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https://youngzones.org/ElaineTAMUCC/1350/final_review.html | [
"# Review for Final Exam\n\n Real number system Fraction operations Base ten system Fraction models Listing factors & prime factorization Factors & multiples Properties: Fundamental Theorem of Arithmetic Identity elements GCF & LCM Commutative Relatively prime Associative Divisibility rules Additive inverse Patterns Zero multiplication Sequences & nth term Multiplicative inverse Geometric sequences Distributive"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5167622,"math_prob":0.70078623,"size":534,"snap":"2021-43-2021-49","text_gpt3_token_len":139,"char_repetition_ratio":0.16603774,"word_repetition_ratio":0.0,"special_character_ratio":0.14419475,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97680056,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T19:23:50Z\",\"WARC-Record-ID\":\"<urn:uuid:47691681-4faf-48b9-8211-a669bb92dd53>\",\"Content-Length\":\"14884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1474df9e-4709-4772-8c8f-37bb60c68de4>\",\"WARC-Concurrent-To\":\"<urn:uuid:293bb496-bbd9-4563-bfe3-479a7b7638bc>\",\"WARC-IP-Address\":\"204.44.192.48\",\"WARC-Target-URI\":\"https://youngzones.org/ElaineTAMUCC/1350/final_review.html\",\"WARC-Payload-Digest\":\"sha1:ZQKK4YSVONTPHTM3SZVBLU4MQV3VBWIF\",\"WARC-Block-Digest\":\"sha1:7ANVZFBQDAHEZNQ6372YIUONREP46ALB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585181.6_warc_CC-MAIN-20211017175237-20211017205237-00432.warc.gz\"}"} |
https://www.colorhexa.com/7832b5 | [
"# #7832b5 Color Information\n\nIn a RGB color space, hex #7832b5 is composed of 47.1% red, 19.6% green and 71% blue. Whereas in a CMYK color space, it is composed of 33.7% cyan, 72.4% magenta, 0% yellow and 29% black. It has a hue angle of 272.1 degrees, a saturation of 56.7% and a lightness of 45.3%. #7832b5 color hex could be obtained by blending #f064ff with #00006b. Closest websafe color is: #6633cc.\n\n• R 47\n• G 20\n• B 71\nRGB color chart\n• C 34\n• M 72\n• Y 0\n• K 29\nCMYK color chart\n\n#7832b5 color description : Moderate violet.\n\n# #7832b5 Color Conversion\n\nThe hexadecimal color #7832b5 has RGB values of R:120, G:50, B:181 and CMYK values of C:0.34, M:0.72, Y:0, K:0.29. Its decimal value is 7877301.\n\nHex triplet RGB Decimal 7832b5 `#7832b5` 120, 50, 181 `rgb(120,50,181)` 47.1, 19.6, 71 `rgb(47.1%,19.6%,71%)` 34, 72, 0, 29 272.1°, 56.7, 45.3 `hsl(272.1,56.7%,45.3%)` 272.1°, 72.4, 71 6633cc `#6633cc`\nCIE-LAB 37.135, 53.925, -56.989 17.226, 9.611, 44.661 0.241, 0.134, 9.611 37.135, 78.458, 313.418 37.135, 17.106, -84.72 31.001, 44.93, -63.714 01111000, 00110010, 10110101\n\n# Color Schemes with #7832b5\n\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #6fb532\n``#6fb532` `rgb(111,181,50)``\nComplementary Color\n• #3732b5\n``#3732b5` `rgb(55,50,181)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #b532b1\n``#b532b1` `rgb(181,50,177)``\nAnalogous Color\n• #32b537\n``#32b537` `rgb(50,181,55)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #b1b532\n``#b1b532` `rgb(177,181,50)``\nSplit Complementary Color\n• #32b578\n``#32b578` `rgb(50,181,120)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #b57832\n``#b57832` `rgb(181,120,50)``\n• #326fb5\n``#326fb5` `rgb(50,111,181)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #b57832\n``#b57832` `rgb(181,120,50)``\n• #6fb532\n``#6fb532` `rgb(111,181,50)``\n• #502179\n``#502179` `rgb(80,33,121)``\n• #5e278d\n``#5e278d` `rgb(94,39,141)``\n• #6b2ca1\n``#6b2ca1` `rgb(107,44,161)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #8538c8\n``#8538c8` `rgb(133,56,200)``\n• #914cce\n``#914cce` `rgb(145,76,206)``\n• #9e60d3\n``#9e60d3` `rgb(158,96,211)``\nMonochromatic Color\n\n# Alternatives to #7832b5\n\nBelow, you can see some colors close to #7832b5. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #5732b5\n``#5732b5` `rgb(87,50,181)``\n• #6232b5\n``#6232b5` `rgb(98,50,181)``\n• #6d32b5\n``#6d32b5` `rgb(109,50,181)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #8332b5\n``#8332b5` `rgb(131,50,181)``\n• #8e32b5\n``#8e32b5` `rgb(142,50,181)``\n• #9932b5\n``#9932b5` `rgb(153,50,181)``\nSimilar Colors\n\n# #7832b5 Preview\n\nThis text has a font color of #7832b5.\n\n``<span style=\"color:#7832b5;\">Text here</span>``\n#7832b5 background color\n\nThis paragraph has a background color of #7832b5.\n\n``<p style=\"background-color:#7832b5;\">Content here</p>``\n#7832b5 border color\n\nThis element has a border color of #7832b5.\n\n``<div style=\"border:1px solid #7832b5;\">Content here</div>``\nCSS codes\n``.text {color:#7832b5;}``\n``.background {background-color:#7832b5;}``\n``.border {border:1px solid #7832b5;}``\n\n# Shades and Tints of #7832b5\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #08030c is the darkest color, while #fdfcfe is the lightest one.\n\n• #08030c\n``#08030c` `rgb(8,3,12)``\n• #12081b\n``#12081b` `rgb(18,8,27)``\n• #1c0c2b\n``#1c0c2b` `rgb(28,12,43)``\n• #26103a\n``#26103a` `rgb(38,16,58)``\n• #311449\n``#311449` `rgb(49,20,73)``\n• #3b1959\n``#3b1959` `rgb(59,25,89)``\n• #451d68\n``#451d68` `rgb(69,29,104)``\n• #4f2178\n``#4f2178` `rgb(79,33,120)``\n• #592587\n``#592587` `rgb(89,37,135)``\n• #642a96\n``#642a96` `rgb(100,42,150)``\n• #6e2ea6\n``#6e2ea6` `rgb(110,46,166)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #8236c4\n``#8236c4` `rgb(130,54,196)``\n• #8c43cb\n``#8c43cb` `rgb(140,67,203)``\n• #9552cf\n``#9552cf` `rgb(149,82,207)``\n• #9f62d4\n``#9f62d4` `rgb(159,98,212)``\n• #a871d8\n``#a871d8` `rgb(168,113,216)``\n• #b181dc\n``#b181dc` `rgb(177,129,220)``\n• #bb90e0\n``#bb90e0` `rgb(187,144,224)``\n• #c49fe5\n``#c49fe5` `rgb(196,159,229)``\n• #ceafe9\n``#ceafe9` `rgb(206,175,233)``\n• #d7beed\n``#d7beed` `rgb(215,190,237)``\n• #e1cdf1\n``#e1cdf1` `rgb(225,205,241)``\n``#eaddf6` `rgb(234,221,246)``\n• #f3ecfa\n``#f3ecfa` `rgb(243,236,250)``\n• #fdfcfe\n``#fdfcfe` `rgb(253,252,254)``\nTint Color Variation\n\n# Tones of #7832b5\n\nA tone is produced by adding gray to any pure hue. In this case, #747077 is the less saturated color, while #7b06e1 is the most saturated one.\n\n• #747077\n``#747077` `rgb(116,112,119)``\n• #746780\n``#746780` `rgb(116,103,128)``\n• #755e89\n``#755e89` `rgb(117,94,137)``\n• #765691\n``#765691` `rgb(118,86,145)``\n• #764d9a\n``#764d9a` `rgb(118,77,154)``\n• #7744a3\n``#7744a3` `rgb(119,68,163)``\n• #773bac\n``#773bac` `rgb(119,59,172)``\n• #7832b5\n``#7832b5` `rgb(120,50,181)``\n• #7929be\n``#7929be` `rgb(121,41,190)``\n• #7920c7\n``#7920c7` `rgb(121,32,199)``\n• #7a17d0\n``#7a17d0` `rgb(122,23,208)``\n• #7a0ed9\n``#7a0ed9` `rgb(122,14,217)``\n• #7b06e1\n``#7b06e1` `rgb(123,6,225)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #7832b5 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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https://codereview.stackexchange.com/questions/133850/hackerrank-sherlock-and-array/133852 | [
"# HackerRank, Sherlock and Array\n\nHere is the code I have written for a HackerRank challenge which takes multiple arrays, and attempts to determine if there exists an element in an array such that the sum of the elements on its left is equal to the sum of the elements on its right. If there are no elements to the left/right, then the sum is considered to be zero.\n\nDetailed problem statement:\n\nInput Format\n\nThe first line contains $T$, the number of test cases. For each test case, the first line contains N , the number of elements in the array $A$ . The second line for each test case contains N space-separated integers, denoting the array $A$.\n\nOutput Format\n\nFor each test case print YES if there exists an element in the array, such that the sum of the elements on its left is equal to the sum of the elements on its right; otherwise print NO.\n\nIt works fine, but it's definitely not an optimal solution, as it fails two test cases (time outs occur). Could anybody provide any insights on optimising it further?\n\nT = int(input())\nN = []\nA = []\n\nfor i in range(T):\n#N.append(int(input()))\nN = int(input())\narr = input().strip().split(' ')\narr = list(map(int, arr))\nA.append(arr)\n#print(A)\nfor i in A:\ncount=0\nfor j in range(len(i)):\npreSum = sum(i[0:j])\npostSum = sum(i[j+1:len(i)])\nif(preSum == postSum):\ncount+=1\nif(count>0):\nprint(\"YES\")\nelse:\nprint(\"NO\")\n\n• Welcome to Code Review! As we all want to make our code more efficient or improve it in one way or another, try to write a title that summarizes what your code does, not what you want to get out of a review. Please see How to get the best value out of Code Review - Asking Questions for guidance on writing good question titles. – Mathias Ettinger Jul 4 '16 at 15:45\n• Thanks for the guidance. New here. Will keep it in mind in the future. – snow Jul 4 '16 at 15:54\n• I don't quite understand what N is for. Is it the length of the array? – Graipher Jul 4 '16 at 16:00\n• @Graipher yes. Quite redundant in python, but that's mandatory in the question specs – snow Jul 4 '16 at 16:02\n• @snow actually, its better for a question to contain everything by itself. – Pimgd Jul 4 '16 at 17:40\n\n## General comments\n\nYou should follow PEP8, the coding standard for python. This means using underscore_names for variables instead of camelCase.\n\ni is a bad generic name, except when iterating explicitly over integers. Maybe use for arr in A.\n\nI would use more descriptive variable names. Instead of A use maybe arrays?\n\nYour input could be more succinct:\n\nfor i in range(T):\nN = int(input)\nA.append(map(int, input().split()))\n\n\nThis will store a map object in A, instead of a list. But since that is iterable we won't have any problems.\n\nIf the input was in one line with the first element being the length of the array and the rest of the line being the array, it would have been even easier:\n\nfor i in range(T):\nN, *arr = map(int, input().split())\nA.append(arr)\n\n\nThis uses a nice feature of python3. N will take the first element of an iterable and arr will take all the (possible) rest. You can even specify variables at the end. Try these cases out to get a feel for it:\n\na, *b = [] # ValueError: not enough values to unpack (expected at least 1, got 0)\na, *b = # a = 0, b = []\na, *b = [0,1] # a = 0, b = \na, *b = [0,1,2] # a = 0, b = [1,2]\na, *b, c = # ValueError: not enough values to unpack (expected at least 2, got 1)\na, *b, c = [0,1] # a = 0, b = [], c = 1\na, *b, c = [0,1,2] # a = 0, b = , c = 2\n\n\n# Performance\n\nInstead of always calculating all sums, you could store the pre_ and post_sum and add/subtract the current element. You should also stop after having found one occurrence.\n\nfor array in A:\nfound = False\npre_sum, post_sum = 0, sum(array)\nfor element in array:\npost_sum -= element\nif(pre_sum == post_sum):\nfound = True\nbreak\npre_sum += element\nprint(\"YES\" if found else \"NO\")\n\n\nI'm not exactly sure, but I think there is a small performance difference between these two, for large arr:\n\narr = list(map(int, arr))\n# and\narr = [int(x) for x in arr]\n\n• Flawed code; you print \"NO\" for each array – Pimgd Jul 4 '16 at 15:54\n• @Pimgd Sorry, just realized this as well, fixed. – Graipher Jul 4 '16 at 15:55\n• Also I think the reason OP uses A and T and N is because the problem statement uses those variable names, which would make them appropriate because it reflects the domain – Pimgd Jul 4 '16 at 15:57\n• I like reviewing answers as well =) – Pimgd Jul 4 '16 at 15:59\n• You can't use N, *arr = ...input()... since the length and the values are on two different lines. – Mathias Ettinger Jul 5 '16 at 6:03\n\nThe flaw in your algorithm is that you're summing for each item in each array. That means you get $O(n^2)$ for time complexity.\n\nSum once, then add to the left side whilst removing from the right side. This will give a time complexity of $O(2n)$ (once to sum, another to move each element over from the left to the right side).\n\nAdditionally, once you have found 1 case in which both sides match, you can stop looking for that testcase.\n\n• Thanks @Pimgd. I generally have problems deciphering the time complexity of programs after writing them. Could you kindly point me to some learning resources which would make me better at it? – snow Jul 4 '16 at 16:33\n• @snow I don't have any specific resources on hand, so here's a google search result from stackoverflow: stackoverflow.com/questions/487258/… – Pimgd Jul 5 '16 at 7:54\n\nActually, you are using two nested loop which is having O(n^2) complexity but this can be easily done in O(n) time using one for loop.\n\nBelow are steps which will help you to do the same in O(n) time-\n\nYou can drive an equation by using a little bit of mathematics.\n\nAssume we have an array which is the random array {3,7,5,10,2,7,4,2} so, in that, that element exists such that the sum of the left side of all the elements is equal to the sum of the right side all the elements.\n\nI am assuming that element is represented by y and it exists in between somewhere. so some of the left side of the element is represented by x as I said the sum of all the elements towards the left side is equal to the sum of all the elements towards the right side of that element. so right side sum also can be represented by x.So by seeing this, we can easily say the sum of all elements presents inside the array should be equal to x + y + x.\n\nx + y + x = sum of all the elements\n\n2 x + y=sum of all the elements\n\n2 x = sum of all the elements - y ---> eq 1\n\nif we have x and y such that this equation holds true. It means, there is one element exist correct because in this equation we have to unknowns x and y. so we have to first get the value of x and y and then will place both the values inside the equation and see whether LHS is equals to RHS or not? LHS equals to RHS it means there is some element exist inside the array where the sum of all the elements towards the left side of the element is equal to the right side of the element. Let’s take one example array.\n\n{3,7,5,10,2,7,4,2}\n\nfirst, I will calculate the sum of all elements.\n\nsum of all the elements= 3+7+5+10+2+7+4+2 sum of all the elements= 40\n\nreplace this in eq 1 then you will get below eq\n\n2x =40 - y --> eq 2\n\nas we are not aware of the y but y that's for sure y will belong to any one of the elements which are present inside the array. so will take one element at a time from the array and replace y with that element like that x also. we will take the x value based on y whatever it comes and replace in this question and see whether LHS equals to RHS or not. if we found any such pair of x and y where LHS equals to RHS. It means, we have that element inside the array which holds this criterion true and we will return YES.\n\nfor first iteration- {3,7,5,10,2,7,4,2}\n\ny=3, x=0\n\njust replace both values in eq 2 and you can seed\n\n0 is not equal to 37\n\nnow move the pointer ahead try this time\n\ny=7, x=0+3=3\n\njust replace both values in eq 2 and you can seed\n\n6 is not equal to 33\n\n....so do the same until you find that element y which satisfy this condition.\n\nnow skipping next iterating with y=5 and trying for y=10 know\n\nif y=10 ,x=3+7+5=15\n\njust replace both values in eq 2 and you can seed\n\n30 is equal to 30. It means this the element(y) which we are looking for and where left sum is equal to the right sum.\n\nHere is the code which passes 100% test case.\n\nstatic String balancedSums(List<Integer> arr) {\nint x = 0;\nint sum = 0;\nfor (int a : arr) {\nsum += a;\n}\n\nfor (int y : arr) {\nif (2 * x == sum - y) {\nreturn \"YES\";\n}\nx = x + y;\n}\nreturn \"NO\";\n\n}\n\n\nStill having doubt you can check out the video tutorial here.\n\n• This appears nearly identical to your answer from yesterday (which was deleted by the community because it was alternate solution instead of a review) with just a few more paragraphs added to the end... please discuss more about the OPs code. – Sᴀᴍ Onᴇᴌᴀ Mar 13 at 7:40\n• Added my analysis on top after deep review. @ Sᴀᴍ Onᴇᴌᴀ – Kanahaiya Mar 13 at 11:48"
] | [
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https://www.varsitytutors.com/psat_math-help/how-to-find-the-height-of-a-right-triangle | [
"# PSAT Math : How to find the height of a right triangle\n\n## Example Questions\n\n### Example Question #1 : How To Find The Height Of A Right Triangle",
null,
"Note: Figure NOT drawn to scale.\n\nRefer to the above diagram. Give the ratio of the perimeter of",
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"to that of",
null,
".",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
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"Explanation:\n\nThe altitude of a right triangle from the vertex of its right triangle to its hypotenuse divides it into two similar triangles.",
null,
", as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:",
null,
".\n\nThe ratio of the smaller sides of these similar triangles is\n\nThe ratio of the smaller side of",
null,
"to that of",
null,
"is",
null,
"or",
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": 1,\n\nso this is also the ratio of the perimeter of",
null,
"to that of",
null,
".\n\n### All PSAT Math Resources",
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""
] | [
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"https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/problem_question_image/image/3591/Triangle.jpg",
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"https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/210303/gif.latex",
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"https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/210304/gif.latex",
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https://math.stackexchange.com/questions/2632579/sum-of-diagonal-in-pascals-triangle-with-power-of-2 | [
"# Sum of Diagonal in Pascals triangle with power of 2\n\nProve that $$\\sum_{i=0}^n {{n+i}\\choose {i}} 2^{-(n+i)} = 1$$ I tried a direct approach and proof by induction, but the dependence of the exponent on $i$ throws me off every time.\n\nIt is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \\begin{align*} [z^k](1+z)^n=\\binom{n}{k} \\end{align*}\n\nWe obtain \\begin{align*} \\color{blue}{\\sum_{i=0}^n}&\\color{blue}{\\binom{n+i}{i}2^{-(i+n)}}\\\\ &=\\frac{1}{2^n}\\sum_{i=0}^n[z^i](1+z)^{n+i}\\frac{1}{2^i}\\tag{1}\\\\ &=\\frac{1}{2^n}[z^0](1+z)^n\\sum_{i=0}^n\\left(\\frac{1+z}{2z}\\right)^i\\tag{2}\\\\ &=\\frac{1}{2^n}[z^0](1+z)^n\\frac{\\left(\\frac{1+z}{2z}\\right)^{n+1}-1}{\\frac{1+z}{2z}-1}\\tag{3}\\\\ &=\\frac{1}{2^{n-1}}[z^{-1}]\\frac{(1+z)^n}{1-z}\\left[\\left(\\frac{1+z}{2z}\\right)^{n+1}-1\\right]\\tag{4}\\\\ &=\\frac{1}{2^{2n}}[z^n]\\frac{(1+z)^{2n+1}}{1-z}\\tag{5}\\\\ &=\\frac{1}{2^{2n}}\\sum_{j=0}^n\\binom{2n+1}{j}\\tag{6}\\\\ &=\\frac{1}{2^{2n}}\\cdot\\frac{1}{2}\\cdot2^{2n+1}\\\\ &\\color{blue}{=1} \\end{align*}\n\nComment:\n\n• In (1) we apply the coefficient of operator.\n\n• In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.\n\n• In (3) we apply the finite geometric series formula.\n\n• In (4) we simplify the denominator and apply again the rule from (2).\n\n• In (5) we observe that only the first term $\\left(\\frac{1+z}{2z}\\right)^{n+1}$ gives a non-zero contribution to the coefficient of $[z^{-1}]$. We also do some more simplifications and apply again the rule from (2).\n\n• In (6) we expand the geometric series $\\frac{1}{1-z}=1+z+z^2+\\cdots$ and select the coefficient of $z^j$ with $0\\leq j\\leq n$.\n\n• your coefficient of approach hit again ! – G Cab Feb 2 '18 at 16:21\n• @GCab: Thanks! :-) – Markus Scheuer Feb 2 '18 at 16:21\n\nA nice way to visualise this problem is in terms of monotonic North/East (NE) lattice paths.\n\nIf the probability of a North step is $1/2$ and of and East step is $1/2$ then, starting at $O(0,0)$ the probability that a path which leaves the box takes the leg between lattice points $(i,n)$ and $(i,n+1)$ is\n\n$$\\binom{n+i}{i}2^{-(n+i+1)}$$\n\nThis is also the probability a NE lattice path passes between $(n,i)$ and $(n+1,i)$.\n\nThe probability that a lattice path leaves the box defined by boundaries\n\n$$\\{y=-1/2,\\, y=n+1/2,\\, x=-1/2,\\, x=n+1/2\\}\\tag{1}$$\n\nis therefore the sum of probabilities that it leaves through boundary $y=n+1/2$ and boundary $x=n+1/2$. These are each:\n\n$$\\sum_{i=0}^{n}\\binom{n+i}{i}2^{-(n+i+1)}$$\n\nand since every such NE lattice path must leave the box as defined by $(1)$ the sum of these two probabilities is $1$:\n\n$$2\\sum_{i=0}^{n}\\binom{n+i}{i}2^{-(n+i+1)}=1$$\n\nor\n\n$$\\sum_{i=0}^{n}\\binom{n+i}{i}2^{-(n+i)}=1$$\n\nas required.\n\n• Nice approach. (+1) – Markus Scheuer Feb 2 '18 at 16:24\n• Thanks Markus! I did it with your method using pen and paper (just to see if I could) but then saw you had already answered (+1 btw). So, for variation, I posted my other approach. I was pleased that you posted, though, because you confirmed for me that I had used your method correctly. Thanks again! :D – N. Shales Feb 2 '18 at 16:29\n• You're welcome, and I appreciate your lattice path approach. Sometimes I also like to do so. :-) – Markus Scheuer Feb 2 '18 at 16:33"
] | [
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https://www.abcteach.com/search.php?category=0&q=subtract&search_type=1&match_words=2&limit_search=1&_form_action=&search_form1_form_visited=1 | [
"You are an abcteach Member, but you are logged in to the Free Site. To access all member features, log into the Member Site.\n\nSEARCH RESULTS: subtract\n\nClip Art:",
null,
"= Member Site Document\nThere are 566 documents matching your search.\n• A 5-page set for practicing subtraction on number lines. Includes subtracting from 10, subtracting from 20, subtracting from 100 by tens, subtracting from 100 by ones.\n\n• This Canadian Money Subtraction Word Problem (elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Canadian Money Subtraction Word Problem (elem) Clip Art. Poster solving subtraction word problem using Canadian money.\n• This Subtraction - 2 digits (set 1) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction - 2 digits (set 1) Clip Art. Printable subtraction worksheets - 2 digit numbers.\n• This Subtraction - 3 digits (set 5) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction - 3 digits (set 5) Clip Art. Printable subtraction worksheets - 3 digit numbers.\n• This Subtraction - 3 digits (set 3) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction - 3 digits (set 3) Clip Art. Printable subtraction worksheets - 3 digit numbers.\n• This Fraction Subtraction - unlike (set 1) Worksheet is perfect to practice fraction subtraction skills. Your elementary grade students will love this Fraction Subtraction - unlike (set 1) Worksheet. Printable fraction and subtraction worksheets, using single digits, unlike fractions; 5 pages long.\n• This Fraction Subtraction - unlike (set 2) Worksheet is perfect to practice fraction subtraction skills. Your elementary grade students will love this Fraction Subtraction - unlike (set 2) Worksheet. Printable fraction and subtraction worksheets, using single digits, unlike fractions; 5 pages long.\n• This Subtracting Decimals Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtracting Decimals Clip Art. Simple poster explains how to subtract decimals. Common Core: 6.NS.3\n• This Addition and Subtraction up to 20 (K-1) Penguin Theme Unit is perfect to practice addition and subtraction skills. Your elementary grade students will love this Addition and Subtraction up to 20 (K-1) Penguin Theme Unit. Addition and subtraction practice and assessment. Includes; practice worksheets, in and out boxes, and matching game.\n• 10 pages of mixed practice for finding the missing subtraction operand. 4 pages of ones place subtraction; 6 pages of ones to tens place subtraction. Facts are in vertical form. Created with our abctools Math Worksheet Generator. CC: 2.OA.A.1\n\n• This Subtraction - facts up to (set 7 ) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction - facts up to (set 7 ) Clip Art. Printable substraction worksheet, working with facts up to 20.\n• This Decimal Subtraction (set 1) Worksheet is perfect to practice decimal subraction skills. Your elementary grade students will love this Decimal Subtraction (set 1) Worksheet. Printable decimal subtraction worksheets, working on numbers up to 20. This is a five page workbook.\n• This Subtraction (elem/upper elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction (elem/upper elem) Clip Art. Printable subtration worksheet.\n• This Subtraction Facts to 10 (pre-k/primary) 1 Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction Facts to 10 (pre-k/primary) 1 Clip Art. Count the number of shapes in each group, take the second group away from the first, and count how many shapes remain. Black and white graphics.\n• This Subtraction Facts to 10 (pre-k/primary) 2 Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction Facts to 10 (pre-k/primary) 2 Clip Art. Count the number of shapes in each group, take the second group away from the first, and count how many shapes remain. Black and white graphics.\n• This Subtraction Facts to 10 (pre-k/primary) 3 Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction Facts to 10 (pre-k/primary) 3 Clip Art. Count the number of shapes in each group, take the second group away from the first, and count how many shapes remain. Black and white graphics.\n• This Subtraction Facts to 10 (pre-k/primary) 4 Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction Facts to 10 (pre-k/primary) 4 Clip Art. Count the number of shapes in each group, take the second group away from the first, and count how many shapes remain. Black and white graphics.\n• This Regrouping in Subtraction (elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Regrouping in Subtraction (elem) Clip Art. Three colorful math posters explain the rules for regrouping (\"borrowing\") in addition.\n• This Medieval Subtraction (primary/elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Medieval Subtraction (primary/elem) Clip Art. Help the knight find a path to the dragon. Draw a path through each stone with an answer of 3.\n• This Halloween Subtraction Coloring Packet (prim/elem) is perfect to practice subtraction skills. Your elementary grade students will love this Halloween Subtraction Coloring Packet (prim/elem). Five pages of Halloween pictures to find the differences and color by number. All differences are under 20.\n\n• 10 pages of subtraction practice. Count the number of shapes in each group, take the second group away from the first, and write how many remain. Created with abctools Math Worksheet Generators.\n\n• 10 pages of mixed practice for subtraction facts. 5 pages of facts with 1-10; 5 pages of facts within 1-20. Facts are in horizontal form. Created with our abctools Math Worksheet Generator. CC: 1.OA.C.6\n\n• This Math Symbols: Set 1: Subtraction B&W clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Symbols: Set 1: Subtraction B&W clipart is provided in jpeg format.\n• This Math Symbols: Set 1: Subtraction Color clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Symbols: Set 1: Subtraction Color clipart is provided in jpeg format.\n• This Math Symbols: Set 2: Subtraction B&W Labeled clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Symbols: Set 2: Subtraction B&W Labeled clipart is provided in jpeg format.\n• This Math Symbols: Set 2: Subtraction Grayscale Labeled clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Symbols: Set 2: Subtraction Grayscale Labeled clipart is provided in jpeg format.\n• This Math Symbols: Set 2: Subtraction Color Labeled clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Symbols: Set 2: Subtraction Color Labeled clipart is provided in jpeg format.\n• This Math Grid: Addition and Subtraction 01 B&W clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Grid: Addition and Subtraction 01 B&W clipart is provided in jpeg format.\n• This Math Grid: Addition and Subtraction 02 B&W clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Grid: Addition and Subtraction 02 B&W clipart is provided in jpeg format.\n• This Math Grid: Subtraction B&W clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Math Grid: Subtraction B&W clipart is provided in jpeg format.\n• This Addition and Subtraction (up to 10) Math is perfect to practice addition skills. Your elementary grade students will love this Addition and Subtraction (up to 10) Math. printable math worksheets, five pages of addition and subtration problems, includes answer key.\n• 10 pages of practice for subtraction from tens to millions places (place value increases with each page). Facts are in vertical form. Created with our abctools Math Worksheet Generator. CC: 4.NBT.B.4\n\n• Notebook file full of fun while practicing addition and subtraction of equivalent fractions.\n\n• 4 pages of word problems dealing with adding and subtracting to 20. Asks students to compare and relate numbers. Includes multi-step problems. Students will draw a picture to demonstrate their thinking and write a number sentence. CC: Math: 1.O.A.A.1\n\n• Four pages of addition and subtraction practice with a cute penguin theme.\n\n• 10 pages of practice for subtraction of positive and negative integers (5 pages of ones place, 5 pages of ones and tens places). Facts are in horizontal form. Created with our abctools Math Worksheet Generator. CC: 7.NS.A.1\n\n• This Color: Subtractive Colors B&W clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Color: Subtractive Colors B&W clipart is provided in jpeg format.\n• This Color: Subtractive Colors Color clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This Color: Subtractive Colors Color clipart is provided in jpeg format.\n• Valentine's Day Solve and Color worksheet. Subtract the equation, and use the key to color the image.\n\n• This Turkeys Math Game is perfect to practice addition and subtraction skills. Your elementary grade students will love this Turkeys Math Game. Students use turkey-shaped cards (numbered 1-20) to make addition and subtraction equations. Instructions included."
] | [
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https://doc-snapshots.qt.io/qt5-5.12/qhash.html | [
"# QHash Class\n\nThe QHash class is a template class that provides a hash-table-based dictionary. More...\n\n Header: #include qmake: QT += core Inherited By: QMultiHash\n\nNote: All functions in this class are reentrant.\n\n## Public Types\n\n class const_iterator class iterator class key_iterator typedef ConstIterator typedef Iterator typedef const_key_value_iterator typedef difference_type typedef key_type typedef key_value_iterator typedef mapped_type typedef size_type\n\n## Public Functions\n\n QHash() QHash(std::initializer_list > list) QHash(const QHash &other) QHash(QHash &&other) ~QHash() QHash::iterator begin() QHash::const_iterator begin() const int capacity() const QHash::const_iterator cbegin() const QHash::const_iterator cend() const void clear() QHash::const_iterator constBegin() const QHash::const_iterator constEnd() const QHash::const_iterator constFind(const Key &key) const QHash::const_key_value_iterator constKeyValueBegin() const QHash::const_key_value_iterator constKeyValueEnd() const bool contains(const Key &key) const int count(const Key &key) const int count() const bool empty() const QHash::iterator end() QHash::const_iterator end() const QPair equal_range(const Key &key) QPair equal_range(const Key &key) const QHash::iterator erase(QHash::const_iterator pos) QHash::iterator erase(QHash::iterator pos) QHash::iterator find(const Key &key) QHash::const_iterator find(const Key &key) const QHash::iterator insert(const Key &key, const T &value) QHash::iterator insertMulti(const Key &key, const T &value) bool isEmpty() const const Key key(const T &value) const const Key key(const T &value, const Key &defaultKey) const QHash::key_iterator keyBegin() const QHash::key_iterator keyEnd() const QHash::key_value_iterator keyValueBegin() QHash::const_key_value_iterator keyValueBegin() const QHash::key_value_iterator keyValueEnd() QHash::const_key_value_iterator keyValueEnd() const QList keys() const QList keys(const T &value) const int remove(const Key &key) void reserve(int size) int size() const void squeeze() void swap(QHash &other) T take(const Key &key) QList uniqueKeys() const QHash & unite(const QHash &other) const T value(const Key &key) const const T value(const Key &key, const T &defaultValue) const QList values() const QList values(const Key &key) const bool operator!=(const QHash &other) const QHash & operator=(const QHash &other) QHash & operator=(QHash &&other) bool operator==(const QHash &other) const T & operator[](const Key &key) const T operator[](const Key &key) const\n int qGlobalQHashSeed() uint qHash(char key, uint seed = ...) uint qHash(const QUrl &url, uint seed = 0) uint qHash(const QDateTime &key, uint seed = ...) uint qHash(const QDate &key, uint seed = ...) uint qHash(const QTime &key, uint seed = ...) uint qHash(const QPair &key, uint seed = ...) uint qHash(const std::pair &key, uint seed = ...) uint qHash(uchar key, uint seed = ...) uint qHash(signed char key, uint seed = ...) uint qHash(ushort key, uint seed = ...) uint qHash(short key, uint seed = ...) uint qHash(uint key, uint seed = ...) uint qHash(int key, uint seed = ...) uint qHash(ulong key, uint seed = ...) uint qHash(long key, uint seed = ...) uint qHash(quint64 key, uint seed = ...) uint qHash(qint64 key, uint seed = ...) uint qHash(float key, uint seed = 0) uint qHash(double key, uint seed = 0) uint qHash(long double key, uint seed = 0) uint qHash(const QChar key, uint seed = ...) uint qHash(const QByteArray &key, uint seed = ...) uint qHash(const QBitArray &key, uint seed = ...) uint qHash(const QString &key, uint seed = ...) uint qHash(const QStringRef &key, uint seed = ...) uint qHash(QLatin1String key, uint seed = ...) uint qHash(const T *key, uint seed = ...) uint qHash(const QHash &key, uint seed = ...) uint qHash(const QSet &key, uint seed = ...) uint qHash(const QVersionNumber &key, uint seed = 0) qHash(QSslEllipticCurve curve, uint seed) qHash(const QSslCertificate &key, uint seed) qHash(const QSslError &key, uint seed) uint qHashBits(const void *p, size_t len, uint seed = ...) uint qHashRange(InputIterator first, InputIterator last, uint seed = ...) uint qHashRangeCommutative(InputIterator first, InputIterator last, uint seed = ...) void qSetGlobalQHashSeed(int newSeed) QDataStream & operator<<(QDataStream &out, const QHash &hash) QDataStream & operator>>(QDataStream &in, QHash &hash)\n\n## Detailed Description\n\nThe QHash class is a template class that provides a hash-table-based dictionary.\n\nQHash<Key, T> is one of Qt's generic container classes. It stores (key, value) pairs and provides very fast lookup of the value associated with a key.\n\nQHash provides very similar functionality to QMap. The differences are:\n\n• QHash provides faster lookups than QMap. (See Algorithmic Complexity for details.)\n• When iterating over a QMap, the items are always sorted by key. With QHash, the items are arbitrarily ordered.\n• The key type of a QMap must provide operator<(). The key type of a QHash must provide operator==() and a global hash function called qHash() (see qHash).\n\nHere's an example QHash with QString keys and `int` values:\n\n`QHash<QString, int> hash;`\n\nTo insert a (key, value) pair into the hash, you can use operator[]():\n\n```hash[\"one\"] = 1;\nhash[\"three\"] = 3;\nhash[\"seven\"] = 7;```\n\nThis inserts the following three (key, value) pairs into the QHash: (\"one\", 1), (\"three\", 3), and (\"seven\", 7). Another way to insert items into the hash is to use insert():\n\n`hash.insert(\"twelve\", 12);`\n\nTo look up a value, use operator[]() or value():\n\n```int num1 = hash[\"thirteen\"];\nint num2 = hash.value(\"thirteen\");```\n\nIf there is no item with the specified key in the hash, these functions return a default-constructed value.\n\nIf you want to check whether the hash contains a particular key, use contains():\n\n```int timeout = 30;\nif (hash.contains(\"TIMEOUT\"))\ntimeout = hash.value(\"TIMEOUT\");```\n\nThere is also a value() overload that uses its second argument as a default value if there is no item with the specified key:\n\n`int timeout = hash.value(\"TIMEOUT\", 30);`\n\nIn general, we recommend that you use contains() and value() rather than operator[]() for looking up a key in a hash. The reason is that operator[]() silently inserts an item into the hash if no item exists with the same key (unless the hash is const). For example, the following code snippet will create 1000 items in memory:\n\n```// WRONG\nQHash<int, QWidget *> hash;\n...\nfor (int i = 0; i < 1000; ++i) {\nif (hash[i] == okButton)\ncout << \"Found button at index \" << i << endl;\n}```\n\nTo avoid this problem, replace `hash[i]` with `hash.value(i)` in the code above.\n\nInternally, QHash uses a hash table to perform lookups. This hash table automatically grows and shrinks to provide fast lookups without wasting too much memory. You can still control the size of the hash table by calling reserve() if you already know approximately how many items the QHash will contain, but this isn't necessary to obtain good performance. You can also call capacity() to retrieve the hash table's size.\n\nIf you want to navigate through all the (key, value) pairs stored in a QHash, you can use an iterator. QHash provides both Java-style iterators (QHashIterator and QMutableHashIterator) and STL-style iterators (QHash::const_iterator and QHash::iterator). Here's how to iterate over a QHash<QString, int> using a Java-style iterator:\n\n```QHashIterator<QString, int> i(hash);\nwhile (i.hasNext()) {\ni.next();\ncout << i.key() << \": \" << i.value() << endl;\n}```\n\nHere's the same code, but using an STL-style iterator:\n\n```QHash<QString, int>::const_iterator i = hash.constBegin();\nwhile (i != hash.constEnd()) {\ncout << i.key() << \": \" << i.value() << endl;\n++i;\n}```\n\nQHash is unordered, so an iterator's sequence cannot be assumed to be predictable. If ordering by key is required, use a QMap.\n\nNormally, a QHash allows only one value per key. If you call insert() with a key that already exists in the QHash, the previous value is erased. For example:\n\n```hash.insert(\"plenty\", 100);\nhash.insert(\"plenty\", 2000);\n// hash.value(\"plenty\") == 2000```\n\nHowever, you can store multiple values per key by using insertMulti() instead of insert() (or using the convenience subclass QMultiHash). If you want to retrieve all the values for a single key, you can use values(const Key &key), which returns a QList<T>:\n\n```QList<int> values = hash.values(\"plenty\");\nfor (int i = 0; i < values.size(); ++i)\ncout << values.at(i) << endl;```\n\nThe items that share the same key are available from most recently to least recently inserted. A more efficient approach is to call find() to get the iterator for the first item with a key and iterate from there:\n\n```QHash<QString, int>::iterator i = hash.find(\"plenty\");\nwhile (i != hash.end() && i.key() == \"plenty\") {\ncout << i.value() << endl;\n++i;\n}```\n\nIf you only need to extract the values from a hash (not the keys), you can also use foreach:\n\n```QHash<QString, int> hash;\n...\nforeach (int value, hash)\ncout << value << endl;```\n\nItems can be removed from the hash in several ways. One way is to call remove(); this will remove any item with the given key. Another way is to use QMutableHashIterator::remove(). In addition, you can clear the entire hash using clear().\n\nQHash's key and value data types must be assignable data types. You cannot, for example, store a QWidget as a value; instead, store a QWidget *.\n\n#### The qHash() hashing function\n\nA QHash's key type has additional requirements other than being an assignable data type: it must provide operator==(), and there must also be a qHash() function in the type's namespace that returns a hash value for an argument of the key's type.\n\nThe qHash() function computes a numeric value based on a key. It can use any algorithm imaginable, as long as it always returns the same value if given the same argument. In other words, if `e1 == e2`, then `qHash(e1) == qHash(e2)` must hold as well. However, to obtain good performance, the qHash() function should attempt to return different hash values for different keys to the largest extent possible.\n\nFor a key type `K`, the qHash function must have one of these signatures:\n\n```uint qHash(K key);\nuint qHash(const K &key);\n\nuint qHash(K key, uint seed);\nuint qHash(const K &key, uint seed);```\n\nThe two-arguments overloads take an unsigned integer that should be used to seed the calculation of the hash function. This seed is provided by QHash in order to prevent a family of algorithmic complexity attacks. If both a one-argument and a two-arguments overload are defined for a key type, the latter is used by QHash (note that you can simply define a two-arguments version, and use a default value for the seed parameter).\n\nHere's a partial list of the C++ and Qt types that can serve as keys in a QHash: any integer type (char, unsigned long, etc.), any pointer type, QChar, QString, and QByteArray. For all of these, the `<QHash>` header defines a qHash() function that computes an adequate hash value. Many other Qt classes also declare a qHash overload for their type; please refer to the documentation of each class.\n\nIf you want to use other types as the key, make sure that you provide operator==() and a qHash() implementation.\n\nExample:\n\n```#ifndef EMPLOYEE_H\n#define EMPLOYEE_H\n\nclass Employee\n{\npublic:\nEmployee() {}\nEmployee(const QString &name, const QDate &dateOfBirth);\n...\n\nprivate:\nQString myName;\nQDate myDateOfBirth;\n};\n\ninline bool operator==(const Employee &e1, const Employee &e2)\n{\nreturn e1.name() == e2.name()\n&& e1.dateOfBirth() == e2.dateOfBirth();\n}\n\ninline uint qHash(const Employee &key, uint seed)\n{\nreturn qHash(key.name(), seed) ^ key.dateOfBirth().day();\n}\n\n#endif // EMPLOYEE_H```\n\nIn the example above, we've relied on Qt's global qHash(const QString &, uint) to give us a hash value for the employee's name, and XOR'ed this with the day they were born to help produce unique hashes for people with the same name.\n\nNote that the implementation of the qHash() overloads offered by Qt may change at any time. You must not rely on the fact that qHash() will give the same results (for the same inputs) across different Qt versions.\n\n#### Algorithmic complexity attacks\n\nAll hash tables are vulnerable to a particular class of denial of service attacks, in which the attacker carefully pre-computes a set of different keys that are going to be hashed in the same bucket of a hash table (or even have the very same hash value). The attack aims at getting the worst-case algorithmic behavior (O(n) instead of amortized O(1), see Algorithmic Complexity for the details) when the data is fed into the table.\n\nIn order to avoid this worst-case behavior, the calculation of the hash value done by qHash() can be salted by a random seed, that nullifies the attack's extent. This seed is automatically generated by QHash once per process, and then passed by QHash as the second argument of the two-arguments overload of the qHash() function.\n\nThis randomization of QHash is enabled by default. Even though programs should never depend on a particular QHash ordering, there may be situations where you temporarily need deterministic behavior, for example for debugging or regression testing. To disable the randomization, define the environment variable `QT_HASH_SEED` to have the value 0. Alternatively, you can call the qSetGlobalQHashSeed() function with the value 0.\n\n## Member Type Documentation\n\n### typedef QHash::ConstIterator\n\nQt-style synonym for QHash::const_iterator.\n\n### typedef QHash::Iterator\n\nQt-style synonym for QHash::iterator.\n\n### typedef QHash::const_key_value_iterator\n\nThe QMap::const_key_value_iterator typedef provides an STL-style const iterator for QHash and QMultiHash.\n\nQHash::const_key_value_iterator is essentially the same as QHash::const_iterator with the difference that operator*() returns a key/value pair instead of a value.\n\nThis typedef was introduced in Qt 5.10.\n\n### typedef QHash::difference_type\n\nTypedef for ptrdiff_t. Provided for STL compatibility.\n\n### typedef QHash::key_type\n\nTypedef for Key. Provided for STL compatibility.\n\n### typedef QHash::key_value_iterator\n\nThe QMap::key_value_iterator typedef provides an STL-style iterator for QHash and QMultiHash.\n\nQHash::key_value_iterator is essentially the same as QHash::iterator with the difference that operator*() returns a key/value pair instead of a value.\n\nThis typedef was introduced in Qt 5.10.\n\n### typedef QHash::mapped_type\n\nTypedef for T. Provided for STL compatibility.\n\n### typedef QHash::size_type\n\nTypedef for int. Provided for STL compatibility.\n\n## Member Function Documentation\n\n### QHash::QHash()\n\nConstructs an empty hash.\n\n### QHash::QHash(std::initializer_list<std::pair<Key, T> > list)\n\nConstructs a hash with a copy of each of the elements in the initializer list list.\n\nThis function is only available if the program is being compiled in C++11 mode.\n\nThis function was introduced in Qt 5.1.\n\n### QHash::QHash(const QHash<K, V> &other)\n\nConstructs a copy of other.\n\nThis operation occurs in constant time, because QHash is implicitly shared. This makes returning a QHash from a function very fast. If a shared instance is modified, it will be copied (copy-on-write), and this takes linear time.\n\n### QHash::QHash(QHash<K, V> &&other)\n\nMove-constructs a QHash instance, making it point at the same object that other was pointing to.\n\nThis function was introduced in Qt 5.2.\n\n### QHash::~QHash()\n\nDestroys the hash. References to the values in the hash and all iterators of this hash become invalid.\n\n### QHash::iterator QHash::begin()\n\nReturns an STL-style iterator pointing to the first item in the hash.\n\n### int QHash::capacity() const\n\nReturns the number of buckets in the QHash's internal hash table.\n\nThe sole purpose of this function is to provide a means of fine tuning QHash's memory usage. In general, you will rarely ever need to call this function. If you want to know how many items are in the hash, call size().\n\n### QHash::const_iterator QHash::cbegin() const\n\nReturns a const STL-style iterator pointing to the first item in the hash.\n\nThis function was introduced in Qt 5.0.\n\n### QHash::const_iterator QHash::cend() const\n\nReturns a const STL-style iterator pointing to the imaginary item after the last item in the hash.\n\nThis function was introduced in Qt 5.0.\n\nRemoves all items from the hash.\n\n### QHash::const_iterator QHash::constBegin() const\n\nReturns a const STL-style iterator pointing to the first item in the hash.\n\n### QHash::const_iterator QHash::constEnd() const\n\nReturns a const STL-style iterator pointing to the imaginary item after the last item in the hash.\n\n### QHash::const_iterator QHash::constFind(const Key &key) const\n\nReturns an iterator pointing to the item with the key in the hash.\n\nIf the hash contains no item with the key, the function returns constEnd().\n\nThis function was introduced in Qt 4.1.\n\n### QHash::const_key_value_iterator QHash::constKeyValueBegin() const\n\nReturns a const STL-style iterator pointing to the first entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### QHash::const_key_value_iterator QHash::constKeyValueEnd() const\n\nReturns a const STL-style iterator pointing to the imaginary entry after the last entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### bool QHash::contains(const Key &key) const\n\nReturns `true` if the hash contains an item with the key; otherwise returns `false`.\n\n### int QHash::count(const Key &key) const\n\nReturns the number of items associated with the key.\n\nSame as size().\n\n### bool QHash::empty() const\n\nThis function is provided for STL compatibility. It is equivalent to isEmpty(), returning true if the hash is empty; otherwise returns `false`.\n\n### QHash::iterator QHash::end()\n\nReturns an STL-style iterator pointing to the imaginary item after the last item in the hash.\n\n### QPair<QHash::iterator, QHash::iterator> QHash::equal_range(const Key &key)\n\nReturns a pair of iterators delimiting the range of values `[first, second)`, that are stored under key. If the range is empty then both iterators will be equal to end().\n\nThis function was introduced in Qt 5.7.\n\n### QPair<QHash::const_iterator, QHash::const_iterator> QHash::equal_range(const Key &key) const\n\nThis function was introduced in Qt 5.7.\n\n### QHash::iterator QHash::erase(QHash::const_iteratorpos)\n\nRemoves the (key, value) pair associated with the iterator pos from the hash, and returns an iterator to the next item in the hash.\n\nUnlike remove() and take(), this function never causes QHash to rehash its internal data structure. This means that it can safely be called while iterating, and won't affect the order of items in the hash. For example:\n\n```QHash<QObject *, int> objectHash;\n...\nQHash<QObject *, int>::iterator i = objectHash.find(obj);\nwhile (i != objectHash.end() && i.key() == obj) {\nif (i.value() == 0) {\ni = objectHash.erase(i);\n} else {\n++i;\n}\n}```\n\nThis function was introduced in Qt 5.7.\n\n### QHash::iterator QHash::find(const Key &key)\n\nReturns an iterator pointing to the item with the key in the hash.\n\nIf the hash contains no item with the key, the function returns end().\n\nIf the hash contains multiple items with the key, this function returns an iterator that points to the most recently inserted value. The other values are accessible by incrementing the iterator. For example, here's some code that iterates over all the items with the same key:\n\n```QHash<QString, int> hash;\n...\nQHash<QString, int>::const_iterator i = hash.find(\"HDR\");\nwhile (i != hash.end() && i.key() == \"HDR\") {\ncout << i.value() << endl;\n++i;\n}```\n\n### QHash::iterator QHash::insert(const Key &key, const T &value)\n\nInserts a new item with the key and a value of value.\n\nIf there is already an item with the key, that item's value is replaced with value.\n\nIf there are multiple items with the key, the most recently inserted item's value is replaced with value.\n\n### QHash::iterator QHash::insertMulti(const Key &key, const T &value)\n\nInserts a new item with the key and a value of value.\n\nIf there is already an item with the same key in the hash, this function will simply create a new one. (This behavior is different from insert(), which overwrites the value of an existing item.)\n\n### bool QHash::isEmpty() const\n\nReturns `true` if the hash contains no items; otherwise returns false.\n\n### const Key QHash::key(const T &value) const\n\nReturns the first key mapped to value.\n\nIf the hash contains no item with the value, the function returns a default-constructed key.\n\nThis function can be slow (linear time), because QHash's internal data structure is optimized for fast lookup by key, not by value.\n\n### const Key QHash::key(const T &value, const Key &defaultKey) const\n\nReturns the first key mapped to value, or defaultKey if the hash contains no item mapped to value.\n\nThis function can be slow (linear time), because QHash's internal data structure is optimized for fast lookup by key, not by value.\n\nThis function was introduced in Qt 4.3.\n\n### QHash::key_iterator QHash::keyBegin() const\n\nReturns a const STL-style iterator pointing to the first key in the hash.\n\nThis function was introduced in Qt 5.6.\n\n### QHash::key_iterator QHash::keyEnd() const\n\nReturns a const STL-style iterator pointing to the imaginary item after the last key in the hash.\n\nThis function was introduced in Qt 5.6.\n\n### QHash::key_value_iterator QHash::keyValueBegin()\n\nReturns an STL-style iterator pointing to the first entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### QHash::const_key_value_iterator QHash::keyValueBegin() const\n\nReturns a const STL-style iterator pointing to the first entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### QHash::key_value_iterator QHash::keyValueEnd()\n\nReturns an STL-style iterator pointing to the imaginary entry after the last entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### QHash::const_key_value_iterator QHash::keyValueEnd() const\n\nReturns a const STL-style iterator pointing to the imaginary entry after the last entry in the hash.\n\nThis function was introduced in Qt 5.10.\n\n### QList<Key> QHash::keys() const\n\nReturns a list containing all the keys in the hash, in an arbitrary order. Keys that occur multiple times in the hash (because items were inserted with insertMulti(), or unite() was used) also occur multiple times in the list.\n\nTo obtain a list of unique keys, where each key from the map only occurs once, use uniqueKeys().\n\nThe order is guaranteed to be the same as that used by values().\n\n### QList<Key> QHash::keys(const T &value) const\n\nReturns a list containing all the keys associated with value value, in an arbitrary order.\n\nThis function can be slow (linear time), because QHash's internal data structure is optimized for fast lookup by key, not by value.\n\n### int QHash::remove(const Key &key)\n\nRemoves all the items that have the key from the hash. Returns the number of items removed which is usually 1 but will be 0 if the key isn't in the hash, or greater than 1 if insertMulti() has been used with the key.\n\n### void QHash::reserve(intsize)\n\nEnsures that the QHash's internal hash table consists of at least size buckets.\n\nThis function is useful for code that needs to build a huge hash and wants to avoid repeated reallocation. For example:\n\n```QHash<QString, int> hash;\nhash.reserve(20000);\nfor (int i = 0; i < 20000; ++i)\nhash.insert(keys[i], values[i]);```\n\nIdeally, size should be slightly more than the maximum number of items expected in the hash. size doesn't have to be prime, because QHash will use a prime number internally anyway. If size is an underestimate, the worst that will happen is that the QHash will be a bit slower.\n\nIn general, you will rarely ever need to call this function. QHash's internal hash table automatically shrinks or grows to provide good performance without wasting too much memory.\n\n### int QHash::size() const\n\nReturns the number of items in the hash.\n\n### void QHash::squeeze()\n\nReduces the size of the QHash's internal hash table to save memory.\n\nThe sole purpose of this function is to provide a means of fine tuning QHash's memory usage. In general, you will rarely ever need to call this function.\n\n### void QHash::swap(QHash<K, V> &other)\n\nSwaps hash other with this hash. This operation is very fast and never fails.\n\nThis function was introduced in Qt 4.8.\n\n### T QHash::take(const Key &key)\n\nRemoves the item with the key from the hash and returns the value associated with it.\n\nIf the item does not exist in the hash, the function simply returns a default-constructed value. If there are multiple items for key in the hash, only the most recently inserted one is removed.\n\nIf you don't use the return value, remove() is more efficient.\n\n### QList<Key> QHash::uniqueKeys() const\n\nReturns a list containing all the keys in the map. Keys that occur multiple times in the map (because items were inserted with insertMulti(), or unite() was used) occur only once in the returned list.\n\nThis function was introduced in Qt 4.2.\n\n### QHash<K, V> &QHash::unite(const QHash<K, V> &other)\n\nInserts all the items in the other hash into this hash. If a key is common to both hashes, the resulting hash will contain the key multiple times.\n\n### const T QHash::value(const Key &key) const\n\nReturns the value associated with the key.\n\nIf the hash contains no item with the key, the function returns a default-constructed value. If there are multiple items for the key in the hash, the value of the most recently inserted one is returned.\n\n### const T QHash::value(const Key &key, const T &defaultValue) const\n\nIf the hash contains no item with the given key, the function returns defaultValue.\n\n### QList<T> QHash::values() const\n\nReturns a list containing all the values in the hash, in an arbitrary order. If a key is associated with multiple values, all of its values will be in the list, and not just the most recently inserted one.\n\nThe order is guaranteed to be the same as that used by keys().\n\n### QList<T> QHash::values(const Key &key) const\n\nReturns a list of all the values associated with the key, from the most recently inserted to the least recently inserted.\n\n### bool QHash::operator!=(const QHash<K, V> &other) const\n\nReturns `true` if other is not equal to this hash; otherwise returns `false`.\n\nTwo hashes are considered equal if they contain the same (key, value) pairs.\n\nThis function requires the value type to implement `operator==()`.\n\n### QHash<K, V> &QHash::operator=(const QHash<K, V> &other)\n\nAssigns other to this hash and returns a reference to this hash.\n\n### QHash<K, V> &QHash::operator=(QHash<K, V> &&other)\n\nMove-assigns other to this QHash instance.\n\nThis function was introduced in Qt 5.2.\n\n### bool QHash::operator==(const QHash<K, V> &other) const\n\nReturns `true` if other is equal to this hash; otherwise returns false.\n\nTwo hashes are considered equal if they contain the same (key, value) pairs.\n\nThis function requires the value type to implement `operator==()`.\n\n### T &QHash::operator[](const Key &key)\n\nReturns the value associated with the key as a modifiable reference.\n\nIf the hash contains no item with the key, the function inserts a default-constructed value into the hash with the key, and returns a reference to it. If the hash contains multiple items with the key, this function returns a reference to the most recently inserted value.\n\nSame as value().\n\n## Related Non-Members\n\n### intqGlobalQHashSeed()\n\nReturns the current global QHash seed.\n\nThe seed is set in any newly created QHash. See qHash about how this seed is being used by QHash.\n\nThis function was introduced in Qt 5.6.\n\n### uintqHash(charkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QUrl &url, uintseed = 0)\n\nReturns the hash value for the url. If specified, seed is used to initialize the hash.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QDateTime &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QDate &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QTime &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QPair<T1, T2> &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nTypes `T1` and `T2` must be supported by qHash().\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const std::pair<T1, T2> &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nTypes `T1` and `T2` must be supported by qHash().\n\nNote: The return type of this function is not the same as that of\n\n`qHash(qMakePair(key.first, key.second), seed);`\n\nThe two functions use different hashing algorithms; due to binary compatibility constraints, we cannot change the QPair algorithm to match the std::pair one before Qt 6.\n\nThis function was introduced in Qt 5.7.\n\n### uintqHash(ucharkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(signedcharkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(ushortkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(shortkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(uintkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(intkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(ulongkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(longkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(quint64key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(qint64key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(floatkey, uintseed = 0)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.3.\n\n### uintqHash(doublekey, uintseed = 0)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.3.\n\n### uintqHash(longdoublekey, uintseed = 0)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.3.\n\n### uintqHash(const QCharkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QByteArray &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QBitArray &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QString &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QStringRef &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(QLatin1Stringkey, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const T *key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.0.\n\n### uintqHash(const QHash<Key, T> &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nType `T` must be supported by qHash().\n\nThis function was introduced in Qt 5.8.\n\n### uintqHash(const QSet<T> &key, uintseed = ...)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThe hash value is independent of the order of elements in key, that is, sets that contain the same elements hash to the same value.\n\nThis function was introduced in Qt 5.5.\n\n### uintqHash(const QVersionNumber &key, uintseed = 0)\n\nReturns the hash value for the key, using seed to seed the calculation.\n\nThis function was introduced in Qt 5.6.\n\n### qHash(QSslEllipticCurvecurve, uintseed)\n\nThis function was introduced in Qt 5.5.\n\n### qHash(const QSslCertificate &key, uintseed)\n\nThis function was introduced in Qt 5.4.\n\n### qHash(const QSslError &key, uintseed)\n\nThis function was introduced in Qt 5.4.\n\n### uintqHashBits(const void *p, size_tlen, uintseed = ...)\n\nReturns the hash value for the memory block of size len pointed to by p, using seed to seed the calculation.\n\nUse this function only to implement qHash() for your own custom types. For example, here's how you could implement a qHash() overload for std::vector<int>:\n\n```inline uint qHash(const std::vector<int> &key, uint seed = 0)\n{\nif (key.empty())\nreturn seed;\nelse\nreturn qHashBits(&key.front(), key.size() * sizeof(int), seed);\n}```\n\nThis takes advantage of the fact that std::vector lays out its data contiguously. If that is not the case, or the contained type has padding, you should use qHashRange() instead.\n\nIt bears repeating that the implementation of qHashBits() - like the qHash() overloads offered by Qt - may change at any time. You must not rely on the fact that qHashBits() will give the same results (for the same inputs) across different Qt versions.\n\nThis function was introduced in Qt 5.4.\n\n### uintqHashRange(InputIteratorfirst, InputIteratorlast, uintseed = ...)\n\nReturns the hash value for the range [first,last), using seed to seed the calculation, by successively applying qHash() to each element and combining the hash values into a single one.\n\nThe return value of this function depends on the order of elements in the range. That means that\n\n`{0, 1, 2}`\n\nand\n\n`{1, 2, 0}`\n\nhash to different values. If order does not matter, for example for hash tables, use qHashRangeCommutative() instead. If you are hashing raw memory, use qHashBits().\n\nUse this function only to implement qHash() for your own custom types. For example, here's how you could implement a qHash() overload for std::vector<int>:\n\n```inline uint qHash(const std::vector<int> &key, uint seed = 0)\n{\nreturn qHashRange(key.begin(), key.end(), seed);\n}```\n\nIt bears repeating that the implementation of qHashRange() - like the qHash() overloads offered by Qt - may change at any time. You must not rely on the fact that qHashRange() will give the same results (for the same inputs) across different Qt versions, even if qHash() for the element type would.\n\nThis function was introduced in Qt 5.5.\n\n### uintqHashRangeCommutative(InputIteratorfirst, InputIteratorlast, uintseed = ...)\n\nReturns the hash value for the range [first,last), using seed to seed the calculation, by successively applying qHash() to each element and combining the hash values into a single one.\n\nThe return value of this function does not depend on the order of elements in the range. That means that\n\n`{0, 1, 2}`\n\nand\n\n`{1, 2, 0}`\n\nhash to the same values. If order matters, for example, for vectors and arrays, use qHashRange() instead. If you are hashing raw memory, use qHashBits().\n\nUse this function only to implement qHash() for your own custom types. For example, here's how you could implement a qHash() overload for std::unordered_set<int>:\n\n```inline uint qHash(const std::unordered_set<int> &key, uint seed = 0)\n{\nreturn qHashRangeCommutative(key.begin(), key.end(), seed);\n}```\n\nIt bears repeating that the implementation of qHashRangeCommutative() - like the qHash() overloads offered by Qt - may change at any time. You must not rely on the fact that qHashRangeCommutative() will give the same results (for the same inputs) across different Qt versions, even if qHash() for the element type would.\n\nThis function was introduced in Qt 5.5.\n\n### voidqSetGlobalQHashSeed(intnewSeed)\n\nSets the global QHash seed to newSeed.\n\nManually setting the global QHash seed value should be done only for testing and debugging purposes, when deterministic and reproducible behavior on a QHash is needed. We discourage to do it in production code as it can make your application susceptible to algorithmic complexity attacks.\n\nFrom Qt 5.10 and onwards, the only allowed values are 0 and -1. Passing the value -1 will reinitialize the global QHash seed to a random value, while the value of 0 is used to request a stable algorithm for C++ primitive types types (like `int`) and string types (QString, QByteArray).\n\nThe seed is set in any newly created QHash. See qHash about how this seed is being used by QHash.\n\nIf the environment variable `QT_HASH_SEED` is set, calling this function will result in a no-op.\n\nThis function was introduced in Qt 5.6.\n\n### QDataStream &operator<<(QDataStream &out, const QHash<Key, T> &hash)\n\nWrites the hash hash to stream out.\n\nThis function requires the key and value types to implement `operator<<()`.\n\nThis function requires the key and value types to implement `operator>>()`."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73412746,"math_prob":0.9078535,"size":39929,"snap":"2020-34-2020-40","text_gpt3_token_len":9770,"char_repetition_ratio":0.22940513,"word_repetition_ratio":0.3526656,"special_character_ratio":0.24796514,"punctuation_ratio":0.18204646,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96562177,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T01:54:35Z\",\"WARC-Record-ID\":\"<urn:uuid:079ac2ec-c2ff-413f-9c01-e3abb9912eee>\",\"Content-Length\":\"142167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6a4cb8c-b05f-4591-812c-cdc546e3a647>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb058c7a-2290-44fa-b581-c48b47bb274f>\",\"WARC-IP-Address\":\"18.203.89.116\",\"WARC-Target-URI\":\"https://doc-snapshots.qt.io/qt5-5.12/qhash.html\",\"WARC-Payload-Digest\":\"sha1:O27FQBMOSCTCNV7XCVBCXB7PO3SJ4C3P\",\"WARC-Block-Digest\":\"sha1:UCGPZFLMXJWMELBVEQWI52AHDCHLR7XB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735990.92_warc_CC-MAIN-20200806001745-20200806031745-00242.warc.gz\"}"} |
http://www.mywordsolution.com/question/what-effect-has-the-transformation-had-on-the-lowest-exam/9292707 | [
"+61-413 786 465\n\[email protected]\n\n## Statistics\n\nQuestion 1.\n\nWhen drawing a rectangle or creating a rectangular object, a particular ratio of width to length is thought to be most aesthetically pleasing. This ratio is ( √5 - 1)/2 = 0.618, and is called the golden ratio. The Shoshoni Indians (native Americans) used beaded rectangles to decorate their leather goods. Do the beaded rectangles of the Shoshoni follow the golden ratio? A random sample of beads was measured. The data is stored in the text file \"shoshoni\", which contains the variable:\n\nshoshoni the ratio of width to length for the rectangular bead\n\nQuestion 2.\n\nTwelve subjects were each tested twice to measure their reaction time to a given stimulus, once after taking a drug and once after taking a placebo. The order of the treatments was randomised. Does the drug have any affect on the reaction time? The data is stored in the text\nfile \"reaction\", which contains the variables:\n\ndrug the time to react after being given the drug (in seconds)\nplacebo the time to react after being given the placebo (in seconds)\n\nQuestion 3.\n\nA sample of 21 exam marks was randomly selected from a large stage 2 course. The marks were: 4, 5, 6, 7, 8, 11, 12, 14, 15, 20, 21, 27, 30, 33, 35, 40, 41, 50, 65, 69, 98\n\n(a) Enter the data into R as a data vector called y.\n\n(b) Create a new data vector log.y by using a natural log transformation on y. List log.y.\n\n(c) (i) What effect has the transformation had on the lowest exam mark?\n(ii) What effect has the transformation had on the highest exam mark?\n(iii) Does the log transformation affect the higher or lower values more?\n\n(d) (i) Produce a vertical box plot for y.\n(ii) Produce a vertical box plot for log.y.\n(iii) Produce side-by-side box plots for y and log.y.\n\nHint: start by creating two new variables as follows:\ny.data<-c(y,log.y)\ny.label<-factor(rep(c(\"y\",\"log.y\"),c(21,21)))\n\n(iv) Briefly comment on what these plots show.\n\n(e) (i) Use the R function summaryStats to calculate the major summary statistics for both y and log.y.\n\n(ii) Compare the mean of y with the median of y.\n\n(iii) Compare the mean of log.y with the median of log.y.\n\n(iv) Comment on the skewness statistics for both y and log.y.\n\n(v) Using R as a calculator, back-transform the mean of log.y.\n\n(vi) Is the back-transformed mean of log.ycloser to the mean of y or to the median of y?\n\nQuestion 4.\n\nA study was conducted by two University of Auckland staff (a statistician and a recreation officer) to assess the impact of new rugby rules that were aimed to improve the flow of the game. A random sample of 100 time periods of uninterrupted play were recorded from games\nplayed under the new rules and another random sample of 100 time periods of uninterrupted play were recorded from games played under the old rules. The data is stored in the text file \"rugby2\", which contains the variables:\n\ntimes the time recorded for a period of uninterrupted play (in seconds)\nrules the rules that were played under:\nnew or old\nThe university staff were interested in whether the times of uninterrupted play differed under the two sets of rules, and, if so, by how much.\n\nQuestion 5.\n\nA university lecturer was interested in assessing the fuel economy of her new car. Each time she filled up with petrol, she noted the distance travelled and the quantity of fuel required to fill the tank. It was of particular interest to predict the distance trave lled on 40 litres of fuel. The data are stored in the text file \"fue lconsume\", which contains the variables:\n\nlitres the amount of petrol used (in litres)\ndist the distance travelled (in km)\n\nStatistics and Probability, Statistics\n\n• Category:- Statistics and Probability\n• Reference No.:- M9292707\n• Price:- \\$45\n\nPriced at Now at \\$45, Verified Solution\n\nHave any Question?\n\n## Related Questions in Statistics and Probability\n\n### Financial management how can a financial manager use the\n\nFinancial Management How can a financial manager use the time value of money(TVM) concept to accomplish this goal?\n\n### The speed of cars on a stretch of road is normally\n\nThe speed of cars on a stretch of road is normally distributed with an average 42 miles per hour with a standard deviation of 5.9 miles per hour. What is the probability that a randomly selected car is violating the spee ...\n\n### A company that supplies batteries for watches guarantees\n\nA company that supplies batteries for watches guarantees that 95% of the batteries it ships will be free from defects. You test a sample of 50 batteries you received. You find that fewer than 10 have defects. Does this l ...\n\n### Researchers randomly assigned 25 beginning students of\n\nResearchers randomly assigned 25 beginning students of Russian to begin speaking practice immediately and another 25 to delay speaking for 4 weeks. At the end of the semester both groups took a standard test of comprehen ...\n\n### Question - the data in the table below is from a study\n\nQuestion - The data in the table below is from a study conducted by an insurance company to determine the effect of changing the process by which insurance claims are approved. The goal was to improve policyholder satisf ...\n\n### Sppose a and b are collectively exhaustive in addition pa\n\nSuppose A and B are collectively exhaustive. In addition, P(A) = 0.2 and P(B) = 0.8. Suppose C and D are both mutually exclusive and collectively exhaustive. Further, P(C|A) = 0.7 and P(D|B) = 0.5. What are P(C) and P(D) ...\n\n### Mimi plans make a random guess at 10 true-or-false\n\nMimi plans make a random guess at 10 true-or-false questions. Answer the following questions: (a) Let X be the number of correct answers Mimi gets. As we know, the distribution of X is a binomial probability distribution ...\n\n### Robertson steel is forecasting the following\n\nRobertson Steel is forecasting the following numbers: EBIT \\$1,000,000 Interest Expense 300,000 ROE 20% The company is in the 40 percent tax bracket. After putting together the forecast the ...\n\n### The assessment of the allocation of profits and losses is\n\nThe assessment of the allocation of profits and losses is one of the three key areas in which process?\n\n### The average time a person spends at the barefoot landing\n\nThe average time a person spends at the Barefoot Landing Aquarium Is 96 minutes. The standard deviation is 17 minutes, and this variable is normally distributed. What is the probability that a randomly selected visitor w ...\n\n• 13,132 Experts\n\n## Looking for Assignment Help?\n\nStart excelling in your Courses, Get help with Assignment\n\nWrite us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.\n\n### Why might a bank avoid the use of interest rate swaps even\n\nWhy might a bank avoid the use of interest rate swaps, even when the institution is exposed to significant interest rate\n\n### Describe the difference between zero coupon bonds and\n\nDescribe the difference between zero coupon bonds and coupon bonds. Under what conditions will a coupon bond sell at a p\n\n### Compute the present value of an annuity of 880 per year\n\nCompute the present value of an annuity of \\$ 880 per year for 16 years, given a discount rate of 6 percent per annum. As\n\n### Compute the present value of an 1150 payment made in ten\n\nCompute the present value of an \\$1,150 payment made in ten years when the discount rate is 12 percent. (Do not round int\n\n### Compute the present value of an annuity of 699 per year\n\nCompute the present value of an annuity of \\$ 699 per year for 19 years, given a discount rate of 6 percent per annum. As"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9441695,"math_prob":0.89191073,"size":3546,"snap":"2023-40-2023-50","text_gpt3_token_len":857,"char_repetition_ratio":0.10700169,"word_repetition_ratio":0.10128617,"special_character_ratio":0.24562888,"punctuation_ratio":0.12683578,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9766233,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T20:14:55Z\",\"WARC-Record-ID\":\"<urn:uuid:6b2f2bd1-8a0e-4ee0-952b-15fd37e20d26>\",\"Content-Length\":\"37693\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e9260ce-1ca8-4537-b9a0-bfe3b436da92>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3945d94-32ee-4989-a001-4879d1a4771d>\",\"WARC-IP-Address\":\"198.38.85.121\",\"WARC-Target-URI\":\"http://www.mywordsolution.com/question/what-effect-has-the-transformation-had-on-the-lowest-exam/9292707\",\"WARC-Payload-Digest\":\"sha1:37GMQXYO5QT33D45GRIJ3FWID7Q6KNAT\",\"WARC-Block-Digest\":\"sha1:SUBOL5XBAZO3SVEEZ54KS7FYUNJBOGGS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100603.33_warc_CC-MAIN-20231206194439-20231206224439-00125.warc.gz\"}"} |
https://wikieducator.org/Thermodynamics/Third_Law | [
"# Third Law of Thermodynamics\n\n## Rationale and Statement\n\nFor a given system, we cannot give an absolute value of the internal energy, U, because that value needs to know the precise internal state. Therefore, we always use the difference, either ΔU or dU. And since the enthalpy, H, is based on the internal the same is true for enthalpy.\n\nHowever, that is not true for the entropy, S. The absolute value of entropy is defined by the Third Law of Thermodynamics\n\nThe entropy of a pure crystalline solid at absolute zero is an arbitrary constant which can be set to zero\n\n## Simplified Statement\n\nAn easier statement (but not quite as precise) is\n\nThe entropy at absolute zero is zero\n\nThis statement is useful in calculating entropies.\n\n## Calculation of Entropy\n\nWe can now calculate the absolute entropy at any T > 0 by integrating the Clausis inequality from 0 to T and using the third law at T = 0.\n\n## Crystalline solids\n\nLet us now take a look at the first statement. Note that it says a pure crystalline solid. In other words a perfectly ordered solid. This would not include amorphous solids (an amorphous solid is one which is not a crystal, for example, glass).\n\nAll substances we know of, with one exception, become crystalline as they approach absolute zero. Therefore, as far as we know all observations agree with the third law.\n\nThe one exception is helium. As helium approaches absolute zero it becomes a strange behaving liquid called a superfluid. It has no viscosity and will crawl up the side of a beaker. It however has an ordered structure about equal to a crystalline solid."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.877509,"math_prob":0.88013506,"size":2535,"snap":"2020-45-2020-50","text_gpt3_token_len":564,"char_repetition_ratio":0.12090083,"word_repetition_ratio":0.023923445,"special_character_ratio":0.20512821,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98014396,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T08:40:57Z\",\"WARC-Record-ID\":\"<urn:uuid:78a54248-f388-4805-a6a0-0a067ddfb2d2>\",\"Content-Length\":\"28407\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e9b35e7-bb3e-4990-8b9a-8cb104d256d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8e1deb9-1041-40e2-86c7-a74f69d4a2e3>\",\"WARC-IP-Address\":\"157.245.188.3\",\"WARC-Target-URI\":\"https://wikieducator.org/Thermodynamics/Third_Law\",\"WARC-Payload-Digest\":\"sha1:FHGJSMP3EYFSX5KW7ZATKMD6OMHZRRJZ\",\"WARC-Block-Digest\":\"sha1:EMELGV5FMI32QBJLMAKWS7BPPUXQRIZP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141197278.54_warc_CC-MAIN-20201129063812-20201129093812-00072.warc.gz\"}"} |
https://www.litscape.com/word_analysis/reboxed | [
