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https://www.vluchtschemas.nl/destinations.html	[ " Flight Schedules countries, flight schedules, flight numbers, flight days.\n\n Main menu Overview all Destinations with airlines and flightday Overview all airlines", null, "Nederlands English Deutsch Español Back\n\n Click on the country or airline for detailed information on flight times. Flight days Summer 2022 Winter 2022-2023 Country Destination Airline Daily Mo Tu We Th Fr Sa Su Daily Mo Tu We Th Fr Sa Su Argentina Buenos Aires (EZE) Air Europa x x British Airways x Iberia x x KLM / Air France x x Lufthansa x x Cordoba (COR) Air Europa x x x x x x Aruba Oranjestad (AUA) KLM / Air France x x TUI fly BE x TUI fly NL x x x x x x Barbados Bridgetown (BGI) KLM / Air France x x x Bolivia Santa Cruz (VVI) Air Europa x x x x Bonaire Kralendijk (BON) Air Belgium x x KLM / Air France x x TUI fly NL x x Brazil Fortaleza (FOR) Air Europa x x Recife (REC) Air Europa x x TAP Portugal x x Rio de Janeiro (GIG) British Airways x x x x x x x x x x KLM / Air France x x Lufthansa x x x x x x x x x x TAP Portugal x x Salvador de Bahia (SSA) Air Europa x x x TAP Portugal x x x x x x Sao Paulo (GRU) Air Europa x x British Airways x x x x x Iberia x x KLM / Air France x x Latam x x Lufthansa x x Swiss Airlines x x TAP Portugal x x Chile Santiago de Chile (SCL) Iberia x x KLM / Air France x x Colombia Bogota (BOG) Air Europa x x Avianca x x Delta Airlines x x Iberia x x KLM / Air France x x Lufthansa x x x x x x x x x x Cartagena (CTG) KLM / Air France x x x x x Cali (CLO) Avianca x x x x x x x x Medellin (MDE) Air Europa x x x x x x x x x Costa Rica Liberia (LIR) KLM / Air France x x x x San Jose (SJO) Aeromexico x x Air Europa x x x British Airways x x x Delta Airlines x x Iberia x x x x x x x x x x x x Iberojet x x x x x KLM / Air France x x Lufthansa x x x x x x Swiss Airlines x x x x x x United Airlines x x Cuba Havana (HAV) Air Caraïbes x x x Air Europa x x x x x x x Condor x x x Iberia x x x x x Iberojet x x x x x x x KLM / Air France x x TUI fly BE x x World2fly x x x x x x Holguin (HOG) Condor x x x Varadero (VRA) Condor x x x x x Eurowings Discover x x Iberojet x TUI fly BE x x x x TUI fly NL x x x x Curacao Willemstad (CUR) Air Belgium x x KLM / Air France x x TUI fly NL x x Dominican Republic Puerto Plata (POP) Condor x x x x Edelweis Air x x Eurowings Discover x Punta Cana (PUJ) Air Belgium x x x x Air Caraïbes x x x x x x x x x Air Europa x x x x x x x Condor x x x x x x x x x Edelweis Air x x x x Eurowings Discover x x x x x x x x x x x Iberojet x x x KLM / Air France x x x x x x x TAP Portugal x x x x x x TUI fly BE x x x x x x TUI fly NL x x x x Wamos Air x x x World2fly x x x x x x Santo Domingo (SDQ) Air Europa x x Condor x x Iberia x x TUI fly BE x x x KLM / Air France x x x x x x x x x Equador Guayaquil (GYE) Air Europa x x x Iberia x x x x x x x KLM / Air France x Quito (UIO) Air Europa x x x x x x Iberia x x x x x KLM / Air France x x Florida Miami (MIA) Air Europa x x x x x x x American Airlines x x British Airways x x Delta Airlines x x Iberia x x KLM / Air France x x Lufthansa x x Swiss Airlines x x TAP Portugal x x TUI fly BE x x United Airlines x x Orlando (MCO) British Airways x x Delta Airlines x x Lufthansa x x United Airlines x x Jamaica Montego Bay (MBJ) British Airways x x Condor x x Eurowings Discover x x x TUI fly BE x x TUI fly NL x x Martinique Fort de France (FDF) Air Belgium Mauricio Mauricio (MRU) Iberojet x Mexico Cancun (CUN) Aeromexico x x Air Europa x x x x x x x x Air Caraïbes x x x Condor x x x x x Delta Airlines x x Eurowings Discover x x x x x Iberojet x x x x x x x x KLM / Air France x x x TAP Portugal x x x x x x x x x x x TUI fly BE x x x x x x TUI fly NL x x x x United Airlines x x Wamos Air x x World2fly x x x x x x Guadalajara (GDL) Aeromexico x x x Los Cabos (SJD) Iberojet x Mexico City (MEX) Aeromexico x x Emirates x x Iberia x x KLM / Air France x x Lufthansa x x United Airlines x x Monterrey (MTY) Aeromexico x x x x x x Panama Panama city (PTY) Air Europa x x x x x x x x Eurowings Discover x x x x x x KLM / Air France x x Iberia x x x x x x Paraguay Asuncion (ASU) Air Europa x x x x x x x x x Peru Lima (LIM) Air Europa x x Iberia x x KLM / Air France x x Puerto Rico San Juan (SJU) Condor Delta Airlines x x Iberia x x x x x x x x x KLM / Air France x Saint Martin Philipsburg (SXM) Air Caraïbes x x x KLM / Air France x x Surinam Paramaribo (PBM) KLM / Air France x x x x x x x x x x Surinam Airways x x x x x x x x Trinidad & Tabago Port of Spain (POS) KLM / Air France x x x Uruguay Montevideo (MVD) Air Europa x x x x x x x x Iberia x x x x x x x Venezuela Caracas (CCS) Air Europa x x x Iberia x x x", null, "Home", null, "", null, "", null, "Menu Flight Schedules", null, "" ]	[ null, "https://www.vluchtschemas.nl/Flaggen/netherlands.gif", null, "https://www.vluchtschemas.nl/figures/boven.gif", null, "https://www.vluchtschemas.nl/Flaggen/netherlands.gif", null, "https://www.vluchtschemas.nl/Flaggen/england.gif", null, "https://www.vluchtschemas.nl/Flaggen/germany.gif", null, "https://www.vluchtschemas.nl/figures/boven.gif", null 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https://igeeksmagz.com/mathsolving-584	[ "# Personal math tutor near me\n\nKeep reading to understand more about Personal math tutor near me and how to use it. Our website can help me with math work.\n\n## The Best Personal math tutor near me\n\nApps can be a great way to help students with their algebra. Let's try the best Personal math tutor near me. Solve for x right triangles by using a Pythagorean formula. This calculator is useful for determining the length of a side of a right triangle, known as the hypotenuse. The Pythagorean relationship between sides x and y is: The ratio or proportion between sides x and y is given by: Substituting this into the above equation gives: or in other words: This can be simplified further as shown below: Therefore, solving for \"x\" right triangles involves applying this formula to any right triangle with lengths equal to 1, 2, and 3. If the hypotenuse (AB) is known then \"x\" can be determined from the equation. For example, if AB = 9 then \"x\" = 9. On the other hand, if AB = 16 then \"x\" = 16. For example, if AB = 12 then \"x\" = 12.\n\nIf an expression cannot be factored, then the process must begin again from scratch. Factoring is the process of breaking down an expression into two separate expressions, one of which has a common factor. . . If an expression cannot be factored, then the process must begin again from scratch. Factoring is done to solve equations when both sides of an equation have a common factor. An easy way to solve this type of equation is by using a combination of variables called a substitution method. A substitution method will take one side of the equation and substitute each variable for its corresponding term on the other side of the equation. The resulting equation will have one fewer term than there are variables in the original equation; this will usually lead to a simplified result with a smaller value for each variable.\n\nGraph equations are a common problem in mathematics. They are used to calculate the position of a point on a graph, for example. The goal is to solve for a specific value in a graph. Here, we will show you how to solve graph equations using Pythagoras' theorem. This method is often referred to as \"dot-to-dot.\" How to solve graph equations using Pythagoras' theorem If you have a triangle with vertices (A, B, and C) and you want to know the length of side AC, then use Pythagoras' theorem to solve for A: [AB=sqrt{AC}] Solution: Substitute the values and simplify: AB=2AC so [A=(-1)^2sqrt{AC}=(-1)sqrt{AC}] Solution: Substitute the values and simplify: A=-(1)AC so [B=(2)^2sqrt{AC}] Solution: Substitute the values and simplify: B=4AC so [C=(-1)^2sqrt{AC}] The rule of Pythagoras states that when solving for distance or ratio between two points, it's best to find their sum or difference first. For example, if you want to know how far 2 cars are apart from each other\n\nSolve with steps is one of the most popular types of puzzles. In this type, you must solve each step in sequence to reach the final solution. Solving with steps puzzles are great for people who want a quick yet challenging brain workout while also providing a sense of accomplishment. If you’re new to solving with steps, start off by simply counting out each step and then visualize yourself making your way through the puzzle. Once you have all the steps down, it will only take a few extra seconds to complete the puzzle. People unfamiliar with solving with steps often end up trying to count out each and every step when they should be focusing on just one or two steps at a time. This can quickly lead to frustration and confusion if you have a lot of information to process at once. Instead, focus on just one or two key steps that you need to remember and try to encode them in your memory as quickly as possible so that you can easily recall them later on.\n\nVery easy to use. Feel like there could be more options for different things. It just seems very minimal and basic, like take photo, solve, explain, done. Which is great, I just want to be able to write equations and such. But overall, def worth 5 stars,", null, "Jayda Kelly\nI love this app. I don't recommend to use this app for everything (since we should use our brain) but I use this at times when I don't understand something. It scans the problem properly, and at the end it has the solution and the strategy. It really helps me with homework. Great job, keep going.", null, "Ysabella Sanchez\nEllipse solver How to do a math problem Solving an algebra problem Solve each system by elimination solver Circle equation solver Free help with math" ]	[ null, "https://igeeksmagz.com/aGU6320a40fd2d67/25388788904_72d2f5ec6f_z-150x150.jpg", null, "https://igeeksmagz.com/aGU6320a40fd2d67/team_3-150x150.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9412072,"math_prob":0.98289526,"size":3892,"snap":"2022-40-2023-06","text_gpt3_token_len":874,"char_repetition_ratio":0.10468107,"word_repetition_ratio":0.117221415,"special_character_ratio":0.22122303,"punctuation_ratio":0.08997429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99823004,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T03:52:00Z\",\"WARC-Record-ID\":\"<urn:uuid:f132db2d-3b2d-4c48-a956-b567c531af34>\",\"Content-Length\":\"25346\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0bb7e3a3-b76c-49bf-ba98-f66627a584d4>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4746630-a7ab-4a29-a1f9-b0d448f97fd9>\",\"WARC-IP-Address\":\"107.167.10.241\",\"WARC-Target-URI\":\"https://igeeksmagz.com/mathsolving-584\",\"WARC-Payload-Digest\":\"sha1:UVBC3LUXF4445FNOC423DXLYNLLRIW6P\",\"WARC-Block-Digest\":\"sha1:X4462UIMGVB3LBUHBGNGPKGBQSPDQF5Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500042.8_warc_CC-MAIN-20230203024018-20230203054018-00649.warc.gz\"}"}
https://forums.atariage.com/topic/350418-sorting-looking-for-a-better-algorithm/page/2/	[ "IGNORED\n\n# Sorting: looking for a better algorithm\n\n## Recommended Posts\n\nThis is a fascinating topic.\n\nSo I looked at my algorithm again, and I was actually mistaken as it's not really a gnome sort either because the latter goes through the entire list once the lowest swap is made in the current loop, whereas in my algorithm I jump directly to the location of the first swap for that loop iteration after the lowest swap for that iteration. Here's an example of a run for a worse case scenario:\n\n5 4 3 2 1\n\n4 5 3 2 1\n\n4 3 5 2 1\n\n3 4 5 2 1\n\n3 4 2 5 1\n\n3 2 4 5 1\n\n2 3 4 5 1\n\n2 3 4 1 5\n\n2 3 1 4 5\n\n2 1 3 4 5\n\n1 2 3 4 5\n\nThat's 10 swaps for that run. Here are the swap numbers for each n from 2 to 7. To clarify, each time a swap situation exists, then 3 additional operations occur in the algorithm, otherwise no additional operations occur. So the number of swaps can be substituted for number of ops = swaps * 3.\n\nn swaps swapsn-swapsn-1\n\n2 1 0\n\n3 3 2\n\n4 6 3\n\n5 10 4\n\n6 15 5\n\n7 21 6\n\nClearly it's a linear progression, so for the worst case scenario the algorithm has a growth of O(n) which is much better than even the combsort or the gnome sort which have a worst case scenario of O(n2)! Am I missing something here?\n\nThe number of swaps for any n here (n>2) is: ( [ i = 3 to n ]SUM ( i - 1 ) ) + 1\n\n•", null, "1\n##### Share on other sites\n\n99'er magazine had an article on sorting methods in TI BASIC. My older brother studied that, implemented them, but it was beyond me then.\n\nLast summer I coded Heapsort in Forth.\n\nTwo interesting properties of Heapsort: First, it doesn't need recursion. It iterates through an array, but skipping forward by powers of 2. Imagine adding up the series 1,2,4,8. How many steps will get you up to N? The answer is log2(N). (More precisely, the sum is N-1.)\n\nIt does this loop on each of N elements. To move X to its destination, it has to compare-swap X with log2(N) other elements.\n\nMaybe that helps reinforce why we say N log N complexity. Outer loop N times, inner loop log2(N) times.\n\nAnd you can see where N log N is faster than Bubble Sort (which might compare every element against each other, NN.)\n\nSecond interesting property, Heapsort always takes N log N steps. No more, no less. I should mention that it takes two passes. Disadvantages.\n\nIn contrast, quicksort has many opportunities to exit early, avoiding a lot of compare-swap. It too is O(N log N)--but imagine that , on average, a lot of elements are left alone after one or two swaps.\n\nConsider that there are special situations where one type of sort is advantageous to your kind of data, or for what you want to get out. And the fastest sort might not be the right choice.\n\nOne situation is that you need a sorted list AND the ability to quickly insert new elements or remove some. While still keeping the list sorted.\n\nAnd not by shoving a giant block of memory one position to the right!\n\nHeapsort , after the first pass, gives you a \"heap\" which is ideal for further insertion/removal. (Guaranteed.) Sometimes that is the special property you need.\n\nBinary trees, in particular red-black trees, are another great choice for insertion/removal.\n\n•", null, "5\n##### Share on other sites\n\n1 hour ago, Vorticon said:\n\nThis is a fascinating topic.\n\nSo I looked at my algorithm again, and I was actually mistaken as it's not really a gnome sort either because the latter goes through the entire list once the lowest swap is made in the current loop, whereas in my algorithm I jump directly to the location of the first swap for that loop iteration after the lowest swap for that iteration. Here's an example of a run for a worse case scenario:\n\n5 4 3 2 1\n\n4 5 3 2 1\n\n4 3 5 2 1\n\n3 4 5 2 1\n\n3 4 2 5 1\n\n3 2 4 5 1\n\n2 3 4 5 1\n\n2 3 4 1 5\n\n2 3 1 4 5\n\n2 1 3 4 5\n\n1 2 3 4 5\n\nThat's 10 swaps for that run. Here are the swap numbers for each n from 2 to 7. To clarify, each time a swap situation exists, then 3 additional operations occur in the algorithm, otherwise no additional operations occur. So the number of swaps can be substituted for number of ops = swaps 3.\n\nn swaps swapsn-swapsn-1\n\n2 1 0\n\n3 3 2\n\n4 6 3\n\n5 10 4\n\n6 15 5\n\n7 21 6\n\nClearly it's a linear progression, so for the worst case scenario the algorithm has a growth of O(n) which is much better than even the combsort or the gnome sort which have a worst case scenario of O(n2)! Am I missing something here?\n\nHow many times do you touch each element in the array?\n\nThat will determine the 'n' factor.\n\nIf your method is O(n) then you only have to scan the array once.\n\nIn other words: a 5 element array would need a 1 to 5 FOR loop run only once to complete the sort.\n\nThat's why my test example had a \"passes\" variable to count how many times I had to run through the array.