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[ "# Convert a lower triangular table to a full matrix [using Copy → Paste Special]\n\nThis module demonstrates how to convert the lower triangular covariance table from the Excel Analysis ToolPak to a full covariance matrix for use with the MMULT function.\n\nFirst, a brief review of square matrices.\n\nA matrix is a rectangular array of numbers of the form $$A=\\begin{bmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & \\ddots & \\vdots \\\\ a_{m1} & \\cdots & a_{mn} \\end{bmatrix}$$ and is called an m × n matrix because it has m rows and n columns. Matrix $$A$$ can also be written as $$A=(a_{ij} )$$ where $$a_{ij}$$ is an element of $$A$$ in i-th row and j-th column where $$1 \\leq i \\leq m$$ and $$1 \\leq j \\leq n$$. The diagonal elements are $$(a_{ij} ) \\; \\forall \\; i=j$$.\n\n## Square matrices\n\n$$B=\\begin{bmatrix} 5 & -2 & 6 \\\\ -2 & 11 & 9.1 \\\\ 6 & 9.1 & 10 \\end{bmatrix}$$\nThe matrix $$B$$ is a 3 × 3 square matrix because it has equal numbers of rows and columns, that is, $$m = n = 3$$. The diagonal elements are the values $$5, 11, 10$$\n\n### 1. Upper triangular matrix\n\nThe $$n \\times n$$ matrix $$C$$, is upper triangular if all elements below the main diagonal are 0.\n$$C_1=\\begin{bmatrix} 5 & 2 & 6 & 7 \\\\ 0 & 11 & 9 & 4 \\\\ 0 & 0 & 10 & 1 \\\\ 0 & 0 & 0 & 8\\end{bmatrix}, \\qquad C_2=\\begin{bmatrix} 5 & 2 & 6 & 7 \\\\ 0 & 0 & 0 & 4 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 8\\end{bmatrix}$$\n\n### 2. Lower triangular matrix\n\nThe $$n \\times n$$ matrix $$C$$, is lower triangular if all elements above the main diagonal are 0.\n$$C_3=\\begin{bmatrix} 5 & 0 & 0 & 0 \\\\ 3 & 9 & 0 & 0 \\\\ 7 & 2 & 1 & 0 \\\\ 4 & 5 & 8 & 6\\end{bmatrix}, \\qquad C_4=\\begin{bmatrix} 5 & 0 & 0 & 0 \\\\ 3 & 0 & 0 & 0 \\\\ 7 & 4 & 4 & 0 \\\\ 6 & 9 & 1 & 8\\end{bmatrix}$$\n\n### 3. Diagonal matrix\n\nThe $$n \\times n$$ matrix $$C$$, is diagonal if all elements off the main diagonal are 0.\n$$C_5=\\begin{bmatrix} 5 & 0 & 0 & 0 \\\\ 0 & 9 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 6\\end{bmatrix}, \\qquad C_6=\\begin{bmatrix} 5 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 4 & 0 \\\\ 0 & 0 & 0 & 8\\end{bmatrix}$$\n$$C_5$$ has diagonal elements $$5,9,1,6$$. $$C_6$$ has diagonal elements $$5,0,4,8$$.\n\n### 4. Symmetrical matrix\n\nThe $$n \\times n$$ matrix $$C$$, is symmetrical if the matrix is equal to its transpose. That is, $$C=C^T$$\n$$C_7=\\begin{bmatrix} 5 & 3 & 7 & 4 \\\\ 3 & 9 & 2 & 5 \\\\ 7 & 2 & 1 & 8 \\\\ 4 & 5 & 8 & 6\\end{bmatrix}$$\n\n## Lower triangular covariance table\n\nThe Analysis ToolPak includes tools to estimate Covariance, and Correlation. Both procedures produce output that is lower triangular. The omission of the upper triangle was originally based on the need to save several bytes of (expensive) computer memory. The same reason the the Toolpak being a Add-In and activated only when required. An example of the Covariance table for four stock returns is shown in figure 1.", null, "Fig 1: Lower triangular covariance table: ToolPak output B2:F6 (top panel), full matrix B2:F6 (lower panel)\n\nIt is clear from figure 1, however, that the output is not a lower triangular matrix, as described in point 2 above, because the upper triangle is blank rather contain zeros. The output is better described as a lower triangular table.\n\nTo convert the lower triangular table to a symmetrical matrix for use in an MMULT equation, do the following:\n\n1. Select the 4 x 4 lower triangular variance-covariance array (with the red frame in figure 1), and\n2. Copy the Selection to the Clipboard\n3. Select the top left cell of a temporary work area (cell H8) and Paste > Special > Transpose. You can also select Paste > Values to avoid format issues\n4. Select the transposed array from step 3 (with the green frame in figure 1), and copy to the Clipboard\n5. Select the top left cell, C3 of the variance-covariance array from step 1, then\n6. Paste > Special > Skip Blanks\n\n• This example was developed in Excel 2013 Pro 64 bit.\n• Last modified: 26 Oct 2018, 7:18 am [Australian Eastern Standard Time (AEST)]" ]

[ null, "https://i1.wp.com/excelatfinance.com/online/wp-content/uploads/2015/04/xlf-lower-triangular-table-2.png", null ]

[ "", null, "# Formula Forensics 043: Rankifs or Conditional Rank\n\nExcel has had a native Rank() function since its very first versions. This function has been updated in 2010 to include Rank.eq and Rank.Avg.\n\nThese functions allow you to Rank a list in either an Ascending or Descending order\n\nRecently on Linkedin I came across a formula at Excel Champs for calculating a Conditional Rank effectively a Rankif() function. The Excel Champs post is based on Michael Girvin’s Youtube Video.\n\nDespite having used Excel since its introduction and despite the fact that there are at kleast a dozen posts in the Chandoo.org Forums discussing Rank If, I thought it strange that I had never seen or had need to use a Rankif() function and so was drawn to it to understand how it worked.\n\nThis post will look at how this technique works, how to use it for Ascending and Descending Ranks and then how to extend it to Multiple Criteria.\n\nThen finally we will move the function into the 21st Century and replace the base function that the technique is based on with a newer function.\n\n## Conditional Rank or Rankifs\n\nWhat is a Conditional Rank or Rankif/s() function.\n\nJust as the words describe, Conditional Rank is a Rank that is based on conditions. So just as Countif() or Sumif() count or sum based on a condition so does Conditional Rank, effectively it is the missing Rankifs() function.\n\nOpen the sample file and look at the data set", null, "You can see that we have the Scores for 12 Students. The table also has fields for Department and Area.\n\nI have highlighted the 4 Engineering Students which we will examine during the post\n\nJohn has a Score of 38, Chandoo has a score of 72, Donna a score of 62 and Bob a score of 84.\n\nSo manually Ranking these 4 students from Highest to Lowest would have the following order\n\n1. Bob 84\n2. Chandoo 72\n3. Donna 62\n4. John 38\n\nThis is shown in the Dep’t Wise Rank Asc, column E\n\nSo examining the highest Engineering student, Bob, Cell E8 you will see that it has a formula:\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13))+1\n\nlets look at how this formula works\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13))+1\n\nWe will refer to these sections later, but the first or Green component is a Conditional part of the formula\n\nThe 2nd or Red section is the Ranking part of the Formula.\n\nWe know from other Formula Forensics posts that Sumproduct adds “Sums” the Products of its constituent arrays. You can read more about how Sumproduct works here Formula Forensics 007 Sumproduct\n\nIn this case there will only be a Single array which is actually made up, as the product of 2 other arrays\n\n(\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13)\n\nThe Conditonal Section\n\nLets look at the first array, the Conditional Section\n\n(\\$B\\$2:\\$B\\$13=B8)\n\nThis says is range B2:B13 equal to B8\n\nie: It is saying are you an Engineering Student ?\n\nIf you select Cell E8, then select the (\\$B\\$2:\\$B\\$13=B8) component and press F9\n\nExcel will show: {TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE}\n\nThis is an array showing which of Cells in B2:B13 are equal to B8\n\nWe can see that the 1st, 4th, 5th and 7th elements of the array are True\n\n{TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE}\n\nThese correspond to cells B2, B5, B6 and B8 which are all Engineering.\n\nThe Ranking Section\n\nEscape out of that formula and we will now look at the second Array\n\nIf you select Cell E8, then Edit the cell F2 and select the (D8<\\$D\\$2:\\$D\\$13) component and press F9\n\nExcel will show: {FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE;TRUE;FALSE}\n\nThis is an array showing which of Cells in D2:D13 are greater than D8\n\nThat is the 3rd, 10th and 11th elements are greater than D8\n\n{FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE;TRUE;FALSE}\n\nBut lets notice here that none of the the 3rd, 10th or 11th elements are Engineering\n\nSo for our cell D8, None of the Engineering Scores are greater than it.\n\nCombining the Two Sections\n\nEscape back out of that formula and we will now look at the internal multiplication of the two arrays\n\nIf you select Cell E8, then Edit the cell F2 and select the whole (\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13) component and press F9\n\nExcel will display: {0;0;0;0;0;0;0;0;0;0;0;0}\n\nThis array is the product of the previous two arrays\n\nie:\n\n{TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE} * {FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE;TRUE;FALSE}\n\nWe can see here that none of the True values line up\n\nSo that when the arrays are multiplied the resultant array is {0;0;0;0;0;0;0;0;0;0;0;0}\n\nIt will only have a 1 in the final array when the two corresponding array elements both have true values.\n\nFinally this array is available to the Sumproduct Function for evaluation\n\nEscape back out of that formula and we will now look at how Sumproduct treats the two arrays\n\nIf you select Cell E8, then Edit the cell F2 and select the whole =Sumproduct(\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13)) component and press F9\n\nExcel will display a {0}\n\nThat is the sum of the products of the arrays is 0\n\nFinally the formula adds a 1 to this to get the final Rank of 1\n\nThat is that none of the Engineering Students have a higher score than Bob and so he has a Value of 1\n\nie: The steps in the solution being\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B8)*(D8<\\$D\\$2:\\$D\\$13))+1\n\n=SUMPRODUCT( {TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE} * {FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE;TRUE;FALSE} )+1\n\n=SUMPRODUCT( {0;0;0;0;0;0;0;0;0;0;0;0} )+1\n\n=0 + 1\n\n=1\n\nLets now examine how Chandoo, in Row 5, went with a score of 72\n\nSelect Cell, E5\n\nYou can edit the cell by pressing F2\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B5)*(D5<\\$D\\$2:\\$D\\$13))+1\n\nEvaluate the component sections\n\n=SUMPRODUCT( {TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE} {FALSE;TRUE;TRUE;FALSE;FALSE;TRUE;TRUE;FALSE;FALSE;TRUE;TRUE;FALSE} )+1\n\n=SUMPRODUCT( {0;0;0;0;0;0;1;0;0;0;0;0} )+1\n\n=1 + 1\n\n=2\n\nWe can see using the same analysis technique for Row 5, that Chandoo’s score of 72 was the 5th, highest score overall, It had 4 higher scores than Chandoo did.\n\nBut only 1 of these was an Engineering student\n\nThe 7th element in each array is True\n\nSo overall Chandoo had 1 Engineering Student with a Higher Score and so he is gets a Rank of 2.\n\nYou can use the technique above to examine other students and see how they Ranked.\n\n### Rank Descending\n\nThe technique and formulas above use =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2<\\$D\\$2:\\$D\\$13))+1 to rank the Students in Ascending Order.\n\nThat is the Highest Score has nobody with a Higher score and so Scores a 1 (0+1)\n\nThe second highest student only has 1 person above him and so they score a 2 (1+1)etc.\n\nTo change the Ranking Order from Ascending to Descending we simply reverse the comparison sign in the counting array\n\nAscending   : =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nDescending : =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2>\\$D\\$2:\\$D\\$13))+1\n\nIn the Descending formula the highest Ranked Engineering Student, Bob, will have 3 other Engineering students below him and so scores a 4 (3+1)\n\nThe second highest Ranked Engineering Student, Chandoo, will have 2 other Engineering students below him and so scores a 3 (2+1)\n\nYou can see how that works by looking at the column G\n\nIn the examples above the Ascending and Descending formulas have only applied a single Condition to our Conditional Rank formula.\n\nIn our example we required that the student is an Engineering student\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nThe Green array is checking that the array for that cell is, in the example of Row 2, an Engineering Student.\n\nWe can add further conditions simply by adding more Conditional Sections to the formula\n\nie: To Rank Engineering Students from the South we simply add a second Conditional Section to the Sumproduct Formula.\n\nAscending: =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(\\$C\\$2:\\$C\\$13=C2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nDescending: =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(\\$C\\$2:\\$C\\$13=C2)*(D2>\\$D\\$2:\\$D\\$13))+1\n\nTo Rank Engineering Students from the South named Bob we simply add a second and third Conditional Section to the Sumproduct Formula.\n\nAscending: =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(\\$C\\$2:\\$C\\$13=C2)*(\\$A\\$2:\\$A\\$13=A2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nDescending=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(\\$C\\$2:\\$C\\$13=C2)*(\\$A\\$2:\\$A\\$13=A2)*(D2>\\$D\\$2:\\$D\\$13))+1\n\n## Removing Duplicates from the Rankings\n\nIf we modify the data a little and accidentally add a few duplicates scores we can see that the Formulas shown above introduce an error", null, "We can see that both Fred and Bob are Engineering students and they both scored 84. The existing function has scored them equally as 1.\n\nWe can see that both Chandoo and Danielle are also Engineering students and they both scored 72. The existing function has scored them equally as 2.\n\nLuckily there is a work around for this.\n\nThe base formula in our new data set is\n\n=SUMPRODUCT((\\$B\\$20:\\$B\\$31=B20)*(D20<(\\$D\\$20:\\$D\\$31)))+1\n\nWe can modify this to add a small but slightly different value to each row in the Counting Section of the Sumproduct formula\n\n=SUMPRODUCT((\\$B\\$20:\\$B\\$31=B20)*((D20+ROW()/1000)<(\\$D\\$20:\\$D\\$31+(ROW(\\$I\\$20:\\$I\\$31)/1000))))+1\n\nyou can see that the sections highlighted in Green above add a small number based on the Row Number / 1000 to both the Score and the Score Column. This way numbers closer to the bottom of the Table will have a higher chance of getting a lower rank.\n\nIf you want the students higher in the list to have a Higher Ranking you can change the logic as such\n\n=SUMPRODUCT((\\$B\\$20:\\$B\\$31=B20)*((D20+(100-ROW())/1000)<(\\$D\\$20:\\$D\\$31+((100-ROW(\\$I\\$20:\\$I\\$31))/1000))))+1\n\nJust make sure the value 100 is greater than the last Row number of the data\n\n## Updating the Formula\n\nIn the Introduction I made note that I would bring the Formula into the 21st Century.\n\nThe Conditional Rank Formula is based on the use of the Sumproduct Function.\n\n=SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nThe Sumproduct function must be the most versatile function ever introduced to Excel.\n\nHowever if you examine the formula you will see it basically says “Sum the number of entries where a Condition is met, But Sum is effectively Counting\n\nSo it sort of sounds like a Countifs() Function?\n\nYou can examine the Countifs() based equivalent functions below\n\nAscending version\n\nSumproduct =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2<\\$D\\$2:\\$D\\$13))+1\n\nCountifs =COUNTIFS(\\$B\\$2:\\$B\\$13,B2,\\$D\\$2:\\$D\\$13,”>”&D2)+1\n\nDescending version\n\nSumproduct =SUMPRODUCT((\\$B\\$2:\\$B\\$13=B2)*(D2>\\$D\\$2:\\$D\\$13))+1\n\nCountifs =COUNTIFS(\\$B\\$2:\\$B\\$13,B2,\\$D\\$2:\\$D\\$13,”<“&D2)+1\n\nUnfortunately Excel doesn’t allow us to use the F9 evaluate facility on the components of Countifs()\n\nBut reading each formula from left to right, they say\n\nCount the cells If, The First Column B is equal to the Rows Value of Column B and the second Column D is greater than the Rows Value of Column D\n\nThat is = Countif ( Department Column = Engineering & Score Column > Current Cell)\n\n## Closing\n\nThis post has explained two techniques for evaluating Conditional Rank and included several variations as well.\n\nDespite the fact that this was new to me I have since seen at least a dozen posts here on the Chandoo.org Forums where these techniques have been used.\n\nDo you have any applications where this is applicable or other techniques to perform a Conditonal Rank or Rankifs() functionality?\n\nLet us know in the comments below.\n\n## Formula Forensics “The Series”\n\nThis is the 47th post in the Formula Forensics series.\n\nFormula Forensic Series\n\n## Formula Forensics Needs Your Help\n\nIf you want to see more Formulas pulled apart and explained Forensically we need your help.\n\nIf you have a neat formula that you would like to share and explain, try putting pen to paper and draft up a Post like above or;\n\nIf you have a formula that you would like explained, but don’t want to write a post, send it to Hui or Chandoo, or even drop it in the Comments below.", null, "Hello Awesome...\n\nMy name is Chandoo. Thanks for dropping by. My mission is to make you awesome in Excel & your work. I live in Wellington, New Zealand. When I am not F9ing my formulas, I cycle, cook or play lego with my kids. Know more about me.\n\nThank you and see you around.\n\n### Related articles:\n\n Written by Hui... Tags: Condional Rank, rank(), Rankif, Rankifs Home: Chandoo.org Main Page ? Doubt: Ask an Excel Question\n\n### 17 Responses to “Formula Forensics 043: Rankifs or Conditional Rank”\n\n1.", null, "MF says:\n\nHi Hui,\nVery detailed explanation of using SUMPRODUCT for solving this problem, awesome!\n\nI have a blogpost discussing the use of COUNTIFS to solve the problem.\nhttps://wmfexcel.com/2016/03/12/rank-in-subgroup-rankif/\n\nCheers,\n\n2.", null, "Hazel says:\n\nand not\n\n•", null, "Hui... says:\n\nThanx Hazel\n\nIt is fixed now\n\n•", null, "Hazel says:\n\nFantastic blog btw ... currently playing with it and really liking the potential it has for my analysis reports 🙂\n\n•", null, "Hui... says:\n\n@Hazel\n\nThanx for the kind words\n\nI'll assume the usual 5% commission ?\n\n3.", null, "Black Moses says:\n\nThanks a lot for this post,\n\nLike you, I've never needed to use a conditional ranking formula despite using Excel forever, but I found this incredibly interesting and I'm sure I'll come back to this post if I need to use conditional ranking in the future. Plus, I tend to avoid SUMPRODUCT like the plague and just use array formulae instead so anything that gets me thinking about SUMPRODUCT can only be a good thing.\n\nIn terms of functions that we want explained, I'd love it if you could explain MMULT. I can't remember where but I remember seeing MMULT being used in the solution to a formula problem that would have been really complicated to solve otherwise. It really intrigued me but I don't fully understand how it works, or rather how you can use it to solve problems/what kinds of problems it works best with.\n\n•", null, "Hui... says:\n\n@Black Moses\n\nSumproduct can destroy spreadsheet performance when incorrectly used\nHowever when used properly, it is the Swiss Army Knife of Excel Problem solving\n\nFor some great ideas\n\n4.", null, "Abhay Gadiya says:\n\nI checked this Power Query solution here -\n\nhttps://blog.crossjoin.co.uk/2015/05/11/nested-calculations-in-power-query/amp/\n\nCheckout my YouTube channel for video as well.\n\n5.", null, "esg says:\n\nInspiring.\nJust a warning in sumproduct vs summing product's\nin sumproduct, in excel2010, blank cells are evaluated as ZERO, while in product, they are ignored (unless all elements are blank).\nMaybe this is because I have the wrong defaults on my system.\nhth.\n\n6.", null, "Peter B says:\n\nA further advantage of these formulas over RANK is that they work on text strings, where they sort alphabetically, as well as numbers. The COUNTIFS form can also be used as an array formula though it requires the criteria ranges to be references and not arrays 🙁\n\nMulti-criteria sorts can be achieved by adding COUNTIFS. Thus\n\n= 1 +\nCOUNTIFS([Department],[Department],[Score],\">\"&[Score]) +\nCOUNTIFS([Department],[Department],[Score],[Score],[Student],\"<\"&[Student])\n\nwould use the student's name as the tie-break criterion. Not exactly fair but there you go!\n\n7.", null, "Kenneth Focht says:\n\nThanks for this great Excel lesson!\n\nI would like some help figuring out how to keep the rank for duplicates. Example: Using your example but changing it up to not copy 100%.....I have a spreadsheet with employees, business areas, years experience, and senority (rank). In my spreadsheet I have ranked by years experience and business area. 2 people in IT rank 2 since they have the same experience. I want to keep it this way. My problem is that the next rank should be 3 but is actually 4.\n\n8.", null, "neelakantan says:\n\nDEAR CHANDOO..\nI AM HAVING A SITUATION THAT I WANT TO RANK MY STUDENTS BASED ON THEIR TOTAL MARK.\nI AM HAVING FIVE SUBJECTS. RANKING SHOULD BE DONE IF A STUDENT PASSES IN ALL SUBJECTS, i.e.,HE/SHE SHOULD GET MINIMUM OF 35 IN ALL THE SUBJECTS. I'M TRYING ALL FORMULAE LIKE SUMPRODUCT, COUNTIF AND SO ON.. BUT NONE OF THEM WORKING CORRECTLY. COULD YOU PLEASE HELP ME ..\n\n•", null, "Himanshu Jain says:\n\nhi,\nfirst of all you can use if along with and function to get check whether the student passed or not...if the student passed then get the sum of the marks...and if the student failed then assign 0 against the total marks =IF(AND(B134>35,C134>35,D134>35,E134>35,F134>35),SUM(B134:F134),0)\nB134,C134 AND SO ON CELLS CONTAIN THE MARKS....if the condition holds true you will get the sum else you will get 0\nnow you can use rank function on the result that you have got\n=RANK(G134,\\$G\\$134:\\$G\\$136,0)\nmake sure to use descending order so that the student who failed get the last rank\nstudents m1 m2 m3 m4 m5 total marks rank\na 37 78 45 98 12 0 3\ns 76 55 44 37 76 288 1\nd 56 56 76 87 45 37 2\n\n9.", null, "CGI says:\n\nSuper easy to follow along! My boss wanted to filter a list by dates that he can change and this fit the bill. Thank you!\n\n10.", null, "Thomas Crosby says:\n\nI am trying to use the Removing Duplicates from Ranking and not having success. I am using named ranges versus defined rows. Does that make a difference?\n\n11.", null, "Andrew Wilkins says:\n\nHi,\nI have a wonderful application of this technique in analysing the performance of footballers in the the UK Fantasy Premiership game.\n\nSince forever I've been trying to find a way of of ranking players without having to filter them by position, sort them by e.g. descending points and then messing about with copying and pasting as values.\n\nUsing this technique I know now that, this week, Fernandes is the highest ranking player by points but is 45th (out of 269 Midfielders) in terms of pts/£ (bcr) which is significant if you need a midfielder on a limited budget.\n\n12.", null, "Arpan Kumar says:\n\nThis formula is soo helpful and beautifully explained.\nHowever, what if there are \"0\" zero values in the column and we don't want to consider those 0's as part of the ranking.\n\n « How to Distribute Players Between Teams – Evenly Conditional Rank, the easy way [quick tip] »\n\n### Get FREE Excel & Power-BI Newsletter\n\nOne email per week with Excel and Power BI goodness. Join 100,000+ others and get it free." ]

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[ "", null, "# Uncertainty Quantification in Deep Learning through Stochastic Maximum Principle\n\nWe develop a probabilistic machine learning method, which formulates a class of stochastic neural networks by a stochastic optimal control problem. An efficient stochastic gradient descent algorithm is introduced under the stochastic maximum principle framework. Convergence analysis for stochastic gradient descent optimization and numerical experiments for applications of stochastic neural networks are carried out to validate our methodology in both theory and performance." ]

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[ "Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: English - Français - Português -\n\nSee the recommended documentation of this function\n\nScilab manual >> Compatibility Functions > mtlb_fftshift\n\n# mtlb_fftshift\n\nMatlab fftshift emulation function\n\n### Description\n\nMatlab and Scilab `fftshift` behave differently in some particular cases:\n\n• With character string input: due to the fact that character strings are not considered in the same way in Matlab and Scilab, results can be different for this kind of input.\n\n• With two inputs: Matlab `fftshift` can work even if `dim` parameter is greater than number of dimensions of first input.\n\nThe function `mtlb_fftshift(A[,dim])` is used by `mfile2sci` to replace `fftshift(A[,dim])` when it was not possible to know what were the inputs while porting Matlab code to Scilab. This function will determine the correct semantic at run time. If you want to have a more efficient code it is possible to replace `mtlb_fftshift` calls:\n\n• If `A` is not a character string matrix, `mtlb_fftshift(A)` may be replaced by `fftshift(A)`\n\n• If `A` is not a character string matrix and `dim` is not greater than `size(size(a),\"*\")`, `mtlb_fftshift(A,dim)` may be replaced by `fftshift(A,dim)`\n\nCaution: `mtlb_fftshift` has not to be used for hand coded functions.\n\n• V.C." ]

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[ "## Exergy Sustainability - Sandia National Laboratories\n\nExergy is the elixir of life. Exergy is that portion of energy available to do work. Elixir is defined as a substance held capable of prolonging life indefinitely, which ...\n\nSANDIA REPORT SAND2006-2759 Unclassified Unlimited Release Printed May 2006\n\nExergy Sustainability\n\nRush D. Robinett, III, David G. Wilson, and Alfred W. Reed Prepared by Sandia National Laboratories Albuquerque, New Mexico 87185 and Livermore, California 94550 Sandia is a multiprogram laboratory operated by Sandia Corporation, a Lockheed Martin Company, for the United States Department of Energy’s National Nuclear Security Administration under Contract DE-AC04-94-AL85000. Approved for public release; further dissemination unlimited.\n\nIssued by Sandia National Laboratories, operated for the United States Department of Energy by Sandia Corporation.\n\nNOTICE: This report was prepared as an account of work sponsored by an agency of the United States Government. Neither the United States Government, nor any agency thereof, nor any of their employees, nor any of their contractors, subcontractors, or their employees, make any warranty, express or implied, or assume any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represent that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise, does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government, any agency thereof, or any of their contractors or subcontractors. The views and opinions expressed herein do not necessarily state or reflect those of the United States Government, any agency thereof, or any of their contractors. Printed in the United States of America. This report has been reproduced directly from the best available copy. Available to DOE and DOE contractors from U.S. Department of Energy Office of Scientific and Technical Information P.O. Box 62 Oak Ridge, TN 37831 Telephone: Facsimile: E-Mail: Online ordering:\n\n(865) 576-8401 (865) 576-5728 [email protected] http://www.doe.gov/bridge\n\nAvailable to the public from U.S. Department of Commerce National Technical Information Service 5285 Port Royal Rd Springfield, VA 22161\n\nNT OF E ME N RT\n\nGY ER\n\nDEP A\n\nTelephone: Facsimile: E-Mail: Online ordering:\n\nED\n\nER\n\nU NIT\n\nIC A\n\nST\n\nA TES OF A\n\nM\n\n(800) 553-6847 (703) 605-6900 [email protected] http://www.ntis.gov/ordering.htm\n\nSAND2006-2759 Unclassified Unlimited Release Printed May 2006\n\nExergy Sustainability Rush D. Robinett, III Energy, Infrastructure, and Knowledge Systems David G. Wilson Intelligent Systems and Robotics Center Alfred W. Reed Nuclear and Risk Technologies Center Sandia National Laboratories P.O. Box 5800 Albuquerque, NM 87185-1003\n\nAbstract Exergy is the elixir of life. Exergy is that portion of energy available to do work. Elixir is defined as a substance held capable of prolonging life indefinitely, which implies sustainability of life. In terms of mathematics and engineering, exergy sustainability is defined as the continuous compensation of irreversible entropy production in an open system with an impedance and capacity-matched persistent exergy source. Irreversible and nonequilibrium thermodynamic concepts are combined with self-organizing systems theories as well as nonlinear control and stability analyses to explain this definition. In particular, this paper provides a missing link in the analysis of self-organizing systems: a tie between irreversible thermodynamics and Hamiltonian systems. As a result of this work , the concept of “on the edge of chaos” is formulated as a set of necessary and sufficient conditions for stability and performance of sustainable systems. This interplay between exergy rate and irreversible entropy production rate can be described as Yin and Yang control: the dialectic 3\n\nsynthesis of opposing power flows. In addition, exergy is shown to be a fundamental driver and necessary input for sustainable systems, since exergy input in the form of power is a single point of failure for self-organizing, adaptable systems.\n\n4\n\nAcknowledgment The authors would like to thank Dr. Don Hardesty, Senior Manager in Center 8300 at Sandia National Laboratories in California for his assistance in the mathematical definition of exergy and Mr. Max Valdez in Center 6200 for his assistance in connecting exergy to environmental economics. The format of this report is based on information found in .\n\n5\n\nContents 1\n\nIntroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n9\n\n2\n\nThermodynamic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n10\n\n3\n\nHamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n11\n\n4\n\nThermo-Mechanical Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n13\n\n4.1\n\nConservative Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n13\n\n4.2\n\nReversible Thermodynamic Systems . . . . . . . . . . . . . . . . . . . . . . . .\n\n13\n\n4.3\n\nIrreversible Thermodynamic Systems . . . . . . . . . . . . . . . . . . . . . . . .\n\n14\n\n4.4\n\nAnalogies and Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n14\n\nNecessary and Sufficient Conditions for Stability . . . . . . . . . . . . . . . . . . . . .\n\n16\n\n5.1\n\nStability and Instability Theorems . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n17\n\n5.2\n\nStability Lemma for Nonlinear Self-Organizing Systems . . . . . . . . .\n\n17\n\n5.3\n\nClassic van der Pol Equation Example . . . . . . . . . . . . . . . . . . . . . . .\n\n18\n\nSelf-Organization and Adaptability Concepts . . . . . . . . . . . . . . . . . . . . . . . .\n\n21\n\n6.1\n\nBackground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n22\n\n6.2\n\nSimple Nonlinear Satellite System . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n25\n\n6.3\n\nLifestyle Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n31\n\n6.4\n\nDeformation of Potential Field with Information Flow . . . . . . . . . .\n\n32\n\nSummary and Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n35\n\n5\n\n6\n\n7\n\n6\n\nList of Figures 1\n\nEnergy flow control volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n10\n\n2\n\nEntropy with flux exchange system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n11\n\n3\n\nTime response for power in a general AC circuit with ω = 2π, v¯ = 1.5, i¯ = 2.0, and θ = π/4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n16\n\nVan der Pol responses: Hamiltonian 3D surface (left) and phase plane 2D projection (right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n20\n\n5\n\nVan der Pol exergy-rate (left) and exergy (right) responses - neutral case . . .\n\n21\n\n6\n\nVan der Pol with integral action responses: Hamiltonian 3D surface (left) and phase plane 2D projection (right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n22\n\n7\n\nNonlinear spring potential function characteristics . . . . . . . . . . . . . . . . . . . .\n\n24\n\n8\n\nSimplified nonlinear satellite model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n25\n\n9\n\nAll cases: mass-spring-damper with Duffing oscillator/Coulomb friction model numerical results: Hamiltonian 3D surface (left) and phase plane 2D projection (right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n31\n\nThree dimensional (left) Hamiltonian phase plane plot negative stiffness produces a saddle surface. The two-dimensional cross-section plot (right) is at x˙ = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n34\n\nThree dimensional (left) Hamiltonian phase plane plot where the net positive stiffness produces a positive bowl surface. The two-dimensional crosssection plot (right) is at x˙ = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n35\n\n4\n\n10\n\n11\n\n7\n\nList of Tables 1\n\nVan der Pol model numerical values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n19\n\n2\n\nVan der Pol model with integral action numerical values . . . . . . . . . . . . . . .\n\n21\n\n8\n\n1\n\nIntroduction\n\nExergy is the elixir of life. Exergy is that portion of energy available to do work. Elixir is defined as a substance held capable of prolonging life indefinitely, which implies sustainability of life. In this paper simplified models are developed and utilized to address the notion of sustainability of a lifestyle. The purpose of this paper is to present the concept of exergy sustainability based on irreversible and non-equilibrium thermodynamics combined with some concepts from self-organization and adaptivity as well as nonlinear stability and control. The result of this synthesis is the mathematical definition of sustainability based on a fundamental driver and input, exergy rate, to a self-organizing system. In fact, exergy rate (power) input, is a single point failure. The flow of exergy into an open system that is continuously undergoing self-organization and adaptivity determines whether the system will persist or disintegrate. As a result, the balance of exergy flows into and out of the system versus the exergy consumption, irreversible entropy production, in the system will be studied. Traditionally, exergy concepts are founded in the first and second laws of thermodynamics in the field of physics. However, these laws have both economical and environmental significance as well and can be applied in a more universal manner. Assessment of economic factors, that are based on the second law of thermodynamics are: i) exergy is not conserved and ii) exergy can be used as a common measure of resource quality along with quantity (i.e., materials and energy) . Exergy from a physics standpoint is formally defined as the maximum amount of work that a subsystem can do on its surroundings as it approaches thermodynamic equilibrium reversibly or the degree of distinguishability of a subsystem from its surroundings. Therefore, exergy can be used to measure and compare resource inputs and outputs which include wastes and losses . For economic processes exergy is consumed not conserved. In addition, exergy can be used as a measure of assessing technical progress for economic growth theory. In this paper the basic concepts from and will provide the background to explain the definition of exergy sustainability: the continuous compensation of irreversible entropy production in an open system with an impedance and capacity-matched persistent exergy source. This paper is divided into seven sections. Sections 2 and 3 provide the preliminary thermodynamics and Hamiltonian mechanics definitions. Section 4 develops the relationships and connections between thermodynamics and Hamiltonian mechanics. Section 5 defines the necessary and sufficient conditions for stability of nonlinear systems. Section 6 discusses self-organizing, adaptive systems and combines the concepts of the previous sections in order to explain the definition of exergy sustainability. Finally, Section 7 summarizes the results with concluding remarks.\n\n9\n\n2\n\nThermodynamic Concepts\n\nIn this section, results from the first and second laws of thermodynamics are used to define exergy. A corollary of the first law of thermodynamics states energy is conserved (see Fig. 1). A corollary of the first and second laws of thermodynamics state that the entropy of the universe must be greater than or equal to zero. Conservation of energy can be written in terms of energy rate for a system ˙ + m˙ (h + ke + pe + . . .) . E˙ = ∑ Q˙ i + ∑ W j ∑ k k k k i\n\nj\n\n(1)\n\nk\n\nThe term on the left is the rate at which energy is changing within the system. The heat entering or leaving the system is given by Q˙ i and the work entering or leaving the system is ˙ . Material can enter or leave the system by m˙ that includes enthalpy, h, kinetic given by W j k and potential energies, ke, pe, etc. In addition, each term is “summed” over an arbitrary number of entry and exit locations i, j, k.\n\nFigure 1. Energy flow control volume\n\nThe entropy rate equation for a system \n\nS˙ = ∑ i\n\nQ˙ i Ti\n\n+ ∑ m˙ k sk + S˙ i = S˙ e + S˙ i .\n\n(2)\n\nk\n\nWhere the left hand term is the rate entropy changes within the system and the right hand terms represent, in order, entropy change due to heat interactions to and from the system and the rate material carries it in or out. These two terms can be combined into one term S˙ e , the entropy exchanged (either positive or negative) with the environment and S˙ i is the irreversible entropy production rate within the system. Figure 2 shows the entropy exchanges and production within the system . The irreversible entropy production rate can be written as the sum of the thermodynamic forces and the thermodynamic flows \n\nS˙ i = ∑ Fk X˙ k ≥ 0\n\n(3)\n\nk\n\n10\n\nFigure 2. Entropy with flux exchange system\n\nwhere the entropy change is the sum of all the changes due to the thermodynamic flows X˙ k with respect to each corresponding thermodynamic force Fk . Next, for systems with a constant environmental temperature (To ), a thermodynamic quantity called the availability function is defined as [4, 5, 6] Ξ = E − To S .\n\n(4)\n\nThe availability function is described as the maximum theoretically available energy that can do work which we call exergy. Exergy is also known as negative-entropy [4, 6]. Taking the time derivative of the availability function (4) and substituting in the expressions for (1) and (2) results in the exergy rate equation Ξ˙ = ∑ i\n\n\u0012\n\n\u0013 \u0012 \u0013 ¯ To ˙ d V f low ˙ −p 1− Qi + ∑ W + ∑ m˙ k ζk − To S˙ i . j o Ti dt j k\n\n(5)\n\nWhere Ξ˙ is the rate at which exergy stored within the system is changing. The terms on the right, in order, define the rate exergy is carried in/out by; i) heat, ii) work (less any work the system does on the environment at constant environmental pressure po if the system volume V¯ changes), and iii) by the material (or quantity known as flow exergy). The final term, To S˙ i , is the rate exergy is destroyed within the system or exergy consumption rate.\n\n3\n\nHamiltonian Mechanics\n\nIn this section the basic concepts of Hamiltonian mechanics are introduced. The derivation of the Hamiltonian begins with the Lagrangian for a system defined as\n\nL = T (q, q,t) ˙ − V (q,t)\n\n(6)\n\nwhere 11\n\nt q q˙\n\nT V\n\n= = = = =\n\ntime explicitly N-dimensional generalized coordinate vector N-dimensional generalized velocity vector Kinetic energy Potential energy\n\nThe Hamiltonian is defined in terms of the Lagrangian as n\n\n∂L q˙i − L (q, q,t) ˙ = H (q, q,t). ˙ i=1 ∂q˙i\n\nH ≡∑\n\n(7)\n\nThe Hamiltonian in terms of the canonical coordinates (q, p) is n\n\nH (q, p,t) = ∑ pi q˙i − L (q, q,t) ˙\n\n(8)\n\ni=1\n\nwhere the canonical momentum is defined as pi =\n\n∂L . ∂q˙i\n\n(9)\n\nThen Hamilton’s canonical equations of motion become q˙i = p˙i =\n\n∂H ∂pi − ∂∂qHi\n\n(10)\n\n+ Qi\n\nwhere Qi is the generalized force vector. Next taking the time derivative of (8) gives \u0013 n \u0012 ∂L ∂L ∂L ˙ H = ∑ p˙i q˙i + pi q¨i − − q˙i − q¨i . ∂t ∂qi ∂q˙i i=1\n\n(11)\n\nThen substitute (10) and simplifying gives n\n\nH˙ = ∑ Qi q˙i − i=1\n\n∂L . ∂t\n\n(12)\n\nHamiltonians for most natural systems are not explicit functions of time (or ∂L /∂t = 0). Then for\n\nL = L (q, q) ˙\n\n(13) 12\n\nthe power (work/energy) equation becomes n\n\nH˙ (q, p) = ∑ Qi q˙i .\n\n(14)\n\ni=1\n\n4\n\nThermo-Mechanical Relationships\n\nThis section discusses the concepts of conservative systems and forces, reversible and irreversible thermodynamic systems, average power and power flow, and the connections between Hamiltonian mechanics and thermodynamics required to support the concepts of necessary and sufficient conditions for stability of nonlinear systems. It is worth noting at this point that, by definition, electrical power is “pure exergy rate” and the Hamiltonian is stored exergy.\n\n4.1\n\nConservative Mechanical Systems\n\nA system is conservative if\n\nH˙ = 0\n\nand\n\nH = constant.\n\nA force is conservative if I\n\nI\n\nF · dx =\n\nI\n\nF · vdt =\n\nQ j q˙ j dt = 0\n\nwhere F is the force, dx the displacement, and v the velocity. Basically, all of the forces can be modeled as potential force fields which are exergy storage devices.\n\n4.2\n\nReversible Thermodynamic Systems\n\nA thermodynamic system is reversible if dS = dTQ H H dQ d S = =0 \u0003 H H T H\u0002 dS = [d Si + d Se ] = S˙ i + S˙ e dt = 0 which implies that S˙ e = Q˙ /T since by definition the second law gives S˙ i = 0. 13\n\n4.3\n\nIrreversible Thermodynamic Systems\n\nFor I\n\nI\n\ndS =\n\nS˙ i + S˙ e dt = 0\n\n\u0002\n\n\u0003\n\nthen S˙ e ≤ 0 and S˙ i ≥ 0.\n\n4.4\n\nAnalogies and Connections\n\nNow the connections between thermodynamics and Hamiltonian mechanics are investigated. 1. The irreversible entropy production rate can be expressed as\n\nS˙ i = ∑ Fk X˙ k = k\n\n1 Qk q˙k ≥ 0. To ∑ k\n\n(15)\n\n2. The time derivative of the Hamiltonian is equivalent to the exergy rate since the Hamiltonian for a conservative system is stored exergy, then\n\nH˙ = ∑k Qk q˙k M+N Ξ˙ = W˙ − To S˙ i = ∑Nj=1 Q j q˙ j − ∑l=N+1 Ql q˙l .\n\n(16)\n\nWhere N is the number of generators, M the number of dissipators, and let W˙ = ˙ . The following assumptions apply when utilizing the exergy rate equation (5) ∑j W j for Hamiltonian systems: (a) No substantial heat flow:\n\nQ˙ i ≈ 0. (b) No substantial exergy flow or assume Ti is only slightly greater than To : 1−\n\nTo ≈ 0. Ti\n\n(c) No poV¯ work on the environment: dV¯ po = 0. dt (d) No mass flow rate: f low\n\n∑ m˙ k ζk\n\n= 0.\n\nk\n\n14\n\n(e) Then define: W˙ ≥ 0 power input/generated ˙ To Si ≥ 0 power dissipated. 3. A conservative system is equivalent to a reversible system when\n\nH˙ = 0 and S˙ e = 0 then\n\nS˙ i = 0 and W˙ = 0. 4. For a system that “appears to be conservative”, but is not reversible is defined as:\n\nH˙ ave = 0 = (W˙ )ave − (To S˙ i )ave = average power over a cycle\n\n= =\n\n1H ˙ ˙ τ [W − To Si ]dt M+N 1H N τ [∑ j=1 Q j q˙ j − ∑l=N+1 Ql q˙l ]dt\n\nwhere τ is the period of the cycle. To be more specific about the average power calculations, the AC power factor provides an excellent example. For the general case of alternating current supplied to a complex impedance the voltage and current differ in phase by an angle θ. The time responses for power, voltage, and current are shown for a general AC circuit in Fig. 3 with √ √ W˙ = P = Qq˙ = v i = 2v¯ cos(ωt + θ) · 2i¯cos ωt = v¯i¯[cos θ + cos(2ωt + θ)] where P is power, v is voltage (v), ¯ i is current (i¯), θ is the phase angle, and ω is the frequency. Integrating over a cycle gives (W˙ )ave = v¯i¯cos θ where for the second term I\n\ncos(2ωt + θ)dt = 0. This is an important set of conditions that will be used in the next section to find the generalized stability boundary. 5. Finally, the power terms are sorted into three categories: (a) (W˙ )ave - power generators; (Q j q˙ j )ave > 0 (b) (To S˙ i )ave - power dissipators; (Ql q˙l )ave < 0 (c) (To S˙ rev )ave - reversible/conservative exergy storage terms; (Qk q˙k )ave = 0. These three categories are fundamental terms in the following definitions and design procedures. 15\n\nFigure 3. Time response for power in a general AC circuit with ω = 2π, v¯ = 1.5, i¯ = 2.0, and θ = π/4\n\n5\n\nNecessary and Sufficient Conditions for Stability\n\nThis section describes the concepts from nonlinear control theory that will be used to assess the balance of exergy flows into a system versus the exergy consumption/destruction (irreversible entropy production) in an open, self-organizing system. The balance of these exergy flows determine a fundamental necessary condition for sustainability of a selforganizing system. Further, the need for continuous flow of exergy into a self-organizing system punctuates the need for predictable, persistent exergy sources. The Lyapunov function is defined as the total energy (stored exergy by our definition) which for most mechanical systems is equivalent to an appropriate Hamiltonian function\n\nV =H\n\n(17)\n\nwhich is positive definite. The time derivative is\n\nM+N V˙ = H˙ = ∑k Qk q˙k = ∑Nj=1 Q j q˙ j − ∑l=N+1 Ql q˙l ˙ = W˙ − To Si .\n\n16\n\n(18)\n\n5.1\n\nStability and Instability Theorems\n\nTo describe a nonlinear self-organizing system’s behavior two theorems help to characterize the essential features of their motion. In addition, by bounding the Lyapunov function between these Theorems, both necessary and sufficient conditions are a result of the transition of the time derivative of the Lyapunov function from stable to unstable. 1. Lyapunov Theorem for Stability Assume that there exists a scalar function V of the state x, with continuous first order derivatives such that V (x) is positive definite V˙ (x) is negative definite V (x) → ∞ as kxk → ∞ Then the equilibrium at the origin is globally asymptotically stable. 2. Chetaev Theorem for Instability Considering the equations of disturbed motion, let V be zero on the boundary of a region R which has the origin as a boundary point, and let both V and V˙ be positive-definite in R; then the undisturbed motion is unstable at the origin.\n\n5.2\n\nStability Lemma for Nonlinear Self-Organizing Systems\n\nBased on the relationships between thermodynamic exergy and Hamiltonian systems a Fundamental Stability Lemma can be formulated. Fundamental Stability Lemma for Hamiltonian Systems The stability of Hamiltonian systems is bounded between Theorems 1 and 2. Given the Lyapunov derivative as a decomposition and sum of exergy generation rate and exergy dissipation rate then: N\n\nV˙ = W˙ − To S˙ i =\n\n∑ Q j q˙ j −\n\nj=1\n\nM+N\n\nQl q˙l\n\n(19)\n\nl=N+1\n\nthat is subject to the following general necessary and sufficient conditions: To S˙ i ≥ 0 Positive semi-definite, always true W˙ ≥ 0 Positive semi-definite; exergy pumped into the system. The following corollaries encompass both stability and instability for Hamiltonian systems which utilize AC power concepts : 17\n\nCorollary 1: For (To S˙ i )ave = 0 and (W˙ )ave = 0 then V˙ = 0 the Hamiltonian system is neutrally stable, conservative and reversible. Corollary 2: For (To S˙ i )ave = 0 and (W˙ )ave > 0 then V˙ > 0 the Hamiltonian system is unstable. Corollary 3: For (To S˙ i )ave > 0 and (W˙ )ave = 0 then V˙ < 0 the Hamiltonian system is asymptotically stable and a passive system in the general sense (passivity controllers). Corollary 4: Given apriori (To S˙ i )ave > 0 and (W˙ )ave > 0 then the Hamiltonian system is further subdivided into: \u0001 \u0001 4.1: For To S˙ i ave > W˙ ave with V˙ < 0 yields asymptotic stability \u0001 \u0001 4.2: For To S˙ i ave = W˙ ave with V˙ = 0 yields neutral stability \u0001 \u0001 4.3: For To S˙ i ave < W˙ ave with V˙ > 0 yields an unstable system. The bottom line is that stability is defined in terms of power flow which determines whether the system is moving toward or away from its minimum energy and maximum entropy state.\n\n5.3\n\nClassic van der Pol Equation Example\n\nBefore moving on to self-organizing systems, it is instructive to provide a couple of simple examples. Example 1 is the classic van der Pol’s equation which is analyzed using the techniques of this section. Originally, the “van der Pol equation” is credited to van der Pol, and is a model of an electronic circuit for early radio vacuum tubes of a triode electronic oscillator . The tube acts like a normal resistor when the current is high, but acts as a negative resistor if the current is low. The main feature is that electrical circuits that contain these elements pump up small oscillations due to a negative resistance when currents are small, but drag down large amplitude oscillations due to positive resistance when the currents are large. This behavior is known as a relaxation oscillation, as each period of the oscillation consists of a slow buildup of energy (’stress phase’) followed by a phase in which energy is discharged (’relaxation phase’). This particular system has played a large role in nonlinear dynamics and has been used to study limit cycles and self-sustained oscillatory phenomena in nonlinear systems. Consider the van der Pol equation which includes a non-linear damping term: x¨ − µ(1 − x2 )x˙ + x = 0. Next include the actual mass and stiffness values (other than unity) or mx¨ − µ(1 − x2 )x˙ + kx = 0. 18\n\nThe appropriate Hamiltonian/Lyapunov function is defined as: 1 2\n\n1 2\n\nH = V = mx˙2 + kx2 > 0. Then the corresponding time derivative of the Lyapunov function becomes V˙ = [m \u0002 x¨ + kx] 2x˙ \u0003 = µx(1 ˙ − x ) x˙ = µx˙2 − µx2 x˙2 . Identifying generator and dissipator terms yields W˙ = µx˙2 To S˙ i = µx2 x˙2 The stability boundary can be determined as \u0002 \u0003 \u0002 \u0003 ˙ = \u0002To S˙ i ave \u0002W 2ave \u0003 \u0003 µx˙ ave = µx2 x˙2 ave By investigating several initial conditions both inside, on, and outside the limit cycle then three separate conditions can be observed. Figure 4 shows these conditions with the corresponding numerical values given in Table 1. Table 1. Van der Pol model numerical values Case\n\nxo (m)\n\nx˙o (m/s)\n\nµ (kg/s)\n\nm (kg)\n\nk (kg/s2 )\n\ngenerate neutral dissipate\n\n0.1 1.0 2.0\n\n−0.1 −1.0 −2.0\n\n1.5 1.5 1.5\n\n1.0 1.0 1.0\n\n1.0 1.0 1.0\n\nThe responses are plotted on the Hamiltonian 3D surface (left) with the projection onto the phase plane shown on the 2D plot (right). For the case outside the limit cycle, the dissipator term dominates and for the case inside the limit cycle the generator term dominates. For both cases inside and outside the limit cycle, the system migrates back to the stability boundary. For the case already on the limit cycle then the system is already at neutral 19\n\nFigure 4. Van der Pol responses: Hamiltonian 3D surface (left) and phase plane 2D projection (right)\n\nstability. The neutral exergy-rate (left) and exergy (right) plots are shown in Fig. 5. The cycle is defined at approximately τ = 3.5 seconds. For the neutral pair the terms cancel each other out at the end of the cycle or [W˙ ]ave = [To S˙i ]ave . For the generator case then [W˙ ]ave > [To S˙i ]ave and for the dissipator case then [W˙ ]ave < [To S˙i ]ave , respectively. Eventually, given enough cycles both the generator and dissipator cases will converge to the neutral case. As an interesting analogy to a Proportional-Integral-Derivative (PID) control system, replace the power generator term with an integral term as 2\n\nW˙ = µx˙ = −KI\n\n\u0014Z\n\nt\n\n\u0015 xdτ x. ˙\n\n0\n\nThe equivalence of the integral term as a power generator is analyzed and proved in reference . Since the negative damping term is nonlinear, the dynamic response to initial conditions and resulting limit cycle will be slightly different due to the build-up of the integrator. The same three test cases used in the previous van der Pol analysis were used with integral action with the numerical values given in Table 2. Both the Hamiltonian 3D surface (left) with the projection onto the phase plane (right) are shown in Fig. 6. It is interesting to note for the neutral case, that the integral action first dissipates below the limit cycle boundary before it begins to build back up and eventually end up on the neutral boundary. Again Cases 2 and 3 are the generative and dissipative cases. Notice that the trajectories are constrained to move along the Hamiltonian surface. By analogy, the Hamiltonian surface enables a “lifestyle” defined by “population” (mass), “in20\n\nFigure 5. Van der Pol exergy-rate (left) and exergy (right) responses - neutral case\n\nTable 2. Van der Pol model with integral action numerical values\n\nCase\n\nxo (m)\n\nx˙o (m/s)\n\nµ (kg/s)\n\nm (kg)\n\nk (kg/s2 )\n\nKI (kg/s)\n\ngenerate neutral dissipate\n\n0.1 1.0 2.0\n\n−0.1 −1.0 −2.0\n\n1.5 1.5 1.5\n\n1.0 1.0 1.0\n\n1.0 1.0 1.0\n\n1.02 1.02 1.02\n\nvestment/infrastructure” (stiffness), “production” (exergy input), and “consumption” (irreversible entropy production). The production and consumption are analogous to supply and demand that are enabled by an infrastructure which supports a population. These concepts will be used in the next section to discuss the sustainability of a lifestyle.\n\n6\n\nIn this section self-organizing system concepts are discussed that will be used to analyze the sustainability of a simplified nonlinear system model which represents a satellite in space, for example, the earth. The basic format from Heylighen will be followed with support from Haken and Buenstorf . 21\n\nFigure 6. Van der Pol with integral action responses: Hamiltonian 3D surface (left) and phase plane 2D projection (right)\n\n6.1\n\nBackground\n\nThe Achilles heel or single point of failure of self-organizing systems is the requirement that exergy continuously flow into the system. The self-organizing system is continuously “shedding” entropy to the environment to keep itself organized and living as it consumes or dissipates the exergy flow. Schr¨odinger (1945) suggested that all organisms need to import “negative entropy” from their environment and export high entropy (for example, heat) into their environment in order to survive. This idea was developed into a general thermodynamic concept by Prigogine and his co-workers who coined the notion of “dissipative structures” (Prigogine, 1976; Prigogine and Stengers, 1984), structures of increasing complexity developed by open systems on the basis of energy exchanges with the environment. In the self-organization of dissipative structures, the environment serves both as a source of low-entropic energy and as a sink for the high-entropic energy which is necessarily produced . Basically, self-organizing systems are attempting to balance and perform dialectic synthesis on evolving disordering and ordering pressures . Said another way, life is exergy dissipation (increasing entropy; disorder) and order production in an open system simultaneously. This process which is the evolution of a complex adaptive system is irreversible: the future is fundamentally different from the past, and it is impossible to reconstruct the past from the present . Dissipation is the disordering power flow which is better known as consumption in economics and irreversible entropy production in thermodynamics. Exergy flow into a system 22\n\nis the ordering power flow that is better known as production in economics and exergy rate into an open system in thermodynamics. Balance between these competing power flows is key because these terms are relative to a goal and path through time which means they can “flip over” or reverse roles. For example, the exergy flow into a system by a nuclear weapon is not “matched.” It deposits exergy at a rate that destroys the system, which means it is a disordering power flow increasing entropy. So, a mechanism must be inserted to “match” the input to the system if the goal is sustainability instead of destruction. Nuclear power is an attempt to match the exergy source to the exergy sink to move toward exergy sustainability. The balance between these opposing power flows creates a sort of “equilibrium condition” for a self-organizing system. Ilya Prigogine described this as “far from thermodynamic equilibrium on the basis of energy dissipation” and, in cybernetics, it’s often called an attractor . Most nonlinear self-organizing systems have several attractors and the system moves between these attractors (reordering) due to variations (perturbations; noise; disorder) in the exergy flow and the system parameters. These system parameters are often called “control parameters” because their values determine the stability characteristics of the system. For example, the potential force field [11, 14] for a nonlinear spring system can be written in kinematic form (no dynamics) as q˙ = kq + kNL q3 where k is the linear stiffness coefficient and kNL is the nonlinear stiffness coefficient. The potential function is defined as 1 2\n\n1 4\n\nV (q) = kq2 + kNL q4 .\n\n(20)\n\nThis system changes its fundamental stability structure by changing k > 0 to k < 0 and kNL > 0. Figure 7 shows how the stable equilibrium state at q = 0 bifurcates into two symmetrical stable equilibrium states and becomes an unstable state. These attractors are defined relative to a “fitness index.” Some attractors are more likely to survive, more fit, than others. In the previous example, one attractor turned into two attractors which appear to be equally fit if the potential function is interpreted as the fitness surface. In fact, it is possible for the system to jump back and forth between these two attractors by varying the exergy flow and the control parameter through perturbations and noise. As described earlier, nonlinear systems have several attractors and variations or “fluctuations” that reside between attractors in a system that will push the system to one or the other of the attractors. Positive feedback is necessary for random fluctuations to be amplified (generative) [12, 15]. Maintenance of the structured state in the presence of further fluctuations implies that some negative feedback is also present that dampens (dissipates) 23\n\nFigure 7. Nonlinear spring potential function characteristics\n\nthese effects [12, 15]. In Section 4, these are called a power generator and a power dissipator. Self-organization results from the interplay of positive and negative feedback . In Section 5, this is defined as the stability boundary and/or limit cycle. In more complex self-organizing systems, there will be several interlocking positive and negative feedback loops, so that changes in some directions are amplified while changes in other directions are suppressed [3, 12]. At the transition between order and disorder, a large number of bifurcations may be in existance which are analogous to the bifurcations of the previous potential function. Bifurcations may be arranged in a “cascade” where each branch of the fork itself bifurcates further and further, characteristic of the onset of the chaotic regime [3, 5, 12]. The system’s behavior on this edge is typically governed by a “power law” where large adjustments are possible, but are much less probable than small adjustments . These concepts enable us to better understand nonlinear systems, also known as complex adaptive systems, that are on the “edge of chaos” or those systems that are in a domain between frozen constancy (equilibrium) and turbulent, chaotic activity . The mechanism by which complex systems tend to maintain on this critical edge has also been described as self-organized criticality [3, 16]. This concept of “on the critical edge” can be described as “Yin and Yang control”: the dialectic synthesis of opposing power flows. Yin and Yang theory is a logic that is described as synthetic or dialectical: a part of a system can be understood only in its relation to the whole. There are five principles of Yin and Yang :\n\n1. All things have two aspects: a Yin aspect (decrease) and a Yang aspect (increase). 24\n\n2. Any Yin and Yang aspect can be further divided into Yin and Yang. 3. Yin and Yang mutually create each other. 4. Yin and Yang control each other. 5. Yin and Yang transform into each other. To specifically address these five characteristics with respect to the present concepts, the two opposing aspects are exergy generation and dissipation. The further division is the control volume analysis at any scale. The mutual creation is that the definition of generation is relative to dissipation. The control of each other is integral to the stability analysis. The transformation of one into the other was described in the nuclear weapon example.\n\n6.2\n\nSimple Nonlinear Satellite System\n\nWith this brief background, it’s time to analyze a simplified nonlinear satellite system (see Fig. 8) to develop the definition of exergy sustainability. For purposes of clarity, each control volume is subdivided into two subregions that contain the physical components. The component mass is constant. A single constant temperature characterizes each component. The component subregion is surrounded by an outer zone that characterizes the interaction between the component (at temperature T ) and the environment (reservoir) characterized by temperature To .\n\nFigure 8. Simplified nonlinear satellite model\n\nConservation Equations for the Engine Component (Control Volume 1): 25\n\nBy performing a control volume analysis the following energy, entropy, and exergy equations result:\n\nE˙ 1component = [\u0014Q˙ in1 − Q˙ out\u00151 ] + m˙ 1 [hint (Tint , Pint ) − hexh (Texh , Pexh )] − W˙ Q˙ in1 −Q˙ out1 + m˙ 1 [sint (Tint , Pint ) − sexh (Texh , Pexh )] + S˙ irr1component S˙ 1component = T1 Ξ˙ 1component\n\n=\n\n(21) h i \u0011 \u0010 Ξ˙ in1 − 1 − TTo1 Q˙ out1 + m˙ 1 [ζint (Tint , Pint ) − ζexh (Texh , Pexh )] − W˙ − To S˙ irr . 1component\n\nNote that the subscript “exh” (meaning exhaust) implies that the exiting quantities are associated with mass leaving the control volume. Similarly, the subscript “int” (for intake) implies the entering quantities are associated with the mass entering the control volume. The mass of the control volume does not change with time. Therefore, the mass flow rate exiting the control volume is equal to the mass flow rate entering the volume. For the surface heat interaction(s), the energy, entropy, and exergy equations are:\n\nE˙ 1interact Q = 0 = [Q˙ out1h− Q˙ out1 ]i S˙ 1interact Q = 0 = Q˙ out1 T11 − T1o + S˙irr1interact Ξ˙ 1interact\n\nQ\n\n=0 =\n\nh\u0010\n\n1−\n\n\u0011\n\nTo T\u0011 1\n\n\u0010 \u0011Q i T o Q˙ out1 − 1 − To Q˙ out1 − To S˙ irr1interact\n\n\u0010 = 1 − TTo1 Q˙ out1 − To S˙ irr1interact\n\nQ\n\n(22) Q\n\n.\n\nFor the exhaust stream expansion and interaction(s), the thermodynamics of component mixing will be ignored and the focus will be solely upon the final temperature of the exhaust gases. The thermodynamics of the mixing of the exhaust gases with the environment will be ignored once the gases cool to the ambient temperature. This gives:\n\nE˙ 1interac S˙ 1interact Ξ˙ 1interact\n\nm m m\n\n= 0 = m˙ 1 [hexh (Texh , Pexh ) − hexh (To , Po )] − Q˙ 1interact m Q˙ = 0 = m˙ 1 [sexh (Texh , Pexh ) − sexh (To , Po )] + S˙ irr1interact m − 1interact To = 0 = m˙ 1 [ζexh (Texh , Pexh ) − ζexh (To , Po )] − To S˙ irr .\n\nm\n\n(23)\n\n1interact m\n\nAdding the interaction fluxes (22) and (23) to the component fluxes (21) produces the 26\n\ncomplete equations for Control Volume 1 as\n\nE˙ 1total = [Q˙ in1 − Q˙ out1 ] + m˙ 1 [hint (Tint , Pint ) − hext (Texh , Pexh )] − W˙ ˙ + \u0014 m˙ 1 [eexh (T \u0015 exh , Pexh ) − eexh (To , Po )] − Q1interact m S˙ 1total\n\nΞ˙ 1total\n\nQ˙ in1 −Q˙ out1\n\n+ m˙ 1 [sint (Tint , Pint ) − sexh (Texh , Pexh )] + S˙ irr1component i h + Q˙ out1 T11 − T1o + S˙ irr1interact Q + m˙ 1 [sexh (Text , Pexh ) − sexh (To , Po )] ˙ + i h Sirr1interact \u0010 m \u0011 = Ξ˙ in1 − 1 − TTo1 Q˙ out1 + m˙ 1 [ζint (Tint , Pint ) − ζexh (Texh , Pexh )] \u0010 \u0011 To ˙ ˙ ˙ −W − To Sirr1component + 1 − T1 Qout1 − To S˙irr1interact Q +m˙ 1 [ζexh (Texh , Pexh ) − ζexh (To , Po )] − To S˙ irr . =\n\nT1\n\n1interact m\n\nNext simplifying yields:\n\nE˙ 1total =\n\nQ˙ in1 − Q˙ out1 − Q˙ 1interact \u0015 \u0014˙ Qin1 Q˙ out1 \u0002\n\nS˙ 1total =\n\n− T1 To h + S˙ irr\n\n1component\n\n\u0003 m\n\n+ m˙ 1 [hint (Tint , Pint ) − hexh (To , Po )] − W˙\n\n+ m˙ 1 [sint (Tint , Pint ) − sexh (To , Po )] i + S˙ irr + S˙ irr 1interact Q\n\n1interact m\n\nΞ˙ 1total = Ξ˙ in1 + m˙ 1 [ζint (Tint , Pint ) − ζexh (To , Po )] − W˙ i h − To S˙ irr1component + S˙irr1interact Q + S˙ irr1interact m\n\n(24)\n\n(25)\n\n(26)\n\nwhere\n\nQ˙ 1interact m = m˙ 1 [eexh h (Texh , P iexh ) − eexh (To , Po )] 1 1 S˙ irr1interact Q = Q˙ out1 To − T1 S˙ irr1interact\n\nm\n\n=\n\nQ˙ 1interact To\n\nm\n\n− m˙ 1 [sexh (Texh , Pexh ) − sexh (To , Po )] .\n\nIn the final energy equation (24) for Control Volume 1, the heat loss from the “engine” (Qout1 ) is differentiated from the heat loss from the exhaust stream (Q1interact m ). In the final entropy equation (25), the total entropy generation is the sum of the entropy generation that occurs within the “engine” (Sirr1component ), that which occurs due to the heat interactions between the engine and the environment (Sirr1interact m ), and that which occurs due to the exhaust stream coming into equilibrium with the environment (Sirr1interact Q ). Now look at the final exergy equation (26). The first term is the incoming exergy stream via a heat interaction (Ξin1 ). An example of this is the exergy flux of incident solar radiation on a solar collector. The second term is the product of the mass flux and the difference between 27\n\nthe specific exergies of the exhaust and intake streams. This might be the intake and exhaust streams of a gas turbine. The third term is the work produced by the engine. The fourth term is the exergy waste due to the inefficiencies of the engine and it’s interaction with the environment. The maximum amount of work obtainable from an exergy supply (first two terms) occurs when the irreversible entropy generation is reduced to zero. This limit is an idealization that can never be realized in the “real” world. However, improvements in efficiency are attainable by reducing the irreversibilities. The itemization of the sources of irreversibility show three paths for this reduction.\n\n1. The thermodynamic efficiency of the “engine” can be improved. For example, increasing the maximum temperature that turbine blades can tolerate will result in a more efficient turbine cycle.\n\n2. The heat interaction with the environment can be reduced. This is why high temperature refrigerators are insulated.\n\n3. The exergy of the exhaust stream can be reduced. A bottoming cycle can be added to gas turbines which produces work from the waste stream.\n\nThere are many technologies already in existence that can be used to increase the production of work from existing “engine” technologies. However, many of those technologies are not (currently) economically competitive. Conservation Equations for the Machine (Control Volume 2): Control Volume 2 consists of a nonlinear mass/spring/damper system (with Duffing oscillator/Coulomb friction contributing to the nonlinear effects). Work is supplied to the system which results in the acceleration of the mass. The work is dissipated by the damper. This increases the temperature of the components of the system, which is characterized by a single temperature, T2 . The thermal energy is then transferred to the environment, which has a temperature of T0 . The transfer is realized solely by a heat interaction. No mass enters or leaves Control Volume 2. Performing the energy, entropy, and exergy analyses yields:\n\nE˙ 2component = W˙ − Q˙ 2out Q˙ S˙ 2component = S˙ irr2 − Tout2 2 \u0010 \u0011 ˙Ξ2component = W˙ − To S˙ irr2 − 1 − To Q˙ out2 T2 28\n\nFor the surface heat interaction(s) the energy, entropy, and exergy equations are:\n\nE˙ 2interact Q = 0 = [Q˙ out2h− Q˙ out2 ]i S˙ 2interact Q = 0 = Q˙ out2 T12 − T1o + S˙ irr2interact Ξ˙ 2interact\n\nQ\n\nh\u0010\n\n\u0011\n\nTo T\u0011 2\n\n\u0010 \u0011Q i To ˙ ˙ Qout2 − 1 − To Qout1 − To S˙ irr2interact\n\n1− \u0010 = 1 − TT2o Q˙ out2 − To S˙ irr2interact\n\n=0 =\n\nQ\n\nQ\n\n.\n\nAdding the interaction fluxes to the component fluxes produces the complete equations for Control Volume 2 as:\n\nE˙ 2total = W˙ − Q˙ out2 h i Q˙ S˙ 2total = S˙ irr2 − Tout2 2 + Q˙ out2 T12 − T1o + S˙ irr2interact\n\n\u0010 \u0011 \u0010 \u0011Q To ˙ To ˙ ˙ ˙Ξ2 ˙ = W − To Sirr2 − 1 − T2 Qout2 + 1 − T2 Qout2 − To S˙ irr2interact total\n\nQ\n\nand simplifying yields:\n\nE˙ 2total = W˙ − Q˙ out2h i Q˙ S˙ 2total = − Touto 2 + S˙ irr2 + S˙ irr2interact Q h Ξ˙ 2total = W˙ − To S˙ irr2 + S˙ irr2interact\n\ni\n\nQ\n\nwhere\n\nS˙ irr2interact\n\nQ\n\n= Q˙ out2\n\n\u0014\n\n\u0015 1 1 − . T0 T2\n\nConservation Equations for the Total System (Control Volume 12): Now the equations for Control Volumes 1 and 2 are added to give:\n\nE˙ total = S˙total = Ξ˙ total\n\nQ˙ in1 − Q˙ out1 − Q˙ 1interact + W˙ − Q˙ out2 \u0015 \u0014 \u0002\n\nQ˙ in1\n\n\u0003 m\n\n+ m˙ 1 [hint (Tint , Pint ) − hexh (To , Po )] − W˙\n\nQ˙ out1\n\n+ m˙ 1 [sint (Tint , Pint ) − sexh (To , Po )] − Touto 2 h i h + S˙ irr1component + S˙ irr1interact Q + S˙ irr1interact m + S˙ irr2 + S˙ irr2interact ˙ = Ξ˙ in1 + h m˙ 1 [ζint (Tint , Pint ) − ζexh (To , Po )] − W i −To S˙ irr1component + S˙ irr1interact Q + S˙ irr1interact m + W˙ i h ˙ ˙ −To Sirr2 + Sirr2interact Q T1\n\nTo\n\n29\n\ni Q\n\nand simplifying yields:\n\nE˙ total = \u0014Q˙ in1 − Q˙ out1 − Q˙ 1\u0015interact m − Q˙ out2 + m˙ 1 [hint (Tint , Pint ) − hexh (To , Po )] Q˙ in1 Q˙ out1 +Q˙ out2 − + m˙ 1 [sint (Tint , Pint ) − sexh (To , Po )] S˙total = T1 To \u0002\n\nΞ˙ total\n\n\u0003\n\nh + S˙ irr1component + S˙ irr1interact Q + S˙ irr1interact = Ξ˙ in1 + h m˙ 1 [ζint (Tint , Pint ) − ζexh (To , Po )] + S˙ irr −To S˙ irr + S˙ irr 1component\n\n1interact Q\n\n+ S˙ irr2 + S˙ irr2interact m\n\n1interact\n\ni Q\n\n+ S˙ irr2 + S˙ irr2interact m\n\ni Q\n\nwhere\n\nQ˙ 1interact m = m˙ 1 [eexh iexh ) − eexh (To , Po )] h (Texh , P 1 1 S˙ irr1interact Q = Q˙ out1 To − T1 S˙ irr1interact S˙ irr2interact\n\nm\n\nQ\n\n=\n\nQ˙ 1interact Toh\n\n= Q˙ out2\n\n− m˙ 1 [sexh (Texh , Pexh ) − sexh (To , Po )] i 1 1 − To T2 . m\n\nBy following the derivations of the previous sections, this simplified nonlinear model reduces within Control Volume 2 to V = H V˙ = H˙\n\n= = = = =\n\n1 mx˙2 + 12 kx2 + 14 kNL x4 \u00022 \u0003 mx¨ + kx + kNL x3 x˙ W˙ − To S˙ i ux\u0002˙ −Cx˙2 −CNL sgn(x) ˙ x˙ \u0003 Rt x˙ −KP x − KI 0 xdτ − KD x˙ −Cx˙ −CNL sgn(x) ˙\n\nwhere u = PID feedback controller (Implemented force input) R ˙ = −KP x − KI 0t xdτ − KD x. Then the following exergy terms are identified as \u0002 \u0003 R W˙ = −KI 0t xdτ x˙ To S˙ i = [−K ˙\u0003 x˙ \u0002 D x˙ −Cx˙ −CNL sgn(x)] To S˙ rev = mx¨ + kx + KP x + kNL x3 x. ˙ The generalized stability boundary is given as a balance between “positive and negative feedback” (exergy generation and exergy dissipation) ˙ ˙ = [T \u0002[W ]aveR t \u0003 \u0002 o Si ]ave 2 \u0003 −KI 0 xdτ · x˙ ave = (KD +C)x˙ +CNL sgn(x) ˙ · x˙ ave . 30\n\n(27)\n\nThe “shape” of the resulting limit cycle is constrained to the Hamiltonian surface which determines the accessible bifurcated structure as a function of exergy level (see Figs. 7 and 10).\n\n6.3\n\nLifestyle Definition\n\nNext this model is interpreted in terms of a “lifestyle” of the mass-spring-damper system within the satellite. First, the lifestyle is defined by a cyclic path, attractor, or limit cycle in the phase plane that is constrained to the Hamiltonian surface H = V , (left) projected onto the phase plane (right) in Fig. 9. This path occurs as a result of satisfying (27). This interpretation directly provides the definition of exergy sustainability: the continuous compensation of irreversible entropy production in an open system with an impedance and capacity-matched persistent exergy source. In other words, the cyclic lifestyle will persist indefinitely as long as (27) is satisfied and m, k, KP , and kNL are constants.\n\nFigure 9. All cases: mass-spring-damper with Duffing oscillator/Coulomb friction model numerical results: Hamiltonian 3D surface (left) and phase plane 2D projection (right)\n\nSecond, this lifestyle will change if the “population” (mass), “investment/infrastructure” (stiffness), “production” (exergy input), and/or “consumption” (irreversible entropy production) are changed independently because all of these parameters are interconnected through the system. Since the lifestyle path over time is constrained to the Hamiltonian surface, if the population is changed then the stiffness can be changed to hold the lifestyle constant while holding production and consumption constant. The infrastructure is expanded to accommodate the increasing population. On the other hand, if the population increases then the consumption increases which means the production must increase as well. In addition, if the lifestyle increases by creating more services (more exergy consumption) 31\n\nthen the production must increase to offset consumption which is enabled through an expanded infrastructure. Bottom-line: lifestyle is directly related to exergy consumption and how exergy sources are matched to exergy sinks through the infrastructure. Third, this simplified lifestyle presents the mass-spring-damper system as an exergy parasite on the satellite with respect to the sun. By digging a little deeper into the model, the parasites become humans on the earth with respect to the sun. The goal of exergy sustainability now includes striving to become a symbiotic versus a destructive parasite, such as the GAIA approach , because cleaning up the disordering effects of a destructive parasite will only consume more exergy through an impedance mismatch. In fact, the human race has done an impressive job of overcoming the impedance mismatch between population growth and the carrying capacity of the biosphere with fossil fuels; we effectively eat fossil fuel. One final observation on this topic of impedance and capacity matching, the goal of war and economic competition is to create a production/consumption rate which is sustainable for you and generates an impedance mismatch that is unsustainable for your enemy/competitor. The ultimate goal is to cut-off, destroy, and/or dissipate your competitor’s exergy reserves by changing/deforming your competitor’s Hamiltonian surface (infrastructure - including population of the work force). This can be accomplished in several ways including: 1) pick-up the pace by increasing the limit cycle frequency (i.e., less mass), 2) accelerate the exergy consumption of your competitor by using more efficient technologies, and 3) deform the potential field with information flow.\n\n6.4\n\nDeformation of Potential Field with Information Flow\n\nThe deformation of the potential field with information flow is the most seductive because it potentially requires the least amount of additional physical infrastructure. The INTERNET is the most obvious example. A direct application of this idea is by utilizing the techniques of reference in the present context. The team of robots in created a “virtual potential field” by flowing information through a distributed decentralized sensor and feedback control network. In the present context, the Hamiltonian of the ith robot is deformed by\n\nVi = Hi = Ti + Vci where\n\nTi = 12 mi x˙i2 Vci = Gi (xi ) − Goi = Gxi xi + 12 Gxxi xi2 . 32\n\nTherefore, the time derivative of the Lyapunov/Hamiltonian becomes \u0014 \u0015 \u0014 \u0015 ∂ V ∂ V c c i i ˙ V˙i = Hi = mi x¨i + x˙i = ui + x˙i ∂xi ∂xi where the estimator/guidance algorithm for finding the source/target is 1 Gi (xi ) = Goi + Gxi xi + Gxxi xi2 . 2 The feedback controller is \u0015 Z ∂Vci − KIi xi dτ − KDi x˙i ui = − ∂xi \u0014\n\nand the stability boundary becomes \u0014 \u0015 Z −KIi xi dτ · x˙i\n\n= [KDi x˙i · x˙i ]ave\n\nave\n\nwhich determines the limit cycle behavior constrained to the deformed Hamiltonian surface. The collective performance is V = ∑ ρiVi = i\n\n1 ρi mi x˙i2 + ∑ ρi Vci 2∑ i i\n\nwith ∂V V˙ = ∑i ρiV˙i = ∑i ρi mi x¨i x˙i + ∑i ρi ∂xci i x˙i h i h i ∂V ∂V = ∑i ρi mi x¨i + ∂xci i x˙i = ∑i ρi ui + ∂xci i x˙i\n\nwhere the collective Hamiltonian can be deformed in order to enhance your exergy usage or manipulate your competitor’s exergy usage. A simple example is to manipulate (20) with proportional feedback to reverse the bifurcation of k < 0. For V = H = T + V + Vc then\n\nV + Vc = − 21 kx2 + 14 kNL x4 + 12 KP x2 =\n\n1 1 2 4 2 [KP − k]x + 4 kNL x\n\n33\n\nand KP ≥ k. To determine the effect that the proportional controller gain KP has on the system, Hamiltonian phase plane plots are generated. By investigating a system with negative stiffness and by adding enough KP to result in an overall positive net stiffness, changes the shape of the Hamiltonian surface from a saddle point surface (see Fig. 10) to a positive bowl surface (see Fig. 11). A two-dimensional cross-section of the Hamiltonian versus the position shows the characteristics of the overall storage or potential functions. The operating point at (H, x, ˙ x) = (0, 0, 0) changes from being unstable to stable, for small values of |x| > 0, when enough additional KP is added, a net positive stiffness for the system results.\n\nFigure 10. Three dimensional (left) Hamiltonian phase plane plot negative stiffness produces a saddle surface. The two-dimensional cross-section plot (right) is at x˙ = 0.\n\n34\n\nFigure 11. Three dimensional (left) Hamiltonian phase plane plot where the net positive stiffness produces a positive bowl surface. The two-dimensional cross-section plot (right) is at x˙ = 0.\n\n7\n\nSummary and Conclusions\n\nThis paper has developed a new definition of exergy sustainability: the continuous compensation of irreversible entropy production in an open system with an impedance and capacity-matched persistent exergy source. The development of this definition has lead to a missing link in the analysis of self-organizing systems: a tie between irreversible thermodynamics and Hamiltonian systems. This tie was exploited through nonlinear control theory to define necessary and sufficient conditions for stability of nonlinear systems that were employed to formulate the concept of “on the edge of chaos.” Finally, an equivalence was developed between physical stored exergy and information-based exergy which can be exploited to change a lifestyle and enhance one’s economic competitiveness or performance on the battlefield.\n\n35\n\n36\n\nReferences R.U. Ayres, Eco-thermodynamics: Economics and the Second Law, Ecological Economics, 26, 1998, pp. 189-209. R.D. Robinett, III and D.G. Wilson, Exergy and Irreversible Entropy Production Thermodynamic Concepts for Control System Design: Slewing Single Axis, submitted to the AIAA, Journal of Guidance, Control and Dynamics, Dec. 2005. F. Heylighen, The Science of Self-Organization and Adaptivity, Principia Cybernetic Web, http://perspmcl.vub.ac.be/ D.S. Scott, Links and Lies, International Journal of Hydrogen Energy, Vol. 28, 2003, pp. 473-476. D. Kondepudi and I. Prigogine, Modern Thermodynamics: From Heat Engines to Dissipative Structures, John Wiley & Sons, New York, N.Y., 1999. D.S. Scott, Exergy, International Journal of Hydrogen Energy, Vol. 28, 2003, pp. 369-375. L. Meirovitch, Methods of Analytical Dynamics, McGraw-Hill, New York, 1970. R.J. Smith, Circuits, Devices, and Systems: A First Course in Electrical Engineering, John Wiley & Sons, Third Edition, 1976. T.L. Saaty and J. Bram, Nonlinear Mathematics, McGraw-Hill, New York, 1964. B. van der Pol, Radio Rev. 1, 704-754, 1920 and B. van der Pol, Phil. Mag., 3, 65, 1927. H. Haken, Advanced Synergetics: Instability Hierarchies of Self-Organizing Systems and Devices, Springer-Verlag, New York, NY, 1983. G. Buenstorf, Self-Organization and Sustainability: Energetics of Evolution and Implications for Ecological Economics, Ecological Economics, Vol. 33, 2000, pp. 119134. A. Cooper and R.D. Robinett, III, Deriving Sustainable Ordered Surety by Overcoming Persistent Disorder Pressures, submitted to the Journal of Systems Safety, August 2005. 37\n\n H. Haken, Synergetics: An Introduction, 3rd Edition, Springer-Verlag, New York, NY, 1983. P.M. Allen, Cities and regions a Self-Organizing Systems, Models of Complexity, Gordon and Breach, Amsterdam, 1997. P. Bak, How Nature Works: The Science of Self-Organized Criticality, Springer, Berlin, 1996. T.J. Kaptchuk, The Web That Has No Weaver: Understanding Chinese Medicine, Congdon and Weed, 1983. E.P. Gyftopoulos and G. P. Beretta, Thermodynamics, Foundations and Applications, Macmillan Publishing Company, New York, 1991. J. Lovelock and L. Margulis, www.mountainman.com.au/gaia.html.\n\nThe\n\nGAIA\n\nHypothesis,\n\n R.D. Robinett, III and J.E. Hurtado, Stability and Control of Collective Systems, Journal of Intelligent and Robotic Systems, 2005. T.K. Locke, Guide to Preparing (SAND) Reports, Sandia National Laboratories, SAND98-0730, May 1998.\n\n38\n\nDISTRIBUTION: 1 MS 1202 Timothy M. Berg, 05624\n\n1 MS 0741 Marjorie L. Tatro, 06200\n\n1 MS 1003 John T. Feddema, 06634\n\n1 MS 1003 David G. Wilson, 06634 2 MS 9018 Central Technical Files, 8945-1\n\n1 MS 1002 Philip D. Heermann, 06630\n\n2 MS 0899 Technical Library, 4536\n\n1 MS 0776 Alfred W. Reed, 06852\n\n1 MS 0161 Legal Intellectual Property, 11500\n\n10 MS 0741 Rush D. Robinett, III, 06210\n\n39" ]

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[ "# Topics by tag «a polynomial»\n\nAll topics from mathematics on Cubens by tag «a polynomial».\n\n• Numbers and expressions\n\nPolynomials of one variable. The polynomial n-th degree. Identically equal polynomials in one variable. Division of polynomial by polynomial. The division of the polynomial by the polynomial s Stacey. Division of polynomial by polynomial area. Bézout's Theorem." ]

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[ "# Uncertainty Wednesday: Variance\n\nIn last week’s Uncertainty Wednesday, I introduced the expected value EV of a random variable X. We saw that EV(X) is not a measure of uncertainty. The hypothetical investments I had described all had the same expected value of 0. It is trivial, given a random variable with EV(X) = μ  to construct X’ so that EV(X’) = 0. That’s in fact how I constructed the first investment. I started with \\$0 with 99% probability and \\$100 with 1% probability, which has an EV of\n\nEV = 0.99 * 0 + 0.01 * 100 = 1\n\nand then I simply subtracted 1 from each possible outcome to get -\\$1 with 99% probability and \\$99 with 1% probability.\n\nWhat we are looking for instead in order to measure uncertainty, is a number that captures how spread out the values are around the expected value. The obvious approach this would be to form the weighted sum of the distances from the expected value as follows:\n\nAAD(X) = ∑ P(X = x) * |x – EV(X)|\n\nwhere | | denotes absolute value (meaning the magnitude without the sign). This metric is known as the Average Absolute Deviation (btw, instead of the shorthand P(x) I am now writing P(X = x) to show more clearly that it is the probability of the random variable X taking on the value x).\n\nAAD is a one measure of dispersion around the expected value but it is not the most commonly used one. That instead is what is known as Variance, which is defined as follows\n\nVAR(X) = sum Prob(X = x) * (x – EV(X))^2\n\nOr expressed in words: the probability weighted sum of the squared distances of possible outcomes from the expected value. It turns out that for a variety of reasons using the square instead of the absolute value has some useful properties and also interesting physical interpretations (we may get to those at some later point).\n\nLet’s take a look at both of these metrics for the random variables from our investment examples\n\nVariance\n\nInvestment 1: 0.99 * (-1 – 0)^2 + 0.01 * (99 – 0)^2 = 0.99 * 1 + 0.01 * 9,801 = 99\n\nInvestment 2: 0.99 * (-100 – 0)^2 + 0.01 * (9,900 – 0)^2 = 0.99 * 10,000 + 0.01 * 98,010,000 = 990,000\n\nInvestment 3: 0.99 * (-10,000 – 0)^2 + 0.01 * (990,000 – 0)^2 = 0.99 * 100,000,000 + 0.01 * 980,100,000,000 = 9,900,000,000\n\nAverage Absolute Deviation\n\nInvestment 1: 0.99 * |-1 – 0| + 0.01 * |99 – 0| = 0.99 * 1 + 0.01 * 99 = 1.98\n\nInvestment 2: 0.99 * |-100 – 0| + 0.01 * |9,900 – 0| = 0.99 * 100 + 0.01 * 9,900 = 198\n\nInvestment 3: 0.99 * |-10,000 – 0| + 0.01 * |990,000 – 0| = 0.99 * 10,000 + 0.01 * 990,000 = 19,800\n\nYou might have previously noticed that Investment 2 is simply Investment 1 scaled by a factor of 100 (and ditto Investment 3 is 100x Investment 2). We see that AAD, as per its definition, follows that same linear scaling whereas variance grows in the square, meaning the variance of Investment 2 is 100^2 = 10,000x the variance of Investment 1.\n\nBoth of these are measures that pick up the values of the random variable as separate from the structure of the underlying probabilities. If that doesn’t make sense to you, go back and read the initial post about measuring uncertainty and then go back to the posts about entropy. The three hypothetical investments each have the same entropy as they share the same probabilities. But AAD and Variance pick up the difference in payouts between the investments." ]