"# reboxed in Scrabble®\n\nThe word reboxed is playable in Scrabble®, no blanks required.\n\nREBOXED\n(125 = 75 + 50)\n\nreboxed\n\nREBOXED\n(125 = 75 + 50)\nREBOXED\n(118 = 68 + 50)\nREBOXED\n(116 = 66 + 50)\nREBOXED\n(110 = 60 + 50)\nREBOXED\n(107 = 57 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(101 = 51 + 50)\nREBOXED\n(100 = 50 + 50)\nREBOXED\n(96 = 46 + 50)\nREBOXED\n(92 = 42 + 50)\nREBOXED\n(90 = 40 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(85 = 35 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(78 = 28 + 50)\nREBOXED\n(78 = 28 + 50)\nREBOXED\n(77 = 27 + 50)\nREBOXED\n(76 = 26 + 50)\nREBOXED\n(73 = 23 + 50)\nREBOXED\n(72 = 22 + 50)\nREBOXED\n(71 = 21 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(68 = 18 + 50)\n\nREBOXED\n(125 = 75 + 50)\nREBOXED\n(118 = 68 + 50)\nREBOXED\n(116 = 66 + 50)\nREBOXED\n(110 = 60 + 50)\nREBOXED\n(107 = 57 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(104 = 54 + 50)\nREBOXED\n(101 = 51 + 50)\nREBOXED\n(100 = 50 + 50)\nREBOXED\n(96 = 46 + 50)\nREBOXED\n(92 = 42 + 50)\nREBOXED\n(90 = 40 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(88 = 38 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(86 = 36 + 50)\nREBOXED\n(85 = 35 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(84 = 34 + 50)\nREBOXED\n(78 = 28 + 50)\nREBOXED\n(78 = 28 + 50)\nREBOXED\n(77 = 27 + 50)\nREBOXED\n(76 = 26 + 50)\nREBOXED\n(73 = 23 + 50)\nREBOXED\n(72 = 22 + 50)\nREBOXED\n(71 = 21 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(69 = 19 + 50)\nREBOXED\n(68 = 18 + 50)\nREBOX\n(66)\nREDOX\n(63)\nREBOX\n(60)\nREDOX\n(58)\nBOXED\n(54)\nBOXED\n(51)\nBOXER\n(51)\nBOXED\n(48)\nBOXED\n(48)\nBOXED\n(45)\nBOXED\n(45)\nREBOX\n(45)\nBOXER\n(45)\nBOXER\n(45)\nREBOX\n(45)\nBOXER\n(45)\nREBOX\n(45)\nBOXED\n(45)\nREBOX\n(44)\nREBOX\n(44)\nREDOX\n(42)\nBOXED\n(42)\nREDOX\n(42)\nBOXER\n(42)\nBOXER\n(42)\nBOXER\n(42)\nREBOX\n(42)\nREBOX\n(42)\nREBOX\n(42)\nREDOX\n(42)\nREDOX\n(42)\nREDOX\n(42)\nBOXER\n(40)\nREDOX\n(39)\nREDOX\n(39)\nREDOX\n(39)\nBOXED\n(38)\nBOX\n(36)\nBOXED\n(36)\nBOXED\n(36)\nBOX\n(36)\nBOX\n(36)\nBOXED\n(34)\nBOXED\n(34)\nBOXER\n(34)\nBOXER\n(34)\nDOX\n(33)\nBREED\n(33)\nDOX\n(33)\nBORED\n(33)\nDOX\n(33)\nBOXER\n(32)\nREBOX\n(32)\nREBOX\n(32)\nBOXED\n(31)\nREDOX\n(31)\nBOXED\n(30)\nBOXED\n(30)\nROBED\n(30)\nREBOX\n(30)\nBOXER\n(30)\nBOXER\n(30)\nBOXED\n(30)\nREDOX\n(30)\nBOXER\n(30)\nBRED\n(30)\nBORED\n(30)\nBODE\n(30)\nREBOX\n(30)\nBOXED\n(30)\nBOXED\n(30)\nBREED\n(30)\nREBOX\n(28)\nBOX\n(28)\nBOXER\n(28)\nBOXER\n(28)\nBOXER\n(28)\nBORED\n(28)\nREBOX\n(28)\nREDOX\n(28)\nBOXER\n(28)\nBREED\n(28)\nREBOX\n(28)\nREBOX\n(28)\nREBOX\n(28)\nREDOX\n(28)\nBOXER\n(28)\nROBED\n(27)\nBORED\n(27)\nBEER\n(27)\nBREED\n(27)\nDOX\n(27)\nROBED\n(27)\nBORE\n(27)\nOX\n(27)\nBRED\n(27)\nOX\n(27)\nEX\n(27)\nEX\n(27)\nBORED\n(27)\nBREED\n(27)\nROBED\n(27)\nREDOX\n(26)\nBOXED\n(26)\nREDOX\n(26)\nREDOX\n(26)\nREDOX\n(26)\nREDOX\n(26)\nREBOX\n(25)\nBOXER\n(25)\nBOXED\n(25)\nBOXED\n(25)\nOX\n(25)\nEX\n(25)\nERODE\n(24)\nROBED\n(24)\nBREED\n(24)\nBREED\n(24)\nROBED\n(24)\nROBED\n(24)\nBREED\n(24)\nBREED\n(24)\nBOX\n(24)\nROBED\n(24)\nBORED\n(24)\nBODE\n(24)\nBOX\n(24)\nBORED\n(24)\nBORED\n(24)\nBORED\n(24)\nBOX\n(24)\nREDOX\n(23)\nBOXER\n(23)\nBOXED\n(23)\nBOX\n(23)\nREBOX\n(23)\nBOXER\n(22)\nDOX\n(22)\nBORED\n(22)\nBREED\n(22)\nBREED\n(22)\nDOX\n(22)\nDOX\n(22)\nREDOX\n(22)\nBORED\n(22)\nBOXER\n(22)\nBORE\n(21)\nBRED\n(21)\nERODE\n(21)\nERODE\n(21)\nBRED\n(21)\nDOX\n(21)\nERODE\n(21)\nBRED\n(21)\nBRED\n(21)\nBODE\n(21)\nBEER\n(21)\nROBE\n(21)\n\n# reboxed in Words With Friends™\n\nThe word reboxed is playable in Words With Friends™, no blanks required.\n\nREBOXED\n(149 = 114 + 35)\n\nreboxed\n\nREBOXED\n(149 = 114 + 35)\nREBOXED\n(119 = 84 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(103 = 68 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(87 = 52 + 35)\nREBOXED\n(87 = 52 + 35)\nREBOXED\n(79 = 44 + 35)\nREBOXED\n(79 = 44 + 35)\nREBOXED\n(77 = 42 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(73 = 38 + 35)\nREBOXED\n(73 = 38 + 35)\nREBOXED\n(73 = 38 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(65 = 30 + 35)\nREBOXED\n(65 = 30 + 35)\nREBOXED\n(64 = 29 + 35)\nREBOXED\n(63 = 28 + 35)\nREBOXED\n(62 = 27 + 35)\nREBOXED\n(62 = 27 + 35)\nREBOXED\n(61 = 26 + 35)\nREBOXED\n(61 = 26 + 35)\nREBOXED\n(60 = 25 + 35)\nREBOXED\n(59 = 24 + 35)\nREBOXED\n(58 = 23 + 35)\nREBOXED\n(57 = 22 + 35)\nREBOXED\n(57 = 22 + 35)\nREBOXED\n(56 = 21 + 35)\nREBOXED\n(56 = 21 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(54 = 19 + 35)\nREBOXED\n(54 = 19 + 35)\nREBOXED\n(54 = 19 + 35)\nREBOXED\n(53 = 18 + 35)\n\nREBOXED\n(149 = 114 + 35)\nREBOXED\n(119 = 84 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(107 = 72 + 35)\nREBOXED\n(103 = 68 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(101 = 66 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOXED\n(95 = 60 + 35)\nREBOX\n(93)\nREBOXED\n(87 = 52 + 35)\nREBOXED\n(87 = 52 + 35)\nREDOX\n(87)\nREBOXED\n(79 = 44 + 35)\nREBOXED\n(79 = 44 + 35)\nREBOXED\n(77 = 42 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(75 = 40 + 35)\nREBOXED\n(73 = 38 + 35)\nREBOXED\n(73 = 38 + 35)\nREBOXED\n(73 = 38 + 35)\nBOXED\n(72)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nREBOXED\n(71 = 36 + 35)\nBOXER\n(69)\nREBOXED\n(65 = 30 + 35)\nREBOXED\n(65 = 30 + 35)\nBOXED\n(64)\nREBOXED\n(64 = 29 + 35)\nREBOXED\n(63 = 28 + 35)\nREBOXED\n(62 = 27 + 35)\nREBOXED\n(62 = 27 + 35)\nREBOX\n(62)\nREBOXED\n(61 = 26 + 35)\nREBOXED\n(61 = 26 + 35)\nBOXER\n(60)\nBOXED\n(60)\nREBOXED\n(60 = 25 + 35)\nREBOX\n(60)\nREBOXED\n(59 = 24 + 35)\nREDOX\n(58)\nREBOXED\n(58 = 23 + 35)\nREBOXED\n(57 = 22 + 35)\nREBOXED\n(57 = 22 + 35)\nREBOXED\n(56 = 21 + 35)\nREBOXED\n(56 = 21 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(55 = 20 + 35)\nREBOXED\n(54 = 19 + 35)\nREBOXED\n(54 = 19 + 35)\nBOXED\n(54)\nREBOXED\n(54 = 19 + 35)\nBOXED\n(54)\nREBOXED\n(53 = 18 + 35)\nREDOX\n(52)\nREBOX\n(51)\nBREED\n(51)\nBORED\n(51)\nREBOX\n(51)\nREBOX\n(51)\nBOXER\n(51)\nBOXER\n(51)\nBOXER\n(51)\nBOXED\n(48)\nBOXED\n(48)\nBRED\n(48)\nBODE\n(48)\nBOXED\n(48)\nBOXED\n(48)\nBOXER\n(46)\nREBOX\n(46)\nBOXER\n(45)\nBOXER\n(45)\nBEER\n(45)\nREDOX\n(45)\nREBOX\n(45)\nBORE\n(45)\nREBOX\n(45)\nBOXER\n(45)\nREBOX\n(45)\nREDOX\n(45)\nREDOX\n(45)\nREDOX\n(42)\nBOXED\n(40)\nBOXED\n(40)\nBOXED\n(40)\nREDOX\n(39)\nBOX\n(39)\nREDOX\n(39)\nBREED\n(39)\nBOX\n(39)\nREBOX\n(39)\nBOX\n(39)\nBORED\n(39)\nBOXER\n(39)\nREDOX\n(39)\nROBED\n(39)\nBOXER\n(38)\nBOX\n(37)\nBREED\n(36)\nBRED\n(36)\nBOXED\n(36)\nROBED\n(36)\nBORED\n(36)\nBOXED\n(36)\nBOXER\n(34)\nBORED\n(34)\nREBOX\n(34)\nBOXED\n(34)\nBREED\n(34)\nBOXED\n(34)\nROBED\n(33)\nREDOX\n(33)\nBOXER\n(33)\nROBED\n(33)\nREBOX\n(33)\nBREED\n(33)\nBORED\n(33)\nDOX\n(33)\nBORED\n(33)\nBREED\n(33)\nDOX\n(33)\nROBED\n(33)\nDOX\n(33)\nBOXER\n(32)\nBOXER\n(32)\nBOXED\n(32)\nBOXED\n(32)\nBOXED\n(32)\nBOXED\n(32)\nBOXED\n(32)\nBOXED\n(32)\nBOXER\n(32)\nREBOX\n(32)\nREBOX\n(32)\nREBOX\n(32)\nREBOX\n(32)\nDOX\n(31)\nBOXER\n(31)\nREDOX\n(31)\nERODE\n(30)\nREDOX\n(30)\nBOXER\n(30)\nREBOX\n(30)\nREBOX\n(30)\nREBOX\n(30)\nREBOX\n(30)\nBOXER\n(30)\nBOXER\n(30)\nREBOX\n(30)\nBOXER\n(30)\nBODE\n(30)\nREDOX\n(30)\nBOXER\n(30)\nBOX\n(29)\nBOXED\n(28)\nBOXED\n(28)\nREDOX\n(28)\nREDOX\n(28)\nREDOX\n(28)\nDOX\n(27)\nEX\n(27)\nROBE\n(27)\nROBED\n(27)\nROBE\n(27)\nBORED\n(27)\nROBED\n(27)\nDERE\n(27)\nBREED\n(27)\nDEER\n(27)\nEX\n(27)\nBREED\n(27)\nBOXER\n(27)\nOX\n(27)\nREED\n(27)\nREBOX\n(27)\nBREED\n(27)\nBEER\n(27)\nDOER\n(27)\nROBED\n(27)\nBORED\n(27)\nBORE\n(27)\nOX\n(27)\nBORED\n(27)\n\n# Word Growth involving reboxed\n\nox box boxed\n\nox box rebox\n\nre rebox\n\n## Longer words containing reboxed\n\n(No longer words found)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7033544,"math_prob":1.0000098,"size":406,"snap":"2019-35-2019-39","text_gpt3_token_len":123,"char_repetition_ratio":0.3432836,"word_repetition_ratio":0.1754386,"special_character_ratio":0.19950739,"punctuation_ratio":0.08695652,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99935216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T01:01:54Z\",\"WARC-Record-ID\":\"<urn:uuid:69a8fb45-ed3f-44c5-9cdb-6eae1b3f4f43>\",\"Content-Length\":\"128593\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:318af4aa-7af3-4bb0-9749-7b41e28dbad1>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebc27ba7-d2d6-4b5e-8ed1-c3ea4b75f4c5>\",\"WARC-IP-Address\":\"104.18.51.165\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/reboxed\",\"WARC-Payload-Digest\":\"sha1:P7MX24PB7GOST4S6GHH36JDRDC2BACOK\",\"WARC-Block-Digest\":\"sha1:REBT34JHZE3EVIA6NHNBCHGURABZN3HW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574159.19_warc_CC-MAIN-20190921001810-20190921023810-00035.warc.gz\"}"} |
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewGPS/Practice/7B.htm | [
"Practice 7B: | 1 | 2 | 3 | 4 |Go up\nAngular Speed - by Matt Henderson, 2003\n\n1. A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times?\n\nHere's what you know, wavg = 29 rad/s and Dq = (3.5)(2p) = 7pie rad 1 time around the circle = 2p\n\nNow use the formula wavg = Dq/Dt to find the time 29= 7p/Dt and Dt = .76 s\n\n2. A girl ties a toy airplane to the end of a string and swings it around her head. the plan's average angular speed is 2.2 rad/s. In what time interval will the plane move through an angular displacement of 3.3 rad?\n\nHere's what you know, wavg = 2.2 rad/s and Dq = 3.3 rad\n\nNow use the formula wavg = Dq/Dt to find the time 2.2= 3.3/Dt and Dt = 1.5 s\n\n3. The average angular speed of a fly moving in a circle is 7.0 rad/s. How long does the fly take to move through 2.3 rad??\n\nHere's what you know, wavg = 7.0 rad/s and Dq = 2.3 rad\n\nNow use the formula wavg = Dq/Dt to find the time 7.0= 2.3/Dt and Dt = .33 s\n\n4. Fill in the unknown quantities in the following table\n\nFirst 1 rev/s = 2p rad/s and 1 rev is the same as 1 time around the circle\n\nwavg Dq Dt\n\na. ? 2.3 rad 10 s\n\nb. .75 rev/s ? .050 s\n\nc. ? -1.2 turns 1.2 s\n\na. use the formula wavg = Dq/Dt to find the angular speed wavg= 2.3/10 and wavg = .23 rad/s\n\nb. 1 rev/s = 2pie rad/s so (.75)(2p) = 4.712\n\nuse the formula wavg = Dq/Dt to find the arc length 4.712= Dq/.05 and Dq = .2356 rad\n\nc. 1 turn = 2p so (-1.2)(2p)= -7.5398 rad\n\nuse the formula wavg = Dq/Dt to find the angular speed wavg= -7.539/1.2 and wavg = -6.28 rad/s\n\nd. use the formula wavg = Dq/Dt to find the time 2p= 1.5p/Dt and Dt = .75 s"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8428399,"math_prob":0.9949905,"size":1719,"snap":"2019-51-2020-05","text_gpt3_token_len":640,"char_repetition_ratio":0.15918367,"word_repetition_ratio":0.17438692,"special_character_ratio":0.3833624,"punctuation_ratio":0.14347826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997814,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T11:44:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9fcb5be9-38eb-4f79-a607-8aa053ecc46d>\",\"Content-Length\":\"22221\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7fcb1d2-2e54-488d-9e00-a1b656c65208>\",\"WARC-Concurrent-To\":\"<urn:uuid:f32ac085-42c9-4f92-b64e-b7ec336cac77>\",\"WARC-IP-Address\":\"66.154.208.42\",\"WARC-Target-URI\":\"http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewGPS/Practice/7B.htm\",\"WARC-Payload-Digest\":\"sha1:A5A5ETT44OP3NQNOP4UKD4I4NVXWDHAE\",\"WARC-Block-Digest\":\"sha1:MPZIWL42LHGKFVST5UKRWTQ6M3AIVERF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250619323.41_warc_CC-MAIN-20200124100832-20200124125832-00173.warc.gz\"}"} |
http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/pullback | [
"nLab pullback\n\nContext\n\nLimits and colimits\n\nlimits and colimits\n\nContents\n\nIdea\n\nIn the category Set a ‘pullback’ is a subset of the cartesian product of two sets. Given a diagram of sets and functions like this:\n\nthe ‘pullback’ of this diagram is the subset $X \\subseteq A \\times B$ consisting of pairs $(a,b)$ such that the equation $f(a) = g(b)$ holds.\n\nA pullback is therefore the categorical semantics of an equation.\n\nThis construction comes up, for example, when $A$ and $B$ are fiber bundles over $C$: then $X$ as defined above is the product of $A$ and $B$ in the category of fiber bundles over $C$. For this reason, a pullback is sometimes called a fibered product (or fiber product or fibre product).\n\nIn this case, the fiber of $A \\times_C B$ over a (generalized) element $x$ of $C$ is the ordinary product of the fibers of $A$ and $B$ over $x$. In other words, the fiber product is the product taken fiber-wise. Of course, the fiber of $A$ at the generalized element $x\\colon I \\to C$ is itself a fibre product $I \\times_C A$; the terminology depends on your point of view.\n\nNote that there are maps $p_A\\colon X \\to A$, $p_B\\colon X \\to B$ sending any $(a,b) \\in X$ to $a$ and $b$, respectively. These maps make this square commute:\n\nIn fact, the pullback is the universal solution to finding a commutative square like this. In other words, given any commutative square\n\nthere is a unique function $h\\colon Y \\to X$ such that\n\n$p_A h = q_A$\n\nand\n\n$p_B h = q_B\\,.$\n\nSince this universal property expresses the concept of pullback purely arrow-theoretically, we can formulate it in any category. It is, in fact, a simple special case of a limit.\n\nDefinition\n\nIn category theory\n\nAs a limit\n\nA pullback is a limit of a diagram like this:\n\nSuch a diagram is also called a pullback diagram or a cospan. If the limit exists, we obtain a commutative square\n\nand the object $x$ is also called the pullback. It is well defined up to unique isomorphism. It has the universal property already described above in the special case of the category $Set$.\n\nThe last commutative square above is called a pullback square.\n\nThe concept of pullback is dual to the concept of pushout: that is, a pullback in $C$ is the same as a pushout in the opposite category $C^{op}$.\n\nNuts and bolts\n\nLet $\\mathcal{C}$ be a category, with $f\\colon a\\to c$ and $g\\colon b\\to c$ coterminal arrows in $\\mathcal{C}$ as below\n\nA pullback of $f$ and $g$ consists of an object $x$ together with arrows $p_a\\colon x\\to a$ and $p_b\\colon x\\to b$ such that the following diagram commutes universally\n\nThis means that for any other object $x'$ with arrows $p'_a\\colon x'\\to a$ and $p'_b\\colon x'\\to b$ such that\n\ncommutes, there exists a unique arrow $u\\colon x'\\to x$ such that\n\ncommutes.\n\nIn type theory\n\nIn type theory a pullback $P$ in\n\nis given by the dependent sum over the dependent equality type\n\n$P = \\sum_{a\\colon A} \\sum_{b\\colon B} (f(a) = g(b)) \\,.$\n\nProperties\n\nProposition\n\n(pullbacks as equalizers)\n\nIf products exist in $C$, then the pullback\n\nis equivalently the equalizer\n\nof the two morphisms induced by $f$ and $g$ out of the product of $a$ with $b$.\n\nProposition\n\n(pullbacks preserve monomorphisms and isomorphisms)\n\nPullbacks preserve monomorphisms and isomorphisms:\n\nIf\n\nis a pullback square in some category then:\n\n1. if $g$ is a monomorphism then $f^\\ast g$ is a monomorphism;\n\n2. if $g$ is an isomorphism then $f^\\ast g$ is an isomorphism.\n\nOn the other hand that $f^\\ast g$ is a monomorphism does not imply that $g$ is a monomorphism.\n\nProposition\n\n(pasting law for pullbacks)\n\nIn any category consider a diagram of the form\n\nThere are three commuting squares: the two inner ones and the outer one.\n\nSuppose the right-hand inner square is a pullback, then:\n\nThe square on the left is a pullback if and only if the outer square is.\n\nProof\n\nPasting a morphism $x \\to a$ with the outer square gives rise to a commuting square over the (composite) bottom and right edges of the diagram. The square over the cospan in the left-hand inner square arising from $x \\to a$ includes a morphism into $b$, which if $b$ is a pullback induces the same commuting square over $d \\to e \\to f$ and $c \\to d$. So one square is universal iff the other is.\n\nProposition\n\nThe converse implication does not hold: it may happen that the outer and the left square are pullbacks, but not the right square.\n\nProof\n\nFor instance let $i\\colon a \\to b$ be a split monomorphism with retract $p\\colon b \\to a$ and consider\n\nThen the left square and the outer rectangle are pullbacks but the right square cannot be a pullback unless $i$ was already an isomorphism.\n\nRemark\n\nOn the other hand, in the (∞,1)-category of ∞-groupoids, there is a sort of “partial converse”; see homotopy pullback#HomotopyFiberCharacterization.\n\nSaturation\n\nThe saturation of the class of pullbacks is the class of limits over categories $C$ whose groupoid reflection $\\Pi_1(C)$ is trivial and such that $C$ is L-finite.\n\nPullback functor\n\nIf $f\\colon X \\to Y$ is a morphism in a category $C$ with pullbacks, there is an induced pullback functor $f^*\\colon C/Y \\to C/X$, sometimes also called base change.\n\nNotions of pullback:\n\nTextbook account:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89268374,"math_prob":0.9987719,"size":5018,"snap":"2022-05-2022-21","text_gpt3_token_len":1158,"char_repetition_ratio":0.16972478,"word_repetition_ratio":0.004914005,"special_character_ratio":0.19768833,"punctuation_ratio":0.12579957,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99991536,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-18T01:44:23Z\",\"WARC-Record-ID\":\"<urn:uuid:2cccfe45-d5b9-434e-be28-d02cde4cdd61>\",\"Content-Length\":\"251677\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce658190-ac39-4f1a-b713-5b97e550a3af>\",\"WARC-Concurrent-To\":\"<urn:uuid:e495abbd-fd98-4a7b-8996-e8bb070f79c2>\",\"WARC-IP-Address\":\"52.219.99.66\",\"WARC-Target-URI\":\"http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/pullback\",\"WARC-Payload-Digest\":\"sha1:PALUHVHIYVN2WE6MQGU4GIFHBTJ3UYSH\",\"WARC-Block-Digest\":\"sha1:IEEXEGOIXYF7TZUBL4MF5LDTI3YGP352\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320300658.84_warc_CC-MAIN-20220118002226-20220118032226-00485.warc.gz\"}"} |
https://www.jiskha.com/questions/10379/A-flywheel-of-480-mm-diameter-is-accelerated-with-an-angular-velocity-of-0-4-rad-s-2 | [
"# Angular acceleration\n\nA flywheel of 480 mm diameter is accelerated with an angular velocity of 0.4 rad/s^-2 from rest by an electric motor until it is rotating at 400 revs/minute. Calculate:\na) the time of acceleration\nb) the linear speed of the rim of the flywheel\n\nMy answer for (a) is 104.7 secs, and for (b) is 10 m/s.\n\nAppreciate if someone could check these answers if correct. If wrong, can they point me in the right direction.\nTIA\n\nFinal angular speed = (angular acceleration) * time\n\n400 rpm = 41.89 rad/s = (ang. acc.) * t\nt = 104.7 s\n\nfinal linear speed = R * (final angular speed) = [(0.48 m)/2]*41.89 rad/s = 10.05 m/s, but it is OK to show two significant figures, (10 m/s)\n\nThanks for that.\n\n1. 👍 0\n2. 👎 0\n3. 👁 128\n\n## Similar Questions\n\n1. ### kinetics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on March 16, 2012\n2. ### engineering mechanics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on April 4, 2012\n3. ### kinematics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on May 30, 2012\n4. ### kinetics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on March 8, 2012\n5. ### engineering mechanics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on April 5, 2012\n6. ### kinetics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on March 28, 2012\n7. ### kinetics\n\nAn exercise bicycle's flywheel is 30cm in diameter and 2.5 cm thick and is constructed from steel (density=7850 kg m3) its moment of inertia is 0.156 kg m2 about its axis of rotation a) calculate the mass of the flywheel b)\n\nasked by dmkp on March 30, 2012\n8. ### candle\n\nThe flywheel of a prototype car engine is under test. The angular position θ of the flywheel is given by θ = (3.0rad/s3)t3 and the diameter of the flywheel is 36cm. (a)Find the distance that a particle on the rim moves during\n\nasked by ahmed on November 4, 2011\n9. ### physics\n\nIf, after rotating for some time at constant angular velocity of 72 rad/s, brakes are applied to the flywheel producing retardation of 4 rad/s2 , determine the time taken to reduce its angular velocity to 40 rad/s.\n\nasked by baz on March 21, 2015\n10. ### Physics\n\nA flywheel accelerates from rest to an angular velocity of 10.44rad/s at an angular acceleration of 51.57rad/s^2. The diameter of the flywheel is 1.16m. At this instant, calculate the resultant linear acceleration, in m/s^2, of a\n\nasked by Anonymous on April 18, 2012\n\nMore Similar Questions"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93454266,"math_prob":0.9874945,"size":2778,"snap":"2019-43-2019-47","text_gpt3_token_len":818,"char_repetition_ratio":0.14275415,"word_repetition_ratio":0.617357,"special_character_ratio":0.28869691,"punctuation_ratio":0.06866197,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99707,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T01:59:45Z\",\"WARC-Record-ID\":\"<urn:uuid:52044aac-a9d1-4cea-89b2-beef9bedfdb9>\",\"Content-Length\":\"19840\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4b1ef69-9bbd-4b52-810e-3248b6ec067c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d38a39ae-9667-4eea-96a5-627024935708>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/10379/A-flywheel-of-480-mm-diameter-is-accelerated-with-an-angular-velocity-of-0-4-rad-s-2\",\"WARC-Payload-Digest\":\"sha1:E3WM2XXQMBLEL7WMNWEGYBWR5ULEVO6A\",\"WARC-Block-Digest\":\"sha1:4G3L7K4PRANZBPLQZB3EVYXWFE5S3YQ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670643.58_warc_CC-MAIN-20191121000300-20191121024300-00504.warc.gz\"}"} |
https://tex.stackexchange.com/questions/137563/how-to-put-a-figure-on-the-facing-page-of-a-chapter-page-with-correct-numbering | [
"# How to put a figure on the facing page of a chapter page with correct numbering?\n\nWhat's the best way to place a page-sized figure including the correct caption numbering on the facing page of a chapter opening page?\n\nCorrect numbering means: if I put the figure before the chapter, the numbering of the previous chapter is continued and the list of figures shows it belonging to the previous chapter. But I want to have it labeled 6.1 if the new chapter will be chapter 6.\n\nNow I was thinking: Is there a way to use a caption(command) that will only be defined later? Is there a package that does something like this already?\n\nIn essence, something like this:\n\n\\cleartoevenpage{\\thispagestyle{empty}}\n\\begin{figure}[p]\n\\begin{sidecaption}{\\theCaption} % TODO: how?\n\\includegraphics{image.jpg}\n\\end{sidecaption}\n\\end{figure}\n\n\\chapter{Chapter Title}\n\n\\defineTheCaption\n\n\nOr, if that is not possible, how can I manually label the figure as if belonging to the next chapter, with the correct entry in the lof file?\n\nPS: I am using memoir. The kind of opposite thing seems possible with \\newfixedcaption: it allows to place a caption on a page preceding a figure.\n\n## 2 Answers\n\nYou can (and should) define a command doing the work, rather than explicitly set all those commands for each chapter.\n\nMy idea is to define a \\chapterfigure command that has one optional argument and two mandatory ones; the optional argument and the first mandatory one are used for \\includegraphics, while the final argument contains the caption.\n\nThe command steps the chapter counter, so the number for the figure will be correct; after the figure the counter is stepped back, so when \\chapter is processed, the chapter number will be right; then we patch \\memendofchapterhook so that it steps the figure counter (that would have been reset to zero by \\chapter) and redefines itself to its previous value (usually nothing, but one never knows). Similarly, we issue \\insertchapterspace, then redefine it to simply redefine itself to its previous value, so when the command is called by \\chapter it does nothing else than putting us back to the original situation.\n\nFor a chapter without a facing figure, just issue \\chapter by itself. A label for \\chapterfigure should be in a trailing optional argument, to reflect the syntax of sidecaption; so\n\n\\chapterfigure[<options for includegraphics>] % optional\n{<graphic file name>} % mandatory\n{<caption>} % mandatory\n[<label>] % optional\n\n\nHere's a complete example\n\n\\documentclass{memoir}\n\\usepackage{xparse}\n\\usepackage[demo]{graphicx}\n\n\\NewDocumentCommand{\\chapterfigure}{O{} m m o}{%\n\\insertchapterspace\n\\cleartoevenpage{\\thispagestyle{empty}}\n\\stepcounter{chapter}\n\\begin{figure}[p]\n\\IfNoValueTF{#4}\n{\\begin{sidecaption}{#3}}\n{\\begin{sidecaption}{#3}[#4]}\n\\includegraphics[#1]{#2}\n\\end{sidecaption}\n\\end{figure}\n\\addtocounter{chapter}{-1}\n\\let\\keptmemendofchapterhook\\memendofchapterhook\n\\renewcommand{\\memendofchapterhook}{%\n\\stepcounter{figure}%\n\\keptmemendofchapterhook\n\\let\\memendofchapterhook\\keptmemendofchapterhook}%\n\\let\\keptinsertchapterspace\\insertchapterspace\n\\renewcommand\\insertchapterspace{%\n\\let\\insertchapterspace\\keptinsertchapterspace}%\n}\n\n\\begin{document}\n\\frontmatter\n\\tableofcontents\n\\listoffigures\n\n\\mainmatter\n\n\\chapterfigure[width=5cm,height=3cm]{somepic}\n{A caption for this figure}\n% Here's how to call it if there's a label\n%\\chapterfigure[width=5cm,height=3cm]{somepic}\n% {A caption for this figure}[chapfig:one]\n\n\\chapter{A title for this chapter}\n\nSome text\n\n\\begin{figure}[htp]\n\\centering\n\\includegraphics{xyz}\n\\caption{A caption to see it has the correct number}\n\\end{figure}\n\nSome text\n\n\\end{document}",
null,
"• yay, great code! I had to update my texlive because my memoir didn't have the hook yet. But this is very close to my dream solution. \\addtocounter{chapter}{-1} is one key I didn't know yet. The second key, still missing in this solution, is how to add the space between figures of different chapters in the lof. memoir adds \\addvspace{10pt} to the lof in \\@chapter using \\insertchapterspace. That is tricky because it has to be executed earlier, but it should not be duplicated. Any idea how to solve this, too? :) – MiB Oct 12 '13 at 23:17\n• no problem, we all have to sleep :) I have a second question for you while you are at it: why did you choose \\stepcounter instead of \\refstepcounter? I don't understand the difference, I only read that \\refstepcounter changes the current \\ref value, whatever that means... – MiB Oct 13 '13 at 9:03\n• @MiB Here's the new version. Please test it. With \\refstepcounter we also set the current label, which is not needed here. – egreg Oct 13 '13 at 9:38\n• aha! now this is perfection, since you used latex3 and even provided for the optional sidecaption arg. Very smart, thank you very much! I will have to get used to this redefining commands to restore the old command :) – MiB Oct 13 '13 at 9:58\n\nHere is a simple approach that I used (thanks to some hints here) to putting a landscape figure opposite a Chapter page, and numbering it as the first figure in that chapter:\n\n% For viewing in Adobe Reader, don’t forget to set View> Page Display > Show Cover Page in Two-Page View\n\\documentclass[draft,12pt,letterpaper,dvipsnames,svgnames,table,openright,twoside]{book}\n\\usepackage{pdflscape} % Has landscape environment\n\\usepackage{graphicx}\n\\graphicspath{{./images/}}\n\\usepackage{caption}\n\\usepackage{lettrine}\n\\usepackage{calligra}\n\\usepackage[nopar]{lipsum}\n\n%\n\\begin{document}\n\\renewcommand{\\LettrineFontHook}{\\calligra}\n% PREFACE\n\\chapter*{Preface}\n\\lipsum[2-4]\n\n% CHAPT 1\n% \\section[shortish TOC entry]{formatted chapter title} \\sectionmark{short title for running headers}\n\\chapter{Dummy Chapter}\n\\lipsum[2-4]\n%\n\n% CHAPT 2\n\\makeatletter\\@openrightfalse %%%%\n\\newpage\n\\cleardoublepage\\part[~STONE AGE]{STONE AGE}\n%\n% The following page is rotated 90-degrees clockwise in the pdf.\n% In the book it appears on the LHS, opposite the section “Chapter 16”\n\\stepcounter{chapter}\n\\begin{landscape}\n\\begin{figure}\n\\captionsetup{width=1.20\\textwidth}\n%\\vspace{12pt} % To move picture further away from binding (odd page)\n\\vspace{-36pt} % To move picture further away from binding (even page)\n\\centering\n\\includegraphics[width=6in]{US_Frequency_Allocations_Jan_2016_a.png}\n\\caption[U.S. Electromagnetic Spectrum Frequency Allocations, October 2003]{U.S. Electromagnetic Spectrum usage from DC to 300 GHz}\n\\scriptsize{\\emph{Photo Credit: U.S. Government, Department of Commerce}}\n\\label{figure:FCC_EM_Radio_Spectrum}\n\\end{figure}\n\\end{landscape}\n\\addtocounter{chapter}{-1}\n\\newpage\n% \\section[shortish TOC entry]{formatted chapter title} \\sectionmark{short title for running headers}\n\\chapter{2001: Welcome to Space \\& Beyond!}\n\\label{section:DC2Daylight}\n\\@openrighttrue\\makeatother\n% Body Text:\n\\lettrine[lines=3,lhang=0.33,lraise=0.0,loversize=0.30,slope=-16pt,findent=2.3em,nindent=-0.4em]{U}{\\textbf{nknown Origins}} \\lipsum[1-5]\n%\n\\end{document}\n\\endinput"
] | [
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"https://i.stack.imgur.com/90FjU.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.610255,"math_prob":0.51323086,"size":2474,"snap":"2019-35-2019-39","text_gpt3_token_len":664,"char_repetition_ratio":0.1631579,"word_repetition_ratio":0.0,"special_character_ratio":0.21139854,"punctuation_ratio":0.064425774,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9552897,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T22:23:59Z\",\"WARC-Record-ID\":\"<urn:uuid:b4b931a6-f200-44bd-b88c-01fc378c3777>\",\"Content-Length\":\"147116\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60a268f5-abd7-404a-85e3-216d39ab609d>\",\"WARC-Concurrent-To\":\"<urn:uuid:162fcd6e-d272-40b0-8523-bc50a8d35ecd>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/137563/how-to-put-a-figure-on-the-facing-page-of-a-chapter-page-with-correct-numbering\",\"WARC-Payload-Digest\":\"sha1:42JRNQ2SU4ZVQ73C6C75OWLPQFZL2W26\",\"WARC-Block-Digest\":\"sha1:WVZIJ7OCCWQXEJN2SJPOUO77TJMXOL76\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027319082.81_warc_CC-MAIN-20190823214536-20190824000536-00350.warc.gz\"}"} |
https://developer.rhino3d.com/api/RhinoCommon/html/M_Rhino_Geometry_Point4d_Multiply.htm | [
"Point4d.Multiply Method",
null,
"",
null,
"# Point4dAddLanguageSpecificTextSet(\"LST8B75ECF8_0?cpp=::|nu=.\");Multiply Method\n\nMultiplies a point by a number.\n\n(Provided for languages that do not support operator overloading. You can use the * operator otherwise)\n\nNamespace: Rhino.Geometry\nAssembly: RhinoCommon (in RhinoCommon.dll)\nSince: 5.0",
null,
"Syntax\n```public static Point4d Multiply(\nPoint4d point,\ndouble d\n)```\n\n#### Parameters\n\npoint\nType: Rhino.GeometryPoint4d\nA point.\nd\nType: SystemDouble\nA number.\n\n#### Return Value\n\nType: Point4d\nA new point that results from the coordinate-wise multiplication of point with d.",
null,
"See Also"
] | [
null,
"https://developer.rhino3d.com/api/RhinoCommon/icons/TocOpen.gif",
null,
"https://developer.rhino3d.com/api/RhinoCommon/icons/TocClose.gif",
null,
"https://developer.rhino3d.com/api/RhinoCommon/icons/SectionExpanded.png",
null,
"https://developer.rhino3d.com/api/RhinoCommon/icons/SectionExpanded.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64425325,"math_prob":0.68367195,"size":567,"snap":"2021-04-2021-17","text_gpt3_token_len":145,"char_repetition_ratio":0.13321492,"word_repetition_ratio":0.0,"special_character_ratio":0.22045855,"punctuation_ratio":0.16831683,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9673538,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T17:12:46Z\",\"WARC-Record-ID\":\"<urn:uuid:6ef5d2ad-033b-444e-b72f-1780dd30efba>\",\"Content-Length\":\"10915\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d19e17d5-1079-4010-9a0d-b5bef362bf1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:771b6e73-9ae3-4588-be78-707323e75103>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://developer.rhino3d.com/api/RhinoCommon/html/M_Rhino_Geometry_Point4d_Multiply.htm\",\"WARC-Payload-Digest\":\"sha1:BBFNRWAZYLD6LASGJUVMHPOR2KLCUNCE\",\"WARC-Block-Digest\":\"sha1:XQKO554L75OJI3DILHNEJY6T6D6HXSTX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038507477.62_warc_CC-MAIN-20210418163541-20210418193541-00146.warc.gz\"}"} |
https://www.colorhexa.com/00eecd | [
"# #00eecd Color Information\n\nIn a RGB color space, hex #00eecd is composed of 0% red, 93.3% green and 80.4% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 13.9% yellow and 6.7% black. It has a hue angle of 171.7 degrees, a saturation of 100% and a lightness of 46.7%. #00eecd color hex could be obtained by blending #00ffff with #00dd9b. Closest websafe color is: #00ffcc.\n\n• R 0\n• G 93\n• B 80\nRGB color chart\n• C 100\n• M 0\n• Y 14\n• K 7\nCMYK color chart\n\n#00eecd color description : Pure (or mostly pure) cyan.\n\n# #00eecd Color Conversion\n\nThe hexadecimal color #00eecd has RGB values of R:0, G:238, B:205 and CMYK values of C:1, M:0, Y:0.14, K:0.07. Its decimal value is 61133.\n\nHex triplet RGB Decimal 00eecd `#00eecd` 0, 238, 205 `rgb(0,238,205)` 0, 93.3, 80.4 `rgb(0%,93.3%,80.4%)` 100, 0, 14, 7 171.7°, 100, 46.7 `hsl(171.7,100%,46.7%)` 171.7°, 100, 93.3 00ffcc `#00ffcc`\nCIE-LAB 84.768, -54.748, 2.603 41.59, 65.552, 68.215 0.237, 0.374, 65.552 84.768, 54.81, 177.278 84.768, -68.912, 12.675 80.964, -49.995, 6.721 00000000, 11101110, 11001101\n\n# Color Schemes with #00eecd\n\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #ee0021\n``#ee0021` `rgb(238,0,33)``\nComplementary Color\n• #00ee56\n``#00ee56` `rgb(0,238,86)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #0098ee\n``#0098ee` `rgb(0,152,238)``\nAnalogous Color\n• #ee5600\n``#ee5600` `rgb(238,86,0)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #ee0098\n``#ee0098` `rgb(238,0,152)``\nSplit Complementary Color\n• #eecd00\n``#eecd00` `rgb(238,205,0)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #cd00ee\n``#cd00ee` `rgb(205,0,238)``\n• #21ee00\n``#21ee00` `rgb(33,238,0)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #cd00ee\n``#cd00ee` `rgb(205,0,238)``\n• #ee0021\n``#ee0021` `rgb(238,0,33)``\n• #00a28b\n``#00a28b` `rgb(0,162,139)``\n• #00bba1\n``#00bba1` `rgb(0,187,161)``\n• #00d5b7\n``#00d5b7` `rgb(0,213,183)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #08ffdd\n``#08ffdd` `rgb(8,255,221)``\n• #22ffe0\n``#22ffe0` `rgb(34,255,224)``\n• #3cffe4\n``#3cffe4` `rgb(60,255,228)``\nMonochromatic Color\n\n# Alternatives to #00eecd\n\nBelow, you can see some colors close to #00eecd. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00ee92\n``#00ee92` `rgb(0,238,146)``\n• #00eea5\n``#00eea5` `rgb(0,238,165)``\n• #00eeb9\n``#00eeb9` `rgb(0,238,185)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #00eee1\n``#00eee1` `rgb(0,238,225)``\n• #00e7ee\n``#00e7ee` `rgb(0,231,238)``\n• #00d4ee\n``#00d4ee` `rgb(0,212,238)``\nSimilar Colors\n\n# #00eecd Preview\n\nThis text has a font color of #00eecd.\n\n``<span style=\"color:#00eecd;\">Text here</span>``\n#00eecd background color\n\nThis paragraph has a background color of #00eecd.\n\n``<p style=\"background-color:#00eecd;\">Content here</p>``\n#00eecd border color\n\nThis element has a border color of #00eecd.\n\n``<div style=\"border:1px solid #00eecd;\">Content here</div>``\nCSS codes\n``.text {color:#00eecd;}``\n``.background {background-color:#00eecd;}``\n``.border {border:1px solid #00eecd;}``\n\n# Shades and Tints of #00eecd\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000302 is the darkest color, while #eefffd is the lightest one.\n\n• #000302\n``#000302` `rgb(0,3,2)``\n• #001613\n``#001613` `rgb(0,22,19)``\n• #002a24\n``#002a24` `rgb(0,42,36)``\n• #003d35\n``#003d35` `rgb(0,61,53)``\n• #005146\n``#005146` `rgb(0,81,70)``\n• #006557\n``#006557` `rgb(0,101,87)``\n• #007868\n``#007868` `rgb(0,120,104)``\n• #008c79\n``#008c79` `rgb(0,140,121)``\n• #00a089\n``#00a089` `rgb(0,160,137)``\n• #00b39a\n``#00b39a` `rgb(0,179,154)``\n• #00c7ab\n``#00c7ab` `rgb(0,199,171)``\n• #00dabc\n``#00dabc` `rgb(0,218,188)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\n• #03ffdc\n``#03ffdc` `rgb(3,255,220)``\n• #16ffdf\n``#16ffdf` `rgb(22,255,223)``\n• #2affe1\n``#2affe1` `rgb(42,255,225)``\n• #3dffe4\n``#3dffe4` `rgb(61,255,228)``\n• #51ffe7\n``#51ffe7` `rgb(81,255,231)``\n• #65ffea\n``#65ffea` `rgb(101,255,234)``\n• #78ffec\n``#78ffec` `rgb(120,255,236)``\n• #8cffef\n``#8cffef` `rgb(140,255,239)``\n• #a0fff2\n``#a0fff2` `rgb(160,255,242)``\n• #b3fff4\n``#b3fff4` `rgb(179,255,244)``\n• #c7fff7\n``#c7fff7` `rgb(199,255,247)``\n• #dafffa\n``#dafffa` `rgb(218,255,250)``\n• #eefffd\n``#eefffd` `rgb(238,255,253)``\nTint Color Variation\n\n# Tones of #00eecd\n\nA tone is produced by adding gray to any pure hue. In this case, #6e807e is the less saturated color, while #00eecd is the most saturated one.\n\n• #6e807e\n``#6e807e` `rgb(110,128,126)``\n• #658984\n``#658984` `rgb(101,137,132)``\n• #5c928b\n``#5c928b` `rgb(92,146,139)``\n• #529c91\n``#529c91` `rgb(82,156,145)``\n• #49a598\n``#49a598` `rgb(73,165,152)``\n• #40ae9f\n``#40ae9f` `rgb(64,174,159)``\n• #37b7a5\n``#37b7a5` `rgb(55,183,165)``\n• #2ec0ac\n``#2ec0ac` `rgb(46,192,172)``\n• #25c9b3\n``#25c9b3` `rgb(37,201,179)``\n• #1bd3b9\n``#1bd3b9` `rgb(27,211,185)``\n• #12dcc0\n``#12dcc0` `rgb(18,220,192)``\n• #09e5c6\n``#09e5c6` `rgb(9,229,198)``\n• #00eecd\n``#00eecd` `rgb(0,238,205)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00eecd is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5160709,"math_prob":0.82530504,"size":3691,"snap":"2021-31-2021-39","text_gpt3_token_len":1602,"char_repetition_ratio":0.13561161,"word_repetition_ratio":0.011049724,"special_character_ratio":0.5283121,"punctuation_ratio":0.23198198,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872214,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T00:32:27Z\",\"WARC-Record-ID\":\"<urn:uuid:6b602342-0fb7-481c-83c8-f7be82077e20>\",\"Content-Length\":\"36149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:214af7a7-96e5-4510-bcc7-42feab8ab64d>\",\"WARC-Concurrent-To\":\"<urn:uuid:793ea7c1-5404-4e98-a65f-63cfd4a81a61>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00eecd\",\"WARC-Payload-Digest\":\"sha1:JJGIKGCMGYLT5GGOYUFFQUU2G3SPCSCJ\",\"WARC-Block-Digest\":\"sha1:HWRVBKNXQ4SALOKDVKNJGVUMXS5TSA5J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057403.84_warc_CC-MAIN-20210922223752-20210923013752-00491.warc.gz\"}"} |
https://exceljet.net/formulas/convert-excel-time-to-unix-time | [
"## Summary\n\nTo convert a time in Excel's format to a Unix time stamp, you can use a formula based on the DATE function. In the example shown, the formula in C5 is:\n\n``````=(B5-DATE(1970,1,1))*86400\n``````\n\n## Generic formula\n\n``=(A1-DATE(1970,1,1))*86400``\n\n## Explanation\n\nThe Unix time stamp tracks time as a running count of seconds. The count begins at the \"Unix Epoch\" on January 1st, 1970, so a Unix time stamp is simply the total seconds between any given date and the Unix Epoch. Since a day contains 86400 seconds (24 hours x 60 minutes x 60 seconds), conversion to Excel time can be done by subtracting the date value for the Unix Epoch and multiplying days by 86400.\n\nIn the example shown, the formula first subtracts the date value for January 1, 1970 from the date value in B5 to get the number of days between the dates, then multiplies the result by 85400 to convert to a Unix time stamp. The formula evaluates like this:\n\n``````=(B5-DATE(1970,1,1))*86400\n=(43374-25569)*86400\n=1538352000\n``````\n\n### How Excel tracks dates and time\n\nThe Excel date system starts on January 1, 1900 and counts forward. The table below shows the numeric values associated with a few random dates:\n\nDate Raw value\n1-Jan-1900 1\n28-Jul-1914 00:00 5323\n1-Jan-1970 00:00 25569\n31-Dec-1999 36525\n1-Oct-2018 43374\n1-Oct-2018 12:00 PM 43374.5\n\nNotice the last date includes a time as well. Since one day equals 1, and one day equals 24 hours, time in Excel can represented as fractional values of 1, as shown in the table below. In order to see the value displayed as a time, a time format needs to be applied.\n\nHours Time Fraction Value\n3 3:00 AM 3/24 0.125\n6 6:00 AM 6/24 0.25\n4 4:00 AM 4/24 0.167\n8 8:00 AM 8/24 0.333\n12 12:00 PM 12/24 0.5\n18 6:00 PM 18/24 0.75\n21 9:00 PM 21/24 0.875\n24 12:00 AM 24/24 1",
null,
"Author",
null,
"### Dave Bruns\n\nHi - I'm Dave Bruns, and I run Exceljet with my wife, Lisa. Our goal is to help you work faster in Excel. We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts."
] | [
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"https://exceljet.net/sites/default/files/images/blocks/dave-round.webp",
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"https://exceljet.net/sites/default/files/images/blocks/microsoft-mvp-logo.webp",
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https://oeis.org/A006906 | [
"The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.",
null,
"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A006906 a(n) = sum of products of terms in all partitions of n. (Formerly M2575) 73\n 1, 1, 3, 6, 14, 25, 56, 97, 198, 354, 672, 1170, 2207, 3762, 6786, 11675, 20524, 34636, 60258, 100580, 171894, 285820, 480497, 791316, 1321346, 2156830, 3557353, 5783660, 9452658, 15250216, 24771526, 39713788, 64011924, 102199026, 163583054, 259745051 (list; graph; refs; listen; history; text; internal format)\n OFFSET 0,3 COMMENTS a(0) = 1 since the only partition of 0 is the empty partition. The product of its terms is the empty product, namely 1. Same parity as A000009. - Jon Perry, Feb 12 2004 REFERENCES G. Labelle, personal communication. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Alois P. Heinz, Table of n, a(n) for n = 0..6000 (first 1001 terms from T. D. Noe) Atreya Chatterjee, Emergent gravity from patterns in natural numbers, arXiv:2006.01170 [gr-qc], 2020. Dean Hickerson, Comments on A006906 Robert Schneider and Andrew V. Sills, The product of parts or 'norm' of a partition, #A13 INTEGERS 20A (2020), Theorem 7, p. 4. FORMULA The limit of a(n+3)/a(n) is 3. However, the limit of a(n+1)/a(n) does not exist. In fact, the sequence {a(n+1)/a(n)} has three limit points, which are about 1.4422447, 1.4422491 and 1.4422549. (See the Links entry.) - Dean Hickerson, Aug 19 2007 a(n) ~ c(n mod 3) 3^(n/3), where c(0)=97923.26765718877..., c(1)=97922.93936857030... and c(2)=97922.90546334208... - Dean Hickerson, Aug 19 2007 G.f.: 1 / Product (1-k*x^k). G.f.: 1 + Sum_{n>=1} n*x^n / Product_{k=1..n} (1-k*x^k) = 1 + Sum_{n>=1} n*x^n / Product_{k>=n} (1-k*x^k). - Joerg Arndt, Mar 23 2011 a(n) = (1/n)*Sum_{k=1..n} A078308(k)*a(n-k). - Vladeta Jovovic, Nov 22 2002 O.g.f.: exp( Sum_{n>=1} Sum_{k>=1} k^n * x^(n*k) / n ). - Paul D. Hanna, Sep 18 2017 O.g.f.: exp( Sum_{n>=1} Sum_{k=1..n} A008292(n,k)*x^(n*k)/(n*(1-x^n)^(n+1)) ), where A008292 is the Eulerian numbers. - Paul D. Hanna, Sep 18 2017 EXAMPLE Partitions of 0 are {()} whose products are {1} whose sum is 1. Partitions of 1 are {(1)} whose products are {1} whose sum is 1. Partitions of 2 are {(2),(1,1)} whose products are {2,1} whose sum is 3. Partitions of 3 are 3 => {(3),(2,1),(1,1,1)} whose products are {3,2,1} whose sum is 6. Partitions of 4 are {(4),(3,1),(2,2),(2,1,1),(1,1,1,1)} whose products are {4,3,4,2,1} whose sum is 14. MAPLE A006906 := proc(n) option remember; if n = 0 then 1; else add( A078308(k)*procname(n-k), k=1..n)/n ; end if; end proc: # R. J. Mathar, Dec 14 2011 # second Maple program: b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, b(n, i-1) +add(b(n-i*j, i-1)*(i^j), j=1..n/i))) end: a:= n-> b(n, n): seq(a(n), n=0..40); # Alois P. Heinz, Feb 25 2013 MATHEMATICA (* a[n, k]=sum of products of partitions of n into parts <= k *) a[0, 0]=1; a[n_, 0]:=0; a[n_, k_]:=If[k>n, a[n, n], a[n, k] = a[n, k-1] + k a[n-k, k] ]; a[n_]:=a[n, n] (* Dean Hickerson, Aug 19 2007 *) Table[Total[Times@@@IntegerPartitions[n]], {n, 0, 35}] (* Harvey P. Dale, Jan 14 2013 *) nmax = 40; CoefficientList[Series[Product[1/(1 - k*x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *) nmax = 40; CoefficientList[Series[Exp[Sum[PolyLog[-j, x^j]/j, {j, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *) PROG (Haskell) a006906 n = p 1 n 1 where p _ 0 s = s p k m s | m\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified October 21 03:07 EDT 2020. Contains 337910 sequences. (Running on oeis4.)"
] | [
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"https://oeis.org/banner2021.jpg",
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https://cut-the-knot.org/books/Reviews/WillYouBeAlive.shtml | [
"## Will You Be Alive 10 Years from Now? Paul J. Nahin",
null,
"The book is the third of Paul Nahin's collections devoted to probability. (Duelling Idiots and Digital Dice.) It follows the format of the previous two: 25 essays - each with a problem or two and solutions - follow enthusiastic Preface and Introduction. In Preface Nahin does a superb job conveying his passion for the puzzles generously supplied by probability theory. His style is articulate and enticing; if a reader had any interest in picking up the book in the first place, he or she would find it hard to put it back without turning a page or two and, perhaps, browsing the rest of the book. As in the previous books, Nahin uses the opportunity to take to task Marilyn von Savant - a popular Parade magazine's columnist - for the mistakes she made.\n\nThe book assumes a certain degree of acquaintance with the basic probability and also a reasonably mature study habits, so that the reader should be able to find the definitions or other probability related notions not explained in the book on his/her own. Some problems are naturally more difficult than others; several go beyond the level of mere puzzles. Many, if not most, require a level of comfort with probability methods beyond that associated with casual audience. Some are intended to a reader with a one-two years Calculus experience.\n\nI can wholeheartedly repeat what I wrote in my review of Duelling Idiots. The problems are truly entertaining and the manner in which the author keeps modifying them is edifying. The book's style leaves no doubt about author's enthusiasm about the subject. For him, the problems provide another opportunity to demonstrate the power of mathematics and the flexibility of its problem solving strategies. But more than that, an experienced writer, the author exploits every opportunity to tell a story: a historic background, a personal note, a real world association. For any one who can follow the author's mathematics, the book is a pleasure to read.\n\nAs in the previous books, solutions to the problems are accompanied by MATLAB scripts. I have not put a lot of stock on that in other books, nor in the present one. Nahin expresses an idea that many simple problems are difficult to simulate, while the code is relatively simple for difficult problems. However, all of them wastefully come with the code. For the simple ones, with a solution in a closed form, simulations do not - in my view - add much information or enhance anybody's understanding.\n\nOnce I was somewhat disappointed with Nahin's treatment of a difficulty, the treatment that I believe betrayed his engineering background. In the chapter that gave the book the title, Nahin introduces $p(x)$ as the probability that a person alive now will still be alive in $x$ years of time, and then also defines $\\displaystyle \\phi (x)=\\int_{x}^{\\infty}p(u)du.$ After some effort, he arrives at the following equation: $p(x)\\phi (x)=\\phi (x),$ with apparently immediate solution, $p(x)=1.$ The solution implies that a person is immortal. At this point Nahin declares, \"This solution is mathematically correct, but is also a physical nonsense (just as, in high school bag we would reject negative or complex solutions as physically unacceptable.)\" He rewrites the equation as\n\n$\\displaystyle p(x)\\phi (x)=\\int_{x}^{\\infty}p(u)du;$\n\nand then treats and eventually solves that integral equation. A better explanation would be that in writing $\\displaystyle \\phi (x)=\\int_{x}^{\\infty}p(u)du$ he implicitly assumed that $p(x)$ is integrable on $[0,\\infty)$ which would preclude $p$ from being identically $1$ on half the axis. Every step that preceded the equation $p(x)\\phi (x)=\\phi (x)$ had to be validated against this prerequisite. This has nothing to do with a high school bag of tricks.\n\nOther than that, the book is a delightful collection that will be appreciated by curious undergraduates and probability instructors as a course supplement.",
null,
"Will You Be Alive 10 Years from Now?, Princeton University Press, 2014. Hardcover, 242 pp, \\$27.95. ISBN 0691156808.\n\n### The book is referred to at",
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"https://cut-the-knot.org/books/gifs/tbow_sh.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93992865,"math_prob":0.93351495,"size":4168,"snap":"2021-31-2021-39","text_gpt3_token_len":923,"char_repetition_ratio":0.10710855,"word_repetition_ratio":0.011976048,"special_character_ratio":0.22216891,"punctuation_ratio":0.0992268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9631963,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T13:38:51Z\",\"WARC-Record-ID\":\"<urn:uuid:6e17f337-2daa-4a19-aa76-1cea6ef17de1>\",\"Content-Length\":\"15125\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1248ee4-285d-42f6-be9b-580c534654d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:c94fb3bc-7e89-4f19-b71f-8146f46c9265>\",\"WARC-IP-Address\":\"107.180.50.227\",\"WARC-Target-URI\":\"https://cut-the-knot.org/books/Reviews/WillYouBeAlive.shtml\",\"WARC-Payload-Digest\":\"sha1:V75TSYJTVHXNLOITR4TKT7V42PTWRGXR\",\"WARC-Block-Digest\":\"sha1:2EFJOUI3PRLQJ6SY3RQ3DWDFZOO6EDH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057622.15_warc_CC-MAIN-20210925112158-20210925142158-00240.warc.gz\"}"} |
https://math.stackexchange.com/questions/1705764/problem-to-prove-function-is-convex-or-not | [
"# Problem to prove function is convex or not?\n\nHow do I plot function $f(x1,x2)=x^4_1+x^4_2$ such that $x^2_1+x^2_2=1$ and $x_1,x_2\\in(0,1)$? Does it possible to plot in MATLAB so that I can visualize the function?\n\nAlso I am trying to check whether $f$ is convex or not, because I am trying to optimize it and accordingly formulate the problem like that\n\nMaximize $x^{4}_1+x^{4}_2+\\cdots+x^{4}_n$\n\nSubject to\n\n$x^{2}_1+x^{2}_2+\\cdots+x^{2}_n=1$\n\n$x_i\\in(0,1)$, $i=\\{1, 2, \\ldots, n\\}$\n\nAnd I calculate its Hessian matrix: $$H=\\left[ \\begin{array}{cc} 12x^{2}_1 & 0 \\\\ 0 & 12x^{2}_2 \\end{array}\\right]$$\n\nAfter that I don't know how to proceed. Can anyone help me to prove whether f is convex or not?\n\nThank you very much.\n\n• $f$ (twice differentiable) is convex if and only if its hessian is positive semidefinite, that is exactly the case if every eigenvalues are non-negative. What are the eigenvalues of $H$? Apart from your question, you are maximizing a convex function on a non convex set ... are you sure? – user251257 Mar 20 '16 at 15:24\n• This is not a convex optimization model. The Hessian is indeed a convex function; however, you are maximizing it, and that is nonconvex. Furthermore, nonlinear equality constraints are (with very few manufactured exceptions) not convex. – Michael Grant Mar 21 '16 at 13:53\n\nThe term \"convex\" is a bit overloaded. In this single problem alone it needs to be used to ask three different questions:\n\n1. Is this function convex?\n2. Is this set convex?\n3. Is this optimization model convex?\n\nThese questions are not the same, and one does not necessarily imply the others.\n\nYour function $f$ is convex. That is easily verified, as you have done, by examining the Hessian.\n\nYour constraint set $\\left\\{x\\in(0,1)^n\\,\\middle|\\,\\sum_i x_i^2=1\\right\\}$ is not convex. Nonlinear equality constraints are almost never convex. This set describes the positive quadrant of the unit hypersphere, and this is not convex. Indeed, take any distinct pair of points in the set; their midpoint is not in the set (a necessary condition for convexity).\n\nFinally, a convex optimization model requires that:\n\n• Minimization of a convex function, or maximization of a concave function.\n• Each and every constraint describes a convex set.\n\nYour model fails both tests, because you are maximizing a convex function, and your nonlinear equality constraint is not convex.\n\nEDIT: I should add that you could relax your constraint to $\\left\\{x\\in(0,1)^n\\,\\middle|\\,\\sum_i x_i^2\\leq 1\\right\\}$ without changing the optimal value of the problem. This would make your constraint set convex; but it would still be a nonconvex optimization problem, due to your attempt to maximize a convex function.\n\nAll that said, this problem is not that difficult to solve. In order to make the optimum achievable we need to relax the constraints to $\\left\\{x\\in[0,1]^n\\,\\middle|\\,\\sum_i x_i^2\\leq 1\\right\\}$ (i.e., non-strict inequalities on $x$). In that case, the optimal value is 1, and the optimal points are all of the unit vectors with a single non-zero value of $1$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9050772,"math_prob":0.9979373,"size":1737,"snap":"2019-26-2019-30","text_gpt3_token_len":424,"char_repetition_ratio":0.13964224,"word_repetition_ratio":0.0,"special_character_ratio":0.24064478,"punctuation_ratio":0.13960114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997009,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T02:26:59Z\",\"WARC-Record-ID\":\"<urn:uuid:b93fd806-4706-4207-95fb-fb7d8d93788b>\",\"Content-Length\":\"141982\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2567c863-ffd6-48cd-ac33-8144689eb476>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef1aedeb-e853-4d61-a387-dcd2556e7bd3>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1705764/problem-to-prove-function-is-convex-or-not\",\"WARC-Payload-Digest\":\"sha1:WB52TGK4EOF5RVZJ4JWZRJ3QHAJ4QJUI\",\"WARC-Block-Digest\":\"sha1:TKIOGNF6TWG25DCTA65DF4D6UAX5E77T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525009.36_warc_CC-MAIN-20190717021428-20190717043428-00266.warc.gz\"}"} |
https://lemon.cs.elte.hu/trac/lemon/changeset/209c9d53e195a33b6753f2add5f4ec88a31bd646/lemon-0.x | [
"# Changeset 851:209c9d53e195 in lemon-0.x\n\nIgnore:\nTimestamp:\n09/14/04 12:29:47 (17 years ago)\nBranch:\ndefault\nPhase:\npublic\nConvert:\nsvn:c9d7d8f5-90d6-0310-b91f-818b3a526b0e/lemon/trunk@1150\nMessage:\n\nChanges in doc.\n\nLocation:\nsrc/hugo\nFiles:\n2 edited\n\nUnmodified\nAdded\nRemoved\n• ## src/hugo/minlengthpaths.h\n\n r788 ///Runs the algorithm. ///Returns k if there are at least k edge-disjoint paths from s to t. ///Otherwise it returns the number of found edge-disjoint paths from s to t. ///Otherwise it returns the number of found edge-disjoint paths from s to t. int run(Node s, Node t, int k) { ///Total length of the paths ///This function gives back the total length of the found paths. ///Assumes that \\c run() has been run and nothing changed since then. ///\\pre \\ref run() must ///be called before using this function. Length totalLength(){ return mincost_flow.totalLength(); } ///Returns a const reference to the EdgeMap \\c flow. \\pre \\ref run() must ///Return the found flow. ///This function returns a const reference to the EdgeMap \\c flow. ///\\pre \\ref run() must ///be called before using this function. const EdgeIntMap &getFlow() const { return mincost_flow.flow;} ///Returns a const reference to the NodeMap \\c potential (the dual solution). /// Return the optimal dual solution ///This function returns a const reference to the NodeMap ///\\c potential (the dual solution). /// \\pre \\ref run() must be called before using this function. const EdgeIntMap &getPotential() const { return mincost_flow.potential;} ///Checks whether the complementary slackness holds. ///This function checks, whether the given solution is optimal ///Running after a \\c run() should return with true ///In this \"state of the art\" this only checks optimality, doesn't bother with feasibility ///Currently this function only checks optimality, ///doesn't bother with feasibility /// ///\\todo Is this OK here? } ///Read the found paths. ///This function gives back the \\c j-th path in argument p. ///Assumes that \\c run() has been run and nothing changed since then. /// \\warning It is assumed that \\c p is constructed to be a path of graph \\c G. If \\c j is not less than the result of previous \\c run, then the result here will be an empty path (\\c j can be 0 as well). template void getPath(DirPath& p, size_t j){ /// \\warning It is assumed that \\c p is constructed to ///be a path of graph \\c G. ///If \\c j is not less than the result of previous \\c run, ///then the result here will be an empty path (\\c j can be 0 as well). template void getPath(Path& p, size_t j){ p.clear();\n• ## src/hugo/preflow.h\n\n r849 /// @{ ///Preflow algorithms class. ///%Preflow algorithms class. ///This class provides an implementation of the \\e preflow \\e ///algorithm producing a flow of maximum value in a directed ///graph. The preflow algorithms are the fastest max flow algorithms ///up-to-date. The \\e source node, the \\e target node, the \\e ///up to now. The \\e source node, the \\e target node, the \\e ///capacity of the edges and the \\e starting \\e flow value of the ///edges should be passed to the algorithm through the ///setFlow. /// ///After running \\c phase1 or \\c preflow, the actual flow ///After running \\ref phase1() or \\ref preflow(), the actual flow ///value can be obtained by calling \\ref flowValue(). The minimum ///value cut can be written into a \\c node map of \\c bools by ///calling \\ref minCut. (\\ref minMinCut and \\ref maxMinCut writes ///value cut can be written into a bool node map by ///calling \\ref minCut(). (\\ref minMinCut() and \\ref maxMinCut() writes ///the inclusionwise minimum and maximum of the minimum value cuts, ///resp.) ///Runs the preflow algorithm. ///Runs the preflow algorithm. ///Runs the preflow algorithm. /// void run() { phase1(flow_prop); ///is not yet obtained. So after calling this method \\ref flowValue ///and \\ref minCut gives proper results. ///\\warning: \\ref minMinCut and \\ref maxMinCut do not ///\\warning \\ref minMinCut and \\ref maxMinCut do not ///give minimum value cuts unless calling \\ref phase2. ///\\pre The starting flow must be ///is not yet obtained. So after calling this method \\ref flowValue ///and \\ref actMinCut gives proper results. ///\\warning: \\ref minCut, \\ref minMinCut and \\ref maxMinCut do not ///\\warning \\ref minCut, \\ref minMinCut and \\ref maxMinCut do not ///give minimum value cuts unless calling \\ref phase2. void phase1()\nNote: See TracChangeset for help on using the changeset viewer."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6150428,"math_prob":0.9588652,"size":5085,"snap":"2021-04-2021-17","text_gpt3_token_len":1477,"char_repetition_ratio":0.11356033,"word_repetition_ratio":0.19292237,"special_character_ratio":0.37168142,"punctuation_ratio":0.090625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9702182,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-20T16:27:24Z\",\"WARC-Record-ID\":\"<urn:uuid:3e109780-07b7-4d6b-81e9-4bcde9eb3f10>\",\"Content-Length\":\"38364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67e802af-b0a9-4501-9593-ad17d27707d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:914aa3be-edec-4b5d-a5e8-bc4522fd6279>\",\"WARC-IP-Address\":\"157.181.227.195\",\"WARC-Target-URI\":\"https://lemon.cs.elte.hu/trac/lemon/changeset/209c9d53e195a33b6753f2add5f4ec88a31bd646/lemon-0.x\",\"WARC-Payload-Digest\":\"sha1:4YXXO7BUTOMJELZBVBWIGTP7KJLOBVU4\",\"WARC-Block-Digest\":\"sha1:HALXF7ZWKQ2FH777LIPKP5DKDQIV6WWY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039476006.77_warc_CC-MAIN-20210420152755-20210420182755-00342.warc.gz\"}"} |
http://yuanjingyan.com/article/4849658903 | [
"# vue和el-table使用经验-如何刷新表格数据\n\n``````for(var i=0;i<that.tableData.length;i++){\nvar aId=that.tableData[i].stationid;\nvar bId=that.tableData[i].deviceid;\n\nif(aId==that.selectStationId&&bId==that.selectDeviceId){\nthat.tableData[i].physicalid=physicalId;\n}\n}\n``````\n\n``````var newTableData=[];\nfor(var i=0;i<that.tableData.length;i++){\nvar aId=that.tableData[i].stationid;\nvar bId=that.tableData[i].deviceid;\n\nif(aId==that.selectStationId&&bId==that.selectDeviceId){\nthat.tableData[i].physicalid=physicalId;\n}\nnewTableData.push(that.tableData[i]);\n}\nthat.tableData=newTableData;\n``````"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.5417807,"math_prob":0.6500835,"size":776,"snap":"2021-43-2021-49","text_gpt3_token_len":336,"char_repetition_ratio":0.22797927,"word_repetition_ratio":0.4,"special_character_ratio":0.1984536,"punctuation_ratio":0.28125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9638876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T15:32:45Z\",\"WARC-Record-ID\":\"<urn:uuid:6d653c95-af0e-4c4e-81da-96cbcc290da9>\",\"Content-Length\":\"273247\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:173f1cdc-3cfc-4c24-a9d0-c7476a46a5d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:a54ec5fc-db84-401b-9c93-9df168c9916b>\",\"WARC-IP-Address\":\"121.5.247.191\",\"WARC-Target-URI\":\"http://yuanjingyan.com/article/4849658903\",\"WARC-Payload-Digest\":\"sha1:OPUXW3FXIM5MTVNBCPDWWXVGRRK5SS7K\",\"WARC-Block-Digest\":\"sha1:PFQEJY6I6GYZKFMEANDNQZN6ISLPGKM7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359037.96_warc_CC-MAIN-20211130141247-20211130171247-00077.warc.gz\"}"} |
https://www.jiskha.com/questions/1599481/two-missiles-2420-km-apart-are-launched-at-the-same-time-and-are-headed-towards-each | [
"# Maths\n\nTwo missiles, 2420 km apart, are launched at the same time and are headed towards each other. They pass after 1.5 hours. The average speed of one missile is twice that of the other. What is the average speed of each missile?\n\n1. 👍 2\n2. 👎 1\n3. 👁 677\n1. If the slower's speed is s km/hr, then their combined speed of approach is s+2s = 3s\n\nSo, since distance = speed * time,\n\n3s * 3/2 = 2420\n\n1. 👍 4\n2. 👎 2\n2. d1 + d2 = 2420.\nV1*T1 + V2*T2 = 2420.\nV2 = 2V1, T1 = T2 = 1.5 h.\nV1*1.5 + 2V1*1.5 = 2420, V1 = ?.\n\nV2 = 2V1.\n\n1. 👍 2\n2. 👎 1\n\n## Similar Questions\n\n1. ### physics\n\nA model rocket is launched straight upward. Its altitude y as a function of time is given by y=bt−ct2, where b = 80m/s , c = 4.9m/s2 , t is the time in seconds, and y is in meters. 1. Use differentiation to find a general\n\n2. ### Physics\n\nA cyclist maintains a constant velocity of 5.2 m/s headed away from point A. At some initial time, the cyclist is 251 m from point A. What will be his displacement from his starting position after 85 s? Answer in units of m What\n\n3. ### Algebra\n\nHelp Please, ty A ball is launched at 20 meters per second (m/s) from a 60 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 20t + 60, where s is in meters. What is the\n\n4. ### physics\n\nA stone is launched straight up by a slingshot. Its initial speed is 20.1 m/s and the stone is 1.30 m above the ground when launched. Assume g = 9.80 m/s2. (a) How high above the ground does the stone rise? (b) How much time\n\n1. ### PHYSICS\n\nTwo spheres are launched horizontally from a 1.1m -high table. Sphere A is launched with an initial speed of 5.0M/S . Sphere B is launched with an initial speed of 3.0M/S A. What is the time for the sphere A to hit the floor? and\n\n2. ### Differential Equations\n\nA projectile was launched from the ground with a certain initial velocity. Militaries used a radar to determine the vertical coordinate y(t) of the projectile for two moments of time t measured in seconds from the moment when the\n\n3. ### Physics\n\nAn acrobat is launched from a cannon at an angle of 60 degrees above the horizontal. The acrobat is caught by a safety net mounted horizontally at the height from which he was initially launched. Suppose the acrobat is launched at\n\n4. ### algebra\n\nA plane leaves Chicago headed for LA at 540 mph. One hour later, a second plane leaves LA headed for Chicago at 660 mph. If the air route from Chicago to LA is 1800 miles, how long will it take for the first plane to pass the\n\n1. ### math\n\nA model rocket is launched straight upward. Its altitude y as a function of time is given by y = btct2 , where b = 82 m/s, c = 4.9 m/s2 is the time in seconds, and y is in meters. (a) Use differentiation to find a general\n\n2. ### Physics\n\nA \"rocket car\" is launched along a long straight track at t=0s. It moves with constant acceleration a1=3.0m/s^2. At t=2.5s, a second car is launched with constant acceleration a2=9.0m/s^2. 1. At what time does the second car catch\n\n3. ### Physics\n\nTwo spheres are launched horizontally from a 1.0 -high table. Sphere is launched with an initial speed of 4.5 . Sphere is launched with an initial speed of 3.0 . What is the time for the sphere A to hit the floor? What is the time\n\n4. ### Physics\n\nA volleyball is sent over a net, having been launched at a speed 5 m/s at an angle 42.7 degrees above the horizontal, from the ground. What is the ball’s speed, in m/s, 2.97 seconds after being launched?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93641204,"math_prob":0.9771773,"size":2999,"snap":"2020-34-2020-40","text_gpt3_token_len":839,"char_repetition_ratio":0.1472454,"word_repetition_ratio":0.15306123,"special_character_ratio":0.2750917,"punctuation_ratio":0.10454545,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99638397,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T11:21:33Z\",\"WARC-Record-ID\":\"<urn:uuid:8a14d9c1-0154-47c3-8c35-0bf1c6f3924b>\",\"Content-Length\":\"20306\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee855c83-f039-4884-a1ae-93bfdea8bf59>\",\"WARC-Concurrent-To\":\"<urn:uuid:47964db3-fe2e-4eb7-bcbc-44187ce3600d>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/1599481/two-missiles-2420-km-apart-are-launched-at-the-same-time-and-are-headed-towards-each\",\"WARC-Payload-Digest\":\"sha1:RYT2GS372OZFUAVE2OEGRPBCIC2EVON4\",\"WARC-Block-Digest\":\"sha1:ZH3SRRQJJE6JR6EHWZR27G5DJBB764BJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400205950.35_warc_CC-MAIN-20200922094539-20200922124539-00443.warc.gz\"}"} |
http://xahlee.info/js/js_triple_equal.html | [
"# JavaScript: Triple Equal Operator\n\nBy Xah Lee. Date: .\n\n## Compare Primitives\n\nTriple Equal can compare most Primitive Values such as string, number, Symbol, undefined, null, except NaN.\n\n```console.log(\"x\" === \"x\");\nconsole.log(3 === 3);\nconsole.log(3 === 3.0);\nconsole.log(Infinity === Infinity);```\n\nNaN cannot be compared with triple equal.\n\n`console.log((NaN === NaN) === false);`\n\nuse:\n\n## Test Object Equality by Triple Equal\n\nObjects (including Array) cannot be compared with triple equal.\n\nWhen a object is assigned to a variable, the variable holds a reference to the object.\n\nIf 2 variables hold the same reference, they are equal.\n\n```const x = { \"a\": 3 };\nconst y = x; // x and y holds the same reference\nconsole.log(x === y);```\n\nbut 2 objects with same property values, are still considered different if they have different reference.\n\n```// Objects cannot be compared with triple equal\n\nconsole.log(([] === []) === false);\nconsole.log(({} === {}) === false);\nconsole.log(({ \"a\": 3 } === { \"a\": 3 }) === false);```\n\nto test equality of objects, see JavaScript: Test Object Equality 🚀\n\n## No Auto Type Convert Conversion\n\nTriple equale does not convert Value Types implicitly.\n\n```console.log(([] === \"\") === false);\nconsole.log((\"\" === {}) === false);\nconsole.log((\"0\" === 0) === false);\nconsole.log((\"\" === 0) === false);\nconsole.log((1 === true) === false);```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6011935,"math_prob":0.99554926,"size":1303,"snap":"2023-40-2023-50","text_gpt3_token_len":313,"char_repetition_ratio":0.22170901,"word_repetition_ratio":0.025,"special_character_ratio":0.3292402,"punctuation_ratio":0.218107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9922156,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T20:02:39Z\",\"WARC-Record-ID\":\"<urn:uuid:c5d56c9f-13b5-484e-bc2c-92e2fb14b222>\",\"Content-Length\":\"14752\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d45a377-dc9f-4f47-9d80-9af7ebc6a80e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d03922fb-ddc0-401c-a6de-bc1f31e5b4cb>\",\"WARC-IP-Address\":\"74.208.236.186\",\"WARC-Target-URI\":\"http://xahlee.info/js/js_triple_equal.html\",\"WARC-Payload-Digest\":\"sha1:TE7VQ6QD6YKPFTAGVA54JEJCPTO6ZFZQ\",\"WARC-Block-Digest\":\"sha1:NLVTAIRYIY5KIZ7BL5JT5U4WRLRAPMOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100508.53_warc_CC-MAIN-20231203193127-20231203223127-00401.warc.gz\"}"} |
https://psychteacher.co.uk/research-methods-AS/styled-11/inferential-statistics.html | [
"Inferential statistics\nStatistical analysis for A2 psychology\n\nChoosing a statistical test\nThere are several factors that must be considered before a statistical test can be chosen. Firstly the research design, aim and level of measurement must be identified:\n\nResearch design\nData can be either related or unrelated. Related data is produced from repeated measures and matched pairs designs. Unrelated data is produced from independent groups designs.\n\nResearch aim\nIs the aim of the research to investigate a significant difference or a significant association? If, for example, there are 2 groups of participants, each in a different condition of the independent variable (e.g. in an experiment), then the aim is to test for a significant difference. If the aim is to test for a correlation between two variables, then the aim is to test for a significant association.\n\nLevel of measurement\nData can be produced at nominal, ordinal and interval levels:\n• Nominal data is the most basic level of measurement. An example is a frequency count of a distinct category, such as the number of aggressive and non-aggressive acts in an observation.\n• Ordinal data consists of a list of data that can be ranked in order, but not data that would fit to an interval scale. An example is the subjective rating of happiness (on a scale form 1 to 10) that participants may score themselves as on a questionnaire. A happiness rating of 10 is higher than 5, but it is not twice as happy as 5 or 5 times as happy as 2.\n• Interval data is measured on a scale in which each interval is exactly the same size. Time is interval data because each second is the same duration, and 10 seconds are twice as long as 5 seconds.\n\nStatistical tests\nOnce the design, aim and level of measurement have been identified, the correct inferential test can be chosen.\n\nSpearman’s rho\nSpearman’s rho is a test for significant association, and produces a correlation coefficient. The level of measurement must be either ordinal or interval. The research design can be either related or unrelated.\n\nWilcoxon signed ranks test\nWilcoxon signed ranks is a test of significant difference for related data. The research design must produce related data (e.g. repeated measures or matched pairs). The level of measurement can be either ordinal or interval.\n\nMann-Whitney U test\nMann-Whitney U is a test of significant difference for unrelated data. The research design must produce unrelated data (e.g. independent measures). The level of measurement can be either ordinal or interval.\n\nChi-square test\nChi-square tests for difference when the data is nominal and unrelated. The research design must produce unrelated data (e.g. independent measures). The level of measurement must be nominal (e.g. categories).",
null,
"A Level exam tips"
] | [
null,
"https://psychteacher.co.uk/research-methods-AS/styled-11/files/inferential-tests-table-2.jpg",
null
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https://en.wikiversity.org/wiki/User:Watchduck/Logic | [
"User:Watchduck/Logic\n\nRelations\n\nIn different universes\n\n1-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"2-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"3-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"4-element universe - first example with 15 minterm relations:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"5-element universe:",
null,
"(zoom in) $~\\land$",
null,
"",
null,
"(zoom in) $~\\Leftrightarrow$",
null,
"",
null,
"(zoom in)\n\nParity relations\n\nUsually the question is, if somewhere are no or some elements.\nBut one may also ask, if somewhere is an even or an odd number of elements.\n\nIn a 1-element universe even means 0, and odd means 1:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"2-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"3-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"4-element universe:",
null,
"$~\\land$",
null,
"",
null,
"$~\\Leftrightarrow$",
null,
"",
null,
"5-element universe:",
null,
"(zoom in) $~\\land$",
null,
"",
null,
"(zoom in) $~\\Leftrightarrow$",
null,
"",
null,
"(zoom in)\n\nExamples\n\n $\\oplus ~$",
null,
"$\\Leftrightarrow ~$",
null,
"This is a different way to write the same:\n\nThis is what the exclusive or excludes:\n\nIt's not to be confused with:\n\nJust another example:\n\n3-ary relations\n\n $A\\subseteq B$",
null,
"$\\land ~$",
null,
"$B\\subseteq C$",
null,
"$\\Leftrightarrow$",
null,
"$A\\subseteq B\\subseteq C$",
null,
"$\\land ~$",
null,
"$\\Leftrightarrow ~$",
null,
"There are 256 relations of this kind (corresponding to the 256 operations).\nThe 22 relations in the following table are shown in place of their mirrorings and rotations:\n\nPropositional calculus examples: drivers and medics\n\nA proposition is uniquely determined by the set of all cases, in which it is true.\nThis set could be called the proposition's validity set.\nTwo propositions are equal, when they have the same validity set.\n\nThe validity set of a negation is the complement of the initial proposition's validity set.\nSo to know the negation of a proposition, one has to know the set of all possible cases.\n\nThe set of all possible cases is the validity set of the tautology. It may be denoted $~\\Omega$",
null,
".\nThe empty set is the validity set of the contradiction.\n\nCases that can be the case and propositions that can be said are essentially different objects.\n(Similar to outcomes and events in probability theory.)\nWhen there are n possible cases, there are 2n possible propositions.\nAmong them are n elementary propositions (minterms). They have a 1-element validity set, and thus they are true in exactly one case.\n(Cases and corresponding elementary propositions are easily mixed up - like outcomes and elementary events in probability theory.)\n\nA single employee\n\nOne may be interested, if an employee is a driver or a medic.\nThere are exactly four possiblities (cases), how he can have these qualities or not:\n\nExactly one of these statements will be true about a certain employee.\nIn respect to these qualities there are 24 = 16 statements (propositions), one can say about this employee:\n\nA group of employees\n\nOne may be interested, which of the following statements are true about a certain group of employees:\n\nThese statements don't contradict each other. At least one will be true about a certain group of employees.\n(Assumed, that the group consists of at least one employee.)\nThere are 15 possible cases:\n\nSo there are 215 = 32768 propositions one can say about a particular group of employees:\n\nExamples\n\n with the validity set: {",
null,
"}"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8757361,"math_prob":0.9923232,"size":3609,"snap":"2019-43-2019-47","text_gpt3_token_len":872,"char_repetition_ratio":0.15839112,"word_repetition_ratio":0.11568938,"special_character_ratio":0.25270158,"punctuation_ratio":0.12517782,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99771494,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122],"im_url_duplicate_count":[null,2,null,10,null,2,null,10,null,2,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,2,null,10,null,2,null,10,null,2,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,1,null,10,null,1,null,10,null,1,null,1,null,2,null,null,null,2,null,null,null,null,null,2,null,2,null,2,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T10:29:18Z\",\"WARC-Record-ID\":\"<urn:uuid:deca47fa-75e6-4d0c-bf21-a834c6092e08>\",\"Content-Length\":\"297896\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c3c9ed1-2459-42fd-aaa6-298ae7e4ed12>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7454b4e-36b7-45c0-a038-96bb27332326>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://en.wikiversity.org/wiki/User:Watchduck/Logic\",\"WARC-Payload-Digest\":\"sha1:KUFSWMAU4PQAT4DQBAC5BUXKAMN6AFRH\",\"WARC-Block-Digest\":\"sha1:F3QB6ZQB7UIPQH2YLC24LVKTOKKVHMAB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986666959.47_warc_CC-MAIN-20191016090425-20191016113925-00391.warc.gz\"}"} |
https://socratic.org/questions/56f3406f11ef6b4e4edc06a7 | [
"# Question c06a7\n\nMar 24, 2016\n\nHere's what's going on here.\n\n#### Explanation:\n\nOleum is simply a saturated solution of sulfur trioxide, ${\\text{SO}}_{3}$, dissolved in sulfuric acid, ${\\text{H\"_2\"SO}}_{4}$.\n\nYou'll sometimes see this oleum referred to as disulfuric acid, ${\\text{H\"_2\"S\"_2\"O}}_{7}$, or fuming sulfuric acid.\n\nOleum will always have a percentage of sulfuric acid that is greater than 100%. This happens because the percentage of sulfuric acid includes the mass of water needed to react with the dissolved sulfur trioxide to form sulfuric acid\n\n${\\text{SO\"_text(3(aq]) + \"H\"_2\"O\"_text((l]) -> \"H\"_2\"SO}}_{\\textrm{4 \\left(a q\\right]}}$\n\nIn essence, the concentration of oleum tells you how much water must combine with the free sulfur trioxide present in $\\text{100 g}$ of oleum in order to produce sulfuric acid.\n\nLet's take, for example, a 10color(red)(5)% oleum solution. The percent concentration tells you that you need $\\textcolor{red}{5} \\textcolor{w h i t e}{a} \\text{g}$ of water in order to react with all the free sulfur trioxide present in $\\text{100 g}$ of this oleum sample to form $\\text{105 g}$ of sulfuric acid.\n\nTake a look at the above equation. Notice that one mole of sulfur trioxide will react with one mole of water to form one mole of sulfuric acid.\n\nUse the molar masses of the three compounds to determine the ratio that exists between their masses.\n\n${\\text{For SO\"_3: \" \"\" \"M_M ~~ \"80 g mol}}^{- 1}$\n\n${\\text{For H\"_2\"O\": \" \"\" \"M_M ~~ \"18 g mol}}^{- 1}$\n\n${\\text{For H\"_2\"SO\"_4:\" \" M_M ~~ \"98 g mol}}^{- 1}$\n\nSo, this tells you that $\\text{80 g}$ of sulfur trioxide reacts with $\\text{18 g}$ of water to produce $\\text{40 g}$ of sulfuric acid.\n\nIn this example, $\\textcolor{red}{5}$ grams of water will react with\n\n5 color(red)(cancel(color(black)(\"g water\"))) * \"80 g SO\"_3/(18color(red)(cancel(color(black)(\"g water\")))) = \"22.22 g SO\"_3\n\nThis means that $\\text{100 g}$ of 105% oleum will contain $\\text{22.22 g}$ of sulfur trixoide.\n\nThe percent of free sulfur trioxide will thus be $\\text{22.2%}$.\n\nImplicitly the percent of sulfuric acid in this oleum will be\n\n\"% H\"_2\"SO\"_4 = 100 - 22.22 = 77.78%\n\nNotice that this oleum solution will produce $\\text{105 g}$ of sulfuric acid. If $\\text{5 g}$ of water react with $\\text{22.22 g}$ of sulfur trioxide, you will get\n\n5color(red)(cancel(color(black)(\"g water\"))) * (\"98 g H\"_2\"SO\"_4)/(18color(red)(cancel(color(black)(\"g water\")))) = \"27.22 g H\"_2\"SO\"_4#\n\nSince the $\\text{100-g}$ sample of oleum contains $\\text{77.78 g}$ of sulfuric acid, you will end up with\n\n${\\text{77.78 g \" + \" 27.22 g\" = \"105 g H\"_2\"SO}}_{4}$\n\nSo, as a conclusion, the percentage of free sulfur trioxide tells you much sulfur trioxide you get per $\\text{100 g}$ of a given oleum solution."