\n\n•", null, "2\n##### Share on other sites\n\n1 minute ago, TheBF said:\n\nHow many times do you touch each element in the array?\n\nThat will determine the 'n' factor.\n\nIf your method is O(n) then you only have to scan the array once.\n\nIn other words: a 5 element array would need a 1 to 5 FOR loop run only once to complete the sort.\n\nThat's why my test example had a \"passes\" variable to count how many times I had to run through the array.\n\nAnd that's exactly what happens. I have only one loop that runs from 1 to n. Inside each iteration, it backtracks until the previous number is smaller than the current one then picks back up where it was initially in the array. Here's the code:\n\n```5000 REM ARRAY SORTING ROUTINE\n5010 FOR I=1 TO R\n5020 IF I=R THEN RETURN\n5030 POINT=I\n5080 V1=DLIST(POINT,1)::V2=DLIST(POINT+1,1)\n5090 IF V1>V2 THEN DLIST(POINT,1)=V2::DLIST(POINT+1,1)=V1 ELSE 5140\n5100 POINT=POINT-1::IF POINT>=1 THEN 5080 ELSE 5140\n5140 NEXT I```\n\n•", null, "2\n##### Share on other sites\n\n18 minutes ago, Vorticon said:\n\nAnd that's exactly what happens. I have only one loop that runs from 1 to n. Inside each iteration, it backtracks until the previous number is smaller than the current one then picks back up where it was initially in the array. Here's the code:\n\n```5000 REM ARRAY SORTING ROUTINE\n5010 FOR I=1 TO R\n5020 IF I=R THEN RETURN\n5030 POINT=I\n5080 V1=DLIST(POINT,1)::V2=DLIST(POINT+1,1)\n5090 IF V1>V2 THEN DLIST(POINT,1)=V2::DLIST(POINT+1,1)=V1 ELSE 5140\n5100 POINT=POINT-1::IF POINT>=1 THEN 5080 ELSE 5140\n5140 NEXT I```\n\nOk.\n\nSo I guess the only thing to do is time it for an array of size 50 and then an array of size 100 and see if it the time increase is double or what the factor actually turns out to be. You may have a new sort algorithm.", null, "##### Share on other sites\n\nOK so I tested 2 arrays each initialized with the worst case scenario, i.e in descending order. One array is 50 elements and the second is 100 elements.\n\nSort times:\n\n50 elements: 57 seconds\n\n100 elements: 229 seconds\n\nDefinitely not quadratic! I wonder if @TheBF can compare this with his other sorts.\n\nHere's the test code. Timing is manual.\n\n```10 CALL CLEAR::OPTION BASE 1\n20 DIM A(50),B(100)\n30 REM FILL ARRAYS\n35 PRINT \"INITIALIZING ARRAYS....\"::PRINT\n40 FOR I=1 TO 100\n60 B(I)=101-I::IF I<51 THEN A(I)=101-I\n80 NEXT I\n85 PRINT \"READY! PRESS A KEY\"::PRINT\n86 CALL KEY(0,K,S)::IF S=0 THEN 86\n90 CALL SORT(A(),50)\n95 PRINT::PRINT \"PRESS A KEY FOR NEXT ARRAY\"::PRINT\n100 CALL KEY(0,K,S)::IF S=0 THEN 100\n110 CALL SORT(B(),100)\n120 CALL KEY(0,K,S)::IF S<>0 THEN STOP\n\n1000 SUB SORT(C(),N)\n1005 PRINT \"SORTING FOR \";N;\" ELEMENTS\"\n1010 FOR I=1 TO N\n1020 IF I=N THEN 1075\n1030 P=I\n1040 V1=C(P)::V2=C(P+1)\n1050 IF V1>V2 THEN C(P)=V2::C(P+1)=V1 ELSE 1070\n1060 P=P-1::IF P>=1 THEN 1040\n1070 NEXT I\n1075 PRINT \"DONE SORTING!\"\n1080 SUBEND```\n\n•", null, "1\n##### Share on other sites\n\n2 hours ago, TheBF said:\n\nHow many times do you touch each element in the array?\n\nThis is the relevant count. The number of compares, regardless of whether each leads to a swap.\n\nConsider the problem as \"how many steps does it take to sort and prove that it is sorted?\" The proof requires comparing a lot more numbers along the way.\n\nAlso, O() notation is an upper bound. Some algorithms require exactly that number of steps--better algorithms in the same O() class might avoid some of the compares or swaps.\n\nGetting an element to its final spot with fewer swaps.\n\nQuicksort and Heapsort are both O(N log N) but Quicksort avoids many swaps on average. While Heapsort always does the upper limit number of swaps!\n\nWe also worry about computer implementations: which takes longer, compares or swaps? Does the computer stall when numbers are not near neighbors? (Computer waits on slow RAM.)\n\n•", null, "1\n##### Share on other sites\n\n57 minutes ago, Vorticon said:\n\n50 elements: 57 seconds\n\n100 elements: 229 seconds\n\nDefinitely not quadratic! I wonder if @TheBF can compare this with his other sorts.\n\n50^2/57 = 43.859\n\n100^2/229 = 43.668\n\nWhich fits 44 * x^2\n\n•", null, "1\n##### Share on other sites\n\n52 minutes ago, FarmerPotato said:\n\n50^2/57 = 43.859\n\n100^2/229 = 43.668\n\nWhich fits 44 * x^2\n\n1 hour ago, Vorticon said:\n\nOK so I tested 2 arrays each initialized with the worst case scenario, i.e in descending order. One array is 50 elements and the second is 100 elements.\n\nSort times:\n\n50 elements: 57 seconds\n\n100 elements: 229 seconds\n\nDefinitely not quadratic! I wonder if @TheBF can compare this with his other sorts.\n\nHere's the test code. Timing is manual.\n\n```10 CALL CLEAR::OPTION BASE 1\n20 DIM A(50),B(100)\n30 REM FILL ARRAYS\n35 PRINT \"INITIALIZING ARRAYS....\"::PRINT\n40 FOR I=1 TO 100\n60 B(I)=101-I::IF I<51 THEN A(I)=101-I\n80 NEXT I\n85 PRINT \"READY! PRESS A KEY\"::PRINT\n86 CALL KEY(0,K,S)::IF S=0 THEN 86\n90 CALL SORT(A(),50)\n95 PRINT::PRINT \"PRESS A KEY FOR NEXT ARRAY\"::PRINT\n100 CALL KEY(0,K,S)::IF S=0 THEN 100\n110 CALL SORT(B(),100)\n120 CALL KEY(0,K,S)::IF S<>0 THEN STOP\n\n1000 SUB SORT(C(),N)\n1005 PRINT \"SORTING FOR \";N;\" ELEMENTS\"\n1010 FOR I=1 TO N\n1020 IF I=N THEN 1075\n1030 P=I\n1040 V1=C(P)::V2=C(P+1)\n1050 IF V1>V2 THEN C(P)=V2::C(P+1)=V1 ELSE 1070\n1060 P=P-1::IF P>=1 THEN 1040\n1070 NEXT I\n1075 PRINT \"DONE SORTING!\"\n1080 SUBEND```\n\nI had to go out to a rehearsal, but I can try comparing this to combsort which I have in BASIC.\n\n-or- I have a sort workbench in Forth so maybe I can translate your algorithm to Forth and then do comparisons on much bigger data.\n\n•", null, "2\n##### Share on other sites\n\n1 hour ago, FarmerPotato said:\n\n50^2/57 = 43.859\n\n100^2/229 = 43.668\n\nWhich fits 44 * x^2\n\nBut processing time is not. In the context of sort performance, if a sort algorithm is O(n2) shouldn't processing time increase quadratically? This is not the case here.\n\n##### Share on other sites\n\nFor the comparison here is COMBSORT in BASIC using the VORTSORT chassis. (I named it)", null, "For the small amount of extra code COMBSORT is a lot of bang for the buck.\n\n```1 ! VORTSORT chassis with COMBSORT\n2 ! 50 elements 9.5 seconds\n3 ! 100 elements 23 seconds\n10 CALL CLEAR::OPTION BASE 1\n20 DIM A(50),B(100)\n30 REM FILL ARRAYS\n35 PRINT \"INITIALIZING ARRAYS....\"::PRINT\n40 FOR I=1 TO 100\n60 B(I)=101-I::IF I<51 THEN A(I)=101-I\n80 NEXT I\n85 PRINT \"READY! PRESS A KEY\"::PRINT\n86 CALL KEY(0,K,S)::IF S=0 THEN 86\n90 CALL COMBSORT(A(),50)\n95 PRINT::PRINT \"PRESS A KEY FOR NEXT ARRAY\"::PRINT\n100 CALL KEY(0,K,S)::IF S=0 THEN 100\n110 CALL COMBSORT(B(),100)\n120 CALL KEY(0,K,S)::IF S<>0 THEN STOP\n\n2000 SUB COMBSORT(C(),N)\n2100 PRINT \"SORTING FOR \";N;\" ELEMENTS\"\n2420 PASSES=0\n2424 GAP=N\n2440 SORTED=1\n2442 GAP=INT(GAP/1.3)\n2450 FOR I=1 TO N-GAP\n2460 IF C(I)<C(I+GAP) THEN 2530\n2470 rem swap the array items\n2480 TEMP=C(I)\n2490 C(I)=C(I+GAP)\n2500 C(I+GAP)=TEMP\n2520 SORTED=0\n2530 NEXT I\n2540 PASSES=PASSES+1\n2550 REM test if sort is finished\n2570 IF GAP=1 THEN 2600\n2580 GOTO 2440\n2600 SUBEND\n```\n\n•", null, "1\n##### Share on other sites\n\nWow. The difference is impressive... So much for Vortsort", null, "I did track the number of comparisons instead for just swaps for arrays 50,100,150 and 200 elements and I got the following:\n\nClearly not linear. And as @FarmerPotato noted, n2/#comparisons is a constant = 2. So yeah, this sort is an O(n2) type under the worst case scenario.\n\nOh well, I guess fame will have to wait... 😝\n\n•", null, "3\n•", null, "2\n##### Share on other sites\n\n19 hours ago, SteveB said:\n\n...still some Quicksort implementation totally freak out when handed an already sorted list ... The language for the implementation should support recursions, I wouldn't do it in TI BASIC. When implementing it in assembly you need to implement the recursion manually. A whole new science is sorting datasets larger than your RAM, which has been a big issue in the early mainframe world... tape-to-tape sorting.\n\nThe Quicksort I wrote once, when I learned about the algorithm, doesn't \"freak out\", but it does run for a longer time when the list is already sorted.\nYou can \"unwrap\" the recursion even when writing it in e.g. Pascal. They claimed where the article was described that it would be more efficient. It may of course depend on how recursion is implemented in the language.\n\nSince I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either.\n\n•", null, "3\n##### Share on other sites\n\n7 hours ago, Vorticon said:\n\nOh well, I guess fame will have to wait... 😝\n\nThere is a reason that there is not \"THE ONE[tm]\" sorting algorithm. It depends very much on data, hardware, programming language ... you just need to find the niche where VORTSORT outperforms the others.\n\nThe O() logic also looks at infinity ... most data we are dealing with is finite. A simple O(N^3) may outperform a O(nlog(n)) on a small dataset on a real machine. And coding time is also important 🙃\n\n•", null, "4\n##### Share on other sites\n\nSince I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either.\n\nWould you happen to have the code for it? I assume this was done using UCSD Pascal which supports recursion.\n\n##### Share on other sites\n\n16 hours ago, FarmerPotato said:\n\nTwo books that I learned from are Algorithms, by Sedgwick, and Numerical Recipes in C.\n\nSedgwick comes in Pascal, C and other flavors.\n\nNumerical Recipes covers just enough sorting, and is a reference on many other mathematical computing topics.\n\nI have used many references for this and other topics of scientific computing (more for data analysis than sorting) over the years—including Numerical Recipes: The Art of Scientific Computing by Press, Teukolsky, Vetterling, and Flannery and various volumes of The Art of Computer Programming by D. E. Knuth, but probably none more than Numerical Recipes.\n\n...lee\n\n•", null, "4\n##### Share on other sites\n\n•", null, "5\n##### Share on other sites\n\nOn 4/18/2023 at 4:22 AM, SteveB said:\n\nAnd coding time is also important\n\nIndeed. Vortsort can be compressed into just 4 lines of code in XB and performance is likely comparable to Insertion or Gnome sort or perhaps even a little better.\n\n```1000 SUB VORTSORT(C(),N)\n1010 FOR I=1 TO N-1::P=I\n1040 V1=C(P)::V2=C(P+1)::IF V1<V2 THEN 1070 ELSE C(P)=V2::C(P+1)=V1::P=P-1::IF P>=1 THEN 1040\n1070 NEXT I::SUBEND```\n\nI'm keeping it for future use.\n\n•", null, "4\n##### Share on other sites\n\n<pedant>\n\nI think the ELSE 1070 at the end of line 1040 can be omitted - it's redundant.\n\n</pedant>", null, "•", null, "4\n##### Share on other sites\n\nThe Quicksort I wrote once, when I learned about the algorithm, doesn't \"freak out\", but it does run for a longer time when the list is already sorted.\nYou can \"unwrap\" the recursion even when writing it in e.g. Pascal. They claimed where the article was described that it would be more efficient. It may of course depend on how recursion is implemented in the language.\n\nSince I managed to sort a thousand integers in about half a second on the TI 99/4A, I was happy. Due to the limited memory capacity of such a small computer there's normally not the need to sort vast amounts of numbers either.\n\nIt would be very interesting to see how a non-recursive Quicksort performs in TI BASIC. The amazing speed of the sort would be offset by the extra BASIC instructions which are very slow.\n\nBelow some size of the data there would be no benefit versus a simpler algorithm I suspect.\n\nI think I have textbook with the non-recursive version in Pascal. I might take a run at it.\n\n•", null, "2\n##### Share on other sites\n\n18 minutes ago, Willsy said:\n\n<pedant>\n\nI think the ELSE 1070 at the end of line 1040 can be omitted - it's redundant.\n\n</pedant>", null, "Yup. Forgot to remove it when I reformatted the code.\n\n•", null, "3\n##### Share on other sites\n\nPosted (edited)\n3 hours ago, Vorticon said:\n\nWould you happen to have the code for it? I assume this was done using UCSD Pascal which supports recursion.\n\nHere it is. The assembly version, that is. I did start with a Pascal version, but don't have that handy now.\n\nQuicksort\nIt's still kind of recursive in prinicple, but not with a full scale recursive routine call.\n\nBy mistake I wrote somewhere that it uses bubblesort to finish up, but it doesn't. It's insertionsort I use to finalize short lists.\n\nThe program is based on a program written in Algol from the beginning.\n\n•", null, "1\n•", null, "1\n##### Share on other sites\n\n7 hours ago, Vorticon said:\n\n```1000 SUB VORTSORT(C(),N)\n1010 FOR I=1 TO N::IF N=I THEN SUBEXIT ELSE P=I\n1040 V1=C(P)::V2=C(P+1)::IF V1<V2 THEN 1070 ELSE C(P)=V2::C(P+1)=V1::P=P-1::IF P>=1 THEN 1040 ELSE 1070\n1070 NEXT I::SUBEND```\n\nI'm keeping it for future use.\n\nAgree this is valuable because it's so short.\n\nyou can eliminate IF I=N\n\nby saying FOR I=1 TO N-1\n\n•", null, "1\n##### Share on other sites\n\nHere it is. The assembly version, that is. I did start with a Pascal version, but don't have that handy now.\n\nQuicksort\nIt's still kind of recursive in prinicple, but not with a full scale recursive routine call.\n\nBy mistake I wrote somewhere that it uses bubblesort to finish up, but it doesn't. It's insertionsort I use to finalize short lists.\n\nThe program is based on a program written in Algol from the beginning.\n\nAnd here is a version that assembles with the default TI Assembler. You will have to un-comment the BLWP line and the two below.\n\nReadme from my time working on it:\n\nThis code is from a demo in UCSD Pascal by @apperson850 on artiage.com.\n\nThe original has been modified to assemble with the stock TI-99 ASSEMBLER. The\nDATA in the Forth's memory, taking the address and size of an array from Forth's\ndata stack and jumping into QUICKSORT.\n\n1000 items can be sorted in under 2 seconds on a stock console.\n\nSpoiler\n``` TITL \"QUICKSORT FOR INTEGERS\"\n\n* Procedure sorting integers in ascending order\n* Called from Pascal host by QSORT(A,N)\n Declared as PROCEDURE QSORT(VAR A:VECTOR* N:INTEGER)* EXTERNAL\n Vector should be an array [..] of integer\n\n Author: @APERSON850 on Atariage.com\n\n* Modified for TI Assembler, to be called from Camel99 Forth\n* 2022 Brian Fox\n\n--------------\n Workspace registers for subroutine at START\n\n R0 Array base pointer\n* R1 Array end pointer\n* R2 L\n* R3 R\n* R4 I\n* R5 J\n* R6 KEY\n* R7 Temporary\n* R8 Temporary\n* R9 Subfile stack pointer\n* R10 Main stackpointer\n* R11 Pascal return address (P-system WS)\n* R12 Not used\n* R13 Calling program's Workspace Pointer\n* R14 Calling program's Program Counter\n* R15 Calling program's Status Register\n-----------------\n\n The actual vector in the Pascal-program could be indexed [1..n]\n* This routine assumes only that n indicates the number of elements, not the\n* last array index.\n\nDEF QSORT\n\nLIMIT EQU 16 Quick- or Insertionsort limit\n\n * Don't need this for Forth *\n BLWP @QSORT\n* AI R10,4 Simulate pop of two words\n B R11 Back to Pascal host\n\nQSORT DATA SORTWS,START Transfer vector for Quicksort\n\nSTART MOV @12(R13),R10 Get stackpointer (R6) from Forth's WP->R10\nLI R9,ENDSTK SUBFILE STACKPOINTER\nMOV R10+,R1 * POP PARAMETER N\nMOV R10+,R0 POP ARRAY POINTER\nDEC R1\nSLA R1,1\nA R0,R1 CALCULATE ARRAY ENDPOINT\n\nMOV R0,R2 L:=1\nMOV R1,R3 R:=N\nMOV R1,R7\nS R0,R7\nCI R7,LIMIT\nJLE INSERT FIGURE OUT IF QUICKSORT IS NEEDED\n\nMNLOOP MOV R2,R7\nSRL R7,1\nMOV R3,R8\nSRL R8,1\nA R8,R7\nANDI R7,>FFFE R7:=INT((L+R)/2)\nMOV R7,R8\nMOV @2(R2),R7\nMOV R8,@2(R2) A[(L+R)/2]:=:A[L+1]\n\nC @2(R2),R3\nJLT NOSWP1\nMOV @2(R2),R8\nMOV R3,@2(R2)\nMOV R8,R3 A[L+1]:=:A[R]\n\nNOSWP1 C R2,R3\nJLT NOSWP2\nMOV R2,R8\nMOV R3,R2\nMOV R8,R3 A[L]:=:A[R]\n\nNOSWP2 C @2(R2),R2\nJLT NOSWP3\nMOV @2(R2),R8\nMOV R2,@2(R2)\nMOV R8,R2 A[L+1]:=:A[L]\n\nNOSWP3 MOV R2,R4\nINCT R4 I:=L+1\nMOV R3,R5 J:=R\nMOV R2,R6 KEY:=A[L]\nJMP INCLOP\n\nINRLOP MOV R4,R8 LOOP UNWRAPPING\nMOV R5,R4\nMOV R8,R5 A[I]:=:A[J]\n\nINCLOP INCT R4 I:=I+1\nC R4,R6\nJLT INCLOP A[I]<KEY\n\nDECLOP DECT R5 J:=J-1\nC R5,R6\nJGT DECLOP A[J]>KEY\n\nC R4,R5\nJLE INRLOP IF I<=J THEN CONTINUE\n\nOUT MOV R2,R8\nMOV R5,R2\nMOV R8,R5 A[L]:=:A[J]\n\nDEL1 MOV R5,R7 Quicksort subfiles?