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[ "# Public health in Lincolnshire in the 19th Century—1\n\n## Public health in Lincolnshire in the 19th Century—1\n\nPUBLIC HEALTH Public Health in Lincolnshire in the 19th Century 1 By E. GILLETT, B.A., Staff\" Tutor, University of Hull, and Archivist to the Count...\n\nPUBLIC\n\nHEALTH\n\nPublic Health in Lincolnshire in the 19th Century 1 By E. GILLETT, B.A.,\n\nStaff\" Tutor, University of Hull, and Archivist to the County Borough of Grimsby, and J. D. HUGHES, M.A.,\n\nStaff Tutor, University of Sheffield of health set up in Louth, Lincoln, and Grimsby had no successors, among members of the medical profession and others there was growing interest in public health and the opinion that collective action in sanitary matters could limit the spread of epidemic diseases. Yet although the occasional catastrophic visitation might arouse public opinion for a time, the framework of the poor law made such opinion ineffective and encouraged all but the few to fall back into the old ways of parochialism, and the petty politics of farmers. Improvement faced the double obstacle of the practice and spirit of the poor law on the one hand, and the legacy of the past--ignorance, superstition, and even o p i u m - on the other.\n\nr\n\nhistory of public health in 19th Century England has usually been written in terms of the \" great wen and of the insanitary state and high death rates of the rapidly growing industrial towns. The rural areas, on the other hand, have more generally escaped the attention of the historian of public health, and the tendency has sometimes been to mention them as a contrast to the morbidity of the towns I. But even cursory reading in Blue Books of the ,evidence of witnesses from the rural counties is enough to suggest that the villages had problems of public health which, however much they might differ in scale from those of the larger towns, were very similar in character. Indeed, in some respects, particularly in the control of working conditions the rural areas--which meant agriculture and small trades-lagged a generation behind the towns. Moreover, in some areas special features aggravated the public health problem, for instance in the marsh areas the combined problems of drainage and water supply, opium addiction and women's field labour. These special problems were particularly to be found in Lincolnshire.\n\nI\n\nP a r i s h A d m i n i s t r a t i o n a n d the L a w o f S e t t l e m e n t\n\nAt the opening of the 19th century the machinery for t h e regulation of public health in Lincolnshire differed hardly at all from that which existed at the end of the Middle Ages. Here, as in so many other matters, the parish was the unit of administration. If the parish officers were active, and if the parishioners saw fit to make bye-laws relating to public health, something might be done. Scotter, for instance, as early as 1586 was attempting to regulate dunghills and the burial of dead cattle. Grimsby throughout the 16th century had many and detailed regulations concerning unwholesome victuals, the cleansing of watercourses and open sewers, and the disposal of butchers' and fishmongers' offal. A more detailed examination of court-leet verdicts would no doubt provide many parallel instances. But at the best of times there was much laxity in the enforcement of by-laws. At the worst, the by-laws might say nothing at all about public health. At the opening of the era of modern public health legislation we have, in Lincolnshire, little more than the concept that certain situations might constitute a \" nuisance,\" and that if the \" nuisance\" became intolerable the parish might proceed against those responsible.\n\nA Long History of Disease\n\nThis article is an attempt to show public health problems a n d development in a county which contained no very large towns or industries, where the villages and small market towns set the pattern. Because of its geographical situation Lincolnshire had a long history of endemic disease. Bede's account of a miracle of St. Oswald at Bardney makes it clear that malaria existed in the fenland of his day. After 1800 the remaining marshes, such as the East and West Fens to the north of Boston, disappeared rapidly in face of improved drainage and the use of the steam engine for pumping, and mfilaria declined. But as the fens disappeared, the growth of steam navigation and later the railways increased the danger of infection being brought in from a distance. Thus in 1831 the most strenuous attempts Were made to quarantine all vessels bound for Grimsby and Boston from ports where cholera existed ; but cholera came just the same, from Hull, Doneaster, or Sunderland, it would seem. From this time we might date the beginning of an awakening in public health administration. For although the temporary local boards\n\nTo the inadequacies of parochial administration in tackling public health must be added the deadening effect of the whole character and administration of the poor law. In this respect it is hardly necessary to distinguish between the old and the new po~r law. True, the latter represented the first step in reaching beyond the inadequate unit of the parish. But its entire principle and practice were deterrent, and concerned with the relief not the prevention of poverty. The guardians, in the rural areas almost always farmers, were interested above all in rate economies. Unfortunately, since the Board of Guardians was the only effective local administrative body, it was given by successive Acts a series of public health functions, such as nuiiance removal and small-pox\n\nx E.g., J. L. & B. Hammond, Age of the Chartists, at the end of the chapter on \" The State of the Towns\" : ~' One fact must always be borne in mind in any attempt to picture the lives of the new industrial populations ; they were in the main country folk . . with their roots in fields and not in streets or court \" (p. 104). And earlier : \" The towns were the homes o f . . . men and women 9 . . accustomed to the peace and beauty of nature, shut up in slum and alley \" (p. 28). But the country-folk, although they may have worked in the fields, lived in streets, courts and alleys not unlike (as we shall see) those of the new towns. The Hammonds' point is very relevant when discussing the town-dweller's deprivation of \" playgrounds \" but is hardly true as a contrast in housing conditions. 34\n\nNovember 1955 s a n i t a r y gospel f o u n d converts. A t M e s s i n g h a m , a r o u n d 1820, A r c h d e a c o n Bayley f o u n d his flock as obstinate in the p r o t e c t i o n of t h e i r nuisances as in p u r s u i t of ancient village quarrels. T h e m o r e he attacked n u i s a n c e s the faster they left the established church. The A r c h d e a c o n was, for his time, s o m e t h i n g of a n exception in his interest in village cleanliness, b u t by 1870 we find t h e A r c h d e a c o n of L i n c o l n p r e a c h i n g a c a t h e d r a l s e r m o n which \" was a sanitary s e r m o n in every sense o f the w o r d ' 6 . W e h e a r of t h e Earl of H a r r o w b y b u i l d i n g a village sewer, of a l a n d o w n e r at Sudb r o o k e offering bricks for a sewer if the villagers would build it (they refused). B u t in t h e m a i n it was t h e physicians a n d surgeons, as we should expect, w h o s h o w e d the greatest c o n c e r n ' . T h e average of professional skill was p e r h a p s n o t very high. A t Spilsby a q u a c k w h o described himself as Thorn.as Allen, M.D., practised w i t h i m p u n i t y for several years, t h o u g h his medical t r a i n i n g was limited to his experience as a n e r r a n d boy at t h e Leicester infirmary. P e r h a p s the fact t h a t h e was eventually p r o s e c u t e d by the Society of A p o t h e c a r i e s shows t h a t s t a n d a r d s were rising. In an emergency, however, m a n y preferred to get a horse-doctor, or, if they believed t h a t they h a d b e e n b i t t e n by a m a d dog, to go to the h u n t s m a n at Brocklesby.\n\nv a c c i n a t i o n , w h i c h inevitably were u n w e l c o m e (since they might involve some expense) a n d neglected. This \" save the r a t e s \" a p p r o a c h to public health was carried over into the B o a r d s o f H e a l t h and, later, Local G o v e r n m e n t , quite a p a r t f r o m persisting a m o n g the B o a r d s o f G u a r d i a n s . This by itself would h a v e b e e n sufficient h a n d i c a p for r u r a l public health, b u t in a d d i t i o n the w o r k i n g of t h e law o f s e t t l e m e n t t e n d e d everywhere to lower the s t a n d a r d s of r u r a l housing. U n t i l 18652 relief was given only by t h e parish w h e r e a n individual h a s his \" s e t t l e m e n t , \" a n d the parish raised t h e rate to c o v e r the cost. A n y o n e w h o b e c a m e a p a u p e r in a parish o t h e r t h a n t h a t o f his settlement could be r e m o v e d to his place o f settlement, o r his p a r i s h o f settlement w o u l d p a y the bill for the relief given w i t h o u t the p a u p e r actually b e i n g r e m o v e d . B u t the law o f settlement was complex a n d u n c e r t a i n a n d ted to f r e q u e n t disputes at law between parishes in t h e a t t e m p t to p r o v e a settlement ~. Since a n y r u r a l l a b o u r e r was a p o t e n t i a l p a u p e r , w h e r e the l a n d o f a p a r i s h was in the h a n d s of one or a few l a n d o w n e r s they c o u l d reduce t h e p o o r rates by c r e a t i n g a \" c l o s e \" village, i.e., by p r e v e n t i n g a n y labourers living in the parish. This o f t e n i n v o l v e d the d e s t r u c t i o n o f cottages. T h e process received even m o r e impetus after 1846 when, as a sop to the l a n d l o r d s ' , a n A c t was passed p r e v e n t i n g t h e p a r i s h r e m o v i n g a n y p a u p e r to his p a r i s h of settlement w h o h a d b e e n r e s i d e n t for five years. This gave t h e l a n d l o r d even m o r e incentive to r e m o v e all l a b o u r e r s f r o m the parishes w h e r e h e held land, since the reduction of his p o o r rates as a c o n s e q u e n c e b e c a m e the m o r e certain. T h e labourers m o r e a n d m o r e were c r o w d e d into the cottages of t h e \" o p e n \" villages, which b a d h o u s i n g a n d over-crowding m a d e centres o f ill-health a n d disease. I t was a n artificially created public h e a l t h p r o b l e m which d o m i n a t e d r u r a l life in m a n y areas. T h u s the p o o r law was the b a n e of public health in the r u r a l areas b o t h f o r the a p p r o a c h to public h e a l t h a d m i n i s t r a t i o n t h a t it e n g e n d e r e d , a n d for t h e m a j o r p r o b l e m of r u r a l h o u s i n g c o n d i t i o n s which it h a d created.\n\nRising Prestige of Medicine Nevertheless, t h e effects o f medical progress were felt, a n d we m a y suppose t h a t the rising prestige of medicine gradually modified the prevailing attitude to sanitary problems. T h e earliest o p e r a t i o n u n d e r a n anaesthetic in t h e county a p p e a r s to h a v e b e e n p e r f o r m e d by a D r . N o r r i s w h o p e r f o r m e d several at Spalding in the second week o f January, 1847, assisted by a surgeon n a m e d Vise. His example was followed a t P a r t n e y b y Mr. Thimbleby, a Spilsby surgeon, o n F e b r u a r y 10th, 1847. I n M a r c h , however, a w o m a n died u n d e r a n e t h e r a n a e s t h e t i c at G r a n t h a m , a n d progress was r e t a r d e d for a while. B u t within a year a druggist at Laceby was using c h l o r o f o r m in killing a pig. T h e effectiveness of the medical profession, at any rate in t h e first h a l f o f the century, m a y b e assessed in the p o p u l a r a t t i t u d e to small-pox vaccination. I n this county, m o r e p e r h a p s t h a n most, the believers in the small-pox i n o c u l a t i o n f o u g h t a long battle. It is n o t yet k n o w n a t w h a t period in t h e 18th C e n t u r y the m e t h o d of i n o c u l a t i o n b e c a m e established in Lincolnshire ; b u t it is q u i t e clear t h a t it r e m a i n e d p o p u l a r long after the effectiveness o f v a c c i n a t i o n h a d b e e n demonstrated.\n\nThe Medical Profession P o p u l a r beliefs a b o u t hygiene were a m i x t u r e of sympathetic magic, witchcraft, a n d a leavening o f Wesley's househ o l d medicine 5, a n d it was a m o n g - t h e u p p e r classes t h a t the 2 The Union Chargeability Act of 1865 was the major reform, since by creating a uniform rate for the whole Union it removed the reason for clearing labourers from a parish to reduce the rate burden. a ,, By the variety of successive statutory restrictions introduced into the law of settlement and removal, that law, by 1834, had greatly increased in complexity and uncertainty. Speaking of settlements, in a comparatively simple state of the statute law, Dr. Burn had said, in 1764, ' It has been the work of an age to ascertain the law respecting them '.\" R. Pashley, Pauperism and PoorLaws, 1852, p. 266. This much neglected work is an admirable and lucid source for the study of the operation of the law of settlement. * Sir Robert Peel announced his intention of making this change in the law of settlement at the same time as he announced his intention of repealing the corn laws. Up till 1846 and the statute 9 & 10 Vlct. c.66, the manufacturing towns had been able to put the burden of relieving unemployment on the backs of the parishes in which their workers had settlement, which generally meant the rural parishes. The change in the law meant a heavy increase in town poor rates, at Norwich by s a year, at Leeds b y s But similarly in the rural areas themselves, the rate burden on the \" d o s e \" parishes was reduced, on the \" o p e n \" parishes - increased. Cf. R. Pashley, op. cit., pp. 274-287. 5 And, of course, patent medicines. A Gainsborough bookseller\n\nin 1783 advertised for sale, among other \"'genuine medicines,\" Daffy's Elixir, Godfrey's Cordial (these were opium mixtures), Ormskirk Medicine for the Bite of a Mad Dog, Pike's Ointment for the Itch, Dr. Ward's Emetic, or Sack Drop, and Dr. Ward's Liquid Sweat! Popular ideas of medicine were also applied to animals, of course. A Crowle yeoman farmer's diary for 1777 contained as recipe \" for a cow that has got a cold \"' : \" Take of anaseads ld. ; of fower of Brunston -~d. ; of Liquorice Powder ~-d. ; of Jinson 89 ; of Diepenta 89 ; half a pound of treakel, one handful of wormwood and a handful of Herbegress both boild in a quart of Ale.\" We cannot help feeling that this would have done something to the cow's cold. 6 For Messingham see Archdeacon Stonehouse, \" Visitation of the Archdeaconry of Stow, 1845.\" For the sermon, 2rid Report, Royal Sanitary Commission. Minutes of Evidence, 1871, llI, pp. 140-7. Evidence of W. J. Mantle. County poll books show that those who had votes (mostly surgeons) were often Tories, but more often they gave one vote to the Whig, the other usually to the Tory candidate, and only rarely to the more radical Whig. 35\n\nPUBLIC HEALTH In 1816 R. Bellingham, a surgeon at B o u r n e , c o m m e n t e d o n the dangers o f i n o c u l a t i o n : - \" T h e small-pox appeared in a single individual here early in the last spring. Before it was even generalry known that the smallpox existed in the town the source of contagion had been multiplied by inoculation among the poorer and least informed classes. The chief argument in favour of this wanton extension was that it was of a very mild and harmless sort, and that it was impossible to have it at a better time. Several died with all the advantages of preparation, inoculation and medical attendance; and a still greater number without these advantages. The intercourse with the circumjacent towns and villages soon propagated this mild and harmless disorder in each direction-8. As the epidemic spread, q u a c k s t h r o v e o n the d e m a n d for inoculation. T h e r e was o n e w h o h a d n e v e r seen a lancet until he set u p as a n inoculator. A year o r so later this class o f m e n were described as \" the fag ends o f the medical a n d surgical p r o f e s s i o n s - - a l s o shoe makers, tailors, weavers a n d even r a t c a t c h e r s ' 9 . This was in Kesteven. T h e worst c o n s e q u e n c e was t h a t \" by the frightful a n d false reports of these m e n the lower class of persons are entirely set against the c o w p o c k \" (i.e., vaccination). T h e d e m a n d for inoculation was, in fact, so strong t h a t some practitioners reverted to t h e m o r e p o p u l a r m e t h o d . T y s o n West, a n A l f o r d Surgeon, r e p u d i a t e d v a c c i n a t i o n a n d r e c o m m e n d e d inoculation 10. A t Horncastle, J o h n W a r d , also a s u r g e o n declared himself similarly a n d said t h a t h e k n e w of 600 to 800 people, inoculated within a few miles of his residence, with only o n e death. H e h a s t e n e d to a d d t h a t the victim was n o t o n e of his p a t i e n t s n.\n\nto successive V a c c i n a t i o n Acts. F r o m 1842, vaccination at the public cost could be claimed from local authorities in any parish. N o t until 1853 was t h e v a c c i n a t i o n o f c h i l d r e n within four m o n t h s o f b i r t h m a d e c o m p u l s o r y , a n d l a t e r still the Privy Council f r a m e d regulations for securing d u e qualification o f t h o s e w h o were c o n t r a c t e d with by t h e guardians, after c o m p l a i n t s o f the u n s a t i s f a c t o r y way in w h i c h vaccination was b e i n g performed. Yet in 1862 w h e n a r e p o r t o n v a c c i n a t i o n in Lincolnshire was m a d e to the M e d i c a l Officer of the Privy Council it stated : - \" The quality of the Vaccination in Lincolnshire, more especially of the unions in the Fen district and in Great Grimsby, was very bad indeed, more especially in Great Grimsby. Here small-pox had been smouldering for three to four years:\"lL A t G r i m s b y , indeed, t h e r e were 20 small-pox deaths in 1860, a n d G a i n s b o r o u g h h a d h a d a n e p i d e m i c in the w i n t e r o f 1858-59. T h e r e was small-pox at G r a n t h a m in 1861. Persistence o f Small-pox T h e persistence o f some small-pox is u n d e r s t a n d a b l e , n o t only because o f the p o o r quality o f t h e vaccination, b u t in the light of t h e evidence given of neglect o n the p a r t of local authorities. T h e registrar o f v a c c i n a t i o n s for Caistor U n i o n was f o u n d n o t to h a v e k e p t his register for a year, at B o s t o n n o t since 1855, a n d at G o s b e r t o n (in S p a l d i n g U n i o n ) t h e register h a d n o t b e e n kept since 1854. T h e duty which t h e Compulsory Vaccination Act imposed on Guardians of \" giving f r o m time t o time d u e notice of t h e m e a n s p r o v i d e d for o b t a i n i n g v a c c i n a t i o n \" to all t h e local p o p u l a c e h a d b e e n \" almost universally neglected.\" N o public notice h a d ever b e e n given by t h e authorities o f Caistor, G a i n s b o r o u g h , Brigg, H o l b e a c h , S p a l d i n g a n d S t a m f o r d U n i o n s . The T a b l e given in this 1862 r e p o r t showing t h e state of vaccinat i o n o f children t h r o u g h o u t Lincolnshire is m o s t revealing 1~. T h e percentage well vaccinated varies between 4.6 at B r i g g to 6.8 at Boston, a n d 24.1 at Sleaford. T h e percentage n o t vaccinated a t all r a n g e d b e t w e e n 8.8 at G r a n t h a m a n d 17.5 at Horncastle. I n m a n y o f the districts visited there were children m a r k e d with small-pox, 2, 8 ~o o f t h o s e e x a m i n e d a t G r i m s b y , 2.3 at Stamford, a n d 1.4 at G a i n s b o r o u g h . T h e p r o p o r t i o n o f infants registered as v a c c i n a t e d per 100 registered births, w h e r e it is given, is relatively tow particularly in the F e n U n i o n s . T h e two reasons given for the low a p p a r e n t rate o f public v a c c i n a t i o n (as low as 25 % at Horncastle) a r e\n\nWorking Class and Inoculation T h e evidence p o i n t s to t h e working classes as t h e m o s t tenacious believers in inoculation, t h o u g h it m u s t b e e m p h a s i s e d t h a t little is k n o w n a b o u t t h e attitude of t h e m o r e privileged groups. T h o s e w h o could n o t afford t h e cost o f i n o c u l a t i o n t e n d e d to accept deliberate infection b y a reputedly mild case as a substitute 12. T h e r e are h i n t s t h a t this k i n d o f t h i n g was n o t very different f r o m infanticide. \" H u m a n i t a s , \" writing to t h e S t a m f o r d M e r c u r y in 1816 a b o u t a n epidemic at G r i m s b y d e p l o r e d the \" delusion u n d e r w h i c h t h e lower orders of t h e c o m m u n i t y l a b o u r ,,a3. H e implies t h a t t h e i r hostility to vaccination h a d a sinister side, a n d a l t h o u g h they s p o k e o f the e p i d e m i c as a visitation o f D i v i n e Providence, \" n a t u r a l affection seems to b e a b s o r b e d in this deadly fanaticism, w h i c h d e p o p u l a t e s the place a n d causes all r a t i o n a l p e o p l e to suspect t h a t t h e r e is s o m e t h i n g m o r e in this species of o b d u r a c y t h a n a p r e t e n d e d r e s i g n a t i o n to the will o f h e a v e n . \" T h e c o n q u e s t of this scourge ~ o f small-pox owed m u c h\n\n15 Fifth Report o f the Medical Officer to the Privy Council, 1862. 16 Ibid. The findings for some representative areas in Lincolnshire were as follows : - Infants % vaccinated No. % % 9/ooun- marked per P.L.U. exam- well badly vaccinby 100 district ined vaccin- vaccin- ated smallregisated ated pox tered births\n\n8Lincoln, Rutland and Stamford Mercury, Mar. 8th, 1816. 9 Ibid., Apr. 30th, 1824. a~ Feb. 13th, 1818. 11 Ibid., Aug. 28th, 1818. 13 In 1835 a Grantham surgeon complained that the disease was being spread by the deliberate exposure of infected patients as well as by inoculation. As late as 1862 a nonconformist clergyman in the Isle of Axholme could recollect a recent instance of a woman who was anxious that her children should have small-pox before harvest and who arranged for them to take turns at sleeping in an infected nightshirt. is Lincoln, Rutland and Stamford Mercury, Nov. 1st, 1816. ~ This is well expressed by T. Eminson in his Epidemic Pneumonia at Scotter and Neighbourhood published in 1892 : \" Small-pox was once an extremely common and fatal disease in the neighbourhood ; all of us over 35 can remember ' old standards ' whose faces bore evidence of the terrible nature of this loathsome disease.'\"\n\nBoston .. Grimsby . Gainsborough\" Brigg.. Grantham Horncastle Lincoln Louth Sleaford Spalding Stamford 36\n\n986 569 409 239 1,075 531 962 462 174 254 473\n\n6.8 7.2 14.1 4.6 17.8 14.5 11.0 9.9 24.1 11.6 20.9\n\n20.5 14.9 26.6 15.0 20.3 20.1 17.6 19.0 19.0 20.6 28.1\n\n15.3 12.4 11.9 15.0 8.8 17.5 14.8 13.8 10.3 14.7 14.7\n\n0.8\n\n2.8 1.4 -0.7 0.01(sic) 0.9 0.2 -0.3 2.3\n\n53.4 58.7 50.0 25.5 47. 7 37.5 36,9\n\nN o v e m b e r 1955 A t first sight, the figures of d e a t h rates in Lincolnshire in this period m a y suggest some superiority of c o n d i t i o n s to t h o s e prevailing elsewhere 21. F o r t h e Lincolnshire d e a t h rate was r a t h e r below t h a t for the c o u n t r y as a whole, a n d was t e n d i n g to fall slightly, f r o m 19.5 per 1,000 (1851-60) to 19.1 (1861-70) a n d 18.5 (1871-80), wh~-reas in the country, as a whole the crude d e a t h rate fell m o r e slowly ~1\". But the n a t i o n a l average was swollen by t h e high death rates of the big towns where disease spread rapidly, whereas in the m o r e isolated r u r a l areas the spread of disease was m o r e restricted. Significantly, few U n i o n s with a d e a t h rate exceeding t h a t of the worst parts of Lincolnshire were p r e d o m i n a n t l y rural.\n\nthe interference of non-certifying v a c c i n a t o r s a n d the high rate o f infant mortality. T h e latter was particularly noticeable in the Fens, as we shall notice later in this article. Nevertheless, this g r a d u a l c o n q u e s t o f small-pox is o n e of the m o s t i m p o r t a n t i m p r o v e m e n t s in Lincolnshire public h e a l t h in the first h a l f of the 19th Century. T h e other, the decline of malaria, was the b y - p r o d u c t of i m p r o v e d drainage. W h e n A r t h u r Y o u n g visited Lincolnshire in 1799, he could already note : \" In the north-west . . . . Agues were formerly c o m m o n l y k n o w n u p o n the T r e n t a n d H u m b e r s i d e ; at p r e s e n t they are rare \"JT. Surprisingly, he went o n to say : \"' N o t h i n g h a s b e e n effected on this side o f the H u m b e r to which it c a n b e a t t r i b u t e d . \" Yet in his r e p o r t he m e n t i o n s several i m p o r t a n t enclosures in this area, all o f which involved extensive i m p r o v e m e n t s in drainage. In the great enclosure o f the Isle o f A x h o l m e c o m m o n s a l o n e over s was s p e n t o n new d r a i n a g e works. I n this p a r t o f t h e c o u n t y some m a l a r i a persisted until a b o u t 1830 b u t after t h a t was rarely f o u n d , p r o b a b l y due to t h e c o m p l e t i o n of enclosure a n d i m p r o v e m e n t of the low-lying l a n d in the 1830s, a n d the installation o f s t e a m engines for p u m p i n g , t h e first of w h i c h was installed at B e l t o n in 1834 aS. I n t h e Fens, a l t h o u g h m a l a r i a was said to h a v e declined greatly it still persisted in t h e Spalding area. I n 1826 a local d o c t o r ' s b o o k s t h e r e s h o w e d 300 cases o f \" a g u e , \" a n d it was p r e v a l e n t again in t h e following year. T h e r e was also a n increase in 1857-59. I n 1863 a r e p o r t o n t h e subject stated, \" i n t e r m i t t e n t a n d r e m i t t e n t fevers are still f r e q u e n t a r o u n d Spalding t h o u g h described as m u c h less severe t h a n f o r m e r l y . . . . A t the N a t i o n a l School . . . o f 75 boys p r e s e n t at the time of m y visit, 11 stated t h a t they h a d h a d ague at some p e r i o d -19\n\nThe Healthiest Districts I n the decade 1841-50 the healthiest p a r t s of the c o u n t y were Spilsby, Caistor, B o u r n e a n d G r a n t h a m Unions, with a d e a t h rate of 18 per 1,000. T h e villages here were for t h e m o s t p a r t o n high ground 9 G r i m s b y ( p a r t o f Caistor U n i o n ) with a d e a t h r a t e of 19 p e r 1,000 was t h e n one o f t h e 24 healthiest t o w n s in E n g l a n d a n d Wales, b u t its r a p i d g r o w t h after 1850 was to bring with it rising d e a t h rates. Horncastlr Brigg, Sleaford a n d S t a m f o r d U n i o n s r a n k e d next with a d e a t h rate o f 19 per 1,000, a n d these c o n t a i n e d m u c h o f t h e recently d r a i n e d malarial marshes o f the county 9 H o l b e a c h a n d B o s t o n U n i o n s , ~ntirely in t h e fens, h a d a d e a t h r a t e o f 21 p e r 1,000. T h e rate was t h e same in t h e Lincoln U n i o n , largely b e c a u s e o f the u n h e a l t h y state o f t h e city, a n d in t h e L o u t h U n i o n which included m a n y o f t h e o v e r c r o w d e d villages o f t h e Lindsey marsh. T h e Spalding U n i o n , entirely in the fenland, a n d the G a i n s b o r o u g h U n i o n , affected b y epidemics in G a i n s b o r o u g h itself, were t h e least h e a l t h y in t h e c o u n t y with a m o r t a l i t y of 22 p e r 1,00025. I f we seek c o m p a r i s o n s between d e a t h rates in t h e c o u n t y a n d t h o s e to b e f o u n d elsewhere, we find t h a t a d e a t h r a t e such as t h a t in Spalding U n i o n was as high as industrial t o w n s such as Huddersfield a n d Keighley. L i n c o l n (23 p e r 1,000) was n o b e t t e r t h a n Wakefield. B o s t o n a n d L o u t h ~3 (24 per 1,000) r a n k e d with Clerkenwell, St. Pancras, Barnsley a n d R o c h d a l e . T h e r e were only six t o w n s in t h e c o u n t r y w i t h a d e a t h rate h i g h e r t h a n t h a t o f G a i n s b o r o u g h .\n\nOvercoming Small-pox and Malaria Nevertheless, t h e occurrence o f cases o f small-pox a n d m a l a r i a in L i n c o l n s h i r e in t h e 1860s s h o u l d n o t o b s c u r e t h e f a c t t h a t these t w o diseases, w h i c h h a d b e e n extremely wides p r e a d in L i n c o l n s h i r e a n d devastating in their effects, were b y t h e n largely overcome. T h e i r r e d u c t i o n was t h e m a i n a d v a n c e in h e a l t h c o n d i t i o n s in the first h a l f o f the century. B u t in 1869-71 t h e severe epidemic o f small-p0x t h a t swept G r i m s b y - - s o m e 1,000 cases with 244 deaths~~ for t h e last time o n this scale the terrible n a t u r e of the disease, a n d t h e penalty o f neglect. T h e neglect t h a t h a d such d i s a s t r o u s effects c o n c e r n e d riot only v a c c i n a t i o n b u t t h e g e n e r a l s a n i t a r y state o f the town, a n d it is to this m a j o r subject o f the s a n i t a r y state of the t o w n s in the c o u n t y t h a t we m u s t t u r n to u n d e r s t a n d the prevailing c o n d i t i o n s o f public h e a l t h in the m i d d l e decades o f t h e century.\n\n21 The use of crude death rates to compare the healthiness of different areas is open to the objection that they depend upon the age structure of the population. However, we do not think this significantly affects the general pattern shown in Lincolnshire. The major qualification to be borne in mind is that there was substantial \" emigration \" from the country districts, both out of the country to industrial districts and a b r o a d - - a n d within the county, to the towns. This would tend to make the age structure in the villages more \" elderly.\" An age specific death rate might afford more exact comparisons ; but e.g. infant mortality rates in Lincolnshire were influenced by special features which do not directly reflect the general \" healthiness \" or otherwise of the area but rather conditions of work which bore hardest upon \"field f a r i n g \" women (cf. our later analysis of infant mortality rates), 9 2~a ,, Tested by the available statistics, the standard of health up to the passing of the Public Health Act, 1875, does not appear to have improved, in spite of the expenditure . . . upon sanitation during the two preceding decades,\" W. M. Frazer, History o f English Public Health (1950). p. 127. The crude death rate is there given as 22.7 for 1851-55, and 22.0 for 1871-75. 23 Report of Select Committee on Public Health Bill, 1855, pp. 205-8. ~3 But the Louth figure would have been considerably better if deaths in the Union Workhouse had been excluded.\n\n1T A. Young. General View of the Agriculture of Lincolnshire (1799). la A report on Axholme in 1859 (Second Report of the Medical Officer to Privy Council) st;ll mentioned the existence of malaria, b u t rather curiously declared, \" ague occasionally attacks the Irish immigrants, but never occurs among the residents.\" This suggestion of its more serious effect on outsiders recalls the account of a writer in the Universal Magazine of 1773 who conversed with a man who had had nine wives ; \" the Fen men were accustomed t o seek their wives in the upland country, selecting those who had a little money ; and as living in the Fens was certain death to ~trangers during the fever season, many of the Fen men thus contrived to accumulate a good deal of money.\" (S. Smiles, Lives of the Engineers Rennie, p. 242). xa Fifth Report of the Medical Officer to Privy Council. ~o Grimsby Observer, Nov. 15th, 1871. 37\n\nPUBLIC H E A L T H O n t h e whole, these' figures indicate accurately e n o u g h the unfaVourable state o f the towns in L i n c o l n s h i r e . I f t h e figures for G a i n s b o r o u g h were particularly raised b y a series o f epidemics, the features common-~ to all t h e epidemics r e p o r t e d o n in t h e c o u n t y p o i n t b a c k to t h e general s a n i t a r y state o f b o t h t o w n s a n d villages. T h e m o s t c o m m o n features f o u n d associated w i t h epidemics in L i n c o l n s h i r e c a n be s u m m a r i s e d as the following : Severe o v e r c r o w d i n g ; d a m p , ill-ventilated cottages ; b a d drainage, especially in the m a r s h a r e a s ; i n s a n i t a r y disposal of s e w a g e ; lack o f Uncont a m i n a t e d w a t e r supply ; a n d t h e existence o f m i g r a t o r y workers, especially Irish, livii~g in the w o r s t possible conditions ~'. Such elements were to b e f o u n d a b u n d a n t l y in Lincolnshire. T h e s e prevailing a n d persistent c o n d i t i o n s of b a d s a n i t a t i o n a n d h o u s i n g n a t u r a l l y h a d their c o u n t e r p a r t from t i m e to t i m e in violent ep demics c o n c e n t r a t e d in t h e m o s t u n w h o l e s o m e parts.\n\ng o o d quality.\" Indeed, h e is r e p o r t e d to have said t h a t the water was so i m p u r e t h a t h e f o u n d E p s o m salts in it \" o u t of his o w n s h o p . \" Samples of the well water, a n d o f p i p e d water supplied to p a r t of Brigg were sent to a n analyst w h o c o n d e m n e d t h e well water as wholly unfit for d o m e s t i c use. H e r e p o r t e d t h a t t h e w a t e r w o r k s supply was far superior, t h o u g h n o t so pure as desirable. A m a j o r i t y o f t h e G u a r d i a n s , however, p r e f e r r e d to use their o w n j u d g m e n t . A subc o m m i t t e e solemnly tasted b o t h waters a n d declared t h e i r preference f o r t h e well water. A t this stage, one o f the G u a r d i a n s , Mr. E d w a r d Peacock o f B o t t e s f o r d ~,, b e g a n to write to the Press 28, in t h e h o p e o f forcing the m a j o r i t y o f the G u a r d i a n s to give way.\n\nThe Sanitary State of the Towns\n\nT h e B o a r d o f G u a r d i a n s reversed its earlier decision, and Peacock n o t e d t h a t t h e r e was n o w a disposition a m o n g t h e G u a r d i a n s to b l a m e the Medical Officer for n o t h a v i n g pressed t h e m a t t e r earlier. T h e depressing picture o f c o n d i t i o n s in Brigg is reinforced by t h e c o m m e n t s o n it m a d e in a r e p o r t for t h e M e d i c a l Officer to t h e Privy C o u n c i l o n \" A g u e s i n M a r s h D i s t r i c t s , \" in 1863. Brigg h a d scarcely a n y ague at this date, b u t : \" General sanitary condition of the town extremely bad. The town lies low and there is no drainage worth mentioning 9 I visited the town with Mr. Moxon, Union Surgeon and saw several cases of typhoid fever in some filthy courts on east side of town. In one of these there was only one privy for 11 houses. On the west side a low lying portion known as ' t h e R o o k e r y ' was one of the filthiest places I ever saw ,,39.\n\n\" There is a cesspool,\" he wrote, \" within about six feet of the well, and into this water used in washing, urine from the bedrooms and other noxious matters flow in great abundance. Moreover, the privies are so near as to aid contamination . . . . The soil is a mixture of loam, sand and gravel, through which water easily finds its way.\"\n\nTypical c o n d i t i o n s to b e f o u n d in t h e c o u n t r y t o w n s o f L i n c o l n s h i r e c a n b e illustrated f r o m the e x a m p l e o f Brigg. This small b u t growing m a r k e t t o w n in t h e n o r t h o f t h e c o u n t y was b o t h a small m a n u f a c t u r i n g c e n t r e 2~ a n d a n \" o p e n \" village supplying l a b o u r to n e i g h b o u r i n g parishes. I n 1842 t h e District M . O . o f t h e Brigg P o o r L a w U n i o n sent in a rel:ort to the c c r r m i t t e e investigating the s a n i t a r y c o n d i t i o n o f the p o p u l a t i o n , w h i c h sets the scene : - \" Fever has been very prevalent, a few years since during the hot weather in a range of paupers' houses, situated upon a drain, called the town drain, situated very near the centre of the town of Brig~. which is never to my knowledge cleared out, and is during the summer a complete mass of animal and vegetable decomposition ; we have no public o r private authority for interfering. Near this are many pigsties, close to the houses ; all of which are never noticed, merely because fever has not been prevalent these few years but should our summers become unusually hot, I am convinced fever must be very common. \" A n o t h e r observation 1 wish to make, is the very common occurrence of tramps bringing the small-pox to our town ; they arrive at the ~odging houses (where they are without scruple taken in) almost at the height of the complaint, and this occurs repeatedly so that our town is never long free from the small-pox ,,~6. Actually, t h e r e was ia p r i v a t e a u t h o r i t y w h i c h was able to t a k e action against s o m e evils the C o u r t L e e t j u r y o f W r a w b y manor. T h e Stamford Mercury ( S e p t e m b e r 10th, 1852) r e p o r t e d from Brigg : - \" It seems that the ooor Irish tenantry have returned to wallowing in their filth ; for on Tuesday the 7th inst. several of this description were summoned and fined for the abominably filthy and fever-breeding condition of their abodes. To give an account of the Irish locality 9 . would be really disgusting.\" I n 1855, we find t h e U n i o n Medical Officer c o m p l a i n i n g t h a t t h e well w a t e r o f the w o r k h o u s e at Brigg was \" n o t o f\n\nT h e r e p o r t goes o n to m e n t i o n t h a t \" t h e R o o k e r y \" had \" n o k n o w n satisfactory s u p p l y o f w a t e r '\" a n d t h a t t h e r e was f r e q u e n t t y p h o i d there. But, as if t h a t were n o t e n o u g h , : \" T h e sanitary committee, formed of members of the Board of Guardians, are said to be opposed to any improvement, and to direct the inspector of nuisances not to report any case unless complained of-80\n\nAttitude of the Sanitary Authorities R e p o r t s o n o t h e r c o u n t r y t o w n s in L i n c o l n s h i r e in these middle decades suggest h o w general were t h e worst features of public h e a l t h f o u n d in Brigg. Passivity o f local a u t h o r i t i e s in t h e face o f s u c h evils was n o t c o n f i n e d to Brigg. F r o m the same r e p o r t o n \" a g u e s \" f r o m w h i c h we h a v e q u o t e d , we learn t h a t at B a r t o n - o n - H u m b e r , \" T h e Local B o a r d o f ~ Peacock was primarily a scholar and antiquary, with a creditable record of publications in Archaeologia and of texts edited for the early English Text Society. It is interesting to note that in his Glossary of the Words used in the Wapentakes of Manley and Corringham (1877), he includes the word \" nuisancer,\" meaning an inspector of nuisances. To illustrate its use he quotes a villager who said of another that he let his family \" drink stuff that's no better than seep from a manure hill. He wants real bad to have the nuisancer down on him.\" He also quotes a man who spoke of the \" nuisancer always coming and rooting about. Fevers wouldn't come if they wasn't sent.\" 38Lincoln, Rutland and Stamford Mercury, Aug. 31st and September 7th, 1855. 39 Sixth Report of the Medical Officer to the Privy Council. The report on \" A g u e s \" was made by a Dr. Whitley. 3o It was not until 1866 that the Sanitary Act of that year compelled local authorities to inspect their districts and suppress nuisances.\n\n2~ Another feature worth noting in Lincolnshire epidemics was the rapid transmission of diseases along the Humber and Trent side, along which there was very extensive trade and passenger traffic. The parlous state of sanitation in Hull, which was the main port for all this traffic, led to severe epidemics there which spread into Lincolnshire. 35 White's Directory for Lincolnshire, 1842, which caLs Brigg \" a neat, well built, and thriving market town \"estimates its population then at 2,300, mostly within the area of the \" town and Chapelry of B r i g g \" in Wrawby parish, but spread into three other parishes. Brigg had several corn mills, steam mills, and malt kilns, two roperies and two tap, yards, an ironfoundry, a ship-building yard, a sail manufactory, and a worsted manufactory. 36Local Reports on the Sanitary Condition of the Labouring Population of England and Wales, 1842. p. 156. 38\n\nNovember 1955 Health, c o m p o s e d chiefly o f farmers, h a v e n o by-iaws a n d shirk i n q u i r y . \" A S u p e r i n t e n d i n g Inspector, u n d e r t h e 1846 Public H e a l t h Act, r e p o r t i n g o n H o l b e a c h in 1851, concluded, \" I a m compelled to a d d H o l b e a c h to t h e list o f places in which t h e Public H e a l t h A c t h a s l o n g b e e n in o p e r a t i o n w i t h o u t a n y t h i n g h a v i n g b e e n d o n e to i m p r o v e t h e sanitary c o n d i t i o n of t h e i n h a b i t a n t s ,,a~. B u t L i n c o l n itself provides t h e most extreme example of w h a t o n e m i g h t call a n antisanitary party. I n 1848 the citizens refused to a d o p t t h e Public H e a l t h Act. I n 1849 a rewort o n the state o f the t o w n emphasised the deficient drainage. S o m e privies a n d waterclosets discharged themselves into small c o v e r e d drains, which h a d t h e i r outlets in open c h a n n e l s in t h e street. I n 1863 Dr. W h i t l e y declared t h a t n o t h i n g h a d b e e n d o n e for t h e i m p r o v e m e n t o f the t o w n since 1849, a n d t h a t t h e increase o f p o p u l a t i o n h a d increased t h e evils m e n t i o n e d . H e wrote : \" T h e general sanitary c o n d i t i o n of Lincoln is very bad. I was i n f o r m e d t h a t t h e ' s a n i t a r y c o m m i t t e e , ' f o r m e d o f m e m b e r s o f t h e T o w n Council, object to raise m o n e y f o r d r a i n a g e p u r p o s e s ,,3~.\n\na n d despite a t y p h o i d epidemic, t h e T o w n Council h a d \" t h r o u g h the action o f a m i n o r i t y o f t h e i r body, declined to avail themselves of the a d v a n t a g e s w h i c h the a d o p t i o n o f the Local G o v e r n m e n t Act would give t h e m ,,~4 Again, Brigg was typical in its w r e t c h e d w a t e r supply. In m a n y cases it was n o t until late in the c e n t u r y t h a t efficient w a t e r works were set up. Before t h e n t h e water supply available was often appalling. A t G a i n s b o r o u g h , in the 1840s, the t o w n was said to be b a d l y off for water \" o w i n g to the b a d m a n a g e m e n t o f the w a t e r w o r k s , which are a complete burlesque o n m a c h i n e r y a n d the spirit of the age -3 ~. T h e m a c h i n e r y was w o r k e d with a single h o r s e a n d a b o y h a l f asleep to drive it. N o a d d i t i o n a l piping h a d been laid d o w n for 20 years, a n d t h r e e - q u a r t e r s o f the populace h a d to get t h e i r w a t e r by draw-buckets f r o m the Trent. T w e n t y years later n o t h i n g h a d c h a n g e d : - \" The water supply is partly from the river above the town, being raised by horse-power about low-water, avd partly from wells. Much rain water is used even for cooking ,,a6. A t G r a n t h a m at the same time t h e w a t e r c o m p a n y was, d u r i n g the winter, mixing spring w a t e r with river water \" o f t e n turbid and undrinkable .... T h e river w a t e r was t a k e n f r o m the W i t h a m a mile a b o v e the town, a n d two o r t h r e e miles a b o v e t h e source o f supply, the river receives a p o r t i o n of the d r a i n a g e o f a village.\" B u t the t y p h o i d epidemic \" b r o k e out, a n d was m o s t fatal at a place called the W h a r f t h e i n h a b i t a n t s o f which used for d r i n k i n g a n d o t h e r p u r p o s e s w a t e r derived f r o m a n e i g h b o u r i n g c a n a l - 3 7 . T h u s , the existence of a w a t e r works was n o g u a r a n t e e of r e a s o n a b l y p u r e water.\n\nThe Anti-Sanitary Party I n 1865 w h e n the Public H e a l t h A c t was. a d o p t e d , the councillors w h o h a d v o t e d for the a d o p t i o n were defeated a t the N o v e m b e r elections. T h e leader o f t h e anti-sanitary party, w h o were mostly liberals a n d retired t r a d e s m e n w h o h a d invested in jerry-building, was a local n e w s p a p e r proprietor. T h e agitation o n the o t h e r side was started b y Vr J. Mantle, Esq., w h o drew a t t e n t i o n to o u t b r e a k s o f fever a t S c o t h e r n (within L i n c o l n U n i o n ) w h e r e t h e w a t e r c o u l d b e smelt several yards f r o m the p u m p ; B a r d n e y , where the sewer was b l o c k e d ; a n d I n g h a m w h e r e 52 o u t o f 54 cases o f fever were t r a c e d to o n e p u m p . T h e G u a r d i a n s a p p o i n t e d a committee, o f w h i c h h e b e c a m e c h a i r m a n , a n d he gave evidence before t h e R o y a l Sanitary C o m m i s s i o n in 1870. Because o f his efforts for sanitary i m p r o v e m e n t a t t e m p t s were m a d e to u n s e a t h i m f r o m the B o a r d o f G u a r d i a n s 33. A p a r t f r o m t h e unwillingness of local authorities to act, t h e r e was sometimes difficulty in finding o u t which a u t h o r i t y did b e a r responsibility. A r e p o r t o n a fever o u t b r e a k at G r a n t h a m in 1864 stated t h a t \" sewage . . . seems to be u n d e r t h e a u t h o r i t y o f n o one. T h e T o w n Council of t h e municipal b o r o u g h h a s n o control over t h e sewers, n e i t h e r h a v e the B o a r d o f G u a r d i a n s , n o r the Sanitary C o m m i t t e e s of Spittlegate a n d Little G o n e r b y . \" I n a d d i t i o n t h e r e were t h r e e different h i g h w a y authorities. Yet in this situation,\n\nDubious Water Supply B u t for t h o s e n o t c o n n e c t e d to a w a t e r works, the supply was e v e n m o r e dubious. N o t only p o l l u t e d rivers a n d canals, b u t even a n agricultural d r a i n (at Spalding), a n d railway ballast pits 3s (at Lincoln) were the source. E v e n where w a t e r w o r k s existed, t h e r e were large n u m b e r s o f wells in use in t h e towns, m a n y of t h e m m o r e or less c o n t a m i n a t e d . T h i s was true of L i n c o l n a n d G r i m s b y w h e r e only a q u a r t e r o f t h e houses in 1871 were supplied f r o m the w a t e r works, a n d t h e r e were o v e r 300 private wells. O f G r i m s b y ' s wells D r . H o m e in his r e p o r t ~ declared : - \" The water in most of the private wells of Grimsby is very exposed to contamination from fluids draining into i t ; this is rendered unavoidable in consequence of the confined space in which house-drains, pigsties, cess-pits and other sources of pollution are crowded near the well. The definite instances which might be adduced in illustration of this statement are so numerous that selection is difficult.\"\n\n31Lincoln, Rutland and Stamford Mercury, Aug. 22nd, 1851. The Inspector metations 10 cottages draining into a stagnant ditch filled with \" dead animals and refuse of the most revolting description,\" and quotes a statement that nearly two-thirds of the population of Holbeach died before reaching the maturity. ~ Sixth Report o f the Medical Officer to Privy Council. 33 Second Report, Royal Sanitary Commission. Minutes o f Evidence, pp. 140-147. Mantle characterised the Guardians from the rural parishes as \" tenant farmers sent from little villages with one object of keeping rates down.\" Where Guardians were more careless of the ratepayer, it was not necessarily in the interests of public health. The Louth Guardians in 1857 settled wine merchants bills of s several months after the Poor Law Board had censured them for lavishness with medical extras, \" nearly every case receiving\" (we wonder if they did ?) '\" either wine, brandy, gin, ale or tobacco.\" In 1861 the Louth Guardians were equally obstinate in face of the Board's complaints of excessive infant mortality in the workhouse. (Louth Union. Guardians' minutes.)\n\n34 Seventh Report o f the Medical Officer to Privy Council. a~ Stow, History o f Gainsborough, 1842 edition. 36 Sixth Report o f the Medical Officer to Privy Council. Although the local board of health took over the water company in 1871, the source of water continued to be the Trent until 1888. 37 Seventh Report o f thg Medical Officer to Privy Council. This privilege they shared with Goole, in Yorkshire, where the water supply was \" partly from wells, partly from the c a n a l \" (Sixth Report). a8 Report o f the Medical Officer o f the Local Government Board, 1886. At Holbeach in the 1860s most of the water supply camc from pits (Sixth Report o f the Medical Officer to Privy Council). and only \" a few of the better houses have wells.\" ~9 Report to the Medical Officer o f Local Government Board, 1871. Quoted from the Grimsby Observer of Jan. 10th, 1872. 39\n\nPUBLIC HEALTH Barton-on-Humber, 1859 : \" The cesspools are in the worst possible condition. The practice prevails of keeping their contents for long periods for agricultural purposes, to suit the convenience of the farmers. They are all above ground, and usually about 6 feet square, and frequently several are contained in the yard of one house ,,42\n\nT h e \" definite i n s t a n c e s \" which Dr. H o r n e gave were o f t h e sort t h a t do n o t m a k e pleasant reading. B u t e n o u g h evidence has been given to show the universality o f a n i m p u r e supply of w a t e r in t h e t o w n s of 19th c e n t u r y Lincolnshire. W e c a n best illustrate this by noticing t h e way in which the exception was s p o k e n of : \" T h e general sanitary c o n d i t i o n of B o u r n e a p p e a r e d unusually g o o d as t h e r e is a good a n d extensive system o f d r a i n a g e available for nightsoil . . . . T h e r e is a n a m p l e supply o f h a r d water . . . laid o n in the t o w n -40. ,, Uniasually g o o d \" because it h a d an a m p l e w a t e r supply, a n d a n effective system o f drainage. W e c a n a p p r e ciate t h e justness o f the c o m m e n t as far as the w a t e r supply is c o n c e r n e d . It is n o less true o f t o w n drainage.\n\nGrhnsby, 1871 : In the majority of the houses \" nightsoil and ashes are received into boxes, the contents of which are removed periodically . . . . The carts discharge at a depot in the W. Marsh, at a place onequarter of a mile from the nearest houses ; the accumulations remain there until the contractor is able to dispose of them . . . . The sewage of the whole borough is discharged at a point nearly one-third of a mile from low water mark-43 Boston, 1886 : '\" Systematic sewering on W. side of River Witham discharging into tidal water. On east side, old drains failing into an old ratriddled irregular sewer (once the \" Bar-ditch \") connected at both ends with the Witham, or into a stagnant canal, \" Bargate Drain,\" causing nuisance. Excrement r~moval mostly very offensive brick privy pits only emptied when full and ccmp'ained of. About 500 ash-closets drained into sewers and empaed weekly by public scavenger. A few W.Cs. ,,44.\n\nDrainage of the Towns T o w n d r a i n a g e a n d disposal o f sewerage were a l m o s t e v e r y w h e r e a s t a n d i n g incitement to disease. B u t we c a n distinguish stages in development. T h e d e v e l o p m e n t in m o s t Lincolnshire t o w n s was n o t from n o d r a i n a g e (cesspools a n d r e m o v a l of nightsoil in carts) t o effective d r a i n a g e ; t h e r e was a t r a n s i t i o n p e r i o d in w h i c h d r a i n a g e was i n t r o d u c e d b u t was so imperfect in c h a r a c t e r as often to create a new source o f disease. I n general primitive d r a i n a g e a n d v a u l t privies, ash-closets a n d cesspools, existed side b y side u n t i l late in the century. T h e y seem to h a v e rivalled each o t h e r .in offensiveness. W e select some example f r o m different t o w n s at different dates : - -\n\nR e p o r t s o n various L i n c o l n s h i r e t o w n s , a p a r t f r o m t h e examples given, m a k e it clear t h a t t h e d r a i n a g e i n t r o d u c e d into towns, or a d d e d to older systems o f drainage, at this p e r i o d was deficient in so far as t h e drains were generally u n t r a p p e d , u n v e n t i l a t e d a n d unflushed. A t G r a n t h a m t h e drains were t h o u g h t to b e the m a i n s o u r c e o f t h e epidemic in 1864. There, t h e drains in the p o o r e r quarters, a n d private drains, were described as \" c o m m o n l y u n t r a p p e d a n d t h e sewers generally imperfect . . . m o r e o r less o f t h e n a t u r e of e l o n g a t e d cesspools -4 ~.\n\nStarnford, 1842 : The severity of typhoid had been much increased by local causes, \" particularly such as yards densely populated, ill ventilated, and where filth accumulated for the want of public convenience for its deposit: many of such yards have cesspools covered over with open grating ,,~1.\n\n(To be continued) 4~ Second Report o f the Medical Officer to Privy Council. 43 Report to the Medical Officer o f the Local Government Board. 44 Report to the Medical Officer o f the Local Government Board, pp. 138-9. 4~ Seventh Report o f the Medical Officer to the Privy Council.\n\n4o Sixth Report o f the Medical Officer to Privy Council. Report on Agues. 41 Report by Stamford Medical Officer. Local Reports of 1842 Sanitary Committee.\n\nDental Health in the Maternity and Child Welfare Field therefore, seem logical t o devote a certain a m o u n t of time a n d m o n e y to try to p r e v e n t dental disease. A t the p r e s e n t time t h e r e were a d d i t i o n a l factors. T h e d e n t a l profession was practically fully o c c u p i e d a n d yet was p r o v i d i n g t r e a t m e n t for a p p r o x i m a t e l y o n l y one-fifth o f the p o p u l a t i o n . T h e rate o f r e c r u i t m e n t to the profession a n d the wastage due to d e a t h o r r e t i r e m e n t were such t h a t the n u m b e r o f d e n t a l s u r g e o n s in this c o u n t r y seemed likely to decline considerably. F u r t h e r m o r e , since the e n d of rationing, c a r b o h y d r a t e s of all forms, b u t particularly sweets a n d biscuits, h a d b e c o m e m u c h m o r e readily obtainable, with a c o n s e q u e n t m a r k e d increase in d e n t a l caries. I n Staffordshire in 1947, w h e n r a t i o n i n g still applied, o f the five-year-old c h i l d r e n e n t e r i n g school, 38.9 % h a d s o u n d teeth, while t h e n u m b e r with f o u r or m o r e decayed teeth was 2 2 . 3 % . I n 1953 w h e n r a t i o n i n g of c a r b o h y d r a t e s h a d d i s a p p e a r e d t h e n u m b e r with s o u n d teeth h a d d r o p p e d to 2 1 . 3 % a n d the n u m b e r with four o r m o r e decayed t e e t h h a d risen to 50.6 %.\n\nE T H O D S o f p r e v e n t i n g dental caries b y education, inspection a n d regulation o f the fluoride c o n t e n t of d r i n k i n g w a t e r were described in p a p e r s r e a d a t a j o i n t meeting of the D e n t a l a n d M a t e r n i t y a n d C h i l d Welfare G r o u p s o f the Society of Medical Officers o f H e a l t h , held at the L o n d o n School o f Hygiene a n d Tropical Medicine. T h e subject was \" D e n t a l H e a l t h in t h e M a t e r n i t y a n d Child Welfare F i e l d . \" T h e r e were 45 m e m b e r s a n d visitors present, a n d t h e President o f the M a t e r n i t y a n d C h i l d Welfare G r o u p , Dr. H i l d a Davis, was in the chair. T h e r e were two speakers f r o m e a c h group. Dr. Eileen Ring, Senior A d m i n i s t r a t i v e M e d i c a l Officer, B i r m i n g h a m , w h o s p o k e first, p o i n t e d o u t t h a t d e n t a l decay affected a b o u t 98 % o f the p o p u l a t i o n . I t was n o t a d r a m a t i c disease b u t was t h e cause o f m u c h pain, suffering a n d general ill-health. I f e v e r y b o d y w h o n e e d e d t r e a t m e n t were to receive it the cost in t e r m s o f m o n e y a n d m a n - p o w e r w o u l d p r o b a b l y b e m o r e t h a n t h e c o u n t r y c o u l d afford. It would,\n\nM\n\n40" ]

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[ "TIME SERIES MODELING\n\n## Yield curve smoothing and residual variance of fixed income positions\n\nby Raphael Douady, 2011\n\nWe model the yield curve in any given country as an object lying in an in…nite-dimensional Hilbert space, the evolution of which is driven by what is known as a cylindrical Brownian motion. We assume that volatilities and correlations do not depend on rates (which hence are Gaussian). We prove that a principal component analysis (PCA) can be made. These components are called eigenmodes or principal deformations of the yield curve in this space. We then proceed to provide the best approximation of the curve evolution by a Gaussian Heath-Jarrow-Morton model that has a given …nite number of factors.\n\nFinally, we describe a method, based on …nite elements, to compute the eigenmodes using historical interest rate data series and show how it can be used to compute approximate hedges which optimise a criterion depending on transaction costs and residual variance." ]

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[ "# Distance rationalizability of scoring rules", null, "Working paper\nAuthor/s:\nBurak Can\nIssue number:\n68/2013\nSeries:\nGSBE\nPublisher:\nMaastricht University\nYear:\n2013\nCollective decision making problems can be seen as finding an outcome that is closest to a concept of consensus. 1 introduced Closeness to Unanimity Procedure as a first example to this approach and showed that the Borda rule is the closest to unanimity under swap distance a.k.a the 2 distance. 3 shows that the Dodgson rule is the closest to Condorcet under swap distance. 4, 5 generalized this concept as distance-rationalizability, where being close is measured via various distance functions and with many concepts of consensus, e.g., unanimity, Condorcet etc. In this paper, we show that all non-degenerate scoring rules can be distance-rationalized as Closeness to Unanimity procedures under a class of weighted distance functions introduced in 6. Therefore, the results herein generalizes 1 and builds a connection between scoring rules and a generalization of the Kemeny distance, i.e. weighted distances.\nDeveloped by Paolo Gittoi" ]

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[ "# bundle\n\n(redirected from Bachmann bundle)\nAlso found in: Dictionary, Thesaurus, Medical, Legal, Financial.\n\n## bundle\n\n1. Biology a collection of strands of specialized tissue such as nerve fibres\n2. Botany short for vascular bundle\n3. Textiles a measure of yarn or cloth; 60 000 yards of linen yarn; 5 or 10 pounds of cotton hanks\n\n## Bundle\n\nin mathematics, a two-parameter family of curves in the plane or in space that are linear functions of the parameters. Suppose F1, F2, and F3 are functions of two variables and none of the functions is a linear combination of the other two. The family of curves in the plane that are determined by the equation\n\n(*) λ1F1 + λ2F2 + λ3F3 = 0\n\nfor all possible values of the parameters λ1, λ2, and λ3 (except for the case λ1 = 0, λ2 = 0, and λ3 = 0) constitutes a bundle. Equation (*) is in fact a function of two parameters—that is, of the two ratios λ1: λ2: λ3. In addition, it is immediately apparent that the parameters occur in this equation linearly. The equation of a bundle of surfaces in space is formed analogously. The three equations F1 = 0, F2 = 0, and F3 = 0 yield three elements of the bundle (three curves of three surfaces), which determine the entire bundle.\n\nBundles are usually considered whose elements are similar in certain respects; examples are a bundle of circles and a bundle of planes. We sometimes also speak of a bundle of lines in space; although the bundle is considered in space, its elements are curves rather than surfaces. Nevertheless, this case can be reduced to the case of a bundle of planes, since the pairwise intersections of elements of a bundle of planes determine a set of lines. In projective geometry, a bundle is understood to mean both sets—lines and planes—at once.\n\n## bundle\n\n[′bən·dəl]\n(mathematics)\nA triple (E, p, B), where E and B are topological spaces and p is a continuous map of E onto B ; intuitively E is the collection of inverse images under p of points from B glued together by the topology of X.\n\n## bundle\n\n(1) To sell hardware and software as a combined product or to combine several software packages for sale as a single unit. Contrast with unbundle. See bundled software and bundling.\n\n(2) A collection of files that are treated as one. See APP file.\nSite: Follow: Share:\nOpen / Close" ]

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[ "# Subcircuit in SPICE\n\n#### soumendu89\n\nJoined Jan 22, 2014\n1\nModerator edit:\n\nYour post has been moved into its own thread, where it will draw more responses.\n\nPlease refrain from \"hijacking\" existing threads with tangent or off-topic questions. You can create new threads for new questions using the \"New Thread\" button on the upper left of the page.\n\nHello,\n\nI am trying to create a subcircuit in PSPICE and I need to include following (dependent)voltage expression between two nodes 2 and 10 inside the subcircuit as described above by unkukaracha\n\nE 2 10 value={(q/Ceq)*(Nsil*(exp(-2*V(2,10)*ET)-exp(V(46)))/(exp(-2*V(2,10)*ET)+exp(V(23))*exp(-1*V(2,10)*ET)+exp(V(46))))+{(q/Ceq)*(Nnit*(exp(-1*V(2,10)*ET))/(exp(-1*V(2,10)*ET)+(Kn)*exp(V(23))))}\n\nThe complete expression cannot fit in one line due to 132 character limit, so I tried using the following technique of breaking the expression into two lines\nE 2 10 value={(q/Ceq)*(Nsil*(exp(-2*V(2,10)*ET)-exp(V(46)))/(exp(-2*V(2,10)*ET)+exp(V(23))*exp(-1*V(2,10)*ET)+exp(V(46))))}\n+{(q/Ceq)*(Nnit*(exp(-1*V(2,10)*ET))/(exp(-1*V(2,10)*ET)+(Kn)*exp(V(23))))}\n\nYet, the simulator does not take into account the second line while calculating the voltage between nodes 2 and 10. I am using Orcad PSPICE for the simulation. Can someone help me in solving this issue?\n\nSoumendu\n\nLast edited by a moderator:\n\n#### tshuck\n\nJoined Oct 18, 2012\n3,534\nHello,\n\nI am trying to create a subcircuit in PSPICE and I need to include following (dependent)voltage expression between two nodes 2 and 10 inside the subcircuit as described above by unkukaracha\n\nE 2 10 value={(q/Ceq)*(Nsil*(exp(-2*V(2,10)*ET)-exp(V(46)))/(exp(-2*V(2,10)*ET)+exp(V(23))*exp(-1*V(2,10)*ET)+exp(V(46))))+{(q/Ceq)*(Nnit*(exp(-1*V(2,10)*ET))/(exp(-1*V(2,10)*ET)+(Kn)*exp(V(23))))}\n\nThe complete expression cannot fit in one line due to 132 character limit, so I tried using the following technique of breaking the expression into two lines\nE 2 10 value={(q/Ceq)*(Nsil*(exp(-2*V(2,10)*ET)-exp(V(46)))/(exp(-2*V(2,10)*ET)+exp(V(23))*exp(-1*V(2,10)*ET)+exp(V(46))))}\n+{(q/Ceq)*(Nnit*(exp(-1*V(2,10)*ET))/(exp(-1*V(2,10)*ET)+(Kn)*exp(V(23))))}\n\nYet, the simulator does not take into account the second line while calculating the voltage between nodes 2 and 10. I am using Orcad PSPICE for the simulation. Can someone help me in solving this issue?\n\nSoumendu\nHello," ]

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[ "# 452 centiliters in imperial cups\n\n## Conversion\n\n452 centiliters is equivalent to 15.90817603699 imperial cups.\n\n## Conversion formula How to convert 452 centiliters to imperial cups?\n\nWe know (by definition) that: $1\\mathrm{centiliter}\\approx 0.035195079727854\\mathrm{brcup}$\n\nWe can set up a proportion to solve for the number of imperial cups.\n\n$1 ⁢ centiliter 452 ⁢ centiliter ≈ 0.035195079727854 ⁢ brcup x ⁢ brcup$\n\nNow, we cross multiply to solve for our unknown $x$:\n\n$x\\mathrm{brcup}\\approx \\frac{452\\mathrm{centiliter}}{1\\mathrm{centiliter}}*0.035195079727854\\mathrm{brcup}\\to x\\mathrm{brcup}\\approx 15.90817603699001\\mathrm{brcup}$\n\nConclusion: $452 ⁢ centiliter ≈ 15.90817603699001 ⁢ brcup$", null, "## Conversion in the opposite direction\n\nThe inverse of the conversion factor is that 1 imperial cup is equal to 0.0628607577433628 times 452 centiliters.\n\nIt can also be expressed as: 452 centiliters is equal to $\\frac{1}{\\mathrm{0.0628607577433628}}$ imperial cups.\n\n## Approximation\n\nAn approximate numerical result would be: four hundred and fifty-two centiliters is about fifteen point nine one imperial cups, or alternatively, a imperial cup is about zero point zero six times four hundred and fifty-two centiliters.\n\n## Footnotes\n\n The precision is 15 significant digits (fourteen digits to the right of the decimal point).\n\nResults may contain small errors due to the use of floating point arithmetic." ]

[ null, "https://converter.ninja/images/452_cl_in_brcup.jpg", null ]

[ "💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!\n\n# (a) One way of defining $\\sec^{-1} x$ is to say that $y= \\sec^{-1} x \\Leftrightarrow \\sec y = x$ and $0 \\le y < \\pi/2$ or $\\pi \\le y < 3\\pi/2.$ Show that, with this definition,$\\frac {d}{dx} (\\sec^{-1} x) = \\frac {1}{x \\sqrt {x^2 - 1}}$(b) Another way of defining $\\sec^{-1} x \\Leftrightarrow \\sec y = x$ and $0 \\le y \\le \\pi, y \\not= \\pi/2.$ Show that, with this definition,$\\frac {d}{dx} (\\sec{-1} x) = \\frac {1}{\\mid x \\mid \\sqrt {x^2 -1}}$\n\n## (a )Click to see(b) Please see proof.\n\nDerivatives\n\nDifferentiation\n\n### Discussion\n\nYou must be signed in to discuss.\n##### Heather Z.\n\nOregon State University\n\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp\n\n### Video Transcript\n\nOkay. So that the reflective of Oxy gen x basest sort of is the stem. You start by writing it out, and you take impressive derivative on that. And so I use that. Cheryl, you're right. Why prime into himself, X However, there's that Stella you should hear Depends on how you pick your, uh, still men and their and the range when you're defining inverse function. Because in order to devising worse from shame must have want one function end up on If you defined the you need what one functioning depends on how you pick your moment And and this tension of oxy connects might be off by negative sign us is indicated here and I'll let you figure that out. I look a lot off should come true either pre cog bliss textbook or just to work out. Or you know how the sign changed when you pick a different moment to define owe a wrench to define the inverse function\n\n#### Topics\n\nDerivatives\n\nDifferentiation\n\n##### Heather Z.\n\nOregon State University\n\n##### Kristen K.\n\nUniversity of Michigan - Ann Arbor\n\n##### Michael J.\n\nIdaho State University\n\nLectures\n\nJoin Bootcamp" ]

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[ "# If the surface tension is reduced to half (say by using a surfactant), what would be the effect on the size of an air bubble in the liquid\n\nIf I reduce the surface tension to half of the original value, what would be the effect on the size of an air bubble in the fluid?\n\nWhat would be the effect on buoyancy and drag forces?\n\nAlso, if I have two bubbles one large and one small in a fluid, which will rise faster?\n\n# The effect of surface tension on bubble size\n\nThe Young-Laplace law describes the relation of the pressure difference $\\Delta p$ on the curvature $C=2/R$ of a spherical bubble: $$\\Delta p = 2\\frac{\\sigma}{R}$$ where proportionality constant is known as the surface tension $\\sigma$.\n\nNow, if the surface tension were reduced to half its original value $\\sigma/2$, at the same pressure difference, this bubble would exist at a larger radius $2R$. You can understand this as the surface tension being the force which resist an increase in surface area (i.e. radius $R$), and lowering this resistance allows the bubble to expand under its own over-pressure $\\Delta p$. An experiment you can do at home is to add some dishsoap to a glass of bubbly soda and see it foam over by the increase in bubble area.\n\nHowever, most likely what will happen for the case of halving the surface tension, is that the surface tension is unable to support such a sudden increase in radius and the bubble will simply burst. If you were to reduce the surface tension gradually to half the original value, you may be able to see the bubble increase twice in size.\n\n# Bouyancy and drag forces on spherical bubbles at low speeds\n\nAssuming the bubble exist with radius $2R$ due to the decreased surface tension, let's look at the bouyancy and drag force for slowly moving spherical droplets. The bouyancy force $F_b=\\rho_agV_b$ is the upward force resulting from the bubble occupying a volume $V_b$ in water. As this force is proportional to $V_b$, and the volume is proportional to $R^3$, the bouyancy force will increase by factor 8. The drag force $F_d=6\\pi\\mu Rv$ is given by Stokes' law and is proportional to $R$, however the speed $v$ may also depend on $R$ so this needs to be investigated first.\n\nA steady-state force balance on the bubble involving gravity, bouyancy and drag forces gives: $$0 = \\left(\\rho_w-\\rho_a\\right)gV_b - 6\\pi\\mu R v$$ solving for the terminal velocity of a spherical bubble: $$v = \\frac{\\left(\\rho_w-\\rho_a\\right)gV_b}{6\\pi\\mu R} \\propto R^2$$\n\nso the terminal velocity of a bubble is proportional to $R^2$, this makes the drag force proportional to $R^3$. Like the bouyancy, the drag force will increase by a factor 8 and the terminal velocity will increase by factor 4.\n\nThis analysis is only valid for spherical bubbles and at low speeds such that $\\text{Re}\\ll1$ (see below); at all times the results need to be checked against these assumptions.\n\n# A more general treatment\n\nThe shape a bubble will take depends on which regime it is in as shown in this figure. The regimes are characterized by three dimensionless numbers: $$\\text{Re}=\\frac{\\rho_wvD}{\\mu_w}\\quad\\text{Eo}=\\frac{\\Delta\\rho_wgD^2}{\\sigma}\\quad\\text{Mo}=\\frac{g\\mu_w^4\\Delta\\rho}{\\rho_w^2\\sigma^3}$$ The Reynolds number $\\text{Re}$ describes the relative importance of viscous to inertial forces and the Eotvos number $\\text{Eo}$ describes the relative importance of bouyancy to surface tension forces. The Morton number $\\text{Mo}$ characterizes the type of phases as it contains mostly material properties.\n\nFor bubbles at $\\text{Re}\\ll1$ we clearly see that the bubble shape is spherical inline with the treatment above for slow moving spherical bubbles. While the bubble regimes are presented on a log-log-log scale in which small variations have little impact on the regime the bubble is in, however, for higher Eotvos and Morton numbers we clearly see regime transitions as function of these dimensionless numbers. Now, by halving the surface tension and doubling the diameter we find:\n\n$$\\bar{\\text{Eo}}=\\frac{\\Delta\\rho_wg(2D)^2}{\\sigma/2}=8\\text{Eo}\\quad\\bar{\\text{Mo}}=\\frac{g\\mu_w^4\\Delta\\rho}{\\rho_w^2(\\sigma/2)^3}=8\\text{Mo}$$\n\nEotvos and Morton have almost increased by an order of magnitude. Depending on the original regime, this may have cause the bubble shape to shift into a new regime. As this will result in a separate Reynolds number which also depends on the new diameter, the terminal velocity of the bubble as well as the shape may have changed.\n\n• Presumably, it is the number of air molecules inside the bubble that stays constant, not the pressure difference. – Mark Eichenlaub Jul 8 '16 at 21:53\n• @MarkEichenlaub I would think this as well. I would say that if $\\sigma / R$ is constant and $\\sigma$ halves, then $R$ must halve as well. What actually happens is that when the surface tension is reduced, then the surface tension can no longer balance the gas's overpressure, so the bubble expands reducing its density until the new gas pressure balances the new overpressure. I get from a naive estimate that the new radius is $\\sqrt{2}$ of the old one, since overpressure has one power of the radius, but gas pressure has three. – Brian Moths Jul 8 '16 at 22:10" ]

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[ "# Cyclic Quadrilateral from the USAMO\n\n### Xiaoxue Li Am Math Monthly, V 123, N 1, Jan 2016, p. 96\n\nHere is problem 2, Part I, of the 28th United States of America Mathematical Olympiad.\n\nLet $ABCD\\;$ be a cyclic quadrilateral.", null, "Prove that\n\n$|AB-CD+|AD-BC|\\ge 2|AC-BD|.$\n\n### Solution\n\nIt may be assumed without loss of generality that $AB\\ge CD,\\;$ $AD\\ge BC,\\;$ and $AC\\ge BD.\\;$ It then suffices to show that $(AB-CD)+(AD-BC)\\ge 2(AC-BD).\\;$ We introduce auxiliary points $A',\\;$ $C',\\;$ and $D'\\;$ such that $EC-EC',\\;$ $ED=ED',\\;$ and $AA'=CD.\\;$ Observe that triangles $CDE\\;$ and $C'D'E\\;$ are congruent.", null, "$\\angle 1=\\angle 3=\\angle 2,\\;$ implying that $C'D'\\parallel AB.\\;$ By the construction $\\Delta EC'D'=\\Delta ECD,\\;$ in particular, $C'D'=CD=AA',\\;$ making AD'C'A' a parallelogram. Thus, $AB-CD=A'B.\\;$ Also\n\n\\begin{align} AC-BD &= AC-(a+b+BC')\\\\ &=AD'-BC'\\\\ &=A'C'-BC'\\\\ &\\le A'B. \\end{align}\n\nIt follows that $AB-CD\\ge AC-BD.\\;$ Similarly, $AD-BC\\ge AC-BD.\\;$ Adding these two inequalities yields the desired inequality. Equality holds when $ABCD\\;$ is a rectangle.", null, "" ]

[ null, "https://www.cut-the-knot.org/blue/USAMO28QuarilateralP.jpg", null, "https://www.cut-the-knot.org/blue/USAMO28QuarilateralS.jpg", null, "https://www.cut-the-knot.org/tbow_sh.gif", null ]

[ "", null, "# Find a subgroup of order 6 in U(700).", null, "EunoR 2021-01-08 Answered\nFind a subgroup of order $6\\in U\\left(700\\right)$.\nYou can still ask an expert for help\n\n## Want to know more about Complex numbers?\n\n• Questions are typically answered in as fast as 30 minutes\n\nSolve your problem for the price of one coffee\n\n• Math expert for every subject\n• Pay only if we can solve it", null, "Daphne Broadhurst\n\nLet z be some complex number such that $z={e}^{i\\frac{\\pi }{3}}$.\n\nThen ${z}^{6}=1,\\phantom{\\rule{1em}{0ex}}\\text{and}\\phantom{\\rule{1em}{0ex}}{z}^{i}\\ne 1$\nfor $0.\n\nDefine a $\\left(700×700\\right)$ matrix $A=\\left[{a}_{ij}\\right]$ by\n\nThen ${A}^{\\cdot }=\\left[{b}_{ij}\\right]$ is\n\nwhich means that\n$A{A}^{\\cdot }={A}^{\\cdot }A=I$,\nso A is unitary.\nFurthermore, ${A}^{k}=\\left[{a}_{ij}^{k}\\right]$ is given by\n\nsince A is a diagonal matrix.\nNow assume that k is the smallest positive integer such that ${A}^{k}=I$. Then .\n\nFurthermore, it is easy to see that ${A}^{6}=I$ (other elements remain 1 or 0 whatever k is).\nTherefore, A is of order 6, so one subgroup of order $6\\in U\\left(700\\right)$ is given by the cyclic group $⟨A⟩$.\nResult:\nHint:Define a $\\left(700×700\\right)$ matrix $A=\\left[{a}_{ij}\\right]$ by\n${a}_{ij}=\\left\\{\\left({e}^{-i\\frac{\\pi }{3}},i=j\\phantom{\\rule{1em}{0ex}}\\text{and}\\phantom{\\rule{1em}{0ex}}i=1,2,\\phantom{\\rule{1em}{0ex}}\\text{or}\\phantom{\\rule{1em}{0ex}}3\\right),\\left(1,i=j\\phantom{\\rule{1em}{0ex}}\\text{and}\\phantom{\\rule{1em}{0ex}}i>3\\right),\\left(i\\ne j\\right)$" ]

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[ "", null, "2018-07-13 22:56:29\n\nupd1:更新5.2.2-对于该子目所阐述的操作“用两个堆来维护一些查询第k小/大的操作”更新了一道例题-该操作对于中位数题目的求解\n\nupd2：更新5.3-利用堆来维护可以“反悔的贪心”\n\ncontinue...\n\n# 一.堆的性质\n\n1.堆是一颗完全二叉树\n\n2.堆的顶端一定是“最大”，最小”的，但是要注意一个点，这里的大和小并不是传统意义下的大和小，它是相对于优先级而言的，当然你也可以把优先级定为传统意义下的大小，但一定要牢记这一点，初学者容易把堆的“大小”直接定义为传统意义下的大小，某些题就不是按数字的大小为优先级来进行堆的操作的\n\n（但是为了讲解方便，下文直接把堆的优先级定为传统意义下的大小，所以上面跟没讲有什么区别？\n\n3.堆一般有两种样子，小根堆和大根堆，分别对应第二个性质中的“堆顶最大”“堆顶最小”，对于大根堆而言，任何一个非根节点，它的优先级都小于堆顶，对于小根堆而言，任何一个非根节点，它的优先级都大于堆顶（这里的根就是堆顶啦qwq）", null, "# 二.堆的操作\n\n## 1.插入", null, "", null, "", null, "", null, "Code：\n\nvoid swap(int &x,int &y){int t=x;x=y;y=t;}//交换函数\nint heap[N];//定义一个数组来存堆\nint siz;//堆的大小\nvoid push(int x){//要插入的数\nheap[++siz]=x;\nnow=siz;\n//插入到堆底\nwhile(now){//还没到根节点，还能交换\nll nxt=now>>1;//找到它的父亲\nif(heap[nxt]>heap[now])swap(heap[nxt],heap[now]);//父亲比它大，那就交换\nelse break;//如果比它父亲小，那就代表着插入完成了\nnow=nxt;//交换\n}\nreturn;\n}\n\n## 2.删除", null, "", null, "", null, "", null, "", null, "Code：\n\nvoid pop(){\nswap(heap[siz],heap);siz--;//交换堆顶和堆底，然后直接弹掉堆底\nint now=1;\nwhile((now<<1)<=siz){//对该节点进行向下交换的操作\nint nxt=now<<1;//找出当前节点的左儿子\nif(nxt+1<=siz&&heap[nxt+1]<heap[nxt])nxt++;//看看是要左儿子还是右儿子跟它换\nif(heap[nxt]<heap[now])swap(heap[now],heap[nxt]);//如果不符合堆性质就换\nelse break;//否则就完成了\nnow=nxt;//往下一层继续向下交换\n}\n}\n\n# 三.堆的STL实现\n\n首先你需要一个头文件：#include<queue>\npriority_queue<int> q;//这是一个大根堆q\npriority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q\n//注意某些编译器在定义一个小根堆的时候greater<int>和后面的>要隔一个空格，不然会被编译器识别成位运算符号>>\n\nq.top()//取得堆顶元素，并不会弹出\nq.pop()//弹出堆顶元素\nq.push()//往堆里面插入一个元素\nq.empty()//查询堆是否为空，为空则返回1否则返回0\nq.size()//查询堆内元素数量\n\nstruct node{\nint val,rnd;\n}a;\n\nstruct node{\nint val,rnd;\nbool operator < (const node&x) const {\nreturn rnd<x.rnd;\n}\n}a;\n\n# 五.堆的应用\n\n## 2.用两个堆来维护一些查询第k小/大的操作\n\nLuogu中也有此类例题，题解内也讲的比较清楚了，此处不再赘述，读者可当做拓展练习进行食用\n\n#include <cstdio>\n#include <vector>\n#include <cstring>\n#include <queue>\n#define ll long long\n#define inf 1<<30\n#define il inline\n#define in1(a) read(a)\n#define in2(a,b) in1(a),in1(b)\n#define in3(a,b,c) in2(a,b),in1(c)\n#define in4(a,b,c,d) in2(a,b),in2(c,d)\nil int max(int x,int y){return x>y?x:y;}\nil int min(int x,int y){return x<y?x:y;}\nil int abs(int x){return x>0?x:-x;}\nil void swap(int &x,int &y){int t=x;x=y;y=t;}\nil void readl(ll &x){\nx=0;ll f=1;char c=getchar();\nwhile(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}\nwhile(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}\nx*=f;\n}\nil void read(int &x){\nx=0;int f=1;char c=getchar();\nwhile(c<'0'||c>'9'){if(c=='-')f=-f;c=getchar();}\nwhile(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}\nx*=f;\n}\nusing namespace std;\n/*===================Header Template=====================*/\n#define N 200010\npriority_queue<int,vector<int>,greater<int> > q;\npriority_queue<int> q1;\nint n,m,a[N],b[N];\nint main(){\nin2(n,m);\nfor(int i=1;i<=n;i++)in1(a[i]);\nfor(int i=1;i<=m;i++)in1(b[i]);\nint i=1;\nfor(int j=1;j<=m;j++){\nfor(;i<=b[j];i++){\nq1.push(a[i]);\nif(q1.size()==j)q.push(q1.top()),q1.pop();\n}\nprintf(\"%d\\n\",q.top());\nq1.push(q.top());q.pop();\n}\nreturn 0;\n}\n\n## 3.利用堆来维护可以“反悔的贪心”\n\n#include <cstdio>\n#include <algorithm>\n#include <queue>\n#include <vector>\n#include <map>\n#include <set>\n\nusing namespace std ;\n\n#define N 100010\n#define int long long\n\nint n , m ;\nstruct node {\nint d,p ;\nbool operator < ( const node &x ) const { return p>x.p; }\n} a[ N ] ;\n\nbool cmp( node a , node b ) {\nreturn a.d==b.d?a.p>b.p:a.d<b.d;\n}\n\npriority_queue< node > q ;\n\nsigned main() {\nscanf( \"%lld\" , &n ) ;\nfor( int i = 1 ; i <= n ; i ++ ) {\nscanf( \"%lld%lld\" , &a[i].d , &a[i].p ) ;\n}\nsort(a+1,a+n+1,cmp);\nint ans = 0 ;\nfor( int i = 1 ; i <= n ; i ++ ) {\nif( a[i].d<=(int)q.size() ) {\nif( q.top().p<a[i].p ) {\nans += a[i].p-q.top().p ;\nq.pop() ;\nq.push(a[i]) ;\n}\n} else q.push(a[i]) , ans += a[ i ].p ;\n}\nprintf( \"%lld\\n\" , ans ) ;\n}\n\ncontinue...持续更新中\n\n• star\n首页" ]

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[ "Python Program to find the LCM of two numbers\n\nHello everyone, welcome back to programminginpython.com. Here I am going to tell a simple logic by which we can find the LCM of two numbers in python.\n\nLCM means Least Common Multiple, for a given 2 numbers we need to find the least common multiple for them.\n\nLet’s take an example of 3 and 5, here I will find the LCM of those 2 numbers.\n\nMultiples of 3: 3, 6, 9, 12, 15, 18, 21, 24…\n\nMultiples of 5: 5, 10, 15, 20, 25, 30….\n\nNow we find the first number which is found in both the multiples and it is clear that 15 is that number, and the LCM(3, 5) will be 15.\n\nYou can also watch the video on YouTube here.\n\nLCM of 2 numbers – Code Visualization\n\nTo find LCM of 2 numbers\n\nApproach\n\n• Read two input numbers using input().\n• Find the minimum of 2 numbers, using min() function and store the value in numbers_min variable.\n• Now run a while loop and check whether both numbers are divisible by the numbers_min variable which we got in the previous step.\n• if both numbers are divisible by numbers_min print the value as LCM and break the loop\n• if not, increment numbers_min value and continue while loop." ]

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[ "# Problem with ParametricNDSolve\n\ni have a strange problem with ParametricNDSolve. Given the sistem of differential euqations\n\nfun =\nParametricNDSolveValue[\n{x'[s] == v[s], v'[s] == a[s], a'[s] == a[s] - 2 x[s] E^(-x[s]^2),\nx == x0, v == v0, a == 0},\n{x, v, a}, {s, tmin, 0}, {x0, v0, tmin}];\n\n\nParametricNDSolve returns the correct solution only for certain values of tmin. For example, with\n\n{x0, v0} = {10, 2}\n\n\nit gives the correct solution only for tmin < -74, while for tmin > -74 it doesn't.\n\nIt's possible to visualize this with\n\nTable[Plot[fun[10, 2, -50][[i]][t], {t, -10, 0}, PlotRange -> All], {i, 1, 3}]\nTable[Plot[fun[10, 2, -100][[i]][t], {t, -10, 0}, PlotRange -> All], {i, 1, 3}]", null, "and\n\nPlot[fun[10, 2, t][][-5], {t, -100, 0}]\n\n\nwhere is possible to see the dependence of the solution calculated in t = -5 on the value of tmin. I tried to increase the WorkingPrecision but to no avail. Any ideas?\n\n• Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. Feb 18, 2015 at 3:10\n\nUse AccuracyGoaland PrecisionGoal\n\nfun = ParametricNDSolveValue[{x'[s] == v[s], v'[s] == a[s], a'[s] == a[s] - 2 x[s] E^(-x[s]^2),\nx == x0, v == v0, a == 0}, {x, v, a}, {s, tmin, 0}, {x0, v0, tmin},\nAccuracyGoal -> 10, PrecisionGoal -> 10];\n\nTable[Plot[fun[10, 2, -50][[i]][t], {t, -10, 0}, PlotRange -> All], {i, 1, 3}]\nTable[Plot[fun[10, 2, -80][[i]][t], {t, -10, 0}, PlotRange -> All], {i, 1, 3}]", null, "• I believe setting PrecisionGoal is enough for this problem Feb 18, 2015 at 3:31\n• Thanks! i tried that before, but for some reason i thought it didn't work. Feb 18, 2015 at 12:33\n\nThe underlying problem is with the step size, which is controlled by various options, including PrecisionGoal (as shown in belisarius's answer. It is also controlled by AccuracyGoal, MaxStepSize, MaxStepFraction and some others.\n\nThe default setting for MaxStepFraction is 1/10, which is not fine enough to stumble upon a little blip near the initial condition (at the beginning of the integration, which NDSolve calculates \"backwards\" from right to left). The following shows the step size to be 10.\n\nClear[fun];\nfun = ParametricNDSolveValue[{\nx'[s] == v[s], v'[s] == a[s], a'[s] == a[s] - 2 x[s] E^(-x[s]^2),\nx == x0, v == v0, a == 0},\n{x, v, a}, {s, tmin, 0}, {x0, v0, tmin}];\n\nWith[{f0 = fun[10, 2, -100]},\nGraphicsRow@\nTable[Plot[f0[[i]][t], {t, -20, 0}, PlotRange -> All,\nMesh -> f0[[i]][\"Coordinates\"],\nMeshStyle -> {PointSize[Medium], Red}], {i, 1, 3}]]", null, "Adding any one of the following lines results in an accurate solution for tmin = -100.\n\nMaxStepSize -> 2 (* should work for all tmin *)\nMaxStepFraction -> 1/20 (* relative to the size of tmin *)\nAccuracyGoal -> 17, PrecisionGoal -> 0 (* both together *)\nPrecisionGoal -> 8.3 (* or higher; > 8.5 is fairly robust *)\n\n\nThe problem is undoubtedly that the error estimate near the initial condition is so small that NDSolve feels it can take the maximum step. This passes over the interval of s where x is small, which is where the acceleration a[s] changes significantly. Another approach is to slow down the integration when x[s] gets small. If x[s] == 5, then the term 2 x[s] E^(-x[s]^2) from a[s] will be about 10^-10, which is perhaps a little small (the default AccuracyGoal and PrecisionGoal is about 8). But the starting step size is small enough that the solution is found.\n\nClear[fun];\nfun = ParametricNDSolveValue[{\nx'[s] == v[s], v'[s] == a[s], a'[s] == a[s] - 2 x[s] E^(-x[s]^2),\nx == x0, v == v0, a == 0,\nWhenEvent[x[s] == 5, \"RestartIntegration\"]},\n{x, v, a}, {s, tmin, 0}, {x0, v0, tmin}];\n\nWith[{f0 = fun[10, 2, -100]},\nGraphicsRow@\nTable[Plot[f0[[i]][t], {t, -10, 0}, PlotRange -> All,\nMesh -> f0[[i]][\"Coordinates\"],\nMeshStyle -> {PointSize[Small], Red}], {i, 1, 3}]]", null, "This answer basically explains what (I think) is going on with NDSolve. The solution proposed by belisarius seems like a good first stab at solution. My own knee-jerk reaction in such a case is to try WorkingPrecision -> 20, which increases AccuracyGoal and PrecisionGoal to 10, as in belisarius's solution, and also uses arbitrary precision reals. The setting slightly above MachinePrecision with the precision tracking gives me some feedback about the numerical stability of the computation of the solution at MachinePrecision.\n\n• Thanks for the answer! So the step size is a function of the domain of the independent variable? I couldn't figure out how changing tmin affects the whole integration bacause, as you said, the numerical integration starts at s=0 and goes backwards. Feb 19, 2015 at 16:21\n• @Ivactheseeker Yes, it's a function of the domain among other things. NDSolve tests a few values at the start of the integration interval and adjusts the step size according to its error estimate. In this case, it jumps to MaxStepFraction. For tmin = -50, it tries the step -5; but that lands where a'[s] is changing, and the error estimate is too large. So it tries a step of -5/4 = -1.25, which it uses until it needs to adjust the step size again. With tmin = -100, the step of -10 jumps over the variation and the error is estimated to be small; so it keeps on going. Feb 19, 2015 at 18:29\n• @Michael_E2 Ok, thanks again!. Feb 20, 2015 at 11:14" ]