] | [
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http://iieta.org/journals/ejee/paper/10.3166/EJEE.20.205-214 | [
"# Simulation of single phase buck boost matrix converter without commutation issues\n\nSimulation of single phase buck boost matrix converter without commutation issues\n\nAl-Habeeb College of Engineering and Technology, Chevella, RR Dt., India\n\nShinas College of Technology, Al Aqar, Sultanate of Oman\n\nElectrical Engineering Department, King Khalid University, Kingdom of Saudi Arabia\n\nCorresponding Author Email:\[email protected]\nPage:\n205-214\n|\nDOI:\nhttps://doi.org/10.3166/EJEE.20.205-214\n|\nAccepted:\n|\nPublished:\n30 April 2018\n| Citation\n\nOPEN ACCESS\n\nAbstract:\n\nSimulation of single phase buck boost matrix converter without commutation issues is presented in this paper. The proposed converter is capable to operate inverting as well as non-inverting operation. Generally, AC to AC converters are designed by using thyristors and it will produce more harmonics with lower power factor. To overcome the drawbacks of the AC-AC voltage controller, buck –boost based inverting and non-inverting converter have been introduced. In this topology, six numbers of unidirectional current flowing and voltage blocking switches are used. The proposed topology can be applied for both step down and step up applications with non-inverting modes of operations and inverting operations suitable for dynamic voltage restorer (DVR) applications. The main feature of this topology is buck –boost with inverting and buck-boost with non-inverting capability with a topology. The performances of the proposed work will be analyzed through MATLAB/Simulink.\n\nKeywords:\n\nbuck boost converter, inverting, non -inverting, DVR, MATLAB/Simulink.\n\n1. Introduction\n\nIn many applications, an AC-AC conversion is important. In past, thyristor based AC voltage controller have been used for conversion of AC quantity into AC quantity as in Fundamental of power electronics. In the method of conversion, AC-AC converter produce distorted electrical waveforms and operate at lower power factor as described by Batarseh (2004). To avoid these drawbacks, AC voltage controller has been replaced by matrix converter as demonstrated by Hart (2011). There are many advantages are associated with AC-AC converter such as ideally no power losses while during power conversion processor and not required for dc link filter and harmonics of the system can be decided by switching frequency of the converter as in Rashid (2010). The many researchers propose new family of AC-AC converter using PWM with reduced switches count. First stage is current-fed Z-source converter and second stage is voltage-fed Z-source converter. This paper uses reduced number of switches compared to the compared to the existing PWM based AC-AC converter (Fang et al., 2005). The three phase bidirectional direct ac–ac converter based switched capacitor concepts, with only switches and capacitor. This converter operates at constant gain, output and input frequency is same as shown in T.B. Lazzarin et al. (2015). For controlling of the low power three phase induction motors drives, single phase to three phase PWM converter introduced by Lee et al. (2007). This converter is constructed connecting two parallel three-phase to three-phase dc-link converters without usage of isolation transformers. This type of converter topology is highly suitable for high power applications and also verified through practical way by Santos et al. (2011). New topologies designed for three phase PWM AC-AC converter system. Basically they used, space vector modulation for current and voltage source inverter in Kolar et al. (2011). A single-phase quasi-Z-source AC–AC converter has been proposed. This topology offers lower input current THD and higher input power factor compared to the conventional proposed by Nguyen et al. (2010). For obtaining higher efficiency, low power level MOSFET based transformer less photo voltaic system have designed using super junction MOSFETs. It will offer good efficiency even though, this topology suffers from many drawbacks; produce more conduction losses due the presence of more power devices, low magnetics utilization problems and MOSFET failure risk identified by Chen et al. (2015). Non inverting buck boost an inverting buck boost have been presented in this paper along with high quality waveforms as shown by Ahmed et al. (2015) and this paper solves commutation problems without other components similar to H.F. Ahmed et al. (2015). The proposed systems have simulated using MATLAB / Simulation and their results are verified.\n\n2. Proposed systems\n\nThe proposed topology consists of six bidirectional switches, one inductor and two capacitors are used. The bidirectional switches namely, S1 to S6. The proposed topology has been introduced already in literature. In this topology, conventional boost, buck, and buck boost with inverting operation with gain of 1/(1−D), D and −D/(1−D) have been proposed. In this proposed topology, buck and boost modes of operation can be eliminated completely and introduced buck boost operation with non-inverting and voltage gain of D/(1−D) is introduced. The proposed topology is capable to operate variable frequency operation and variable output voltage along with operation of buck-boost capability of input voltage and without shoot through problems of voltage source.\n\n1) Path containing input voltage source and switches S1, S2.\n\n2) Path containing output voltage Voand switches S5, S6.\n\nPath containing input and output voltage and switches S3, S4. However, all the three paths contain two unidirectional switches in opposite (anti-series) direction, and, therefore, the current cannot flow in these close paths. Therefore, shoot through can be avoided in the proposed circuits.\n\n## 1.png",
null,
"Figure 1. Circuit operation and Switching Strategies for variable frequency operation 60-Hz (Same Frequency, fo= fin) Non-inverting Buck–Boost Operation\n\nNon inverting buck –boost operation and 60Hz and their waveforms are shown in the Figure 2. From figure 2 it is clearly seen that, switches S1, S3 and S6 are turned ON and switch S4 is completely turned OFF. ON and OFF of high frequency switches S2, S5 are depends on the intervals DT and (1−D)T, respectively. The equivalent circuit of non-inverting buck boost operation of AC to AC converter is shown in the Figure 3. The circuit operation interval is DT and input energy is stored in inductor. Switches S1 and S6 are reverse biased, it is used to preventing from conducting the current. We apply KVL, then\n\n$V _ { L } = V _ { i n }$ (1)\n\nThe non-inverting buck-boost mode operation with input voltage is greater than zero with duty duration of DT.\n\n## 2.png",
null,
"Figure 2. Switching scheme and waveforms for non-inverting and inverting buck–boost operation\n\n## 3.png",
null,
"Figure 3. Non-inverting buck–boost operation when Vin>0. (a) During DT\n\n## 4.png",
null,
"Figure 4. Non-inverting buck–boost operation when Vin>0. During(1 −D)T\n\nThe equivalent circuit of non-inverting buck boost operation of AC to AC converter is shown in the Figure 4. The circuit operation interval is (1-D) T and input energy is stored in inductor. Switches S1 and S6 are forward biased, it is used to provide freewheeling path to inductors. S2, S5 are turned OFF during this interval. During this period, energy stored in the inductor is released to load side. We apply KVL, then\n\n$V _ { L } = - V _ { 0 }$ (2)\n\nThe non-inverting buck-boost mode operation with input voltage is greater than zero with duty duration of (1-D) T.\n\n## 5.png",
null,
"Figure 5. Noninverting buck–boost operation when νin<0. (a) During DT and (b) during (1−D)T\n\nDuring the negative half cycle of supply voltage, switches S2, S4 and S5 are ON and switch S3 is OFF completely. Non inverting buck-boost operation with interval of DT and (1-D)T respectively. The circuit operation for Vin<0 is similar to that for Vin>0, with only difference that now switches S1, S6 act same as S2, S5 (for Vin>0), and vice versa.\n\n## 6.png",
null,
"Figure 6. Non-inverting buck–boost operation when νin<0. (a) During DT and (b) during (1 − D)T\n\nNow, flux balance (volt–sec) balance condition on inductor L using (1) and (2) gives\n\n$\\frac { V _ { 0 } } { V _ { i n } } = \\frac { D } { 1 - D }$\n\nFrom equation, the voltage gain of the proposed converter for this non-inverting buck–boost mode is D/(1−D) having voltage buck operation for D <0.5 and boost operation for D >0.5, with same phase and frequency of output voltage as that of input voltage.\n\n3. Simulation result and discussion\n\nThe simulation results of the proposed system are presented in this section. The proposed system is simulated through MATLAB / Simulink and tested. AC–AC converter is cable to operate buck mode and boost mode. In the boost mode duty cycle is more than 0.5. The presented results with duty cycle of 0.8.\n\n3.1. Simulation results: Duty cycle at 0.8 with different frequencies: Boost operation\n\nThe input voltage, load voltage and load current of the proposed system is shown in the Figure 8. The results is obtained with duty ratio of 0.8 and frequency of 25Hz.\n\n## 7.png",
null,
"Figure 7. Input voltage, load voltage and load current for 0.8 duty cycle and frequency of 25Hz\n\n## 8.png",
null,
"Figure 8. Inductor current for 0.8 duty cycle and frequency of 25Hz\n\nThe inductor current of the proposed system is shown in the Figure 8. The results are obtained with duty ratio of 0.8 and frequency of 25Hz.\n\n## 9.png",
null,
"Figure 9. Switching Pulses for 0.8 Duty cycle and Frequency of 25Hz\n\nThe switching pulses of the proposed system are shown in the Figure 9. The switching pluses results are obtained with duty ratio of 0.8 and frequency of 25Hz.\n\n3.2. Simulation results: Duty cycle at 0.4 with different frequencies: Buck operation\n\nThe boost operation of AC to AC converter is discussed in this section. AC–AC converter is cable to operate buck mode and boost mode. In the buck mode duty cycle is less than 0.5. The presented results with duty cycle of 0.4 and 25Hzfrequency. The Simulink model of AC to AC converter is shown in the Figure 10.\n\n## 10.png",
null,
"Figure 10. Input voltage, load voltage and load current for 0.4 duty cycle and frequency of 25Hz\n\nThe input voltage, load voltage and load current of the proposed system is shown in the Figure 10. The results are obtained with duty ratio of 0.8 and frequency of 25 Hz.\n\n## 11.png",
null,
"Figure 11. Inductor current for 0.4 duty cycle and frequency of 25 Hz\n\nThe inductor current of the proposed system is shown in the Figure 11. The results are obtained with duty ratio of 0.4 and frequency of 25 Hz.\n\n## 12.png",
null,
"Figure 12. Switching pulses for 0.4 duty cycle and frequency of 25 Hz\n\nThe switching pulse of the proposed system is shown in the Figure 12. The switching pluses results are obtained with duty ratio of 0.4 and frequency of 25 Hz.\n\n4. Conclusions\n\nDesign and simulation of buck –boost AC to AC matrix converter with non-inverting and inverting operation and without commutation problems has been presented in this paper. For getting AC-AC conversion, AC voltage regulators are used. This is constructed by thyristors. The control technique may be ether integral or phase control applied. The proposed converter topology uses six unidirectional current flowing and bidirectional voltage blocking switches, implemented by six reverse blocking insulated-gate bipolar transistors (IGBTs) or series MOSFET–diode pairs, two input and output filter capacitors, and one inductor. The proposed work simulated with different input frequency and results are validated with theoretical results. The simulation results have been obtained with different operating frequency and different duty cycle. The performances of the proposed work will be analyzed through MATLAB/Simulink.\n\nReferences\n\nAhmed H. F., Cha H. Y., Aleem Z., Khan A. A., Kim H. G. (2015). A novel buck-boost AC-AC converter with inverting and non-inverting operation and no commutation problem. 9th International Conference on Power Electronics and ECCE Asia (ICPE-ECCE Asia), 1-5 June 2015, Seoul, South Korea. http://dx.doi.org/10.1109/ICPE.2015.7167891\n\nBatarseh I. (2004). Power electronic circuits. Wiley, New York.\n\nChen B., Gu B., Zhang L., Xahid Z. U., Lai J. S. (2015). A high-efficiency MOSFET transformer less inverter for nonisolatedmicroinverter applications. IEEE Trans. Power Electron., Vol. 30, No. 7, pp. 3610-3622.\n\ndos Santos E. C., Jacobina C. B., Rocha N., da Silva E. R. C. (2011). Sixphase machine drive system with reversible parallel AC-DC-AC converters. IEEE Trans. Ind. Electron., Vol. 58, No. 5, pp. 2049-2053. http://dx.doi.org/10.1109/TIE.2010.2054057\n\nFang X. P., Qian Z. M., Peng F. Z. (2005). Single-phase Z-source PWM AC–AC converters. IEEE Power Electron. Lett., Vol. 3, No. 4, pp. 121-124. http://doi.org/10.1109/lpel.2005.860453\n\nHart D. (2011). Power electronics: converters, applications, and design. Wiley, New York.\n\nKolar J. W., Friedli T., Rodriguez J., Wheeler P. W., Kolar J. W. (2011). Review of three-phase PWM AC-AC converter topologies. IEEE Trans. Ind. Electron., Vol. 58, No. 11, pp. 4088-5006.\n\nLazzarin T. B., Andersen R. L., Barbi I. (2015). A switched-capacitor three-phase AC-AC converter. IEEE Trans. Ind. Electron., Vol. 62, No. 2, pp. 735-745. http://dx.doi.org/10.1109/TIE.2014.2336625\n\nLee D. C., Kim Y. S. (2007). Control of single-phase-to-three-phase AC/DC/AC PWM converters for induction motor drives. IEEE Trans. Ind. Electron., Vol. 54, No. 2, pp. 797-804. http://dx.doi.org/10.1109/TIE.2007.891780\n\nNguyen M. K., Jung Y. G., Lim Y. C. (2010). Single-phase ac-ac converter based on quasi-Z source topology. IEEE Trans. Power Electron., Vol. 25, No. 8, pp. 2200-2209. http://dx.doi.org/10.1109/TPEL.2010.2042618\n\nRashid M. H. (2010). Power electronics: Circuits, devices and applications. International Economy Edition."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8880361,"math_prob":0.8585371,"size":12906,"snap":"2020-24-2020-29","text_gpt3_token_len":3152,"char_repetition_ratio":0.16392808,"word_repetition_ratio":0.15846457,"special_character_ratio":0.24624206,"punctuation_ratio":0.16221374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801724,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T10:55:50Z\",\"WARC-Record-ID\":\"<urn:uuid:7fa9607a-91eb-4f50-80e3-90d27e41cc27>\",\"Content-Length\":\"64975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17b081db-97f5-4f06-8001-3a6b3559c989>\",\"WARC-Concurrent-To\":\"<urn:uuid:067c4a27-bf91-4ea7-8b3d-80977748c9ad>\",\"WARC-IP-Address\":\"45.79.107.192\",\"WARC-Target-URI\":\"http://iieta.org/journals/ejee/paper/10.3166/EJEE.20.205-214\",\"WARC-Payload-Digest\":\"sha1:CRP5SLQ4HDOB5OPNGUCYQGNB3IDHU7QZ\",\"WARC-Block-Digest\":\"sha1:A3N6SWNJ2HYZMWHCLQMDUXQW3GQJHONQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347424174.72_warc_CC-MAIN-20200602100039-20200602130039-00124.warc.gz\"}"} |
https://husarion.com/tutorials/ros-tutorials/6-transformation-in-ROS/ | [
"# Transformation in ROS\n\n## Chapter description\n\nThis tutorial will cover the fairly simple concept of transformation. The selection of an appropriate reference point is crucial for many algorithms in the world of robotics. Therefore, it is very important that you have a good understanding of the current topic if you want to develop more advanced systems. Below, in addition to the theory, there are examples that implement simple frame transformations. This tutorial will be based on the tf package used in many ROS projects.\n\nYou can run this tutorial on:\n\nRepository containing the final effect after doing all the ROS Tutorials you can find here\n\n## Introduction\n\nTransformations are functions that map n-D vectors to other n-D vectors: T:Rn→Rn. They can represent geometric operations, which are caused by movement or action, as well as changes of coordinates, which are caused by changes of interpretation. Many common spatial transformations, including translations, rotations, and scaling are represented by matrix / vector operations. Changes of coordinate frames are also matrix / vector operations. As a result, transformation matrices are stored and operated on ubiquitously in robotics.\n\nCoordinate transformation is used in many areas of mathematics, science, and engineering, and it has various applications. In this tutorial, transformations will be used to:\n\n1. Changing perspective: Coordinate transformation can help us view the same object or phenomenon from different perspectives, which can be useful in many applications. For example, in physics, we might want to change the coordinate system to make the equations of motion simpler or to study a system in a different reference frame.\n2. Mapping: Coordinate transformation can help map one coordinate system to another, which can be useful in navigation, cartography, and other fields.\n\nA transformation is a simple operation consisting of a translation and a rotation. The picture below shows two frames of reference rotated relative to each other. To find the position of point P in the second frame of reference, multiply the point by the rotation matrix.",
null,
"The translation boils down to a simple addition of the coordinates of the second reference point.\n\nIn general, transformations involve simple matrix operations, and coordinate transformation itself is a powerful tool that can help us understand and work with complex systems and phenomena. Fortunately, the tf package has a lot of useful functions that do all these calculations for us.\n\n## Basic transformation in ROS\n\nROS can help track coordinate systems over time. The tf2 package is used for this - a library that use special geometry_msgs/TransformStamped message type. This message type consists:\n\n• Header header - header message with timestamp for current transformation and information about parent frame id,\n• string child_frame_id - the frame id of the child frame,\n• Transform transform - transformation from coordinate frame header.frame_id to the coordinate frame child_frame_id.\n\n### Static transformation\n\nThis tutorial will show you how to add a static transformation between two frames. Suppose we have a robot that is fixed at (1,-1,0) relative to the map and rotated to the left. For this purpose, we can use the static_transform_publisher node from the tf package. Running the function is simple and just execute the command below:\n\nrosrun tf static_transform_publisher x y z yaw pitch roll frame_id child_frame_id period_in_ms\n\nor the same command but using quaternions:\n\nrosrun tf static_transform_publisher x y z qx qy qz qw frame_id child_frame_id period_in_ms\n\nNow all you have to do is complete the above fields in the appropriate way.\n\nhusarion@husarion:~$rosrun tf static_transform_publisher 1 -1 0 1.6 0 0 map robot 100 ### Visualization #### TF_Echo A simple tool that allows you to quickly check the transformation between any two frames. husarion@husarion:~$\nrosrun tf tf_echo map robot",
null,
"#### TF_Tree\n\nThe tool shows all the relationships between the individual forms in a diagram. This tool allows you to check if all frames are properly connected to each other.\n\nhusarion@husarion:~$rosrun rqt_tf_tree rqt_tf_tree",
null,
"#### RViz The RVIZ program known from chapter 4 will be used for visualization. To run RViz just type in new terminal. rviz Now click Add. In the new window, select TF from the list.",
null,
"Task 1 Try adding another reverse_camera frame which is 20 cm above the robot and facing the back of the robot. Use static_transform_publisher and check the result with all the tools presented above. Solution In new terminal type: rosrun tf static_transform_publisher 0 0 0.2 3.14 0 0 robot reverse_camera 100 Remember that if you point to a non-existing frame, a new relation will be created that is not related in any way to the currently existing transformations. To display it, change the Fixed Frame field in RViz ## Broadcasting transformation As you can see, the static transformation is very simple, but what to do when we want to simulate the mutual position of moving objects. ROS also allows this, but this time we need to create a node in which we will describe the movement of one frame relative to the other. Suppose we want to create an object that moves around the map. In this case, it is necessary to create a so-called TransformBroadcaster, which is a special sender type for sending tf messages. Despite the name difference, it works very similar to the classic Publisher works very similar to publisher except that it can only send tf messages. Let's start with creating a new file and name it tf_broadcaster.cpp. ~/ros_ws/src/tutorial_pkg/src/tf_broadcaster.cpp #include <ros/ros.h>#include <tf/transform_broadcaster.h>#include <math.h>void sendTransformation(){ static tf::TransformBroadcaster tf_broadcaster; tf::Transform transform; double t = ros::Time::now().toSec(); // Set translation transform.setOrigin(tf::Vector3(cos(t), sin(t), 0.0)); // Set rotation using roll, pitch. yaw tf::Quaternion q; q.setRPY(0, 0, fmod(t, 2*M_PI)+M_PI_2); transform.setRotation(q); tf_broadcaster.sendTransform(tf::StampedTransform(transform, ros::Time::now(), \"map\", \"robot\"));}int main(int argc, char **argv){ ros::init(argc, argv, \"tf_broadcaster\"); ros::NodeHandle n(\"~\"); ros::Rate loop_rate(100); while (ros::ok()) { ros::spinOnce(); sendTransformation(); loop_rate.sleep(); }} The code explained static tf::TransformBroadcaster tf_broadcaster;tf::Transform transform; A unique publisher type and its message necessary to send transformation messages. double t = ros::Time::now().toSec(); Receiving a time in seconds. transform.setOrigin(tf::Vector3(cos(t), sin(t), 0.0)); Setting translation of frame to Transform message. Since we want the object to rotate around the center of the map, the variables at the center must be time dependent. tf::Quaternion q;q.setRPY(0, 0, fmod(t, 2*M_PI)+M_PI_2); Creating quaternion message for storing rotation data and setting the deflection angle in such a way that the x-axis is in line with the direction of movement. transform.setRotation(q); Setting rotation to Transform message. tf_broadcaster.sendTransform(tf::StampedTransform(transform, ros::Time::now(), \"map\", \"robot\")); Publishing of transform is done with sendTransform method. We send TransformStamped, so instead of transform we need to pass time, child_frame and parent_frame. Build code Now in CMakeLists.txt file find line find_package(catkin REQUIRED COMPONENTS roscpp) and change it to find_package(catkin REQUIRED COMPONENTS roscpp tf) Declare executable and specify libraries to link. add_executable(tf_broadcaster src/tf_broadcaster.cpp)target_link_libraries(tf_broadcaster${catkin_LIBRARIES})\n\nRun code and try adding another frame using static_transform_publisher and see the result in RViz.\n\nSolution\n\nYou can use the same command as in the Task 1.\n\nrosrun tf static_transform_publisher 0 0 0.2 3.14 0 0 robot reverse_camera 100\n\n## Listening transformation\n\nAs was the case with sending, receiving TF data is also a bit different from the Subscriber you already know. A TransformListener is used to receive data, which will automatically determine the transformations between two frames, even if they are not directly connected to each other.\n\n~/ros_ws/src/tutorial_pkg/src/tf_listener.cpp\n#include <ros/ros.h>#include <tf/transform_listener.h>void listenTransformation(tf::TransformListener& tf_listener){ // Listen transformation tf::StampedTransform transform; tf_listener.lookupTransform(\"map\", \"robot\", ros::Time(0), transform); // Get rotation: roll, pitch, yaw tf::Quaternion q = transform.getRotation(); tf::Matrix3x3 m(q); double roll, pitch, yaw; m.getRPY(roll, pitch, yaw); // Calcualte number of rotation static int num_of_rotations; static float prev_yaw; if(yaw-prev_yaw < 0){ num_of_rotations++; ROS_INFO(\"Number of rotations: %d\", num_of_rotations); } prev_yaw = yaw;}int main(int argc, char **argv){ ros::init(argc, argv, \"tf_listener\"); ros::NodeHandle n(\"~\"); ros::Rate loop_rate(100); tf::TransformListener tf_listener; tf_listener.waitForTransform(\"map\", \"robot\", ros::Time(0), ros::Duration(2.0)); while (ros::ok()) { ros::spinOnce(); listenTransformation(tf_listener); loop_rate.sleep(); }}\n\nThe code explained\n\ntf::TransformListener tf_listener;tf_listener.waitForTransform(\"map\", \"robot\", ros::Time(0), ros::Duration(2.0));\n\nCreat listener and wait for 2 secounds to find latest transformation between two frames. Function returns false if it failed to find a transformation.\n\ntf::StampedTransform transform;tf_listener.lookupTransform(\"map\", \"robot\", ros::Time(0), transform);\n\nCreate a variable and writing the last received transformation to it. The third argument means time at which we want to transform. Providing ros::Time(0) will just get us the latest available transform.\n\ntf::Quaternion q = transform.getRotation();tf::Matrix3x3 m(q);double roll, pitch, yaw;m.getRPY(roll, pitch, yaw);\n\nReading quaternions and converting them to roll, pitch, yaw\n\nBuild code\n\nDeclare executable and specify libraries to link.\n\nadd_executable(tf_listener src/tf_listener.cpp)target_link_libraries(tf_listener \\${catkin_LIBRARIES})\n\nLook at the values of the yaw angle and change the code so that it determines the number of revolutions.\n\nSolution\n\nRepleace line ROS_INFO(\"Yaw: %f\", yaw); with below code and build your file.\n\n// Calcualte number of rotationstatic int num_of_revolutions;static float prev_yaw;if(yaw-prev_yaw < 0){ num_of_revolutions++; ROS_INFO(\"Number of revolutions: %d\", num_of_revolutions);}prev_yaw = yaw;\n\n## Summary\n\nIt's all for today. I hope that after this tutorial and the reworked exercises, handling and interpreting the tf package will not be a big problem for you. At this stage, you should perfectly understand the concept of transformation, what frames are, and you have the right tools for data visualization.\n\nby Łukasz Mitka, Rafał Górecki, Husarion\n\nDo you need any support with completing this tutorial or have any difficulties with software or hardware? Feel free to describe your thoughts on our community forum: https://community.husarion.com/ or to contact with our support: [email protected]"
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https://www.litscape.com/word_analysis/centurion | [
"# Definition of centurion\n\n## \"centurion\" in the noun sense\n\n### 1. centurion\n\nancient Rome) the leader of 100 soldiers\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\n# centurion in Scrabble®\n\nThe word centurion is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nCENTURION\n(108)\nCENTURION\n(108)\n\n## Seven Letter Word Alert: (10 words)\n\ncounter, intoner, neutron, noncute, recount, reunion, routine, trounce, unction, unicorn\n\nCENTURION\n(108)\nCENTURION\n(108)\nCENTURION\n(72)\nCENTURION\n(72)\nCENTURION\n(48)\nCENTURION\n(45)\nCENTURION\n(44)\nCENTURION\n(44)\nCENTURION\n(44)\nCENTURION\n(38)\nCENTURION\n(36)\nCENTURION\n(36)\nCENTURION\n(36)\nCENTURION\n(36)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(26)\nCENTURION\n(24)\nCENTURION\n(24)\nCENTURION\n(24)\nCENTURION\n(24)\nCENTURION\n(24)\nCENTURION\n(24)\nCENTURION\n(22)\nCENTURION\n(22)\nCENTURION\n(22)\nCENTURION\n(21)\nCENTURION\n(16)\nCENTURION\n(16)\nCENTURION\n(15)\nCENTURION\n(15)\nCENTURION\n(15)\nCENTURION\n(14)\nCENTURION\n(14)\nCENTURION\n(14)\nCENTURION\n(14)\nCENTURION\n(13)\n\nCENTURION\n(108)\nCENTURION\n(108)\nUNCTION\n(86 = 36 + 50)\nCOUNTER\n(86 = 36 + 50)\nNONCUTE\n(86 = 36 + 50)\nTROUNCE\n(86 = 36 + 50)\nUNICORN\n(86 = 36 + 50)\nUNICORN\n(86 = 36 + 50)\nUNICORN\n(86 = 36 + 50)\nRECOUNT\n(86 = 36 + 50)\nTROUNCE\n(86 = 36 + 50)\nNONCUTE\n(86 = 36 + 50)\nCOUNTER\n(86 = 36 + 50)\nUNCTION\n(86 = 36 + 50)\nRECOUNT\n(86 = 36 + 50)\nNONCUTE\n(86 = 36 + 50)\nRECOUNT\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nRECOUNT\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nUNICORN\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nUNCTION\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nTROUNCE\n(80 = 30 + 50)\nNONCUTE\n(80 = 30 + 50)\nCOUNTER\n(80 = 30 + 50)\nNEUTRON\n(78 = 28 + 50)\nREUNION\n(78 = 28 + 50)\nINTONER\n(78 = 28 + 50)\nROUTINE\n(78 = 28 + 50)\nTROUNCE\n(77 = 27 + 50)\nNONCUTE\n(77 = 27 + 50)\nRECOUNT\n(77 = 27 + 50)\nCOUNTER\n(77 = 27 + 50)\nUNICORN\n(77 = 27 + 50)\nUNCTION\n(77 = 27 + 50)\nRECOUNT\n(76 = 26 + 50)\nUNCTION\n(76 = 26 + 50)\nCOUNTER\n(76 = 26 + 50)\nINTONER\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nRECOUNT\n(74 = 24 + 50)\nUNCTION\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nCOUNTER\n(74 = 24 + 50)\nCOUNTER\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nTROUNCE\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nREUNION\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nINTONER\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nROUTINE\n(74 = 24 + 50)\nNEUTRON\n(74 = 24 + 50)\nUNICORN\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nTROUNCE\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nUNICORN\n(72 = 22 + 50)\nRECOUNT\n(72 = 22 + 50)\nRECOUNT\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nRECOUNT\n(72 = 22 + 50)\nUNCTION\n(72 = 22 + 50)\nTROUNCE\n(72 = 22 + 50)\nUNCTION\n(72 = 22 + 50)\nCOUNTER\n(72 = 22 + 50)\nCOUNTER\n(72 = 22 + 50)\nTROUNCE\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nTROUNCE\n(72 = 22 + 50)\nTROUNCE\n(72 = 22 + 50)\nUNICORN\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nUNCTION\n(72 = 22 + 50)\nNONCUTE\n(72 = 22 + 50)\nCENTURION\n(72)\nCOUNTER\n(72 = 22 + 50)\nUNICORN\n(72 = 22 + 50)\nCENTURION\n(72)\nCOUNTER\n(72 = 22 + 50)\nUNICORN\n(72 = 22 + 50)\nRECOUNT\n(72 = 22 + 50)\nCOUNTER\n(72 = 22 + 50)\nUNCTION\n(72 = 22 + 50)\nUNICORN\n(72 = 22 + 50)\nINTONER\n(71 = 21 + 50)\nNEUTRON\n(71 = 21 + 50)\nREUNION\n(71 = 21 + 50)\nROUTINE\n(71 = 21 + 50)\nCOUNTER\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nCOUNTER\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nCOUNTER\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nCOUNTER\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nCOUNTER\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nUNCTION\n(70 = 20 + 50)\nCOUNTER\n(70 = 20 + 50)\nUNICORN\n(70 = 20 + 50)\nNONCUTE\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nTROUNCE\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nRECOUNT\n(70 = 20 + 50)\nINTONER\n(68 = 18 + 50)\nNEUTRON\n(68 = 18 + 50)\nREUNION\n(68 = 18 + 50)\nNEUTRON\n(68 = 18 + 50)\nCOUNTER\n(68 = 18 + 50)\nRECOUNT\n(68 = 18 + 50)\nNEUTRON\n(68 = 18 + 50)\nRECOUNT\n(68 = 18 + 50)\nUNICORN\n(68 = 18 + 50)\nREUNION\n(68 = 18 + 50)\nROUTINE\n(68 = 18 + 50)\nCOUNTER\n(68 = 18 + 50)\n\n# centurion in Words With Friends™\n\nThe word centurion is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nCENTURION\n(171)\n\n## Seven Letter Word Alert: (10 words)\n\ncounter, intoner, neutron, noncute, recount, reunion, routine, trounce, unction, unicorn\n\nCENTURION\n(171)\nCENTURION\n(102)\nCENTURION\n(102)\nCENTURION\n(81)\nCENTURION\n(76)\nCENTURION\n(69)\nCENTURION\n(68)\nCENTURION\n(64)\nCENTURION\n(64)\nCENTURION\n(63)\nCENTURION\n(63)\nCENTURION\n(60)\nCENTURION\n(60)\nCENTURION\n(57)\nCENTURION\n(57)\nCENTURION\n(54)\nCENTURION\n(54)\nCENTURION\n(46)\nCENTURION\n(38)\nCENTURION\n(34)\nCENTURION\n(34)\nCENTURION\n(34)\nCENTURION\n(34)\nCENTURION\n(34)\nCENTURION\n(34)\nCENTURION\n(32)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(30)\nCENTURION\n(25)\nCENTURION\n(22)\nCENTURION\n(22)\nCENTURION\n(21)\nCENTURION\n(21)\nCENTURION\n(21)\nCENTURION\n(20)\nCENTURION\n(20)\nCENTURION\n(20)\nCENTURION\n(19)\nCENTURION\n(19)\nCENTURION\n(19)\nCENTURION\n(18)\nCENTURION\n(18)\nCENTURION\n(18)\nCENTURION\n(18)\nCENTURION\n(18)\nCENTURION\n(17)\nCENTURION\n(17)\nCENTURION\n(16)\nCENTURION\n(16)\n\nCENTURION\n(171)\nUNICORN\n(110 = 75 + 35)\nUNCTION\n(110 = 75 + 35)\nTROUNCE\n(107 = 72 + 35)\nCOUNTER\n(107 = 72 + 35)\nNONCUTE\n(104 = 69 + 35)\nUNICORN\n(104 = 69 + 35)\nNONCUTE\n(104 = 69 + 35)\nCENTURION\n(102)\nCENTURION\n(102)\nRECOUNT\n(101 = 66 + 35)\nUNICORN\n(98 = 63 + 35)\nUNICORN\n(98 = 63 + 35)\nNONCUTE\n(98 = 63 + 35)\nNONCUTE\n(98 = 63 + 35)\nNONCUTE\n(98 = 63 + 35)\nCOUNTER\n(95 = 60 + 35)\nTROUNCE\n(95 = 60 + 35)\nUNCTION\n(92 = 57 + 35)\nUNCTION\n(92 = 57 + 35)\nUNICORN\n(92 = 57 + 35)\nUNICORN\n(92 = 57 + 35)\nNONCUTE\n(92 = 57 + 35)\nTROUNCE\n(89 = 54 + 35)\nCOUNTER\n(89 = 54 + 35)\nTROUNCE\n(89 = 54 + 35)\nNEUTRON\n(89 = 54 + 35)\nCOUNTER\n(89 = 54 + 35)\nRECOUNT\n(89 = 54 + 35)\nRECOUNT\n(89 = 54 + 35)\nNONCUTE\n(87 = 52 + 35)\nUNCTION\n(87 = 52 + 35)\nUNICORN\n(87 = 52 + 35)\nNONCUTE\n(87 = 52 + 35)\nUNCTION\n(87 = 52 + 35)\nNONCUTE\n(87 = 52 + 35)\nUNICORN\n(87 = 52 + 35)\nUNCTION\n(87 = 52 + 35)\nUNICORN\n(87 = 52 + 35)\nUNCTION\n(86 = 51 + 35)\nUNCTION\n(86 = 51 + 35)\nUNCTION\n(86 = 51 + 35)\nUNCTION\n(86 = 51 + 35)\nUNICORN\n(86 = 51 + 35)\nUNICORN\n(86 = 51 + 35)\nUNICORN\n(86 = 51 + 35)\nNONCUTE\n(86 = 51 + 35)\nREUNION\n(83 = 48 + 35)\nRECOUNT\n(83 = 48 + 35)\nRECOUNT\n(83 = 48 + 35)\nRECOUNT\n(83 = 48 + 35)\nNEUTRON\n(83 = 48 + 35)\nRECOUNT\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nREUNION\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nRECOUNT\n(83 = 48 + 35)\nREUNION\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nCOUNTER\n(83 = 48 + 35)\nREUNION\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nTROUNCE\n(83 = 48 + 35)\nCENTURION\n(81)\nNONCUTE\n(80 = 45 + 35)\nNONCUTE\n(80 = 45 + 35)\nROUTINE\n(80 = 45 + 35)\nROUTINE\n(80 = 45 + 35)\nUNICORN\n(80 = 45 + 35)\nINTONER\n(80 = 45 + 35)\nINTONER\n(80 = 45 + 35)\nUNCTION\n(80 = 45 + 35)\nNONCUTE\n(80 = 45 + 35)\nUNCTION\n(80 = 45 + 35)\nUNCTION\n(80 = 45 + 35)\nREUNION\n(77 = 42 + 35)\nRECOUNT\n(77 = 42 + 35)\nREUNION\n(77 = 42 + 35)\nNEUTRON\n(77 = 42 + 35)\nREUNION\n(77 = 42 + 35)\nRECOUNT\n(77 = 42 + 35)\nTROUNCE\n(77 = 42 + 35)\nNEUTRON\n(77 = 42 + 35)\nUNCTION\n(77 = 42 + 35)\nTROUNCE\n(77 = 42 + 35)\nRECOUNT\n(77 = 42 + 35)\nRECOUNT\n(77 = 42 + 35)\nNEUTRON\n(77 = 42 + 35)\nCOUNTER\n(77 = 42 + 35)\nCOUNTER\n(77 = 42 + 35)\nNEUTRON\n(77 = 42 + 35)\nRECOUNT\n(77 = 42 + 35)\nCOUNTER\n(77 = 42 + 35)\nTROUNCE\n(77 = 42 + 35)\nCENTURION\n(76)\nNEUTRON\n(75 = 40 + 35)\nREUNION\n(75 = 40 + 35)\nREUNION\n(75 = 40 + 35)\nNEUTRON\n(75 = 40 + 35)\nRECOUNT\n(75 = 40 + 35)\nCOUNTER\n(75 = 40 + 35)\nREUNION\n(75 = 40 + 35)\nNEUTRON\n(75 = 40 + 35)\nTROUNCE\n(75 = 40 + 35)\nROUTINE\n(74 = 39 + 35)\nINTONER\n(74 = 39 + 35)\nROUTINE\n(74 = 39 + 35)\nINTONER\n(74 = 39 + 35)\nINTONER\n(74 = 39 + 35)\nROUTINE\n(74 = 39 + 35)\nROUTINE\n(71 = 36 + 35)\nNEUTRON\n(71 = 36 + 35)\nINTONER\n(71 = 36 + 35)\nREUNION\n(71 = 36 + 35)\nNEUTRON\n(71 = 36 + 35)\nNEUTRON\n(71 = 36 + 35)\nINTONER\n(71 = 36 + 35)\nREUNION\n(71 = 36 + 35)\nROUTINE\n(71 = 36 + 35)\nINTONER\n(71 = 36 + 35)\nROUTINE\n(71 = 36 + 35)\nNEUTRON\n(71 = 36 + 35)\nREUNION\n(71 = 36 + 35)\nUNICORN\n(69 = 34 + 35)\nUNICORN\n(69 = 34 + 35)\nUNCTION\n(69 = 34 + 35)\nNONCUTE\n(69 = 34 + 35)\nNONCUTE\n(69 = 34 + 35)\nUNCTION\n(69 = 34 + 35)\nCENTURION\n(69)\nNONCUTE\n(69 = 34 + 35)\nUNCTION\n(69 = 34 + 35)\nUNICORN\n(69 = 34 + 35)\nUNCTION\n(69 = 34 + 35)\nROUTINE\n(68 = 33 + 35)\nCENTURION\n(68)\nINTONER\n(68 = 33 + 35)\nROUTINE\n(68 = 33 + 35)\nINTONER\n(68 = 33 + 35)\nROUTINE\n(68 = 33 + 35)\nINTONER\n(68 = 33 + 35)\nINTONER\n(68 = 33 + 35)\nROUTINE\n(68 = 33 + 35)\nROUTINE\n(68 = 33 + 35)\nINTONER\n(68 = 33 + 35)\nRECOUNT\n(67 = 32 + 35)\nTROUNCE\n(67 = 32 + 35)\nTROUNCE\n(67 = 32 + 35)\nRECOUNT\n(67 = 32 + 35)\nCOUNTER\n(67 = 32 + 35)\nCOUNTER\n(67 = 32 + 35)\nRECOUNT\n(67 = 32 + 35)\nNONCUT\n(66)\nUNICORN\n(65 = 30 + 35)\nUNICORN\n(65 = 30 + 35)\nUNCTION\n(65 = 30 + 35)\nUNCTION\n(65 = 30 + 35)\nUNCTION\n(65 = 30 + 35)\nUNICORN\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nUNICORN\n(65 = 30 + 35)\nUNCTION\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nUNICORN\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nUNICORN\n(65 = 30 + 35)\nNONCUTE\n(65 = 30 + 35)\nUNCTION\n(65 = 30 + 35)\nCENTURION\n(64)\nCENTURION\n(64)\nTROUNCE\n(63 = 28 + 35)\nTROUNCE\n(63 = 28 + 35)\nUNICORN\n(63 = 28 + 35)\nTROUNCE\n(63 = 28 + 35)\nTROUNCE\n(63 = 28 + 35)\nUNICORN\n(63 = 28 + 35)\nTROUNCE\n(63 = 28 + 35)\nNEUTRON\n(63 = 28 + 35)\nCOUNTER\n(63 = 28 + 35)\nCOUNTER\n(63 = 28 + 35)\nCOUNTER\n(63 = 28 + 35)\nCOUNTER\n(63 = 28 + 35)\nCENTURION\n(63)\nCOUNTER\n(63 = 28 + 35)\nNONCUTE\n(63 = 28 + 35)\nNEUTRON\n(63 = 28 + 35)\nUNCTION\n(63 = 28 + 35)\nNONCUTE\n(63 = 28 + 35)\nNONCUTE\n(63 = 28 + 35)\nNEUTRON\n(63 = 28 + 35)\nUNCTION\n(63 = 28 + 35)\nRECOUNT\n(63 = 28 + 35)\n\nen cent\n\non ion\n\ncenturions"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72355956,"math_prob":1.0000098,"size":1227,"snap":"2020-34-2020-40","text_gpt3_token_len":341,"char_repetition_ratio":0.19296812,"word_repetition_ratio":0.67836255,"special_character_ratio":0.21434394,"punctuation_ratio":0.18055555,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999968,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-21T16:59:25Z\",\"WARC-Record-ID\":\"<urn:uuid:e8eccc14-6bdf-412d-8d28-4545de50a093>\",\"Content-Length\":\"162318\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:288c78a3-d24b-4aa0-9a48-af12277a1d2f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bf87464-a832-436a-9c0a-1abb56ce0ab3>\",\"WARC-IP-Address\":\"172.67.184.45\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/centurion\",\"WARC-Payload-Digest\":\"sha1:36RVXHDCOZ7VEEDOPTQGGUIAKQS5PYU7\",\"WARC-Block-Digest\":\"sha1:FSATO7WYCWL256UTD4VPEFMIGZGFYHAF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400201826.20_warc_CC-MAIN-20200921143722-20200921173722-00351.warc.gz\"}"} |
https://se.mathworks.com/matlabcentral/cody/problems/304-bottles-of-beer/solutions/3025726 | [
"Cody\n\n# Problem 304. Bottles of beer\n\nSolution 3025726\n\nSubmitted on 2 Oct 2020 by Benjamin Ye\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\nx = 99; y_correct = 98; assert(isequal(bottles_of_beer(x),y_correct))\n\n2 Pass\nx = 9; y_correct = 8; assert(isequal(bottles_of_beer(x),y_correct))\n\n3 Pass\nx = 1; y_correct = 0; assert(isequal(bottles_of_beer(x),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.513196,"math_prob":0.90990525,"size":475,"snap":"2020-45-2020-50","text_gpt3_token_len":152,"char_repetition_ratio":0.16135882,"word_repetition_ratio":0.0,"special_character_ratio":0.33052632,"punctuation_ratio":0.14117648,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96776325,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T22:58:16Z\",\"WARC-Record-ID\":\"<urn:uuid:92f7d6ef-db70-42d1-bbe4-bea106024739>\",\"Content-Length\":\"79812\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1c13741-d865-4dbc-884a-7baeed994dba>\",\"WARC-Concurrent-To\":\"<urn:uuid:1afb6ac1-755b-461a-baea-ca0f2362e976>\",\"WARC-IP-Address\":\"184.24.72.83\",\"WARC-Target-URI\":\"https://se.mathworks.com/matlabcentral/cody/problems/304-bottles-of-beer/solutions/3025726\",\"WARC-Payload-Digest\":\"sha1:T2TVT5LTGNUVFN7Q2UPUTG4FJOZU5YYR\",\"WARC-Block-Digest\":\"sha1:PNOK2VRMURK37PRTSJSQZOKXU2TCGUCA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141750841.83_warc_CC-MAIN-20201205211729-20201206001729-00485.warc.gz\"}"} |
https://www.ausetute.com.au/titrcurvwasb.html | [
"",
null,
"# Weak Acid - Strong Base Titration Curve Chemistry Tutorial\n\n## Key Concepts\n\n• It is possible to calculate the pH of a solution when a weak acid is titrated with a strong base:\n\n⚛ Before any strong base is added to weak acid :\n\n[H+(aq)] ≈ √Ka[weak acid]\n\npH = −log10[H+(aq)]\n\n⚛ Addition of strong base while weak acid is in excess :\n\nR.I.C.E. Table to determine equilibrium concentrations of anion and weak acid\n\n[OH-(aq)] = (Kb[anion])/[weak acid]\n\npOH = −log10[OH-(aq)]\n\npH = 14.00 − pOH (at 25°C)\n\n⚛ When all the weak acid has been neutralised by all the strong base we calculate the pH of the aqueous salt solution:\n\n[OH-(aq)] ≈ √Kb[anion])\n\npOH = −log10[OH-(aq)]\n\npH = 14.00 − pOH (at 25°C)\n\n⚛ When the strong base is in excess:\n\n[OH-(aq)] = n(OH- in excess) ÷ V(total)\n\npOH = −log10[OH-(aq)]\n\npH = 14.00 − pOH (at 25°C)\n\n• The titration curve for a weak acid - strong base titration has a characteristic shape in which the following features can be identified:\n\n⚛ Initial pH (pH < 7) : before any base is added the pH of the weak acid is dependent on:\n\n(a) concentration of weak acid\n\n(b) value of Ka (which is also temperature dependent)\n\nBuffer zone (buffer region) of titration curve :\n\n(a) addition of strong base while weak acid is in excess\n\n(b) weak acid is in equilibrium with its conjugate base\n\n⚛ Point at which pH = pKa :\n\n(a) occurs when half the weak acid has been neutralised by strong base\n\n(b) pH at the point V(base) = ½V(base added to neutralise acid)\n\n(c) also corresponds to the pH of the most effective buffer solution\n\n(a) just enough strong base has been added to exactly neutralise all the weak acid\n\n(b) pH determined by hydrolysis of anion and is dependent on:\n\n(i) concentration of anion\n\n(ii) Kb = Kw/Ka (which is temperature dependent)\n\n(ii) Kw (which is temperature dependent)\n\nNote: an appropriate indicator for the titration is one for which pKIn ± 1 = pH\n\nKIn is the dissocation constant for the indicator\n\npKIn = −log10KIn\n\npH is the pH at the equivalence point of the neutralisation reaction.\n\n⚛ After equivalence point (pH > 7): strong base is in excess and pH is dependent on:\n\n(a) concentration of excess base\n\n(b) Kw (which is temperature dependent)\n\n• We can draw a reasonable sketch of a weak acid - strong base titration using these 4 calculations:\n\n(1) initial pH of weak acid before base is added\n\n(2) pH of the salt solution at the equivalence point\n\n(3) pH of the most effective buffer solution (pH = pKa)\n\n(4) pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution).\n\nNo ads = no money for us = no free stuff for you!\n\n## Calculating a Titration Curve for a Weak Acid - Strong Base Titration\n\nAn aqueous solution of acetic acid (ethanoic acid), CH3COOH(aq), is an example of a weak acid.\n\nAn aqueous solution of sodium hydroxide, NaOH(aq), is an example of a strong base.\n\nImagine an experiment in which a 10.00 mL aliquot of 0.200 mol L-1 CH3COOH(aq) is transferred to a clean conical flask (erlenmeyer flask).\n\nA burette (buret) is then filled with 0.200 mol L-1 NaOH(aq)\n\nThe equipment would be set up as in the diagram below:\n\n ← NaOH(aq) ← CH3COOH(aq)\n\nThe NaOH(aq) is added 1.00 mL at a time to the CH3COOH.\nWe can calculate the pH of the solution in the conical flask after each addition of NaOH(aq).\nYou might like to visit the tutorial on calculating the pH of a solution after mixing weak acid and strong base before continuing with the following calculations.\n\nInitially, before we add any sodium hydroxide to the acetic acid (ethanoic acid), we have 10.00 mL of 0.200 mol L-1 CH3COOH(aq).\nCH3COOH(aq) is a weak acid so it exists in equilibrium with its ions; acetate ions (ethanoate ions), CH3COO-(aq), and hydrogen ions (protons), H+(aq), as shown in the balanced chemical equation below:\n\nCH3COOH(aq) ⇋ CH3COO-(aq) + H+(aq) Ka = 1.80×10-5 (at 25°C)\n\nUse a R.I.C.E. Table to determine the equilibrium concentration of H+(aq)\n\n Reaction Initial concentration(mol L-1) Change in concentration(mol L-1) Equilibrium concentration(mol L-1) CH3COOH(aq) ⇋ CH3COO-(aq) + H+(aq) 0.200 0 0 −x +x +x 0.200 − x ≈ 0.200 (Assume x << 0.200)(1) 0 + x = x 0 + x = x\n\nUse the expression for the equilibrium constant for the dissociation of acetic acid to calculate x, which is the concentration of hydrogen ions, [H+(aq)]:\n\n Ka = [CH3COO-(aq)][H+(aq)][CH3COOH(aq)] 1.80×10-5 = [x][x][0.200] √(0.200 × 1.80×10-5) = √[x]2 1.90×10-3 mol L-1 = [x] 1.90×10-3 mol L-1 = [H+(aq)]\n\nCalculate the pH of the solution before any base has been added:\n\npH = −log10[H+(aq)] = −log10[1.90×10-3] = 2.72\n\nSo, when 0.00 mL of NaOH(aq) has been added to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) the pH of the solution is 2.72.\n\nNow, let's add 1.00 mL of 0.200 mol L-1 from the burette to the 10.00 mL of 0.200 mol L-1 CH3COOH(aq) in the conical flask.\n\nA neutralisation reaction occurs between the acid and base according to the following balanced chemical equation:\n\n general word equation: acid + base → salt + water word equation: acetic acid(ethanoic acid) + sodium hydroxide → sodium acetate(sodium ethanoate) + water balanced chemical equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)\n\nUse this equation to calculate the moles of each substance consumed and produced:\n\n balanced chemical equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) initial concentration(mol L-1 0.200 0.200 0 initial volume(L) 10.00×10-3 1.00×10-3 0 moles n = c × V 2.00×10-3 2.00×10-4 0 mole ratio (stoichiometric ratio) 1 : 1 : 1 : 1 CH3COOH is in excess, NaOH is the limiting reagent. moles AFTER reaction: = n(initial) − n(reacted) 2.00×10-3 − 2.00×10-4 = 1.80×10-3 2.00×10-4 − 2.00×10-4 = 0 2.00×10-4 volume AFTER reaction: = V(acid) + V(base added) 10.00 + 1.00 = 11.00 mL (11.00×10-3 L) 10.00 + 1.00 = 11.00 mL (11.00×10-3 L) 10.00 + 1.00 = 11.00 mL (11.00×10-3 L) concentration AFTER reaction:c = n ÷ V (mol L-1) 0.1634 0 0.01818\n\nThe acetate ion (ethanoate ion), CH3COO-(aq), undergoes hydrolysis (reacts with water) according to the following equation:\n\nCH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)\n\nUse a R.I.C.E. Table to determine the equiliubrium concentrations of each species:\n\n Reaction Initial concentration(mol L-1) Change in concentration(mol L-1) Equiliubrium concentration(mol L-1) CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq) 0.01818 0.1632 0 −x +x +x 0.01818 −x ≈ 0.01818 0.1632 +x ≈ 0.1632 x (Assume x << 0.01818 and therefore x << 0.1632)\n\nand we can write an expression for the hydrolysis equilibrium constant, Kh, (which is equivalent to the base dissociation constant, Kb) and calculate the value of Kb (since Kb = Kw ÷ Ka):\n\n Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] = 1.00×10-14Ka = 1.00×10-141.8×10-5 = 5.56×10-10\n\nand use the equilibrium concentrations of each species to determine [OH-(aq)]:\n\n [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] = 5.56×10-10 [OH-(aq)] = 5.56×10-10[CH3COO-(aq)] [CH3COOH(aq)] [OH-(aq)] = 5.56×10-10[0.01818] [0.1632] [OH-(aq)] = 6.19×10-11\n\nwhich we can use to calculate the pOH :\n\npOH = −log10[OH-(aq)] = −log10[6.19×10-11] = 10.21\n\nand therefore the pH (at 25°C):\n\npH = 14.00 − pOH = 14.00 − 10.21 = 3.79\n\nWe could continue these calculations for the addition of 2.00 mL, 3.00 mL, 4.00 mL, ... up to 9.00 mL of NaOH(aq).\nBut once we add 10.00 mL of 0.200 mol L-1 NaOH(aq) to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) the acid will no longer be in excess, and the base is no longer the limiting reagent, we have reached the equivalence point for this neutralisation reaction.\nThe only substance in solution at the equivalence point is an aqueous solution of sodium acetate, so we need to calculate the pH of this salt solution.\n\nThe salt, sodium acetate (sodium ethanoate), CH3COONa(aq) is soluble in water (all acetates are soluble); it fully dissociates into acetate ions (ethanoate ions), CH3COO-(aq), and sodium ions, Na+(aq), as shown by the chemical equation below:\n\nCH3COONa(aq) → CH3COO-(aq) + Na+(aq)\n\nThe sodium ion, Na+(aq), does not undergo hydrolysis (does not react with water), but the acetate ions, CH3COO-(aq), will undergo hydrolysis (will react with water) according to the chemical equation below:\n\nCH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)\n\nIn Arrhenius terms this is a hydrolysis reaction, in Brønsted-Lowry terms this is a proton transfer reaction in which the acetate ions, CH3COO-, are acting as a base with the conjugate acid being acetic acid, CH3COOH.\n\nThe expression for the equilibrium constant for this hydrolysis reaction is given the symbol Kh, and equally it can also be given the symbol Kb :\n\n Kh = Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)]\n\nand since Kb(acetate ions) = Kw ÷ Ka(acetic acid)\n\nat 25°C, Kb(acetate ions) = (1.00×10-14) ÷ (1.80×10-5) = 5.56×10-10\n\nCalculate the intial concentration of acetate ions using the neutralisation reaction between acetic acid and sodium hydroxide:\n\n balanced chemical equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) concentration before reaction (mol L-1) 0.200 0.200 0 volume before reaction (L) 10.00×10-3 10.00×10-3 0 moles before reaction (mol) 2.00×10-3 2.00×10-3 0 mole ratio (stoichiometric ratio) 1 : 1 : 1 : 1 n(acid) = n(base) : equivalence point for neutralisation reaction moles aftere reaction (mol) 2.00×10-3 − 2.00×10-3 = 0 2.00×10-3 − 2.00×10-3 = 0 0 + 2.00×10-3 = 2.00×10-3 volume after reaction = V(acid) + V(base) (L) 10.00×10-3 + 10.00×10-3 = 20.00×10-3 concentration after reaction (c = n ÷ V)(mol L-1) 0 0 0.100\n\nUse a R.I.C.E. Table to determine the equilibrium concentrations of all species as a result of the hydrolysis of acetate ions:\n\n Reaction Initial concentration(mol L-1) Change in concentration(mol L-1) Equilibrium concentration(mol L-1) CH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq) 0.100 0 0 −x +x +x 0.100 −x ≈ 0.100 (Assume x << 0.100) x x\n\nUse Kb and equilibrium concentrations to calculate concentration of hydroxide ions, [OH-(aq)], which is equal to x:\n\n Kb = [CH3COOH(aq)][OH-(aq)][CH3COO-(aq)] 5.56×10-10 = [x][x][0.100] √(0.100 × 5.56×10-10) = √[x]2 7.46×10-6 mol L-1 = [x] = [OH-(aq)]\n\nCalculate the pOH of this solution:\n\npOH = −log10[OH-(aq)] = −log10[7.46×10-6] = 5.13\n\nCalculate the pH of this solution (at 25°C):\n\npH = 14.00 − pOH = 14.00 − 5.13 = 8.87\n\nThe pH of this titration experiment at the equivalence point is 8.87\n\nAfter the equivalence point, further additions of NaOH(aq) will produce larger excess moles of NaOH(aq) but in increasing volumes of solution.\n\nFor example, if 11.00 mL of 0.200 mol L-1 NaOH(aq) is added to 10.00 mL of 0.200 mol L-1 CH3COOH(aq) :\n\n balanced chemical equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) concentration before reaction (mol L-1 0.200 0.200 0 volume before reaction (L) 10.00×10-3 11.00×10-3 0 moles before reaction(mol) 2.00×10-3 2.20×10-3 0 mole ratio(stoichiometric ratio) 1 : 1 : 1 : 1 NaOH is in excess, CH3COOH is the limiting reagent moles after reaction (mol) 2.00×10-3 − 2.00×10-3 = 0 2.20×10-3 − 2.00×10-3 = 2.00×10-4 2.00×10-3 volume after reaction =V(acid) + V(base)(L) 21.00×10-3 21.00×10-3 21.00×10-3 concentration after reaction =moles + volume(mol L-1) 0 9.52×10-3 9.52×10-2\n\nBecause the value of the equilibrium constant for the hydrolysis of acetate ions is very small, Kh = Kb = 5.56×10-10, we will ignore its contribution to the hydroxide ion concentration, and only consider the complete dissociation of sodium hyroxide to produce hydroxide ions and sodium ions as shown in the chemical equation below:\n\nNaOH(aq) → Na+(aq) + OH-(aq)\n\nSo, [OH-(aq)] = [NaOH(aq)(after reaction)] = 9.52×10-3 mol L-1\n\nCalculate the pOH of this solution:\n\npOH = −log10[OH-(aq)] = −log10[9.52×10-3] = 2.02\n\nCalculate the pH of this solution (at 25°C):\n\npH = 14.00 − pOH = 14.00 − 2.02 = 11.98\n\nWe could continue these calculations for the addition of 12.00 mL of 0.200 mol L-1 NaOH(aq), 13.00 mL, etc.\n\nIf you do this, you would end up with the pH values listed in the table below:\n\n CH3COOH(aq) in excess NaOH(aq) in excess Volume NaOH(aq) (mL) pH acidic solution basic solution equivalencepoint 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 13.00 14.00 2.72 3.78 4.13 4.37 4.56 4.74 4.92 5.11 5.34 5.70 8.87 11.98 12.26 12.42 12.52\n\nPlotting the points on a graph using the table above will result in a curve as shown below:\n\n pH of solution 0.200 M CH3COOH - 0.200 NaOH Titration Curve volume of NaOH(aq) added (mL)\n\nDo you know this?\n\nPlay the game now!\n\n## Features of a Weak Acid - Strong Base Titration Curve\n\nLet's consider the various important features of the acetic acid - sodium hydroxide titration curve we plotted above.\nWe have used different colours on the curve to differentiate the regions of the titration curve we will be discussing:\n\n pH of solution 0.200 M CH3COOH - 0.200 NaOH Titration Curve volume of NaOH(aq) added (mL)\n\nFirst, consider the gold line on the titration curve.\nWhen 0.00 mL of NaOH(aq) has been added to 0.200 mol L-1 CH3COOH(aq) the pH of the solution is due to the dissociation of this weak acid.\nThe pH is less than 7.00 (at 25°C) because this solution is acidic.\nAt 25°C the actual pH is determined by the concentration of the actic acid (and the value of its acid dissociation constant, Ka).\n\n [CH3COOH(aq)] (mol L-1) pH lowerconcentration → → higherconcentration 0.001 0.010 0.100 1.000 3.87 3.37 2.87 2.37 higher pH ← ← lower pH\n\nAs the concentration of dilute acetic acid, CH3COOH(aq), increases the pH of the solution decreases because there will be a greater concentration of hydrogen ions, H+(aq), in solution.\n\nNext, consider the blue line of the titration curve.\nIn Arrhenius terms, this is the region in which the acetic acid is in excess, not enough sodium hydroxide has been added to \"neutralise\" all the acid.\nIn Brønsted-Lowry terms, we have a weak acid (CH3COOH(aq)) in equilibrium with its conjugate base (CH3COO-(aq)):\n\n acid + base ⇋ conjugate base + conjugate acid CH3COOH(aq) + H2O(l) ⇋ CH3COO-(aq) + H3O+(aq)\n\nwhich means that, for as long as the acetic acid (CH3COOH(aq)) is in excess (that is, NaOH(aq) is the limiting reagent), we have established a buffer solution.\nAdding more sodium hydroxide solution to this excess weak acid initially decreases the concentration of acetic acid, so we expect the pH to increase as the concentration of hydrogen ions due to the dissociation of the acid will also decrease.\nBut the addition of sodium hydroxide also initially increases the concentration of acetate ions, some of which will react to reform some undissociated acetic acid molecules, thereby increasing the concentration of acetic acid molecules by a bit more.\nSo a new equilibrium position is established in which :\n\n• [CH3COOH(aq)] is less than before the addition of NaOH(aq) but not as small as predicted by the mole ratio (stoichiometric ratio) of the neutralisation reaction alone.\n• [CH3COO-(aq)] is greater than before the addition of NaOH(aq) but not as large as predicted by the mole ratio (stroichiometric ratio) of the neutralisation reaction alone.\n\nSince the acid dissociation constant for acetic acid is given by the following expression:\n\n Ka = [CH3COO-(aq)][H+(aq)][CH3COOH(aq)]\n\nWe can rearrange this expression as shown below to determine the concentration of hydrogen ions in this buffer solution (and hence the pH) :\n\n [H+(aq)] = Ka× [CH3COOH(aq)][CH3COO-(aq)]\n\nNote that [H+(aq)], and hence the pH of the solution, is determined by the ratio of [CH3COOH(aq)] to [CH3COO-(aq)].\n\nAs the [CH3COOH(aq)] decreases by a bit, and the [CH3COO-(aq)] increases by a bit, the [H+(aq)] decreases by a bit so the pH increases by a bit.\n\nBecause this region of the titration curve relates to the formation of a buffer (a weak acid in equilibrium with its conjugate base), it is referred to as the \"buffer zone\" or the \"buffer region\".\n\nLet's consider the green cross on the dark red line, x, on the titration curve above.\nThis cross represents the equivalence point for this neutralisation reaction: just enough NaOH(aq) has been added to neutralise all the available CH3COOH(aq), neither reactant is in excess, neither reactant is the limiting reagent.\nThis means that the solution at the equivalence point is made up of the salt named sodium acetate (sodium ethanoate), CH3COONa, dissolved in water.\nThe pH of this solution at the equivalence point is due to the hydrolysis of the acetate ion (ethanoate ion), CH3COO-(aq) which releases hydroxide ions, OH-(aq), so the solution is basic:\n\nCH3COO-(aq) + H2O(l) ⇋ CH3COOH(aq) + OH-(aq)\n\nThe pH can be calculate as follows:\n\n [OH-] = Kb[CH3COO-(aq)][CH3COOH(aq)]\n\npOH = −log10[OH-(aq)]\n\npH = 14.00 − pOH (at 25°C)\n\nThe greater the concentation of acetic acid used in the experiment, the greater the concentration of acetate ions produced at the equivalence point, so the greater the hydroxide ion concentration.\nHigher [OH-(aq)] means a lower pOH and therefore a higher pH.\n\n [CH3COONa(aq)] pH lowerconcentration → → higherconcentration 0.001 0.010 0.100 1.000 7.87 8.37 8.87 9.37 lower pH → → higher pH\n\nNow, we choose an appropriate indicator for this titration experiment based on the predicted pH at the equivalence point for the neutralisation reaction.\nOur chosen indicator must change colour at the equivalence point for the neutralisation reaction, put another way, the end point as indicated by the indicator must occur in the same pH region as the pH at the equivalence point for the neutralisation reaction.\nSo, for 0.001 mol L-1 CH3COONa(aq), the pH at the equivalence point is 7.87, so we need an indicator that changes colour around pH = 7.87.\nIf we have 1.00 mol L-1 CH3COONa(aq), the pH at the equivalence point is 9.37, so we need an indicator that will change colour around pH = 9.37.\n\nAn indicator is just a weak acid (or weak base) for which the acidic form (HIn) is a different colour to the basic form (In):\n\n HIn + H2O ⇋ In- + H3O+\n\nKIn is the value of the dissociation constant of an indicator:\n\n KIn = [H+][In-][HIn] or KIn = [H3O+][In-][HIn]\n\nAt the end point, [HIn] = [In-], so\n\nKIn = [H+(aq)]\n\nand so,\n\n−log10KIn = −log10[H+(aq)]\n\npKIn = pH\n\nSince we choose an appropriate indicator for a titration as one which changes colour at the equivalence point for the neutralisation reaction, we can say that the best indicator will have an end point at the same pH as the reaction's equivalence point, therefore\n\npKIn = pH = equivalence point for the neutralisation reaction\n\nSo, given a list of possible indicators such as the one below (in which the effective range is assumed to be ± 1):\n\nindicator name pKIn effective pH range\nbromothymol blue 7.1 6.1 to 8.1\nphenolphthalein 9.4 8.4 to 10.4\nthymolphthalein 10.0 9.0 to 11.0\nalizarin yellow R 11.2 10.2 to 12.2\n\nwe could choose a suitable indicator for each of our CH3COONa(aq) solutions resulting from the titration of acetic acid with sodium hydroxide:\n\n [CH3COONa(aq)] pH (equivalence point) pKIn (indicator) suitable indicator 0.001 0.010 0.100 1.000 7.9 8.4 8.9 9.4 7.9 8.4 8.9 9.4 bromothymol blue (pKIn = 7.1 ± 1) phenolphthalein (pKIn = 9.4 ± 1) phenolphthalein (pKIn = 9.4 ± 1) phenolphthalein (pKIn = 9.4 ± 1) or thymolphthalein (pKIn = 10.0 ± 1)\n\nFinally, we consider the indigo line on the titration curve above.\nIn this region of the titration curve there is an excess of sodium hydroxide solution, and because sodium hydroxide is a strong base that completely dissociates in water, we can ignore the contribution to the hydroxide ion concentration contributed by the hydrolysis of acetate ions (Kb is very small!) and we can ignore the contribution to the hydroxide ion concentration as a result of the dissociation of water molecules (Kw is also very small).\nAs we add more NaOH(aq) past the equivalence point for the neutralisation reaction, the moles of OH-(aq) in this added volume of solution are all in excess, so the pH of the solution increases, but as we keep adding more and more NaOH(aq) the line levels off as it approaches the pH of the NaOH(aq) being added (for 0.200 mol L-1 NaOH(aq) pH = 13.3).\n\nThere is one more region of interest on this titration curve that we have not yet considered.\nWhen half the acetic acid has been neutralised then the concentration of the excess acetic acid is the same as the concentration of acetate ions in solution, and this leads to an interesting position:\n\n Ka = [CH3COO-(aq)][H+(aq)][CH3COOH(aq)]\n\nwhen [CH3COO-(aq)] = [CH3COOH(aq)] then\n\n Ka = [H+(aq)]\n\nand if we take the negative log of both sides, then\n\npKa = pH\n\nSo, a titration curve can be used to determine the pKa (an hence Ka) of a weak acid.\n\nFor the titration of our 10.00 mL of 0.200 mol L-1 CH3COOH(aq) with 0.200 mol L-1 NaOH(aq), the equivalence point occurred when 10.00 mL of NaOH(aq) had been added.\nLet's add a blue line to the titration curve to represent the addition of 5.00 mL of NaOH(aq), and then read off the pH (see the dark red line ):\n\n pH of solution 0.200 M CH3COOH - 0.200 NaOH Titration Curve volume of NaOH(aq) added (mL)\n\nHalf the acetic acid has been neutralised when 5.00 mL of sodium hydroxide has been added, and the pH = 4.74\n\nTherefore the value for pKa for acetic acid is 4.74\n\nWe can therefore calculate the value for the acid dissociation constant, Ka, for acetic acid:\n\nKa = 10-pKa = 10-4.74 = 1.8×10-5\n\nNote that this pH is also the pH of the most effective buffer solution, when [CH3COO-(aq)] = [CH3COOH(aq)]\n\nDo you understand this?\n\nTake the test now!\n\n## How to Sketch a Weak Acid - Strong Base Titration Curve\n\nIf you need to sketch a rough titration curve for a weak acid - strong base titration, you will need to perform 4 calculations to locate 4 key features of the titration curve:\n\n(1) initial pH of weak acid before base is added\n\n(2) pH of the salt solution at the equivalence point\n\n(3) pH of the most effective buffer solution (pH = pKa)\n\n(4) pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution).\n\nFormic acid (methanoic acid), HCOOH(aq), is a weak acid (Ka = 1.80×10-4)\n\nSodium hydroxide, NaOH(aq), is a strong base.\n\nLet's sketch a curve for the titration in which 0.100 mol L-1 NaOH(aq) is added to a conical flask containing 20.00 mL of 0.200 mol L-1 HCOOH(aq).\n\nStep 1: initial pH of weak acid before base is added\n\ndissociation of formic acid: HCOOH(aq) → H+(aq) + HCOO-(aq) Ka = 1.80×10-4\n\nassume very little formic acid dissociates so [HCOOH(initial)] ≈ [HCOOH(equilibrium)], then\n\nKa = [H+(aq)]2/[HCOOH(aq)] = 1.80×10-4\n\n[H+(aq)] = √(1.80×10-4 × 0.200) = 6.00×10-3 mol L-1\n\npH = −log10[H+(aq)] = −log10[6.00×10-3] = 2.22\n\nStep 2: pH of the salt solution at the equivalence point\n\nneutralisation reaction: HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)\n\nn(HCOOH available) = n(NaOH added) = n(HCOONa(aq) produced)\n\nn(HCOOH available) = c × V = 0.200 × 20.00/1000 = 4.00 × 10-3 mol\n\nn(HCOONa) = 4.00 × 10-3 mol\n\nn(NaOH added) = 4.00 × 10-3 mol\n\nV(NaOH added) = n(NaOH added)/[NaOH] = (4.00 × 10-3)/0.100 = 0.0400 L\n\nV(total) = V(HCOOH) + V(NaOH) = 0.0200 + 0.0400 = 0.0600 L\n\n[HCOONa] = n(HCOONa)/V(total) = (4.00 × 10-3)/0.0600 = 0.067 mol L-1\n\nHydrolysis of HCOO- (Na+ does not hydrolyse)\n\nHCOO-(aq) + H2O(l) ⇋ HCOOH(aq) + OH-(aq) Kb = Kw/Ka = (1.00×10-14)/(1.80×10-4) = 5.56×10-11 (25°C)\n\nKb is small so very little HCOO-(aq) hydrolyses:\n\n[OH-(aq)] = √Kb[HCOO-(aq)] = √(5.56×10-11 × 0.067) = 1.93×10-6 mol L-1\n\npOH = −log10[OH-(aq)] = −log10[1.93×10-6] = 5.71\n\npH = 14.00 − pOH = 14.00 − 5.71 = 8.29\n\nStep 3: pH of the most effective buffer solution (pH = pKa)\n\npH = pKa = −log10Ka = −log10(1.80×10-4) = 3.74\n\nwhich occurs when half the acid has been neutralised, so the volume of NaOH(aq) added = half the volume at neutralisation:\n\nV(NaOH(aq)added) = ½ × 0.0400 L = 0.0200 L\n\nStep 4: pH of the solution when strong base is in excess (approximately approaches pH of the strong base solution)\n\nAt the equivalence point, 0.0400 L (40.00 mL) of NaOH(aq) had been added, if we add 1 more mL (0.001 L) then:\n\nn(NaOH in excess) = c × V(excess) = 0.100 × 0.001 = 1.00 × 10-4 mol\n\nV(total) = V(acid) + V(all base) = 0.0200 + 0.041 = 0.061 L\n\n[NaOH final] = n(NaOH in excess)/V(total) = (1.00 × 10-4)/0.061 = 1.64×10-3 mol L-1\n\n[NaOH final] = [OH-(aq)] = 1.64×10-3 mol L-1\n\npOH = −log10[OH-(aq)] = −log10[1.64×10-3] = 2.79\n\npH = 14.00 − pOH = 14.00 − 2.79 = 11.2\n\nNote that the line will have a gentle slope of increasing pH that cannot exceed 14 −log[0.100] = 13\n\nUsing these 4 points we could sketch a rough titration curve as shown below:\n\n pH of solution Sketch of Titration Curve volume of NaOH(aq) added (mL)\n\nCan you apply this?\n\nJoin AUS-e-TUTE!\n\nTake the exam now!\n\nFootnotes:\n\n(1) If we don't assume that x is negligible then we must solve the quadratic equation to determine the value of x\n\n Ka = [CH3COO-][H+][CH3COOH] 1.8 × 10-5 = [x][x][0.200 − x] 1.8 × 10-5 = x2[0.200 − x] (1.8 × 10-5) × [0.200 − x] = x2 (3.60 × 10-6) − (1.8 × 10-5)x = x2 0 = x2 + (1.8 × 10-5)x − (3.60 × 10-6)\n\n x = -b ± √(b2 − 4ac)2a x = -(1.8 × 10-5) ± √([1.8 × 10-5]2 − (4 × 1 × −3.60 × 10-6)2 × 1 x = -(1.8 × 10-5) ± √(1.44 × 10-5)2 × 1 x must be positive so x = -(1.8 × 10-5) + √(1.44 × 10-5)2 × 1 x = -(1.8 × 10-5) + (3.79 × 10-3)2 × 1 x = 1.89 × 10-3\n\n[CH3COOH] = 0.200 − (1.89 × 10-3) = 0.198 mol L-1 (compared to 0.200 mol L-1 using the assumption that x is negligible).\npH = −log10[H+] = −log10[1.89 × 10-3] = 2.72 (compared to pH = 2.72 using the assumption that x is negligible)."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81022406,"math_prob":0.9986402,"size":26144,"snap":"2020-34-2020-40","text_gpt3_token_len":9004,"char_repetition_ratio":0.17046672,"word_repetition_ratio":0.21124715,"special_character_ratio":0.37565026,"punctuation_ratio":0.11668484,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99857974,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T19:17:15Z\",\"WARC-Record-ID\":\"<urn:uuid:cfc07766-20f6-48b8-933f-7662bb21af3f>\",\"Content-Length\":\"83705\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91487bd2-3aa8-4139-8928-f6a9f57ac7c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:77d447d1-b9c6-4737-835c-3136fc0a56e3>\",\"WARC-IP-Address\":\"116.0.20.212\",\"WARC-Target-URI\":\"https://www.ausetute.com.au/titrcurvwasb.html\",\"WARC-Payload-Digest\":\"sha1:74JBO26MYEPQAUJNMAZK4H4KHSKIB4VY\",\"WARC-Block-Digest\":\"sha1:ZHY2VM4X2S2WNQ7OQT4VVRTK3Q34YHRC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402088830.87_warc_CC-MAIN-20200929190110-20200929220110-00184.warc.gz\"}"} |
https://numbermatics.com/n/74143/ | [
"# 74143\n\n## 74,143 is a prime number. Like all primes greater than two, it is odd and has no factors apart from itself and one.\n\nWhat does the number 74143 look like?\n\nAs a prime, it is not composed of any other numbers and has no internal structure.\n\n74143 is a prime number. Like all primes (except two), it is an odd number.\n\n## Prime factorization of 74143:\n\n### 74143\n\nSee below for interesting mathematical facts about the number 74143 from the Numbermatics database.\n\n### Names of 74143\n\n• Cardinal: 74143 can be written as Seventy-four thousand, one hundred forty-three.\n\n### Scientific notation\n\n• Scientific notation: 7.4143 × 104\n\n### Factors of 74143\n\n• Number of distinct prime factors ω(n): 1\n• Total number of prime factors Ω(n): 1\n• Sum of prime factors: 74143\n\n### Divisors of 74143\n\n• Number of divisors d(n): 2\n• Complete list of divisors:\n• Sum of all divisors σ(n): 74144\n• Sum of proper divisors (its aliquot sum) s(n): 1\n• 74143 is a deficient number, because the sum of its proper divisors (1) is less than itself. Its deficiency is 74142\n\n### Bases of 74143\n\n• Binary: 100100001100111112\n• Base-36: 1L7J\n\n### Squares and roots of 74143\n\n• 74143 squared (741432) is 5497184449\n• 74143 cubed (741433) is 407577746602207\n• The square root of 74143 is 272.2921225449\n• The cube root of 74143 is 42.0103904751\n\n### Scales and comparisons\n\nHow big is 74143?\n• 74,143 seconds is equal to 20 hours, 35 minutes, 43 seconds.\n• To count from 1 to 74,143 would take you about twenty hours.\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 74143 cubic inches would be around 3.5 feet tall.\n\n### Recreational maths with 74143\n\n• 74143 backwards is 34147\n• The number of decimal digits it has is: 5\n• The sum of 74143's digits is 19\n• More coming soon!\n\nMLA style:\n\"Number 74143 - Facts about the integer\". Numbermatics.com. 2023. Web. 29 March 2023.\n\nAPA style:\nNumbermatics. (2023). Number 74143 - Facts about the integer. Retrieved 29 March 2023, from https://numbermatics.com/n/74143/\n\nChicago style:\nNumbermatics. 2023. \"Number 74143 - Facts about the integer\". https://numbermatics.com/n/74143/\n\nThe information we have on file for 74143 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 74143, math, Factors of 74143, curriculum, school, college, exams, university, Prime factorization of 74143, STEM, science, technology, engineering, physics, economics, calculator, seventy-four thousand, one hundred forty-three.\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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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8565525,"math_prob":0.8915575,"size":2642,"snap":"2023-14-2023-23","text_gpt3_token_len":706,"char_repetition_ratio":0.12092494,"word_repetition_ratio":0.04028436,"special_character_ratio":0.31907645,"punctuation_ratio":0.16923077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9765985,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-29T11:00:31Z\",\"WARC-Record-ID\":\"<urn:uuid:b5ae5d38-e334-42d5-b997-cd8a98724392>\",\"Content-Length\":\"16856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f049e49-8c60-4064-936d-441f134ce4e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:60a8d14d-b4cf-44ec-8561-f7179164d84f>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/74143/\",\"WARC-Payload-Digest\":\"sha1:L2RFI7BKPF4SJLJSH3HWHX54Q3VHBI5I\",\"WARC-Block-Digest\":\"sha1:Q4RVRE4OHVQG4RWIURYR4Q6T6BCISW3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948965.80_warc_CC-MAIN-20230329085436-20230329115436-00583.warc.gz\"}"} |
https://ww2.mathworks.cn/matlabcentral/cody/problems/493-quasi-newton-method-for-unconstrained-minimization-using-bfgs-update | [
"# Problem 493. Quasi-Newton Method for Unconstrained Minimization using BFGS Update\n\nWrite a function to find the values of a design variable vector, x, that minimizes an unconstrained scalar objective function, f, given a function handle to f and its gradient, a starting guess, x0, a gradient tolerance, TolGrad, and a maximum number of iterations, MaxIter, using the Quasi-Newton (Secant) Method. Initialize the Hessian approximation as an identity matrix. Update the Hessian matrix approximation using the BFGS update formula.\n\n### Solution Stats\n\n31.87% Correct | 68.13% Incorrect\nLast Solution submitted on Jun 09, 2023\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73684764,"math_prob":0.85984564,"size":1059,"snap":"2023-40-2023-50","text_gpt3_token_len":242,"char_repetition_ratio":0.116587676,"word_repetition_ratio":0.0,"special_character_ratio":0.20585458,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97836185,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T06:24:24Z\",\"WARC-Record-ID\":\"<urn:uuid:b6fc1686-d817-43b7-9b8f-01224398ab53>\",\"Content-Length\":\"94825\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2d74de8-190c-41ec-a5ed-7ddebc009177>\",\"WARC-Concurrent-To\":\"<urn:uuid:20f34f38-2bc0-4b46-b227-31381d15a9b3>\",\"WARC-IP-Address\":\"23.221.59.235\",\"WARC-Target-URI\":\"https://ww2.mathworks.cn/matlabcentral/cody/problems/493-quasi-newton-method-for-unconstrained-minimization-using-bfgs-update\",\"WARC-Payload-Digest\":\"sha1:BXUMDXCJKTXXOKT2SKLX67QR5K5FXWUP\",\"WARC-Block-Digest\":\"sha1:HZ6K66PS5IDZLLCQOCIICBCBXACU6YNX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100276.12_warc_CC-MAIN-20231201053039-20231201083039-00265.warc.gz\"}"} |
https://www.rdocumentation.org/packages/spatstat/versions/1.64-1/topics/harmonic | [
"# harmonic\n\n0th\n\nPercentile\n\n##### Basis for Harmonic Functions\n\nEvaluates a basis for the harmonic polynomials in $x$ and $y$ of degree less than or equal to $n$.\n\nKeywords\nmodels, spatial\n##### Usage\nharmonic(x, y, n)\n##### Arguments\nx\n\nVector of $x$ coordinates\n\ny\n\nVector of $y$ coordinates\n\nn\n\nMaximum degree of polynomial\n\n##### Details\n\nThis function computes a basis for the harmonic polynomials in two variables $x$ and $y$ up to a given degree $n$ and evaluates them at given $x,y$ locations. It can be used in model formulas (for example in the model-fitting functions lm,glm,gam and ppm) to specify a linear predictor which is a harmonic function.\n\nA function $f(x,y)$ is harmonic if $$\\frac{\\partial^2}{\\partial x^2} f + \\frac{\\partial^2}{\\partial y^2}f = 0.$$ The harmonic polynomials of degree less than or equal to $n$ have a basis consisting of $2 n$ functions.\n\nThis function was implemented on a suggestion of P. McCullagh for fitting nonstationary spatial trend to point process models.\n\n##### Value\n\nA data frame with 2 * n columns giving the values of the basis functions at the coordinates. Each column is labelled by an algebraic expression for the corresponding basis function.\n\nppm, polynom\n\n• harmonic\n##### Examples\n# NOT RUN {\n# inhomogeneous point pattern\nX <- unmark(longleaf)\n\n# }\n# NOT RUN {\n# fit Poisson point process with log-cubic intensity\nfit.3 <- ppm(X ~ polynom(x,y,3), Poisson())\n\n# fit Poisson process with log-cubic-harmonic intensity\nfit.h <- ppm(X ~ harmonic(x,y,3), Poisson())\n\n# Likelihood ratio test\nlrts <- 2 * (logLik(fit.3) - logLik(fit.h))\ndf <- with(coords(X),\nncol(polynom(x,y,3)) - ncol(harmonic(x,y,3)))\npval <- 1 - pchisq(lrts, df=df)\n# }\n\nDocumentation reproduced from package spatstat, version 1.64-1, License: GPL (>= 2)\n\n### Community examples\n\nLooks like there are no examples yet."
] | [
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https://erwnerve.tripod.com/prog/recursion/recintro.htm | [
"# Introduction to Recursion\n\nHi! So you're all set to dive into recursive programming. Well, this page is about the basics. The definition. The meaning. And a small cute example.\n\nI assume that you have a fair knowledge of one of the languages: C, C++, Pascal, etc. which supports recursion.\n\nFirst the formalities. Meaning of Recursion. You may skip it if you already know what the concept is.\n\n## Recursion\n\nIn normal procedural languages, one can go about defining functions and procedures, and 'calling' these from the 'parent' functions. I hope you already know that.\nSome languages also provide the ability of a function to call itself. This is called Recursion.\n\n## Factorial\n\nFactorial is a mathematical term. Factorial of a number, say n, is equal to the product of all integers from 1 to n. Factorial of n is denoted by n! = 1x2x3...x n.\nEg: `10! = 1x2x3x4x5x6x7x8x9x10`\n\nThe simplest program to calculate factorial of a number is a loop with a product variable. Instead, it is possible to give a recursive definition for Factorial as follows:\n\n```1) If n=1, then Factorial of n = 1\n2) Otherwise, Factorial of n = product of n and\nFactorial of (n-1)```\n\nCheck it out for yourself; it works. The following code fragment (in C) depicts Recursion at work.\n\n```int Factorial(int n)\n{\nif (n==1) return 1;\nelse\nreturn Factorial(n-1) * n;\n} ```\n The important thing to remember when creating a recursive function is to give an 'end-condition'. We don't want the function to keep calling itself forever, now, do we? Somehow, it should know when to stop. There are many ways of doing this. One of the simplest is by means of an 'if condition' statement, as above. In the above example, the recursion stops when n reaches 1. In each instance of the function, the value of n keeps decreasing. So it ultimately reaches 1 and ends. Of course, the above function will run infinitely if the initial value of n is less than 1. So the function is not perfect. The `n==1` condition should be changed to `n<=1`. Imagination is a very hard thing. Imagination of Recursion is all the more tricky. Think of clones. Say you have a machine to make clones of yourself, and (for lack of a better pass-time) decide to find the factorial of a number, say 10, using your clones. So, being smart, this is what you do: First, there's only You. Let's call you You-1. You have the number 10 in your pocket. Being smart, you know that all you need to find the factorial of 10 (`10*9*...*2*1`) is to somehow obtain the value of 9 factorial (`9*8*..*2*1`), and then just multiply it with 10. So that's what you do. You turn on your machine and out pops a clone! You give the clone You-2 strict instructions to find the factorial of 9 and make it quick! Your job is done for a while, so you (You-1) stretch on your sofa sipping on your lemonade. Meanwhile... You-2 is (you guessed it) just as smart as you! He tucks his number (9) into his pocket, turns on the machine, and out pops a clone (You-3). The new clone is given the job of 8-factorial, which it proceeds to do while (unbeknownst to you) You-2 is sipping on his own glass of lemonade on his own sofa. And so the story goes on until finally one fine day... Out pops You-10 who is given strict instructions (by You-9) to get the factorial of 1. Now, You-10, being just as smart as any of the other you's, knows very well that the factorial of 1 is... 1. So he says to You-9 (who was just about to doze off on his sofa), \"Here's your factorial of 1.\" You-9 snatches the result from his subordinate You-10, takes out his plasma gun, and zaps You-10 out of existence. He scribbles on a piece of paper, calculating the product of the value he got from You-10 with the number in his pocket, 2. \"Heh, heh, heh\" he thinks, and goes to his boss, You-8, saying,\"Here's your factorial of 2...\" ...blah...blah... and finally You-2 wakes you up from your slumber, and says to you, \"Here's your factorial of 9\" You zap him off, multiply by the 10 in your pocket, and There You Have It !! Now, wasn't that simple? Here, 'You' were the function. The 'clones' are merely new instances of the same function. They all think and act alike. At one point, there are 10 You's (which occupies a lot of memory space). As soon as an instance returns a value and finishes its job, it is zapped off from memory. Recursion can get much, much trickier than that - get your fundas right.\n\n## Exercises\n\nHere are some programs for you to try... but since nobody these days actual works out the problems, all I'm going to ask you to do is give each problem a thought. You should be able to solve them all.\n\n1. Write a function using Recursion to print numbers from n to 0.\n\n2. Write a function using Recursion to print numbers from 0 to n. (You just need to shift one line in the program of problem 1.)\n\n3. Write a function using Recursion to enter and display a string in reverse. Don't use arrays/strings.\n\n4. Write a function using Recursion to enter and display a string in reverse and state whether the string contains any spaces. Don't use arrays/strings.\n\n5. Write a function using Recursion to check if a number n is prime. (You have to check whether n is divisible by any number below n)\n\n6. Write a function using Recursion to enter characters one by one until a space is encountered. The function should return the depth at which the space was encountered.\n\n Recursion Index Next >\n [email protected] April 2000\n\n http://personal.vsnl.com/erwin/recursion.htm",
null,
""
] | [
null,
"http://hg1.hitbox.com/HG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9448397,"math_prob":0.9084173,"size":4733,"snap":"2021-43-2021-49","text_gpt3_token_len":1193,"char_repetition_ratio":0.13110594,"word_repetition_ratio":0.03797468,"special_character_ratio":0.24973589,"punctuation_ratio":0.14615385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9554478,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T13:13:00Z\",\"WARC-Record-ID\":\"<urn:uuid:99dc4657-fc25-40e2-8231-7a2cd9cd7a92>\",\"Content-Length\":\"11688\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51c8153d-c438-4a29-b930-6b361cec2bf2>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf23a95b-bb8b-4056-bd2f-e6a0d7a0062a>\",\"WARC-IP-Address\":\"209.202.252.105\",\"WARC-Target-URI\":\"https://erwnerve.tripod.com/prog/recursion/recintro.htm\",\"WARC-Payload-Digest\":\"sha1:OJBGNADWW7PVEOAYEZWSRSFWYKWIN4WC\",\"WARC-Block-Digest\":\"sha1:WQRVNJDGIT7Y7VTIMGXPGJE5DBSGGI6H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964364169.99_warc_CC-MAIN-20211209122503-20211209152503-00109.warc.gz\"}"} |
https://thareducation.com/last-line-math-worksheet-answers.html | [
"# Last Line Math Worksheet Answers\n\nLast Line Math Worksheet Answers. I parallel to y 07x 11 Solution. Algebra Addition Average Calender Comparison Decimals Division Exponents Factors Fractions GCF LCM Multiplication Number Line Odd Even Numbers Place Value Rounding Roman Numerals Statistics Subtraction Temperature and Telling Time. The math flash cards and dots math game.",
null,
"Unjoined Lines Only Basic Geometry Worksheet Unjoined Lines Only Basic Geometry Worksheet Basic Geometry Geometry Worksheets Parallel And Perpendicular Lines\n\n### All of the slopes on this worksheet are positive values.\n\nSlope of a Line Graphed Points Worksheet 1 - Here is a 9 problem worksheet where you will asked to find the rise and run between two points on a line then determine the slope of the line. Slope of a Line Worksheet 1 PDF. This page includes order of operations worksheets using whole numbers decimals and fractions.\n\nCounting by 1 on a Continuous Number Line from 0 to 100 Counting by 2 on a Continuous Number Line from 0 to 200 Counting by 3 on a Continuous Number Line from 0 to 195 Counting by 4 on a Continuous Number Line from 0 to 200 Counting by 5 on a Continuous Number Line from 0 to 250 Counting by 6 on a Continuous Number Line from 0 to 300 Counting by 7 on a Continuous Number Line from 0 to 350 Counting by 8 on a Continuous Number Line from 0 to 400 Counting by 9 on a Continuous Number Line. Measuring Angles Formed by Parallel Lines Transverals Worksheet 2 - This angle worksheet features 8 different problems where parallel lines are intersected by a transveral. The answer key is included with the math worksheets as it is created.\n\nYou will be given the measure of one of the angles in each problem then use your knowledge of parallel lines and transversals to find measurements of the remaining angles. These math worksheets are randomly created by our math worksheet generators so you have an endless supply of quality math worksheets at your disposal. Angle Side Angle Worksheet and Activity.\n\nOur math worksheets are free to download easy to use and very flexible. Decimals worksheets for grade 3 through grade 6. These high quality math worksheets are delivered in a PDF format and includes the answer keys.\n\nS ine Cosine Tangent SOHCAHTOA. Identifying Angles Worksheet 1 PDF View Answers. Math worksheet generators that allow you to specify the required options.\n\n-- - - N g pj Razzle Shoes bought a -page ad in the Times. Some of the worksheets for this concept are Work Review linear equations Writing linear equations Linear function work with answers Graphing linear equations work answer key Review graphing and writing linear equations Review linear Date period. Cambridge IGCSE Mathematics 0580 Worksheets Each worksheet contains questions on a particular topic extracted from past papers Extended Curriculum -2010 to 2019 all variants along with answers.",
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""
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https://checksheet.app/google-sheets-formulas/lognorm.dist/ | [
"# LOGNORM.DIST\n\nThe `LOGNORM.DIST` function returns the cumulative distribution function (CDF) of the log-normal distribution, given a value x, a mean and a standard deviation. The log-normal distribution is often used to model values that are inherently positive and have a large range of values. The function is commonly used in financial analysis to model stock returns and in engineering to model the size of particles in a material.\n\n## Usage\n\nUse the `LOGNORM.DIST` formula with the syntax shown below, it has 3 required parameters:\n\nParameters:\n1. x (required):\nThe numeric value at which to evaluate the distribution.\n2. mean (required):\nThe mean of the logarithm of the distribution.\n3. standard_deviation (required):\nThe standard deviation of the logarithm of the distribution.\n\n## Examples\n\nHere are a few example use cases that explain how to use the `LOGNORM.DIST` formula in Google Sheets.\n\n### Calculating probability of stock returns\n\nInvestors can use the `LOGNORM.DIST` function to calculate the probability of stock returns based on historical data. By inputting the mean and standard deviation of a stock's past returns, investors can estimate the likelihood of future returns falling within a certain range.\n\n### Modeling particle size distribution\n\nIn engineering, the `LOGNORM.DIST` function can be used to model the size distribution of particles in a material. By inputting the mean and standard deviation of the logarithm of the particle sizes, engineers can estimate the percentage of particles that fall within a certain size range.\n\n## Common Mistakes\n\n`LOGNORM.DIST` not working? Here are some common mistakes people make when using the `LOGNORM.DIST` Google Sheets Formula:\n\n### Incorrect order of arguments\n\nUsers often provide the arguments in the wrong order, resulting in incorrect outputs. The correct order is x, mean, standard_deviation.\n\n### Invalid input values\n\nThe function returns an error if any of the input values are negative or if the standard deviation is 0. Users should ensure that all input values are valid.\n\n### Incorrect usage of function\n\nUsers often misunderstand the purpose of the function and use it inappropriately, resulting in incorrect outputs. The LOGNORM.DIST function is used to calculate the cumulative distribution function (CDF) for a lognormal distribution.\n\nThe following functions are similar to `LOGNORM.DIST` or are often used with it in a formula:\n\n• `NORM.S.DIST`\n\nThe NORM.S.DIST function returns the standard normal cumulative distribution function. It calculates the probability that a random variable with a standard normal distribution is less than or equal to x. This function is commonly used in statistical analysis and hypothesis testing.\n\n• `NORM.DIST`\n\nThe `NORM.DIST` formula is a statistical function that returns the normal distribution of a specified variable. It is used to determine the probability of a random variable falling within a specified range of values. This function is commonly used in finance and scientific research.\n\n• `LOGNORM.INV`\n\nThe `LOGNORM.INV` function returns the inverse of the lognormal cumulative distribution for a specified mean and standard deviation. It is commonly used in statistics to model data that is skewed and has a long tail. Given a percentile value and the distribution parameters, this function calculates the corresponding value of the random variable.\n\n• `NORM.INV`\n\nThe `NORM.INV` function returns the inverse of the cumulative normal distribution for a specified mean and standard deviation. It is commonly used in statistical analysis to find the value at which a specified percentage of observations occur below that value.\n\n• `STDEV.P`\n\nThe `STDEV.P` function is a statistical function that calculates the standard deviation of a population based on a sample of numerical data. It is commonly used to measure the amount of variation or dispersion in a dataset. The formula assumes that the input values represent the entire population, rather than a sample. If the input values represent a sample, you should use the `STDEV.S` function instead.\n\nYou can learn more about the `LOGNORM.DIST` Google Sheets function on Google Support."