\nS R2,R7 R7:=J-L\nMOV R3,R8\nS R4,R8\nINCT R8 R8:=R-I+1\nCI R7,LIMIT\nJH DEL2\nCI R8,LIMIT\nJH DEL2\n\nCI R9,ENDSTK LVL=0?\nJEQ INSERT No more Quicksorting at all?\n\nMOV R9+,R2 POP L\nMOV R9+,R3 POP R\nJMP MNLOOP\n\nDEL2 C R7,R8 Determine what is small and large subfile\nJL ELSE2\n\nMOV R2,@LSFL\nMOV R5,@LSFR\nDECT @LSFR\nMOV R4,@SSFL\nMOV R3,@SSFR\nJMP DEL3\n\nELSE2 MOV R4,@LSFL\nMOV R3,@LSFR\nMOV R2,@SSFL\nMOV R5,@SSFR\nDECT @SSFR\n\nDEL3 CI R7,LIMIT Is small subfile big enough to be sorted by\nJLE THEN3 Quicksort?\nCI R8,LIMIT\nJH ELSE3\n\nTHEN3 MOV @LSFL,R2 Don't Quicksort small subfile, only large\nMOV @LSFR,R3\nJMP MNLOOP\n\nELSE3 DECT R9 Stack large subfile\nMOV @LSFR,R9 PUSH R\nDECT R9\nMOV @LSFL,R9 PUSH L\nMOV @SSFL,R2 Sort small subfile\nMOV @SSFR,R3\nJMP MNLOOP\n\n\n Insertion Sort finishing up\n\nINSERT MOV R1,R4\nDECT R4 I:=N-1\nC R4,R0\nJL LEAVE Check if any looping at al\n\nFORI C @2(R4),R4\nJGT NEXTI If next is greater than this, it's OK\n\nMOV R4,R6 KEY:=A[I]\nMOV R4,R5\nINCT R5 J:=I+1\n\nWHILE MOV R5,@-2(R5) A[J-1]:=A[J]\nINCT R5 J:=J+1\nC R5,R1\nJH ENDWHL J>N?\nC R5,R6 A[J]<KEY?\nJLT WHILE\nENDWHL MOV R6,@-2(R5) A[J-1]:=KEY\nNEXTI DECT R4\nC R4,R0 Check if passed array base point\nJHE FORI\n\nLEAVE RTWP * Return Forth workspace\n* AI R6,4 * remove parameters from Forth stack\nB @R10 * Branch to Forth interpreter\n\n--------------\n DATA AREA\n\nSORTWS BSS >20 Workspace for sorting routine\nSUBSTK BSS >40 Internal subfile stack\nENDSTK EQU SUBSTK+>40 End of that stack\n\n variable used by quicksort\nLSFL DATA 0 Large SubFile Left pointer\nLSFR DATA 0 Large SubFile Right pointer\nSSFL DATA 0 Small SubFile Left pointer\nSSFR DATA 0 Small SubFile Right pointer\n\nEND\n```\n\n•", null, "1\n##### Share on other sites\n\n1 hour ago, FarmerPotato said:\n\nAgree this is valuable because it's so short.\n\nyou can eliminate IF I=N\n\nby saying FOR I=1 TO N-1\n\nYes! Every little trim helps.\n\n•", null, "1\n\n## Join the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.\nNote: Your post will require moderator approval before it will be visible.", null, "× Pasted as rich text. Paste as plain text instead\n\nOnly 75 emoji are allowed." ]	[ null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/emoticons/atariage_icon_smile.gif", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/emoticons/atariage_icon_smile.gif", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/emoticons/atariage_icon_lol.gif", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_haha.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/emoticons/atariage_icon_winking.gif", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/emoticons/atariage_icon_winking.gif", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_thanks.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/reactions/react_like.png", null, "https://content.invisioncic.com/r322239/set_resources_6/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90156996,"math_prob":0.825864,"size":1545,"snap":"2023-14-2023-23","text_gpt3_token_len":469,"char_repetition_ratio":0.11161584,"word_repetition_ratio":0.05105105,"special_character_ratio":0.30938512,"punctuation_ratio":0.06741573,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746126,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,4,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T17:03:50Z\",\"WARC-Record-ID\":\"<urn:uuid:5222e8bc-c36a-4dfd-a9fa-1f824f684bcc>\",\"Content-Length\":\"449330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b80b8f65-067b-4040-a60f-d71500b38cfe>\",\"WARC-Concurrent-To\":\"<urn:uuid:0829e892-89d9-434a-b6d8-79e89e9f93ef>\",\"WARC-IP-Address\":\"13.32.208.21\",\"WARC-Target-URI\":\"https://forums.atariage.com/topic/350418-sorting-looking-for-a-better-algorithm/page/2/\",\"WARC-Payload-Digest\":\"sha1:63MBQPRBFLUNLVNMH5FHOKPUCOAMRKUH\",\"WARC-Block-Digest\":\"sha1:BDVKGQETIVWQS6G2CJMZWY3ZARHGZ4HC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652149.61_warc_CC-MAIN-20230605153700-20230605183700-00621.warc.gz\"}"}
https://whatmaster.com/what-is-a-geometric-figure/	[ "CONCEPTS\n\n# What is a geometric figure?\n\nWe explain what geometric figures are and the ways in which they can be classified. In addition, some examples of these figures.\n\n1. ### What is a geometric figure?\n\nA geometric figure is the visual and functional representation of a non-empty and closed set of points in a geometric plane. That is, figures that delimit flat surfaces through a set of lines (sides) that join their points in a specific way. Depending on the order and number of these lines we will talk about one figure or another.\n\nGeometric figures are the work of geometry , a branch of mathematics that studies representational planes and the relationships between the forms we can imagine in them. These are, therefore, abstract objects, according to which our perspective and our way of spatially understanding the universe around us are determined.\n\nYou can classify the geometric figures according to their shape and number of sides, but also based on the number of dimensions represented, being able to talk about:\n\n• Dimensional figures (0 dimensions) . It basically refers to the point.\n• Linear figures (1 dimension) . These are the lines and the curves, that is, lines with some orientation and determined path.\n• Flat figures (2 dimensions) . Polygons, planes and surfaces, which lack depth but have a measurable length and width.\n• Volumetric figures (3 dimensions) . Three-dimensional figures add depth and perspective to the matter, and can be considered geometric bodies, such as polyhedra and solids in revolution.\n• N-dimensional figures (n-dimensions) . These are theoretical abstractions endowed with n appreciable dimensions.\n\nWe should note that to define the geometric figures , abstractions such as the point, the line and the plane are often used , which are in turn considered geometry figures.\n\n1. ### Examples of geometric figures", null, "Squares necessarily have four equal sides.\n\nSome examples of geometric figures are:\n\n• Triangles . Flat figures characterized by having three sides, that is, three lines in contact forming three vertices. Depending on the type of angle they build, they can be equilateral triangles (three equal sides), isosceles (two equal and one different) or scalenes (all unequal).\n• Squares . These flat figures are always identical in proportion but not in size, having four sides necessarily of the same length. Its four angles will then be right angles (90 °).\n• Rhombus s . Similar to the square, they have four identical sides in contact, but none constitute right angles, but acute and two obtuse.\n• Circumferences . It is a flat and closed curve on itself, in which any chosen point of the line is at the same identical distance from the center (or axis). It could be called a perfect circle.\n• Ellipses . Closed curves similar to the circumference, but with two axes or centers instead of one, generating a flattened or elongated spheroid, depending on whether it revolves around its minor or major axis, respectively.\n• Pyramids . Three-dimensional geometric bodies formed by a quadrangular base and four isosceles triangles that act as sides." ]	[ null, "https://whatmaster.com/wp-content/uploads/2020/02/03-193.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92865074,"math_prob":0.9822912,"size":3035,"snap":"2021-21-2021-25","text_gpt3_token_len":618,"char_repetition_ratio":0.14879578,"word_repetition_ratio":0.004008016,"special_character_ratio":0.20296541,"punctuation_ratio":0.122486286,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962667,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T21:19:15Z\",\"WARC-Record-ID\":\"<urn:uuid:89cfc51a-5ff0-4398-a601-617a56a97114>\",\"Content-Length\":\"131478\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9fbbb2d1-a94c-44c4-9e85-a6f4ae4ddbc2>\",\"WARC-Concurrent-To\":\"<urn:uuid:5885d43f-8e3b-41bb-8e45-f4f92d66a72d>\",\"WARC-IP-Address\":\"74.208.236.160\",\"WARC-Target-URI\":\"https://whatmaster.com/what-is-a-geometric-figure/\",\"WARC-Payload-Digest\":\"sha1:DC2QR5ACG3IW2NUUWRKUL6TW25K7RUEQ\",\"WARC-Block-Digest\":\"sha1:JB6PNNDB66LQLEDYDM4GBOJWVKMCJUNI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487586390.4_warc_CC-MAIN-20210612193058-20210612223058-00083.warc.gz\"}"}
https://www.turito.com/ask-a-doubt/Mathematics-lin-creates-a-pattern-using-the-white-and-black-tiles-the-number-of-white-tiles-present-in-the-next-sha-qe11e76	[ "Mathematics\nEasy\n\nQuestion\n\n# Lin creates a pattern using the white and black tiles.", null, "The number of white tiles present in the next shape of Lin’s pattern is _____.\n\n## 15 16 20 25", null, "Hint:\n\n## The correct answer is: 16\n\n### The number of tiles is increasing with the number of shape. The rule used to create the shape is the number of tiles is given by a square. And the number is increasing by 1.The first shape requires 12 = 1 white tiles.The second shape requires 22 = 4 white tiles.The third shape requires 32 = 9 white tiles.So, the next figure will be square of next number that is 4.Therefore, 4th shape requires 42 = 16 white tiles.\n\nWe have to be careful about the sequence of the shape. We have to increase the number of tiles in proper order.\n\n### Related Questions to study", null, "", null, "#### With Turito Foundation.", null, "#### Get an Expert Advice From Turito.", null, "", null, "" ]	[ null, "https://mycourses.turito.com/tokenpluginfile.php/c161933dbfaab094c54655ab71e9b8f0/1/question/questiontext/888095/1/1207599/Picture65.png", null, "https://www.turito.com/images/idea_green.svg", null, "https://www.turito.com/_next/image", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9079809,"math_prob":0.95818377,"size":1042,"snap":"2023-40-2023-50","text_gpt3_token_len":257,"char_repetition_ratio":0.20905587,"word_repetition_ratio":0.029850746,"special_character_ratio":0.2581574,"punctuation_ratio":0.09954751,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9653665,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T13:42:58Z\",\"WARC-Record-ID\":\"<urn:uuid:481f8b91-17c3-4f9c-af19-180d59037417>\",\"Content-Length\":\"74648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51ea5497-fba3-4482-b8b7-8193e05e61c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:da5d41ae-0f64-4546-aadf-59c48497e52c>\",\"WARC-IP-Address\":\"3.108.140.208\",\"WARC-Target-URI\":\"https://www.turito.com/ask-a-doubt/Mathematics-lin-creates-a-pattern-using-the-white-and-black-tiles-the-number-of-white-tiles-present-in-the-next-sha-qe11e76\",\"WARC-Payload-Digest\":\"sha1:XYTWDTMO7J3YDT3PXK5Y62TE4DDEK7CZ\",\"WARC-Block-Digest\":\"sha1:LTL7AMM3YZ4KDXCXVTCWNI4SW4QZJYJV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00293.warc.gz\"}"}
https://coding-gym.org/tags/subarray/	[ "## eTonno influencers accuracy\n\nSolutions A brute-force solution (considering the set of all possible transactions) is clearly too expensive. To grasp the optimal solution, consider this example: 1 1, 7, 2, 3, 6, 7, 6, 7 To obtain the maximum profit, the - possibly - most intuitive set of transactions is: 1 (buy=1, sell=7) + (buy=2, sell=7) + (buy=6, sell=7) = 6 + 5 + 1 = 12 Look at the increasing sub-series 2, 3, 6, 7. [Read More]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7413337,"math_prob":0.9823136,"size":431,"snap":"2022-40-2023-06","text_gpt3_token_len":144,"char_repetition_ratio":0.12412178,"word_repetition_ratio":0.0,"special_character_ratio":0.34106728,"punctuation_ratio":0.19587629,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9847828,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T21:35:53Z\",\"WARC-Record-ID\":\"<urn:uuid:1e514eae-0bf8-4d4b-a3e3-3ce0c134b81c>\",\"Content-Length\":\"8702\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ff10defc-c4c4-4b55-9b1f-a41abc7c8d16>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0576060-bf53-44b6-bb2d-0b548d19c089>\",\"WARC-IP-Address\":\"35.185.44.232\",\"WARC-Target-URI\":\"https://coding-gym.org/tags/subarray/\",\"WARC-Payload-Digest\":\"sha1:5MJOTZCQT7RCAYVOUOFA5PZMVYYT433A\",\"WARC-Block-Digest\":\"sha1:J7HZZRU5YBCSCFJKO4IDNPW76L45FTG3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337668.62_warc_CC-MAIN-20221005203530-20221005233530-00337.warc.gz\"}"}
https://www.baeldung.com/cs/graphs-max-number-of-edges	[ "## 1. Overview\n\nIn this tutorial, we’ll discuss how to calculate the maximum number of edges in a directed graph.\n\n## 2. A Simple Definition of Directed Graph\n\nIn graph theory, graphs can be categorized generally as a directed or an undirected graph. In this section, we’ll focus our discussion on a directed graph.\n\nLet’s start with a simple definition. A graph", null, "is a directed graph if all the edges in the graph", null, "have direction. The vertices and edges in", null, "should be connected, and all the edges are directed from one specific vertex to another.\n\nThe main difference between a directed and an undirected graph is reachability. Let’s explain this statement with an example:", null, "We’ve taken a graph", null, ". The vertex set", null, "contains five vertices:", null, ". The edge set", null, "of", null, "contains six edges:", null, ".\n\nNow as we discussed, in a directed graph all the edges have a specific direction. For example, edge", null, "can only go from vertex", null, "to", null, ". Unlike an undirected graph, now we can’t reach the vertex", null, "from", null, "via the edge", null, ".\n\nHence in a directed graph, reachability is limited and a user can specify the directions of the edges as per the requirement.\n\n## 3. When a Directed Graph Contains Maximum Edges?\n\nIn this section, we’ll discuss some conditions that a directed graph needs to hold in order to contain the maximum number of edges.\n\nFirstly, there should be at most one edge from a specific vertex to another vertex. This ensures all the vertices are connected and hence the graph contains the maximum number of edges. In short, a directed graph needs to be a complete graph in order to contain the maximum number of edges.\n\nIn graph theory, there are many variants of a directed graph. To make it simple, we’re considering a standard directed graph. So in our directed graph, we’ll not consider any self-loops or parallel edges.\n\n## 4. A General Formula for Maximum Edges Calculation\n\nIn this section, we’ll present a general formula to calculate the maximum number of edges that a directed graph can contain.\n\nLet’s assume an undirected graph with", null, "vertices. Further, we’re also assuming that the graph has a maximum number of edges. In such a case, from the starting vertex, we can draw", null, "edges in the graph. Continuing this way, from the next vertex we can draw", null, "edges.\n\nHence the maximum number of edges in an undirected graph is:", null, "Now, in an undirected graph, all the edges are bidirectional. We can convert an undirected graph into a directed graph by replacing each edge with two directed edges. Hence the revised formula for the maximum number of edges in a directed graph:", null, "## 5. An Example\n\nIn this section, we’ll take some directed graph and calculate the maximum number of edges according to the formula we derived:", null, "Now, we already discussed some conditions and assumptions for a directed graph such that it contains the maximum number of edges. Let’s verify first whether this graph contains the maximum number of edges or not.\n\nFirst, let’s check if it is a complete directed graph or not. In a complete directed graph, all the vertices are reachable from one another. In the above graph, we can see all the vertices are reachable from one another.