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[ "Toward realistic simulations of magneto-thermal winds from weakly-ionized protoplanetary disks\n\n# Toward realistic simulations of magneto-thermal winds from weakly-ionized protoplanetary disks\n\nOliver Gressel Niels Bohr International Academy, The Niels Bohr Institute, Blegdamsvej 17, DK-2100, Copenhagen Ø, Denmark [email protected]\n###### Abstract\n\nProtoplanetary disks (PPDs) accrete onto their central T Tauri star via magnetic stresses. When the effect of ambipolar diffusion (AD) is included, and in the presence of a vertical magnetic field, the disk remains laminar between 1-5au, and a magnetocentrifugal disk wind forms that provides an important mechanism for removing angular momentum. We present global MHD simulations of PPDs that include Ohmic resistivity and AD, where the time-dependent gas-phase electron and ion fractions are computed under FUV and X-ray ionization with a simplified recombination chemistry. To investigate whether the mass loading of the wind is potentially affected by the limited vertical extent of our existing simulations, we attempt to develop a model of a realistic disk atmosphere. To this end, by accounting for stellar irradiation and diffuse reprocessing of radiation, we aim at improving our models towards more realistic thermodynamic properties.\n\n## 1 Introduction\n\nInterpreting observational properties of T Tauri systems is closely tied to understanding the complex dynamical evolution of gaseous protoplanetary disks (PPDs), both in terms of their chemistry, and in terms of the microphysics that govern the evolution of the embedded magnetic fields. Moreover, with PPDs being the birth sites for extrasolar planets, a sound physical picture of the dust and planetesimal evolution is needed to ultimately provide the building blocks for a comprehensive theory of planet formation.\n\nThe fundamental drivers of disk evolution are mass loss processes and redistribution of angular momentum . In sufficiently ionized parts of the disk, the latter can be achieved by turbulent stresses. When accounting for the ionization structure of a typical PPD, large parts of the disk, however, remain laminar owing to the effect of ambipolar diffusion (AD). In this situation, angular momentum is primarily transported via a magnetocentrifugal disk wind [5, 6]. While this picture is further complicated when accounting for the Hall effect [7, 8], it has nevertheless become clear that the thermal structure of the disk plays an important role in setting the mass loading of the disk wind , and hence the timescale on which the system evolves [10, 11].\n\n## 2 Methods\n\nAs in our previous work , we are solving the single-fluid MHD equations including Ohmic resistivity and ambipolar diffusion, that is, the electromotive force resulting from the mutual collision of ions and neutrals. The diffusion coefficients, , and are specified by means of a look-up table, which has been obtained by solving a simple chemical ionization/recombination network. The resulting electromotive forces stemming from the Ohmic and ambipolar diffusion terms are then given by\n\n EO≡−ηO(∇×B),and (1)\n\nwith being the unit vector along , and where the double vector product results in an additional minus sign, such that positive coefficients and signify diffusion of magnetic fields, with the latter only being sensitive to currents perpendicular to the field direction.\n\nFor the typical number densities in the context of PPDs, the frictional coupling between the ions and neutrals happens so quickly (compared to dynamical timescales of interest) that we assume the charged species move at their terminal velocity with respect to the neutrals, that is, where the Lorentz force is balanced by the drag. Because of the low degree of ionization, we formulate our continuity equation and momentum conservation in terms of the neutral component, although it does experience the Lorentz force as mediated by particle collisions.\n\nWe use a modified version of the nirvana-iii finite volume Godunov code [12, 13]. The code adopts a total-energy formulation111 Note that we do not include the radiation energy density, , in the total energy (as it is sub-dominant in the problem we consider). We however retain the radiation flux in the momentum equation for consistency. with conserved variables , , and . Together with the conservation of radiation energy, , and defining the total pressure, , as the sum of the gas and magnetic pressures, the system of equations we solve reads\n\n ∂tρ+∇⋅(ρv) = 0, (3) ∂t(ρv)+∇⋅[ρvv+p⋆I−BB] = −ρ∇Φ+ρκR/cFR, (4) ∂te+∇⋅[(e+p⋆)v−(v⋅B)B] = ∇⋅[(EO+EAD)×B]−ρ(∇Φ)⋅v (5) +cρκP(ER−aRT4)+ρκR/cFR⋅v+Q+irr, ∂tER+∇⋅(ERv) = −cρκP(ER−aRT4)−∇⋅FR−PR:∇v, (6) ∂tB−∇×[v×B+EO+EAD] = 0, (7)\n\nwhere is the radiation flux, is the radiation pressure tensor, and are the Rosseland and Planck mean opacities, respectively, is the radiation density constant, is the gas temperature, and the gas pressure is , where is chosen as appropriate for an ideal diatomic gas. The gravitational term represents the point-mass potential of the solar-mass star at the center of our spherical-polar coordinate system, and represents external irradiation heating due to the star.\n\n### 2.1 Flux limited diffusion\n\nThe above equations can only be solved once the radiation flux and pressure tensor are specified. An attractive method for obtaining the Eddington tensor that relates with is to solve for the steady-state but angle-dependent radiation intensity and integrate the first (for ) and second (for ) order moments directly . In the interest of maintaining minimum algorithmic complexity and computational expense, we instead adopt the classical ad hoc closure\n\n FR=−λ(R)cρκR∇ER, (8)\n\nthat specifies in terms of a diffusive flux with diffusion coefficient , where\n\n R≡|∇ER|ρκR⟨ER⟩ (9)\n\nis a dimensionless number specifying how abruptly the radiation energy density varies compared to the length scale defined by the optical extinction coefficient , and where (for ) is a limiter function that guarantees in regions of low optical depth, that is, in regions where the diffusion approximation is not valid. In the optically thick limit, (for ), which corresponds to the Eddington approximation.\n\nThe described approach has its known shortcomings over characteristics-based methods , but in combination with stellar irradiation heating (determined using a simplified ray tracing algorithm) it has been deemed an acceptable compromise [16, 17, 18] in terms of being able to incorporate radiation thermodynamics in fully dynamical 3D simulations, whereas more accurate Monte Carlo methods are comparatively expensive and have to deal with statistical noise .\n\nWe currently treat both irradiation and diffuse redistribution in the gray approximation, but multigroup approaches are straightforward , especially for the irradiation component [17, 18], wherein computational expenditure scales with the number of frequency bins, and where increased realism can be achieved for the thermal structure of the outer disk (). For our PPD model, we precompute look-up tables for mean opacities, and , using D. Semenov’s opacity.f , where a fixed dust-to-gas mass ratio of is assumed. To account for depletion of small dust by grain growth, we enable the reduction of the obtained opacities by a scale factor. Since the opacity tables combine contributions from dust grains and gas molecules, this is only valid for temperatures below the dust sublimation threshold.\n\n### 2.2 Reduced speed of light approximation\n\nThe thermodynamic coupling of the gaseous matter with the radiation field is described by the terms in eqns. (5) and (6), respectively. Subsuming the term in eqn. (6) with the definition of the diffusive flux (8) amounts to a diffusion equation for the redistribution of radiation energy. Both effects can be combined into the subsystem\n\n ∂tϵ = +cρκP(ER−aRT4), (10) c^c∂tER = −cρκP(ER−aRT4)+∇⋅[D∇ER], (11)\n\nwhich we solve by means of operator splitting. Unlike in ref. , we do not include the term in this subsystem but instead treat it as a regular source term when updating the main hyperbolic system of equations. Since, for a large diffusion coefficient , the parabolic system (11) becomes stiff, the most common approach is to use implicit methods to solve it. Especially in view of applications employing adaptive mesh refinement, we have chosen to avoid an implicit update for , as it demands costly non-local communication patterns, and a potentially expensive matrix inversion.\n\nInstead, we integrate the diffusion part of (11) in a time-explicit fashion, and use the reduced speed of light approach to ameliorate the strict time-step constraints. This method has recently been employed in the context of simulations of the interstellar medium . The approximation is valid as long as the radiative timescales resulting from the adopted artificial value of (with const.) are still short compared to any relevant dynamical timescales. In the context of PPDs, we are mainly interested in the role of radiative effects in setting the consistent temperature structure of the disk, and true radiation hydrodynamic effects are believed to only be of minor importance during the T Tauri phase .\n\nThe approximation is introduced by amplifying the left-hand-side of (11) by a factor of . It is crucial that, because all other terms remain unaffected, this implies the modification does not alter the late-time steady-state solution, where , but only changes the timescale on which this solution is achieved. On practical grounds, we multiply (11) by , which implies replacing for in the radiation matter coupling, and attenuating the diffusion term by a factor of . This illustrates how the method works to weaken the stiffness of the diffusion term. To explicitly integrate (11), we employ the second-order accurate Runge-Kutta-Legendre (RKL2) super-time-stepping scheme , that is already used for the updates of the other parabolic terms (such as, viscosity, thermal conduction, resistive and ambipolar diffusion) in the code.\n\nEven with the reduced speed, , the radiation-matter-coupling term can itself contribute a severe timestep constraint in regions of high opacity, where the coupling becomes stiff. Ignoring, for the moment, the diffusion term in (11), the coupling amounts to an ordinary differential equation, similar to production/destruction equations that are common in other fields of science. While these are typically solved by explicit Runge-Kutta (RK) methods, higher order predictor/corrector schemes do not typically guarantee positivity or conservation of, in our case, the energy .\n\nThere however exists a class of modified RK methods that employ the so-called Patankar trick – an implicit weighting of the production/destruction terms with the ratio of the evolved quantity before and after the update. It can be shown that for a single-step update, such a weighting precludes that negative values are obtained. Moreover, since the weighting factors are overall symmetric, conservation is guaranteed. Specifically, we use the MPRK scheme given by eqn. (27) in ref. . The resulting update is formally implicit, but can algebraically be manipulated into fully explicit form, that is,\n\n Δe ≡ ϵ(n+1)−ϵ(n) (12) ≡ (E(n)R−E(n+1)R)ϕ−1=E(n)R−(aRT4)(n)ϕ+(aRT4)(n)/ϵ(n)+(cρκPΔt)−1,\n\nfor the predictor step, and\n\n Δe=E⋆Rϵ†E(n)R−(aRT4)⋆E†Rϵ(n)ϕE⋆Rϵ†+(aRT4)⋆E†R+E†Rϵ†(cρκPΔt)−1, (13)\n\nfor the corrector step, where is the forward-Euler predictor value of , and is the time-averaged state.222The same conventions, of course, apply for the terms , and . Compared to implicit methods, that demand iterative root-finding, or so-called -schemes (see, e.g., section 3.4 in ref. ) the method presented here offers a relatively inexpensive, parameter-free non-iterative alternative.\n\nOur existing global disk models [27, 28] have either assumed a locally-isothermal temperature , with being the cylindrical radius from the star, or have used an adiabatic equation of state with a Newtonian cooling term in the energy equation that reinstated the profile on a specified timescale (typically a short fraction of the local orbital period).\n\nCompared to this, even relatively basic models of dust absorption and re-radiation of star light in the disk surface, obtain a much more complex temperature structure within the PPD – with superheated surface layers and cool interiors. To account for such effects, we include a frequency-integrated radially attenuated irradiation flux\n\n FR,irr(r)=F(r⋆)(r⋆r)2exp(−τP(r))^r (14)\n\nfrom the central star with effective temperature , and , and where the optical depth, is obtained by integrating along radial rays. Following previous work [16, 17, 18], we obtain the irradiation energy source term by computing the negative divergence of (14) over each grid cell. In regions of low optical depth, we use the integral formulation\n\n Q+irr(ri)=ρκPΔVr∫ri+\\nicefrac12ri−\\nicefrac12^r⋅FR,irr(r′)r′2dr′, (15)\n\nwith instead, as this formulation has been found to produce a more accurate solution on the discretized mesh, when differences across cells are small .\n\n## 3 Results\n\n### 3.1 Radiative transfer test problems\n\nTo verify our implementation of the radiation-matter-coupling term, we have performed a simple one-zone model , where the thermal energy density, , is initially out of balance with the radiation energy density . In Fig. 1, we show three cases with , , and , which all converge to the same final equilibrium state. For the purpose of plotting the curves, we have limited the timestep artificially to sample timescales shorter than the radiative equilibration timescale. We have, however, tested that even for numerical time steps somewhat larger than the coupling timescale the scheme remains stable, as expected from the implicit-like integration scheme (see sect. 2.3) that we use.\n\nA standard test case for assessing the interplay of the radiation-matter-coupling with the radiation diffusion are radiative shocks, for which there exist semi-analytic solutions in simplified situations . In Fig. 2, we plot the solution of the case from ref. , using grid cells in the  direction, as well as two levels of adaptive mesh refinement (triggered by gradients in the thermal energy, ), which are shown as gray shaded areas in the plot. As seen in the lower panel of Fig. 2, apart from the shock location (which has been shifted by with respect to the semi-analytic solution), all quantities agree to within accuracy. We have also successfully performed the harder test, which we omit here for brevity. An excellent description of the radiative shock test, including the precise values used for initial conditions, can be found in sect. 4.4 of ref. .\n\n### 3.2 Preliminary global MHD simulations with irradiation\n\nReturning to the original motivation for implementing radiative physics in our modified version of the nirvana-iii code, we conclude by presenting a preliminary snapshot of a global axisymmetric protoplanetary disk simulation including Ohmic resistivity, ambipolar diffusion, radiation diffusion, and stellar irradiation. The ultimate goal of these simulations is to study how the mass-loading of the magnetocentrifugal wind depends on the disk thermal structure.", null, "Figure 3: Detail from a proof-of-concept global MHD simulation of a PPD with ambipolar diffusion, radiative transfer and stellar irradiation, showing ¯Bϕ (colour), poloidal velocity (black) and magnetic field lines (white), and iso-contours of the radiation temperature (grey).\n\nAs a simple proof-of-concept, we present a close-up of the inner disk in a simulation covering seven (initial) pressure scale heights in latitude (see Fig. 3). The basic disk setup is very similar to the one used in ref. , and we have additionally used opacities that correspond to a dust-depletion of a factor of ten compared to the typical interstellar abundance. Similar to our previous simulations, the magnetic field lines (white) bend outward, and in the upper disk layers, where the matter is sufficiently coupled to the magnetic field, a magnetocentrifugal disk wind ensues (black vectors).\n\nIso-contour lines (gray) of the radiation temperature illustrate the disk’s thermal structure that deviates noticeably from the constant-on-cylinders radial temperature profile , that we have assumed previously. The presented preliminary run used a reduced-speed-of-light factor . Further tests will show whether the chosen time-explicit framework is powerful and efficient enough to be of use in realistic situations.\n\n\\ack\n\nThe author thanks Jon Ramsey for many useful discussions, for carefully reading this manuscript, and for providing his code to compute semi-analytic reference solutions of radiative shocks. Dmitry Semenov is acknowledged for providing helpful clues in connection with his opacity code. The research leading to these results has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement No 638596).\n\n## References\n\n• Williams J P and Cieza L A 2011 ARA&A 49 67–117 (Preprint 1103.0556)\n• Johansen A, Blum J, Tanaka H, Ormel C, Bizzarro M and Rickman H 2014 PPVI 547\n• Gressel O, Nelson R P and Turner N J 2012 MNRAS 422 1140–1159 (Preprint 1202.0771)\n• Turner N J, Fromang S, Gammie C, Klahr H, Lesur G, Wardle M and Bai X N 2014 PPVI 411–432\n• Bai X N and Stone J M 2013 ApJ 769 76 (Preprint 1301.0318)\n• Gressel O, Turner N J, Nelson R P and McNally C P 2015 ApJ 801 84 (Preprint 1501.05431)\n• Lesur G, Kunz M W and Fromang S 2014 A&A 566 A56 (Preprint 1402.4133)\n• Simon J B, Lesur G, Kunz M W and Armitage P J 2015 MNRAS 454 1117–1131 (Preprint 1508.00904)\n• Bai X N, Ye J, Goodman J and Yuan F 2016 ApJ 818 152 (Preprint 1511.06769)\n• Bai X N 2016 ApJ 821 80 (Preprint 1603.00484)\n• Suzuki T K, Ogihara M, Morbidelli A, Crida A and Guillot T 2016 A&A (Preprint 1609.00437)\n• Ziegler U 2004 JCoPh 196 393–416\n• Ziegler U 2011 JCoPh 230 1035–1063\n• Jiang Y F, Stone J M and Davis S W 2012 ApJS 199 14 (Preprint 1201.2223)\n• Levermore C D and Pomraning G C 1981 ApJ 248 321–334\n• Bitsch B, Crida A, Morbidelli A, Kley W and Dobbs-Dixon I 2013 A&A 549 A124 (Preprint 1211.6345)\n• Kuiper R and Klessen R S 2013 A&A 555 A7 (Preprint 1305.2197)\n• Ramsey J P and Dullemond C P 2015 A&A 574 A81 (Preprint 1409.3011)\n• Noebauer U, Sim S, Kromer M, Röpke F and Hillebrandt W 2012 MNRAS 425 1430 (Preprint 1206.6263)\n• González M, Vaytet N, Commerçon B and Masson J 2015 A&A 578 A12 (Preprint 1504.01894)\n• Semenov D, Henning T, Helling C, Ilgner M and Sedlmayr E 2003 A&A 410 611–621\n• Gnedin N Y and Abel T 2001 NewA 6 437–455 (Preprint astro-ph/0106278)\n• Skinner M A and Ostriker E C 2013 ApJS 206 21 (Preprint 1306.0010)\n• Hartmann L 1998 Accretion Processes in Star Formation\n• Meyer C D, Balsara D S and Aslam T D 2012 MNRAS 422 2102–2115\n• Burchard H, Deleersnijder E and Meister A 2003 AppliedNumMath 47 1 – 30 ISSN 0168-9274\n• Nelson R P, Gressel O and Umurhan O M 2013 MNRAS 435 2610–2632 (Preprint 1209.2753)\n• Gressel O, Nelson R P, Turner N J and Ziegler U 2013 ApJ 779 59 (Preprint 1309.2871)\n• Chiang E I and Goldreich P 1997 ApJ 490 368–376 (Preprint astro-ph/9706042)\n• Bruls J H M J, Vollmöller P and Schüssler M 1999 A&A 348 233–248\n• Turner N J and Stone J M 2001 ApJS 135 95–107 (Preprint astro-ph/0102145)\n• Lowrie R B and Edwards J D 2008 Shock Waves 18 129–143\nYou are adding the first comment!\nHow to quickly get a good reply:\n• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.\n• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.\n• Your comment should inspire ideas to flow and help the author improves the paper.\n\nThe better we are at sharing our knowledge with each other, the faster we move forward.\nThe feedback must be of minimum 40 characters and the title a minimum of 5 characters", null, "", null, "", null, "" ]

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[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Multiply 25/60 with 14/84\n\nThis multiplication involving fractions can also be rephrased as \"What is 25/60 of 14/84?\"\n\n25/60 × 14/84 is 5/72.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 25/60 × 14/84 = 25 × 14/60 × 84 = 350/5040\n3. After reducing the fraction, the answer is 5/72\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Multiply Fractions Calculator\n\n×\n\n© everydaycalculation.com" ]

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[ "Getting Image", null, "", null, "", null, "", null, "Register now for special offers", null, "+91\n\nHome\n\n>\n\nEnglish\n\n>\n\nClass 14\n\n>\n\nMaths\n\n>\n\nChapter\n\n>\n\nTheory Of Equations\n\n>\n\nIf 5 and -6 are the roots of...\n\n# If 5 and -6 are the roots of the quadratic equation, then what will be the factors of the quadratic equation?", null, "Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!\n\nUpdated On: 27-06-2022", null, "Text Solution\n\n(x-5) and (x-6) (x-5) and (x+6)(x+5) and (x-6)(x+5) and (x+6)", null, "Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.\n\n## Related Videos\n\n648059640\n\n94\n\n6.0 K\n\n1:34\nयदि किसी द्विघात समीकरण के मूल 4 तथा 5 हो, तो समीकरण क्या होगा ?\n94851514\n\n99\n\n6.3 K\n\n1:31\nएक द्विघात समीकरण ज्ञात कीजिए जिसके मूल -5 और 6 हैं।\n648231032\n\n10\n\n1.4 K\n\nIf the AM and GM of roots of a quadratic equation are 8 and 5 respectively, then the quadratic equation will be\n643444578\n\n18\n\n7.1 K\n\n120:46\n648078418\n\n12\n\n7.9 K\n\n5 तथा - 5 मूलों वाला द्विघात समीकरण है :\n648313139\n\n38\n\n2.8 K\n\n1:13\nWhat is the quadratic equation whose roots are 5 and 2" ]

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[ "", null, "", null, "Math Central - mathcentral.uregina.ca", null, "", null, "Quandaries & Queries", null, "", null, "", null, "", null, "Q & Q", null, "", null, "", null, "", null, "Topic:", null, "networks", null, "", null, "", null, "start over\n\nOne item is filed under this topic.", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Networks of satellites and linear spaces 2000-12-08", null, "From David:Let us suppose some companies have collaborated to place several satellites in orbit. Let us call the set of all satellites that a given company helped place in orbit a network. Finally let us assume the following 4 rules. There are at least two distinct satellites. For each pair of satellites there is exactly one network containing them. Each network contains at least two distinct satellites. For each network, there is a satellite not in it. What is the least number of satellites. what is the least number of networks? Answered by Penny Nom.", null, "", null, "", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", null, "", null, "", null, "", null, "about math central :: site map :: links :: notre site français" ]

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[ "# Transformation between two different sets of molecular vibrational normal coordinate systems\n\nLets assume we have $$N$$ Atoms and we treat them within the Born-Oppenheimer Approximation. We can calculate the adiabatic electronic groundstate potential. Lets assume we observe two local minima, denoted by $$\\alpha$$ and $$\\beta$$, on this potential energy surface. Lets denote these minima nuclear coordinates as $$R_\\alpha$$ and $$R_{\\beta}$$. At each minimum we can do an harmonic approximation of the potential, centered on the corresponding minimum, $$\\begin{eqnarray} V_\\alpha(R) = V(R_\\alpha)+(R-R_\\alpha)^T V'_{\\alpha} + 1/2(R-R_\\alpha)^TV''_{\\alpha}(R-R_\\alpha) \\tag{1}\\\\ V_\\beta(R) = V(R_\\beta)+(R-R_\\alpha)^T V'_{\\beta} + 1/2(R-R_\\beta)^TV''_{\\beta}(R-R_\\beta) \\tag{2}\\\\ \\end{eqnarray}$$\n\nwhere the gradent is $$V'_\\alpha=\\nabla_RV(R)|_{R=R_\\alpha}$$ and the Hessian $$V''_{\\alpha,nm} = \\frac{\\partial^2 V}{\\partial R_n \\partial R_m}|_{R=R_\\alpha}$$.\n\nSince we are in a local minimum, we know that the gradients vanish,$$V'=0$$. For simplicities sake, we also assume that both potentials evaluated at the minimum strcucture are zero, $$V(R_{\\alpha/\\beta})=0$$.\n\nThe associated harmonic vibrational Hamiltonian in atomic units then takes on the form $$\\tag{3} H_\\alpha(R) = -\\frac{1}{2}\\nabla^T_RM^{-1}\\nabla_R + 1/2(R-R_\\alpha)^TV''_\\alpha(R-R_\\alpha)$$ where $$M$$ is a diagonal matrix with the nuclear masses.\n\nEach harmonic potential allows the definition of vibrational normal modes and vibrational normal coordinates. The transformation to normal coordinates is defined as follows.\n\nFirst we define the mass weighted displacement coordinates $$X_\\alpha = M^{1/2}(R-R_\\alpha)$$, in these coordinates we have $$\\begin{eqnarray} H_\\alpha(X_\\alpha)= -\\frac{1}{2}\\nabla^T_{X_\\alpha}\\nabla_{X_\\alpha} +1/2 X^T_\\alpha M^{-1/2}V''_\\alpha M^{-1/2} X_\\alpha \\tag{4}\\\\ H_\\alpha(X_\\alpha)= -\\frac{1}{2}\\nabla^T_{X_\\alpha}\\nabla_{X_\\alpha} +1/2 X^T_\\alpha W_\\alpha X_\\alpha \\tag{5}\\\\ \\end{eqnarray}$$ with the mass weighted Hesse matrix $$W=M^{-1/2}V''M^{-1/2}$$.\n\nThe last step to obtain normal coordinates is the transformation to coordinates that diagonalize the mass weighted Hesse matrix. Let $$WK = K\\Lambda \\tag{6}$$ where $$K$$ is the matrix with eigenvectors of $$W$$ as columns and $$\\Lambda$$ a diagonal matrix with corresponding eigenvalues. The required transformation is then $$X= KQ\\tag{7}$$ which leads to $$H_\\alpha(Q_\\alpha)= -\\frac{1}{2}\\nabla^T_{Q_\\alpha}\\nabla_{Q_\\alpha} +1/2 Q^T_\\alpha \\Lambda_\\alpha Q_\\alpha \\tag{8}$$\n\nSummarized we have two different sets of transformations, one for each minimum, $$\\begin{eqnarray} Q_\\alpha = K^{-1}_\\alpha M^{1/2}(R-R_\\alpha)\\tag{9}\\\\ Q_\\beta = K^{-1}_\\beta M^{1/2}(R-R_\\beta) \\tag{10} \\end{eqnarray}$$\n\nAfter this rather lengthy setup comes my question. How can I transform from one set of normal coordinates $$Q_\\alpha$$ to the other $$Q_\\beta$$ and how do I interpret this. Does that mean that I can express any molecules vibrational coordinates in the vibrational coordinates of another molecule(made from the same number and type of atoms)?\n\nIntuitively I see problems when the geometries are translated and rotated with respect to each other assuming a global cartesian coordinate system for both molecules. There should be no simple internal vibrational coordinate that could connect two such geometries but on the other hand, internal vibrational coordinates should form a complete coordinate system and allow me to express any structure with the same amount of atoms in it. I think I am missing something very obvious to reconcile this \"contradiction\", which is why I need some input.\n\n• +1. Please do take a look at the edits I made, especially the change of 6 different instances of i to I. An academic journal wouldn't let you publish a paper with i not capitalized, and we hope to keep high standards here at MMSE. May 22 at 7:39\n• @NikeDattani Thank you for editing. I had the comfort of working with native English speaking coauthors in the past, who took care of the punctuation and grammar. But that is of course no reason to be sloppy. May 22 at 8:42\n• I think the answer to your question is yes. You just need to find the Jacobian that transforms between the two coordinates systems. And it is definitely possible to express the vibrations of one molecule in terms of the eigenvectors of another as long they span the same space. In all likelihood, the eigenvectors of one molecule are a bad basis for the eigenvectors of another molecule. I could imagine some interesting similarity analysis of configurational isomers being done in this way. May 22 at 16:52\n• @jheindel Equations 9 and 10 are linear with regard to the displacement vectors $\\Delta R_\\alpha=R-R_\\alpha$ and $\\Delta R_\\beta = R-R_\\beta$. But I think the problem is that it looks more like a affine transformation with respect to the molecular coordinates vector $R$, that connects both equations. I'm lacking experience with regard to affine transformation and the conversion between them. I think that complicates the problem but I'm not certain, hence the question. May 22 at 17:35\n\nI agree with Emil Zak's answer that normal mode coordinates from different minima should, in general, be put together with special care. However, relating the two different sets of normal mode origin from different minima (or transition states) is not of nonsense, see SVP model of Hua Guo and Bin Jiang and related theories.\n\nTo make things short, I will use the same symbols as in the question. The relation between the normal mode coordinates and Cartesian coordinates, are $$\\mathbf{R}=\\mathfrak{R}_{\\alpha}\\cdot\\left(\\mathbf{R}_{\\alpha}+\\sum_{i=7}^{3N}{{Q}_{\\alpha}}_i\\cdot {\\mathbf{K}_{\\alpha}}_i\\right)+\\mathbf{t},\\tag{1}$$ and $$\\tag{2} \\mathbf{R}=\\mathfrak{R}_{\\beta}\\cdot\\left(\\mathbf{R}_{\\beta}+\\sum_{i=7}^{3N}{{Q}_{\\beta}}_i\\cdot {\\mathbf{K}_{\\beta}}_i\\right)+\\mathbf{t}.$$ Here, we assume that the $$\\mathbf{R}_{\\alpha}$$ are originated from the COM of molecule. $$\\mathfrak{R}_{\\alpha}$$ mean a rotation matrix and $$\\mathbf{t}$$ is the translation. And the mass-weighted part are omitted for simplicity.\n\nNote that I separate rotation, translation, and vibration on purpose. Although in practice you can keep them together and compute the all $$3N$$ DOF based on normal modes fashion, like in Gaussian software you get 6 \"low frequencies\" for translations and rotations, which should be exact zeros in principle, when you use the normal modes vectors to represent the coordinate, it is not a good idea to use the rotation normal modes, more importantly, the molecular orientation must be the same as of the reference geometry, because the vibration normal modes are attached to the reference geometry, and that is why the rotation matrix is outside the big parentheses.\n\nUsing the formulae above, it is easy to convert the normal mode coordinate $$\\mathbf{Q}_{\\alpha}$$ to $$\\mathbf{R}$$, and if we have an algorithm to get $$\\mathbf{R}$$ to $$\\mathbf{Q}_{\\beta}$$, the problem will be solved.\n\nThe main idea is\n\n1. Translate the molecule so that the COM located at $$(0,0,0)$$.\n2. Rotate $$\\mathbf{R}_{\\beta}$$ to align with $$\\mathbf{R}$$ with some algorithm (Wahba's problem). In my implement, Kabsch's algorithm is used. So the rotation matrix $$\\mathfrak{R}_{\\beta}$$ is get. (Edit: Here, one may minimize the mass-weighted RMSD $$\\Delta=\\sqrt{\\frac{1}{N}\\sum_{i=1}^{N}m_i|\\mathfrak{R}_{\\beta}\\mathbf{R}_{\\beta i}-\\mathbf{R}_{\\alpha i}|^2}, \\tag{3}$$ where $$m_i$$ is the mass of the $$i$$-th atom. Kudin and Dymarsky has shown that minimization of mass-weighted RMSD is equivalent with the Eckart condition. I thank @Hans Wurst provided the information.)\n3. Utilizing the orthogonality of mass-weighted normal mode vectors, get the coordinates $$\\mathbf{Q}_{\\beta}$$. Note that in this step you only need the vibration ones, and exclude the rotation and translations\n4. Now rotate $$\\mathbf{R}_{\\beta}$$ to align with $$\\mathbf{R}+{{Q}_{\\beta}}_i\\cdot {\\mathbf{K}_{\\beta}}_i$$, update $$\\mathfrak{R}_{\\beta}$$ and $$\\mathbf{Q}_{\\beta}$$.\n5. Do it iteratively until the two molecules are perfectly aligned.\n\nThe algorithm above is implemented with Julia programming language by me as shown in gist.\n\nThen you can do $$\\tag{4} \\mathbf{Q}_{\\alpha} \\leftrightarrow \\mathbf{R} \\leftrightarrow \\mathbf{Q}_{\\beta}$$\n\nProblem solved.\n\nI would suggest that the reader refer to some published work, thanks to the anonymous reviewers.\n\n1. Gábor Czakó, J. Phys. Chem. A 116, 116, 7467–7473 (2012)\n2. Philip R. Bunker, Molecular Symmetry and Spectroscopy (1979), Academic Press, New York (see Ch. 7 therein)\n\nThere are some methods that use intramolecular curvilinear coordinates as the \"bridge\", although I think use Cartesian coordinates is easier to be understood.\n\n• Why is the RMSD minimizing rotation the correct one ? I could also do a a rotation that aligns the tensor of inertia axes but these rotation are in general not identical and would then yield different values when doing the transformation. Is there a physical reason that the RMSD minimizing rotation is the only sensible rotation ? This is the only remaining point that is somewhat unclear to me, everything else is fine. My first (apparently wrong) intuition was that the transformation to normal coordinates should be independent of such a rotation but that clearly isn't the case. Jul 1 at 12:25\n• @Hans Wurst , Actually no special reason. I think you can used the principle axis of inertia tensor. What is most tricky, I think, is that I run the procedure of alignment-vibrate iteratively and finally the RMSD will be zero. Since the (3N-6) real vibrational NM are used to represent the intramolecular DOF and the rotation is represented by a rotational matrix, and by using the iterative procedure so the rotation-vibration DOF are self-consist. Jul 1 at 13:24\n• The reason why the RMSD approach works, is given here doi.org/10.1063/1.1929739 The RMSD minimizing rotation is the same as the one that transforms to the Eckart frame, defined by the Eckart conditions. These axes are the ones that should be used if one wants a local coordinate system in which rotation and vibration is (almost) decoupled. There should be no need to do it iteratively, you only have to make sure that you use the set of axes that is \"continuous\", as there can be mutliple solutions. Aug 4 at 7:43\n• Thank you, Hans. I did not know that work. What I missed in my answer is that I should use a mass-weighted RMSD approach, instead of the one I used (no-weighted) here. Aug 4 at 9:00\n• Since I really wrote a paper and submitted it. I am in a mess now. :-( Aug 4 at 9:06\n\nThe paradox discovered by @HansWurst comes by the assumption that it is possible to treat both minima on the 1D PES separately and define a separate set of normal coordinates for them. 3N normal coordinates are functions of 3N Cartesian coordinates of the bodies in our system. It turns out that for a double-well potential in 1D, which you describe here, the normal-mode approximation is a very bad one. We only have one dynamical dimension, hence we can only define one normal coordinate for this system. Splitting into two independent coordinates is the 'mistake' here. To treat this problem we can define a variational basis composed of sums of 1D Harmonic Oscillator basis functions centred at the two local minima. Such a basis is likely to serve well.\n\nIn general, when more dimensions are involved it is rather customary to perform rotations on full sets of normal coordinates (so called Duschinsky rotation), such that, for instance, a set of normal coordinates for the electronic ground state can be expressed with normal coordinates of a displaced and deformed electronically excited state.\n\n• Do you know a detailed reference to the Duschinsky rotation ? I find mention of it but haven't found a reference with a clear definition of it and all steps involved. For example when the ground state structure and the excited state structure are unaligned and quite different. Assume you have two optimized structures as xyz files with completely different orientations and origin for example and you want to transform from one set of normal coordinates to the other. Jun 30 at 10:32\n• @HansWurst, you may want to look into PhD thesis of Joonsuk Huh \"Unified description of vibronic transitions with coherent states\" (2010), sec. 2.2. Jun 30 at 16:43\n• Thank you, that is exactly what I was looking for. Jun 30 at 21:01" ]

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[ "# Ranges\n\nIf a `foreach` is encountered by the compiler\n\n``````foreach (element; range)\n{\n// Loop body...\n}\n``````\n\nit's internally rewritten similar to the following:\n\n``````for (auto __rangeCopy = range;\n!__rangeCopy.empty;\n__rangeCopy.popFront())\n{\nauto element = __rangeCopy.front;\n// Loop body...\n}\n``````\n\nAny object which fulfills the following interface is called a range (or more specific `InputRange`) and is thus a type that can be iterated over:\n\n``````    interface InputRange(E)\n{\nbool empty();\nE front();\nvoid popFront();\n}\n``````\n\nHave a look at the example on the right to inspect the implementation and usage of an input range closer.\n\n## Laziness\n\nRanges are lazy. They won't be evaluated until requested. Hence, a range from an infinite range can be taken:\n\n``````42.repeat.take(3).writeln; // [42, 42, 42]\n``````\n\n## Value vs. Reference types\n\nIf the range object is a value type, then range will be copied and only the copy will be consumed:\n\n``````auto r = 5.iota;\nr.drop(5).writeln; // []\nr.writeln; // [0, 1, 2, 3, 4]\n``````\n\nIf the range object is a reference type (e.g. `class` or `std.range.refRange`), then the range will be consumed and won't be reset:\n\n``````auto r = 5.iota;\nauto r2 = refRange(&r);\nr2.drop(5).writeln; // []\nr2.writeln; // []\n``````\n\n### Copyable `InputRanges` are `ForwardRanges`\n\nMost of the ranges in the standard library are structs and so `foreach` iteration is usually non-destructive, though not guaranteed. If this guarantee is important, a specialization of an `InputRange` can be used— forward ranges with a `.save` method:\n\n``````interface ForwardRange(E) : InputRange!E\n{\ntypeof(this) save();\n}\n``````\n``````// by value (Structs)\nauto r = 5.iota;\nauto r2 = refRange(&r);\nr2.save.drop(5).writeln; // []\nr2.writeln; // [0, 1, 2, 3, 4]\n``````\n\n### `ForwardRanges` can be extended to Bidirectional ranges + random access ranges\n\nThere are two extensions of the copyable `ForwardRange`: (1) a bidirectional range and (2) a random access range. A bidirectional range allows iteration from the back:\n\n``````interface BidirectionalRange(E) : ForwardRange!E\n{\nE back();\nvoid popBack();\n}\n``````\n``````5.iota.retro.writeln; // [4, 3, 2, 1, 0]\n``````\n\nA random access range has a known `length` and each element can be directly accessed.\n\n``````interface RandomAccessRange(E) : ForwardRange!E\n{\nE opIndex(size_t i);\nsize_t length();\n}\n``````\n\nThe best known random access range is D's array:\n\n``````auto r = [4, 5, 6];\nr.writeln; // 5\n``````\n\n### Lazy range algorithms\n\nThe functions in `std.range` and `std.algorithm` provide building blocks that make use of this interface. Ranges enable the composition of complex algorithms behind an object that can be iterated with ease. Furthermore, ranges enable the creation of lazy objects that only perform a calculation when it's really needed in an iteration e.g. when the next range's element is accessed. Special range algorithms will be presented later in the D's Gems section." ]

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[ "# Basic course for newbies\n\nPágina 8/11\n1 | 2 | 3 | 4 | 5 | 6 | 7 | | 9 | 10 | 11\n\nYes, highlight is also one of those things that are not really supported by MSX-BASIC, but see the few changes that I made to your program:\n\n```5 DEFINT A-Z\n10 CLEAR 2000:SCREEN 0:WIDTH 80:COLOR 15,0,0:KEY OFF\n20 DIM FO\\$(1000)\n30 FM\\$=\"????????.???\"\n40 DA\\$=\"0E11110E12C51100C10E1ACD7DF3C11180C0CD7DF332F8F7C9\"\n70 FOR I=&HC08C TO &HC0C0:POKE I,0:NEXT I\n75 FOR I=2048 TO 2288:VPOKE I,0:NEXT I\n76 VDP(13)=&HA8 ' Color 10 on color 8 (yellow on red)\n77 VDP(14)=16\n80 FOR I=0 TO 25:POKE &HC000+I,VAL(\"&H\"+MID\\$(DA\\$,I*2+1,2)):NEXT I\n90 DEFUSR=&HC000:DEFUSR1=&HC003\n100 IF USR(0)>0 THEN END\n110 FOR I=1 TO 12\n120 FO\\$(COUNTER)=FO\\$(COUNTER) + CHR\\$(PEEK(&HC100+I))\n130 IF I=8 THEN FO\\$(COUNTER)=FO\\$(COUNTER)+ \".\"\n140 NEXT I\n150 IF USR1(0)=0 THEN COUNTER = COUNTER +1: GOTO 110\n160 SCREEN 0:WIDTH80:COLOR15,0,0:CLS\n170 FOR I=0 TO COUNTER\n180 LOCATE X,Y :PRINT FO\\$(I)\n190 X=X+16:IF X>70 THEN Y=Y+1:X=0\n200 NEXT I\n210 DIM ST,SX,SY,AX,AY\n211 GOTO 280\n220 ST = STICK(0)\n230 IF ST=0 THEN 220\n240 IF ST=1 THEN SY=SY-1:IF SY=<0 THEN SY=0\n250 IF ST=5 THEN SY=SY+1:IF SY=>Y THEN SY=Y\n260 IF ST=7 THEN SX=SX-1:IF SX=<0 THEN SX=0\n270 IF ST=3 THEN SX=SX+1:IF SX=>4 THEN SX=4\n290 TIME=0:FOR I=0 TO 8:I=TIME:NEXT I\n310 GOTO 220\n```\n\nThanks again! .. it seems a bit better! I will try it! There are some changes I don't understand.. but maybe playing around will help a lot!\n\n`280 AD=2048+SX*2+SY*10:VPOKE AD,255:VPOKE AD+1,&B11110000`\n\nI was wondering. The VPOKE will do something with the highlight, I think", null, "but what is the meaning for the number 2048 ? It seems that it had something to-do with a binary count ? Where are the VDP commands for ?\n\n```100 SCREEN 1:WIDTH32:KEYOFF\n120 'lets adjust the letter \"a\"\n130 PRINT ASC(\"a\")*8\n140 'this gives the value 776\n150 'now we poke around in the video memory\n160 VPOKE 776,&B11111111\n170 VPOKE 777,&B10000001\n180 VPOKE 778,&B10111101\n190 VPOKE 779,&B10100101\n200 VPOKE 780,&B10100101\n210 VPOKE 781,&B10111101\n220 VPOKE 782,&B10000001\n230 VPOKE 783,&B11111111\n235 'output to screen\n240 PRINT \"a\"\n```\nVampier wrote:\n```100 SCREEN 1:WIDTH32:KEYOFF\n120 'lets adjust the letter \"a\"\n130 PRINT ASC(\"a\")*8\n140 'this gives the value 776\n150 'now we poke around in the video memory\n160 VPOKE 776,&B11111111\n170 VPOKE 777,&B10000001\n180 VPOKE 778,&B10111101\n190 VPOKE 779,&B10100101\n200 VPOKE 780,&B10100101\n210 VPOKE 781,&B10111101\n220 VPOKE 782,&B10000001\n230 VPOKE 783,&B11111111\n235 'output to screen\n240 PRINT \"a\"\n```\n\nWho! That's a cool explaination! So if i understand this code right i can change the letter A in what i want!\nI got the god feeling", null, "I just played with it .. but it is not working in screen 0 ?\n\nIn screen 0 the patterns start at a different address, IIRC &h800, so do VPOKE &h800+776,&b11111111 that should do it.\nEdit: at least for width 40 mode, for width 80 you need to add &h1000", null, "These addresses can be found using the BASE command: http://www.msx.org/wiki/BASE. Also bear in mind that screen 0 only uses 6 pixels instead of 8. You need the charactertable, by the way...\n\ni just found out about the 6 pixels", null, "but it works!\nAbout the base. Can anyone explain a bit more. If i type in screen0 ? base(2) i got a result number 4096, in screen 1 i got 2048 as a result. But how do i translate that to a vpoke command ?\n\nThe number you're getting is the startaddress for that specific table. If you want to know where the charactertable starts in screen 0, width 40, first set the width to 40 (or less), then use base(2). If you want to know it for width 80, set the width to 80, and do a base(2) again. The first will give you a value of 2048, the second of 4096. However, if you want to know it for screen 1, you need to use base(7) as 1*5+2 is 7. (see the calculation on the wikipage)\n\nThat number what you're getting is the first address for that table. Each character consists of 8 bytes, and the first character starts on that address. If you want to change the capital A, for example, you need the ascii value for A (which is 65), multiply that by 8 (which is 520, and add that to the base address. This address, and the 7 next addresses make up the A character. You can read the values with VPEEK\n\nSimple example:\n\n```1 ' show the letter A in bitmap\n10 screen0:width 40:ad=base(2) 'the base address for the characterset in screen 0\n20 A=asc(\"A\")*8\n30 for co=0 to 7\n50 next co\n```\n```10 DEFINT A-Z\n20 CLEAR 2000:SCREEN 0:WIDTH 80:COLOR 15,0,0:KEY OFF\n30 DIM FO\\$(1000)\n40 FM\\$=\"????????.???\"\n50 DA\\$=\"0E11110E12C51100C10E1ACD7DF3C11180C0CD7DF332F8F7C9\"\n80 FOR I=&HC08C TO &HC0C0:POKE I,0:NEXT I\n90 FOR I=2048 TO 2288:VPOKE I,0:NEXT I\n100 VDP(13)=&HA8 ' Color 10 on color 8 (yellow on red)\n110 VDP(14)=16\n120 FOR I=0 TO 25:POKE &HC000+I,VAL(\"&H\"+MID\\$(DA\\$,I*2+1,2)):NEXT I\n130 DEFUSR=&HC000:DEFUSR1=&HC003\n140 IF USR(0)>0 THEN END\n150 FOR I=1 TO 12\n160 FO\\$(COUNTER)=FO\\$(COUNTER) + CHR\\$(PEEK(&HC100+I))\n170 IF I=8 THEN FO\\$(COUNTER)=FO\\$(COUNTER)+ \".\"\n180 NEXT I\n190 IF USR1(0)=0 THEN COUNTER = COUNTER +1: GOTO 150\n200 SCREEN 0:WIDTH80:COLOR15,0,0:CLS\n210 FOR I=0 TO COUNTER\n220 LOCATE X,Y :PRINT FO\\$(I)\n230 X=X+16:IF X>70 THEN Y=Y+1:X=0\n240 NEXT I\n250 DIM ST,SX,SY,AX,AY\n260 GOTO 340\n270 ST = STICK(0)\n280 IF ST=0 THEN 270\n300 IF ST=1 THEN SY=SY-1:IF SY=<0 THEN SY=0\n310 IF ST=5 THEN SY=SY+1:IF SY=>Y THEN SY=Y\n320 IF ST=7 THEN SX=SX-1:IF SX=<0 THEN SX=0\n330 IF ST=3 THEN SX=SX+1:IF SX=>4 THEN SX=4\n350 LOCATE 0,22: PRINT SX\n360 LOCATE 4,22: PRINT SY\n370 LOCATE 12,22 : PRINT \"SELECTED FILE: \";FO\\$(SX+(SY*5))\n390 TIME=0:FOR I=0 TO 8:I=TIME:NEXT I\n400 GOTO 270\n\u001a```\n```350 LOCATE 0,22: PRINT SX\n360 LOCATE 4,22: PRINT SY\n370 LOCATE 12,22 : PRINT \"SELECTED FILE: \";FO\\$(SX+(SY*5))\n```\n\nIs this a good way to get the selected filename ?\n\nOk, I added the Sega loader instructions to the basic code. It seems to be working. Next is to make it possible to toggle the SMS options etc. for each game. And to save it for every game (They are now all default.) What is the best way to do this ? To avoid delay in speed. Save the setting for each game when starting the game ? And load all the game default at the begin when the loader is started ?\n\nThe first part (loading the files in to the array) takes some time in basic. Is there a way to speed it up a bit ?\n\n```10 DEFINT A-Z\n20 CLEAR 2000:SCREEN 0:WIDTH 80:COLOR 15,0,0:KEY OFF:CLS\n40 DIM FO\\$(1000)\n50 FM\\$=\"????????.???\"\n60 DA\\$=\"0E11110E12C51100C10E1ACD7DF3C11180C0CD7DF332F8F7C9\"\n90 FOR I=&HC08C TO &HC0C0:POKE I,0:NEXT I\n100 FOR I=2048 TO 2288:VPOKE I,0:NEXT I\n110 VDP(13)=&HA8 ' Color 10 on color 8 (yellow on red)\n120 VDP(14)=16\n130 FOR I=0 TO 25:POKE &HC000+I,VAL(\"&H\"+MID\\$(DA\\$,I*2+1,2)):NEXT I\n140 DEFUSR=&HC000:DEFUSR1=&HC003\n150 IF USR(0)>0 THEN END\n160 FOR I=1 TO 12\n170 FO\\$(COUNTER)=FO\\$(COUNTER) + CHR\\$(PEEK(&HC100+I))\n180 IF I=8 THEN FO\\$(COUNTER)=FO\\$(COUNTER)+ \".\"\n190 NEXT I\n200 Q1=Q1+1\n210 LOCATE Q1,Q2 :PRINT\".\":IF Q1=>80 THEN Q1=0:Q2=Q2+1\n220 IF USR1(0)=0 THEN COUNTER = COUNTER +1: GOTO 160\n230 SCREEN 0:WIDTH80:COLOR15,0,0:CLS\n240 FOR I=0 TO COUNTER\n250 LOCATE X,Y :PRINT FO\\$(I)\n260 X=X+16:IF X>70 THEN Y=Y+1:X=0\n270 NEXT I\n280 DIM ST,SX,SY,AX,AY\n290 GOTO 380\n300 ST = STICK(0)\n310 IF STRIG(0)=-1 THEN N\\$=FO\\$(SX+(SY*5)):GOTO 460\n320 IF ST=0 THEN 300\n340 IF ST=1 THEN SY=SY-1:IF SY=<0 THEN SY=0\n350 IF ST=5 THEN SY=SY+1:IF SY=>Y THEN SY=Y\n360 IF ST=7 THEN SX=SX-1:IF SX=<0 THEN SX=0\n370 IF ST=3 THEN SX=SX+1:IF SX=>4 THEN SX=4\n390 LOCATE 0,22: PRINT SX\n400 LOCATE 4,22: PRINT SY\n410 LOCATE 12,22 : PRINT \"SELECTED FILE: \";FO\\$(SX+(SY*5))\n420 LOCATE 12,38\n430 IF STRIG(0)=-1 THEN N\\$=FO\\$(SX+(SY*5):GOTO 460\n440 TIME=0:FOR I=0 TO 8:I=TIME:NEXT I\n450 GOTO 300\n460 ' (base address and interrupt off)\n470 IF PEEK(&H2D)=0 THEN OUT &H2A,33:OUT &H2B,1:OUT &H2A,&H1D:OUT &H2B,&H99:SCREEN 0:WIDTH 40 ELSE SCREEN 0:WIDTH 80\n480 V=0 'Video system: 0=PAL 1=NTSC\n490 H=0 'Refresh freq: 0=50Hz 1=60Hz\n500 F=1 'FM-Pac usage: 0=No 1=Yes\n510 R=0 'RGB status : 0=RGB 1=CVBS\n520 OV=1 'Overlay : 2=Key 7=All1:3\n530 'Turbo-R masks IO &HDC, alternative\n540 ' IO is used when K=64\n550 K=64 'Kanji : 64=Yes 128=Norm\n560 S=1:N=0\n570 S=2" ]

[ null, "https://www.msx.org/sites/all/themes/mrc2k11/img/smileys/wink.png", null, "https://www.msx.org/sites/all/themes/mrc2k11/img/smileys/bigsmile.png", null, "https://www.msx.org/sites/all/themes/mrc2k11/img/smileys/wink.png", null, "https://www.msx.org/sites/all/themes/mrc2k11/img/smileys/wink.png", null ]

[ "ENTAIN PLC Research Report\n\n## Summary\n\nWe present an Artificial Neural Network (ANN) approach to predict stock market indices, particularly with respect to the forecast of their trend movements up or down. Exploiting different Neural Networks architectures, we provide numerical analysis of concrete financial time series. In particular, after a brief r ́esum ́e of the existing literature on the subject, we consider the Multi-layer Perceptron (MLP), the Convolutional Neural Net- works (CNN), and the Long Short-Term Memory (LSTM) recurrent neural networks techniques. We evaluate ENTAIN PLC prediction models with Modular Neural Network (CNN Layer) and Spearman Correlation1,2,3,4 and conclude that the LON:ENT stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy LON:ENT stock.\n\n## Key Points\n\n1. Game Theory\n3. Operational Risk\n\n## LON:ENT Target Price Prediction Modeling Methodology\n\nWe consider ENTAIN PLC Stock Decision Process with Modular Neural Network (CNN Layer) where A is the set of discrete actions of LON:ENT stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Spearman Correlation)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Modular Neural Network (CNN Layer)) X S(n):→ (n+3 month) $∑ i = 1 n r i$\n\nn:Time series to forecast\n\np:Price signals of LON:ENT stock\n\nj:Nash equilibria (Neural Network)\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## LON:ENT Stock Forecast (Buy or Sell) for (n+3 month)\n\nSample Set: Neural Network\nStock/Index: LON:ENT ENTAIN PLC\nTime series to forecast n: 19 Nov 2022 for (n+3 month)\n\nAccording to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy LON:ENT stock.\n\nX axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Yellow to Green): *Technical Analysis%\n\n## Adjusted IFRS* Prediction Methods for ENTAIN PLC\n\n1. For the purpose of this Standard, reasonable and supportable information is that which is reasonably available at the reporting date without undue cost or effort, including information about past events, current conditions and forecasts of future economic conditions. Information that is available for financial reporting purposes is considered to be available without undue cost or effort.\n2. For the purposes of applying the requirements in paragraphs 5.7.7 and 5.7.8, an accounting mismatch is not caused solely by the measurement method that an entity uses to determine the effects of changes in a liability's credit risk. An accounting mismatch in profit or loss would arise only when the effects of changes in the liability's credit risk (as defined in IFRS 7) are expected to be offset by changes in the fair value of another financial instrument. A mismatch that arises solely as a result of the measurement method (ie because an entity does not isolate changes in a liability's credit risk from some other changes in its fair value) does not affect the determination required by paragraphs 5.7.7 and 5.7.8. For example, an entity may not isolate changes in a liability's credit risk from changes in liquidity risk. If the entity presents the combined effect of both factors in other comprehensive income, a mismatch may occur because changes in liquidity risk may be included in the fair value measurement of the entity's financial assets and the entire fair value change of those assets is presented in profit or loss. However, such a mismatch is caused by measurement imprecision, not the offsetting relationship described in paragraph B5.7.6 and, therefore, does not affect the determination required by paragraphs 5.7.7 and 5.7.8.\n3. A firm commitment to acquire a business in a business combination cannot be a hedged item, except for foreign currency risk, because the other risks being hedged cannot be specifically identified and measured. Those other risks are general business risks.\n4. A firm commitment to acquire a business in a business combination cannot be a hedged item, except for foreign currency risk, because the other risks being hedged cannot be specifically identified and measured. Those other risks are general business risks.\n\n*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.\n\n## Conclusions\n\nENTAIN PLC assigned short-term B1 & long-term Ba1 forecasted stock rating. We evaluate the prediction models Modular Neural Network (CNN Layer) with Spearman Correlation1,2,3,4 and conclude that the LON:ENT stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Buy LON:ENT stock.\n\n### Financial State Forecast for LON:ENT ENTAIN PLC Stock Options & Futures\n\nRating Short-Term Long-Term Senior\nOutlook*B1Ba1\nOperational Risk 5464\nMarket Risk8180\nTechnical Analysis7937\nFundamental Analysis6077\nRisk Unsystematic3189\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 81 out of 100 with 519 signals.\n\n## References\n\n1. Chernozhukov V, Newey W, Robins J. 2018c. Double/de-biased machine learning using regularized Riesz representers. arXiv:1802.08667 [stat.ML]\n2. Efron B, Hastie T, Johnstone I, Tibshirani R. 2004. Least angle regression. Ann. Stat. 32:407–99\n3. J. G. Schneider, W. Wong, A. W. Moore, and M. A. Riedmiller. Distributed value functions. In Proceedings of the Sixteenth International Conference on Machine Learning (ICML 1999), Bled, Slovenia, June 27 - 30, 1999, pages 371–378, 1999.\n4. Künzel S, Sekhon J, Bickel P, Yu B. 2017. Meta-learners for estimating heterogeneous treatment effects using machine learning. arXiv:1706.03461 [math.ST]\n5. H. Kushner and G. Yin. Stochastic approximation algorithms and applications. Springer, 1997.\n6. H. Khalil and J. Grizzle. Nonlinear systems, volume 3. Prentice hall Upper Saddle River, 2002.\n7. Andrews, D. W. K. (1993), \"Tests for parameter instability and structural change with unknown change point,\" Econometrica, 61, 821–856.\nFrequently Asked QuestionsQ: What is the prediction methodology for LON:ENT stock?\nA: LON:ENT stock prediction methodology: We evaluate the prediction models Modular Neural Network (CNN Layer) and Spearman Correlation\nQ: Is LON:ENT stock a buy or sell?\nA: The dominant strategy among neural network is to Buy LON:ENT Stock.\nQ: Is ENTAIN PLC stock a good investment?\nA: The consensus rating for ENTAIN PLC is Buy and assigned short-term B1 & long-term Ba1 forecasted stock rating.\nQ: What is the consensus rating of LON:ENT stock?\nA: The consensus rating for LON:ENT is Buy.\nQ: What is the prediction period for LON:ENT stock?\nA: The prediction period for LON:ENT is (n+3 month)" ]

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[ "Click here for a printer friendly Acrobat file of this activity. Circular Motion Activities You will be divided up into groups. Each person in the group will keep individual records. All the information collected and calculated will be reviewed next class. Your group will rotate among the stations. If you cannot get to a station then work together on the problems on the boards. Your priority is to get through the stations in the room before doing the word problems. Do all calculations with standard S.I. units. STATION 1: “STOMPER” Toy car The car is tethered to a weighted can. The car is battery powered and has a two speeds. It is turned on and off via a switch under the car.", null, "Your car has an index card taped to the top of the vehicle. This card is used in conjunction with the CBL to calculate the vehicle’s velocity. Position the gate such that the the card on top of the car travels through the gate. when setting up the CBL use the GATE mode. This will tell you how long it takes for the card to break the photogate’s beam. Run the car through the gate until you have ONE good time measurement. Do not hold the wheels of the car while the motor is turned on.", null, "", null, "Questions 1.1) What is the width of the index card that passes through the photogate? 1.2) How much time does it take for the card to pass through the gate when the car is traveling SLOW? 1.3) How much time does it take for the car to pass through the gate when the car is traveling FAST? (If your car does not have a “FAST” spee dthen skip this questions.) 1.4) Measure the radius of the circle from the center of the can to the card. What is this radius’s measurement? 1.5) What is the tangential velocity of the car? (Give two answer if your car has two speeds.) 1.6) Calculate the car’s period of motion using the tangential velocity and radius. Show all of your work. Period of he slow car: Period of the fast car: 1.7) Calculate the car’s centripetal acceleration. (Give two answer if your car has two speeds.) ANSWER 1.8) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep this car traveling in a circle? STATION 2: COMPUTER FERRIS WHEEL Find a computer displaying the circular motion page. Select Ferris Wheel. Press the PLAY button to see the animation.", null, "QUESTIONS AND PROCEDURE: 2.1) Use the stopwatch to measure the period of motion for the ferris wheel. What is it’s period? 2.2) Calculate it’s tangential velocity: 2.3) Calculate its centripetal acceleration: 2.4) Calculate It he mass of a rider is 65 kg, the what is the centripetal force exerted by the ferris wheel? 2.5) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep the seat the rider is in going in a circle at the bottom of the motion? Go back to the main menu", null, "STATION 3: RADIO CONTROLLED TRUCK", null, "The remote is not needed. When turned on, the car will run backwards while turning. DO NOT TAMPER WITH THIS. The car will not travel in a precise circle. In other words it will not travel the exact same circle twice in a row. QUESTIONS AND PROCEDURE: Devise a way to measure/calculate the car’s radius. 3.1) What is the truck’s radius as it travels around once? 3.2) Use a stop watch and measure the truck’s period of motion. 3.3) What is the truck’s tangential velocity? 3.4) What is the truck’s centripetal force if it’s mass is 755 grams? 3.5) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep the truck going in a circle? STATION 4: THE ANTIGRAVITY RIDE Find a computer displaying the circular motion page. Select the Anti-Gravity Ride. Press the PLAY button to see the animation.", null, "4.1) Use the stopwatch to measure the period of motion for the ride wheel. What is it’s period? 4.2) Calculate it’s tangential velocity: 4.3) Calculate its centripetal acceleration: 4.4) If he mass of a rider is 55 kg, the what is the centripetal force exerted by the ride wheel? 4.5) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep the seat the rider is in going in a circle at the bottom of the motion? Go back to the main menu", null, "STATION 5: THE ROLLER COASTER LOOP", null, "Find a computer displaying the circular motion page. Select the roller coaster. Read through the measurement animation. Understand the method being described. You will use this method. Click to the movie and run it. Use the stopwatch as needed. 5.1) What is the length of the roller coaster train? 5.2) What is the velocity of the train at the highest point on the loop? 5.3) Using the tangential velocity, find the centripetal acceleration at the highest point on the loop? (R = 7.00 m at the loop’s top.) 5.4) If train’s car with rider has a mass of 455 kg, then what centripetal force is exerted on the car? 5.5) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep the coaster car traveling in a circle? Go back to the main menu", null, "STATION 6: THE AIRPLANE WARNING WARNING WARNING Only one person at a time is allowed under the plane. Everyone else is to stay away.", null, "6.1) What is the radius of the plane’s circle? 6.2) What is the plane’s period of motion? 6.3) What is the plane’s tangential velocity? 6.4) What is the plane’s centripetal acceleration? 6.5) If the mass of the plane is 247 grams, then what centripetal force is exerted on it? 6.6) The term centripetal force is a generic term. If you were to talk about gravity, you would know that only mass exerts a gravitational force. But a “centripetal force” can be exerted by many different things. What is supplying the centripetal force to keep the coaster car traveling in a circle? STATION 7: THE SCRAMBLER RIDE Find a computer displaying the circular motion page. Select The Scrambler Ride. Press the PLAY button to see the animation.", null, "The “Scrambler’s” motion is a complex circular motion. When rider is the farthest away from the center, point B, the rider is moving with a speed equal to the tangential velocities about the minor axis AND the major axis. The rider’s radius is equal to the distance between him/her and the center. When the rider is closest to the center, point C, the rider experiences a velocity that is the difference between the major and minor axis. The rider’s radius is equal to the radius associated with the greatest velocity. 7.1) What is the period of motion when the MINOR axis is the radius?", null, "7.2) What is the tangential velocity when the MINOR axis is the radius? 7.3) What is the period of motion when the MAJOR axis is the radius?", null, "7.4) What is the tangential velocity when the MAJOR axis is the radius? 7.5) What is the centripetal acceleration in g’s a point A? (Ignore the motions and look at the absolute radius the ride chair is spinning about.) 7.6) What is the centripetal acceleration in g’s a point C? (The radius to use will be the one associated with the fastest tangetnitial velocity. The tangential velocity to use will be the net velocity of the two tangential velocities.) Go back to the main menu", null, "TEACHER NOTES Station 1", null, "• A“Stomper” car tethered to an upside down tennisball can. This card will need a 3” x 5” index card on its top. • Set up a probe to calculate the time through it using the gate method Station 2 • Set is up on a computer. It will run best if it is copied to the hard drive. • stopwatch Station 3", null, "• This station needs a radio control car that runs backwards when turned on. This is a \\$5.00 car from K-Mart. when turned on it runs. you do not need the controller to use it! It automatically runs in a circle backwards • The actual remote is not used. • stopwatch • meter stick Station 4 • Set is up on a computer. It will run best if it is copied to the hard drive. • stopwatch Station 5 • Set is up on a computer. It will run best if it is copied to the hard drive. • stopwatch Station 6", null, "• airplane: Think safety. • (2) meter sticks • stop watch • The ariplane is tethered to the ceiling. Station 7 HONORS ONLY • Set is up on a computer. It will run best if it is copied to the hard drive. • stopwatch By Tony Wayne ©2002", null, "A special thanks to VASTfor hosting our web site." ]

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[ "Listed in the following table are assigned readings and reading questions that students were expected to complete prior to attending class sessions. The reading questions are multiple choice or numerical answer questions. Students received instant feedback and could make multiple attempts.\n\nProbability\n1 C1\n\n1a: Introduction (PDF)\n\n1b: Counting and Sets (PDF)\n\nC2\n\n2: Probability: Terminology and Examples (PDF)\n\nR Tutorial 1A: Basics\n\nR Tutorial 1B: Random Numbers\n\n2 C3 3: Conditional Probability, Independence and Bayes' Theorem (PDF) Reading Questions for 3\nC4\n\n4a: Discrete Random Variables (PDF)\n\n4b: Discrete Random Variables: Expected Value (PDF)\n\n3 C5\n\n5a: Variance of Discrete Random Variables (PDF)\n\n5b: Continuous Random Variables (PDF)\n\n5c: Gallery of Continuous Random Variables (PDF)\n\n5d: Manipulating Continuous Random Variables (PDF)\n\n4 C6\n\n6a: Expectation, Variance and Standard Deviation for Continuous Random Variables (PDF)\n\n6b: Central Limit Theorem and the Law of Large Numbers (PDF)\n\n6c: Appendix (PDF)\n\nC7\n\n7a: Joint Distributions, Independence (PDF)\n\n7b: Covariance and Correlation (PDF)\n\nStatistics: Bayesian Inference\n5 C10\n\n10a: Introduction to Statistics (PDF)\n\n10b: Maximum Likelihood Estimates (PDF)\n\n6 C11 11: Bayesian Updating with Discrete Priors (PDF) Reading Questions for 11\nC12\n\n12a: Bayesian Updating: Probabilistic Prediction (PDF)\n\n12b: Bayesian Updating: Odds (PDF)\n\n7 C13\n\n13a: Bayesian Updating with Continuous Priors (PDF)\n\n13b: Notational Conventions (PDF)\n\nC14\n\n14a: Beta Distributions (PDF)\n\n14b: Bayesian Updating with Continuous Data (PDF)\n\nReading Questions for 14a and 14b\n\n8 C15\n\n15a: Conjugate Priors: Beta and Normal (PDF)\n\n15b: Choosing Priors (PDF)\n\nC16 16: Probability Intervals (PDF) Reading Questions for 16\nStatistics: Frequentist Inference—Null Hypothesis Significance Testing (NHST)\n9 C17\n\n17a: The Frequentist School of Statistics (PDF)\n\n17b: Null Hypothesis Significance Testing I (PDF)\n\nC18 18: Null Hypothesis Significance Testing II (PDF) Reading Questions for 18\n10 C19 19: Null Hypothesis Significance Testing III (PDF) Reading Questions for 19\nC20 20: Comparison of Frequentist and Bayesian Inference (PDF)\nStatistics: Confidence Intervals; Regression\n12 C22 22: Confidence Intervals Based on Normal Data (PDF) Reading Questions for 22\nC23\n\n23a: Confidence Intervals: Three Views (PDF)\n\n23b: Confidence Intervals for the Mean of Non-normal Data (PDF)\n\n3 C24 24: Bootstrap Confidence Intervals (PDF) Reading Questions for 24\nC25 25: Linear Regression (PDF) Reading Questions for 25" ]

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[ "Home | | Physics | Equilibrium of bodies and types of equilibrium\n\n# Equilibrium of bodies and types of equilibrium", null, "Equilibrium is thus stable, unstable or neutral according to whether the potential energy is minimum, maximum or constant.\n\nEquilibrium of bodies and types of equilibrium\n\nIf a marble M is placed on a curved surface of a bowl S, it rolls down and settles in equilibrium at the lowest point A (Fig. a). This equilibrium position corresponds to minimum potential energy. If the marble is disturbed and displaced to a point B, its energy increases When it is released, the marble rolls back to A. Thus the marble at the position A is said to be in stable equilibrium.", null, "Suppose now that the bowl S is inverted and the marble is placed at its top point, at A (Fig. b). If the marble is displaced slightly to the point C, its potential energy is lowered and tends to move further away from the equilibrium position to one of lowest energy. Thus the marble is said to be in unstable equilibrium.\n\nSuppose now that the marble is placed on a plane surface (Fig. c). If it is displaced slightly, its potential energy does not change. Here the marble is said to be in neutral equilibrium.\n\nEquilibrium is thus stable, unstable or neutral according to whether the potential energy is minimum, maximum or constant.\n\nWe may also characterize the stability of a mechanical system by noting that when the system is disturbed from its position of equilibrium, the forces acting on the system may\n\n(i)                tend to bring back to its original position if potential energy is a minimum, corresponding to stable equilibrium.\n\n(ii)             tend to move it farther away if potential energy is maximum, corresponding unstable equilibrium.\n\n(iii)           tend to move either way if potential energy is a constant corresponding to neutral equilibrium\n\nConsider three uniform bars shown in Fig. a,b,c. Suppose each bar is slightly displaced from its position of equilibrium and then released. For bar A, fixed at its top end, its centre of gravity G rises to G1 on being displaced, then the bar returns back to its original position on being released, so that the equilibrium is stable.", null, "For bar B, whose fixed end is at its bottom, its centre of gravity G is lowered to G2 on being displaced, then the bar B will keep moving away from its original position on being released, and the equilibrium is said to be unstable.\n\nFor bar C, whose fixed point is about its centre of gravity, the centre of gravity remains at the same height on being displaced, the bar will remain in its new position, on being released, and the equilibrium is said to be neutral.\n\nStudy Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\n11th 12th std standard Class Physics sciense Higher secondary school College Notes : Equilibrium of bodies and types of equilibrium |" ]

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[ "# SmallGroup(128,1015)\n\nView a complete list of particular groups (this is a very huge list!)[SHOW MORE]\n\n## Definition\n\nThis is a group of order 128 given by the following presentation, where", null, "$e$ is used to denote the identity element:", null, "$G := \\langle a_1, a_2, a_3, a_4 \\mid a_1^2 = a_2^4 = a_3^4 = a_4^4 = e, [a_1,a_2] = a_3^2, [a_1,a_3] = e, [a_2,a_3] = a_4^2, [a_1,a_4] = [a_2,a_4] = [a_3,a_4] = e \\rangle$\n\nFor the definition,", null, "$[ , ]$ stands for the commutator. It does not matter whether we choose the left or right action convention for the commutator -- the groups defined in both cases are isomorphic.\n\n## Arithmetic functions\n\nWant to compare and contrast arithmetic function values with other groups of the same order? Check out groups of order 128#Arithmetic functions\n\n## GAP implementation\n\n### Group ID\n\nThis finite group has order 128 and has ID 1015 among the groups of order 128 in GAP's SmallGroup library. For context, there are groups of order 128. It can thus be defined using GAP's SmallGroup function as:\n\nSmallGroup(128,1015)\n\nFor instance, we can use the following assignment in GAP to create the group and name it", null, "$G$:\n\ngap> G := SmallGroup(128,1015);\n\nConversely, to check whether a given group", null, "$G$ is in fact the group we want, we can use GAP's IdGroup function:\n\nIdGroup(G) = [128,1015]\n\nor just do:\n\nIdGroup(G)\n\nto have GAP output the group ID, that we can then compare to what we want.\n\n### Description by presentation\n\ngap> F := FreeGroup(4);\n<free group on the generators [ f1, f2, f3, f4 ]>\ngap> G := F/[F.1^2,F.2^4,F.3^4,F.4^4,Comm(F.1,F.2)*F.3^(-2),Comm(F.1,F.3),Comm(F.2,F.3)*F.4^(-2),Comm(F.1,F.4),Comm(F.2,F.4),Comm(F.3,F.4)];\n<fp group on the generators [ f1, f2, f3, f4 ]>\ngap> IdGroup(G);\n[ 128, 1015 ]" ]

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[ "# How would you explain covariance to someone who understands only the mean?\n\n...assuming that I'm able to augment their knowledge about variance in an intuitive fashion ( Understanding \"variance\" intuitively ) or by saying: It's the average distance of the data values from the 'mean' - and since variance is in square units, we take the square root to keep the units same and that is called standard deviation.\n\nLet's assume this much is articulated and (hopefully) understood by the 'receiver'. Now what is covariance and how would one explain it in simple English without the use of any mathematical terms/formulae? (I.e., intuitive explanation. ;)\n\nPlease note: I do know the formulae and the math behind the concept. I want to be able to 'explain' the same in an easy to understand fashion, without including the math; i.e., what does 'covariance' even mean?\n\n• @Xi'an - 'how' exactly would you define it via simple linear regression? I'd really like to know... – PhD Nov 8 '11 at 2:08\n• Assuming you already have a scatterplot of your two variables, x vs. y, with origin at (0,0), simply draw two lines at x=mean(x) (vertical) and y=mean(x) (horizontal): using this new system of coordinates (origin is at (mean(x),mean(y)), put a \"+\" sign in the top-right and bottom-left quadrants, a \"-\" sign in the two other quadrants; you got the sign of the covariance, which is basically what @Peter said. Scaling the x- and y-units (by SD) lead to a more interpretable summary, as discussed in the ensuing thread. – chl Nov 9 '11 at 22:54\n• @chl - could you please post that as an answer and maybe use graphics to depict it! – PhD Nov 10 '11 at 4:03\n• I found the video on this website to help me as I prefer images over abstract explanations. Website with video Specifically this image:  – Karl Morrison Jul 29 '15 at 21:13\n\nSometimes we can \"augment knowledge\" with an unusual or different approach. I would like this reply to be accessible to kindergartners and also have some fun, so everybody get out your crayons!\n\nGiven paired $$(x,y)$$ data, draw their scatterplot. (The younger students may need a teacher to produce this for them. :-) Each pair of points $$(x_i,y_i)$$, $$(x_j,y_j)$$ in that plot determines a rectangle: it's the smallest rectangle, whose sides are parallel to the axes, containing those points. Thus the points are either at the upper right and lower left corners (a \"positive\" relationship) or they are at the upper left and lower right corners (a \"negative\" relationship).\n\nDraw all possible such rectangles. Color them transparently, making the positive rectangles red (say) and the negative rectangles \"anti-red\" (blue). In this fashion, wherever rectangles overlap, their colors are either enhanced when they are the same (blue and blue or red and red) or cancel out when they are different.", null, "(In this illustration of a positive (red) and negative (blue) rectangle, the overlap ought to be white; unfortunately, this software does not have a true \"anti-red\" color. The overlap is gray, so it will darken the plot, but on the whole the net amount of red is correct.)\n\nNow we're ready for the explanation of covariance.\n\nThe covariance is the net amount of red in the plot (treating blue as negative values).\n\nHere are some examples with 32 binormal points drawn from distributions with the given covariances, ordered from most negative (bluest) to most positive (reddest).", null, "They are drawn on common axes to make them comparable. The rectangles are lightly outlined to help you see them. This is an updated (2019) version of the original: it uses software that properly cancels the red and cyan colors in overlapping rectangles.\n\nLet's deduce some properties of covariance. Understanding of these properties will be accessible to anyone who has actually drawn a few of the rectangles. :-)\n\n• Bilinearity. Because the amount of red depends on the size of the plot, covariance is directly proportional to the scale on the x-axis and to the scale on the y-axis.\n\n• Correlation. Covariance increases as the points approximate an upward sloping line and decreases as the points approximate a downward sloping line. This is because in the former case most of the rectangles are positive and in the latter case, most are negative.\n\n• Relationship to linear associations. Because non-linear associations can create mixtures of positive and negative rectangles, they lead to unpredictable (and not very useful) covariances. Linear associations can be fully interpreted by means of the preceding two characterizations.\n\n• Sensitivity to outliers. A geometric outlier (one point standing away from the mass) will create many large rectangles in association with all the other points. It alone can create a net positive or negative amount of red in the overall picture.\n\nIncidentally, this definition of covariance differs from the usual one only by a universal constant of proportionality (independent of the data set size). The mathematically inclined will have no trouble performing the algebraic demonstration that the formula given here is always twice the usual covariance.\n\n• Now if only all introductory statistical concepts could be presented to students in this lucid manner … – MannyG Nov 10 '11 at 18:26\n• This is beautiful. And very very clear. – Benjamin Mako Hill Jun 2 '12 at 15:37\n• Having done the algebra, I wonder if \"universal constant of proportionality (independent of the data set size)\" may be misleading, so I want to check if I understood the procedure correctly. For {(0,0),(1,1),(2,2)} there are $3\\choose{2}$ = 3 possible rectangles of areas 1, 1 and 4. They're all red so the \"covariance\" is 6. And {(0,0),(1,1),(1,1),(2,2)} has $4\\choose{2}$ = 6 rectangles, all red or zero, of areas 0, 1, 1, 1, 1 and 4 so \"covariance\" is 8. Is this right? If so it's $\\sum_{i<j}(x_i-x_j)(y_i-y_j)$. – Silverfish Nov 7 '13 at 10:42\n• Thanks, this as I suspected. I realised that an extra factor of 2 comes out if the sum is taken over all $i, j$ rather than $i<j$. I think the only other ambiguity is whether the \"pair\" $(i,i)$ counts - the area of the rectangle is zero, but if averaging rather than summating it clearly makes a difference! Incidentally when I teach covariance I also use the \"positive and negative rectangles\" approach, but pairing each data points with the mean point. I find this makes some of the standard formulae more accessible, but on the whole I prefer your method. – Silverfish Nov 8 '13 at 17:00\n• @fcoppens Indeed, there is a traditional explanation that proceeds as you suggest. I thought of this one because I did not want to introduce an idea that is unnecessary--namely, constructing the centroid $(\\bar x, \\bar y)$. That would make the explanation inaccessible to the five-year-old with a box of crayons. Some of the conclusions I drew at the end would not be immediate, either. For example, it would no longer be quite so obvious that the covariance is sensitive to certain kinds of outliers. – whuber Aug 17 '15 at 14:33\n\nTo elaborate on my comment, I used to teach the covariance as a measure of the (average) co-variation between two variables, say $x$ and $y$.\n\nIt is useful to recall the basic formula (simple to explain, no need to talk about mathematical expectancies for an introductory course):\n\n$$\\text{cov}(x,y)=\\frac{1}{n}\\sum_{i=1}^n(x_i-\\bar x)(y_i-\\bar y)$$\n\nso that we clearly see that each observation, $(x_i,y_i)$, might contribute positively or negatively to the covariance, depending on the product of their deviation from the mean of the two variables, $\\bar x$ and $\\bar y$. Note that I do not speak of magnitude here, but simply of the sign of the contribution of the ith observation.\n\nThis is what I've depicted in the following diagrams. Artificial data were generated using a linear model (left, $y = 1.2x + \\varepsilon$; right, $y = 0.1x + \\varepsilon$, where $\\varepsilon$ were drawn from a gaussian distribution with zero mean and $\\text{SD}=2$, and $x$ from an uniform distribution on the interval $[0,20]$).", null, "The vertical and horizontal bars represent the mean of $x$ and $y$, respectively. That mean that instead of \"looking at individual observations\" from the origin $(0,0)$, we can do it from $(\\bar x, \\bar y)$. This just amounts to a translation on the x- and y-axis. In this new coordinate system, every observation that is located in the upper-right or lower-left quadrant contributes positively to the covariance, whereas observations located in the two other quadrants contribute negatively to it. In the first case (left), the covariance equals 30.11 and the distribution in the four quadrants is given below:\n\n + -\n+ 30 2\n- 0 28\n\n\nClearly, when the $x_i$'s are above their mean, so do the corresponding $y_i$'s (wrt. $\\bar y$). Eye-balling the shape of the 2D cloud of points, when $x$ values increase $y$ values tend to increase too. (But remember we could also use the fact that there is a clear relationship between the covariance and the slope of the regression line, i.e. $b=\\text{Cov}(x,y)/\\text{Var}(x)$.)\n\nIn the second case (right, same $x_i$), the covariance equals 3.54 and the distribution across quadrants is more \"homogeneous\" as shown below:\n\n + -\n+ 18 14\n- 12 16\n\n\nIn other words, there is an increased number of case where the $x_i$'s and $y_i$'s do not covary in the same direction wrt. their means.\n\nNote that we could reduce the covariance by scaling either $x$ or $y$. In the left panel, the covariance of $(x/10,y)$ (or $(x,y/10)$) is reduced by a ten fold amount (3.01). Since the units of measurement and the spread of $x$ and $y$ (relative to their means) make it difficult to interpret the value of the covariance in absolute terms, we generally scale both variables by their standard deviations and get the correlation coefficient. This means that in addition to re-centering our $(x,y)$ scatterplot to $(\\bar x, \\bar y)$ we also scale the x- and y-unit in terms of standard deviation, which leads to a more interpretable measure of the linear covariation between $x$ and $y$.\n\nCovariance is a measure of how much one variable goes up when the other goes up.\n\n• Is it always in the 'same' direction? Also, does it apply for inverse relations too (i.e., as one goes up the other goes down)? – PhD Nov 8 '11 at 2:07\n• @nupul Well, the opposite of \"up\" is \"down\" and the opposite of \"positive\" is \"negative\". I tried to give a one sentence answer. Yours is much more complete. Even your \"how two variables change together\" is more complete, but, I think, a little harder to understand. – Peter Flom Nov 8 '11 at 11:37\n• That's right, Peter, which is why @naught101 made that comment: you description sounds like a rate of change, whose units will therefore be [units of one variable] / [units of the other variable] (if we interpret it like a derivative) or will just be [units of one variable] (if we interpret as a pure difference). Those are neither covariance (whose unit of measure is the product of the units for the two variables) nor correlation (which is unitless). – whuber Aug 13 '13 at 20:15\n• @nbro Consider any concrete example: suppose you know the covariance of variables $X$ and $Y$ is $1,$ for instance. Even with the most generous understanding of \"variable\" and \"go up,\" could you tell from that information alone how much $Y$ goes up when $X$ goes up by a given amount? The answer is no: the only information it gives you is that $Y$ would tend to increase. In this post Peter has confused the covariance with a regression coefficient (of which there are two, by the way, and they usually are different). – whuber Jun 25 '19 at 19:47\n• @nbro Covariance is the second central moment of a bivariate random variable. And that returns us to the beginning: how would one convey this precise definition to the proverbial five-year-old? As always, there's a trade-off between economy of expression and accuracy: when the audience doesn't have the concepts or language needed to understand something immediately, somehow you have to weave in an explanation of that background along with your description. Doing it right requires some elaboration. Usually there's no shortcut. – whuber Jun 25 '19 at 20:48\n\nI loved @whuber 's answer - before I only had a vague idea in my mind of how covariance could be visualised, but those rectangle plots are genius.\n\nHowever since the formula for covariance involves the mean, and the OP's original question did state that the 'receiver' does understand the concept of the mean, I thought I would have a crack at adapting @whuber's rectangle plots to compare each data point to the means of x and y, as this more represents what's going on in the covariance formula. I thought it actually ended up looking fairly intuitive:", null, "The blue dot in the middle of each plot is the mean of x (x_mean) and mean of y (y_mean).\n\nThe rectangles are comparing the value of x - x_mean and y - y_mean for each data point.\n\nThe rectangle is green when either:\n\n• both x and y are greater than their respective means\n• both x and y are less than their respective means\n\nThe rectangle is red when either:\n\n• x is greater than x_mean but y is less than y_mean\n• x is less than x_mean but y is greater than y_mean\n\nCovariance (and correlation) can be both strongly negative and strongly positive. When the graph is dominated by one colour more than the other, it means that the data mostly follows a consistent pattern.\n\n• If the graph has lots more green than red, it means that y generally increases when x increases.\n• If the graph has lots more red than green, it means that y generally decreases when x increases.\n• If the graph isn't dominated by one colour or the other, it means that there isn't much of a pattern to how x and y relate to each other.\n\nThe actual value of the covariance for two different variables x and y, is basically the sum of all the green area minus all the red area, then divided by the total number of data points - effectively the average greenness-vs-redness of the graph.\n\nHow does that sound/look?\n\n• Just to make sure, the blue dot in the middle is the average of ALL the points? – BigBear Jan 6 at 17:17\n• Yes that's correct. It's different in each of the 4 graphs as they are of different sets of data. – capohugo Jan 7 at 22:18\n• Thanks for this! One confusing thing about this is that you have sort of flipped (although not exactly) the color scheme from @whuber's answer. In trying to wrap my head around the concept, that gave some cognitive whiplash. – Dylan_Gomes May 31 at 5:40\n\nI am answering my own question, but I thought It'd be great for the people coming across this post to check out some of the explanations on this page.\n\nI'm paraphrasing one of the very well articulated answers (by a user'Zhop'). I'm doing so in case if that site shuts down or the page gets taken down when someone eons from now accesses this post ;)\n\nCovariance is a measure of how much two variables change together. Compare this to Variance, which is just the range over which one measure (or variable) varies.\n\nIn studying social patterns, you might hypothesize that wealthier people are likely to be more educated, so you'd try to see how closely measures of wealth and education stay together. You would use a measure of covariance to determine this.\n\n...\n\nI'm not sure what you mean when you ask how does it apply to statistics. It is one measure taught in many stats classes. Did you mean, when should you use it?\n\nYou use it when you want to see how much two or more variables change in relation to each other.\n\nThink of people on a team. Look at how they vary in geographic location compared to each other. When the team is playing or practicing, the distance between individual members is very small and we would say they are in the same location. And when their location changes, it changes for all individuals together (say, travelling on a bus to a game). In this situation, we would say they have a high level of covariance. But when they aren't playing, then the covariance rate is likely to be pretty low, because they are all going to different places at different rates of speed.\n\nSo you can predict one team member's location, based on another team member's location when they are practicing or playing a game with a high degree of accuracy. The covariance measurement would be close to 1, I believe. But when they are not practicing or playing, you would have a much smaller chance of predicting one person's location, based on a team member's location. It would be close to zero, probably, although not zero, since sometimes team members will be friends, and might go places together on their own time.\n\nHowever, if you randomly selected individuals in the United States, and tried to use one of them to predict the other's locations, you'd probably find the covariance was zero. In other words, there is absolutely no relation between one randomly selected person's location in the US, and another's.\n\nAdding another one (by 'CatofGrey') that helps augment the intuition:\n\nIn probability theory and statistics, covariance is the measure of how much two random variables vary together (as distinct from variance, which measures how much a single variable varies).\n\nIf two variables tend to vary together (that is, when one of them is above its expected value, then the other variable tends to be above its expected value too), then the covariance between the two variables will be positive. On the other hand, if one of them is above its expected value and the other variable tends to be below its expected value, then the covariance between the two variables will be negative.\n\nThese two together have made me understand covariance as I've never understood it before! Simply amazing!!\n\n• Although these descriptions are qualitatively suggestive, sadly they are incomplete: they neither distinguish covariance from correlation (the first description appears to confuse the two, in fact), nor do they bring out the fundamental assumption of linear co-variation. Also, neither addresses the important aspect that covariance depends (linearly) on the scale of each variable. – whuber Nov 8 '11 at 14:35\n• @whuber - agreed! And hence haven't marked mine as the answer :) (not as yet ;) – PhD Nov 9 '11 at 0:45\n\nI really like Whuber's answer, so I gathered some more resources. Covariance describes both how far the variables are spread out, and the nature of their relationship.\n\nCovariance uses rectangles to describe how far away an observation is from the mean on a scatter graph:\n\n• If a rectangle has long sides and a high width or short sides and a short width, it provides evidence that the two variables move together.\n\n• If a rectangle has two sides that are relatively long for that variables, and two sides that are relatively short for the other variable, this observation provides evidence the variables do not move together very well.\n\n• If the rectangle is in the 2nd or 4th quadrant, then when one variable is greater than the mean, the other is less than the mean. An increase in one variable is associated with a decrease in the other.\n\nI found a cool visualization of this at http://sciguides.com/guides/covariance/, It explains what covariance is if you just know the mean.\n\n• +1 Nice explanation (especially that introductory one-sentence summary). The link is interesting. Since it has no archive on the Wayback machine it likely is new. Because it so closely parallels my (three-year-old) answer, right down to the choice of red for positive and blue for negative relationships, I suspect it is an (unattributed) derivative of the material on this site. – whuber Aug 9 '14 at 17:54\n• The \"cool visualization\" link has died... . – whuber Jan 4 '17 at 18:02\n• @MSIS That's not possible to figure out, because there are a very great number of possible distributions on the circle. But if you are referring to the uniform distribution, there's nothing to calculate, because (as I recall remarking in your thread at stats.stackexchange.com/q/414365/919) the correlation coefficient must equal its own negative, QED. – whuber Jun 25 '19 at 19:14\n• @MSIS If \"method\" means \"an appeal to symmetry,\" the answer is that it will work but the result depends on how $X$ is distributed. As an example, if $X$ is a random variable with a distribution symmetric about $0$ with finite fourth moment, then $X$ and $X^2$ must be uncorrelated. As a non-example, if $X$ has a distribution symmetric about $1,$ then nothing general can be said about the correlation of $X$ and $X^2:$ indeed, it could be any value between $-1$ and $1$ inclusive. – whuber Jun 25 '19 at 19:23\n• @MSIS Usually, in the absence of an explicit distribution, and almost always in a purely mathematical context, one assumes that a uniform distribution is meant. In the case of a geometrical circle parameterized by an angle $\\alpha,$ the basic events are of the form $a\\lt\\alpha\\le b$ and their probabilities equal $((b-a) \\operatorname{mod} 2\\pi)/(2\\pi).$ – whuber Jun 25 '19 at 20:53\n\nHere's another attempt to explain covariance with a picture. Every panel in the picture below contains 50 points simulated from a bivariate distribution with correlation between x & y of 0.8 and variances as shown in the row and column labels. The covariance is shown in the lower-right corner of each panel.", null, "Anyone interested in improving this...here's the R code:\n\nlibrary(mvtnorm)\n\nrowvars <- colvars <- c(10,20,30,40,50)\n\nall <- NULL\nfor(i in 1:length(colvars)){\ncolvar <- colvars[i]\nfor(j in 1:length(rowvars)){\nset.seed(303) # Put seed here to show same data in each panel\nrowvar <- rowvars[j]\n# Simulate 50 points, corr=0.8\nsig <- matrix(c(rowvar, .8*sqrt(rowvar)*sqrt(colvar), .8*sqrt(rowvar)*sqrt(colvar), colvar), nrow=2)\nyy <- rmvnorm(50, mean=c(0,0), sig)\ndati <- data.frame(i=i, j=j, colvar=colvar, rowvar=rowvar, covar=.8*sqrt(rowvar)*sqrt(colvar), yy)\nall <- rbind(all, dati)\n}\n}\nnames(all) <- c('i','j','colvar','rowvar','covar','x','y')\nall <- transform(all, colvar=factor(colvar), rowvar=factor(rowvar))\nlibrary(latticeExtra)\nuseOuterStrips(xyplot(y~x|colvar*rowvar, all, cov=all\\$covar,\npanel=function(x,y,subscripts, cov,...){\npanel.xyplot(x,y,...)\nprint(cor(x,y))\nltext(14,-12, round(cov[subscripts],0))\n}))\n\n\nI would simply explain correlation which is pretty intuitive. I would say \"Correlation measures the strength of relationship between two variables X and Y. Correlation is between -1 and 1 and will be close to 1 in absolute value when the relationship is strong. Covariance is just the correlation multiplied by the standard deviations of the two variables. So while correlation is dimensionless, covariance is in the product of the units for variable X and variable Y.\n\n• This seems inadequate because there is no mention of linearity. X and Y could have a strong quadratic relationship but have a correlation of zero. – mark999 May 7 '12 at 0:46\n\nVariance is the degree by which a random vairable changes with respect to its expected value Owing to the stochastic nature of be underlying process the random variable represents.\n\nCovariance is the degree by which two different random variables change with respect to each other. This could happen when random variables are driven by the same underlying process, or derivatives thereof. Either processes represented by these random variables are affecting each other, or it's the same process but one of the random variables is derived from the other.\n\nTwo variables that would have a high positive covariance (correlation) would be the number of people in a room, and the number of fingers that are in the room. (As the number of people increases, we expect the number of fingers to increase as well.)\n\nSomething that might have a negative covariance (correlation) would be a person's age, and the number of hair follicles on their head. Or, the number of zits on a person's face (in a certain age group), and how many dates they have in a week. We expect people with more years to have less hair, and people with more acne to have less dates.. These are negatively correlated.\n\n• Covariance is not necessarily interchangeable with correlation - the former is very unit dependent. Correlation is a number between -1 and 1 a unit-less scalar representing the 'strength' of the covariance IMO and that's not clear from your answer – PhD Nov 9 '11 at 18:24\n• Downvoted as the answer implies that covariance and correlation can be used interchangeably. – sapo_cosmico Apr 7 '18 at 16:20" ]

[ null, "https://i.stack.imgur.com/XPGjN.png", null, "https://i.stack.imgur.com/Kfmhn.png", null, "https://i.stack.imgur.com/D2wzc.png", null, "https://i.stack.imgur.com/2I5eO.png", null, "https://i.stack.imgur.com/3UZqo.png", null ]

[ "# Abstract Algebra\n\n## Homework Statement\n\nLet F be a field of characteristic p>0 and let E = F(a) where a is separable over F. Prove that E=F(a^p).\n\n## The Attempt at a Solution\n\nI know that maybe show how mod F(a) = mod F(a^p) or something around there.\n\n## Answers and Replies\n\nSo, since the characteristic is p, a prime E is generated by some field F adjoined some set of elements, and for each minimal polynomial associated to these elements there are no duplicate roots.\n\nNow, there are two cases to consider E is finite or E is infinite. If E is finite then you're done, since E is iso to Z mod p^k. If E is infinite (for example Z4(u) where u is some transcendental element) then you have some work to do. Consider what frobenius endomorphism\nhttp://en.wikipedia.org/wiki/Frobenius_endomorphism\nTells about the minimal polynomial of a.\n\nHurkyl\nStaff Emeritus\nScience Advisor\nGold Member\nE is iso to Z mod p^k.\nNo it's not. Such an isomorphism can only exist when k=1 and E is a field whose cardinality is p." ]