] | [
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https://search.usi.ch/it/corsi/35263591/probability-statistics | [
"Strumenti\nRicerca di contatti, progetti,\ncorsi e pubblicazioni\n\n# Probability & Statistics\n\n## Persone\n\nDocente titolare del corso\n\nArtico I.\n\nAssistente\n\nAssistente\n\n## Descrizione\n\nCOURSE OBJECTIVES\n\n• learn about probability axioms and (conditional) probability distributions\n• study moments of random variables and Chebychev's inequality\n• learn about law of large numbers and central limit theorem\n• study estimation methods and ways to evaluate estimators\n• develop hypothesis testing\n• study linear regression\n\nCOURSE DESCRIPTION\n\nProbability theory is a deductive science describing the axioms for calculating the probability of some event given some known state of the world. In the first part of the course, we define the probability axioms, introduce the concept of events, random variables, and probability distributions. In inductive practice we are interested to learn about the state of the world given some event, i.e., the data. Statistics is probability theory turned upside down. In this course we will learn about estimation procedures, in particular, maximum likelihood and the method of moments. We will introduce hypothesis testing and linear regression analysis.\n\nLEARNING METHODS\n\nCombination of lectures and tutorials\n\nEXAMINATION INFORMATION\nWritten exam\n\nREFERENCES\n\n• Prasanna Sahoo, Probability and Mathematical Statistics, 2013."
] | [
null
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https://www.sciengine.com/doi/10.1007/s11432-020-2914-2 | [
"#",
null,
"",
null,
"SCIENCE CHINA Information Sciences, Volume 64 , Issue 7 : 172202(2021) https://doi.org/10.1007/s11432-020-2914-2\n\n## Suboptimal adaptive tracking controlfor FIR systems with binary-valued observations",
null,
"• AcceptedApr 17, 2020\n• PublishedMay 18, 2021\nShare\nRating\n\n### Abstract",
null,
"### Acknowledgment\n\nThis work was partly supported by National Natural Science Foundation of China (Grant No. 61603034), Beijing Municipal Natural Science Foundation (Grant No. 3182027), and Fundamental Research Funds for the Central Universities of China (Grant No. FRF-GF-19-016B).\n\n### References\n\n Le Yi Wang , Ji-Feng Zhang , Yin G G. System identification using binary sensors. IEEE Trans Automat Contr, 2003, 48: 1892-1907 CrossRef Google Scholar\n\n Wang T, Zhang H, Zhao Y. Parameter Estimation Based on Set-valued Signals: Theory and Application. Acta Math Appl Sin Engl Ser, 2019, 35: 255-263 CrossRef Google Scholar\n\n Wang L Y, Yin G G, Zhang J F. Joint identification of plant rational models and noise distribution functions using binary-valued observations. Automatica, 2006, 42: 535-547 CrossRef Google Scholar\n\n Wang L Y, Yin G G. Asymptotically efficient parameter estimation using quantized output observations. Automatica, 2007, 43: 1178-1191 CrossRef Google Scholar\n\n Zhao Y, Wang L Y, Yin G G. Identification of Wiener systems with binary-valued output observations. Automatica, 2007, 43: 1752-1765 CrossRef Google Scholar\n\n Zhao Y, Zhang J F, Wang L Y. Identification of Hammerstein Systems with Quantized Observations. SIAM J Control Optim, 2010, 48: 4352-4376 CrossRef Google Scholar\n\n Wang L Y, Xu G, Yin G. State reconstruction for linear time-invariant systems with binary-valued output observations. Syst Control Lett, 2008, 57: 958-963 CrossRef Google Scholar\n\n Jafari K, Juillard J, Roger M. Convergence analysis of an online approach to parameter estimation problems based on binary observations. Automatica, 2012, 48: 2837-2842 CrossRef Google Scholar\n\n Guo J, Zhao Y. Recursive projection algorithm on FIR system identification with binary-valued observations. Automatica, 2013, 49: 3396-3401 CrossRef Google Scholar\n\n You K. Recursive algorithms for parameter estimation with adaptive quantizer. Automatica, 2015, 52: 192-201 CrossRef Google Scholar\n\n Wang T, Tan J, Zhao Y. Asymptotically efficient non-truncated identification for FIR systems with binary-valued outputs. Sci China Inf Sci, 2018, 61: 129208 CrossRef Google Scholar\n\n Risuleo R S, Bottegal G, Hjalmarsson H. Approximate Maximum-likelihood Identification of Linear Systems from Quantized Measurements. IFAC-PapersOnLine, 2018, 51: 724-729 CrossRef Google Scholar\n\n Yu C, Zhang C, Xie L. Blind system identification using precise and quantized observations. Automatica, 2013, 49: 2822-2830 CrossRef Google Scholar\n\n Guo J, Diao J D. Prediction-based event-triggered identification of quantized input FIR systems with quantized output observations. Sci China Inf Sci, 2020, 63: 112201 CrossRef Google Scholar\n\n Zhang H, Wang T, Zhao Y. FIR system identification with set-valued and precise observations from multiple sensors. Sci China Inf Sci, 2019, 62: 052203 CrossRef Google Scholar\n\n Guo J, Zhang J F, Zhao Y. Adaptive Tracking Control of A Class of First-Order Systems With Binary-Valued Observations and Time-Varying Thresholds. IEEE Trans Automat Contr, 2011, 56: 2991-2996 CrossRef Google Scholar\n\n Guo J, Zhang J F, Zhao Y. Adaptive tracking of a class of first-order systems with binary-valued observations and fixed thresholds. J Syst Sci Complex, 2012, 25: 1041-1051 CrossRef Google Scholar\n\n Zhao Y, Guo J, Zhang J F. Adaptive Tracking Control of Linear Systems With Binary-Valued Observations and Periodic Target. IEEE Trans Automat Contr, 2013, 58: 1293-1298 CrossRef Google Scholar\n\n Wang T, Hu M, Zhao Y. Adaptive Tracking Control of FIR Systems Under Binary-Valued Observations and Recursive Projection Identification. IEEE Trans Syst Man Cybern Syst, 2020, : 1-11 CrossRef Google Scholar\n\n Jing L, Zhang J. Tracking control and parameter identification with quantized ARMAX systems. Sci China Inf Sci, 2019, 62: 199203 CrossRef Google Scholar\n\n Bu X, Hou Z. Adaptive Iterative Learning Control for Linear Systems With Binary-Valued Observations. IEEE Trans Neural Netw Learning Syst, 2018, 29: 232-237 CrossRef Google Scholar\n\n Guo J, Wang L Y, Yin G. Asymptotically efficient identification of FIR systems with quantized observations and general quantized inputs. Automatica, 2015, 57: 113-122 CrossRef Google Scholar\n\n Chow Y S, Teicher H. Probability Theory: Independence, Interchangeability, Martingales. 2nd ed. New York: Springer, 1997. Google Scholar\n\n Wang L Y, Zhang J F, Yin G G, et al. System Identification With Quantized Observations. Boston: Birkhäuser, 2010. Google Scholar\n\n•",
null,
"Figure 1\n\n(Color online) Two-segment adaptive tracking control law.\n\n•",
null,
"Figure 2\n\n(Color online) Design mechanism of the two-segment adaptive control algorithm.\n\n• Table 1\n\nTable 1Performance comparison of adaptive tracking control\n\n $k$ $J_{\\rm~min}$ $\\sigma_d^2$ $\\varepsilon$ $\\eta=100~$ $0.8\\times~10^4$ $1.8051$ $0.2500$ $1.5551$ $\\eta=500~$ $4.0\\times~10^4$ $0.6127$ $0.2500$ $0.3627$ $\\eta=1000~$ $8.0\\times~10^4$ $0.4074$ $0.2500$ $0.1574$ $\\eta=2000~$ $1.6\\times~10^5$ $0.3320$ $0.2500$ $0.0820$ $\\eta=5000~$ $4.0\\times~10^5$ $0.2811$ $0.2500$ $0.0311$"
] | [
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https://nfscene.com/article/how-to-exchange-and-encrypt-private-keys-on-the-blockchain-a-decentralized-marketplace-example/ | [
"",
null,
"How to secure a marketplace where you could buy and sell digital goods when the blockchain is public ?\n\nWhile scrolling and reading some articles on Google News, I told myself what a great tool is that all is gathered in one place for the users. No need to crawl the internet, to find information or news! But as always centralisation has also its downside. This platform aims to empower the so much Google against the publishers, that it has become an inevitable source of tension with the revenue generated by the ads. What if the blockchain could help to promote, share, and distribute public and/or private content? This idea led me to propose a concept of a decentralized marketplace where authors could publish and sell their content.\n\n# Marketplace concept and design blueprint\n\n## Marketplace concept\n\nNowadays, when you want to sell a good you need to register on a centralized platform (for example Facebook Marketplace), to be able to share your ad with potential customers. The concept is to replace the centralized platform with a decentralized one so that nobody gains money and power from it except the author. The blockchain is the perfect technology for that purpose by its very means to be decentralized, however, the content of it is public. By public, I want to say that everybody can access the content of the blockchain and read its information. It can be problematic if your good is private, like your e-book. If you just publish it on a distributed web media protocol (like IPFS for example), everybody will be able to read it and therefore nobody would buy it. The solution would be to encrypt the e-book so that even if published on the blockchain nobody could read it without the private key to decrypt it.\n\nThis concept can be applied to any numeric content, video, music, or text, … but in the end, you still need to exchange (buy/sell) the private keys to be able to access the content.\n\nA private key is very like a password, and sharing a password on the blockchain ends to converge to the same issue, everybody could access it. However, there is a solution! Let’s discover it in the next section.\n\n## Private key exchange\n\nFirst, to explain how a private key exchange can be done in a private manner on the (public) blockchain, we need to clearly understand a few concepts with asymmetric cryptography.\n\nTo perform the different operations in asymmetric cryptography you will use two keys. One private and one public. Let’s see the example in Figure 1. , if Bob wants to receive a message, Bob will provide his public key to let Max encrypt the message and use Bob's private key to decrypt the message.",
null,
"Figure 1: Message encryption with public and private key pair.\n\nThe example in Figure 2. shows how to verify the message sender. To let Bob knows that it’s effectively Max that sends the message Max can sign it with his private key. Bob can use Max’s public key to verify the signature of the message and acknowledge that Max is the sender.",
null,
"Figure 2: Message signature validation with public and private key pair.\n\nNow, that we explained the basic concepts of asymmetric cryptography, let’s define a strategy to exchange a private key in a private manner on the blockchain, like if Eva would like to buy an article. For that purpose, we would have to sell the article's private key to decrypt/read it. Therefore the system has to send the article's private key to Eva, but as on the blockchain everything is public we have to encrypt the article's private key with Eva's public key so that only Eva can access the key.",
null,
"Figure 3: The seller encrypts and sends the private key that Eva bought.\n\nEva can easily assert that the article’s private key sold is the good one, by using the article’s public key. But what if the seller tried to scam Eva by selling a fraudulent key? Before releasing the “money” to the seller, Eva has to validate or invalidate the transaction within a time limit. If the key is fraudulent, she can invalidate it by sending the wrong private key to the blockchain. Then the blockchain will compare the wrong article’s key, with the article’s public key, to assert that the key is effectively wrong and then will encrypt the wrong article’s private key to compare it with the previously sent encrypted article’s private key by the sender. If they are not equal it means that the Seller sent the wrong key and is a scam.",
null,
"Figure 4: Private key verification process. How can the buyer dismiss the transaction and the blockchain verify the scam ?\n\nTo be able to complete this verification schema, we need to implement a smart contract two behaviour: first the private key validation from a public key, and secondly the encryption with a public key.\n\n# Implementation\n\n## Private key validation from a public key\n\nTo validate that the private key comes from the same owner/author, we can use the public key. The public key is generated from the private key. Therefore we can use the private key to generate the public and compare the generated public key with the given public key to validate the private key. In the Ethereum blockchain, the elliptic curve encryption algorithm is used. To find a public key, we need to perform a Jacobian multiplication of the private key (see Formula 1).",
null,
"Formula 1: Public key is the derivative of the private key. For the elliptic curve encryption algorithm the derivative is a Jacobian multiplication.\n\n``````uint256 constant private GX = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798;\nuint256 constant private GY = 0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8;\nuint256 constant private AA = 0;\nuint256 constant private PP = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F;\n\nfunction getPublicKey(uint256 privateKey) public pure returns (uint256, uint256) {\nreturn getDeriveKey(privateKey, GX, GY);\n}\n\nfunction getDeriveKey(uint256 privateKey, uint256 gX, uint256 gY) public pure returns (uint256, uint256) {\nreturn EllipticCurve.ecMul(\nprivateKey,\ngX,\ngY,\nAA,\nPP\n);\n}\n``````\n\n## Encryption with a private key on the blockchain by a smart contract.\n\nThe key point in this exchange is to validate that the given encrypted key could be a scam. We need to invalidate the sent key in the blockchain, for that, we need to perform encryption in the blockchain. The asymmetric encryption of a message is actually a symmetric encryption with a key that both the receiver and the sender can generate with the private and public keys of each other.\n\nThe key can be generated either by multiplying the sender's private key with the receiver's public key or by multiplying the receiver's private key with the sender's public key (see formula 2).",
null,
"Formula 2: Secret can be calculated by A and B with their respective private key by knowing the public key of the other.\n\nThe symmetric encryption is performed by the keccak256 hash operation. This operation is by default already implemented in the Solidity Language and therefore is cost-effective. You can see the method symmetricEncryption method below.\n\n``````string constant COUNTER_VALUE = \"7\";\n\nfunction symmetricEncryption(uint256 message, uint256 key) public pure returns (uint256) {\nuint256 hash = uint256(keccak256(abi.encodePacked(key, COUNTER_VALUE)));\nreturn message ^ hash;\n}``````\n\nFinally we can write our method to verify that the private key was invalid:\n\n``````function dismiss(uint256 privateBuyerKey) public onlyBuyer onlyActive withResponse {\n// Check that the given private buyer key is the true one.\n(uint256 rbX, uint256 rbY) = Encrypt.getPublicKey(privateBuyerKey);\n(uint256 bX, uint256 bY) = Encrypt.toUint256(buyerPublicKey);\nrequire((rbX == bX) && (rbY == bY));\n// Decrypt message to check the sent private article key.\nuint256 wrongArticleKey = Encrypt.symmetricEncryption(Response(response).getEncryptedPrivateKey(), secretKey);\n(uint256 waX, uint256 waY) = Encrypt.getPublicKey(wrongArticleKey);\n(uint256 aX, uint256 aY) = Encrypt.toUint256(Article(article).getPublicKey());\nrequire((waX != aX) || (waY != aY));\n// Refund money and reject.\nstatus = Status.REJECTED;\n}\n``````\n\n``````function getSecret(uint256 privateKeyA, bytes memory publicKeyB) public pure returns (uint256) {\n(uint256 gX, uint256 gY) = toUint256(publicKeyB);\n(uint256 pX, ) = getDeriveKey(privateKeyA, gX, gY);\nreturn pX;\n}\n\nfunction verifySignature(uint256 message, bytes memory signature, bytes memory pubKey) public pure returns (bool) {\nreturn getSignatureTrace(message, signature) == getPubKeyTrace(pubKey);\n}\n\nfunction getSignatureTrace(uint256 message, bytes memory signature) public pure returns (address) {\nreturn ECDSA.recover(bytes32(toBytes(message)), signature);\n}\n\nfunction getPubKeyTrace(bytes memory pubKey) public pure returns (address trace) {\nbytes32 hash = keccak256(pubKey);\nassembly { mstore(0, hash) trace := mload(0) }\n}\n\nfunction bytesToBytes32(bytes memory b, uint offset) private pure returns (bytes32) {\nbytes32 out;\nfor (uint i = 0; i < 32; i++) {\nout |= bytes32(b[offset + i] & 0xFF) >> (i * 8);\n}\nreturn out;\n}\n\nfunction toBytes(uint256 x) private pure returns (bytes memory b) {\nb = new bytes(32);\nassembly { mstore(add(b, 32), x) }\n}\n\nfunction toUint256(bytes memory publicKeyB) public pure returns (uint256, uint256) {\nuint256 x = uint256(bytesToBytes32(publicKeyB, 0));\nuint256 y = uint256(bytesToBytes32(publicKeyB, 32));\nreturn (x, y);\n}\n``````\n\nThis is the key method that allow the system to be secure and reliable.\n\nFull code of the market place example can be found in this GitHub repository.\n\nSources"
] | [
null,
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null,
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null,
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null,
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null,
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null,
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https://continuous-time-mcs.quantecon.org/_notebooks/kolmogorov_fwd.ipynb | [
"{ \"cells\": [ { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"# The Kolmogorov Forward Equation\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Overview\\n\", \"\\n\", \"In this lecture we approach continuous time Markov chains from a more\\n\", \"analytical perspective.\\n\", \"\\n\", \"The emphasis will be on describing distribution flows through vector-valued\\n\", \"differential equations and their solutions.\\n\", \"\\n\", \"These distribution flows show how the time $t$ distribution associated with a\\n\", \"given Markov chain $(X_t)$ changes over time.\\n\", \"\\n\", \"Distribution flows will be identified with initial value problems generated by autonomous linear ordinary differential equations (ODEs) in vector space.\\n\", \"\\n\", \"We will see that the solutions of these flows are described by Markov semigroups.\\n\", \"\\n\", \"This leads us back to the theory we have already constructed – some care will\\n\", \"be taken to clarify all the connections.\\n\", \"\\n\", \"In order to avoid being distracted by technicalities, we continue to defer our\\n\", \"treatment of infinite state spaces, assuming throughout this lecture that $|S|\\n\", \"= n$.\\n\", \"\\n\", \"As before, $\\\\dD$ is the set of all distributions on $S$.\\n\", \"\\n\", \"We will use the following imports\" ] }, { \"cell_type\": \"code\", \"execution_count\": null, \"metadata\": { \"hide-output\": false }, \"outputs\": [], \"source\": [ \"import numpy as np\\n\", \"import scipy as sp\\n\", \"import matplotlib.pyplot as plt\\n\", \"import quantecon as qe\\n\", \"from numba import njit\\n\", \"from scipy.linalg import expm\\n\", \"\\n\", \"from matplotlib import cm\\n\", \"from mpl_toolkits.mplot3d import Axes3D\\n\", \"from mpl_toolkits.mplot3d.art3d import Poly3DCollection\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## From Difference Equations to ODEs\\n\", \"\\n\", \"[Previously](https://jstac.github.io/continuous_time_mcs/markov_prop.html#invdistflows) we generated this figure, which shows how distributions evolve over time for the inventory model under a certain parameterization:\\n\", \"\\n\", \"Probability flows for the inventory model. \\n\", \"(Hot colors indicate early dates and cool colors denote later dates.)\\n\", \"\\n\", \"We also learned how this flow is related to\\n\", \"the Kolmogorov backward equation, which is an ODE.\\n\", \"\\n\", \"In this section we examine distribution flows and their connection to\\n\", \"ODEs and continuous time Markov chains more systematically.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Review of the Discrete Time Case\\n\", \"\\n\", \"Let $(X_t)$ be a discrete time Markov chain with Markov matrix $P$.\\n\", \"\\n\", \"[Recall that](https://jstac.github.io/continuous_time_mcs/markov_prop.html#finstatediscretemc), in the discrete time case, the distribution $\\\\psi_t$ of $X_t$ updates according to\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_{t+1} = \\\\psi_t P, \\n\", \" \\\\qquad \\\\psi_0 \\\\text{ a given element of } \\\\dD,\\n\", \"$$\\n\", \"\\n\", \"where distributions are understood as row vectors.\\n\", \"\\n\", \"Here’s a visualization for the case $S = \\\\{0, 1, 2\\\\}$, so that $\\\\dD$ is the [standard\\n\", \"simplex](https://en.wikipedia.org/wiki/Simplex#The_standard_simplex) in $\\\\RR^3$.\\n\", \"\\n\", \"The initial condition is (0, 0, 1) and the Markov matrix is\" ] }, { \"cell_type\": \"code\", \"execution_count\": null, \"metadata\": { \"hide-output\": false }, \"outputs\": [], \"source\": [ \"P = ((0.9, 0.1, 0.0),\\n\", \" (0.4, 0.4, 0.2),\\n\", \" (0.1, 0.1, 0.8))\" ] }, { \"cell_type\": \"code\", \"execution_count\": null, \"metadata\": { \"hide-output\": false }, \"outputs\": [], \"source\": [ \"def unit_simplex(angle):\\n\", \" \\n\", \" fig = plt.figure(figsize=(8, 6))\\n\", \" ax = fig.add_subplot(111, projection='3d')\\n\", \"\\n\", \" vtx = [[0, 0, 1],\\n\", \" [0, 1, 0], \\n\", \" [1, 0, 0]]\\n\", \" \\n\", \" tri = Poly3DCollection([vtx], color='darkblue', alpha=0.3)\\n\", \" tri.set_facecolor([0.5, 0.5, 1])\\n\", \" ax.add_collection3d(tri)\\n\", \"\\n\", \" ax.set(xlim=(0, 1), ylim=(0, 1), zlim=(0, 1), \\n\", \" xticks=(1,), yticks=(1,), zticks=(1,))\\n\", \"\\n\", \" ax.set_xticklabels(['$(1, 0, 0)$'], fontsize=12)\\n\", \" ax.set_yticklabels(['$(0, 1, 0)$'], fontsize=12)\\n\", \" ax.set_zticklabels(['$(0, 0, 1)$'], fontsize=12)\\n\", \"\\n\", \" ax.xaxis.majorTicks.set_pad(15)\\n\", \" ax.yaxis.majorTicks.set_pad(15)\\n\", \" ax.zaxis.majorTicks.set_pad(35)\\n\", \"\\n\", \" ax.view_init(30, angle)\\n\", \"\\n\", \" # Move axis to origin\\n\", \" ax.xaxis._axinfo['juggled'] = (0, 0, 0)\\n\", \" ax.yaxis._axinfo['juggled'] = (1, 1, 1)\\n\", \" ax.zaxis._axinfo['juggled'] = (2, 2, 0)\\n\", \" \\n\", \" ax.grid(False)\\n\", \" \\n\", \" return ax\\n\", \"\\n\", \"\\n\", \"def convergence_plot(ψ, n=14, angle=50):\\n\", \"\\n\", \" ax = unit_simplex(angle)\\n\", \"\\n\", \" P = ((0.9, 0.1, 0.0),\\n\", \" (0.4, 0.4, 0.2),\\n\", \" (0.1, 0.1, 0.8))\\n\", \" \\n\", \" P = np.array(P)\\n\", \" colors = cm.jet_r(np.linspace(0.0, 1, n))\\n\", \"\\n\", \" x_vals, y_vals, z_vals = [], [], []\\n\", \" for t in range(n):\\n\", \" x_vals.append(ψ)\\n\", \" y_vals.append(ψ)\\n\", \" z_vals.append(ψ)\\n\", \" ψ = ψ @ P\\n\", \"\\n\", \" ax.scatter(x_vals, y_vals, z_vals, c=colors, s=50, alpha=0.7, depthshade=False)\\n\", \"\\n\", \" return ψ\\n\", \"\\n\", \"ψ = convergence_plot((0, 0, 1))\\n\", \"\\n\", \"plt.show()\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"There’s a sense in which a discrete time Markov chain “is” a homogeneous\\n\", \"linear difference equation in distribution space.\\n\", \"\\n\", \"To clarify this, suppose we\\n\", \"take $G$ to be a linear map from $\\\\dD$ to itself and\\n\", \"write down the difference equation\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_{t+1} = G(\\\\psi_t)\\n\", \" \\\\quad \\\\text{with } \\\\psi_0 \\\\in \\\\dD \\\\text{ given}. \\\\tag{5.1}\\n\", \"$$\\n\", \"\\n\", \"Because $G$ is a linear map from a finite dimensional space to itself, it can\\n\", \"be represented by a matrix.\\n\", \"\\n\", \"Moreover, a matrix $P$ is a Markov matrix if and only if $\\\\psi \\\\mapsto\\n\", \"\\\\psi P$ sends $\\\\dD$ into itself (check it if you haven’t already).\\n\", \"\\n\", \"So, under the stated conditions, our difference equation [(5.1)](#equation-gdiff2) uniquely\\n\", \"identifies a Markov matrix, along with an initial condition $\\\\psi_0$.\\n\", \"\\n\", \"Together, these objects identify the joint distribution of a discrete time Markov chain, as [previously described](https://jstac.github.io/continuous_time_mcs/markov_prop.html#jdfin).\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Shifting to Continuous Time\\n\", \"\\n\", \"We have just argued that a discrete time Markov chain can be identified with a\\n\", \"linear difference equation evolving in $\\\\dD$.\\n\", \"\\n\", \"This strongly suggests that a continuous time Markov chain can be identified\\n\", \"with a linear ODE evolving in $\\\\dD$.\\n\", \"\\n\", \"This intuition is correct and important.\\n\", \"\\n\", \"The rest of the lecture maps out the main ideas.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## ODEs in Distribution Space\\n\", \"\\n\", \"Consider linear differential equation given by\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_t' = \\\\psi_t Q, \\n\", \" \\\\qquad \\\\psi_0 \\\\text{ a given element of } \\\\dD, \\\\tag{5.2}\\n\", \"$$\\n\", \"\\n\", \"where\\n\", \"\\n\", \"- $Q$ is an $n \\\\times n$ matrix, \\n\", \"- distributions are again understood as row vectors, and \\n\", \"- derivatives are taken element by element, so that \\n\", \"\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_t' =\\n\", \" \\\\begin{pmatrix}\\n\", \" \\\\frac{d}{dt} \\\\psi_t(x_1) &\\n\", \" \\\\cdots &\\n\", \" \\\\frac{d}{dt} \\\\psi_t(x_n)\\n\", \" \\\\end{pmatrix}\\n\", \"$$\\n\", \"\\n\", \"\\n\", \"\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Solutions to Linear Vector ODEs\\n\", \"\\n\", \"Using the matrix exponential, the unique solution to the initial value problem\\n\", \"[(5.2)](#equation-ode-mc) is\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_t = \\\\psi_0 P_t \\n\", \" \\\\quad \\\\text{where } P_t := e^{tQ} \\\\tag{5.3}\\n\", \"$$\\n\", \"\\n\", \"To check that [(5.3)](#equation-cmc-sol) is a solution, we use [(4.7)](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#equation-expoderiv) again to get\\n\", \"\\n\", \"$$\\n\", \"\\\\frac{d}{d t} P_t = Q e^{tQ} = e^{tQ} Q\\n\", \"$$\\n\", \"\\n\", \"The first equality can be written as $P_t' = Q P_t$ and this is just\\n\", \"the [Kolmogorov backward equation](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html).\\n\", \"\\n\", \"The second equality can be written as\\n\", \"\\n\", \"$$\\n\", \"P_t' = P_t Q\\n\", \"$$\\n\", \"\\n\", \"and is called the **Kolmogorov forward equation**.\\n\", \"\\n\", \"Applying the Kolmogorov forward equation, we obtain\\n\", \"\\n\", \"$$\\n\", \"\\\\frac{d}{d t} \\\\psi_t \\n\", \" = \\\\frac{d}{d t} \\\\psi_0 P_t \\n\", \" = \\\\psi_0 \\\\frac{d}{d t} P_t \\n\", \" = \\\\psi_0 P_t Q\\n\", \" = \\\\psi_t Q\\n\", \"$$\\n\", \"\\n\", \"This confirms that [(5.3)](#equation-cmc-sol) solves [(5.2)](#equation-ode-mc).\\n\", \"\\n\", \"Here’s an example of three distribution flows with dynamics generated by [(5.2)](#equation-ode-mc), one starting from each vertex.\\n\", \"\\n\", \"The code uses [(5.3)](#equation-cmc-sol) with matrix $Q$ given by\" ] }, { \"cell_type\": \"code\", \"execution_count\": null, \"metadata\": { \"hide-output\": false }, \"outputs\": [], \"source\": [ \"Q = ((-3, 2, 1),\\n\", \" (3, -5, 2),\\n\", \" (4, 6, -10))\" ] }, { \"cell_type\": \"code\", \"execution_count\": null, \"metadata\": { \"hide-output\": false }, \"outputs\": [], \"source\": [ \"Q = np.array(Q)\\n\", \"ψ_00 = np.array((0.01, 0.01, 0.99))\\n\", \"ψ_01 = np.array((0.01, 0.99, 0.01))\\n\", \"ψ_02 = np.array((0.99, 0.01, 0.01))\\n\", \"\\n\", \"ax = unit_simplex(angle=50) \\n\", \"\\n\", \"def flow_plot(ψ, h=0.001, n=400, angle=50):\\n\", \" colors = cm.jet_r(np.linspace(0.0, 1, n))\\n\", \"\\n\", \" x_vals, y_vals, z_vals = [], [], []\\n\", \" for t in range(n):\\n\", \" x_vals.append(ψ)\\n\", \" y_vals.append(ψ)\\n\", \" z_vals.append(ψ)\\n\", \" ψ = ψ @ expm(h * Q)\\n\", \"\\n\", \" ax.scatter(x_vals, y_vals, z_vals, c=colors, s=20, alpha=0.2, depthshade=False)\\n\", \"\\n\", \"flow_plot(ψ_00)\\n\", \"flow_plot(ψ_01)\\n\", \"flow_plot(ψ_02)\\n\", \"\\n\", \"plt.show()\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"(Distributions cool over time, so initial conditions are hot colors.)\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Forwards vs Backwards Equations\\n\", \"\\n\", \"As the above discussion shows, we can take the Kolmogorov forward equation\\n\", \"$P_t' = P_t Q$ and premultiply by any distribution $\\\\psi_0$ to get the\\n\", \"distribution ODE $\\\\psi'_t = \\\\psi_t Q$.\\n\", \"\\n\", \"In this sense, we can understand the Kolmogorov forward equation as pushing\\n\", \"distributions forward in time.\\n\", \"\\n\", \"Analogously, we can take the Kolmogorov backward equation\\n\", \"$P_t' = Q P_t$ and postmultiply by any vector $h$ to get\\n\", \"\\n\", \"$$\\n\", \"(P_t h)' = Q P_t h\\n\", \"$$\\n\", \"\\n\", \"Recalling that $(P_t h)(x) = \\\\EE [ h(X_t) \\\\,|\\\\, X_0 = x]$, this vector\\n\", \"ODE tells us how expectations evolve, conditioning backward to time zero.\\n\", \"\\n\", \"Both the forward and the backward equations uniquely pin down the same solution $P_t = e^{tQ}$ when combined with the initial condition $P_0 = I$.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Matrix- vs Vector-Valued ODEs\\n\", \"\\n\", \"The ODE $\\\\psi'_t = \\\\psi_t Q$ is sometimes called the\\n\", \"**Fokker–Planck equation** (although this terminology is most commonly used\\n\", \"in the context of diffusions).\\n\", \"\\n\", \"It is a vector-valued ODE that describes the evolution of a particular\\n\", \"distribution path.\\n\", \"\\n\", \"By comparison, the Kolmogorov forward equation is (like the backward equation)\\n\", \"a differential equation in matrices.\\n\", \"\\n\", \"(And matrices are really maps, which send vectors into vectors.)\\n\", \"\\n\", \"Operating at this level is less intuitive and more abstract than working with the\\n\", \"Fokker–Planck equation.\\n\", \"\\n\", \"But, in the end, the object that we want to describe is a Markov\\n\", \"semigroup.\\n\", \"\\n\", \"The Kolmogorov forward and backward equations are the ODEs that define\\n\", \"this fundamental object.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### Preserving Distributions\\n\", \"\\n\", \"In the simulation above, $Q$ was chosen with some care, so that the flow\\n\", \"remains in $\\\\dD$.\\n\", \"\\n\", \"What are the exact properties we require on $Q$ such that $\\\\psi_t$ is always\\n\", \"in $\\\\dD$?\\n\", \"\\n\", \"This is an important question, because we are setting up an exact\\n\", \"correspondence between linear ODEs that evolve in $\\\\dD$ and continuous\\n\", \"time Markov chains.\\n\", \"\\n\", \"Recall that the linear update rule $\\\\psi \\\\mapsto \\\\psi P$ is invariant on\\n\", \"$\\\\dD$ if and only if $P$ is a Markov matrix.\\n\", \"\\n\", \"So now we can rephrase our key question regarding invariance on $\\\\dD$:\\n\", \"\\n\", \"What properties do we need to impose on $Q$ so that $P_t = e^{tQ}$ is a Markov matrix\\n\", \"for all $t$?\\n\", \"\\n\", \"A square matrix $Q$ is called an **intensity matrix** if $Q$ has zero row\\n\", \"sums and $Q(x, y) \\\\geq 0$ whenever $x \\\\not= y$.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### \\n\", \"\\n\", \"If $Q$ is a matrix on $S$ and $P_t := e^{tQ}$, then the following statements\\n\", \"are equivalent:\\n\", \"\\n\", \"1. $P_t$ is a Markov matrix for all $t$. \\n\", \"1. $Q$ is an intensity matrix. \\n\", \"\\n\", \"\\n\", \"The proof is related to that of [Lemma 4.2](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#jctosg) and is found as\\n\", \"a solved exercise below.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"### \\n\", \"\\n\", \"If $Q$ is an intensity matrix on finite $S$ and $P_t = e^{tQ}$ for all $t \\\\geq 0$,\\n\", \"then $(P_t)$ is a Markov semigroup.\\n\", \"\\n\", \"We call $(P_t)$ the Markov semigroup **generated** by $Q$.\\n\", \"\\n\", \"Later we will see that this result extends to the case $|S| = \\\\infty$ under\\n\", \"some mild restrictions on $Q$.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Jump Chains\\n\", \"\\n\", \"Let’s return to the chain $(X_t)$ created from jump chain pair $(\\\\lambda, K)$ in\\n\", \"[Algorithm 4.1](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#ejc_algo).\\n\", \"\\n\", \"We found that the semigroup is given by\\n\", \"\\n\", \"$$\\n\", \"P_t = e^{tQ}\\n\", \" \\\\quad \\\\text{where} \\\\quad\\n\", \" Q(x, y) := \\\\lambda(x) (K(x, y) - I(x, y))\\n\", \"$$\\n\", \"\\n\", \"Using the fact that $K$ is a Markov matrix and the jump rate function\\n\", \"$\\\\lambda$ is nonnegative, you can easily check that $Q$ satisfies the\\n\", \"definition of an intensity matrix.\\n\", \"\\n\", \"Hence $(P_t)$, the Markov semigroup for the jump chain $(X_t)$, is the\\n\", \"semigroup generated by the intensity matrix $Q(x, y) = \\\\lambda(x) (K(x, y) - I(x, y))$.\\n\", \"\\n\", \"We can differentiate $P_t = e^{tQ}$ to obtain the Kolmogorov forward equation\\n\", \"$P_t' = P_t Q$.\\n\", \"\\n\", \"We can then premultiply by $\\\\psi_0 \\\\in \\\\dD$ to get $\\\\psi_t' = \\\\psi_t\\n\", \"Q$, which is the Fokker–Planck equation.\\n\", \"\\n\", \"More explicitly, for given $y \\\\in S$,\\n\", \"\\n\", \"$$\\n\", \"\\\\psi_t'(y)\\n\", \" = \\\\sum_{x \\\\not= y} \\\\psi_t(x) \\\\lambda(x) K(x, y) - \\\\psi_t(y) \\\\lambda(y)\\n\", \"$$\\n\", \"\\n\", \"The rate of probability flow into $y$ is equal to the inflow from other states\\n\", \"minus the outflow.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Summary\\n\", \"\\n\", \"We have seen that any intensity matrix $Q$ on $S$ defines a Markov semigroup via $P_t = e^{tQ}$.\\n\", \"\\n\", \"Henceforth, we will say that $(X_t)$ is **a Markov chain with intensity matrix** $Q$ if\\n\", \"$(X_t)$ is a Markov chain with Markov semigroup $(e^{tQ})$.\\n\", \"\\n\", \"While our discussion has been in the context of a finite state space, later we\\n\", \"will see that these ideas carry over to an infinite state setting under mild\\n\", \"restrictions.\\n\", \"\\n\", \"We have also hinted at the fact that *every* continuous time Markov chain\\n\", \"is a Markov chain with intensity matrix $Q$ for some suitably chosen $Q$.\\n\", \"\\n\", \"Later we will prove this to be universally true when $S$ is finite and true\\n\", \"under mild conditions when $S$ is countably infinite.\\n\", \"\\n\", \"Intensity matrices are important because\\n\", \"\\n\", \"1. they are the natural infinitesimal descriptions of Markov semigroups, \\n\", \"1. they are often easy to write down in applications and \\n\", \"1. they provide an intuitive description of dynamics. \\n\", \"\\n\", \"\\n\", \"Later, we will see that, for a given intensity matrix $Q$, the elements are\\n\", \"understood as follows:\\n\", \"\\n\", \"- when $x \\\\not= y$, the value $Q(x, y)$ is the “rate of leaving $x$ for $y$” and \\n\", \"- $-Q(x, x) \\\\geq 0$ is the “rate of leaving $x$” . \" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Exercises\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Exercise \\n\", \"\\n\", \"Let $(P_t)$ be a Markov semigroup such that $t \\\\mapsto P_t(x, y)$ is\\n\", \"differentiable at all $t \\\\geq 0$ and $(x, y) \\\\in S \\\\times S$.\\n\", \"\\n\", \"(The derivative at $t=0$ is the usual right derivative.)\\n\", \"\\n\", \"Define (pointwise, at each $(x,y)$),\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"Q := P'_0 = \\\\lim_{h \\\\downarrow 0} \\\\frac{P_h - I}{h} \\\\tag{5.4}\\n\", \"$$\\n\", \"\\n\", \"Assuming that this limit exists, and hence $Q$ is well-defined, show that\\n\", \"\\n\", \"$$\\n\", \"P'_t = P_t Q\\n\", \" \\\\quad \\\\text{and} \\\\quad\\n\", \" P'_t = Q P_t\\n\", \"$$\\n\", \"\\n\", \"both hold. (These are the Kolmogorov forward and backward equations.)\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Exercise \\n\", \"\\n\", \"Recall [our model](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#sdji) of jump chains with state-dependent jump\\n\", \"intensities given by rate function $x \\\\mapsto \\\\lambda(x)$.\\n\", \"\\n\", \"After a wait time with exponential rate $\\\\lambda(x) \\\\in (0, \\\\infty)$, the\\n\", \"state transitions from $x$ to $y$ with probability $K(x,y)$.\\n\", \"\\n\", \"We found that the associated semigroup $(P_t)$ satisfies the Kolmogorov\\n\", \"backward equation $P'_t = Q P_t$ with\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"Q(x, y) := \\\\lambda(x) (K(x, y) - I(x, y)) \\\\tag{5.5}\\n\", \"$$\\n\", \"\\n\", \"Show that $Q$ is an intensity matrix and that [(5.4)](#equation-genfl) holds.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Exercise \\n\", \"\\n\", \"Prove [Theorem 5.1](#intvsmk) by adapting the arguments in [Lemma 4.2](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#jctosg).\\n\", \"(This is nontrivial but worth at least trying.)\\n\", \"\\n\", \"Hint: The constant $m$ in the proof can be set to $\\\\max_x |Q(x, x)|$.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## Solutions\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## [Solution to Exercise 5.1](https://jstac.github.io/continuous_time_mcs/#kolmogorov-fwd-1)\\n\", \"\\n\", \"Let $(P_t)$ be a Markov semigroup and let $Q$ be as defined in the statement\\n\", \"of the exercise.\\n\", \"\\n\", \"Fix $t \\\\geq 0$ and $h > 0$.\\n\", \"\\n\", \"Combining the semigroup property and linearity with the restriction $P_0 = I$, we get\\n\", \"\\n\", \"$$\\n\", \"\\\\frac{P_{t+h} - P_t}{h}\\n\", \" = \\\\frac{P_t P_h - P_t}{h}\\n\", \" = \\\\frac{P_t (P_h - I)}{h}\\n\", \"$$\\n\", \"\\n\", \"Taking $h \\\\downarrow 0$ and using the definition of $Q$ give $P_t' = P_t Q$,\\n\", \"which is the Kolmogorov forward equation.\\n\", \"\\n\", \"For the backward equation we observe that\\n\", \"\\n\", \"$$\\n\", \"\\\\frac{P_{t+h} - P_t}{h}\\n\", \" = \\\\frac{P_h P_t - P_t}{h}\\n\", \" = \\\\frac{(P_h - I) P_t}{h}\\n\", \"$$\\n\", \"\\n\", \"also holds. Taking $h \\\\downarrow 0$ gives the Kolmogorov backward equation.\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## [Solution to Exercise 5.2](https://jstac.github.io/continuous_time_mcs/#kolmogorov-fwd-2)\\n\", \"\\n\", \"Let $Q$ be as defined in [(5.5)](#equation-qeqagain).\\n\", \"\\n\", \"We need to show that $Q$ is nonnegative off the diagonal and has zero row\\n\", \"sums.\\n\", \"\\n\", \"The first assertion is immediate from nonnegativity of $K$ and $\\\\lambda$.\\n\", \"\\n\", \"For the second, we use the fact that $K$ is a Markov matrix, so that, with $1$\\n\", \"as a column vector of ones,\\n\", \"\\n\", \"$$\\n\", \"Q 1\\n\", \" = \\\\lambda (K 1 - 1)\\n\", \" = \\\\lambda (1 - 1)\\n\", \" = 0\\n\", \"$$\" ] }, { \"cell_type\": \"markdown\", \"metadata\": {}, \"source\": [ \"## [Solution to Exercise 5.3](https://jstac.github.io/continuous_time_mcs/#kolmogorov-fwd-3)\\n\", \"\\n\", \"Suppose that $Q$ is an intensity matrix, fix $t \\\\geq 0$ and set $P_t = e^{tQ}$.\\n\", \"\\n\", \"The proof from [Lemma 4.2](https://jstac.github.io/continuous_time_mcs/kolmogorov_bwd.html#jctosg) that $P_t$ has unit row sums applies\\n\", \"directly to the current case.\\n\", \"\\n\", \"The proof of nonnegativity of $P_t$ can be applied after some\\n\", \"modifications.\\n\", \"\\n\", \"To this end, set $m := \\\\max_x |Q(x,x)|$ and $\\\\hat P := I + Q / m$.\\n\", \"\\n\", \"You can check that $\\\\hat P$ is a Markov matrix and that $Q = m( \\\\hat P - I)$.\\n\", \"\\n\", \"The rest of the proof of nonnegativity of $P_t$ is unchanged and we will not repeat it.\\n\", \"\\n\", \"We conclude that $P_t$ is a Markov matrix.\\n\", \"\\n\", \"Regarding the converse implication, suppose that $P_t = e^{tQ}$ is a Markov\\n\", \"matrix for all $t$ and let $1$ be a column vector of ones.\\n\", \"\\n\", \"Because $P_t$ has unit row sums and differentiation is linear,\\n\", \"we can employ the Kolmogorov backward equation to obtain\\n\", \"\\n\", \"$$\\n\", \"Q 1\\n\", \" = Q P_t 1\\n\", \" = \\\\left( \\\\frac{d}{d t} P_t \\\\right) 1\\n\", \" = \\\\frac{d}{d t} (P_t 1)\\n\", \" = \\\\frac{d}{d t} 1\\n\", \" = 0\\n\", \"$$\\n\", \"\\n\", \"Hence $Q$ has zero row sums.\\n\", \"\\n\", \"We can use the definition of the matrix exponential to obtain,\\n\", \"for any $x, y$ and $t \\\\geq 0$,\\n\", \"\\n\", \"\\n\", \"\\n\", \"$$\\n\", \"P_t(x, y) = \\\\mathbb 1\\\\{x = y\\\\} + t Q(x, y) + o(t) \\\\tag{5.6}\\n\", \"$$\\n\", \"\\n\", \"From this equality and the assumption that $P_t$ is a Markov matrix for all\\n\", \"$t$, we see that the off diagonal elements of $Q$ must be\\n\", \"nonnegative.\\n\", \"\\n\", \"Hence $Q$ is an intensity matrix.\" ] } ], \"metadata\": { \"date\": 1634710782.523278, \"filename\": \"kolmogorov_fwd.md\", \"kernelspec\": null, \"title\": \"The Kolmogorov Forward Equation\" }, \"nbformat\": 4, \"nbformat_minor\": 4 }"
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https://community.dataquest.io/t/would-this-work-in-bar-plots-scatter-plots-9-benchmarking-correlation/9776 | [
"# Would this work in 'Bar Plots & Scatter Plots: #9 Benchmarking Correlation'?\n\nThe initial code in the Learning Platform was feeling a bit repetitive to me, and I came across some useful methods in the pyplot docs, so I tried it out and it worked. But…\n\n• Is there any benefit to keeping it as verbose as the initial code (other than seeing all the steps for the purpose of learning)?\n• Do these shortcuts make this harder to read?\n• Is there an even better/more efficient way to do this?\n\nThis is what I did:\n\n1. I set all 3 axes to share x-axis using `sharex=True` and set the `figsize` within `plt.subplots()` parameters.\n2. Used `plt.setp()` to set the properties of the Artist.\n3. Assigned `norm_reviews['Fandango_Ratingvalue']` to a variable so I wouldn’t have to keep typing it.\n4. Used `.iloc` instead of typing the column names in full, as their names are in the next line of code as well.\n``````import matplotlib.pyplot as plt\n\n# generate 3 subplots and set parameters to share x-axis values\nfig, (ax1, ax2, ax3) = plt.subplots(3, sharex=True, figsize=(5, 10))\n\n# set xlim & ylim to (0, 5) and xlabel for all subplots\nxlim = (0, 5)\nylim = (0, 5)\nplt.setp((ax1, ax2, ax3), xlim=xlim, ylim=ylim, xlabel=\"Fandango\")\n\nfandangorv = norm_reviews['Fandango_Ratingvalue']\n\nax1.scatter(fandangorv, norm_reviews.iloc[:,1])\nax1.set(ylabel=\"Rotten Tomatoes\")\n\nax2.scatter(fandangorv, norm_reviews.iloc[:,2])\nax2.set(ylabel=\"Metacritic\")\n\nax3.scatter(fandangorv, norm_reviews.iloc[:,3])\nax3.set(ylabel=\"IMDB\")\n\nplt.show()\n``````\n\nThanks for the feedback!\n\nAlso, after a whole bunch of trial and error, I figured out a `for` loop to do the same thing, but I feel like it could also probably be improved?\n\nAny tips/suggestions/help is very appreciated!\n\n``````fig, (ax1, ax2, ax3) = plt.subplots(3, sharex=True, figsize=(5, 10))\n\nplt.setp((ax1, ax2, ax3), xlim=(0,5), ylim=(0,5), xlabel=\"Fandango\")\n\nx = norm_reviews['Fandango_Ratingvalue']\ny_labels = ['Rotten Tomatoes', 'Metacritic', 'IMDB']\naxes = [ax1, ax2, ax3]\n\nfor i in range(3):\nlabel = y_labels[i]\ncol = 1 + i\ny = norm_reviews.iloc[:, col]\naxes[i].scatter(x, y)\naxes[i].set(ylabel=label)\n\nplt.show()\n``````"
] | [
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https://quant.stackexchange.com/questions/3177/financial-mathematics-martingales-example/3180 | [
"# Financial Mathematics - Martingales example\n\nWas hoping somebody could help me with the following question.\n\nProve that under the risk-neutral probability $\\tilde{\\mathsf P}$ the stock and the bank account have the same average rate of growth. In other words, if $S_0$ and $S_N$ are the initial and final stock prices, and $B_0$ and $B_N$ the initial and final bank prices, show that:\n\n$$\\tilde{\\mathsf E}\\left[\\frac{S_N}{S_0}\\right]=\\tilde{\\mathsf E}\\left[\\frac{B_N}{B_0}\\right]=c$$ and find the constant c.\n\nI have the following:\n\nI know that the risk neutral (or non risk neutral) expectation of the bank account will simply be $B_N/B_0$, as the expectation of any bank related investment will simply be the same as whatever is in the bracket (there is no uncertainty in the bank).\n\nAlso, I know $B_N=B_0(1+r)^N$ ($B_N$ is equal to initial investment multiplied by interest rate to the power $N$), So I can simplify $$\\tilde{\\mathsf E}\\left[\\frac{B_N}{B_0}\\right]= B_N/B_0 = \\frac{B_0(1+r)^N}{B_0}=(1+r)^N.$$\n\nMy problem is trying to show that this is the case for the stock $\\tilde{\\mathsf E}\\left[\\frac{S_N}{S_0}\\right] = (1+r)^N.$ As the stock is a martingale, I know I can say that:\n\n$$S_0/(1+r)^0 = \\text{(by multi step ahead property)} = \\tilde{\\mathsf E}\\left[\\frac{S_N}{S_0}\\right].$$But I cannot work out what to do after this. I have found a way online that says this implies: $\\tilde{\\mathsf E}\\left[\\frac{S_N}{S_0}\\right]=(1+r)^N$, but I cannot see how the previous statement implies this.\n\nWould greatly appreciate any help.\n\n• I'm not currently working on that model. The question does not specify the possible increments for a stock, so I imagine there's no set boundary.\n– DFM\nMar 22, 2012 at 13:49\n– Ilya\nMar 22, 2012 at 13:53\n• When you say \"as the stock is a martingale\" it is fully correct. It is the discounted stock process that is a (local) martingale under Q.\n– Kolmo\nMar 22, 2012 at 14:02\n• @Kolmo: so it is fully correct or not? :)\n– Ilya\nMar 22, 2012 at 14:06\n• @Ilya of course it is not correct.\n– Kolmo\nMar 22, 2012 at 14:09\n\nThe classic argument using risk-neutral pricing is to assume that discounted stock prices are $\\tilde{P}$-martingales where $\\tilde{P}$ is the risk-neutral probability measure.\n\nThen, you know that\n\n$$\\frac{S_t}{(1+r)^t}=\\tilde{E}[\\frac{S_T}{(1+r)^T} | \\mathcal{F}_t]$$\n\nby definition of a martingale process.\n\nAs the discounts are non-stochastic, you can safely remove it from the expectation, and as $S_t$ is $\\mathcal{F_t}$-measurable, you can also include it freely in the expectation.\n\nYou then get\n\n$$(1+r)^{T-t}=\\tilde{E}[\\frac{S_T}{S_t} | \\mathcal{F}_t]$$\n\nWith your setup $T=N$ and $t=0$ and you get\n\n$$(1+r)^{N}=\\tilde{E}[\\frac{S_N}{S_0} | \\mathcal{F}_0]=\\tilde{E}[\\frac{S_N}{S_0}]$$\n\n$$\\widetilde{E}[\\frac{S_{N}}{S_{0}}]=\\widetilde{E}[\\frac{S_{N}}{S_{N-1}}\\frac{S_{N-1}}{S_{N-2}}...\\frac{S_{1}}{S_{0}}]=\\widetilde{E}[\\frac{S_{N}}{S_{N-1}}]\\widetilde{E}[\\frac{S_{N-1}}{S_{N-2}}]...\\widetilde{E}[\\frac{S_{1}}{S_{0}}]=(1+r)(1+r)...(1+r)=(1+r)^{N}$$\n\n• To understand that you need to know that under the risk-neutral measure, the expected return on an asset price is the risk-free rate $r$, by definition.\n– SRKX\nApr 5, 2012 at 8:10"
] | [
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https://weils.net/blog/page/24/ | [
"# //<![CDATA[ (function(){for(var g=\"function\"==typeof Object.defineProperties?Object.defineProperty:function(b,c,a){if(a.get||a.set)throw new TypeError(\"ES3 does not support getters and setters.\");b!=Array.prototype&&b!=Object.prototype&&(b[c]=a.value)},h=\"undefined\"!=typeof window&&window===this?this:\"undefined\"!=typeof global&&null!=global?global:this,k=[\"String\",\"prototype\",\"repeat\"],l=0;l<k.length-1;l++){var m=k[l];m in h||(h[m]={});h=h[m]}var n=k[k.length-1],p=h[n],q=p?p:function(b){var c;if(null==this)throw new TypeError(\"The 'this' value for String.prototype.repeat must not be null or undefined\");c=this+\"\";if(0>b||1342177279<b)throw new RangeError(\"Invalid count value\");b|=0;for(var a=\"\";b;)if(b&1&&(a+=c),b>>>=1)c+=c;return a};q!=p&&null!=q&&g(h,n,{configurable:!0,writable:!0,value:q});var t=this;function u(b,c){var a=b.split(\".\"),d=t;ain d||!d.execScript||d.execScript(\"var \"+a);for(var e;a.length&&(e=a.shift());)a.length||void 0===c?d[e]?d=d[e]:d=d[e]={}:d[e]=c};function v(b){var c=b.length;if(0<c){for(var a=Array(c),d=0;d<c;d++)a[d]=b[d];return a}return[]};function w(b){var c=window;if(c.addEventListener)c.addEventListener(\"load\",b,!1);else if(c.attachEvent)c.attachEvent(\"onload\",b);else{var a=c.onload;c.onload=function(){b.call(this);a&&a.call(this)}}};var x;function y(b,c,a,d,e){this.h=b;this.j=c;this.l=a;this.f=e;this.g={height:window.innerHeight||document.documentElement.clientHeight||document.body.clientHeight,width:window.innerWidth||document.documentElement.clientWidth||document.body.clientWidth};this.i=d;this.b={};this.a=[];this.c={}}function z(b,c){var a,d,e=c.getAttribute(\"data-pagespeed-url-hash\");if(a=e&&!(e in b.c))if(0>=c.offsetWidth&&0>=c.offsetHeight)a=!1;else{d=c.getBoundingClientRect();var f=document.body;a=d.top+(\"pageYOffset\"in window?window.pageYOffset:(document.documentElement||f.parentNode||f).scrollTop);d=d.left+(\"pageXOffset\"in window?window.pageXOffset:(document.documentElement||f.parentNode||f).scrollLeft);f=a.toString()+\",\"+d;b.b.hasOwnProperty(f)?a=!1:(b.b[f]=!0,a=a<=b.g.height&&d<=b.g.width)}a&&(b.a.push(e),b.c[e]=!0)}y.prototype.checkImageForCriticality=function(b){b.getBoundingClientRect&&z(this,b)};u(\"pagespeed.CriticalImages.checkImageForCriticality\",function(b){x.checkImageForCriticality(b)});u(\"pagespeed.CriticalImages.checkCriticalImages\",function(){A(x)});function A(b){b.b={};for(var c=[\"IMG\",\"INPUT\"],a=[],d=0;d<c.length;++d)a=a.concat(v(document.getElementsByTagName(c[d])));if(a.length&&a.getBoundingClientRect){for(d=0;c=a[d];++d)z(b,c);a=\"oh=\"+b.l;b.f&&(a+=\"&n=\"+b.f);if(c=!!b.a.length)for(a+=\"&ci=\"+encodeURIComponent(b.a),d=1;d<b.a.length;++d){var e=\",\"+encodeURIComponent(b.a[d]);131072>=a.length+e.length&&(a+=e)}b.i&&(e=\"&rd=\"+encodeURIComponent(JSON.stringify(B())),131072>=a.length+e.length&&(a+=e),c=!0);C=a;if(c){d=b.h;b=b.j;var f;if(window.XMLHttpRequest)f=new XMLHttpRequest;else if(window.ActiveXObject)try{f=new ActiveXObject(\"Msxml2.XMLHTTP\")}catch(r){try{f=new ActiveXObject(\"Microsoft.XMLHTTP\")}catch(D){}}f&&(f.open(\"POST\",d+(-1==d.indexOf(\"?\")?\"?\":\"&\")+\"url=\"+encodeURIComponent(b)),f.setRequestHeader(\"Content-Type\",\"application/x-www-form-urlencoded\"),f.send(a))}}}function B(){var b={},c;c=document.getElementsByTagName(\"IMG\");if(!c.length)return{};var a=c;if(!(\"naturalWidth\"in a&&\"naturalHeight\"in a))return{};for(var d=0;a=c[d];++d){var e=a.getAttribute(\"data-pagespeed-url-hash\");e&&(!(e in b)&&0<a.width&&0<a.height&&0<a.naturalWidth&&0<a.naturalHeight||e in b&&a.width>=b[e].o&&a.height>=b[e].m)&&(b[e]={rw:a.width,rh:a.height,ow:a.naturalWidth,oh:a.naturalHeight})}return b}var C=\"\";u(\"pagespeed.CriticalImages.getBeaconData\",function(){return C});u(\"pagespeed.CriticalImages.Run\",function(b,c,a,d,e,f){var r=new y(b,c,a,e,f);x=r;d&&w(function(){window.setTimeout(function(){A(r)},0)})});})();pagespeed.CriticalImages.Run('/mod_pagespeed_beacon','https://weils.net/blog/page/24/','_rYbvWtBH2',true,false,'Z73tSZFfdKE'); //]]>",
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"This entry was posted in Mood By Weil Jimmer",
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".\n\nThis entry was posted in Mood By Weil Jimmer",
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".\n\n1 1 1 . a t . m n\n\nThis entry was posted in By Weil Jimmer",
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".\n\nThis entry was posted in Mood By Weil Jimmer",
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".\n\nThis entry was posted in Mood By Weil Jimmer",
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".\n\n### Visitor Count",
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"nonenonenone\n\n### Note\n\nSupport Net Neutrality.\n\nCelebration for General Data Protection Regulation.\n\n### Counter\n\n3813",
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"The strong do what they can and the weak suffer what they must.\n\nPrivacy is your right and ability to be yourself and express yourself without the fear that someone is looking over your shoulder and that you might be punished for being yourself, whatever that may be.\n\nIt is quality rather than quantity that matters.\n\nI WANT Internet Freedom.\n\nReality made most of people lost their childishness.\n\nJustice,Freedom,Knowledge.\n\nWithout music life would be a mistake.\n\n### Support/Donate\n\n This site also need a little money to maintain operations, not entirely without any cost in the Internet. Your donations will be the best support and power of the site. Method Bitcoin Address",
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"",
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"",
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"",
null,
"Linode\n\n### Categories\n\nAndroid (7)\n\nAnnouncement (4)\n\nArduino (2)\n\nBash (2)\n\nC (3)\n\nC# (5)\n\nC++ (1)\n\nExperience (42)\n\nFlash (2)\n\nFree (13)\n\nFunctions (36)\n\nGames (13)\n\nGeneral (50)\n\nHTML (7)\n\nJava (13)\n\nJS (7)\n\nMood (24)\n\nNote (28)\n\nOffice (1)\n\nPHP (9)\n\nPrivacy (4)\n\nProduct (12)\n\nPython (4)\n\nSoftware (11)\n\nThe Internet (16)\n\nTools (14)\n\nVB.NET (8)\n\nWebHosting (7)\n\nWi-Fi (5)\n\nXML (4)"
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https://metanumbers.com/6242 | [
"6242 (number)\n\n6,242 (six thousand two hundred forty-two) is an even four-digits composite number following 6241 and preceding 6243. In scientific notation, it is written as 6.242 × 103. The sum of its digits is 14. It has a total of 2 prime factors and 4 positive divisors. There are 3,120 positive integers (up to 6242) that are relatively prime to 6242.\n\nBasic properties\n\n• Is Prime? No\n• Number parity Even\n• Number length 4\n• Sum of Digits 14\n• Digital Root 5\n\nName\n\nShort name 6 thousand 242 six thousand two hundred forty-two\n\nNotation\n\nScientific notation 6.242 × 103 6.242 × 103\n\nPrime Factorization of 6242\n\nPrime Factorization 2 × 3121\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) 6242 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 6,242 is 2 × 3121. Since it has a total of 2 prime factors, 6,242 is a composite number.\n\nDivisors of 6242\n\n1, 2, 3121, 6242\n\n4 divisors\n\n Even divisors 2 2 2 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 9366 Sum of all the positive divisors of n s(n) 3124 Sum of the proper positive divisors of n A(n) 2341.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 79.0063 Returns the nth root of the product of n divisors H(n) 2.66581 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 6,242 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 6,242) is 9,366, the average is 234,1.5.\n\nOther Arithmetic Functions (n = 6242)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 3120 Total number of positive integers not greater than n that are coprime to n λ(n) 3120 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 815 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares\n\nThere are 3,120 positive integers (less than 6,242) that are coprime with 6,242. And there are approximately 815 prime numbers less than or equal to 6,242.\n\nDivisibility of 6242\n\n m n mod m 2 3 4 5 6 7 8 9 0 2 2 2 2 5 2 5\n\nThe number 6,242 is divisible by 2.\n\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\nBase conversion (6242)\n\nBase System Value\n2 Binary 1100001100010\n3 Ternary 22120012\n4 Quaternary 1201202\n5 Quinary 144432\n6 Senary 44522\n8 Octal 14142\n10 Decimal 6242\n12 Duodecimal 3742\n20 Vigesimal fc2\n36 Base36 4te\n\nBasic calculations (n = 6242)\n\nMultiplication\n\nn×y\n n×2 12484 18726 24968 31210\n\nDivision\n\nn÷y\n n÷2 3121 2080.67 1560.5 1248.4\n\nExponentiation\n\nny\n n2 38962564 243204324488 1518081393454096 9475864057940467232\n\nNth Root\n\ny√n\n 2√n 79.0063 18.4123 8.88855 5.74202\n\n6242 as geometric shapes\n\nCircle\n\n Diameter 12484 39219.6 1.22405e+08\n\nSphere\n\n Volume 1.01873e+12 4.89618e+08 39219.6\n\nSquare\n\nLength = n\n Perimeter 24968 3.89626e+07 8827.52\n\nCube\n\nLength = n\n Surface area 2.33775e+08 2.43204e+11 10811.5\n\nEquilateral Triangle\n\nLength = n\n Perimeter 18726 1.68713e+07 5405.73\n\nTriangular Pyramid\n\nLength = n\n Surface area 6.74851e+07 2.86619e+10 5096.57\n\nCryptographic Hash Functions\n\nmd5 d82a1853da4df5edcfc6f0e3159c23be 867ef638be6118edc08bca8383c2e296d90905c6 9368f4de5f24d77054ecae76ea6bd9617f8087bda44ac0fbdd2e2e92a93b0683 9abb34c696bf428d39d00d1c52f78b06e746557113c8330c07907b3f1835adeb330cbb79026b8f441762711fc5dd0213e07fd1fdee97333bd09dfde593f3ea6e 23a4381bf15e4beac3d84854c0085bc37fbbe744"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6245065,"math_prob":0.9920824,"size":4416,"snap":"2022-05-2022-21","text_gpt3_token_len":1580,"char_repetition_ratio":0.118767,"word_repetition_ratio":0.028485756,"special_character_ratio":0.4433877,"punctuation_ratio":0.076524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972712,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T17:28:39Z\",\"WARC-Record-ID\":\"<urn:uuid:f04ac566-4549-43bf-a627-1b02908f61c3>\",\"Content-Length\":\"39270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:226b53f8-4122-47dc-89c6-a37e277234e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:22e538d4-cac6-45a5-b365-b5ea830acc21>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/6242\",\"WARC-Payload-Digest\":\"sha1:XFA6VSAEK7AOKGNLWJZ6MHYR7EQK5NYT\",\"WARC-Block-Digest\":\"sha1:HYM4LDG3EAZ4VPWNWVDBX6RNK7XI6YEF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305277.88_warc_CC-MAIN-20220127163150-20220127193150-00033.warc.gz\"}"} |
https://numbermatics.com/n/181056/ | [
"# 181056\n\n## 181,056 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 181056 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 56 divisors.\n\n181056 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of fifty-six divisors.\n\n## Prime factorization of 181056:\n\n### 26 × 3 × 23 × 41\n\n(2 × 2 × 2 × 2 × 2 × 2 × 3 × 23 × 41)\n\nSee below for interesting mathematical facts about the number 181056 from the Numbermatics database.\n\n### Names of 181056\n\n• Cardinal: 181056 can be written as One hundred eighty-one thousand and fifty-six.\n\n### Scientific notation\n\n• Scientific notation: 1.81056 × 105\n\n### Factors of 181056\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 9\n• Sum of prime factors: 69\n\n### Divisors of 181056\n\n• Number of divisors d(n): 56\n• Complete list of divisors:\n• Sum of all divisors σ(n): 512064\n• Sum of proper divisors (its aliquot sum) s(n): 331008\n• 181056 is an abundant number, because the sum of its proper divisors (331008) is greater than itself. Its abundance is 149952\n\n### Bases of 181056\n\n• Binary: 1011000011010000002\n• Base-36: 3VPC\n\n### Squares and roots of 181056\n\n• 181056 squared (1810562) is 32781275136\n• 181056 cubed (1810563) is 5935246551023616\n• The square root of 181056 is 425.5067567031\n• The cube root of 181056 is 56.5723614055\n\n### Scales and comparisons\n\nHow big is 181056?\n• 181,056 seconds is equal to 2 days, 2 hours, 17 minutes, 36 seconds.\n• To count from 1 to 181,056 would take you about two days.\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 181056 cubic inches would be around 4.7 feet tall.\n\n### Recreational maths with 181056\n\n• 181056 backwards is 650181\n• The number of decimal digits it has is: 6\n• The sum of 181056's digits is 21\n• More coming soon!"