\n\nSecondly, in our directed graph, there shouldn’t be any parallel edges or self-loop. Our example directed graph satisfies this condition too. Now let’s proceed with the edge calculation. Note that each edge here is bidirectional. Hence, each edge is counted as two independent directed edges.\n\nThe maximum number of edges =", null, "and the above graph has all the edges it can contain.\n\nLet’s take another graph:", null, "Does this graph contain the maximum number of edges? Let’s check.\n\nTo verify this, we need to check if all the vertices can reach from one another. If we take a deep loop in the graph, we can see a lot of vertices can’t reach each other via a single edge.\n\nTherefore, we can conclude that the given directed graph doesn’t contain the maximum number of edges. According to our formula, this graph has the capacity to contain maximum of", null, "edges. But the graph has 16 edges in this example.\n\n## 6. Conclusion\n\nIn this tutorial, we’ve discussed how to calculate the maximum number of edges in a directed graph.\n\nWe’ve presented a general formula for calculating the maximum number of edges in a directed graph and verified our formula with the help of a couple of examples." ]	[ null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-eaa54ad1d5903544229dbbebdf92afbd_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-1e40206e25474f738eeb7ca968031abf_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-1e40206e25474f738eeb7ca968031abf_l3.svg", null, "https://www.baeldung.com/wp-content/uploads/sites/4/2020/06/graph-1.png", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-e3192da0128dfabe5fce82166bdc373c_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-54e215a7a583b4f357a5a627420bcf2f_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-505498cd4401f369536779b85c9f4206_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-638a7387bd72763290cc777a9b509c38_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-e3192da0128dfabe5fce82166bdc373c_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-3ffabe3c897ee301a3d3ad0b12041d64_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-ac91793e30799352150fdae8a6ae5d48_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-c13da9eae23428ebdd0fed62ec5a2124_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-93741ff3d67e852e96df8314f03552f6_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-c13da9eae23428ebdd0fed62ec5a2124_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-93741ff3d67e852e96df8314f03552f6_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-ac91793e30799352150fdae8a6ae5d48_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-7354bae77b50b7d1faed3e8ea7a3511a_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-b8d591aa3d3a6700d7bc61014596c9fb_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-91c3bf97740f19afb5e1c465cfc5510d_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-03fd9865d9c031490aea97d5172a89da_l3.svg", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-833956ebdd46ec4302a307b272da5b79_l3.svg", null, "https://www.baeldung.com/wp-content/uploads/sites/4/2020/06/graph-2.png", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-be79e9c203cad64e02e64c142b220f53_l3.svg", null, "https://www.baeldung.com/wp-content/uploads/sites/4/2020/06/Capture.png", null, "https://www.baeldung.com/wp-content/ql-cache/quicklatex.com-76c2f9fc5a2660966460ba939a42d02c_l3.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9472039,"math_prob":0.983411,"size":4249,"snap":"2023-40-2023-50","text_gpt3_token_len":881,"char_repetition_ratio":0.23674911,"word_repetition_ratio":0.08819538,"special_character_ratio":0.20475405,"punctuation_ratio":0.11214953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996699,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,4,null,null,null,null,null,2,null,null,null,null,null,2,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,6,null,2,null,2,null,2,null,4,null,2,null,4,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T18:13:10Z\",\"WARC-Record-ID\":\"<urn:uuid:e6b5b6e7-d5ea-433b-af73-3b18b697145a>\",\"Content-Length\":\"73142\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30229368-bd6b-44f4-b69c-3e3fa0ce746a>\",\"WARC-Concurrent-To\":\"<urn:uuid:006269ea-de9e-4b24-81b9-a190dea3bdc6>\",\"WARC-IP-Address\":\"172.66.40.248\",\"WARC-Target-URI\":\"https://www.baeldung.com/cs/graphs-max-number-of-edges\",\"WARC-Payload-Digest\":\"sha1:7GMKAPHTNUVWE3CUDRBBKBHWTI7PH6S7\",\"WARC-Block-Digest\":\"sha1:APVAM7ZBDZHDJQU5OPXKHDH4DTFOYKEE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511002.91_warc_CC-MAIN-20231002164819-20231002194819-00773.warc.gz\"}"}
https://quantitysystem.wordpress.com/2010/04/15/work-and-torque-solution/	[ "I am really sorry\n\nI didn’t notice that I haven’t posted how did I solve the dilemma of Torque and Work Problem\n\nhowever the solution was there in my discussion of quantity System http://quantitysystem.codeplex.com/Thread/View.aspx?ThreadId=23672\n\nand also on this post of my personal blog http://blog.lostparticles.net/?p=28\n\n# Problem Core: Length\n\nSimply I defined TWO Length Types\n\n## Normal Length (NL)\n\nThis is the normal length that we refer to it in our daily life\n\nThis is the radius length that have an origin point in center of circle\n\n## Quantities\n\nWork: Force * Normal Length\n\nAngle: Normal Length / Radial Length\n\nIn quantity system I made the L dimension as NL+RL\n\nCAN YOU SEE ANGLE\n\nI made it explicitly a quantity that is NOT dimensionless\n\nand THIS SOLVED ALL my problems of this Problem\n\nI won’t argue much let us test 🙂\n\nTorque * Angle = Work\n\nFRL (NL/RL) = F * NL <== see what I mean\n\nTorque * Angular Speed = Power\n\nFRL (NL/RLT) = FNL/T <== where T is the time.\n\nso you may wonder about Angle and Solid Angle\n\nin SI they are all dimensionless but in Quantity System YOU CAN’T consider them like this\n\n(By the way I’ve break the checking for these two quantities to be summable with dimensionless numbers – just to keep the fundamentals as it is although I am not convinced and I may remove it in future but damn it I need support from any physics guy)\n\nANGLE : NL/RL\n\nSolid Angle: NL^2/RL^2 because its area over area\n\nFrequency = 1/T\n\nAngular Velocity = (NL/RL)/T\n\ndo you remember the conversion between RPM to frequency\n\nRPM (Revolution Per Minute) is a Angle / Time\n\nlets test the law\n\nOmega = 2pifrequency\n\npi: radian value which is NL/RL\n\nOmega = (NL/RL) * 1/T = (NL/RL)/T which angular velocity\n\n# Discoveries\n\n## PI value\n\nPI is ratio of any circle‘s circumference to its diameter WHICH MEANS (NL/RL)\n\nthis is the same value as the ratio of a circle’s area to the square of its radius WHICH MEANS (NL^2/RL^2)\n\nwhich corresponds to radian unit and stradian unit for angle and solid angle quantities.\n\n## Reynolds Number\n\nI am not holding back ( Reynolds number is a dimensionless number that measure inertial forces to viscous forces)\n\nWHY we always differentiating between Flow in Pipes and Flow on Flat plate\n\nlet’s see with normal length in quantity system\n\nQs> rho = 3[Density]\nDensity: 3 <kg/m^3>\nQs> v=0.5<m/s>\nSpeed: 0.5 <m/s>\nQs> l=4<m>\nLength: 4 m\nQs> mue = 2<Pa.s>\nViscosity: 2 <Pa.s>\nQs> rhovl/mue\nDimensionlessQuantity: 3 <kg/m.s^2.Pa>\n\nit shows it is a dimensionless quantity\n\nok let us force my theory about flow in pipes\n\nthe pipe have a diameter and Reynolds number is calculated by rhovDIAMETER/viscosity\n\nso let me add d as a diameter and solve again\n\nQs> d=0.5<m!>\n\nQs> rhovd/mue\n\nDerivedQuantity: 0.375 <kg/m.s^2.Pa>\n\nNOTE: Adding ‘!’ after the length unit will mark the quantity as Radial Length quantity.\n\nERROR it shouldn’t be DerivedQuantity at all it should be Dimensionlesss\n\nso there is another term that should be fixed. Do you know which term ???\n\nlets try the velocity with vr=0.5<m!/s>\n\nQs> vr=0.5<m!/s>\n\nDerivedQuantity: 0.5 <m/s>\n\nQs> rhovrd/mue\n\nDerivedQuantity: 0.375 <kg/m.s^2.Pa>\n\nALSO ERROR\n\nok let us try the density\n\nQs> rhor = 3<kg/m!^3>\n\nDerivedQuantity: 3 <kg/m^3>\n\nQs> rhorvd/mue\n\nSolidAngle: 0.375 <kg/m.s^2.Pa>\n\nCAN YOU SEE THE SOLID ANGLE Quantity\n\ndo you remember the above argue about Solid Angle is dimensionless number\n\ncan I pretend now that reynolds number for pipes is CORRECT ???\n\nthe new density which is kg/m!^3\n\nThis is driving me nuts\n\n## Pump Affinity Laws\n\nAlso pump equations led to the same SolidAngle Quantity not DimensionLess one\n\n# Epilogue\n\nI’ve said what I’ve discovered till now about this problem I hope that may be someday someone explain to me or convince me that Angle is really a Dimensionless number after all of this and that my assumption about Torque and Work is Wrong.\n\nFrom → Uncategorized\n\nOne Comment" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8694817,"math_prob":0.95537186,"size":3733,"snap":"2020-45-2020-50","text_gpt3_token_len":1002,"char_repetition_ratio":0.11558058,"word_repetition_ratio":0.0,"special_character_ratio":0.23921779,"punctuation_ratio":0.070866145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99598795,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T14:29:32Z\",\"WARC-Record-ID\":\"<urn:uuid:52f19000-d217-499f-9877-a2e3fc8fc7cc>\",\"Content-Length\":\"66820\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d829dfbe-9bb7-498b-92c1-20ce43668ff3>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8d27c05-fab5-4c0c-a081-2e5e5d45131c>\",\"WARC-IP-Address\":\"192.0.78.12\",\"WARC-Target-URI\":\"https://quantitysystem.wordpress.com/2010/04/15/work-and-torque-solution/\",\"WARC-Payload-Digest\":\"sha1:HATUJEK376C4S5N6RRC5DWKW7B2GSLIP\",\"WARC-Block-Digest\":\"sha1:3FXA3PYUEWXXK7UQZ6A2E7XZHUSWOXWM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141188800.15_warc_CC-MAIN-20201126142720-20201126172720-00506.warc.gz\"}"}
https://members.kidpid.com/ask/reply/49836/	[ "community around the world.\n\nActivity Discussion Math MATHS Reply To: MATHS\n\n• ### Aseem\n\nMember\nJune 4, 2023 at 8:44 am\n2\n\nTo simplify the expression (2x + 3y) – (4x – 2y), we need to distribute the negative sign to the terms inside the parentheses. Here are the steps to simplify it:\n\nStep 1: Distribute the negative sign to the terms inside the second set of parentheses:\n\n(2x + 3y) – 4x + 2y.\n\nStep 2: Combine like terms. In this case, we can combine the x terms and the y terms separately:\n\n2x – 4x + 3y + 2y.\n\nStep 3: Simplify the x terms by combining the coefficients:\n\n(2 – 4)x + 3y + 2y.\n\nStep 4: Simplify the y terms by combining the coefficients:\n\n-2x + 5y.\n\nTherefore, the simplified expression for (2x + 3y) – (4x – 2y) is -2x + 5y.\n\n+" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7227888,"math_prob":0.9954003,"size":706,"snap":"2023-40-2023-50","text_gpt3_token_len":237,"char_repetition_ratio":0.14814815,"word_repetition_ratio":0.11510792,"special_character_ratio":0.3371105,"punctuation_ratio":0.1632653,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9945438,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T19:20:17Z\",\"WARC-Record-ID\":\"<urn:uuid:7a47393d-c72b-4c97-b303-c336c81632fe>\",\"Content-Length\":\"112330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3bd08c6-6235-48e3-afb5-c31a66e199a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9439a94-9403-422f-9bc0-31627e51b529>\",\"WARC-IP-Address\":\"68.178.152.103\",\"WARC-Target-URI\":\"https://members.kidpid.com/ask/reply/49836/\",\"WARC-Payload-Digest\":\"sha1:GFOFEYKCRVV3VILOL6BC6IQ4FW4JUFKT\",\"WARC-Block-Digest\":\"sha1:TZQOYODACFWQRO6HBKEGVD3DDII5NUS5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511406.34_warc_CC-MAIN-20231004184208-20231004214208-00070.warc.gz\"}"}
https://electronics.stackexchange.com/questions/538187/how-are-two-definitions-of-baud-rate-same	[ "# How are two definitions of baud rate same?\n\nThe bit rate V baud rate has been a confusing topic for me lately. I feel like I am clear about the difference between these two.\n\nBut going through the definitions of baud rate, I find two different definitions in use:\n\n1. A baud rate is the number of times a signal in a communications channel changes state or varies. For example, a 2400 baud rate means that the channel can change states up to 2400 times per second. The term “change state” means that it can change from 0 to 1 or from 1 to 0 up to X (in this case, 2400) times per second. (This is the definition from my teacher's notes.)\n1. Baud rate as the no. of symbols transmitted per second. Consider a symbol \"A\" which is encoded as \"0100 0001\" in ASCII format. If I send this symbol within 1 second, I have bitrate of 8 bits per second.\n\nFor baud rate: According to first definition, baud rate is 1 baud per second.\n\nBut according to second definition, the baud rate is 3 bauds per second since the signal changes its state three times (0 -> 1 -> 00000 -> 1).\n\nSo how do these two definitions link together? I can't see any match between these two definitions with \"my\" level of knowledge. Thank you!!\n\nAnd how is the calculation done in following image? I understand the calculation for the bit rate but how is baud rate calculated?", null, "• The letter 'A' is not a symbol in this context. In this context, a symbol is a quantum of information that can be transmitted on the communication channel. Dec 19 '20 at 15:20\n• @ElliotAlderson would you elaborate please? Dec 19 '20 at 15:24\n• Sounds like a non-standard definition in the second one. \"Symbol\" sounds vague. I agree with Elliot's comment. It should be referring to the least indivisible information or something in the context you read. The first one is standard definition though. Dec 19 '20 at 15:49\n• @Mitu Raj The terms symbol and symbol rate are not vague and have been in use for at least 40 years. Dec 19 '20 at 16:37\n• I would be curious to see any reliable source which correctly defines what is \"Symbol\" in information theory. As of the above definition of baudrate, does a Symbol means the same as \"number of transitions of a signal\" in a second (cz that's what baudrate means)? I don't think so. It has different meanings as per how the information is encoded. Dec 19 '20 at 19:31\n\nData transmission has always some bits of overhead for giving the payload data a frame so that decoding it is possible with some certainty. Thus sync signals and also parity signals are there. On bus systems further multiplexing (routing) information is part of it. In modern systems there is even some parts used for data priorization (effectively part of the routing duty) and for media quality measurement such as DSL signal optimisation.\n\nFor RS232 you have for example the start and stop bits, and the optional parity bits - applied to each character bit (might it be 5 to 8 bits long). For Ethernet you have the whole IP header that encapsulates your payload data in a package. Further there are For CAN bus there is an address field and some checksums along with other bits. For bit-serial busses like I2C, SPI and alike you often will find read-write bits, address bits and also start/stop bits along with checksums and even feedback/confirmation/error bits.\n\nRationale: What's sent on the channel is more than just the data bits you want to be sent but also some extra data making it possible for the receiver to really decode your data bits. For this some extra bits are added and later on removed again. This extra bits are counted on the channel's baud rate but they have to be subtracted to estimate the maximum transfer data rate that you as a user will e able to achieve.\n\nWhat else? Parallel busses have more than one wire and thus are able to transmit multiple bits per transmission step. Further the control flow is realized as separate wires. Thus an 8 wire + control flow wire interface can have 8 times the data rate than the baud rate.