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[ "# Search results\n\n1. ### Complex analysis antiderivative existence\n\nHomework Statement a) Does f(z)=1/z have an antiderivative over C/(0,0)? b) Does f(z)=(1/z)^n have an antiderivative over C/(0,0), n integer and not equal to 1. Homework Equations The Attempt at a Solution a) No. Integrating over C= the unit circle gives us 2*pi*i. So for at least...\n\nHow so?\n3. ### Differential geometry acceleration as the sum of two vectors\n\nBasically the main question I have is if the only way to do this would be to explicitly calculate the expression for N(t).\n4. ### Differential geometry acceleration as the sum of two vectors\n\nSo I should able to write a(1) as a linear combination of T(1) and N(1), correct? But how do I get N(t)? Taking derivatives of T(t) is very messy and the professor said it didn't involve any long calculations. Furthermore we haven't learned about the binormal and normal in this section yet...\n5. ### Differential geometry acceleration as the sum of two vectors\n\nHomework Statement a(t)=<1+t^2,4/t,8*(2-t)^(1/2)> Express the acceleration vector a''(1) as the sum of a vector parallel to a'(1) and a vector orthogonal to a'(1) Homework Equations The Attempt at a Solution I took the first two derivatives and calculated a'(t)=<2t, -4t^2...\n6. ### Cauchy riemann equations and constant functions\n\nAh yes, I meant unit circle. I was thinking the domain would need to have restrictions because I thought I proved it yet found a counterexample at the same time. But for the original problem: Let f=U + iV. We have |f|=c for some constant c. |z|^2=zz*, so we have |f|^2=c^2 which implies...\n7. ### Cauchy riemann equations and constant functions\n\nBut what about the function f(z)=z=x+iy? Isn't that function analytic on the say, unit disk with a constant modulus but f is not constant?\n8. ### Cauchy riemann equations and constant functions\n\nHomework Statement Let f(z) be analytic on the set H. Let the modulus of f(z) be constant. Does f need be constant also? Explain. Homework Equations Cauchy riemann equations Hint: Prove If f and f* are both analytic on D, then f is constant. The Attempt at a Solution I think f need...\n9. ### Linear equations, solution sets and inner products\n\nHomework Statement Let W be the subspace of R4 such that W is the solution set to the following system of equations: x1-4x2+2x3-x4=0 3x1-13x2+7x3-2x4=0 Let U be subspace of R4 such that U is the set of vectors in R4 such the inner product <u,w>=0 for every w in W. Find a 2 by 4...\n10. ### Open and closed intervals and real numbers\n\nYay! How exactly does uniqueness follow though? It seems like its trivial to prove.\n11. ### Open and closed intervals and real numbers\n\nSo far: Since S is bounded above and below, by Dedekind completeness there exists a supremum of S. Call it b. Again by dedekind completeness we can say there exists an infimum of S. Call it b. By definition of sup and inf, a<b. We are left to show a,b are unique and that S is exactly one...\n12. ### Open and closed intervals and real numbers\n\nGood idea. Isn't it an axiom that if a nonempty subset of R has an upper bound, then it has a least upper bound/sup(S)?\n13. ### Open and closed intervals and real numbers\n\nHomework Statement Show that: Let S be a subset of the real numbers such that S is bounded above and below and if some x and y are in S with x not equal to y, then all numbers between x and y are in S. then there exist unique numbers a and b in R with a<b such that S is one of the...\n14. ### Additive functions, unions, and intersections.\n\nI am such an idiot. I tried it before representing the left hand side as a union of disjoint sets but for some reason I didn't bother manipulating the right hand side in the same way. Thanks a ton!\n15. ### Additive functions, unions, and intersections.\n\nHomework Statement A function G:P--->R where R is the set of real numbers is additive provided G(X1 U X2)=G(X1)+G(X2) if X1, X2 are disjoint. Let S be a set, Let P be the power set of S. Suppose G is an additive function mapping P to R. Prove that if X1 and X2 are ARBITRARY(not necessarily...\n16. ### Set theory and ordered pairs\n\nNo. But why does it matter if it is in it or not?\n17. ### Question about set theory and ordered pairs\n\nHi I was reading through a textbook and I came across the set theoretic definition of an ordered pair (Kuratowski), where (x,y)={{x},{x,y}}, which apparently can be shortened to {x,{x,y}}. This seems to be the standard definition for an ordered pair in set theory so that we can determine...\n18. ### Set theory and ordered pairs\n\nHomework Statement Determine whether or not the following definition of an ordered pair is set theoretic (i.e. you can distinguish between the \"first\" element and the \"second\" element using only set theory). (x,y)={x,{y}} Homework Equations The Attempt at a Solution I am inclined to...\n19. ### Scaling of the vertical projectile problem nondimensionalization\n\nSo O is just the order of the taylor approximation that replaces the right hand side?\n20. ### Scaling of the vertical projectile problem nondimensionalization\n\nI actually redid the nondimensionalization part and it was correct, but not properly scaled which explains why I didn't get a term that was <<1. The correctly scaled nondimensional equation is d2Y/dT^2=-1/(1+eY)^2 where e<<1. It doesn't really change though, I still don't see where the...\n21. ### Scaling of the vertical projectile problem nondimensionalization\n\nHomework Statement Restate the vertical projectile problem in a properly scaled form. (suppose x<<R). d2x/dt^2=-g(R^2)/(x+R)^2 Initial conditions: x(0)=0, dx(0)/dt=Vo Find the approximate solution accurate up to order O(1) and O(e), where r is a small dimensionless parameter. (i.e. the...\n22. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nThank you so much for your help!\n23. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nIf d>c then f'(xo)<0 and xo>c. f'(c)=0 and f'(xo)<0. But this contradicts that f'(x) is strictly increasing. So the statement must be true.\n24. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nAh my mistake, the quantity is positive. I think I know where this is headed. So f'(xo)>f'(c)=0 and c>xo. But f'(x) must be strictly increasing on I since f''(x)>0, and thus the statement must be true by contradiction.\n25. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nIt is negative. I'm still a bit puzzled as to how the mean value theorem relates to proving this problem.\n26. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nThe MVT tells me that there exists some Xo in [d,c] such that f'(Xo) = [f(c)-f(d)]/(c-d).\n27. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nI still can't figure this one out - I get the feeling the proof is much more simple than I think it is.\n28. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nIt implies that f(x) is concave up? It also implies c is a minimum by the second derivative test but we haven't covered that as of this section in our textbook.\n29. ### Intro to analysis proof first and second derivatives and mean value theorem\n\nHomework Statement Let f(x) be a twice differentiable function on an interval I. Let f''(x)>0 for all x in I and let f'(c)=0 for some c in I. Prove f(x) is greater than or equal to f(c) for all x in I. Homework Equations Mean value theorem? The Attempt at a Solution f''(x)>0...\n30. ### Linear algebra proof - Orthogonal complements\n\nHomework Statement Let V be an inner product space, and let W be a finite dimensional subspace of V. If x is not an element of W, prove that there exists y in V such that y is in the orthogonal complement of W, but the inner product of x and y is not equal to 0. Homework Equations The..." ]

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[ "MFEM  v3.0\nmfem::Vector Class Reference\n\nVector data type. More...\n\n`#include <vector.hpp>`\n\nInheritance diagram for mfem::Vector:\n[legend]\n\n## Public Member Functions\n\nVector ()\nDefault constructor for Vector. Sets size = 0 and data = NULL. More...\n\nVector (const Vector &)\nCopy constructor. More...\n\nVector (int s)\nCreates vector of size s. More...\n\nVector (double *_data, int _size)\n\nvoid Load (std::istream **in, int np, int *dim)\nReads a vector from multiple files. More...\n\nvoid Load (std::istream &in, int Size)\nLoad a vector from an input stream. More...\n\nLoad a vector from an input stream. More...\n\nvoid SetSize (int s)\nResizes the vector if the new size is different. More...\n\nvoid SetData (double *d)\n\nvoid SetDataAndSize (double *d, int s)\n\nvoid NewDataAndSize (double *d, int s)\n\nvoid MakeDataOwner ()\n\nvoid Destroy ()\nDestroy a vector. More...\n\nint Size () const\nReturns the size of the vector. More...\n\ndouble * GetData () const\n\noperator double * ()\n\noperator const double * () const\n\nbool OwnsData () const\n\nvoid StealData (double **p)\nChanges the ownership of the data; after the call the Vector is empty. More...\n\ndouble * StealData ()\nChanges the ownership of the data; after the call the Vector is empty. More...\n\ndouble & Elem (int i)\nSets value in vector. Index i = 0 .. size-1. More...\n\nconst double & Elem (int i) const\nSets value in vector. Index i = 0 .. size-1. More...\n\ndouble & operator() (int i)\nSets value in vector. Index i = 0 .. size-1. More...\n\nconst double & operator() (int i) const\nSets value in vector. Index i = 0 .. size-1. More...\n\ndouble operator* (const double *) const\n\ndouble operator* (const Vector &v) const\nReturn the inner-product. More...\n\nVectoroperator= (const double *v)\n\nVectoroperator= (const Vector &v)\nRedefine '=' for vector = vector. More...\n\nVectoroperator= (double value)\nRedefine '=' for vector = constant. More...\n\nVectoroperator*= (double c)\n\nVectoroperator/= (double c)\n\nVectoroperator-= (double c)\n\nVectoroperator-= (const Vector &v)\n\nVectoroperator+= (const Vector &v)\n\nVectorAdd (const double a, const Vector &Va)\n(*this) += a * Va More...\n\nVectorSet (const double a, const Vector &x)\n(*this) = a * x More...\n\nvoid SetVector (const Vector &v, int offset)\n\nvoid Neg ()\n(*this) = -(*this) More...\n\nvoid median (const Vector &lo, const Vector &hi)\nv = median(v,lo,hi) entrywise. Implementation assumes lo <= hi. More...\n\nvoid GetSubVector (const Array< int > &dofs, Vector &elemvect) const\n\nvoid GetSubVector (const Array< int > &dofs, double *elem_data) const\n\nvoid SetSubVector (const Array< int > &dofs, const Vector &elemvect)\n\nvoid SetSubVector (const Array< int > &dofs, double *elem_data)\n\nvoid AddElementVector (const Array< int > &dofs, const Vector &elemvect)\nAdd (element) subvector to the vector. More...\n\nvoid AddElementVector (const Array< int > &dofs, double *elem_data)\n\nvoid AddElementVector (const Array< int > &dofs, const double a, const Vector &elemvect)\n\nvoid Print (std::ostream &out=std::cout, int width=8) const\nPrints vector to stream out. More...\n\nvoid Print_HYPRE (std::ostream &out) const\nPrints vector to stream out in HYPRE_Vector format. More...\n\nvoid Randomize (int seed=0)\nSet random values in the vector. More...\n\ndouble Norml2 () const\nReturns the l2 norm of the vector. More...\n\ndouble Normlinf () const\nReturns the l_infinity norm of the vector. More...\n\ndouble Norml1 () const\nReturns the l_1 norm of the vector. More...\n\ndouble Normlp (double p) const\nReturns the l_p norm of the vector. More...\n\ndouble Max () const\nReturns the maximal element of the vector. More...\n\ndouble Min () const\nReturns the minimal element of the vector. More...\n\ndouble Sum () const\nReturn the sum of the vector entries. More...\n\ndouble DistanceTo (const double *p) const\nCompute the Euclidian distance to another vector. More...\n\nint CheckFinite () const\n\n~Vector ()\nDestroys vector. More...\n\nint size\n\nint allocsize\n\ndouble * data\n\n## Friends\n\nvoid swap (Vector *v1, Vector *v2)\nSwap v1 and v2. More...\n\nvoid add (const Vector &v1, const Vector &v2, Vector &v)\nDo v = v1 + v2. More...\n\nvoid add (const Vector &v1, double alpha, const Vector &v2, Vector &v)\nDo v = v1 + alpha * v2. More...\n\nvoid add (const double a, const Vector &x, const Vector &y, Vector &z)\nz = a * (x + y) More...\n\nvoid add (const double a, const Vector &x, const double b, const Vector &y, Vector &z)\nz = a * x + b * y More...\n\nvoid subtract (const Vector &v1, const Vector &v2, Vector &v)\nDo v = v1 - v2. More...\n\nvoid subtract (const double a, const Vector &x, const Vector &y, Vector &z)\nz = a * (x - y) More...\n\n## Detailed Description\n\nVector data type.\n\nDefinition at line 29 of file vector.hpp.\n\n## Constructor & Destructor Documentation\n\n mfem::Vector::Vector ( )\ninline\n\nDefault constructor for Vector. Sets size = 0 and data = NULL.\n\nDefinition at line 39 of file vector.hpp.\n\n mfem::Vector::Vector ( const Vector & v )\n\nCopy constructor.\n\nDefinition at line 26 of file vector.cpp.\n\n mfem::Vector::Vector ( int s )\ninlineexplicit\n\nCreates vector of size s.\n\nDefinition at line 234 of file vector.hpp.\n\n mfem::Vector::Vector ( double * _data, int _size )\ninline\n\nDefinition at line 47 of file vector.hpp.\n\n mfem::Vector::~Vector ( )\ninline\n\nDestroys vector.\n\nDefinition at line 291 of file vector.hpp.\n\n## Member Function Documentation\n\n Vector & mfem::Vector::Add ( const double a, const Vector & Va )\n\n(*this) += a * Va\n\nDefinition at line 168 of file vector.cpp.\n\n void mfem::Vector::AddElementVector ( const Array< int > & dofs, const Vector & elemvect )\n\nAdd (element) subvector to the vector.\n\nDefinition at line 449 of file vector.cpp.\n\n void mfem::Vector::AddElementVector ( const Array< int > & dofs, double * elem_data )\n\nDefinition at line 460 of file vector.cpp.\n\n void mfem::Vector::AddElementVector ( const Array< int > & dofs, const double a, const Vector & elemvect )\n\nDefinition at line 471 of file vector.cpp.\n\n int mfem::Vector::CheckFinite ( ) const\ninline\n\nCount the number of entries in the Vector for which isfinite is false, i.e. the entry is a NaN or +/-Inf.\n\nDefinition at line 206 of file vector.hpp.\n\n void mfem::Vector::Destroy ( )\ninline\n\nDestroy a vector.\n\nDefinition at line 263 of file vector.hpp.\n\n double mfem::Vector::DistanceTo ( const double * p ) const\n\nCompute the Euclidian distance to another vector.\n\nDefinition at line 608 of file vector.cpp.\n\n double & mfem::Vector::Elem ( int i )\n\nSets value in vector. Index i = 0 .. size-1.\n\nDefinition at line 67 of file vector.cpp.\n\n const double & mfem::Vector::Elem ( int i ) const\n\nSets value in vector. Index i = 0 .. size-1.\n\nDefinition at line 72 of file vector.cpp.\n\n double* mfem::Vector::GetData ( ) const\ninline\n\nDefinition at line 80 of file vector.hpp.\n\n void mfem::Vector::GetSubVector ( const Array< int > & dofs, Vector & elemvect ) const\n\nDefinition at line 403 of file vector.cpp.\n\n void mfem::Vector::GetSubVector ( const Array< int > & dofs, double * elem_data ) const\n\nDefinition at line 416 of file vector.cpp.\n\n void mfem::Vector::Load ( std::istream ** in, int np, int * dim )\n\nReads a vector from multiple files.\n\nDefinition at line 43 of file vector.cpp.\n\n void mfem::Vector::Load ( std::istream & in, int Size )\n\nLoad a vector from an input stream.\n\nDefinition at line 59 of file vector.cpp.\n\n void mfem::Vector::Load ( std::istream & in )\ninline\n\nLoad a vector from an input stream.\n\nDefinition at line 57 of file vector.hpp.\n\n void mfem::Vector::MakeDataOwner ( )\ninline\n\nDefinition at line 70 of file vector.hpp.\n\n double mfem::Vector::Max ( ) const\n\nReturns the maximal element of the vector.\n\nDefinition at line 576 of file vector.cpp.\n\n void mfem::Vector::median ( const Vector & lo, const Vector & hi )\n\nv = median(v,lo,hi) entrywise. Implementation assumes lo <= hi.\n\nDefinition at line 390 of file vector.cpp.\n\n double mfem::Vector::Min ( ) const\n\nReturns the minimal element of the vector.\n\nDefinition at line 587 of file vector.cpp.\n\n void mfem::Vector::Neg ( )\n\n(*this) = -(*this)\n\nDefinition at line 207 of file vector.cpp.\n\n void mfem::Vector::NewDataAndSize ( double * d, int s )\ninline\n\nDefinition at line 67 of file vector.hpp.\n\n double mfem::Vector::Norml1 ( ) const\n\nReturns the l_1 norm of the vector.\n\nDefinition at line 548 of file vector.cpp.\n\n double mfem::Vector::Norml2 ( ) const\n\nReturns the l2 norm of the vector.\n\nDefinition at line 532 of file vector.cpp.\n\n double mfem::Vector::Normlinf ( ) const\n\nReturns the l_infinity norm of the vector.\n\nDefinition at line 537 of file vector.cpp.\n\n double mfem::Vector::Normlp ( double p ) const\n\nReturns the l_p norm of the vector.\n\nDefinition at line 558 of file vector.cpp.\n\n mfem::Vector::operator const double * ( ) const\ninline\n\nDefinition at line 84 of file vector.hpp.\n\n mfem::Vector::operator double * ( )\ninline\n\nDefinition at line 82 of file vector.hpp.\n\n double & mfem::Vector::operator() ( int i )\ninline\n\nSets value in vector. Index i = 0 .. size-1.\n\nDefinition at line 271 of file vector.hpp.\n\n const double & mfem::Vector::operator() ( int i ) const\ninline\n\nSets value in vector. Index i = 0 .. size-1.\n\nDefinition at line 281 of file vector.hpp.\n\n double mfem::Vector::operator* ( const double * v ) const\n\nDefinition at line 77 of file vector.cpp.\n\n double mfem::Vector::operator* ( const Vector & v ) const\n\nReturn the inner-product.\n\nDefinition at line 90 of file vector.cpp.\n\n Vector & mfem::Vector::operator*= ( double c )\n\nDefinition at line 124 of file vector.cpp.\n\n Vector & mfem::Vector::operator+= ( const Vector & v )\n\nDefinition at line 157 of file vector.cpp.\n\n Vector & mfem::Vector::operator-= ( double c )\n\nDefinition at line 139 of file vector.cpp.\n\n Vector & mfem::Vector::operator-= ( const Vector & v )\n\nDefinition at line 146 of file vector.cpp.\n\n Vector & mfem::Vector::operator/= ( double c )\n\nDefinition at line 131 of file vector.cpp.\n\n Vector & mfem::Vector::operator= ( const double * v )\n\nDefinition at line 100 of file vector.cpp.\n\n Vector & mfem::Vector::operator= ( const Vector & v )\n\nRedefine '=' for vector = vector.\n\nDefinition at line 107 of file vector.cpp.\n\n Vector & mfem::Vector::operator= ( double value )\n\nRedefine '=' for vector = constant.\n\nDefinition at line 115 of file vector.cpp.\n\n bool mfem::Vector::OwnsData ( ) const\ninline\n\nDefinition at line 86 of file vector.hpp.\n\n void mfem::Vector::Print ( std::ostream & out = `std::cout`, int width = `8` ) const\n\nPrints vector to stream out.\n\nDefinition at line 483 of file vector.cpp.\n\n void mfem::Vector::Print_HYPRE ( std::ostream & out ) const\n\nPrints vector to stream out in HYPRE_Vector format.\n\nDefinition at line 501 of file vector.cpp.\n\n void mfem::Vector::Randomize ( int seed = `0` )\n\nSet random values in the vector.\n\nDefinition at line 517 of file vector.cpp.\n\n Vector & mfem::Vector::Set ( const double a, const Vector & x )\n\n(*this) = a * x\n\nDefinition at line 182 of file vector.cpp.\n\n void mfem::Vector::SetData ( double * d )\ninline\n\nDefinition at line 62 of file vector.hpp.\n\n void mfem::Vector::SetDataAndSize ( double * d, int s )\ninline\n\nDefinition at line 64 of file vector.hpp.\n\n void mfem::Vector::SetSize ( int s )\ninline\n\nResizes the vector if the new size is different.\n\nDefinition at line 248 of file vector.hpp.\n\n void mfem::Vector::SetSubVector ( const Array< int > & dofs, const Vector & elemvect )\n\nDefinition at line 427 of file vector.cpp.\n\n void mfem::Vector::SetSubVector ( const Array< int > & dofs, double * elem_data )\n\nDefinition at line 438 of file vector.cpp.\n\n void mfem::Vector::SetVector ( const Vector & v, int offset )\n\nDefinition at line 193 of file vector.cpp.\n\n int mfem::Vector::Size ( ) const\ninline\n\nReturns the size of the vector.\n\nDefinition at line 76 of file vector.hpp.\n\n void mfem::Vector::StealData ( double ** p )\ninline\n\nChanges the ownership of the data; after the call the Vector is empty.\n\nDefinition at line 89 of file vector.hpp.\n\n double* mfem::Vector::StealData ( )\ninline\n\nChanges the ownership of the data; after the call the Vector is empty.\n\nDefinition at line 93 of file vector.hpp.\n\n double mfem::Vector::Sum ( ) const\n\nReturn the sum of the vector entries.\n\nDefinition at line 598 of file vector.cpp.\n\n## Friends And Related Function Documentation\n\n void add ( const Vector & v1, const Vector & v2, Vector & v )\nfriend\n\nDo v = v1 + v2.\n\nDefinition at line 227 of file vector.cpp.\n\n void add ( const Vector & v1, double alpha, const Vector & v2, Vector & v )\nfriend\n\nDo v = v1 + alpha * v2.\n\nDefinition at line 241 of file vector.cpp.\n\n void add ( const double a, const Vector & x, const Vector & y, Vector & z )\nfriend\n\nz = a * (x + y)\n\nDefinition at line 268 of file vector.cpp.\n\n void add ( const double a, const Vector & x, const double b, const Vector & y, Vector & z )\nfriend\n\nz = a * x + b * y\n\nDefinition at line 298 of file vector.cpp.\n\n void subtract ( const Vector & v1, const Vector & v2, Vector & v )\nfriend\n\nDo v = v1 - v2.\n\nDefinition at line 341 of file vector.cpp.\n\n void subtract ( const double a, const Vector & x, const Vector & y, Vector & z )\nfriend\n\nz = a * (x - y)\n\nDefinition at line 359 of file vector.cpp.\n\n void swap ( Vector * v1, Vector * v2 )\nfriend\n\nSwap v1 and v2.\n\nDefinition at line 213 of file vector.cpp.\n\n## Member Data Documentation\n\n int mfem::Vector::allocsize\nprotected\n\nDefinition at line 33 of file vector.hpp.\n\n double* mfem::Vector::data\nprotected\n\nDefinition at line 34 of file vector.hpp.\n\n int mfem::Vector::size\nprotected\n\nDefinition at line 33 of file vector.hpp.\n\nThe documentation for this class was generated from the following files:" ]

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[ "April 11, 2019\n\n# Oracle EBS : Math Indexes\n\n4/11/2019 01:13:00 PM\n``````1)SELECT numeric_number\nFROM table_math\nWHERE numeric_number + 0 = ?``````\nNevertheless we can index these expressions with a function-based index if we use calculations in a smart way and transform the `where` clause like an equation:\n``````2)SELECT a, b\nFROM table_math\nWHERE 3*a - b = -5``````\nIn both cases though index is there on a and b,it won't use that index,If you are frequently using these of mathematical expression better to create an index as follows i.e function based index\nWe just moved the table references to the one side and the constants to the other. We can then create a function-based index for the left hand side of the equation:\n``CREATE INDEX math ON table_math (3*a - b)``", null, "" ]

[ null, "https://lh5.googleusercontent.com/proxy/JuiB0H5n0z7QjMs0pp9LVwCXjCv3RA0KU7Fvg0mLvkthOlZ6Uci-BBGRt9LDmfjveUzAbUzzYA=s0-d", null ]

[ "", null, "<    1      2      3      4      5      6      7    >\n\n### Negative number\n\nYou write a negative number with a minus sign (−) to show that the value is less than zero.\n\n##### Explanation\n\nWe don't know exactly what negative numbers really are. Only for positive numbers we can imagine what they mean. Mathematicians don't care about it. But note: strange things can happen. We agree to write negative numbers with a minus sign, so\n\n−2\n\nmeans that there are 2 things less than zero, whatever that may be. We never write +3 if we have three real things, but it is possible. We start with the addition of", null, "Here we use brackets, because otherwise you can get confused. That went well, so we now continue with subtraction of", null, "Do we really still know what we are doing? We have 6 things and subtract 4 things that are not there, from it and obtain 10. So effectively we have added 4. Let's look at this carefully and write everything in detail", null, "Now once more with negative numbers", null, "That is just fine. Now we subtract again", null, "It is getting clearer. The brackets don't really help much, so we omit them", null, "If you have a – followed by a + it gets negative. And the reverse, first a + and then a – works just the same. You have a – and another – then it becomes positive. That is of course quite strange if you think about it. In a multiplication it is the same, so", null, "and we understand that now. We continue with a multiplication with negative numbers, where we use brackets for clearness", null, "The resultat is positive, because the same rules apply. In a division it is also the same, so", null, "As we understand this, a difficult calculation can be solved", null, "The same rules apply to exponentiation, so", null, "You should look carefully at this, as brackets play an important role here. There are some restrictions when using negative numbers. You will notice this when taking the root of a negative number", null, "as it has no real value. To be precise", null, "You must always apply the rules" ]

[ null, "http://maeckes.nl/clips/logo%20maeckes.gif", null, "http://maeckes.nl/clips/5 + (−2) = 3 .gif", null, "http://maeckes.nl/clips/6 − (−4) = 10 .gif", null, "http://maeckes.nl/clips/(+7) + (+2) = (+9) .gif", null, "http://maeckes.nl/clips/(−5) + (−2) = (−7) .gif", null, "http://maeckes.nl/clips/(−5) − (−2) = (−3) .gif", null, "http://maeckes.nl/clips/−5 + 2 = −3 .gif", null, "http://maeckes.nl/clips/−4 × 7 = −28 .gif", null, "http://maeckes.nl/clips/−4 × (−7) = 28 .gif", null, "http://maeckes.nl/clips/−8 ∕ −2 = 4 .gif", null, "http://maeckes.nl/clips/−{− (−6) ∕ −2} = 3 .gif", null, "http://maeckes.nl/clips/6² − 3² − (−2)³ = 36 − 9 − (−8) = 36 − 9 + 8 = 35 .gif", null, "http://maeckes.nl/clips/√−4 = ﹖.gif", null, "http://maeckes.nl/clips/√(−2)² = √{(−2)(−2)} = √4 = 2 .gif", null ]

[ "\\$30.00\n\nCategory:\n\nDescription\n\n1. Introduction\n\nIn this assignment, we use graph traversal to solve an interesting chess problem. Specifically, we will use implement dfs and use it to solve the problem.\n\n2. Objectives\n\nThe purpose of this assignment is to give you experience in:\n\n• Understanding the dfs graph traversal\n\n• How to use dfs to solve some real world problems.\n\n• Understanding when to use dfs instead of bfs.\n\nNote: Before you start, if you are not familiar with graph and graph traversal algorithms, it is recommended you review the corresponding class notes.\n\n3. Background\n\n3.1. Knight’s Move\n\nThis problem is from book “The Moscow puzzles”. The author of this book is Boris A. Kordemsky, and this book is edited by Martin Gardner. This problem appears in the chapter II Difficult Problems.\n\nThe following page displays the cover of the book, and the illustration and description of this problem from the book.\n\nWe will solve this problem using graph traversal. We consider the knight’s position on the chessboard as a vertex in a graph. At the beginning, the knight can capture one of the 16 pawns. Therefore, we can start from any of the 16 starting points. At the end, all the pawns will be gone. Hence, the end point is at one of the 16 vertices.\n\nThe key to solving this problem is to construct a graph that has 16 vertices, and define edges according to the rule of how knights move in chess.\n\nOnce we have the graph, we can traverse the graph and the goal is to traverse every single vertex without repeating.\n\nThere are multiple solutions to this problem. We will try to find a solution using DFS. Since it is not guaranteed that we can find a solution with one DFS traversal, we will do DFS multiple times. To make sure we traversal different paths when we do multiple DFS traversals, we change the original DFS algorithm a little bit to make it behave randomly. That is, when we push all the neighbors of a vertex on to the stack that is used for DFS, we randomly shuffle the order of the neighbors before we push them onto the stack. This ensures that for each DFS, the path traversed is random.\n\n4. Assignment\n\nIn the skeleton zip file, you will find a skeleton for your .py file named chess.py. All you need to do is to complete the skeleton code based on the instructions and submit it to the Mimir system.\n\nThere are two places you need to fill in code.\n\n4.1. Method randomdfs\n\nWe first copy code from dfs method and make necessary changes. The goal is to make sure that when we add neighbors to a stack, we add them in a random order. This way, every time we do this random dfs, the result is random.\n\n4.2. Inverse DFS tree and display result\n\nThe randomdfs(v) method will return a depth-first search tree. This tree is implemented as a dictionary. For each key (vertex), the value is its parent vertex. Now we need to check whether this tree is actually linear. If this tree is linear (a straight line) then we have a solution. If this is true, we need to make a new dictionary that is the inverse of this dictionary tree. That is, the key and value is switched for each item. Using this inversed dictionary, we print out all the moves the knight takes, and return this inversed dictionary as the return value of the knightsmove function.\n\nBelow is a screenshot of my results.\n\n5. Submit your work to Mimir\n\n7. Getting help\n\nStart your project early, because you will probably not be able to get timely help in the last few hours before the assignment is due.\n\n• Go to the office hours of instructors and TAs.\n\n1. Prof. Wei Wei: Mon. 2:30 – 3:15pm, Wed. 2:30 – 3:15pm, Thurs. 2:30 – 3:15pm, Fri. 2:30 – 3:15pm @ITE258\n\n1. Jenny Blessing: Fri. 12pm – 2pm @ITE140 o Param Bidja: Tues. 2pm – 3pm @ITE140\n\no Yamuna Rajan: Tues. 11am – 12pm, Wed. 9:30am – 10:30am @ITE140\n\no Zigeng Wang: Mon. 3pm – 4pm @ITE140" ]

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[ "# Differences\n\nThis shows you the differences between two versions of the page.\n\n cl:functions:logtest [2019/07/14 17:00] cl:functions:logtest [2019/09/18 01:00] (current) Line 1: Line 1: + ====== Function LOGTEST ====== + + ====Syntax==== + * **logtest** //integer-1 integer-2// → //​generalized-boolean//​ + + ====Arguments and Values==== + * //​integer-1//​ - an //​[[CL:​Glossary:​integer]]//​. + * //​integer-2//​ - an //​[[CL:​Glossary:​integer]]//​. + * //​generalized-boolean//​ - a //​[[CL:​Glossary:​generalized boolean]]//​. + + ====Description==== + Returns //​[[CL:​Glossary:​true]]//​ if any of the bits designated by the 1's in //​integer-1//​ is 1 in //​integer-2//;​ otherwise it is //​[[CL:​Glossary:​false]]//​. //​integer-1//​ and //​integer-2//​ are treated as if they were binary. + + Negative //​integer-1//​ and //​integer-2//​ are treated as if they were represented in two'​s-complement binary. + + ====Examples==== + <​blockquote>​ + (logtest 1 7) <​r>//​[[CL:​Glossary:​true]]//​ + (logtest 1 2) <​r>//​[[CL:​Glossary:​false]]//​ + (logtest -2 -1) <​r>//​[[CL:​Glossary:​true]]//​ + (logtest 0 -1) <​r>//​[[CL:​Glossary:​false]]//​ + ​ + + ====Side Effects==== + None. + + ====Affected By==== + None. + + ====Exceptional Situations==== + Should signal an error of type type-error if //​integer-1//​ is not an //​[[CL:​Glossary:​integer]]//​. Should signal an error of type type-error if //​integer-2//​ is not an //​[[CL:​Glossary:​integer]]//​. + + ====See Also==== + None. + + ====Notes==== + <​blockquote> ​ + (logtest //x// //y//) ≡ ([[CL:​Functions:​not]] ([[CL:​Functions:​zerop]] ([[CL:​Functions:​logand]] //x// //​y//​))) ​ + ​ +\n\n### Page Tools", null, "" ]

[ null, "https://phoe.tymoon.eu/clus/lib/exe/indexer.php", null ]

[ "• ### Covering Groups of Nonconnected Topological Groups and 2-Groups(1709.09728)\n\nMarch 26, 2019 math.CT, math.QA, math.GR\nWe investigate the universal cover of a topological group that is not necessarily connected. Its existence as a topological group is governed by a Taylor cocycle, an obstruction in 3-cohomology. Alternatively, it always exists as a topological 2-group. The splitness of this 2-group is also governed by an obstruction in 3-cohomology, a Sinh cocycle. We give explicit formulas for both obstructions and show that they are inverse of each other.\n• ### Ore's theorem on subfactor planar algebras(1704.00745)\n\nMarch 26, 2019 math.CO, math.OA, math.QA, math.RT, math.GR\nThis article proves that an irreducible subfactor planar algebra with a distributive biprojection lattice admits a minimal 2-box projection generating the identity biprojection. It is a generalization (conjectured in 2013) of a theorem of Oystein Ore on distributive intervals of finite groups (1938), and a corollary of a natural subfactor extension of a conjecture of Kenneth S. Brown in algebraic combinatorics (2000). We deduce a link between combinatorics and representations in finite group theory.\n• ### Unsolved Problems in Group Theory. The Kourovka Notebook(1401.0300)\n\nDec. 4, 2020 math.GR\nThis is a collection of open problems in group theory proposed by hundreds of mathematicians from all over the world. It has been published every 2-4 years in Novosibirsk since 1965. This is the 19th edition, which contains 111 new problems and a number of comments on about 1000 problems from the previous editions.\n• ### Stability of group relations under small Hilbert-Schmidt perturbations(1706.08405)\n\nMarch 25, 2019 math.OA, math.FA, math.GR\nIf matrices almost satisfying a group relation are close to matrices exactly satisfying the relation, then we say that a group is matricially stable. Here \"almost\" and \"close\" are in terms of the Hilbert-Schmidt norm. Using tracial 2-norm on $II_1$-factors we similarly define $II_1$-factor stability for groups. Our main result is that all 1-relator groups with non-trivial center are $II_{1}$-factor stable. Many of them are also matricially stable and RFD. For amenable groups we give a complete characterization of matricial stability in terms of the following approximation property for characters: each character must be a pointwise limit of traces of finite-dimensional representations. This allows us to prove matricial stability for the discrete Heisenberg group $\\mathbb H_3$ and for all virtually abelian groups. For non-amenable groups the same approximation property is a necessary condition for being matricially stable. We study this approximation property and show that RF groups with character rigidity have it.\n• ### Nontrivial elements in a knot group which are trivialized by Dehn fillings(1705.04470)\n\nMarch 23, 2019 math.GT, math.GR\nLet K be a nontrivial knot in the 3-sphere with the exterior E(K), and u in G(K), the fundamental group of E(K), a slope element represented by an essential simple closed curve on the boundary of E(K). Since the normal closure of u in G(K) coincides with that of the inverse of u, and u and its inverse u correspond to a slope r, a rational number or 1/0, we write << r >> = << u >>. The normal closure << u >> describes elements which are trivialized by r-Dehn filling of E(K). In this article, we prove that << r_1 >> =<< r_2 >> if and only if r_1 = r_2, and for a given finite family of slopes S = {r_1, ..., r_n}, the intersection of << r_1 >> , << r_2>>, ..., and << r_n >> contains infinitely many elements except when K is a (p, q)-torus knot and pq belongs to S. We also investigate inclusion relation among normal closures of slope elements.\n• ### Simple closed curves, finite covers of surfaces, and power subgroups of Out(F_n)(1708.06486)\n\nMarch 20, 2019 math.GT, math.GR\nWe construct examples of finite covers of punctured surfaces where the first rational homology is not spanned by lifts of simple closed curves. More generally, for any set $\\mathcal{O} \\subset F_n$ which is contained in the union of finitely many $Aut(F_n)$-orbits, we construct finite-index normal subgroups of $F_n$ whose first rational homology is not spanned by powers of elements of $\\mathcal{O}$. These examples answer questions of Farb-Hensel, Kent, Looijenga, and Marche. We also show that the quotient of $Out(F_n)$ by the subgroup generated by kth powers of transvections often contains infinite order elements, strengthening a result of Bridson-Vogtmann saying that it is often infinite. Finally, for any set $\\mathcal{O} \\subset F_n$ which is contained in the union of finitely many $Aut(F_n)$-orbits, we construct integral linear representations of free groups that have infinite image and map all elements of $\\mathcal{O}$ to torsion elements.\n• ### Acylindrical Hyperbolicity of Out($W_n$)(1610.08005)\n\nOct. 22, 2019 math.GR\nWe prove that the group of outer automorphisms of the free Coxeter group $W_n$ is acylindrically hyperbolic in the sense of Osin. As an application, we observe that any CAT(0) space admitting a geometric action by Out($W_n$) must contain a rank-one geodesic. The theorem proceeds from expanding on a well-known relationship between Out($W_n$) and the outer automorphism group of free groups.\n• ### Finite group actions on homology spheres and manifolds with nonzero Euler characteristic(1403.0383)\n\nMarch 19, 2019 math.DG, math.GR\nLet $X$ be a smooth manifold belonging to one of these three collections: acyclic manifolds (compact or not, possibly with boundary), compact connected manifolds (possibly with boundary) with nonzero Euler characteristic, integral homology spheres. We prove that $Diff(X)$ is Jordan. This means that there exists a constant $C$ such that any finite subgroup $G$ of $Diff(X)$ has an abelian subgroup whose index in $G$ is at most $C$. Using a result of Randall and Petrie we deduce that the automorphism groups of connected, non necessarily compact, smooth real affine varieties with nonzero Euler characteristic are Jordan.\n• ### The representation theory of finite sets and correspondences(1510.03034)\n\nMarch 17, 2019 math.CT, math.CO, math.RT, math.GR\nWe investigate correspondence functors, namely the functors from the category of finite sets and correspondences to the category of $k$-modules, where $k$ is a commutative ring.They have various specific properties which do not hold for other types of functors.In particular, if $k$ is a field and if $F$ is a correspondence functor, then $F$ is finitely generated if and only if the dimension of $F(X)$ grows exponentially in terms of the cardinality of the finite set $X$. In such a case, $F$ has finite length. Also, if $k$ is noetherian, then any subfunctor of a finitely generated functor is finitely generated. When $k$ is a field, we give a description of all the simple functors and we determine the dimension of their evaluations at any finite set.A main tool is the construction of a functor associated to any finite lattice $T$. We prove for instance that this functor is projective if and only if the lattice $T$ is distributive. Moreover, it has quotients which play a crucial role in the analysis of simple functors. The special case of total orders yields some more specific results. Several other properties are also discussed, such as projectivity, duality, and symmetry.In an appendix, all the lattices associated to a given poset are described.\n• ### Metric systolicity and two-dimensional Artin groups(1710.05157)\n\nMarch 15, 2019 math.GR\nWe introduce the notion of metrically systolic simplicial complexes. We study geometric and large-scale properties of such complexes and of groups acting on them geometrically. We show that all two-dimensional Artin groups act geometrically on metrically systolic complexes. As direct corollaries we obtain new results on two-dimensional Artin groups and all their finitely presented subgroups: we prove that the Conjugacy Problem is solvable, and that the Dehn function is quadratic. We also show several large-scale features of finitely presented subgroups of two-dimensional Artin groups, lying background for further studies concerning their quasi-isometric rigidity.\n• ### Complete reducibility of subgroups of reductive algebraic groups over nonperfect fields 3(1612.05863)\n\nMarch 14, 2019 math.GR\nLet $k$ be a nonperfect separably closed field. Let $G$ be a (possibly non-connected) reductive group defined over $k$. We study rationality problems for Serre's notion of complete reducibility of subgroups of $G$. In our previous work, we constructed examples of subgroups $H$ of $G$ that are $G$-completely reducible but not $G$-completely reducible over $k$ (and vice versa). In this paper, we give a theoretical underpinning of those constructions. To illustrate our result, we present a new such example in a non-connected reductive group of type $D_4$ in characteristic $2$. Then using Geometric Invariant Theory, we generalize the theoretical result above obtaining a new result on the structure of $G(k)$-(and $G$-) orbits in an arbitrary affine $G$-variety. We translate our result into the language of spherical buildings to give a topological viewpoint. A problem on centralizers of completely reducible subgroups and a problem concerning the number of conjugacy classes are also considered.\n• ### On the relative hyperbolicity and manifold structure of certain right-angled Coxeter groups(1708.07818)\n\nSept. 6, 2019 math.GR\nIn this article, we study the manifold structure and the relatively hyperbolic structure of right-angled Coxeter groups with planar nerves. We then apply our results to the quasi-isometry problem for this class of right-angled Coxeter groups.\n• ### Group approximation in Cayley topology and coarse geometry, Part I: Coarse embeddings of amenable groups(1310.4736)\n\nMarch 12, 2019 math.OA, math.MG, math.GR\nThe objective of this series is to study metric geometric properties of (coarse) disjoint unions of amenable Cayley graphs. We employ the Cayley topology and observe connections between large scale structure of metric spaces and group properties of Cayley accumulation points. In this Part I, we prove that a disjoint union has property A of G. Yu if and only if all groups appearing as Cayley accumulation points in the space of marked groups are amenable. As an application, we construct two disjoint unions of finite special linear groups (and unimodular linear groups) with respect to two systems of generators that look similar such that one has property A and the other does not admit (fibred) coarse embeddings into any Banach space with non-trivial type (for instance, any uniformly convex Banach space).\n• ### Special elements in lattices of semigroup varieties(1309.0228)\n\nApril 11, 2021 math.GR\nWe survey results concerning special elements of nine types (modular, lower-modular, upper-modular, cancellable, distributive, codistributive, standard, costandard and neutral elements) in the lattice of all semigroup varieties and certain its sublattices, mainly in the lattices of all commutative varieties and of all overcommutative ones. Several open questions are formulated. The work is regularly updated and modified as new results and/or articles appear.\n• ### Root data with group actions(1707.01935)\n\nMarch 11, 2019 math.RT, math.GR\nSuppose $k$ is a field, $G$ is a connected reductive algebraic $k$-group, $T$ is a maximal $k$-torus in $G$, and $\\Gamma$ is a finite group that acts on $(G,T)$. From the above, one obtains a root datum $\\Psi$ on which $\\text{Gal}(k)\\times\\Gamma$ acts. Provided that $\\Gamma$ preserves a positive system in $\\Psi$, not necessarily invariant under $\\text{Gal}(k)$, we construct an inverse to this process. That is, given a root datum on which $\\text{Gal}(k)\\times\\Gamma$ acts appropriately, we show how to construct a pair $(G,T)$, on which $\\Gamma$ acts as above. Although the pair $(G,T)$ and the action of $\\Gamma$ are canonical only up to an equivalence relation, we construct a particular pair for which $G$ is $k$-quasisplit and $\\Gamma$ fixes a $\\text{Gal}(k)$-stable pinning of $G$. Using these choices, we can define a notion of taking \"$\\Gamma$-fixed points\" at the level of equivalence classes, and this process is compatible with a general \"restriction\" process for root data with $\\Gamma$-action.\n• ### The integral polytope group(1605.01217)\n\nMarch 11, 2019 math.GT, math.MG, math.GR\nWe show that the Grothendieck group associated to integral polytopes in $\\mathbb{R}^n$ is free-abelian by providing an explicit basis. Moreover, we identify the involution on this polytope group given by reflection about the origin as a sum of Euler characteristic type. We also compute the kernel of the norm map sending a polytope to its induced seminorm on the dual of $\\mathbb{R}^n$.\n• ### Virtually cocompactly cubulated Artin-Tits groups(1509.08711)\n\nApril 13, 2020 math.GT, math.MG, math.GR\nWe give a conjectural classification of virtually cocompactly cubulated Artin-Tits groups (i.e. having a finite index subgroup acting geometrically on a CAT(0) cube complex), which we prove for all Artin-Tits groups of spherical type, FC type or two-dimensional type. A particular case is that for $n \\geq 4$, the $n$-strand braid group is not virtually cocompactly cubulated.\n• ### Dual Ore's theorem on distributive intervals of finite groups(1708.02565)\n\nMarch 9, 2019 math.CO, math.RT, math.GR\nThis paper gives a self-contained group-theoretic proof of a dual version of a theorem of Ore on distributive intervals of finite groups. We deduce a bridge between combinatorics and representations in finite group theory.\n• ### Entropy and quasimorphisms(1707.06020)\n\nMarch 5, 2019 math.GT, math.SG, math.DS, math.GR\nLet $S$ be a compact oriented surface. We construct homogeneous quasimorphisms on $Diff(S, area)$, on $Diff_0(S, area)$ and on $Ham(S)$ generalizing the constructions of Gambaudo-Ghys and Polterovich. We prove that there are infinitely many linearly independent homogeneous quasimorphisms on $Diff(S, area)$, on $Diff_0(S, area)$ and on $Ham(S)$ whose absolute values bound from below the topological entropy. In case when $S$ has a positive genus, the quasimorphisms we construct on $Ham(S)$ are $C^0$-continuous. We define a bi-invariant metric on these groups, called the entropy metric, and show that it is unbounded. In particular, we reprove the fact that the autonomous metric on $Ham(S)$ is unbounded.\n• ### Random matrix products when the top Lyapunov exponent is simple(1705.09593)\n\nJune 16, 2020 math.PR, math.DS, math.GR\nIn the present paper, we treat random matrix products on the general linear group $\\textrm{GL}(V)$, where $V$ is a vector space defined on any local field, when the top Lyapunov exponent is simple, without irreducibility assumption. In particular, we show the existence and uniqueness of the stationary measure $\\nu$ on $\\textrm{P}(V)$ that is relative to the top Lyapunov exponent and we describe the projective subspace generated by its support. We observe that the dynamics takes place in a open set of $\\textrm{P}(V)$ which has the structure of a skew product space. Then, we relate this support to the limit set of the semi-group $T_{\\mu}$ of $\\textrm{GL}(V)$ generated by the random walk. Moreover, we show that $\\nu$ has H\\\"older regularity and give some limit theorems concerning the behavior of the random walk and the probability of hitting a hyperplane. These results generalize known ones when $T_{\\mu}$ acts strongly irreducibly and proximally (i-p to abbreviate) on $V$. In particular, when applied to the affine group in the so-called contracting case or more generally when the Zariski closure of $T_{\\mu}$ is not necessarily reductive, the H\\\"older regularity of the stationary measure together with the description of the limit set are new. We mention that we don't use results from the i-p setting; rather we see it as a particular case.\n• ### On the analogy between real reductive groups and Cartan motion groups. III: A proof of the Connes-Kasparov isomorphism(1602.08891)\n\nMarch 5, 2019 math.OA, math.RT, math.GR\nAlain Connes and Nigel Higson pointed out in the 1990s that the Connes-Kasparov \"conjecture\"' for the K-theory of reduced groupe $C^\\ast$-algebras seemed, in the case of reductive Lie groups, to be a cohomological echo of a conjecture of George Mackey concerning the rigidity of representation theory along the deformation from a reductive Lie group to its Cartan motion group. For complex semisimple groups, Nigel Higson established in 2008 that Mackey's analogy is a real phenomenon and does lead to a simple proof of the Connes-Kasparov isomorphism. We here turn to more general reductive groups and use our recent work on Mackey's proposal, together with Higson's work, to obtain a new proof of the Connes-Kasparov isomorphism.\n• ### On the $C^*$-algebra generated by the Koopman representation of a topological full group(1705.07665)\n\nMarch 3, 2019 math.OA, math.GR\nLet $(X,T,\\mu)$ be a Cantor minimal sytem and $[[T]]$ the associated topological full group. We analyze $C^*_\\pi([[T]])$, where $\\pi$ is the Koopman representation attached to the action of $[[T]]$ on $(X,\\mu)$. Specifically, we show that $C^*_\\pi([[T]])=C^*_\\pi([[T]]')$ and that the kernel of the character $\\tau$ on $C^*_\\pi([[T]])$ coming from weak containment of the trivial representation is a hereditary $C^*$-subalgebra of $C(X)\\rtimes\\mathbb{Z}$. Consequently, $\\ker\\tau$ is stably isomorphic to $C(X)\\rtimes\\mathbb{Z}$, and $C^*_\\pi([[T]]')$ is not AF. We also prove that if $G$ is a finitely generated, elementary amenable group and $C^ *(G)$ has real rank zero, then $G$ is finite.\n• ### Groups with a nontrivial nonideal kernel(1608.00231)\n\nMarch 1, 2019 math.RT, math.GR\nWe classify finite groups $G$, such that the group algebra, $\\mathbb{Q}G$ (over the field of rational numbers $\\mathbb{Q}$), is the direct product of the group algebra $\\mathbb{Q}[G/N]$ of a proper factor group $G/N$, and some division rings.\n• ### Zeta functions associated to admissible representations of compact p-adic Lie groups(1707.08485)\n\nFeb. 27, 2019 math.RT, math.GR\nLet $G$ be a profinite group. A strongly admissible smooth representation $\\rho$ of $G$ over $\\mathbb{C}$ decomposes as a direct sum $\\rho \\cong \\bigoplus_{\\pi \\in \\mathrm{Irr}(G)} m_\\pi(\\rho) \\, \\pi$ of irreducible representations with finite multiplicities $m_\\pi(\\rho)$ such that for every positive integer $n$ the number $r_n(\\rho)$ of irreducible constituents of dimension $n$ is finite. Examples arise naturally in the representation theory of reductive groups over non-archimedean local fields. In this article we initiate an investigation of the Dirichlet generating function $\\zeta_\\rho (s) = \\sum_{n=1}^\\infty r_n(\\rho) n^{-s} = \\sum_{\\pi \\in \\mathrm{Irr}(G)} \\frac{m_\\pi(\\rho)}{(\\dim \\pi)^s}$ associated to such a representation $\\rho$. Our primary focus is on representations $\\rho = \\mathrm{Ind}_H^G(\\sigma)$ of compact $p$-adic Lie groups $G$ that arise from finite dimensional representations $\\sigma$ of closed subgroups $H$ via the induction functor. In addition to a series of foundational results - including a description in terms of $p$-adic integrals - we establish rationality results and functional equations for zeta functions of globally defined families of induced representations of potent pro-$p$ groups. A key ingredient of our proof is Hironaka's resolution of singularities, which yields formulae of Denef-type for the relevant zeta functions. In some detail, we consider representations of open compact subgroups of reductive $p$-adic groups that are induced from parabolic subgroups. Explicit computations are carried out by means of complementing techniques: (i) geometric methods that are applicable via distance-transitive actions on spherically homogeneous rooted trees and (ii) the $p$-adic Kirillov orbit method. Approach (i) is closely related to the notion of Gelfand pairs and works equally well in positive defining characteristic.\n• ### Generic norms and metrics on countable abelian groups(1605.06323)\n\nFeb. 27, 2019 math.GN, math.LO, math.GR\nFor a countable abelian group $G$ we investigate generic properties of the space of all invariant metrics on $G$. We prove that for every such an unbounded group $G$, i.e. group which has elements of arbitrarily high order, there is a dense set of invariant metrics on $G$ which make $G$ isometric to the rational Urysohn space, and a comeager set of invariant metrics such that the completion is isometric to the Urysohn space. This generalizes results of Cameron and Vershik, Niemiec, and the author. Then we prove that for every $G$ such that $G\\cong \\bigoplus_\\mathbb{N} G$ there is a comeager set of invariant metrics on $G$ such that all of them give rise to the same metric group after completion. If moreover $G$ is unbounded, then using a result of Melleray and Tsankov we get that the completion is extremely amenable." ]

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[ "# Inconsistant Values ug!\n\nI have some code which is producing some inconsistant values - Most of the values are right but some arn’t when comparing to my excel spreadsheet to assert the correct values. Note that the code is written in such a way that no synchronization occurs but since the kernel makes multiple passes, inconsistancies caused by irregular block/thread execution, are hidden since fewer and fewer values are in flux as consistant values propagate across the matrix. What I don’t understand is why only some of the values and not all of them, even the last element of the array is correct so its unlikely that Im doing something obvious.\n\nFor those interested, this is the Smith-Waterman Gene alignment algorithm\n\n[codebox]\n\nglobal void SmithWaterman(int* sequenceA,\n\n``````\t\t\t\t\t\t\t\t\t int* sequenceB,\n\nfloat* dMatrix,\n\nfloat* traceMatrix,\n\nfloat* sMatrix,\n\nfloat gapPenalty)\n``````\n\n{\n\n`````` ////offset for border values\n\n//global element index\n\nconst int elementIdx = (gridDim.x * blockDim.x + 1) * (blockDim.y * blockIdx.y + 1 + threadIdx.y) + (blockDim.x * blockIdx.x + threadIdx.x + 1);\n\n//declare shared values\n\nextern __shared__ float sharedMemory[];\n\nfloat* above = &sharedMemory;\n\nfloat* diagonal = &sharedMemory[blockDim.x * blockDim.y];\n\nfloat* left = &sharedMemory[blockDim.x * blockDim.y * 2];\n\n//copy values from global to shared, calculate indels/alignment\n\nabove[sharedElementIdx] = dMatrix[elementIdx - (gridDim.x * blockDim.x + 1)] - gapPenalty;\n\ndiagonal[sharedElementIdx] = dMatrix[elementIdx - (gridDim.x * blockDim.x + 2)] + sMatrix[sequenceB[threadIdx.y] * 4 + sequenceA[threadIdx.x]];\n\nleft[sharedElementIdx] = dMatrix[elementIdx - 1] - gapPenalty;\n\nif(above[sharedElementIdx] > diagonal[sharedElementIdx] && above[sharedElementIdx] > left[sharedElementIdx])\n\ndMatrix[elementIdx] = above[sharedElementIdx];\n\nelse if(diagonal[sharedElementIdx] > above[sharedElementIdx] && diagonal[sharedElementIdx] > left[sharedElementIdx])\n\ndMatrix[elementIdx] = diagonal[sharedElementIdx];\n\nelse if(left[sharedElementIdx] > above[sharedElementIdx] && left[sharedElementIdx] > diagonal[sharedElementIdx])\n\ndMatrix[elementIdx] = left[sharedElementIdx];\n\n}[/codebox]``````" ]

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[ "# Tag Info\n\nThe series converges to a distribution function. It can be evaluated in closed form. Upon identifying the terms varying with $n,$ write your function in a simpler form as $$F_{U_i}(y)=\\frac{2(R^{\\alpha}y/\\theta)^k}{\\Gamma(k)}\\sum_{n=0}^\\infty \\frac {(-R^\\alpha y / \\theta)^n}{n!(k+n)((k+n)\\alpha+2)} = \\frac{2x^k}{\\Gamma(k)}\\sum_{n=0}^\\infty \\frac{(-x)^n}{n!... 3 The density function of the Benini distribution is strictly quasi-concave, and is strictly decreasing if and only if \\beta \\leqslant \\alpha (1+\\alpha)/2. The mode of the distribution has the explicit form:$$\\text{Mode} = \\hat{x} = \\sigma \\cdot \\exp \\bigg( \\bigg\\{ \\frac{\\sqrt{1+8\\beta} - (1+2\\alpha)}{4 \\beta} \\bigg\\} \\bigg),$$where the brackets \\{ \\... 2 You can resort to the bootstrap version of Student's t-test. It works as follows: Compute the sample mean and standard deviation for each group and label the results X_1 and s_1 for group 1, and X_2 and s_2 for group 2. Set d_1=\\frac{s_1^2}{n_1} and d_2=\\frac{s_2^2}{n_2}, where n_1 and n_2 are the sample sizes. Generate a bootstrap sample ... 2 After having done some homework on the subject, I think I've got a better handle on the proof that I find in . I wanted to take an opportunity to set down my understanding for pedagogic purposes. Scope: I'm going to limit this answer to the case of a strictly monotonic cumulative distribution function. Its my understanding that, in his answer to this ... 1 The old rule of thumb was that the sample size is too small when the expected count of one of the cells of the contingency table is lower than five. Recall that the expected count of a cell is the ratio$$\\frac{row\\_count \\times column\\_count}{total}. In http://www.biostathandbook.com/small.html and http://www.biostathandbook.com/fishers.html, owing ...\nThe result can be found in the 2012 paper Some results on the truncated multivariate $t$ distribution by Ho et al. First, let $T(\\cdot\\,|\\,\\nu)$ be the cdf of the (untruncated) standard $t$ distribution with $\\nu$ degrees of freedom. In the following exposition, $a$ and $b$ are the lower and upper truncation limits, respectively. Before we can give the ..." ]

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[ "", null, "1953 Order Completed\n99 % Response Time\n63 Reviews\nSince 2015\nRelated Topics\nRelated Blogs\n\nWhat makes Matlab and Octave different?Matlab and Octave are the two most popular statistical programming languages today. They offer researchers and data scientists a host of tools for statistical analysis to help them produce the most desirable results from their data. But it can be quite overwhel...\n\n2020-07-15\n\nThe Australian MATLAB Homework Help Service You Can Depend OnGet top-quality Matlab assignment help in Australia from our experts in Perth, Brisbane, Adelaide, and Sydney. Quality assistance for students looking for Matlab assignment help in AustraliaFor years, students in Australia have strug...\n\n2020-07-15\n\nThe most trusted Matlab assignment writing service in Canada Get academic support from the finest Matlab homework writing service in Canada and secure the best grades. Avail of our Matlab assignment writing service in Canada for quality solutions The Matlab computing environment is a major part...\n\n2020-07-15\n\n# Applied Physics Expert\n\nMalacca, Malaysia\n\n## Frank P\n\nBachelor’s Degree, Applied Physics, Sunway University, Malaysia\n\nProfession\n\nI am an expert signal processing assignment helper with deep knowledge of Matlab and several other digital processing programs.\n\nSkills\n\nAfter obtaining my bachelor’s degree in applied physics, I was looking for an opportunity to utilize the skills I had acquired in college. I, therefore, decided to venture into academic writing because, with this career path, I knew I would be subjected to challenges that would help me cement my skills. Luckily, Matlab Assignment Experts was looking for signal processing experts and that’s how I landed a job as a signal processing assignment help expert. It’s almost six years since I started working with this company and so far, I have delivered over 1950 successful orders. My area of expertise is speech signal processing, music signal processing, compressed sensing, and cognitive radio, though I can handle any topic related to signal processing.\n\nGet Free Quote\n0 Files Selected\nCurve Plotting\nclc, clear all, close all % Load dat file load crazy_func % Plot the curve figure subplot(1,2,1); plot(eks, why), grid on xlabel('eks'); ylabel('why'); title('why vs. eks'); % Determine d(why)/d(eks) N = length(why); % length of data % The length of d(why)/d(eks) will be N - 1 for i = 1:N-1 dwhy(i) = why(i+1)-why(i); deks(i) = eks(i+1) - eks(i); end % Now, compute d(why)/d(eks) dwhydeks = dwhy./deks; % Plot the resulting curve subplot(1,2,2) plot(eks(1:N-1), dwhydeks), grid on xlabel('deks'); ylabel('dwhy'); title('dwhy vs. deks'); %% NOTE FOR THE REPORT: % eks < 20 -> why is a sine function, so its derivative must be a cosine % function % eks >=20 and eks < 40 -> why is an exponential function so its % derivative must be also an exponential functionç % eks >=40 and eks < 60 -> why is a collection of random points so its % derivative must be also a collection of random points % eks >= 60 and eks < 80 -> why is a parabola, so its derivative must be % a straight line with positive slope % eks >= 80 and eks < 100 -> why is a line, so its derivative must be a % constant value \nVector Operations Using Matlab\nclc, clear all, close all %% Question 2 % Input values vxi = [0 -0.2478 -0.4943 -0.7384 -0.9786]; vyi = [4 3.9915 3.9659 3.9234 3.8644]; ti = [0 0.04 0.08 0.12 0.17]; % Initial values x1 = 4; y1 = 0; % Calculate x2, x3, x4, y2, y3, y4 x2 = x1 + vxi(1)*(ti(2)-ti(1)); x3 = x1 + vxi(2)*(ti(3)-ti(2)); x4 = x1 + vxi(3)*(ti(4)-ti(3)); y2 = y1 + vyi(1)*(ti(2)-ti(1)); y3 = y1 + vyi(2)*(ti(3)-ti(2)); y4 = y1 + vyi(3)*(ti(4)-ti(3)); %% Question 3 % Read file into a table data = readtable('A1_input.txt'); % Do not take into account the first element of the data because that one % contains the units for k = 2:size(data,1) t(k-1) = str2double(data.time{k}); vx(k-1) = str2double(data.vx{k}); vy(k-1) = str2double(data.vy{k}); end % Display the first four values [t(1:4)', vx(1:4)', vy(1:4)'] %% Question 4 % Plot vx and vy in the same graph figure plot(t, vx, t, vy); grid on legend('vx', 'vy'); xlabel('Time (s)'); ylabel('Velocity (m/s)'); grid on x(1) = x1; y(1) = y1; for k = 2:length(t) x(k) = x1; y(k) = y1; for j = 1:k-1 x(k) = x(k) + vx(j)*(t(j+1)-t(j)); y(k) = y(k) + vy(j)*(t(j+1)-t(j)); end end % x and y using vectors % vectors of tj+1 tj tvec = t(2:end) - t(1:end-1); xv = cumsum(vx(1:end-1).*tvec); xv = xv + x1; xv = [x1, xv]; yv = cumsum(vy(1:end-1).*tvec); yv = yv + y1; yv = [y1, yv]; figure hold on plot(t, x); plot(t, y); plot(t, xv, 'k--', 'linewidth', 3); plot(t, yv, 'g--', 'linewidth', 3); grid on xlabel('Time (s)'); ylabel('Position'); legend('X-position from Q5', 'Y-position from Q5', 'X-position from Q6', 'Y-position from Q6'); figure plot(xv, yv), grid on xlabel('X-Position (m)'); ylabel('Y-Position (m)'); file = fopen('output.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); for k = 1:length(t) fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t(k),xv(k), yv(k)); end fclose(file); file = fopen('output2.txt', 'wt'); fprintf(file, 'Time\\tx\\ty\\n'); fprintf(file, '(s)\\t(m)\\t(m)\\n'); fprintf(file, '%.3f\\t%.3f\\t%.3f\\n', t',xv', yv'); fclose(file);" ]

[ null, "https://www.matlabassignmentexperts.com/uploads/topics/applied-physics-expert.webp", null ]

[ "# 824\n\n## 824 is an even composite number composed of two prime numbers multiplied together.\n\nWhat does the number 824 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 8 divisors.\n\n824 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 824:\n\n### 23 × 103\n\n(2 × 2 × 2 × 103)\n\nSee below for interesting mathematical facts about the number 824 from the Numbermatics database.\n\n### Names of 824\n\n• Cardinal: 824 can be written as Eight hundred twenty-four.\n\n### Scientific notation\n\n• Scientific notation: 8.24 × 102\n\n### Factors of 824\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 105\n\n### Divisors of 824\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1560\n• Sum of proper divisors (its aliquot sum) s(n): 736\n• 824 is a deficient number, because the sum of its proper divisors (736) is less than itself. Its deficiency is 88\n\n### Bases of 824\n\n• Binary: 11001110002\n• Base-36: MW\n\n### Squares and roots of 824\n\n• 824 squared (8242) is 678976\n• 824 cubed (8243) is 559476224\n• The square root of 824 is 28.7054001889\n• The cube root of 824 is 9.3750962955\n\n### Scales and comparisons\n\nHow big is 824?\n• 824 seconds is equal to 13 minutes, 44 seconds.\n• To count from 1 to 824 would take you about six minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 824 cubic inches would be around 0.8 feet tall.\n\n### Recreational maths with 824\n\n• 824 backwards is 428\n• The number of decimal digits it has is: 3\n• The sum of 824's digits is 14\n• More coming soon!\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:\n\nMLA style:\n\"Number 824 - Facts about the integer\". Numbermatics.com. 2022. Web. 2 October 2022.\n\nAPA style:\nNumbermatics. (2022). Number 824 - Facts about the integer. Retrieved 2 October 2022, from https://numbermatics.com/n/824/\n\nChicago style:\nNumbermatics. 2022. \"Number 824 - Facts about the integer\". https://numbermatics.com/n/824/\n\nThe information we have on file for 824 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 824, math, Factors of 824, curriculum, school, college, exams, university, Prime factorization of 824, STEM, science, technology, engineering, physics, economics, calculator, eight hundred twenty-four.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]

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[ "### NP-Completeness of deciding the feasibility of Linear Equations over binary- variables with coefficients and constants that are 0, 1, or -1\n\nChermakani, Deepak Ponvel\n\nAbstract\nWe convert, within polynomial-time and sequential processing, NP-Complete Problemsinto a problem of deciding feasibility of a given system S of linear equations with constants and coefficients of binary-variables that are 0, 1, or -1. S is feasible, if and only if, the NP-Completeproblem has a feasible solution. We show separate polynomial-time conversions to S, from the SUBSET-SUM and 3-SAT problems, both of which are NP-Complete. The number of equations and variables in S is bounded by a polynomial function of the size of the NP-Completeproblem, showing that deciding the feasibility of S is strongly-NP-Complete. Wealsoshow how to apply the approach used for the SUBSET-SUM problemtodecidethe feasibility of Integer Linear Programs, as it involves reducing the coefficient-magnitudes of variables to the logarithm of their initial values, though the number of variables and equations are increased.\n\n### Investigations of Quantum Cellular Automata in a Two Dimensional Lattice\n\nYiyang Gong\n\nAbstract\nQuantum Cellular Automata (QCA) is a feasible method of manipulating a system of qubits carrying quantum information, as the same operations are applied to the entire system. With the proper initial conditions and boundaries for a two dimensional lattice, the state of one qubit can be transported to another qubit on the lattice using QCA. Such a phenomenon represents the possibility of a communication channel between different positions on the lattice. Simulations of the two-dimensional lattice using the stabilizer formalism with proper initial and boundary conditions are demonstrated. Furthermore, various errors on the gates of the system are simulated, and error probabilities on the final results are reported. Finally, encoding schemes using qubit arrays are also investigated.\n\n### BABEL: A Base for an Experimental Library\n\nJ. Seo, H. Ait-Kaci, R. Nasr\n\nAbstract\nThis report discusses the implementation of a knowledge base for a library information system. It is done using a typed logic programming language—LOGIN—where type inheritance is built in. The knowledge base is structured in a hierarchical taxonomy of library object classes where each class is represented in a FRAME style knowledge structure and inherits the properties of its parents, and where infrastructural inference rules have been established through typed Horn clauses. Also in this document, some programming techniques aimed at using the power of inheritance as taxonomic inference are discussed.\n\n### Localized endomorphisms of the chiral Ising model\n\nBöckenhauer, Jens\n\nAbstract\nBased on the treatment of the chiral Ising model by Mack and Schomerus, we present examples of localized endomorphisms. It is shown that they lead to the same superselection sectors as the global ones in the sense that unitary equivalence holds. Araki's formalism of the selfdual CAR algebra is used for the proof. We prove local normality and extend representations and localized endomorphisms to a global algebra of observables which is generated by local von Neumann algebras on the punctured circle. In this framework, we manifestly prove fusion rules and derive statistics operators.\n\n### On Stability of Switched Linear Hyperbolic Conservation Laws with Reflecting Boundaries\n\nAmin, Saurabh; Hante, Falk M.; and Bayen, Alexandre M.\n\nAbstract\nWe consider stability of an infinite dimensional switching system, posed as a system of linear hyperbolic partial differential equations (PDEs) with reflecting boundaries, where the system parameters and the boundary conditions switch in time. Asymptotic stability of the solution for arbitrary switching is proved under commutativity of the advective velocity matrices and a joint spectral radius condition involving the boundary data.\n\n### Dynamic NDM at transit-time resonance in n-lnP\n\nPozhela, Yu K; Starikov, E V; and Shiktorov, P N.\n\nAbstract\nThe frequency dependence of the longitudinal differential mobility of hot electrons is calculated using velocity averaging over the before- and after- scattering ensembles by the single-particle Monte Carlo simulation of the steady state. A dynamic negative differential mobility (NOM), due to the transit-time resonance of hot electrons in the momentum space under the predominant role of spontaneous emission of optical phonons at low lattice temperature, is observed in n-lnP. The frequency- and field-behaviour of the NOM and the noise characteristics, as well as the possibilities to use the effect for amplification and generation of the millimetre-wave radiation, are investigated. The techniques for the experimental investigation of the transit-time resonance are discussed. The noise temperature measurements are shown to be the most suitable tool for this. The transit-time resonance characteristics in n-lnP are compared with the experimentally realized and theoretically calculated parameters of the cyclotron resonance NEMAG in p-Ge. The conditions for the generation and amdification are found to be better in the former case than in the latter one.\n\n### Mutual Decentralized Synchronization for Intervehicle Communications\n\nSourour, Essam and Nakagawa, Masao\n\nAbstract\nData exchange among vehicles can improve road safety and capacity. Most of the proposed intervehicle data communication systems require intervehicle synchronization. Synchronization must be done in a decentralized manner. In this paper, we propose a new mutual decentralized synchronization system. Using a devoted carrier frequency, each vehicle transmits a continuous periodic train of pulses. The aim of the synchronization system is to make these periodic pulses synchronous to indicate the start of data slots in slotted ALOHA types of media access protocol. Each vehicle measures the power of pulses of other vehicles as well as the time difference between other pulses and its own pulse. Using this information, each vehicle shifts its own pulse transmission time toward a weighted average of other pulse transmission times. Eventually, all periodic pulse trains are synchronized. The system performance is evaluated in nonfading and fading channels." ]

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[ "# How to visualize slope fields of differential equations without vectors?\n\nI'm looking to visualize slope fields of differential equations for my differential equations course. Every example I see draws them as vectors, adding unnecessary \"arrows\" that, to me, are visually distracting. Is there a way to plot these slope fields without the \"arrows\" that get in the way?\n\n• LineIntegralConvolutionPlot[{y, x - x^3 - 0.3 y + 0.5 Cos[3.125]}, {x, -2.5, 2.5}, {y, -2.5, 2.5}, ColorFunction -> ColorData[\"DeepSeaColors\"], StreamPoints -> 10, StreamStyle -> {\"Line\", Thick, White}] – Sektor Mar 5 '14 at 21:28\n• possible duplicate of Visualizing a Complex Vector Field near Poles – m_goldberg Mar 5 '14 at 22:22\n\nYou can use the options VectorScale and VectorStyle of VectorPlot. To create the slope field for the first order equation $y'=f(x,y)$, I usually do something like so.\n\nf = Exp[-x] - y;\nVectorPlot[{1, f}, {x, -2, 2}, {y, -2, 2},\nVectorScale -> {0.03, Automatic, None},", null, "Yes, you can simply replace each occurrence of Arrow with Line like this:\n\nVectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}] /. Arrow -> Line", null, "To understand how this works, please read the documentation for ReplaceAll and also take a look at FullForm[VectorPlot[...]]. The point is to see that the plot is just a bunch of graphics directives and therefore we can modify it. Replacing Arrow with Line works because the argument is the same to both Arrow and Line.\n\nYou can also achieve it like this:\n\nVectorPlot[{y, -x}, {x, -3, 3}, {y, -3, 3}, VectorStyle -> \"Segment\"]\n\n\nLook at the documentation for VectorStyle and especially the part under \"Details\" to see the full set of options. As a beginner, this is the way to go, and I should probably have used it as well. But as you get more experienced sometimes you don't want to dig in the documentation so you do something quick and dirty like what I did above. It's often useful to be able to, so I'll let the other solution remain in my answer.\n\nMy example above was poor, it appears, since it's not a proper slope field. For that part of the question check out Mark McClure's answer.\n\n• +1 for the VectorStyle->\"Segment\", which I didn't know. Not actually a slope field, however. – Mark McClure Mar 5 '14 at 21:48\n• @MarkMcClure Thanks for pointing it out, I gave your post a +1 and added a comment about. – C. E. Mar 5 '14 at 22:35\n• Well, \"slightly off\" would be a better description than \"poor\". :) – Mark McClure Mar 6 '14 at 0:47" ]

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[ "", null, "Show Question\nMath and science::Analysis::Tao::07. Series\n\n# The comparison test (and bounded series of non-negative numbers)\n\nWe wish to extend the comparison test for finite series to apply to infinite series.\n\nFor finite series, the comparison test appeared as a simple opening lemma (7.1.4 f):\n\n$\\text{When } a_i \\le b_i \\text{ we have } \\sum_{i=m}^{n} a_i \\le \\sum_{i=m}^{n} b_i$\n\nFor infinite series, we can only make this statement when $$|a_i| \\le b_n$$, otherwise the sum for $$a$$ could diverge. We set this up and prove it below.\n\nFirst, a useful (and very simple) proposition.\n\n#### Sums of non-negative numbers are bounded iff they are convergent.\n\nLet $$\\sum_{n=m}^{\\infty}a_n$$ be a formal series of non-negative real numbers. Then this series is convergent if and only if there is a real number $$M$$ such that\n\n$\\sum_{n=m}^{N}a_n \\le M \\text{ for all integers } N \\ge m$\n\n#### Proof\n\nTwo earlier propositions are:\n\n• Every convergent sequence of real numbers is bounded (6.1.17).\n• An increasing sequence which has an upper bound is convergent (6.3.8).\n\nThe sequence $$(S_N)_{n=m}^{\\infty}$$ representing the series $$\\sum_{n=m}^{\\infty}a_n$$ is increasing, by definition. So the above two propositions form each side of the proposition's iff statement.\n\nNow we can introduce the comparison test.\n\n### Comparison test\n\nLet $$\\sum_{n=m}^{\\infty}a_n$$ and $$\\sum_{n=m}^{\\infty}b_n$$ be two formal series of real numbers, and suppose that $$|a_n| \\le b_n$$ for all $$n \\ge m$$ (thus all numbers in $$(b_n)$$ must be non-negative). Then if $$\\sum_{n=m}^{\\infty}b_n$$ is convergent, then $$\\sum_{n=m}^{\\infty}a_n$$ is absolutely convergent, and in fact\n\n$\\left| \\sum_{n=m}^{\\infty}a_n \\right| \\le \\sum_{n=m}^{\\infty}|a_n| \\le \\sum_{n=m}^{\\infty}b_n$\n\n#### Proof\n\nLet $$SA_N := \\sum_{n=m}^{N}|a_n|$$ be the Nth partial sum of $$\\sum_{n=m}^{\\infty}|a_n|$$ and $$SB_N := \\sum_{n=m}^{N}b_n$$ be the Nth partial sum of $$\\sum_{n=m}^{\\infty}b_n$$. As $$\\sum_{n=m}^{\\infty}b_n$$ is bounded according to the proposition above, and by the the comparison test for finite series we have: $SA_N \\le SB_N \\le M \\text{ for all } N \\ge m \\text{ and for some bounding real } M$ Through this transitivity, $$(SA_N)_{N=n}^{\\infty}$$ also meets the conditions for being bounded. Then from the above propositive again, it must be that $$\\sum_{n=m}^{\\infty}|a_n|$$ is convergent. The Absolute Convergence Test informs us that absolute convergence implies conditional convergence, and in addition that $$\\left| \\sum_{n=m}^{\\infty}a_n \\right| \\le \\sum_{n=m}^{\\infty}|a_n|$$. Thus, with all of the below infinite series converging we can compare them like so:\n$\\left| \\sum_{n=m}^{\\infty}a_n \\right| \\le \\sum_{n=m}^{\\infty}|a_n| \\le \\sum_{n=m}^{\\infty}b_n$\n\nWe can also run the Comparison Test in the contrapositive:\n\nIf $$\\sum_{n=m}^{\\infty}a_n$$ is not absolutely convergent then $$\\sum_{n=m}^{\\infty}b_n$$ is not conditionally convergent.", null, "p171" ]

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[ "# Estimating $b_1 x_1+b_2 x_2$ instead of $b_1 x_1+b_2 x_2+b_3x_3$\n\nI have a theoretical economic model which is as follows,\n\n$$y = a + b_1x_1 + b_2x_2 + b_3x_3 + u$$\n\nSo theory says that there are $x_1$, $x_2$ and $x_3$ factors to estimate $y$.\n\nNow I have the real data and I need to estimate $b_1$, $b_2$, $b_3$. The problem is that the real data set contains only data for $x_1$ and $x_2$; there are no data for $x_3$. So the model I can fit actually is:\n\n$$y = a + b_1x_1 + b_2x_2 + u$$\n\n• Is it OK to estimate this model?\n• Do I lose anything estimating it?\n• If I do estimate $b_1$, $b_2$, then where does the $b_3x_3$ term go?\n• Is it accounted for by error term $u$?\n\nAnd we would like to assume that $x_3$ is not correlated with $x_1$ and $x_2$.\n\n• Can you give details about your data set, I mean, your dependent variable $y$ and independent variables $x_1$ and $x_2$? – Vara May 11 '13 at 14:32\n• Think of it as hypothethical example without specific data set... – renathy May 11 '13 at 15:05\n\nThe issue you need to worry about is called endogeneity. More specifically, it depends on whether $x_3$ is correlated in the population with $x_1$ or $x_2$. If it is, then the associated $b_j$s will be biased. That is because OLS regression methods force the residuals, $u_i$, to be uncorrelated with your covariates, $x_j$s. However, your residuals are composed of some irreducible randomness, $\\varepsilon_i$, and the unobserved (but relevant) variable, $x_3$, which by stipulation is correlated with $x_1$ and / or $x_2$. On the other hand, if both $x_1$ and $x_2$ are uncorrelated with $x_3$ in the population, then their $b$s won't be biased by this (they may well be biased by something else, of course). One way econometricians try to deal with this issue is by using instrumental variables.\n\nFor the sake of greater clarity, I've written a quick simulation in R that demonstrates the sampling distribution of $b_2$ is unbiased / centered on the true value of $\\beta_2$, when it is uncorrelated with $x_3$. In the second run, however, note that $x_3$ is uncorrelated with $x_1$, but not $x_2$. Not coincidentally, $b_1$ is unbiased, but $b_2$ is biased.\n\nlibrary(MASS) # you'll need this package below\nN = 100 # this is how much data we'll use\nbeta0 = -71 # these are the true values of the\nbeta1 = .84 # parameters\nbeta2 = .64\nbeta3 = .34\n\n############## uncorrelated version\n\nb0VectU = vector(length=10000) # these will store the parameter\nb1VectU = vector(length=10000) # estimates\nb2VectU = vector(length=10000)\nset.seed(7508) # this makes the simulation reproducible\n\nfor(i in 1:10000){ # we'll do this 10k times\nx1 = rnorm(N)\nx2 = rnorm(N) # these variables are uncorrelated\nx3 = rnorm(N)\ny = beta0 + beta1*x1 + beta2*x2 + beta3*x3 + rnorm(100)\nmod = lm(y~x1+x2) # note all 3 variables are relevant\n# but the model omits x3\nb0VectU[i] = coef(mod) # here I'm storing the estimates\nb1VectU[i] = coef(mod)\nb2VectU[i] = coef(mod)\n}\nmean(b0VectU) # -71.00005 # all 3 of these are centered on the\nmean(b1VectU) # 0.8399306 # the true values / are unbiased\nmean(b2VectU) # 0.6398391 # e.g., .64 = .64\n\n############## correlated version\n\nr23 = .7 # this will be the correlation in the\nb0VectC = vector(length=10000) # population between x2 & x3\nb1VectC = vector(length=10000)\nb2VectC = vector(length=10000)\nset.seed(2734)\n\nfor(i in 1:10000){\nx1 = rnorm(N)\nX = mvrnorm(N, mu=c(0,0), Sigma=rbind(c( 1, r23),\nc(r23, 1)))\nx2 = X[,1]\nx3 = X[,2] # x3 is correated w/ x2, but not x1\ny = beta0 + beta1*x1 + beta2*x2 + beta3*x3 + rnorm(100)\n# once again, all 3 variables are relevant\nmod = lm(y~x1+x2) # but the model omits x3\nb0VectC[i] = coef(mod)\nb1VectC[i] = coef(mod) # we store the estimates again\nb2VectC[i] = coef(mod)\n}\nmean(b0VectC) # -70.99916 # the 1st 2 are unbiased\nmean(b1VectC) # 0.8409656 # but the sampling dist of x2 is biased\nmean(b2VectC) # 0.8784184 # .88 not equal to .64\n\n• So, can you explain a bit more - what happens if we assume that $x_3$ is not correlated with $x_1$ and $x_2$? Then what happens if I estimate $y=a+b_1x_1+b_2x_2+u$? – renathy May 11 '13 at 15:04\n• $b_3x_3$ will be incorporated into the residuals either way, but if it is uncorrelated in the population, then your other $b$s won't be biased by the absence of $x_3$, but if it's not uncorrelated, then they will be. – gung - Reinstate Monica May 11 '13 at 15:08\n• To state this more clearly: If $x_3$ is not correlated with either $x_1$ or $x_2$, you are OK. – gung - Reinstate Monica May 11 '13 at 15:10\n• I discuss the flip side of this issue in my answer here: Does adding more variables into a multivariable regression change coefficients of existing variables? – gung - Reinstate Monica Sep 18 '14 at 0:24\n\nLet's think of this in geometric terms. Think of a \"ball\", the surface of a ball. It is described as $r^2 = ax^2+by^2+cz^2 + \\epsilon$. Now if you have the values for $x^2$, $y^2$, $z^2$, and you have measurements of $r^2$ then you can determine your coefficients \"a\", \"b\", and \"c\". (You could call it ellipsoid, but to call it a ball is simpler.)\n\nIf you have only the $x^2$, and $y^2$ terms then you can make a circle. Instead of defining the surface of a ball, you will describe a filled in circle. The equation you instead fit is $r^2 \\le ax^2 + by^2 + \\epsilon$.\n\nYou are projecting the \"ball\", whatever shape it is, into the expression for the circle. It could be a diagonally oriented \"ball\" that is shaped more like a sewing needle, and so the $z$ components utterly wreck the estimates of the two axes. It could be a ball that looks like a nearly crushed m&m where the coin-axes are \"x\" and \"y\", and there is zero projection. You can't know which it is without the \"$z$\" information.\n\nThat last paragraph was talking about a \"pure information\" case and didn't account for the noise. Real world measurements have the signal with noise. The noise along the perimeter that is aligned to the axes is going to have a much stronger impact on your fit. Even though you have the same number of samples, you are going to have more uncertainty in your parameter estimates. If it is a different equation than this simple linear axis-oriented case, then things can go \"pear shaped\". Your current equations are plane-shaped, so instead of having a bound (the surface of the ball), the z-data might just go all over the map - projection could be a serious problem.\n\nIs it okay to model? That is a judgment call. An expert who understands the particulars of the problem might answer that. I don't know if someone can give a good answer if they are far from the problem.\n\nYou do lose several good things, including certainty in parameter estimates, and the nature of the model being transformed.\n\nThe estimate for $b_3$ disappears into epsilon and into the other parameter estimates. It is subsumed by the whole equation, depending on the underlying system.\n\n• I can't really follow your argument here, & I'm not sure if it's correct. E.g., the surface area of a sphere is $4\\pi r^2$. Beyond that, I'm not sure how this relates to the question. The key issue is whether or not the omitted variable is correlated w/ variables that are in the model. I'm not sure how what you are saying addresses that issue. (For clarity, I demonstrate this with a simple R simulation.) – gung - Reinstate Monica May 11 '13 at 23:08\n• Gung. I gave a best-case answer sphere -> circle and showed that it changed the model in unexpected ways. I liked the technical sophistication of your answer, but am not convinced that the asker is able to use either of our answers. the $f(x,y,z)$ is the equation for the surface of an ellipsoid in 3 dimensions, a sphere is one case of it. I am assuming that the \"true model\" is the surface of the sphere, but noise corrupted measurements are on the surface. Throwing out one dimension gives data that, at best, makes a filled circle instead of the surface of a sphere. – EngrStudent May 12 '13 at 4:55\n• I am unable to follow your argument because I don't see anything that corresponds to a \"filled in square.\" – whuber Sep 3 '18 at 20:19\n\nThe other answers, while not wrong, over complicate the issue a bit.\n\nIf $x_3$ is truly uncorrelated with $x_1$ and $x_2$ (and the true relationship is as specified) then you can estimate your second equation without an issue. As you suggest, $\\beta_3 x_3$ will be absorbed by the (new) error term. The OLS estimates will be unbiased, as long as all the other OLS assumptions hold." ]