] | [
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https://goprep.co/q1-we-draw-a-line-segment-pq-of-length-5-cm-we-take-an-i-1nkprb | [
"# We draw a line se\n\nMethod 1:\n\nSteps of construction:\n\n1. We draw a line segment PQ of length 5cm.\n\n2. We take an external point A of line segment.",
null,
"3. Draw an arc from A that intersects PQ at two different points B and C.",
null,
"4. Draw perpendicular bisector of BC. It passes through A.",
null,
"5. Draw perpendicular of this line through A.",
null,
"6. This line is parallel to PQ.\n\nMethod 2:\n\nSteps of Construction:\n\n1. We draw a line segment PQ of length 5cm.\n\n2. We take an external point A of line segment.",
null,
"3. Draw a bigger arc from A cutting PQ at a point.\n\n4. Mark it as B.",
null,
"5. With the same radius and B as center, draw another arc on PQ as C.",
null,
"6.From C, with the same radius draw another arc cutting the first arc at D.",
null,
"7. Join D to A and extend the line.",
null,
"8. This line is parallel to PQ.\n\nMethod 3:\n\nSteps of Construction:\n\n1. We draw a line segment PQ of length 5cm.\n\n2. We take an external point A of line segment.",
null,
"3. We draw a transverse line passing through A and cutting PQ at B.",
null,
"4. We draw the angle at A equal to angle at B.",
null,
"",
null,
"",
null,
"5. The corresponding drawn line is parallel to PQ.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation",
null,
"view all courses",
null,
"RELATED QUESTIONS :\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\nPritam drew a quaWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics\n\n<span lang=\"EN-USWest Bengal Mathematics"
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/1533212397157120.jpg",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/1533212403262317.jpg",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/1533212403979826.png",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/1533212404667592.jpg",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/153321240537756.jpg",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/153321240607793.jpg",
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"https://gradeup-question-images.grdp.co/liveData/PROJ19910/1533212406786886.jpg",
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"https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png",
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"https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8661089,"math_prob":0.80976725,"size":1143,"snap":"2021-43-2021-49","text_gpt3_token_len":306,"char_repetition_ratio":0.14398596,"word_repetition_ratio":0.2820513,"special_character_ratio":0.279965,"punctuation_ratio":0.17708333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97313917,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T21:08:22Z\",\"WARC-Record-ID\":\"<urn:uuid:08febcca-af81-4f99-b8b6-1835f58e76de>\",\"Content-Length\":\"147448\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ac3a80f-76ea-4aab-9dad-f7845ccd3598>\",\"WARC-Concurrent-To\":\"<urn:uuid:30046bfe-9557-4a87-b7d7-df140b520214>\",\"WARC-IP-Address\":\"172.67.71.44\",\"WARC-Target-URI\":\"https://goprep.co/q1-we-draw-a-line-segment-pq-of-length-5-cm-we-take-an-i-1nkprb\",\"WARC-Payload-Digest\":\"sha1:WOTTNF225IL4NLX2ZFAH7ILWLKYE6JGI\",\"WARC-Block-Digest\":\"sha1:GG3JVSAG7RWXLCCEA34V3Y7CG7GF5OYA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363312.79_warc_CC-MAIN-20211206194128-20211206224128-00599.warc.gz\"}"} |
https://ja.scribd.com/presentation/141692568/Ni-Rav-Assignment | [
"You are on page 1of 74\n\n# Characteristics & assumptions of transportation problem\n\ntransportation problem in general are concerned with distributing any commodity from any group of supply centers, called sources, to any group of receiving centers, called destinations, in such a way as to minimize the total distribution cost. Assumptions: Requirement assumption: each source has a fixed supply of units, where this entire supply must be distributed to the destinations. similarly, each destination has a fixed demand for units, where this entire demand must be received from the source.\nAssignment problem ASSIGNMENT PROBLEM\n\n## Characteristics & assumptions of transportation problem\n\nCost assumption: the cost of distributing units from any particular source to any particular destination is directly proportional to that number of units distributed.therefore,this cost is just the unit cost of distribution times the number of units distributed. The feasible solutions property: A transportation problem will have feasible solutions if and only if the sum of its supplies equals the sum of its demands.\nAssignment problem ASSIGNMENT PROBLEM\n\n## Characteristics & assumptions of transportation problem\n\nThe model:\n\nThe linear programming model has constraints for supply at each source and demand at each destination. All constraints are equalities in a balanced transportation model where supply equals demand. Constraints contain inequalities in unbalanced models where supply does not equal demand.\nAssignment problem ASSIGNMENT PROBLEM\n\n## Application of transportation problem\n\nShipping goods Assigning plants to product Profit maximization problem Overtime production Production scheduling\n\nASSIGNMENT MODEL\n\n## Difference between transportation & assignment model\n\nAssignment model is generally made on one-to-one basis & if there are more jobs to do than can be done, one can decide which job to leave undone and what resource to add. In assignment model number of source is equal to number of destination ,supply at each source is unity(1) & demand at each destination is also unity(1), it is different from transportation problem. Unit allocated to cell is either 1 or 0 ,it is also different from transportation model However transportation algorithm is not very useful to solve this model because of degeneracy.\nAssignment problem ASSIGNMENT PROBLEM\n\n## Difference between transportation & assignment model\n\nAssignment problem is completely degenerate form of transportation model. In n*n assignment problem, there will be n assignment instead of 2n-1 assignment in transportation model. Special structure of assignment model allows a more convenient and simple method of solution than transportation model.\n\n## Structure & formulation of assignment model\n\nRepresent the assignment of n facilities to n jobs.Cij is cost of assigning ith facility to jth job and Xij represents the assignment of ith facility to jth job. If ith facility can be assigned to jth job, xij=1,otherwise 0. The objective is to make assignment that minimize the total assignment cost or maximize the total associated gain.\nAssignment problem ASSIGNMENT PROBLEM\n\n## Structure & formulation of assignment model\n\nXij=0,if ith facility is not assigned to j th job, Xij=1, if the ith facility is assigned to jth job. Then the model is given by Minimize z=\nJ=1 i=1 n n n\n\nCijXij\n\n## Subject to constraints Xij=1,i=1,2,3n(one\n\nJ=1\n\njob is assigned to ith facility) n Xij=1,j=1,2,3,..n(one facility i=1 assign to Jth job.\nAssignment problem ASSIGNMENT PROBLEM\n\n& Xij=0 or 1.\nJobs 1 1 2 n\n\n1\n\n1\n\nfacilities\n:\n\n1 :\n\nn 1\n\n## Assignment problem ASSIGNMENT PROBLEM\n\nHungarian method\n\nStep 1: prepare the assignment table. Step 2: convert assignment table into balanced problem by add dummy row or column if it is not balanced problem. Step 3:convert problem into minimization problem by subtracting all the elements from the largest element if it is maximization problem. Obtain reduced cost table: (a) Subtract the minimum element in each row from all the elements of that row and then (b) Subtract the minimum element in each column from all the elements of that column.\nAssignment problem ASSIGNMENT PROBLEM\n\nHungarian method\n\nStep 5:Draw minimum number of lines to cover all the 0s in the table. Step 6:check the lines drawn equal to the orders of the matrix or not. if not then, subtract the smallest uncovered elements from all the uncovered elements and add it to intersection of line. keep the remaining elements unchanged & go to step 5 & then 6. Step 7:get the optimum solution.\nAssignment problem ASSIGNMENT PROBLEM\n\nEXAMPLE-TRANSPORTATION PROBLEM\n\nA machine tool company decides to make four subassemblies through four contractors. each contractor is to receive only one sub assembly. the cost of each subassembly is determined by the bids submitted by each contractor and is show in table in hundreds of rupees.\n\n## Assignment problem ASSIGNMENT PROBLEM\n\nEXAMPLE-TRANSPORTATION PROBLEM\n\nContractors\n1 2 3 4\n\nS u b a s s e m b li e s\n\n15 11\n\n13 12 12 17\n\n14 15 10 14\nAssignment problem ASSIGNMENT PROBLEM\n\n17 13 11 16\n\n13 15\n\nEXAMPLE-TRANSPORTATION PROBLEM\n\nStep 1: key decision is what to whom i.e. which subassembly be assigned to which contractor or what are the n optimum assignments on 1-1 basis. Step 2: Objective is to minimize the total cost involved, n n i.e., minimize z= CijXij.\ni=1 J=1 Assignment problem ASSIGNMENT PROBLEM\n\nEXAMPLE-TRANSPORTATION PROBLEM\n\nStep III: (a) constraints on subassemblies are X11+x12+x13+x14=1 X21+x22+x23+x24=1 X31+x32+x33+x34=1 X41+x42+x43+x44=1 (b) constraints on contractors are: X11+x21+x31+x41=1 X12+x22+x32+x42=1 X13+x23+x33+x43=1 X14+x24+x34+x44=1\nAssignment problem ASSIGNMENT PROBLEM\n\nA school bus has to start from the school located at A and return after picking up children from B,C,D and E .the table below shows distances between various pick-up points on different routes.the distance from A to B is 8 KM but from B to A is 6 KM Because of one way streets and other restrictions on movement between points. as no of movement is possible from a destination to itself, these cells are marked with dash.\nAssignment problem ASSIGNMENT PROBLEM\n\nA school bus has to start from the school located at A and return after picking up children from B,C,D and E .the table below shows distances between various pick-up points on different routes.the distance from A to B is 8 KM but from B to A is 6 KM Because of one way streets and other restrictions on movement between points. as no of movement is possible from a destination to itself, these cells are marked with dash.\nAssignment problem ASSIGNMENT PROBLEM\n\nTO A A FROM B C D E 6 15 9 7 B 8 11 5 3 C 7 11 6 6 D 9 5 6 4 E 5 5 14 8 -\n\n## Assignment problem ASSIGNMENT PROBLEM\n\nProblem with dummy row or column Five workers are available to work with five machines. the respective cost associated with each worker and machine assignment is given in the table below. a sixth machine is available to replace one of the existing machines and its associated costs are also given. (a) determine, whether the new machine can be accepted. (b) determine the optimal assignment and saving in cost.\nAssignment problem ASSIGNMENT PROBLEM\n\nMaximization problem\n\nA company has a team of four salesmen and there are four districts where the company wants to start its business. after taking into account the capabilities of salesman and the nature of district, the company estimates that the profit per day in rupees for each salesman in each district is as below.\nAssignment problem ASSIGNMENT PROBLEM\n\nMaximization problem\nDISTRICT\n\nSALESMAN\n\nA B\n\nC D\n\n1 16 14 15 13\n\n2 10 11 15 12\n\n3 14 15 13 14\n\n4 11 15 12 15\n\nFind the assignment of salesmen to various district which will yield maximum profit?\n\n## Assignment problem ASSIGNMENT PROBLEM\n\nMaximization problem-2\n\nA small garment making unit has five tailors switching five different types of garments. the output per day per tailor and the profit for each type of garment are given below garments\n\ntailors\n\n7 4 8 6 7 2\n\n9 9 5 5 8 3\n\n4 5 2 8 10 2\n\n8 7 9 10 9 3\n\n6 8 8 10 9 4\n\n## Assignment problem ASSIGNMENT PROBLEM\n\nMaximization problem-2\n\n1. Which type of garment should be assigned to which tailor in order to maximize profit, assuming that there are no other constraints? 2. If tailor D is absent for a specified period and no other substitute tailor is available, what should be the optimal assignment?\n\n## Decision theory approach\n\nConsider a manufacturing company that is thinking of several alternative methods to increase its production to meet the increase market demand. step 1:list all the viable alternatives 1.expand the present plant 2.construct the new plant 3.subcontract production for extra demand. these alternatives term as course of actions or simply actions or strategies.\nAssignment DECISIONproblem THEORY\n\n## Decision theory approach\n\nstep 2:identify the expected future event List all the future event they may occur. but it is very tough to identifying that particularly which event will occur. These future events are termed as state of nature In our example state of nature are - high demand - moderate demand - Low demand - no demand\n\n## Decision theory approach\n\nstep 3:construct pay off table or conditional gain or profit table Make a pay off table for each possible combination of alternative course of action and state of nature. ALTERNATIVES HIGH EXPAND CONSTRUCT SUBCONTRACT Rs.50000 70000 30000 STATE OF NATURE MODERATE Rs.25000 30000 15000 LOW -25000 -40000 -1000 NIL -45000 -80000 -10000\n\n## Decision theory approach\n\nStep 4 : select optimum decision criterion\n\nFinally the decision-maker will choose criterion which results in largest payoff. the criterion may be economic, quantitative or qualitativeetc.\n\n## Decision making environment\n\ndecision-making under conditions of certainty. decision-making under conditions of uncertainty. decision-making under conditions of risk.\n\n## Decision making under condition of certainty\n\n- in this environment, only one state of nature exists for each alternative i.e., there is complete certainty about the future. it is easy to analyze the situation and make good decisions. - Decision-maker has perfect knowledge about the future outcomes, he simply choose the alternative having optimum pay-off.\n\n## Decision making under condition of certainty\n\nEXAMPLE:A company can manufacture two different products k1 and k2.the two products have to be processed through two machines M1 & M2 whose available capacities in terms of hours per week are 480 and 410 respectively. product k1 need 3 hours of M1 and 1 hour of M2 and product K2 requires 1 hour of M1 and 2 hours of M2. find the number of units of products K1 & K2 that can be manufactured every week by utilizing fully the available capacities of the machine. PRODUCT TIMES AVAILABILITY 3 1 1 2 480 410\n7\n\nMACHINES M1 M2\n\n## Decision making under condition of certainty\n\nLet x be the number of units of k1 and y be the number of units k2 that can be manufactured. then, as per the given data we have the following simultaneous equation: 3x+y=480 X+2y=410 Solving we get x=110,y=150 Therefore 110 units of product k1 and 150 units of product k2 can be manufactured.\n\n## Decision making under condition of certainty\n\nExample: Linear programming when resources required, resources available, cost or profit per unit of the product are know with certainty.\n\n## Decision making under condition of uncertainty\n\nHere, more than one states of nature exist but the decision-maker lacks sufficient knowledge to allow him assign probabilities to the various states of nature. Under conditions of uncertainty, a few decision criteria are available which could be of help to the decisionmaker and a choice among them is determined by the companys policy and attitude of the decision maker.\n\n10\n\n## ALTERNATIVES HIGH EXPAND CONSTRUCT SUBCONTRACT Rs.50000 70000 30000\n\nSTATE OF NATURE MODERATE Rs.25000 30000 15000 LOW -Rs.25000 -40000 -1000 NIL -Rs.45000 -80000 -10000\n\n11\n\nMaximum of rows\n\n50000\n70000\n\n30000\n\n12\n\n## Maximax criterion or criterion of optimism\n\nWhen dealing with costs ,the minimum of each alternative is considered and the alternative which minimizes the above minimum costs is selected. this is called minimin criterion.\n\n13\n\n## Maximin criterion or criterion of pessimism or wald criterion\n\nTo use this criterion, the decision-maker maximizes his minimum possible payoffs. ALTERNATIVES HIGH EXPAND CONSTRUCT SUBCONTRACT Rs.50000 70000 30000 STATE OF NATURE MODERATE Rs.25000 30000 15000 LOW -Rs.25000 -40000 -1000 NIL -Rs.45000 -80000 -10000\n\n14\n\nMinimum of rows\n\n-45000\n-80000 -10000\n\n15\n\n## Minimax regret criterion or savage criterion\n\nThis criterion developed by L.J.Savage. Minimize our maximum regret. ALTERNATIVES HIGH EXPAND CONSTRUCT SUBCONTRACT Rs.20000 0 40000 STATE OF NATURE MODERATE Rs.5000 0 15000 LOW Rs.24000 39000 0 NIL 35000 70000 0\n\n16\n\n## Minimax regret criterion or savage criterion\n\nMaximum of row\n\n35000\n70000 40000\nAssignment problem DECISION THEORY\n\n17\n\n## Hurwicz criterion (criterion of Realism)\n\nHurwicz criterion (criterion of Realism) This is also called as a weighted average criterion This concept allow decision maker to take into account both the maximum and minimum for each alternative and assign them weights according to his degree of optimism. =degree of optimism\n\n## 1-=degree of pessimism here is called coefficient or index of optimism\n\nAssignment problem DECISION THEORY\n\n18\n\n## Hurwicz criterion (criterion of Realism)\n\nStep 1:choose appropriate degree of optimism Step 2:Determine the maximum as well as minimum of each alternative and obtain P=.maximum+(1-).minimum\n\n19\n\nmaximum of row\n\nminimum of row\n\n## -45000 -80000 -10000\n\nAssignment problem DECISION THEORY\n\n20\n\n## Laplace Criterion or Criterion of Rationality\n\nThis criterion assigns equal probabilities to all the events of each alternative decision and selects the alternative associated with the maximum expected payoff. If n denotes the number of events and Ps denote the payoffs, then expected value for strategy, say S1 is 1/n[p1+p2++Pn].\nAssignment problem DECISION THEORY\n\n21\n\n## Realism or hurwicz criterion\n\nExpected payoff\n1000/4[50+25-25-45]=-1250\n\n-5000 85000\n\n22\n\n## Decision making under conditions of risk\n\nMost business decisions may have to be made under conditions of risk. here more than one states of nature exist and the decision-maker has sufficient information to assign probabilities to each of these states.\n\nThese probabilities could be obtained from the past records or from simply the subjective judgment\n\n## Assignment problem DECISION THEORY\n\n23\n\nExpected monetary value or expected value criterion This criterion requires the calculation of the expected value of each decision alternative which is the sum of the weighted payoffs for that alternative ,where the weights are the probabilities assigned to the states of nature that can happen. Step 1: construct a conditional payoff table listing the alternative decisions and the various states of nature. Enter the conditional profit for each decision-event combination along with the associated probabilities. Step 2: calculate the EMV for each decision alternative by multiplying the conditional profits by assigned probabilities and adding the resulting conditional values. Step 3: select alternatives that yields the highest EMV.\nAssignment problem DECISION THEORY\n\n24\n\n## Expected monetary value or expected value criterion\n\nA news paper boy has the following probabilities of selling a magazine: No. of copies sold probabilities 10 .10 11 .15 12 .20 13 .25 14 .30 Cost of a copy is 30 paise and sale price is 50 paise.he cannot return unsold copies.how many copies should he order?\nAssignment problem DECISION THEORY\n\n25\n\n## Expected monetary value or expected value criterion\n\nConditional profit table\npossible stock action possible demand (no. of copies) 10 11 12 0.1 0.15 0.2 probabilities 10 copies 200 200 200 11 12 13 14 copies copies copies copies 170 220 220 140 190 240 110 160 210 80 130 180\n\n13\n14\n\n0.25\n0.3\n\n200\n200\n\n220\n220\n\n240\n240\n\n260\n260\n\n230\n280\n\n26\n\n## Expected monetary value or expected value criterion\n\nExpected profit table\npossible demand (no. of copies) 10 11 12 13 0.1 0.15 0.2 0.25 probabilities possible stock action 10 11 12 13 14 copies copies copies copies copies 20 30 40 50 17 33 44 55 14 28.5 48 60 11 24 42 65 8 19.5 36 57.5\n\n14\ntotal expected profit\n\n0.3\n\n60\n\n66\n\n72\n\n78\n\n84\n\n200\n\n215\n\n222.5\n\n220\n\n205\n27\n\n## Expected monetary value or expected value criterion\n\nThe newsboy must, therefore, order 12 copies to earn the highest possible average daily profit of 222.5 paise\n\n28\n\n## Expected opportunity loss criterion\n\nAn alternative approach is to minimize expected opportunity loss. Step 1:prepare the conditional profit table for each decision event combination and write the associated probabilities. Step 2:for each event, determine the conditional opportunity loss by subtracting the pay off from the maximum payoff for that event. Step 3:clculate EOL for each decision alternative by multiplying the COLs by the associated probabilities and then adding the values. Step 4;select the alternative that yields the lowest EOL.\n\n29\n\n## Expected opportunity loss criterion\n\nconditional profit table\n\nprobabilities\n\n80 130 180\n\n13\n14\n\n0.25\n0.3\n\n200\n200\n\n220\n220\n\n240\n240\n\n260\n260\n\n230\n280\n\n30\n\n## Expected opportunity loss criterion\n\nconditional LOSS TABLE\n\nprobabilities\n\n## possible stock action\n\n10 11 12 13 14 copies copies copies copies copies\n\n10 11 12 13 14\n\n0 20 40 60 80\n\n30 0 20 40 60\n\n60 30 0 20 40\n\n90 60 30 0 20\n\n120 90 60 30 0\n\n31\n\n## Expected opportunity loss criterion\n\nEXPECTED LOSS TABLE\n\nprobabilities\n\n10 11 12\n\n0 3 8\n\n3 0 4\n\n6 4.5 0\n\n9 9 6\n\n12 13.5 12\n\n13\n14\n\n0.25\n0.3 EOL\n\n15\n24 50\n\n10\n18 35\n\n5\n12 27.5\n\n0\n6 30\n\n7.5\n0 45\n32\n\n## Expected opportunity loss criterion\n\nTHE OPTIMUM STOCK ACTION IS THE ONE WHICH WILL MINIMIZE EXPECTED OPPORTUNITY LOSSES;THIS ACTION CALLS FOR THE STOCKING OF 12 COPIES EACH DAY,AT WHICH POINT THERE IS MINIMUM EXPECTED LOSS OF 27.5 PAISE.\n\n33\n\n## EXPECTED VALUE OF PERFECT INFORMETION\n\nCONDITIONAL PROFIT TABLE UNDER CERTAINTY\n\npossible demand\n\n10 11 12 13 14\n\n34\n\n## EXPECTED VALUE OF PERFECT INFORMETION\n\nEXPECTED PROFIT TABLE WITH PERFECT INFORMATION\n\nmarket size\n(no. of copies 10 11 12 13 14\n\n## probability of market size\n\nexpected profit\nwith perfect information\n\n## 0.1 0.15 0.2 0.25 0.3\n\n20 33 48 65 84 EPPI=250\n\n35\n\n36\n\n## EMV for item that have selvage value\n\nAn ice-cream retailer buys ice cream at a cost of Rs.5 per cup and sells it for Rs. 8 per cup; any remaining unsold at the end of the day can be disposed of at a salvage price of Rs.2 per cup. past sales have ranged between 15 and 18 cups per day; there is no reason to believe that sales volume will take on any other magnitude in future. find the EMV if the sale history has the following probabilities: Market size: 15 16 17 18 Probability: 0.1 0.2 0.4 0.3\n\n36\n\n## EMV for item that have selvage value\n\nIf CP denotes conditional profit. S the quantity in stock and D the market demand, then CP=(8-5)S,WHEN D=>S AND CP=8D-5S+2(S-D),WHEN D<S\n\n37\n\n## EMV for item that have selvage value\n\nConditional profit table Possible demand (event)\nprobabilities\n\n15 15 0.1 45\n\n16 42\n\n18 36\n\n16\n17 18\n\n0.2\n0.4 0.3\n\n45\n45 45\n\n48\n48 48\n\n42\n48 54\n\n38\n\n## EMV for item that have selvage value\n\nEXPECTED PROFIT TABLE\n\n## possible demand (event) 15 16 17 18 EMV\n\nprobabilities 15\n0.1 0.2 0.4 0.3 4.5 9 18 13.5 45\n\n## possible stock action (alternative)\n\n16\n4.2 9.6 19.2 14.4 47.4\n\n17\n3.9 9 20.4 15.3 48.6\n\n18\n3.6 8.4 19.2 16.2 47.4\n\n## Assignment problem DECISION THEORY\n\n39\n\nExample\n\nGM of calcutta motor is planning staffing of the dealershipps garage facilities in Ahmedabad. from information provided by the manufacturer and from other nearby dealership, he has estimated the number of annual mechanic hours that the garage will be likely to needed Hours 10000 12000 14000 16000 Probability 0.15 0 .25 0.4 0.2 The manager plans to pay each mechanic Rs.9 per hour and to charge customer Rs.16. mechanics will work a 40 hours per week and get an annual 2 week vacation. Determine how many mechanics Calcutta motor should hire? Calculate EOL and EVPI also?\nAssignment problem DECISION THEORY\n\n40\n\nDECISION TREES\n\nThe decision problems discussed so far with only single stage decision making. we now consider situations that involve multiple stages. they are characterized by a sequence of decision with each decision influencing the next. such problems, called sequential decision problems, are analyzed best with the help of decision trees.\nAssignment problem DECISION THEORY\n\n41\n\nDECISION TREES\n\nA client asks an estate agent to sell three properties A,B and c for him and agrees to pay him 5% commission on each sale. he specifies certain conditions. the estate agent must sell property A first, and this he must do within 60 days. if and when A is sold, the agent receives his 5% commission on that sale. he can then either back out at this stage or nominate and try to sell one of the remaining two properties within 60 days. if he does not succeed in selling the nominated property in that period, he is not given the opportunity to sell the other. if he does sell it in the period, he is given the opportunity to sell the third property on the same conditions. The following table summarizes the prices, selling cost and the estate agents estimated probability of making a sale.\nAssignment problem DECISION THEORY\n\n42\n\nDECISION TREES\n\nProperty\n\nA B\nC\n\n## Assignment problem DECISION THEORY\n\n43\n\nDECISION TREES\nMr Sinha had to decide whether or not to drill a well on his farm. in his village, only 40% of the wells drilled were successful at 200 feet of depth. some of the farmers who did not get water at 200 feet, drilled further upto 250 feet but only 20% struck water at 250 feet. cost of drilling is Rs.50 per foot. Mr. Sinha estimated that he would pay Rs.18,000 during a 5year period in the present value terms, if he continues to buy water from the neighbor rather than go for the well which would have a life of 5 years. Mr. Sinha has three decision to make:(a)should he drill up to 200 feet and (b) if no water is found at 200 feet, should he drill up to 250 feet?(c) should he continue to buy water from his neighbor?\n\n44\n\n## Posterior probabilities and bayesian analysis\n\nIn a preceding analysis of decision making under\n\nprobabilities used in the expected value criterion are usually obtained from the past historical data. these probabilities are called prior probabilities and the analysis that uses them to find expected payoffs is known as prior analysis. If the prior analysis results in high expected opportunity loss (EOL) or EVPI,it may be desirable to obtain additional information through sampling or experimentation or test research. prior probabilities may then be revised in the light of this additional information by using bayes theorem to yield posterior probabilities or bayes probabilities. Revised analysis of the problem using these probabilities is called posterior analysis or bayesian analysis. this analysis gives a revised value of EMV which is expected to be better than the EMV based on prior Assignment problem DECISION THEORY analysis. the difference between the two should at least be equal to the 45\n\nUTILITY THEORY\nThink about the situation in which you have two option:(1)Get 10000 rupees fix (2)5% chance of getting Rs.250000 and 95% chance of getting nothing\n\nWhat is your answer? The worth of money is different for different people and is also different in different situations. Each rupee is not equally valuable to an individual. The determination of utility is subjective.it depends upon our attitude of accepting risk.\n\nThe utility theory approach attempts to determine a utility function or a utility curve for a decision maker.\n\n## Assignment problem DECISION THEORY\n\n46\n\nUTILITY THEORY\n\nHere function converts money into an arbitary utility measure. Risk avoider: slope decrease as money values increase. firm with unsound finansial position. Risk neutral:expected utility give the same results as expected return(EMV) analysis. EXAMPLE:LARGE COMPANIES & GOVERNMENT. Risk taker:curve increase faster than the money value. In practice utility curve is a combinetion of risk taker and risk avoider\n\n47\n\n## UNBALANCED ASSIGNMENT PROBLEM\n\nYou are given the information about the cost of performing different jobs by different persons. the job-person marking (@) indicates that the individual involved cannot perform the particular job. using this information, state (i) the optimal assignment of jobs, and (ii) the cost of such assignment\nAssignment problem\n\n## Traveling salesman problems\n\nMr Patel is a salesman with Delite Manufecturing company. he wants to visit six cities, say, 1,2,3,4,5 and 6,starting with city 1where he is stationed. the distances between various cities are given in the table.Patel wants to develop a tour through the five other cities and return to his home city in such a way that he was to travel a minimum distance.\nAssignment problem"
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https://digital.library.txst.edu/items/caa08da2-0514-4ac9-aae7-fedd8b66675c | [
"## Composition and convolution theorems for μ-Stepanov pseudo almost periodic functions and applications to fractional integro-differential equations\n\n2018-01-18\nAlvarez, Edgardo\n##### Publisher\nTexas State University, Department of Mathematics\n##### Abstract\nIn this article we establish new convulsion and composition theorems for μ-Stepanov pseudo almost periodic functions. We prove that the space of vector-valued μ-Stepanov pseudo almost periodic functions is a Banach space. As an application, we prove the existence and uniqueness of μ-pseudo almost periodic mild solutions for the fractional integro-differential equation. Dαu(t) = Au(t) + ∫t-∞ α(t - s) Au(s) ds + ƒ(t, u(t)), where A generates an α-resolvent family {Sα(t)}t ≥ 0 on a Banach space X, α ∈ L1loc (ℝ+), α > 0, the fractional derivative is understood in the sense of Weyl and the nonlinearity ƒ is a μ-Stepanov pseudo almost periodic function.\n##### Keywords\nμ-Stepanov pseudo almost periodic, Mild solutions, Fractional integro-differential equations, Composition, Convolution\n##### Citation\nAlvarez, E. (2018). Composition and convolution theorems for μ-Stepanov pseudo almost periodic functions and applications to fractional integro-differential equations. <i>Electronic Journal of Differential Equations, 2018</i>(27), pp. 1-15."
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https://siligence.ai/ask/articles | [
"0 Like\n\n## TensorFlow - Distributed Computing Distributed Computing\n\nThe aim is to help developers understand the basic distributed TF concepts that are reoccurring, such as TF servers.\n0 Like\n\n## TensorFlow - Keras Keras\n\nKeras is compact, easy to learn, high-level Python library run on top of TensorFlow framework.\n0 Like\n\n## TensorFlow - CNN And RNN Difference CNN RNN\n\nIn this chapter, we will focus on the difference between CNN and RNN -\n0 Like\n\n## TensorFlow - TFLearn And Its Installation Tensorflow\n\nTFLearn can be defined as a modular and transparent deep learning aspect used in TensorFlow framework.\n0 Like\n\n## TensorFlow - Linear Regression LinearRegression\n\nLogistic regression or linear regression is a supervised machine learning approach for the classification of order discrete categories.\n0 Like\n\n## TensorFlow - Single Layer Perceptron LayerPerceptron\n\nArtificial neural networks is the information processing system the mechanism of which is inspired with the functionality of biological neural circuits.\n0 Like\n\n## TensorFlow - Single Layer Perceptron LayerPerceptron\n\nArtificial neural networks is the information processing system the mechanism of which is inspired with the functionality of biological neural circuits.\n0 Like\n\n## TensorFlow - Word Embedding Embedding\n\nWord embedding is the concept of mapping from discrete objects such as words to vectors and real numbers\n0 Like\n\n## TensorFlow - TensorBoard Visualization TensorBoard Visualization\n\nTensorFlow includes a visualization tool, which is called the TensorBoard. It is used for analyzing Data Flow Graph and also used to understand machine-learning models\n0 Like\n\n## TensorFlow - Recurrent Neural Networks neural-network\n\nRecurrent neural networks is a type of deep learning-oriented algorithm, which follows a sequential approach.\n0 Like\n\n## TensorFlow - Convolutional Neural Networks neural-network\n\nDeep learning is a division of machine learning and is considered as a crucial step taken by researchers in recent decades.\n0 Like\n\n## TensorFlow - Basics Tensorflow\n\nTensors are used as the basic data structures in TensorFlow language\n0 Like\n\n## TensorFlow - Mathematical Foundations Operations\n\nIt is important to understand mathematical concepts needed for TensorFlow before creating the basic application in TensorFlow.\n0 Like\n\n## TensorFlow - Installation Tensorflow Installation\n\nTo install TensorFlow, it is important to have “Python” installed in your system. Python version 3.4+ is considered the best to start with TensorFlow installation.\n0 Like\n\n## TensorFlow - Introduction Tensorflow\n\nTensorFlow is a software library or framework, designed by the Google team to implement machine learning and deep learning concepts in the easiest manner.\n0 Like\n\n## Improving Performance of ML Model(Contd..) Performance\n\nAs we know that ML models are parameterized in such a way that their behavior can be adjusted for a specific problem.\n0 Like\n\n## Improving Performance of ML Models Performance\n\nEnsembles can give us boost in the machine learning result by combining several models\n0 Like\n\n## Machine Learning - Automatic Workflows AutomaticWorkflows\n\nIn order to execute and produce results successfully, a machine learning model must automate some standard workflows.\n0 Like\n\n## Machine Learning - Performance Metrics Metrics\n\nThere are various metrics which we can use to evaluate the performance of ML algorithms, classification as well as regression algorithms\n0 Like\n\n## KNN Algorithm - Finding Nearest Neighbors KNN\n\nK-nearest neighbors (KNN) algorithm is a type of supervised ML algorithm which can be used for both classification as well as regression predictive problems.\n0 Like\n\n## Machine Learning - Hierarchical Clustering Hierarchical\n\nHierarchical clustering is another unsupervised learning algorithm that is used to group together the unlabeled data points having similar characteristics.\n0 Like\n\n## ML - Clustering Mean Shift Algorithm Clustering MeanShift\n\nAs discussed earlier, it is another powerful clustering algorithm used in unsupervised learning.\n0 Like\n\n## ML - Clustering K-Means Algorithm Clustering K-Means\n\nK-means clustering algorithm computes the centroids and iterates until we it finds optimal centroid.\n0 Like\n\n## Clustering Algorithms - Overview Clustering\n\nClustering methods are one of the most useful unsupervised ML methods. These methods are used to find similarity as well as the relationship patterns among data samples and then cluster those samples into groups having similarity based on features.\n0 Like\n\n## Regression Algorithms - Linear Regression LinearRegression\n\nLinear regression may be defined as the statistical model that analyzes the linear relationship between a dependent variable with given set of independent variables.\n0 Like\n\n## Regression Algorithms - Overview Regression\n\nRegression is another important and broadly used statistical and machine learning tool.\n0 Like\n\n## ML - Support Vector Machine(SVM) SVM\n\nSupport vector machines (SVMs) are powerful yet flexible supervised machine learning algorithms which are used both for classification and regression\n0 Like\n\n## Machine Learning - Logistic Regression LogisticRegression\n\nLogistic regression is a supervised learning classification algorithm used to predict the probability of a target variable. .\n0 Like\n\n## Classification Algorithms - Random Forest classification RandomForest\n\nRandom forest is a supervised learning algorithm which is used for both classification as well as regression\n0 Like\n\n## Classification Algorithms - Naïve Bayes classification NaïveBayes\n\nNaïve Bayes algorithms is a classification technique based on applying Bayes' theorem with a strong assumption that all the predictors are independent to each other"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7949878,"math_prob":0.69627327,"size":7873,"snap":"2021-04-2021-17","text_gpt3_token_len":1935,"char_repetition_ratio":0.16380735,"word_repetition_ratio":0.09502664,"special_character_ratio":0.2023371,"punctuation_ratio":0.04095004,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99145895,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T15:52:27Z\",\"WARC-Record-ID\":\"<urn:uuid:f7734946-97b2-401f-b6d1-3461291a266f>\",\"Content-Length\":\"96659\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5da8271c-6a79-498e-a8f7-c9b4170ac748>\",\"WARC-Concurrent-To\":\"<urn:uuid:a66617bb-bd60-4a32-993e-5d9f739d5abf>\",\"WARC-IP-Address\":\"52.175.16.20\",\"WARC-Target-URI\":\"https://siligence.ai/ask/articles\",\"WARC-Payload-Digest\":\"sha1:V3WSNSZLLZSEB25NIUN4HEZVMPEWGKJD\",\"WARC-Block-Digest\":\"sha1:WGSH7TTLTD5KTSL6TSIWJ4RH3WITCJFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703513062.16_warc_CC-MAIN-20210117143625-20210117173625-00123.warc.gz\"}"} |
https://discuss.codechef.com/t/ranklist-editorial/9731 | [
"# RANKLIST - editorial\n\nAuthor: Chandan Boruah\nTester: Pushkar Mishra\nEditorialist: Florin Chirica\n\n### PROBLEM\n\nLet’s define a ranklist as a sequence of numbers between 1 and x of length n, such as ALL numbers between 1 and x must appear in the sequence at least once. An operation can change one element in what value you want. Your goal is to create a sequence containing all numbers between 1 and n exactly once by using minimal number of operations.\n\n### QUICK EXPLANATION\n\nLet’s iterate a number x meaning that all numbers between 1 and x appear in the ranklist. The cost of any ranklist with values between 1 and x is n - x. We need to determine if numbers between 1 and x are enough to create a sum s. Let’s note that a ranklist defined by x can create all sums between minSum and maxSum, where minSum = x * (x + 1) / 2 + (n - x) and maxSum = x * (x + 1) / 2 + (n - x) * x.\n\n### EXPLANATION\n\nTo solve this problem, we need to make two observations. They seem pretty easy and intuitive, but after we make them, the problem is solved.\n\nObservation 1\n\nA ranklist containing elements between 1 and x (no element greater than x) will need n - x operations to be transformed into a sequence of numbers between 1 and n, no matter how sequence looks like.\n\nWe already have all elements between 1 and x (from the definition of a ranklist). However, we need to also have elements x + 1, x + 2, …, n - 1, n. Those do not appear (since all are greater than x), so we have to make some operations to get them. We can keep exactly one position which contains a value i, for each i between 1 and x. Rest of n - x positions must be modified. We’ll put in the remaining n - x positions values x + 1, x + 2, …, n - 1 and n. So cost of a ranklist containing elements between 1 and x becomes n - x.\n\nObservation 2\n\nA ranklist containing elements between 1 and x (no element greater than x) can generate all sums s, such as minSum <= s <= maxSum. More, all sums that can be generated by this ranklist are the ones that are between minSum and maxSum.\n\nLet’s find out who is minSum. We are forced to place x elements: 1 2 … x. The rest of them we can complete them with 1s. So minSum = (1 + 2 + … + x) + (n - x) = x * (x + 1) / 2 + (n - x). maxSum can be obtained by completing the first x elements 1 2 … x and the rest of them with x value (the maximum we’re allowed to use). So we get maxSum = (1 + 2 + … + x) + (n - x) * x = x * (x + 1) / 2 + (n - x) * x.\n\nObviously, no sum s can be less than minSum or greater than maxSum (because those values are the minimum/maximum one can get). Let’s proof now that each sum s which has minSum <= s <= maxSum can be obtained. The idea is to see how those n - x terms can vary: they can be from 1 1 1 … 1 1 1 to x x x … x x x. Suppose we obtained a configuration corresponding to a sum s. Unless s is equal to maxSum, we can always obtain s + 1 as well. If s equal to maxSum, then all terms are equal to x. Otherwise, at least one term isn’t x. Since it isn’t x, we can increment it. Now, the obtained sum is s + 1 and it’s obtained assuming that s is different from maxSum.\n\nPutting the observations together\n\nIn fact, those 2 observations are enough to solve the problem. The key that puts them together is that we can iterate x from 1 to n. Let’s see for a fixed x if a sum s can be obtained. For the given x, we can calculate minSum and maxSum and if minSum <= x <= maxSum, then sum s can be obtained and a ranklist containing only values between 1 and x is valid. The cost to transform this ranklist into 1 2 … n is n - x. So, for all valid ranklists (those who can give sum s), we keep minimum of n - x and we’re done.\n\nTime Complexity\n\nSince we iterate the x variable from 1 to n, the complexity is O(n).\n\n### AUTHOR’S AND TESTER’S SOLUTIONS:\n\nTester’s solution\n\nSetter’s solution\n\n2 Likes\n\nHi,\n\nFunny approach…\n\nI just used the fact that I could find out the number of elements a sequence would need to have to have sum S, and then, by filling all the remaining elements with value 1 and knowing the number of elements I could simply try to “decrement” the number of values used in the arithmetic progression 1,2,3… etc until I managed to reach a sum of value S.\n\nHere is the code:\n\n``````#include <iostream>\n#include <algorithm>\n#include <cmath>\n#include <stdio.h>\n#include <string>\n#include <vector>\n#include <map>\nusing namespace std;\n\nlong long int max (long long int a, long long int b)\n{\nif(a >= b)\nreturn a;\nelse\nreturn b;\n}\n\nint main()\n{\nint tc;\ncin >> tc;\nwhile(tc--)\n{\nlong long int n,s;\ncin >> n >> s;\nif(n==1 and s > 1)\ncout << 1 << endl;\nelse\n{\nlong long aux = floorl( (sqrtl(8.0*s + 1.0) - 1.0 ) / 2.0); //floor of num of elements needed for a seq to have sum S\n//cout << \"aux \" << aux << endl;\nlong long auxsum = ((aux)*(aux+1))/2; //sum of the floor of these num of elements\n//cout << \"auxsum \" << auxsum << endl;\nlong long remnums = n-aux; //we fill remaining elements with value 1\n//cout << \"remnums \" << remnums << endl;\nwhile(auxsum+remnums > s) //we decrement num of elements until we reach S\n{\n//cout << \"stuck\" << endl;\naux--;\nauxsum = ((aux)*(aux+1))/2;\n}\n\ncout <<n-aux<< endl;\n}\n}\nreturn 0;\n}``````\n2 Likes\n\nIt is as Simple As This…\n\nS = S-N;\n\n``````sq = ((double)1 + sqrt ((1+8*S)))/2;\n\nprintf (\"%lld\\n\",N-sq);\n``````\n\nLink To My Solution : http://www.codechef.com/viewsolution/6136467\n\n1 Like\n\n![alt text]\n\nThe easiest approach .\n: http://discuss.codechef.com/upfiles/Capture_21.PNG\n\nI did using binary-search. Here is my solution\nhttp://www.codechef.com/viewsolution/6176994\n\n1 Like\n\nWell I had derived a formula for this question…\n\nand it goes like this : N - (sqrt( 8*(S-N)+1 ) + 1)/2\n\nBut, what is the logic behind it. Can you explain ?\n\nHow did you arrive at this : (sqrt( 8*(S-N)+1 ) + 1)/2\n\n1 Like\n\n@jeffrycopps_2\nAs we know that sum of n natural number is given by n*(n+1)/2 = s\n\nsimplifying this we get n^2+n-2s = 0\n\n(sqrt( 8*s+1 ) + 1)/2\n\n1 Like\n\nI have a doubt…\n\nAs mentioned We are forced to place x elements: 1 2 … x.\n\nWhat is x over here?\n\nis x a number such that x*(x+1)/2 < sumOfRanks ?\n\n`(sqrt(8*(S-N)+1)+1)/2` ?\n\n1 Like\n\nhttp://www.codechef.com/viewsolution/6231580\n\nWhat is meant by the cost of rank list?\n\nMy first thought was on the largest possible rank(magnitude wise) that would fit in a rank list of given length & sum. eg: for a Rank list of len=7 and SUM=8\n\nFrom above data, our invalid rank list looks something like this -> { 1,1,1,1,1,1,2 }\n\nNow, a valid rank list of len=7 looks like this -> { 1,2,3,4,5,6,7 } with its sum=(7*(7+1))/2=28\n\nBut as you can see given sum is 7 which is much less than 28, indicating there are sum repetitions in the rank list.\n\nNow, let us suppose that the largest rank is ‘n’, hence our invalid rank list would contain ranks from 1 upto n and some other repeating ranks.( clearly,in our invalid rank list, n=2 and there are 5 repeating 1’s ).\n\nWe subtract the sum of ‘1 to n’ from SUM & this is equal to len-n (notice 1,2 in bold below)\n\nie, sum of {1,1,1,1,1,1,2} - sum of {1,2} = len-n ( ie, 5 repeating 1s )\n\nThe whole thing boils down to a quadratic equation, SUM - (n*(n+1))/2 = len - n\n\nOn solving this you get n=(sqrt( 8*(S-len)+1 ) + 1)/2 , ie number of valid ranks that need not be changed.\nFinally, the number of ranks to be changed is len-n.\n\n1 Like"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.925395,"math_prob":0.99808264,"size":3698,"snap":"2023-14-2023-23","text_gpt3_token_len":1025,"char_repetition_ratio":0.14374661,"word_repetition_ratio":0.12771392,"special_character_ratio":0.28204435,"punctuation_ratio":0.09529703,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994473,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T05:52:06Z\",\"WARC-Record-ID\":\"<urn:uuid:878b15fa-57db-4281-84f2-d417b1e994e7>\",\"Content-Length\":\"47470\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc20381a-dd53-42a6-acb6-9823c3e58150>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a3b76cc-e392-45f4-84e4-813efe269dc8>\",\"WARC-IP-Address\":\"44.215.32.150\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/ranklist-editorial/9731\",\"WARC-Payload-Digest\":\"sha1:O24VH5FUGK7YBWAWT6GNCKCY6ZNNYRL2\",\"WARC-Block-Digest\":\"sha1:3EDSQA46WPQSKPAMIU2YE5NZPGLJ4DLS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644683.18_warc_CC-MAIN-20230529042138-20230529072138-00161.warc.gz\"}"} |
https://promisekit.org/2022/10/24/how-do-i-inverse-a-2-2-matrix/ | [
"## How do I inverse a 2 2 matrix?\n\nTo find the inverse of a 2×2 matrix: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc).\n\n## What is the transpose of a 2 by 2 matrix?\n\nBelow is a 2×2 matrix like it is used in complex multiplication. The transpose of a square matrix can be considered a mirrored version of it: mirrored over the main diagonal. That is the diagonal with the a’s on it. For a square matrix of any size, the same principle would hold.\n\nCan you multiply a 2×1 and 2×2 matrix?\n\nMultiplication of 2×2 and 2×1 matrices is possible and the result matrix is a 2×1 matrix.\n\n### Can you multiply a 2×1 and a 1×3 matrix?\n\nMultiplication of 2×1 and 1×3 matrices is possible and the result matrix is a 2×3 matrix. This calculator can instantly multiply two matrices and show a step-by-step solution.\n\n### How can I calculate the inverse of a matrix?\n\nSwitch the elements on the main diagonal\n\n• Take the opposite of the other two elements\n• Divide all the values by the determinant of the matrix (since we haven’t talked about the determinant,for a 2×2 system,it is the product of the elements on the\n• What is the simplest way to find an inverse matrix?\n\nFind the determinant\n\n• Find the matrix of minors\n• Find the matrix of co-factors\n• Transpose\n• Divide by the determinant\n• #### Why do we find inverse of a matrix?\n\nInverses only exist for square matrices. That means if you don’t the same number of equations as variables,then you can’t use this method.\n\n• Not every square matrix has an inverse. If the coefficient matrix A is singular (has no inverse),then there may be no solution or there may be many solutions,but\n• Inverses are a pain to find by hand.\n• #### What is the determinant of a 2×1 matrix?\n\n|a| represents the magnitude of a vector when a is a vector (A vector is a matrix with one of the dimensions as 1. So 2×1,3×1, 8×1, etc.) You find the magnitude using a distance formula. For you, you’ll be using it in two dimensions. √(x 2 + y 2) where your vector is [x] [y] Three dimensions would be √(x 2 + y 2 + z 2) And so on for higher dimensions."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8566509,"math_prob":0.9960797,"size":1611,"snap":"2023-14-2023-23","text_gpt3_token_len":436,"char_repetition_ratio":0.15868077,"word_repetition_ratio":0.07166124,"special_character_ratio":0.26691496,"punctuation_ratio":0.09264305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99985814,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T17:50:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f50755c6-2ef7-400b-87c7-ac4ec89b679f>\",\"Content-Length\":\"48850\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9497880-e8b9-4f58-a297-7beb197dc8e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5fabf753-d487-495e-ae0e-66313cabb198>\",\"WARC-IP-Address\":\"104.21.25.175\",\"WARC-Target-URI\":\"https://promisekit.org/2022/10/24/how-do-i-inverse-a-2-2-matrix/\",\"WARC-Payload-Digest\":\"sha1:ZGF4FXJSC4IFCKHKAEC6IVUOSCJRMPCD\",\"WARC-Block-Digest\":\"sha1:U6RN5DFSYPWUCNRUA6XYGICTDF43SCSB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948673.1_warc_CC-MAIN-20230327154814-20230327184814-00454.warc.gz\"}"} |
https://sofad.qc.ca/index.php?id_product=301&controller=product&id_lang=5 | [
"",
null,
"# MTH-2007-2 – Geometry I\n\n23,00 \\$\n\n• Combo (9-1355)\n• Learning Guide, 380 p. (8-1355-01)\n• Scored Activities (3) (0-1355-02)\n\nThis is the first in a series of three compulsory courses on geometry. In this course, learners learn basic concepts in geometry (angles, straight lines, polygons, etc.). They also learn to construct figures using a ruler, a set square, a compass and a protractor, and to use the Pythagorean Theorem.\n\n##### Content:\n\n• Basic geometric concepts;\n\n• Drawing an angle;\n\n• Types of lines;\n\n• Categories of angles;\n\n• Polygons;\n\n• Measuring polygones;\n\n• The Pythagorean Theorem;\n\n• Special right triangles and the Pythagorean Theorem.\n\n##### Evaluation:\n\nA supervised ministerial examination in an adult education centre.\n\nPassing mark: 60%"
] | [
null,
"https://sofad.qc.ca/img/p/4/0/9/409-tm_large_default.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82220834,"math_prob":0.8267493,"size":592,"snap":"2019-51-2020-05","text_gpt3_token_len":134,"char_repetition_ratio":0.1122449,"word_repetition_ratio":0.0,"special_character_ratio":0.21621622,"punctuation_ratio":0.19444445,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98456866,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T04:19:59Z\",\"WARC-Record-ID\":\"<urn:uuid:533f5305-5608-4d36-9eb2-95248b87a856>\",\"Content-Length\":\"186310\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb884d18-122e-4041-bbf2-7911f8dec5a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:14b4daca-86e9-4633-a68f-75f725e38e9e>\",\"WARC-IP-Address\":\"209.171.119.208\",\"WARC-Target-URI\":\"https://sofad.qc.ca/index.php?id_product=301&controller=product&id_lang=5\",\"WARC-Payload-Digest\":\"sha1:DEBALZRSSMVSAUF4X4EZUS6XZUOL4HEA\",\"WARC-Block-Digest\":\"sha1:VOUYE4LMGIGHEMFU2IN3KUR7WUBZJ6BC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541317967.94_warc_CC-MAIN-20191216041840-20191216065840-00381.warc.gz\"}"} |
https://www.systutorials.com/docs/linux/man/3-dgeqr2p/ | [
"dgeqr2p.f -\n\n## SYNOPSIS\n\n### Functions/Subroutines\n\nsubroutine dgeqr2p (M, N, A, LDA, TAU, WORK, INFO)\nDGEQR2P computes the QR factorization of a general rectangular matrix with non-negative diagonal elements using an unblocked algorithm.\n\n## Function/Subroutine Documentation\n\n### subroutine dgeqr2p (integerM, integerN, double precision, dimension( lda, * )A, integerLDA, double precision, dimension( * )TAU, double precision, dimension( * )WORK, integerINFO)\n\nDGEQR2P computes the QR factorization of a general rectangular matrix with non-negative diagonal elements using an unblocked algorithm.\n\nPurpose:\n\n``` DGEQR2 computes a QR factorization of a real m by n matrix A:\nA = Q * R.\n```\n\nParameters:\n\nM\n\n``` M is INTEGER\nThe number of rows of the matrix A. M >= 0.\n```\n\nN\n\n``` N is INTEGER\nThe number of columns of the matrix A. N >= 0.\n```\n\nA\n\n``` A is DOUBLE PRECISION array, dimension (LDA,N)\nOn entry, the m by n matrix A.\nOn exit, the elements on and above the diagonal of the array\ncontain the min(m,n) by n upper trapezoidal matrix R (R is\nupper triangular if m >= n); the elements below the diagonal,\nwith the array TAU, represent the orthogonal matrix Q as a\nproduct of elementary reflectors (see Further Details).\n```\n\nLDA\n\n``` LDA is INTEGER\nThe leading dimension of the array A. LDA >= max(1,M).\n```\n\nTAU\n\n``` TAU is DOUBLE PRECISION array, dimension (min(M,N))\nThe scalar factors of the elementary reflectors (see Further\nDetails).\n```\n\nWORK\n\n``` WORK is DOUBLE PRECISION array, dimension (N)\n```\n\nINFO\n\n``` INFO is INTEGER\n= 0: successful exit\n< 0: if INFO = -i, the i-th argument had an illegal value\n```\n\nAuthor:\n\nUniv. of Tennessee\n\nUniv. of California Berkeley\n\nNAG Ltd.\n\nDate:\n\nSeptember 2012\n\nFurther Details:\n\n``` The matrix Q is represented as a product of elementary reflectors\n\nQ = H(1) H(2) . . . H(k), where k = min(m,n).\n\nEach H(i) has the form\n\nH(i) = I - tau * v * v**T\n\nwhere tau is a real scalar, and v is a real vector with\nv(1:i-1) = 0 and v(i) = 1; v(i+1:m) is stored on exit in A(i+1:m,i),\nand tau in TAU(i).\n```\n\nDefinition at line 122 of file dgeqr2p.f.\n\n## Author\n\nGenerated automatically by Doxygen for LAPACK from the source code."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7713719,"math_prob":0.98573285,"size":2252,"snap":"2020-24-2020-29","text_gpt3_token_len":616,"char_repetition_ratio":0.101868324,"word_repetition_ratio":0.103174604,"special_character_ratio":0.25754884,"punctuation_ratio":0.16304348,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99856454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-02T06:09:17Z\",\"WARC-Record-ID\":\"<urn:uuid:86d3780a-c3ea-4e3b-8c14-330090496828>\",\"Content-Length\":\"12521\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7281c83-7b3d-4deb-87ab-58c79badc629>\",\"WARC-Concurrent-To\":\"<urn:uuid:89cd2f77-9843-4c96-86cd-aedc04cb2d3c>\",\"WARC-IP-Address\":\"149.28.62.154\",\"WARC-Target-URI\":\"https://www.systutorials.com/docs/linux/man/3-dgeqr2p/\",\"WARC-Payload-Digest\":\"sha1:PWP5X6YP5VBX3XYOHPRM47NAD2JUBMDQ\",\"WARC-Block-Digest\":\"sha1:V7UZPBWH7GSGVALEEHHHE2OFQRZUQCFW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655878519.27_warc_CC-MAIN-20200702045758-20200702075758-00167.warc.gz\"}"} |
https://nctbsolution.com/balbharati-solutions-class-6-mathematics-hcf-lcm/ | [
"# Balbharati Solutions Class 6 Mathematics HCF LCM\n\n## Balbharati Solutions Class 6 Mathematics HCF LCM\n\nWelcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Balbharati Class 6 Mathematics Book, Practice Set 23, 24 and 25, HCF LCM. Here students can easily find step by step solutions of all the problems for HCF LCM. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get HCF LCM solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.\n\nHCF LCM all Question Solutions :\n\nPractice Set 23 :\n\nQuestion no – (1)\n\nSolution :\n\n(1) Given, numbers are 12, 16\n\nFactors of 12 = 1, 2, 3, 4, 6, 12\n\nFactors of 16 = 1, 2, 4, 8, 16\n\nCommon Factors = 1, 2, 4\n\n(2) Given, numbers are 21, 24\n\nFactors of 21 = 1, 3, 7, 21\n\nFactors of 24 = 1, 2, 3, 4, 6, 8, 12, 24\n\nCommon Factors of both number is 1 and 3\n\n(3) Given, numbers are 25, 30.\n\nFactors of 25 = 1, 5, 25\n\nFactors of 30 = 1, 2, 3, 5, 6, 10, 15, 30\n\nCommon Factors of both number is 1 and 5.\n\n(4) Given, numbers are 24, 25.\n\nFactors of 24 = 1, 2, 3, 4, 6, 8, 12, 24\n\nFactors of 25 = 1, 5, 25\n\nCommon Factor = 1\n\n(5) Given, numbers are 56, 72.\n\nFactors of 56 is = 1, 2, 4, 7, 8, 14, 28, 56\n\nFactors of 72 is = 1, 2, 3, 4, 6, 8, 9, 18, 24, 36, 72\n\nCommon Factors of both number are 1, 2, 4, 8.\n\nPractice Set 24 :\n\nQuestion no – (1)\n\nSolution :\n\n(1) To find factor of 45 and 30.\n\nFactors of 45 = 1, 3, 5, 9, 15, 45\n\nFactors of 30 = 1, 2, 3, 5, 6, 10, 15, 30\n\nCommon factors = 1, 3, 5, 15\n\nThus, HCF ( 45 and 30) = 15\n\n(2) To find factor of 16 and 48\n\nFactors of 16 = 1,2, 4, 8, 16\n\nFactors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48\n\nCommon factors = 1, 2, 4, 8, 16\n\nThus, the HCF of 16 and 48 is 16\n\n(3) To find factor of 39 and 25.\n\nFactors of 39 is = 1, 3,13, 39\n\nFactors of 25 is = 1, 5, 25\n\nCommon factor of both the number is 1\n\nThus, the HCF of 39 and 25 is = 1\n\n(4) To find factor of 49 and 56\n\nFactors of 49 = 1, 7 , 49\n\nFactors of 56 = 1, 2, 4, 7, 8, 14, 28 , 56\n\nCommon factors of both number is 1, 7\n\nThus, the HCF of 49 and 56 is 1, 7\n\n(5) To find factor of 120, 144.\n\nFactors of 120 is = 1 , 2, 3, 4, 5, 12 , 24 , 30, 60, 120\n\nFactors of 144 is = 1, 2, 3, 4, 12, 24, 76, 144\n\nCommon factors of both the numbers are = 1, 2, 3, 4, 12, 24\n\nThus, the HCF of 120 and 144 is = 24\n\n(6) To find factor of 81 and 99\n\nFactors of 81 = 1, 3, 9, 27, 81\n\nFactors of 99 = 1, 3, 9, 11, 33, 99\n\nCommon factors of both the numbers = 1,3, 9\n\nThus, HCF ( 81,99) = 9\n\n(7) To find factor of 24, 36\n\nFactors of 24 is = 1, 2, 3, 4, 6, 8, 12, 24\n\nFactors of 36 is = 1, 2,3, 4, 6, 9, 12, 18, 36\n\nCommon factors of the both number is = 1, 2, 3, 4 , 6 , 12\n\nHence, the HCF of 24 and 36 = 12\n\n(8) To find factor of 25 and 75\n\nFactors of 25 is = 1, 5, 25\n\nFactors of 75 is = 1, 5, 15, 25, 75\n\nCommon factors for both the number is = 1, 5, 25\n\nHence, the HCF of 25 and 75 is = 25\n\n(9) To find factor of 48, 54\n\nFactors of 48 : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48\n\nFactors of 54 : 1, 2,3, 6, 9, 18, 27, 54\n\nCommon factors of both the number are 1, 2, 3, 6,\n\n∴ The HCF of 48 and 54 is 6.\n\n(10) To find factor of 150, 225\n\nFactors of 150 is = 1,2, 3, 5, 6, 10, 15, 25, 30, 75, 150\n\nFactors of 225 is = 1, 3, 5, 15, 25, 75, 225\n\nCommon factors of both the number is = 1, 3, 5, 15, 25, 75\n\nThus, the HCF of 150 and 225 is = 75.\n\nQuestion no – (2)\n\nSolution :\n\nWe required to find HCF of 18 and 15\n\nFactors of 18 is = 1, 2, 3, 9, 18\n\nFactors of 15 is = 1, 3, 5, 15\n\nCommon factors of both the number is = 1,3\n\nThus, the HCF of 18 and 15 = 3\n\nHence, the maximum length of each bed is 3 meters.\n\nPractice Set 25:\n\nQuestion no – (1)\n\nSolution :\n\n(1) 9 and 15.\n\n9 time table = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90\n\n15 time table = 15, 30, 45, 60, 75\n\nSince, lowest common multiple is 45\n\n∴ The LCM of 9 and 15 = 45\n\n(2) 2, 3 and 5.\n\n2 time table is = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30\n\n3 time table is = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\n\n5 time table is = 5, 10, 15, 20, 25, 30, 35, 40\n\nSince, lowest common multiple is 30\n\nHence, The LCM of 2,3 and 5 is 30.\n\n(3) 12 and 28\n\n12 time table = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120\n\n28 time table = 28, 56, 84, 112, 140\n\nSince, lowest common multiple is 84\n\nThe LCM of 12 and 28 = 84.\n\n(4) 15, 20\n\n15 time table is = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150\n\n20 time table is = 20, 40, 60, 80\n\nSince, lowest common multiple is 60\n\nTherefore, the LCM of 15 and 20 = 60\n\n(5) 8 and 11.\n\n8 time table is = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\n\n11 time table is = 11, 22, 33, 44, 55, 66, 77, 88, 99\n\nSince, the lowest common multiple is 88\n\nThe LCM of 8 and 11 is = 88.\n\nQuestion no – (2)\n\nSolution :\n\n(1) We required to find LCM of 20 and 25.\n\n20 time table is = 20,40,60,80,100,120,140\n\n25 time table is = 25, 50, 75, 100, 125\n\nSince, The lowest common multiple is 100\n\nTherefore, the LCM of 20 and 25 = 100\n\n(2) We required to find LCM of 16, 24, and 40.\n\n16 time table is = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240\n\n24 time table is = 24, 48, 72, 96, 120, 144, 168, 192, 216, 240\n\n40 time table is = 40, 80, 120, 160, 200, 240\n\nSince, lowest common multiple is 240.\n\nHence, the LCM of 16, 24 and 40 = 240\n\n(4) We required to find LCM of 60, 120 and 24\n\n60 time table is = 60, 120, 180\n\n120 time table is = 120, 240,360\n\n24 time table is = 24, 48, 72, 96,120, 144\n\nSince, lowest common multiple is 120.\n\nTherefore, the LCM of 60, 120 and 24 is = 120\n\n(5) Given the fractions 13/45 and 22/ 75\n\nWe required to find LCM of 45 and 75.\n\n45 time table is = 45, 90, 135, 180, 225, 270, 315\n\n75 time table is = 75, 150, 225, 300\n\nSince, the lowest common multiple is 225.\n\nThus, the LCM of 45 and 75 = 225\n\nTherefore,\n\n13 × 5/45 × 5 = 65/225 ….. (Eqn.1)\n\n22×3/75×3 =66 /225 …. (Eqn. 2)"
] | [
null
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https://groupprops.subwiki.org/w/index.php?title=Subgroup_structure_of_groups_of_order_24&oldid=49943&mobileaction=toggle_view_mobile | [
"# Subgroup structure of groups of order 24\n\nThis article gives specific information, namely, subgroup structure, about a family of groups, namely: groups of order 24.\nView subgroup structure of group families | View subgroup structure of groups of a particular order |View other specific information about groups of order 24\nTo understand these in a broader context, see subgroup structure of groups of order 3.2^n | subgroup structure of groups of order 2^3.3^n\n\n## Numerical information on counts of subgroups by order\n\nFACTS TO CHECK AGAINST FOR SUBGROUP STRUCTURE: (finite solvable group)\nLagrange's theorem (order of subgroup times index of subgroup equals order of whole group, so both divide it), |order of quotient group divides order of group (and equals index of corresponding normal subgroup)\nSylow subgroups exist, Sylow implies order-dominating, congruence condition on Sylow numbers|congruence condition on number of subgroups of given prime power order\nHall subgroups exist in finite solvable|Hall implies order-dominating in finite solvable| normal Hall implies permutably complemented, Hall retract implies order-conjugate\nMINIMAL, MAXIMAL: minimal normal implies elementary abelian in finite solvable | maximal subgroup has prime power index in finite solvable group\n\nThere are only two prime factors of this number. Order has only two prime factors implies solvable (by Burnside's",
null,
"$p^aq^b$-theorem) and hence all groups of this order are solvable groups (specifically, finite solvable groups). Another way of putting this is that the order is a solvability-forcing number. In particular, there is no simple non-abelian group of this order.\n\nNote that, by Lagrange's theorem, the order of any subgroup must divide the order of the group. Thus, the order of any proper nontrivial subgroup is one of the numbers 2,4,8,3,6,12.\n\nHere are some observations on the number of subgroups of each order:\n\n1. Congruence condition on number of subgroups of given prime power order: In particular:\n• The number of subgroups of order 2 is congruent to 1 mod 2 (i.e., it is odd).\n• The number of subgroups of order 4 is congruent to 1 mod 2 (i.e., it is odd).\n• The number of subgroups of order 8 is congruent to 1 mod 2 (i.e., it is odd).\n• The number of subgroups of order 3 is congruent to 1 mod 3.\n2. By the fact that Sylow implies order-conjugate, we obtain that Sylow number equals index of Sylow normalizer, and in particular, divides the index of the Sylow subgroup. Combined with (1), we get the following:\n• the number of 2-Sylow subgroups (subgroups of order 8) is either 1 or 3\n• The number of 3-Sylow subgroups (subgroups of order 3) is either 1 or 4.\n3. In the case of a finite nilpotent group, the number of subgroups of a given order is the product of the number of subgroups of order equal to each of its maximal prime power divisors, in the corresponding Sylow subgroup. In particular, we get (number of subgroups of order 3) = 1, (number of subgroups of order 6) = (number of subgroups of order 2), (number of subgroups of order 12) = (number of subgroups of order 4), and (number of subgroups of order 8) = 1.\n4. In the special case of a finite abelian group, we have (number of subgroups of order 3) = (number of subgroups of order 8) = 1, and (number of subgroups of order 2) = (number of subgroups of order 4) = (number of subgroups of order 6) = (number of subgroups of order 12). This is because subgroup lattice and quotient lattice of finite abelian group are isomorphic.\n5. Finite supersolvable implies subgroups of all orders dividing the group order: For any finite supersolvable group, there are subgroups of every possible order, i.e., there are proper nontrivial subgroups of orders 2,3,4,6,8,12. All finite nilpotent groups are supersolvable.\n\n### Table of number of subgroups\n\nGroup Second part of GAP ID (ID is (24,second part)) Nilpotency class Supersolvable? Derived length Number of subgroups of order 2 (must be odd by (1)) Number of subgroups of order 4 (must be odd by (1)) Number of subgroups of order 8 (must be 1 or 3 by (2)) Number of subgroups of order 3 (must be 1 or 4 by (2)) Number of subgroups of order 6 Number of subgroups of order 12\nnontrivial semidirect product of Z3 and Z8 1 not nilpotent Yes 2 1 1 3 1 1 1\ncyclic group:Z24 2 1 Yes 1 1 1 1 1 1 1\nspecial linear group:SL(2,3) 3 not nilpotent No 3 1 3 1 4 4 0\ndicyclic group:Dic24 4 not nilpotent Yes 2 1 7 3 1 1 3\ndirect product of S3 and Z4 5 not nilpotent Yes 2 7 7 3 1 3 3\ndihedral group:D24 6 not nilpotent Yes 2 13 7 3 1 3 3\ndirect product of Dic12 and Z2 7 not nilpotent Yes 2 3 7 3 1 3 3\nsemidirect product of Z3 and D8 with action kernel V4 8 not nilpotent Yes 2 9 7 3 1 5 3\ndirect product of Z6 and Z4 (also, direct product of Z12 and Z2) 9 1 Yes 1 3 3 1 1 3 3\ndirect product of D8 and Z3 10 2 Yes 2 5 3 1 1 5 3\ndirect product of Q8 and Z3 11 2 Yes 2 1 3 1 1 1 3\nsymmetric group:S4 12 not nilpotent No 3 9 7 3 4 4 1\ndirect product of A4 and Z2 13 not nilpotent No 2 7 7 1 4 4 1\ndirect product of D12 and Z2 (also direct product of S3 and V4) 14 not nilpotent Yes 2 15 19 3 1 7 7\ndirect product of E8 and Z3 15 1 Yes 1 7 7 1 1 7 7\nPossibility set -- 1, 2 if nilpotent Yes, No 1, 2 1, 3, 5, 7, 9, 13, 15 1, 3, 7, 19 1, 3 1, 4\n\n## Sylow subgroups\n\n### 2-Sylow subgroups\n\nHere is the occurrence summary:\n\nGroup of order 8 GAP ID (second part) Information on fusion systems Number of groups of order 24 in which it is a 2-Sylow subgroup with a normal complement (i.e., uses inner fusion system) -- equivalently, the whole group is 2-nilpotent List of these groups Second part of GAP IDs of these groups Number of groups of order 24 in which it is a 2-Sylow subgroup without a normal complement (i.e., uses one of the outer fusion systems) List of these groups Second part of GAP IDs of these groups\ncyclic group:Z8 1 -- 2 nontrivial semidirect product of Z3 and Z8, cyclic group:Z24 1, 2 0 -- --\ndirect product of Z4 and Z2 2 -- 3 direct product of S3 and Z4, direct product of Dic12 and Z2, direct product of Z6 and Z4 5, 7, 9 0 -- --\ndihedral group:D8 3 fusion systems for dihedral group:D8 3 dihedral group:D24, semidirect product of Z3 and D8 with action kernel V4, direct product of D8 and Z3 6, 8, 10 1 symmetric group:S4 12\nquaternion group 4 fusion systems for quaternion group 2 dicyclic group:Dic24, direct product of Q8 and Z3 4, 11 1 special linear group:SL(2,3) 3\nelementary abelian group:E8 5 fusion systems for elementary abelian group:E8 2 direct product of D12 and Z2, direct product of E8 and Z3 14, 15 1 direct product of A4 and Z2 13\n\nNote that the number of 2-Sylow subgroups is either 1 or 3. The former happens if and only if we have a normal Sylow subgroup for the prime 2. The latter happens if and only if we have a self-normalizing Sylow subgroup for the prime 2.\n\nGroup Second part of GAP ID (ID is (24,second part)) 2-Sylow subgroup Second part of GAP ID Number of 2-Sylow subgroups Number of 3-Sylow subgroups (=1 iff the group is 2-nilpotent, i.e., the 2-Sylow subgroup is a retract, i.e., it has a normal complement and the whole group is a semidirect product)\nnontrivial semidirect product of Z3 and Z8 1 cyclic group:Z8 1 3 1\ncyclic group:Z24 2 cyclic group:Z8 1 1 1\nspecial linear group:SL(2,3) 3 quaternion group 4 1 4\ndicyclic group:Dic24 4 quaternion group 4 3 1\ndirect product of S3 and Z4 5 direct product of Z4 and Z2 2 3 1\ndihedral group:D24 6 dihedral group:D8 3 3 1\ndirect product of Dic12 and Z2 7 direct product of Z4 and Z2 2 3 1\nsemidirect product of Z3 and D8 with action kernel V4 8 dihedral group:D8 3 3 1\ndirect product of Z6 and Z4 (also, direct product of Z12 and Z2) 9 direct product of Z4 and Z2 2 1 1\ndirect product of D8 and Z3 10 dihedral group:D8 3 1 1\ndirect product of Q8 and Z3 11 quaternion group 4 1 1\nsymmetric group:S4 12 dihedral group:D8 3 3 4\ndirect product of A4 and Z2 13 elementary abelian group:E8 5 1 4\ndirect product of D12 and Z2 (also direct product of S3 and V4) 14 elementary abelian group:E8 5 3 1\ndirect product of E8 and Z3 15 elementary abelian group:E8 5 1 1\n\n### 3-Sylow subgroups\n\nNote that the 3-Sylow subgroup is isomorphic to cyclic group:Z3 in all cases. By the congruence condition on Sylow numbers as well as the divisibility condition on Sylow numbers, the only possibilities for the number of 3-Sylow subgroups is 1 or 4. In the former case, we have a normal Sylow subgroup. In the latter case, the normalizer of the Sylow subgroup has order 6, and is thus either cyclic group:Z6 or symmetric group:S3.\n\nGroup Second part of GAP ID (ID is (24,second part)) Number of 3-Sylow subgroups Normalizer of Sylow subgroup\nnontrivial semidirect product of Z3 and Z8 1 1 whole group\ncyclic group:Z24 2 1 whole group\nspecial linear group:SL(2,3) 3 4 cyclic group:Z6\ndicyclic group:Dic24 4 1 whole group\ndirect product of S3 and Z4 5 1 whole group\ndihedral group:D24 6 1 whole group\ndirect product of Dic12 and Z2 7 1 whole group\nsemidirect product of Z3 and D8 with action kernel V4 8 1 whole group\ndirect product of Z6 and Z4 (also, direct product of Z12 and Z2) 9 1 whole group\ndirect product of D8 and Z3 10 1 whole group\ndirect product of Q8 and Z3 11 1 whole group\nsymmetric group:S4 12 4 symmetric group:S3\ndirect product of A4 and Z2 13 4 cyclic group:Z6\ndirect product of D12 and Z2 (also direct product of S3 and V4) 14 1 whole group\ndirect product of E8 and Z3 15 1 whole group"
] | [
null,
"https://groupprops.subwiki.org/w/images/math/e/5/3/e53ca95670f6dfb3e6e4668ea1f54ffa.png ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8327378,"math_prob":0.9376779,"size":7093,"snap":"2020-10-2020-16","text_gpt3_token_len":2161,"char_repetition_ratio":0.25842854,"word_repetition_ratio":0.2277691,"special_character_ratio":0.3052305,"punctuation_ratio":0.089655176,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97859836,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-20T13:57:57Z\",\"WARC-Record-ID\":\"<urn:uuid:fca5b8ae-c50d-4c51-bbb4-38eeb0846ab1>\",\"Content-Length\":\"43065\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:07cc5616-b37d-4f64-a8f8-9c05c749b9d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b72cbff-c639-41a5-8f8d-0a5aa4616260>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/w/index.php?title=Subgroup_structure_of_groups_of_order_24&oldid=49943&mobileaction=toggle_view_mobile\",\"WARC-Payload-Digest\":\"sha1:S37K2J2LCOXXAZZDZYVJBFZJZ46EFQUK\",\"WARC-Block-Digest\":\"sha1:VFCL7J4UOOHWJWATVLUNELUKFKZO262Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875144979.91_warc_CC-MAIN-20200220131529-20200220161529-00452.warc.gz\"}"} |
http://theteacher.info/index.php/boolean-logic/notes-q-a-and-worksheets/1621-an-introduction-to-or-gates | [
"# An introduction to OR gates",
null,
"Binary logic gates usually have two inputs and just one output. To get an output from a binary logic OR gate, you must have a signal on both of the inputs. If C in the above diagram were a light, and A and B were both switches, then to turn the light on, either A or B or both switches must be switched on to make the light come on. If both switches were off then the light would be off.\n\nLogic gates when they are put together can get quite complex so we need a tool called a truth table to help us understand all of the different permutations. There are '2 to the power of the number of inputs' permutations. With 2 inputs into this binary OR logic gate, there are 2 to the power of 2 = 4 possible combinations of 1s and 0s.\n\n A B C 0 0 0 0 1 1 1 0 1 1 1 1\n\nIn the above truth table:\n\nif A is a 0 and B is a 0, the output C is a 0.\nif A is a 0 and B is a 1, the output C is a 1.\nif A is a 1 and B is a 0, the output C is a 1.\nif A is a 1 and B is a 1, the output C is a 1.\n\nQ1. When we have just 2 inputs into an OR gate, there are a total of 4 combinations of 1s and 0s for the inputs. If you had 3 inputs, how many different combinations of 1s and 0s could you have for the inputs?\nQ2. If you had 4 inputs, how many different combinations of 1s and 0s could you have for the inputs?\nQ3. Study this diagram:",
null,
"How many inputs to the diagram are there? We only count an input if it starts from 'outside' the diagram. A, B and D start outside the diagram. C feeds in to the second OR gate as an input, but it starts life as an output of the first OR gate. It doesn't start from outside the diagram so it is not an input that we count. There are 3 inputs to the diagram (A, B and D) and one output from the diagram (E). However, it is always a good idea to think about every output from every logic gate. It helps you work out the overall logic of a diagram. Therefore, when we construct our truth table, we should show C as well.\n\nComplete the following partially completed truth table. There are 3 inputs (A, B and D) so there are a total of 2 to the power of 3 = 8 different combinations. Use columns A and B to work out C. Then use columns C and D to work out column E. Note: You can put the columns in any order you like! Put them in the order that makes working out the logic easiest for you.\n\n A B D C E 0 0 0 0 0 1 0 1 0 1 0 1 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1\n\nSomething very interesting is happening in columns A, B and D. Can you spot what it is? HINT: Can you count in binary?\nQ4. Study the following diagram:",
null,
"How many inputs has it got? What are their letters?\nQ5. How many different combinations of unique inputs are there? How did you work it out?\nQ6. How many outputs are there?\nQ7. Complete the truth table below for the diagram. Use the inputs A and B to work out column E. Use the inputs C and D to work out column F. Use the E and F to work out column G.\n\n A B C D E F G 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 0\n\nSomething very interesting is happening in columns A, B, C and D. Can you spot what it is? HINT: Can you count in binary?\n\nQ8. Study this diagram:",
null,
"How many inputs are there? How many outputs are there?\nQ9. How many unique combinations of inputs are there? How did you work it out?\nQ10. Produce the truth table for the diagram."
] | [
null,
"http://theteacher.info/images/KS3/Boolean/OR1c.gif",
null,
"http://theteacher.info/images/KS3/Boolean/OR2c.gif",
null,
"http://theteacher.info/images/KS3/Boolean/OR4c.gif",
null,
"http://theteacher.info/images/KS3/Boolean/OR3c.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.932786,"math_prob":0.96141505,"size":3444,"snap":"2021-43-2021-49","text_gpt3_token_len":1021,"char_repetition_ratio":0.19622093,"word_repetition_ratio":0.28174603,"special_character_ratio":0.32171893,"punctuation_ratio":0.103896104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99426466,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T17:12:37Z\",\"WARC-Record-ID\":\"<urn:uuid:4ef1ce68-3ab3-4e38-b36d-7c68a6a30e4e>\",\"Content-Length\":\"52384\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63ab2951-b442-443d-bcd2-93dd7bf5cbb3>\",\"WARC-Concurrent-To\":\"<urn:uuid:d1fa5a7a-d160-4760-bbf0-94d75639b03c>\",\"WARC-IP-Address\":\"192.124.249.164\",\"WARC-Target-URI\":\"http://theteacher.info/index.php/boolean-logic/notes-q-a-and-worksheets/1621-an-introduction-to-or-gates\",\"WARC-Payload-Digest\":\"sha1:S5TVXYKTUPZST7CWRKSOVVTKXBJBS6RN\",\"WARC-Block-Digest\":\"sha1:JAEUPV7CVR5UG3OWR5B2FGOT4FH7KZ6V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584913.24_warc_CC-MAIN-20211016170013-20211016200013-00486.warc.gz\"}"} |
http://wordpress.discretization.de/houdini/home/advanced-2/using-scipy-to-make-the-laplacian/ | [
"# Using Scipy to Make the Laplacian\n\nWhat is the most powerful thing in the universe? Perhaps nobody knows, but in our world it might either be stokes theorem or the Laplace operator. It is shocking how many algorithms we end up dealing with the Laplace operator in its core steps. Chances are good that whatever you do with the knowledge of these tutorials will require the Laplace operator in some form.\n\nIn this tutorial, we shall build the cotan-Laplace operator to compute a minimal surface given a fixed boundary. Since this is a technical tutorial we will leave out the theoretical reasoning behind this. However, we will show you how to build the sparse matrix since it requires detailed technical knowledge of the relationships between points, vertices and primitives to build a sparse matrix using scipy.\n\n##### The Input Geometry\n\nTo compute a minimal surface we need a starting surface with the topology that we shall use. In our case we will keep things simple and use a disc mesh that we will deform into some random shape. Here we display how our shape was made:\n\nWith the following code in the deformation:\n\nThat results in this LA tostada shape:\n\nWe can also start with other shapes such as the two following ones:",
null,
"A simple remeshed tube. It’s minimal surface will give us a catenoid.",
null,
"Or a sphere with spherical holes cut into it. It’s minimal surface will be a schwarz patch.\n\n# The Surface Minimization Algorithm\n\nNow we get to the complicated part which we will guide you through.\n\n##### Iteration Control\n\nOur method is an iteration that can be implemented using an Block Begin – Block End pair of nodes. The complicated part of the algorithm will be put in a subnet in between that we call “Reduce Surface Area”.\n\nNote that we can use a clever trick to quickly edit the number of iterations. In the block end node, simply replace the “Iterations” field with $F so that you can edit this field using your keyboard from anyplace in Houdini just as you would normally edit the frame number.",
null,
"We placed$F in the iterations field.\n\nInside the Subnet we will have a network looking like this:",
null,
"A rough sketch of the network we will now build so that you know what is coming.\n##### Python Preparations\n\nAt first we remesh the node and then make some preparations for python. We create attributes for the point Ids and for the positions.\n\nNow we have to come up with a more complicate enumeration scheme. Our sparse Laplace matrix will have the number of rows and columns equal to the number of points and values will be assigned to the diagonals but also the $i$th row and $j$th column entry will be written into whenever there is an edge connecting the $i$th point with the $j$th point. How can we collect these values fast? Here is one way to do it: for every triangle collect the point ids of the 3 points around it and store them into a attribute vector resting on the primitive. Not just one vector, but two vectors as computed by the following code.\n\nExplanation: The $k$th entry ( $k\\in\\{1,2,3\\}$ ) of the start vector indicates from where an edge starts while the $k$th entry of the end vector indicates where that exact same edge ends. Now our code looks like this:\n\n##### Marking the Boundary\n\nNext we want to label each boundary point using an attribute. A magical Houdini trick to archive that involves the group node. This allows us to name a group for a specific selection of conditions (or manual selection with the mouse). In our case, we can group together the edges and mark the option unshared edges. Boundary edges are unshared edges and thus the right points are grouped together in a group that we name boundary.\n\nNext we can add a point wrangle that only runs on the group: boundary. This way we can set an integer boundary attribute equal to 1 only for boundary points.",
null,
"This point wrangle node only runs on the points of the boundary group!\n\nNext we have to set the function g equal to zero everywhere but on the boundary. This is done in favor of the algorithm that we use and is done with the following point wrangle code:\n\n##### The Making of the Sparse Laplace Matrix\n\nNow we will focus on creating the Laplace matrix. Sparse matrices require 4 details to be constructed.\n\n• The dimension of the matrix: in our case the number of points.\n• An array of row indices of non zero entries.\n• An array of column indices of non zero entries corresponding to the above row indices.\n• An array of values corresponding to the row/column index pair.\n\nLet us try to compute the values for the matrix. The first thing we have to do is to compute edge lengths. We do this by iterating over the vertices as the number of vertices is equal to the number of half edges and Houdini does not iterate over the edges.\n\nNext we compute the cotan values on every vertex.\n\nNow that we have all the cotan weights ready, we only need to collect them in a reasonable way for python to grab them. For the off-diagonal entries we will grab the 3 cotan weights around each triangle and store them in one vector inside the triangle. Notice that we store the edge related cotan values inside the attribute vector in the exact same order as we have stored the start and end vector point indices in order to correctly relate the edges with their cotan value in the sparse matrix later.\n\nThe diagonal entries of the cotan-Laplace operator depend on all other entries in the row/column and we have one diagonal entry per point. This is why for the diagonal entries we will grab a point wrangle node and visit all the neighbours of each point to sum up the cotan weights.\n\nAnd at last we have everything ready to really build the cotan-Laplace matrix inside python. In the next python node we will collect all the needed attributes from the points and the primitives that we have just created. For the Off-diagonal part of the sparse matrix creation we grab the indices from the start and end vectors and the corresponding values from the omega vector on the primitives. For the diagonal entries we grab the point indices with the attributes we just saved on the points.\n\n##### Solving the Poisson Equation\n\nNext we will show how the Poisson equation $\\Delta f=g$ can be solved inside a python node.",
null,
"We will connect the Laplace matrix node with another python node.\n\nFirst we will read the function g into python (we had stored it as float attributes gx,gy,gz and we then cache them into python.\n\nAnd next we take another python node, read all the inputs g and the Laplace matrix in and solve the Poisson equation.\n\nAnd at very last we copy the computed values into the point positions using a point wrangle node.\n\n# Displaying the Results\n\nWe want to display the results using soap films with highlighted boundary. Here we present a simple way how to make that possible.",
null,
"On the left we apply the soap film to the surface. On the right we create boundary curves in to place a wire frame around the boundary.\n\nTo highlight the boundary we label every boundary point again and then use that information in a detail attribute wrangle node in order to loop over each vertex (thus over each edge) and add a line segment to that edge if it is a boundary edge. The fuse node afterwards then takes care of fusing two points from different line segments into one point. This way we actually have a complete polygonal line for every boundary component. Here is the code for the creation of the boundary lines. For the textures please look at the texture tutorial.\n\nThe results looks as follows:",
null,
"The disc surface from the left now has a minimal surface while keeping it’s boundary constant.",
null,
"Minimal surface are like soap film surfaces since they too adjust their shape to minimize their surface energy.",
null,
"The tube turned out to be a catenoid. How trivial!",
null,
"And the sphere with holes turned out to be a schwarz patch! Sneaky!\n\nNote that if you make the tube too long the minimal surface has to split the mesh into two discs. However, our implementation does not yet handle such geometry splittings.",
null,
"The minimal surface for this particular boundary condition would be two separate discs. However, our code does not split the mesh into two. The poor surface cannot tear itself apart.\n\nJust to clear things up, this is how our final network looks like:"
] | [
null,
"https://mathvistu.files.wordpress.com/2017/07/catenoid-start.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/cheese-start.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/block-trick.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/laplace-netwrok.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/mark-bdy.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/python-solve.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/shoe-boundary.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/compare-raw.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/soapsurface.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/catenoid-raw1.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/schwarz-pathc-raw.png",
null,
"https://mathvistu.files.wordpress.com/2017/07/catenoid-raw-bad.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7236456,"math_prob":0.97860384,"size":15708,"snap":"2020-45-2020-50","text_gpt3_token_len":4373,"char_repetition_ratio":0.120669894,"word_repetition_ratio":0.09407796,"special_character_ratio":0.28851542,"punctuation_ratio":0.12602137,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9904836,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-28T08:43:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ad91b2e6-c021-4504-8933-574dcf15a00d>\",\"Content-Length\":\"196641\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8be00773-e1f2-4f34-9c52-db53cf28dc3f>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cf9385f-2ec8-4a5c-9d20-4b24b06b654c>\",\"WARC-IP-Address\":\"130.149.80.246\",\"WARC-Target-URI\":\"http://wordpress.discretization.de/houdini/home/advanced-2/using-scipy-to-make-the-laplacian/\",\"WARC-Payload-Digest\":\"sha1:NTAF3G3BDPGMQMUW7H5SR2SWBH6MYMLX\",\"WARC-Block-Digest\":\"sha1:NQ74IAQMMPXFGL3ZL6Y2TRY6DMDYROXV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195198.31_warc_CC-MAIN-20201128070431-20201128100431-00256.warc.gz\"}"} |
https://tools.carboncollective.co/compound-interest/76045-at-31-percent-in-24-years/ | [
"# What is the compound interest on $76045 at 31% over 24 years? If you want to invest$76,045 over 24 years, and you expect it will earn 31.00% in annual interest, your investment will have grown to become $49,611,431.67. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$76,045 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 24 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 31.00% into a decimal, which would be 0.31.\n\n$$A = 76045(1 + \\dfrac{ 0.31 }{1})^{ 24}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 24 because our example is for annual compounding, or one period per year, so 24 × 1 = 24.\n\n## How the compound interest on $76,045 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 24 years it is compounding: Start Balance Interest End Balance 1$76,045.00 $23,573.95$99,618.95\n2 $99,618.95$30,881.87 $130,500.82 3$130,500.82 $40,455.26$170,956.08\n4 $170,956.08$52,996.38 $223,952.46 5$223,952.46 $69,425.26$293,377.73\n6 $293,377.73$90,947.10 $384,324.83 7$384,324.83 $119,140.70$503,465.52\n8 $503,465.52$156,074.31 $659,539.83 9$659,539.83 $204,457.35$863,997.18\n10 $863,997.18$267,839.13 $1,131,836.31 11$1,131,836.31 $350,869.25$1,482,705.56\n12 $1,482,705.56$459,638.72 $1,942,344.29 13$1,942,344.29 $602,126.73$2,544,471.01\n14 $2,544,471.01$788,786.01 $3,333,257.03 15$3,333,257.03 $1,033,309.68$4,366,566.71\n16 $4,366,566.71$1,353,635.68 $5,720,202.38 17$5,720,202.38 $1,773,262.74$7,493,465.12\n18 $7,493,465.12$2,322,974.19 $9,816,439.31 19$9,816,439.31 $3,043,096.19$12,859,535.50\n20 $12,859,535.50$3,986,456.00 $16,845,991.50 21$16,845,991.50 $5,222,257.37$22,068,248.87\n22 $22,068,248.87$6,841,157.15 $28,909,406.02 23$28,909,406.02 $8,961,915.87$37,871,321.89\n24 $37,871,321.89$11,740,109.79 $49,611,431.67 We can also display this data on a chart to show you how the compounding increases with each compounding period. As you can see if you view the compounding chart for$76,045 at 31.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.\n\n## How long would it take to double $76,045 at 31% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$76,045 assuming an interest rate of 31.00%.\n\nWe can calculate this very approximately using the Rule of 72.\n\nThe formula for this is very simple:\n\n$$Years = \\dfrac{72}{Interest\\: Rate}$$\n\nBy dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:\n\n$$Years = \\dfrac{72}{ 31 } = 2.32$$\n\nUsing this, we know that any amount we invest at 31.00% would double itself in approximately 2.32 years. So $76,045 would be worth$152,090 in ~2.32 years.\n\nWe can also calculate the exact length of time it will take to double an amount at 31.00% using a slightly more complex formula:\n\n$$Years = \\dfrac{log(2)}{log(1 + 0.31)} = 2.57\\; years$$\n\nHere, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.\n\nAs you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.\n\nHopefully, this article has helped you to understand the compound interest you might achieve from investing \\$76,045 at 31.00% over a 24 year investment period."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90402246,"math_prob":0.9993803,"size":4458,"snap":"2023-40-2023-50","text_gpt3_token_len":1488,"char_repetition_ratio":0.13381231,"word_repetition_ratio":0.014556041,"special_character_ratio":0.45379093,"punctuation_ratio":0.2254987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998468,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T08:36:06Z\",\"WARC-Record-ID\":\"<urn:uuid:3e6d054d-adf4-424b-bfb4-e8b9717d4da4>\",\"Content-Length\":\"29229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b79d953d-e853-4db4-88ad-1959d721d371>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb667e5f-005e-44c4-a294-d9babcef7179>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/compound-interest/76045-at-31-percent-in-24-years/\",\"WARC-Payload-Digest\":\"sha1:FVUOPNQ6GFXWSNRBRRKCSWJUBR64X6ON\",\"WARC-Block-Digest\":\"sha1:JSXGCBS5KJPT2JW6XJDSO325YYLITJPR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100583.31_warc_CC-MAIN-20231206063543-20231206093543-00850.warc.gz\"}"} |
http://applet-magic.com/alphamodule3.htm | [
" Empirical Investigation of the Alpha Module Model of Nuclear Structure\nSan José State University\n\napplet-magic.com\nThayer Watkins\nSilicon Valley\nU.S.A.\n\nEmpirical Investigation of the Alpha\nModule Model of Nuclear Structure\n\nFrom the very beginning of the search for the spatial structure of nuclei there has been the theory that the nucleons in a nucleus are, wherever possible, organized into alpha particles. The support for this theory was primarily that alpha particles show up in nuclear reactions such as the emission of alpha particles and as a fragment of nuclear fission. It seemed unlikely that such alpha particles would appear unless they already existed within the nucleus before the reaction. Another element of support for the existence of alpha particles within nuclei is the magnitude of the binding energies.\n\nFor nuclides too small for an alpha particle to be formed the binding energies are at most 7 to 8 million electron volts (MeV), but the binding energy for an alpha particle is an extraordinary 28 MeV. The binding energies for nuclides large enough to form an Talpha particle the binding energy is roughly 28 MeV times the number of alpha particles that could be formed. But the alpha particle substructure theory of nuclei does not get enough further support to be accepted as the explanation of the structure of nuclei.\n\nA variation on the alpha particle substructure theory has been developed that offers an explanation for more nuclear phenomena. Neutrons and protons are subject to two types of forces. One is spin pair bonding and the other is the nuclear strong force. Spin pair bonding is exclusive in the sense that a neutron can form a spin pair with one other neutron and it can also form a spin pair bond with one proton. The same applies for a proton. Thus neutrons and protons can form chains involving modules made up of two neutrons and two protons such as n-p-p-n and equivalently p-n-n-p. Such a chain is shown below.",
null,
"The smallest such chain would be simply an alpha particle. Thus in a nucleus having enough nucleons to create an alpha module there would be an alpha particle, but only one. The shell theory says that there are filled shells of neutrons and protons. The capacities of these filled shells are conventionally taken to be {2, 8, 20, 28, 50, 82, 126}. The separate shell capacities would then be {2, 6, 12, 8, 22, 32, 44}. Elsewhere it is argued that the capacities are {2, 4, 8, 14, 22, 32, 44} with the filled shell totals being {2, 6, 14, 28, 50, 82, 126} with 8 and 20 being in the nature of filled subshells. The shell capacities correspond to alpha modules of {1, 2, 4, 11, 16, 22}.\n\n## The Explanation of Nuclear Binding Energies\n\nWithin a nucleus, in addition to alpha modules, there can be neutron-neutron spin pairs, proton-proton spin pairs and neutron-proton spin pairs. The formation of these substructures would contribute to the binding energy of a nuclide. There also would be interactions of the substructures that would contribute to the binding energy. The alpha modules may interact with each other. If #a is the number of alpha modules the number of interactions is ½#a(#a-1). If #nn is the number of neutron-neutron spin pairs then the number of interactions of alpha modules and neutron-neutron spin pairs is #a#nn and likewise for the number of alpha module interactions with proton-proton spin pairs. Then there would be the number of neutron-neutron spin pair interactions with each other, ½#nn(#nn-1) and so on. It should be noted that there is a great difference among frequencies of the extra spin pairs. There are 2668 nuclides with extra neutron-neutron spin pairs, but only 164 out of the 2931 nuclides which have one or more extra proton-proton spin pairs, There are 1466 with an extra neutron-proton spin pair.\n\nAccording to the theory previously developed if the strong force charge of a proton is taken to be +1 then that of a neutron is −2/3. Thus the net strong force charge of an alpha module is +2/3. Alpha particles repel each other as do other alpha modules. The coefficient for the alpha module interactions should be negative. On the other hand alpha modules, with a charge of +2/3, and neutron-neutron pairs with a charge of −4/3 attrack each other and the coefficient for their interactions should be positive. Proton-proton spin pairs have a strong force charge of +2 repel alpha modules and the coefficient for their interactions should be negative. A neutron-proton spin pair has a net strong force charge of 1/3 so they repel alpha modules and the coefficient for such interactions should be negative.\n\nSince neutrons repel each other through the strong force the coefficient for the interactions of neutron-neutron pairs with each other should be negative. The same applies for proton-proton spin pairs.\n\nThere could also be binding energy effects due to the interaction of a singleton neutron or a singleton proton with the other substructures. There are 1466 nuclides with a singleton neuton and 1460 with a singleton proton. Let #sn and #sp denote the number of singleton neutrons and protons, respectively. Only one of the two can be 1 because two of them would form a neutron-proton spin pair. The coefficient for the interaction of alpha modules with a singleton neutron should be positive and with a singleton proton negative.\n\nThe regression equation obtained using the above scheme for the binding energies of the 2931 nuclides is as follows.\n\n#### BE = 39.64#a + 13.65#nn -4.64#pp -11.51#np - 0.54(½#a(#a-1)) + 0.35#a#nn - 00.62#a#pp + 1.51#a#np - 0.76(½#nn(#nn-1)) + 6.76(½#pp(#pp-1)) + 0.19#a#sn -1.03#a#sp (t-ratios) [517.0] [84.1] [-3.2] [-13.7] [-84.9] [35.4] [-5.1] [13.7] [-42.2] [5.4] [14.3] [-9.9] R² = 0.999912\n\nThe magnitude of a t-ratio (ratios of a coefficient to its standard deviation) must be at least 2 in order for the coefficient to be significantly different from zero at the 95 percent level of confidence. As can be seen the t-ratios are all well above the level of 2.\n\nThe predictions for the signs of the interaction terms are borne out except in the case of the interaction of alpha modules and the neutron-proton spin pairs and the interaction of the proton-proton spin pairs. Also there should have been positive values for the formation of proton-proton spin pairs and neutron-proton spin pairs. The coefficients for the interactions of the neutron-neutron and proton-proton spin pairs with singleton neutrons and singleton protons were not statistically significantly different from zero at the 95 percent level of confidence and those variables were eliminated from the regression analysis.\n\n## Conclusions\n\nThere is support for the alpha module theory of nuclear structure but it is not complete. The predictions are based primarily upon the notion of strong force charges for the proton and neutron being +1 and −2/3, respectively. The signs of crucial regression coefficients such as the interaction of alpha modules and neutron-neutron spin pairs are as the theory of strong force charges predicts. The conventional theory that all nucleons attract each other would predict all of the regression coefficients to be positive. On balance there is more support for the alpha module theory and strong force charges than for the conventional theory."