\n\nThe key word \"symbols\" is a term found in modem technology. With a certain signal shape more than one bit is encoded. This can be for example multiple tone signals send in parallel (DTMF) or other types of modulation (FSK, PSK, QAM, ...). Thus meaning the tone frequency and the switching speed between sent symbols (no longer bits) is quite decoupled. (Of course there is the Nyquist theorem of limiting a channel to some maximum raw data transmission capability.) So if you have for example 64-value quadrature amplitude modulation - then you will have a bit equivalent to those 64 values of useable channel capacity that is sent and received on each of the baud-rate steps. 64 values would mean 6 bits of raw data per baud-rate step. (Still even with this a few values might be reserved in order to achieve various types of bit and byte border synchronization as well as for channel duties.)\n\n• Channel overhead is not part of the question (although it needs to be considered at link level). Dec 19 '20 at 17:00\n• Good point, Peter. The question was not that clear for me if this item in the stack has even awareness - it definitely contributes in a larger amount of cases to the difference between user seen capacity and baud rate. - Further paragraphs on parallel busses and on /symbol/ based transmissions (or modulated signals) were there already. Dec 20 '20 at 9:32\n\nThe letter 'A' is not a symbol in this context. A symbol is a recognizable distinct value on the communication channel; it is a quantum of transmittable information.\n\nFor a binary communication channel, a symbol is either a 0 or a 1. If you are talking about conventional UART kinds of transmission, then every transmitted byte of information will also need a start symbol and a stop symbol. So, you have to transmit 10 symbols in order to transmit 8 bits of information. The maximum information (bit) rate is 8/10 of the maximum baud rate.\n\nSome kinds of communication can send multiple bits per symbol. In the old fashioned touch-tone dual-tone multiple-frequency (DTMF) telephone systems, pressing a button on the keypad caused a sound to be transmitted. Each sound was a symbol. Each sound comprised two audio frequencies, one frequency corresponding to a row on the keypad and the other frequency corresponding to the column. With four rows and four columns, each sound corresponded to 1 of 16 keys. So, each sound conveyed 4 bits of information.\n\n• Actually, each button press transmitted two tones, hence the name \"dual tone\". Dec 19 '20 at 16:09\n• So in this, a symbol would be 1 or 0? If that is the case there would be 8 symbols. So, 8 baud per second. Then how is it equal to 3 state changes in the signal? Dec 19 '20 at 16:10\n• @Hearth You are right, I'll change that to something else. Dec 19 '20 at 16:39\n• @G-aura-V You are comparing apples and cabbages. You talk about two completely different communication systems, one that encodes a single bit per baud and another that encodes three bits per baud. Dec 19 '20 at 16:41\n• @Elliot Alderon I didn't get you. Which system in the question encodes three bits per baud? Dec 20 '20 at 3:37\n\nbit rate = (bits/symbol)*(symbols/second)\n\nIf you are going to transmit three bits at a time you have to use eight different frequencies (as in the case of FSK). So, each symbol will have three bits and the symbol rate (which we call baud rate) will be decided prior to transmission. This is your 'second' definition. For some modulation schemes such as ASK, BPSK, and PSK, the bit rate equals the baud rate." ]	[ null, "https://i.stack.imgur.com/4ZfD3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93958485,"math_prob":0.90499586,"size":1288,"snap":"2021-43-2021-49","text_gpt3_token_len":306,"char_repetition_ratio":0.16666667,"word_repetition_ratio":0.0,"special_character_ratio":0.25543478,"punctuation_ratio":0.09811321,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95398897,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-27T20:58:22Z\",\"WARC-Record-ID\":\"<urn:uuid:d3a428cf-d6cf-44d0-9441-c63708cc440f>\",\"Content-Length\":\"167937\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:33369717-d2f7-4746-875b-c699a6597f37>\",\"WARC-Concurrent-To\":\"<urn:uuid:c6aaef65-0dbc-44a0-ba48-7d2f55549b4d>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/538187/how-are-two-definitions-of-baud-rate-same\",\"WARC-Payload-Digest\":\"sha1:POVCAZANA5C3B6GPEYMYVIRGQC725W3T\",\"WARC-Block-Digest\":\"sha1:YNMZAT3X5RGUN5RG42FQZYP2G2NHFNS5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358233.7_warc_CC-MAIN-20211127193525-20211127223525-00007.warc.gz\"}"}
https://math.stackexchange.com/questions/1814721/solutions-of-lfloor-4x-rfloor-lfloor-3x-rfloor-1	[ "Solutions of $\\lfloor 4x\\rfloor+\\lfloor 3x\\rfloor=1$\n\nFind all solutions of $$\\lfloor 4x\\rfloor+\\lfloor 3x\\rfloor=1$$\n\nI have no idea as to how to go about this question. I would be grateful if somebody would please show me how to solve such questions.\n\nMany thanks!\n\n• Yes Sir, it is the floor function. – user342209 Jun 5 '16 at 16:24\n\nIf $x\\le0$, then the LHS is non-positive.\n\nSo $x$ has to be positive. Since both terms on the LHS are integers with $\\lfloor 3x\\rfloor\\le \\lfloor 4x\\rfloor$, we have $$\\lfloor 4x\\rfloor=1\\quad \\text{and}\\quad \\lfloor 3x\\rfloor =0$$ from which $\\color{red}{1/4\\le x\\lt 1/3}$ follows.\n\n• Could you please elaborate a bit? I couldn't understand the part about $\\lfloor 3x\\rfloor\\le \\lfloor 4x\\rfloor$. How do we compare between $\\lfloor ax\\rfloor$ and $\\lfloor bx\\rfloor$ for any general $x$ (ie positive or negative values of $x$), where $a,b$ are constants? – user342209 Jun 5 '16 at 16:45\n• @user342209 : For $x\\gt 0$, we have $3x\\lt 4x$, and so $\\lfloor 3x\\rfloor\\le \\lfloor 4x\\rfloor$. Does this help or not? – mathlove Jun 5 '16 at 16:47\n• @user342209: In general, if $ax\\le bx$, then $\\lfloor ax\\rfloor\\le\\lfloor bx\\rfloor$. – mathlove Jun 5 '16 at 16:55\n• Beat me to it. I was typing on my phone. – user223391 Jun 5 '16 at 16:57\n• @mathlove It helps a lot! – user342209 Jun 5 '16 at 17:06\n\nHint: $$\\lfloor4x \\rfloor+\\lfloor3x \\rfloor=1$$ $$4x-1<\\lfloor4x \\rfloor\\le4x$$ $$3x-1<\\lfloor3x \\rfloor\\le3x$$ Then $$3x+4x-2<\\lfloor4x \\rfloor+\\lfloor3x \\rfloor\\le3x+4x$$ $$7x-2\\le1<7x$$ $$\\frac 17< x\\le\\frac37$$\n\nCase 1) $\\frac 17< x<\\frac14$\n\nCase 2) $\\frac 14\\le x<\\frac13$\n\nCase 3) $\\frac 13\\le x<\\frac37$\n\nAnswer: $$\\frac 14\\le x<\\frac13$$\n\n• Sir, isn't $[x]\\le x$? Then how can $4x\\le\\lfloor4x \\rfloor<4x+1$? – user342209 Jun 5 '16 at 16:32\n• You probably mean $4x-1$ and $4x$. – user223391 Jun 5 '16 at 16:36\n\nLet $a+b=1$ where $a,b\\in \\Bbb{Z}$.\n\nNote if $x <0$ then $\\lfloor 3x\\rfloor<0$ and likewise with $4x$. So $a,b\\geq 0$.\n\nSo the equation boils down to $a=1$ and $b=0$ or vice versa. Note that $\\lfloor 4x\\rfloor\\geq \\lfloor 3x\\rfloor$ so you need to find $x$ so that $\\lfloor 3x\\rfloor=0$ and $\\lfloor 4x \\rfloor = 1$.\n\nYou want $0<3x<1$ and $1\\leq 4x< 2$. That is, $1/4 \\leq x < 1/3$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69032824,"math_prob":0.99996924,"size":1428,"snap":"2019-26-2019-30","text_gpt3_token_len":593,"char_repetition_ratio":0.18820225,"word_repetition_ratio":0.30331755,"special_character_ratio":0.37114847,"punctuation_ratio":0.07491857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99995136,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T15:23:49Z\",\"WARC-Record-ID\":\"<urn:uuid:c42fd27c-99c6-417c-9c26-dadf91b19505>\",\"Content-Length\":\"156706\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e704ef23-f007-48e9-9235-74ba5a5020d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9c95090-e394-4e3d-a2c5-7dcacf104c35>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1814721/solutions-of-lfloor-4x-rfloor-lfloor-3x-rfloor-1\",\"WARC-Payload-Digest\":\"sha1:F3AGAXWRB77TV22W43PBU5U6ICGM3P25\",\"WARC-Block-Digest\":\"sha1:WHQ4Y3J22A3XYV2KLJQUNNIPTNGBFNGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999261.43_warc_CC-MAIN-20190620145650-20190620171650-00153.warc.gz\"}"}
https://www.designcise.com/web/tutorial/how-to-get-key-value-pair-from-url-query-string-in-php	[ "How To Get Key/Value Pair From URL Query String In PHP?\n\nA look at different ways to retrieve the query string and parameter values therein\n\n• By Daniyal Hamid\n• October 15, 2015\n\nIn this article we look at different ways in PHP to retrieve the query string and its parameter values from a URL. We'll be using the following string for all the examples in this article:\n\nhttp://www.designcise.com/?key1=value1&key2=value2\n\nRetrieving The Query String\n\nQuery String Of Current URL:\n\nUsing \\$_SERVER['QUERY_STRING'] in PHP we can get the full query string from the current URL. For example:\n\necho \\$_SERVER['QUERY_STRING'];\n\n// Output: key1=value1&key2=value2\n\nQuery String From A String\n\nTo retrieve the query string from a string we can use the parse_url function. For example:\n\necho parse_url('http://www.designcise.com/?key1=value1&key2=value2', PHP_URL_QUERY);\n\n// Output: key1=value1&key2=value2\n\nQuery String Parameters As An Array\n\nUsing The \\$_GET Array:\n\nUsing \\$_GET array in PHP we can directly access any key/value pair in the current URL. For example:\n\necho \\$_GET['key1']; // Output: value1\n\nParsing A Query String\n\nWhen we have all the key/value pairs in a single string, we can use the parse_str function to convert the key/value pairs to an array:\n\n// method #1\nparse_str(\\$_SERVER['QUERY_STRING'], \\$output);\n\n// method #2\nparse_str(parse_url('http://www.designcise.com/?key1=value1&key2=value2', PHP_URL_QUERY), \\$output);\necho print_r(\\$output, TRUE);\n\n// Output:\nArray (\n['key1'] => \"value1\"\n['key2'] => \"value2\"\n)\n\nHandling Duplicate Keys In A Query String:\n\nUsing either of the two methods explained above, the output of duplicate keys in a query string would be a child array for that key with numerically indexed values. For demonstration purposes, consider the output of the following url:\n\nhttp://www.designcise.com/?key[]=value1&key[]=value2\n\n// Output:\nArray (\n['key'] => Array (\n => test1\n => test2\n)\n)\n\nThings To Remember:\n\nWhen using the methods explained above, remember the following are true:\n\n• Using \\$_GET would be the fastest way to access a key/value pair in the current url. Wherever possible, this should be the preferred choice.\n• Both parse_str and \\$_GET automatically urldecode values, which means any %## encoding is decoded in the given string. Plus symbols (i.e. +) are decoded to a space character. If this is not the behavior you're looking for, then you may use \\$_SERVER['QUERY_STRING'] to retrieve the entire query string in its original form or call urlencode on individual values.\n• Duplicate key names followed by square brackets (for example, ?key[]=value1&key[]=value2) create a child array for that key with numerically indexed values." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5945473,"math_prob":0.7432149,"size":2477,"snap":"2019-43-2019-47","text_gpt3_token_len":604,"char_repetition_ratio":0.12252325,"word_repetition_ratio":0.056179777,"special_character_ratio":0.25878078,"punctuation_ratio":0.14254385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9522122,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T16:57:04Z\",\"WARC-Record-ID\":\"<urn:uuid:1b7dc6f5-4495-479c-8703-fe6359ac9fd1>\",\"Content-Length\":\"18132\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:167ab67d-4e04-4aed-b8be-163ac7bc15f6>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cd7d837-ac50-41d3-94d0-de27e9f97f74>\",\"WARC-IP-Address\":\"213.32.120.154\",\"WARC-Target-URI\":\"https://www.designcise.com/web/tutorial/how-to-get-key-value-pair-from-url-query-string-in-php\",\"WARC-Payload-Digest\":\"sha1:FDA2EO6JIZYP3T53W3SUVVZB5ZX3CCD6\",\"WARC-Block-Digest\":\"sha1:2IH7UV4QWQ5OZJ2YF6FZKLMQZH7AWCOL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986669057.0_warc_CC-MAIN-20191016163146-20191016190646-00378.warc.gz\"}"}
https://www.fmz.com/strategy/380219	[ "# Candle Strength\n\nAuthor: Zer3192, Date: 2022-08-27 11:57:02\nTags:\n\n```/backtest\nstart: 2021-05-08 00:00:00\nend: 2022-05-07 23:59:00\nperiod: 4h\nbasePeriod: 15m\nexchanges: [{\"eid\":\"Futures_Binance\",\"currency\":\"BTC_USDT\"}]\n/\n// This source code is subject to the terms of the Mozilla Public License 2.0 at https://mozilla.org/MPL/2.0/\n\n//@version=5\nindicator(\"Candle Strength\",overlay = true )\n\n//round function for rounding up to n decimal places\n//Thanks to Proper Round Function - QuantNomad\n\nroundn(x, n) =>\nmult = 1\nif n != 0\nfor i = 1 to math.abs(n) by 1\nmult = 10\nmult\n\nn >= 0 ? math.round(x mult) / mult : math.round(x / mult) * mult\n\n//calculating strength\ngreen = roundn(((close-low)/(high-low) 100), 2)\nred = roundn(((high-close)/(high-low)100), 2)\n\nif (close>open)\nl = label.new(x = bar_index,y = close,text=str.tostring(green) + \" %\",color=color.green,textcolor = color.white,style=label.style_label_up)\nlabel.set_yloc(l,yloc.belowbar)\nstrategy.entry(\"Enter Long\", strategy.long)\nelse if (open > close)\nl2 = label.new(x = bar_index,y = close,text=str.tostring(red)+ \" %\",color=color.red,textcolor=color.white,style=label.style_label_down)\nlabel.set_yloc(l2,yloc = yloc.abovebar)\nstrategy.entry(\"Enter Short\", strategy.short)\n\n```\n\nMore" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5073403,"math_prob":0.9112602,"size":1280,"snap":"2023-40-2023-50","text_gpt3_token_len":407,"char_repetition_ratio":0.09874608,"word_repetition_ratio":0.013071896,"special_character_ratio":0.35078126,"punctuation_ratio":0.23791821,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9764429,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T00:03:06Z\",\"WARC-Record-ID\":\"<urn:uuid:3c03f342-1afa-43df-ab06-ae9bea4e37e2>\",\"Content-Length\":\"8226\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf3db980-0220-462f-a70e-94b54c208fa0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a2ce15c-67ac-431c-810a-e54b8f42076e>\",\"WARC-IP-Address\":\"34.92.158.77\",\"WARC-Target-URI\":\"https://www.fmz.com/strategy/380219\",\"WARC-Payload-Digest\":\"sha1:3E33F7FBXYF3BXN3PVWGPY4DDQPFLT7B\",\"WARC-Block-Digest\":\"sha1:SLFMKTNMMMYLS4RFBTE62ANWD45WOAUG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100535.26_warc_CC-MAIN-20231204214708-20231205004708-00571.warc.gz\"}"}