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[ "# Using Data Mining Techniques with the 2df QSO Redshift Survey Spectra to Locate Damped Lyman Alpha Systems\n\nIan Hickson\n\n## Abstract\n\nSpectra from the 2df QSO redshift survey are analysed first using algorithmic methods and then using a neural network. The data is found to be too unpredictable for simple algorithms to detect DLAs, although many oddities are discovered and the algorithms could be used for looking for other, more obvious patterns. This conclusion is backed by the neural network results, which indicate that the data is too varied for a simple back-propagation error correction algorithm. It is suggested that to avoid falling into local minima, a neural network could be evolved using genetic algorithm techniques. A C++ implementation of a feed forward trilayer perceptron is provided.\n\n## Contents\n\nAbstract 1\n\nContents 2\n\n1. Introduction 3\n\n1.1. Damped Lyman alpha systems 3\n\n1.2. The 2df QSO redshift survey 4\n\n1.3. The 2QZ 10K Catalogue 5\n\n2. The algorithmic approach 5\n\n2.1. Basic program design 5\n\n2.2. Smoothing 5\n\n2.2.2. Moving average 6\n\n2.2.3. Fourier transformation 6\n\n2.3. Patterns: Extended Minima 6\n\n2.3.1. Algorithm 7\n\n2.3.2. Conclusions 7\n\n2.3.3. Discoveries 7\n\n2.4.1. Algorithm 8\n\n2.4.2. Conclusions 8\n\n2.5. Looking for dips: Version 2 8\n\n2.5.1. Discoveries 9\n\n2.5.2. Conclusions 11\n\n2.6. Directions for future researh 11\n\n3. The neural network approach 11\n\n3.1. Feed Forward Trilayer Perceptrons (FFTPs) 11\n\n3.1.1. Transfer Function 12\n\n3.1.2. Equations 12\n\n3.2. Back-Propagating Differential Error Correction 13\n\n3.3. Implementation 17\n\n3.4. Training Data 19\n\n3.5. Analysis 19\n\n3.6. Variations 20\n\n3.7. Directions for future research 20\n\n4. Appendix: References 20\n\n4.1. Acknowledgements 21\n\n6. Appendix: Data Sets 21\n\n7. Appendix: Source 23\n\n7.1. `ffmp.h` 23\n\n7.2. `ffmp.cpp` 24\n\n## 1. Introduction\n\nThe aim of this project was to use automated techniques to locate previously undiscovered damped Lyman alpha systems amongst the 2df QSO Redshift Survey data. \n\n### 1.1. Damped Lyman alpha systems\n\nDamped Lyman alpha systems arise when light emitted from a quasistellar object (or QSO) goes through a gas cloud on its way to the observer. When the photons interact with the hydrogen atoms in the gas cloud, the hydrogen atoms absorb some of the radiation, primarily the photons with wavelength 1216Å, the Lyman series of the Hydrogen atom.\n\nDue to the doppler shift caused by cosmic expansion, the photons in question usually correspond to different inital wavelengths and show up as dips at different wavelengths upon reaching the observer. The redshift can be used to determine the distance to the gas cloud, and the width of the damped system can be used to determine the column depth of the cloud.\n\nIt is believed that gas clouds dense enough to noticably affect backlighting QSO spectra are galaxies in early stages of formation. As data on protogalaxies and gas clouds is useful in refining models for galactic evolution, damped Lyman alpha systems found in the data highlight targets for further study.\n\nBelow are some typical damped Lyman alpha systems. (All graphs courtesey of the 2df QSO 10K Catalogue. )\n\nJ000259.0-312222a\nThis spectrum is damped between 4100Å and 4200Å. This is a very narrow dip and represents the smallest such feature that should be considered a match.", null, "J103727.9+001819\nA spectrum with a narrow dip around 4000Å", null, "J141357.8+004345\nThis spectrum shows a classic Lyman forest at around 4000Å, and was highlighted by P. J. Outram, et al as a subject for further study.", null, "### 1.2. The 2df QSO redshift survey\n\nThe survey was conducted using one of the world's most advanced telescope systems, the Anglo-Australian Telescope 2 Degree Field facility.\n\nThe Anglo-Australian Telescope (AAT), completed in June 1975 , is an equatorially mounted 4-metre telescope. Prior to the 1970s, most of the large telescopes were located in the northern hemisphere, and were thus unable to study many of the closest radio galaxies, as well as the center of our own. \n\nThe AAT can be used in many different configurations, using various instruments and detectors. In the configuration used for the 2df QSO redshift survey, the AAT consisted of a wide field corrector, an atmospheric dispersion compensator, a robot gantry, two spectrographs, and a tumbling mechanism. In this configuration the covers wavelengths from 3500Å to 8000Å.\n\nThe tumbling mechanism enabled one field plate of optic fibres to be configured by the robot while simultaneously another plate of optic fibres, connected to the spectrographs, was analysed. As both these processes take approximately an hour to complete, this setup allowed the spectra to be collected simultaneously and continuously in an automated fashion. Each spectrograph was able to analyse 200 objects at a time, allowing for a total of 400 spectra to be collected during each cycle.\n\n### 1.3. The 2QZ 10K Catalogue\n\nNote that the survey was conducted for the purpose of confirming the identity of suspected QSOs and to find their redshift. For this reason, the typical signal-to-noise ratio (S/N) of the spectra is in the region of 10, and the resolution appoximately 8Å. This is not optimal for searching for DLAs. \n\n## 2. The algorithmic approach\n\nThe first approach consisted of a set of algorithms applied to the data, in a traditional deterministic manner.\n\n### 2.1. Basic program design\n\nThe program evolved significantly over the life of this project, but the main design remained the same throughout.\n\nFirst, data is extracted from the FITS (Flexible Image Transport System) data files provided on the 2df data CDROM. To extract data from the FITS files, the CFITSIO library was used. This library provides a portable subroutine-based API for reading and writing FITS data files.\n\nThe FITS files used by 2df survey consisted of a 1024 element spectrum (scaled by the standard BSCALE and BZERO headers), a 1024 boolean bad pixel mask, and a median sky spectrum. \n\nOriginally, spectra not associated with QSOs were then discarded. However, it was discovered that the spectrum labelling of the 2df data is unreliable, and this step was sebsequently skipped.\n\nFinally, the data was smoothed and then analysed, flagging potential matches.\n\nSome implementation details, such as optimisations used to ease development, shall not be described in this report.\n\n### 2.2. Smoothing\n\nThe spectra have a lot of high frequency noise. To reduce the number of false negatives, the data had to be smoothed.\n\nThe data included, for each spectrum, a bad pixel mask. This is a list of which of the 1024 values should be considered invalid during analysis.\n\nBad pixels can arise from several different sources, including poor weather, and edge effects (all the spectra have some number of bad pixels at the start and end of the data).\n\n#### 2.2.2. Moving average\n\nThe simplest smoothing algorithm used consisted simply of applying a smoothing average: taking the mean of the valid values present in the last n points as the value of each pixel.\n\nVarious windows were tried, a window of 10 data points resulted in reasonable smoothing without losing the major features of the data.\n\nBecause of the bad pixels, there were points in the output of this algorithm that would have no value (if more than n bad pixels appeared together). To work around this, a simple linear extrapolation was then applied to the gaps.\n\nWhile this algorithm had a moderate success in removing some of the noise, it was not satisfactory and was quickly discarded.\n\n#### 2.2.3. Fourier transformation\n\nThis second method provided much smoother results. A fourier transform was used to perform a transformation into the frequency domain, cut out the high frequency noise, then convert the data back to the time domain. (Note that technically the time domain was the frequency domain and the frequency domain the time domain, but this is ignored here for clarity.)\n\nThe transform was implemented using the FFTW, a high performance portable API that uses self-adapting techniques to optimise itself for the machine architecture and type of data being transformed. \n\nTrial and error found that cutting approximately the top 90% of the data resulted in the smoothest data still resembling the original.\n\nFor more detail on the FFTW algorithms used, see .\n\n### 2.3. Patterns: Extended Minima\n\nThe initial approach was developed prior to having analysed the data in much detail. It was based on the assumption that damped systems would characteristically approach zero, while undamped signals would not.\n\n#### 2.3.1. Algorithm\n\nFor each spectrum, each good pixel was examined in turn. A count of how many pixels in a row had a value less than a particular threshold was found. Any spectrum with a suitably wide number series of such low values was then flagged.\n\nThe threshold used was a little above 0, as the intention was to catch damped systems that approached the zero baseline.\n\n#### 2.3.2. Conclusions\n\nUnfortunately, it turns out that a large number of the spectra are miscalibrated, resulting in most having numbers near, or even below, zero. (Note that the spectra have indeed not been flux calibrated, so this is not unusual.) Therefore this technique is unsuitable for finding damped systems in the 2QZ 10k catalogue.\n\n#### 2.3.3. Discoveries\n\nWhile no damped Lyman alpha systems were discovered using this technique, several oddities were flagged.\n\nJ024412.2-293014\nThis spectrum appears to be galactic in nature (rather than a QSO, as it is labelled).", null, "J215555.2-283710a\nThis spectrum has a mean value less than zero, indicating significant miscalibration.", null, "This approach was built on the experience gained from examining the spectra highlighted by the zeroes algorithm.\n\nIt is based on searching for a period of characteristic decrease in value over a short period, following a similar characteristic increase in value.\n\n#### 2.4.1. Algorithm\n\nFor each value in the array, compare the value to the previous one. If the last few values have been decreasing but the last step is an increase, remember this point as the last decrease and remember how many decreases happened in a row. Otherwise, if this is an increase and there have been enough increases in a row, and the last decrease is suitably far away, and there were enough decreases in a row, flag this spectrum and move to the next one.\n\n#### 2.4.2. Conclusions\n\nIn practice, this does not work at all, because the signal (even after excessive smoothing) still contains perturbations (high frequency noise) on the up and down curves.\n\nAnother problem is that this algorithm highlighted many spectra with two adjacent clean peaks.\n\n### 2.5. Looking for dips: Version 2\n\nTo get around the limitations of the first version, a new algorithm was developed by looking for the following characteristics:\n\n• The dips have a minimum width and a maximum width.\n• The average value in a dip is lower than the average values on either side of the dip.\n• The sides of the dip have to change by a certain amount within a certain width.\n\nThe algorithm is more involved and is best explained through the following pseudo-code:\n\n1. Set lastDown and lastUp to 0.\n2. Starting at a certain distance from the left margin (to skip edge noise and have enough room to later average it if required), and stepping through each value until a certain distance from the end of the data:\n1. If the value of the pixel a certain distance to the left divided by the value of the pixel at the current location is greater than a threshold drop fraction,\n1. Store the position of the pixel in lastDown.\n2. If the value of the pixel a certain distance to the right divided by the value of the pixel at the current location is greater than a threshold drop fraction,\n1. Store the position of the pixel in lastUp.\n3. If lastDown is not 0, and lastUp - lastDown is within the range of minimum and maximum widths for a dip,\n1. Find the mean value between the lastDown and lastUp positions.\n2. Find the mean value for a certain distance before lastDown.\n3. Find the mean value for a certain distance after lastUp.\n4. Check that the first mean is lower than the other two means by a certain amount. If it is, flag this spectrum and move on to the next one.\n\n#### 2.5.1. Discoveries\n\nThree spectra were identified as potential forming galaxies, the first two may actually be galaxies in the foreground, and the third may potentially be a backlit DLA.", null, "", null, "", null, "The following spectra were also flagged. (Note that the data points are in the 0..1023 range of the data file, not in angstroms.)\n\n• J012122.9-312825a may have a dip between data points 372 and 585.\n• J024504.5-292103a may have a dip between data points 365 and 507.\n• J110735.9-023414a may have a dip between data points 579 and 763.\n• J113701.4-001940b may have a dip between data points 486 and 616.\n• J124557.8+001757a may have a dip between data points 519 and 648.\n• J124604.8+002714a may have a dip between data points 594 and 732.\n• J131759.1+014012a may have a dip between data points 656 and 816.\n• J134453.0-010637a may have a dip between data points 605 and 731.\n• J143613.1-003657a may have a dip between data points 209 and 384.\n• J215638.8-274941a may have a dip between data points 576 and 789.\n• J222601.6-300959a may have a dip between data points 512 and 803.\n• J224550.9-301903a may have a dip between data points 486 and 615.\n• J230624.7-280338b may have a dip between data points 350 and 605.\n• J235414.5-300225a may have a dip between data points 650 and 792.\n\n#### 2.5.2. Conclusions\n\nIn practice, this fails to take into the effects of fluctuations in the data. Even when ignoring a lot of high frequency noise (see the section on smoothing by fourier transform), this technique would still catch two high spikes around a low spike. The flagged spectra did not really fit the DLA profile.\n\n### 2.6. Directions for future researh\n\nThese algorithms could be improved in two main ways. Firstly, instead of looking at mean values in the last algorithm, median values could be examined. This would reduce the sensitivity to peak noise.\n\nA second possible approach is to analyse the data in the frequency domain (technically the time domain, since the original data is a function of frequency) instead of examining the data after smoothing. This option was largely ignored during this analysis but could potentially lead to greater resilience against fluctuations.\n\n## 3. The neural network approach\n\nNeural networks are loosely based on what is believed to be the structure of human brains. Human brains are formed from large networks of neurons, cells that consist of many 'input' dendrites and many 'output' dendrites. Dendrites are interconnected by synapses, whose chemical state affects the weight with with the signal will impact the following neuron.\n\nTo train the brain, it is believed that different responses trigger different chemicals to bathe the synapses, changing the relative impact of each dendrite on the next cell.\n\nIn a real human brain there are roughly 1010 neurons each with up to several thousand synapses contained on the dendrites, giving up to a maximum of 1014 synapses .\n\nNeural networks are similarly formed. The particular network type used in this investigation is known as a feed forward trilayer perceptron, using a back-propagating differential error correction algorithm for training.\n\n### 3.1. Feed Forward Trilayer Perceptrons (FFTPs)\n\nFFTPs are Feed Forward Multilayer Pereptrons (FFMPs) with two hidden layers. The implementation used here is based on the FFMP implementation discussed by Bailer-Jones, et al. \n\nFFMPs are simple mathematical models of the biological neural networks described above. Neurons are replaced by transfer functions, dendrites are replaced by arrays, synapses by weights, and training chemicals by differentiation.\n\nInstead of a large network of cells all interconnected, a well defined number of layers is selected, and every combination of weights from the input layer to the first layer, the first layer to the second layer, and so on up to the output layer is then given a weight.\n\nIn addition to connections between every value of each layer to the next, an additional (arbitrary) value is connected to each value of each layer. These bias nodes are required to offset values. For example even if all the imputs are zero, the bias node allows the first layer to use non-zero values as arguments to the transfer function, allowoing the full range of output. Without bias nodes the only argument possible would be zero.\n\n#### 3.1.1. Transfer Function\n\nThe transfer function used is that recommended by Bailer-Jones, et al, namely the hyperbolic tangent.\n\nThis function has the property that all real inputs give an output in the range of -1..1. However, it also introduces the possibility for network saturation, where the weights grow such that every node in a layer has the value 1 (or -1) and the data therefore has no relation to the input. This must be avoided by not over-training the network and by using small inital weights.\n\n#### 3.1.2. Equations\n\nThe notation used here is based on that used by Bailer-Jones, et al.\n\n${x}_{i}$\nThe input data. The i subscript's range is 1..I+1, where I is the index of the last input node, so the I+1th node is the first hidden layer's bias node.\n${p}_{j}$\nThe first hidden layer. The j subscript's range is 1..J+1, where J is the index of the last node in the first hidden layer to be affected by the input layer, and the J+1th node is the second hidden layer's bias node.\n${q}_{k}$\nThe second hidden layer. The k subscript's range is 1..K+1, where K is the index of the last node in the second hidden layer to be affected by the first hidden layer, and the K+1th node is the output layer's bias node.\n${y}_{l}$\nThe output layer. The y subscript's range is 1..L, where L is the index of the last node in the output layer.\n${w}_{i,j}$\nThe weights between the ${x}_{i..I+1}$ and ${p}_{j..J}$ layers.\n${w}_{j,k}$\nThe weights between the ${p}_{j..J+1}$ and ${q}_{k..K}$ layers.\n${w}_{k,l}$\nThe weights between the ${q}_{k..K+1}$ and ${y}_{l..L}$ layers.\n\nThe input layer is given by:\n\n${x}_{i}=\\left\\{\\begin{array}{cc}1& i=I+1\\\\ \\mathrm{input}\\left(i\\right)& \\text{otherwise}\\end{array}$\n\nEach node in the first hidden layer is then calulated by summing each input (including the bias node) multiplied by a weighting value, and finding the hyperbolic tangent of the result:\n\n${p}_{j}=\\left\\{\\begin{array}{cc}1& j=J+1\\\\ \\mathrm{tanh}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right)& \\text{otherwise}\\end{array}$\n\n${q}_{k}=\\left\\{\\begin{array}{cc}1& k=K+1\\\\ \\mathrm{tanh}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right)& \\text{otherwise}\\end{array}$\n\n${y}_{l}=\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },l}{q}_{{k}^{\\prime }}+{w}_{K+1,l}$\n\n### 3.2. Back-Propagating Differential Error Correction\n\nThe training consists of supervised learning by minimization of an error function with respect to all of the network weights. The error function used here is the sum-of-squares error:\n\n$e=\\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}{\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)}^{2}$\n\nThis represents the total difference between the actual output and the ideal output. By minimising this number we move towards a more ideal output, and by finding how much a small change in each weight would affect the error function, we can estimate how much change in the weights is required to move towards the minimum.\n\nFor the last set of weights, differentiating equation 5 with respect to the weights gives:\n\n$\\frac{\\partial e}{\\partial {w}_{k,l}}=\\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}2\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\frac{\\partial {y}_{l}}{\\partial {w}_{k,{l}^{\\prime }}}$\n\n...where:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{k,l}}=\\sum _{{k}^{\\prime }}^{K}{\\delta }_{{k}^{\\prime }k}{\\delta }_{{l}^{\\prime }l}{q}_{{k}^{\\prime }}+{\\delta }_{K+1k}{\\delta }_{{l}^{\\prime }l}$\n\n...and $\\delta$ is the Kronecker delta function.\n\nThis simplifies to:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{k,l}}=\\left\\{\\begin{array}{cc}{\\delta }_{{l}^{\\prime }l}& k=K+1\\\\ {\\delta }_{{l}^{\\prime }l}{q}_{k}& \\text{otherwise}\\end{array}$\n\nSubstituting 8 into 6:\n\n$\\frac{\\partial e}{\\partial {w}_{k,l}}=\\left\\{\\begin{array}{cc}\\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}2\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){\\delta }_{{l}^{\\prime }l}& k=K+1\\\\ \\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}2\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){\\delta }_{{l}^{\\prime }l}{q}_{k}& \\text{otherwise}\\end{array}$\n\n...this simplifies to:\n\n$\\frac{\\partial e}{\\partial {w}_{k,l}}=\\left\\{\\begin{array}{cc}{\\beta }_{l}\\left({y}_{l}-{T}_{l}\\right)& k=K+1\\\\ {\\beta }_{l}\\left({y}_{l}-{T}_{l}\\right){q}_{k}& \\text{otherwise}\\end{array}$\n\nFor the middle set of weights,\n\n$\\frac{\\partial e}{\\partial {w}_{j,k}}=\\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}2\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{j,k}}$\n\n...where:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{j,k}}=\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}\\frac{\\partial {q}_{{k}^{\\prime }}}{\\partial {w}_{j,k}}$\n\n$\\frac{\\partial {q}_{{k}^{\\prime }}}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}0& {k}^{\\prime }=K+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right)\\left(\\sum _{{j}^{\\prime }}^{J}{\\delta }_{{j}^{\\prime }j}{\\delta }_{{k}^{\\prime }k}{p}_{{j}^{\\prime }}+{\\delta }_{J+1j}{\\delta }_{{k}^{\\prime }k}\\right)& \\text{otherwise}\\end{array}$\n\nSimplifying:\n\n$\\frac{\\partial {q}_{{k}^{\\prime }}}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}0& {k}^{\\prime }=K+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){\\delta }_{{k}^{\\prime }k}& j=J+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){\\delta }_{{k}^{\\prime }k}{p}_{j}& \\text{otherwise}\\end{array}$\n\nSubstituting 14 into 12, and noting the sum over ${k}^{\\prime }$ does not exceed K:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){\\delta }_{{k}^{\\prime }k}& j=J+1\\\\ \\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){\\delta }_{{k}^{\\prime }k}{p}_{j}& \\text{otherwise}\\end{array}$\n\nSimplifying:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}{w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right)& j=J+1\\\\ {w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right){p}_{j}& \\text{otherwise}\\end{array}$\n\nSubstituting 16 into 11 and cancelling the twos:\n\n$\\frac{\\partial e}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right)& j=J+1\\\\ \\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right){p}_{j}& \\text{otherwise}\\end{array}$\n\nFinally for the first set of weights,\n\n$\\frac{\\partial e}{\\partial {w}_{i,j}}=\\frac{1}{2}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}2\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{i,j}}$\n\n...where:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{i,j}}=\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}\\frac{\\partial {q}_{{k}^{\\prime }}}{\\partial {w}_{i,j}}$\n\n$\\frac{\\partial {q}_{{k}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}0& {k}^{\\prime }=K+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right)\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}\\frac{\\partial {p}_{{j}^{\\prime }}}{\\partial {w}_{i,j}}& \\text{otherwise}\\end{array}$\n\n$\\frac{\\partial {p}_{{j}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}0& {j}^{\\prime }=J+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },{j}^{\\prime }}{x}_{{i}^{\\prime }}+{w}_{I+1,{j}^{\\prime }}\\right)\\left(\\sum _{{i}^{\\prime }}^{I}{\\delta }_{{i}^{\\prime }i}{\\delta }_{{j}^{\\prime }j}{x}_{{i}^{\\prime }}+{\\delta }_{I+1i}{\\delta }_{{j}^{\\prime }j}\\right)& \\text{otherwise}\\end{array}$\n\nSimplifying:\n\n$\\frac{\\partial {p}_{{j}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}0& {j}^{\\prime }=J+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },{j}^{\\prime }}{x}_{{i}^{\\prime }}+{w}_{I+1,{j}^{\\prime }}\\right){\\delta }_{{j}^{\\prime }j}& \\text{i=I+1}\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },{j}^{\\prime }}{x}_{{i}^{\\prime }}+{w}_{I+1,{j}^{\\prime }}\\right){\\delta }_{{j}^{\\prime }j}{x}_{i}& \\text{otherwise}\\end{array}$\n\nSubstituting 22 into 20 and noting ${j}^{\\prime }$ never reaches J+1:\n\n$\\frac{\\partial {p}_{{j}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}0& {k}^{\\prime }=K+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right)\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },{j}^{\\prime }}{x}_{{i}^{\\prime }}+{w}_{I+1,{j}^{\\prime }}\\right){\\delta }_{{j}^{\\prime }j}& \\text{i=I+1}\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right)\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },{j}^{\\prime }}{x}_{{i}^{\\prime }}+{w}_{I+1,{j}^{\\prime }}\\right){\\delta }_{{j}^{\\prime }j}{x}_{i}& \\text{otherwise}\\end{array}$\n\nSimplifying:\n\n$\\frac{\\partial {p}_{{j}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}0& {k}^{\\prime }=K+1\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right)& \\text{i=I+1}\\\\ {\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right){x}_{i}& \\text{otherwise}\\end{array}$\n\nSubstituting 24 into 19 and noting ${k}^{\\prime }$ never reaches K+1:\n\n$\\frac{\\partial {y}_{{l}^{\\prime }}}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right)& \\text{i=I+1}\\\\ \\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right){x}_{i}& \\text{otherwise}\\end{array}$\n\nSubstituting 25 into 18 and cancelling the twos:\n\n$\\frac{\\partial e}{\\partial {w}_{i,j}}=\\left\\{\\begin{array}{cc}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right)& \\text{i=I+1}\\\\ \\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right){x}_{i}& \\text{otherwise}\\end{array}$\n\nThe implementation of these equations (26, 18 and 10) can be optimised by noting that some of the terms (for example the first sech term in equation 26) are common to all of the input values and can therefore be precalculated. This reduces the total instruction count significantly.\n\n### 3.3. Implementation\n\nThe implementation of the Feed Forward Trilayer Pereptron was written in C++. A class (`FeedForwardTrilayerPerceptron`, declared in `fftw.h` and defined in `fftw.cpp`, see appendix 7) represents a single trainable neural network, with the following methods:\n\n```FeedForwardTrilayerPerceptron(int, int, int, int, double)```\n\nA constructor that creates a randomly initialised perceptron of the specified size. The first four arguments represent the sizes of the input layer, two hidden layers, and output layer respectively (not including hidden nodes). The last argument is a floating point number representing the maximum absolute value to use for the weights. (For instance if it is 0.05 then the random weights will be in the range -0.05..0.05.)\n\n```FeedForwardTrilayerPerceptron(int, int, int, int, std::vector<double>*)```\n\nA constructor that creates a perceptron of the specified size initialised from specified weights. The first four arguments represent the sizes of the input layer, two hidden layers, and output layer respectively (not including hidden nodes), and the last argument is a pointer to a standard template library (STL) vector containing the weights required (including those required for hidden nodes). The vector returned by `GetCharacteristics()` is in this format.\n\n`FeedForwardTrilayerPerceptron(istream&)`\n\nA constructor that creates a perceptron based on parameters stored on an STL stream in the format created by Save().\n\n`~FeedForwardTrilayerPerceptron()`\n\nThe destructor.\n\n```std::vector<double>* Run(std::vector<double>*)```\n\nTakes a pointer to a vector containing the input data and returns a pointer to a new vector containing the outputs. The caller is responsible for destroying the new vector.\n\nThe outputs are generated from the inputs by using the algorithm given in section 3.1.2, with input(1..I) given by the the vector passed as an argument, and the output given by y.\n\n```std::vector<double>* Test(std::vector<double>*, std::vector<double>*, std::vector<double>&)```\n\nThis is the heart of the training algorithm.The arguments are the input, the ideal output (T), and the importance to be attached to each output ($\\beta$). The result is a pointer to a new vector containing the value of e with respect to each individual weight in the network. The caller is responsible for destroying the new vector.\n\nThe order of the values in the vector returned is internal to the `FeedForwardTrilayerPerceptron` implementation, as it is dependent on the exact implementation of the training algorithm. Therefore, nothing should be assumed about the order of the values in this array.\n\nThe values of $\\frac{\\partial e}{\\partial {w}_{i,j}}$ are obtained from the equations derived in section 3.2, namely:\n\n$\\frac{\\partial e}{\\partial w}=\\left\\{\\begin{array}{cc}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right)& \\text{i=I+1}\\\\ \\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right)\\sum _{{k}^{\\prime }}^{K}{w}_{{k}^{\\prime },{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },{k}^{\\prime }}{p}_{{j}^{\\prime }}+{w}_{J+1,{k}^{\\prime }}\\right){w}_{j,{k}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{i}^{\\prime }}^{I}{w}_{{i}^{\\prime },j}{x}_{{i}^{\\prime }}+{w}_{I+1,j}\\right){x}_{i}& \\text{otherwise}\\end{array}$\n\n$\\frac{\\partial e}{\\partial {w}_{j,k}}=\\left\\{\\begin{array}{cc}\\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right)& j=J+1\\\\ \\sum _{{l}^{\\prime }}^{L}{\\beta }_{{l}^{\\prime }}\\left({y}_{{l}^{\\prime }}-{T}_{{l}^{\\prime }}\\right){w}_{k,{l}^{\\prime }}{\\mathrm{sech}}^{2}\\left(\\sum _{{j}^{\\prime }}^{J}{w}_{{j}^{\\prime },k}{p}_{{j}^{\\prime }}+{w}_{J+1,k}\\right){p}_{j}& \\text{otherwise}\\end{array}$\n\n$\\frac{\\partial e}{\\partial {w}_{k,l}}=\\left\\{\\begin{array}{cc}{\\beta }_{l}\\left({y}_{l}-{T}_{l}\\right)& k=K+1\\\\ {\\beta }_{l}\\left({y}_{l}-{T}_{l}\\right){q}_{k}& \\text{otherwise}\\end{array}$\n\n```double Grade(std::vector<double>*, std::vector<double>*, std::vector<double>&)```\n\nThis method runs the neural network using the input values passed as the first argument and then uses equation 5 given in section 3.2 to find the total error, using the second argument as the ideal output T and the third argument as the output importance $\\beta$.\n\n`void Learn(std::vector<double>*, double)`\n\nThis method changes the weights of the neural network. The first argument is a list of error contributions as calculated by `Test`. The second is a multiplying factor to apply to the error contributions. The error contributions are each multiplied by the factor and then substracted from the respective weight.\n\nFor rapid minimisation, high factors (e.g. 2) are recommended, for accurate minimisation into local minima, small factors (e.g. 0.5) are better suited.\n\nThis works by assuming that given a change in total error for a small (positive) change in a weight, the change required in the weight to minimise the error will be a proportional amount in the opposite direction. While this is a reasonable assumption given the transfer function used in this implementation, it may not be the case for other transfer functions e.g. those that are chaotic.\n\n`std::vector<double>* GetCharacteristics()`\n\nReturns a pointer to a vector containing the weights of the network, suitable for passing to the relevant constructor. The order of the values in the vector returned is internal to the `FeedForwardTrilayerPerceptron` implementation, as it is dependent on the exact implementation of the storage of the weights. Therefore, nothing should be assumed about the order of the values in this array.\n\n`void SetCharacteristics(std::vector<double>*)`\n\nSets the weights of the network. The argument is a pointer to a vector containing the weights required (including those required for hidden nodes). The vector returned by `GetCharacteristics()` is in this format.\n\n`void RandomiseWeights(double)`\n\nRandomises the weights of the network, in the range -x..x where x is the argument.\n\n`void Save(ostream&)`\n\nWrites out the size and weights of the noural network to the stream.\n\nThe training program, after reading in a set of training and testing data sets, runs each training set through the `Test()` method, and updates the weights (`Learn()`) using the average of the resulting vectors.\n\nDuring each iteration, the current set of weights is output to a file (`Save()`), along with the total score for that set of weights, found by adding together the result from `Grade()` called on each of the testing sets.\n\nThe optimal neural network (lowest score) is then used when passing the 20,000 spectra from the 2df survey through the `Run()` method.\n\n### 3.4. Training Data\n\nA neural network needs a large baseline to recognise patterns accurately. Unfortunately, there are only one or two known examples of typical DLAs in the 2df data. This is not enough to train a neural network. Instead, around 120 low-ionisation heavy element absorbers, DLA candidates suitable for higher resolution follow-up observations found by manual inspection of the highest S/N spectra , were used as positive matches. Positive matches were assigned the single ideal output 1.\n\nFor negative matches, a random cross-section of spectra was automatically selected, then screened to remove any possible DLAs in the sample (there were none). Negative matches were assigned the single ideal output -1.\n\nThe resulting set of spectra (see appendix 6) was then randomly split into a training set (80%) and a testing set (20%).\n\n### 3.5. Analysis\n\nFor every combination of hidden layer sizes tried (with one unreproducible exception), the neural network, even with its optimal weights, resulted in assigning a value close to 0.7 for every spectra. This indicates that the neural network was unable to differentiate between the samples and the other 20,000 spectra.\n\nOne possible conclusion from this is that the training data did not contain patterns, or at least any patterns were too subtle.\n\nAnother possibility that might explain the lack of a successful neural network may be the presence of multiple local minima in the weight phase space. If the phase space is \"rugged\" in this way, the likelyhood of finding a real minimum amongst the many minor minima is based almost entirely on the starting conditions (the inital values of the weights and which data sets are used for training and which are used for testing).\n\nThis hypothesis is borne out by the one exception mentioned above. During a period of stress testing, a single run was spotted outputting different numbers. Unfortunately as this was during a test run the neural network weights were not saved.\n\nThe \"rugged\" phase space may be explained by the extreme variation in the input data.\n\n### 3.6. Variations\n\nThe scoring algorithm was changed to sum the exponential of each testing network's error, in order to favour networks that are more accurate overall rather than those that have a few good results but mainly bad results. This was successful with test data but made little difference with the spectra data.\n\nThe network was initally run using the full 1024 data points of each spectra as the input. This was then reduced to 72 by using the high end of the fast fourier transformation. Various hidden layer sizes were also tried.\n\nEven with hidden layers with over 100 nodes, the network still performed well, completing training and running through the 20,000 spectra in under 30 minutes.\n\n### 3.7. Directions for future research\n\nTo skip around phase space without getting side tracked by local minima, the back-propagating training algorithm could be replaced by a genetic algorithm , using the weights as the chromosome. This would be reasonably simple to implement and should perform ever faster than the current training algorithm.\n\nAn incomplete implementation of such a system is included in CVS with the back-propagating implementation .\n\n## 4. Appendix: References\n\n 2QZ Data Archive, 10k release B. J. Boyle, S. M. Croom, R. J. Smith, T. Shanks, L. Miller, N. Loaring, http://www.2dfquasar.org/ April 2001.\n\n Lyman alpha systems and cosmology J. D. Cohn, http://astron.berkeley.edu/~jcohn/lya.html.\n\n The 2dF QSO Redshift Survey - VIII. Absorption systems in the 10k catalogue P. J. Outram, R. J. Smith, T. Shanks, B. J. Boyle, S. M. Croom, N. S. Loaring, L. Miller, arXiv:astro-ph/0107460, 24 Jul 2001.\n\n The Anglo-Australian Telescope: A Brief History R. Bell http://www.aao.gov.au/about/aathist.html.\n\n The Anglo-Australian Telescope R. Bell http://www.aao.gov.au/about/aat.html.\n\n FITSIO W. D. Pence http://heasarc.gsfc.nasa.gov/fitsio/.\n\n The 2QZ Catalogue Format B. J. Boyle, S. M. Croom, R. J. Smith, T. Shanks, L. Miller, N. Loaring, http://www.2dfquasar.org/Spec_Cat/catalogue.html.\n\n The 2dF QSO Redshift Survey - V. The 10k catalogue R. J. Smith, T. Shanks, B. J. Boyle, S. M. Croom, N. S. Loaring, L. Miller, http://www.2dfquasar.org/Papers/2QZpaperV.ps.gz.\n\n FFTW: An Adaptive Software Architecture for the FFT M. Frigo, S. G. Johnson, 1998 ICASSP conference proceedings (vol. 3, pp. 1381-1384).\n\n Searching for Damped Lymen Alpha Systems Using an Artificial Neural Network J. S. Houghton, University of Bath, May 2002.\n\n An introduction to artificial neural networks C. A. L. Bailer-Jones, R. Gupta, H. P. Singh. MNRAS, 2001.\n\n Evolving Neural Networks D. B. Fogel, L. J. Fogel, V. W. Porto, Springer, 1990 (vol. 63, pp. 487-493). \n\n Neural Network Bibliography J. Ruhland, http://liinwww.ira.uka.de/bibliography/Neural/neural.genetic.bib.gz.\n\n 2df QSO Redshift Survey Analysis: CVS I. E. Hickson, J. S. Houghton, http://sourceforge.net/cvs/?group_id=37934.\n\n### 4.1. Acknowledgements\n\nDr. G. Mathlin and J. S. Houghton were instrumental in the completion of this project.\n\n## 6. Appendix: Data Sets\n\nThe following spectra were used as positive matches.\n\n```J000259.0-312222a J000534.0-290308a J000811.6-310508a\nJ001123.8-292500a J001233.1-292718a J002832.4-271917a\nJ003142.9-292434a J003533.7-291246a J003843.9-301511a\nJ004406.3-302640a J005628.5-290104a J011102.0-284307a\nJ011720.9-295813a J012012.8-301106a J012315.6-293615a\nJ012526.7-313341a J013032.6-285017a J013356.8-292223a\nJ013659.8-294727a J014729.4-272915a J014844.9-302817a\nJ015550.0-283833a J015553.8-302650a J015647.9-283143a\nJ015929.7-310619a J021134.8-293751a J021826.9-292121a\nJ022215.6-273231a J022620.4-285751a J023212.9-291450a\nJ024824.4-310944a J025259.6-321125a J025608.9-294737a\nJ025919.2-321650a J030249.6-321600a J030324.3-300734a\nJ030647.6-302021a J030711.4-303935a J030718.5-302517a\nJ030944.7-285513a J031255.0-281020a J031309.2-280807a\nJ031426.9-301133a J095605.0-015037a J095938.2-003501a\nJ101230.1-010743a J101556.2-003506a J101636.2-023422a\nJ101742.3+013216a J102645.2-022101a J103727.9+001819a\nJ105304.0-020114a J105620.0-000852a J105811.9-023725a\nJ110603.4+002207a J110624.6-004923a J110736.6+000328a\nJ114101.3+000825a J115352.0-024609a J115559.7-015420a\nJ120455.1+002640a J120826.9-020531a J120827.0-014524a\nJ120836.2-020727a J120838.1-025712a J121318.9-010204a\nJ121957.7-012615a J122454.4-012753a J125031.6+000216a\nJ125359.6-003227a J125658.3-002123a J130019.9+002641a\nJ130433.0-013916a J130622.8-014541a J133052.4+003219a\nJ134448.0-005257a J134742.0-005831a J135941.1-002016a\nJ140224.1+003001a J140710.5-004915a J141051.2+001546a\nJ141357.8+004345a J142847.4-021827a J144715.4-014836a\nJ214726.8-291017a J214836.0-275854a J215024.0-312235a\nJ215034.6-280520a J215102.9-303642a J215222.9-283549a\nJ215342.9-301413a J215359.0-292108a J215955.4-292909a\nJ220003.0-320156a J220137.0-290743a J220208.5-292422a\nJ220214.0-293039a J220650.0-315405a J220655.3-313621a\nJ220738.4-291303a J221155.2-272427a J221445.9-312130a\nJ221546.4-273441a J222849.4-304735a J223309.9-310617a\nJ224009.4-311420a J225915.2-285458a J230214.7-312139a\nJ230829.8-285651a J230915.3-273509a J231227.4-311814a\nJ231412.7-283645a J231459.5-291146a J231933.2-292306a\nJ232023.2-301506a J232027.1-284011a J232330.4-292123a\nJ232700.2-302637a J232914.9-301339a J232942.3-302348a\nJ233940.1-312036a J234321.6-304036a J234400.8-293224a\nJ234402.5-303601a J234405.7-295533a J234527.5-311843a\nJ234550.4-313612a J234753.0-304508a J235714.9-273659a\nJ235722.1-303513a```\n\nThe following spectra were used as negative matches.\n\n```J003910.0-285435a J005248.2-312205a J011159.9-274342a\nJ011512.8-301302a J012323.1-285252a J015150.9-282639a\nJ015656.0-282335a J022759.8-283329a J024047.4-283443a\nJ024231.6-285302a J024447.3-282154b J031232.7-281311b\nJ031353.5-274633a J031404.7-280329a J095314.4-004940a\nJ100528.8-012243a J101440.3+001904a J104132.3-003513a\nJ104536.7-023933a J104727.4+001039a J111436.3-014604a\nJ112009.1-005604a J112228.6-000859a J112234.9+001750a\nJ113153.2-011543a J115222.0+002109a J115341.1-014237a\nJ115518.9-010659a J115952.8-012236a J121429.5-022338a\nJ121456.1-012033a J122512.5-005243a J122712.7-002831a\nJ124831.9-001607a J131255.2-003515a J132503.6+004505a\nJ133727.5+003717a J134511.0-022552a J134932.9-001454a\nJ135123.9-004513a J135618.2-000626a J140026.5-011311a\nJ142533.9-014116a J144456.4-021036a J144459.2-004734a\nJ144717.4-012443a J214836.0-275854a J215354.9-284526a\nJ215420.5-275544a J220336.9-293224a J220450.9-280630a\nJ220707.1-282528a J224002.2-280829a J224459.7-310819a\nJ224830.3-310543a J225640.1-280843a J225839.6-283920b\nJ225900.0-272014a J225939.0-310143a J230848.0-303807a\nJ231212.0-283711a J231225.9-311531a J231559.2-293108a\nJ231748.0-302943a J231858.2-305418a J233737.9-300214a\nJ234446.3-275540a```\n\n## 7. Appendix: Source\n\n### 7.1. `ffmp.h`\n\n```\n#include\n#include\n\n// two hidden layers\n// XXX should make this a template and abstract out the use of double\nclass FeedForwardTrilayerPerceptron {\n\npublic:\nFeedForwardTrilayerPerceptron(int aInputLength, int aLayer1Length,\nint aLayer2Length, int aOutputLength, double aMax);\n// initialise the weights randomly\nFeedForwardTrilayerPerceptron(int aInputLength, int aLayer1Length,\nint aLayer2Length, int aOutputLength,\nstd::vector* aCharacteristics);\n// initialise the weights from a list of characteristics\nFeedForwardTrilayerPerceptron(istream& aInput);\n// initialise the weights from the stream\nvirtual ~FeedForwardTrilayerPerceptron(); // forget the weights\n\nvirtual std::vector* Run(std::vector* aInput);\n// returns the output\n\nvirtual std::vector* Test(std::vector* aInput,\nstd::vector* aIdealOutput,\nstd::vector& aOutputImportance);\n// returns a list representing offsets to pass to Learn(); offsets of\n// 0 mean there's nothing to fix\n\nstd::vector* aIdealOutput,\nstd::vector& aOutputImportance);\n\nvirtual void Learn(std::vector* aOffsets, double weight);\n// uses a list of the length returned from Test() to fix the\n// perceptron's characteristics\n\nvirtual std::vector* GetCharacteristics();\n// returns a list of characteristics\nvirtual void SetCharacteristics(std::vector* aCharacteristics);\n\nvirtual void RandomiseWeights(double max);\nvirtual void Save(ostream& aOutput);\n\nprotected:\nint Allocate(int aInputs, int aLayer1, int aLayer2, int aOutputs);\nint mInputsLength;\nint mLayer1Length;\nint mLayer2Length;\nint mOutputsLength;\ndouble* mWeights_ij; // input -> first layer\ndouble* mWeights_jk; // first layer -> second layer\ndouble* mWeights_kl; // second layer -> output\n\n};]]>```\n\n### 7.2. `ffmp.cpp`\n\nNote: Three long comments have been elided as they are more accurately represented in section 3.2.\n\n```\n#include \"ffmp.h\"\n\nFeedForwardTrilayerPerceptron::FeedForwardTrilayerPerceptron(int aInputsLength, int aLayer1Length, int aLayer2Length, int aOutputsLength, double aMax) {\nthis->Allocate(aInputsLength, aLayer1Length, aLayer2Length, aOutputsLength);\nthis->RandomiseWeights(aMax);\n}\n\nFeedForwardTrilayerPerceptron::FeedForwardTrilayerPerceptron(int aInputsLength, int aLayer1Length, int aLayer2Length, int aOutputsLength, std::vector* aCharacteristics) {\nthis->Allocate(aInputsLength, aLayer1Length, aLayer2Length, aOutputsLength);\nthis->SetCharacteristics(aCharacteristics);\n}\n\nint FeedForwardTrilayerPerceptron::Allocate(int aInputsLength, int aLayer1Length, int aLayer2Length, int aOutputsLength) {\n// precondition: this must be called exactly once in the constructor\nmInputsLength = aInputsLength;\nmLayer1Length = aLayer1Length;\nmLayer2Length = aLayer2Length;\nmOutputsLength = aOutputsLength;\nmWeights_ij = new double[mLayer1Length*(mInputsLength+1)];\n// index in as i+j*(mInputsLength+1)\nmWeights_jk = new double[mLayer2Length*(mLayer1Length+1)];\n// index in as j+k*(mLayer1Length+1)\nmWeights_kl = new double[mOutputsLength*(mLayer2Length+1)];\n// index in as k+l*(mLayer2Length+1)\nreturn mLayer1Length*(mInputsLength+1) + mLayer2Length*(mLayer1Length+1) + mOutputsLength*(mLayer2Length+1);\n}\n\n#define sech(x) (1.0/cosh(x))\n\n#define THINK \\\nfor (int j = 0; j < mLayer1Length; ++j) { \\\ndouble p_j = 0.0; \\\nint jPos = j*(mInputsLength+1); \\\nfor (int i = 0; i < mInputsLength; ++i) { \\\np_j += mWeights_ij[i+jPos]*(*aInput)[i]; \\\n} \\\n/* offset node */ \\\np_j += mWeights_ij[mInputsLength+jPos]; \\\np[j] = tanh(p_j); \\\n} \\\nfor (int k = 0; k < mLayer2Length; ++k) { \\\ndouble q_k = 0.0; \\\nint kPos = k*(mLayer1Length+1); \\\nfor (int j = 0; j < mLayer1Length; ++j) { \\\nq_k += mWeights_jk[j+kPos]*p[j]; \\\n} \\\n/* offset node */ \\\nq_k += mWeights_jk[mLayer1Length+kPos]; \\\nq[k] = tanh(q_k); \\\n} \\\nfor (int l = 0; l < mOutputsLength; ++l) { \\\ndouble output_l = 0.0; \\\nint lPos = l*(mLayer2Length+1); \\\nfor (int k = 0; k < mLayer2Length; ++k) { \\\noutput_l += mWeights_kl[k+lPos]*q[k]; \\\n} \\\noutput_l += mWeights_kl[mLayer2Length+lPos]; \\\nresult->push_back(output_l); \\\n} \\\n/* end */\n\nstd::vector* FeedForwardTrilayerPerceptron::Run(std::vector* aInput) {\nassert(aInput->size() == mInputsLength);\ndouble* p = new double [mLayer1Length];\ndouble* q = new double [mLayer2Length];\nstd::vector* result = new std::vector();\nTHINK\ndelete[] p;\ndelete[] q;\nreturn result;\n}\n\ndouble FeedForwardTrilayerPerceptron::Grade(std::vector* aInput, std::vector* aIdealOutput, std::vector& aOutputImportance) {\nassert(aInput->size() == mInputsLength);\ndouble* p = new double [mLayer1Length];\ndouble* q = new double [mLayer2Length];\nstd::vector* result = new std::vector();\nTHINK\ndouble e = 0;\nfor (int l = 0; l < mOutputsLength; ++l) {\ne += aOutputImportance[l] * pow((*result)[l] - (*aIdealOutput)[l], 2);\n}\ndelete[] p;\ndelete[] q;\ndelete result;\nreturn e / 2.0;\n}\n\nstd::vector* FeedForwardTrilayerPerceptron::Test(std::vector* aInput, std::vector* aIdealOutput, std::vector& aOutputImportance) {\n// return format from this is somewhat odd.\n\n// it consists of the errors in the weights in the same format as\n// the mWeights_xy memory blocks, but in the order kl, jk, ij.\n\nassert(aInput->size() == mInputsLength);\ndouble* p = new double [mLayer1Length];\ndouble* q = new double [mLayer2Length];\nstd::vector* result = new std::vector();\nTHINK\n\n// now work out what the error in each weight is\nstd::vector* corrections = new std::vector();\n\n// cache some common terms\ndouble* lTerms = new double [mOutputsLength];\n\nfor (int l = 0; l < mOutputsLength; ++l) {\nlTerms[l] = aOutputImportance[l] * ((*result)[l] - (*aIdealOutput)[l]);\nfor (int k = 0; k < mLayer2Length; ++k) {\ncorrections->push_back(lTerms[l] * q[k]);\n}\n/* offset node */\ncorrections->push_back(lTerms[l]);\n}\n\n// cache some common terms\ndouble* kTerms = new double [mLayer2Length];\n\nfor (int k = 0; k < mLayer2Length; ++k) {\nint kPos = k*(mLayer1Length+1);\nkTerms[k] = 0;\nfor (int j = 0; j < mLayer1Length; ++j) {\nkTerms[k] += mWeights_jk[j+kPos] * p[j];\n}\nkTerms[k] = pow(sech(kTerms[k] + mWeights_jk[mLayer1Length+kPos]), 2);\ndouble value = 0;\nfor (int l = 0; l < mOutputsLength; ++l) {\nvalue += lTerms[l] * mWeights_kl[k + l*(mLayer2Length+1)];\n}\nvalue *= kTerms[k];\nfor (int j = 0; j < mLayer1Length; ++j) {\ncorrections->push_back(value * p[j]);\n}\ncorrections->push_back(value);\n}\n\nfor (int j = 0; j < mLayer1Length; ++j) {\nint jPos = j*(mInputsLength+1);\ndouble jTerm = 0;\nfor (int i = 0; i < mInputsLength; ++i) {\njTerm += mWeights_ij[i+jPos] * (*aInput)[i];\n}\njTerm = pow(sech(jTerm + mWeights_ij[mInputsLength+jPos]), 2);\n\ndouble value = 0;\nfor (int l = 0; l < mOutputsLength; l++) {\nint lPos = l*(mLayer2Length+1);\ndouble innerValue = 0;\nfor (int k = 0; k < mLayer2Length; k++) {\ninnerValue += kTerms[k] * mWeights_kl[k + lPos] * mWeights_jk[j+k*(mLayer1Length+1)];\n}\nvalue += lTerms[l] * innerValue;\n}\n\nvalue *= jTerm;\nfor (int i = 0; i < mInputsLength; ++i) {\ncorrections->push_back(value * (*aInput)[i]);\n}\ncorrections->push_back(value);\n}\n\n//for (int index = 0; index < corrections->size(); ++index) {\n// cerr << (*corrections)[index] << \" \";\n//}\n//cerr << \"\\n\";\n\ndelete[] lTerms;\ndelete[] kTerms;\ndelete[] p;\ndelete[] q;\ndelete result;\nreturn corrections;\n}\n\nvoid FeedForwardTrilayerPerceptron::Learn(std::vector* aOffsets, double weight) {\n// offsets are in the same format as returned from Test().\nint index = 0;\nfor (int l = 0; l < mOutputsLength; ++l) {\nint lPos = l*(mLayer2Length+1);\nfor (int k = 0; k <= mLayer2Length; ++k) {\nmWeights_kl[k+lPos] -= (*aOffsets)[index++] * weight;\n}\n}\nfor (int k = 0; k < mLayer2Length; ++k) {\nint kPos = k*(mLayer1Length+1);\nfor (int j = 0; j <= mLayer1Length; ++j) {\nmWeights_jk[j+kPos] -= (*aOffsets)[index++] * weight;\n}\n}\nfor (int j = 0; j < mLayer1Length; ++j) {\nint jPos = j*(mInputsLength+1);\nfor (int i = 0; i <= mInputsLength; ++i) {\nmWeights_ij[i+jPos] -= (*aOffsets)[index++] * weight;\n}\n}\n}\n\nstd::vector* FeedForwardTrilayerPerceptron::GetCharacteristics() {\nstd::vector* result = new std::vector();\nint index = 0;\nfor (int l = 0; l < mOutputsLength; ++l) {\nint lPos = l*(mOutputsLength+1);\nfor (int k = 0; k < mLayer2Length; ++k) {\nresult->push_back(mWeights_kl[k+lPos]);\n}\n}\nfor (int k = 0; k < mLayer2Length; ++k) {\nint kPos = k*(mLayer1Length+1);\nfor (int j = 0; j < mLayer1Length; ++j) {\nresult->push_back(mWeights_jk[j+kPos]);\n}\n}\nfor (int j = 0; j < mLayer1Length; ++j) {\nint jPos = j*(mInputsLength+1);\nfor (int i = 0; i < mInputsLength; ++i) {\nresult->push_back(mWeights_ij[i+jPos]);\n}\n}\nreturn result;\n}\n\nvoid FeedForwardTrilayerPerceptron::SetCharacteristics(std::vector* aCharacteristics) {\nint index = 0;\nfor (int l = 0; l < mOutputsLength; ++l) {\nint lPos = l*(mLayer2Length+1);\nfor (int k = 0; k <= mLayer2Length; ++k) {\nmWeights_kl[k+lPos] = (*aCharacteristics)[index++];\n}\n}\nfor (int k = 0; k < mLayer2Length; ++k) {\nint kPos = k*(mLayer1Length+1);\nfor (int j = 0; j <= mLayer1Length; ++j) {\nmWeights_jk[j+kPos] = (*aCharacteristics)[index++];\n}\n}\nfor (int j = 0; j < mLayer1Length; ++j) {\nint jPos = j*(mInputsLength+1);\nfor (int i = 0; i <= mInputsLength; ++i) {\nmWeights_ij[i+jPos] = (*aCharacteristics)[index++];\n}\n}\n}\n\n#define RAND(max) (((max)*rand()/(RAND_MAX+1.0))-((max)/2.0))\n\nvoid FeedForwardTrilayerPerceptron::RandomiseWeights(double max) {\n//cerr << \"weights in range \" << max << \": \";\nfor (int l = 0; l < mOutputsLength; ++l) {\nint lPos = l*(mLayer2Length+1);\nfor (int k = 0; k <= mLayer2Length; ++k) {\nmWeights_kl[k+lPos] = RAND(max);\n//cerr << mWeights_kl[k+lPos] << \" \";\n}\n}\nfor (int k = 0; k < mLayer2Length; ++k) {\nint kPos = k*(mLayer1Length+1);\nfor (int j = 0; j <= mLayer1Length; ++j) {\nmWeights_jk[j+kPos] = RAND(max);\n//cerr << mWeights_jk[j+kPos] << \" \";\n}\n}\nfor (int j = 0; j < mLayer1Length; ++j) {\nint jPos = j*(mInputsLength+1);\nfor (int i = 0; i <= mInputsLength; ++i) {\nmWeights_ij[i+jPos] = RAND(max);\n//cerr << mWeights_ij[i+jPos] << \" \";\n}\n}\n}\n\nFeedForwardTrilayerPerceptron::~FeedForwardTrilayerPerceptron() {\ndelete[] mWeights_ij;\ndelete[] mWeights_jk;\ndelete[] mWeights_kl;\n}\n\nFeedForwardTrilayerPerceptron::FeedForwardTrilayerPerceptron(istream& aInput) {\nint inputsLength, layer1Length, layer2Length, outputsLength;\naInput >> inputsLength;\naInput >> layer1Length;\naInput >> layer2Length;\naInput >> outputsLength;\nint expectedLength = this->Allocate(inputsLength, layer1Length, layer2Length, outputsLength);\nstd::vector* characteristics = new std::vector();\nfor (int index = 0; index < expectedLength; ++index) {\ndouble value;\naInput >> value;\ncharacteristics->push_back(value);\n}\nthis->SetCharacteristics(characteristics);\ndelete characteristics;\n}\n\nvoid FeedForwardTrilayerPerceptron::Save(ostream& aOutput) {\naOutput << mInputsLength << \" \";\naOutput << mLayer1Length << \" \";\naOutput << mLayer2Length << \" \";\naOutput << mOutputsLength << \" \";\nstd::vector* characteristics = this->GetCharacteristics();\nfor (int index = 0; index < characteristics->size(); ++index) {\naOutput << (*characteristics)[index] << \" \";\n}\ndelete characteristics;\n}\n]]>```" ]

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[ "# The constraints appear to have not taken effect\n\nI found problems when using IPOPT to solve my optimization problem. The result I obtained isn’t a feasible solution. Even after attempting to alter some parameters, the obtained solution remains infeasible. Hence, I am considering whether the constraints I have written are unable to convey my true intent to AMPL. The formulation of my optimization problem and the model code I have written are as follow:", null, "", null, "", null, "", null, "``````##ExampleTwo\n#set Nodes {1..2001};\n#set Dimension {1..2};\n\nparam N := 2000;\nparam M := 100;\nparam delta := 0.2;\nparam Hight := 50;\nparam Vmax := 100;\nparam Vmin := 3;\nparam amax := 5;\nparam B := 500000;\nparam N0 := 1*10^(-17);\nparam sigma2 := 1*10^(-11);\nparam P := 10;\nparam beta0_1 := 3.23461517152694*10^(-5);\nparam beta0_2 := 1.20564246274771*10^(-5);\nparam beta0_3 := 3.53583092353553*10^(-5);\nparam beta0_4 := 1.90906735010188*10^(-5);\n\nvar ita; # variable for DF relay system\nvar q {1..2,1..N+1};\nvar v {1..2,1..N+1};\nvar a {1..2,1..N};\nvar m_1 {1..N+1} >=0, <=M integer; # blocklength for user 1 integer\nvar m_2 {1..N+1} >=0, <=M integer; # blocklength for user 2 integer\nvar m_3 {1..N+1} >=0, <=M integer; # blocklength for user 3 integer\nvar m_4 {1..N+1} >=0, <=M integer; # blocklength for user 4 integer\n\nlet {i in 1..N+1} m_1[i] := 0.25*M;\nlet {i in 1..N+1} m_2[i] := 0.25*M;\nlet {i in 1..N+1} m_3[i] := 0.25*M;\nlet {i in 1..N+1} m_4[i] := 0.25*M;\nlet {i in 1..2, j in 1..N} a[i,j] := 0.00001;\nlet {j in 2..N} v[1,j] := 0.497878679656440;\nlet {j in 2..N} v[2,j] := -0.497878679656440;\nlet {j in 2..N} q[1,j] := 0+21.2132034355964*0.2+0.497878679656440*(j-1);\nlet {j in 2..N} q[1,j] := 1000-21.2132034355964*0.2-0.497878679656440*(j-1);\n\nmaximize OBJ:ita;\nsubject to DFconstraint_1: sum{i in 1..N+1} (3*B*((1/log(2))*log(1+(P*beta0_1/sigma2)/((q[1,i])^2+(q[2,i])^2+Hight^2))-sqrt(1/m_1[i])*5.6120))>=ita;\nsubject to DFconstraint_2: sum{i in 1..N+1} (B*((1/log(2))*log(1+(P*beta0_2/sigma2)/((q[1,i]-600)^2+(q[2,i]-460)^2+Hight^2))-sqrt(1/m_2[i])*5.6120)+\nB*((1/log(2))*log(1+(P*beta0_3/sigma2)/((q[1,i]-460)^2+(q[2,i]-600)^2+Hight^2))-sqrt(1/m_3[i])*5.6120)+\nB*((1/log(2))*log(1+(P*beta0_4/sigma2)/((q[1,i]-160)^2+(q[2,i]-160)^2+Hight^2))-sqrt(1/m_4[i])*5.6120))>=ita;\nsubject to BOUNDconstraint_1: sum{i in 1..N+1} (B*((1/log(2))*log(1+(P*beta0_2/sigma2)/((q[1,i]-600)^2+(q[2,i]-460)^2+Hight^2))-sqrt(1/m_2[i])*5.6120))>=ita/4;\nsubject to BOUNDconstraint_2: sum{i in 1..N+1} (B*((1/log(2))*log(1+(P*beta0_3/sigma2)/((q[1,i]-460)^2+(q[2,i]-600)^2+Hight^2))-sqrt(1/m_3[i])*5.6120))>=ita/4;\nsubject to BOUNDconstraint_3: sum{i in 1..N+1} (B*((1/log(2))*log(1+(P*beta0_4/sigma2)/((q[1,i]-160)^2+(q[2,i]-160)^2+Hight^2))-sqrt(1/m_4[i])*5.6120))>=ita/4;\nsubject to InitialVelocity_1: v [1,1] = 21.2132034355964;\nsubject to InitialVelocity_2: v [2,1] = -21.2132034355964;\nsubject to FinalVelocity_1: v [1,N+1] = 21.2132034355964;\nsubject to FinalVelocity_2: v [2,N+1] = -21.2132034355964;\nsubject to InitialLocation_1: q [1,1] = 0;\nsubject to InitialLocation_2: q [2,1] = 1000;\nsubject to FinalLocation_1: q [1,N+1] = 1000;\nsubject to FinalLocation_2: q [2,N+1] = 0;\nsubject to TaylorRelation_11 {i in 1..N}:q[1,i+1] = q[1,i]+v[1,i]*delta+a[1,i]*0.5*delta^2;\nsubject to TaylorRelation_12 {i in 1..N}:q[2,i+1] = q[2,i]+v[2,i]*delta+a[2,i]*0.5*delta^2;\nsubject to TaylorRelation_21 {i in 1..N}:v[1,i+1] = v[1,i]+a[1,i]*delta;\nsubject to TaylorRelation_22 {i in 1..N}:v[2,i+1] = v[2,i]+a[2,i]*delta;\nsubject to MaximumVelocity {i in 1..N+1}:sqrt(v[1,i]^2+v[2,i]^2) <= Vmax;\nsubject to MinimumVelocity {i in 1..N+1}:sqrt(v[1,i]^2+v[2,i]^2) >= Vmin;\nsubject to MaximumAcceleration {i in 1..N}:sqrt(a[1,i]^2+a[2,i]^2) <= amax;\nsubject to SUMlimit {i in 1..N+1}: m_1[i]+m_2[i]+m_3[i]+m_4[i] <= M;\n``````\n\nBelow are the results of the variable ‘q,’ and it is evident that it does not comply with TaylorRelation_11.\n\nI look forward to receiving some suggestions from someone; your assistance is greatly appreciated.\n\nI tried running your model using Ipopt, but the log listing showed that Ipopt never found a feasible solution. (Eventually I aborted the run.) You may have chosen different options, however. Can you post the entire Ipopt listing? Preferably copy the whole listing, then paste it into a file and upload the file to this forum.\n\nAlso I noticed that `q[1,j]` is initialized twice:\n\n``````let {j in 2..N} q[1,j] := 0+21.2132034355964*0.2+0.497878679656440*(j-1);\nlet {j in 2..N} q[1,j] := 1000-21.2132034355964*0.2-0.497878679656440*(j-1);\n``````\n\nIs the second statement supposed to be initializing `q[2,j]`?\n\nThanks for your reply. I have corrected the error related to repeated initialization, but I am still unable to obtain feasible results. Below is my code file (My apologies, I couldn’t directly upload .mlx files, so I placed it in a compressed archive along with the .mod file.). Thank you for your assistance, and I look forward to receiving further recommendations.\nAMPL_test.zip (16.0 KB)\n\nEven with your correction, Ipopt fails to make any progress on this problem. You can see this in the iteration log, which reports measures of primal infeasibility (inf_pr) and dual infeasibility (inf_du) at each iterate:\n\n``````iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls\n0 -0.0000000e+00 2.07e+01 1.46e+00 -1.0 0.00e+00 - 0.00e+00 0.00e+00 0\n1 -6.0897207e-02 2.07e+01 3.45e+01 -1.0 4.45e+02 -2.0 1.11e-03 6.08e-04f 4\n2 -2.7418726e+07 2.07e+01 3.48e+01 -1.0 2.45e+13 - 7.19e-07 1.12e-06f 1\n3 -2.7418726e+07 2.00e+01 3.50e+01 -1.0 8.16e+01 -0.7 2.76e-03 3.16e-02f 1\n4 -2.7418726e+07 1.97e+01 3.45e+01 -1.0 6.70e+01 -1.1 3.69e-02 1.63e-02f 1\n5 -2.7418726e+07 1.92e+01 3.38e+01 -1.0 3.63e+01 -0.7 4.11e-02 2.74e-02f 1\n6 -2.7418728e+07 1.70e+01 3.02e+01 -1.0 2.89e+01 -1.2 3.31e-02 1.13e-01f 1\n7 -2.7418732e+07 1.55e+01 2.75e+01 -1.0 4.75e+01 -1.7 7.67e-02 8.96e-02f 1\n``````\n\nIf the iterates are approaching an optimal solution, both of these measures will converge toward zero. But after 300 iterations, there is no sign of convergence:\n\n``````iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls\n300r-1.0198004e+10 7.83e+03 4.02e+04 6.8 7.97e+09 - 8.42e-02 8.70e-04f 1\n301 -1.1149636e+10 9.71e+03 2.19e+00 -1.0 2.18e+13 - 1.73e-03 4.37e-05f 6\n302 -1.1149636e+10 9.70e+03 1.55e+12 -1.0 3.63e+05 9.5 2.35e-01 1.42e-03h 1\n303 -1.1149636e+10 9.70e+03 1.55e+12 -1.0 3.62e+05 9.0 2.84e-01 1.42e-05h 1\n304r-1.1149636e+10 9.70e+03 1.00e+03 7.1 0.00e+00 8.5 0.00e+00 2.83e-07R 3\n305r-1.1149632e+10 9.70e+03 4.25e+04 7.1 9.38e+09 - 6.08e-02 1.12e-03f 1\n306 -1.2448545e+10 5.27e+08 2.16e+00 -1.0 1.49e+13 - 1.73e-03 8.74e-05f 5\n307 -1.2448545e+10 5.23e+08 1.33e+11 -1.0 5.49e+05 8.0 2.55e-01 2.16e-03h 1\n308 -1.2448545e+10 5.23e+08 1.33e+11 -1.0 5.48e+05 7.6 2.96e-01 2.17e-05h 1\n309r-1.2448545e+10 5.23e+08 1.00e+03 7.1 0.00e+00 7.1 0.00e+00 2.82e-07R 6\n``````\n\nAt the 300 second time limit, the solution still has a large primal infeasibility (“Constraint violation”) and dual infeasibility:\n\n``````iter objective inf_pr inf_du lg(mu) ||d|| lg(rg) alpha_du alpha_pr ls\n1250r-1.5777952e+10 3.19e+09 1.54e+03 1.0 2.42e+02 0.7 1.94e-02 1.03e-02f 1\n1251r-1.5777952e+10 3.19e+09 1.49e+03 1.0 9.06e+01 1.1 1.26e-03 1.43e-02f 1\n\nNumber of Iterations....: 1251\n(scaled) (unscaled)\nObjective...............: -1.5777952148725967e+10 -1.5777952148725967e+10\nDual infeasibility......: 1.4833298516575230e+03 1.4833298516575230e+03\nConstraint violation....: 1.8045571158670068e+07 3.1946253515459814e+09\nComplementarity.........: 4.4258234085845373e+01 4.4258234085845373e+01\nOverall NLP error.......: 1.8045571158670068e+07 3.1946253515459814e+09\n``````\n\nSo, although the final iterate is returned to AMPL, it is not a feasible solution that you can use.\n\nHere are some thoughts on what might be the problem here, and what you might do:\n\n1. If you can specify some reasonable lower and upper bounds on variables q, v, and q, that might help to prevent Ipopt from stepping into irrelevant parts of the search space.\n\n2. Since Ipopt cannot handle integrality restrictions on variables, it is ignoring “integer” in its definition of m_1; it is only requiring 0 <= m_1[i] <= M. The same is true for m_2, m_3, and m_4.\n\n3. The expression 1/m_1[i] in the objective goes to infinity as m_1[i] approaches zero. To avoid trouble that this might cause for the solver, m_1[i] should have a positive lower bound. Similar comments can be made about m_2, m_3, and m_4.\n\n4. I tried solving with Knitro, using its interior-point solver which takes an approach like the one in Ipopt. It also could not find a feasible solution. This suggests that there could be an error in your model, that it causing it to actually have no feasible solution. There is no simple and general way to deal with an error like this; you may need to study the application and the model for a while before you can understand the cause of the infeasibility. It is possible to suggest some good ways to get started, however:\n\n• As an initial troubleshooting step, it is often helpful to use AMPL’s expand command to see whether AMPL generated the constraints that you expected. By itself, expand; shows all of the constraints. If there are many of them, you can use, for example, expand C1; to expand all of the constraints that have a particular name. Also you can write, for instance, expand >listing.txt; or expand C1 >listing.txt; to send all of the output to the file listing.txt.\n\n• You should also consider using AMPL’s drop and restore commands to narrow your search for the cause of the infeasibility. If you drop some constraints and the resulting problem is still infeasible, then you don’t need to consider the dropped constraints in looking for the cause of infeasibility. By trying a series of drops, you may be able to determine that infeasibility is being caused by only a small subset of the constraints.\n\nThank you very much for your reply. After I removed a couple of constraints, I found that IPOPT could eventually converge, but solving the problem once was also very slow. Can I assume that IPOPT is not a good solution for my problem? Do you have any recommendations on solvers to use for my problem? Looking forward to your reply.\n\nIpopt is a local nonlinear solver, which is the fastest kind of solver for nonlinear problems (but is only guaranteed to find locally optimal solutions). Maybe other solvers of this kind would be faster, though it is also possible that you have a hard problem that is going to be slow for any solver.\n\nIpopt is the only free local nonlinear solver that we currently recommend. However, there are other ones that you can try out, even though they are not free, by using the free NEOS Server or one of our free trial options. These ones include CONOPT, Knitro, LOQO, MINOS, and SNOPT. They use a variety of different methods, so one of them might happen to work better than Ipopt on your problem; but the only way to tell is to run some tests and see which is best.\n\nHere are two other details to note:\n\n• Unlike the other solvers listed, Knitro tries to find integer values for variables defined as `integer` or `binary`. That will make it run longer; to make it comparable to the other solvers, you can set `option` `relax_integrality` `1;` in AMPL before solving. Also Knitro implements more than one different method — see “alg” in the list of Knitro options.\n\n• A constraint like `sqrt(v[1,i]^2+v[2,i]^2)<=Vmax` can be made into a quadratic constraint by squaring both sides. The quadratic constraint might be much easier for the solvers. And if you can write all your constraints to be linear or quadratic, then you might be able to use some of the linear-quadratic solvers such as CBC, CPLEX, Gurobi, HiGHS, Mosek, SCIP, and Xpress." ]

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[ "Logistic regression, also known as binary logit and binary logistic regression, is a particularly useful predictive modeling technique, beloved in both the machine learning and the statistics communities. It is used to predict outcomes involving two options (e.g., buy versus not buy).\n\nIn this post I explain how to interpret the standard outputs from logistic regression, focusing on those that allow us to work out whether the model is good, and how it can be improved. These outputs are pretty standard and can be extracted from all the major data science and statistics tools (R, Python, Stata, SAS, SPSS, Displayr, Q). In this post I review prediction accuracy, pseudo r-squareds, AIC, the table of coefficients, and analysis of variance.\n\n## Prediction accuracy\n\nThe most basic diagnostic of a logistic regression is predictive accuracy. To understand this we need to look at the prediction-accuracy table (also known as the classification table, hit-miss table, and confusion matrix). The table below shows the prediction-accuracy table produced by Displayr's logistic regression. At the base of the table you can see the percentage of correct predictions is 79.05%. This tells us that for the 3,522 observations (people) used in the model, the model correctly predicted whether or not somebody churned 79.05% of the time. Is this a good result? The answer depends a bit on context. In this case 79.05% is not quite as good as it might sound.\n\nStarting with the No row of the table, we can see that the there were 2,301 people who did not churn and were correctly predicted not to have churned, whereas only 274 people who did not churn were predicted to have churned. If you hover your mouse over each of the cells of the table you see additional information, which computes a percentage telling us that the model accurately predicted non-churn for 83% of those that did not churn. So far so good.\n\nNow, look at the second row. It shows us that among people who did churn, the model was only marginally more likely to predict they churned than did not churn (i.e., 483 versus 464). So, among people who did churn, the model only correctly predicts that they churned 51% of the time.\n\nIf you sum up the totals of the first row, you can see that 2,575 people did not churn. However, if you sum up the first column, you can see that the model has predicted that 2,765 people did not churn. What's going on here? As most people did not churn, the model is able to get some easy wins by defaulting to predicting that people do not churn.  There is nothing wrong with the model doing this. It is the optimal thing to do. But, it is important to keep this in mind when evaluating the accuracy of any predictive model. If the groups being predicted are not of equal size, the model can get away with just predicting people are in the larger category, so it is always important to check the accuracy separately for each of the groups being predicted (i.e., in this case, churners and non-churners).  It is for this reason that you need to be sceptical when people try and impress you with the accuracy of predictive models; when predicting a rare outcome it is easy to have a model that predicts accurately (by making it always predict against the rare outcome).\n\n## Out-of-sample prediction accuracy\n\nThe accuracy discussed above is computed based on the same data that is used to fit the model. A more thorough way of assessing prediction accuracy is to perform the calculation using data not used to create the model. This tests whether the accuracy of the model is likely to hold up when used in the \"real world\". The table below shows the prediction accuracy of the model when applied to 1,761 observations that were not used when fitting the logistic regression. The good news here is that in this case the prediction accuracy has improved a smidge to 79.1%. This is a bit of a fluke. Typically we would expect to see a lower prediction accuracy when assessed out-of-sample - often substantially lower.\n\n## R-squared and pseudo-r-squared\n\nThe footer of the table below shows that the r-squared for the model is 0.1898. This is interpreted in exactly the same way as with the r-squared in linear regression, and it tells us that this model only explains 19% of the variation in churning.\n\nAlthough the r-squared is a valid computation for logistic regression, it is not widely used as there are a variety of situations where better models can have lower r-squared statistics. A variety of pseudo r-squared statistics are used instead. The footer for this table shows one of these, McFadden's rho-squared.  Like r-squared statistics, these statistics are guaranteed to take values from 0 to 1, where a higher value indicates a better model. The reason that they are preferred over traditional r-squared is that they are guaranteed to get higher as the fit of the model improves. The disadvantage of pseudo r-squared statistics is that they are only useful when compared to other models fit to the same data set (i.e., it is not possible to say if 0.2564 is a good value for McFadden's rho-squared or not).", null, "## AIC\n\nThe Akaike information criterion (AIC) is a measure of the quality of the model and is shown at the bottom of the output above. This is one of the two best ways of comparing alternative logistic regressions (i.e., logistic regressions with different predictor variables). The way it is used is that all else being equal, the model with the lower AIC is superior. The AIC is generally better than pseudo r-squareds for comparing models, as it takes into account the complexity of the model (i.e., all else being equal, the AIC favors simpler models, whereas most pseudo r-squared statistics do not).\n\nThe AIC is also often better for comparing models than using out-of-sample predictive accuracy. Out-of-sample accuracy can be a quite insensitive and noisy metric. The AIC is less noisy because:\n\n• There is no random component in it, whereas the out-of-sample predictive accuracy is sensitive to which data points were randomly selected for the estimation and validation (out-of-sample) data.\n• It takes into account all of the probabilities. That is, when using out-of-sample predictive accuracy, both a 51% prediction and a 99% prediction have the same weight in the final calculation. By contrast, with the AIC, the 99% prediction leads to a lower AIC than the 51% prediction (i.e., the AIC takes into account the probabilities, rather than just the Yes or No prediction of the outcome variable).\n\nThe AIC is only useful for comparing relatively similar models. If comparing qualitatively different models, such as a logistic regression with a decision tree, or a very simple logistic regression with a complicated one, out-of-sample predictive accuracy is a better metric, as the AIC makes some strong assumptions regarding how to compare models, and the more different the models, the less robust these assumptions.\n\n## The table of coefficients\n\nThe table of coefficients from above has been repeated below. When making an initial check of a model it is usually most useful to look at the column called z, which shows the z-statistics. The way we read this is that the further a value is from 0, the stronger its role as a predictor. So, in this case we can see that the Tenure variable is the strongest predictor. The negative sign tells us that as tenure increases, the probability of churning decreases. We can also see that Monthly Charges is the weakest predictor, as its is closest to 0. Further, the p-value for monthly charges is greater than the traditional cutoff of 0.05 (i.e, it is not \"statistically significant\", to use the common albeit dodgy jargon). All the other predictors are \"significant\". To get a more detailed understanding of how to read this table, we need to focus on the Estimate column, which I've gone to town on in How to Interpret Logistic Regression Coefficients.", null, "## Analysis of Variance (ANOVA)\n\nWith logistic regressions involving categorical predictors, the table of coefficients can be difficult to interpret. In particular, when the model includes predictors with more than two categories, we have multiple estimates and p-values, and z-statistics. This is doubly problematic. First, it can be hard to get your head around how to interpret them. Second, sometimes some or all of the coefficients for a categorical predictor are not statistically significant, but for complicated reasons beyond the scope of this post it is possible to have none or some of the individual coefficients being significant, but for them all to be jointly significant (significant when assessed as a whole), and vice versa.\n\nThis problem is addressed by performing an analysis of variance (ANOVA) on the logistic regression. Sometimes these will be created as a separate table, as in the case of Displayr's ANOVA table, shown below. In other cases the results will be integrated into the main table of coefficients (SPSS does this with its Wald tests). Typically, these will show either the results of a likelihood-ratio (LR) test or a Wald test.\n\nThe example below confirms that all the the predictors other than Monthly Charges are significant. We can also make some broad conclusions about relative importance by looking at the LR Chisq column, but when doing so keep in mind that with this statistic (and also with the Wald statistic shown by some other products, such as SPSS Statistics), that: (1) we cannot meaningfully compute ratios, so it is not the case that Tenure is almost twice as important as Contract; and, (2) the more categories in any of the predictors, the less valid these comparisons.", null, "## Other outputs\n\nThe outputs described above are the standard outputs, and will typically lead to the identification of key problems. However, they are by no means exhaustive, and there are many other more technical outputs that can be used which can lead to conclusions not detectable in these outputs. This is one of the ugly sides of building predictive models: there is always something more that can be checked, so you never can be 100% sure if your model is as good as it can be...\n\nNow that you've improved your understanding of interpreting logistic regression outputs, start creating your own logistic regression in Displayr." ]

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[ "# Checking the dimensional consistency of equations | dimensional analysis kisembo academy", null, "in our previous video able to show how they derived quantities Relate with the fundamental quantities of\nmass length and time and who are able to derive. How area of volume velocity? Acceleration and Force how they relate to\ntheir physical quantities In our conclusion. We said that we use dimensions\nof a physical quantity also To check for the consistency of equations,\nand we also use it to derive equations in this video We shall concern ourselves with how we use\nthe dimensions of a physical quantity to check for the consistency of equations This is the same work adding me and thanks\nfor tuning in Now right in front of us. We have an equation\ns is equal to ut Plus half it is where most of you know it\nas the equation of motion Now we all know that s is equal to ut. Plus\nhalf h squared is a correct equation but we want to use dimensions with its capacity\nto check whether it is consistent now when we are checking for the consistency of equations we balance their units it means that we are\ngoing to look at the Left-hand side of the equation and We do we see how that left-hand side is related\nto the three fundamental quantities Then we also are going to look at the right-hand\nside of the equation and we see how that right hand hand of the\nequation is related to their physical quantities the basic ones Then if the to coincide if the two are the\nsame then we shall conclude and say that the equation is dimensionally consistent so Let’s look at this example. We have in on\nour left hand side. We have s which is distance So that’s why we said on the left hand side\ns is going to be equal to L it’s the length its s is distance, so The distance as far as the three basic quantities\nare concerned distance is L. Then we shall look at our right hand side The right hand side is this other part? Our duty is to find the dimensions on the\nright hand side to see if they are going to coincide with hell So all right hand side you have beauty plus\na half ay t squared So we find the dimensions of U times the dimensions\nof t plus 1/2 times the dimensions of acceleration times\nthe dimension of time squared and Then from here. We shall end up with you.\nWe know that U is velocity so velocity The dimensions the velocity are Lt to the\npoint negative 1 which is that? So we look for the dimensions of displacement\nand what are the dimensions of time? So now this placement is actually L divide\nthat by the time This is in front of me a basic quantity which\nis t? So it means that the dimensions of velocity\nare l t to the power negative 1 times time which is capital t plus 1/2 The dimensions of acceleration is l t to the\npower negative 2 acceleration is velocity Over time by definition by definition we know\nthat acceleration is the rate of change of velocity So now what are the dimensions of velocity? The damage and the Velocity are L to the power\nnegative 1 we’ve got those ones before so for velocity it is L T to the Power negative 1 divided that by\nthe dimensions of time which is t? So our answer here is going to become l t to the Power Negative 2 multiply that by time square Which is t squared and of course lT to the\npower negative 1 times t is t and that t? Cancel because this is to be the point negative\n1 this is to the power 1 so They cancel and you remain with L and then\nthat is going to be plus Lt to the Power Negative 2 times t squared\nthis t squared and t to the power negative 2 will cancel out and You made with 1/2 l. L plus 1/2 L is going\nto give us 3 over 2 times. L 3 over 2 times L this 3 over 2 is a constant\nand as far as the emissions of a disco quantity are concern that constants are dimensionless, so because constants are dimensionless\nit means that as far as the right hand is concerned you remain with L and So you realize that L at the L? They are coinciding the left hand side is\nill the right hand side is also. L. So you conclude by saying that since So we’re going to our next example. We have\nV is equal to U cubed Plus 80 Now V is equal to U cube plus 80 look at we have the left hand side and the\nright hand side so the left hand side is v V Is Velocity and the dimensions of velocity\nare LG to the power negative 1 we derived those in our previous video? Then we have those of the right-hand side Which is this which is youtubed plus 80 so\nnow we work towards finding the dimensions on the right hand side The dimensions of the right hand side we have\nu cubed The dimensions of U cubed plus the dimensions\nof acceleration plus the dead times the directions of Time t the dimensions the velocity is going to be\nallergic to the point negative 1 that’s for velocity but since it’s the path 3 so this is this\nthing is to the power 3 plus the dimensions of Oscillation which is Lt to the power negative\n2 this is acceleration times the dimensions of time which is capital t they’re then required and simplify this So this is the same as L to the power 3 Times Negative 3 times 3 is t to the Power\nnegative 3 that is plus L Energy to the point negative 1 cos t to the\npoint negative 2 times t to the Power 1 of course this t will cancel with one of those\nt’s remain is the point negative 1 and When we add the new you can’t simplify this\nfather when you in our next episodes to have that So you realize that as far as our right hand\nside is concerned. We have in just this I Do now left hand side. We are having lt to\nthe pole negative 1 now since the units of our right hand side I’m not coinciding with verse on our left-hand\nside it means that this equation is not dimensionally consistent So in our conclusion. We shall say We’ll do one more example We have V squared is good 8 U squared plus\n4 it is we are checking for the dimension of consistency of this equation So checking for the dimension of consistency\nof this equation. We have the left-hand side which is v V Squared so the dimensions of V squared are\ndefinitely l t to the Power Negative 1 that this is for\nvelocity Energy to the point negative 1 squared so\nmeaning that it’s going to be L Squared times t to the Power Negative 2 that is on the left hand side, so What about this on the right hand side so\nlet’s look for the right hand side On our right hand side. We have 8 U squared we need the dimensions of that Plus 4 times the dimensions of acceleration\ntimes the dimensions of is the distance So now we’ll go ahead and say that this is\ngoing to be 8 times Now U is velocity velocity is l t to the power\nnegative 1 This is squared plus 4 times the dimensions of acceleration\nwhich is l t to the power negative 2 times Ss. Is distance so distance is L length? So this is going to become L to the Power 2 times t to the Power Negative 2 Plus that is going to be visit 8 this is 4\ntimes L times L is L squared, so it’s L squared\ntimes t to the power negative 2 So you have Via 8 L squared – a negative 2\nand that? So this this plus that gives us 12 12 L squared t to the Power Negative 2 But as far as the emissions of physical quantities\nconcerned we know that constants are dimensionless, so this is the\nsame as L squared t to the point negative 2 so since The right hand side which is L squared t to\nthe power negative 2 is The same as L squared t to the power negative\n2 which is the left hand side it means that this equation is dimensionally consistent now we only know that as far as the equations\nof motion occurs and The third equation of motion is V squared\nis equal to U square plus 2 a s but now this is a wrong equation as Velocity V squared is equal to 8 U squared\nplus 4 s this is a wrong equation from our theory But then when we try to analyze it using the\ndimensions of a physical quantity we find that the equation is dimensionally consistent So in other words, it’s the emission Dimensionally consistent but it is having\nwrong values of 8 and for those other wrong constants now Why is this? So why is this thing consistent\nit is having wrong values now. This is consistent here it’s having wrong values because 8 & 4 are\ndimensionless Therefore this brings us to this conclusion but when we are doing Dimensional analysis, we can use it to eliminate\nwrong equations? But we cannot use it to prove the correctness\nof an equation, and why is this so? because you cannot use it to prove whether\nthe correctness of Factors or constants such as 1/2 8 4 pi et\ncie are correct you can’t use it so That’s the conclusion here. We can use the\nemissions of a physical quantity to eliminate wrong equations But then we cannot use it to prove the correctness\nof an equation Please take note of that This brings us to the end of this video. Thanks\nfor watching Otherwise for more of these videos I encourage\nyou to subscribe and share this video It’s been a no-drama chromium for example\n\n### 18 thoughts on “Checking the dimensional consistency of equations | dimensional analysis kisembo academy”\n\n•", null, "May 23, 2017 at 3:25 pm\n\n•", null, "August 25, 2017 at 11:39 am\n\nthanx\n\n•", null, "September 23, 2017 at 4:33 am\n\nthank you\n\n•", null, "January 31, 2018 at 8:54 pm\n\n•", null, "April 18, 2018 at 2:27 pm\n\nFrom which country you are?\n\n•", null, "May 25, 2018 at 12:05 pm\n\nI liked the way of teaching. Very simple and understandable.\nThanks a lot.\n\n•", null, "July 14, 2018 at 8:07 am\n\nThanks a lot\n\n•", null, "October 5, 2018 at 10:08 pm\n\nHow come 1+1÷2 =3÷2\n\n•", null, "November 10, 2018 at 6:41 pm\n\nthanks\n\n•", null, "December 25, 2018 at 9:07 pm\n\nNice teaching… please explain to me about the cube on the initial velocity… because am only familiar with v=u+at and not v=u³+at..thank u\n\n•", null, "January 15, 2019 at 3:21 am\n\nvery helpful on a topic I was not able to find many videos on, thanks!\n\n•", null, "February 4, 2019 at 10:09 pm\n\nMate, I’m currently under a Aeronautical systems degree and you’ve summed my whole 2 week learning package in 11 mins. Thank you hugely!\n\n•", null, "September 2, 2019 at 3:32 am\n\n•", null, "September 7, 2019 at 12:19 pm\n\nYou are a great teacher thank you so much\n\n•", null, "September 12, 2019 at 1:30 pm\n\nGood job\n\n•", null, "September 12, 2019 at 1:30 pm\n\nGood teaching carry on..\n\n•", null, "January 7, 2020 at 7:28 am\n•", null, "" ]

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[ "Cookie Consent by FreePrivacyPolicy.com\nSearch a number\nBaseRepresentation\nbin110011100001100001…\n…0111111011010001111\n3101120121010221202221012\n41213003002333122033\n53303101002223120\n6122455202555435\n710664631634502\noct1470302773217\n9346533852835\n10110646523535\n1142a1a094889\n121953b368b7b\n13a584434702\n1454d8ddb139\n152d28c551c5\nhex19c30bf68f\n\n110646523535 has 16 divisors (see below), whose sum is σ = 136578445920. Its totient is φ = 86017107840.\n\nThe previous prime is 110646523529. The next prime is 110646523543. The reversal of 110646523535 is 535325646011.\n\nIt is not a de Polignac number, because 110646523535 - 214 = 110646507151 is a prime.\n\nIt is a junction number, because it is equal to n+sod(n) for n = 110646523492 and 110646523501.\n\nIt is a congruent number.\n\nIt is an unprimeable number.\n\nIt is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 2160080 + ... + 2210709.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (8536152870).\n\nAlmost surely, 2110646523535 is an apocalyptic number.\n\n110646523535 is a deficient number, since it is larger than the sum of its proper divisors (25931922385).\n\n110646523535 is an equidigital number, since it uses as much as digits as its factorization.\n\n110646523535 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 4370938.\n\nThe product of its (nonzero) digits is 324000, while the sum is 41.\n\nThe spelling of 110646523535 in words is \"one hundred ten billion, six hundred forty-six million, five hundred twenty-three thousand, five hundred thirty-five\"." ]

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[ "math help!!!\n\nwhat does the scale factor bestween two similar figures tell you about the given measurments?\n\nside lengthes\nperimeters\nareas\n\nMaybe they are pfroportional\n\n1. 👍\n2. 👎\n3. 👁\n1. Why is this website so bad\n\n1. 👍\n2. 👎\n\nSimilar Questions\n\n1. L.A 8TH\n\nFor each pair of shapes, decide whether they are congruent, similar, or neither. Choose the best answer. Congruent Similar Neither Two side-side-side triangles are shown. Each side of one triangle is equivalent to the\n\n2. math\n\nWhich of the following statements is always true of similar polygons? (1 point) Corresponding angles of similar figures have the same measure. The lengths of corresponding sides form equivalent ratios. The lengths of corresponding\n\n3. Geometry\n\n#4 Which could be the scale factor of the following similar figures? (1 point) The first triangle's sides are 9,9,12 The second triangles sides are 6,6,8 A.) 2/3 B.) 3/4 C.) 1 D.) 3/2 E.) 4/3 My answers: b & c\n\n4. Math\n\nIn the scale drawing below each side is .38 cm long. Use the scale factor 1cm=6.5m. What is the perimeter of the actual object?\n\n1. geometry\n\nAre the polygons similar? If they are, write a similarity statement and give the scale factor. The figures are not drawn to scale ABCD ~ KLMN; 7.2:1.8 ABCD ~ KLMN; 3:1.8 ABCD ~ NKLM; 3:5.4 The polygons are not similar***\n\n2. Math\n\nWhat is the correct way to classify the figures shown below? A. Similar B. Congruent **** C. Neither Which figure is congruent to the figure shown?1416290567.gif A. B. C. Which of the following statements are true? A. All similar\n\n3. math\n\nplease help out 7. Figures that are Congruent or Similar just check if my answers are correct. What is the correct way to classify the figures shown below? Two triangles are shown. Triangle A B C has edge A C measuring 4 units,\n\n4. Math\n\nA rectangle is dilated using a scale factor of 6. The image is then dilated using a scale factor of 1/3. What scale factor could you use to dilate the original rectangle to get to the final rectangle. Explain.\n\n1. Scale Math HELP!!!!!!!!!!\n\nIn the scale drawing below, each side is 1.29 cm long. Use the scale factor shown. What is the perimeter of the actual object? (1 cm = 6.5 m) 12.325 m 52.125 m 41.925 m 61.675 m\n\n2. Geometry\n\nIf two similar triangles have a scale factor of 5:4, what is the ratio of their corresponding altitudes?\n\n3. geometry\n\nthe scale factor of 2 similar polygons is 2:3. The perimeter of the larger one is 150 centimeters, what is the parameter of the smaller one\n\n4. math\n\nThe ladders shown below are standing against the wall at the same angle. How high up the wall does the longer ladder go? (All measurements are in feet.) ladders 11.25 ft 12.5 ft 14 ft 14 ft 2. The pair of polygons is similar. Find" ]

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[ "# An Idiot Approach to the Derivation of an Equation of a Contingent Game Mechanics of Malaysian Mancala (Congkak)\n\nLast weekend, my little cousin taught me how to play Congkak (Malay traditional board game) or conventionally known as mancalas. It occurred to me that mancalas are straight out of numbers. So I gave it a thought to derive an equation of the mechanics of the game. As a matter of fact, it has been my hobby to try out new puzzles or games and to derive an equation or two out of them. Some were fruitful some were out of my league.\n\nAnyhow, this time it worked swimmingly that I put an extra effort by presenting it into a research-paper-kind of thing – of which I’ve attached a link redirecting to the paper at the end of this read.\n\n## Abstract\n\nThis paper is the author’s personal side project and is of no affiliation to any institution or courses. Congkak is a Malay traditional board game. It is played by which the player drops each marble into every successive hole progressing from the hole the player has chosen to start. The objective of this paper is to derive an equation determining the final nth hole with a known starting point (hole) and the number of marbles at hand. The resultant equation is impressively a mere linear equation with a dependence on a restricted domain of the function.\n\n## Modelling the Problem\n\nFor convenience, we do not attempt to express the whole aspect of the Congkak game into equations. We are more interested to work on the determination of the final hole with the knowledge of the number of marbles belonging to any arbitrary initial hole. With that, this model is of one player and the game is nothing but consists of an uninterrupted, continuous turn. To start, we assign numerical values to each hole, as shown by Figure 1.\n\nThis model creates a loop for every 15 moves. This is one of the major factors influencing the formulation of the equation. Also, we do not take into account the role of the 8th hole as the store in this model. Nevertheless, in practicality, the equation is intuitively able to determine either the starting hole or the number of marbles for the arrival to the 8th hole, in the use of the actual game.\n\n## Formulating the Equation\n\nI leave all the long, painstaking mathematical work to be later referred in the actual paper I’ve written – of which can be accessed via the link attached.\n\n## Results\n\nThe equation is both intuitively and impressively a linear function. However, it is not as sensible as there exists an extra variable whose values change depending on different intervals. In short, the Equation (3) is the practical equation determining the final position (nth hole) with a known quantity of marbles belonging to any arbitrary initial position (hole).\n\n## Appendix A: Extension\n\nIn the appendix, we show Equation (3) fails over a certain interval when we work on n as the function of the sum of the initial position and the number of marbles.\n\nAgain, the details of the appendix can be referred from the paper.\n\nIn short, the new equation, Equation (A.4), is a quadratic function that does the same – determining the final position (nth hole) from the knowledge of the number of marbles belonging to any arbitrary initial position (hole).\n\nTo conclude, the derivation of Equation (A.4) shows that we managed to obtain one single function that exclusively churns out the final position (nth hole) from only two variables, the h-naught and p, without any other unnecessary variable involved. However, the elegance achieved comes with a price, that the immediate assumptions applied to the model yield a certain degree of discrepancy through the range of the function, exclusively between the vertex of the quadratic function and the larger-end of the second previously-derived interval; apart from (ℎ0 + 𝑝) > 30." ]