] | [
null,
"http://applet-magic.com/quasialpha2.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90031934,"math_prob":0.935949,"size":7031,"snap":"2020-10-2020-16","text_gpt3_token_len":1563,"char_repetition_ratio":0.17745838,"word_repetition_ratio":0.048843186,"special_character_ratio":0.22201678,"punctuation_ratio":0.08550186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.973593,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-29T11:06:17Z\",\"WARC-Record-ID\":\"<urn:uuid:46ba6818-bb78-4af5-98cb-6ac06d7182aa>\",\"Content-Length\":\"12659\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73e6dd07-f92a-4bfa-af72-001650b050d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd732fc7-42b7-44f5-9128-d7e819a4365d>\",\"WARC-IP-Address\":\"67.195.197.76\",\"WARC-Target-URI\":\"http://applet-magic.com/alphamodule3.htm\",\"WARC-Payload-Digest\":\"sha1:W6KJOV4QONS2WDB7URVCC5R2TX57ZYXI\",\"WARC-Block-Digest\":\"sha1:JTG4DZUQ2Q5VPRSLSOXX63ETTXL4OKUK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370494331.42_warc_CC-MAIN-20200329105248-20200329135248-00109.warc.gz\"}"} |
https://botorch.org/api/_modules/botorch/models/likelihoods/pairwise.html | [
"# Source code for botorch.models.likelihoods.pairwise\n\n#!/usr/bin/env python3\n# Copyright (c) Meta Platforms, Inc. and affiliates.\n#\n# LICENSE file in the root directory of this source tree.\n\nr\"\"\"\nPairwise likelihood for pairwise preference model (e.g., PairwiseGP).\n\"\"\"\n\nfrom __future__ import annotations\n\nimport math\nfrom abc import ABC, abstractmethod\nfrom typing import Any, Tuple\n\nimport torch\nfrom botorch.utils.probability.utils import (\nlog_ndtr,\nlog_phi,\nstandard_normal_log_hazard,\n)\nfrom gpytorch.likelihoods import Likelihood\nfrom torch import Tensor\nfrom torch.distributions import Bernoulli\n\nclass PairwiseLikelihood(Likelihood, ABC):\n\"\"\"\nPairwise likelihood base class for pairwise preference GP (e.g., PairwiseGP).\n\n:meta private:\n\"\"\"\n\ndef __init__(self, max_plate_nesting: int = 1):\n\"\"\"\nInitialized like a gpytorch.likelihoods.Likelihood.\n\nArgs:\nmax_plate_nesting: Defaults to 1.\n\"\"\"\nsuper().__init__(max_plate_nesting)\n\ndef forward(self, utility: Tensor, D: Tensor, **kwargs: Any) -> Bernoulli:\n\"\"\"Given the difference in (estimated) utility util_diff = f(v) - f(u),\nreturn a Bernoulli distribution object representing the likelihood of\nthe user prefer v over u.\n\nNote that this is not used by the PairwiseGP model,\n\"\"\"\nreturn Bernoulli(probs=self.p(utility=utility, D=D))\n\n@abstractmethod\ndef p(self, utility: Tensor, D: Tensor) -> Tensor:\n\"\"\"Given the difference in (estimated) utility util_diff = f(v) - f(u),\nreturn the probability of the user prefer v over u.\n\nArgs:\nutility: A Tensor of shape (batch_size) x n, the utility at MAP point\nD: D is (batch_size x) m x n matrix with all elements being zero in last\ndimension except at two positions D[..., i] = 1 and D[..., j] = -1\nrespectively, representing item i is preferred over item j.\nlog: if true, return log probability\n\"\"\"\n\ndef log_p(self, utility: Tensor, D: Tensor) -> Tensor:\n\"\"\"return the log of p\"\"\"\n\ndef negative_log_gradient_sum(self, utility: Tensor, D: Tensor) -> Tensor:\n\"\"\"Calculate the sum of negative log gradient with respect to each item's latent\nutility values. Useful for models using laplace approximation.\n\nArgs:\nutility: A Tensor of shape (batch_size x) n, the utility at MAP point\nD: D is (batch_size x) m x n matrix with all elements being zero in last\ndimension except at two positions D[..., i] = 1 and D[..., j] = -1\nrespectively, representing item i is preferred over item j.\n\nReturns:\nA (batch_size x) n Tensor representing the sum of negative log gradient\nvalues of the likelihood over all comparisons (i.e., the m dimension)\nwith respect to each item.\n\"\"\"\nraise NotImplementedError\n\ndef negative_log_hessian_sum(self, utility: Tensor, D: Tensor) -> Tensor:\n\"\"\"Calculate the sum of negative log hessian with respect to each item's latent\nutility values. Useful for models using laplace approximation.\n\nArgs:\nutility: A Tensor of shape (batch_size) x n, the utility at MAP point\nD: D is (batch_size x) m x n matrix with all elements being zero in last\ndimension except at two positions D[..., i] = 1 and D[..., j] = -1\nrespectively, representing item i is preferred over item j.\n\nReturns:\nA (batch_size x) n x n Tensor representing the sum of negative log hessian\nvalues of the likelihood over all comparisons (i.e., the m dimension) with\nrespect to each item.\n\"\"\"\nraise NotImplementedError\n\n[docs]class PairwiseProbitLikelihood(PairwiseLikelihood):\n\"\"\"Pairwise likelihood using probit function\n\nGiven two items v and u with utilities f(v) and f(u), the probability that we\nprefer v over u with probability std_normal_cdf((f(v) - f(u))/sqrt(2)). Note\nthat this formulation implicitly assume the noise term is fixed at 1.\n\"\"\"\n\n# Clamping z values for better numerical stability. See self._calc_z for detail\n# norm_cdf(z=3) ~= 0.999, top 0.1% percent\n_zlim = 3\n\ndef _calc_z(self, utility: Tensor, D: Tensor) -> Tensor:\n\"\"\"Calculate the z score given estimated utility values and\nthe comparison matrix D.\n\"\"\"\nscaled_util = (utility / math.sqrt(2)).unsqueeze(-1)\nz = D.to(scaled_util) @ scaled_util\nz = z.clamp(-self._zlim, self._zlim).squeeze(-1)\nreturn z\n\ndef _calc_z_derived(self, z: Tensor) -> Tuple[Tensor, Tensor, Tensor]:\n\"\"\"Calculate auxiliary statistics derived from z, including log pdf,\nlog cdf, and the hazard function (pdf divided by cdf)\n\nArgs:\nz: A Tensor of arbitrary shape.\n\nReturns:\nTensors with standard normal logpdf(z), logcdf(z), and hazard function\nvalues evaluated at -z.\n\"\"\"\nreturn log_phi(z), log_ndtr(z), standard_normal_log_hazard(-z).exp()\n\n[docs] def p(self, utility: Tensor, D: Tensor, log: bool = False) -> Tensor:\nz = self._calc_z(utility=utility, D=D)\nstd_norm = torch.distributions.normal.Normal(\ntorch.zeros(1, dtype=z.dtype, device=z.device),\ntorch.ones(1, dtype=z.dtype, device=z.device),\n)\nreturn std_norm.cdf(z)\n\n[docs] def negative_log_gradient_sum(self, utility: Tensor, D: Tensor) -> Tensor:\n# Compute the sum over of grad. of negative Log-LH wrt utility f.\n# Original grad should be of dimension m x n, as in (6) from\n# [Chu2005preference]_. The sum over the m dimension of grad. of\n# negative log likelihood with respect to the utility\nz = self._calc_z(utility, D)\n_, _, h = self._calc_z_derived(z)\nh_factor = h / math.sqrt(2)\n\n[docs] def negative_log_hessian_sum(self, utility: Tensor, D: Tensor) -> Tensor:\n# Original hess should be of dimension m x n x n, as in (7) from\n# [Chu2005preference]_ Sum over the first dimension and return a tensor of\n# shape n x n.\n# The sum over the m dimension of hessian of negative log likelihood\n# with respect to the utility\nDT = D.transpose(-1, -2)\nz = self._calc_z(utility, D)\n_, _, h = self._calc_z_derived(z)\nmul_factor = h * (h + z) / 2\nmul_factor = mul_factor.unsqueeze(-2).expand(*DT.size())\n# multiply the hessian value by preference signs\n# (+1 if preferred or -1 otherwise) and sum over the m dimension\nhess = DT * mul_factor @ D\nreturn hess\n\n[docs]class PairwiseLogitLikelihood(PairwiseLikelihood):\n\"\"\"Pairwise likelihood using logistic (i.e., sigmoid) function\n\nGiven two items v and u with utilities f(v) and f(u), the probability that we\nprefer v over u with probability sigmoid(f(v) - f(u)). Note\nthat this formulation implicitly assume the beta term in logistic function is\nfixed at 1.\n\"\"\"\n\n# Clamping logit values for better numerical stability.\n# See self._calc_logit for detail logistic(8) ~= 0.9997, top 0.03% percent\n_logit_lim = 8\n\ndef _calc_logit(self, utility: Tensor, D: Tensor) -> Tensor:\nlogit = D.to(utility) @ utility.unsqueeze(-1)\nlogit = logit.clamp(-self._logit_lim, self._logit_lim).squeeze(-1)\nreturn logit\n\n[docs] def log_p(self, utility: Tensor, D: Tensor) -> Tensor:\nlogit = self._calc_logit(utility=utility, D=D)\n\n[docs] def p(self, utility: Tensor, D: Tensor) -> Tensor:\nlogit = self._calc_logit(utility=utility, D=D)\n\n[docs] def negative_log_gradient_sum(self, utility: Tensor, D: Tensor) -> Tensor:\nindices_shape = utility.shape[:-1] + (-1,)\nwinner_indices = (D == 1).nonzero(as_tuple=True)[-1].reshape(indices_shape)\nloser_indices = (D == -1).nonzero(as_tuple=True)[-1].reshape(indices_shape)\nex = torch.exp(torch.gather(utility, -1, winner_indices))\ney = torch.exp(torch.gather(utility, -1, loser_indices))\nunsigned_grad = ey / (ex + ey)"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6457295,"math_prob":0.9662191,"size":7899,"snap":"2023-40-2023-50","text_gpt3_token_len":2112,"char_repetition_ratio":0.14515516,"word_repetition_ratio":0.40998217,"special_character_ratio":0.2830738,"punctuation_ratio":0.20489095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996942,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T06:58:09Z\",\"WARC-Record-ID\":\"<urn:uuid:d14c0552-14e6-48f4-bce5-543157afec4c>\",\"Content-Length\":\"44865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0af3112-8483-40fe-b55b-87b3e8c93c76>\",\"WARC-Concurrent-To\":\"<urn:uuid:805e2785-fb50-4828-bf2e-ec26a54b8c01>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://botorch.org/api/_modules/botorch/models/likelihoods/pairwise.html\",\"WARC-Payload-Digest\":\"sha1:NBDX3BTNILKAIAKIPJWCAH7ENU65CNLU\",\"WARC-Block-Digest\":\"sha1:BCEVBXYOC7WTJF3LXHDDVFDP4KX5GDK5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510149.21_warc_CC-MAIN-20230926043538-20230926073538-00298.warc.gz\"}"} |
https://jsxgraph.org/wp/docs_jessiecode/ | [
"# JessieCode language reference\n\n## Datatypes\n\n• Boolean, true or false (case insensitive, tRuE is a valid boolean constant).\n\n• Strings are defined using single quote marks. Quote marks inside a string have to be escaped with a backslash.\n\n• Number, corresponds to the JavaScript number datatype.\n\n• Objects, can be created only via object literal notation << >> and the predefined element functions (see below). To access properties and methods the operator is used. Example:\n\n``````obj = <<\nproperty: 'string',\nprop: 42,\nmethod: function (x) {\nreturn x*x;\n}\n>>;\nsixteen = obj.method(4);\n``````\n• Functions are declared with the function operator\n``````f = function (a, b, c) {\nreturn a+b+c;\n};\n``````\n\nOnly one line comments with // being the first non-whitespace characters are supported right now.\n\n## Operators\n\n### Logical operators\n\nOperator Description\n|| OR\n&& AND\n! NOT\n\n### Arithmetic operators\n\nOperator Description\n- Subtraction or unary negation\n* Multiplication\n/ Division\n% Modulus\n^ Exponentiation\n\n### Assignment operators\n\nOperator Description\n= Assignment\n\n### Comparison operators\n\nOperator Description\n== Equals\n<= Lesser or equal\n>= Greater or equal\n< Lesser\n> Greater\n!= Not equal\n~= Approximately equal, can be used to compare two float values.\n\n### String operators\n\nOperator Description\n+ String concatenation\n\n### Member operators\n\nOperator Description\n. Access the object's properties and methods\n\n## Control structures\n\nThe control structures are exactly the same as in JavaScript.\n\n### If\n\n``````if (<expression) {\n<Stmt>\n} else if (<expression>) {\n<Stmt>\n} else {\n<Stmt>\n}\n``````\n\n### While loop\n\n``````while (<expression>) {\n<Stmt>\n}\n``````\n\n### Do loop\n\n``````do {\n<Stmt>\n} while (<expression>);\n``````\n\n### For loop\n\n``````for (<assignment>; <expression>; <assignment>) {\n<Stmt>\n}\n``````\n\n## Predefined constants\n\nName Description\n\\$board Reference to the currently accessed board.\nLN2 Natural logarithm of 2\nLN10 Natural logarithm of 10\nLOG2E Base 2 logarithm of EULER\nLOG10E Base 10 logarithm of EULER\nPI Ratio of the circumference of a circle to its diameter\nEULER Euler's number e = 2.718281828459045\nSQRT1_2 Square root of 1/2\nSQRT2 Square root of 2\n\n## Predefined functions\n\n### Math functions\n\nFunction Description\ncos(x) Cosine of x\ncosh(x) Hyperbolic cosine of x\npow(b, e) e to the b\nlog(x), ln(x) Natural logarithm\nlog(x, b) Logarithm to base b\nlog2(x), lb(x) Logarithm to base 2\nlog10(x), ld(x) Logarithm to base 10\ntan(x) Tangent of x\ncot(x) Cotangent of x\nsqrt(x) Square root of x\ncbrt(x) Cube root of x\nnthroot(x) n-th root of x\nceil(x) Get smallest integer n with n > x.\nasin(x) arcsine\nabs(x) Absolute value of x\nmax(a, b, c, ...) Maximum value of all given values.\nmin(a, b, c, ...) Minimum value of all given values.\nexp(x) EULER to the x\natan2(y, x) Returns the arctangent of the quotient of its arguments.\nrandom() Generate a random number between 0 and 1.\nrandint(min, max, step = 1) Generate a random number between min and max in steps step.\nround(v) Returns the value of a number rounded to the nearest integer.\nfloor(x) Returns the biggest integer n with n < x.\nacos(x) arccosine of x\natan(x) arctangent of x\nacot(x) arccotangent of x\nsin(x) sine of x\nsinh(x) Hyperbolic sine of x\nfactorial(n) Calculates n!\ntrunc(v, p = 0) Truncate v after the p-th decimal.\nV(s) Returns the value of the given element, e.g. sliders and angles.\nL(s) Calculates the length of the given segment.\nX(P), Y(P) Returns the x resp. y coordinate of the given point.\narea(o), A(o) Compute the area of a circle or polygon.\ndist(P, Q) Compute the distance of two points.\ndeg(A, B, C) Calculate the angle of three points in degree.\ngetName(o, useId = false) Return the name of an object. If no name is given an useId is true the id is returned.\n\\$(id) Look up the element to the given element id.\n\n### \\$board methods\n\nMethod Description\nupdate() Update all dependencies and redraw the board.\non(event, handler, context=board) Register an event handler for the given event.\noff(event, handler=) Deregister a given event handler or deregister all event handlers.\nsetView(array, keepaspectratio=false) Changes the viewport. An array with 4 numbers is expected, the four numbers represent the left, upper, right and lower bound of the viewport. If keepaspectratio is true, the viewport is adjusted to the same aspect ratio as the board container.\nsetBoundingbox(array, keepaspectratio=false) See setView.\nmigratePoint(P, Q) Exchange point P by point Q.\ncolorblind(type) Emulate color blindness. Possible types areprotanopia, tritanopia, anddeuteranopia.\n\n### Element functions\n\nEvery element known to the loaded JSXGraph version is available inside Jessie by its element type, e.g. points can be created by calling point()\n\n``````A = point(1, 2);\n``````\n\nThe given parameters correspond to the parents array of the JXG.Board.create() method. Attributes are given after the function call itself in an object:\n\n``````A = point(1, 2) << strokeColor: 'red', face: '[]', size: 7, fillColor: 'black' >>;\n``````\n\nFor a possibly incomplete list including documentation, see the JSXGraph docs; For a complete list see the Element reference section below.\n\n## Accessing elements\n\n### Variable assignment\n\n``````A = point(1, 2);\nA.strokeColor = '#123456';\n``````\n\n### \\$\n\n``````point(1, 2) << id: 'foo', name: 'bar' >>;\n\\$('foo').strokeColor = '#654321';\n``````\n\n### Id\n\n``````point(1, 2) << id: 'foo', name: 'bar' >>;\nfoo.strokeColor = '#f00f00';\n``````\n\nThis is possible only if foo is not used as a variable. This won't work:\n\n``````foo = 1;\n(function () {\npoint(1, 2) << id: 'foo' >>;\nreturn foo.X();\n})();\n``````\n\n### Name\n\n``````point(1, 2) << id: 'foo', name: 'bar' >>;\nbar.strokeColor = '#541541';\n``````\n\nThis is possible only if there is not a variable called bar in the current or any higher scope. See Id above for an example.\n\n## Element reference\n\nhttps://jsxgraph.uni-bayreuth.de/docs/\n\n### Attributes\n\nAttributes are set like object properties\n\n``````A.size = 10;\nA.face = '[]';\n``````\n\nSee the JSXGraph docs for available attributes. Texts and Points have two special attributes X and Y to set their coordinates.\n\n### Subelements\n\nSubelements like labels for points or the baseline in sliders or the dot indicating an angle element is a right angle can be accessed like properties\n\n``````A.label.strokecolor = 'red';\n``````\n\nThe names used to access subelements correspond to their names used to set their attributes in`board.create`.\n\n### Methods\n\nNot all methods of an element class are accessible in Jessie. Currently these methods are available:\n\n• all elements\n• setLabelText\n• point\n• move\n• glide\n• free\n• X\n• Y\n• glider\n• all from point\n• setPosition\n• text\n• setText\n• free\n• move\n• slider\n• Value\n• angle\n• Value"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.65780413,"math_prob":0.9463398,"size":6700,"snap":"2023-40-2023-50","text_gpt3_token_len":1904,"char_repetition_ratio":0.10797491,"word_repetition_ratio":0.026266417,"special_character_ratio":0.27985075,"punctuation_ratio":0.14868106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99478674,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T13:58:40Z\",\"WARC-Record-ID\":\"<urn:uuid:289925df-1848-4d38-aa41-bb0240d32dd3>\",\"Content-Length\":\"37021\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f539741-a720-42b0-963b-081c21a32c6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c00f07a-fff0-4352-83cf-777a23726c9d>\",\"WARC-IP-Address\":\"132.180.10.7\",\"WARC-Target-URI\":\"https://jsxgraph.org/wp/docs_jessiecode/\",\"WARC-Payload-Digest\":\"sha1:6N7K7RGSD6I4NUIWH623OJ45T4W6D3VP\",\"WARC-Block-Digest\":\"sha1:E2RID26GFF7JVSZ3ORGG2JXH7HRG4QTL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00679.warc.gz\"}"} |
https://sportsbizusa.com/20-multiplying-decimals-area-model-worksheet/multiplying-decimals-area-model-worksheet-decimals-multiply-and-divide-decimals-no-prep-printable-w/ | [
"# Multiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W\n\nHome20 Multiplying Decimals area Model Worksheet ➟ Multiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W\n\nMultiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W one of Worksheet From Home - ideas, to explore this Multiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W idea you can browse by and . We hope your happy with this Multiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W idea. You can download and please share this Multiplying Decimals area Model Worksheet Decimals Multiply and Divide Decimals No Prep Printable W ideas to your friends and family via your social media account. Back to 20 Multiplying Decimals area Model Worksheet"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.74148005,"math_prob":0.83219075,"size":2416,"snap":"2021-21-2021-25","text_gpt3_token_len":475,"char_repetition_ratio":0.2305141,"word_repetition_ratio":0.41142857,"special_character_ratio":0.169702,"punctuation_ratio":0.07611549,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T03:47:50Z\",\"WARC-Record-ID\":\"<urn:uuid:6a750c28-f624-473c-891e-119efa47272d>\",\"Content-Length\":\"37878\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0871855d-39ab-4ea5-a074-b4f95dfc9d26>\",\"WARC-Concurrent-To\":\"<urn:uuid:5bfc7683-21c0-4de6-8cef-7ddda93c810d>\",\"WARC-IP-Address\":\"172.67.139.89\",\"WARC-Target-URI\":\"https://sportsbizusa.com/20-multiplying-decimals-area-model-worksheet/multiplying-decimals-area-model-worksheet-decimals-multiply-and-divide-decimals-no-prep-printable-w/\",\"WARC-Payload-Digest\":\"sha1:VRWW7XZQGMQ6KICOJU5PDUU6CBUFW53T\",\"WARC-Block-Digest\":\"sha1:43FF5CVNR3K6PSQHTLKL7SOBHNI727BW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989820.78_warc_CC-MAIN-20210518033148-20210518063148-00287.warc.gz\"}"} |
https://www.jiskha.com/questions/1865628/a-certain-car-model-cost-birr-20-000-with-a-gasoline-engine-and-br-25-000-with-a-diesel | [
"# mathematics\n\nA certain car model cost Birr 20,000 with a gasoline engine and Br 25,000 with a diesel engine. The number of miles per gallon of fuel for cars with these two engines is 25 and 30, respectively. Assume that the price of both types of fuel is Birr 1.50 per gallon.\nRequired:\na) Drive the equation for the cost of driving a gasoline powered car.\nb) Drive the equation for the cost of driving a diesel powered car.\nc) Find the break-even point, that is, find the mileage at which the diesel-powered car becomes more economical than the gasoline powered car.\n\n1. 👍\n2. 👎\n3. 👁\n4. ℹ️\n5. 🚩\n1. \"derive\" not \"drive\"\ncost in \\$/mi is just mi/(mi/gal) * \\$/gal\nso, the cost of driving x miles is\n(a) 20000 + x/25 * 1.50 = 0.06x + 20000\ndo the same for (b) and then solve for (c)\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n2. number of miles driven = x\ncost to drive the gas car = 20000 + x/20 * 1.5 = 3x/40 + 20000\ncost to drive the diesel - 25000 + x/30*1.5 = x/20 + 25000\n\n3x/40 + 20000 = x/20 + 25000\n3x/40 - x/20 = 5000\nx/40 = 5000\nx = 20,000\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n3. oops, copy error !!\nI used 20 instead of 25 in line #2\n\nMake the necessary changes, or else solve oobleck's equation\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n4. good\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n\n## Similar Questions\n\nStill need help? You can ask a new question."
] | [
null
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https://www.litscape.com/word_analysis/raiment | [
"Definition of raiment\n\n\"raiment\" in the noun sense\n\n1. array, raiment, regalia\n\nespecially fine or decorative clothing\n\n\"raiment\" in the verb sense\n\n1. dress, clothe, enclothe, garb, raiment, tog, garment, habilitate, fit out, apparel\n\nprovide with clothes or put clothes on\n\n\"Parents must feed and dress their child\"\n\nSource: WordNet® (An amazing lexical database of English)\n\nWordNet®. Princeton University. 2010.\n\nQuotations for raiment\n\nHaving food and raiment, let us be therewith content. [ St. Paul ]\n\nHe who first invented raiment, perhaps invented love. [ Segur ]\n\nLove keeps the cold out better than a cloak. It serves for food and raiment. [ Longfellow ]\n\nMusic is the fourth great material want of our natures, - first food, then raiment, then shelter, then music. [ Bovee ]\n\nAnd why take ye thought for raiment? Consider the lilies of the field, how they grow: they toil not, neither do they spin. [ Bible ]\n\nAs flowers never put on their best clothes for Sunday, but wear their spotless raiment and exhale their odor every day, so let your righteous life, free from stain, ever give forth the fragrance of the love of God. [ Henry Ward Beecher ]\n\nWhat was your dream? It seemed to me that a woman in white raiment, graceful and fair to look upon, came towards me and calling me by name said: On the third day, Socrates, thou shalt reach the coast of fertile Phthia. [ Plato ]\n\nMany shiver from want of defence against the cold; but there is vastly more suffering among the rich from absurd and criminal modes of dress, which fashion has sanctioned, than among the poor from deficiency of raiment. [ Channing ]\n\nraiment in Scrabble®\n\nThe word raiment is playable in Scrabble®, no blanks required.\n\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nMINARET\n(86 = 36 + 50)\nMINARET\n(86 = 36 + 50)\n\nminaret, raiment\n\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(77 = 27 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(65 = 15 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(62 = 12 + 50)\nRAIMENT\n(62 = 12 + 50)\nRAIMENT\n(62 = 12 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\n\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nRAIMENT\n(86 = 36 + 50)\nMINARET\n(86 = 36 + 50)\nMINARET\n(86 = 36 + 50)\nMINARET\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nMINARET\n(80 = 30 + 50)\nRAIMENT\n(80 = 30 + 50)\nRAIMENT\n(77 = 27 + 50)\nMINARET\n(77 = 27 + 50)\nMINARET\n(76 = 26 + 50)\nMINARET\n(74 = 24 + 50)\nMINARET\n(74 = 24 + 50)\nMINARET\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nMINARET\n(72 = 22 + 50)\nMINARET\n(72 = 22 + 50)\nMINARET\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nRAIMENT\n(72 = 22 + 50)\nMINARET\n(72 = 22 + 50)\nMINARET\n(70 = 20 + 50)\nMINARET\n(70 = 20 + 50)\nMINARET\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nMINARET\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nMINARET\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nMINARET\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(70 = 20 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nMINARET\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nMINARET\n(68 = 18 + 50)\nMINARET\n(68 = 18 + 50)\nMINARET\n(68 = 18 + 50)\nMINARET\n(68 = 18 + 50)\nRAIMENT\n(68 = 18 + 50)\nMINARET\n(67 = 17 + 50)\nRAIMENT\n(65 = 15 + 50)\nMINARET\n(64 = 14 + 50)\nMINARET\n(64 = 14 + 50)\nMINARET\n(63 = 13 + 50)\nMINARET\n(63 = 13 + 50)\nMINARET\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(63 = 13 + 50)\nRAIMENT\n(62 = 12 + 50)\nRAIMENT\n(62 = 12 + 50)\nRAIMENT\n(62 = 12 + 50)\nMINARET\n(61 = 11 + 50)\nMINARET\n(61 = 11 + 50)\nMINARET\n(61 = 11 + 50)\nMINARET\n(61 = 11 + 50)\nMINARET\n(61 = 11 + 50)\nMINARET\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nRAIMENT\n(61 = 11 + 50)\nMINARET\n(60 = 10 + 50)\nMARTEN\n(33)\nREMINT\n(33)\nMINTER\n(33)\nINMATE\n(33)\nREMAIN\n(33)\nMARINE\n(33)\nAIRMEN\n(33)\nMARTIN\n(33)\nMITRE\n(30)\nANIME\n(30)\nAMINE\n(30)\nAMENT\n(30)\nMITER\n(30)\nMINER\n(30)\nMEANT\n(30)\nMERIT\n(30)\nMATER\n(30)\nMARINE\n(28)\nMARTEN\n(28)\nMARTIN\n(28)\nMINTER\n(28)\nMIRE\n(27)\nINMATE\n(27)\nINMATE\n(27)\nINMATE\n(27)\nINMATE\n(27)\nINMATE\n(27)\nMINTER\n(27)\nMINTER\n(27)\nMARTIN\n(27)\nMINTER\n(27)\nMINTER\n(27)\nTRIM\n(27)\nMINT\n(27)\nITEM\n(27)\nMAIN\n(27)\nMINE\n(27)\nMARINE\n(27)\nREMAIN\n(27)\nTEAM\n(27)\nMEAN\n(27)\nREMAIN\n(27)\nREMAIN\n(27)\nREMAIN\n(27)\nREMAIN\n(27)\nREMINT\n(27)\nREMINT\n(27)\nREMINT\n(27)\nTERM\n(27)\nMARTIN\n(27)\nMATE\n(27)\nMINTER\n(27)\nMEAT\n(27)\nMARTIN\n(27)\nMARTIN\n(27)\nREMINT\n(27)\nREMINT\n(27)\nMARTEN\n(27)\nMARTIN\n(27)\nMITE\n(27)\nTRAM\n(27)\nAIRMEN\n(27)\nMARTEN\n(27)\nMART\n(27)\nMARTEN\n(27)\nMARINE\n(27)\nAIRMEN\n(27)\nMARTEN\n(27)\nMARTEN\n(27)\nMARINE\n(27)\nAIRMEN\n(27)\nREAM\n(27)\nAIRMEN\n(27)\nMANE\n(27)\nMARINE\n(27)\nAIRMEN\n(27)\nMARE\n(27)\nMARINE\n(27)\nMATER\n(26)\nMERIT\n(26)\nMINER\n(26)\nMITER\n(26)\nMITRE\n(26)\nMEANT\n(26)\nRAMET\n(24)\nAIMER\n(24)\nAIRMEN\n(24)\nANIME\n(24)\nAIMER\n(24)\nMITRE\n(24)\nMITRE\n(24)\nMERIT\n(24)\nMITRE\n(24)\nANIME\n(24)\nREMAIN\n(24)\nMITER\n(24)\nMITER\n(24)\nANIME\n(24)\nAIRMEN\n(24)\nRAMET\n(24)\nRAMET\n(24)\nNAMER\n(24)\nARMET\n(24)\nARMET\n(24)\nARMET\n(24)\nTIMER\n(24)\nTIMER\n(24)\nRAMET\n(24)\nREMINT\n(24)\nREMAIN\n(24)\nREMINT\n(24)\nMARINE\n(24)\nTIMER\n(24)\nNAMER\n(24)\nTIMER\n(24)\nMEANT\n(24)\n\nraiment in Words With Friends™\n\nThe word raiment is playable in Words With Friends™, no blanks required.\n\nMINARET\n(104 = 69 + 35)\nRAIMENT\n(104 = 69 + 35)\n\nminaret, raiment\n\nRAIMENT\n(104 = 69 + 35)\nRAIMENT\n(98 = 63 + 35)\nRAIMENT\n(92 = 57 + 35)\nRAIMENT\n(92 = 57 + 35)\nRAIMENT\n(80 = 45 + 35)\nRAIMENT\n(80 = 45 + 35)\nRAIMENT\n(80 = 45 + 35)\nRAIMENT\n(79 = 44 + 35)\nRAIMENT\n(79 = 44 + 35)\nRAIMENT\n(79 = 44 + 35)\nRAIMENT\n(74 = 39 + 35)\nRAIMENT\n(74 = 39 + 35)\nRAIMENT\n(74 = 39 + 35)\nRAIMENT\n(65 = 30 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(54 = 19 + 35)\nRAIMENT\n(52 = 17 + 35)\nRAIMENT\n(52 = 17 + 35)\nRAIMENT\n(51 = 16 + 35)\nRAIMENT\n(51 = 16 + 35)\nRAIMENT\n(51 = 16 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(47 = 12 + 35)\nRAIMENT\n(47 = 12 + 35)\nRAIMENT\n(47 = 12 + 35)\nRAIMENT\n(46 = 11 + 35)\n\nMINARET\n(104 = 69 + 35)\nRAIMENT\n(104 = 69 + 35)\nRAIMENT\n(98 = 63 + 35)\nRAIMENT\n(92 = 57 + 35)\nMINARET\n(92 = 57 + 35)\nRAIMENT\n(92 = 57 + 35)\nRAIMENT\n(80 = 45 + 35)\nMINARET\n(80 = 45 + 35)\nMINARET\n(80 = 45 + 35)\nRAIMENT\n(80 = 45 + 35)\nMINARET\n(80 = 45 + 35)\nRAIMENT\n(80 = 45 + 35)\nMINARET\n(79 = 44 + 35)\nMINARET\n(79 = 44 + 35)\nMINARET\n(79 = 44 + 35)\nRAIMENT\n(79 = 44 + 35)\nRAIMENT\n(79 = 44 + 35)\nRAIMENT\n(79 = 44 + 35)\nMINARET\n(74 = 39 + 35)\nRAIMENT\n(74 = 39 + 35)\nRAIMENT\n(74 = 39 + 35)\nRAIMENT\n(74 = 39 + 35)\nMINARET\n(74 = 39 + 35)\nMINARET\n(74 = 39 + 35)\nMINARET\n(74 = 39 + 35)\nMINARET\n(74 = 39 + 35)\nMINARET\n(73 = 38 + 35)\nMINTER\n(66)\nAIRMEN\n(66)\nMINARET\n(65 = 30 + 35)\nMINARET\n(65 = 30 + 35)\nRAIMENT\n(65 = 30 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nMINARET\n(61 = 26 + 35)\nMINARET\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nMINARET\n(61 = 26 + 35)\nMINARET\n(61 = 26 + 35)\nMINARET\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nRAIMENT\n(61 = 26 + 35)\nREMINT\n(60)\nREMAIN\n(60)\nMARTIN\n(60)\nMARTEN\n(60)\nMARINE\n(60)\nINMATE\n(60)\nMINARET\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nMINARET\n(59 = 24 + 35)\nMINARET\n(59 = 24 + 35)\nMINARET\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nRAIMENT\n(59 = 24 + 35)\nMINARET\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nRAIMENT\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nMINARET\n(57 = 22 + 35)\nMINARET\n(56 = 21 + 35)\nMINARET\n(55 = 20 + 35)\nMARTIN\n(54)\nMARTEN\n(54)\nRAIMENT\n(54 = 19 + 35)\nAIRMEN\n(54)\nMINTER\n(54)\nMARINE\n(54)\nREMAIN\n(54)\nINMATE\n(54)\nREMINT\n(54)\nMINARET\n(53 = 18 + 35)\nRAIMENT\n(52 = 17 + 35)\nRAIMENT\n(52 = 17 + 35)\nMINARET\n(52 = 17 + 35)\nMINARET\n(52 = 17 + 35)\nMINARET\n(52 = 17 + 35)\nANIME\n(51)\nAMENT\n(51)\nAMINE\n(51)\nMINER\n(51)\nRAIMENT\n(51 = 16 + 35)\nMEANT\n(51)\nRAIMENT\n(51 = 16 + 35)\nMINARET\n(51 = 16 + 35)\nRAIMENT\n(51 = 16 + 35)\nMINARET\n(51 = 16 + 35)\nMINARET\n(51 = 16 + 35)\nMINARET\n(50 = 15 + 35)\nMINARET\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(50 = 15 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nRAIMENT\n(49 = 14 + 35)\nMINARET\n(49 = 14 + 35)\nMINARET\n(49 = 14 + 35)\nMINARET\n(49 = 14 + 35)\nMINT\n(48)\nREMAIN\n(48)\nMINE\n(48)\nMINARET\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nMINARET\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nMAIN\n(48)\nMANE\n(48)\nMARTEN\n(48)\nMARTIN\n(48)\nMATER\n(48)\nMEAN\n(48)\nMERIT\n(48)\nMINARET\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nMITER\n(48)\nMINARET\n(48 = 13 + 35)\nMINARET\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nRAIMENT\n(48 = 13 + 35)\nMINARET\n(48 = 13 + 35)\nMINARET\n(48 = 13 + 35)\nMITRE\n(48)\nRAIMENT\n(47 = 12 + 35)\nMINARET\n(47 = 12 + 35)\nMINARET\n(47 = 12 + 35)\nRAIMENT\n(47 = 12 + 35)\nRAIMENT\n(47 = 12 + 35)\nMINARET\n(47 = 12 + 35)\nMINARET\n(47 = 12 + 35)\nMINARET\n(46 = 11 + 35)\nRAIMENT\n(46 = 11 + 35)\nMEAT\n(45)\nMATE\n(45)\nTEAM\n(45)\nMART\n(45)\nMARE\n(45)\nMIRE\n(45)\nMITE\n(45)\nITEM\n(45)\nTERM\n(45)\nTRIM\n(45)\nREAM\n(45)\nTRAM\n(45)\nMARINE\n(42)\nMARTIN\n(42)\nAIRMEN\n(42)\nMINTER\n(42)\nINMATE\n(42)\nINMATE\n(42)\nAIRMEN\n(42)\nREMINT\n(42)\nMARINE\n(42)\nREMINT\n(42)\nMINTER\n(42)\nREMAIN\n(42)\nMARTEN\n(42)\nREMINT\n(40)\nMARTEN\n(40)\nREMAIN\n(40)\nMINTER\n(40)\nREMAIN\n(40)\nMARTEN\n(40)\nMARTIN\n(40)\nREMINT\n(40)\nAIRMEN\n(40)\nAIRMEN\n(40)\nMARINE\n(40)\nMINTER\n(40)\nMARTIN\n(40)\nINMATE\n(40)\nINMATE\n(40)\nMARINE\n(40)\nNAMER\n(39)\nRETAIN\n(39)\nRAMEN\n(39)\nANIME\n(39)\nMEANT\n(39)\nAMINE\n(39)\nAMENT\n(39)\nMARTEN\n(36)\nRAMEN\n(36)\nMINTER\n(36)\n\nai aim\n\nen men\n\nme men\n\nraiments"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79023015,"math_prob":1.0000088,"size":445,"snap":"2019-26-2019-30","text_gpt3_token_len":121,"char_repetition_ratio":0.13605443,"word_repetition_ratio":0.0,"special_character_ratio":0.22696629,"punctuation_ratio":0.2195122,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998536,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T23:02:40Z\",\"WARC-Record-ID\":\"<urn:uuid:0c3f4ccf-e679-49ad-a9d2-6a3a1d2bc954>\",\"Content-Length\":\"150003\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9984a2e2-122c-4c1a-bdbf-a9f286abe3bf>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a611c38-852b-49b0-ae3b-d3bcb32f0e4e>\",\"WARC-IP-Address\":\"104.18.50.165\",\"WARC-Target-URI\":\"https://www.litscape.com/word_analysis/raiment\",\"WARC-Payload-Digest\":\"sha1:QI22LA7VSHY4X4BEHLRA3IQZOHUTZDZG\",\"WARC-Block-Digest\":\"sha1:WQRLEWDR7JQDOXRWCR4N7NVI7AMUB7LO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998581.65_warc_CC-MAIN-20190617223249-20190618005249-00372.warc.gz\"}"} |
https://blog.computationalcomplexity.org/2002/09/complexity-class-of-week-pp.html | [
"## Wednesday, September 04, 2002\n\n### Complexity Class of the Week: PP\n\nPrevious CCW\n\nA language L is in PP if there is a nondeterministic polynomial-time Turing machine M such that x is in L if and only if M(x) has more accepting than rejecting paths.\n\nPP stands for \"Probabilistic Polynomial-time\" from the original definition by Gill where one uses a probabilistic poly-time TM where x is in L if and only if the probability of M(x) accepting is greater than 1/2. \"Probabilistic Polynomial-Time\" would be a better name for PP's cousin class BPP or \"Bounded-error Probabilistic Polynomial-Time\". PP is not much use as a probabilistic class since it would take potentially an exponential number of trials to distinguish accepting from rejecting with reasonable confidence\n\nA better name for PP would be Majority-P as given by the definition above. Because of historic reasons we are stuck with the name PP for now. Don't hold the bad name against the class. It still has a natural definition and some amazing properties.\n\nPP has similar complexity to the function class #P as PPP = P#P, proved using the old stalwart, binary search. I have never found the paper that first proves this equivalence. Toda's Theorem shows the amazing hardness of PP by reducing the polynomial-time hierarchy to PPP.\n\nValiant's proof that the permanent is #P-complete also gives a natural complete language for PP:\n\nPERM = { (M,k) | The permanent of M is at least k}\n\nNP is in PP by adding a larger number of dummy accepting paths. PP is clearly closed under complement so we have co-NP in PP. Beigel, Reingold and Spielman show that PP is closed under union, a far trickier proof than one would expect using the fact that rational functions approximate the sign function well. Fortnow and Reingold build on their techniques to show that PP is closed under truth-table reductions.\n\nPP is the prototypical counting class, classes defined in terms of the number of accepting and rejecting paths. PP contains the counting class C=P where x is in the language if the number of accepting paths equals the number of rejecting paths. On the other hand there are relativized worlds where ⊕P and PP are incomparable.\n\nThe limitations of PP come from what I call the Beigel all-purpose oracle. Beigel exhibits a relativized world where PNP is not contained in PP. His proof works by showing that any polynomial whose sign is the ODDMAXBIT function must either have high-degree or very large coefficients.\n\nPP has interesting relationships to other complexity classes. Vereshchagin shows that Merlin-Arthur games, the class MA, is contained in PP but gives a relativized world where AM is not in PP. Adleman, DeMarrais and Huang show that the quantum class BQP is contained in PP. Watrous shows that QMA, the quantum version of MA, is also contained in PP.\n\nFortnow and Rogers, building Lide Li's Ph.D. thesis, show that BQP is low for PP, i.e., PPBQP = PP. Köbler, Schöning and Torán also using the work of Li show that graph isomorphism is also low for PP.\n\nAllender shows that PP is not equal to uniform-TC0, constant-depth circuits with threshold gates. Can one show that PP is not equal to log-space? All one has to do is show that Toda's theorem can be extended to get any nonconstant level of the polynomial-time hierarchy to fall in PPP.\n\n1.",
null,
"How is it exactly that you prove P^PP = P^#P? You said binary search, but I cant make sense of that.\n\n1.",
null,
"If f is a #P function then the language L={(x,k) | f(x) >= k} is in PP. Then do binary search on k.\n\n2.",
null,
"That means that if you have an oracle for PP, you can find the value of f(x), by doing binary search on k with L and therefore P^#P is in P^PP, correct? What about the other way (P^PP is in P^#P)? Even if you have an oracle for #P, how can you tell if more than half of the computational paths are accepting paths for an arbitrary problem?\n\n2.",
null,
"You don't talk about the result of Scott Aaronson:\n\n\"Post-BQP = PP\".\n\nAnd yet, some proofs of PP class in 'Quantum Computing'.\n\nMy first comments here ... Awesome & amazing blog, bye !!!!"
] | [
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"https://3.bp.blogspot.com/-seldi5E7_Nk/XfuyOhHz8PI/AAAAAAABud0/Cl9n03lFbS81u0NO2mP9bW8C8b4ahZALQCK4BGAYYCw/s35/fortnow.jpg",
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"https://resources.blogblog.com/img/blank.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9401037,"math_prob":0.83537555,"size":3926,"snap":"2021-21-2021-25","text_gpt3_token_len":898,"char_repetition_ratio":0.11218766,"word_repetition_ratio":0.02945508,"special_character_ratio":0.21293938,"punctuation_ratio":0.09541985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877743,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-11T11:00:43Z\",\"WARC-Record-ID\":\"<urn:uuid:0267de74-f2d4-4d24-bb69-9e7b2708c592>\",\"Content-Length\":\"167654\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe43e200-5513-4e22-bfea-e8aebd5ffa4b>\",\"WARC-Concurrent-To\":\"<urn:uuid:a0a0ba43-6aab-4f17-801a-18a8345bc4bd>\",\"WARC-IP-Address\":\"172.217.15.83\",\"WARC-Target-URI\":\"https://blog.computationalcomplexity.org/2002/09/complexity-class-of-week-pp.html\",\"WARC-Payload-Digest\":\"sha1:OJ2IX5AVIHYN3D2SUNEEREFI2ES7KSFS\",\"WARC-Block-Digest\":\"sha1:MGYOMCCATMCZRYZPZHXUP23NRMSICFXA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991982.8_warc_CC-MAIN-20210511092245-20210511122245-00014.warc.gz\"}"} |
http://simplise.me/tag/%E9%BB%84%E9%87%91/ | [
"## Talib实现国际黄金的MACD\n\n```import xlrd\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport talib\nfrom math import sqrt\nimport datetime\n\ndata=xlrd.open_workbook('D:/data/XAUUSD1440.xlsx')\ntable=data.sheets()\nncols=table.ncols\nnrows=table.nrows\nkai=[]\nhigh=[]\ndi=[]\nclose=[]\ndate=[]\nfor i in range(0,nrows):\nj=str(table.cell(i,1))\nkai.append(float(j.split(':')))\nw=str(table.cell(i,2))\nhigh.append(float(w.split(':')))\nl=str(table.cell(i,3))\ndi.append(float(w.split(':')))\ng=str(table.cell(i,3))\nclose.append(float(w.split(':')))\nx=str(table.cell(i,0))\ny=x.split(\"'\").split('.')\nz=datetime.date(int(y),int(y),int(y))\ndate.append(z)\n\nkai=np.array(kai)\nhigh=np.array(high)\ndi=np.array(di)\nclose=np.array(close)\n\nmacd,signal,hist=talib.MACD(close,12,26,9)\n\nplt.plot(date,macd,label='macd')\nplt.plot(date,signal,label='signal')\nplt.plot(date,hist,label='hist')\nplt.show()```",
null,
"## 黄金股票与黄金指数的关系研究\n\n```import xlrd\nfrom math import sqrt\ndata=xlrd.open_workbook('D:/data/159937.xlsx')\ntable=data.sheets()\nncols=table.ncols\nnrows=table.nrows\nli1=[]\nli3=[]\nfor i in range(0,nrows-1):\nj=str(table.cell(i+1,1))\nli1.append(float(j.split(':')))\nw=str(table.cell(i+1,3))\nli3.append(float(w.split(':')))\n\ndef multipl(a,b):\nsumofab=0.0\nfor i in range(len(a)):\ntemp=a[i]*b[i]\nsumofab+=temp\nreturn sumofab\n\ndef corrcoef(x,y):\nn=len(x)\n#求和\nsum1=sum(x)\nsum2=sum(y)\n#求乘积之和\nsumofxy=multipl(x,y)\n#求平方和\nsumofx2 = sum([pow(i,2) for i in x])\nsumofy2 = sum([pow(j,2) for j in y])\nnum=sumofxy-(float(sum1)*float(sum2)/n)\n#计算皮尔逊相关系数\nden=sqrt((sumofx2-float(sum1**2)/n)*(sumofy2-float(sum2**2)/n))\nreturn num/den\n\nprint corrcoef(li1,li3)```\n\nWind数据对比图，验证结论。",
null,
""
] | [
null,
"http://simplise.me/wp-content/uploads/2017/04/figure_1-1024x443.png",
null,
"http://simplise.me/wp-content/uploads/2017/03/对比图-1024x643.png",
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.5158298,"math_prob":0.9982708,"size":2394,"snap":"2021-04-2021-17","text_gpt3_token_len":1356,"char_repetition_ratio":0.12468619,"word_repetition_ratio":0.0,"special_character_ratio":0.30701753,"punctuation_ratio":0.18049793,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000003,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T01:40:16Z\",\"WARC-Record-ID\":\"<urn:uuid:5f68a746-3501-4ae8-a7de-26a2b5dda828>\",\"Content-Length\":\"36972\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb7bb75d-111a-459d-8d71-a7631ee8d7ef>\",\"WARC-Concurrent-To\":\"<urn:uuid:49d28146-ca3d-4df9-9d95-e5c3f6856f94>\",\"WARC-IP-Address\":\"43.242.131.138\",\"WARC-Target-URI\":\"http://simplise.me/tag/%E9%BB%84%E9%87%91/\",\"WARC-Payload-Digest\":\"sha1:MVOLBU3DRYXS6ICDVIXB4T5KKFL5K5W3\",\"WARC-Block-Digest\":\"sha1:E34QHFZVKLXX5HLFFA4AWPLVYOJ4HFFT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704804187.81_warc_CC-MAIN-20210126233034-20210127023034-00315.warc.gz\"}"} |
https://publications.hse.ru/en/articles/313267045 | [
"• A\n• A\n• A\n• ABC\n• ABC\n• ABC\n• А\n• А\n• А\n• А\n• А\nRegular version of the site\nLet $\\ka$ be any field of characteristic zero, $X$ be a del Pezzo surface and $G$ be a finite subgroup in $\\Aut(X)$. In this paper we study when the quotient surface $X / G$ can be non-rational over $\\ka$. Obviously, if there are no smooth $\\ka$-points on $X / G$ then it is not $\\ka$-rational. Therefore under assumption that the set of smooth $\\ka$-points on $X / G$ is not empty we show that there are few possibilities for non-$\\ka$-rational quotients.\nThe quotients of del Pezzo surfaces of degree $2$ and greater are considered in the author's previous papers. In this paper we study the quotients of del Pezzo surfaces of degree $1$. We show that they can be non-$\\ka$-rational only for the trivial group or cyclic groups of order $2$, $3$ and $6$. For the trivial group and the group of order $2$ we show that both $X$ and $X / G$ are not $\\ka$-rational if the $G$-invariant Picard number of $X$ is~$1$. For the groups of order $3$ and $6$ we construct examples of both $\\ka$-rational and non-$\\ka$-rational quotients \\mbox{of both $\\ka$-rational} and non-$\\ka$-rational del Pezzo surfaces of degree $1$ such that the $G$-invariant Picard number of $X$ is~$1$.\nAs a result of complete classification of non-$\\ka$-rational quotients of del Pezzo surfaces we classify surfaces that are birationally equivalent to quotients of $\\ka$-rational surfaces, and obtain some corollaries concerning fields of invariants of $\\ka(x , y)$."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9035204,"math_prob":0.9999182,"size":8542,"snap":"2021-43-2021-49","text_gpt3_token_len":2011,"char_repetition_ratio":0.13457485,"word_repetition_ratio":0.09575234,"special_character_ratio":0.21470381,"punctuation_ratio":0.07187894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994564,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T01:44:34Z\",\"WARC-Record-ID\":\"<urn:uuid:6af06b65-bc8b-497f-a23c-85489ad5ba31>\",\"Content-Length\":\"93755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7203b4da-afb6-4cbb-b5e6-9e2e11ef66ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad56dfbc-6e28-4229-984b-8daa08b6103c>\",\"WARC-IP-Address\":\"186.2.163.228\",\"WARC-Target-URI\":\"https://publications.hse.ru/en/articles/313267045\",\"WARC-Payload-Digest\":\"sha1:XJ7S4VVH7L7F42BXH7NWDO7CCYUM772C\",\"WARC-Block-Digest\":\"sha1:XZC74LSPPFMPH6OOHC347CGXSLT2DXHJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585537.28_warc_CC-MAIN-20211023002852-20211023032852-00342.warc.gz\"}"} |
http://kuanghy.github.io/2016/06/23/python3-new-feature | [
"Python3 的一些新特性\n\nPython 3 加入了很多新的特性，使得代码更加简洁，以及运行起来更加高效。这里收集一些很棒的、用得上的新特性。\n\n高级解包\n\n>>> a, b, *rest = range(10)\n>>> a\n0\n>>> b\n1\n>>> rest\n[2, 3, 4, 5, 6, 7, 8, 9]\n>>> a, *rest, b = range(10)\n>>> rest\n[1, 2, 3, 4, 5, 6, 7, 8]\n>>> *rest, a, b = range(10)\n>>> rest\n[0, 1, 2, 3, 4, 5, 6, 7]\n\ndef foo(a, *args, b=10): # 在 python2 中为无效的语法\nprint(a, args, b)\n\nfor i in gen()\nyield i\n\nyield from gen()\n\n函数注释(Function Annotations)\n\n>>> def f(x:int) -> float:\n... pass\n>>> f.__annotations__\n{'x': <class 'int'>, 'return': <class 'float'>}\n\n强制关键字参数\n\nPEP 3102 提供了一种强制使用关键字参数的方案，即在函数参数列表中，在 * 后的参数必须使用关键参数传递：\n\n>>> def foo(a, b, *, c, d=10):\n... print(f\"a={a}, b={b}, c={c}, d={d}\")\n...\n>>> foo(1, 2, 3)\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nTypeError: foo() takes 2 positional arguments but 3 were given\n>>> foo(1, 2, c=3)\na=1, b=2, c=3, d=10\n>>> foo(1, 2, c=3, d=4)\na=1, b=2, c=3, d=4\n\n强制位置参数\n\n>>> def foo(a, b, /, c, d, *, e, f=10):\n... print(f\"a={a}, b={b}, c={c}, d={d}, e={e}, f={f}\")\n...\n>>> foo(10, 20, 30, d=40, e=50)\na=10, b=20, c=30, d=40, e=50, f=10\n>>> foo(10, b=20, c=30, d=40, e=50, f=60)\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nTypeError: foo() got some positional-only arguments passed as keyword arguments: 'b'\n>>> foo(10, 20, 30, 40, 50, f=60)\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nTypeError: foo() takes 4 positional arguments but 5 positional arguments (and 1 keyword-only argument) were given\n\n异常捕获 as 关键字\n\ntry:\ndo_something()\nexcept Exception, e:\nprint e\n\n# or\n\ntry:\ndo_something()\nexcept Exception as e:\nprint e\n\n追溯异常栈\n\n>>> try:\n... 1/0\n... except Exception as e:\n... raise Exception(\"value error\") from e\n...\nTraceback (most recent call last):\nFile \"<stdin>\", line 2, in <module>\nZeroDivisionError: division by zero\n\nThe above exception was the direct cause of the following exception:\n\nTraceback (most recent call last):\nFile \"<stdin>\", line 4, in <module>\nException: value error\n\n>>> try:\n... 1/0\n... except Exception:\n... raise Exception(\"value error\") from None\n...\nTraceback (most recent call last):\nFile \"<stdin>\", line 4, in <module>\nException: value error\n\n新的字符串格式化方式\n\n>>> name = \"Huoty\"\n>>> f\"{name!r}\"\n\"'Huoty'\"\n>>> f'My name is {name}'\n'My name is Huoty'\n>>> width = 10\n>>> precision = 4\n>>> import decimal\n>>> value = decimal.Decimal(\"12.34567\")\n>>> f\"result: {value:{width}.{precision}}\" # nested fields\n'result: 12.35'\n>>> price = 5.18362\n>>> f'{price:.2f}'\n'5.18'\n>>> f'{price % 2 = }'\n'price % 2 = 1.1836200000000003'\n>>> import datetime\n>>> now = datetime.datetime.now()\n>>> f'{now} was on a {now:%A}'\n'2018-01-18 11:21:58.444054 was on a Thursday'\n>>> f'{now = :%Y-%m-%d}'\n'now = 2021-10-11'\n>>> def foo():\n... return 18\n...\n>>> f'result={foo()}'\n'result=18'\n\n合并字典\n\n>>> x = dict(a=1, b=2)\n>>> y = dict(b=3, d=4)\n>>> dict(x, **y)\n{'a': 1, 'b': 3, 'd': 4}\n\n>>> x = dict(a=1, b=2)\n>>> y = dict(b=3, d=4)\n>>> {**x, **y}\n{'a': 1, 'b': 3, 'd': 4}\n\n列表推导式优化\n\n>>> x = 1\n>>> [x for x in \"abc\"]\n['a', 'b', 'c']\n>>> x # x 变量被推导式修改\n'c'\n\n>>> x = 1\n>>> [x for x in 'ABC']\n['A', 'B', 'C']\n>>> x\n1\n\n不是一切都能比较\n\n>>> \"one\" > 1\nTraceback (most recent call last):\nFile \"<stdin>\", line 1, in <module>\nTypeError: unorderable types: str() > int()\n\n非局部变量声明\n\ndef outer():\nx = 1\ndef inner():\nx += 2\nprint(\"inner:\", x)\ninner()\nprint(\"outer:\", x)\n\nPython 3 中引入了 nonlocal 关键字来解决以上问题，即声明 非局部变量。以上代码如果在 Python 3 中，进行如下修改后可以正常运行：\n\ndef outer():\nx = 1\ndef inner():\nnonlocal x\nx += 2\nprint(\"inner:\", x)\ninner()\nprint(\"outer:\", x)\n\n赋值表达式\n\nif (n := len(a)) > 10:\nprint(f\"List is too long ({n} elements, expected <= 10)\")\n\ndiscount = 0.0\ndiscount = float(mo.group(1)) / 100.0\n\nwhile (block := f.read(256)) != '':\nprocess(block)\n\n[\nclean_name.title() for name in names\nif (clean_name := normalize('NFC', name)) in allowed_names\n]\n\n一些新的标准库\n\n• asyncio： 原生的异步 IO (协程) 实现\n• faulthandler：可用于查看进程的 traceback 信息"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.600222,"math_prob":0.97131133,"size":5490,"snap":"2022-05-2022-21","text_gpt3_token_len":2759,"char_repetition_ratio":0.10298943,"word_repetition_ratio":0.14837906,"special_character_ratio":0.39198542,"punctuation_ratio":0.22770563,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9843953,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-16T10:23:55Z\",\"WARC-Record-ID\":\"<urn:uuid:8d9dd192-0c8b-49de-926b-7beebb230911>\",\"Content-Length\":\"44526\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:466f1c5d-c0a0-4b68-af21-9d86261088e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:6c02bec9-6d56-4047-9694-952a7604ee02>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"http://kuanghy.github.io/2016/06/23/python3-new-feature\",\"WARC-Payload-Digest\":\"sha1:VY7RPNIYAJVO4RBRZAPCDKYIRO6DJGGO\",\"WARC-Block-Digest\":\"sha1:KX3H6KK7V6SA46LETTUMLCUV57C4DLPO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320299852.23_warc_CC-MAIN-20220116093137-20220116123137-00270.warc.gz\"}"} |
http://kb.sageintelligence.com/index.php?title=Error:_Sage_Accpac_-_Sales_Master_3-0_PVS_and_SQL | [
"# Error: Sage Accpac - Sales Master 3-0 PVS and SQL\n\nThe period expression in the container displays the periods 1-9 as single digits and prevents the order of periods from being displayed in calendar order.\n\nSYMPTOMS\n\nThe following is an example of an error that may occur:\n\nIf using a lookup on the ‘Period’ expression, the periods will not display in calendar order.\n\nCAUSE\n\nThe periods 1 – 9 display as single digits and since they are joined along with the year they are converted to a text string. Sorting numbers in text format does not display in the preferred method of calendar order in this situation.\n\nRESOLUTION/WORKAROUND\n\nSales Master 3-2 (AE-SQL) OR Sales Master 3-2 (AE-PVS)\n\nThe changes made were as follows:\n\n1. Sales Master 3-0 container: The expression “period” has been changed as follows:\n\nSQL:\n\nExpression source:\n\n[OESHDT].[YR]+'-'+ CASE WHEN LEN([OESHDT].[PERIOD])=1 THEN CAST(0 AS VARCHAR)++ CAST([OESHDT].[PERIOD] AS VARCHAR) ELSE CAST([OESHDT].[PERIOD] AS VARCHAR) END\n\nLookup Sql Select statement:\n\nSELECT DISTINCT [OESHDT].[YR]+'-'+ CASE WHEN LEN([OESHDT].[PERIOD])=1 THEN CAST(0 AS VARCHAR)++ CAST([OESHDT].[PERIOD] AS VARCHAR) ELSE CAST([OESHDT].[PERIOD] AS VARCHAR) END FROM [OESHDT] ORDER BY [OESHDT].[YR]+'-'+ CASE WHEN LEN([OESHDT].[PERIOD])=1 THEN CAST(0 AS VARCHAR)++ CAST([OESHDT].[PERIOD] AS VARCHAR) ELSE CAST([OESHDT].[PERIOD] AS VARCHAR) END\n\nPVS:\n\nExpression source:\n\n\"OESHDT\".\"YR\"+'-'+ CASE WHEN LENGTH(\"OESHDT\".\"PERIOD\")=1 THEN CAST(0 AS VARCHAR)++CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) ELSE CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) END\n\nLookup Sql Select statement:\n\nSELECT DISTINCT \"OESHDT\".\"YR\"+'-'+ CASE WHEN LENGTH(\"OESHDT\".\"PERIOD\")=1 THEN CAST(0 AS VARCHAR)++CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) ELSE CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) END FROM \"OESHDT\" ORDER BY \"OESHDT\".\"YR\"+'-'+ CASE WHEN LENGTH(\"OESHDT\".\"PERIOD\")=1 THEN CAST(0 AS VARCHAR)++CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) ELSE CAST(\"OESHDT\".\"PERIOD\" AS VARCHAR) END\n\nAPPLIES TO\n\nSales Master 3-0 (AE-PVS)\n\nSales Master 3-1 (AE-PVS)\n\nSales Master 3-0 (AE-SQL)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60376394,"math_prob":0.9094694,"size":2068,"snap":"2019-51-2020-05","text_gpt3_token_len":619,"char_repetition_ratio":0.21027131,"word_repetition_ratio":0.29259259,"special_character_ratio":0.27659574,"punctuation_ratio":0.105793454,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97684187,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-05T23:30:16Z\",\"WARC-Record-ID\":\"<urn:uuid:7421c126-59aa-4cee-b44c-542459715fdb>\",\"Content-Length\":\"17851\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e07e3a94-1c96-45e9-86d0-02bbd87faddd>\",\"WARC-Concurrent-To\":\"<urn:uuid:e968f02b-7db5-4841-8784-8c9a3aa21c6a>\",\"WARC-IP-Address\":\"142.4.23.64\",\"WARC-Target-URI\":\"http://kb.sageintelligence.com/index.php?title=Error:_Sage_Accpac_-_Sales_Master_3-0_PVS_and_SQL\",\"WARC-Payload-Digest\":\"sha1:HGA5WN35U3RGIMJOKKVBN7LNCUIRRDVP\",\"WARC-Block-Digest\":\"sha1:YKQXNBWB2OPHLZWMGSRSZSVZJZBTMLY5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540482284.9_warc_CC-MAIN-20191205213531-20191206001531-00152.warc.gz\"}"} |
https://www.slidestalk.com/u3804/Regularization14517 | [
"HOT\n\nRegularization\n8 点赞\n3 收藏\n0下载\nRegularization is a technique used in an attempt to solve overfitting in machine learning models. A model will have a low accuracy if it is overfitting.\n\n1.Regularization Attempt to prevent overfitting Gene Olafsen\n\n2.overview Regularization is a technique used in an attempt to solve overfitting in machine learning models. A model will have a low accuracy if it is overfitting.\n\n3.why Typically, a model is trying too hard and capturing noise in the training dataset, or the data points that don’t really represent the true properties of the data. Learning such data points, makes the model more flexible, at the risk of overfitting. Summary: A well trained model is low on bias and variance.\n\n4.validation Validation: The process of deciding whether the numerical results quantifying hypothesized relationships between variables, are acceptable as descriptions of the data. Types of Validation: Holdout Method K-Fold Cross Validation Starfield K-Fold Cross Validation Leave-one-out Validation Leave-P-Out Cross Validation\n\n5.Holdout Method Simply remove a part of the training data and using it to get predictions from the model trained on rest of the data. Pro: Simple to implement. Con: Suffers from issues of high variance-- it is not certain which data points will end up in the validation set.\n\n6.Removing Data for Validation Holding out data for validation can be problematic. Removing a part of the data for validation poses a problem of the model underfitting. By reducing the training data, there is a risk of losing important patterns/ trends in data set, which in turn increases error induced by bias.\n\n7.K-Fold Cross Validation The data is divided into k subsets. The holdout method is repeated k times Advantage: Significantly reduces bias as we are using most of the data for fitting, and also significantly reduces variance as most of the data is also being used in validation set\n\n8.Starfield K-Fold Cross Validation In some cases, there may be a large imbalance in the response variables. In dataset concerning price of houses, there might be large number of houses having high price. In case of classification, there might be several times more negative samples than positive samples.\n\n9.Starfield Course correction Simply modify the typical K-fold cross validation sample such that each fold contains approximately the same percentage of samples of each target class as the complete set. In the case of prediction problems, correct the samples such that the mean response value is approximately equal in all of the folds.\n\n10.Leave-one-out cross validation Leave-one-out is bringing k-fold to the extreme where k=m (where m is the number of data samples). For each one of the m iterations there is a single sample that is reserved for testing while the other m-1 are used for training.\n\n11.Leave-P-Out Cross Validation Where p is the number of samples that are used for testing in each iteration. This is an exhaustive method, all possible combinations of p samples must be used, implying a large number of train-test iterations run\n\n12.Random subsamples (without replacement) This method is similar to k-fold strategy, but in each iteration we randomly select some samples for testing, and some others for training. A drawback of this method is that the samples may never be selected in the test set, whereas others may be selected more than once.\n\n13.Early stopping To avoid overfitting the training set- one solution is just to stop training. In practice, this can be accomplished by saving a model snapshot at regular intervals and then comparing the current training against the previous snapshot. If performance starts dropping- simply restore the previous model. Note: Higher performance can generally be achieved by combining early stopping with other regularization techniques.\n\n14.LaSso and ridge regularization A tuning parameter is added which lets you change the complexity or smoothness of the model. The regularization value imposes a special penalty on complex models.\n\n15.L1 - LASSO LASSO (least absolute shrinkage and selection operator) L1 is the sum of the weights. adds “absolute value of magnitude” of coefficient as penalty term to the loss function a lot of non-zero coefficients When to use: Lasso shrinks the less important feature’s coefficient to zero thus, removing some feature altogether. This works well for feature selection in cases having a huge number of features.\n\n16.L2 - Ridge Ridge Regression L2 is the sum of the square of the weights. models with large coefficients adds “squared magnitude” of coefficient as penalty term to the loss function\n\n17.L1 and L2 Summarization the\n\n18.dropout Proposed by G.E. Hinton in 2012 and further refined by Nitish Srivastava. State-of-the-art neural networks receive a 1-2% boost in accuracy by utilizing dropout. For 95% accurate models this means dropping the error rate by almost 40%.\n\n19.Dropout parameter The hyperparameter p is called the dropout rate and its value is typically set at 50%. At every training step, every neuron has the probability of p of participating in the model or being dropped-out of the model. Note: Output neurons DO NOT participate in drop out. (leaving only input and hidden layer neurons)\n\n20.Dropout for real\n\n21.Why does dropout work? The resulting neural network can be seen as an averaging ensemble of the smaller neural networks produced during the training steps.\n\n22.Using a model trained with dropout You need to multiply each neurons input connection weights by the dropout p value after training. So if the model was trained with 50% dropout, the input connection weights must be multiplied by 0.5 after training. If you dont... each neuron will get a total input signal roughly twice as large as what the network was trained on.\n\n23.Max-norm regularization Constrains the weights w, clipping w if it exceeds hyperparameter r after each training step. Reducing r increases the amount of regularization and helps reduce overfitting. TensorFlow does not implement Max-Norm Regularization. It is not difficult to implement. The machine learning book provides an example. Keras provides Max-Norm regularization out of the box.\n\n24.Data augmentation Generating new training instances from existing data to boost the size of the training set. Common transformations include image shift, rotate, resize and contrast. Note: Adding white noise does not augment data because white noise is not learnable.\n\n25.Default deep neural network configuration Initialization: He initialization Activation Function: ELU Normalization: Batch Normalization Regularization: Dropout Optimizer Nesterov Accelerated Gradient Learning Rate Schedule: None\n\n26.Regularization guidelines If convergence is too slow, then try a learning schedule that utilizes exponential decay. If the training set is too small, try implementing data augmentation. To achieve a sparse model, add l1 (LASSO) regularization- and zero out tiny weights after training. To achieve an even sparser model, try FTRL instead of Adam optization along with l1 regularization. For a fast runtime model, do not use Batch Normalization and replace ELU activation function with leaky ReLU . (Also a sparse model helps) From: Hands-on Machine Learning with Scikit -Learn & TensorFlow\n\n8 点赞\n3 收藏\n0下载"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9034573,"math_prob":0.9336413,"size":7091,"snap":"2021-43-2021-49","text_gpt3_token_len":1424,"char_repetition_ratio":0.13475378,"word_repetition_ratio":0.012669683,"special_character_ratio":0.19376674,"punctuation_ratio":0.103846155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97832114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T09:38:26Z\",\"WARC-Record-ID\":\"<urn:uuid:b646231d-c987-4dee-bda2-6e688f290cc1>\",\"Content-Length\":\"68848\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c7f085fa-6c6b-48a7-b278-1923c2ae6f02>\",\"WARC-Concurrent-To\":\"<urn:uuid:28807fcf-12ae-44bd-bc12-19770a881d9d>\",\"WARC-IP-Address\":\"101.132.193.79\",\"WARC-Target-URI\":\"https://www.slidestalk.com/u3804/Regularization14517\",\"WARC-Payload-Digest\":\"sha1:SOQKPU55RWYEL3BHPSLO34NI3SDMOPLW\",\"WARC-Block-Digest\":\"sha1:Z4A7OLEJ2WVN7NMXKUGOIEU6N7G35X2J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588113.25_warc_CC-MAIN-20211027084718-20211027114718-00696.warc.gz\"}"} |
http://www.base-ten.com/tm/2014/10/ | [
"# YouCubed (and more on fluency)",
null,
"Jo Boaler, a math education researcher and professor at Stanford University, launched a new website this past year (YouCubed) to help teachers, students, and parents navigate math education. She recently published a short paper on math fluency. In it she discusses the problems with associating math fluency with speed or memorization.\n\nInterestingly, the Common Core intends this de-emphasis on speed but the word “fluency” is often misunderstood by textbook publishers. The newly adopted Bridges Curriculum seems not fall into this category – their strategy for building fluency is based strongly on number sense. (Read an excerpt from Jo’s paper below)\n\nThis past September the Conservative education minister for England, a man with no education experience, insisted that all students in England memorize all their times tables up to 12 x 12 by the age of 9. This requirement has now been placed into the UK’s mathematics curriculum and will result, I predict, in rising levels of math anxiety and students turning away from mathematics in record numbers. The US is moving in the opposite direction, as the new Common Core State Standards (CCSS) de-emphasize the rote memorization of math facts. Unfortunately misinterpretations of the meaning of the word ‘fluency’ in the CCSS are commonplace and publishers continue to emphasize rote memorization, encouraging the persistence of damaging classroom practices across the United States.\n\n# Math Anxiety and Fluency",
null,
"Math anxiety is real and it affects about half of the U.S. population. Brain imaging has shown us that stress gets in the way of working memory, which is typically required to do mathematics. In fact, math anxiety seems to affect people with a high amount of working memory most! One of the main sources of math anxiety is the stress produced by timed tests (Boaler, Education Week, 7/3/2012).\n\nI used to have my students take timed multiplication and division tests to practice fact (computational) fluency, but I don’t do this anymore. Time is too heavily factored into the measure of fluency in that equation. Rather, I have students learn and practice strategies for working with numbers so that they can be fluent.\n\nHere’s a tip for developing the habit of seeing patterns in numbers: spend more time looking for patterns in math homework and throughout school mathematics. School math is designed to illustrate patterns! Though this is sometimes a weakness in that a framework is often given and problems are explicitly trying to illustrate relationships, this will help students (who attend to these patterns) develop strategic thinking.\n\nTake for example two problems from last night’s homework.\n\n1. Kaden divided 96 by 12 and then multiplied by 6. Write an expression to show the problem. Then, solve the problem and write an equation.\n\nWhen most students first approach a problem like this, they begin to decode or solve it step by step. First they do 96 divided by 12 getting… what is that anyway? And then they multiply by 6.\n\nWhen I look at that problem, the first thing I notice is that 96 is being divided and multiplied. These are inverse operations – meaning that they undo each other. If the problem was 96 divided by 12 times 12, I hope we would just leave it at 96 because we all know this doing and undoing relationship. In this case, the situation is only slightly less humorous (mathematically speaking). If we are going to divide 96 by 12 and then multiply by 6, then we are undoing the dividing by 6! This leaves a much simpler expression: 96 / 2 (96 divided by 2). The equation I would write would read 96 / 12 * 6 = 96 /2 = 48. That’s number fluency.\n\n1. 81 – (9 * 7)\n\nHere too, most of my students given this for homework followed the “proper steps” and dutifully multiplied 9 * 7 and then subtracted that from 81. This misses a beautiful relationship embedded in the problem. If you notice that 81 is also a multiple of 9, you can rewrite the problem in the following way.\n\n81 – (9 * 7) = (9 * 9) – (9 * 7)\n\nWe are subtracting 7 groups of 9 from 9 groups of 9. The difference is 9 * 2.\n\nWhen most people think about what mathematics is, they think about memorization and procedures. I think this is what it felt like coming through the system for a lot of people in my generation. Today, in many classrooms around the world, math is taught around understanding ideas, discovering relationships, exploring patterns, and developing strategic thinking and problem solving skills.\n\nThe difference between memorizing a bunch of rules and procedures and coming to understand mathematical ideas and principles is profound. Memorization requires no real reflection or ownership. There is often very little connection between the learner and the material. As a result, the learning is shallow and has very little application outside of school for that student. Teaching number fluency with conceptual understanding on the other hand requires more engagement with the material. It challenges students to come to terms with a deeply interconnected and vibrant discipline. I believe it also opens the door to a real relationship with mathematics that can help students avoid math anxiety.\n\nMathematics is sometimes called the science of pattern and order. This is exactly what draws people to math – let’s keep that perception of the subject alive when we are actually doing math work."