http://aspalliance.com/36_Creating_Excel_XP_spreadsheets_by_using_NET_and_XML	[ "", null, "", null, "", null, "", null, "", null, "Print", null, "Add To Favorites", null, "Email To Friend", null, "Rate This Article Creating Excel XP spreadsheets by using .NET and XML\n page 1 of 1\nPublished: 26 Sep 2003\nUnedited - Community Contributed\nAbstract\nExcel XP has a cool feature related to XML as the product can easily read and write properly-structured XML documents, run web queries against properly-structured XML data sources and save entire workbooks as XML spreadsheets. This article demonstrates one view to the utilization of this feature in .NET, namely creating a basic spreadsheet that has formulas.", null, "by\nFeedback\nViews (Total / Last 10 Days): 17809/ 39\n\nThe article\n\nSource XML\n\n<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<sales>\n<product>\n<name>CD</name>\n<price>20</price>\n<sold>15</sold>\n</product>\n<product>\n<name>Book</name>\n<price>50</price>\n<sold>7</sold>\n</product>\n<product>\n<name>Pen</name>\n<price>2</price>\n<sold>18</sold>\n</product>\n</sales>\n\nTarget XML\n\nThis is an example of the format Excel can open straight as a file from file system or as a web page and then produce a working spreadsheet with formulas. Note that this is a reduced version to keep the example concise.\nIf you save an Excel spreadsheet as XML and take a look at it, there would be content related to worksheet's options, protection, document's properties and so on. However, those are optional elements in the XML format and therefore stripped out of this example.\n\n<?xml version=\"1.0\" encoding=\"utf-8\"?>\n\n<Styles>\n<Style ss:ID=\"s1\">\n<Font x:Family=\"Swiss\" ss:Bold=\"1\" />\n</Style>\n</Styles>\n\n<Worksheet ss:Name=\"Sheet1\">\n<Table>\n<Row>\n<Cell>\n<Data ss:Type=\"String\">Demonstration spreadsheet created with .NET</Data>\n</Cell>\n</Row>\n\n<Row ss:Index=\"3\">\n<Cell ss:StyleID=\"s1\">\n<Data ss:Type=\"String\">Product</Data>\n</Cell>\n<Cell ss:StyleID=\"s1\">\n<Data ss:Type=\"String\">Sold</Data>\n</Cell>\n<Cell ss:StyleID=\"s1\">\n<Data ss:Type=\"String\">Price</Data>\n</Cell>\n<Cell ss:StyleID=\"s1\">\n<Data ss:Type=\"String\">Sum</Data>\n</Cell>\n</Row>\n<Row>\n<Cell>\n<ss:Data ss:Type=\"String\">CD</ss:Data>\n</Cell>\n<Cell>\n<ss:Data ss:Type=\"Number\">15</ss:Data>\n</Cell>\n<Cell>\n<Data ss:Type=\"Number\">20</Data>\n</Cell>\n<Cell ss:Formula=\"=RC[-2]RC[-1]\">\n</Cell>\n</Row>\n<Row>\n<Cell>\n<ss:Data ss:Type=\"String\">Book</ss:Data>\n</Cell>\n<Cell>\n<ss:Data ss:Type=\"Number\">7</ss:Data>\n</Cell>\n<Cell>\n<Data ss:Type=\"Number\">50</Data>\n</Cell>\n<Cell ss:Formula=\"=RC[-2]RC[-1]\">\n</Cell>\n</Row>\n<Row>\n<Cell>\n<ss:Data ss:Type=\"String\">Pen</ss:Data>\n</Cell>\n<Cell>\n<ss:Data ss:Type=\"Number\">18</ss:Data>\n</Cell>\n<Cell>\n<Data ss:Type=\"Number\">2</Data>\n</Cell>\n<Cell ss:Formula=\"=RC[-2]*RC[-1]\">\n</Cell>\n</Row>\n<Row>\n<Cell ss:Index=\"3\" ss:StyleID=\"s1\">\n<Data ss:Type=\"String\">Sum</Data>\n</Cell>\n<Cell ss:Formula=\"=R[-3]C+R[-2]C+R[-1]C\">\n</Cell>\n</Row>\n</Table>\n</Worksheet>\n</Workbook>\n\n<Style ss:ID=\"s1\">\n<Font x:Family=\"Swiss\" ss:Bold=\"1\" />\n</Style>\n\nThis declares a style that can be used at the spreadsheet by referring to ss:StyleID=\"s1\"\n\n<Worksheet ss:Name=\"Sheet1\">\n\nDeclares a worksheet with name “Sheet 1”.\n\n<Data ss:Type=\"String\">\n\nSpecifies the data type of the content of the cell.\n\n<Row ss:Index=\"3\">\n\nSpecifies that the row is placed at the third row at the spreadsheet.\n\n<Cell ss:Index=\"3\"…>\n\nSpecifies that the cell is placed at the third cell in this row at the spreadsheet.\n\n<Cell ss:Formula=\"=R[-3]C+R[-2]C+R[-1]C\"></Cell>\n\nSpecifies the formula to be used to produce the content for the cell. The syntax is the basic RC syntax that uses relative locations. R means row and C means cell.\n\nIf you want to know more about the exact syntax, see resources provided at the end of this article.\n\nSo?\n\nWell, it should be obvious that creating these Excel compatible XML files is very easy task in .NET. You could create the XML for example using XSLT or DOM and output it from web page using built-in Xml Control. Consider also the benefits if these docs i.e. Excel spreadsheets would be transmitted using XML web services. Also according to Microsoft the Office 2003 will have better support for XML. The result XML provided earlier was successfully tested with beta version of Excel 2003 by Thomas Johansen, http://authors.aspalliance.com/aylar .\n\nAs much as it is cool to have this kind feature, there are also restrictions. You can't save charts, OLE objects, VBA code etc. special features. I've included sample codes for the XSLT approach. It contains the source XML, the XSLT style sheet with extension C# class and the page to test the transformation and to produce the result XML. If you try to open the page from your server using Excel XP, you get fully working spreadsheet that has formulas.\n\nResources\n\nXML in Excel and the Spreadsheet component (MSDN)\n\nTransform XML Files when importing into Microsoft Excel 2002 (MSDN)", null, "", null, "", null, "", null, "", null, "" ]	[ null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/icons_11x8/icon-print.gif", null, "http://aspalliance.com/images/icons_11x8/icon-favorite.gif", null, "http://aspalliance.com/images/icons_11x8/icon-envelope.gif", null, "http://aspalliance.com/images/icons_11x8/icon-rate.gif", null, "http://aspalliance.com/images/photos/8cbd3d4f-1d72-4777-9409-46f875fcc8f8_80_80.jpg", null, "http://aspalliance.com/images/gear_16x16.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null, "http://aspalliance.com/images/clear.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63701177,"math_prob":0.47267985,"size":4586,"snap":"2020-34-2020-40","text_gpt3_token_len":1315,"char_repetition_ratio":0.16302924,"word_repetition_ratio":0.03797468,"special_character_ratio":0.28892282,"punctuation_ratio":0.13943355,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95936954,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-06T07:24:28Z\",\"WARC-Record-ID\":\"<urn:uuid:81ab437c-6295-47f4-9c83-3010aa99fd6c>\",\"Content-Length\":\"84173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a9c6c41-8847-49aa-85f8-f1789eb9cead>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3303ab1-e2ea-4f7a-9316-a168d624d895>\",\"WARC-IP-Address\":\"76.74.234.215\",\"WARC-Target-URI\":\"http://aspalliance.com/36_Creating_Excel_XP_spreadsheets_by_using_NET_and_XML\",\"WARC-Payload-Digest\":\"sha1:TJQUMP5NYBDBOVVZEFQUWVWBZWZQ5MXW\",\"WARC-Block-Digest\":\"sha1:BIR5OOR7RYOUMMLCFBUGA2E35TF4MR26\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439736883.40_warc_CC-MAIN-20200806061804-20200806091804-00389.warc.gz\"}"}
https://numbermatics.com/n/920736/	[ "# 920736\n\n## 920,736 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 920736 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 72 divisors.\n\n920736 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of seventy-two divisors.\n\n## Prime factorization of 920736:\n\n### 25 × 32 × 23 × 139\n\n(2 × 2 × 2 × 2 × 2 × 3 × 3 × 23 × 139)\n\nSee below for interesting mathematical facts about the number 920736 from the Numbermatics database.\n\n### Names of 920736\n\n• Cardinal: 920736 can be written as Nine hundred twenty thousand, seven hundred thirty-six.\n\n### Scientific notation\n\n• Scientific notation: 9.20736 × 105\n\n### Factors of 920736\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 9\n• Sum of prime factors: 167\n\n### Divisors of 920736\n\n• Number of divisors d(n): 72\n• Complete list of divisors:\n• Sum of all divisors σ(n): 2751840\n• Sum of proper divisors (its aliquot sum) s(n): 1831104\n• 920736 is an abundant number, because the sum of its proper divisors (1831104) is greater than itself. Its abundance is 910368\n\n### Bases of 920736\n\n• Binary: 111000001100101000002\n• Base-36: JQG0\n\n### Squares and roots of 920736\n\n• 920736 squared (9207362) is 847754781696\n• 920736 cubed (9207363) is 780558346679648256\n• The square root of 920736 is 959.5498944821\n• The cube root of 920736 is 97.2848114109\n\n### Scales and comparisons\n\nHow big is 920736?\n• 920,736 seconds is equal to 1 week, 3 days, 15 hours, 45 minutes, 36 seconds.\n• To count from 1 to 920,736 would take you about three 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 920736 cubic inches would be around 8.1 feet tall.\n\n### Recreational maths with 920736\n\n• 920736 backwards is 637029\n• The number of decimal digits it has is: 6\n• The sum of 920736's digits is 27\n• More coming soon!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8160243,"math_prob":0.9603531,"size":3414,"snap":"2021-31-2021-39","text_gpt3_token_len":1013,"char_repetition_ratio":0.11994135,"word_repetition_ratio":0.034843206,"special_character_ratio":0.4004101,"punctuation_ratio":0.14114115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99316543,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T19:54:36Z\",\"WARC-Record-ID\":\"<urn:uuid:38748c6c-99e8-49f7-9e87-3796c1c4063d>\",\"Content-Length\":\"21898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d14c55d4-0821-4946-a0f6-cc8b26313625>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2d4074c-db4d-468a-9178-c4b6f12af07c>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/920736/\",\"WARC-Payload-Digest\":\"sha1:TWUNUYAUVBIWS6NB4JMMEJMQD2OGDGUG\",\"WARC-Block-Digest\":\"sha1:RYJDIK2PWSXJY7C46ZOBA2YPQM2WAJTM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060882.17_warc_CC-MAIN-20210928184203-20210928214203-00513.warc.gz\"}"}
https://piping-designer.com/index.php/disciplines/electrical/2517-wavelength-velocity	[ "# Wavelength Velocity\n\nWritten by Jerry Ratzlaff on . Posted in Electrical Engineering\n\n## Wavelength Velocity formula\n\n $$\\large{ v_w = \\frac { f } { \\lambda } }$$\n\n### Where:\n\n Units English Metric $$\\large{ v_w }$$ = wavelength velocity $$\\large{\\frac{ft}{sec}}$$ $$\\large{\\frac{m}{s}}$$ $$\\large{ f }$$ = frequency $$\\large{Hz}$$ $$\\large{Hz}$$ $$\\large{ \\lambda }$$ (Greek symbol \\lambda) = wavelength $$\\large{ft}$$ $$\\large{m}$$", null, "" ]	[ null, "https://piping-designer.com/images/Piping%20Designer%20Gallery/P-D_Logo_1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5519477,"math_prob":1.0000091,"size":368,"snap":"2022-40-2023-06","text_gpt3_token_len":129,"char_repetition_ratio":0.27747253,"word_repetition_ratio":0.0,"special_character_ratio":0.41847825,"punctuation_ratio":0.022727273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000072,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T18:56:29Z\",\"WARC-Record-ID\":\"<urn:uuid:9419bbbf-3b82-4085-b891-8243d7a2c96c>\",\"Content-Length\":\"27082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:46b23d93-bf52-4102-ae8d-f01e762d5568>\",\"WARC-Concurrent-To\":\"<urn:uuid:f0a313d4-1981-4d60-8f48-2a20506da993>\",\"WARC-IP-Address\":\"200.225.40.42\",\"WARC-Target-URI\":\"https://piping-designer.com/index.php/disciplines/electrical/2517-wavelength-velocity\",\"WARC-Payload-Digest\":\"sha1:2BSGD7EZLMJ73P6B5F364OPZKKNBONYG\",\"WARC-Block-Digest\":\"sha1:VTZCXGCJBKA5O3WFV4MPH3ELV4HIU7WD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335276.85_warc_CC-MAIN-20220928180732-20220928210732-00720.warc.gz\"}"}
https://texar-pytorch.readthedocs.io/en/latest/_modules/texar/torch/run/metric/summary.html	[ "# Source code for texar.torch.run.metric.summary\n\n```# Copyright 2019 The Texar Authors. All Rights Reserved.\n#\n# you may not use this file except in compliance with the License.\n# You may obtain a copy of the License at\n#\n#\n# Unless required by applicable law or agreed to in writing, software\n# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.\n# See the License for the specific language governing permissions and\n\"\"\"\nExecutor metrics for summaries.\n\"\"\"\n\nfrom collections import deque\nfrom typing import Any, Deque, Optional, Sequence\nimport weakref\n\nimport numpy as np\nfrom torch.optim.optimizer import Optimizer\n\nfrom texar.torch.run.metric.base_metric import StreamingMetric\n\n__all__ = [\n\"Average\",\n\"AveragePerplexity\",\n\"RunningAverage\",\n\"LR\",\n]\n\n[docs]class Average(StreamingMetric[float, float]):\nr\"\"\"The average of a specific predicted value.\n\nAverage is a :class:`~texar.torch.run.metric.StreamingMetric`, requires only\npredicted values. Average values are unbounded :class:`float` numbers. By\ndefault, lower values are better, but the behavior can be configured.\n\nKeyword Args:\npred_name (str): Name of the predicted value. This will be used as the\nkey to the dictionary returned by the model. Defaults to ``\"loss\"``.\nhigher_is_better (bool, optional): If specified, the\n:attr:`higher_is_better` attribute for the instance is overwritten\nby the specified value. Defaults to `False`.\n\"\"\"\nhigher_is_better = False\nrequires_label = False\n\nsum: float\n\ndef __init__(self, , pred_name: str = \"loss\",\nhigher_is_better: bool = False):\n# pylint: disable=useless-super-delegation\nsuper().__init__(pred_name=pred_name, higher_is_better=higher_is_better)\n\ndef reset(self) -> None:\nsuper().reset()\nself.sum = 0.0\n\ndef add(self, predicted: Sequence[float], _) -> None:\nself.count += len(predicted)\nself.sum += sum(predicted)\n\ndef value(self) -> float:\nif self.count == 0:\nreturn 0.0\nreturn self.sum / self.count\n\n[docs]class AveragePerplexity(Average):\n# TODO: Create a `WeightedAverage` class that takes `(value, weight)`\nhigher_is_better = False\n\ndef add(self, predicted: Sequence[float], _) -> None:\n\n[docs]class RunningAverage(StreamingMetric[float, float]):\nr\"\"\"The running average of a specific predicted value, i.e., the average\ncomputed over the most recent :attr:`queue_size` values.\n\nRunning average is a :class:`~texar.torch.run.metric.StreamingMetric`,\nrequires only predicted values. Running average values are unbounded\n:class:`float` numbers. By default, lower values are better, but the\nbehavior can be configured.\n\nKeyword Args:\nqueue_size (int): Size of the queue to keep history values. The running\naverage is computed over the most recent :attr:`queue_size` values.\npred_name (str): Name of the predicted value. This will be used as the\nkey to the dictionary returned by the model. Defaults to ``\"loss\"``.\nhigher_is_better (bool, optional): If specified, the\n:attr:`higher_is_better` attribute for the instance is overwritten\nby the specified value.\n\"\"\"\nhigher_is_better = False\nrequires_label = False\n\nhistory: Deque[float]\nsum: float\n\ndef __init__(self, queue_size: int, , pred_name: str = \"loss\",\nhigher_is_better: bool = False):\nsuper().__init__(pred_name=pred_name, higher_is_better=higher_is_better)\nif not isinstance(queue_size, int) or queue_size <= 0:\nraise ValueError(\"'queue_size' must be a position integer\")\nself.queue_size = queue_size\n\ndef reset(self) -> None:\nsuper().reset()\nself.sum = 0.0\nself.history = deque()\n\ndef add(self, predicted: Sequence[float], _) -> None:\nif len(predicted) >= self.queue_size:\nself.history = deque(predicted[-self.queue_size:])\nself.sum = sum(self.history)\nelse:\nfor _ in range(len(predicted) -\n(self.queue_size - len(self.history))):\nself.sum -= self.history.popleft()\nself.sum += sum(predicted)\nself.history.extend(predicted)\n\ndef value(self) -> float:\nif len(self.history) == 0:\nreturn 0.0\nreturn self.sum / len(self.history)\n\n[docs]class LR(StreamingMetric[Any, float]):\nr\"\"\"The learning rate (LR) of the given optimizer. This is not exactly a\nmetric, but rather a convenience object to print learning rates in log.\n\nLR is a :class:`~texar.torch.run.metric.StreamingMetric`, requires neither\npredicted values nor labels. LR values are unbounded :class:`float` numbers,\nwith no clear definition of \"better\". Comparison of two learning rates are\nnot meaningful.\n\nKeyword Args:\noptimizer: The optimizer instance.\nparam_group (int, optional): Index of the parameter group to obtain the\nlearning rate from. Defaults to 0. You don't need to specify this if\nthe optimizer contains only one parameter group (e.g., constructed\nusing :python:`optim_class(model.parameters())`.\n\"\"\"\n\nrequires_pred = False\nrequires_label = False\n\ndef __init__(self, optimizer: Optimizer, param_group: int = 0):\nsuper().__init__(pred_name=None)\nself.optimizer = weakref.ref(optimizer)\nself.group = param_group\n\npass\n\ndef value(self) -> float:\nreturn self.optimizer().param_groups[self.group]['lr'] # type: ignore\n\ndef better(self, cur: float, prev: float) -> Optional[bool]:\n# pylint: disable=unused-argument\n# Always return `None` to indicate values are uncomparable.\nreturn None\n\ndef __getstate__(self):\n# There's no point in pickling an `LR` metric; just ignore it.\nreturn None\n\ndef __getnewargs__(self):\n# But when unpickling, we need to make sure we can construct something.\n# This requires passing a dummy `optimizer` to which a weakref can be\n# constructed. In this case, we use an arbitrary built-in class.\nreturn (int,)\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.51193094,"math_prob":0.8313412,"size":5702,"snap":"2020-45-2020-50","text_gpt3_token_len":1389,"char_repetition_ratio":0.13180064,"word_repetition_ratio":0.21798365,"special_character_ratio":0.2569274,"punctuation_ratio":0.23258004,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9740546,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T07:57:44Z\",\"WARC-Record-ID\":\"<urn:uuid:c89c5399-b70c-441d-89e2-7e97957cc0ac>\",\"Content-Length\":\"75808\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9052407-1749-4403-a13c-4f5383e49ab8>\",\"WARC-Concurrent-To\":\"<urn:uuid:66188171-3111-492b-b100-844b0a09fcc5>\",\"WARC-IP-Address\":\"104.17.33.82\",\"WARC-Target-URI\":\"https://texar-pytorch.readthedocs.io/en/latest/_modules/texar/torch/run/metric/summary.html\",\"WARC-Payload-Digest\":\"sha1:RYFB3NRVPLOO5PIUW6VPFWFACWFRDWJX\",\"WARC-Block-Digest\":\"sha1:2HYJQRBZRCJYUOE3TYDZJ5VRB72ZZK3C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107888402.81_warc_CC-MAIN-20201025070924-20201025100924-00272.warc.gz\"}"}