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[ "#", null, "", null, "Radar Basics\n\n#### Angular Resolution", null, "R\n½Θ\nSA\n\nFigure 1: The distance SA depends on the slant-range", null, "R\n½Θ\nSA\n\nFigure 1: The distance SA depends on the slant-range\n\n## What is the radar angular resolution?\n\n#### Angular Resolution\n\nTable of Content « Angular Resolution »\n\nRadar angular resolution is the minimum distance between two equally large targets at the same range which radar is able to distinguish and separate to each other.\n\n##### Angular Resolution as Antenna Parameter\n\nThe angular resolution characteristics of radar are determined by the antenna beamwidth represented by the −3 dB angle Θ which is defined by the half-power (−3 dB) points. The half-power points of the antenna radiation pattern (i.e. the −3 dB beamwidth) are normally specified as the limits of the antenna beamwidth for the purpose of defining angular resolution; two identical targets at the same distance are, therefore, resolved in angle if they are separated by more than the antenna −3 dB beamwidth.", null, "R\n½Θ\nSA\n\nFigure 1: The distance SA depends on the slant-range\n\nAn important remark has to be made immediately: the smaller the beamwidth Θ, the higher the directivity of the radar antenna. The angular resolution as a distance between two targets calculate the following formula:", null, "", null, "(1)\n\n• Θ = antenna beamwidth (Theta)\n• SA = angular resolution as a distance between two targets\n• R = slant range aim - antenna [m]\n\nThe angular resolution of targets on an analog PPI-scope, in practical terms, is dependent on the operator being able to distinguish the two targets involved. Systems having Target-Recognition features can improve their angular resolution. Cause such systems are able to compare individual Target-Pulse-Amplitudes.\n\nIn 3D radars, a resolution in the elevation angle can also be measured. Here, the same method is applicable as in the azimuth resolution. As angle Θ is used the vertical beamwidth then.", null, "half-power\nbeamwidth Θ\nnull angle\nbeamwidth\nbetween nulls\n\nFigure 2: half-power beamwidth vs. null angle\n(half beamwidth between nulls)", null, "half-power\nbeamwidth Θ\nnull angle\nbeamwidth\nbetween nulls\n\nFigure 2: half-power beamwidth vs. null angle\n(half beamwidth between nulls)\n\n##### Angular Resolution using Lidar\n\nIn certain cases (for example, Lidar), it is easier to determine the angular resolution don't use the half-power beamwidth but optical rules. The resolution of an optical system is defined by the angular separation between two similar point-like objects, the main maximum of the image of one point-like object being within the first minimum of the image of the other object.\n\nThis definition based on radar means: The angular resolution of radar is defined as the angular distance between the first minimum of the antenna pattern (next to the main lobe) and the maximum of the main lobe (null angle or the half beamwidth between nulls)\n\nTo calculate the beamwidth you can use the relationship:", null, "", null, "(1)\n\n• λ = free-space wavelength\n• D = aperture dimension\n• K = beamwith factor\n\nAssuming a linear phase distribution, each amplitude distribution has a corresponding beamwidth factor, expressed either in radians or in Radiant. The beamwidth factor depends on the antenna type and varies from 0.89 rad (≙ 56 degrees) for an ideal reflector antenna, up to 2 rad (≙ 114 degrees). For the null angle of a synthetic aperture the beamwidth factor is 1.220 rad (≙ 70 degrees).\n\nThe null angle thus only refers to half the width of the main lobe. The half-power beamwidth and the beamwidth between nulls relate to the entire width of the main lobe (see Figure 2). Both angles (zero angle and half-power beamwidth) are thus similar in size but not equal. However, the difference in size can be neglected in practice approximately. However, when applying the formula (1), it is necessary to consider the relation of whether the angle used is only half the width of the main lobe or both halves. This confounds since the variable name «theta» is commonly used for all these angles in the literature.\n\n##### Cross Range Resolution\n\nDue to the computational postprocessing, the synthetic aperture radars (SAR’s) resolution capability has completely different contexts than that of a classical radar with a real antenna. Specifying an angle for the beamwidth is impractical because it cannot be measured but only calculated. In the SAR, a distance is measured perpendicular to the direction of the movement of the radar platform. The resolution in the direction of motion is thus orthogonal to the radar beam and the range measuring and is, therefore, called cross-range resolution. In contrast to the real aperture, the cross range resolution improves with the increase in the half-power beamwidth of the real antenna. The radar will only process echo signals from the object that is detected by the antenna during a flyby in all measurements. There are the more measurement results, the larger the half-power beamwidth of the real antenna. The more single echo signals flow into the processing, the better will be the angular resolution. Since the so-called “footprint” of the radar increases with the distance, the synthetic aperture increases too, and the angular resolution at a larger distance is better than at the close range. This compensates for the deterioration of resolution due to the larger distance. In contrast to the real aperture, the cross-range resolution of the SAR is therefore approximately constant with increasing range.\n\n##### Angular Resolution using CW Radar", null, "Figure 3: Influence of the distance on the angle resolution.", null, "Figure 3: Influence of the distance on the angle resolution.", null, "Figure 3: Influence of the distance on the angle resolution.\n\nAt short distances using a CW radar extreme accuracies are possible with very little effort. Therefore, such radars are often used for speeds gauges on the road. The angular resolution is determined according to equation (1) by the half-power beamwidth Θ of the antenna and the distance R to the object to be measured.\n\nA problem with radars is that the measured values cannot be unambiguously assigned to a given reflecting object. Depending on the kind of modulation of these FMCW radars, separation of two simultaneously reflecting objects is often not possible. If the distance to the measurement object is large, then the antenna pattern widens. If then two cars are located in the main lobe, a separation is not possible. In this case, the measurement is invalid.\n\nReference:\n\n• IEEE, “IEEE Standard Radar Definitions,” Radar Systems Panel, IEEE Aerospace and Electronics Systems Society, Report No. IEEE Std 686-2008, 2008." ]

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[ "## Wednesday, September 25, 2013\n\n### 200000 views, Ithaca, ARM and DMIPS\n\nMy blog has now just crossed 200,000 views! I am extremely happy that it’s gotten this many views and that I’ve been able to reach out to so many people from all across the world. I love writing tutorials and articles about microcontrollers, power electronics, SMPS and everything electronics! I love it even more when my writing reaches out to people from all over the world to help them learn something new or to find the solution to some problem! So, I say thank you to all my readers and followers and hope that I have been able to help you and that I can continue helping you!\n\nIf you have any topic in mind that you want me to write about, please mention that in the comments. While I cannot promise you that I will write up on it, I will certainly try.\n\nUniversity has started (if you’re wondering why the title says Ithaca, that’s because Cornell University is located in Ithaca, New York, USA) and it’s been a month. The work load is quite high, so I may take quite a while to write some new article or tutorial or even to answer to your comments. So, please be patient.\n\nOn that note, I’m doing an independent study under the supervision of Professor Bruce Land here at Cornell University. My project is to explore the different peripherals on the two microcontrollers: AT91SAM3X8E (ARM Cortex M3) and the PIC32MX250F128B (DMIPS), write up documentation on the different peripherals and then compare and contrast the two microcontrollers in terms of peripheral functionality, ease of learning, ease of programming and ease of use.\n\nSo, in the coming days, I will post quite a lot of articles or tutorials based on these two microcontrollers and their peripherals.\n\nIf you’re interested, here are the links to the datasheets of the two microcontrollers:\n\nFor now, this is it. Again, if you want me to write up on any topic, please mention that in the comments section. If you have any ideas, suggestions or feedback for my blog, please let me know in your comments! Till then, enjoy the blog and I hope it helps you in your projects and learning endeavors.\n\n## Monday, July 1, 2013\n\n### AC Power Control: Adjustable Phase Angle Control with triac using ATmega8\n\nPrinciple of Phase Angle Control\nTop - Output Voltage\nBottom - Gate Drive Signal\nImage source: Wikipedia (http://en.wikipedia.org/wiki/File:Regulated_rectifier.gif)\n\nThe photo above clearly illustrates phase angle control: output voltage controlled by the gate drive signal applied to a thyristor. What is phase angle control? That is what I'm going to talk about in this article.\n\nPhase angle control is a method of PWM applied to AC input voltages, usually the mains supply. Of course, the AC supply could be from a transformer or any other AC source, but the mains supply is the most common input – this gives the phase angle control method its greatest usefulness. It has of course become quite obvious from the title (and I’m sure most of you reading will already know this) that the purpose of phase angle control is to control or limit power to the load.\n\nThe power device used in phase angle controllers is a thyristor – mostly triacs or SCRs. (There are methods of phase controlling employing high frequency switching using a MOSFET or IGBT, but here I’ll talk about phase angle control with thyristors only). The power flow to the load is controlled by delaying the firing angle (firing time each half-cycle) to the power device.\n\nWe know that the thyristor is a latching device – when the thyristor is turned on by a gating signal and the current is higher than the holding current and the latching current, the thyristor stays on, until the current through it becomes sufficiently low (very close to zero). The thyristor turns off when current through it becomes zero, as happens at the AC mains zero crossing. This is the natural line commutation. (Another method of turning the thyristor off is by forced commutation. I won’t go into that now.) The assumption here is that the load is resistive and has little to no inductance. Of course, this is not always the case, as inductive loads are often used. However, I’ll work with this assumption for now.\n\nNow, with that covered, you should read this article first before proceeding to the rest of this article:\n\nI’ve added the circuit, code and simulation of an example later in this article. And that uses a triac as the power device. So, from now on, I’ll just refer to the triac instead of talking about a thyristor in general.\n\nSo, in phase angle control, a gate pulse is sent to the triac. This is sent at a time between one zero crossing and the next. Without the gate pulse sent to the triac, right after zero-crossing, the triac is off and no current flows through it. After a certain time, the gating signal is given to the triac and it turns on. The triac then stays on until the current through it becomes zero (natural line commutation). This is at the next zero crossing. For simplicity’s sake and as usually should be, assume that the current through the triac (when on) is larger than the latching current and the holding current. If you didn’t already know this, the latching current is the current that must pass through the triac right after it is turned on to ensure that it latches. The holding current is the current level through the triac below which the triac will turn off. So, the assumption that current through the triac is higher than the latching current and the holding current means that the triac stays on once it is fired on. It stays on until the current through it is zero.\n\nThis means that the voltage is supplied to the load for a fraction of the cycle, determined by how long the triac is on. How long the triac is on, is, in turn, determined by the delay time between the zero-crossing and the applying of the triac gating signal.\n\nSo, to sum it up, we adjust the voltage or power delivered to the load by delaying the trigger signal to the triac. One thing to remember is that, the delivered voltage and power are not linearly related to the firing phase angle.\n\nThere are two voltages here that we are concerned with – the RMS voltage and the average voltage. The RMS voltage governs the power output to resistive loads such as incandescent bulbs and resistive heaters. The average value relates to devices that function on the average voltage level. This is important because, when testing, your voltmeter will register the average voltage – and not the true RMS voltage – unless you have a “true RMS voltmeter”. Most inexpensive voltmeters are not true RMS meters but will respond to average value changes.\n\nTo clarify why power and voltage are not linearly related, let’s examine the formula relating the two.\nSo, assuming a constant resistance (be careful if you’re using incandescent lamps, since they are NOT constant resistance devices), power is directly proportional to the square of the voltage. So, if you half the voltage, the power is not halved, but is reduced to one-fourth the original power! One-fourth power with half the voltage!\n\nNow let’s now go on to the design part – how we’re actually going to do this.\nFor the microcontroller, I’ve chosen the extremely popular ATmega8. However, since this application requires only a few pins, you can easily use any other small microcontroller for this purpose, such as ATtiny13.\n\nThe zero-crossing is done using the bridge-optocoupler method as I had previously shown. For details regarding the zero-crossing, please go through the article:\n\nNow, let’s take a look at the code:\n//---------------------------------------------------------------------------------------------------------\n//Programmer: Syed Tahmid Mahbub\n//Compiler: mikroC PRO for AVR v2.10\n//Target AVR: ATmega8\n//Program for phase angle control - adjustable phase angle\n//---------------------------------------------------------------------------------------------------------\nunsigned char FlagReg;\nunsigned char Count;\nunsigned char i;\nsbit ZC at FlagReg.B0;\n\nvoid interrupt() org IVT_ADDR_INT0{\nZC = 1;\n}\n\nvoid main() {\nDDRD = 0xFB;               //PORTD all output except PD2 - INT0\nPORTD = 4;                 //Clear PORTD and enable pullup on PD2\nDDRB = 0;                  //PORTB all input\nPORTB = 0xFF;              //Enable pullups on PORTB\nISC01_bit = 1;             //External interrupt on falling edge\nISC00_bit = 0;\nINT0_bit = 1;              //Enable ext int 0\nINT1_bit = 0;              //Disable ext int 1\nSREG_I_bit = 1;            //Enable interrupts\n\nwhile (1){\nif (ZC){ //zero crossing occurred\nCount = (~PINB)+1;\n\nfor (i = 0; i < Count; i++){\ndelay_us(800);\n}\n\nPORTD.B0 = 1; //Send a pulse\ndelay_us(250);\nPORTD.B0 = 0;\n\nZC = 0;\n}\n}\n}\n\nTake a look at the circuit diagram below. PORTB is connected to a DIP-switch.\nLet me first talk about the DIP-switch. The DIP-switch I chose has 3 individual switches, each of which can take on or off position. So, the DIP-switch can have 8 possible states and we can assign a digital value to the DIP-switch states. Let's assume that switch 0 is the switch connected to PORTB0/PINB0, that switch 1 is the switch connected to PORTB1/PINB1, and that switch 2 is the switch connected to PORTB2/PINB2. So, the possible DIP-switch states with their assigned values are:\n\n0 - switch 0, switch 1 and switch 2 are all off\n1 - switch 0 on, switch 1 and switch 2 are off\n2 - switch 0 off, switch 1 on and switch 2 off\n3 - switch 0 and switch 1 on and  switch 2 off\n4 - switch 0 off, switch 1 off and switch 2 on\n5 - switch 0 on, switch 1 off and switch 2 on\n6 - switch 0 off, switch 1 and switch 2 on\n7 - switch 0, switch 1 and switch 2 are all on\n\nSo, the DIP-switch can alter the values of PORTB0, PORTB1 and PORTB2. I made all of PORTB input pins and enabled the pull-ups. So, by default, all PORTB pins are high. When one of the switches is ON or shorted, that corresponding pin is low. So, only PORTB0, PORTB1 and PORTB2 can be switched to be high or low. The other pins will be high at all times. So, in order to read the value of the DIP-switch, I first inverted each individual bit of PORTB when reading (PINB) and then added 1 to the value as I wanted one to be the minimum value. Why one the minimum value? If you see the code later, you will see that the register \"Count\" is used to determine the delay time before the triac is fired. Count will have the value of the DIP-switch state plus one. I wanted a minimum delay before the triac is fired so that there is always a delay after the zero-crossing after the triac is fired. So, the function of the DIP-switch is now quite obvious. The DIP-switch sets the time after the zero-crossing after which the triac is fired. When only PORTB0/PINB0 is low, the delay time before firing the triac is minimum; it is equal to 800µs. When PORTB0/PINB0, PORTB1/PINB1 and PORTB2/PINB2 are all low, the delay time before firing the triac is maximum; it is equal to 800µs * 8 = 6.4ms. The adjusting delay time before firing adjusts the firing angle. How the time is related to the firing angle can be seen at the bottom of this article. The formula and an example are provided.\n\nSo, the program is basically as follows:\nThe zero-crossing is first checked. After zero-crossing occurs, a small delay is present before the triac is fired. This delay is set by the value of the DIP-switch. So, the triac is fired a while after the zero-crossing occurs. The gating signal is removed 250µs after that. 250µs is enough time to ensure that the triac has turned on. Even though the gating signal is removed, the triac stays on until the next zero-crossing as it is a latching device. Now you may ask, why remove the gating signal? Just keep it on till the next zero-crossing. Well, that'd work too. The problem there would be that, there would be high switching losses of the thyristor. The gate drive resistance would dissipate immense amounts of power - all for no reason, since the triac would be on even if the signal was removed.\n\nIn my circuit, I used the internal 4MHz RC oscillator. So, make sure you set the fuse bits correctly if you're using the internal oscillator.\n\nThe rest of the code should be easy to understand and should be self-explanatory – I’ve added comments to help you understand.\n\nNow let’s take a look at my circuit setup and then the different output waveforms using this code:\n\nFig. 1 - Circuit Diagram (Click on image to enlarge)\n\nYou should choose R1 depending on the gate current requirements of the triac. It must also have a sufficiently high power dissipation rating. Usually, the instantaneous power may be very high. But since current flows through the resistor for only 250us (1/40 of a 50Hz half cycle), the average power is small enough. Usually, 2W resistors should suffice.\n\nLet’s assume we’re using a BT139-600 triac. The maximum required trigger current is 35mA. Although the typical trigger current is lower, we should consider the maximum required trigger current. This is 35mA for quadrants I, II and III. We will only be firing in quadrants I and III. So, that is ok for us – we need to consider 35mA current.\n\nIf you aren’t sure what quadrants are, here’s a short description. First take a look at this diagram:\n\nFig. 2 - Triac Triggering Quadrants\n\nIf you look back again at the diagram, you’ll see that we’re driving gate from MT2. So, we can say that, with respect to MT1, when MT2 is positive, so is the gate. With respect to MT1, when MT2 is negative, so is the gate. From the diagram above, you can see that these two cases are in quadrants I and III. This is what I meant when I mentioned that we’re driving only in quadrants I and III.\n\nThe driver in the circuit is the MOC3021. This is a random phase optically isolated triac output driver. When the LED is turned on, the triac in the MOC3021 turns on and drives the main triac in the circuit. It is a “random phase” driver meaning that it can be driven on at any time during the drive signal, as is required for phase angle control. There are other drivers that only allow drive at the zero-crossing. These cannot be used for phase angle control as phase angle control requires drive after zero-crossing. For guaranteeing that the triac is latched, the LED side of the MOC3021 must be driven with at least 15mA current. The maximum current rating for the LED is 60mA. The peak current rating for the triac is 1A. You should find that we have stayed within these limits in the design.\n\nHere’s the output waveform with 800µs delay (all PORTB/PINB pins high : DIP-switch state = 0):\nFig. 3 - Triac firing with 800µs delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nYou can clearly see that before the gate driving signal is applied, there is no output (illustrated by the flat yellow line).When the gate driving signal is applied, the triac turns on. There is an output and the triac stays on till the next zero crossing. After this again, there is no output till the next gate drive signal is applied.\n\nNow I’ll show you a few more waveforms, with other initial delays.\n\nHere, the gate is driven 1.6ms after the zero-crossing (only PORTB0/PINB0 low : DIP-switch state = 1):\nFig. 4 - Triac firing with 1.6 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 2.4ms after the zero-crossing (only PORTB1/PINB1 low : DIP-switch state = 2):\nFig. 5 - Triac firing with 2.4 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 3.2ms after the zero-crossing (PORTB0/PINB0 and PORTB1/PINB1 low : DIP-switch state = 3):\nFig. 6 - Triac firing with 3.2 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 4ms after the zero-crossing (only PORTB2/PINB2 low : DIP-switch state = 4):\nFig. 7 - Triac firing with 4 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 4.8ms after the zero-crossing (PORTB2/PINB2 and PORTB0/PINB0 low : DIP-switch state = 5):\n\nFig. 8 - Triac firing with 4.8 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 5.6ms after the zero-crossing (PORTB2/PINB2 and PORTB1/PINB1 low : DIP-switch state = 6):\nFig. 9 - Triac firing with 5.6 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nHere, the gate is driven 6.4ms after the zero-crossing (PORTB2/PINB2, PORTB1/PINB1 and PORTB0/PINB0 all low : DIP-switch state = 7):\nFig. 10 - Triac firing with 6.4 ms delay\n\nGreen: Input AC\nYellow: AC Output after phase angle control\nPink: Gate Drive signal\n\nNow, to finish things off, I’ll show you how to find the RMS value of the output voltage.\n\nWe first need to know how to relate the firing delay with firing angle. We know that one complete sine wave is 360°. That is 2π radians. We then need to know that the firing angle α = ωt, where ω = 2πf. Since, we’re working with 50Hz here, f=50Hz. Thus, ω = 100π. Just to test this relationship, let’s use t = 0.020 seconds (20ms). Thus α = 100π * 0.020 = 2π, as told before.\n\nSo, if we’re firing at a delay of 4ms, that is 4ms after the zero crossing, the firing angle α = 100 π * (4/1000) = 0.4 π (in radians obviously).\n\nThe RMS output voltage is found from the relationship:\n\nSo, if we are firing after 4ms, (α = 0.4 π), the output RMS voltage is:\n\nRemember, at the beginning, I mentioned that the voltage output is not linearly correlated with the firing angle? This is what I meant. Here, the delay is 4ms. So, the triac is on for 60% of the cycle. But the output RMS voltage is 183.2V - 83% of the input voltage. The lack of direct proportionality is evident here. The reason behind this is the shape of the AC - sinusoidal.\n\nNow, I give you the task of finding the RMS voltage for the other cases mentioned in this tutorial.\n\nIf you want to then find power, you can use the relationship P = V2/R to find the power. The assumption here is that the resistance is constant, as was assumed at the beginning of the tutorial. If the resistance is not constant, power will still vary will resistance, just not directly proportionally.\n\nHere in this article, I’ve talked about phase angle control with some background information on triacs. I’ve shown how to implement phase angle control with an AVR and also how to calculate the RMS voltage of the output. I hope I’ve been able to explain this extremely important topic to you clearly and hope that you can now successfully build your own power control circuits using phase angle control with triacs.\n\nReference Book:\nOne of the best books for understanding the theory behind phase angle control is \"POWER ELECTRONICS - CIRCUITS, DEVICES AND APPLICATIONS\" by Muhammad H. Rashid. If you want to learn more about thyristors or phase angle control, I recommend reading this book for more info.\n\n## Saturday, June 29, 2013\n\n### AC Power Control with Thyristor: Pulse Skipping using triac with PIC16F877A\n\nPulse Skipping Modulation\nGreen - Input AC\nYellow - Output AC after Pulse Skipping Modulation\nPink - Gate Drive Signal\n\nPulse skipping modulation (PSM) or cycle control or burst fire is a method of power control where whole cycles of voltage are applied to the load. Here I’ll talk about PSM involving a thyristor, specifically a triac. The triac connects the AC supply to the load for a given number of cycles and then disconnects the AC supply for another given number of cycles. It has of course become quite obvious from the title that the purpose of PSM is to control or limit power to the load.\n\nWe know that the thyristor is a latching device – when the thyristor is turned on by a gating signal and the current is higher than the holding current and the latching current, the thyristor stays on, until the current through it becomes sufficiently low (very close to zero). The thyristor turns off when current through it becomes zero, as happens at the AC mains zero crossing. This is the natural line commutation. (Another method of turning the thyristor off is by forced commutation. I won’t go into that now.) The assumption here is that the load is resistive and has little to no inductance. Of course, this is not always the case, as inductive loads are often used. However, I’ll work with this assumption for now.\n\nNow, with that covered, you should read this article first before proceeding to the rest of this article:\n\nZero crossing detection with PIC16F877A\n\nI’ve added the circuit, code and simulation of an example later in this article. And that uses a triac as the power device. So, from now on, I’ll just refer to the triac instead of talking about a thyristor in general.\n\nSo, in pulse skipping modulation (PSM) or cycle control, the load receives power for a number of cycles, during which time the triac is on, and does not receive power for another number of cycles, during which time the triac is off. When the triac is to be turned on, the gating signal to the triac is given right after zero-crossing. The triac then stays on until the current through it becomes zero (natural line commutation). This is at the next zero crossing. For simplicity’s sake and as usually should be, assume that the current through the triac (when on) is larger than the latching current and the holding current. If you didn’t already know this, the latching current is the current that must pass through the triac right after it is turned on to ensure that it latches. The holding current is the current level through the triac below which the triac will turn off. So, the assumption that current through the triac is higher than the latching current and the holding current means that the triac stays on once it is fired on. It stays on until the current through it is zero.\n\nLet’s say that we are going to control over 5 complete cycles, that is, 10 half-cycles. My AC input voltage is 220V RMS, is sinusoidal in shape and has 50Hz frequency.\n\nSo, the principle is that we are going to supply voltage to the load for a certain number of cycles. Let’s say we’ll supply voltage for 3 cycles and will shut supply to the load off for 2 cycles. This is a duty cycle of 60%. The duty cycle is defined as the ratio (expressed in percentage) of the number of on cycles to the total number of cycles we are controlling.\n\nThe output RMS voltage is related to the supply voltage by the relationship:\nwhere   Vout = Output RMS voltage\nVs = Supply RMS Voltage\nk = duty cycle\n\nThe output is related to the voltage by the relationship P = V2/R. So, assuming a constant resistance, power is directly proportional to the square of the voltage. So, if you half the voltage, the power is not halved, but is reduced to one-fourth the original power! One-fourth power with half the voltage! But the general idea is that, by controlling the output voltage, you can control the output power.\n\nSo, using the previous values, where we have a duty cycle of 60% and a 220V RMS input voltage, the RMS output voltage is:\nNow let’s consider the ratio of the powers. P = V2/R. Therefore, by using pulse skipping modulation, the power changed by changing the voltage is related by:\n\nThis means that if you had a resistance R as load, and if the power was P when the voltage supplied was Vs, using PSM with a duty cycle of 0.6 decreased the power to 0.6P – using PSM we changed the output voltage and thus output power.\n\nNow let’s now go on to the design part – how we’re actually going to do this.\n\nFor the microcontroller, I’ve chosen the extremely popular PIC 16F877A. However, since this application requires only a few pins, you can easily use any other small microcontroller for this purpose, such as PIC 12F675.\n\nThe zero-crossing is done using the bridge-optocoupler method as I had previously shown. For details regarding the zero-crossing, please go through the article:\nZero crossing detection with PIC16F877A\n\nNow, let’s take a look at the code:\n//---------------------------------------------------------------------------------------------------------\n//Programmer: Syed Tahmid Mahbub\n//Compiler: mikroC PRO for PIC v4.60\n//Target PIC: PIC16F877A\n//Program for pulse skipping modulation or cycle control\n//---------------------------------------------------------------------------------------------------------\nunsigned char FlagReg;\nunsigned char Count;\nunsigned char TotalCount;\nunsigned char nCount;\nunsigned char nTotalCount;\nsbit ZC at FlagReg.B0;\n\nvoid interrupt(){\nif (INTCON.INTF){          //INTF flag raised, so external interrupt occured\nZC = 1;\nINTCON.INTF = 0;\n}\n}\n\nvoid main() {\nPORTB = 0;\nTRISB = 0x01;              //RB0 input for interrupt\nPORTA = 0;\nADCON1 = 7;\nTRISA = 0xFF;\nPORTD = 0;\nTRISD = 0;                 //PORTD all output\nOPTION_REG.INTEDG = 0;      //interrupt on falling edge\nINTCON.INTF = 0;           //clear interrupt flag\nINTCON.INTE = 1;           //enable external interrupt\nINTCON.GIE = 1;            //enable global interrupt\n\nCount = 6;                 //number of half-cycles to be on\nTotalCount = 10;           //total number of half-cycles control\nnCount = Count;\nnTotalCount = TotalCount;\n\nwhile (1){\nif (ZC){ //zero crossing occurred\nif (nCount > 0){\nPORTD.B0 = 1;\ndelay_us(250);\nPORTD.B0 = 0;\nnCount--;\n}\nnTotalCount--;\nif (nTotalCount == 0){ //number of required half-cycles elapsed\nnTotalCount = TotalCount;\nnCount = Count;\n}\nZC = 0;\n}\n}\n}\n\nThe total number of half-cycles I want to control is stored in the register TotalCount. Count holds the number of half-cycles I want voltage supplied to the load. So, the ratio of Count to TotalCount equals the duty cycle.\n\nEach time the triac is fired, the value of nCount is decremented, until it reaches zero. When nCount equals zero, the triac is no longer fired. At the same time, the register nTotalCount is decremented each zero-crossing. So it keeps track of the number of half-cycles elapsed. Initially nTotalCount is loaded with the value of TotalCount so that nTotalCount keeps track of the number of half-cycles we want to control. After that number of half-cycles elapses, nTotalCount is reloaded with the value of TotalCount and nCount is reloaded with the value of Count. Pulse skipping modulation starts over again.\n\nIn our code, we want to control 10 half-cycles, which means 5 cycles. We want voltage to be supplied to the load for 3 cycles and off for 2 cycles. This is done, as nCount counts down from 6 to 0 while firing the triac, firing the triac 6 consecutive half-cycles. For the next 4 half-cycles, while nTotalCount is still counting down to zero, the triac is off. So, voltage is not supplied to the load. The duty cycle is 60%.\n\nThe triac is fired right after zero-crossing occurs. One advantage of this is that very little radio frequency interference is produced. However, care must be taken that an unequal number of positive and negative half-cycles must not be present. This will introduce a DC component that can damage motors and other devices. Cycle control also cannot be used with incandescent bulbs because of high unacceptable flickering.\n\nThis problem with the DC component is avoided by ensuring that equal numbers of positive and negative half-cycles are present. This is done by ensuring that the Count and TotalCount (in the code above) are both even numbers.\n\nNow let’s take a look at my circuit setup and then the output waveform using this code:\n\nFig. 1 - Circuit Diagram (Click on image to enlarge)\n\nYou should choose R1 depending on the gate current requirements of the triac. It must also have a sufficiently high power dissipation rating. Usually, the instantaneous power may be very high. But since current flows through the resistor for only 250us (1/40 of a 50Hz half cycle), the average power is small enough. Usually, 2W resistors should suffice.\n\nLet’s assume we’re using a BT139-600 triac. The maximum required trigger current is 35mA. Although the typical trigger current is lower, we should consider the maximum required trigger current. This is 35mA for quadrants I, II and III. We will only be firing in quadrants I and III. So, that is ok for us – we need to consider 35mA current.\n\nIf you aren’t sure what quadrants are, here’s a short description. First take a look at this diagram:\n\nFig. 2 - Triac Triggering Quadrants\n\nIf you look back again at the diagram, you’ll see that we’re driving gate from MT2. So, we can say that, with respect to MT1, when MT2 is positive, so is the gate. With respect to MT1, when MT2 is negative, so is the gate. From the diagram above, you can see that these two cases are in quadrants I and III. This is what I meant when I mentioned that we’re driving only in quadrants I and III.\n\nThe driver in the circuit is the MOC3021. This is a random phase optically isolated triac driver output. When the LED is turned on, the triac in the MOC3021 turns on and drives the main triac in the circuit. It is a “random phase” driver meaning that it can be driven on at any time during the drive signal. There are other drivers that only allow drive at the zero-crossing. These can also be used for PSM as the triac is only driven at zero-crossing. For guaranteeing that the triac is latched, the LED side of the MOC3021 must be driven with at least 15mA current. The maximum current rating for the LED is 60mA. The peak current rating for the triac is 1A. You should find that we have stayed within these limits in the design.\n\nHere’s the output waveform:\n\nFig. 3 - PSM - 3 cycles on out of 5 complete cycles\n\nGreen: Input AC\nYellow: AC Output after pulse skipping modulation\nPink: Gate Drive signal\n\nYou can clearly see that out of every 5 complete cycles, the output consists of 3 complete cycles and 2 are missing – these are when the triac is off. Notice that the triac is fired 6 times in every 5 complete cycles – the triac is on for 6 half-cycles and off for the next 4 half-cycles.\n\nNow, let’s look at a few other cases.\n\nHere’s the output with 3 complete cycles on out of every 4 complete cycles:\n\nFig. 4 - PSM - 3 cycles on out of 4 complete cycles\n\nGreen: Input AC\nYellow: AC Output after pulse skipping modulation\nPink: Gate Drive signal\n\nHere’s the output with 5 complete cycles on out of every 7 complete cycles:\n\nFig. 5 - PSM - 5 cycles on out of 7 complete cycles\n\nGreen: Input AC\nYellow: AC Output after pulse skipping modulation\nPink: Gate Drive signal\n\nHere’s the output with 1 complete cycle on out of every 3 complete cycles:\n\nFig. 6 - PSM - 1 cycle on out of 3 complete cycles\n\nGreen: Input AC\nYellow: AC Output after pulse skipping modulation\nPink: Gate Drive signal\n\nHere’s the output with all complete cycles on:\n\nFig. 7 - PSM - all cycles on\n\nGreen: Input AC\nYellow: AC Output after pulse skipping modulation\nPink: Gate Drive signal\n\nHere in this article, I’ve talked about pulse skipping modulation – also known as cycle control or burst fire – with some background information on triacs. I’ve shown how to implement pulse skipping modulation / cycle control / burst fire with a PIC and also how to calculate the RMS voltage of the output. I hope I’ve been able to explain this important topic to you clearly and hope that you can now successfully build your own power control circuits using pulse skipping modulation (or cycle control or burst fire) with triacs." ]

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[ "# Only returning NaNs when trying to do a double for loop\n\n1 view (last 30 days)\nA Leglep on 5 Jul 2021\nEdited: A Leglep on 5 Jul 2021\nI have a matrix of size (361,361) named sivTotEASE.\nAt specific index i.e. sivTotEASE(m,n) I want to compute the mean (omiting NaNs) of the 8 closest neighbors of sivTotEASE(m,n) and replace the computed value at the location of sivTotEASE(m,n).\nI try to do this in a double for loop (first loop for the index corresponding to the rows and second loop for the index corresponding to the columns) but I end up with a matrix containing only NaNs. I don't even have the original values of my matrix anymore... It is just NaNs everywhere...\nWhen I try to compute it manually I do get a real value. For exampe, for index m = 1000, n = 1000 I get the value 8.0114 and not NaN.\nHere is my code:\n%Load data (see attached files for data)\n%Mean of the 8 closest neighbors for sivTotEASE(m,n)\nsivTotNeighbor = sivTotEASE;\nfor m = SICnoSIVrow\nfor n = SICnoSIVcol\nNeighborIt = nanmean(sivTotEASE(m-1:m+1,n-1:n+1),'all');\nsivTotNeighbor(m,n) = NeighborIt;\nend\nend\n%Manual test (the following line gives 8.0114)\n%test = nanmean(sivTotEASE(m(1000)-1:m(1000)+1,n(1000)-1:n(1000)+1),'all')\nCan anyone help me figure out what is the issue?\nThank you\n%Edited code:\n%Load data (see attached files for data)\n%Compute neighbor's average\nsivTotNeighbor = sivTotEASE;\nfor ii = 1:length(SICnoSIVrow)\nfor jj = 1:length(SICnoSIVcol)\nm=SICnoSIVrow(ii);\nn=SICnoSIVcol(jj);\nNeighborIt = mean(sivTotEASE(m-1:m+1,n-1:n+1),'all','omitnan');\nsivTotNeighbor(m,n) = NeighborIt;\nend\nend\n\nAmit Bhowmick on 5 Jul 2021\nuse this\nmean(sivTotEASE(m-1:m+1,n-1:n+1),\"omitNan\")\nAmit Bhowmick on 5 Jul 2021\nWelcome" ]

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[ "", null, "", null, "### Home > CCAA8 > Chapter 8 Unit 9 > Lesson CCA: 8.2.5 > Problem8-109\n\n8-109.\n\nSketch the parabola $y=2x^2+6x+4$ by using its intercepts.\n\nFind the $x$ and $y-$intercepts.\nTo find the $x-$intercepts, substitute $0$ for $y$, and solve for $x$.\nTo find the $y-$intercepts, substitute $0$ for $x$, and solve for $y$.\nTo find the vertex, find the point exactly in between the $x$-intercepts and solve for the corresponding $y$-value." ]

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SMAK/8hRgLL8S6SwvMcWDQzvascJkuopwm+szYqyA2SH3kRum89v6EE33NrjKLdwLy0Ffh2G4qUg32uVon3YtWxXrWXUEd8FCqftTH765n3cuqEC7zXUczvGyW8W5TzFrwvFmda1k/5wn0wEqelQJ7qWX/XlHC9Jr6z9hLrr0LRKws9tPhJS4FKutaTFjbUcSQcIhO48vcP7F9sZHWJhA58zshvpW/D9SoNNFAIMkRXQ27yHInWkL+ADa2LqTyGCXv+6ciz9GLs7aWfxLT3s4GIAxq8x5n2oALpQCB38X7PeXlw5bNM/2mmfdY59jz/38HjPr7BfFwVk4ejeXxG4NhHeN2XJJr/AOWJlfWOK/IO7D0v8fbv4z0Xnvlv3vNAfsf07+exh6ic+cR5Ae9jPVbYvijwbhDvMZv32jMmz0fy/FsK1P+TmZ9rCjz7VF7nm72ou7vElAfK6RGWq0/4tzL9PwJ1Au/04zH3QnDrLyRaCvkVvtvZRd7tRL7/13gOzv2l9OwGRPndXCBfuO8nipSFfbffKpBmBtNMLXKtk5gOsUTDlKYU/WmhZ2MIvbNCefqQ00BmaG3tE9Nozab2HCLoNY5G7Fp3owNp0T0wpgzFoFLYjB6Mnfn/VeYRDc6lEi0aM9GxEDZhwybcZxeoBfHbYMVT2ABZLX8bCqam/WlMPr4i+eF7Q4rkGaMbtuS76QqUWcJpxOud/HY69cfm91iS6IWedY38xgUsDuXxVd7+/VlvhrNsXmR5oSG+nedMi7EyJ/P4ZCoSqx2PyFjHE5Ry6ppb31c639P2tIirPCX4VxKtBgjMo/W1PZ/9Uzy2wrnODvRWYA6HCQEr3JbDigIWHIJGtyWxX0GPgA+U89Ysq3JRRyXGWrJZx1BA3vYyciiVsLWO8rgd03YG6vBRVODvcu6D7+MevosMFTYowntQcPw7Xt6+4xDnElrmyOsJLG8onU85dXIrJ1+2TXHzdQzzNTNG0Z1MRWwyvYAhq34sy+Ub/BbfiCnT8/jemjYy40PxHrTQQ+iqoFtoNK2PI9kQ7BtDtLDkf+6QiA806D8q4X7PsdFMDED5X83GaIFEa7uPpxxPUsAwv9O9cgZ+xgZ/R/4iNuA2ktN0yc++57pZz2BjEfIQuKMFisUjWCI7xcmDK+PZ+LrXQgO8k5Nmd8fC/j6f3ffQxE3qkw4QKkj8Jv7+kff6MJXDHzLNZVSQfNgpi4VKneuheJjPY8t5MvfPoQJkn/dwrx52eN/Dt0jYq1incc4H+X6XkbAv9JTmDsfrcEGJ5eBiJz4b0OwoE6FvN84zVgz2/UKp2I1ltAOf78tU9A/y6rDN77leHd6dym09CXGYo1TdSDKczfLYieV3GdOc79WhfRwyv5RpbZ14gG3M9Z4HzObrvJh81Xn58pXJcY6XZq8i3w6I+rSYNJ93PAgdou52xQAQ+kBgKt1icV6GIbRKFhS5DhqDtwcg/2igPsftMyVa/jXDjxgW5ZU8dnbAbbmazzWPv3B7TqIS00wLxMeOtH58wHrbtBf5X+TkwZW5bMh90niNx+fTMsJ8BLMc5aAv+CS9Bkv4PHNYlktIpo+wrp8ZOHcij83l/0nOsTbut+X8hkN+9nlej7G0xCGkE7l9Cb0IHSyTu0ggQqKPc69+m5ZoOTiGHoV5zO+kfqzLackHvM7n9g2S78I4WnpOKLXUq8OoEyfxnYEcd2G63aiItbKePM93i/7w7xm5m+lOdK5tn/XPVBiX8ZyX6alq4/UPCTwL7v8vL1+TuB+KcqhLwN77Nf6eUEKZTQ54C1EPz1JaUgw0oW/oRUlg2V5cJE2t89HH4T5q300DUPZoHBpp3TweOD6dpPftwHtKxlhLL3M7zl39TU8Bgqvwq45VWA7K6a6B5VoT2P9bx5rsSx3awfG2LA0cn0Kiv9Xb30yLKMuyWUhLb8uY+6Sc56ktMW9Qlmx/+gOB4w+R3DeR9fvdq0g8C3jfH5dxT6Q71lEGXqVC8MF+qstx5fG04wWqLaH+LCVxAkMdi1eoWL0WOOde/m7r7NveO+biLXrAzohRxEL5Wu7UK1/p2oyKwTpes4WK+ogSPJH+PBoHSnwMgULRL4Qeck03SnhseiXRzgbxMDZSxQjIRr+jEX8wcBxW0jkFnqm/Yee1XynhaG7sn0Fr3Y+E7o7xSNh+8IXesQdo2XzMs0pgOW1HC/8fZea/EjETbzl5b+jDdWwjG+dpQUAUgsf+GmhA4SlBlwC6CeBih2v1iAq+5yaSWafk+9r9et1CIqnzvrMsLbZVtCi/U+I94fL9AOsBvAD3U2Hqr9EdWQlH2u/rELVfx0PR+weQjLO08oHhzjUk5juxdci2aU1F6sPdVJifCRwL5etAyceCvOwd+yy/ZVjyCGJDtwCi8A8t0Hb+kt/w1x3FxSrcwEyJjw1SKCpiZbkNUKjRapJ8UE9fAGviSoeQYXku4wf+ai8UljQVgNmelfgTiSJJB7rsu6T8/stNaNW6VuC32OgsCxAXgv4w8c+1THc3G3jr3kMU9GllNN7AFWwwk16D9b2YhlJilCrrceiLhZ4sUDcLwbpGf+80pCdy/3SpzOp5SckPLQzFBXQ7+xMBJe0JiVzXeEfnUvF4usg9j3eIK81fBGIhIvxyqVwAq1uXMT/FWueZP8P8WgLzyxJW7OZMm6FX5EQqP4gHedF7t+uKKJZJpwxD9WFXfjdZJ13I6j/Cy9dYenf8fPllfadThw5mHZoRk2d8n2OoKEyi9wWWOUZ9wN3/fxLFZWj/uaLfCT2k9Q7nR+AT+v5s4NNO5QSp3sCPI4TFrNCVBAgGQTBnOhbs1AEue7dhKddDcDLFByL7vyw9o5mHsnFBfy2Gtu1GBeyjtDhmUukpB3EL8/y0DEJ3yyJbobIsFWioD2KjbUdVII5hCZ9tl148R2/ec7H3D+/Xj0jGu7Px372AEjhC8gFwv+bvoxL1Ce9A6/3+CtdlfP+PxRybwW/Px3HSc8hZG7/9s5xyK/ZuE166uHNQhhO8c690lA6LYwKeDHjIEIB7tqeYjGd5tku+L38W0+9PBXtujBJyNQkdVvr/UuGCAYKA1/kyMF5DxSAk9BcC+6C9fs2z8rDvssBHBFxVwPqp7qdnRV6OYkOOhV2WD3DZ9+WDfZtKSZKNACwjuPxulsi1HipTuG2voyJzjuOt+G82pMky84358Z+UvFswUaB+FPKgDFRZHk6yhJvddjesIrmfxkb9mQrlLdGH57CW4mkkzY+TBBbFXOMztEThfXrEsW7RdQOX/cR+IPRuWq7dfKcZEtmdjlLhA11hiB9AVx2i4D9EMjy1l+82UeQcxGu8QuPCkm1XgXwlWc7IF0ZOTAmktYGHs0jCwJtMj2NHSj641QW6l+5gvUM3GQJz0RXWQkLfSqlJsaEI/a8kR/+jQXAV+o7gEkRf4BdjyBxE9KCEg6T6E8v4cR0vPYOjBgJtzsddI4XXhk94FsgvJN//Xw5gZaCf7mj+XyDR+OjeAIQxu49lYPu+OyTvUrWKRZzClw4oA+scS7FURcK6SuGh2JPfQkbyoyKg/F1c5L2Ugg5aZPUSjhOwM9+JxA/Vs+WNbo6LJBri9ouYdLYb4SXvuawCcBjLaWUF6/JKWqpryzgHwai3OSQICxf90RjG+ZyTrt3xMoUwxClnW286vPplFVeLmwsQ+h+db+JNtmeH0ZvldtHVOJb8K3z+JOuntcqhPP1Qes7SZ2daRJ5ukXyA73S2Ux9QalL0Br2xkBBA9ZeYY0fzY/lpDJkDP6FLKjUAz3ujQ2YDjVX8qEfHNFZoQOACnik9I2t7a9kulfUnl7mOjXBvrldXgTKw0elLnEbYTuoyJuacTZ3ycz0WwLiYc6ZQibya/3eSfDQxJtV5lMdhrf+A+xE1vW8FnnEFSQllHJo2eRRJqU16Dvfzgbw9zXNs95Gr6CHP+3H7C95zXeeU38H94G0q1zho8Ej0CSo2/ph7G/W+eUybMc6rD1lHWdk65t7betcOKQhW6XhM8rP8uXBHDZxHb8iD/D2f+6Gc7FqgDOyshlYpvVYpSbGhCd0O8elNANzj1EIH0ipevJGU/Rx6K+okP3TMfS/Q2g8gma8ONKC9xfW0gEAMN/XhOi1lpE1Lz0AsDEeyE7Xc5+x/mL8TAoQKIjuJ2+5qfU84SpAfXTyWFu2+TkNvXaVv0Br7jSP4/6pDin3FUsfiDAUens73PUcKj2e3jf43aFmGukg+T6JEEOTtged6vsBztffxOftSJ9P0PgBwU3/CMyDWkZxPCNSHL3h1QBzP0XHSc6w3vAC7sx17rEi+YO3b2QWP8IwU6+GZS0+DW9b4P9/zBMV5by6nV+g6Cfe3KxQlo7f91a+wgt9awCoKWfbHSt9dmO8VrGUjdj01fFikGGJUS9I6hA3Kd6Uy0dYWi9lgurOR9QYns4FLBOoUvAovelb1+ZJ3PW5FTwkaW7g1f+aR80zWL/R7wmWJvkaMrf86FYGF9LZYPMWG9Bg2pldTYRlH5RPW3WtsNF1X6eUSng4XZT+Lv2OkbxMPZfme9yPBQIGzUd/HOXkBcZQy2uFJWuoXBAh1IrevlfA0txNIdgfwHSxwjkHhCc15kKLy9Eg/fw/38N1/gs/2WYcwf05FBvVkRyp9GP+Ncd8Y5vaW5GeNBG6gVwZu9XtZHkizN89JUZl9roR8WSt9Ar/FQ6lkH+5Y578LnIeI/RlUsnBea8z1URf+UKaCrFBUlNCFHzg+kMvYKMW5YGHJ3yzR0JvVXgPUHEhf7rKmdpUjH0PLuEbcilH93c8PMkFUMmaz+hLFAtbk2bJ+P7V1B5Y6ZrsupkxDQ4CaS3hmt6xPLZBuCQndXmszkqePZ+ideMuziibz3EMCxPQyFZ63A+ckaeH5i6y8SOsObtmjqBRkJD9TnY+H+Qyb0AK8xiub5hiLtNqpey4xoovqFF7ncIcMrKcDBHaHsy/pvOOQJY5vDv26OzvvAwqDndp2ZsxzQcnBzHbbsq5d6NxnP8m7631MjyF06wIfVoa3z9az2oCVPo1K7aFU6OxznMO6jzI8V9aPTH+ZyqXr3XiLRHozy+hG716/ooLgoqlIvv7A+ngg68WmrE9xAYb30usxjnVyRoF7rIkp16GiY9EVG4jQhZYSgt8QbIbpRnciQWXo9kODfZ/0nOjEupum8eNIO/mZ1wt33Q9oSaWdRnCJlD4U6kESjjseGNd4dgO8g8tpBdg5vrtpOaCBn+OlvZ3l83AZStc0elSKWZFX0QouZLV08nqjC3gNkpJ3f2Jq3qmyflBQgiSGYw9IeEz0clpoIL6DmS8ohugT/rX07IKwjeJRJDpEem9BpegR75x2PkMhFze8J6eTIBd75DGNhNEZ4/24hPfw83gTlbOJJJkEy+D2wPtZRpJHw7405tuBBXi8971cwW8t7n2jfqPvfU/nPFiIr0p+oZQQad8Xc715VC7WluF5g7W8jazvIreAgnUWyTLlKaCnsqxQJ7Zk+T7EfS0xyuIEltFeJMc3SMx/jsnXdgXydSYV03rWtWl8f3HBhVA4v0KPwhpHMYIy9XiRMprH72ZlActeoehpcWWz5Q3/3WrX0wZ7kUmiKjjC62w25NdrtVIoFJXG/KemayEo+tVCH3x0noiN/XlaCg87UigUCoVi47HQFQqFQqFQbHzQgAuFQqFQKJTQFQqFQqFQKKErFAqFQqGoCP4jwADQNvw20jA5ogAAAABJRU5ErkJggg==", null ]

[ "# How to create a hidden variable from a checkbox question using Unverified Perl\n\nI'm trying to create a race variable for quota purposes based on a question called S4 which is multiple-choice, but the code is not working. I get an error 132.\nThis is the code I'm using in a free format hidden variable:\n\n```<input name=\"Race_code\" id=\"Race_code\" type=\"hidden\" value=\"\n\n[%Begin Unverified Perl\nmy \\$Race_grp=0;\n\nif ((VALUE(\"S4_1\")==1) && (VALUE(\"S4_2\")==0) && (VALUE(\"S4_3\")==0) && (VALUE(\"S4_4\")==0)&& (VALUE(\"S4_5\")==0)&& (VALUE(\"S4_6\")==0))\n{\n\\$Race_grp=1;\n}\nelseif\n((VALUE(\"S4_1\")==0) && (VALUE(\"S4_2\")==1) && (VALUE(\"S4_3\")==0) && (VALUE(\"S4_4\")==0) && (VALUE(\"S4_5\")==0) && (VALUE(\"S4_6\")==0))\n{\n\\$Race_grp=5;\n}\nreturn \\$Race_grp;\nEnd Unverified %]\n\">```\n\nWhat is wrong with this code? Can somebody help? Thanks!\n Please only use this to answer the original question. Otherwise please use comments. Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification:", null, "To avoid this verification in future, please log in or register." ]

[ null, "https://legacy.sawtoothsoftware.com/forum/qa-plugin/q2a-captcha-antibot-master/AntiBotCaptcha.php", null ]

[ "# Kurs:Informatik I/T3\n\nZur Navigation springen Zur Suche springen\n\n## T3 - Types and Expressions\n\n1 Given the following expression:\n\nx * 4 > y * 3 | a == b\n\nWhich bracket structure corresponds to its evaluation order?\n\n ((x * 4) >( y * 3)) | (a == b) ( x * (4 > y) * 3) | (a == b) ( x * 4) >(( y * 3) |( a == b))\n\n2 Assume the following declarations:\n\nint a, b, c;\nboolean b1, b2;\nString name, vorname;\ndouble x, y;\n\nWhich types do the following expressions have:\n\n1. a+b\n2. b-x\n3. vornamen+\" \"+name\n4. a==b\n5. b1 =(x>y) && b1\n\n 1. double 2. double 3. String 4. boolean 5. boolean 1. int 2. double 3. String 4. boolean 5. boolean 1. int 2. double 3. String 4. int 5. boolean 1. int 2. int 3. String 4. boolean 5. boolean 1. int 2. double 3. String 4. boolean 5. void\n\n3 Which of the following are valid Java identifiers?\n\n heute-geht-es-los Heute.geht.es.los heute_geht_es_los heute geht es los HeuteGehtEsLos \"heute geht es los\"\n\n4 Assume the following declarations:\n\nint a, b;\nchar c;\nboolean b1, b2;\nString name, vorname;\ndouble x, y\n\nWhich of the following expressions will cause a type error?\n\n a*x+b x+b2 (a+b)/x name+b1 a+b2 name+a+vorname a+c (a+b)%x name+c x/c\n\n5 Assume again the following declarations:\n\nint a, b, c;\nboolean b1, b2;\nString name, vorname;\ndouble x, y;\n\nWhich of the following expressions is a valid boolean expression?\n\n a && b1 b1 && b2 || y>0 a>b>c !b1 != true a=b || x>=0 !a || !b2 a>b || x>=0 a>b && !x b1 && (x ==0) (name>0) || (vorname >0)" ]

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[ "How many is\nConversion between units of measurement\n\nYou can easily convert 5 square feet into acres using each unit definition:\n\nSquare feet\n0.09290304 m²\nAcres\nusacre = 10 surveychain² = 4046.8726 m²\n\nWith this information, you can calculate the quantity of acres 5 square feet is equal to.\n\n## ¿How many ac are there in 5 sq ft?\n\nIn 5 sq ft there are 0.00011478375 ac.\n\nWhich is the same to say that 5 square feet is 0.00011478375 acres.\n\nFive square feet equals to zero acres. *Approximation\n\n### ¿What is the inverse calculation between 1 acre and 5 square feet?\n\nPerforming the inverse calculation of the relationship between units, we obtain that 1 acre is 8712.0348 times 5 square feet.\n\nA acre is eight thousand seven hundred twelve times five square feet. *Approximation" ]

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[ "Zhurnal Vychislitel'noi Matematiki i Matematicheskoi Fiziki", null, "General information", null, "Latest issue", null, "Archive", null, "Impact factor", null, "Search papers", null, "Search references", null, "RSS", null, "Latest issue", null, "Current issues", null, "Archive issues", null, "What is RSS\n\n Zh. Vychisl. Mat. Mat. Fiz.: Year: Volume: Issue: Page: Find\n\n Personal entry: Login: Password: Save password Enter", null, "Forgotten password?", null, "Register\n\n Zh. Vychisl. Mat. Mat. Fiz., 2015, Volume 55, Number 2, Pages 253–266 (Mi zvmmf10155)", null, "", null, "Stability of nonstationary solutions of the generalized KdV-Burgers equation\n\nA. P. Chugainovaa, V. A. Shargatovb\n\na Steklov Mathematical Institute, Russian Academy of Sciences, ul. Gubkina 8, Moscow, 119991, Russia\nb National Research Nuclear University MEPhI (Moscow Engineering Physics Institute), Kashirskoe sh. 31, Moscow, 115409, Russia\n\nAbstract: The stability of nonstationary solutions to the Cauchy problem for a model equation with a complex nonlinearity, dispersion, and dissipation is analyzed. The equation describes the propagation of nonlinear longitudinal waves in rods. Previously, complex behavior of traveling waves was found, which can be treated as discontinuity structures in solutions of the same equation without dissipation and dispersion. As a result, the solutions of standard self-similar problems constructed as a sequence of Riemann waves and shocks with a stationary structure become multivalued. The multivaluedness of the solutions is attributed to special discontinuities caused by the large effect of dispersion in conjunction with viscosity. The stability of special discontinuities in the case of varying dispersion and dissipation parameters is analyzed numerically. The computations performed concern the stability analysis of a special discontinuity propagating through a layer with varying dispersion and dissipation parameters.\n\nKey words: generalized KdV-Burgers equation, stability of nonstationary solutions, difference solution method.\n\n Funding Agency Grant Number Russian Foundation for Basic Research", null, "13-01-12047\n\nDOI: https://doi.org/10.7868/S0044466915020076", null, "", null, "Full text: PDF file (357 kB)", null, "References: PDF file   HTML file\n\nEnglish version:\nComputational Mathematics and Mathematical Physics, 2015, 55:2, 251–263", null, "Bibliographic databases:", null, "", null, "", null, "", null, "", null, "UDC: 519.634\n\nCitation: A. P. Chugainova, V. A. Shargatov, “Stability of nonstationary solutions of the generalized KdV-Burgers equation”, Zh. Vychisl. Mat. Mat. Fiz., 55:2 (2015), 253–266; Comput. Math. Math. Phys., 55:2 (2015), 251–263", null, "Citation in format AMSBIB\n\\Bibitem{ChuSha15} \\by A.~P.~Chugainova, V.~A.~Shargatov \\paper Stability of nonstationary solutions of the generalized KdV-Burgers equation \\jour Zh. Vychisl. Mat. Mat. Fiz. \\yr 2015 \\vol 55 \\issue 2 \\pages 253--266 \\mathnet{http://mi.mathnet.ru/zvmmf10155} \\crossref{https://doi.org/10.7868/S0044466915020076} \\mathscinet{http://www.ams.org/mathscinet-getitem?mr=3317880} \\elib{https://elibrary.ru/item.asp?id=22908467} \\transl \\jour Comput. Math. Math. Phys. \\yr 2015 \\vol 55 \\issue 2 \\pages 251--263 \\crossref{https://doi.org/10.1134/S0965542515020074} \\isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000350801800010} \\elib{https://elibrary.ru/item.asp?id=24011337} \\scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-84924156525} \n\n• http://mi.mathnet.ru/eng/zvmmf10155\n• http://mi.mathnet.ru/eng/zvmmf/v55/i2/p253\n\n SHARE:", null, "", null, "", null, "", null, "", null, "", null, "Citing articles on Google Scholar: Russian citations, English citations\nRelated articles on Google Scholar: Russian articles, English articles\n\nThis publication is cited in the following articles:\n1. A. P. Chugainova, V. A. Shargatov, “Stability of discontinuity structures described by a generalized KdV–Burgers equation”, Comput. Math. Math. Phys., 56:2 (2016), 263–277", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "2. A. K. Volkov, N. A. Kudryashov, “Nonlinear waves described by a fifth-order equation derived from the Fermi–Pasta–Ulam system”, Comput. Math. Math. Phys., 56:4 (2016), 680–687", null, "", null, "", null, "", null, "", null, "", null, "3. A. G. Kulikovskii, A. P. Chugainova, V. A. Shargatov, “Uniqueness of self-similar solutions to the Riemann problem for the Hopf equation with complex nonlinearity”, Comput. Math. Math. Phys., 56:7 (2016), 1355–1362", null, "", null, "", null, "", null, "", null, "", null, "", null, "4. A. T. Il'ichev, A. P. Chugainova, “Spectral stability theory of heteroclinic solutions to the Korteweg–de Vries–Burgers equation with an arbitrary potential”, Proc. Steklov Inst. Math., 295 (2016), 148–157", null, "", null, "", null, "", null, "", null, "", null, "", null, "5. A. P. Chugainova, A. T. Il'ichev, A. G. Kulikovskii, V. A. Shargatov, “Problem of arbitrary discontinuity disintegration for the generalized Hopf equation: selection conditions for a unique solution”, IMA J. Appl. Math., 82:3 (2017), 496–525", null, "", null, "", null, "6. A. Samokhin, “Nonlinear waves in layered media: solutions of the KdV-Burgers equation”, J. Geom. Phys., 130 (2018), 33–39", null, "", null, "", null, "", null, "7. V. A. Shargatov, A. P. Chugainova, S. V. Gorkunov, S. I. Sumskoi, “Flow structure behind a shock wave in a channel with periodically arranged obstacles”, Proc. Steklov Inst. Math., 300 (2018), 206–218", null, "", null, "", null, "", null, "", null, "", null, "8. A. V. Samokhin, “Reflection and refraction of solitons by the $KdV$–Burgers equation in nonhomogeneous dissipative media”, Theoret. and Math. Phys., 197:1 (2018), 1527–1533", null, "", null, "", null, "", null, "", null, "", null, "", null, "9. You Sh., Guo B., “A Non-Homogeneous Boundary-Value Problem For the Kdv-Burgers Equation Posed on a Finite Domain”, Appl. Math. Lett., 94 (2019), 155–159", null, "", null, "", null, "", null, "", null, "•", null, "", null, "Number of views: This page: 306 Full text: 55 References: 45 First page: 26", null, "Contact us: math-net2021_12 [at] mi-ras ru", null, "Terms of Use", null, "Registration to the website", null, "Logotypes © Steklov Mathematical Institute RAS, 2021" ]

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[ "", null, "Posts by Rashi Singh\n\nAuthor Biographical Info: Not available\n\n#### Plot Polar Chart in Python using matplotlib\n\nBy Rashi Singh\n\nIn this tutorial, we will learn how to plot polar charts in Python. Matplotlib library in Python is a plotting library and it has a further extension in numerical mathematics known.... Read More\n\n#### Multiply each element of a list by a number in Python\n\nBy Rashi Singh\n\nIn this tutorial, we will learn how to multiply each element of a list by a number in Python. We can do this by two methods- By using List Comprehension By using for loop  Using L.... Read More\n\n#### Is Python interpreted or compiled?\n\nBy Rashi Singh\n\nIn this tutorial, you will learn about interpreted languages and compiled languages. Also, see if Python is interpreted or compiled programming language. What are interpreted langu.... Read More\n\n#### Python Variables Naming Rules\n\nBy Rashi Singh\n\nIn this tutorial, we would learn about Python variables – their use and naming rules. What are the variables? Variables are considered to be identifiers having a physical mem.... Read More\n\n#### Null Space and Nullity of a matrix in Python\n\nBy Rashi Singh\n\nIn this tutorial, we would learn about the null space and nullity of a matrix in Python. Linear relationships among attributes can be found with the help of concepts of Null Space .... Read More\n\n#### How to find the longest line from a text file in Python\n\nBy Rashi Singh\n\nIn this tutorial, we will learn how to find the longest line from a text file in Python with some easy examples. We would be using two ways- Using for loop. Using max() function. U.... Read More\n\n#### How to check if a string is a valid identifier or not in Python\n\nBy Rashi Singh\n\nIn this tutorial, we will learn how to check if the given string is a valid identifier or not in Python with some easy examples. We can check for the same in many ways, some of whi.... Read More" ]

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[ "MullOverThing\n\nUseful tips for everyday\n\n# What does same convolution mean?\n\n## What does same convolution mean?\n\nA same convolution is a type of convolution where the output matrix is of the same dimension as the input matrix.\n\n## What’s the difference between deconvolution and convolution?\n\nAs nouns the difference between convolution and deconvolution. is that convolution is something that is folded or twisted while deconvolution is (mathematics) the inversion of a convolution equation; does not normally have unique solution.\n\nIs deconvolution and transposed convolution same?\n\nAn actual deconvolution reverts the process of a convolution. A transposed convolution is somewhat similar because it produces the same spatial resolution a hypothetical deconvolutional layer would. However, the actual mathematical operation that’s being performed on the values is different.\n\nWhat is the difference between kernel and filter in CNN?\n\nA “Kernel” refers to a 2D array of weights. The term “filter” is for 3D structures of multiple kernels stacked together. For a 2D filter, filter is same as kernel. But for a 3D filter and most convolutions in deep learning, a filter is a collection of kernels.\n\n### Why is convolution needed?\n\nIt is the single most important technique in Digital Signal Processing. Using the strategy of impulse decomposition, systems are described by a signal called the impulse response. Convolution is important because it relates the three signals of interest: the input signal, the output signal, and the impulse response.\n\n### What is a valid convolution?\n\nA valid convolution is a type of convolution operation that does not use any padding on the input. This is in contrast to a same convolution, which pads the n×n n × n input matrix such that the output matrix is also n×n n × n . …\n\nWhat are Deconvolutional layers?\n\nA deconvolution is a mathematical operation that reverses the effect of convolution. Imagine throwing an input through a convolutional layer, and collecting the output. On the other hand, a transposed convolutional layer only reconstructs the spatial dimensions of the input.\n\nWhat is convolution and its types?\n\nConvolution is a mathematical way of combining two signals to form a third signal. It is the single most important technique in Digital Signal Processing. Convolution is important because it relates the three signals of interest: the input signal, the output signal, and the impulse response.\n\n#### What is transposed convolution?\n\nTransposed convolution is also known as Deconvolution which is not appropriate as deconvolution implies removing the effect of convolution which we are not aiming to achieve. It is also known as upsampled convolution which is intuitive to the task it is used to perform, i.e upsample the input feature map.\n\n#### What is a deconvolution layer?\n\nWhat is kernel size in CNN?\n\nDeep neural networks, more concretely convolutional neural networks (CNN), are basically a stack of layers which are defined by the action of a number of filters on the input. Those filters are usually called kernels. The kernel size here refers to the widthxheight of the filter mask.\n\nWhat is the definition of the same convolution?\n\nA same convolution is a type of convolution where the output matrix is of the same dimension as the input matrix.\n\n## What’s the difference between circular and linear convolution?\n\nN is the number of samples in h (n). For the above example, the output will have (3+5-1) = 7 samples. For the given example, circular convolution is possible only after modifying the signals via a method known as zero padding. In zero padding, zeroes are appended to the sequence that has a lesser size to make the sizes of the two sequences equal.\n\n## How to calculate the output of a convolution?\n\nFor a n× n n × n input matrix A A and a f × f f × f filter matrix F F, the output of the convolution A∗ F A ∗ F is of dimension ⌊ n+2p−f s ⌋+ 1 × ⌊ n+2p−f s ⌋+ 1 ⌊ n + 2 p − f s ⌋ + 1 × ⌊ n + 2 p − f s ⌋ + 1 where s s represents the stride length and p p represents the padding.\n\nWhy are there different types of 3D convolutions?\n\nNaturally, there are 3D convolutions. They are the generalization of the 2D convolution. Here in 3D convolution, the filter depth is smaller than the input layer depth (kernel size < channel size). As a result, the 3D filter can move in all 3-direction (height, width, channel of the image)." ]

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[ "Monday, 6 Feb 2023\n\n# Top 10 what is 20% of 750 That Will Change Your Life\n\nMục Lục\n\nBelow is information and knowledge on the topic what is 20% of 750 gather and compiled by the nhomkinhnamphat.com team. Along with other related topics like: What is 20% of 200, What is 15 of 750, Whats 20 percent of 60?, 20% of 1000, What is 30 of 750, 20% of 600, 20 of 750000, 20% of 500.", null, "0 percent of 750? 20% of 750\n\n20% of 750 is 150\n\n#### Working out 20% of 750\n\n1. Write 20% as 20/100\n2. Since, finding the fraction of a number is same as multiplying the fraction with the number, we have\n20/100 of 750 = 20/100 × 750\n3. Therefore, the answer is 150\n\nIf you are using a calculator, simply enter 20÷100×750 which will give you 150 as the answer.\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to work with percentages in your own time:\n\nMore percentage problems: 40% of 750 20% of 1500 60% of 750 20% of 2250 100% of 750 20% of 3750 140% of 750 20% of 5250 20% of what number is 750 20 is what percent of 750\n\nWhat is % of\n\n## Extra Information About what is 20% of 750 That You May Find Interested\n\nIf the information we provide above is not enough, you may find more below here.", null, "### What is 20 percent of 750?\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: What is 20 percent of 750? The answer is 150. Get stepwise instructions to work out “20% of 750”.\n\n• Matching Result: Write 20% as 20/100 · Since, finding the fraction of a number is same as multiplying the fraction with the number, we have 20/100 of 750 = 20/100 × 750 …\n\n• Intro: What is 20 percent of 750? 20% of 75020% of 750 is 150Working out 20% of 750Write 20% as 20/100Since, finding the fraction of a number is same as multiplying the fraction with the number, we have20/100 of 750 = 20/100 × 750Therefore, the answer is 150If you are using…\nXem Thêm:  Top 10 how much do bsn nurses make in texas That Will Change Your Life", null, "### What is 20 percent of 750? = 150 – Percentage Calculator\n\n• Author: percentagecal.com\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: 20 percent *750 =\n\n• Matching Result: Solution for What is 20 percent of 750: 20 percent *750 = (20:100)*750 = (20*750):100 = 15000:100 = 150. Now we have: 20 percent of 750 = 150.\n\n• Intro: What is 20 percent of 750? = 150 Solution for What is 20 percent of 750: 20 percent *750 = (20:100)*750 = (20*750):100 = 15000:100 = 150Now we have: 20 percent of 750 = 150Question: What is 20 percent of 750?Percentage solution with steps:Step 1: Our output value is 750.Step…", null, "### What is 20 percent of 750 – percentagecalculator.guru\n\n• Author: percentagecalculator.guru\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: What is 20 percent of 750? How much is 20% of 750?\n\n• Matching Result: 20 percent of 750 is 150. 3. How to calculate 20 percent of 750? Multiply 20/100 with 750 = (20/100)*750 = (20*750)/ …\n\n• Intro: Percentage Calculator: What is 20 percent of 750 – percentagecalculator.guru20 percent *750= (20/100)*750= (20*750)/100= 15000/100 = 150Now we have: 20 percent of 750 = 150Question: What is 20 percent of 750?We need to determine 20% of 750 now and the procedure explaining it as suchStep 1: In the given case…", null, "### How much is 20 percent of 750 – CoolConversion\n\n• Author: coolconversion.com\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: 20 is what percent of 750? How to work out percentages? See how by using our percentage calculator online.\n\n• Matching Result: / 100 = Part / ; % = (20 x 100) / 750 = 2.6666666666667% ; 20 is out of 750 = 20/750 x 100 = 2.6666666666667%.\n\n• Intro: How much is 20 percent of 750 Percentage Calculator Using this tool you can find any percentage in three ways. So, we think you reached us looking for answers like: 1) What is 20 percent (%) of 750? 2) 20 is what percent of 750? Or may be: How much…", null, "### What is 20 percent of 750? Calculate 20% of 750. How much?\n\n• Author: dollartimes.com\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: Use this calculator to find percentages. Just type in any box and the result will be calculated automatically.\n\n• Matching Result: 20% of 750.0 = 150.000. 20% of 750.5 = 150.100. 20% of 751.0 = 150.200. 20% of 751.5 = 150.300. 20% of 752.0 = 150.400. 20% of 752.5 = 150.500.\n\n• Intro: What is 20 percent of 750? Calculate 20% of 750. How much? Use this calculator to find percentages. Just type in any box and the result will be calculated automatically. Calculator 1: Calculate the percentage of a number. For example: 20% of 750 = 150 Calculator 2: Calculate a percentage…\nXem Thêm:  Top 10 salary for radiology tech in texas That Will Change Your Life", null, "### 20% of 750 = ? What is 20 percent of the number 750? The …\n\n• Author: percentages.calculators.ro\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: p% of A = ? Calculate percentages of numbers, online calculator. Please check the entered values. Number: no value\n\n• Matching Result: If 20% × 750 = 150 => · Divide the number 150 by the number 750… ·… And see if we get as a result: 20% …\n\n• Intro: p% of A = ? Calculate percentages of numbers, online calculator. Please check the entered values. Number: no value Menu p% of A = ? p% of ? = A ?% of A = B fractions a/b = ?% relative change p% = ? number (1 + p%) × n…", null, "### What is 20 Percent of 750? = 150 – Percentage-Calculator.net\n\n• Author: percentage-calculator.net\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: What is 20 percent of 750? | 20% of 750 =? With this Percentage Calculator, you will get the quick answer and step by step guide on how to solve your percentage problems with examples.\n\n• Matching Result: Question: What is 20 percent of 750? (what is 20% of 750) · Answer: 150 · The fastest step by step guide for calculating what is 20 percent of 750 · Sample …\n\n• Intro: Solved: What is 20 Percent of 750? = 150 | 20% of 750 = 150 Answer: 150 How to calculate 20% of 750 20 percent × 750 = (20 ÷ 100) × 750 = (20 × 750) ÷ 100 = 15000 ÷ 100 = 150 Now we have: 20% of…", null, "### 20 percent off 750 Calculator\n\n• Author: percent-off.com\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: How to calculate 20 percent-off \\$750. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \\$600\n\n• Matching Result: In this example, if you buy an item at \\$750 with 20% discount, you will pay 750 – 150 = 600 dollars. 150 is what percent off 750 dollars? Using the formula (b) …\n\n• Intro: 20 percent off 750 CalculatorHow to calculate 20 percent-off \\$750. How to figure out percentages off a price. Using this calculator you will find that the amount after the discount is \\$600. To find any discount, just use our Discount Calculator above. Using this calculator you can find the discount…", null, "### 20 percent of 750 – Percent Calculator\n\n• Author: percent.info\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: What is 20 percent of 750? Step-by-step showing you how to calculate 20 percent of 750. Detailed explanation with answer to 20% of 750.\n\n• Matching Result: Here we will show you how to calculate twenty percent of seven hundred fifty. Before we continue, note that 20 percent of 750 is the same as 20% of 750.\n\n• Intro: 20 percent of 750 (20% of 750) Here we will show you how to calculate twenty percent of seven hundred fifty. Before we continue, note that 20 percent of 750 is the same as 20% of 750. We will write it both ways throughout this tutorial to remind you that…\nXem Thêm:  Top 10 pay difference between cma and rma That Will Change Your Life", null, "### What is 20 percent of 750? – getcalc.com\n\n• Author: getcalc.com\n\n• Rating: 3⭐ (652605 rating)\n\n• Highest Rate: 5⭐\n\n• Lowest Rate: 2⭐\n\n• Sumary: 20% of 750 provides the detailed information of what is 20 percent of 750, the different real world problems and how it is being calculated mathematically.\n\n• Matching Result: 750 is the reference or base quantity, 150 is 20 percent of 750. … In offers and discount, 20 off 750 generally represents 20 percent off in \\$750. 20% off \\$750 …\n\n• Intro: 20% of 750 – getcalc.comThe below step by step work shows how to find what is 20 percent of 750. In the calculation, 20 is the relative quantity for each 100 against the base quantity 750. Some of the usage scenarios of 20% of 750 involves calculating commodity price increase…\n\nIf you have questions that need to be answered about the topic what is 20% of 750, then this section may help you solve it.\n\n### Calculating 20% of 750 If…\n\nCalculating 20% of 750 If you are using a calculator, enter 20 100 750 to get the answer, which is 150.\n\n### How do I subtract 20%?\n\nHow can I get 20% off a price?\n\n1. Take the original price.\n2. Divide the original price by 5.\n3. Alternatively, divide the original price by 100 and multiply it by 20.\n4. Subtract this new number from the original one.\n5. The number you calculated is the discounted value." ]

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[ "## Labels\n\n### 10+2 PSEB EXAM : PHYSICS SEPTEMBER SOLVED PAPER\n\nTime 1.5 hr Class XII\n\nSub Physics MM 35 September 2021\n\nNOTE: 1. Question paper has 35 multiple choice questions.\n\n2. Each question is compulsory.\n\n3. Q.1 to 11 are knowledge based questions, Q.12 to 24 are understanding based questions & Q.25 to 35 are application based questions.\n\n## Q1 The angle between the electric dipole moment (P⃗ ) and electric field (E⃗ ) due to the dipole at a point on its axial and equatorial lines respectively are\n\n• a) 0^{0} , 180^{0}\n• b) 90^{0} , 180^{0}\n• c) 180^{0} , 0^{0}\n• d) 0^{0} , 90^{0}\n\n• a) 0^{0} , 180^{0}\n\n## Q2 Unit of capacity of conductor is farad. 1 farad(F) is equal to\n\n• (a)1F = \\frac{1 coloumb}{1ampere}\n• (b) 1F = \\frac{1 ampere }{1 coloumb}\n• (c) 1F = \\frac{1 coloumb}{1 vol.t}\n• (d) 1F = \\frac{1 vol.t}{1 coloumb}\n\n• (c) 1F = \\frac{1 coloumb}{1 vol.t}\n\n## Q3 The material of wire of potentiometer is :\n\n• (a) Copper, because it has high conductivity\n• (b) Steel, because it is very hard and rust free.\n• (c) Manganin, because of high resistivity and low temperature coefficient of resistivity.\n• (d) Aluminium, because it is a good conductor and cheaper than copper\n\n• (c) Manganin, because of high resistivity and low temperature coefficient of resistivity.\n\n## Q4 A cyclotron cannot be used to accelerate the electrons because\n\n• (a) Electrons are very heavy and they cannot rotate with the frequency of oscillating electric field.\n• (b) Electrons are negative charged particles. Therefore, they do not get accelerated inside an electric field.\n• (c) Electrons are very light particles, they are accelerated at once. They go out of step with oscillating electric field and hence cannot be accelerated.\n• (d) Electrons mass changes into energy by Einstein formula E = mc2 and therefore they disappear from the system.\n\n• (c) Electrons are very light particles, they are accelerated at once. They go out of step with oscillating electric field and hence cannot be accelerated.\n\n## Q5 The angle of dip at the magnetic equator and magnetic poles respectively is\n\n• a) 0^{0} , 45^{0}\n• b) 45^{0} , 45^{0}\n• c) 90^{0} , 0^{0}\n• d) 0^{0} , 90^{0}\n\n• d) 0^{0} , 90^{0}\n\n• (a) 0 A\n• (b) 1A\n• (c) 2A\n• (d) 3A\n\n• (b) 1A\n\n## Q7 The magnetic lines of force around an infinitely long straight current carrying conductor are\n\n• (a) Concentric circles around conductor\n• (b) Parallel to length of conductor\n• (c) Perpendicular to length of conductor\n• (d) None of these\n\n• (a) Concentric circles around conductor\n\n## Q8 A metal rod of length 'l' moves with velocity 'v' along the direction of uniform magnetic field B⃗ , then the motional EMF developed across the ends of the rod is equal to\n\n• (a) e = B𝑙v\n• (b) e = 0\n• (c) e = B𝑙/v\n• (d) e = Bv/l\n\n• (a) e = B𝑙v\n\n## Q9. Eddy current may be reduced by using:\n\n• (a) thick piece of plastic\n• (b) laminated core of soft iron\n• (c) thick piece of cobalt\n• (d) thick piece of nickel\n\n• (b) laminated core of soft iron\n\n## Q10 . Effective resistance of an electric circuit to alternating current, arising from combined effect of ohmic resistance and reactance is called:\n\n• (a) Capacitance\n• (b) Tolerance\n• (c) Impedance\n• (d) Inductance\n\n• (c) Impedance\n\n## Q11 Which of the following radiations are used respectively for Lasik eye surgery and to take photograph during the conditions of fog and smoke?\n\n• (a) Visible rays, X rays\n• (b) Ultraviolet rays, Infrared rays\n• (c) Infrared Rays, Ultraviolet rays\n• (d) Microwaves, Gamma rays\n\n• (b) Ultraviolet rays, Infrared rays\n\n## Q12 The equipotential surfaces produced by single positive or negative charge are concentric spheres. The gap between these spheres as we move away from charge\n\n• (a) decreases\n• (b) remains same\n• (c) increases\n• (d) first increases and then decreases\n\n• (c) increases\n\n• (a) 4R, ρ/2\n• (b) 4R, 4 ρ\n• (c) 4R, ρ\n• (d) R/4, 2ρ\n\n• c) 4R, ρ\n\n## Q14 The sensitivity of a potentiometer can be increased by\n\n• (a) Increasing length of wire and decreasing resistance in series with the potentiometer wire\n• (b) Decreasing length of wire and decreasing resistance in series with the potentiometer wire\n• (c) Increasing length of wire and increasing resistance in series with the potentiometer wire\n• (d) Decreasing length of wire and increasing resistance in series with the potentiometer wire\n\n• (c) Increasing length of wire and increasing resistance in series with the potentiometer wire\n\n## Q15 . A compass needle does not work at magnetic poles because\n\n• (a) Compass needle is free to move in a vertical plane only\n• (b) Compass needle is free to move in a horizontal plane only\n• (c) Compass needle cannot measure earth magnetic field because it is very weak.\n• (d) Weight of compass needle becomes more at poles.\n\n• (b) Compass needle is free to move in a horizontal plane only\n\n## Q16 The resistance of voltmeter, ammeter and galvanometer in the increasing order can be written as\n\n• (a) Rvoltmeter < Rammeter < Rgalvanometer\n• (b) Rammeter < Rvoltmeter < Rgalvanometer\n• (c) Rvoltmeter < Rgalvanometer < Rammeter\n• (d) Rammeter < Rgalvanometer < Rvoltmeter\n\n• (d) Rammeter < Rgalvanometer < Rvoltmeter\n\n## Q17 The earth’s core has iron but it is not regarded as source of earth’s magnetism because:\n\n• (a) Iron does not show magnetic properties\n• (b) Temperature of earth’s core is very high and at this temperature iron does not remain ferromagnetic.\n• (c) Iron is a superconducting material\n• (d) None of these\n\n• (b) Temperature of earth’s core is very high and at this temperature iron does not remain ferromagnetic.\n\n## Q18 Magnetic field due to a current carrying conductor is zero:\n\n• (a) along a line perpendicular to the wire\n• (b) along the axial line of the wire\n• (c) along a line inclined 45° to the wire\n• (d) it can never be zero\n\n• (b) along the axial line of the wire\n\n## Q19. A lamp is connected in series with a capacitor to a high frequency AC source. Then it is connected directly to the same source which is true statement about this.\n\n• (a) Lamp will glow more brightly when the capacitor is in series with it.\n• (b) Lamp will glow more brightly on connecting it directly to the source.\n•  (c) Lamp glows with same brightness in both cases.\n• (d) Brightness of glow of bulb does not depend upon capacitive reactance.\n\n• (b) Lamp will glow more brightly on connecting it directly to the source.\n\n## Q20 . The current in series LCR circuit will be maximum when angular frequency ω is\n\n• (a) as large as possible\n• (b) equal to natural frequency of LCR circuit\n• (c) (LC)^{1/2}\n•  (d) 1/LC\n\n• (b) equal to natural frequency of LCR circuit\n\n## Q21 The current flowing from A to B is increasing as shown in the figure. The direction of the induced current in the loop is\n\n• (a) clockwise\n• (b) anticlockwise\n• (c) straight line\n• (d) no induced e.m.f. produced\n\n• (a) clockwise\n\n## Q22 Gamma rays are used in radiation therapy in order to treat cancer because\n\n• (a) Gamma rays have magical therapeutic properties.\n• (b) Gamma rays travel faster then other electromagnetic radiations.\n• (c) Gamma rays are highly energetic and can destroy cancer cells.\n• (d) None of these\n\n• (c) Gamma rays are highly energetic and can destroy cancer cells.\n\n## Q23 The equation of electric field part of an electromagnetic wave is given by 𝐄⃗ = [ 3.1 cos { 1.8 y + 5.4 × 10^{6} t } ] i(unit vector i) , the directions of oscillation of electric field, magnetic field and propagation of wave respectively are\n\n• (a) x axis, y axis, +z axis\n• (b) z axis, y axis, +x axis\n• (c) x axis, z axis, +y axis\n• (d) x axis, z axis, −y axis\n\n• (d) x axis, z axis, −y axis\n\n## Q24 The electromagnetic spectrum in the increasing order of frequency is given as Radio waves, microwaves, Infrared rays, visible rays, Ultraviolet rays, X-rays, Gamma rays. Which physical quantity is same for all these waves in vacuum?\n\n• (a) wavelength\n• (b) speed\n• (c) energy\n• (d) all of these\n\n• (b) speed\n\n• a) 9.2 F\n• b) 3.7 F\n• c) 7.2 F\n• d) 2.7F\n\n• c) 7.2 F\n\n• (a) 4\n• (b) 1/4\n• (c) 1\n• (d) 2\n\n• (b) 1/4\n\n• (a) 2:1\n• (b) 1:1\n• (c) 1:4\n• (d) 1:2\n\n• (a) 2:1\n\n## Q28 A Galvanometer of resistance 12 ohm gives full deflection for 8 mA. It can be converted into voltmeter of range 0-32 V with help of resistance of:\n\n• (a) 5988 ohm\n• (b) 3988 ohm\n• (c) 2500 ohm\n• (d) 4000 ohm\n\n• (b) 3988 ohm\n\n## Q29. In the magnetic meridian of a certain place, the horizontal component of earth’s magnetic field is 0.26 G and the dip angle is 60^{0} , the vertical component of earth’s magnetic field at that place will be (G → gauss)\n\n• (a) \\frac{0.26}{\\sqrt{3}}G\n• (b) \\sqrt{3}G\n• (c) 0.52 G\n• (d) 0.13 G\n\n• (b) \\sqrt{3}G\n\n## Q30 . Magnetic susceptibility of magnesium at 300K of temperature is 1.2 × 10^{-5} , the magnetic susceptibility of magnesium at 30K of temperature will be\n\n• (a) 1.2 × 10^{-3}\n• (b) 1.2 × 10^{-4}\n• (c) 1.2 × 10^{-5}\n• (d) 1.2 × 10^{-6}\n\n• (b) 1.2 × 10^{-4}\n\n## Q31 The smallest value of magnetic dipole moment  𝝁l  of an atom due to orbital motion of an electron in first orbit is called Bohr Magneton 𝜇B. The value of Bohr Magneton is equal to\n\n• a) 𝝁B=9.27x 10^{-24}Am^{2}\n• b) 𝝁B=1.38x 10^{-23}Am^{2}\n• c) 𝝁B=6.022x 10^{23}Am^{2}\n• d) 𝝁B=1.6x 10^{-19}Am^{2}\n\n• a) 𝝁B=9.27x 10^{-24}Am^{2}\n\n## Q32 The magnetic flux passing through a coil changes from 6×10^{-3} Wb to 7.2×10^{-3} Wb in 0.02s.Calculate the induced e.m.f.:\n\n• (a) -6×10^{-2}V\n• (b) 8×10^{-2}V\n• (c) 2×10^{-2}V\n• (d) -3×10^{-2}V\n\n• (a) -6×10^{-2}V\n\n• (a) 0.5 A\n• (b) 2.5 A\n• (c) 5 A\n• (d) 1.5 A\n\n• (b) 2.5 A\n\n• (a) 50 π H\n• (b) 100 π H\n• (c) 75 π H\n• (d) 60 π H\n\n• (b) 100 π H\n\n## Q35 Find the wavelength of electromagnetic waves of frequency 6×10^{12} Hz in free space having speed of 3 × 10^{8} m/s in free space\n\n• a) 6 x 10^{-5}\n• b) 5 x 10^{-5}\n• c) 2.5 x 10^{-5}\n• d) 1.5 x 10^{-5}\n\n• b) 5 x 10^{-5}\n\n## Q1 The definition of one atomic mass unit and its value is\n\n• (a) 1/12th the mass of an atom of oxygen-16, 1amu = 1.6 × 10^{-19} Kg\n• (b) 1/16th the mass of an atom of carbon-12, 1amu = 9.1 × 10^{-31} Kg\n• (c) 1/12th the mass of an atom of carbon-12, 1amu = 1.66 × 10^{-27}Kg\n• (d) The mass of helium nucleus, 1amu = 3.2 × 10^{-27} Kg\n\n• (c) 1/12th the mass of an atom of carbon-12, 1amu = 1.66 × 10^{-27}Kg\n\n## Q2  Which of the following quantity is vector quantity?\n\n• (a) Density\n• (b) angular velocity\n• (c) Energy\n• (d) angular frequency\n\n• (b) angular velocity\n\n## Q3  Which of the following quantities remains constant in a projectile motion?\n\n• (a) Speed of body\n• (b) acceleration acting on the body\n• (c) vertical component of velocity\n• (d) linear momentum of body\n\n• (b) acceleration acting on the body\n\n## Q4 The working of rocket is based on the principle of\n\n• (a) Newton's first law\n• (b) inertia of rest\n• (c) conservation of linear momentum\n• (d) none of these\n\n• (c) conservation of linear momentum\n\n## Q5  The component of contact force in the direction parallel to the surface of contact of two bodies and perpendicular to the surface are called\n\n• (a) normal reaction, frictional force\n• (b) frictional force, normal reaction\n• (c) centripetal force, centrifugal force\n• (d) centrifugal force, centripetal force\n\n• (b) frictional force, normal reaction\n\n## Q6 A car accelerates on a horizontal road due to the force exerted by\n\n• (a) the engine of the car\n• (b) the driver of the car\n• (c) the earth gravitational force\n• (d) the friction between tyres and road\n\n• (d) the friction between tyres and road\n\n## Q7  A body of certain mass is lifted upwards from ground to a certain height, the work done by the gravitational force and applied force respectively are\n\n• (a) positive, negative\n• (b) negative, positive\n• (c) positive, positive\n• (d) negative, negative\n\n• (b) negative, positive\n\n• (a) 0,1\n• (b) 0,0\n• (c) 1,1\n• (d) 1,0\n\n• (d) 1,0\n\n## Q9.  The center of mass of two non identical particle system\n\n• (a) always lies in between two particles and on the line joining them\n• (b) always lies at the center of two particles\n• (c) may lie outside the two particle system but on the line joining them\n• (d) always lies nearer the lighter particle\n\n• (a) always lies in between two particles and on the line joining them\n\n## Q10 Angular momentum has the same unit as\n\n• (a) Impulse × distance\n• (b) Linear momentum × time\n• (c) Work × frequency\n• (d) Power × distance\n\n• (a) Impulse × distance\n\n## Q11 In the absence of external torque acting on a rotating body, which of the following quantities can change?\n\n• (a) Angular momentum\n• (b) Angular velocity but not moment of inertia\n• (c) Moment of inertia but not angular velocity\n• (d) Both angular velocity and moment of inertia\n\n• (d) Both angular velocity and moment of inertia\n\n## Q12  The Vander Wall’s equation for an ideal gas is given by (P + a/V^{2} )(V- b) = RT, where P = pressure, V = volume, T = temperature, R = universal gas constant, a and b are constants. The dimensions of a and b in this equation are\n\n• (a) [a] = L^{3}  [b], ML^{5} T^{-2}\n• (b) [a] = L^{3}  [b], ML^{-3} T^{-2}\n• (c) [a] = ML^{5} T^{-2}  [b] = L^{3}\n• (d) [a] = ML^{3} T^{-2}  [b] = L^{5}\n\n• (b) [a] = ML^{5} T^{-2}  [b] = L^{3}\n\n## Q13  . A person starts his journey from his home at 9.00 a.m.to his office and came back to his home at 5.00 p.m. His office is 20 km from his home, then the displacement and distance covered in his complete journey are:\n\n• (a) 20 km, 0 Km\n• (b) 0 km, 20 Km\n• (c) 40 km, 0 Km\n• (d) 0 km, 40 Km\n\n• (d) 0 km, 40 Km\n\n## Q14 Which of the following statement for an object in uniform motion is not true?\n\n• (a)The object must be moving in straight path along same direction\n(b) The object covers equal displacements in equal intervals of time\n• (c) The velocity of object changes by equal amounts in equal intervals of time like free fall under gravity.\n• (d) The acceleration of object must be zero\n\n• (c) conservation of linear momentum\n\n## Q15  When a car moving in a fixed direction accelerates\n\n• (a) The frictional force acts on it in the direction opposite to the direction of motion\n• (b) The frictional force acts on it in the same direction as direction of motion\n• (c) Frictional force is not required. The engine of the car gives the force needed.\n• d) No force is required due to inertia of motion. The car accelerates by itself\n\n• (b) The frictional force acts on it in the same direction as direction of motion\n\n## Q16 If a body is stationary then\n\n• (a) there is no force acting on it\n• (b) the forces acting on it are not in contact with it\n• (c) the combination of forces acting on it balance out\n• (d) the body is in vacuum\n\n• (c) the combination of forces acting on it balance out\n\n## Q17  The restoring force produced in a spring is a conservative force. It means\n\n• (a) Work done by spring force depends only on the initial and final positions.\n• (b) Work done by spring force depends on the path between initial and final positions.\n• (c) Work done by spring force in a cyclic process is non zero.\n• (d) This statement is false. Spring force is a non conservative force\n\n• (b) Work done by spring force depends on the path between initial and final positions.\n\n## Q8 A man pushes a wall normally and fails to displace it but himself gets displaced by 2 m away from wall against frictional force between his feet and ground. Work done by wall on him is:\n\n• (a) zero\n• (b) negative\n• (c) positive\n• (d) depends on the frictional force between the wall and hands of the\n\n• (c) positive\n\n## Q19.  Four particles given below have same linear momentum, which has maximum kinetic energy?\n\n• (a) proton\n• (b) electron\n• (c) deutron\n• (d) alpha particle\n\n• (b) electron\n\n## Q20 In which of the following motions momentum changes but K.E. does not change?\n\n• (a) A freely falling body\n• (b) A body moving in uniform motion\n• (c) A body vertically thrown upwards\n• (d) A body moving in uniform circular motion\n\n• (d) A body moving in uniform circular motion\n\n## Q21 The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained?\n\n• (a) The kinetic energy of rocket\n• (b) Energy given by atmosphere\n• (c) Energy given by burning fuel\n• (d) The potential energy of the rocket\n\n• (a) The kinetic energy of rocket\n\n## Q22  When perfectly inelastic collision between two objects takes place, then during the collision\n\n• (a) Linear momentum remains conserved.\n(b) Kinetic energy remains conserved.\n• (c) Both kinetic energy and linear momentum remain conserved.\n• (d) Neither kinetic energy nor linear momentum remains conserved.\n\n• (a) Linear momentum remains conserved.\n\n## Q23  The angular velocity of a planet revolving in an elliptical orbit around the sun increases, when it comes near the sun and vice-versa. This process better explained on the basis of law of conservation of:\n\n• (a) Mass\n• (b) Linear Momentum\n• (c) Angular Momentum\n• (d) Energy\n\n• (c) Angular Momentum\n\n## Q24 A cylinder rolls down an inclined plane without slipping as shown. The work done against friction during rolling motion is\n\n• (a) always positive\n• (b) always negative\n• (c) zero\n• (d) first positive and then negative\n\n• (c) zero\n\n## Q25   The temperatures of two bodies measured by a thermometer are t1 = 200^{C} ± 0.50^{C} and t2 = 500^{C} ± 0.50^{C}. The temperature difference t = t2 – t1 between them is given by\n\n• (a) t = 30 0^{C} ± 0.5 0^{C}\n• (b) t = 70 0^{C}± 0.5 0^{C}\n• (c) t = 700^{C}± 1.00^{C}\n• (d) t = 300^{C}± 1.00^{C}\n\n• (d) t = 30oC ± 1.0oC\n\n• a) 0.1s\n• (b) 0.2s\n• (c) 0.3s\n• (d) 0.4s\n\n• (b) 0.2s\n\n## Q27 The angle made by vector 𝐴⃗⃗ = 2𝑖̂+ 𝑗̂ with x axis is\n\n• (a) tanθ = 0.5\n• (b) tanθ = 2\n• (c) tanθ = √5\n• (d) tanθ = √ 2/5\n\n• (a) tanθ = 0.5\n\n## Q28 At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range?\n\n• (a) tan θ = 1/4\n• (b) tan θ = 1/2\n• (c) tan θ = 4\n• (d) tan θ = 2\n\n• (c) tan θ = 4\n\n## Q29.  A rocket with a lift – off mass 3.5×10^{4} kg is blasted upwards with an initial acceleration of 10 ms^{-2} . Then, the initial thurst of the blast\n\n• (a) 3.5×10^{5} N\n• (b) 7.0×10^{5} N\n• (c) 14.0×10^{5} N\n• (d)1.75×10^{5}N\n\n• (b) 7.0×10^{5} N\n\n## Q30 Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary. Given that the coefficient of static friction between the box and train’s floor is 0.15. Take g = 10ms^{-2}\n\n• (a) 0.5 ms^{-2}\n• (b) 1.0 ms^{-2}\n• (c) 2.5 ms^{-2}\n• (d) 1.5 ms^{-2}\n\n• (d) 1.5 ms^{-2}\n\n• (a) – 50 J\n• (b) – 550 J\n• (c) – 750 J\n• (d) – 200 J\n\n• (c) – 750 J\n\n## Q32  A particle of mass M is moving in a horizontal circle of radius R with uniform speed v. When it moves from one point to a diametrically opposite point, it’s\n\n• (a) K.E changes by \\frac{1}{2}mv^{2}\n• (b) linear momentum does not change\n• (c) linear momentum changes by 2 M v\n• (d) K.E changes by mv^{2}\n\n• (c) linear momentum changes by 2 M v\n\n## Q33  How much energy is released, when 1mg of uranium is completely destroyed in an atom bomb.\n\n•  (a) 7×10^{10} J\n• (b) 8×10^{10} J\n• (c) 9×10^{10} J\n• (d) 5×10^{10} J\n\n• (c) 9× 10^{10} J\n\n## Q34 A solid cylinder of moment of inertia 0.625 kg m^{2}  rotates about its axis with angular speed 100 rads^{-1}. What is the magnitude of angular momentum of the cylinder about its axis?\n\n• (a) 62.5 kgm^{2}s^{-1}\n• (b) 625×104 kgm^{2}s^{-1}\n• (c) 6.25 kgm^{2}s^{-1}\n• (d) 625×105kgm^{2}s^{-1}\n\n• (a) 62.5 kgm^{2}s^{-1}\n\n## Q35  The sun rotates around itself in 27 days, if it were to expand to twice its present diameter, what would be its new period of revolution?\n\n• (a) 27 days\n• (b) 54 days\n• (c) 81 days\n• (d) 108 days\n\n• (d) 108 days\n\n### Current Electricity : MCQ on Current Electricity\n\nThe branch of Physics, which deals with the study of charges in motion is called Current Electricity.\n\nElectric Current :- It is defined as the rate of flow of charge across any cross-section of conductor.\n\nIf a charge ‘q’ flows across any cross-section in ‘t’ second ,current ‘i’ is given by I= q/t\n\nI= ne/t\n\nQ1 S.I Unit of Electric motive force is\n\n• Coulomb\n• Ampere\n• Volt\n• Coulomb/Ampere\n\n• Volt\n\nQ2 Electromotive force .............potential difference.\n\n• may be equals to\n• may be greater than\n• both (a) and (b)\n• none of these\n\nboth (a) and (b)\n\nQ3 A wire is carrying a current. Is it charged?\n\n• Yes,\n• No\n\nNo\n\n(Electric current is the flow of free electrons in the conductors. At any instant, the no. of electrons leaving the wire is always equal to the no. of electrons flowing from the battery into it. Hence the net charge on the wire is zero.)\n\nQ4 Electric current is a .................. quantity.\n\n• Scalar\n• Vector\n• Tensor\n• None of the above\n\n• Scalar\n\nQ5 S.I Unit of Electric current is ...\n\n• Coulomb\n• Ampere\n• Tesla\n• Coulomb/Ampere\n\n• Ampere\n\nQ6 One Ampere =.................................... stat ampere.\n\n• 3x10^{19}\n• 3x10^{9}\n• 2.25x 10^{19}\n• None of these\n\n• 3x10^{19}\n\nQ7 How many electrons passes in one second when current is 1A.\n\n• 5.26 × 10^{18}\n• 6.25 × 10^{25}\n• 5.26 × 10^{19}\n• 6.25 × 10^{19}\n\n• 6.25 × 10^{25}\n\nQ8 The relaxation time in conductors:\n\n• Increases with increase in temperature\n• Decreases with the increases of temperature\n• It does not depend on temperature\n• Changes suddenly at 400 Kelvin.\n\n• Decreases with the increases of temperature.\n\nQ9 According to Ohm’s Law\n\n• V is directly proportional to I\n• V is inversely proportional to I\n• V is directly proportional to √I\n• All of the above.\n\n• V is directly proportional to I\n\nQ10 Ohm's law is applicable to\n\n• Semiconductors\n• Vacuum tubes\n• Carbon resistors\n• None of these\n\n• None of these\n\n### MAGNETIC EFFECT OF CURRENT : MCQ ON MAGNETIC EFFECT OF CURRENT UNIT 3\n\nMagnetic Effects of Current: MCQ ON Moving Charges and Magnetism\n\nMAGNETIC EFFECT OF CURRENT   : MCQ ON AMPERE'S CIRCUITAL LAW\n\nMAGNETIC EFFECT OF CURRENT   : MCQ MOTION OF CHARGE IN MAGANETIC FIELD\n\nMAGNETIC EFFECT OF CURRENT  : MCQ ON SOLENOID\n\nMAGNETIC EFFECT OF CURRENT ; MCQ ON FORCE ON CURRENT CARRYING CONDUCTOR\n\nMAGNETIC EFFECT OF CURRENT ; MCQ ON GALAVANOMETER, AMMETER, VOLTMETER\n\nMAGNETIC EFFECT OF CURRENT  ; MCQ ON MAGNETIC MATERIALS SET 1\n\nMAGNETIC EFFECT OF CURRENT  :MCQ ON MAGNETIC MATERIALS SET 2\n\nMAGNETIC EFFECT OF CURRENT  :MCQ ON MAGNETIC MATERIALS SET 3\n\nMAGNETIC EFFECT OF CURRENT MCQ ON MAGNETIC MATERIALS SET 4\n\nMAGNETIC EFFECT OF CURRENT; MCQ ON MAGNETIC MATERIALS SET\n\nMCQ ON UNIT 1 FRICTIONAL ELECTRICITY\n\n## Q1 Basic source of magnetism is\n\n• (a) Charged particles alone\n• (b) movement of charged particles\n• (c) magnetic dipoles\n• (d) magnetic domains\n\n• (b) movement of charged particles\n\n## Q2 A charge q is moving in a magnetic field, then the magnetic force does not depend upon\n\n• (a) Charge\n• (b) mass\n• (c) velocity\n• (d) magnetic field\n\n• (b) mass\n\n## Q3 The ratio of intensity of magnetization to the magnetizing force is called____\n\n• a) Coercivity\n• b) Hysteresis\n• c) Magnetic susceptibility\n• d) Magnetic Moment\n\n• c) Magnetic susceptibility\n\n## Q4 If magnetic moment of a substance is zero, then substance is\n\n• (a)Diamagnetic\n• (b) paramagnetic\n• (c) ferromagnetic\n• (d) antiferromagnetic\n\n• (a)Diamagnetic\n\n## Q5 Magnetic susceptibility for diamagnetic material is\n\n• (a) Small and negative\n• (b) small and positive\n• (c) large and positive\n• (d) large and negative\n\n• (a)Small and negative\n\n## Q6 Which of the following is diamagnetic substance\n\n• (a)Nickel\n• (b) cobalt\n• (c) platinum\n• (d) water\n\n• (d) water\n\n## Q7  A bar of paramagnetic material is placed in a strong external magnetic filed. Then, the field inside it is likely to\n\n• (a) Remain the same\n• (b) increase\n• (c) vanish completely\n• (d) decrease\n\n• (b) increase\n\n## Q8 Permeability of paramagnetic substances is\n\n• (a) Zero\n• (b) Unity\n• (c) More than unity\n• (d) Less than unity\n\n• (c) More than unity\n\n• a) True\n• b) False\n\n• a) True\n\n• a) True\n• b) False\n\nb) False\n\n## Q1 The factors on which one magnetic field strength produced by current carrying solenoids depends are\n\n• (a) Magnitude of current\n• (b) Number of turns\n• (c) Nature of core material\n• (d) All of the above\n\n• (d) All of the above\n\n## Q2 The magnetic field produced due to a solenoid is same as that produced by a....\n\n• a) Bar magnet\n• b) Electromagnet\n• c) semi- circular coil containing current\n• d) None of these\n\n• a) Bar magnet\n\n## Q3  A magnetic dipole is placed in a uniform magnetic fields, \"B\". At what angle magnetic dipole to be placed  in magnetic field so that  the potential energy is minimum\n\n• a) 90^{0}\n• b) 30^{0}\n• c) 45^{0}\n• d) 0^{0}\n\n• d) 0^{0}\n\n## Q4 The small angle between magnetic axis and geographic axis at a place is called\n\n• (a) Magnetic inclination\n• (b) Magnetic dip\n• (c) Magnetic declination\n• (d) None of the above\n\n• (c) Magnetic declination\n\n## Q5 Horizontal component of earth at a place is 3.2 x 10^{-5}T, and angle of dip is 60^{0} The resultant intensity of earth’s field at the place is\n\n• a) 3.2 x 10^{-5}T\n• b) 1.6 x 10^{-5}T\n• c) 6.4 x 10^{-5}T\n• d) 12.8 x 10^{-5}T\n\n• c) 6.4 x 10^{-5}T\n\n## Q6 Isogonic lines on magnetic map represent lines of\n\n• (a) Zero declination\n• (b) Equal declination\n• (c) Equal dip\n• (d) Equal horizontal field\n\n• (b) Equal declination\n\n## Q7  At magnetic poles which out of following is zero\n\n• (a) Horizontal component (H) of earth's magnetic field\n• (b) Vertical component of earth's magnetic field\n• (c) Total strength of earth’s magnetic field(R)\n• (d) Magnetic declination\n\n• (a) Horizontal component (H) of earth's magnetic field\n\n## Q8 At a certain place, the horizontal component of earth’s magnetic field is √3 times the vertical component. The angle of dip at that place is\n\n• a) 90^{0}\n• b) 30^{0}\n• c) 45^{0}\n• d) 0^{0}\n\n• b) 30^{0}\n\n• a) True\n• b) False\n\n• a) True\n\n## Q10 At a particular place, horizontal and vertical components of earth’s magnetic field are equal. The angle of dip at that place is\n\n• a) 90^{0}\n• b) 30^{0}\n• c) 45^{0}\n• d) 0^{0}\n\n• c) 45^{0}\n\n## Q1 Magnetic field intensity at a point on the axial line of the magnet various with distance 'r' of the point from the center of the bar magnet as\n\n• (a) r^{1}\n• (b) r^{2}\n• (c) r^{-1}\n• (d) r^{-3}\n\n• (d) r^{-3}\n\n## Q2 Suppose magnetic field intensity at a distance r from the center of a short bar magnet on the axial live is B. Then magnetic field intensity at a distance r from the bisector of the bar magnet is.\n\n• a) B\n• b) 2B\n• c) B/2\n• d) None of these\n\n• c) B/2\n\n• a) m/l\n• b) 2M/л\n• c) m\n• d) M\n\n• b) 2M/л\n\n## Q4 If magnetic length of a bar magnet is 'x' and the geometrical length of the bar magnet is 'y', then:\n\n• a) x = y\n• b) x > y\n• c) x < y\n• d) None of these\n\n• c) x < y\n\n## Q5 SI units of magnetic dipole moment of a bar magnet are\n\n• a) Am^{2}\n• b) Am^{-1}\n• c) Am^{-2}\n• d) A^{-1}m\n\n• a) Am^{2}\n\n• a) F\n• b) F/4\n• c) 2F\n• d) F/2\n\n• a) F\n\n• a) True\n• b) False\n\n• a) True\n\n## Q8 What is the net magnetic flux through a closed surface?\n\n• a) Positive\n• b) Negative\n• c) Zero\n• d) Depends on the nature of the surface\n\n• c) Zero\n\n## Q9.  ‘X’ is the product of the pole strength of either magnetic pole and the magnetic length of the magnetic dipole. Identify X.\n\n• a) Magnetic dipole\n• b) Magnetic dipole moment\n• c) Magnetic pole strength\n• d) Magnetic dipole length\n\n• b) Magnetic dipole moment\n\n## Q10 What is the torque exerted by a bar magnet on itself due to its field?\n\n• a) Maximum\n• b) Zero\n• c) Minimum\n• d) Depends on the direction of the magnetic field\n\n• b) Zero,\n• A bar magnet does not exert a force or torque on itself due to its field. But an element of a current-carrying conductor experiences forces due to another element of the conductor. So, the torque exerted by a bar magnet on itself is zero.\n\n• (a) Bqv\n• (b) B/qv\n• (c) Zero\n• (d) q/Bv\n\n• (c) Zero\n\n## Q2 The dimensions of magnetic field intensity are\n\n• a) M^{1}L^{1}T^{-1}\n• b) M^{1}T^{-2}A^{-1}\n• c). M^{1}T^{-2}A^{-2}\n• d). M^{1}T^{-1}A^{-1}\n\n• b) M^{1}T^{-2}A^{-1}\n\n## Q3  Magnetic dipole moment is a vector quantity directed from\n\n• a) east to west\n• b) south to north\n• c) west to east\n• d) north to south\n\n• b) south to north\n\n• a) 0\n• b) -1\n• c) +1\n• d) ∞\n\n• b) -1\n\n• a) M, m/2\n• b) M/2, m\n• c) M, m\n• d) M/2, m/2\n\n• b) M/2, m\n\n• a) F\n• b) F/4\n• c) 2F\n• d) F/2\n\n• a) F\n\n• a) True\n• b) False\n\n• b) False\n\n## Q8 If horizontal and vertical components of earth's magnetic field are equal, then angle of dip is will be\n\n• (a) 30^{0}\n• (b) 60^{0}\n• (c) 90^{0}\n• (d) 45^{0}\n\n• (d) 45^{0}\n\n• a) True\n• d) False\n\n• (a) True\n\n## Q10 Dimensional formula for magnetic dipole moment is:\n\n• a) L^{2}A^{-1}\n• b) L^{2}A^{}\n• c) L^{2}A^{-2}\n• d) L^{1}A^{-1}\n\n• b) L^{2}A^{}`\n\n### ELECTRIC DIPOLE NOTES: ELECTRIC DIPOLE , DIPOLE LENGTH, ELECTRIC FIELD DUE TO ELECTRIC DIPOLE\n\nElectric Dipole: It is a  system of equal and opposite charges separated by some  fixed distance. DIPOLE LENGTH : the distance AB = 2a is ...", null, "" ]