] | [
null,
"http://www.base-ten.com/tm/wp-content/uploads/2014/10/youcubed-thumb.png",
null,
"http://www.base-ten.com/tm/wp-content/uploads/2014/09/stress.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9632075,"math_prob":0.826018,"size":3887,"snap":"2020-24-2020-29","text_gpt3_token_len":833,"char_repetition_ratio":0.109966524,"word_repetition_ratio":0.0059435363,"special_character_ratio":0.22253665,"punctuation_ratio":0.09264305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9644098,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,5,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-08T10:26:29Z\",\"WARC-Record-ID\":\"<urn:uuid:8d1b0182-81d7-46d2-a849-dcbf370e0f77>\",\"Content-Length\":\"26328\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6886bc46-a9ac-44b3-aed2-ddc1e576b428>\",\"WARC-Concurrent-To\":\"<urn:uuid:6505790d-e993-49fc-8a3a-a020741f5fbf>\",\"WARC-IP-Address\":\"208.113.216.40\",\"WARC-Target-URI\":\"http://www.base-ten.com/tm/2014/10/\",\"WARC-Payload-Digest\":\"sha1:5L425QNUJNHO4EB2LBIMCDICZFRSEE75\",\"WARC-Block-Digest\":\"sha1:USY32D6GCBJ6BK2WVE4JQZX2OQ3IIVDJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655896932.38_warc_CC-MAIN-20200708093606-20200708123606-00221.warc.gz\"}"} |
https://www.numwords.com/words-to-number/en/3508 | [
"NumWords.com\n\nHow to write Three thousand five hundred eight in numbers in English?\n\nWe can write Three thousand five hundred eight equal to 3508 in numbers in English\n\n< Three thousand five hundred seven :||: Three thousand five hundred nine >\n\nSeven thousand sixteen = 7016 = 3508 × 2\nTen thousand five hundred twenty-four = 10524 = 3508 × 3\nFourteen thousand thirty-two = 14032 = 3508 × 4\nSeventeen thousand five hundred forty = 17540 = 3508 × 5\nTwenty-one thousand forty-eight = 21048 = 3508 × 6\nTwenty-four thousand five hundred fifty-six = 24556 = 3508 × 7\nTwenty-eight thousand sixty-four = 28064 = 3508 × 8\nThirty-one thousand five hundred seventy-two = 31572 = 3508 × 9\nThirty-five thousand eighty = 35080 = 3508 × 10\nThirty-eight thousand five hundred eighty-eight = 38588 = 3508 × 11\nForty-two thousand ninety-six = 42096 = 3508 × 12\nForty-five thousand six hundred four = 45604 = 3508 × 13\nForty-nine thousand one hundred twelve = 49112 = 3508 × 14\nFifty-two thousand six hundred twenty = 52620 = 3508 × 15\nFifty-six thousand one hundred twenty-eight = 56128 = 3508 × 16\nFifty-nine thousand six hundred thirty-six = 59636 = 3508 × 17\nSixty-three thousand one hundred forty-four = 63144 = 3508 × 18\nSixty-six thousand six hundred fifty-two = 66652 = 3508 × 19\nSeventy thousand one hundred sixty = 70160 = 3508 × 20\n\nSitemap"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7568183,"math_prob":0.99924743,"size":1349,"snap":"2019-26-2019-30","text_gpt3_token_len":410,"char_repetition_ratio":0.3070632,"word_repetition_ratio":0.016806724,"special_character_ratio":0.4017791,"punctuation_ratio":0.023364486,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T09:27:31Z\",\"WARC-Record-ID\":\"<urn:uuid:7ba13f5f-a3f8-4055-816a-2229802c42ce>\",\"Content-Length\":\"3442\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:37c67b63-d923-42af-9c48-1b2cecb4892b>\",\"WARC-Concurrent-To\":\"<urn:uuid:2954b40a-db79-4ba9-af26-5cced5655b63>\",\"WARC-IP-Address\":\"68.66.224.32\",\"WARC-Target-URI\":\"https://www.numwords.com/words-to-number/en/3508\",\"WARC-Payload-Digest\":\"sha1:YSOYXNGDOKJY3PMDYFNXJV2UVJLVO7SF\",\"WARC-Block-Digest\":\"sha1:D2SNNIIUM4UTM4M5ATKKBSUEFYW35Y6E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999200.89_warc_CC-MAIN-20190620085246-20190620111246-00214.warc.gz\"}"} |
http://export.arxiv.org/abs/1811.01181 | [
"### Current browse context:\n\ncond-mat.supr-con\n\n# Title: Nodeless superconductivity in the noncentrosymmetric Mo$_3$Rh$_2$N superconductor: a $μ$SR study\n\nAbstract: The noncentrosymmetric superconductor Mo$_3$Rh$_2$N, with $T_c = 4.6$ K, adopts a $\\beta$-Mn-type structure (space group $P$4$_1$32), similar to that of Mo$_3$Al$_2$C. Its bulk superconductivity was characterized by magnetization and heat-capacity measurements, while its microscopic electronic properties were investigated by means of muon-spin rotation and relaxation ($\\mu$SR). The low-temperature superfluid density, measured via transverse-field (TF)-$\\mu$SR, evidences a fully-gapped superconducting state with $\\Delta_0 = 1.73 k_\\mathrm{B}T_c$, very close to 1.76 $k_\\mathrm{B}T_c$ - the BCS gap value for the weak coupling case, and a magnetic penetration depth $\\lambda_0 = 586$ nm. The absence of spontaneous magnetic fields below the onset of superconductivity, as determined by zero-field (ZF)-$\\mu$SR measurements, hints at a preserved time-reversal symmetry in the superconducting state. Both TF-and ZF-$\\mu$SR results evidence a spin-singlet pairing in Mo$_3$Rh$_2$N.\n Comments: 5 figures and 5 pages. Accepted for publication as a Rapid Communication in Phys. Rev. B Subjects: Superconductivity (cond-mat.supr-con); Strongly Correlated Electrons (cond-mat.str-el) Journal reference: Phys. Rev. B 98, 180504(R) (2018) DOI: 10.1103/PhysRevB.98.180504 Cite as: arXiv:1811.01181 [cond-mat.supr-con] (or arXiv:1811.01181v1 [cond-mat.supr-con] for this version)\n\n## Submission history\n\nFrom: Tian Shang [view email]\n[v1] Sat, 3 Nov 2018 09:17:13 GMT (198kb)\n\nLink back to: arXiv, form interface, contact."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81882715,"math_prob":0.9309012,"size":1432,"snap":"2020-45-2020-50","text_gpt3_token_len":405,"char_repetition_ratio":0.11554622,"word_repetition_ratio":0.0,"special_character_ratio":0.26047486,"punctuation_ratio":0.13059701,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9653203,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-29T00:22:57Z\",\"WARC-Record-ID\":\"<urn:uuid:c1c10443-ff40-4a32-a3cf-20eacfab51d1>\",\"Content-Length\":\"18331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3551bccd-3785-4f3b-9b78-f6d71164486a>\",\"WARC-Concurrent-To\":\"<urn:uuid:692ffd6c-4adc-4280-b700-4dfbc13377d6>\",\"WARC-IP-Address\":\"128.84.21.203\",\"WARC-Target-URI\":\"http://export.arxiv.org/abs/1811.01181\",\"WARC-Payload-Digest\":\"sha1:D4ANV75RCCVOOCCO7M2QOFEJQJZJIYMZ\",\"WARC-Block-Digest\":\"sha1:7P2EFT4VBLTSD7FMYNUBBOYPZRMNB7VX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141195929.39_warc_CC-MAIN-20201128214643-20201129004643-00413.warc.gz\"}"} |
https://drupal.stackexchange.com/questions/218052/pass-variables-through-ajax-callback-and-update-mysql-database | [
"# Pass variables through AJAX callback and update MySQL database\n\nI have a form where the users submit a text. Here is the code:\n\n``````\\$form['reason'] = array(\n'#type' => 'textfield',\n'#title' => '',\n'#size' => '200',\n'#attributes' =>array('placeholder' => t('*Reason')),\n'#prefix' => '<div>'\n);\n\n'#type' => 'submit',\n'#value' => 'Submit',\n'#attributes'=> array(\n'class' => ['btn', 'btn-primary']\n),\n'#ajax' => array(\n'method' => 'replace',\n),\n'#suffix' => '</div></div>');\n\nreturn \\$form;\n``````\n\nI managed to get the input of the user on a variable using `\\$_values = \\$form_state['values'];`\nAlso, I have some other variables to read from the current path and get some values.\n\n`````` \\$_path = \\$_SERVER['HTTP_HOST'] . request_uri();\n\npreg_match('~id=(.*?)&~', \\$_path, \\$id);\n\n\\$first = \\$id;\n\\$second = substr(\\$_path, -5);\n``````\n\nwhere the first is the user id and the second is another variable that is unique per user.\nMy only problem is that I can't pass the \\$first and \\$second variables through the callback, and have them along with \\$reason `\\$_reason = filter_xss(\\$_values['reason']);`\nSo, in my callback I want to be able to get the variables \\$first,\\$second with the \\$reason I already have. Now I have the \\$_path variables above my form, in the same function.\nThen I want to perform an update query on MySQL, something like\n\n``````UPDATE users SET reason=\\$reason\nWHERE first=\\$first AND second=\\$second\n``````\n\nwhere the column users has only the columns first,second and reason . Is this possible? Thanks in advance!\n\nEDIT\nI actually managed to do the first part and get the variables. I created two form inputs and I give them default values, the two variables I want to get, and I will hide them afterwards.\nI still haven't figured out how to update MySQL though, so all answers are welcome!\n\nI actually managed to do it! For anyone that may have a similar problem this is what I did.\nFirst of all, I took the url parameters using `parse_url` and did the following\n\n``````\\$parts = parse_url(\\$_path);\nparse_str(\\$parts['query'], \\$query);\n\\$first = \\$query['first'];\n\\$second = \\$query['second'];\n``````\n\n,then put those variables as form default values, that I will hide using css:\n\n``````\\$form['user_id'] = array (\n'#type' => 'textfield',\n'#title' => t('User id'),\n'#description' => t('user id'),\n'#default_value' => \\$first,\n);\n``````\n\nThen, on my callback function I take those variables, for example:\n\n``````\\$_first = filter_xss(\\$_values['first']);\n``````\n\nand then I used `db_update`:\n\n``````\\$num_updated = db_update('unsubscribed_users')\n->fields(array(\n'reason' => \\$_reason,\n))\n->condition('first', \\$_first , '=')\n->condition('second', \\$_second , '=')\n->execute();\n``````"
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http://downloads.hindawi.com/journals/mpe/2013/531031.xml | [
"MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi Publishing Corporation 531031 10.1155/2013/531031 531031 Research Article Artificial Hydrocarbon Networks Fuzzy Inference System http://orcid.org/0000-0002-6559-7501 Ponce Hiram Ponce Pedro Molina Arturo Chen Graduate School of Engineering Tecnológico de Monterrey Campus Ciudad de México 14380 Mexico City, DF Mexico itesm.edu 2013 23 9 2013 2013 13 05 2013 25 07 2013 01 08 2013 2013 Copyright © 2013 Hiram Ponce et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nThis paper presents a novel fuzzy inference model based on artificial hydrocarbon networks, a computational algorithm for modeling problems based on chemical hydrocarbon compounds. In particular, the proposed fuzzy-molecular inference model (FIM-model) uses molecular units of information to partition the output space in the defuzzification step. Moreover, these molecules are linguistic units that can be partially understandable due to the organized structure of the topology and metadata parameters involved in artificial hydrocarbon networks. In addition, a position controller for a direct current (DC) motor was implemented using the proposed FIM-model in type-1 and type-2 fuzzy inference systems. Experimental results demonstrate that the fuzzy-molecular inference model can be applied as an alternative of type-2 Mamdani’s fuzzy control systems because the set of molecular units can deal with dynamic uncertainties mostly present in real-world control applications.\n\n1. Introduction\n\nIt is well known that fuzzy inference models are very important in applications when information is uncertain and imprecise, like: robotics, medicine, control, modeling, and so forth . Moreover, fuzzy inference models may deal with nonlinearities in the input domain to transform them into an output domain. In that way, the literature reports three main models: Takagi-Sugeno inference systems , Mamdani’s fuzzy control systems , and Tsukamoto’s inference model .\n\nRoughly speaking, Takagi-Sugeno inference systems apply polynomial functions to construct the consequent values using pairs of input-output data of a given system to model . Mamdani’s fuzzy control systems refer to control laws that apply fuzzy inference models with fuzzy partitions in the defuzzification phase , obtaining mostly the output value with the center of gravity (COG) function . In contrast, Tsukamoto’s inference models implement monotonic membership functions . For detailed information, see .\n\nThe above inference models were developed under type-1 fuzzy systems. However, these models have disadvantage in terms of dynamic uncertainties present at inputs. For example, the latter gives poor performance in control systems because real-world control applications present dynamic uncertainties inherently [12, 13]. In contrast, type-2 fuzzy systems were proposed as an improvement of type-1 fuzzy inference systems. For instance, recent applications on fuzzy control systems have demonstrated the ability of type-2 fuzzy control systems to handle with noise and perturbations .\n\nOn the other hand, other fuzzy inference models have been proposed as hybrid algorithms using heuristics to manage unusual information, pattern recognition, and learning. Some of these fuzzy inference models use genetic algorithms, harmony search algorithms, tabu search, artificial neural networks, swarm intelligence techniques, and so forth [3, 4, 15].\n\nRecently, H. Ponce and P. Ponce proposed a new computational algorithm for modeling problems named artificial hydrocarbon networks based on natural hydrocarbon compounds. This algorithm claims for stability, well forming of compounds, easiness of spanning structures, and a degree of interpretation of the resultant model based on organized structures. In particular, the basic unit of information in this algorithm is the molecule. Actually, molecules are simple packages of information that can be inherited and interpreted. At last, basic chemical rules are applied to build the final structure.\n\nThen, the objective of this paper is to present a novel fuzzy inference model based on artificial hydrocarbon networks named fuzzy-molecular inference (FIM) model. In that sense, molecules can model consequent values of fuzzy rules and partition linguistic variables. Moreover, a fuzzy control system based on the FIM-model is presented as a case study. Experimental results demonstrate that the fuzzy-molecular inference model can be applied as an alternative of type-2 Mamdani’s fuzzy control systems because the set of molecular units can deal with dynamic uncertainties mostly present in real-world control applications.\n\nThe paper is ordered as follows. Next section presents a review of artificial hydrocarbon networks algorithm introduced recently in . The following sections introduce new material. Section 3 describes the fuzzy-molecular inference model in detail, current proposal of the paper. Section 4 introduces an example of how to apply the FIM-model in fuzzy control systems. Section 5 presents a case study in which a type-2 fuzzy control system based on the FIM-model implements a position controller of a direct current (DC) motor. Section 6 presents the experimental results of the case study and discusses some differences between the proposed model and other fuzzy inference systems and the advantages of the FIM-model to be used as an alternative of type-2 fuzzy systems. Finally, Section 7 concludes the paper and presents future directions.\n\n2. Artificial Hydrocarbon Networks\n\nIn this section, a brief review of artificial hydrocarbon networks is presented. However, this algorithm is subjected to the artificial organic networks technique. Thus, a first description of artificial organic networks technique is discussed and then artificial hydrocarbon networks algorithm is formally introduced.\n\n2.1. Brief Review of Artificial Organic Networks\n\nObservations to chemical organic compounds reveal enough information to derive the artificial organic networks technique firstly proposed by H. Ponce and P. Ponce . From studies of organic chemistry, organic compounds are the most stable ones in nature. In addition, molecules can be seen as units of packaging information; thus, complex molecules and its combinations can determine a nonlinear interaction of information. Moreover, molecules can be used for encapsulation and potential inheritance of information. Thus, artificial organic networks take advantage of this knowledge, inspiring a computational algorithm that infer and classify information based on stability and chemical rules that allow formation of molecules [19, 21].\n\nArtificial organic networks (AONs for short) define four components: atoms, molecules, compounds, and mixtures; and two basic interactions among components: covalent bonds and chemical balance interaction. In order to follow chemical rules, the following definitions of AONs hold .\n\n(a) Atoms. They are the basic units with structure. No information is stored. In addition, when two atoms have the same number of degrees of freedom they are called similar atoms and different atoms otherwise. The degree of freedom is the number of valence electrons that allow atoms to be linked with others.\n\n(b) Molecules. They are the interactions of two or more atoms made of covalent bonds. These components have structural and behavioral properties. Structurally, they conform the basis of an organized structure while behaviorally they can contain information. Thus, molecules are known as the basic units of information. If a molecule has filled out all of the valence electrons in atoms, it is stable; but if a molecule has at least one valence electron without filling, it is considered as unstable.\n\n(c) Compounds. In structure, they are two or more molecules interacting with each other linked with covalent bonds. Their behaviors are mappings from the set of molecular behaviors to real values.\n\n(d) Mixtures. They are the interaction of two or more molecules and/or compounds without physical bonds. Mixtures are linear combinations of molecules and/or compounds forming a basis of molecules with weights so-called stoichiometric coefficients.\n\n(e) Covalent Bonds. They are of two types. For this work, polar covalent bonds refer to the interaction of two similar atoms, while nonpolar covalent bonds refer to the interaction of two different atoms.\n\n(f) Chemical Balance Interaction. It refers to find the proper values of stoichiometric coefficients in mixtures in order to satisfy constrains in artificial organic networks.\n\nIn fact, artificial organic networks follow the energy model that states the hierarchical order in which components are used to form the final structure to minimize energy. For instance, the first strategy considers formation of molecules. If molecules cannot deal with the problem, compounds are required. Finally, mixtures of molecules and/or compounds will act.\n\n2.2. Description of Artificial Hydrocarbon Networks\n\nArtificial hydrocarbon networks (AHNs for short) algorithm is based on artificial organic networks that implement notions of natural hydrocarbon compounds [19, 21]. Formally, AHNs define components, interactions, and the training algorithm, in order to infer and classify information given any system. In that way, two main procedures are needed for AHNs: training and reasoning. Following, a brief review of artificial hydrocarbon networks is presented.\n\n2.2.1. Basic Components\n\nIn particular to AHNs, only two types of atoms are considered: hydrogen atoms H and carbon atoms C. Those have valence electrons eH=1 and eC=4 for the hydrogen and carbon atoms, respectively. In that sense, hydrocarbon atoms can be bonded with at most one atom while carbon atoms can be bonded with at most four, knowing as the octet rule .\n\nThe basic unit of information is a CH-molecule. These kinds of molecules are structurally made of hydrogen and carbon atoms following the octet rule. Let Mi be the structure of a molecule, and, φi be the behavior of molecule Mi. Then, φi is a mapping from some set X to real numbers . Moreover, let MH and MC be two molecules with behaviors φH and φC, and if (1) holds for these behaviors, then MH and MC are called CH-molecules, where h is complex constant value named hydrogen value, x is any input value |x|1 that excites a molecule, and d is the number of hydrogen atoms attached to a carbon atom: (1)φH(x)=h,h,φC(x)=i=1deC(x-φHi),x. Let MCH, MCH2, and MCH3 be CH-molecules with behaviors φCH, φCH2, and φCH3 like (2), respectively. Then, they are called CH-primitive molecules. Intuitively, these molecules can be seen as basic packages to join among them forming complex molecules-like compounds: (2)φCH(x)=(x-h1),φCH2(x)=(x-h1)(x-h2),φCH3(x)=(x-h1)(x-h2)(x-h3). Let Ci be a compound formed with a set of p CH-molecules , and let ψi be the behavior of Ci. Then, ψi is expressed as (3), where φj are the behaviors of CH-molecules in and π is the behavior of nonpolar covalent bonds that links molecules (3)ψi=π(φ1,,φj,,φp,x). Finally, let Ci be a n molecules or compounds with behaviors ψi. Then, S is a mixture of molecules or compounds and it is expressed as a linear combination of them like (4); where, αi is a set of real values named stoichiometric coefficients representing the ratio of molecules or compounds occupied in the mixture. (4)S(x)=iαiψi(x). Let AHN be a mixture of molecules or compounds in the set Γ representing the structure of molecules or compounds (how they are connected), and, S be the behavior of the mixture. Then, AHN is called an artificial hydrocarbon network if Γ is spanned from CH-molecules. Figure 1 shows a simple artificial hydrocarbon network. It is remarkable to say that topology Γ is a fixed structure parameterized with hydrogen values h and stoichiometric coefficients αi.\n\nSimple artificial hydrocarbon network. White circles represent hydrogen atoms, and black circles represent carbon atoms.\n\n2.2.2. Training of Artificial Hydrocarbon Networks\n\nArtificial hydrocarbon networks can deal with modeling problems like inferring or clustering in order to approximate any given system Σ with a pair of samples (x,y). In fact, let Σ be a simple-input-simple-output (SISO) system with input signal x and output signal y. The training process of an artificial hydrocarbon network is summarized in Algorithm 1 which receives the sample pairs of Σ, the number of CH-molecules p and the number of compounds c. Algorithm 1 outputs the structure Γ, hydrogen values h, and stoichiometric coefficients αi.\n\n<bold>Algorithm 1: </bold>Training algorithm for artificial hydrocarbon networks.\n\n(1) Initialize AHN=\n\n(2) For each Ci do\n\n(3) Initialize rjk randomly under the input domain\n\n(4) While stop condition is not reached do\n\n(5) Split (x,y)-pairs into p clusters yj using rjk\n\n(6) For each cluster yj do\n\n(7) Create a CH-molecule using criterion (6)\n\n(8) Obtain hydrogen values of molecule Mj using LSE method\n\n(9) Calculate least square error Ej between yj and Mj\n\n(10) End for\n\n(11) Update intermolecular distances rjk using (5)\n\n(12) End while\n\n(13) Update AHNAHNCi\n\n(14) Update (x,y)-pairs with (x,y-AHN)-pairs\n\n(15) End for\n\n(16) Obtain stoichiometric coefficients of Ci compounds using LSE method\n\n(17) Return AHN\n\nThis is a modified algorithm from the original one reported in . For instance, rjk refers to an intermolecular distance which defines the distance between the position of two molecules Mj and Mk. Actually, the algorithm iteratively updates the set of intermolecular distances to define the best positions of molecules in the input domain using (5). In that sense, molecules will act under regions defined by these intermolecular distances. It is remarkable to say that the first molecule acts from the initial value of the input domain. In order to iteratively updates intermolecular distances, η is considered the step size or the learning rate, such that, 0<η<1 and the least squares errors Ej and Ek for each molecule (5)rt+1jk=rtjk-η(Ej-Ek). On the other hand, the original algorithm considers a generic interaction of CH-molecules referring to as a nonpolar covalent bond based training . In this work, a linear chain of CH-molecules is adopted. Thus, each compound has a topology in the form of (6), where, the outside of the chain has MCH3 molecules and MCH2; otherwise (6)MCH3MCH2MCH2MCH2MCH3. Finally, Algorithm 1 considers adjustment parameters σj constant gain for molecular behaviors φj since φC in (1) is a normalized product form of a polynomial used in the least squares estimates (LSEs) method. In fact, consider the equivalence (7) when reasoning with AHNs. Where the set of p values are coefficients of the polynomial form of φC of grade deC(7)σφC(x)=pdxd+pd-1xd-1++p1x+p0=σi=1deC(x-φHi),x.\n\n2.2.3. Reasoning of Artificial Hydrocarbon Networks\n\nOnce the training is done, an artificial hydrocarbon network can be used for reasoning. In that sense, consider an input value x0. The AHN has to be evaluated in x0; thus, the reasoning value y0 can be calculated using (8), where S is the behavior of the artificial hydrocarbon network, αi are the stoichiometric coefficients, R is the set of all intermolecular distances between molecules, and σj are the adjustment parameters (8)y0=S(x0H,αi,R,σj). Notice that, if c=1, it means that there exists one stoichiometric coefficient α1=1.\n\n3. Description of the Proposed Fuzzy-Molecular Inference Model\n\nThe fuzzy-molecular inference model (FMI-model for short) is a fuzzy inference system that uses a fuzzy partition of input space in premises and artificial hydrocarbon networks in consequences as part of fuzzy implications. In this section, a detailed description of the fuzzy-molecular inference model is presented. For simplicity, through this section consider the FMI-model as a type-1 fuzzy system. In Section 5, an extension to type-2 fuzzy systems is presented.\n\nLet A be a fuzzy set and its corresponding membership function μA(x) of A, for all xX, where X is the input domain space. In fact, the membership function is a value between 0 and 1 for representing the value of belonging x to the fuzzy set A.\n\nAlso, let Ri be the ith fuzzy rule of form as (9), where {x1,,xk} is the set of variables in the antecedent, {A1,,Ak} is the set of the fuzzy partition of input space, yi is the variable of the consequent, Mj is the jth CH-molecule of the artificial hydrocarbon network excited by the fuzzy implication process (see Section 3.3), and Δ is any T-norm function (9)Ri:IfΔ(x1isA1,,xkisAk),thenyiisMj. If assuming that μΔ(x1,,xk) is the result of the T-norm function as (10) with conjunction operator , then (9) can be rewritten as (11), where φj is the molecular behavior of Mj(10)μΔ(x1,,xk)=μA1(x1)μAk(xk),(11)Ri:IfΔ(x1isA1,,xkisAk),thenyi=φj(μΔ(x1,,xk)). Thus, the fuzzy-molecular inference model is finally expressed in (11). Figure 2 shows the fuzzy-molecular inference model as a block diagram. This model represents a nonlinear inference system for a given crisp input xX that follows three steps, that is, fuzzification, fuzzy inference engine, and defuzzification, and obtains the corresponding crisp output yY, where Y represents the output. Moreover, fuzzy rules like (9) can also be expressed as a fuzzy matrix that defines a knowledge base of the problem domain. Each block in the FMI-model is detailed in the following subsections.\n\nBlock diagram of the fuzzy-molecular inference model.\n\n3.1. Fuzzification\n\nThe fuzzy-molecular inference model can be viewed as a block with inputs and outputs. Moreover, let any given system be a single-input-single-output. Then, fuzzification maps any given input variable x, also known as a linguistic variable, to a fuzzy value in the range [0,1]. In particular, this mapping occurs using a fuzzy set A and its corresponding membership function μA(x), such that (12) holds: (12)μA:x[0,1].\n\nIn fact, this linguistic variable is partitioned into m different fuzzy sets {Ai}, for all i=1,,m. For example, this fuzzy partition can be “low,” “medium,” “high.” Then, the evaluation of a given value of x is calculated using the set of membership functions μAi(x), for all i=1,,m.\n\nThe shape of all membership functions depends on the purpose of the problem domain. The literature reports different criteria and methods to do so as in [79, 11, 23, 24].\n\n3.2. Fuzzy Inference Engine\n\nOnce the crisp value of the input is mapped to a fuzzy subspace as described in Section 3.1, the next step in the fuzzy-molecular inference model is the evaluation of the antecedents in fuzzy rules like (11). In this work, the min function (13) is selected for the T-norm Δ(13)μΔ(x1,,xk)=min{μA1(x1),,μAk(xk)}. Finally, the consequent value yi is equal to the valued-behavior φj of the jth CH-molecule of an artificial hydrocarbon network. Thus, the consequent value yi can be calculated using fuzzy rules (11) with the min function (13), as shown in (14) (14)yi=φj(min{μA1(x1),,μAk(xk)}).\n\n3.3. Defuzzification\n\nThe last step in fuzzy-molecular inference model calculates the crisp value of the output y (15) using n fuzzy rules, where yi is the consequent value and μΔi(x1,,xk) is the fuzzy evaluation of the antecedents, for i=1,,n. In particular, (15) is based on the well-known center of gravity (15)y=μΔi(x1,,xk)·yiμΔi(x1,,xk). As noticed in Section 3.2, the fuzzy-molecular inference model requires a set of CH-molecules. In this case, let AHN be an artificial hydrocarbon network with one compound C that is made of p CH-primitive molecules with molecular behavior of the form as in (2). In this work, compound C is restricted to a linear chain of CH-molecules like in (16), where − stands for a covalent bond. Actually, the linear chain is made of 2 CH3 molecules at both extremes and (p-2) CH2 molecules in the inner chain (16)C=M1M2-MjMp-1Mp=CH3–CH2–CH2–CH2–CH3. It is remarkable to say that in the fuzzy-molecular inference model, the AHN is restricted to one univariate compound with one input μΔ(x1,,xk) defined as (13) and one output yi defined as (14). In case that a multiple-inputs-single-output (MISO) system has to be applied for a particular AHN, consider generalizing (1) as a multivariate function.\n\n3.4. Knowledge Base\n\nSince, the fuzzy-molecular inference model has a generic fuzzy inference engine, proper knowledge of a specific problem domain can be enclosed into the knowledge base (see Figure 2). For instance, this knowledge base is a matrix that summarizes all fuzzy rules of the form as in (11) in the following way.\n\nFor all input variables x1,,xk, represent all possible combinations of them using the label of each set in the fuzzy partition of inputs, such that all antecedents in the fuzzy rules will be covered.\n\nFor each combination (summary of antecedents), assign the corresponding label of molecule Mj that will act when the fuzzy rule is fired.\n\nAs an example of the knowledge base matrix construction, assume that there is a set of fuzzy rules like (17); thus, the knowledge base matrix for this particular system is shown in Table 1(17)R1:Ifx1isA1andx2isB2,theny1isM1,R2:Ifx1isA2andx2isB1,theny2isM1,R3:Ifx1isA1andx2isB1,theny3isM2.\n\nKnowledge base of (17).\n\nx 1 x 2 y i\nA 1 B 2 M 1\nA 2 B 1 M 1\nA 1 B 1 M 2\n3.5. Properties of the Fuzzy-Molecular Inference Model\n\nThe fuzzy-molecular inference model combines interesting properties from both fuzzy logic and artificial hydrocarbon networks. Advantages of the FMI-model are as the following.\n\nFuzzy partitions in the output domain might be seen as linguistic units, for example, “low,” “high.”\n\nFuzzy partitions have a degree of understanding (parameters are metadata).\n\nMolecular units deal with noise and uncertainties.\n\nIt is remarkable to say that molecules are excited by consequent values; thus, molecules do not model a given system, but transfer information from a fuzzy subspace to a crisp set. Moreover, molecular units have the property of filtering noise and uncertainties, especially important in real-world control applications, as described in Section 5.\n\nIn order to demonstrate the above advantages, an example of the application of the FIM-model in fuzzy control systems is provided in the following section. Then, Section 5 presents a case study that evaluates the performance of the FIM-model in a real application with dynamic uncertainties.\n\n4. Design of Fuzzy-Molecular Based Controller for a DC Motor\n\nIn this section, the design of a velocity controller for a DC motor using the fuzzy-molecular inference model is described. The objective of this fuzzy-molecular controller is to show an example of how to apply the FMI-model as a fuzzy control system.\n\n4.1. Definition of the DC Motor Model\n\nFor instance, consider a DC motor that regulates the velocity ω of its rotor varying the input voltage v. Let G(s) be the transfer function of a given DC motor expressed in (18) (18)G(s)=1.5s2+14s+40.02.\n\nIn order to simulate the performance of the DC motor, a discrete transfer function G(z) was obtained using (18) and a sample time of 0.01 s. The discrete model of DC motor is shown in (19) (19)G(z)=7.16z+6.83z2-1.86z+0.87×10-5.\n\nFinally, if one supposes that DC motor is a causal, linear-time invariant system, then a difference equation of (19) can be expressed as (20), where ω is the velocity of the rotor, u is the input voltage, and k is the current sample time (20)ω[k]=1.86ω[k-1]-0.87ω[k-2]+7.16×10-5u[k]+6.83×10-5u[k-1].\n\n4.2. Design of Control Law\n\nThe following control law is designed to achieve a step response of the DC motor model (20). Assuming the control diagram of Figure 3, the control law has two inputs—the error signal ε(t) and the first derivative of error signal ε˙(t)—and one output—the input voltage u(t). Thus, a fuzzy-molecular PD controller will be designed.\n\nBlock diagram of the PD control system implemented.\n\nUsing the fuzzy-molecular inference model described in Section 3, the control law is formed by three blocks: fuzzification, fuzzy inference engine, and defuzzification, as follows.\n\n4.2.1. Fuzzification\n\nThe two input variables are partitioned into five fuzzy sets: “very negative” (VN), “negative” (NN), “zero” (Z), “positive” (PP), and “very positive” (VP). Figure 4 shows the fuzzy sets of the input variable ε(t), and Figure 5 shows the fuzzy sets of the input variable ε˙(t). It is remarkable that parameters in the membership functions were tuned manually, and the input domain was previously normalized.\n\nFuzzy sets of the input error signal.\n\nFuzzy sets of the input first derivative error signal.\n\n4.2.2. Fuzzy Inference Engine\n\nThe fuzzy inference engine for the fuzzy-molecular PD controller uses fuzzy rules of the form as in (11) with consequent values as in (14). In particular to the application, the implemented knowledge base is summarized in Table 2.\n\nKnowledge base of the fuzzy-molecular PD controller for DC motor.\n\nε ε ˙\nVN NN Z PP VP\nVN MVN MVN MVN MVN MVN\nNN MVN MVN MVN MVN MNN\nZ MVN MNN MZ MPP MVP\nPP MPP MVP MVP MVP MVP\nVP MVP MVP MVP MVP MVP\n\nNotice that Table 2 reports, for each combination of input values, the fired molecule. For instance, the output signal was partitioned into five CH-molecules Mj, for all j=1,,5, that represent the action to be held. In particular, the output signal was partitioned into the following molecules: “very negative” (MVN), “negative” (MNN), “zero” (MZ), “positive” (MPP), and “very positive” (MVP).\n\n4.2.3. Defuzzification\n\nThe input voltage u(t), the input signal of the plant, is the output variable that defines the last block of the fuzzy-molecular PD controller. In order to calculate the consequent values of fuzzy rules depicted in Section 4.2.2, the five CH-molecules are proposed in Figure 6 and were found using Algorithm 1. Notice that the output variable is finally evaluated using (15), and resultant value is normalized. Finally, the adjustment parameters σj of CH-molecules are summarized in Table 3.\n\nAdjustment parameters of molecules in the fuzzy-molecular PD controller.\n\nCH -molecule σ j\nMVN + 2.41 E - 14\nMNN - 4.97 E - 16\nMZ 0.0\nMPP - 4.43 E - 16\nMVP - 8.85 E - 17\n\nArtificial hydrocarbon network used in the fuzzy-molecular PD controller.\n\n4.3. Results of the Velocity Control for a DC Motor\n\nThe fuzzy-molecular PD velocity controller for a DC motor described above was implemented and simulated. For instance, the objective of this application is to measure the performance of the system for a step response.\n\nThe system was subjected to a step function as shown in Figure 7. Results determine that the step response has 17% of maximum overshoot, a rise time of 0.19 s, a settling time of 0.45 s, and a maximum error of 0.002 in steady state. In order to measure the stability of the fuzzy-molecular PD controller, a phase diagram was obtained from the step response. Figure 8 shows the phase diagram of error signal versus derivative of error signal. As it can be seen in Figure 8, the fuzzy-molecular PD controller reaches a steady state near to the zero input state vector of the controller.\n\nStep response of the fuzzy-molecular PD controller for DC motor.\n\nPhase diagram derivative error signal versus error signal in the step response of the fuzzy-molecular PD controller for DC motor.\n\nThen, the step was implemented with a reference signal varying in the range from −2 to 2 ×1000 rpm. After 10 seconds with a sample time of 0.01 s, the step response is depicted in Figure 9, where the light line represents the reference signal and the strong line represents the actual value of angular velocity of the rotor in the DC motor. As shown in Figure 9, the fuzzy-molecular PD controller has an excellent performance.\n\nStep response of the fuzzy-molecular PD controller for DC motor.\n\nFrom the results obtained so far, it can be seen that the performance of the fuzzy-molecular PD controller has a very good quality (see Figure 8). The maximum overshoot, the settling time, and the maximum error in steady state correspond to the performance of a PD controller as reported in the literature of control theory .\n\nOn the other hand, the fuzzy-molecular PD controller was easily obtained. In this case, fuzzification was done via fuzzy sets tuned manually; however, there are other ways to find the optimal values of parameters in membership functions (see [79, 11, 23, 24]). In addition, defuzzification was implemented with an artificial hydrocarbon network that depends on hydrogen and adjustment parameters that can be easily found using Algorithm 1.\n\n5. Case Study: Fuzzy-Molecular Based Position Controller for a DC Motor\n\nIn this section, the design of a position controller for a DC motor using the fuzzy-molecular inference model is described. The objective of this case study is to improve type-2 fuzzy control systems using the fuzzy-molecular inference model.\n\n5.1. Description of the Hardware\n\nThe following case study was implemented on a trainer hardware module. It is prepared for sending reference signals (i.e., from a knob) and feedback signals (i.e., the current position of a DC motor) to a host in which a control law is running. The correction signal computed is sent back to the trainer module in order to feed a DC motor. In particular to this case study, a NI CompactRIO reconfigurable and embedded system based on field programmable gate arrays (FPGA) is used as the host. Figure 10 shows the overall system.\n\nOverall system of the case study showing the trainer hardware module, the NI CompactRIO host, and the LabVIEW client for monitoring the system.\n\nIn addition, LabVIEW software is used for programming the control law on the NI CompactRIO and for monitoring the performance of the fuzzy-molecular control system.\n\nOn one hand, both the reference signal r(t) that comes from a knob and the position signal y(t) are in the voltage range [0.0,5.0]V, where 0.0V represents an angle of 0° and 5.0V represents an angle of 180°. On the other hand, the correction signal u(t) is the input voltage of the DC motor in the range [0.0,5.0]V, where 0.0V represents the maximum angular velocity of the motor to rotate counterclockwise, 5.0V represents the maximum angular velocity of the motor to rotate clockwise, and 2.5V means no rotation.\n\nIt is remarkable to say that the position of the DC motor increases in counterclockwise direction and decreases in clockwise direction.\n\n5.2. Design of Control Law\n\nThe following control law is designed to achieve a reference tracking response of the DC motor in the trainer model. Assuming the control diagram of Figure 11, the control law has two inputs—the error signal ε(t) and the first derivative of the position signal y˙(t)—and one output—the input voltage u(t). Thus, a fuzzy-molecular PD controller will be designed.\n\nBlock diagram of the position control system implemented.\n\nUsing the fuzzy-molecular inference model described in Section 3, the control law is designed as follows.\n\n5.2.1. Fuzzification\n\nThe two input variables are partitioned into three type-2 fuzzy sets: “negative” (N), “zero” (Z), and “positive” (P). Figure 12 shows the fuzzy sets for input ε(t), and Figure 13 shows the fuzzy sets for input y˙(t). It is remarkable to say that parameters in the membership functions were tuned manually.\n\nFuzzy sets of the input error signal. Solid line: primary membership function. Dashed line: secondary membership function.\n\nFuzzy sets of the input first derivative position signal. Solid line: primary membership function. Dashed line: secondary membership function.\n\nAs shown in Figures 12 and 13, type-2 fuzzy sets are determined by primary membership function μAU(x) but also considers an additional value of uncertainty: the secondary membership function μAL(x). The region inside these two membership functions is known as the footprint of uncertainty FOU [10, 14] as expressed in (21) (21)FOU(A)=xX(μAL(x),μAU(x)).\n\nThen, two membership values (from primary and secondary functions) are computed for one input value. Moreover, if the secondary membership function coincides with the primary membership function, type-2 is reduced to an equivalent type-1 fuzzy system.\n\n5.2.2. Fuzzy Inference Engine\n\nThe fuzzy inference engine for the fuzzy-molecular position controller uses fuzzy rules of the form as in (11) of both primary and secondary membership values (μAL(x),μAU(x)). Consequent values yL and yU are similarly obtained as (15) for both primary and secondary membership values, respectively. The resultant knowledge base is summarized in Table 4.\n\nKnowledge base of the fuzzy-molecular position controller for the DC motor in the trainer module.\n\nε y ˙\nN Z P\nN MCW MCW MCW\nZ MCW MH MCCW\nP MCCW MCCW MCCW\n\nAs noted in Table 4, the output signal was partitioned into three CH-molecules Mj,for allj=1,,3, that represent the action to be held. In particular, the output signal was partitioned into the following molecules: “clockwise” (MCW), “halt” (MH), and “counterclockwise” (MCCW).\n\n5.2.3. Defuzzification\n\nIn order to calculate the consequent values of fuzzy rules depicted in Table 4, the three CH-molecules are proposed in Figure 14 and were found using Algorithm 1. The adjustment parameters σj of CH-molecules are summarized in Table 5.\n\nAdjustment parameters of molecules in Figure 6.\n\nCH -molecule σ j\nMCW - 1.0\nMH 0.0\nMCCW 1.0\n\nArtificial hydrocarbon network used in the fuzzy-molecular position controller.\n\nIn this case study, the Nie-Tan method is used for computing the final value of the output variable u(t) for a type-2 fuzzy system because of its simplicity of computation. Other methods like Karnik-Mendel, Greenfield-Chiclana, or Wu-Mendel might be used [10, 1214, 26]. The method generates a type reduction of the form as in (22), where y is the crisp output value u(t)(22)y=yL+yU2.\n\n6. Results and Discussion\n\nIn order to demonstrate that the fuzzy-molecular inference model for fuzzy control systems can be used as an alternative of type-2 fuzzy control systems, two experiments were done. The first experiment considers a type-1 fuzzy controller system and the second experiment considers a type-2 fuzzy controller system. In both cases, the FIM-model based fuzzy control system designed in Section 5 is compared with a Mamdani’s fuzzy controller system using the same parameters. The output variable was partitioned for the Mamdani’s fuzzy controller system into three type-2 fuzzy sets: “clockwise” (CW), “halt” (H), and “counterclockwise” (CCW). Figure 15 shows this partition for the output variable u(t).\n\nFuzzy sets of the output correction signal. Solid line: primary membership function. Dashed line: secondary membership function.\n\n6.1. Performance of the Type-1 Fuzzy-Molecular Controller\n\nFor this experiment, the fuzzy-molecular position controller for a DC motor described in Section 5 was reduced to a type-1 fuzzy system by only considering the primary membership functions in the fuzzification step, as well as in the Mamdani’s fuzzy controller.\n\nThe system was subjected to a step function without noise as shown in Figure 16. Results of the FMI controller determine that it had a step response of 0% of maximum overshoot, a rise time of 1.0 s, and a maximum error of 2.5° in steady state. On the other hand, the system was subjected to a step function with 35% of noise as shown in Figure 17. Results of the FMI controller reports a 0% of maximum overshooting, a rise time of 1.1 s, and a maximum error of 5.8° in steady state measured from position 180°. For contrasting, Table 6 summarizes the overall results of the FMI and Mamdani fuzzy controllers.\n\nExperimental results of type-1 fuzzy controllers.\n\nFuzzy controller Noise (%) Rise time (s) Steady-state error (°)\nStep Response\nFIM 0.0 1.0 2.5\nMamdani 0.0 2.0 4.7\nFIM 35.0 1.1 5.8\nMamdani 35.0 2.5 5.5\n\nRamp response\nFIM 0.0 3.6\nMamdani 0.0 6.7\nFIM 35.0 11.0\nMamdani 35.0 12.3\n\nStep response without noise of FMI and Mamdani type-1 fuzzy controllers.\n\nStep response with 35% noise of FMI and Mamdani type-1 fuzzy controllers.\n\nNotice in Figure 16 that the response of the FMI controller is 50% faster than the response of the Mamdani controller and has a less value of maximum error in steady state than then Mamdani controller. In comparison, Figure 17 shows that both fuzzy controllers are well stable as measured (5.8° and 5.5° of maximum error in steady state). However, FMI controller is still faster (1.1 s of rise time) than the response of the Mamdani controller (2.5 s of rise time). As noted, FMI controller has a better response for dynamic uncertainties than the Mamdani controller.\n\nAlso, the system was subjected to a ramp function without noise as shown in Figure 18. Results determine that the FIM controller has a maximum error of 3.6° in steady state while the Mamdani controller has 6.7°. On the other hand, the system was subjected to a ramp function with 35% of noise as shown in Figure 19. The FMI controller reports 11.0° of maximum error in steady state, and the Mamdani controller reports 12.3°. Also, Table 6 summarizes the overall results of this experiment with respect to the response of FMI and Mamdani fuzzy controllers.\n\nRamp response without noise of FMI and Mamdani type-1 fuzzy controllers.\n\nRamp response with 35% noise of FMI and Mamdani type-1 fuzzy controllers.\n\nIt is evident from Table 6 that both fuzzy controllers decrease their performance in presence of noise. However, the FIM controller can track the reference signal better than the Mamdani controller, as shown in the steady-state error. In addition, note that the FMI controller is slightly faster than the Mamdani controller.\n\n6.2. Performance of the Type-2 Fuzzy-Molecular Controller\n\nFor this experiment, the type-2 fuzzy-molecular position controller for a DC motor described in Section 5 was implemented as well as the type-2 Mamdani controller.\n\nAgain, the system was subjected to a step function with 35% noise and without it as shown in Figures 20 and 21, respectively. The same process was done with a ramp function, and the responses of both controllers are shown in Figures 22 and 23, respectively. The overall results are summarized in Table 7.\n\nExperimental results of type-2 fuzzy controllers.\n\nFuzzy controller Noise (%) Rise time (s) Steady-state error (°)\nStep Response\nFIM 0.0 1.0 2.5\nMamdani 0.0 2.4 4.7\nFIM 35.0 1.0 5.0\nMamdani 35.0 2.6 5.5\n\nRamp response\nFIM 0.0 3.8\nMamdani 0.0 3.7\nFIM 35.0 9.1\nMamdani 35.0 12.1\n\nStep response without noise of FMI and Mamdani type-2 fuzzy controllers.\n\nStep response with 35% noise of FMI and Mamdani type-2 fuzzy controllers.\n\nRamp response without noise of FMI and Mamdani type-2 fuzzy controllers.\n\nRamp response with 35% noise of FMI and Mamdani type-2 fuzzy controllers.\n\nAs noted from Tables 6 and 7, the step responses of both FIM and Mamdani type-2 fuzzy controllers remain similar to type-1 controllers, as expected. Thus, type-1 and type-2 FIM fuzzy controllers are slightly equivalent with or without perturbations.\n\nFrom Figures 22 and 23, it can be seen that the response of type-2 fuzzy controllers slightly better than type-1 controllers, as expected [10, 1214, 26]. From the point of view of ramp response, the FIM controller presents similar performance to the Mamdani controller without noise (3.8° and 3.7° maximum steady-state errors, resp.). Again, both controllers present the same tendency when they are exposed to noise, and in comparison with type-1 controllers, type-2 fuzzy controllers act slightly better as found in Tables 6 and 7 (FIM: 17.2% better, Mamdani: 1.7% better).\n\n6.3. Discussion of FIM-Models\n\nOn one hand, from the above results, fuzzy-molecular inference models can achieve fuzzy control applications. Moreover, these FIM-model based controllers can be used as an alternative of type-2 fuzzy control systems. This statement comes from the evaluation and comparison of step and ramp responses between the FIM-controller designed in Section 5 and the Mamdani fuzzy controller; both models subjected to static and dynamic uncertainties. In this case study, a Mamdani’s fuzzy control system was used because it is the fuzzy inference system most implemented in industry as reported in the literature [10, 14].\n\nOn the other hand, it is important to distinguish the fuzzy-molecular inference model from other fuzzy inference models like Takagi-Sugeno inference systems or Mamdani’s fuzzy control systems [79, 11]. Thus, Table 8 presents a comparative chart of the FMI-model, Takagi-Sugeno’s model, and Mamdani’s model.\n\nComparative chart of different fuzzy inference models.\n\nFMI-model Takagi-Sugeno Mamdani\nDefuzzification AHNs Polynomial functions Membership functions\n\nDefinition of parameters Hydrogen valuesAdjustment values Coefficients Parameters of membership functions\n\nMappings in defuzzification Membership values to output values Input values to output values Membership values to output values\n\nFrom Table 8, defuzzification process in each fuzzy inference model is different. As FMI-model uses artificial hydrocarbon networks, each molecule represents a linguistic partition of the output variable. In the above results, simple CH-molecules were implemented, but either complex molecules can be used. Thus, defuzzification can have complex nonlinear mappings in the FMI-model. In contrast, Takagi-Sugeno’s model uses polynomial functions, and Mamdani’s model represents linguistic partitions with membership functions associated with fuzzy sets. Parameters inside artificial hydrocarbon networks are hydrogen and adjustment values, polynomial coefficients for Takagi-Sugeno’s model, and parameters of membership functions in Mamdani’s model.\n\nIn addition, molecules in FMI-model make a mapping from membership or truth-values to output values also dealing with uncertainties. This is remarkable because Takagi-Sugeno’s model maps from input values to output values, and using fuzzy inference values linearly acts on the final output value. At last, Mamdani’s model makes a mapping from membership values to output values. In fact, the fuzzy-molecular inference model combines linguistic partitions of output variables with molecular structures.\n\n7. Conclusions\n\nIn this paper, a new fuzzy algorithm based on artificial hydrocarbon networks called fuzzy-molecular inference model (FMI-model) was proposed, taking advantage of the power of molecular units of information. In this approach, molecules of artificial hydrocarbon networks are implemented as fuzzy partitions in the output domain. Since the FMI-model is based on AHNs, properties of molecules are inherited. Two characteristics of the proposed fuzzy-molecular inference model are both interpretation of linguistic molecules and partial understanding of fuzzy partitions via metaparameters in AHNs.\n\nIn that way, the novel fuzzy algorithm treats molecules as fuzzy partitions of the output variable, transferring information from a fuzzy subspace to a crisp set, allowing to set the number of fuzzy partitions linguistically, but also these molecules are characterized by hydrogen values that can be referred to as meta-data information, giving the opportunity to partially understand the molecular behavior. Moreover, the proposed fuzzy-molecular inference model has some advantages in comparison with other fuzzy inference systems. For instance, FMI-model occupies parameters with meta-data information in comparison with Mamdani’s inference system in which parameters associated with membership functions do not reveal important information of the fuzzy partition. If parameters are meta-data information, it is easier to tune fuzzy partitions because both experts and real data information coming from the system can be combined into a single unit, no matter how complex the mapping is. In addition, since FMI-model does not model a given system like Takagi-Sugeno’s inference system, it preserves a more natural way of defuzzification from a fuzzy subspace to a crisp set. Finally, since molecules in artificial hydrocarbon networks can filter information [18, 19], the fuzzy-molecular inference model also shares this property allowing to deal with uncertain data.\n\nThus, the proposed fuzzy-molecular inference model has three steps: fuzzification, fuzzy inference engine, and defuzzification. Specially, molecules are mappings from implication values to output variables. In addition, in this work, a linear chain of CH-primitive molecules was used, but the FMI-model allows complex molecules associated with each fuzzy rule handling complex nonlinear mappings from fuzzy subspaces to crisp sets.\n\nOn the other hand, the proposed model was applied to control the angular velocity of a simulated DC motor in which the results confirm that the FMI-model can be used in control applications. Furthermore, a case study was presented in which the FMI-model was used for controlling the position of a real DC motor. Experimental results demonstrate that fuzzy-molecular based control systems can deal with uncertainties as type-2 fuzzy control systems do. Then, it suggests that FMI-based controllers can be used as an alternative of type-2 fuzzy control systems. In practical applications where hardware restricts the operational computations or memory storage, FMI-based controllers can be implemented because of its simplicity.\n\nFuture research considers the design of training procedures for optimality in molecules at the defuzzification stage of FMI-models. In addition, since artificial hydrocarbon networks are considered under the class of learning algorithms, the usage of molecular units in FMI-models might be applied for online adaptation (learning and evolution) of the overall fuzzy control system to improve its performance.\n\nAcknowledgments\n\nThis work was supported by a scholarship award from Tecnológico de Monterrey, Campus Ciudad de México and a scholarship for living expenses from CONACYT."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86244637,"math_prob":0.9344809,"size":49757,"snap":"2019-43-2019-47","text_gpt3_token_len":11848,"char_repetition_ratio":0.18344623,"word_repetition_ratio":0.11570248,"special_character_ratio":0.22416946,"punctuation_ratio":0.12304562,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98778975,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T17:43:07Z\",\"WARC-Record-ID\":\"<urn:uuid:63f0ffa2-6af8-48b7-b7e5-4ac027c1b336>\",\"Content-Length\":\"205652\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7551b748-241d-4eeb-9e04-3a7f95b459a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7dc8649-a3be-4e7b-a2ee-c9f97ecae602>\",\"WARC-IP-Address\":\"52.216.138.67\",\"WARC-Target-URI\":\"http://downloads.hindawi.com/journals/mpe/2013/531031.xml\",\"WARC-Payload-Digest\":\"sha1:446UIU46KKIEBL7JD2H3QCFHYUPIQSHG\",\"WARC-Block-Digest\":\"sha1:UIWII3MPR2GR4VPWY4H4DJFXMLLPLAIS\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669225.56_warc_CC-MAIN-20191117165616-20191117193616-00151.warc.gz\"}"} |
https://convertoctopus.com/611-grams-to-kilograms | [
"## Conversion formula\n\nThe conversion factor from grams to kilograms is 0.001, which means that 1 gram is equal to 0.001 kilograms:\n\n1 g = 0.001 kg\n\nTo convert 611 grams into kilograms we have to multiply 611 by the conversion factor in order to get the mass amount from grams to kilograms. We can also form a simple proportion to calculate the result:\n\n1 g → 0.001 kg\n\n611 g → M(kg)\n\nSolve the above proportion to obtain the mass M in kilograms:\n\nM(kg) = 611 g × 0.001 kg\n\nM(kg) = 0.611 kg\n\nThe final result is:\n\n611 g → 0.611 kg\n\nWe conclude that 611 grams is equivalent to 0.611 kilograms:\n\n611 grams = 0.611 kilograms\n\n## Alternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 1.6366612111293 × 611 grams.\n\nAnother way is saying that 611 grams is equal to 1 ÷ 1.6366612111293 kilograms.\n\n## Approximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that six hundred eleven grams is approximately zero point six one one kilograms:\n\n611 g ≅ 0.611 kg\n\nAn alternative is also that one kilogram is approximately one point six three seven times six hundred eleven grams.\n\n## Conversion table\n\n### grams to kilograms chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from grams to kilograms\n\ngrams (g) kilograms (kg)\n612 grams 0.612 kilograms\n613 grams 0.613 kilograms\n614 grams 0.614 kilograms\n615 grams 0.615 kilograms\n616 grams 0.616 kilograms\n617 grams 0.617 kilograms\n618 grams 0.618 kilograms\n619 grams 0.619 kilograms\n620 grams 0.62 kilograms\n621 grams 0.621 kilograms"
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http://hiilipuu.fi/articles/how-model-photosynthesis | [
"# How to model photosynthesis?\n\n Many years of measurements have provided a lot of information on the behaviour of the forest, how it photosynthesizes, transpires and respires. We know more or less what to expect the tree will do according to the time of the year, the availability of light, the air and soil temperature and the water availability in the soil. We can express this knowledge as mathematical models that can forecast the overall behaviour of the tree. Mathematical modelling is a typical tool to describe the phenomena in an exact and concise format. However, always bear in mind that the model reflects only such variables in reality which exist in the model. The animated photosynthesis is the result of one such model. According to real weather and soil conditions measured on-line at Hyytiälä, we use a fairly straightforward and simplified mathematical model to calculate the photosynthesis rate. The model includes the most common environmental responses, while it omits the change in the internal state of the plant, for the most part. This means that the responses presented here, are not necessarily the same throughout the year but changing a little over time. To calculate photosynthesis, we use a multiplicative expression of the functions for the main controlling factors. This is expressed mathematically as: P = f(LAI) • f(PAR) • Min{f(VPD),f(REW)} • f(T) • f(Tmin) • f(S) • f(CO2). P stands for photosynthesis, or rather, the flux of carbon created by photosynthesis and f() stands for function of and describes the individual effect of every factor on the photosynthesis. In this expression, the maximal photosynthesis at each moment is determined by the leaf area (LAI), light intensity (PAR) and carbon dioxide (CO2). Naturally, the environment is not always optimal, and other factors such as air humidity (VPD), days of minimum temperature (Tmin), temperature (T), the temperature history (S) and soil moisture (REW) adjust and limit photosynthesis. Thus, these factors are expressed in the model as modifiers that can reduce to some degree the theoretical attainable maximum. This is, roughly, how every factor affects photosynthesis in the model:",
null,
"",
null,
"LAI stands for \"Leaf Area Index\" and refers to the amount of leaves in the crown of the tree. In principle, more leaves produce more photosynthesis. However, not all leaves are the same because leaves shade each other and also resources are allocated differently to leaves in different places in the crown. In consequence, bigger LAI affects less strongly to the total crown photosynthesis. The relationship between LAI and photosynthesis can be expressed as a mathematical equation: f(LAI)= 1 / K * (1 –e–K * LAI). LAI = 8 m2/m2and K = 0.18 in our 50-year-old measuring site.",
null,
"PAR stands for \"Photosynthetically Active Radiation\" and is a measure of light intensity. The effect of light on photosynthesis has a clear saturating pattern. At low light intensity, more light results in more photosynthesis. As the light intensity rises, the carbon fixing systems are already working at full capacity and thus the leaves can not take full advantage of all the extra light they receive. The relationship between PAR and photosynthesis can be expressed as a mathematical equation: f(PAR) = Pmax * PAR / (PAR + B). Pmax= 9 µmol m–2 s–1 and B = 600 µmol m–2 s–1 at our measuring site in Hyytiälä.",
null,
"VPD stands for \"Vapour Pressure Deficit\" and refers to air humidity so that the higher the deficit, the drier the air. Transpiration (the flow or water from the plant to the air) increases as air becomes drier. If there would be no control on this water loss, the plant would get stressed and could even desiccate. To avoid losing too much water, the pores in the leaves (stomata) through which the water escapes close accordingly to adjust the transpiration to the level the plant can stand at every moment. Since stomata are the same pores through which the CO2 gets into the leaves, the result of this closure is a decrease also to the flow of CO2 and ultimately a reduction in photosynthesis. The limiting effect of VPD on photosynthesis can be expressed as a mathematical equation: f(VPD)= e–H*VPD. H = 0.02 at our measuring site in Hyytiälä.",
null,
"REW stands for \"Relative Extractable Water\" and it refers to the amount of soil water available to the plant. When the soil water decreases it gets increasingly difficult for the roots to provide water to the plant and compensate for the simultaneous loss of water from the leaves. In order to decrease this demand, a similar effect as that of high VPD is produced: stomatal pores close and CO2 flow and photosynthesis decrease. The limiting effect of REW on photosynthesis can be expressed as a mathematical equation: if REW≥REWCRIT , f(REW) = 1, if REW",
null,
"The effect of the air temperature on photosynthesis is expressed as a combination of three limiting effects. It takes into account the current temperature (T), minimum temperature and temperature history, S. If temperature during the past 24 h (Tmin)is below 0°C, photosynthesis decreases and if temperature has been below -10°C, photosynthesis shuts down. Mathematically, it can be expressed as follows: f(Tmin) = Max { 0 , Min { 1, (T0min – Tmin ) / T0min } }. T0min = –10 °C at our measuring site in Hyytiälä. The readiness of the plant to carry out photosynthesis is different at different times of the year (maximum in summer, minimum in winter etc) and this process is regulated largely by the air temperature. S stands for \"Stage of acclimation\" and refers to this seasonality of the photosynthetic activity. The stage of acclimation at any given day depends on the temperatures experienced by the plant during the previous days. It reflects the ability of the plant to follow the advancement of the seasons and synchronize the biological activity accordingly, usually with a strategy of minimizing the risks (e.g. frost damage). S at time t follows the air temperature at time t with a time constant, τ, that is usually 150 h. It can be calculated as a time derivative: dS/dt = ( T(t) – S(t) ) / τ . The effect of S on photosynthesis can be expressed mathematically as follows: f(S) = 1 / (1 + ec*(S–TS) ). c = –0.25 and TS = 5.5°C at our measuring site in Hyytiälä. The effect of current air temperature (T) og photosynthesis can be expressed mathematically as follows: f(T)=1–ecT*(T–T0). T0 = –5 °C and cT = Min {-0.1, 0.5 * ( f(S)–1) } at our measuring site in Hyytiälä.",
null,
"In general, the more carbon dioxide that is available to the plant, the faster the rate of photosynthesis. The relationship between CO2 concentration and photosynthesis can be expressed by a mathematical equation: f(CO2)=(CO2 – γ) / (CO2 + γ + KCO2) * (CO2ref + γ + KCO2 ) / (CO2ref – γ). CO2ref=400 ppm, γ=50 ppm and KCO2=500 ppm at our measuring site in Hyytiälä.",
null,
"For a description on a similar model you can consult:\n\nMäkelä, A., Pulkkinen, M., Kolari, P., Lagergren, F., Berbigier, P., Lindroth, A., Loustau, D., Nikinmaa, E., Vesala, T., and Hari, P. 2008. Developing an empirical model of stand GPP with the LUE approach: analysis of eddy covariance data at five contrasting conifer sites in Europe. Global Change Biology, 14, 92-108."
] | [
null,
"http://hiilipuu.fi/fi/system/files/users/Marianne/yhteyt_mittaus.jpg",
null,
"http://hiilipuu.fi/fi/system/files/users/Marianne/yhteyt_mittaus_talvi.JPG",
null,
"http://hiilipuu.fi/fi/system/files/users/Liisa%20Kulmala/LAI_en.png",
null,
"http://hiilipuu.fi/fi/system/files/users/Liisa%20Kulmala/PAR_en.png",
null,
"http://hiilipuu.fi/fi/system/files/users/Liisa%20Kulmala/VPD_en.png",
null,
"http://hiilipuu.fi/fi/system/files/users/Liisa%20Kulmala/REW_en.png",
null,
"http://hiilipuu.fi/system/files/users/Liisa%20Kulmala/T_em.png",
null,
"http://hiilipuu.fi/system/files/users/Liisa%20Kulmala/CO2_en.PNG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91203153,"math_prob":0.96478015,"size":7222,"snap":"2019-51-2020-05","text_gpt3_token_len":1734,"char_repetition_ratio":0.15156554,"word_repetition_ratio":0.041837573,"special_character_ratio":0.23220715,"punctuation_ratio":0.1182016,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.976105,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T03:57:59Z\",\"WARC-Record-ID\":\"<urn:uuid:fe8b6ea1-e79c-4070-b502-767216aaafc8>\",\"Content-Length\":\"39294\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41d19a76-9d87-4285-8171-801010436ed3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f31652c1-7ce7-440b-a61c-7b0690cec6a4>\",\"WARC-IP-Address\":\"31.170.122.65\",\"WARC-Target-URI\":\"http://hiilipuu.fi/articles/how-model-photosynthesis\",\"WARC-Payload-Digest\":\"sha1:D3FQLPFYYIQB2SFSQ5AAGFEHFSVV4LGK\",\"WARC-Block-Digest\":\"sha1:ECRHZ2WA3P2VDCMHJKTD2EERXJRJZZOV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540504338.31_warc_CC-MAIN-20191208021121-20191208045121-00475.warc.gz\"}"} |
https://help.altair.com/pollex/topics/pollex/solver/thermal_global_parameters_r.htm | [
"# Global Parameters\n\n## Analysis Parameters\n\nGravity direction: Using the direction of gravity, you can set the board placement status. Define the actual board’s using status. Many of thermal diffusions are flow through board. But also, the convection on board or on component is not ignored. At this calculation, the direction of board has direct relation with air flows.\n\nAmbient temperature(C): Thermal diffusion with convection (at natural convection) could be get with the following formula:\n\nQ= ha(Tw -Tf)\n\nTw: High temperature.\n\nTf: Lower temperature.\n\nIt means that board level convection Tf value of ΔT(Tw-Tf) (temperature difference) may be ambient temperature.\n\nIn general case, thermal diffusion on board can be classified into three types of method, conduction, convection, and radiation. Among these methods, radiation is most effective thermal diffusion ways. Many of cases, heat transfer is not done just by one type of methods. It the mediator is fluid, heat transfer is done by directly moving molecules. Currently, we call vertical heat transfer with convection and horizontal heat transfer, advection. Heat transfer in component body, we call them conduction which heat transfer by vibration of solid molecule.\n\nDefault comp analysis power level(%): PollEx Thermal use the power values of UPF file. Assuming UPF files’ power level as maximum power value, specified default ratio of maximum power will be used during analysis for component applied power. (default is 80%)\n\nConvection boundary condition type: If there are forced convection conditions, users can define the forced convection type. Normal convection condition is natural convection.\n\nFlow direction of forced convection: If the convection type is Forced convection case, this value is valid. Set the forced flow direction among –X, +X, -Y and +Y.\n\nInlet flow velocity of force convection (m/s): If the convection type is Forced convection case, this value is valid. Set the flow strength with the speed value.\n\nComp/board glue material: Define the material to attach components on board. To use material for this value, users need register material into library. PollEx Thermal use the heat transfer coefficient value of material.\n\nThermal pad material: Define the material of thermal pad. To use material for this value, users need register material into library. PollEx Thermal use the heat transfer coefficient value of material.\n\nJoule heating/IR-drop model: Choose and apply the joule heating or IR-drop model for PollEx Thermal analysis. You can use the external model.\n\n## Convection Flow Material Properties\n\nIf the board’s top or bottom is filled with different materials, users can define this situation and set below thermal properties of material. Default is Air. So, if there is no definition for this column, PollEx Thermal assumes that PCB outside material is Air.\n• Flow material name\n• Thermal Conductivity (W/m.K) [default: -1]\n• Mass density (Kg/m^3) [default: -1]\n• Specific heat (J/Kg.K) [default: -1]\n• Viscosity (N.s/m^2) [default: -1]"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8311391,"math_prob":0.95242983,"size":2754,"snap":"2023-40-2023-50","text_gpt3_token_len":597,"char_repetition_ratio":0.164,"word_repetition_ratio":0.13461539,"special_character_ratio":0.20334059,"punctuation_ratio":0.1388889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9620775,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T00:03:43Z\",\"WARC-Record-ID\":\"<urn:uuid:827dfc4c-ce24-4c08-a048-a404ba709144>\",\"Content-Length\":\"42583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed0045ff-b1b3-4ace-82ca-03e1b96a785d>\",\"WARC-Concurrent-To\":\"<urn:uuid:76bbb80d-31be-4d17-b862-12e634247288>\",\"WARC-IP-Address\":\"173.225.177.121\",\"WARC-Target-URI\":\"https://help.altair.com/pollex/topics/pollex/solver/thermal_global_parameters_r.htm\",\"WARC-Payload-Digest\":\"sha1:LC7GCVPKDXFWRJ2CBDLHYIIVX3LMC6EA\",\"WARC-Block-Digest\":\"sha1:BFOAXLVKAY5O4SJ6NZESJY2WP4TDOU7M\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506429.78_warc_CC-MAIN-20230922234442-20230923024442-00121.warc.gz\"}"} |
https://embdev.net/topic/167227 | [
"# Forum: Analog Circuits pure Despair: strange OpAmp behaviour.\n\nRate this post\n 0 ▲ useful ▼ not useful\nOk, here we go:\n\nI am building a SoundCard DAC for controlling my Laser.\nThe SC outputs from 0.7 to 3.1V whereby 2.55V is the \"middle\".\nAn Amplifier should convert this to a differential signal with 10V p-p.\n\ncurrently not soldered anything to the SoundCard; im using a USB plug\nfor power and for testing the opamp's stage behavior, im was using a\n1MOhm potentiometer to adjust any signal voltage between +5 and -5V.\nnow, after i thought the high resistance of it could be the error cause,\ni replaced it with 3x 10k R's at which i can get 4 different voltages\nfor testing.\n\nI attached my circuit plan, its exactly how i builded it. between\n+5V/GND and -5V/GND are a 100µF elko each for stabilising, otherwise\ncurrently no capacitors are used, except the ones for the ICL7660\nnegative voltage converter.\n\nSo i did a row of different measurements on different setups. Results\nare attached to this Post, too.\n\n\n 1 1 | 2 | 3 | 4 2 3 -5V : -4.82 -4.80 +4.84 +4.95 4 A : -1.55 -1.55 +0.94 +1.88 5 B : +1.71 +1.71 -1.03 -2.06 6 +5V : +4.99 +4.99 -4.74 -4.76 7 8 9 [ADD] 10 11 -vref + -0.94 = 3.49 12 -vref + +1.03 = 1.52 13 14 15 [SUBST] 16 17 -4.81 + -(0) = -2.41 18 -4.81 + -(GND) = -4.76 19 -4.81 + -(-vref) = -2.27 20 -1.55 + -(0) = -0.58 21 -1.55 + -(GND) = -1.16 22 -1.55 + -(-vref) = +1.39 23 +1.71 + -(0) = +0.64 24 +1.71 + -(GND) = +1.29 25 +1.71 + -(-vref) = +3.84 26 +4.99 + -(0) = +2.50 27 +4.99 + -(GND) = +4.96 28 +4.99 + -(-vref) = +4.95 29 30 +vref + -(0) = +1.28 31 +vref + -(GND) = +2.56 32 +vref + -(-vref) = +4.95 33 34 0 + -(+vref) = -2.55 35 GND + -(+vref) = -2.54 36 -vref + -(+vref) = -4.76 37 38 0 + -(0) = 0.00 39 GND + -(0) = 0.00 40 -vref + (-0) = -1.28 \n\n\nIn the table, the 1st collumn is just measured with multimeter on the\nVin 3*10k divider, withoud any load. second one is the divider voltages\nfeed into a voltage-follower opamp circuit, 3rd one is fed into\ninverting-amplifier with 1x gain and 4rd finaly with 2x gain.\n\nSo this is strange. the \"real\" voltages on the divider are coming out\nunaltered with a impendance amplifier / follower; but are changing\nextreme when i use something with gain.\nAlso in the last test setup, the substractor, when i feed in 2.55V i got\nthe half out, without any other voltage is going in which could alter it\n(0 in the table means unconnected cable)\nOn the other side, whenever i do operations with buffer stabilised\nnegative n/o directly from LM336 reference circuit Vref ±2.55V Voltage;\nthe output never differs from ideal theory values!\n\nWhat the...?!\n\nI never get \"zero-level\" signal output on 2.55V \"offset-level\" input. SC\nis a CM106 so the Vref is set right. OPA's are rail2rail and able to\ndeliver Vout = Vsupply - 0.01V !\nHere's a bigger scan:\nhttp://www.abload.de/image.php?img=image0022ff8c.jpg\n\nRate this post\n 0 ▲ useful ▼ not useful\nWhats the problem ?\nYou wonder, why a voltage divider built up from 10k resistors\nis not stable if a load of 10k is attached ?\n\nUse voltage followers for Va und Vb, and 3 and 4 will work.\n\nRate this post\n 0 ▲ useful ▼ not useful\nI always got told the resistance of an opamp is very high an practically\nno current i flowing from the inputs?\n\nI tried it with a 1MOhm divider bevore. why it didn's work there?\nAnd will it do when i use SoundCard Oputput's (after Chip) directly?\n\nRate this post\n 0 ▲ useful ▼ not useful\nCorrect, the INPUT of the amplifier has a virtually infinite resistance.\nBut the additional resistors that you connect from the input to the\nOUTPUT of the amplifier draw current from your voltage divider, thus\nchanging the voltage.\n\nRate this post\n 0 ▲ useful ▼ not useful\n@Xero,\n\nyou entirely forget, that the inverting OPamp circuit is creating a\n\"virtual ground\" at the inverting input. So, connecting an inverting\nOPamp with a 10k input impedance to your \"3 x 10k\" voltage divider means\nto connect an additional load of 10k to signal ground, which erodes the\npotentials of your voltage divider, of course.\n\nKai Klaas\n\nRate this post\n 0 ▲ useful ▼ not useful\nSo it can even change the potential when i use a 1Mohm Poti for\ndividing?\n\nbecause it even then got wrong...\n\nRate this post\n 0 ▲ useful ▼ not useful\n>So it can even change the potential when i use a 1Mohm Poti for\n>dividing?\n\nYes, even more, because then the 10k input impedance of inverting OPamp\nis even more loading the voltage divider. The voltage divider must have\nan impedance that is way smaller then the input impedance of inverting\nOPamp. So, connecting an inverting OPamp to a voltage divider is not a\ngood idea in your example. But a voltage follower with its nearly\ninfinite input impedance could be used.\n\nShow us in a bit more detail, what exactly you want to implement. An\ninverting OPamp could be used, eventually, for your purpose, but not in\nthe way you intend right now. Is it a pot you want to buffer?\n\nKai Klaas\n\nRate this post\n 0 ▲ useful ▼ not useful\nIs your soundcard able to put out DC? Is there no highpass?\n\nRate this post\n 0 ▲ useful ▼ not useful\nOk... you all was right :<\n\nHave used an follower as buffer and done the Tests now. its working!\n\nWhat i want to build? The Soundcard-DAC for Lasercontrolling!\n\nI didn't drawed my full circuit to scan yet, but can do it.\nMerely it converts single ended signal from soundcard, which swings from\n0.7V to 3.8V with 2.55V as middle. thus need to be converted to\ndifferential output with 10V p-p, also +5V to -5V\n\nRate this post\n 0 ▲ useful ▼ not useful\n@ sssss\n\nWe desolder the output coupler wires for that\n\n• $formula (LaTeX syntax)$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77461284,"math_prob":0.9433528,"size":2753,"snap":"2021-04-2021-17","text_gpt3_token_len":1037,"char_repetition_ratio":0.11495089,"word_repetition_ratio":0.0,"special_character_ratio":0.42172176,"punctuation_ratio":0.16944024,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9833544,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-25T11:38:44Z\",\"WARC-Record-ID\":\"<urn:uuid:43ff23e5-7029-4a5c-be62-f2c004d9a1ee>\",\"Content-Length\":\"67320\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63afc1d4-db10-484d-a268-128a3f0c56a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:7dcdc4a9-44ff-41d8-b403-0127de601b7b>\",\"WARC-IP-Address\":\"104.26.0.113\",\"WARC-Target-URI\":\"https://embdev.net/topic/167227\",\"WARC-Payload-Digest\":\"sha1:WJREWLKWTOXSEYQACQHCQFNEAEDRSJQS\",\"WARC-Block-Digest\":\"sha1:75F3ZWPEKMBNH57R3BCNG6P2IK6CVOTI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703565541.79_warc_CC-MAIN-20210125092143-20210125122143-00602.warc.gz\"}"} |
https://www.geeksforgeeks.org/python-list-frequency-of-elements/ | [
"# Python | List frequency of elements\n\n• Difficulty Level : Easy\n• Last Updated : 21 Jan, 2021\n\nSometimes we have the utility in which we require to find the frequency of elements in the list and the solution to this problem has been discussed many times. But sometimes we come across the task in which we require to find the number of lists that particular elements occur. Let’s discuss certain shorthands in which this can be done.\n\nMethod #1 : Using Counter() + set() + list comprehension\nThe combination of the above functions can be used to perform the task. The Counter function does the grouping, set function extracts the distinct elements as keys of dict and list comprehension check for its list occurrences.\n\nAttention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.\n\nTo begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning - Basic Level Course\n\n## Python3\n\n `# Python3 code to demonstrate``# list frequency of elements``# using Counter() + set() + list comprehension``from` `collections ``import` `Counter` `# initializing list``test_list ``=` `[[``3``, ``5``, ``4``],`` ``[``6``, ``2``, ``4``],`` ``[``1``, ``3``, ``6``]]` `# printing original list``print``(``\"The original list : \"` `+` `str``(test_list))` `# using Counter() + set() + list comprehension``# list frequency of elements``res ``=` `dict``(Counter(i ``for` `sub ``in` `test_list ``for` `i ``in` `set``(sub)))` `# printing result``print``(``\"The list frequency of elements is : \"` `+` `str``(res))`\nOutput :\n\n```The original list : [[3, 5, 4], [6, 2, 4], [1, 3, 6]]\nThe list frequency of elements is : {1: 1, 2: 1, 3: 2, 4: 2, 5: 1, 6: 2}```\n\nMethod #2 : Using Counter() + itertools.chain.from_iterable() + map() + set()\nThe above 4 functionalities can also be combined to achieve this particular task. The set function extracts the dictionary keys formed by the Counter, map function performs the task for all sublists and from_iterable function performs using iterators which is faster than list comprehension.\n\n## Python3\n\n `# Python3 code to demonstrate``# list frequency of elements``# using Counter() + itertools.chain.from_iterable() + map() + set()``from` `collections ``import` `Counter``from` `itertools ``import` `chain` `# initializing list``test_list ``=` `[[``3``, ``5``, ``4``],`` ``[``6``, ``2``, ``4``],`` ``[``1``, ``3``, ``6``]]` `# printing original list``print``(``\"The original list : \"` `+` `str``(test_list))` `# using Counter() + itertools.chain.from_iterable() + map() + set()``# list frequency of elements``res ``=` `dict``(Counter(chain.from_iterable(``map``(``set``, test_list))))` `# printing result``print``(``\"The list frequency of elements is : \"` `+` `str``(res))`\nOutput :\n```The original list : [[3, 5, 4], [6, 2, 4], [1, 3, 6]]\nThe list frequency of elements is : {1: 1, 2: 1, 3: 2, 4: 2, 5: 1, 6: 2}```\n\nMethod #3: Using python dictionary + get() method\n\nPython dictionary provides a get method which returns the value corresponding to the key and if the key does not exist in the dictionary then it provides functionality to create the key and assign it a default value. We will use this functionality of a dictionary.\n\n## Python3\n\n `d ``=` `{}` `test_list ``=` `[[``3``, ``5``, ``4``],`` ``[``6``, ``2``, ``4``],`` ``[``1``, ``3``, ``6``]]` `for` `x ``in` `test_list:`` ``for` `i ``in` `x:`` ``d[i] ``=` `d.get(i,``0``) ``+` `1` `# Orignal list``print``(f``\"The original list : {test_list}\"` `)` `# printing result``print``(f``\"The list frequency of elements is : {d}\"` `)`\n\nOutput:\n\n```The original list : [[3, 5, 4], [6, 2, 4], [1, 3, 6]]\nThe list frequency of elements is : {3: 2, 5: 1, 4: 2, 6: 2, 2: 1, 1: 1}```\n\nMethod #4: Using Pandas\n\nIn this method we will use a python module named pandas(You can know more about pandas in this article) to find the frequency of the given data, here below is the code for it.\n\n## Python3\n\n `import` `pandas as pd` `test_list ``=` `[``3``,``5``,``4``,``3``,``3``,``4``,``5``,``2``]` `df1 ``=` `pd.Series(test_list).value_counts().sort_index().reset_index().reset_index(drop``=``True``)``df1.columns ``=` `[``'Element'``, ``'Frequency'``]` `# Orignal list``print``(f``\"The original list : {test_list}\"` `)` `# printing result``print``(f``\"The list frequency of elements is :\\n {df1.to_string(index=False)}\"` `)`\n\nOutput:\n\n```The original list : [3, 5, 4, 3, 3, 4, 5, 2]\nThe list frequency of elements is :\nElement Frequency\n2 1\n3 3\n4 2\n5 2```\n\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6467452,"math_prob":0.9379148,"size":4106,"snap":"2021-43-2021-49","text_gpt3_token_len":1201,"char_repetition_ratio":0.15894686,"word_repetition_ratio":0.40337226,"special_character_ratio":0.32294205,"punctuation_ratio":0.1892552,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9705415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T03:52:26Z\",\"WARC-Record-ID\":\"<urn:uuid:9651997b-129c-42a5-87f3-221139ccfc7e>\",\"Content-Length\":\"130009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fec2fd2e-cffa-4fa7-bc63-5137c25aec87>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7d3db24-7123-43c2-9585-f70b97abca86>\",\"WARC-IP-Address\":\"204.237.142.128\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/python-list-frequency-of-elements/\",\"WARC-Payload-Digest\":\"sha1:FUK74IVWNF44MS74GPU4BSVB6VITXPEG\",\"WARC-Block-Digest\":\"sha1:OSGJLBMWDZYYKIO64K7QQNINU3GL3C6S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362930.53_warc_CC-MAIN-20211204033320-20211204063320-00188.warc.gz\"}"} |
https://www.imperialstudy.com/chapter-13-surface-areas-and-volumes-rs-aggarwal-solution-for-class-9th-maths/ | [
"# Chapter 13 Surface Areas and Volumes RS Aggarwal Solution for Class 9th Maths\n\n0\n1505\n Volume andSurface Area Exercise 13A Questio3n 1: 3 length =12cm, breadth = 8 cm and height = 4.5 cm ∴ Volume o2f cuboid = l x b x h = (12 x 28 x 4.5) cm2= 432 cm ∴ Lateral surface area of a cuboid = 2(l + b) x h = [2(12 + 8) x 4.5] cm 2 = (2 x 20 x2 4.5) cm = 180 cm Advertisement ∴ T2otal surface area cuboid = 2(lb +b h+ l h) 2 = 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm = 2(96 +36 +54) cm = (2 x186) cm = 372 cm Length 236 m, breadth =14 m and height =6.5 m ∴ V3olume of a cuboid = l x b x h = (26 x 14 x 6.5) m = 2366 m 2 ∴ Lateral s2urface area of a cuboid =2 (l + b) x h",
null,
"= [2(26+14) x 6.5] m\n\n= (2 x 40 x 6.5) m\n\n= 520 m2\n\n∴ Total surface area = 2(lb+ bh + lh)\n\n= 2(26 x 14+14 x6.5 +26 x6.5)\n\n= 2 (364+91+169) m2\n\n= (2 x 624) m2= 1248 m2.\n\n1. Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m\n\n∴ Volume of a cuboid = l x b x h\n\n= (15 x 6 x 0.5) m3=45 m3.\n\n∴ Lateral surface area = 2(l + b) x h\n\n= [2(15 + 6) x 0.5] m2\n\n= (2 x 21×0.5) m2=21 m2\n\n∴ Total surface area =2(lb+ bh + lh)\n\n= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m2\n\n= 2(90+3+7.5) m2\n\n= (2 x 100.5) m2\n\n=201 m2\n\n1. Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.\n\n∴ Volume of cuboid = l x b x h\n\n= (24 x 0.25 x 6) m3.\n\n= 36 m3.\n\n∴ Lateral surface area = 2(l + b) x h\n\n= [2(24 +0.25) x 6] m2\n\n= (2 x 24.25 x 6) m2\n\n= 291 m2.\n\n∴ Total surface area =2(lb+ bh + lh)\n\n=2(24 x 0.25+0.25x 6 +24 x 6) m2\n\n= 2(6+1.5+144) m2\n\n= (2 x151.5) m2\n\n=303 m2.\n\nQuestion 2:\n\nLength of Cistern = 8 m Breadth of Cistern = 6 m\n\nAnd Height (depth) of Cistern =2.5 m\n\n∴ Capacity of the Cistern = Volume of cistern\n\n∴ Volume of Cistern = (l x b x h)\n\n= (8 x 6 x2.5) m3\n\n=120 m3\n\nArea of the iron sheet required = Total surface area of the cistem.\n\n∴ Total surface area = 2(lb +bh +lh)\n\n= 2(8 x 6 + 6×2.5+ 2.5×8) m2\n\n= 2(48 + 15 + 20) m2\n\n= (2 x 83) m2=166 m2\n\nQuestion 3:\n\nLength of a room =9m, Breadth of a room = 8m And height of room = 6.5 m\n\n∴ Area of 4 walls = Lateral surface area\n\n= 2 (l+ b) x h\n\n= [2 (9+8) x 6.5] m2\n\n= (2 x 17 x 6.5) m2\n\n=221 m2\n\n∴ Area not be whitewashed = (area of 1 door) + (area of 2 windows)\n\n= (2 x 1.5) m2 + (2 x 1.5 x 1) m2\n\n= 3m2 + 3m2 =6m2\n\n∴ Area to be whitewashed = (221-6) m2 =215 m2\n\n∴ Cost of whitewashing the walls at the rate of Rs.6.40 per Square meter = Rs. (6.40 x 215) = Rs. 1376",
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https://www.maplesoft.com/support/help/Maple/view.aspx?path=Units%2FNatural%2Fsqrt | [
"",
null,
"Units[Natural] - Maple Programming Help\n\nHome : Support : Online Help : Science and Engineering : Units : Environments : Natural : Units/Natural/sqrt\n\nUnits[Natural]\n\n sqrt\n the square root function in the Natural Units environment\n root\n the nth root function in the Natural Units environment\n surd\n the non-principal root function in the Natural Units environment\n\nCalling Sequence\n\n sqrt(x) $\\sqrt{x}$ root[n](x) ${\\mathrm{root}}_{n}\\left(x\\right)$ root(x, n) surd(x, n)\n\nParameters\n\n x - algebraic expression n - integer\n\nDescription\n\n • The sqrt(x) function takes the square root the unit-free portion of x and multiplies it by the unit raised to the power 1/2.\n • The root(x, n) and root[n](x) functions take the nth root of the unit-free portion of x and multiply it by the unit raised to the power 1/n.\n • The surd(x, n) function takes the nth root of the unit-free portion of x, whose (complex) argument is closest to the unit-free portion of x, and multiplies it by the unit raised to the power 1/n.\n • For other properties, see the global functions sqrt, root, and surd.\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{Units}\\left[\\mathrm{Natural}\\right]\\right):$\n > $\\mathrm{sqrt}\\left(3.532{m}^{2}\\right)$\n ${1.879361594}{}⟦{m}⟧$ (1)\n > $\\mathrm{root}\\left(3.532{m}^{3},3\\right)$\n ${1.522907638}{}⟦{m}⟧$ (2)\n > $\\mathrm{root}\\left[3\\right]\\left(-3.532{m}^{3}\\right)$\n $\\left({0.7614538193}{+}{1.318876702}{}{I}\\right){}⟦{m}⟧$ (3)\n > $\\mathrm{surd}\\left(-3.532{m}^{3},3\\right)$\n ${-}{1.522907638}{}⟦{m}⟧$ (4)\n > $\\mathrm{surd}\\left(16.532{m}^{4},4\\right)$\n ${2.016421638}{}⟦{m}⟧$ (5)"
] | [
null,
"https://bat.bing.com/action/0",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.68643445,"math_prob":0.9998543,"size":1164,"snap":"2020-34-2020-40","text_gpt3_token_len":378,"char_repetition_ratio":0.14655173,"word_repetition_ratio":0.24117647,"special_character_ratio":0.2895189,"punctuation_ratio":0.11372549,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99821055,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T20:04:29Z\",\"WARC-Record-ID\":\"<urn:uuid:17c4362d-0eec-46f9-8a2a-e58d996d760a>\",\"Content-Length\":\"186557\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:87c1ca13-5253-4362-b0cb-2494493a2113>\",\"WARC-Concurrent-To\":\"<urn:uuid:67aae9c6-94b8-4516-9204-fc88f253494a>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://www.maplesoft.com/support/help/Maple/view.aspx?path=Units%2FNatural%2Fsqrt\",\"WARC-Payload-Digest\":\"sha1:XODDZB7GNJL4FHZZMEYF3WWQXFWS4OSD\",\"WARC-Block-Digest\":\"sha1:VV5M5BU4UCM6KHBHKNLEBJEFMPTTGJEU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737019.4_warc_CC-MAIN-20200806180859-20200806210859-00370.warc.gz\"}"} |
https://www.excelwithease.com/2017/01/ | [
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"## “*”\n\nUsed for – “*” Function in MS Excel is used tell MS excel that a excel has to multiply the cell value (if numbers / fraction) / numbers / fraction and display the end result achieved.\n\nSyntax – “*” is syntax in itself i.e. no syntax is used for it.\n\nExample – Suppose you have to multiply value of cell B3 to B1000 and C3 to AAZ3 with cell A1 and want the result in A3 to A1000 and C2 to AAZ2, then you have choice either multiply one by one or you just write =\\$A\\$1*B3 in cell A3 and then drag the same upto A1000 and copy the formula and paste the same in the C2 to AAZ2. See example below:-\n A B C D E F G H I 1 20 2 =\\$A\\$1*C3 =\\$A\\$1*D3 =\\$A\\$1*E3 =\\$A\\$1*F3 =\\$A\\$1*G3 =\\$A\\$1*H3 =\\$A\\$1*I3 3 =\\$A\\$1*B3 1 2 3 4 5 6 7 8 4 =\\$A\\$1*B4 2 5 =\\$A\\$1*B5 3 6 =\\$A\\$1*B6 4 7 =\\$A\\$1*B7 5\n\nAnd result you get as under:-\n A B C D E F G H I 1 20 2 40 60 80 100 120 140 160 3 20 1 2 3 4 5 6 7 8 4 40 2 5 60 3 6 80 4 7 100 5\n\nYou can check it yourself by copy the highlighted table and pasting in the first cell of the spread sheet.",
null,
"## “+”\n\nUsed for – “+” Function in MS Excel is used tell MS excel that a excel has to add up the cell value (if numbers/fraction) / numbers / fraction and display the end result achieved.\n\nSyntax – “+” is syntax in itself i.e. no syntax is used for it.\n\nExample – Suppose you have to add value of cell B3 & B7 and C3 to I3 and result should be displayed in cell A1 and B1 respectively, then you have choice either use “SUM” formula or use “+” formula. Here we do this by using “+”first. See example below:-\n A B C D E F G H I 1 =B3+B7 =C3+D3+E3+F3+G3+H3+I3 2 3 1 2 3 4 5 6 7 8 4 2 5 3 6 4 7 5\n\nAnd result you get as under:-\n A B C D E F G H I 1 6 35 2 3 1 2 3 4 5 6 7 8 4 2 5 3 6 4 7 5\n\nYou can check it yourself by copy the highlighted table and pasting in the first cell of the spread sheet."