https://asmedc.silverchair.com/vibrationacoustics/article-abstract/108/3/362/418462/On-Statistical-Analysis-of-Gear-Dynamic-Loads?searchresult=1	[ "Statistical analysis of the gear dynamic load is carried out using piecewise constant mesh stiffness approximation. The dynamics of the spur gear system is modeled as a nonlinear, nonstationary process, and the gear transmission error which acts as a random input to the gear system is generated by passing a Gaussian white noise process through a time invariant shaping filter. The equivalent discrete time state equation and the mean and covariance propagation equations are then written for the augmented system. Then starting from known initial conditions these propagation equations are used to compute the statistics of the steady state response and hence those of the dynamic load. A procedure is presented for the selection of proper initial conditions so as to reach the steady state condition faster, thereby reducing the computational time required. The variations in the statistics of the dynamic load with respect to changes in contact position, random error magnitude, and operating speed are also investigated with the help of a numerical example. The results show that the approach presented in this study provides truer results than the statistical linearization approach used by Tobe et al. . Moreover, the proposed procedure has the advantage that it can be applied to higher-order systems with complex mesh stiffness and torque fluctuations and to systems with symmetrical or nonsymmetrical nonlinearities.\n\nThis content is only available via PDF.\nYou do not currently have access to this content." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83485633,"math_prob":0.927875,"size":3991,"snap":"2022-40-2023-06","text_gpt3_token_len":946,"char_repetition_ratio":0.09982443,"word_repetition_ratio":0.15745394,"special_character_ratio":0.22074668,"punctuation_ratio":0.16644114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96064955,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T21:19:00Z\",\"WARC-Record-ID\":\"<urn:uuid:a7cb3e62-fc30-47b1-93f6-8e0f9233a178>\",\"Content-Length\":\"112661\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6fd9357a-7a5f-4310-b146-28ce00c8d58b>\",\"WARC-Concurrent-To\":\"<urn:uuid:53f2e7c8-3fac-4214-9079-8a40f38c342c>\",\"WARC-IP-Address\":\"52.179.114.94\",\"WARC-Target-URI\":\"https://asmedc.silverchair.com/vibrationacoustics/article-abstract/108/3/362/418462/On-Statistical-Analysis-of-Gear-Dynamic-Loads?searchresult=1\",\"WARC-Payload-Digest\":\"sha1:SMJQRXM73QOTPRVJRRL3F2SFOUXMNX6M\",\"WARC-Block-Digest\":\"sha1:5VRWON2CPI34M57HKCH4MZGSLGW6SURX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337432.78_warc_CC-MAIN-20221003200326-20221003230326-00028.warc.gz\"}"}
http://downloads.hindawi.com/journals/mpe/2015/109325.xml	[ "MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi Publishing Corporation 10.1155/2015/109325 109325 Research Article On Two-Level State-Dependent Routing Polling Systems with Mixed Service Zheng Guan 1 Zhijun Yang 2 Wenhua Qian 1 Min He 1 Liu Ruihua 1 School of Information Science and Technology Yunnan University Kunming 650091 China ynu.edu.cn 2 Educational and Scientific Institute Educational Department of Yunnan Province Kunming 650223 China 2015 18102015 2015 19 03 2015 01 09 2015 18102015 2015 Copyright © 2015 Guan Zheng 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\nBased on priority differentiation and efficiency of the system, we consider an N+1 queues’ single-server two-level polling system which consists of one key queue and N normal queues. The novel contribution of the present paper is that we consider that the server just polls active queues with customers waiting in the queue. Furthermore, key queue is served with exhaustive service and normal queues are served with 1-limited service in a parallel scheduling. For this model, we derive an expression for the probability generating function of the joint queue length distribution at polling epochs. Based on these results, we derive the explicit closed-form expressions for the mean waiting time. Numerical examples demonstrate that theoretical and simulation results are identical and the new system is efficient both at key queue and normal queues.\n\n1. Introduction\n\nIn this paper, we study a class of N+1 queues’ polling systems that consists of one key queue, Qh, and N normal queues, Q1,Q2,,QN, which are attended by a single server. Studies on the polling systems have attracted extensive attentions in the last years due to their vast area of applications in communication network, production, and transportation. Excellent surveys on polling systems analysis and their applications may be found in . However, many studies in the literatures assume that the server visit the queues in a fixed, cyclic order. This might not be a realistic assumption, as queues might have different priority level; queues with high priority should be visit more frequently than the lower ones; sometime queues might be empty and then there is no need to visit. As such, we study the case where the server just visits active queues with customers. Note that as a consequence, after skipping the empty queues, server could provide more visit opportunity to active queues with customers. Furthermore, parallel process of service period and switch-over period allows a successive service between two active queues without the duration of switch-over time. To provide priority differentiation service, queues are separated as one key queue and N normal queues. Two-level route order and mixed service scheme are used to provide high priority to key queue.\n\nIt is observed that in the wide body of literature on polling system hardly can any studies be found that take the consideration of queue state-dependent routing and service priority simultaneously. The reason for this may lie in the fact that the analysis of state-dependent routing polling model is much more complex than that of cyclic polling model, especially in priority differentiated model. In particular, waiting time and queue length analysis of two-level priorities polling systems can be found in , in which the server visits queues in a two-level route; that is, the server polls key queue with exhaustive scheme after each gated service to normal queue . This work is extended in with assigning 1-limited service discipline to normal queues. More recently, Yang et al. set the exhaustive service for normal queue and gated service for key queue to ensure fairness but just acquire the first moment performance of the system as mean queue length at the polling epoch and the mean cyclic time . The parallel discipline is used to improve the delay performance in , in which when the current polling queue has customers in storage the server will process service while switching to the successive queue simultaneously and begins to serve the successor once it finishes the service of the current one. This scheme could improve polling efficiency in high traffic cases. However, the parallel mechanism will be invalid when there is no customer in the queue. In low traffic cases, useless polling to idle queue becomes an obvious liability in cyclic polling model. Routing depends on the event whether a queue is empty or it is not helpful to this problem . In this paper, we consider the special setting to a two-level mixed service polling model, where the key queue is served exhaustively while normal queues are served in 1-limited mechanism. Furthermore, the server no longer checks all the stations in a fixed order; only active stations with transfer requirements could be served and then the switch-over period and service period are processed paralleled. This mechanism increases the system utilization and reduces the mean waiting time.\n\nAlthough the exhaustive service discipline in principle fits the branching property, the present model involves 1-limited service discipline, which does not satisfy the above-mentioned branching property. The explicit analysis of nonbranching service disciplines is mostly in special setting, such as [10, 11] studied on two-queue polling systems and studied on symmetric 1-limited model. In this paper, we follow the special setting in and analyze the mean waiting time of the present model under the assumption on the symmetrical characteristic among normal queues, as will be described in greater detail in Section 2.\n\nInitially, we follow an approach similar to the analysis of , which uses a recursive iteration of a functional equation, for the probability generating function (PGF) of the joint queue-length distribution at moments the server starts a visit period.\n\nThe main contributions of this paper can be summarized as follows. Firstly, we extend the parallel two-level poling system in by using queue state-dependent routing, in which only active queues with customers could be visited by server. This scheme is helpful to avoid the consumptions induced by idle visit. Secondly, under the assumption of a stable system, we obtain the explicit expressions for the PGF for the joint queue length distribution at polling epochs as a starting point of key queue and normal queue separately. Thirdly, we achieve the exact closed-form expression of the mean waiting time under the assumption on the symmetrical characteristic of normal queue.\n\nThe rest of the paper is structured as follows. In Section 2, we give a formal description of the polling model that we study and we introduce the necessary notation. Based on this, in Section 3, we derive the expressions for the mean waiting time of the present model under the assumption of a semisymmetric (symmetrical characteristic of normal queue) stable system, by taking a functional equation for the PGF for the joint queue length distribution at polling epochs as a starting point. In Section 4, numerical results obtained with the proposed analytical models are shown and their very good agreement with realistic simulation results is discussed. Finally, concluding remarks and directions for future research are given in the end.\n\n2. Model Description\n\nConsider a discrete time (timeline is divided into time slot) polling system consisting of N (N2) infinite-buffer queues Q1,Q2,,QN, and Qh. The single server visits active queues in a two-level state-dependent routing order and serves the customers with mixed service discipline.\n\nIn the arrival process, type-j (j=1,2,,N,h) customers arrive at Qj according to an independent Poisson arrival process. The generating function of arrival process in queue j is Aj(zj), with the variance of σλj2=Aj1+λj-λj2 and the arrival rate of λj=Aj1. The total arrival rate is i=1Nλi+λh.\n\nIn the service process, we assume that customers in queue j (j=1,2,,N,h) receive individual service. The service time of a customer at each queue is independent of each other. Their generating function is Bj(zj), with the variance of σβj2=Bj1+βj-βj2 and the mean value βj=Bj1. We propose a two-level server routing make the high priority queue be visited more frequently than others and add mix-service discipline to ensure the high priority of Qh. The load offered to Qj is ρj=λjβj, and the total offered load is equal to i=1Nρi+ρh.\n\nState-Dependent Routing. Queues are partitioned as active queue and idle queue by their buffer condition. Only active queues with customers waiting in the buffer could be visited by the server in order. Idle queue with empty buffer would be skipped in the current polling round.\n\nTwo-Level Polling. The server visits queues governed by a two-level routing. In the first polling level, the server polls between the high priority queue Qh and an active normal queue; in the second level, for each time after the exhaustive service at Qh, one normal active queue is visited in a cyclic order; that is, the server routing in this model is 1hihi+1hN.\n\nIn the switch-over process, a parallel mechanism is used. When the server polls an active queue at time with customers in its buffer, the server will provide service and inquire the next active queue simultaneously and then switch to serve the successor immediately without the switch-over time once it has finished the current service. Combined with the state-dependent routing scheme, over the course of a visit period, the server serves the active queues and normal queue in sequence continuously until the entire system is empty; there will be no consumption of switch-over time anymore in the present model. More especially, we assume the server consume one time slot to confirm the system state when the system is entirely empty.\n\nMix-Service Discipline. Exhaustive discipline is specified for the key queue and 1-limited discipline for normal queues, so that the entire customers in the key queue could be served in the present server round, while those who are in normal queues might need several rounds when there are more than one customer in the buffer. Let Fh denote the duration of a service period for the customers arrive during arbitrary time slot in Qh. This service period consists of the services of its ancestral customers arriving during the exact slot and the services of the offspring line of the ancestral customers . The generating function of Fh is denoted by Fh(zh)=E[zhFh]. Such a functional equation has already been derived in as Fh(zh)=Ah(Bh(zhF(zh))).\n\nIn the remainder of this paper, we are interested in the queue length distributions at the polling epoch of Qi and Qh. Let ξj(n) denote the number of customers present at Qj at tn when the server starts a visit period at Qi, and let ξj(n) denote the number of customers present at Qj at tn when the server starts a visit period at Qh successively with the service of Qi. The joint distribution of ξj(n+1) and ξj(n) is represented by the N-dimensional PGF Gi+1(z1,,zN,zh) and Gih(z1,,zN,zh).\n\nWe analyze the system under stability conditions (i=1Nρi+ρh<1) . Normal queues in the present model are served in a 1-limited manner, which does not satisfy the well-known branching property in polling systems. Therefore, more specifically, in the analyses of mean waiting time, we assume the normal queues are symmetric; that is, normal queues have the same customer arrival rate and service rate.\n\nIn this section, we derive explicit expression for the joint queue length distribution. In Section 3.1, we first obtain expressions for Gi+1(z1,,zN,zh) and Gih(z1,,zN,zh), the joint queue length PGF at the polling epoch at Qi+1 and Qh. These results ultimately lead in Section 3.2 to the first and second moment of the PGF, and obtain the expressions for E[Wi] and E[Wh], the mean waiting time of type-i and type-h customers that arrive at an arbitrary point in time.\n\n3.1. Joint Queue Length Distribution at Polling Epoch\n\nAssuming that the server begin the service of Qi at tn, define a random variable ξj(n) as the number of type-j (j=1,2,,N,h) customers at time tn. Then the status of the entire polling model at time tn can be represented as {ξ1(n),,ξN(n),ξh(n)}. Denote ξj(n+k) as the number of type-j customers at tn+k, the polling epoch of Qi+k. The status of the entire polling model at time tn+k can be represented as {ξ1(n+k),,ξN(n+k),ξh(n+k)} while ξi(n) is the number of type-j customers in at time tn, at which the server begins providing service to Qh and the status of the entire polling model at time tn can be represented as {ξ1(n),,ξN(n),ξh(n)}. Under the necessary and sufficient condition for the stability of the system i=1Nρi+ρh<1, the probability distribution is defined as(1)limnPξjn=xj;j=1,,N,h=πix1,,xN,xh,limnpξjn=yj;j=1,,N,h=πihy1,,yN,yh.The generating functions at tn and tn are(2)Giz1,,zN,zh=x1=0xN=0xh=0z1x1zNxnzhxhπix1,,xN,xhi=1,2,,N,Gihz1,,zN,zh=y1=0yN=0yh=0z1y1zNynzhyhπihy1,,yN,yhi=1,2,,N.