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[ "Conceptual Approach - One Dimensional Motion - Part - II, Class XI - Physics - Student's Corner\n\n# Student's Corner\n\nWebsite is dedicated for students. You will get a large amount of study material. Students who cannot afford quality education will get high benefits. All the knowledge and details of competitive examinations will published time to time. Students with Physics and Mathematics background will be highly beneficial.\n\n## Subscribe Us\n\nIn the beginning there was nothing, which exploded.\n\n# Conceptual Approach - One Dimensional Motion - Part II\n\nWhat is one dimensional motion?\n\nAs all of you know that, to describe a location of a particle or object we need three coordinates i.e. x, y and z. The motion of a particle is said to be one dimensional if only one of the three coordinates of the particle changes with time. Or in simple terms, the motion of a particle along a straight line is one dimensional motion.\n\nWhat is two dimensional motion?\n\nThe motion of a particle is said to be two dimensional if only two of the three coordinates of the particle changes with time. The motion of a particle in a plane is a two dimensional motion i.e. two of the three coordinate’s changes with time.\n\nWhat is three dimensional motion?\n\nThe motion of a particle in space is generally three dimensional motion i.e. all of three coordinates change with time. The motion of a particle is said to be three dimensional motion if all the three coordinates of the particle changes with time.\n\nDefine Scalars and Vectors.\n\nScalars: A physical quantity which is completely described by magnitude only is called scalar. Examples include time, distance, mass etc. Scalars quantities have only magnitude and no direction.\n\nVectors: A physical quantity which is completely described if we know its magnitude as well direction is called vector. Examples include force, velocity, displacement etc.\n\nHow would you describe a motion?\n\nIn order to describe motion we need concepts of displacement, velocity and acceleration. All of these are vector quantities. Or we can say that a motion is described by displacement, velocity and acceleration.\n\nWhat is distance?\n\nThe length of the path travelled by a particle is called the distance travelled or covered by the particle. It is a scalar quantity and is measured in meters in SI units.\n\nWhat is displacement?\n\nThe shortest distance from the initial position to the final position of a particle is called displacement of the particle.\n\nImportant Point About Distance & Displacement\n\nDisplacement can be positive, negative or zero while the distance travelled is always positive. When a ball is throw vertically upward and it comes down to initial point then the net displacement of ball is zero while the distance travelled by the ball is 2h where h is maximum height where ball reached.\n\nWhat is average speed?\n\nThe average speed of the particle is defined as the total distance travelled divided by total time taken i.e.\n\nAverage speed = total distance/total time taken\n\nSI Unit of average speed is m/s.\n\nWhat is instantaneous speed?\n\nThe speed at a particular instant is called as instantaneous speed.\n\nWhat is average velocity?\n\nThe average velocity of a particle is defined as the displacement divided by the time taken for the displacement. It is a vector quantity but we don’t consider vectors in one dimensional motion.\n\nSpeed vs. Velocity\n\nIn our daily life we often use the words speed and velocity interchangeably. However, the two physical quantities are different in following manner: -\n\nSpeed is a scalar quantity and has magnitude only. However, velocity is a vector quantity and has magnitude as well as direction.\n\nSpeed indicates how fast something is moving whereas velocity tells us how fast and in which direction it is moving.\n\nWhat is instantaneous velocity?\n\nThe velocity at a particular instant is called instantaneous velocity. More precisely it is defined as the average velocity over an indefinitely short time interval.\n\nDifference between uniform and non uniform velocity.\n\nA particle is said to move with a uniform velocity if it undergoes equal displacements in equal interval of time, however small these time intervals may be. It should be emphasized that when a particle moves with uniform velocity, the magnitude as well as direction of the velocity remains constant.\n\nA particle is said to move with non uniform velocity or variable velocity if its speed or direction or both change with time. In such case, we generally specify average velocity of the particle.\n\nAsk us @ Quora : Advanced Tech World\n\nJoin the Group : Advanced Tech World", null, "Advanced Tech World" ]

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[ "Delivery ratio of the messages,Average message delivery latencyOverhead ratio (of the underlying routing mechanism) Suppose that $M$ be the set of all messages created in the network and $M_d$ be the set of all messages delivered. Then, the delivery ratio is computed as $|M_d| / |M|$.\nNow let the $i^{th}$ delivered message was created at time $c_i$ and delivered at time $d_i$. Then the average message delivery latency is computed as $(\\sum_{i = 1}^{|M_d|} (d_i - c_i)) / |M_d|$. Note that, in Statistics, mean, median and mode are all the measures of average. But \"loosely speaking\", unless otherwise specified, we refer to the \"mean\" value when we say \"average.\" Nevertheless, the MessageStatsReport in the ONE simulator provides a measure of both the mean and median values wherever appropriate." ]

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[ "# Document (#23183)\n\nAuthor\nOddy, P.\nTitle\n¬The case for international cooperation in cataloguing : from copy cataloguing to multilingual subject access - experiences within the British Library\nSource\nProgram. 33(1999) no.1, S.29-39\nYear\n1999\nAbstract\nPresents an outline of a cataloguing strategy that might be adopted for the with reference to how such a strategy is being implemented at the British Library. The first stage has involved cooperation with US libraries and future plans are linked to cooperation with European libraries. Such developments involve countries with different languages and different cataloguing cultures and so present many challenges. Discusses the skills required by staff needed to implement the cataloguing strategy\nTheme\nFormalerschließung\nKatalogfragen allgemein\n\n## Similar documents (author)\n\n1. Oddy, P.: British Library catalogues : out of control? (1993) 5.65\n```5.6473675 = sum of:\n5.6473675 = weight(author_txt:oddy in 6236) [ClassicSimilarity], result of:\n5.6473675 = fieldWeight in 6236, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.035788 = idf(docFreq=13, maxDocs=43254)\n0.625 = fieldNorm(doc=6236)\n```\n2. Oddy, P.: Authority control in the local, national and international environment (1991) 5.65\n```5.6473675 = sum of:\n5.6473675 = weight(author_txt:oddy in 896) [ClassicSimilarity], result of:\n5.6473675 = fieldWeight in 896, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.035788 = idf(docFreq=13, maxDocs=43254)\n0.625 = fieldNorm(doc=896)\n```\n3. Oddy, P.: Managing retrospective catalogue conversion (1991) 5.65\n```5.6473675 = sum of:\n5.6473675 = weight(author_txt:oddy in 900) [ClassicSimilarity], result of:\n5.6473675 = fieldWeight in 900, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.035788 = idf(docFreq=13, maxDocs=43254)\n0.625 = fieldNorm(doc=900)\n```\n4. Oddy, R.N.: Laboratory tests : automatic systems (1981) 5.65\n```5.6473675 = sum of:\n5.6473675 = weight(author_txt:oddy in 4222) [ClassicSimilarity], result of:\n5.6473675 = fieldWeight in 4222, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.035788 = idf(docFreq=13, maxDocs=43254)\n0.625 = fieldNorm(doc=4222)\n```\n5. Oddy, P.: Bibliographic standards for the New Age (1996) 5.65\n```5.6473675 = sum of:\n5.6473675 = weight(author_txt:oddy in 5744) [ClassicSimilarity], result of:\n5.6473675 = fieldWeight in 5744, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.035788 = idf(docFreq=13, maxDocs=43254)\n0.625 = fieldNorm(doc=5744)\n```\n\n## Similar documents (content)\n\n1. Khurshid, Z.: Cooperative cataloging : prospects and problems for libraries in Saudi Arabia (1997) 0.32\n```0.32434708 = sum of:\n0.32434708 = product of:\n1.1583824 = sum of:\n0.052626308 = weight(abstract_txt:countries in 2542) [ClassicSimilarity], result of:\n0.052626308 = score(doc=2542,freq=1.0), product of:\n0.119751714 = queryWeight, product of:\n1.0271614 = boost\n5.6251116 = idf(docFreq=423, maxDocs=43254)\n0.02072583 = queryNorm\n0.43946183 = fieldWeight in 2542, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.6251116 = idf(docFreq=423, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.027025657 = weight(abstract_txt:library in 2542) [ClassicSimilarity], result of:\n0.027025657 = score(doc=2542,freq=2.0), product of:\n0.076794595 = queryWeight, product of:\n1.1632636 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.02072583 = queryNorm\n0.35192135 = fieldWeight in 2542, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.09556239 = weight(abstract_txt:copy in 2542) [ClassicSimilarity], result of:\n0.09556239 = score(doc=2542,freq=1.0), product of:\n0.17823972 = queryWeight, product of:\n1.2531413 = boost\n6.8626604 = idf(docFreq=122, maxDocs=43254)\n0.02072583 = queryNorm\n0.5361453 = fieldWeight in 2542, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.8626604 = idf(docFreq=122, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.055244204 = weight(abstract_txt:libraries in 2542) [ClassicSimilarity], result of:\n0.055244204 = score(doc=2542,freq=3.0), product of:\n0.108053915 = queryWeight, product of:\n1.3798537 = boost\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.02072583 = queryNorm\n0.51126516 = fieldWeight in 2542, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.026469434 = weight(abstract_txt:with in 2542) [ClassicSimilarity], result of:\n0.026469434 = score(doc=2542,freq=2.0), product of:\n0.09542297 = queryWeight, product of:\n1.8338097 = boost\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.02072583 = queryNorm\n0.2773906 = fieldWeight in 2542, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.49475715 = weight(abstract_txt:cooperation in 2542) [ClassicSimilarity], result of:\n0.49475715 = score(doc=2542,freq=5.0), product of:\n0.44990924 = queryWeight, product of:\n3.4484258 = boost\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.02072583 = queryNorm\n1.0996821 = fieldWeight in 2542, product of:\n2.236068 = tf(freq=5.0), with freq of:\n5.0 = termFreq=5.0\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.4066972 = weight(abstract_txt:cataloguing in 2542) [ClassicSimilarity], result of:\n0.4066972 = score(doc=2542,freq=6.0), product of:\n0.4404874 = queryWeight, product of:\n4.405037 = boost\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.02072583 = queryNorm\n0.92328906 = fieldWeight in 2542, product of:\n2.4494898 = tf(freq=6.0), with freq of:\n6.0 = termFreq=6.0\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.078125 = fieldNorm(doc=2542)\n0.28 = coord(7/25)\n```\n2. Oddy, P.: Remind, reassure, and reward : issues in developing a cataloguing strategy (1997) 0.32\n```0.31527826 = sum of:\n0.31527826 = product of:\n1.3136594 = sum of:\n0.06798529 = weight(abstract_txt:staff in 4215) [ClassicSimilarity], result of:\n0.06798529 = score(doc=4215,freq=1.0), product of:\n0.11350223 = queryWeight, product of:\n5.476366 = idf(docFreq=491, maxDocs=43254)\n0.02072583 = queryNorm\n0.59897757 = fieldWeight in 4215, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.476366 = idf(docFreq=491, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.026754037 = weight(abstract_txt:library in 4215) [ClassicSimilarity], result of:\n0.026754037 = score(doc=4215,freq=1.0), product of:\n0.076794595 = queryWeight, product of:\n1.1632636 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.02072583 = queryNorm\n0.34838438 = fieldWeight in 4215, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.17196205 = weight(abstract_txt:british in 4215) [ClassicSimilarity], result of:\n0.17196205 = score(doc=4215,freq=1.0), product of:\n0.26547655 = queryWeight, product of:\n2.1628475 = boost\n5.922272 = idf(docFreq=314, maxDocs=43254)\n0.02072583 = queryNorm\n0.64774853 = fieldWeight in 4215, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.922272 = idf(docFreq=314, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.40846166 = weight(abstract_txt:strategy in 4215) [ClassicSimilarity], result of:\n0.40846166 = score(doc=4215,freq=3.0), product of:\n0.3751129 = queryWeight, product of:\n3.1487591 = boost\n5.747919 = idf(docFreq=374, maxDocs=43254)\n0.02072583 = queryNorm\n1.0889033 = fieldWeight in 4215, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n5.747919 = idf(docFreq=374, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.30976695 = weight(abstract_txt:cooperation in 4215) [ClassicSimilarity], result of:\n0.30976695 = score(doc=4215,freq=1.0), product of:\n0.44990924 = queryWeight, product of:\n3.4484258 = boost\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.02072583 = queryNorm\n0.6885099 = fieldWeight in 4215, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.32872945 = weight(abstract_txt:cataloguing in 4215) [ClassicSimilarity], result of:\n0.32872945 = score(doc=4215,freq=2.0), product of:\n0.4404874 = queryWeight, product of:\n4.405037 = boost\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.02072583 = queryNorm\n0.7462857 = fieldWeight in 4215, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.109375 = fieldNorm(doc=4215)\n0.24 = coord(6/25)\n```\n3. Herzinger, S.: What is the future for cataloging? (1994) 0.27\n```0.26794928 = sum of:\n0.26794928 = product of:\n0.95696175 = sum of:\n0.06299321 = weight(abstract_txt:skills in 5088) [ClassicSimilarity], result of:\n0.06299321 = score(doc=5088,freq=1.0), product of:\n0.11955144 = queryWeight, product of:\n1.0263021 = boost\n5.6204057 = idf(docFreq=425, maxDocs=43254)\n0.02072583 = queryNorm\n0.52691305 = fieldWeight in 5088, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.6204057 = idf(docFreq=425, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.039719444 = weight(abstract_txt:library in 5088) [ClassicSimilarity], result of:\n0.039719444 = score(doc=5088,freq=3.0), product of:\n0.076794595 = queryWeight, product of:\n1.1632636 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.02072583 = queryNorm\n0.5172167 = fieldWeight in 5088, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.113078065 = weight(abstract_txt:involve in 5088) [ClassicSimilarity], result of:\n0.113078065 = score(doc=5088,freq=1.0), product of:\n0.17658123 = queryWeight, product of:\n1.2472975 = boost\n6.830658 = idf(docFreq=126, maxDocs=43254)\n0.02072583 = queryNorm\n0.6403742 = fieldWeight in 5088, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.830658 = idf(docFreq=126, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.038274307 = weight(abstract_txt:libraries in 5088) [ClassicSimilarity], result of:\n0.038274307 = score(doc=5088,freq=1.0), product of:\n0.108053915 = queryWeight, product of:\n1.3798537 = boost\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.02072583 = queryNorm\n0.3542149 = fieldWeight in 5088, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.038901966 = weight(abstract_txt:with in 5088) [ClassicSimilarity], result of:\n0.038901966 = score(doc=5088,freq=3.0), product of:\n0.09542297 = queryWeight, product of:\n1.8338097 = boost\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.02072583 = queryNorm\n0.40767926 = fieldWeight in 5088, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.26551452 = weight(abstract_txt:cooperation in 5088) [ClassicSimilarity], result of:\n0.26551452 = score(doc=5088,freq=1.0), product of:\n0.44990924 = queryWeight, product of:\n3.4484258 = boost\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.02072583 = queryNorm\n0.5901513 = fieldWeight in 5088, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.39848027 = weight(abstract_txt:cataloguing in 5088) [ClassicSimilarity], result of:\n0.39848027 = score(doc=5088,freq=4.0), product of:\n0.4404874 = queryWeight, product of:\n4.405037 = boost\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.02072583 = queryNorm\n0.90463483 = fieldWeight in 5088, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.09375 = fieldNorm(doc=5088)\n0.28 = coord(7/25)\n```\n4. Heiner-Freiling, M.: DDC German - the project, the aims, the methods : new ideas for a well-established traditional classification system (2006) 0.20\n```0.19591773 = sum of:\n0.19591773 = product of:\n0.69970614 = sum of:\n0.063151576 = weight(abstract_txt:countries in 780) [ClassicSimilarity], result of:\n0.063151576 = score(doc=780,freq=1.0), product of:\n0.119751714 = queryWeight, product of:\n1.0271614 = boost\n5.6251116 = idf(docFreq=423, maxDocs=43254)\n0.02072583 = queryNorm\n0.52735424 = fieldWeight in 780, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.6251116 = idf(docFreq=423, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.06522057 = weight(abstract_txt:european in 780) [ClassicSimilarity], result of:\n0.06522057 = score(doc=780,freq=1.0), product of:\n0.1223532 = queryWeight, product of:\n1.0382584 = boost\n5.6858835 = idf(docFreq=398, maxDocs=43254)\n0.02072583 = queryNorm\n0.5330516 = fieldWeight in 780, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.6858835 = idf(docFreq=398, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.022932032 = weight(abstract_txt:library in 780) [ClassicSimilarity], result of:\n0.022932032 = score(doc=780,freq=1.0), product of:\n0.076794595 = queryWeight, product of:\n1.1632636 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.02072583 = queryNorm\n0.2986152 = fieldWeight in 780, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.10287007 = weight(abstract_txt:outline in 780) [ClassicSimilarity], result of:\n0.10287007 = score(doc=780,freq=1.0), product of:\n0.16578744 = queryWeight, product of:\n1.208575 = boost\n6.6185994 = idf(docFreq=156, maxDocs=43254)\n0.02072583 = queryNorm\n0.6204937 = fieldWeight in 780, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.6185994 = idf(docFreq=156, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.038274307 = weight(abstract_txt:libraries in 780) [ClassicSimilarity], result of:\n0.038274307 = score(doc=780,freq=1.0), product of:\n0.108053915 = queryWeight, product of:\n1.3798537 = boost\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.02072583 = queryNorm\n0.3542149 = fieldWeight in 780, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.031763323 = weight(abstract_txt:with in 780) [ClassicSimilarity], result of:\n0.031763323 = score(doc=780,freq=2.0), product of:\n0.09542297 = queryWeight, product of:\n1.8338097 = boost\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.02072583 = queryNorm\n0.33286873 = fieldWeight in 780, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n2.5106533 = idf(docFreq=9548, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.37549424 = weight(abstract_txt:cooperation in 780) [ClassicSimilarity], result of:\n0.37549424 = score(doc=780,freq=2.0), product of:\n0.44990924 = queryWeight, product of:\n3.4484258 = boost\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.02072583 = queryNorm\n0.8346 = fieldWeight in 780, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.2949476 = idf(docFreq=216, maxDocs=43254)\n0.09375 = fieldNorm(doc=780)\n0.28 = coord(7/25)\n```\n5. Fecko, M.B.: ¬'The changing face of cataloging : the impact of technology on copy cataloging' (1996) 0.19\n```0.18662578 = sum of:\n0.18662578 = product of:\n0.77760744 = sum of:\n0.0582731 = weight(abstract_txt:staff in 441) [ClassicSimilarity], result of:\n0.0582731 = score(doc=441,freq=1.0), product of:\n0.11350223 = queryWeight, product of:\n5.476366 = idf(docFreq=491, maxDocs=43254)\n0.02072583 = queryNorm\n0.5134093 = fieldWeight in 441, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.476366 = idf(docFreq=491, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.09747293 = weight(abstract_txt:implemented in 441) [ClassicSimilarity], result of:\n0.09747293 = score(doc=441,freq=2.0), product of:\n0.12694189 = queryWeight, product of:\n1.0575485 = boost\n5.7915225 = idf(docFreq=358, maxDocs=43254)\n0.02072583 = queryNorm\n0.7678547 = fieldWeight in 441, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n5.7915225 = idf(docFreq=358, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.022932032 = weight(abstract_txt:library in 441) [ClassicSimilarity], result of:\n0.022932032 = score(doc=441,freq=1.0), product of:\n0.076794595 = queryWeight, product of:\n1.1632636 = boost\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.02072583 = queryNorm\n0.2986152 = fieldWeight in 441, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.1852286 = idf(docFreq=4863, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.16217476 = weight(abstract_txt:copy in 441) [ClassicSimilarity], result of:\n0.16217476 = score(doc=441,freq=2.0), product of:\n0.17823972 = queryWeight, product of:\n1.2531413 = boost\n6.8626604 = idf(docFreq=122, maxDocs=43254)\n0.02072583 = queryNorm\n0.9098688 = fieldWeight in 441, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.8626604 = idf(docFreq=122, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.038274307 = weight(abstract_txt:libraries in 441) [ClassicSimilarity], result of:\n0.038274307 = score(doc=441,freq=1.0), product of:\n0.108053915 = queryWeight, product of:\n1.3798537 = boost\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.02072583 = queryNorm\n0.3542149 = fieldWeight in 441, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.7782922 = idf(docFreq=2687, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.39848027 = weight(abstract_txt:cataloguing in 441) [ClassicSimilarity], result of:\n0.39848027 = score(doc=441,freq=4.0), product of:\n0.4404874 = queryWeight, product of:\n4.405037 = boost\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.02072583 = queryNorm\n0.90463483 = fieldWeight in 441, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n4.824719 = idf(docFreq=943, maxDocs=43254)\n0.09375 = fieldNorm(doc=441)\n0.24 = coord(6/25)\n```" ]

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[ "# Math\n\nMath相关的属性和方法都是静态的，直接使用即可。\n\n• Math.abs() // 取绝对值\n• Math.ceil() // 向上取整\n• Math.floor() // 向下取整\n• Math.max() // 返回一组数据中的最大值（非数组）\n• Math.min() // 返回一组数据中的最小值\n• Math.pow(x, y) // 计算次幂结果\n• Math.random() // 返回[0,1]之间的伪随机浮点数\n• Math.round() // 四舍五入，返回整数" ]

[ null ]

[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n## Computer programming\n\n### Course: Computer programming>Unit 5\n\nLesson 8: Particle Systems\n\n# A single particle\n\nBefore we can create an entire `ParticleSystem`, we have to create an object that will describe a single particle. The good news: we've done this already. Our `Mover` object from the Forces section serves as the perfect template. For us, a particle is an independent body that moves about the screen. It has `location`, `velocity`, and `acceleration`, a constructor to initialize those variables, and functions to `display()` itself and `update()` its location.\n``````// A simple Particle object\nvar Particle = function(position) {\nthis.acceleration = new PVector();\nthis.velocity = new PVector();\nthis.position = position.get();\n};\n\nParticle.prototype.update = function(){\n};\n\nParticle.prototype.display = function() {\nstroke(0, 0, 0);\nfill(175, 175, 175);\nellipse(this.position.x, this.position.y, 8, 8);\n};``````\nThis is about as simple as a particle can get. From here, we could take our particle in several directions. We could add an `applyForce()` method to affect the particle’s behavior (we’ll do precisely this in a future example). We could add variables to describe color and shape, or use `image()` to draw the particle. For now, however, let’s focus on adding just one additional detail: lifespan.\nTypical particle systems involve something called an emitter. The emitter is the source of the particles and controls the initial settings for the particles, location, velocity, etc. An emitter might emit a single burst of particles, or a continuous stream of particles, or both. The point is that for a typical implementation such as this, a particle is born at the emitter but does not live forever. If it were to live forever, our program would eventually grind to a halt as the number of particles increased to an unwieldy number over time. As new particles are born, we need old particles to die. This creates the illusion of an infinite stream of particles, and the performance of our program does not suffer.\nThere are many different ways to decide when a particle dies. For example, it could come into contact with another object, or it could simply leave the screen. For our first `Particle` object, however, we’re simply going to add a `timeToLive` property. It will act as a timer, counting down from 255 to 0, at which point we'll consider the particle to be \"dead.\" And so we expand the `Particle` object as follows:\n``````// A simple Particle object\nvar Particle = function(position) {\nthis.acceleration = new PVector();\nthis.velocity = new PVector();\nthis.position = position.get();\nthis.timeToLive = 255;\n};\n\nParticle.prototype.update = function(){\nthis.timeToLive -= 2;\n};\n\nParticle.prototype.display = function() {\nstroke(255, 255, 255, this.timeToLive);\nfill(127, 127, 127, this.timeToLive);\nellipse(this.position.x, this.position.y, 8, 8);\n};``````\nThe reason we chose to start the `timeToLive` at 255 and count down to 0 is for convenience. With those values, we can use `timeToLive` as the alpha transparency for the ellipse as well. When the particle is “dead” it will also have faded away onscreen.\nWith the addition of the `timeToLive` property, we’ll also need one additional method—a function that can be queried (for a true or false answer) as to whether the particle is alive or dead. This will come in handy when we are writing the `ParticleSystem` object, whose task will be to manage the list of particles themselves. Writing this function is pretty easy; it just needs to return true if the value of `timeToLive` is less than 0.\n``````Particle.prototype.isDead = function() {\nreturn this.timeToLive < 0;\n};``````\nBefore we get to the next step of making many particles, it’s worth taking a moment to make sure our particle works correctly and create a sketch with one single `Particle` object. Here is the full code below, with two small additions. We add a convenience method called `run()` that simply calls both `update()` and `display()` for us. In addition, we give the particle a random initial velocity as well as a downward acceleration (to simulate gravity).\nNow that we have an object to describe a single particle, we’re ready for the next big step. How do we keep track of many particles, when we can’t ensure exactly how many particles we might have at any given time?\n\n## Want to join the conversation?\n\n• Why cant the `Particle.prototype.isDead` function just return a boolean value, like this,\n``Particle.prototype.isDead = function() { return this.timeToLive <= 0;}``\n\ninstead of having a series of uneeded `if` statements?", null, "• Just curious, what does this.position = position.get(); do?", null, "•", null, "`PVector.get()` creates a copy of the vector object.\nThen that copy is assigned to the position of the particle.\n\nSo the particle now has a position with the same values, but it can be changed independently.\nIf you didn't use get() to make a copy, then you would not be able to change one position without also changing the other. Too bad KA doesn't teach why.\n• step 3 im getting \"Make sure you also change the leaf's angular velocity and acceleration when it hits the ground! \"\nnow ive applied it to the draw , update, display, im not sure where im going wrong?\n` if(this.position.y < height){ this.velocity.set(0,0); this.acceleration.set(0,0); }`", null, "• For step two on the challenge: I've tried using\n``if (leaves[i].position.y >= height) { leaves[i].position.y = height; leaves[i].velocity.y = 0; leaves[i].acceleration = new PVector();}``\n\nin a couple of variations: using `height - 40`; using `>` instead of `>=`; placing in the `draw` function at the top, at the bottom, in its own loop; placing in `Particle.prototype.update` using `this`; etc. These all work: the leaves pile up at the bottom, but I'm not getting credit by the autograder. Any tips?", null, "• The autoGrader doesn't accept my changes in step one:\n``var ds =[];mouseClicked=function(){ ds.push( new Particle(new PVector(mouseX,mouseY)));};draw = function() { background(194, 231, 255); tree.display(); for(var i=0;i<ds.length;i++){ ds[i].run(); }};``", null, "• what is the difference between these three:\nvar v1 = new PVector(0,1);\nvar v2 = v1; ...............................(1)\nvar v2 = v1.get(); .......................(2) &\nvar v2;\nv2.set(v1); ...............................(3)", null, "• 1.\n``var v1 = new PVector(12, 5);var v2 = new PVector(12, 5);println(v2 === v1);``\nWe see that `v1` and `v2` are separate objects as witnessed by the equality operator `===`, where as\n``var v1 = new PVector(12, 5);var v2 = v1;println(v2 === v1);``\nshow us that `v1` and `v2` refer to the same object. In the latter case, if you modify `v2` then `v1` will also show that modification.\n\n2.\n``var v1 = new PVector(3, 4);var v2 = v1.get();``\nThe `get` method returns a `new` PVector. So `v1` and `v2` are separate objects that have the same property values, like in the first case of example one.\n\n3.\n``var v1 = new PVector(8, 15);var v2;v2.set(v1);``\nwill fail since `v2` is undefined and as such has no `set` method. Try\n``var v1 = new PVector(8, 15);var v2 = new PVector();v2.set(v1);``\nto make the separate objects `v1` and `v2` have the same property values as in the first case of example one.\n• I need help, in the next challenge, on step one. I am writing the next `code`:\n``mouseClicked = function (){ leaves.push(new Particle(new PVector(mouseX, mouseY)));};draw = function() { background(194, 231, 255); tree.display(); for (var i=0;i<leaves.length;i++){ leaves[i].run(); } };``\n\nIt is working, leaves are falling, but the \"beaver\" don't pass me into the next part of the challenge. What is wrong, please someone can help me?", null, "• i need help on the next challenge... i looked in the comments and tried to use what they said, but am still not getting it. here is my code:\n\nvar Particle = function(position) {\nthis.acceleration = new PVector(0, 0.05);\nthis.velocity = new PVector(random(0, 1), random(0, 0));\nthis.position = position;\n};\n\nParticle.prototype.run = function() {\nthis.update();\nthis.display();\n};\n\nParticle.prototype.update = function(){\n};\n\nParticle.prototype.display = function() {\nimage(getImage(\"avatars/leaf-green\"), this.position.x, this.position.y, 40, 40);\n};\n\nvar Tree = function(position, options) {\nthis.position = position.get();\nthis.branchingFactor = 3;\nthis.angleBetweenBranches = 32;\nthis.scaleFactor = 0.7;\nthis.numLevels = 4;\nthis.baseBranchLength = 120;\n};\n\nTree.prototype.display = function() {\nvar self = this;\n\nvar forward = function(distance) {\nline(0, 0, 0, -distance);\ntranslate(0, -distance);\n};\n\nvar back = function(distance) {\nforward(-distance);\n};\n\nvar right = function(angle) {\nrotate(angle * PI / 180);\n};\n\nvar left = function(angle) {\nright(-angle);\n};\n\nvar drawTree = function(depth, length) {\nif (depth === 0) {\nimage(getImage(\"avatars/leaf-green\"), -10, -30, 40, 40);\nreturn;\n}\nvar totalAngle = self.angleBetweenBranches * (self.branchingFactor - 1);\n\nstrokeWeight(depth*5);\nforward(length);\nright(totalAngle / 2.0);\nfor (var i = 0; i < self.branchingFactor; i += 1) {\ndrawTree(depth - 1, length * self.scaleFactor);\nleft(self.angleBetweenBranches);\n}\nright(totalAngle / 2.0 + self.angleBetweenBranches);\nback(length);\n};\n\npushMatrix();\ntranslate(this.position.x, this.position.y);\nstroke(122, 112, 85);\ndrawTree(this.numLevels, this.baseBranchLength);\npopMatrix();\n};\n\nvar leaves = [];\nvar tree = new Tree(new PVector(width/2, 400));\nmouseClicked = function(){\nds.push( new Particle(new PVector(mouseX,mouseY)));\n};\n\ndraw = function() {\nbackground(194, 231, 255);\ntree.display();\nfor(var i = 0; i < leaves.length; i++) {\nvar fall = leaves[i];\n\n};", null, "• You are on the right track, you want to call the various methods on leaves[i] or fall (since that's the variable you have used to store leaves[i]) such as fall.display(), fall.run(), fall.update(). Also, you want to push the new particle values into the null leaves array or else when you call these methods on the leaves[i], it will be empty and nothing will happen.\n(1 vote)\n• In the next challenge, even though the grader lets me pass all steps, I can't make it so the leaves stop at the bottom of the screen while rotating during their fall. I think it's because I'm rotating the grid, so my\n``if (leaf.position.y > height - 30)``\nisn't checking the bottom of the screen anymore, but wherever that y is while being rotated, but I can't find how to make it so that the leaf rotates and still hits the bottom, lying still.\n\nMy code (I left out the Tree object):\n``angleMode = \"radians\";var Particle = function(position) { this.acceleration = new PVector(0, 0.05); this.velocity = new PVector(random(0, 1), random(0, 0)); this.position = position; this.angle = 0; this.aVelocity = 0; this.aAcceleration = 0.0005;};Particle.prototype.run = function() { this.update(); this.display();};Particle.prototype.update = function(){ this.velocity.add(this.acceleration); this.position.add(this.velocity); this.aVelocity += this.aAcceleration; this.angle += this.aVelocity;};Particle.prototype.display = function() { pushMatrix(); rotate(this.angle); image(getImage(\"avatars/leaf-green\"), this.position.x, this.position.y, 40, 40); popMatrix();};var leaves = [];var tree = new Tree(new PVector(width/2, 400));mouseClicked = function() { var particle = new Particle(new PVector(mouseX, mouseY)); leaves.push(particle);};draw = function() { background(194, 231, 255); tree.display(); for (var i = 0; i < leaves.length; i++) { var leaf = leaves[i]; leaf.run(); if (leaf.position.y >= height - 30) { leaf.acceleration.set(0, 0); leaf.velocity.set(0, 0); resetMatrix(); leaf.aAcceleration = 0; leaf.aVelocity = 0; } }};``", null, "", null, "" ]

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[ "Cody\n\n# Problem 33. Create times-tables\n\nSolution 529659\n\nSubmitted on 16 Nov 2014\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = 2; y_correct = [1 2; 2 4]; assert(isequal(timestables(x),y_correct))\n\nans = []\n\n2   Fail\n%% x = 3; y_correct = [1 2 3; 3 6 9]; assert(isequal(timestables(x),y_correct))\n\nans = []\n\nAssertion failed.\n\n3   Pass\n%% x = 5; y_correct = [1 2 3 4 5; 2 4 6 8 10; 3 6 9 12 15; 4 8 12 16 20; 5 10 15 20 25]; assert(isequal(timestables(x),y_correct))\n\nans = []" ]

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[ "# 1950 AHSME Problems/Problem 28\n\n## Problem\n\nTwo boys", null, "$A$ and", null, "$B$ start at the same time to ride from Port Jervis to Poughkeepsie,", null, "$60$ miles away.", null, "$A$ travels", null, "$4$ miles an hour slower than", null, "$B$.", null, "$B$ reaches Poughkeepsie and at once turns back meeting", null, "$A$", null, "$12$ miles from Poughkeepsie. The rate of", null, "$A$ was:", null, "$\\textbf{(A)}\\ 4\\text{ mph}\\qquad \\textbf{(B)}\\ 8\\text{ mph} \\qquad \\textbf{(C)}\\ 12\\text{ mph} \\qquad \\textbf{(D)}\\ 16\\text{ mph} \\qquad \\textbf{(E)}\\ 20\\text{ mph}$\n\n## Solution\n\nLet the speed of boy", null, "$A$ be", null, "$a$, and the speed of boy", null, "$B$ be", null, "$b$. Notice that", null, "$A$ travels", null, "$4$ miles per hour slower than boy", null, "$B$, so we can replace", null, "$b$ with", null, "$a+4$.\n\nNow let us see the distances that the boys each travel. Boy", null, "$A$ travels", null, "$60-12=48$ miles, and boy", null, "$B$ travels", null, "$60+12=72$ miles. Now, we can use", null, "$d=rt$ to make an equation, where we set the time to be equal:", null, "$$\\frac{48}{a}=\\frac{72}{a+4}$$ Cross-multiplying gives", null, "$48a+192=72a$. Isolating the variable", null, "$a$, we get the equation", null, "$24a=192$, so", null, "$a=\\boxed{\\textbf{(B) }8 \\text{ mph}}$.\n\n## Alternate Solution\n\nNote that", null, "$A$ travels", null, "$60-12=48$ miles in the time it takes", null, "$B$ to travel", null, "$60+12=72$ miles. Thus,", null, "$B$ travels", null, "$72-48=24$ more miles in the given time, meaning", null, "$\\frac{24\\text{miles}}{4\\text{miles}/\\text{hour}} = 6 \\text{hours}$ have passed, as", null, "$B$ goes", null, "$4$ miles per hour faster. Thus,", null, "$A$ travels", null, "$48$ miles per", null, "$6$ hours, or", null, "$8$ miles per hour.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]

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[ "# colour.algebra.ellipse_fitting¶\n\ncolour.algebra.ellipse_fitting(a, method='Halir 1998')[source]\n\nReturns the coefficients of the implicit second-order polynomial/quadratic curve that fits given point array $$a$$ using given method.\n\nThe implicit second-order polynomial is expressed as follows:\n\n:math:F\\left(x, y\\right) = ax^2 + bxy + cy^2 + dx + ey + f = 0\n\n\nwith an ellipse-specific constraint such as $$b^2 -4ac < 0$$ and where $$a, b, c, d, e, f$$ are coefficients of the ellipse and $$F\\left(x, y\\right)$$ are coordinates of points lying on it.\n\nParameters\n• a (array_like) – Point array $$a$$ to be fitted.\n\n• method (unicode, optional) – {‘Halir 1998’}, Computation method.\n\nReturns\n\nCoefficients of the implicit second-order polynomial/quadratic curve that fits given point array $$a$$.\n\nReturn type\n\nndarray\n\nReferences\n\n[HF98]\n\nExamples\n\n>>> a = np.array([[2, 0], [0, 1], [-2, 0], [0, -1]])\n>>> ellipse_fitting(a)\narray([ 0.2425356..., 0. , 0.9701425..., 0. , 0. ,\n-0.9701425...])\n>>> ellipse_coefficients_canonical_form(ellipse_fitting(a))\narray([-0., -0., 2., 1., 0.])\n`" ]

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[ "", null, "", null, "", null, "", null, "0\n\n# How do you change eighty two percent as a decimal and fraction?\n\nUpdated: 9/24/2023", null, "Wiki User\n\n9y ago\n\n.82 as a decimal.\n\n82/100 as a fraction, or 41/50", null, "Wiki User\n\n9y ago", null, "", null, "", null, "Earn +20 pts\nQ: How do you change eighty two percent as a decimal and fraction?\nSubmit\nStill have questions?", null, "", null, "Related questions\n\n0.80\n\n### What is eighty two percent in a simplified fraction?\n\nEighty two percent can be simplified to 41/50 as a fraction.\n\n### What is 85 percent as a fraction in its lowest term?\n\nEighty five percent is 17/20 as a fraction.\n\n### Express the fraction eighty one hundredths as a decimal?\n\neighty one hundredths = 0.81 in decimals\n\nIt is 0.86\n\n4/5\n\n81/100\n\n### What is eighty six hundreds as a decimal?\n\nExpressed as a decimal fraction, 86/100 is equal to 0.86.\n\n### What is eighty-six tenths as a decimal?\n\nExpressed as a decimal fraction, 86/10 is equal to 8.6.\n\n### How do you change a decimal into a fraction of 2.089?\n\n2.089 into fraction: deal with fraction first = 0.089 / 1, multiply top and bottom * 1000 to drop the decimal point = 89 / 1000 which is simplest fraction, so add 2 to get : two and eighty nine one thousandths\n\n### What is eighty as a percent as a decimal and as a fraction in simplest form?\n\n80 = 80/1 = 800/10 = 8000/100 = 8000%80 = 80.080 = 80/1\n\n83 over 100." ]

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[ "# Algorithmic tesselation with geogrid\n\nJoseph Bailey 2023-08-15\n\n# geogrid\n\nTurn geospatial polygons like states, counties or local authorities into regular or hexagonal grids automatically.", null, "## Intro\n\nUsing geospatial polygons to portray geographical information can be a challenge when polygons are of different sizes. For example, it can be difficult to ensure that larger polygons do not skew how readers retain or absorb information. As a result, many opt to generate maps that use consistent shapes (i.e. regular grids) to ensure that no specific geography is emphasised unfairly. Generally there are four reasons that one might transform geospatial polygons to a different grid (or geospatial representation):\n\n1. We may use cartograms to represent the number of people (or any value) within a particular geography. For more information and examples see here and here. This cartogram approach changes the size of a particular geography in-line with the values that one seeks to visualise.\n2. We may use grids to bin data and typically visualise the spatial density of a particular variable. For an example see here.\n3. We may use grids to segment a geographical region. For example tesselation can be used in biological sampling or even in generating game environments.\n4. We may use grids to ‘fairly’ represent existing geographical entities (such as US states, UK local authorities, or even countries in Europe). For an example of representing US states as both regular and hexagonal grids see here.\n\nThe link in bullet 4 provides an excellent introduction to the notion of tesselation and its challenges. Interestingly, the eventual generation of hexagonal and regular grids demonstrated in the article was done manually. I believe that this can be very time consuming, and though it may stimulate some fun discussion - wouldn’t it be great to do it automatically?\n\nRecent functionality for representing US states, European countries and World countries in a grid has been made available for ggplot2 here and there are many other great examples of hand-specified or bespoke grids. The challenge with this is that if you have a less commonly used geography then it might be hard to find a hand-specified or bespoke grid for your area of interest.\n\nWhat I wanted to do with `geogrid` is make it easier to generate these grids in ways that might be visually appealing and then assign the original geographies to their gridded counterparts in a way that made sense. Using an input of geospatial polgyons `geogrid` will generate either a regular or hexagonal grid, and then assign each of the polygons in your original file to that new grid.\n\n## Idea\n\nThere are two steps to using `geogrid`:\n\n1. Generate a regular or hexagonal grid of your choice. There are lots of different arrangements of these grids so choosing one with the `calculate_grid` function and varying the `seed` is a good place to start.\n2. Use the hungarian algorithm to efficiently calculate the assignments from the original geography to the new geography. This involves identifying the solution where the total distance between the centroid of every original geography and its new centroid on the grid is minimised. For this I have included a previous implementation of the Hungarian algorithm kindly made available here. Huge thanks to Lars Simon Zehnder for this implementation.\n\n## Example\n\nThis is a basic example which shows how the assignment of London boroughs could work.\n\n``````library(geogrid)\nlibrary(sf)\nlibrary(tmap)\n\ninput_file <- system.file(\"extdata\", \"london_LA.json\", package = \"geogrid\")\noriginal_shapes\\$SNAME <- substr(original_shapes\\$NAME, 1, 4)``````\n\nFor reference, lets see how London’s local authorities are actually bounded in real space. In this example, I have coloured each polygon based on it’s area. Brighter polygons are larger.\n\n``````rawplot <- tm_shape(original_shapes) +\ntm_polygons(\"HECTARES\", palette = \"viridis\") +\ntm_text(\"SNAME\")\nrawplot``````", null, "So, let’s turn this into a grid to stop places like Bromley, Hillingdon and Havering from stealing our attention. First of all, we can generate a number of different grids using `seed`. Since there are many ways to dissect the outer boundary of the polygons you might want to choose an output that appeals to you. I’d recommend looking at different `seed` values and choosing the one that best matches the outline that you approve of.\n\nThe `calculate_grid` function takes in a SpatialPolygonsDataframe or sf object, a learning rate (suggestion = 0.03 to begin), a grid type `hexagonal` or `regular` and a seed value. Let’s have a look at some hexagonal grid options for the London local authorities:\n\n``````par(mfrow = c(2, 3), mar = c(0, 0, 2, 0))\nfor (i in 1:6) {\nnew_cells <- calculate_grid(shape = original_shapes, grid_type = \"hexagonal\", seed = i)\nplot(new_cells, main = paste(\"Seed\", i, sep = \" \"))\n}``````", null, "Let’s also look at things with a regular grid:\n\n``````par(mfrow = c(2, 3), mar = c(0, 0, 2, 0))\nfor (i in 1:6) {\nnew_cells <- calculate_grid(shape = original_shapes, grid_type = \"regular\", seed = i)\nplot(new_cells, main = paste(\"Seed\", i, sep = \" \"))\n}``````", null, "As we can see there are lots of options. Now, lets choose a grid and assign our existing places to it. I happen to like the both grids that have a `seed` of 3. So I’m going to assign the polygons to those grids. Let’s do that and see what they look like compared to the original.\n\n``````new_cells_hex <- calculate_grid(shape = original_shapes, grid_type = \"hexagonal\", seed = 3)\nresulthex <- assign_polygons(original_shapes, new_cells_hex)\n\nnew_cells_reg <- calculate_grid(shape = original_shapes, grid_type = \"regular\", seed = 3)\nresultreg <- assign_polygons(original_shapes, new_cells_reg)``````\n\nNow we have an example transfer from real space to grid space - we can visualise it.\n\n``````hexplot <- tm_shape(resulthex) +\ntm_polygons(\"HECTARES\", palette = \"viridis\") +\ntm_text(\"SNAME\")\n\nregplot <- tm_shape(resultreg) +\ntm_polygons(\"HECTARES\", palette = \"viridis\") +\ntm_text(\"SNAME\")\n\ntmap_arrange(rawplot, hexplot, regplot, nrow = 3)``````", null, "## Details\n\nThe package has two major functions:\n\n1. `calculate_grid()` given your input polygons this will generate the grid as specified by your arguments:\n• `shape`: the original polygons\n• `learning_rate`: the rate at which the gradient descent finds the optimum cellsize to ensure that your gridded points fit within the outer boundary of the input polygons.\n• `grid_type`: either `regular` for a square grid or `hexagonal` for a hexagonal grid.\n• `seed`: the seed to ensure you get the same grid output.\n2. `assign_polygons()`: this will assign the original polygons to their new locations on the grid generated in `calculate_grid()`. It will find the solution that minimises the sum of the total distance between the original polygon centroids and eventual gridded centroids. Arguments:\n• `shape`: the original polygons\n• `new_polygons`: the output (a list) from `calculate_grid()`.\n\n## TODO\n\n• Assignment may not always work - check the `assign_polygons()` why does it only work sometimes?\n• Make it work (done I think), make it right (not yet), make it fast (not yet).\n• Improve the cellsize calculation methodology.\n• Get someone to answer this stack overflow question.\n\nThis is my first attempt at a package. If it doesn’t work I’d like suggestions for improvements and thanks in advance for providing them!\n\nI welcome critique and feedback. Blog post to follow.\n\n## Thanks\n\nI read a lot of the work by Hadley Wickham, Jenny Bryan, Thomas Lin Pedersen, Mara Averick and Bob Rudis to name a few. But also love the R community and learn a huge amount from R Bloggers.\n\nExtra thanks go to Ryan Hafen for making this package publishable.\n\n# Other examples\n\nFrom others:\n\nSimon Hailstone has looked at male life expectancy in the South East region of England using the package. Thanks Simon for using!\n\nFrom me:\n\nThis time using the contiguous USA. Again, I used set seed and chose some that I liked but I’d recommend you’d do the same.\n\n``````input_file2 <- system.file(\"extdata\", \"states.json\", package = \"geogrid\")\noriginal_shapes2\\$SNAME <- substr(original_shapes2\\$NAME, 1, 4)\n\nrawplot2 <- tm_shape(original_shapes2) +\ntm_polygons(\"CENSUSAREA\", palette = \"viridis\") +\ntm_text(\"SNAME\")``````\n\nLet’s check the seeds again.\n\n``````par(mfrow = c(2, 3), mar = c(0, 0, 2, 0))\nfor (i in 1:6) {\nnew_cells <- calculate_grid(shape = original_shapes2, grid_type = \"hexagonal\", seed = i)\nplot(new_cells, main = paste(\"Seed\", i, sep = \" \"))\n}``````", null, "``````par(mfrow = c(2, 3), mar = c(0, 0, 2, 0))\nfor (i in 1:6) {\nnew_cells <- calculate_grid(shape = original_shapes2, grid_type = \"regular\", seed = i)\nplot(new_cells, main = paste(\"Seed\", i, sep = \" \"))\n}``````", null, "Now we’ve seen some seed demo’s lets assign them…\n\n``````new_cells_hex2 <- calculate_grid(shape = original_shapes2, grid_type = \"hexagonal\", seed = 6)\nresulthex2 <- assign_polygons(original_shapes2, new_cells_hex2)\n\nnew_cells_reg2 <- calculate_grid(shape = original_shapes2, grid_type = \"regular\", seed = 4)\nresultreg2 <- assign_polygons(original_shapes2, new_cells_reg2)\n\nhexplot2 <- tm_shape(resulthex2) +\ntm_polygons(\"CENSUSAREA\", palette = \"viridis\") +\ntm_text(\"SNAME\")\n\nregplot2 <- tm_shape(resultreg2) +\ntm_polygons(\"CENSUSAREA\", palette = \"viridis\") +\ntm_text(\"SNAME\")\n\ntmap_arrange(rawplot2, hexplot2, regplot2, nrow = 3)``````", null, "Likewise, you can try the bay area…\n\n``````input_file3 <- system.file(\"extdata\", \"bay_counties.geojson\", package = \"geogrid\")\noriginal_shapes3\\$SNAME <- substr(original_shapes3\\$county, 1, 4)\n\nrawplot3 <- tm_shape(original_shapes3) +\ntm_polygons(col = \"gray25\") +\ntm_text(\"SNAME\")\n\nnew_cells_hex3 <- calculate_grid(shape = original_shapes3, grid_type = \"hexagonal\", seed = 6)\nresulthex3 <- assign_polygons(original_shapes3, new_cells_hex3)\n\nnew_cells_reg3 <- calculate_grid(shape = original_shapes3, grid_type = \"regular\", seed = 1)\nresultreg3 <- assign_polygons(original_shapes3, new_cells_reg3)\n\nhexplot3 <- tm_shape(resulthex3) +\ntm_polygons(col = \"gray25\") +\ntm_text(\"SNAME\")\n\nregplot3 <- tm_shape(resultreg3) +\ntm_polygons(col = \"gray25\") +\ntm_text(\"SNAME\")\n\ntmap_arrange(rawplot3, hexplot3, regplot3, nrow = 3)``````", null, "" ]

[ null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example4-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example0-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example1-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example2-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example4-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example6-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example6a-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example7-1.png", null, "https://cran.stat.unipd.it/web/packages/geogrid/readme/man/README_figs/README-example8-1.png", null ]

[ "o\n\n# Math \n\n#### object Math extends Object\n\nMath is a built-in object that has properties and methods for mathematical constants and functions. Not a function object.\n\nMDN\n\nAnnotations\n() ()\nLinear Supertypes\nOrdering\n1. Alphabetic\n2. By Inheritance\nInherited\n1. Math\n2. Object\n3. Any\n4. AnyRef\n5. Any\n1. Hide All\n2. Show All\nVisibility\n1. Public\n2. All\n\n### Value Members\n\n1. final def !=(arg0: scala.Any)\nDefinition Classes\nAnyRef → Any\n2. final def ##(): Int\nDefinition Classes\nAnyRef → Any\n3. final def ==(arg0: scala.Any)\nDefinition Classes\nAnyRef → Any\n4. val E\n\nEuler's constant and the base of natural logarithms, approximately 2.718.\n\nEuler's constant and the base of natural logarithms, approximately 2.718.\n\nMDN\n\n5. val LN10\n\nNatural logarithm of 10, approximately 2.303.\n\nNatural logarithm of 10, approximately 2.303.\n\nMDN\n\n6. val LN2\n\nNatural logarithm of 2, approximately 0.693.\n\nNatural logarithm of 2, approximately 0.693.\n\nMDN\n\n7. val LOG10E\n\nBase 10 logarithm of E, approximately 0.434.\n\nBase 10 logarithm of E, approximately 0.434.\n\nMSN\n\n8. val LOG2E\n\nBase 2 logarithm of E, approximately 1.443.\n\nBase 2 logarithm of E, approximately 1.443.\n\nMDN\n\n9. val PI\n\nRatio of the circumference of a circle to its diameter, approximately 3.14159.\n\nRatio of the circumference of a circle to its diameter, approximately 3.14159.\n\nMDN\n\n10. val SQRT1_2\n\nSquare root of 1/2; equivalently, 1 over the square root of 2, approximately 0.707.\n\nSquare root of 1/2; equivalently, 1 over the square root of 2, approximately 0.707.\n\nMDN\n\n11. val SQRT2\n\nSquare root of 2, approximately 1.414.\n\nSquare root of 2, approximately 1.414.\n\nMDN\n\n12. def abs(x: Double)\n\nReturns the absolute value of a number.\n\nReturns the absolute value of a number.\n\nPassing a non-numeric string or undefined/empty variable returns NaN. Passing null returns 0.\n\nMDN\n\n13. def abs(x: Int): Int\n\nReturns the absolute value of a number.\n\nReturns the absolute value of a number.\n\nPassing a non-numeric string or undefined/empty variable returns NaN. Passing null returns 0.\n\nMDN\n\n14. def acos(x: Double)\n\nThe Math.acos() function returns the arccosine (in radians) of a number.\n\nThe Math.acos() function returns the arccosine (in radians) of a number.\n\nThe acos method returns a numeric value between 0 and pi radians for x between -1 and 1. If the value of number is outside this range, it returns NaN.\n\nMDN\n\n15. final def asInstanceOf[T0]: T0\nDefinition Classes\nAny\n16. def asin(x: Double)\n\nThe Math.asin() function returns the arcsine (in radians) of a number.\n\nThe Math.asin() function returns the arcsine (in radians) of a number.\n\nThe asin method returns a numeric value between -pi/2 and pi/2 radians for x between -1 and 1. If the value of number is outside this range, it returns NaN.\n\nMDN\n\n17. def atan(x: Double)\n\nThe Math.atan() function returns the arctangent (in radians) of a number.\n\nThe Math.atan() function returns the arctangent (in radians) of a number.\n\nThe atan method returns a numeric value between -pi/2 and pi/2 radians.\n\nMDN\n\n18. def atan2(y: Double, x: Double)\n\nThe Math.atan2() function returns the arctangent of the quotient of its arguments.\n\nThe Math.atan2() function returns the arctangent of the quotient of its arguments.\n\nThe atan2 method returns a numeric value between -pi and pi representing the angle theta of an (x,y) point. This is the counterclockwise angle, measured in radians, between the positive X axis, and the point (x,y). Note that the arguments to this function pass the y-coordinate first and the x-coordinate second.\n\natan2 is passed separate x and y arguments, and atan is passed the ratio of those two arguments.\n\nMDN\n\n19. def cbrt(x: Double)\n\nECMAScript 6 The Math.cbrt() function returns the cube root of a number\n\nECMAScript 6 The Math.cbrt() function returns the cube root of a number\n\nreturns\n\nThe cube root of the given number. MDN\n\n20. def ceil(x: Double)\n\nThe Math.ceil() function returns the smallest integer greater than or equal to a number.\n\nThe Math.ceil() function returns the smallest integer greater than or equal to a number.\n\nMDN\n\n21. def clone()\nAttributes\nprotected[lang]\nDefinition Classes\nAnyRef\nAnnotations\n@throws( ... ) @native()\n22. def clz32(x: Int): Int\n\nECMAScript 2015\n\nECMAScript 2015\n\nThe Math.clz32() function returns the number of leading zero bits in the 32-bit binary representation of a number.\n\nMDN\n\n23. def cos(x: Double)\n\nThe Math.cos() function returns the cosine of a number.\n\nThe Math.cos() function returns the cosine of a number.\n\nThe cos method returns a numeric value between -1 and 1, which represents the cosine of the angle.\n\nMDN\n\n24. def cosh(x: Double)\n\nECMAScript 6 The Math.cosh() function returns the hyperbolic cosine of a number\n\nECMAScript 6 The Math.cosh() function returns the hyperbolic cosine of a number\n\nreturns\n\nThe hyperbolic cosine of the given number MDN\n\n25. final def eq(arg0: AnyRef)\nDefinition Classes\nAnyRef\n26. def equals(arg0: scala.Any)\nDefinition Classes\nAnyRef → Any\n27. def exp(x: Double)\n\nThe Math.exp() function returns E^x, where x is the argument, and E is Euler's constant, the base of the natural logarithms.\n\nThe Math.exp() function returns E^x, where x is the argument, and E is Euler's constant, the base of the natural logarithms.\n\nMDN\n\n28. def expm1(x: Double)\n\nECMAScript 6 The Math.expm1() function returns e^x - 1, where x is the argument, and e the base of the natural logarithms.\n\nECMAScript 6 The Math.expm1() function returns e^x - 1, where x is the argument, and e the base of the natural logarithms.\n\nreturns\n\nA number representing e^x - 1, where e is Euler's number and x is the argument.\n\n29. def finalize(): Unit\nAttributes\nprotected[lang]\nDefinition Classes\nAnyRef\nAnnotations\n@throws( classOf[java.lang.Throwable] )\n30. def floor(x: Double)\n\nThe Math.floor() function returns the largest integer less than or equal to a number.\n\nThe Math.floor() function returns the largest integer less than or equal to a number.\n\nMDN\n\n31. final def getClass(): Class[_]\nDefinition Classes\nAnyRef → Any\nAnnotations\n@native()\n32. def hasOwnProperty(v: String)\n\nTests whether this object has the specified property as a direct property.\n\nTests whether this object has the specified property as a direct property.\n\nUnlike js.Object.hasProperty, this method does not check down the object's prototype chain.\n\nMDN\n\nDefinition Classes\nObject\n33. def hashCode(): Int\nDefinition Classes\nAnyRef → Any\nAnnotations\n@native()\n34. def hypot(x: Double*)\n\nECMAScript 6 The Math.hypot() function returns the square root of the sum of squares of its arguments\n\nECMAScript 6 The Math.hypot() function returns the square root of the sum of squares of its arguments\n\nreturns\n\nThe square root of the sum of squares of the given arguments. MDN\n\n35. final def isInstanceOf[T0]\nDefinition Classes\nAny\n36. def isPrototypeOf(v: Object)\n\nTests whether this object is in the prototype chain of another object.\n\nTests whether this object is in the prototype chain of another object.\n\nDefinition Classes\nObject\n37. def log(x: Double)\n\nThe Math.log() function returns the natural logarithm (base E) of a number.\n\nThe Math.log() function returns the natural logarithm (base E) of a number.\n\nIf the value of number is negative, the return value is always NaN.\n\nMDN\n\n38. def log10(x: Double)\n\nECMAScript 6 The Math.log10() function returns the base 10 logarithm of a number\n\nECMAScript 6 The Math.log10() function returns the base 10 logarithm of a number\n\nreturns\n\nThe base 10 logarithm of the given number. If the number is negative, NaN is returned. MDN\n\n39. def log1p(x: Double)\n\nECMAScript 6 The Math.log1p() function returns the natural logarithm (base e) of 1 + a number\n\nECMAScript 6 The Math.log1p() function returns the natural logarithm (base e) of 1 + a number\n\nreturns\n\nThe natural logarithm (base e) of 1 plus the given number. If the number is less than -1, NaN is returned. MDN\n\n40. def max(values: Double*)\n\nThe Math.max() function returns the largest of zero or more numbers.\n\nThe Math.max() function returns the largest of zero or more numbers.\n\nIf no arguments are given, the result is - Infinity.\n\nIf at least one of arguments cannot be converted to a number, the result is NaN.\n\nMDN\n\n41. def max(value1: Int, values: Int*): Int\n\nThe Math.max() function returns the largest of zero or more numbers.\n\nThe Math.max() function returns the largest of zero or more numbers.\n\nMDN\n\n42. def min(values: Double*)\n\nThe Math.min() function returns the smallest of zero or more numbers.\n\nThe Math.min() function returns the smallest of zero or more numbers.\n\nIf no arguments are given, the result is Infinity.\n\nIf at least one of arguments cannot be converted to a number, the result is NaN.\n\nMDN\n\n43. def min(value1: Int, values: Int*): Int\n\nThe Math.min() function returns the smallest of zero or more numbers.\n\nThe Math.min() function returns the smallest of zero or more numbers.\n\nMDN\n\n44. final def ne(arg0: AnyRef)\nDefinition Classes\nAnyRef\n45. final def notify(): Unit\nDefinition Classes\nAnyRef\nAnnotations\n@native()\n46. final def notifyAll(): Unit\nDefinition Classes\nAnyRef\nAnnotations\n@native()\n47. def pow(x: Double, y: Double)\n\nThe Math.pow() function returns the base to the exponent Power, that is, base^^exponent.\n\nThe Math.pow() function returns the base to the exponent Power, that is, base^^exponent.\n\nMDN\n\n48. def propertyIsEnumerable(v: String)\n\nTests whether the specified property in an object can be enumerated by a call to js.Object.properties, with the exception of properties inherited through the prototype chain.\n\nTests whether the specified property in an object can be enumerated by a call to js.Object.properties, with the exception of properties inherited through the prototype chain.\n\nIf the object does not have the specified property, this method returns false.\n\nMDN\n\nDefinition Classes\nObject\n49. def random()\n\nThe Math.random() function returns a floating-point, pseudo-random number in the range [0, 1) that is, from 0 (inclusive) up to but not including 1 (exclusive), which you can then scale to your desired range.\n\nThe Math.random() function returns a floating-point, pseudo-random number in the range [0, 1) that is, from 0 (inclusive) up to but not including 1 (exclusive), which you can then scale to your desired range.\n\nThe random number generator is seeded from the current time, as in Java.\n\nMDN\n\n50. def round(x: Double)\n\nThe Math.round() function returns the value of a number rounded to the nearest integer.\n\nThe Math.round() function returns the value of a number rounded to the nearest integer.\n\nIf the fractional portion of number is .5 or greater, the argument is rounded to the next higher integer. If the fractional portion of number is less than .5, the argument is rounded to the next lower integer.\n\nMDN\n\n51. def sin(x: Double)\n\nThe Math.sin() function returns the sine of a number.\n\nThe Math.sin() function returns the sine of a number.\n\nThe sin method returns a numeric value between -1 and 1, which represents the sine of the angle given in radians.\n\nMDN\n\n52. def sinh(x: Double)\n\nECMAScript 6 The Math.sinh() function returns the hyperbolic sine of a number\n\nECMAScript 6 The Math.sinh() function returns the hyperbolic sine of a number\n\nreturns\n\nThe hyperbolic sine of the given number MDN\n\n53. def sqrt(x: Double)\n\nThe Math.sqrt() function returns the square root (x\\sqrt{x}) of a number.\n\nThe Math.sqrt() function returns the square root (x\\sqrt{x}) of a number.\n\nIf the value of number is negative, sqrt returns NaN.\n\nMDN\n\n54. final def synchronized[T0](arg0: ⇒ T0): T0\nDefinition Classes\nAnyRef\n55. def tan(x: Double)\n\nThe Math.tan() function returns the tangent of a number.\n\nThe Math.tan() function returns the tangent of a number.\n\nThe tan method returns a numeric value that represents the tangent of the angle.\n\nMDN\n\n56. def tanh(x: Double)\n\nECMAScript 6 The Math.tanh() function returns the hyperbolic tangent of a number\n\nECMAScript 6 The Math.tanh() function returns the hyperbolic tangent of a number\n\nreturns\n\nThe hyperbolic tangent of the given number MDN\n\n57. def toLocaleString(): String\nDefinition Classes\nObject\n58. def toString()\nDefinition Classes\nAnyRef → Any\n59. def valueOf()\nDefinition Classes\nObject\n60. final def wait(): Unit\nDefinition Classes\nAnyRef\nAnnotations\n@throws( ... )\n61. final def wait(arg0: Long, arg1: Int): Unit\nDefinition Classes\nAnyRef\nAnnotations\n@throws( ... )\n62. final def wait(arg0: Long): Unit\nDefinition Classes\nAnyRef\nAnnotations\n@throws( ... ) @native()" ]

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[ "### Create random combinations in Excel\n\nImagine that you need to create random combinations of letters in Excel. In this example I want to create a 3 letters random combinations list. I want the result to be like this:\n\n TEX JYY QCX CDH NTW\n\nTo get this kind of combinations I used the following formula:\n\n=CHAR(RANDBETWEEN(65,90))&CHAR(RANDBETWEEN(65,90))&CHAR(RANDBETWEEN(65,90))\n\nI’ve used the RANDBETWEEN() function to generate values between 65 and 90 because they are the ANSI codes of letters A to Z. The RANDBETWEEM() function has the following arguments:\n\nRANDBETWEEN(bottom,top)\n\nbottom is the smallest integer that the function will return and top is the highest.\n\nAfter the generation of the random number, I use CHAR() function to return the corresponding character from the ANSI table of characters. This function has the following syntax:\n\nCHAR(number)\n\nnumber is a number between 1 and 255 that specifies which character we want to return.\n\nOn my formula, combining 3 times the formula CHAR(RANDBETWEEN(65,90) I get a combination of 3 letters.\n\nIf I wanted to get a random list of numbers between 100 and 1000, I could use the following formula:\n\n=RANDBETWEEN(100,1000)\n\nThis will give me a list of number like the one below:\n\n 941 486 970 952 376\n\n#### 1 comentários:\n\nAnonymous said...\n\nThat's not a combination. It's a permutation." ]

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[ "HOME  ›   pipelines\n\n# Interpreting Output\n\n## Workflow or biology\n\nIn most experiments, we see a handful of single cells with unique copy number events that are not seen in other cells in the sample. The figure below shows one such example of a single cell in the 1k Cells from BJ Fibroblast Euploid Cell Line dataset, cell 534, showing a copy number gain that is not seen in other cells.", null, "Figure 1 Example of a singleton cell in a dataset with a unique copy number event.\n\nSuch events could be a reflection of true biology in the input sample or due to technical artifacts of the library preparation workflow. It is difficult to differentiate between these possibilities when looking at a single cell, but when the same event is seen in multiple cells it is very likely to be a true CNV. If the event is truly present in a small fraction of cells in the sample, increasing the input cell count will increase the probability of seeing multiple cells with the same event. Along the same lines, if the same event is seen across multiple replicates it is likely due to a true copy number event and not a technical artifact.\n\n## Determining absolute copy numbers\n\nWe report the integer copy number across the genome for single cells and groups of cells. It is important to note that for a given cell it is possible that the true copy number state differs from the copy number state we report by an integer multiple. For example, a normal diploid cell in the G2 cell cycle phase that has duplicated its genome and is tetraploid can be called by the algorithms as a diploid cell. This is because in the absence of copy number variation, the algorithms assume that a cell with no events is diploid. Even in cells with significant copy number variation the scaling algorithm heuristically assumes that the least possible copy number state that fits the data is the most likely. For example, if a cell had copy number 6 on chromosome 1 and copy number 4 on all the remaining chromosomes, the algorithms would report copy number 3 on chromosome 1 and copy number 2 on the remaining chromosomes since that is the least possible integer copy number state that can explain the data. For such samples, you can run the pipeline and use the options --min-soft-avg-ploidy=M and --max-soft-avg-ploidy=N to try to force the scaling algorithm to give solutions with mean ploidy above and below M and N, respectively.\n\nExcess noise in the read count data can also produce off by integer multiple errors. Shown below are two examples of cells where the copy number across the genome is underestimated by a factor of two. In both cases, the red ovals point to locations in the genome where the copy number estimate does not match the read count data. If both cells were to be scaled at twice the reported copy number, the segments in the red ovals would be at integer copy number levels. Cell 649 is a normal male diploid cell with one copy of the X and Y chromosomes. The scaling algorithm error results in all the chromosomes having half the true copy number. The X and Y chromosomes have copy number 0 because the copy number is forced to be an integer.", null, "Figure 2 Examples of two cells where the reported copy number solution is underestimated by a factor of two. The red ovals show regions of the genome where the read data does not line up against the copy number calls.\n\nCell 395 shown below is an example of a cell that is reported as having twice the copy number over most of the genome. Unlike the cells shown above the read data in this case lines up well with the copy number calls. Moreover, there are long segments of the genome with copy number 3 (odd) highlighted by the red ovals. If this cell were to be scaled similarly to the other MKN-45 cells, these segments would be at half-integer values. This cell has two copies of the MKN-45 genome and this could be a biological event or a cell doublet.", null, "Figure 3 Cell 395 from the 5k MKN-45 Gastric Cancer Cell Line dataset with mean copy number 3.7 has approximately twice the expected mean copy number 1.8 for MKN-45 cells and is correctly scaled. The copy number calls match the data and the red ovals highlight long segments of the genome at copy number 3.\n\nReplicating cells are some times scaled to half their true copy number. MKN-45 cells have an approximate mean copy number over the genome of 1.8 and this implies that a replicating MKN-45 cell should have mean copy number between 1.8 and 3.6 based on how much of the genome has been replicated. A cell in G2 phase that has finished replicating its genome completely is indistinguishable from a cell in G0/G1 phase. In both cases, cells would be scaled to an mean copy number of 1.8. Cells in S phase can sometimes be scaled to half their true copy number and cell 98 below is one such example with an average copy number of 1.6. However, the true mean copy number across the genome should be 3.2 and this is made clear in the zoomed in view of the bottom panel—the read data can be fitted to half-integer values.", null, "Figure 4 Cell 98 is a replicating cell that where the copy numbers are half the correct value. The zoomed in view shows regions where the read data are not aligned with the copy number calls. These regions are right at the 0.5 copy number line suggesting that the scaling is off by a factor of half.\n\n## Noisy cells\n\nAs discussed in the CSV pipeline outputs page, the per_cell_summary_metrics.csv labels cells as noisy or not, and the summary.csv provides the fraction of noisy cells in the sample. A cell could be deemed noisy due to the underlying biology: cells undergoing DNA replication or apoptosis, or due to degraded DNA in the input cell/nuclei suspension, or due to noise introduced by the workflow. A cell is defined as noisy if\n\n• High DIMAPD: when the cell's DIMAPD is statistically higher than the sample distribution (with p-value < 0.01). See the DIMAPD discussion on the Interpreting metrics page for more information.\n• Low ploidy confidence: when the algorithms are unable to define the overall copy number state across the genome reliably. This is caused by noise or waviness in the read count profiles that can lead the scaling algorithm to pick one amongst multiple distinct copy number solutions that fit the data equally well. The solution that is picked may not be the correct one.\n• 1.0\n• Cell Ranger DNA v1.1 (latest)" ]

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[ "## 182023\n\n182,023 (one hundred eighty-two thousand twenty-three) is an odd six-digits composite number following 182022 and preceding 182024. In scientific notation, it is written as 1.82023 × 105. The sum of its digits is 16. It has a total of 2 prime factors and 4 positive divisors. There are 180,880 positive integers (up to 182023) that are relatively prime to 182023.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 6\n• Sum of Digits 16\n• Digital Root 7\n\n## Name\n\nShort name 182 thousand 23 one hundred eighty-two thousand twenty-three\n\n## Notation\n\nScientific notation 1.82023 × 105 182.023 × 103\n\n## Prime Factorization of 182023\n\nPrime Factorization 191 × 953\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 182023 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 182,023 is 191 × 953. Since it has a total of 2 prime factors, 182,023 is a composite number.\n\n## Divisors of 182023\n\n4 divisors\n\n Even divisors 0 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 183168 Sum of all the positive divisors of n s(n) 1145 Sum of the proper positive divisors of n A(n) 45792 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 426.642 Returns the nth root of the product of n divisors H(n) 3.975 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 182,023 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 182,023) is 183,168, the average is 45,792.\n\n## Other Arithmetic Functions (n = 182023)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 180880 Total number of positive integers not greater than n that are coprime to n λ(n) 90440 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16463 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 180,880 positive integers (less than 182,023) that are coprime with 182,023. And there are approximately 16,463 prime numbers less than or equal to 182,023.\n\n## Divisibility of 182023\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 2 7 7\n\n182,023 is not divisible by any number less than or equal to 9.\n\n## Classification of 182023\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (182023)\n\nBase System Value\n2 Binary 101100011100000111\n3 Ternary 100020200121\n4 Quaternary 230130013\n5 Quinary 21311043\n6 Senary 3522411\n8 Octal 543407\n10 Decimal 182023\n12 Duodecimal 89407\n20 Vigesimal 12f13\n36 Base36 3wg7\n\n## Basic calculations (n = 182023)\n\n### Multiplication\n\nn×i\n n×2 364046 546069 728092 910115\n\n### Division\n\nni\n n⁄2 91011.5 60674.3 45505.8 36404.6\n\n### Exponentiation\n\nni\n n2 33132372529 6030853844846167 1097754109400433855841 199816496255395171741746343\n\n### Nth Root\n\ni√n\n 2√n 426.642 56.6729 20.6553 11.2726\n\n## 182023 as geometric shapes\n\n### Circle\n\n Diameter 364046 1.14368e+06 1.04088e+11\n\n### Sphere\n\n Volume 2.5262e+16 4.16354e+11 1.14368e+06\n\n### Square\n\nLength = n\n Perimeter 728092 3.31324e+10 257419\n\n### Cube\n\nLength = n\n Surface area 1.98794e+11 6.03085e+15 315273\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 546069 1.43467e+10 157637\n\n### Triangular Pyramid\n\nLength = n\n Surface area 5.7387e+10 7.10743e+14 148621\n\n## Cryptographic Hash Functions\n\nmd5 90f4442c2b329ab4ce326b7798772c63 063d4587254e3f7b257906ef238e99355eb10f0e 87c972362ac4edf0fa07b79811dbc4d4ddcae83c67a52b930b445b79a5df17c3 f4604bb15a6d5da85a9706a831e83dc9074a03d7e86203638271d374aa728657b12860844bdbf7f7d4a045df66089d007fb37d4f23e45ede2d9f3651fb09bd2e 71e7cffc1c90f3974d985b935e170151caa41b36" ]

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[ "## Conversion formula\n\nThe conversion factor from feet per second to knots is 0.59248380129641, which means that 1 foot per second is equal to 0.59248380129641 knots:\n\n1 ft/s = 0.59248380129641 kt\n\nTo convert 1018 feet per second into knots we have to multiply 1018 by the conversion factor in order to get the velocity amount from feet per second to knots. We can also form a simple proportion to calculate the result:\n\n1 ft/s → 0.59248380129641 kt\n\n1018 ft/s → V(kt)\n\nSolve the above proportion to obtain the velocity V in knots:\n\nV(kt) = 1018 ft/s × 0.59248380129641 kt\n\nV(kt) = 603.14850971974 kt\n\nThe final result is:\n\n1018 ft/s → 603.14850971974 kt\n\nWe conclude that 1018 feet per second is equivalent to 603.14850971974 knots:\n\n1018 feet per second = 603.14850971974 knots\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.0016579664608052 × 1018 feet per second.\n\nAnother way is saying that 1018 feet per second is equal to 1 ÷ 0.0016579664608052 knots.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that one thousand eighteen feet per second is approximately six hundred three point one four nine knots:\n\n1018 ft/s ≅ 603.149 kt\n\nAn alternative is also that one knot is approximately zero point zero zero two times one thousand eighteen feet per second.\n\n## Conversion table\n\n### feet per second to knots chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from feet per second to knots\n\nfeet per second (ft/s) knots (kt)\n1019 feet per second 603.741 knots\n1020 feet per second 604.333 knots\n1021 feet per second 604.926 knots\n1022 feet per second 605.518 knots\n1023 feet per second 606.111 knots\n1024 feet per second 606.703 knots\n1025 feet per second 607.296 knots\n1026 feet per second 607.888 knots\n1027 feet per second 608.481 knots\n1028 feet per second 609.073 knots" ]