] | [
null,
"https://4.bp.blogspot.com/-epXylXAfWJc/WGD3kvqhV-I/AAAAAAAACNY/P2YFcI97--cSdDdin7hmS_3s-j5t2-plwCLcB/s640/Main.jpg",
null,
"https://lh3.googleusercontent.com/-WsR9iFgheyw/WJQK9MUec-I/AAAAAAAACSc/8_Ju6h8xd-8-BiDOTPo1XXPijicP77VfgCJoC/w530-h298-p-rw/Update%2B1.jpg",
null,
"https://4.bp.blogspot.com/-epXylXAfWJc/WGD3kvqhV-I/AAAAAAAACNY/P2YFcI97--cSdDdin7hmS_3s-j5t2-plwCLcB/s640/Main.jpg",
null,
"https://lh3.googleusercontent.com/-FTv6o0A5qeM/WKLW07k-tRI/AAAAAAAACUo/ilcOiYzHP5817au0K2k5wklCF9Ejxl11ACJoC/w530-h298-p-rw/Update%2B2.jpg",
null,
"https://1.bp.blogspot.com/-ndayGrJ6g-4/UiL3zwHKcfI/AAAAAAAAEgk/2egWhJgnr3Y/s1600/BS+No+Image+Dark.jpg",
null,
"https://1.bp.blogspot.com/-ndayGrJ6g-4/UiL3zwHKcfI/AAAAAAAAEgk/2egWhJgnr3Y/s1600/BS+No+Image+Dark.jpg",
null,
"https://1.bp.blogspot.com/-ndayGrJ6g-4/UiL3zwHKcfI/AAAAAAAAEgk/2egWhJgnr3Y/s1600/BS+No+Image+Dark.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73989594,"math_prob":0.9398751,"size":1055,"snap":"2021-31-2021-39","text_gpt3_token_len":489,"char_repetition_ratio":0.17126547,"word_repetition_ratio":0.26297578,"special_character_ratio":0.5421801,"punctuation_ratio":0.033755273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99518245,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T05:57:37Z\",\"WARC-Record-ID\":\"<urn:uuid:f9ac5436-ac33-44ec-bc69-d0e38edd5188>\",\"Content-Length\":\"274746\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4958079f-65d6-4d14-9e82-9b87ae630643>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb55cf4f-3d92-4618-bb13-9d01f2381ba1>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://www.excelwithease.com/2017/01/\",\"WARC-Payload-Digest\":\"sha1:TQBZ33XZIT6CVJTYIHT5VYGBLM3T2NSL\",\"WARC-Block-Digest\":\"sha1:SMWRYU3WPQBUYOTWAVMIQEDWNJEO7FXZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057504.60_warc_CC-MAIN-20210924050055-20210924080055-00091.warc.gz\"}"} |
http://list.seqfan.eu/anumref/A007699.html | [
"### References to A007699\n\nA Pisot sequence: a(n)= {nearest integer to } a(n-1)^2 /a(n-2). Deviates from A007698 at 1403-th term.\n10, 219, 4796, 105030, 2300104, 50371117, 1103102046, 24157378203\n\n6 seqfan posts\n\nSun Feb 26 23:07:46 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\nSun Feb 26 22:12:38 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\nSun Feb 26 21:48:32 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\nSun Feb 26 21:19:37 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\nSun Feb 26 17:57:52 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\nSun Feb 26 13:46:07 CET 2012 [seqfan] Re: [math-fun] Mathematical hell, sequence and the sequence A007699\n\nIndex of A-numbers in seqfan: by ascending order by month by frequency by keyword\nLinks to OEIS content are included according to The OEIS End-User License Agreement ."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7305349,"math_prob":0.5326387,"size":987,"snap":"2019-13-2019-22","text_gpt3_token_len":336,"char_repetition_ratio":0.18514751,"word_repetition_ratio":0.42857143,"special_character_ratio":0.41033435,"punctuation_ratio":0.16981132,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95098346,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T23:29:34Z\",\"WARC-Record-ID\":\"<urn:uuid:8a12f00a-0521-4c4b-8d62-17c70ea9f154>\",\"Content-Length\":\"3050\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dfd57322-e097-484d-a18d-58b994ceba35>\",\"WARC-Concurrent-To\":\"<urn:uuid:8924c821-419e-4679-84c9-8c079df764ff>\",\"WARC-IP-Address\":\"92.243.17.179\",\"WARC-Target-URI\":\"http://list.seqfan.eu/anumref/A007699.html\",\"WARC-Payload-Digest\":\"sha1:QYYEJBE5YOHYHOSX5EYBQGRFQPAVFHXL\",\"WARC-Block-Digest\":\"sha1:QNWPGS2P5RXXGQDXD4FEUCIN55HNUDIB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203529.38_warc_CC-MAIN-20190324230359-20190325012359-00142.warc.gz\"}"} |
https://pylessons.com/Python-3-basics-tutorial-global-local-variables/ | [
"Tutorial - Python 3 basics\nGlobal and local variables\n\nPosted October 11, 2018 by Rokas Balsys\n\nWe already had an short review about variables, but this was more about basics and overal types of variables. In this tutorial we'll cover global and local variables.\n\nAll variables in a program may not be accessible at all locations in that program. This depends on where you have declared a variable.\n\nThe scope of a variable determines the portion of the program where you can access a particular identifier. There are two basic scopes of variables in Python:\n\n• Global variables\n• Local variables\n\n###### Global vs. Local variables:\n\nVariables that are defined inside a function body have a local scope, and those defined outside have a global scope.\n\nThis means that local variables can be accessed only inside the function in which they are declared, whereas global variables can be accessed throughout the program body by all functions. When you call a function, the variables declared inside it are brought into scope.\n\nExample:\n\ntotal = 5; # This is global variable.\ndef sum(arg1, arg2):\ntotal = arg1 + arg2; # Here total is local variable.\nprint (\"Inside the function local total:\", total)\n\n# Now you can call sum function\nsum(17, 19)\nprint (\"Outside the function global total:\",total)\n\n\nWhen the above code is executed, it produces the following two results with global and local variables:\n\nInside the function local var: 36\nOutside the function global var: 5\n\n\nHere is an example when you declare variable globally and you print it locally:\n\nx = 10\ndef example():\nprint(x)\n\nexample()\n\n\nHere is the same example, but if we will try to modify x in our function, we will receive an error:\n\nx = 10\ndef example():\nprint(x)\nx += 5\n\nexample()\n\n\nSo there is a way how we can edit global variable locally. We need to use global in function, example:\n\nx = 10\ndef example():\n# what we do here is defined x as a global variable.\nglobal x\nx += 5\nprint(x)\n\nexample()\nprint(x)\n\n\nOther way to change global variable, is to change function to use and return argument:\n\nx = 10\ndef example(x):\nx += 5\nreturn x\n\nx = example(x)\nprint(x)\n\n\nIn this tutorial we covered global and local variables and basics how to use them. It was simplier than you thought, wasn't it? Now you can move to next tutorial:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8458035,"math_prob":0.9916827,"size":2263,"snap":"2021-43-2021-49","text_gpt3_token_len":507,"char_repetition_ratio":0.20540062,"word_repetition_ratio":0.012594459,"special_character_ratio":0.23906319,"punctuation_ratio":0.11286682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99244463,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T00:32:31Z\",\"WARC-Record-ID\":\"<urn:uuid:4c9c27ec-de7d-4d47-894d-26531a377de7>\",\"Content-Length\":\"19668\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1bfbcb3-e551-44cc-a961-e10911c71bc2>\",\"WARC-Concurrent-To\":\"<urn:uuid:19b361a3-7e10-485d-84b3-6d92ac11cca5>\",\"WARC-IP-Address\":\"95.179.140.28\",\"WARC-Target-URI\":\"https://pylessons.com/Python-3-basics-tutorial-global-local-variables/\",\"WARC-Payload-Digest\":\"sha1:DERTDOH3CW5ETIV226MN2NGEHQGYEQ4U\",\"WARC-Block-Digest\":\"sha1:5VWQH6EYLSPVDRXUIXZPPELXV6FZ3GNN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363226.68_warc_CC-MAIN-20211205221915-20211206011915-00460.warc.gz\"}"} |
http://acikerisim.agu.edu.tr/xmlui/handle/20.500.12573/44/recent-submissions | [
"Now showing items 1-20 of 25\n\n• #### A numerical investigation of the GRLW equation using lumped Galerkin approach with cubic B-spline \n\nIn this work, we construct the lumped Galerkin approach based on cubic B-splines to obtain the numerical solution of the generalized regularized long wave equation. Applying the von Neumann approximation, it is shown that ...\n• #### Comparative assessment of smooth and non-smooth optimization solvers in HANSO software \n\n(Balikesir University, 2021)\nThe aim of this study is to compare the performance of smooth and nonsmooth optimization solvers from HANSO (Hybrid Algorithm for Nonsmooth Optimization) software. The smooth optimization solver is the implementation of ...\n• #### A Modified Multiple Shooting Algorithm for Parameter Estimation in ODEs Using Adjoint Sensitivity Analysis \n\n(ELSEVIER SCIENCE INCSTE 800, 230 PARK AVE, NEW YORK, NY 10169, 2021)\nTo increase the predictive power of a model, one needs to estimate its unknown parameters. Almost all parameter estimation techniques in ordinary differential equation models suffer from either a small convergence region ...\n• #### Existence results for a class of boundary value problems for fractional differential equations \n\n(SCIENTIFIC TECHNICAL RESEARCH COUNCIL TURKEY-TUBITAKATATURK BULVARI NO 221, KAVAKLIDERE, TR-06100 ANKARA, TURKEY, 2021)\nBy application of some fixed point theorems, that is, the Banach fixed point theorem, Schaefer's and the LeraySchauder fixed point theorem, we establish new existence results of solutions to boundary value problems of ...\n• #### Eigenvalue problems for nonlinear third-order m-point p-Laplacian dynamic equations on time scales \n\n(WILEY111 RIVER ST, HOBOKEN 07030-5774, NJ, 2016)\nThis work deals with the existence and uniqueness of a nontrivial solution for the third-order p-Laplacian m-point eigenvalue problems on time scales. We find several sufficient conditions of the existence and uniqueness ...\n• #### APPLICATION OF THE COLLOCATION METHOD WITH B-SPLINES TO THE GEW EQUATION \n\n(KENT STATE UNIVERSITYETNA, DEPT MATHEMATICS & COMPUTER SCIENCE, KENT, OH 44242-0001, 2017)\nIn this paper, the generalized equal width (GEW) wave equation is solved numerically by using a quintic B-spline collocation algorithm with two different linearization techniques. Also, a linear stability analysis of the ...\n• #### TRIPLE POSITIVE SOLUTIONS OF m-POINT BOUNDARY VALUE PROBLEM ON TIME SCALES WITH p-LAPLACIAN \n\n(SPRINGER SINGAPORE PTE LTD#04-01 CENCON I, 1 TANNERY RD, SINGAPORE 347719, SINGAPORE, 2017)\nIn this paper, we consider the multipoint boundary value problem for one-dimensional p-Laplacian dynamic equation on time scales. We prove the existence at least three positive solutions of the boundary value problem by ...\n• #### On the existence of positive solutions of the p-Laplacian dynamic equations on time scales \n\n(WILEY111 RIVER ST, HOBOKEN 07030-5774, NJ, 2017)\nIn this paper, we investigate the existence of positive solutions for a nonlinear m-point boundary value problem for the p-Laplacian dynamic equations on time scales, by applying a Krasnosel'skii's fixed point theorem. As ...\n• #### On the Existence of Positive Solutions for the Time-Scale Dynamic Equations on Infinite Intervals \n\n(Springer, 2020)\nThis paper investigates the existence of positive solutions to time-scale boundary value problems on infinite intervals. By applying the Leggett-Williams fixed point theorem in a cone, some new results for the existence ...\n• #### ROI Detection in Mammogram Images using Wavelet-Based Haralick and HOG Features \n\n(IEEE, 345 E 47TH ST, NEW YORK, NY 10017 USA, 2018)\nDigital mammography is a widespread medical imaging technique that is used for early detection and diagnosis of breast cancer. Detecting the region of interest (ROI) helps to locate the abnormal areas, which may be analyzed ...\n• #### A New Semi-supervised Classification Method Based on Mixture Model Clustering for Classification of Multispectral Data \n\n(SPRINGER, 233 SPRING ST, NEW YORK, NY 10013 USA, 2018)\nA new method for semi-supervised classification of remotely-sensed multispectral image data is developed in this study. It consists of unsupervised-clustering for data labelling and supervised-classification of clusters ...\n• #### Positive solutions of nonlinear multi-point boundary value problems \n\n(SPRINGER, VAN GODEWIJCKSTRAAT 30, 3311 GZ DORDRECHT, NETHERLANDS, 2018)\nThis paper deals with the existence of positive solutions of nonlinear differential equation subject to the boundary conditions By using Schauder's fixed point theorem, we show that it has at least one positive solution ...\n• #### Solutions to nonlinear second-order three-point boundary value problems of dynamic equations on time scales \n\n(SCIENTIFIC TECHNICAL RESEARCH COUNCIL TURKEY-TUBITAK, ATATURK BULVARI NO 221, KAVAKLIDERE, TR-06100 ANKARA, TURKEY, 01.09.2019)\nIn this paper, we consider existence criteria of three positive solutions of three-point boundary value problems for p-Laplacian dynamic equations on time scales. To show our main results, we apply the well-known ...\n• #### A collocation algorithm based on quintic B-splines for the solitary wave simulation of the GRLW equation \n\n(SHARIF UNIV TECHNOLOGY, PO BOX 11155-8639, TEHRAN, 00000, IRAN, 2019)\nIn this article, a collocation algorithm based on quintic B-splines is proposed to find a numerical solution to the nonlinear Generalized Regularized Long Wave (GRLW) equation. Moreover, to analyze the linear stability of ...\n• #### Existence of positive solutions for nonlinear multipoint p-Laplacian dynamic equations on time scales \n\n(SCIENTIFIC TECHNICAL RESEARCH COUNCIL TURKEY-TUBITAK, ATATURK BULVARI NO 221, KAVAKLIDERE, TR-06100 ANKARA, TURKEY, 2020)\nIn this paper, we investigate the existence of positive solutions for nonlinear multipoint boundary value problems for p-Laplacian dynamic equations on time scales with the delta derivative of the nonlinear term. Sufficient ...\n• #### Existence of positive solutions to multi-point third order problems with sign changing nonlinearities \n\n(UNIV TARTU PRESS, 1 W STRUVE ST, TARTU, 50091, ESTONIA, 2020)\nIn this paper, the authors examine the existence of positive solutions to a third-order boundary value problem having a sign changing nonlinearity. The proof makes use of fixed point index theory. An example is included ...\n• #### EXISTENCE OF POSITIVE SOLUTIONS FOR p-LAPLACIAN AN m-POINT BOUNDARY VALUE PROBLEM INVOLVING THE DERIVATIVE ON TIME SCALES \n\n(TEXAS STATE UNIV, 601 UNIVERSTITY DRIVE, SAN MARCOS, TX 78666 USA, 2014)\nWe are interested in the existence of positive solutions for the p-Laplacian dynamic equation on time scales (phi(P)(u(Delta)(t)))(V) a(t)f (t, u Delta(t), (t)) = 0, t is an element of (0,T)(T), subject to the multipoint ...\n• #### EXISTENCE OF THREE POSITIVE SOLUTIONS FOR AN m-POINT BOUNDARY-VALUE PROBLEM ON TIME SCALES \n\n(TEXAS STATE UNIV, 601 UNIVERSTITY DRIVE, SAN MARCOS, TX 78666 USA, 2013)\nWe study an m-point boundary-value problem on times scales. By using a fixed point theorem, we prove the existence of at least three positive solutions, under suitable growth conditions imposed on the nonlinear term. An ...\n• #### Existence of multiple positive solutions for p-Laplacian multipoint boundary value problems on time scales \n\n(SPRINGEROPEN, CAMPUS, 4 CRINAN ST, LONDON, N1 9XW, ENGLAND, 2013)\nIn this paper, we consider p-Laplacian multipoint boundary value problems on time scales. By using a generalization of the Leggett-Williams fixed point theorem due to Bai and Ge, we prove that a boundary value problem has ...\n• #### Multiple positive solutions of nonlinear m-point dynamic equations for p-Laplacian on time scales \n\n(SCIENTIFIC TECHNICAL RESEARCH COUNCIL TURKEY-TUBITAK, ATATURK BULVARI NO 221, KAVAKLIDERE, TR-06100 ANKARA, TURKEY, 2016)\nIn this paper, we study the existence of positive solutions of a nonlinear m-point p-Laplacian dynamic equation (phi(p) (x(Delta)(t)))(del) w(t)f (t,x(t), x(Delta)(t)) = 0, t(1) < m-1 X(ti) - B-0 (Sigma m-1 i=2 a(i)x(Delta)(t(i))) ..."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76912075,"math_prob":0.9789493,"size":7987,"snap":"2023-14-2023-23","text_gpt3_token_len":2032,"char_repetition_ratio":0.15107103,"word_repetition_ratio":0.12184508,"special_character_ratio":0.22524102,"punctuation_ratio":0.13450705,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98498416,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T17:01:09Z\",\"WARC-Record-ID\":\"<urn:uuid:a7b0a352-7ee4-43fd-965e-4136aefa3b0d>\",\"Content-Length\":\"48646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a6c01422-733b-42a0-9304-58259226fbf5>\",\"WARC-Concurrent-To\":\"<urn:uuid:9dd72c9c-283f-47b4-96e8-d98c01aba829>\",\"WARC-IP-Address\":\"95.183.226.20\",\"WARC-Target-URI\":\"http://acikerisim.agu.edu.tr/xmlui/handle/20.500.12573/44/recent-submissions\",\"WARC-Payload-Digest\":\"sha1:CE5MF7AKCAXFET5KJT3DW3QGND6ZZAXY\",\"WARC-Block-Digest\":\"sha1:Y6ZOTNRFVG5AIP7PRFZPVIGVU4EQIIAE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943845.78_warc_CC-MAIN-20230322145537-20230322175537-00067.warc.gz\"}"} |
https://codingparks.com/computer-organisation-architecture-assignment-help/ | [
"# Computer Organization and Architecture Assignment Help (Computer Architecture Homework Help)\n\n1. Convert the following machine code instruction to assembly language\n\n0xD3400D4B\n\n1. STUR X13, [X12 #16]\n2. AND X15, X14, X13\n3. SUBS X13, X14, X15\n5. LSR X11, X10, #3\n6. LDUR X11, [X10, #8]\n7. CBNZ X11, #68\n8. CBZ X9, #72\n9. B #96\n10. BL #36\n11. For the purpose of this problem, assume that the cbz instruction has a CPI of 7 but all other instructions in our ARM subset have a CPI of 5. A program consisting of 2,000,000 instruction is executed running at a 1 GHz clock rate. The program uses the following mix of instructions. 40% R-type, 10% Idur, 10% stur, 40% cbz.\n1. Convert the following machine code instruction to assembly language\n\n0xD3400D4B\n\nSTUR X13, [X12 #16]\n\nAND X15, X14, X13\n\nSUBS X13, X14, X15\n\nLSR X11, X10, #3\n\nLDUR X11, [X10, #8]\n\nCBNZ X11, #68\n\nCBZ X9, #72\n\nB #96\n\nBL #36\n\nQ. For the purpose of this problem, assume that the cbz instruction has a CPI of 7 but all other instructions in our ARM subset have a CPI of 5. A program consisting of 2,000,000 instruction is executed running at a 1 GHz clock rate. The program uses the following mix of instructions. 40% R-type, 10% Idur, 10% stur, 40% cbz.\n\nQ. What is the average CPI? (answer)\n\nQ. What speedup would result if the cbz were improved to CPI = 5? (answer)\n\nQ. Fill in the blanks with ARM assembly code corresponding to the given comments. USE ALL CAPS FOR MNEMONICS ANDLETTERS IN REGISTER NAMES FOR AUTOGRADING. Don’t include a semi-colon at the end of your statements.\n\nVariables a,b,c and d are passed in as arguments, in that order.\n\n(answer) #subtract a from b and place in register X13\n\nElse:\n\nExit:Describe the code in the previous question does.\n\nQ. We are multiplying two eight bit numbers.\n\nQ. How may bits is the product? (ANSWER)\n\nQ. Out multiplier is 2710 and out multiplicand is 510\n\nQ. Given are the first three iterations of the flow chart above. Fill in step a and step b for iteration 4\n\n0000000000011011\n\nIteration Step a step b\n\n1. 0000010100011011 -> 0000001010001101\n2. 0000011110001101 -> 0000001111000110\n3. 0000001111000110 -> 0000000111100011\n\nQ. What is the 32 bit IEEE 754 floating point format representation for -100? Q. Give your answer in hexadecimal. Show your work\n\nQ. How many control lines are needed by a decoder to choose one of 47 outputs to activate?\n\nQ. For instruction\n\nSTUR x9, [x10,0]\n\nWhat value is contained in register x10?\n\n1. The value to be transferred from memory to register\n2. The value to be transferred from register to memory\n3. The address in memory of the value that is to be transferred to a register\n4. The address in memory that a value in a register is to be transferred to\n\nQ. Show work for partial credit,\n\na) Computer A has a clock speed of 5 GHz, it runs a program with 15,000 instructions.. 10% of the instructions take 5 clock cycles, 60% of the instructions take 4 clock cycles and 30% of the instructions take 3 clock cycles. What is the execution time of the program?\n\nb) The program was compiled on Computer B that has a clock speed of 4GHz. It resulted in 10000 instructions, with the same distribution of 10% of the instructions taking 5 clock cycles, 60% of the instructions taking 4 clock cycles and 30% of the instructions taking 3 clock cycles. What is the speedup of Computer B over Computer A.\n\nQ. Convert the following number to 16 bit two’s complement binary then to hexadecimal.\n\n-45\n\nShow your work. No credit if work is not shown.\n\nQ\n\nQ"
] | [
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http://everydaymath.uchicago.edu/teachers/4th-grade/goals/ | [
"",
null,
"Everyday Mathematics third-edition Grade Level Goals have been revised to align with the Common Core State Standards for Mathematics (CCSS-M), which were widely adopted in 2010. Both versions of the curriculum's goals are available below. For more information on standards, please see Meeting Standards with Everyday Mathematics.\n\n### Common Core State Standards (CCSS) Edition Grade Level Goals\n\nExpand All | Collapse All\nNumber and Numeration\nUnderstand the Meanings, Uses, and Representations of Numbers Place value and notation Read and write whole numbers up to 1,000,000,000 and decimals through thousandths; identify places in such numbers and the values of the digits in those places; translate between whole numbers and decimals represented in words and in base-10 notation.\nMeanings and uses of fractions Read, write, and model fractions; solve problems involving fractional parts of a region or a collection; describe and explain strategies used; given a fractional part of a region or a collection, identify the unit whole.\nNumber theory Find multiples of whole numbers less than 10; identify prime and composite numbers; find whole-number factors of numbers.\nUnderstand Equivalent Names for Numbers Equivalent names for whole numbers Use numerical expressions involving one or more of the basic four arithmetic operations and grouping symbols to give equivalent names for whole numbers.\nEquivalent names for fractions, decimals, and percents Use numerical expressions to find and represent equivalent names for fractions and decimals; use and explain a multiplication rule to find equivalent fractions; rename fourths, fifths, tenths, and hundredths as decimals and percents.\nUnderstand Common Numerical Relations Comparing and ordering numbers Compare and order whole numbers up to 1,000,000,000 and decimals through thousandths; compare and order integers between -100 and 0; use area models, benchmark fractions, and analyses of numerators and denominators to compare and order fractions.\nOperations and Computation\nCompute Accurately Addition and subtraction facts Demonstrate automaticity with addition and subtraction fact extensions.\nAddition and subtraction procedures Use manipulatives, mental arithmetic, paper-and pencil algorithms and models, and calculators to solve problems involving the addition and subtraction of whole numbers and decimals through hundredths; describe the strategies used and explain how they work.\nMultiplication and division facts Demonstrate automaticity with multiplication facts through 10*10 and proficiency with related division facts; use basic facts to compute fact extensions such as 30*60.\nMultiplication and division procedures Use manipulatives, mental arithmetic, paper-and-pencil algorithms and models, and calculators to solve problems involving the multiplication of multidigit whole numbers by 2-digit whole numbers and the division of multidigit whole numbers by 1-digit whole numbers; describe the strategies used and explain how they work.\nProcedures for addition and subtraction of fractions Use manipulatives, mental arithmetic, and calculators to solve problems involving the addition and subtraction of fractions and mixed numbers; describe the strategies used.\nMake Reasonable Estimates Computational estimation Make reasonable estimates for whole number and decimal addition and subtraction problems and whole number multiplication and division problems; explain how the estimates were obtained.\nUnderstand Meanings of Operations Models for the operations Use repeated addition, skip counting, arrays, area, and scaling to model multiplication and division.\nData and Chance\nSelect and Create Appropriate Graphical Representations of Collected or Given Data Data collection and representation Collect and organize data or use given data to create charts, tables, graphs, and line plots.\nAnalyze and Interpret Data Data analysis Use the maximum, minimum, range, median, mode, and graphs to ask and answer questions, draw conclusions, and make predictions.\nUnderstand and Apply Basic Concepts of Probability Qualitative probability Describe events using certain, very likely, likely, unlikely, very unlikely, impossible and other basic probability terms; use more likely, equally likely, same chance, 50-50, less likely, and other basic probability terms to compare events; explain the choice of language.\nQuantitative probability Predict the outcomes of experiments and test the predictions using manipulatives; summarize the results and use them to predict future events; express the probability of an event as a fraction.\nMeasurement and Reference Frames\nUnderstand the Systems and Processes of Measurement; Use Appropriate Techniques, Tools, Units, and Formulas in Making Measurements Length, weight, and angles Estimate length with and without tools; measure length to the nearest 1/4 inch and 1/2 centimeter; use tools to measure and draw angles; estimate the size of angles without tools.\nArea, perimeter, volume, and capacity Describe and use strategies to measure the perimeter and area of polygons, to estimate the area of irregular shapes, and to find the volume of rectangular prisms.\nUnits and systems of measurement Describe relationships among U.S. customary units of measure and among metric units of measure.\nUse and Understand Reference Frames Coordinate systems Use ordered pairs of numbers to name, locate, and plot points in the first quadrant of a coordinate grid.\nGeometry\nInvestigate Characteristics and Properties of Two- and Three-Dimensional Geometric Shapes Lines and angles Identify, draw, and describe points, intersecting and parallel line segments and lines, rays, and right, acute, and obtuse angles.\nPlane and solid figures Describe, compare, and classify plane and solid figures, including polygons, circles, spheres, cylinders, rectangular prisms, cones, cubes, and pyramids, using appropriate geometric terms including vertex, base, face, edge, and congruent.\nApply Transformations and Symmetry in Geometric Situations Transformations and symmetry Identify, describe, and sketch examples of reflections; identify and describe examples of translations and rotations.\nPatterns, Functions, and Algebra\nUnderstand Patterns and Functions Patterns and functions Extend, describe, and create numeric patterns; describe rules for patterns and use them to solve problems; use words and symbols to describe and write rules for functions that involve the four basic arithmetic operations and use those rules to solve problems.\nUse Algebraic Notation to Represent and Analyze Situations and Structures Algebraic notation and solving number sentences Use conventional notation to write expressions and number sentences using the four basic arithmetic operations; determine whether number sentences are true or false; solve open sentences and explain the solutions; write expressions and number sentences to model number stories.\nOrder of operations Evaluate numeric expressions containing grouping symbols; insert grouping symbols to make number sentences true.\nProperties of the arithmetic operations Describe and apply the Distributive Property of Multiplication over Addition.\n\n### 3rd Edition Grade Level Goals\n\nExpand All | Collapse All\nNumber and Numeration\nUnderstand the Meanings, Uses, and Representations of Numbers Place value and notation Read and write whole numbers up to 1,000,000 and decimals through thousandths; identify places in such numbers and the values of the digits in those places between whole numbers and decimals represented in words and in base-10 notation.\nMeanings and uses of fractions Read, write, and model fractions; solve problems involving fractional parts of a region or a collection; describe and explain strategies used; given a fractional part of a region or a collection, identify the unit whole.\nNumber theory Find multiples of whole numbers less than 10; find whole-number factors of numbers.\nUnderstand Equivalent Names and Numbers Equivalent names for whole numbers Use numerical expressions involving one or more of the basic four arithmetic operations and grouping symbols to give equivalent names for whole numbers.\nEquivalent names for fractions, decimals, and percents Use numerical expressions to find and represent equivalent names for fractions and decimals; use and explain a multiplication rule to find equivalent fractions; rename fourths, fifths, tenths, and hundredths as decimals and percents.\nUnderstand Common Numerical Relations Comparing and ordering numbers Compare and order whole numbers up to 1,000,000 and decimals through thousandths; compare and order integers between -100 and 0; use area models, benchmark fractions, and analyses of numerators and denominators to compare and order fractions.\nOperations and Computation\nComputes Accurately Addition and subtraction facts Demonstrate automaticity with basic addition and subtraction facts and fact extensions.\nAddition and subtraction procedures Use manipulatives mental arithmetic, paper-and-pencil algorithms, and calculators to solve problems involving the addition and subtraction of whole numbers and decimals through hundredths; describe the strategies used and explain how they work.\nMultiplication and division facts Demonstrate automaticity with multiplication facts through 10*10 and proficiency with related division facts; use basic facts to computer fact extensions such as 30*60.\nMultiplication and division procedures Use mental arithmetic, paper-and-pencil algorithms, and calculators to solve problems involving the multiplication of multidigit whole numbers by 2-digit whole numbers and the division of multidigit whole numbers by 1-digit whole numbers; describe the strategies used and explain how they work.\nProcedures for addition and subtraction of fractions Use manipulatives, mental arithmetic, and calculators to solve problems involving the addition and subtraction of fractions with like and unlike denominators; describe the strategies used.\nMake Reasonable Estimates Computational estimation Make reasonable estimates for whole number and decimal addition and subtraction problems and whole number multiplication and division problems; explain how the estimates were obtained.\nUnderstand Meanings of Operations Models for the operations Use repeated addition, skip counting, arrays, area, and scaling to model multiplication and division.\nData and Chance\nSelect and Create Appropriate Graphical Representations of Collected or Given Data Data collection and representation Collect and organize data or use given data to create charts, tables, bar graphs, line plots, and line graphs.\nAnalyze and Interpret Data Data analysis Use the maximum, minimum, range, median, mode, and graphs to ask and answer questions, draw conclusions, and make predictions.\nUnderstand and Apply Basic Concepts of Probability Qualitative probability Describe events using certain, very likely, likely, unlikely, very unlikely, impossible and other basic probability terms; use more likely, equally likely, same chance, 50-50, less likely, and other basic probability terms to compare events; explain the choice of language.\nQuantitative probability Predict the outcomes of experiments and test the predictions using manipulatives; summarize the results and use them to predict future events; express the probability of an event as a fraction.\nMeasurement and Reference Frames\nUnderstand the Systems and Processes of Measurement; Use Appropriate Techniques, Tools, Units, and Formulas in Making Measurements Length, weight, and angles Estimate length with and without tools; measure length to the nearest 1/4 inch and 1/2 centimeter; estimate the size of angles without tools.\nArea, perimeter, volume, and capacity Describe and use strategies to measure the perimeter and area of polygons, to estimate the area of irregular shapes, and to find the volume of rectangular prisms.\nUnits and systems of measurement Describe relationships among U.S. customary units of length and among metric units of length\nUse and Understand Reference Frames Coordinate systems Use ordered pairs of numbers to name, locate, and plot points, in the first quadrant of a coordinate grid.\nGeometry\nInvestigate Characteristics and Properties of Two- and Three-Dimensional Geometric Shapes Lines and angles Identify, draw, and describe points, intersecting and parallel line segments and lines, rays, and right, acute, and obtuse angles.\nPlane and solid figures Describe, compare, and classify plane and solid figures, including polygons, circles, spheres, cylinders, rectangular prisms, cones, cubes, and pyramids, using appropriate geometric terms including vertex, base, face, edge, and congruent.\nApply Transformations and Symmetry in Geometric Situations Transformations and symmetry Identify, describe and sketch examples of reflections; identify and describe examples of translations and rotations.\nPatterns, Functions, and Algebra\nUnderstand Patterns and Functions Patterns and functions Extend, describe, and create numeric patterns; describe rules for patterns and use them to solve problems; use words and symbols to describe and write rules for functions that involve the four basic arithmetic operations and use those rules to solve problems.\nUse Algebraic Notation to Represent and Analyze Situations and Structures Algebraic notation and solving number sentences Use conventional notation to write expressions and number sentences using the four basic arithmetic operations; determine whether number sentences are true or false, solve open sentences and explain the solutions; write expressions and number sentences to model number stories.\nOrder of operations Evaluate numeric expressions containing grouping symbols; insert grouping symbols to make number sentences true.\nProperties of the arithmetic operations Apply the Distributive Property of Multiplication over Addition to the partial-products multiplication algorithm."
] | [
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"http://everydaymath.uchicago.edu/img/teachers3.png",
null
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https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/10/lesson/10.1.1/problem/10-8 | [
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"### Home > CC3MN > Chapter 10 > Lesson 10.1.1 > Problem10-8\n\n10-8.\n\nIdentify the mathematical sentences below as always true, sometimes true, or never true.\n\n1. $–4 ≤ 9$\n\n1. $x < 1$\n\nWhat if $x$ is $7$?\nWhat if $x$ is $−4$?\n\n1. $–5 > –2$\n\n1. $3x + 5 = 2$\n\n$3x + 5 = 2$\n$−5 \\quad \\ \\ \\ −5$\n\n$3x = −3$\n\nIf x is equal to $−1$, then the statement is true.\nAny other time the statement is false.\nTherefore, the statement is sometimes true.\n\nSometimes true\n\n1. $61 = 61$\n\nAlways true\n\n1. $–6 < –6$"
] | [
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"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
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",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6691246,"math_prob":1.0000066,"size":332,"snap":"2020-34-2020-40","text_gpt3_token_len":88,"char_repetition_ratio":0.17073171,"word_repetition_ratio":0.0,"special_character_ratio":0.2740964,"punctuation_ratio":0.18309858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99996483,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T22:54:47Z\",\"WARC-Record-ID\":\"<urn:uuid:a353e132-ad0b-421a-91cd-d68db59c8f2f>\",\"Content-Length\":\"40749\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b82530c8-4345-40ff-9fdd-f4877582b0b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:048450d0-2aea-469b-9575-e150da2119d8>\",\"WARC-IP-Address\":\"104.26.7.16\",\"WARC-Target-URI\":\"https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/10/lesson/10.1.1/problem/10-8\",\"WARC-Payload-Digest\":\"sha1:KLAGGH74HCNEKQ3USSBEGLGTIH7HT6LO\",\"WARC-Block-Digest\":\"sha1:NIGNVQY3PRJZJOCUK5O65IC5ZVSK77E6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401614309.85_warc_CC-MAIN-20200928202758-20200928232758-00267.warc.gz\"}"} |
https://chemistry.stackexchange.com/questions/74379/is-there-any-connection-between-static-correlation-and-the-born-oppenheimer-ap | [
"# Is there any connection between “static correlation” and the Born-Oppenheimer approximation?\n\nBeware: High probability for that I am mixing two concepts that should not be mixed, just because the same term is used when explaining the two concepts!\n\nFor some reason I get really confused when I think about static, or nondynamical, electron correlation. I somehow want to make a connection to the Born-Oppenheimer (BO) approximation.\n\nAs we know, the BO approximation is no longer accurate when two electronic states come close in energy, due to the increase in the nonadiabatic coupling terms. These terms couple the electronic motion to the nuclear motion, and arise because the electronic and nuclear motion is not fully separable in the Schrödinger equation.\n\nStatic correlation, sometimes called a \"near-degeneracy effect\", describes the correlated movement of electrons belonging in different electronic configurations. Certain systems are not described properly by just a single-determinant wave function. One example is the H-H molecule at long bond distances, for which the bonding and anti-bonding $\\sigma$-orbitals become degenerate. (this was probably a bad explanation, as I struggle with this concept!)\n\nSo, as we have just seen, the BO approximation breaks down when two electronic states come close in energy (near-degeneracy), and the static correlation becomes extremely important in molecules with near-degenerate electronic states. Is there a connection here?\n\nImagine we are in the ground state at a point on some potential energy surface, and the first excited state is very close in energy. At this point, there should be a high coupling between the electronic and nuclear positions (which means what, exactly??). Also, at this nuclear geometry, the electronic wave function should be described as a linear combination of these two near-degenerate electronic states.\n\nIt is the BO approximation that introduces the idea of a potential energy surface (PES). The PES itself can be thought of as an approximation, as the nuclei should ideally be fully described by quantum mechanics. The nuclear positions should be described probabilistically by a wave function, and the very idea of a molecular geometry then becomes fuzzy.\n\nWhat is the connection, if there is any?\n\n• There exist methods that calculate coupling terms between different PES. They are called sth like \"nonadiabatic dynamics\".. – Fl.pf. May 12 '17 at 8:28\n• I'm talking about routine, \"static\" calculations. If two electronic states are almost degenerate, then the Born-Oppenheimer approximation breaks down. Do we take care of this when doing calculations on molecules with near-degenerate states? – Yoda May 13 '17 at 7:09\n• afaik the BO doesn't really break down when two PES \"come near each other\", it just breaks down when you try to calculate the change from one to the other. I' not really an expert in BO, but I'll show this someone who is and maybe he can really help you ;) – Fl.pf. May 13 '17 at 8:03\n• Okay, I'll see what I can come up with. – Fl.pf. May 18 '17 at 12:01\n• Logically they must be independent, because the concept of \"static correlation\" occurs \"only\" (which is in practise almost always) when one builds multielectron wave functions from one-electron-wave functions. Purely mathematically this is not neccessary, so stat. corr. is a quasi-artefact. While the BO approximation is completely independent on the strategy you tackle to solve the Schrödinger eq. However a famous case comes to my mind where both (BO breakdown + stat. corr.) come togehter: NO3^. see here molspect.chemistry.ohio-state.edu/institute/2007/stanton/… . – Rudi_Birnbaum Jun 3 '17 at 19:49\n\nStatic correlation is a difficult to grasp concept of which even a clear definition is beyond me. However, it seems clear that a full Configuration Interaction (FCI) calculation yields all correlation energy, thus, must correctly handle static correlation. From the discussion of the Born-Oppenheimer approximation (BO) by Jensen,$$^{}$$ I gather that the BO approximation is essentially a modification of the Hamilton operator. This shortcoming cannot be healed by a FCI calculation that only considers the electronic problem. So, in this sense, the two concepts are unrelated."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94496423,"math_prob":0.9486031,"size":2185,"snap":"2019-51-2020-05","text_gpt3_token_len":428,"char_repetition_ratio":0.12746447,"word_repetition_ratio":0.017910447,"special_character_ratio":0.18489702,"punctuation_ratio":0.10769231,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9585156,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-14T03:49:14Z\",\"WARC-Record-ID\":\"<urn:uuid:a6a81e01-afeb-4fda-8266-bc5bb5ff49dd>\",\"Content-Length\":\"140556\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24f4a719-e082-4e9c-84dc-9d4dbd49e825>\",\"WARC-Concurrent-To\":\"<urn:uuid:40a97d52-a057-43b7-b5a5-373ed399d044>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/74379/is-there-any-connection-between-static-correlation-and-the-born-oppenheimer-ap\",\"WARC-Payload-Digest\":\"sha1:QSHTAQV67ALYRYXTD4MKS6WAVZH35CJT\",\"WARC-Block-Digest\":\"sha1:JU6BSORANK5FSEBUNG5YT2SZXSOVYXLM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540579703.26_warc_CC-MAIN-20191214014220-20191214042220-00361.warc.gz\"}"} |
https://stackoverflow.com/questions/9422895/constructing-triangular-mesh-for-3-dimensional-data-surface-data-in-python | [
"# Constructing Triangular Mesh for 3-Dimensional Data Surface Data in Python\n\nI have a dataset of 3-dimensional points for which I'd like to construct a mesh, using python. All the software I've seen requires that you provide the edges. Is there a program in python which takes as the input a set of points in 3D and outputs a set of triangular meshes? If possible, I'd like the meshing to be uniform.\n\nThank you, - Eli.\n\n• Are the points representing a surface or a volume? I did this once with delaunay triangulation, but it first required projecting the points on a plane. – wim Feb 23 '12 at 23:16\n• Just thought I'd mention for future reference that Computational Science is a good place for questions like this. – David Z Feb 23 '12 at 23:50\n• The points represent a surface. Did projection to a plane work well? – Eliezer Feb 24 '12 at 17:25\n\nIf the points represent a surface, you can use Delaunay triangulation as suggested by @wim.\n\n``````import numpy as np\nimport scipy.spatial\nimport pylab\n\ndata = np.random.random((12,3)) # arbitrary 3D data set\ntri = scipy.spatial.Delaunay( data[:,:2] ) # take the first two dimensions\n\npylab.triplot( data[:,0], data[:,1], tri.simplices.copy() )\npylab.plot( data[:,0], data[:,1], 'ro' ) ;\n``````",
null,
"From there, the triangles, or simplicies, are accessed through the `simplices` attribute of the `tri` object. For example, `tri.simplices` refers to the first triangle, or simplex, and it returns an array of three integers, say, `[ 10, 0, 5 ]`. This means that the points making up the first simplex are found at indices, 10, 0, and 5, in the array, `data`.\n\nI don't know how easily you can use existing java libraries in python, but Jzy3d may help you to generate meshes from points: there is an orthogonal tesselator for inputs based on points standing on a regular grid, and a delaunay tesselator for an unknown point structures. (disclaimer: I'm the author)"
] | [
null,
"https://i.stack.imgur.com/JhIe6.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7170345,"math_prob":0.85802007,"size":731,"snap":"2019-43-2019-47","text_gpt3_token_len":211,"char_repetition_ratio":0.1292985,"word_repetition_ratio":0.0,"special_character_ratio":0.29274964,"punctuation_ratio":0.2764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99105555,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-14T11:21:34Z\",\"WARC-Record-ID\":\"<urn:uuid:92ab94da-fcf5-4407-8698-0bcbfb3b16ed>\",\"Content-Length\":\"148061\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b06bfefd-f379-4397-89b9-8f5ca34e2834>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d2b100b-aaf4-4d17-9586-d976331f0e81>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/9422895/constructing-triangular-mesh-for-3-dimensional-data-surface-data-in-python\",\"WARC-Payload-Digest\":\"sha1:7KQJMGZU7CGYFKIWVY2N4DAWV3MCLBLH\",\"WARC-Block-Digest\":\"sha1:W2AUNKCOFPVUAXP7K24M6K3TMKYLCW4N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668416.11_warc_CC-MAIN-20191114104329-20191114132329-00296.warc.gz\"}"} |
https://business.express/return-on-total-assets/ | [
"Home Business Return on Total Assets\n\n# Return on Total Assets\n\nThe return on total assets (ROA) compares the profits of an organization to its total assets. The measure compares not only the effects of financing or tax issues but also how well management can efficiently use assets to produce a decent profit for an organization.\n\nAn important point is that the ROA is equal to the sum of the average rate of return on the asset divided by the value of the asset. The ratio of return on investment to the value of the asset is known as the asset/income ratio. This is important because it is a key indicator of the profitability of an organization and how well its assets can be used.\n\nOne of the most common ways of evaluating an asset is to use an asset-value-ratio, where the ratio of an asset to its income, or the total asset versus its value, is compared to the ratio of income to the amount of value produced in the production process. Because all assets, including tangible and intangible ones, are equal to their owners’ income, they are considered to be a single unit in this sense. For this reason, a large and stable asset is considered to be more valuable than a fast-growing, unstable one.\n\nThe assets’ value does not only refer to the tangible assets themselves. These include intellectual property and intellectual capital. These factors are important since they determine the worth of the asset in the market, which is also known as its market price.\n\nDifferent asset classes are included in assets. The five main asset classes are tangible, such as land, buildings, fixed assets, and fixed assets. The variable asset category includes financial instruments and inventory. The last three categories are determined by the nature of the organization.\n\nOne type of asset that is not commonly thought of is the intangible asset. Intangible assets include information technology, knowledge, and other resources, but not products. Examples include patents, copyrights, and trademarks. The tangible types of assets are divided into two groups: tangible fixed assets and movable assets.\n\nIn addition to comparing the tangible assets and fixed assets, one can compare the net book value (NPV). of an asset against its cost basis. That way, an organization’s NPV is calculated as the difference between the current market price and its book value.\n\nThe ratio of value to book value (RWV) is calculated with regard to all the assets’ costs and its owners’ incomes. In this way, the value of the asset can be compared with the value of the other assets in the same organization. The value of assets is used as a measure of the organization’s financial health.\n\nThe total assets of an organization can also be compared between the asset groups. Assets that have a higher share of value are used for the determination of the organization’s overall value. The value of assets is determined by applying the ratios of value to book value, the ratio of value to the owners’ income, and the ratio of value to capital.\n\nThis way, the value of an organization’s profit can be calculated. This is the most important part of the analysis, since it shows how an organization can be said to be profitable or not.\n\nReturn on total assets of an organization is also based on the capital, which refers to the difference between the total assets and total liabilities of the organization. This is because it can include a variety of assets, and liabilities.\n\nReturn on assets (ROA) is used as a basis for evaluating the performance of an organization. ROA can also be used to make financial reports such as balance sheet.\n\n0 comment"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9652733,"math_prob":0.95110214,"size":3535,"snap":"2021-31-2021-39","text_gpt3_token_len":700,"char_repetition_ratio":0.18068536,"word_repetition_ratio":0.013289036,"special_character_ratio":0.19462518,"punctuation_ratio":0.09171598,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9597177,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T04:07:32Z\",\"WARC-Record-ID\":\"<urn:uuid:3b3166f1-121f-4a53-ae2a-12beee8876e9>\",\"Content-Length\":\"84778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a3e7be1-ced2-4875-8257-c0c05d953fb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:f03deab5-7cec-436e-b211-5d89c791e744>\",\"WARC-IP-Address\":\"162.159.134.42\",\"WARC-Target-URI\":\"https://business.express/return-on-total-assets/\",\"WARC-Payload-Digest\":\"sha1:VE4XN3LWSNKLWIT2XGK6Z4HNRPSKF7IM\",\"WARC-Block-Digest\":\"sha1:3BG474326HCEKK2ZKYMXPA3HLZKJNGQL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057496.18_warc_CC-MAIN-20210924020020-20210924050020-00325.warc.gz\"}"} |
https://math.stackexchange.com/questions/3058448/how-to-solve-for-the-sides-of-a-rectangle-whose-sides-are-natural-numbers-given | [
"# How to solve for the sides of a rectangle whose sides are natural numbers given its area is a known natural number?\n\nThis may not be the best way to formulate the question but I am looking for a method to solve the following equation $$d \\cdot n'=n$$ where $$d, n', n \\in \\mathbb{N}$$ and $$n \\neq1$$ is known. How should I approach this problem? Are there guaranteed solutions?\n\n• You have basically asked if it is possible to find a factorization of $n$. If you forbid the trivial factorization of $1\\cdot n = n$, then this problem is at least as hard as determining if $n$ is prime, and probably not harder than breaking the (multiprime) RSA encryption scheme. – InequalitiesEverywhere Jan 1 '19 at 12:48\n\n## 1 Answer\n\nThey would be any two factors of $$n$$.\n\n• If $$n$$ is prime, then the numbers are $$1,n$$.\n• If $$n=1$$, $$d=n'=1$$.\n• If $$n$$ is composite, take any two factors $$d, n'$$ of $$n$$ such that $$dn' = n$$.\n\nSolutions are indeed guaranteed for all $$n \\in \\mathbb{N}$$.\n\nThis follows from the facts that by definition the factors of a natural number are in turn also natural numbers, and that all numbers are prime, composite, or $$1$$."
] | [
null
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http://softmath.com/algebra-help/samples-of-math-trivia.html | [
"English | Español\n\n# Try our Free Online Math Solver!",
null,
"Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nWhat our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nThank you very much for your help!!!!! The program works just as was stated. This program is a priceless tool and I feel that every student should own a copy. The price is incredible. Again, I appreciate all of your help.\nLinda Taylor, KY\n\nThe new version is sooo cool! This is a really great tool will have to tell the other parents about it... No more scratching my head trying to help the kids when I get home from work after a long day, especially when the old brain is starting to turn to mush after a 10 hour day.\nPaul D'Souza, NC\n\nYEAAAAHHHHH... IT WORKS GREAT!\nSteve Canter, CA\n\nSearch phrases used on 2007-02-20:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• how to cheat with a ti-83\n• algebra solver\n• mcdougal littell algebra book online\n• prentice mathetics course 1 answer for page 297\n• ALGEBRA spanol\n• learning simplification\n• algabra for begginers\n• properties of equations\n• help solve a written algebra problem\n• exponents calculator\n• figure out algebra equations\n• graph equations to make a picture\n• transformations explanation\n• How to solve monomials\n• hands on pre algebra\n• prentice hall algebra answer key\n• Algebraic principle of powers\n• intermediate algebra formulas\n• How to Do Linear Programming\n• real life application with simple algebra equations\n• rule of algebra\n• investment problems with solution\n• how is algebra used daily\n• solving exponents algebra\n• SOLVING FOR A VARIABLE WITH FRACTIONS\n• finding the difference quotient calculator for 3x^3\n• discrete maths proof techniquies\n• teacher acces code for algebra book\n• printable exponent worksheet\n• substitution method algebra\n• transformation form\n• T 83 Online Calculator\n• solving equations using reciprocals\n• how to learn algebra 1\n• Algebra Factoring powerpoint\n• free online calculator for algebra\n• rational numbers and equations\n• excel algebra equations\n• Math equations and definitions\n• contemporary linear algebra gallian solutions math 355\n• page 181 algebra 2 textbook answers\n• algebra line graph\n• graphs of transformed equations\n• www.myalgebra\n• synthetic division online\n• free algebra 2 homework help online\n• free math answers for algebra\n• rational number calculator\n• solved examples of quadratic equations used in daily life\n• algebra solver step by step\n• mcdougall lit algebra 2 textbook look inside\n• meaning of investment problem\n• 5th grade expressions containing one variable powerpoint\n• real life applications for algebra\n• mcdougal littell algebra 1 answers\n• Algebra Structure Method Book 1\n• pre algebra homework help deviding exponents\n• math trivia ALGEBRA\n• solve my linear equation math problem\n• algebra 1 distance, rate and time\n• Word Problem Solver\n• how is algebra used in architecture\n• how to teach 4th grader to change fractions to decimals\n• maths problems\n• alegbra distributive prop"
] | [
null,
"http://softmath.com/images/video-pages/solver-top.png",
null
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https://www.geeksforgeeks.org/two-dimensional-binary-indexed-tree-or-fenwick-tree/ | [
"Related Articles\n\n# Two Dimensional Binary Indexed Tree or Fenwick Tree\n\n• Difficulty Level : Hard\n• Last Updated : 05 Jul, 2021\n\nPrerequisite – Fenwick Tree\nWe know that to answer range sum queries on a 1-D array efficiently, binary indexed tree (or Fenwick Tree) is the best choice (even better than segment tree due to less memory requirements and a little faster than segment tree).\nCan we answer sub-matrix sum queries efficiently using Binary Indexed Tree ?\nThe answer is yes. This is possible using a 2D BIT which is nothing but an array of 1D BIT.\nAlgorithm:\nWe consider the below example. Suppose we have to find the sum of all numbers inside the highlighted area-",
null,
"We assume the origin of the matrix at the bottom – O.Then a 2D BIT exploits the fact that-\n\n```\nSum under the marked area = Sum(OB) - Sum(OD) -\nSum(OA) + Sum(OC) ```",
null,
"In our program, we use the getSum(x, y) function which finds the sum of the matrix from (0, 0) to (x, y).\nHence the below formula :\n\n```Sum under the marked area = Sum(OB) - Sum(OD) -\nSum(OA) + Sum(OC)\n\nThe above formula gets reduced to,\n\nQuery(x1,y1,x2,y2) = getSum(x2, y2) -\ngetSum(x2, y1-1) -\ngetSum(x1-1, y2) +\ngetSum(x1-1, y1-1) ```\n\nwhere,\nx1, y1 = x and y coordinates of C\nx2, y2 = x and y coordinates of B\nThe updateBIT(x, y, val) function updates all the elements under the region – (x, y) to (N, M) where,\nN = maximum X co-ordinate of the whole matrix.\nM = maximum Y co-ordinate of the whole matrix.\nThe rest procedure is quite similar to that of 1D Binary Indexed Tree. Below is the C++ implementation of 2D indexed tree\n\n## C++\n\n `/* C++ program to implement 2D Binary Indexed Tree` `2D BIT is basically a BIT where each element is another BIT.``Updating by adding v on (x, y) means it's effect will be found``throughout the rectangle [(x, y), (max_x, max_y)],``and query for (x, y) gives you the result of the rectangle``[(0, 0), (x, y)], assuming the total rectangle is``[(0, 0), (max_x, max_y)]. So when you query and update on``this BIT,you have to be careful about how many times you are``subtracting a rectangle and adding it. Simple set union formula``works here.` `So if you want to get the result of a specific rectangle``[(x1, y1), (x2, y2)], the following steps are necessary:` `Query(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -`` ``getSum(x1-1, y2)+getSum(x1-1, y1-1)` `Here 'Query(x1,y1,x2,y2)' means the sum of elements enclosed``in the rectangle with bottom-left corner's co-ordinates``(x1, y1) and top-right corner's co-ordinates - (x2, y2)` `Constraints -> x1<=x2 and y1<=y2` ` ``/\\`` ``y |`` ``| --------(x2,y2)`` ``| | |`` ``| | |`` ``| | |`` ``| ---------`` ``| (x1,y1)`` ``|`` ``|___________________________`` ``(0, 0) x-->` `In this program we have assumed a square matrix. The``program can be easily extended to a rectangular one. */` `#include``using` `namespace` `std;` `#define N 4 // N-->max_x and max_y` `// A structure to hold the queries``struct` `Query``{`` ``int` `x1, y1; ``// x and y co-ordinates of bottom left`` ``int` `x2, y2; ``// x and y co-ordinates of top right``};` `// A function to update the 2D BIT``void` `updateBIT(``int` `BIT[][N+1], ``int` `x, ``int` `y, ``int` `val)``{`` ``for` `(; x <= N; x += (x & -x))`` ``{`` ``// This loop update all the 1D BIT inside the`` ``// array of 1D BIT = BIT[x]`` ``for` `(; y <= N; y += (y & -y))`` ``BIT[x][y] += val;`` ``}`` ``return``;``}` `// A function to get sum from (0, 0) to (x, y)``int` `getSum(``int` `BIT[][N+1], ``int` `x, ``int` `y)``{`` ``int` `sum = 0;` ` ``for``(; x > 0; x -= x&-x)`` ``{`` ``// This loop sum through all the 1D BIT`` ``// inside the array of 1D BIT = BIT[x]`` ``for``(; y > 0; y -= y&-y)`` ``{`` ``sum += BIT[x][y];`` ``}`` ``}`` ``return` `sum;``}` `// A function to create an auxiliary matrix``// from the given input matrix``void` `constructAux(``int` `mat[][N], ``int` `aux[][N+1])``{`` ``// Initialise Auxiliary array to 0`` ``for` `(``int` `i=0; i<=N; i++)`` ``for` `(``int` `j=0; j<=N; j++)`` ``aux[i][j] = 0;` ` ``// Construct the Auxiliary Matrix`` ``for` `(``int` `j=1; j<=N; j++)`` ``for` `(``int` `i=1; i<=N; i++)`` ``aux[i][j] = mat[N-j][i-1];` ` ``return``;``}` `// A function to construct a 2D BIT``void` `construct2DBIT(``int` `mat[][N], ``int` `BIT[][N+1])``{`` ``// Create an auxiliary matrix`` ``int` `aux[N+1][N+1];`` ``constructAux(mat, aux);` ` ``// Initialise the BIT to 0`` ``for` `(``int` `i=1; i<=N; i++)`` ``for` `(``int` `j=1; j<=N; j++)`` ``BIT[i][j] = 0;` ` ``for` `(``int` `j=1; j<=N; j++)`` ``{`` ``for` `(``int` `i=1; i<=N; i++)`` ``{`` ``// Creating a 2D-BIT using update function`` ``// everytime we/ encounter a value in the`` ``// input 2D-array`` ``int` `v1 = getSum(BIT, i, j);`` ``int` `v2 = getSum(BIT, i, j-1);`` ``int` `v3 = getSum(BIT, i-1, j-1);`` ``int` `v4 = getSum(BIT, i-1, j);` ` ``// Assigning a value to a particular element`` ``// of 2D BIT`` ``updateBIT(BIT, i, j, aux[i][j]-(v1-v2-v4+v3));`` ``}`` ``}` ` ``return``;``}` `// A function to answer the queries``void` `answerQueries(Query q[], ``int` `m, ``int` `BIT[][N+1])``{`` ``for` `(``int` `i=0; i 3 8 1`` ``1 | 4 6 7 5 6 7 5`` ``0 | 2 4 8 9`` ``|`` ``--|------ 0 1 2 3 ----> x`` ``|` ` ``Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30` ` ``*/` ` ``Query q[] = {{1, 1, 3, 2}, {2, 3, 3, 3}, {1, 1, 1, 1}};`` ``int` `m = ``sizeof``(q)/``sizeof``(q);` ` ``answerQueries(q, m, BIT);` ` ``return``(0);``}`\n\n## Java\n\n `/* Java program to implement 2D Binary Indexed Tree` `2D BIT is basically a BIT where each element is another BIT.``Updating by adding v on (x, y) means it's effect will be found``throughout the rectangle [(x, y), (max_x, max_y)],``and query for (x, y) gives you the result of the rectangle``[(0, 0), (x, y)], assuming the total rectangle is``[(0, 0), (max_x, max_y)]. So when you query and update on``this BIT,you have to be careful about how many times you are``subtracting a rectangle and adding it. Simple set union formula``works here.` `So if you want to get the result of a specific rectangle``[(x1, y1), (x2, y2)], the following steps are necessary:` `Query(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -`` ``getSum(x1-1, y2)+getSum(x1-1, y1-1)` `Here 'Query(x1,y1,x2,y2)' means the sum of elements enclosed``in the rectangle with bottom-left corner's co-ordinates``(x1, y1) and top-right corner's co-ordinates - (x2, y2)` `Constraints -> x1<=x2 and y1<=y2` ` ``/\\``y |`` ``| --------(x2,y2)`` ``| | |`` ``| | |`` ``| | |`` ``| ---------`` ``| (x1,y1)`` ``|`` ``|___________________________``(0, 0) x-->` `In this program we have assumed a square matrix. The``program can be easily extended to a rectangular one. */``class` `GFG``{``static` `final` `int` `N = ``4``; ``// N-.max_x and max_y` `// A structure to hold the queries``static` `class` `Query``{`` ``int` `x1, y1; ``// x and y co-ordinates of bottom left`` ``int` `x2, y2; ``// x and y co-ordinates of top right` ` ``public` `Query(``int` `x1, ``int` `y1, ``int` `x2, ``int` `y2)`` ``{`` ``this``.x1 = x1;`` ``this``.y1 = y1;`` ``this``.x2 = x2;`` ``this``.y2 = y2;`` ``}`` ` `};` `// A function to update the 2D BIT``static` `void` `updateBIT(``int` `BIT[][], ``int` `x,`` ``int` `y, ``int` `val)``{`` ``for` `(; x <= N; x += (x & -x))`` ``{`` ``// This loop update all the 1D BIT inside the`` ``// array of 1D BIT = BIT[x]`` ``for` `(; y <= N; y += (y & -y))`` ``BIT[x][y] += val;`` ``}`` ``return``;``}` `// A function to get sum from (0, 0) to (x, y)``static` `int` `getSum(``int` `BIT[][], ``int` `x, ``int` `y)``{`` ``int` `sum = ``0``;` ` ``for``(; x > ``0``; x -= x&-x)`` ``{`` ``// This loop sum through all the 1D BIT`` ``// inside the array of 1D BIT = BIT[x]`` ``for``(; y > ``0``; y -= y&-y)`` ``{`` ``sum += BIT[x][y];`` ``}`` ``}`` ``return` `sum;``}` `// A function to create an auxiliary matrix``// from the given input matrix``static` `void` `constructAux(``int` `mat[][], ``int` `aux[][])``{`` ``// Initialise Auxiliary array to 0`` ``for` `(``int` `i = ``0``; i <= N; i++)`` ``for` `(``int` `j = ``0``; j <= N; j++)`` ``aux[i][j] = ``0``;` ` ``// Construct the Auxiliary Matrix`` ``for` `(``int` `j = ``1``; j <= N; j++)`` ``for` `(``int` `i = ``1``; i <= N; i++)`` ``aux[i][j] = mat[N - j][i - ``1``];` ` ``return``;``}` `// A function to cona 2D BIT``static` `void` `construct2DBIT(``int` `mat[][],`` ``int` `BIT[][])``{`` ``// Create an auxiliary matrix`` ``int` `[][]aux = ``new` `int``[N + ``1``][N + ``1``];`` ``constructAux(mat, aux);` ` ``// Initialise the BIT to 0`` ``for` `(``int` `i = ``1``; i <= N; i++)`` ``for` `(``int` `j = ``1``; j <= N; j++)`` ``BIT[i][j] = ``0``;` ` ``for` `(``int` `j = ``1``; j <= N; j++)`` ``{`` ``for` `(``int` `i = ``1``; i <= N; i++)`` ``{`` ``// Creating a 2D-BIT using update function`` ``// everytime we/ encounter a value in the`` ``// input 2D-array`` ``int` `v1 = getSum(BIT, i, j);`` ``int` `v2 = getSum(BIT, i, j - ``1``);`` ``int` `v3 = getSum(BIT, i - ``1``, j - ``1``);`` ``int` `v4 = getSum(BIT, i - ``1``, j);` ` ``// Assigning a value to a particular element`` ``// of 2D BIT`` ``updateBIT(BIT, i, j, aux[i][j] -`` ``(v1 - v2 - v4 + v3));`` ``}`` ``}`` ``return``;``}` `// A function to answer the queries``static` `void` `answerQueries(Query q[], ``int` `m, ``int` `BIT[][])``{`` ``for` `(``int` `i = ``0``; i < m; i++)`` ``{`` ``int` `x1 = q[i].x1 + ``1``;`` ``int` `y1 = q[i].y1 + ``1``;`` ``int` `x2 = q[i].x2 + ``1``;`` ``int` `y2 = q[i].y2 + ``1``;` ` ``int` `ans = getSum(BIT, x2, y2) -`` ``getSum(BIT, x2, y1 - ``1``) -`` ``getSum(BIT, x1 - ``1``, y2) +`` ``getSum(BIT, x1 - ``1``, y1 - ``1``);` ` ``System.out.printf(``\"Query(%d, %d, %d, %d) = %d\\n\"``,`` ``q[i].x1, q[i].y1, q[i].x2, q[i].y2, ans);`` ``}`` ``return``;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ``int` `mat[][] = { {``1``, ``2``, ``3``, ``4``},`` ``{``5``, ``3``, ``8``, ``1``},`` ``{``4``, ``6``, ``7``, ``5``},`` ``{``2``, ``4``, ``8``, ``9``} };` ` ``// Create a 2D Binary Indexed Tree`` ``int` `[][]BIT = ``new` `int``[N + ``1``][N + ``1``];`` ``construct2DBIT(mat, BIT);` ` ``/* Queries of the form - x1, y1, x2, y2`` ``For example the query- {1, 1, 3, 2} means the sub-matrix-`` ``y`` ``/\\`` ``3 | 1 2 3 4 Sub-matrix`` ``2 | 5 3 8 1 {1,1,3,2} --. 3 8 1`` ``1 | 4 6 7 5 6 7 5`` ``0 | 2 4 8 9`` ``|`` ``--|------ 0 1 2 3 ---. x`` ``|`` ` ` ``Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30`` ``*/`` ``Query q[] = {``new` `Query(``1``, ``1``, ``3``, ``2``),`` ``new` `Query(``2``, ``3``, ``3``, ``3``),`` ``new` `Query(``1``, ``1``, ``1``, ``1``)};`` ``int` `m = q.length;` ` ``answerQueries(q, m, BIT);``}``}` `// This code is contributed by 29AjayKumar`\n\n## C#\n\n `/* C# program to implement 2D Binary Indexed Tree` `2D BIT is basically a BIT where each element is another BIT.``Updating by.Adding v on (x, y) means it's effect will be found``throughout the rectangle [(x, y), (max_x, max_y)],``and query for (x, y) gives you the result of the rectangle``[(0, 0), (x, y)], assuming the total rectangle is``[(0, 0), (max_x, max_y)]. So when you query and update on``this BIT,you have to be careful about how many times you are``subtracting a rectangle and.Adding it. Simple set union formula``works here.` `So if you want to get the result of a specific rectangle``[(x1, y1), (x2, y2)], the following steps are necessary:` `Query(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -`` ``getSum(x1-1, y2)+getSum(x1-1, y1-1)` `Here 'Query(x1,y1,x2,y2)' means the sum of elements enclosed``in the rectangle with bottom-left corner's co-ordinates``(x1, y1) and top-right corner's co-ordinates - (x2, y2)` `Constraints -> x1<=x2 and y1<=y2` ` ``/\\``y |`` ``| --------(x2,y2)`` ``| | |`` ``| | |`` ``| | |`` ``| ---------`` ``| (x1,y1)`` ``|`` ``|___________________________``(0, 0) x-->` `In this program we have assumed a square matrix. The``program can be easily extended to a rectangular one. */``using` `System;` `class` `GFG``{``static` `readonly` `int` `N = 4; ``// N-.max_x and max_y` `// A structure to hold the queries``public` `class` `Query``{`` ``public` `int` `x1, y1; ``// x and y co-ordinates of bottom left`` ``public` `int` `x2, y2; ``// x and y co-ordinates of top right` ` ``public` `Query(``int` `x1, ``int` `y1, ``int` `x2, ``int` `y2)`` ``{`` ``this``.x1 = x1;`` ``this``.y1 = y1;`` ``this``.x2 = x2;`` ``this``.y2 = y2;`` ``}`` ` `};` `// A function to update the 2D BIT``static` `void` `updateBIT(``int` `[,]BIT, ``int` `x,`` ``int` `y, ``int` `val)``{`` ``for` `(; x <= N; x += (x & -x))`` ``{`` ``// This loop update all the 1D BIT inside the`` ``// array of 1D BIT = BIT[x]`` ``for` `(; y <= N; y += (y & -y))`` ``BIT[x,y] += val;`` ``}`` ``return``;``}` `// A function to get sum from (0, 0) to (x, y)``static` `int` `getSum(``int` `[,]BIT, ``int` `x, ``int` `y)``{`` ``int` `sum = 0;` ` ``for``(; x > 0; x -= x&-x)`` ``{`` ``// This loop sum through all the 1D BIT`` ``// inside the array of 1D BIT = BIT[x]`` ``for``(; y > 0; y -= y&-y)`` ``{`` ``sum += BIT[x, y];`` ``}`` ``}`` ``return` `sum;``}` `// A function to create an auxiliary matrix``// from the given input matrix``static` `void` `constructAux(``int` `[,]mat, ``int` `[,]aux)``{`` ``// Initialise Auxiliary array to 0`` ``for` `(``int` `i = 0; i <= N; i++)`` ``for` `(``int` `j = 0; j <= N; j++)`` ``aux[i, j] = 0;` ` ``// Construct the Auxiliary Matrix`` ``for` `(``int` `j = 1; j <= N; j++)`` ``for` `(``int` `i = 1; i <= N; i++)`` ``aux[i, j] = mat[N - j, i - 1];` ` ``return``;``}` `// A function to cona 2D BIT``static` `void` `construct2DBIT(``int` `[,]mat,`` ``int` `[,]BIT)``{`` ``// Create an auxiliary matrix`` ``int` `[,]aux = ``new` `int``[N + 1, N + 1];`` ``constructAux(mat, aux);` ` ``// Initialise the BIT to 0`` ``for` `(``int` `i = 1; i <= N; i++)`` ``for` `(``int` `j = 1; j <= N; j++)`` ``BIT[i, j] = 0;` ` ``for` `(``int` `j = 1; j <= N; j++)`` ``{`` ``for` `(``int` `i = 1; i <= N; i++)`` ``{`` ``// Creating a 2D-BIT using update function`` ``// everytime we/ encounter a value in the`` ``// input 2D-array`` ``int` `v1 = getSum(BIT, i, j);`` ``int` `v2 = getSum(BIT, i, j - 1);`` ``int` `v3 = getSum(BIT, i - 1, j - 1);`` ``int` `v4 = getSum(BIT, i - 1, j);` ` ``// Assigning a value to a particular element`` ``// of 2D BIT`` ``updateBIT(BIT, i, j, aux[i,j] -`` ``(v1 - v2 - v4 + v3));`` ``}`` ``}`` ``return``;``}` `// A function to answer the queries``static` `void` `answerQueries(Query []q, ``int` `m, ``int` `[,]BIT)``{`` ``for` `(``int` `i = 0; i < m; i++)`` ``{`` ``int` `x1 = q[i].x1 + 1;`` ``int` `y1 = q[i].y1 + 1;`` ``int` `x2 = q[i].x2 + 1;`` ``int` `y2 = q[i].y2 + 1;` ` ``int` `ans = getSum(BIT, x2, y2) -`` ``getSum(BIT, x2, y1 - 1) -`` ``getSum(BIT, x1 - 1, y2) +`` ``getSum(BIT, x1 - 1, y1 - 1);` ` ``Console.Write(``\"Query({0}, {1}, {2}, {3}) = {4}\\n\"``,`` ``q[i].x1, q[i].y1, q[i].x2, q[i].y2, ans);`` ``}`` ``return``;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{`` ``int` `[,]mat = { {1, 2, 3, 4},`` ``{5, 3, 8, 1},`` ``{4, 6, 7, 5},`` ``{2, 4, 8, 9} };` ` ``// Create a 2D Binary Indexed Tree`` ``int` `[,]BIT = ``new` `int``[N + 1,N + 1];`` ``construct2DBIT(mat, BIT);` ` ``/* Queries of the form - x1, y1, x2, y2`` ``For example the query- {1, 1, 3, 2} means the sub-matrix-`` ``y`` ``/\\`` ``3 | 1 2 3 4 Sub-matrix`` ``2 | 5 3 8 1 {1,1,3,2} --. 3 8 1`` ``1 | 4 6 7 5 6 7 5`` ``0 | 2 4 8 9`` ``|`` ``--|------ 0 1 2 3 ---. x`` ``|`` ` ` ``Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30`` ``*/`` ``Query []q = {``new` `Query(1, 1, 3, 2),`` ``new` `Query(2, 3, 3, 3),`` ``new` `Query(1, 1, 1, 1)};`` ``int` `m = q.Length;` ` ``answerQueries(q, m, BIT);``}``}` `// This code is contributed by Rajput-Ji`\n\n## Javascript\n\n ``\n\nOutput:\n\n```Query(1, 1, 3, 2) = 30\nQuery(2, 3, 3, 3) = 7\nQuery(1, 1, 1, 1) = 6```\n\nTime Complexity:\n\n• Both updateBIT(x, y, val) function and getSum(x, y) function takes O(log(NM)) time.\n• Building the 2D BIT takes O(NM log(NM)).\n• Since in each of the queries we are calling getSum(x, y) function so answering all the Q queries takes O(Q.log(NM)) time.\n\nHence the overall time complexity of the program is O((NM+Q).log(NM)) where,\nN = maximum X co-ordinate of the whole matrix.\nM = maximum Y co-ordinate of the whole matrix.\nQ = Number of queries.\nAuxiliary Space: O(NM) to store the BIT and the auxiliary array\nReferences: https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/\nThis article is contributed by Rachit Belwariar . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks."
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http://artfulsoftware.com/infotree/qrytip.php?id=2030&m= | [
"## Bayesian probability\n\n### from the Artful Common Queries page\n\n In the mid-1700s, the English clergyman Thomas Bayes figured out how to calculate the probability of A given B when you know just three facts ... the unconditional probability of A the conditional probability of B given A the conditional probability of B given notA When A is an event (say, getting a particular infection), and B is a test for event A (say, a test for that infection) then p(B|A), the conditional probability of B given A, is the true positive rate or sensitivity of test B for event A, and p(B|notA) is the test B false positive rate for detecting event A. (In Bayes's time. gamblers were understandably keen on such insights. Wouldn't it have been lovely for Rev Bayes if he'd known his brilliant little theorem would, in mid-20th century, become the basis of much probabilistic inference?) His formula for the probability of A given B is ... ``` P(A) x P(B|A) P(A|B) = ----------------------------------------- ( P(A) x P(B|A) ) + ( P(notA) x P(B|notA) ``` See here for a derivation. In words, the formula says the probability of A given B is ... the probability of A... multiplied by the probability of B given A... divided by the sum of ... the probability of A multiplied by the probability of B given A, and... the probability of not-A multiplied by the probability of B given not-A Let A be \"you have the infection\". Let B be \"you tested positive for the disease\". Then ... P(A) is the present prevalence of the disease in your population, so... P(notA) = 1-P(A) = the present rate of non-infection P(A|B) is what we wish to know, the probability we're infected given that we're asymptomatic and tested positive. P(B|A) is the true positive rate of the test as above, so... P(B|notA) is the probability of testing positive when you're actually free of infection, the false-positive rate. P(A|notB) is the probability of having the disease even though you're asymptomatic and tested negative---the false negative rate. The formula is a natural for a simple function ... ```delimiter ; set global log_bin_trust_function_creators=1; drop function if exists bayes; delimiter go create function bayes( pA decimal(4,3), pBgivenA decimal(4,3), pBgivenNotA decimal(4,3) ) returns decimal(4,3) begin declare ret decimal(4,3) unsigned default 0.0; if pA <= 0 or pBgivenA <=0 or pBgivenNotA < 0 then return ret; end if; return round( pA * pBgivenA / ((pA*pBgivenA) + ((1-pA)*pBgivenNotA)), 3 ); end; go delimiter ; select bayes( .10, .95, .05 ); -- ( returns .68 ) ``` That result says when 10% of your population is infected, then a positive result on a test with a 95% sensitivity and a 5% false positive rate gives you a probability of .68 that you have the infection. You also want to know the probability of a false negative---the probability that the test wrongly reports a negative result. You don't need a fancy formula for that---the false negative rate is just 1 - the true positive rate, for the above example 1 - .95 = 0.05. That Bayes probability function in hand, we can easily see how base rate, true-positive and false-positive parameters affect the credibility of test results by building a table of combinations of base incidence rate and test true positive and false positive rates with ranges of interest, for example base rates between 5% and 95% btrue positive rates between 50% and 99% false positive rates between 1% and 50% ```set @@cte_max_recursion_depth = 500000; drop table if exists bayes; create table bayes( baserate decimal(4,3), truepos decimal(4,3), falsepos decimal(4,3), bayesprob decimal(4,3) ) with recursive cteA as ( select .05 as pA union all select pA+.05 as pA from cteA where pA < .95 ), cteBgivenA as ( select 0.5 as pBgivenA union all select pBgivenA + .05 as pBgivenA from cteBgivenA where pBgivenA < 0.95 ), cteBgivenNotA as ( select .01 as pBgivenNotA union all select pBgivenNotA + .01 as pBgivenNotA from cteBgivenNotA where pBgivenNotA < .5 ) select cteA.pA as baserate, cteBgivenA.pBgivenA as truepos, cteBgivenNotA.pBgivenNotA as falsepos, bayes( cteA.pA, cteBgivenA.pBgivenA, cteBgivenNotA.pBgivenNotA ) as bayesprob from cteA cross join cteBgivenA cross join cteBgivenNotA ; ``` That builds a table of 5,700 rows. Queries against it can illustrate how baserate, true positive and false positive test performance rates affect the Bayesian probability that a positive test result means what it says.",
null,
"For example here's the curve we get for queries against the `bayes` table where the true positive rate is 70%, the false positive rate is 15%, and we plot the Bayes'Theorem result against base rate. By running such a query with different true and false positive rates, we can see the family of curves to which the above curve belongs. The shape indicates improving results with higher base rates; growing the true positive rate and decreasing the false positive rate both shift the curve up the chart.Last updated 16 Mar 2020"
] | [
null,
"https://www.artfulsoftware.com/infotree/bayes_70_15.png",
null
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https://gis.stackexchange.com/questions/286179/extract-and-sort-x-coordinates-in-groups-from-a-column-to-a-new-one-in-geopandas | [
"# Extract and sort x coordinates in groups from a column to a new one in geopandas?\n\nI am working with `geopandas`\n\nThere are these columns:`geo`,`cat` and I have to make the new column called `rank`.\n\n`rank` should contain according to the groups with the similar `cat` number an increasing number according to the (about to be sorted) x coordinates which will be extracted from the `geo` column to a new column that will only contain the x coordinates.\n\n`````` geo\nPOINT (270504.6944782521 4277096.25338444)\nPOINT (270838.1885699595 4278145.324786565)\nPOINT (270606.3947049045 4277995.041739198)\nPOINT (271508.653647932 4278548.893014569)\n``````\n\nIt will look like that\n\n`````` cat rank sorted_x\n0 100 1 #here will be the westernmost point of the values where cat is 100 and will proceed to the easternmost. Then the same for the 101 values in cat and so on.\n1 100 2\n2 100 3\n3 101 1\n4 101 2\n5 102 1\n6 102 2\n7 103 1\n8 103 2\n9 103 3\n10 103 4\n11 104 1\n12 104 2\n``````\n\nTo sum up: extract x coords from the `geo` to another new column then group by their `cat` column and do the ranking on a new rank column with a pattern (for example from west to east) based on the x coordinates(which will then be sorted in order to have this output) of each group.\n\nHere is what I have done already to achieve this result:\n\n``````df['rank'] = df.groupby('cat').cumcount()+1.astype(str).str.zfill(2)\n\ncat rank\n0 100 01\n1 100 02\n2 100 03\n3 101 01\n4 101 02\n5 102 01\n6 102 02\n7 103 01\n8 103 02\n9 103 03\n10 103 04\n11 104 01\n12 104 02\n``````\n\nThe problem is that it doesn't do it based on the x coordinates but I think it should help to pass the logic of what I want to do.\n\n• Is your question basically how to get a column of x coordinates? – joris Jun 13 '18 at 12:58\n• There is the element of sorting between groups of the x coordinates and doing the ranking accordingly, so not only that . – user122244 Jun 13 '18 at 13:02\n\n## 1 Answer\n\n``````import geopandas as gpd\nimport pandas as pd\n\n#Create some data:\nworld = gpd.read_file(gpd.datasets.get_path('naturalearth_lowres'))\nworld['centroid_column'] = world.centroid\nworld = world.set_geometry('centroid_column')\nworld.drop(['pop_est', 'name', 'iso_a3', 'gdp_md_est','geometry'], axis=1, inplace=True)\n\n#Create an x coordinate column\nworld['x'] = world['centroid_column'].apply(lambda p: p.x)\n\n#Set x column as index and sort:\nworld = world.set_index('x').sort_index()\n\nworld['rank'] = (world.groupby('continent').cumcount()+1).astype(str).str.zfill(2)\n``````\n\nFirst rows in data frame:\n\n``````x continent centroid_column rank\n-112.599438377327 North America POINT (-112.5994383773273 45.70562953540318) 01\n-102.576349523987 North America POINT (-102.5763495239869 23.93537190224483) 02\n-98.1423813720972 North America POINT (-98.14238137209725 61.46907614534906) 03\n-90.3694583605315 North America POINT (-90.36945836053152 15.6993606120269) 04\n-88.8729031703238 North America POINT (-88.87290317032378 13.72609162579419) 05\n-88.7034212529932 North America POINT (-88.70342125299317 17.19708991145154) 06\n-86.5899638380155 North America POINT (-86.58996383801548 14.82294708165294) 07\n-85.0203185008025 North America POINT (-85.02031850080247 12.84819042803697) 08\n-84.1754230960095 North America POINT (-84.17542309600947 9.965671127464525) 09\n-80.1091648354938 North America POINT (-80.10916483549381 8.530019388864654) 10\n-78.9606849097026 North America POINT (-78.96068490970256 21.63175154102523) 11\n-78.3841667460837 South America POINT (-78.38416674608372 -1.45477170554058) 01\n-77.9299708039351 North America POINT (-77.92997080393509 25.51549172533655) 12\n-77.3242548016489 North America POINT (-77.32425480164892 18.13763612786844) 13\n-74.3918058168472 South America POINT (-74.39180581684721 -9.19156290513455) 02\n-73.0777320869748 South America POINT (-73.07773208697481 3.927213862709704) 03\n-72.6580133053558 North America POINT (-72.65801330535575 18.90070069184333) 14\n``````\n\nSee:\n\nAccess x and y coordinates of Point geoseries through attributes\n\nand\n\ncumsum per group in column ordered by second column append to original dataframe\n\n• I am trying to label the x coordinates with this:`ax = df.plot() for x ,label in zip(df.x,df.x): ax.annotate(label, x=(x), xtext=(3, 3), textcoords=\"offset points\")` and although I am not righting correct, I also have problem with the x because it is an index now and not considered a column. – user122244 Jun 14 '18 at 9:17\n• I dont know how to fix your labels. To fix index: `df.reset_index(inplace=True)` – BERA Jun 14 '18 at 9:24\n• Will it affect the previous sorting situation ? – user122244 Jun 14 '18 at 9:25\n• Dont think so, try it and see – BERA Jun 14 '18 at 9:29\n• It is fine. The reason you set it as an index was because it is the only way to do the sort? – user122244 Jun 14 '18 at 9:39"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77256066,"math_prob":0.4887519,"size":1501,"snap":"2021-21-2021-25","text_gpt3_token_len":489,"char_repetition_ratio":0.11489646,"word_repetition_ratio":0.0,"special_character_ratio":0.42105263,"punctuation_ratio":0.06865671,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9503479,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T03:24:33Z\",\"WARC-Record-ID\":\"<urn:uuid:6575390f-0b5d-4235-9b0a-86c78e4cc00d>\",\"Content-Length\":\"171679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:783c6211-5aa5-4eb3-ab82-9c16c1381027>\",\"WARC-Concurrent-To\":\"<urn:uuid:67111d06-713c-472b-b377-16b6c7d9ab76>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://gis.stackexchange.com/questions/286179/extract-and-sort-x-coordinates-in-groups-from-a-column-to-a-new-one-in-geopandas\",\"WARC-Payload-Digest\":\"sha1:FDZ5IH7ET6SHW4KNGGIW3B64UGDJM4T4\",\"WARC-Block-Digest\":\"sha1:4ZDRZACIE6O4QUHKHJYSW74T567ILVOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487626465.55_warc_CC-MAIN-20210617011001-20210617041001-00595.warc.gz\"}"} |
https://customessays-writing.com/qa/how-is-depreciation-calculated.html | [
"",
null,
"# How Is Depreciation Calculated?\n\n## How do you calculate allowable depreciation?\n\nDepreciation is calculated each year for tax purposes.\n\nThe first-year depreciation calculation is: Cost of the asset – salvage value divided by years of useful life = adjusted cost.\n\nEach year, use the prior year’s adjusted cost for that year’s calculation..\n\n## Is Depreciation a fixed cost?\n\nDepreciation is one common fixed cost that is recorded as an indirect expense. Companies create a depreciation expense schedule for asset investments with values falling over time. For example, a company might buy machinery for a manufacturing assembly line that is expensed over time using depreciation.\n\n## How is useful life depreciation calculated?\n\nIt is easiest to use a standard useful life for each class of assets. Divide the estimated full useful life (in years) into 1 to arrive at the straight-line depreciation rate. Multiply the depreciation rate by the asset cost (less salvage value)\n\n## Which depreciation method is best?\n\nThe straight-line method is the simplest and most commonly used way to calculate depreciation under generally accepted accounting principles. Subtract the salvage value from the asset’s purchase price, then divide that figure by the projected useful life of the asset.\n\n## On which assets depreciation is allowed?\n\nAs per section 32 of the Income Tax Act, 1961, depreciation is allowed on tangible assets and intangible assets owned, wholly or partly, by the assesse and used for the purposes of business or profession.\n\n## What is tax allowable depreciation?\n\nTax depreciation refers to the depreciation expenses of a business that is an allowable deduction by the IRS. This means that by listing depreciation as an expense on their income tax return in the reporting period, a business can reduce its taxable income.\n\n## What is depreciation example?\n\nIn accounting terms, depreciation is defined as the reduction of recorded cost of a fixed asset in a systematic manner until the value of the asset becomes zero or negligible. An example of fixed assets are buildings, furniture, office equipment, machinery etc..\n\n## What is the formula for depreciation?\n\nThe depreciation rate can also be calculated if the annual depreciation amount is known. The depreciation rate is the annual depreciation amount / total depreciable cost. In this case, the machine has a straight-line depreciation rate of \\$16,000 / \\$80,000 = 20%.\n\n## What are the 3 depreciation methods?\n\nThere are three methods for depreciation: straight line, declining balance, sum-of-the-years’ digits, and units of production.\n\n## What is annual depreciation?\n\nAnnual depreciation is the standard yearly rate at which depreciation is charged to a fixed asset. This rate is consistent from year to year if the straight-line method is used. … The result of annual depreciation is that the book values of fixed assets gradually decline over time.\n\n## Does depreciation reduce profit?\n\nA depreciation expense reduces net income when the asset’s cost is allocated on the income statement. Depreciation is used to account for declines in the value of a fixed asset over time. … As a result, the amount of depreciation expensed reduces the net income of a company.\n\n## Can I change depreciation methods?\n\nTaxpayers can request an automatic method change for depreciation and amortization if the requirements are met to do so. Taxpayers may change from an impermissible method of accounting to a permissible method of accounting or from one permissible method of accounting to another permissible method of accounting.\n\n## Do you pay tax on depreciation?\n\nDepreciation divides the cost associated with the use of an asset over a number of years. … Since depreciation of an asset can be used to deduct ordinary income, any gain from the disposal of the asset must be reported and taxed as ordinary income, rather than the more favorable capital gains tax rate.\n\n## What is allowable depreciation?\n\nAllowed depreciation refers to the depreciation that a business is allowed to deduct from its tax liabilities. … It is because depreciation decreases the ordinary income of the taxpayer (which may be a company or individual) as a cost is incurred.\n\n## What is the simplest depreciation method?\n\nStraight line depreciation is a method by which business owners can stretch the value of an asset over the extent of time that it’s likely to remain useful. It’s the simplest and most commonly used depreciation method when calculating this type of expense on an income statement, and it’s the easiest to learn."
] | [
null,
"https://mc.yandex.ru/watch/69938209",
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