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[ "Introduction of Electronics - science lessons for life\n\n## Thursday, January 5, 2017\n\nElectronics\nIntroduction\nElectronics has made a huge impact on our daily lives. We use many electronic devices in our day to day activities. Mobile phone, computers, televisions and radios are some examples for such electronic device\nMaterials that conduct electricity are known as electrical conductors. Conductors (copper, aluminium, iron, lead etc.) and mixed conductors (brass, nychrome, manganin) are examples of these. Materials that do not conduct electricity (ebonite, polythene, plastic, dry wood, asbestos, glass etc) are known as electrical insulators.\n\nThe reason behind the ability to conduct electricity is the ability of some of the electrons in the atoms of such materials to move freely within the conductor. Electrons in the outer shells of conductors act in this manner since they are not tightly bound to the nucleus. Since inter-atomic bonds (covalent bonds) between the atoms of insulators are strong, there are very few electrons that are free to move.\n\nMeanwhile, some materials conduct a small amount of electricity. Such materials are known as semiconductors. Materials such as silicon (Si) and germanium (Ge) in their crystalline form show such properties. These elements belong to the fourth group in the periodic table and have four electrons in their outermost shell. Such elements form crystal lattice structures by sharing the four electrons in their outermost shell to make covalent bonds with four nearby atoms and thereby acquiring a stable electronic configuration having eight electrons in the outermost shell.\n\nHowever, these bonds are rather weak and can be broken from the thermal energy available even at room temperature, releasing electrons.\n\nFigure A shows the covalent bonds of the silicon lattice at 0 K. All the bonds are complete at this temperature. Figure B shows that some bonds have been broken releasing some free electrons at a temperature higher than 0 K. An electron deficiency can be observed at the positions that the free electrons occupied previously. Such positions with an electron deficiency are known as holes. Due to the positively charged protons in the nucleus, a hole gives rise to a positive charge that has not been neutralized (In a neutral atom, the number of protons in the nucleus is equal to the number of and electrons). Therefore a hole is equivalent to a positive charge.\n\nFigure A\nA silicon lattice at 0 K\n\nFigure B\nA silicon lattice at temperature above 0 K\n\nIn semiconductors, not only electrons contribute to the conduction of electricity. When an electron in an adjacent atom jumps to an atom with a hole having a positive charge, the position of the hole can change. By changing the position of a hole from one atom to another in this manner, holes can move around in the lattice and contribute in conducting a current. Electrons act as negative charge carriers\n\nwhile holes act as positive charge carriers.\n\nTherefore, when an electric potential difference is applied across a semiconductor, holes move from the positive to the negative potential while electrons move from the negative to the positive potential and the (conventional) current flows from the positive to the negative potential.\n\n• In metallic conductors, the charge carriers that conduct electricity are the negatively charged electrons.\n• In semiconductors, the negatively charged electrons as well as the positively charged holes act as the charge carriers that contribute in the conduction of electricity.\n• Since a hole is generated in the breaking of a bond to release an electron, the number of carrier electrons present in a semiconductor is equal to the number of holes.\n• Therefore the semiconductor lattice is electrically neutral.\n\nIntrinsic Semiconductors\nPure semiconductor materials such as silicon (Si) and germanium (Ge) that exist in crystaline form as mentioned above are known as intrinsic semiconductors.\n\nEffect of Temperature on the Conduction of Electricity\nSince the random motion of free electrons increases as the temperature is increased, a rise in the temperature inhibits the current flow. Therefore, a temperature rise in conductors causes a decrease in the conductivity (increase in the resistivity). However in semiconductors, a rise in temperature breaks bonds generating more holes and free electrons causing an increase in the conductivity (decrease in the resistivity).\n\nExtrinsic Semiconductors\nLet us consider what happens when a minute amount of the element phosphorous (P) is mixed (doped) to an intrinsic semiconductor such as Si. Phosphorous is an element in group V of the periodic table and has five electrons in the outermost shell. A phosphorous atom makes the number of electrons in its outermost shell eight by acquiring four electrons from four nearby silicon atoms around it. In the process, one of the five electrons is left behind without taking part in forming a\nbond. This electron has the opportunity to move about freely in the lattice.\n\nFigure C\nA Si lattice doped by phosphorous\n\nFigure C shows how a phosphorous atom forms bonds with silicon atoms. The\nelectron left behind increases the conductivity of the lattice. Since negatively\ncharged electrons are introduced to the lattice as charge carriers, the semiconductor\nis known as a negative type or n-type semiconductor. Semiconductors whose\ncarriers have been increased by doping it with another element are known as\nextrinsic semiconductors. By doping an intrinsic semiconductor with other elements in group V such as arsenic (As) and antimony (Sb) also, n-type extrinsic semiconductors can be formed. Since electrons are donated to the lattice by group V elements, they are known as donor atoms.\n\nIf Si which is an intrinsic semiconductor is doped with an element in group III such as boron (B), the boron atom forms bonds with nearby silicon atoms. However, since there are only three electrons in the outermost shell of the boron atom, there is a deficiency of one electron in order to form four bonds. Figure D shows how the atoms and bonds are configured in this case.\n\nFigure D\nA Si lattice doped by boron\n\nA hole exists at the point where the electron is deficient in the boron atom to form a bond. Since holes can conduct electricity as positive charges, the conductivity of silicon increases. As a hole is equivalent to a positive charge, such extrinsic semiconductors are known as positive or p-type semiconductors. By doping an intrinsic semiconductor with other elements in group III such as aluminium (Al), gallium (Ga) and indium (In) instead of B also p-type extrinsic semiconductors\ncan be formed. Since holes that can receive electrons are produced by group III elements, they are known as accepter atoms.\n\nsource by internet and books" ]

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[ "numpy.diagflat¶ numpy.diagflat (v, k=0) [source] ¶ Create a two-dimensional array with the flattened input as a diagonal. If a is 2-D, returns the diagonal of a with the given offset, i.e., the collection of elements of the form a[i, i+offset]. So, for this we are using numpy.diagonal() function of NumPy library. Flip the entries in each row in the left/right direction. diag = [ mat[i][i] for i in range(len(mat)) ] or even. A view of m with the columns reversed. In NumPy dimensions are called axes. You can rate examples to help us improve the quality of examples. To get the leading diagonal you could do. Python diagonal - 30 examples found. See the more detailed documentation for numpy.diagonal if you use this function to extract a diagonal and wish to write to the resulting array; whether it returns a copy or a view depends on what version of numpy you are using. If a has more than two dimensions, then the … numpy.diagonal¶ numpy.diagonal (a, offset=0, axis1=0, axis2=1) [source] ¶ Return specified diagonals. Parameters v array_like. For example, if we have matrix of 2×2 [ [1, 2], [2, 4]] then answer will be (4*1)-(2*2) = 0. Diagonal to set; 0, the default, corresponds to the “main” diagonal, a positive (negative) k giving the number of the diagonal above (below) the main. These are the top rated real world Python examples of numpy.diagonal extracted from open source projects. 2: diagonal(): diagonal function in numpy returns upper left o right diagonal elements. That axis has 3 elements in it, so we say it has a length of 3. Let’s see the program for getting all 2D diagonals of a 3D NumPy array. Numpy linalg det() Numpy linalg det() is used to get the determinant of a square matrix. It is a table of elements (usually numbers), all of the same type, indexed by a tuple of positive integers. But what is the determinant of a Matrix: It is calculated from the subtraction of the product of the two diagonal elements (left diagonal – right diagonal). Input data, which is flattened and set as the k-th diagonal of the output.. k int, optional. If a is 2-D, returns the diagonal of a with the given offset, i.e., the collection of elements of the form a[i, i+offset]. This function return specified diagonals from an n-dimensional array. The sub-arrays whose main diagonals we just obtained; note that each corresponds to fixing the right-most (column) axis, and that the diagonals are “packed” in rows. numpy.fliplr¶ numpy.fliplr (m) [source] ¶ Flip array in the left/right direction. numpy.diag¶ numpy.diag (v, k=0) [source] ¶ Extract a diagonal or construct a diagonal array. numpy.diagonal numpy.diagonal(a, offset=0, axis1=0, axis2=1) [source] Return specified diagonals. diag = [ row[i] for i,row in enumerate(mat) ] And play similar games for other diagonals. numpy.fill_diagonal — NumPy v1.19 Manual, Value to be written on the diagonal, its type must be compatible with that of the array a. wrapbool. numpy.diagonal(a, offset=0, axis1=0, axis2=1) [source] Return specified diagonals. For example, for the counter-diagonal (top-right to bottom-left) you would do something like: diag = [ row[-i-1] for i,row in enumerate(mat) ] Input array, must be at least 2-D. Returns f ndarray. For example, the coordinates of a point in 3D space [1, 2, 1] has one axis. NumPy’s main object is the homogeneous multidimensional array. If a is 2 -D and not a matrix, a 1 -D array of the same type as a containing the diagonal is returned. Parameters m array_like. Columns are preserved, but appear in a different order than before. ¶ Return specified diagonals 3D space [ 1, 2, 1 ] has axis! A 3D numpy array diagonal of the same type, indexed by tuple. Numbers ), all of the output.. k int, optional ¶! For i, row in the left/right direction, the coordinates of point. Examples of numpy.diagonal extracted from open source projects diagonals of a 3D numpy array help us improve quality! Create a two-dimensional array with the flattened input as a diagonal array, must be at least returns! ( len ( mat ) ] and play similar games for other.. 2-D. returns f ndarray elements ( usually numbers ), all of output. Elements ( usually numbers ), all of the output.. k int, optional has..., which is flattened and set as the k-th diagonal of the output.. k int, optional numbers. For i in range ( len ( mat ) ] or even numpy array so we say it has length... Examples of numpy.diagonal extracted from open source projects [ 1, 2, 1 ] has axis... ] Return specified diagonals from an n-dimensional array ) is used to get the of... A two-dimensional array with the flattened input as a diagonal numpy returns upper left right... Positive integers output.. k int, optional for i in range ( len ( mat ) ) ] even! Function in numpy returns upper left o right diagonal elements v, ). Tuple of positive integers help us improve the quality of examples by a tuple of positive integers this function specified. 1, 2, 1 ] has one axis an n-dimensional array we say it has a length of.... You can rate examples to help us improve the quality of examples diagonal elements array... Returns f ndarray we say it has a length of 3 array, must be least! Of the output.. k int, optional numpy library numpy array output.. k int, optional in (! To help us improve the quality of examples linalg det ( ): diagonal function in numpy returns left., indexed by a tuple of positive integers source ] Return specified diagonals (... Has a length of 3 length of 3 numpy.diagonal numpy.diagonal ( a, offset=0, axis1=0, axis2=1 [. 3D space [ 1, 2, 1 ] has one axis in enumerate ( )! For other diagonals different order than before rated real world Python examples of extracted. Numpy linalg det ( ) function of numpy library rated real world Python of... Mat ) ) ] and play similar games for other diagonals in the left/right direction order... Be at least 2-D. returns f ndarray o right diagonal elements numpy.diagonal (... Input data, which is flattened and set as the k-th diagonal of the output.. k int optional..., for this we are using numpy.diagonal ( ) is used to get the determinant of square... The entries in each row in the left/right direction ] [ i ] [ i ] for i range... So we say it has a length of 3 examples to help us improve the of. The determinant of a point in 3D space [ 1, 2, 1 ] has one axis or.... ¶ Create a two-dimensional array with the flattened input as a diagonal the program for getting all 2D diagonals a! Indexed by a tuple of positive integers len ( mat ) ) ] or even Return specified from... For this we are using numpy.diagonal ( a, offset=0, axis1=0, axis2=1 ) [ source ] ¶ specified! Examples to help us improve the quality of examples are using numpy.diagonal ( ) is used to get the of... Input data, which is flattened and set as the k-th diagonal of the output.. k int,.... Of the output.. k int, optional rate examples to help us the... In it, so we say it has a length of 3 offset=0, axis1=0, )! Row [ i ] for i, row in the left/right direction the program for getting all 2D diagonals a..." ]

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[ "## gvSIG 2.0: Scripting, exploit your gvSIG (III): Generate a polygon from a course\n\nSome time ago a friend of the project, Gustavo Agüero, published a post on his blog in which he explained how gvSIG generate a polygon from a bearing (or route or course). We found a very interesting exercise and with its help we implemented a script to undertake the same steps in gvSIG 2.0. The result is explained in this post.\n\nWe want to clarify that all the merit of this exercise is to Gustavo and the errors are because of our ignorance and for not properly follow his instructions. If you see some error, please, comment it and we will correct it 😉\n\nWe start with a bearing or course that presents the data to the following structure LINE, AZIMUT, DISTANCE, being AZIMUT the angle direction in the sense counted clockwise from the geographic north (download rumbo.jpg).\n\nThe first thing to do is to save this file to a format that would be able to read from a script, this file we have created is a csv file that we named rumbo.csv (download rumbo.csv).\n\nOnce we have our file we have to think what we need to get its content into a shapefile. Obviously we’ll need a shapefile where to save the results and we’ll need to create the features to represent the data for including them on the shapefile. To create these features we’ll need to transform the polar coordinates (azimuth, distance) in rectangular coordinates (X, Y).\n\nSome comments on that: Gustavo in his post recommend to make a representation of the data. This was very helpful to us.\n\nIn the script we will create a LINE shapefile and another one for POLYGONS. In the LINE type, we will keep heading data, data from coordinates transformation, and the resulting coordinates. In the POLYGONS shapefile will keep an identifier of the feature and a text field.\n\nWe start from a desktop gvSIG 2.0 (in my case I’ve used is build 2060 RC1) with the latest version of the scripting extension installed (currently the number 36).\n\nOnce we open the script editor (menu bar Tools / Scripting / Scripting Composer) we create a new script and start to write our script.\n\nThe first thing we’re going to do is to create the output layers using the function createShape from the gvsig module. The syntax of this function tells us that you need a data definition for the features, a route that is the place where to store the shapefile we are acreating, the projection of the shapefile, and the type of geometry that will contain. The definition of the data will be created using the function createSchema from the same gvsig module.\n\nThe source code, including comment would be:\n\n```import gvsig\nimport geom\n\ndef main():\n\n'''\nCreate a polygon layer with the following data model:\n- 'ID', Integer\n- 'OBSERVACIONES', string\n- 'GEOMETRY', Geometry\n\nCreate a line layer with the following data model:\n- \"ID\", Integer\n- \"LINEA\", string\n- \"MINUTOS\",long\n- \"DISTANCIA\", double\n- \"X\", double\n- \"Y\", double\n- \"GEOMETRY\", Geometry\n'''\n\n#Set up the projection, it is the same for both layers\nCRS=\"EPSG:32617\"\n\n#Create the object that represent the data model for the polygonal shapefile\nschema_poligonos = gvsig.createSchema()\n\n#Insert the filed from the data model\nschema_poligonos.append('ID','INTEGER', size=7, default=0)\nschema_poligonos.append('OBSERVACIONES','STRING', size=200, default='Sin modificar')\nschema_poligonos.append('GEOMETRY', 'GEOMETRY')\n\n#Set up the layer path. Remember changing it!!!\nruta='/tmp/rumbo-poligonos.shp'\n\n#Create the shapefile\nshape_poligonos = gvsig.createShape(\nschema_poligonos,\nruta,\nCRS=CRS,\ngeometryType=geom.SURFACE\n)\n\n#Create the line shapefile\n\n#Create the object that represent the data model for the line shapefile\nschema_lineas = gvsig.createSchema()\n\n#Insert the field from the data model\nschema_lineas.append('ID','INTEGER', size=7, default=0)\nschema_lineas.append('LINEA','STRING',size=50,default='')\nschema_lineas.append('MINUTOS','LONG', size=7, default=0)\nschema_lineas.append('SEGUNDOS','LONG', size=7, default=0)\nschema_lineas.append('DISTANCIA','DOUBLE', size=20, default=0.0, precision=6)\nschema_lineas.append('X','DOUBLE', size=20, default=0.0, precision=6)\nschema_lineas.append('Y','DOUBLE', size=20, default=0.0, precision=6)\nschema_lineas.append('GEOMETRY', 'GEOMETRY')\n\n#Set up the layer path. Remember changing it!!\nruta='/tmp/rumbo-lineas.shp'\n\n#Create the shapefile\nshape_line = gvsig.createShape(\nschema_lineas,\nruta,\nCRS=CRS,\ngeometryType=geom.MULTILINE\n)```\n\nOk, we already have our output layers, now we are going to see how to get the data from csv file we created. Python has a module that allows csv csv file handling very comfortable ( python csv ). The source code for reading the csv file would be:\n\n```import csv\nimport os.path\n\n#Set up the layer path for the CSV file. Remember changing it!!!\ncsv_file = '/tmp/rumbo.csv'\n\n#Check that the file exists on the provided path\nif not os.path.exists(csv_file):\nprint \"Error, el archivo no existe\"\nreturn\n\n#Open the file on read only mode\ninput = open(csv_file,'r')\n\n#Create a reader object from the CSV module\n\n#Print the line\nprint ', '.join(row)```\n\nWe already know how to create shapefiles and read csv files, the next step would be to create the features, so we must transform the polar coordinates to rectangular and it is at this point that for me things get a little dark so forgive the errors that you can find.\nTo make the coordinate transformation we’ll convert the decimal angle in radian angle. Gustavo in his post how to generate a polygonal (in Spanish) between the files to download you can find a PDF that provides an explanation of the calculations he made, if anybody has doubts I recommend to read it. I only want to clarify that we ‘ll use the python module python math to use math functions sine and cosine, needed to perform the transformation. In addition, our course uses relative coordinates, so we need a starting point and the code to calculate the new coordinates from the above ones. Our starting point is ( X= 361820.959424, Y = 1107908.627000). The source code would be:\n\n```\"\"\"\nConvert decimal degrees, minutes, seconds to radian\n\"\"\"\nimport math\n#Define the “pi” value\nPI = 3.1415926\n\nx_ant = 361820.959424\ny_ant = 1107908.627000\n\n#Apply the equation and obtain the angle in radians\nangulo = (degrees+(minutes/60.0)+(seconds/3600.0))*PI/180.0\n\n#Convert the polar coordinates into rectangular ones\n\n#Add relative coordinates values to get the next coordinate\nx = x_new + x_ant\ny = y_new + y_ant```\n\nThe last piece of our puzzle is to learn how to create the features and their geometries. The layers that we created earlier shape_line and shape_polygons have a method called append that allows us to add the features to the layer. This method accepts a dictionary,\npython dict that uses as key fields the ones we defined in the data structure of the layer (defined when we created it). The values are the ones that should have the feature. The geometries will be created using the geometric module geom from the scripting extension, in particular the createGeometry function, that needs the type of geometry to be created and the dimensions of the geometry. The code that we could use is:\n\n```geometry_multiline = geom.createGeometry(MULTILINE, D2)\n\nvalues = dict()\nvalues[\"ID\"] = id #Calculated field for the registered number\nvalues[\"LINEA\"] = linea_id #First data of the course\nvalues[\"MINUTOS\"] = minutos #decimal minutes from the course\nvalues[\"SEGUNDOS\"] = segundos #decimal seconds from the course\nvalues[\"DISTANCIA\"] = radio #Distance from the course\nvalues[\"X\"] = x #X coordinate calculated\nvalues[\"Y\"] = y #Y coordinate calculated\n\nshape_line.append(values)```\n\nAt this point we have seen everything needed to get what we want from our original course, we only need to assemble the puzzle and the final result would be this:" ]

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[ "# (Python API) Importing, Setting Pivot Point, Locating and Rotating\n\nI’m trying to import a model, set its location it and rotate the whole model using the quaternion mode.\n\nHere’ s my code:\n\nbpy.ops.object.select_all(action='SELECT')\nbpy.ops.object.delete()\nbpy.ops.import_scene.obj(filepath='Bench_1.obj', axis_forward='-Z', axis_up='Y')\nobject = bpy.context.selected_objects\nfor obj in bpy.context.selected_objects:\nif obj.type == 'MESH':\nobj.location.x = 5\nobj.location.z = 5\nobj.location.y = 5\nobj.rotation_mode = \"QUATERNION\"\nobj.rotation_quaternion.x = 1\nobj.rotation_quaternion.y = 1\nobj.rotation_quaternion.z = 0\nobj.rotation_quaternion.w = 1\nbpy.context.scene.update()\n\n\nIt works fine, but in this case model’s pivot point is right in the center.", null, "Some other models may have pivot point far away from geometry, so I can’t neither locate, nor rotate them properly. I decided to put an extra line in my code that will always set pivot point in the center of any imported model:\n\nbpy.ops.object.select_all(action='SELECT')\nbpy.ops.object.delete()\nbpy.ops.import_scene.obj(filepath='Bench_1.obj', axis_forward='-Z', axis_up='Y')\nobject = bpy.context.selected_objects\nbpy.ops.object.origin_set(type='ORIGIN_CENTER_OF_MASS') # EXTRA LINE HERE!\nfor obj in bpy.context.selected_objects:\nif obj.type == 'MESH':\nobj.location.x = 5\nobj.location.z = 5\nobj.location.y = 5\nobj.rotation_mode = \"QUATERNION\"\nobj.rotation_quaternion.x = 1\nobj.rotation_quaternion.y = 1\nobj.rotation_quaternion.z = 0\nobj.rotation_quaternion.w = 1\nbpy.context.scene.update()\n\n\nAfter this update my ‘normal’ models become fractured:", null, "What am I doing wrong?\n\nI suspect that all of the points in the geometry are relative to the origin, but that likely does not explain the problem you are experiencing. I found this function on the web, it seems to move the pivot point without the problems you are experiencing.\n\nimport bpy\nfrom mathutils import Vector, Matrix\n\ndef set_origin(ob, global_origin=Vector()):\nmw = ob.matrix_world\no = mw.inverted() @ Vector(global_origin)\nob.data.transform(Matrix.Translation(-o))\nmw.translation = global_origin\n\n# call the function:\nset_origin(bpy.context.object)" ]

[ null, "https://i.stack.imgur.com/cahuL.png", null, "https://i.stack.imgur.com/ylpf8.png", null ]

[ "# Notation related to probability spaces\n\nLet $A$ and $B$ be events in the probability space $(\\Omega, \\mathcal{F}, \\mathcal{P})$. How would I write that\n\nevents $A$ and $B$ both occur\n\nand\n\nevent $A$ occurs with probability $0.5$ and the event $B$ does not occur\n\nand\n\nneither of the events occurs?\n\n## 1 Answer\n\nAn event $A$ (a subset of $\\Omega$ that is also a member of $\\mathcal F$) is said to occur if and only if the outcome of the experiment is an element that belongs to $A$. The outcome $\\omega$ of the experiment is always one element of the sample space $\\Omega$, but the occurrence of this outcome $\\omega$ leads to many events occurring because $\\omega$ is a member of numerous subsets of $\\Omega$ that are in $\\mathcal F$. In particular, regardless of what $\\omega$ is, one and only one of the events $A$ and $A^c$ must occur because $\\omega$ necessarily belongs to exactly one of $A$ and $A^c$; it cannot belong to both, and it cannot be excluded from both. If $\\mathcal F$ is a finite collection of subsets (there must be $2^n$ of them for some integer $n$), then half the events occur and half don't, regardless of what $\\omega$ is. In some sense, this is also true when $\\mathcal F$ has infinitely many subsets: there is a one-to-one correspondence $A \\leftrightarrow A^c$ between those events that occur and those that don't. $\\Omega$ always occurs and so is called the sure or certain event, while its complement $\\emptyset$ never occurs ans so is called the impossible event (also the null event by some, but there are many, including myself, who deprecate this usage).\n\nWith that as prologue, if you are told that events $A$ and $B$ both have occurred, then the outcome $\\omega$ must be a member of both $A$ and $B$, right? What is the notation for the set of elements that are members of both $A$ and $B$?\n\n• So $\\omega\\in A\\cap B$ for both occuring at the same time? Thanks for the thorough answer, it was helpful. – mmh Sep 17 '15 at 19:31" ]

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[ "A digit addition generator of an integer n is any integer x that satisfy the equation x + s(x) = n, with s(x) being the sum of the digits of x. (We will work under base 10 for convenience.)\n\nFor example, a digit addition generator for 29 would be 19, because 19 + (1 + 9) = 29. Some numbers have more than one generator. An example might be 216, which has generators of 198 and 207.\n\nYour objective is to generate the sequence a_n where a_i is the lowest digit addition generator of every non-negative integer i, and anything other than a non-negative integer if there is none for i. The non-negative terms in your result should match the sequence A096234. You may find this paper related to the challenge.\n\nFewest bytes win; standard rules apply.\n\n• If there is no solution, is it acceptable to loop forever without output, or do we need to terminate?\n– xnor\nJun 15 at 8:31\n• @xnor That's okay. Jun 15 at 8:41\n• Presumably the expected output for 0 is 0? Jun 15 at 14:09\n• @Shaggy Yes. 0 + (0) is indeed 0 Jun 15 at 16:12\n• Just making sure that x<=n and not just x<n. Perhaps you could add n=0 to the challenge as a test case, along with a few others, as a few of us did get tripped up by it. Jun 15 at 22:31\n\n# VyxalgM, 28 bitsv2, 3.5 bytes\n\n'∑+?=\n\n\nTry it Online!\n\nOutputs an empty list for no result. I think (spoiler I did) I ninja'd someone in the process of writing this lol.\n\n## Explained\n\n'∑+?=\n' # filter the range [0, n] by:\n∑+ # sum of digits + x\n?= # equals input?\n# g flag prints minimum.\n\n• Lol, I've been ninja'd Jun 15 at 9:14\n\n# 05AB1E, 7 bytes\n\nÝ.ΔDªOQ\n\n\nOutputs -1 if there is no result.\n\nExplanation:\n\nÝ # Push a list in the range [0, (implicit) input-integer]\n.Δ # Find the first value of this list that's truthy for, or -1 if none are:\nD # Duplicate the current value\nª # Convert the first value to a list of digits,\n# and append the duplicated number to this list\nO # Sum the list together\nQ # Check whether it's equal to the (implicit) input-integer\n# (after which the result is output implicitly)\n\n• That was quick! Jun 15 at 8:40\n• @TomEpsilon Could have been quicker if I saw the challenge 10 minutes earlier. ;) Nice first challenge btw, and welcome to CGCC! Jun 15 at 8:41\n\n# PARI/GP, 44 bytes\n\nf(n,i)=if(i>n,x,i+sumdigits(i)-n,f(n,i+1),i)\n\n\nAttempt This Online!\n\nReturns x when there is no solution.\n\n# Japt, 8 bytes\n\nOutputs -1 for no result.\n\nôÈ+ìxÃbU\n\n\nTry it\n\n(Eventually) outputs undefined for no result.\n\n@¶X+ìx}a\n\n\nTry it\n\nôÈ+ìxÃbU :Implicit input of integer U\nô :Range [0,U]\nÈ :Pass each through the following function\nì : Convert to digit array\nà :End function\nbU :First 0-based index of U\n\n@¶X+ìx}a :Implicit input of integer U\n@ :Function taking an integer X as argument\n¶ : Test U for equality with\nì : Convert X to digit array\n} :End function\na :Get the first integer >=0 that returns true\n\n• Yup, your second version is what I came up with as well. (This is Jacob, I changed my username) Jun 15 at 12:22\n• 6 bytes +ìx ¥N -æ Jun 15 at 12:43\n• Came up with the exact same while I was out to lunch, @noodleman :) Do you want to run with it before I edit it into mine? Jun 15 at 13:44\n• Scratch that, @noodleman; it fails for 0. Jun 15 at 14:08\n• I think I can add the x flag, which does nothing for regular outputs and throws an error for undefined, i.e. empty output, and \"\" == 0. I'm going to post my version now. Jun 15 at 15:37\n\n# APL (Dyalog Extended), 21 bytes\n\nAnoymous tacit prefix function. Requires 0-indexing (⎕IO←0).\n\n⍸⊢<\\⍤=0,⍨⍳+1⊥¨⍎¨∘⍕¨∘⍳\n\n\nTry it online!\n\n⍸ where do we find that\n\n⊢ the argument\n\n<\\ is the first (lit. cum. right-ass. less red.)\n\n⍤ that:\n\n= equals\n\n0,⍨ zero appended to\n\n⍳ the range\n\n+ plus\n\n1⊥ the sum (lit. base-1 eval.)\n\n¨ of each of\n\n⍎ the evaluation\n\n¨ of each\n\n⍤ of:\n\n⍕ the stringification\n\n¨ of each\n\n⍤ of:\n\n⍳ the range\n\n• I’m sorry, but what does “ lit. cum. right-ass. less red.” mean? Jun 15 at 11:25\n• @TomEpsilon literally cumulative right-associative less reduction\nJun 15 at 12:02\n\n# C (clang), 58 bytes\n\nloops forever if there's no solution\n\nd,s;f(*i,n){for(*i=s=0;n-s;)for(s=d=++*i;d;d/=10)s+=d%10;}\n\n\nTry it online!\n\n# C (clang), 70 bytes\n\nFor comparison, this returns negative values when there's no solution.\n\ng,o,l;f(*i,n){for(*i=g=-n;g++;*i=n+l?*i:-g)for(l=o=g;o;o/=10)l+=o%10;}\n\n\nTry it online!\n\n# C++ (gcc), 71 $$\\\\cdots\\$$ 66 64 bytes\n\n[](int&n){for(int m=n,d=n=0,t;m-d;)for(d=t=++n;t;t/=10)d+=t%10;}\n\n\nTry it online!\n\nSaved 6 bytes thanks to c--!!!\n\n• @c-- Nice one - thanks! :D Jun 16 at 10:06\n• @c-- Very good - thanks! :D Jun 22 at 20:40\n\n# JavaScript, 49 bytes\n\nLoops forever, in theory, if there's no result but, in practice, throws a recursion error.\n\nn=>(g=x=>n-eval([x,...x+].join+)?g(++x):x)0\n\n\nTry it online!\n\n# Raku, 34 bytes\n\n{first 0..$^n: {$_+.comb.sum==$n}} Try it online! Returns the first element of the range 0 through the input number $^n/$n such that the element $_ plus the sum of its digits .comb.sum is equal to the input number. If there is no such number, Nil is returned.\n\n# R, 45 46 bytes\n\nEdit: +1 byte to correctly handle input of zero, but then -1 byte thanks pajonk for removal of useless accidental space\n\n\\(n){while(n-F-sum(F%/%10^(0:F)%%10))F=F+1;+F}\n\n\nAttempt This Online!\n\nTests increasing integers until it finds a solution, so loops forever if no solution exists.\n\n• You have an extra space at the start of you code... Jun 15 at 17:57\n• @pajonk - Ah, thanks v much for spotting that! Jun 15 at 18:16\n\n# Pyth, 8 bytes\n\nx|msajdT\n\n\nTry it online!\n\nReturns -1 for no result.\n\n### Explanation\n\nx|msajdTdQQQ # implicitly add dQQQ\n# implicitly assign Q = eval(input())\nm Q # map lambda d over range(Q):\njdT # list the digits of d\na d # append d to this list\ns # sum the list\n| Q # short circuiting or, does nothing unless the input is 0, then [] -> 0\nx Q # find the first index of Q in the mapped list (or xors with 0 for 0), returns -1 if not found\n\n\n• Input 0 also yields -1, which is untrue (should have been 0). I apologize if this might cause non-alphabetical issues. Jun 15 at 16:54\n• @TomEpsilon My bad, did not realize I needed to account for 0. And indeed it has :( but even worse it cost me a byte :((( Jun 15 at 21:09\n\n# Arturo,  40  35 bytes\n\n$->x[0while[x<>+∑digits<=<=]->1+] Try it! Causes an infinite loop when there is no result. $->x[ ; a function taking an integer x\n0 ; push 0 to stack\nwhile[x<> ; while x doesn't equal...\n<=<= ; duplicate top of stack twice\n+∑digits ; digit sum then add\n] ; end while condition\n->1+ ; increment top of stack (while body)\n] ; end function\n\n\n# Elixir, 74 72 bytes\n\nimport Enum\nc=& &1-?0\nr=&find(1..&1,fn x->&1==x+sum map'#{x}',c end)||-1\n\n\nTry it online!\n\n# Wolfram Language (Mathematica), 895349 44 bytes\n\n#&@@Pick[r=Range@#,r+Tr/@IntegerDigits@r,#]&\n\n\nTry it online!\n\nThanks to @ovs and @att!\n\n• That was hell of an optimization. Jun 15 at 8:56\n• Sure it can be more golfed Jun 15 at 8:58\n• 49 bytes with Range[n=#] and Tr instead of Total.\n– ovs\nJun 15 at 9:54\n• Cases variants typically are shorter than Select variants when you need the outer argument in the selection function, but here Pick does better still: 43 bytes\n– att\nJun 16 at 1:00\n\n# Julia, 42 37 bytes\n\n~n=argmax(x->x+sum(digits(x))==n,0:n)\n\n\nAttempt This Online!\n\n-5 bytes thanks to MarcMush: use argmax instead of findfirst\n\n• Your TIO doesn't seem to work. Jun 15 at 22:32\n• I added an example that prints. Before, I had only been using @assert statements, which will silently pass if the tests succeed. Jun 16 at 1:27\n• with argmax, 41 bytes or 37 bytes with julia 1.7+ Jun 16 at 18:34\n\n# Java, 85 bytes\n\nn->{for(int i=-1;i++<n;)if((i+\"\").chars().map(d->d-48).sum()+i==n)return i;return-1;}\n\n\nOutputs -1 if there is no result.\n\nTry it online.\n\nExplanation:\n\nn->{ // Method with integer as both parameter and return-type\nfor(int i=-1;i++<n;) // Loop i in the range (-1,n]:\nif((i+\"\") // Convert i to a String\n.chars() // Then to an IntStream of codepoints\n.map(d->d-48) // Then to an IntStream of digits\n.sum() // And sum those digits together\n+i // Add integer i\n==n) // And if it's equal to input n:\nreturn i; // Return i as result\nreturn-1;} // If no result is found, return -1 instead\n\n\nUsing a recursive function which throws a StackOverflowError when there is no result is also 85 bytes:\n\nn->f(n,0)int f(int n,int i){return(i+\"\").chars().map(d->d-48).sum()+i==n?i:f(n,i+1);}\n\n\nTry it online.\n\n# Charcoal, 13 bytes\n\nＮθＩ⌊Φ⊕θ⁼θ⁺ιΣι\n\n\nTry it online! Link is to verbose version of code. Takes i as an input and outputs None if no generator exists. Explanation: Brute force approach.\n\nＮθ First input as an integer\nθ First input\n⊕ Incremented\nΦ Filter on implicit range\nι Current value\n⁺ Plus\nι Current value\nΣ Digit sum\n⁼ Equals\nθ First input\n⌊ Take the minimum\nＩ Cast to string\nImplicitly print\n\n\n23 bytes for a less inefficient version:\n\nＮθＩ⌊Φ…·⁻θ×⁹Ｌθθ⁼θ⁺ιΣ⌈⟦⁰ι\n\n\nTry it online! Link is to verbose version of code. Explanation: Starts checking for generators starting from i minus 9 times the number of digits of i.\n\n28 bytes for a more efficient version:\n\nＮθＩ⌊ΦＥＬθ⁻⁻θ﹪×⁵θ⁹×⁹ι⁼θ⁺ιΣ⌈⟦⁰ι\n\n\nTry it online! Link is to verbose version of code. Explanation: As above, but only checks for generators which when doubled are equivalent to i modulo 9 (because i equals the generator plus its digit sum but both the generator and the digit sum are equivalent modulo 9) although it actually computes the maximum possible generator by subtracting 5i reduced modulo 9 from i.\n\n# C# (.NET Core), 73 72 bytes\n\nn=>{for(int i=-1;i++<n;)if((i+\"\").Sum(d=>d-'0')+i==n)return i;return-1;}\n\n\nTry it online!\n\nThis is a C# port of this answer. All the kudos goes to Kevin Cruijssen.\n\nI really like that .NET's sum is taking a lambda to avoid the common map followed by sum!\n\n• You can remove the space at return-1; like in my Java answer for -1 byte. :) Jun 16 at 8:45\n• @KevinCruijssen forgot this one. Thank you! Jun 16 at 17:05\n\n# Python 3, 63 bytes\n\ng=lambda n:min(x for x in range(n)if x+sum(map(int,str(x)))==n)\n\n\nTry it online!\n\n• Welcome to Code Golf, and nice first answer! Jun 20 at 21:15\n\n# Python 3, 56 bytes\n\nf=lambda n,o=0:(o+sum(map(int,str(o)))==n)*o or f(n,o+1)\n\n\nTry it online!\n\n• Welcome to Code Golf, and nice answer! Jun 20 at 23:23\n• o*(...)or f(... saves a byte. Jul 17 at 15:24\n\n# Fig, $$\\11\\log_{256}(96)\\approx\\$$ 9.054 bytes\n\n[KFax'=#x+S\n\n\nTry it online!\n\n[ # The first item from\na # the range from 1 to\nx # the input,\nF ' # filtered by:\nS # the digit sum of n\n+ # plus n\n= # is equal to\n#x # the program's input,\nK # sorted (smallest to largest).\n\n\n# Swift, 84  80 bytes\n\n{n in(0...n).map{($0,\"\\($0)\".reduce($0){$0+$1.hexDigitValue!})}.first{$1==n}!.0}\n\n\nTakes an Int in and returns an Int.\n\nTry it online!\n\nUngolfed:\n\n{ (n: Int) -> Int in\n(0...n) // a sequence of all integers from 0 to n, inclusive\n// the parens are needed to avoid parsing as 0...(n.map)\n.map { i in // transform each one\n( // return a tuple of:\ni, // 0. the integer itself\n\"\\(i)\".reduce(i) { $0 +$1.hexDigitValue! } // 1. the digit sum plus the number\n// technically, wholeNumberValue would be more correct than hexDigitValue\n)\n}\n.first { $1 == n }! // find the first one where the sum is the input .0 // and return the number that gave that sum } # Perl 5-MList::Util=sum -pa, 29 bytes $_=0;$_++while\"@F\"-sum$_,/./g\n\n\nTry it online!\n\n# Desmos, 66 bytes\n\nI=[0...n]\nf(n)=I[[i+∑_{k=0}^imod(floor(i/10^k),10)fori=I]=n].min\n\n\nReturns undefined if no solution exists.\n\nTry It On Desmos!\n\nTry It On Desmos! - Prettified\n\n# Ruby, 46 bytes\n\n1.step{|c|p (1..c).find{|x|x+x.digits.sum==c}}\n\n\nTry it online!\n\n# Thunno 2M, 5 bytes\n\næDS+=\n\n\nAttempt This Online!\n\n#### Explanation\n\næDS+= # Implicit input\næ # Filter [1..input] by:\nD # Duplicate\nS # Sum digits\n= # Equals input?\n# Take the minimum\n# Implicit output\n\n\na n=find((==n).ap(+)(sum.map digitToInt.show))[1..n]\n\nn=>(0 to n).find(i=>(i.toString.map(_.asDigit).sum+i)==n).getOrElse(-1)" ]

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[ "# PERBANDINGAN METODE RUNTUN WAKTU FUZZY-CHEN DAN FUZZY-MARKOV CHAIN UNTUK MERAMALKAN DATA INFLASI DI INDONESIA\n\nNURKHASANAH, LINTANG AFDIANTI (2015) PERBANDINGAN METODE RUNTUN WAKTU FUZZY-CHEN DAN FUZZY-MARKOV CHAIN UNTUK MERAMALKAN DATA INFLASI DI INDONESIA. Undergraduate thesis, FSM Universitas Diponegoro.", null, "", null, "Preview\nPDF\n3148Kb\n\n## Abstract\n\nInflation data are financial time series data which often violate assumption if it is modeled with ARIMA Box-Jenkins classic method. Therefore, to forecast inflation are used forecast method which hasn’t requirement classic assumptions, like as fuzzy time series method. Fuzzy time series is a method of predicting data that use principles of fuzzy as basis. Many researches has been developed about this method, such as fuzzy time series developed by Chen (1996) and fuzzy time series-Markov chain developed by Tsaur (2012). In this case, both methods are used to predict inflation data in Indonesia. Result of predicting from both methods are compared with MSE value to in sample data. Method of fuzzy time series-Chen get MSE value 0,656, whereas method of fuzzy time series-Markov chain get MSE value 0,216. Because of this reason, method of fuzzy time series-Markov chain get smallest MSE value. So, this method as the best method. Furthermore, to evaluate the best of predicting model used MAPE value to out sample data. The MAPE value in method of fuzzy time series-Markov chain is 6,610%. As conclusion, model of fuzzy time series Markov chain have best performance. Keywords : Fuzzy time series, Markov chain , MSE, MAPE.\n\nItem Type: Thesis (Undergraduate) H Social Sciences > HA Statistics Faculty of Science and Mathematics > Department of Statistics 47305 Mr Hasbi Yasin 05 Jan 2016 17:42 05 Jan 2016 17:42\n\nRepository Staff Only: item control page" ]

[ null, "http://eprints.undip.ac.id/style/images/fileicons/application_pdf.png", null, "http://eprints.undip.ac.id/47305/thumbnails/1/preview.png", null ]

[ "## 14403\n\n14,403 (fourteen thousand four hundred three) is an odd five-digits composite number following 14402 and preceding 14404. In scientific notation, it is written as 1.4403 × 104. The sum of its digits is 12. It has a total of 2 prime factors and 4 positive divisors. There are 9,600 positive integers (up to 14403) that are relatively prime to 14403.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 12\n• Digital Root 3\n\n## Name\n\nShort name 14 thousand 403 fourteen thousand four hundred three\n\n## Notation\n\nScientific notation 1.4403 × 104 14.403 × 103\n\n## Prime Factorization of 14403\n\nPrime Factorization 3 × 4801\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 14403 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 14,403 is 3 × 4801. Since it has a total of 2 prime factors, 14,403 is a composite number.\n\n## Divisors of 14403\n\n1, 3, 4801, 14403\n\n4 divisors\n\n Even divisors 0 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 19208 Sum of all the positive divisors of n s(n) 4805 Sum of the proper positive divisors of n A(n) 4802 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 120.012 Returns the nth root of the product of n divisors H(n) 2.99938 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 14,403 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 14,403) is 19,208, the average is 4,802.\n\n## Other Arithmetic Functions (n = 14403)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 9600 Total number of positive integers not greater than n that are coprime to n λ(n) 4800 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1695 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 9,600 positive integers (less than 14,403) that are coprime with 14,403. And there are approximately 1,695 prime numbers less than or equal to 14,403.\n\n## Divisibility of 14403\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 4 3 3\n\nThe number 14,403 is divisible by 3.\n\n## Classification of 14403\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (14403)\n\nBase System Value\n2 Binary 11100001000011\n3 Ternary 201202110\n4 Quaternary 3201003\n5 Quinary 430103\n6 Senary 150403\n8 Octal 34103\n10 Decimal 14403\n12 Duodecimal 8403\n20 Vigesimal 1g03\n36 Base36 b43\n\n## Basic calculations (n = 14403)\n\n### Multiplication\n\nn×i\n n×2 28806 43209 57612 72015\n\n### Division\n\nni\n n⁄2 7201.5 4801 3600.75 2880.6\n\n### Exponentiation\n\nni\n n2 207446409 2987850628827 43034012606995281 619818883578553032243\n\n### Nth Root\n\ni√n\n 2√n 120.012 24.3305 10.955 6.7872\n\n## 14403 as geometric shapes\n\n### Circle\n\n Diameter 28806 90496.7 6.51712e+08\n\n### Sphere\n\n Volume 1.25155e+13 2.60685e+09 90496.7\n\n### Square\n\nLength = n\n Perimeter 57612 2.07446e+08 20368.9\n\n### Cube\n\nLength = n\n Surface area 1.24468e+09 2.98785e+12 24946.7\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 43209 8.98269e+07 12473.4\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.59308e+08 3.52122e+11 11760\n\n## Cryptographic Hash Functions\n\nmd5 2258803b6e4f1ef992229c3cef66d75d d03c5953c221eed40a203389e08b4f38f4171620 679eae51449ffd6af50d93f77307060461ab896fb64492a1e315b1855e02fa73 353c4ae6e980a01df54282b89e18e6b20f92d8d58944372d5663f3c8af97214fbd41636fd32db5ee3063d70c007ef6cb3c0eefff99542e18bdede49049f5bdc4 14f880ba5909f91a3f6295189574afcce1e19d7e" ]

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[ "Cody\n\n# Problem 1943. Signed Magnitude\n\nSolution 335681\n\nSubmitted on 18 Oct 2013 by FRANCISCO JAVIER\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = [1 2 -12]; y_correct = -12; assert(isequal(signed_mag(x),y_correct))\n\n2   Pass\n%% x = [1 10 20]; y_correct = 20; assert(isequal(signed_mag(x),y_correct))\n\n3   Pass\n%% x = [0 0.1 -0.2]; y_correct = -0.2; assert(isequal(signed_mag(x),y_correct))" ]

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[ "## Contact\n\nDepartment of Mathematics\nThe City College of New York\n160 Convent Avenue\nNew York, NY 10031\n\nPhone: (212) 650-5346\nFax: (212) 650-6294\[email protected]\n\n# Courses\n\nFor course information (like syllabi, past exams, etc.) see the pages of individual courses.\n\nMath 15000: Mathematics for the Contemporary World\nMath 17300: Introduction to Probability and Statistics\nMath 17700: Introduction to Biostatistics\nMath 18000: Quantitative Reasoning\nMath 18500: Basic Ideas in Mathematics\nMath 19000: College Algebra and Trigonometry\nMath 19500: Precalculus\nMath 20100: Calculus I\nMath 20300: Calculus III\nMath 20500: Elements of Calculus\nMath 20900: Elements of Calculus and Statistics\nMath 21200: Calculus II with Introduction to Multivariable Functions\nMath 21300: Calculus III with Vector Analysis\nMath 30200: Honors II\nMath 30300: Honors III\nMath 30400: Honors IV\nMath 30800: Bridge to Advanced Mathematics\nMath 31000: Independent Study\nMath 31100: Selected Topics in Mathematics\nMath 31103: Special Topics : Honors Linear Algebra\nMath 31200: Selected Topics in Mathematics\nMath 31300: Selected Topics in Mathematics\nMath 31400: Selected Topics in Mathematics\nMath 31500: Selected Topics in Mathematics\nMath 31600: Selected Topics in Mathematics\nMath 31700: Selected Topics in Mathematics\nMath 31900: Selected Topics in Mathematics\nMath 32000: Selected Topics in Mathematics\nMath 32800: Methods of Numerical Analysis\nMath 34200: History of Mathematics\nMath 34500: Theory of Numbers\nMath 34600: Elements of Linear Algebra\nMath 34700: Elements of Modern Algebra\nMath 36000: Introduction to Modern Geometry\nMath 36500: Elements of Combinatorics\nMath 36600: Introduction to Applied Mathematical Computation\nMath 37500: Elements of Probability Theory\nMath 37600: Mathematical Statistics\nMath 37700: Applied Statistics and Probability\nMath 38100: Discrete Models of Financial Mathematics\nMath 38200: Continuous Time Models in Financial Mathematics\nMath 39100: Methods of Differential Equations\nMath 39104: Methods of Differential Equations\nMath 39200: Linear Algebra and Vector Analysis for Engineers\nMath 39204: Linear Algebra and Vector Analysis for Engineers\nMath 39300: Introduction to Applied Fourier Analysis\nMath 39500: Complex Variables for Scientists and Engineers\nMath 41100: Topics in Pure Mathematics\nMath 41103: Topics in Pure Mathematics\nMath 41200: Topics in Applied Mathematics\nMath 41203: Topics in Applied Mathematics\nMath 41300: Topics in Probability and Statistics\nMath 41303: Topics in Probability and Statistics\nMath 43200: Theory of Functions of a Complex Variable\nMath 43400: Theory of Functions of a Real Variable\nMath 43500: Partial Differential Equations I\nMath 44300: Set Theory\nMath 44400: Mathematical Logic\nMath 44500: Dynamical Systems\nMath 44600: Linear Algebra\nMath 44900: Introduction to Modern Algebra\nMath 46100: Differential Geometry\nMath 46300: Topology\nMath 46400: Number Theory\nMath 46700: Mathematical Modeling\nMath 46800: Combinatorial Analysis\nMath 47700: Stochastic Processes I\nMath A1100: Topics in Pure Mathematics\nMath A1103: Topics in Pure Mathematics\nMath A1200: Topics in Applied Mathematics\nMath A1203: Topics in Applied Mathematics\nMath A1300: Topics in Probability and Statistics\nMath A1303: Topics in Probability and Statistics\nMath A3200: Theory of Functions of a Complex Variable I\nMath A3400: Theory of Functions of a Real Variable I\nMath A3500: Partial Differential Equations I\nMath A4300: Set Theory\nMath A4400: Mathematical Logic\nMath A4500: Dynamical Systems\nMath A4600: Linear Algebra\nMath A4900: Modern Algebra I\nMath A6100: Differential Geometry\nMath A6300: Topology I\nMath A6400: Number Theory\nMath A6800: Combinatorial Analysis\nMath A7700: Stochastic Processes I\nMath A9801: Independent Study\nMath A9802: Independent Study\nMath A9803: Independent Study\nMath A9804: Independent Study\nMath B1100: Selected Topics in Pure Mathematics\nMath B1200: Selected Topics in Applied Mathematics\nMath B1300: Selected Topics in Probability and Statistics\nMath B3200: Theory of Functions of a Complex Variable II\nMath B3400: Theory of Functions of a Real Variable II\nMath B3500: Partial Differential Equations II\nMath B4500: Dynamical Systems II\nMath B4900: Modern Algebra II\nMath B6300: Topology II\nMath B7700: Stochastic Processes II\nMath B7800: Advanced Topics in Statistics\nMath B9801: Independent Study\nMath B9802: Independent Study\nMath B9803: Independent Study\nMath B9804: Independent Study" ]

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[ "# The error in MOST logic studybooks (not in ZFC)\n\n0 users think it is non-scientific; after 15 it will be moderated.\n\nAfter a long discussion at Reddit where many counter and counter-counter arguments appeared, I was pointed an error:\n\nS(... S ...) is not defined. I meant to expand S inside such a statement before applying the outward S(), but I’ve realized that S is not defined solely by functional symbols but by a predicate, so “expand S” does not make sense.\n\nI need to confess that the article is worth withdrawal due to a fundamental error.\n\n## An error in most logic studybooks\n\nVictor Porton, no affiliation, [email protected]\n\nAbstract: In most logic study books “extension by definition” is erroneous.\n\nThis error is almost as serious as if we found a contradiction in ZF. There are no contradictions in ZF, but ZF is always used together with such definitions. So, for practical purposes it is as if the error were in ZF itself.\n\nThat’s a security hazard: Nuclear powerplants for instance are made using logic.\n\nIn this article I show how to produce a contradiction using ZF and extension by definition. I also show how to generalize this error to Peano arithmetic (or even predicate calculus).\n\nI show which logical systems are free of this paradox.\n\n# Gödel encoding\n\nEvery logical formula can be written as a string of characters. For example the formula that there exists a natural number whose square is below five:\n\n∃x∈ℕ:(x^2 < 5)\n\ncan be represented as characters,\n\n∃, x, ∈, ℕ, :, (, x,^, 2, >, 5, )\n\nAs Pythagorus said “everything is a number”. Every logical formula can be encoded as a number: For each logical formula we produce a number, and having this number, can reconstruct this formula back.\n\nGödel did it this way (see ):\n\nGödel used a system based on prime factorization. He first assigned a unique natural number to each basic symbol in the formal language of arithmetic with which he was dealing.\n\nTo encode an entire formula, which is a sequence of symbols, Gödel used the following system. Given a sequence", null, "of positive integers, the Gödel encoding of the sequence is the product of the first n primes raised to their corresponding values in the sequence:\n\nenc(x1,…,xn) = 2^x1 · … pn^xn", null, "According to the fundamental theorem of arithmetic, any number (and, in particular, a number obtained in this way) can be uniquely factored into prime factors, so it is possible to recover the original sequence from its Gödel number (for any given number n of symbols to be encoded).\n\nSo, every formula can be represented as a string or even as a single natural number.\n\n# The error, informally\n\nI will denote the set described by a set comprehension (more precisely, its numeric encoding, such as Gödel’s encoding) X as S(X).\n\nLet M be the set comprehension\n\n{ X∊set comprehensions | X not in S(X) }.\n\nTherefore:\n\nM in S(M) ⇔ M not in S(M).\n\n# Extension by definition is erroneous\n\nSo, we have a logical paradox in our basic logic. What’s wrong?\n\nI claim that the wrong construct is extension by definition, a construct appearing almost in every logic studybook without proof. As per Wikipedia :\n\nLet", null, "be a first-order theory (with equality) and f(y, x1, …, xn) a formula of T such that y, x1, …, xn are distinct and include the variables free in f(y, x1, …, xn). Assume that we can prove\n\n∀x1…∀xn∃!y:f(y, x1, …, xn).\n\nin T, i.e. for all x1, …, xn there exists a unique y such that f(y, x1, …, xn).\n\nFurther Wikipedia (and most studybooks) recommends:\n\nForm a new first-order theory T’ from T by adding a new n-ary function symbol g", null, ", the logical axioms featuring the symbol g and the new axiom\n\n∀x1…∀xn∃!y:g(f(y, x1, …, xn), x1, …, xn).\n\nI will prove that following this rule (“extension by definition”) and ZF we can enter into a contradiction.\n\nThere is an obvious recursive way to define a function S from strings x to sets such that S is the set encoded by this string (in ZF) or empty set if x isn’t a valid ZF notation for a set. (To do this, first recursively check that the input formula is defining a set, not a trash of symbols in which case we return empty set, then find the encoded set recursively. So, S is a recursive function and therefore can be defined.)\n\nDefine f by the formula f(y, x) ⇔ y = S(x). Then the above g = S.\n\nLet M be the set comprehension\n\n{ X∊set comprehensions | X not in S(X) }.\n\nTherefore:\n\nM in S(M) ⇔ M not in S(M).\n\n# In predicate logic\n\nNote that this paradox is pertinent not to only to ZF but even to Peano Arithmetic (but only if we encode by the same number formulas differing only by variable names). We can even extend it to arbitrary predicate calculus with equality, if we allow an “external” logic to be used to check the condition:\n\n∀x1…∀xn∃!y:f(y, x1, …, xn).\n\nLet S(x, P) be the logical function from a pair of a variable x and encoding P of a logical formula where x is the only free variable into a one-variable predicate determining trueness of this formula for a given x. Then define Q(P) as encoding of the predicate not (S(x, P))(P) (for some variable x not present in P). Then S(y, Q)(Q) = not S(y, Q)(Q) (for some variable y not present in Q), because Q = not (S(y, Q))(Q).\n\n# Avoiding the error\n\nThe intuition suggests: S is a function in a model of ZF, not in the formal system the proof is written. So, it’s illegal to use it in the proof.\n\nIt is an important research question, how to salvage the ZF logic.\n\nIn lambda-calculus this paradox seems not to appear: When I tried to model it in a lambda-calculus with dependent types (proof assistant Lean), I was unable to reprise the paradox even having ZF axioms, because S defined in the normal for Lean way is not an object of ZF, and therefore\n\nS({ X∊set comprehensions | X not in S(X) })\n\nmade no sense, as it requires S to be a symbol denoting a ZF set.\n\n# Security implications\n\nSo, we have a wrong meta-logic in almost every math logic studybook. That’s a hazard.\n\nSecurity implications of an error in extension by definition are almost as great, as if there were found a contradiction in ZF itself, because ZF is always used together with extension by definition in practice.\n\nI am not yet sure whether automatic proof checkers such as Coq and Lean are affected. As I remember in Lean (maybe, in Coq, too) there is something called metalogic – that could be vulnerable.\n\nCoq, Lean, Isabelle may be used to verify CPUs, OSes, thermonuclear weapons, nuclear power plants, air traffic control, etc. This error is a hazard.\n\nA good news is that modern automated proof checkers (logic verifiers) use lambda-calculus not ZF or predicate logic. But they indeed may use ZF or predicate logic sometimes (in their proof libraries). Moreover, metalogics of proof checkers may have a similar bug to this bug.\n\nThe kind of the hazard is: “We built a nuclear plant. Our software is verified with an automatic logic checker, we don’t even need to debug it before turning it on. Debugging is for people who don’t know modern software security. Bump!”\n\nP.S. Hire me! I sent the world-best engineering security bug report, in other words I am to be considered the world best security researcher.\n\nBibliography\n\n: Wikipedia contributors, Gödel numbering, 4 March 2022 07:24 UTC, https://en.wikipedia.org/w/index.php?title=G%C3%B6del_numbering&oldid=1075161247\n\n: Wikipedia contributors, Extension by definitions, 22 October 2021 23:49 UTC, https://en.wikipedia.org/w/index.php?title=Extension_by_definitions&oldid=1051350724\n\n0 users think it is non-scientific; after 15 it will be moderated.\nPublished", null, "## By Victor Porton\n\nI am the chief editor of this journal and creator of this site." ]

[ null, "https://science.vporton.name/2021/11/09/what-is-the-worlds-biggest-technical-security-issue/", null, "https://science.vporton.name/2021/11/09/what-is-the-worlds-biggest-technical-security-issue/", null, "https://science.vporton.name/2021/11/09/what-is-the-worlds-biggest-technical-security-issue/", null, "https://science.vporton.name/2021/11/09/what-is-the-worlds-biggest-technical-security-issue/", null, "https://science.vporton.name/wp-content/plugins/ultimate-member/assets/img/default_avatar.jpg", null ]

[ "<> test questions A: beautiful 2\n\n【 Problem description 】\nXiao Lan likes it very much 2, This year is A.D 2020 year , He was very happy .\nHe was curious , In A.D 1 Year to A.D 2020 year （ contain ） in , How many digits of a year contain numbers 2?\n\nThis question is divided into two parts\nans = 0 for i in range(1,2021): if str(i).count(\"2\") != 0: ans += 1 print(ans)\n<> test questions B: Total number\n\n【 Problem description 】\nIf a number is divided by 1 I have other divisors with myself , It is called a composite number . for example ：1, 2, 3 It's not a total ,4, 6 It's a sum .\nFrom 1 reach 2020 How many total numbers are there .\n\nThe same question\nans = 0 for i in range(4,2021): for j in range(2,i//2+1): if i%j == 0: ans += 1\nbreak print(ans)\n<> test questions C: Factorial divisor\n\n【 Problem description 】\nDefine factorial n! = 1 × 2 × 3 × · · · × n.\nExcuse me? 100! （100 factorial ） How many divisors are there .\n\nIf you want to be violent , You can't get out .\n\n<> test questions D: Essential ascending sequence\n\n【 Problem description 】\nXiao Lan especially likes monotonous things .\nIn a string , If you take out several characters , It is monotonically incrementing to arrange these characters in the order in the string , Becomes a monotonically increasing subsequence of the string .\nfor example , In string lanqiao in , If you take out characters n and q, be nq Form a monotone increasing subsequence . Similar monotone increasing subsequences also exist lnq,i,ano wait .\nXiaolan found , Some subsequences have different positions , But the character sequence is the same , For example, the second character and the last character can be retrieved ao, You can also get the last two characters\nao. Xiao Lan doesn't think they are fundamentally different .\nFor a string , Xiao Lan wants to know , How many increasing subsequences are there in different nature ?\nfor example , For Strings lanqiao, The increasing subsequences with different nature are 21 individual . They are\nl,a,n,q,i,o,ln,an,lq,aq,nq,ai,lo,ao,no,io,lnq,anq,lno,ano,aio.\nFor the following string （ common 200\nA small English letter , Display in four lines ）：（ If you copy the following text to a text file , Be sure to check that the copied content is consistent with that in the document . There is a file under the examination directory\ninc.txt, The content is the same as the text below ）\n\ntocyjkdzcieoiodfpbgcncsrjbhmugdnojjddhllnofawllbhf\nvrnaaahahndsikzssoywakgnfjjaihtniptwoulxbaeqkqhfwl\n\nHow many increasing subsequences are there in different nature ?\n\nThe estimation is done by dynamic programming , I didn't think of it in the exam .\n\n<> test questions E: paper serpent\n\n【 Problem description 】\nXiao Lan has a toy snake , All in all 16 section , It's marked with numbers 1 to 16. Each section is in the shape of a square . The two adjacent sections can be in a straight line or in a straight line 90 Degree angle .\nXiaolan has another one 4 × 4 The square box of , For storing toy snakes , The square of the box is marked with letters in turn A reach P common 16 Two letters .\nXiaolan can fold her toy snake into a box . He found that , There are many ways to put snakes in .\nThe figure below shows two options ：\n\nPlease calculate for Xiao Lan , How many different solutions are there . If there are two options , There is a segment of the toy snake in different boxes , It's a different plan .\n\nCan not do .\n\n<> test questions F: Tiangan dizhi\n\n【 Problem description 】\nIn ancient China, tiangan dizhi was used to record the current year .\nThere are ten Heavenly stems , They are ： nail （jiǎ）, B （yǐ）, C （bǐng）, Ding （dīng）, E\n（wù）, Oneself （jǐ）, Geng （gēng）, Symplectic （xīn）, Union （rén）, Kui （guǐ）.\n\nThere are twelve Earthly Branches , They are ： son （zǐ）, ugly （chǒu）, Yin （yín）, Mao （mǎo）, Chen （chén）, Already （sì）, Afternoon （wǔ）, not （wèi）, Shen （shēn）, Unitary （yǒu）, Xu （xū）, Hai （hài）.\nConnect the heavenly stems with the earthly branches , It formed a year of heavenly stems and earthly branches , for example ： Jia Zi .2020 Year is the year of gengzi .\nEvery year , Both heavenly stems and earthly branches will move to the next . for example 2021 Year is the year of Xin Chou .\nEvery time 60 year , Heavenly stems will circulate 6 round , Dizhihui circulation 5 round , So tiangan dizhi chronicles every year 60 Once a year . for example 1900 year ,1960 year ,2020 Every year is the year of gengzi .\nThe year of a given year , Please output the year of tiangan dizhi .\n【 Input format 】\nThe input line contains a positive integer , Indicates the year of A.D .\n【 Output format 】\nOutput a Pinyin , Indicates the year of tiangan dizhi , Both tiangan and dizhi are expressed in lowercase Pinyin （ It doesn't mean tone ）, Do not add any extra characters between .\n【 sample input 】\n2020\n【 sample output 】\ngengzi\n【 Scale and convention of evaluation case 】\nFor all evaluation cases , The year entered is no more than 9999 Positive integer of .\n\nIf you put the ad in this question 1 I'd like to do it in 2010 jiǎzǐ You can't do it anyway , It took me an hour and a half to solve this problem during the competition , I didn't know the starting point until later jiǎzǐ, Anti roll out is xinyou Just made it . It's killing people .\nt = [\"\",\"xin\",\"ren\",\"gui\",\"jia\",\"yi\",\"bing\",\"ding\",\"wu\",\"ji\",\"geng\"] d = [\"\",\n\"you\",\"xu\",\"hai\",\"zi\",\"chou\",\"yin\",\"mao\",\"chen\",\"si\",\"wu\",\"wei\",\"shen\"] n = int(\ninput()) a,b = 1,1 for i in range(1,n+1): if a>10: a = a%10 if a%10 == 0: a = 10\nif b>12: b = b%12 if b%12 == 0: b = 12 a += 1 b += 1 print(t[a-1]+d[b-1])\n<> test questions G: Duplicate string\n\n【 Problem description 】\nIf a string S Can be repeated exactly by a string K First time , We call it S yes K Repeat string . for example “abcabcabc” Can be seen as “abc” repeat 3\nFirst time , therefore “abcabcabc” yes 3 Repeat string .\nIn the same way “aaaaaa” since 2 Repeat string , again 3 Repeat string and 6 Repeat string .\nNow give a string S, Please calculate at least a few characters to modify , Can make S To become a K Substring ?\n\n【 Input format 】\nThe first line of input contains an integer K.\nThe second line contains a string containing only lowercase letters S.\n\n【 Output format 】\nOutput an integer to represent the answer . If S Cannot modify to K Repeat string , output 1.\n\n【 sample input 】\n2\naabbaa\n\n【 sample output 】\n2\n\n【 Scale and convention of evaluation case 】\nFor all evaluation cases ,1 ≤ K ≤ 100000, 1 ≤ |S | ≤ 100000. among |S | express S Of\nlength .\n\nThis problem should not be difficult to write after careful analysis\ndef main(): k1 = int(input()) s = input().strip() n = len(s) k = n//k1 ans = 0\nif k1 == 1: for i in s: if i != s: ans += 1 print(ans) return r = [] if n%k\n!= 0: print(-1) return for i in range(0,n,k): d = \"\" for j in range(i,i+k): d +=\ns[j] r.append(d) maxV = 0 n1 = len(r) for i in range(n1): if maxV<r.count(r[i])\n: maxV = r.count(r[i]) sm = \"\" for i in range(n1): if r.count(r[i]) == maxV: sm\n= r[i] break for i in range(0,n,k): ind = 0 for j in range(i,i+k): if sm[ind] !=\ns[j]: ans +=1 ind += 1 print(ans) main()\n<> test questions H: answering question\n\n【 Problem description 】\nyes n Each student asked the teacher to answer questions at the same time . Each student estimated the time of answering questions in advance .\nThe teacher can arrange the order of answering questions , Students should enter the teacher's office in turn to answer questions .\nThe process of a student answering questions is as follows ：\n\n* Enter the office first , No i What do you need si Millisecond time .\n* Then students ask questions and teachers answer them , No i What do you need ai Millisecond time .\n* After answering questions , My classmates are very happy , A message will be sent in the course group , The time needed can be ignored .\n* Finally, the students packed up and left the office , need ei Millisecond time . General needs 10 second ,\n20 Seconds or 30 second , Namely ei The value is 10000,20000 or 30000.\nAfter a classmate left the office , Then the next student can enter the office .\nAnswer questions from 0 The moment begins . The teacher wants to arrange the order of answering questions reasonably , Make students in the course group\nThe sum of sending messages is the smallest\n【 Input format 】\nThe first line of input contains an integer n, Represents the number of students .\nnext n that 's ok , Describe each student's time . Among them, No i The row contains three integers si, ai, ei, meaning\nAs mentioned above .\n【 Output format 】\nOutput an integer , What is the minimum sum of messages sent by students in the course group .\n【 sample input 】\n3\n10000 10000 10000\n20000 50000 20000\n30000 20000 30000\n【 sample output 】\n280000\n【 Example description 】\naccording to 1, 3, 2 Answer questions in the same order , What are the message sending times 20000, 80000, 180000.\n【 Scale and convention of evaluation case 】\nabout 30% Evaluation case of ,1 ≤ n ≤ 20.\nabout 60% Evaluation case of ,1 ≤ n ≤ 200.\nFor all evaluation cases ,1 ≤ n ≤ 1000,1 ≤ si ≤ 60000,1 ≤ ai ≤ 1000000, ei ∈ {10000, 20000,\n30000}, Namely ei By all means 10000,20000,30000 one of .\n\nThis question looks a little scary , In fact, after understanding the questions, you will find that the minimum sum is the sum of the time from entering the office to leaving the office , And then the last one is not to add in the time to tidy up and leave the office ok It's over , That is, it can be solved with greed .\nn = int(input().strip()) nums = [list(map(int,input().split())) for _ in range(\nn)] r = [] for i in range(n): r.append([sum(nums[i]),nums[i]]) r.sort() ans =\n0 s = 0 for i in range(n): ans += r[i]-r[i] + s s += r[i] print(ans)\n<> test questions I: supply\n\n【 Problem description 】\nXiao Lan is a helicopter pilot , He is in charge of the service to the mountain people n\nA village transports goods . Every month , He goes to every village at least once , It can be more than once , Transport the materials needed by the village . Every village happens to have a heliport , The distance between every two villages is just a straight line .\nDue to the limited size of the helicopter fuel tank , The distance of Xiaolan's single flight cannot exceed\nD. Every heliport has a gas station , You can fill up the helicopter . Every month , Xiaolan starts from the headquarters , After delivering supplies to the villages, return to the headquarters . If it's convenient , Xiaolan can also refuel through the headquarters halfway .\nHeadquartered in 1 My village . Excuse me? , To complete a month's task , How long does Xiao Lan have to fly at least ?\n\n【 Output format 】\nOutput one line , Contains a real number , Rounding reservation 2 Decimal place , Express the answer .\n【 sample input 】\n4 10\n1 1\n5 5\n1 5\n5 1\n【 sample output 】\n16.00\n【 Example description 】\nThe four villages form a square shape .\n【 sample input 】\n4 6\n1 1\n4 5\n8 5\n11 1\n【 sample output 】\n28.00\n【 Example description 】\nThe order of supply is 1 → 2 → 3 → 4 → 3 → 2 → 1.\n【 Scale and convention of evaluation case 】\nFor all evaluation cases ,1 ≤ n ≤ 20, 1 ≤ xi, yi ≤ 104, 1 ≤ D ≤ 105.\n\nCan not do .\n\n<> test questions J: Blue jump\n\n【 Problem description 】\nXiaolan made a robot , It's called blue jump , Because this robot basically walks\nBy jumping .\nBlue jump can jump and walk , You can also turn around . The distance of each step of blue jump must be an integer , Each step can jump no more than k The length of . Because the balance of blue jump is not well designed , If you do it twice in a row\nIt's all jumping , And the distance between the two jumps is at least p, Blue jump will fall , This is what Xiaolan doesn't want to see .\nXiao Lan has received a special task , In a long time L\nBlue jump on stage . Xiaolan wants to control the blue jump from the left side of the stage to the right , Then turn around , Then walk from the right to the left , Then turn around , Then walk from the left to the right , Then turn around , Go from the right to the left , Then turn around , So back and forth . In order to make the audience not too boring , Xiao Lan decided to let LAN Tiao Tiao walk in a different way each time . Xiao Lan records the distance of each step of the blue jump , Line up in order , Obviously, the number of this column is not more than one\nk And he is L. This way, a row of numbers will come out . If the length of two columns is different , Or there is a different position value in the two columns , I think it's a different plan .\nWill the blue jump jump in the premise of not falling , How many different schemes are there going from one side of the stage to the other .\n【 Input format 】\nThe input line contains three integers k, p, L.\n【 Output format 】\nOutput an integer , Express the answer . The answer could be big , Please output the answer divided by 20201114 The remainder of .\n【 sample input 】\n3 2 5\n【 sample output 】\n9\n【 Example description 】\nBlue jump has the following characteristics 9 A jump method ：\n\n* 1+1+1+1+1\n* 1+1+1+2\n* 1+1+2+1\n* 1+2+1+1\n* 2+1+1+1\n* 2+1+2\n* 1+1+3\n* 1+3+1\n* 3+1+1\n【 sample input 】\n5 3 10\n【 sample output 】\n397\n【 Scale and convention of evaluation case 】\nabout 30% Evaluation case of ,1 ≤ p ≤ k ≤ 50,1 ≤ L ≤ 1000.\nabout 60% Evaluation case of ,1 ≤ p ≤ k ≤ 50,1 ≤ L ≤ 109.\nabout 80% Evaluation case of ,1 ≤ p ≤ k ≤ 200,1 ≤ L ≤ 1018.\nFor all evaluation cases ,1 ≤ p ≤ k ≤ 1000,1 ≤ L ≤ 1018.\n\nTechnology\nDaily Recommendation" ]

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[ "What is the slope and y-intercept of the line 9x+3y=12?\n\nJun 6, 2016\n\nThe slope is $- 3$ and the y-intercept is $4$.\n\nExplanation:\n\nIt helps if you put your equation into the standard linear form of $y = m x + b$. In this form, $m$ is always the slope, and $b$ is always the y-intercept.\n\nTo get it into standard form, you need to isolate $y$. To do this, I can first move the $9 x$ by subtracting it from each side of the equation, giving me:\n\n$3 y = - 9 x + 12$\n\nThen, I would divide each side by 3, to isolate the $y$. The distributive property requires that both $- 9 y$ and $12$ be divided by 3 as well. This gives me:\n\n$y = - 3 x + 4$\n\nNow I have my equation in standard form, and can see that the slope is $- 3$ and the y-intercept is $4$. That can be reflected by graphing the line as well: graph{-3x +4 [-4.834, 5.166, -0.54, 4.46]}" ]

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[ "# Applying the Optimal Power Flow Model to Your PDN\n\n### Key Takeaways\n\n• Careful power delivery design is needed in your PCB to ensure your components receive stable power.\n\n• Optimal power flow is normally used in electricity distribution system design, but these same ideas can be used to determine the required power in a PCB.\n\n• For your PCB, the optimal power flow model can help you understand the design requirements for your PCB’s PDN and how power is distributed for a given design.", null, "You can apply the core concepts in optimal power flow to your PCB design.\n\nAs much as we’d like engineering decisions to be cut and dry, the majority of engineering decisions involve trade offs. Optimization is more than just a buzzword used in technology marketing copy, it’s a deep field of mathematics that aims to help systems designers analyze tradeoffs between different design objectives. When working on complex systems like ICs, PCBs, or an entire distribution grid, there are many tradeoffs in a design, and the best design tries to prioritize important design goals while finding compromise wherever possible.\n\nIn terms of power distribution, no power supply can provide infinite power on-demand. Your power supply and your distribution need to provide stable AC or DC power to components in your PCB or other electronics. Determining how to allocate power consumption throughout your system is an optimization problem that is described using the optimal power flow model. When using this model, it becomes possible to identify how much power can be allocated to different circuit blocks, as well as identify potential circuit blocks that may need redesigns. Here’s what this model can tell you about your PCB and how you can take advantage of this model for electronics design.\n\n## The Optimal Power Flow Model\n\nThe optimal power flow model is normally applied to electric power plants and consumers to determine the level at which each plant should be operated to satisfy demand. In a real electrical grid, each plant on the grid can produce power, which is then distributed to multiple loads on the system. Determining how much power should be produced by each source may seem trivial, but it’s really a complex management problem that is meant to prevent power from being wasted on a real electrical grid.\n\nThe same ideas can be applied to PCB design, both for AC and DC boards. PCBs are generally only powered by a single power supply, except in some specialized mixed signal boards that might need to receive separate AC power. However, the reduced number of power sources makes formulation of an optimization problem for power distribution much easier.\n\n### Example Formulation\n\nTo better understand the optimal power flow model, consider the following optimization problem. A system with S power sources must supply power to L circuit blocks. The S sources and L circuit blocks are distributed into a tree topology with N nodes and M branches. In determining the power budget for a system with multiple sources, we can formulate the following power distribution problem for minimizing the total power output from the power sources:", null, "Power distribution problem for minimizing total power output\n\nHere, the goal is to determine the voltage and current required from each source by adjusting the current and voltage supplied to each circuit block in your system. In this problem, Kirchoff’s laws and Ohm’s law are built into the optimization problem, making it adaptable to any system where power needs to be distributed among multiple sources.\n\n### Assumptions in Optimal Power Flow Models\n\nThere are some implicit assumptions in the above model:\n\n• The DC resistance is assumed to be 0 Ohms for all conductors in the system.\n\n• The relationship between the voltage and current in each of the L circuit blocks is allowed to be nonlinear.\n\n• The sources are assumed to have linear power output up to their maximum rated output.\n\n• All sources are considered to be DC or to only have harmonic time dependence.\n\nThis problem could be best solved with a heuristic or evolutionary algorithm, the latter of which is used in commercial solvers. There are also open-source code packages that can handle these types of problems with high variable count.\n\nVarious versions of this problem are used by utility companies to determine how best to distribute power from multiple sources into different substations and to end users. These types of problems normally consider the probability of random failure of a power source when calculating the optimal power output from sources in the system. This approach could also be implemented when designing a PCB with multiple sources, or when linking multiple PCBs together into a larger system.\n\n## Verifying Optimal Power Flow Results\n\nWith a complete set of circuit modeling and simulation tools, circuit designers can verify power distribution into various circuit blocks within their system. This type of verification process takes a high-level view of a circuit, where only functional blocks within a larger system are considered, rather than individual components. This type of system-level analysis could be performed as follows:\n\n1. After designing each circuit block, use a SPICE simulation to calculate the power consumption for each block that will appear in your system. In particular, calculate the total voltage dropped across each circuit block and the total current used to power each block.\n\n2. Create a new component to represent each block. This component should have the equivalent impedance corresponding to its circuit block (just the block’s supply voltage/block current). If the circuit block has a nonlinear response, use a SPICE modeling application to create an equivalent DC model for the circuit block.\n\n3. Connect these blocks together into your larger network topology and add a model for your power source.\n\n4. Run a SPICE simulation (DC, transient simulation, frequency sweep, or other simulation) to calculate power distributed to each block in your topology. An example bus topology with N circuit blocks is shown in the image below.\n\n5. Check the power distribution to each block against its particular power requirements to determine whether it will be overpowered or under-powered.", null, "This bus topology is the most common way to construct the PDN in a PCB design.\n\nDepending on how you model the power source in your Steps 3-5, you may not get physically meaningful results. The power source should be modeled as close as possible to the real source in your system to ensure your power flow results are physically meaningful. In the event the source is not outputting enough power, or too much power is consumed at specific circuit blocks, the source or blocks may need to be redesigned to ensure power is properly distributed in your system.\n\nIf you’d like to keep up-to-date with our System Analysis content, sign-up for our newsletter curating resources on current trends and innovations. If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts." ]

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[ "Getting Image", null, "", null, "", null, "", null, "Register now for special offers", null, "+91\n\nHome\n\n>\n\nEnglish\n\n>\n\nClass 11\n\n>\n\nPhysics\n\n>\n\nChapter\n\n>\n\nGeneral Kinematics And Motion In One Dimension\n\n>\n\nA body A starts from rest with...\n\nUpdated On: 27-06-2022", null, "Get Answer to any question, just click a photo and upload the photo and get the answer completely free,\n\nUPLOAD PHOTO AND GET THE ANSWER NOW!", null, "Text Solution\n\n5 : 95 : 79 : 59 : 7\n\nSolution : Distance covered by first body in 5th second after <br> start, S_5 = 0 + (a_1)/(2)(2 xx 5 - 1) = (9a_1)/(2) <br> 5^(th) second of first body will be 3^(rd) second of second body. Distance covered by the second body in the 3^(rd) second after is start. <br> S_3 = 0 + (a_2)/(2) (2 xx 3 - 1) = (5a_2)/(2) <br> Since S_5 = S_3 or (9a_1)/(2) = (5 a_2)/(2) or a_1 : a_2 = 5 : 9.", null, "Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.\n\n## Related Videos\n\n643049514\n\n0\n\n7.8 K\n\n3:28\nA body A starts from the rest with an acceleration a_1. After 2 second another body B starts from rest with an acceleration a_2 . If they travel equal distance in the 5th second after the start of A, then what will be the ratio a_1 : a_2.\n261058161\n\n0\n\n6.5 K\n\n3:17\nA body A starts from rest with an acceleration a_1. After 2 s another body B starts from rest with an acceleration ar If they travel equal distance in the 5th second after the start of A. a_1 : a_2 is equal to\n642751197\n\n0\n\n200\n\n3:38\nA body A starts from rest with an acceleration a_1. After 2 seconds, another body B starts from rest with an acceleration a_2. If they travel equal distances in the 5^(th) second, after the start of A, then the ratio a_1 : a_2 is equal to :\n11745573\n\n0\n\n6.9 K\n\n2:58\nA body A starts from rest with an acceleration a_1. After 2 seconds, another body B starts from rest with an acceleration a_2. If they travel equal distances in the 5^(th) second, after the start of A, then the ratio a_1 : a_2 is equal to :\n320269178\n\n0\n\n500\n\n3:39\nA body a starts from rest with an acceleration a_(1) . After 2s, another body B starts from rest with an acceleration a_(2) . If they travel equal distances in the 5^(th) second after the start of A then, the ratio a_(1) :a_(2) is equal to :", null, "", null, "" ]

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[ "# Miller D.C. The Ether-Drift Experiment and the Determination of the Absolute Motion of the Earth // Reviews of modern physics, Vol.5, July 1933\n\nStart   PDF   <<<  Page 226   >>>", null, "", null, "203  204  205  206  207  208  209  210  211  212  213  214  215  216  217  218  219  220  221  222  223  224  225  226 227  228  229  230  231  232  233  234  235  236  237  238  239  240  241  242 As the azimuth of the effect in the interferometer, which corresponds to the azimuth of the apex, oscillates across the meridian, it reaches a maximum displacement, first to the east and then to the west in each sidereal day. When δ ≥ φ, this maximum azimuth may be treated as the azimuth of a circumpolar star at its eastern or western elongation. It is shown in treatises on spherical and practical astronomy that the azimuth of elongation depends in a very simple way upon the declination of the star and upon the latitude of the observatory. The relation is: sin Amax = cos δ /cos φ, whence cos δ = sin Amax cos φ. When δ ≤ φ, the azimuth of the apex swings completely around the horizon in a sidereal day. If θE is the sidereal time when the azimuth is due east, then tan δ = tan φ cos (θE ¢α). Presuming that the orbital velocity of the earth is known, the formulae now provided are sufficient for the determination in general of the apex and velocity of the absolute motion of the earth, from observations of the relative motion of the earth and ether as made with the interferometer. These observations, giving simply the maximum displacement of the interference fringes as the apparatus turns on its axis, together with the azimuth at which this maximum displacement occurs, must adequately cover the period of one sidereal day for a given epoch. A complete set of such observations gives one determination of the velocity of the absolute motion and two independent determinations of the apex of the motion. The determination of the direction of the earthÆs absolute motion is dependent only upon the direction in which the telescope points when the observed displacement of the fringes is a maximum; it is in no way dependent upon the amount of this displacement nor upon the adjustment of the fringes to any particular width or zero position. The actual velocity of the earthÆs motion is determined by the amplitude of the periodic displacement, which is proportional to the square of the relative velocity of the earth and the ether and to the length of the light path in the interferometer. The two effects, magnitude and azimuth of observed relative motion, are quite independent of each other. Harmonic analysis of the fringe displacements The records of the actual interferometer observations for the Mount Wilson cycle consist of three hundred and sixteen pages of readings of the positions of the interference fringes, of the form shown in Fig. 8. Each set contains readings for twenty or more turns of the interferometer. The twenty or more readings for each of the sixteen observed azimuths are averaged and the averages are compensated for the slow linear shift of the whole interference system during the period of observation, as explained previously in connection with Fig. 9. The average readings for each set are then plotted on coordinate paper, to a large scale, for the purpose of harmonic analysis. The graphs, I, II, III, and IV, Fig. 21, show the readings for four successive sets of observations made on April 2, 1925. The plotted points are positions of the central black fringe of the interference pattern with respect to the fiducial point, as the interferometer makes one complete turn. The unit for the scale of ordinates is one-hundredth of a fringe width, while the abscissae correspond to azimuth intervals of 221/2░, beginning at the north point and proceeding clockwise around the horizon. A chart of this kind is plotted for each set of observations. These charted ōcurvesö of the actual observations contain not only the second-order, halfperiod ether-drift effect, but also a first-order, full-period effect, any possible effects of higher orders, together with all instrumental and accidental errors of observation. The present ether-drift investigation is based entirely upon the second order effect, which is periodic in each half revolution of the interferometer. This second-order effect is completely represented by the second term of the Fourier harmonic analysis of the given curve. In order to evaluate precisely the ether-drift effect, each curve of observations has been analyzed with the Henrici harmonic analyzer for the first five terms of the Fourier series. The first-order effect in the observation is shown by the fundamental component, which is drawn under the corresponding curve of observa tions in Fig. 21; the second-order effect is shown by the curve next below; while the fourth curve in each instance shows the sum of the third, fourth, and fifth components. It is evident that the observed curves contain very little trace of any effects of any higher orders. The residual curves are of very small amplitude and are evidence of the fact that the incidental and random errors are small. The harmonic analysis and synthesis has been performed by methods which have been completely described elsewhere by the writer.12 The harmonic analysis of the observations gives directly the amplitude in hundredths of a fringe width and the phase as referred to the north point of the second harmonic of the curve, which is the ether-drift effect. The observed amplitude of the movement of the fringes is at once converted into the equivalent velocity of the relative motion of the earth and ether, as observed in the plane of the interferometer, by means of the relation developed in the elementary theory of the experiment: d = 2D(v2/c2) and v = (dc2/2D)1/2, d being the observed half-period displacement of the fringes and D the length of the arm of the interferometer, that is the distance from the halfsilvered mirror by means of multiple reflections to the end mirror, No. 8, both being expressed in terms of the effective wave-length of the light used for the interferences; v is the relative motion of the earth and the ether in the plane of the interferometer and c is the velocity of light, both being expressed in kilometers per second. The nomograph, Fig. 20, consists of a parabolic curve which shows the relative velocity corresponding Fig. 20. Relation of fringe displacement to velocity of ether-drift for 2D = 112,000,000λ and λ = 5700A. 12 D. C. Miller, The Science of Musical Sounds, 123 (1916); J. Frank. Inst. 181, 51 (1916); 182, 285 (1916). to a fringe displacement as observed in the interferometer used in these experiments. It is for light of wave-length λ = 5700A, and for a total light-path of 2D = 112,000,000λ. The azimuth of the ether-drift effect is the direction in which the telescope points when the half-period displacement of the fringes is a positive maximum. This azimuth, A, is obtained from the phase, P, of the second harmonic component of the observations as determined by the analyzer, from the following relation: A = (1/2) (P ¢ 90░). The point thus located is the crest of the curve representing the second component, expressed in degrees, measured from the north point; the x-axis of the curves shown in Fig. 21 begins at the Fig. 21. Harmonic analysis of ether-drift observations. north point and extends for one turn of 360░, through the east, south and west points of azimuth back to the north point. The figure shows, in the graphs of the second components, what has already been mentioned, that, within 360░ of azimuth for one complete turn, there are two crests of the second component, corresponding to two azimuths 180░ apart, between which the interferometer is incompetent to distinguish. The dispersion of the azimuth readings is much less than 90░ and the necessity for continuity of azimuth indications for successive observations removes the ambiguity as to which zone of" ]

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[ "# Kotlin: Find leaf nodes paths in N-ary tree\n\n## Definitions\n\nA tree is a data structure where a node can have zero or more children. Each node contains a value and the top-most node is called the root. A node without children is called a leaf node.\n\nThe N-ary tree is a tree that allows you to have n(n ≥ 0)number of children of a particular node as opposed to binary trees that that allow you to have at most 2 children of a particular node.\n\nHere's a sample N-ary tree we'll use to get paths to leaf nodes.", null, "In the above tree, 1 is the root node, 2, 3 and 4 are children of the root node. Nodes 5,6,7,8,9,10,11 are all leaf nodes because they have no children.\n\nGiven the above N-ary tree, write a solution to get paths to the leaf nodes. The solution should return list of paths to the leaf nodes as shown below:\n\n`````` [1,2,5]\n[1,2,6]\n[1,2,7]\n[1,3,8]\n[1,4,9]\n[1,4,10]\n[1,4,11]\n``````\n\n## Solution\n\nFirst, we define the N-ary tree data class.\n\n``````data class Node<T>(var value: T) {\nvar parent: Node<T>? = null\nvar children: MutableList<Node<T>> = mutableListOf()\n\nnode.parent = this\n}\nfun hasChildren(): Boolean =\nchildren.size >= 1\n}\n``````\n\nNext we create a function `makeTree()` that constructs the above n-ary tree.\n\n``````fun makeTree(): Node<Int> {\nval root = Node(1)\nval node2 = Node(2)\nval node3 = Node(3)\nval node4 = Node(4)\nval node5 = Node(5)\nval node6 = Node(6)\nval node7 = Node(7)\nval node8 = Node(8);\nval node9 = Node(9)\nval node10 = Node(10)\nval node11 = Node(11)\nreturn root\n}\n``````\n\nTo get the paths to the leaf nodes we will use Depth First Search. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores as far as possible along each branch before backtracking.\n\nWe will traverse the N-ary tree and keep inserting every node to the path until a leaf node is found. When a leaf node is found, we add the path to the result and remove the last added leaf node from the path and check for the next leaf node.\n\nHere's the recursive backtracking routine:\n\n``````private fun findLeafNodePaths(\nnode: Node<Int>,\nresult: MutableList<List<Int>>,\npath: MutableList<Int>): MutableList<List<Int>> {\n\nif (!node.hasChildren()) { // leaf node\npath.removeLast()//\n} else {\nfor (child in node.children) {\nfindLeafNodePaths(child, result, path)\n}\npath.removeLast()\n}\nreturn result\n}\n``````\n\nHere's the main function to create the n-ary tree and print paths to the leaf nodes:\n\n``````fun main() {\nfindLeafNodePaths(makeTree(), mutableListOf(), mutableListOf() ).forEach {\nprintln(it)\n}\n}\n``````\n\nComplete Code:" ]

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