\n\nAccording to the proposed mechanism, the system variables have the following equations. When the server begins the service on Qi+1 at tn+1, we have(3)ξjn+1=ξjn+ηjνhjh0j=h.vj(n) is the service time in Qj and ηk(vj) is the number of arrivals to Qk during vj(n).\n\nThe server just finishes the service of Qh in an exhaustive manner and starts the polling on Qi+1 at tn+1. Such a functional equation of exhaustive service has already been derived in . Applying these results to our case, we obtain(4)Gi+1z1,z2,,zN,zh=limnEj=1Nzjξin+1zhξhn+1=Gihz1,z2,,zN,Bhj=1NAjzjFhj=1NAjzj.\n\nThe expression can be interpreted as follows. At the start of the visit period at Qi+1, type-i customers are those at the polling epoch of Qh plus the new customers arriving at each queue during the service period of the Qh in exhaustive scheme, and no type-h customer resumes at that moment.\n\nWhen the server begins the service on Qh at tn, we have(5)ξjn=ξjn+ηjνi,jih,ξin+ηiνi-1,j=iξin0,ηjνi,j=h,ξjn=ξjn,jih,0,j=iξin=0,0,j=hνj(n) is the service time in Qj, and ηk(νj) is the number of arrivals to Qk during νj(n).\n\nIn our case, for normal queues, the server just polls the active queues with customers in parallel 1-limited manner. To gain more insight in the state-dependent service discipline, let Pi denote the queue length at the service epoch in an M/G/1 queue with the same arrival process and service-time distribution as Qi. We assume that the k customers have waited in Qi at the start of the busy period with probability pk[0,1), k=0pk=1. Then we can acquire the queue length generating function at the service epoch as Pizi=Aizik=0pkzik, where Ai(zi) is the PGF of the arrival process as defined in Section 2. Specifically, the server does not provide service when the queue length is zero, so we assume that k customers resumed after the end of the busy time in 1-limited service with the probability of pk[0,1), and pk=pk+1 for k=0,1,. Consequently, the probability space could be rebuilt as(6)Pizi=BiAiziAizip0+k=0pkzik=BiAizik=0pkzik-p0zi0zi+p0zi0.\n\nWith the definition of Pi(zi), we have(7)Pizi=BiAiziPizi-Pizizi=0zi+Pizizi=0.Applying these results to our case, we obtain(8)Gihz1,,zN,zh=limnEj=1Nzjξinzhξhn=1ziBij=1NAjzjAhzhGiz1,,zN,zh-Giz1,,zN,zhzi=0+Giz1,,zN,zhzi=0-Giz1,,zN,zhz1,,zN,zh=0+j=1NAjzjAhzhGiz1,,zN,zhz1,,zN,zh=0.\n\nThe expression can be interpreted as follows. At the start of the visit period at Qh, in the case that the former Qi is active, one type-i customer would have been served at tn and new customers arrived at each queue during the service period of the exact type-i customer. The server would skip Qi to Qi+1 when Qi is empty; in that case, the distribution of the number of customers in the systems is represented by the generating function Giz1,,zi,,zN-1,zhzi=0, with the exception as the system is entirely empty, which is represented by the generating function Giz1,,zi,,zN-1,zhz1,,zN-1,zh=0. When the system is entirely empty, the server will stop providing service for one time slot until new customers arrive during this time slot, and this number of customers is represented by the last partition of the addition formula.\n\n3.2. Expression for the Mean Waiting Time\n\nNow we have derived expressions for the PGF Gi+1z1,,zi,,zN,zh and Gihz1,,zi,,zN,zh pertaining to the queue length at polling epoch of Qi+1 and Qh, we use these results to obtain E[Wi], the mean waiting time of type-i normal customers, and E[Wh], the mean waiting time of type-h high priority customers.\n\n3.2.1. The First and Second Moment of <inline-formula><mml:math xmlns:mml=\"http://www.w3.org/1998/Math/MathML\" id=\"M146\"><mml:msub><mml:mrow><mml:mi>G</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi><mml:mo>+</mml:mo><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mfenced separators=\"\|\"><mml:mrow><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mn mathvariant=\"normal\">1</mml:mn></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:mo>…</mml:mo><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:mo>…</mml:mo><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>N</mml:mi></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>h</mml:mi></mml:mrow></mml:msub></mml:mrow></mml:mfenced></mml:math></inline-formula> and <inline-formula><mml:math xmlns:mml=\"http://www.w3.org/1998/Math/MathML\" id=\"M147\"><mml:msub><mml:mrow><mml:mi>G</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi><mml:mi>h</mml:mi></mml:mrow></mml:msub><mml:mfenced separators=\"\|\"><mml:mrow><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mn mathvariant=\"normal\">1</mml:mn></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:mo>…</mml:mo><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:mo>…</mml:mo><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>N</mml:mi></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>z</mml:mi></mml:mrow><mml:mrow><mml:mi>h</mml:mi></mml:mrow></mml:msub></mml:mrow></mml:mfenced></mml:math></inline-formula>\n\nTo start the analysis of mean waiting time of type-j customers, we need to calculate the generating functions and its derivation at the point z=1, z is the abbreviation of the (1×N+1) vector of (z1,,zi,,zN,zh), and 1 is the (1×N+1) vector with 1.\n\nG i + 1 ( z ) is the PGF of the joint queue length at the polling epoch of Qi, so we have(9)Giz1,,zi,,zN,zh=x1=0xi=0xN=0xh=0z1x1zixizNxnzhxhPξ1n=x1,,ξin=xi,,ξNn=xN,ξhn=xh=x1=0xi=0xN=0xh=0z1x1zixizNxnzhxhPξ1n=x1,,ξNn=xN,ξhn=xhξin=xiPξin=xi.\n\nTaking the kth derivative with respect to zi yields(10)kGiz1,z2,,zi,,zN,zhzik=x1=0xN=0xh=0z1x1zixi-kzNxNzhxhxi!xi-k!Pξ1n=x1,,ξNn=xN,ξhn=xhξin=xiPξin=xi.\n\nSetting zi=0 yields(11)kGiz1,z2,,zi,,zN,zhzikzi=0=x1=0xN=0xh=0z1x11zNxNzhxhk!Pξ1n=x1,,ξNn=xN,ξhn=xhξin=kPξin=k=k!Pξin=kx1=0xN=0xh=0z1x11zNxNzhxhPξ1n=x1,,ξNn=xN,ξhn=xhξin=k=k!Pξin=kEz1ξ1n1zNξNnzhξhnξin=k.Rearranging terms and setting k=0, we have(12)Giz1,z2,,zi,,zN,zhzi=0=Pξin=0Ez1ξ1n1zNξNnzhξhnξin=0,Gi1i=Pξin=0.Extending this result we have(13)Gi0=Pξ1n=0,,ξin=0,,ξNn=0,ξhn=0.0 is the (1×N+1) vector with 0, and 1j is the (1×N+1) vector with 0 in jth position and 1 in all other entries.\n\nDefine the first derivative of Gi(z) and Gih(z) at z=1 as (14)gij=limz1,,zi,,zN,zh1Gizzj,gihj=limz1,,zi,,zN,zh1Gihzzj,j,k=1,2,,N,h.(15)gihj=βiλj1-Gi1i+gij+λjGi0(16)gihi=βiλi-11-Gi1i+gii+λiGi0(17)gihh=βiλh1-Gi1i+λhGi0(18)gi+1i=gihi+gihhβhλi1+Fh1(19)gi+1j=gihj+gihhβhλj1+Fh1.Calculate j=1Ngj+1k yields(20)1-Gi1i=NλiGi01-ρh-Nρ.Define the second derivative of Gi(z) and Gih(z) at z=1 as(21)gij,k=limz1,,zi,,zN,zh12Gizzjzkgi0j,k=limz1,,zi-1,zi+1,,zN,zh12Gizzi=0zjzkgi00j,k=limz1,,zN,zh02Gizz1,,zN,zh=0zjzkgihj,k=limz1,,zN,zh12Gihzzjzki=1,2,,Nj,k=1,2,,N,h.\n\nSubstitute (4) and (8) into the above second derivative formulas.\n\nWe assume the N normal queues are symmetrical; that is, λi=λ, βi=β, i=1,2,,N. Then simplifying these we get the second derivative of Gi(z) and Gih(z) at z=1 as follows:(22)gihh,h=B1λh21-Gi1i+βAh11-Gi1i+Ah1Gi0.(23)gii=1-Gi1i211-ρh-Nρ1-ρhρhA1λ1-ρh2+Ahβh2+λhBh+NB1λ2+NβA1+λ1-ρh+λhλBh11-ρh-Nρ+1-Gi1i.\n\nRemark 1.\n\nThough gi(i) is the first derivative at z=1Gi(z) in definition, it is clear that it contains the second moment parameter as Aj(1) and Bj(1). So, gi(i) is a second moment parameter for the system performance.\n\n3.2.2. Analysis of <inline-formula><mml:math xmlns:mml=\"http://www.w3.org/1998/Math/MathML\" id=\"M197\"><mml:mi>E</mml:mi><mml:mo stretchy=\"false\">[</mml:mo><mml:msub><mml:mrow><mml:mi>W</mml:mi></mml:mrow><mml:mrow><mml:mi>h</mml:mi></mml:mrow></mml:msub><mml:mo stretchy=\"false\">]</mml:mo></mml:math></inline-formula> and <inline-formula><mml:math xmlns:mml=\"http://www.w3.org/1998/Math/MathML\" id=\"M198\"><mml:mi>E</mml:mi><mml:mo stretchy=\"false\">[</mml:mo><mml:msub><mml:mrow><mml:mi>W</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi></mml:mrow></mml:msub><mml:mo stretchy=\"false\">]</mml:mo></mml:math></inline-formula>\n\nDefine Wh and Wi as the waiting time of type-h and type-i customers, which denotes the time from the epoch when a customer arrives at the queue to the time it is served. In the present model, high priority type-h customers are served in the exhaustive service and normal type-i customers are served in 1-limited service. Based on the related research works in , the mean waiting time of type-h customers E[Wh] and the type-i customers E[Wi] can be calculated as follows:(24)EWh=gihh,h2λhgihh-Ah12λh21+ρh+λhBh121-ρh,(25)EWi=1λ1-Gi1igii-1λ-A12λ2.\n\nTaking (17), (22) in (24) in the above expressions, we have(26)EWh=121-ρh1-ρh+NλB1+λhBh1-12λh21+ρhAh1.\n\nTaking (17), (22), and (23) in (25) in the above expressions, we have(27)EWi=12λ11-ρh-Nρ1-ρhρhA1λi1-ρh2+Ahβh2+λhBh+NB1λ2+NβA1+λ1-ρh+λhλBh11-ρh-Nρ-A12λ2.\n\n4. Numerical Study\n\nIn this section we study the accuracy of the theoretical analysis and compare the mean waiting time of the present model with two existing two-level polling models. Consider an N+1 queues’ model with one high priority queue Qh and N normal queues Qi (i=1,,N) defined as follows: the service times of all customers are exponentially distributed with mean β in Qi and βh in Qh. The arrival processes are Poisson process with rate λ in Qi and λh in Qh. The relative parameter values are listed in Table 1, in which {a:k:b} means the parameter is varied between a and b in steps of k.\n\nTest bed used to compare the mean waiting time.\n\n Parameter Number of normal queues Arrival rate Service time Switch over time Notation value N λ λ h β β h γ Figure 1(a) 1 : 1 : 9 0.04 0.1 2 2 Figure 1(b) 4 0.02 0.1 : 0.05 : 0.4 1 2 — Figure 1(c) 4 0.02 : 0.02 : 0.18 0.1 1 2 — Figure 1(d) 4 0.02 0.1 2 1 : 1 : 9 — Figure 1(e) 4 0.02 0.1 1 : 1 : 10 2 — Figure 2 4 0.01 : 0.01 : 0.09 0.1 2 2 1\n\nFrom Figure 1, we can clearly see that, firstly, the theoretical value and the simulation result coincided with each other. Secondly, when the total offered load grew with the arrival rate, service time, and the number of queues, with the mean waiting time increasing distinctly in Qi, while the performances in Qh are much better, both queue and mean waiting time are much lower than normal queues, and the growth in Qh with the total offered load presents much more smoothly.\n\nTheoretical and simulation values of E[Wh] and E[Wi] from different values of the load increasing with the increasing of the number of normal queues. (a) is the total offered load increasing with the growth of the number of normal queues. (b) is the total offered load increasing with the growth of the arrival rate of Qh. (c) is the total offered load increasing with the growth of the arrival rate of Qi. (d) is the total offered load increasing with the growth of the service time of Qh. (e) is the total offered load increasing with the growth of the service time of Qi.\n\nIt is worth considering whether the state-dependent mechanism improves the performance of the system comparing with the existing two-level polling systems. In order to answer this question, we compare a classical two-level system with switch-over time , abbreviated as classical system and a parallel two-level system , abbreviated as parallel system in Figure 2. The service discipline in the comparisons is 1-limited service for normal queues and exhaustive service for the key queue. Overall models have the same test bed as shown in Table 1. We just vary the working mechanism.\n\nComparing of mean waiting time among the classical two-level system , the parallel two-level system , and the state-dependent two-level system. (a) is the theoretical value comparison of E[Wi] with the growth of the arrival rate in Qi. (b) is the theoretical value comparison of E[Wh] with the growth of the arrival rate in Qi.\n\nFigure 2 shows the mean waiting time of normal queues in (a) and mean waiting time of key queue in (b). Comparing with the forgoing, the state-dependent system achieves a better performance in delay guarantee and stability. It is clear in Figure 2(a), for lower load, in most of the cases, that there is no customer in the buffers; thus a switch-over time is necessary when the server switches between Qi and Qh in the classical and parallel system, while the empty queues would be skipped in the present model. Therefore, customers in the state-dependent system achieve a lower mean waiting time, which is under 20% of the forgoing. In the heavy traffic, the server could not provide service in the necessary switch-over time for the classical system; consequently, it becomes unstable when the arrival rate of Qi grows over 0.06 in this case. The parallel system and the state-dependent system have better performance in system stability; especially in state-dependent system, the mean waiting time of the normal customers has less than 50% of which in the parallel system. A conclusion can be drawn from a comparison between Figures 2(a) and 2(b), which is that for all three two-level models the mean waiting time of the customers in key queue is significantly lower than that in normal queues, and as illustrated in Figure 2(b), the mean waiting time for h-type customers in state-dependent system is lower than that of the others.\n\n5. Conclusion\n\nWhen comparing the model of the present paper with the existing literature, the contribution of the present paper is twofold. One of the most striking differences is the queues which are partitioned as active queue and idle queue by their buffer condition, and only active queues with customers waiting in the buffer could be visited by the server in a two-level order. As illustrated in the numerical example, both i-type customers in normal queues and h-type customers in key queue acquire better delay performance than those in systems without queue-stated differentiation. Another notable contribution of the paper is that we achieve the closed-form exact expressions of the mean waiting time for customers in normal queues and key queue, under the assumption of the symmetric of normal queues. The total unknowns in these equations are all first moments of random variables and, thus, no correlation terms are required.\n\nConflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of this paper.\n\nAcknowledgments\n\nThis work was supported by a Grant from the National Science Foundation of China (nos. 61463051 and 61463054), the National Science Foundation of Yunnan Province (no. 2012FD002), and the Science Foundation of Yunnan Provincial Department (no. 2014Z010).\n\nTakagi H. Analysis of polling systems Performance Evaluation 1985 5 3 206 Levy H. Sidi M. Polling systems: applications, modeling, and optimization IEEE Transactions on Communications 1990 38 10 1750 1760 10.1109/26.61446 2-s2.0-0025502587 Boon M. A. A. van der Mei R. D. Winands E. M. M. Applications of polling systems Surveys in Operations Research and Management Science 2011 16 2 67 82 10.1016/j.sorms.2011.01.001 2-s2.0-79957844270 Vishnevskii V. M. Semenova O. M. Mathematical models to study the polling systems Automation and Remote Control 2006 67 2 173 220 Yang Z.-J. Zhao D.-F. Ding H.-W. Zhao Y.-F. Research on two-class priority based polling system Acta Electronica Sinica 2009 37 7 1452 1456 2-s2.0-69249127140 Liu Q. Zhao D. Zhou D. An analytic model for enhancing IEEE 802.11 point coordination function media access control protocol European Transactions on Telecommunications 2011 22 6 332 338 10.1002/ett.1482 2-s2.0-81355138818 Yang Z.-J. Ding H.-W. Chen C.-L. Research on E(x) characteristics of two-class polling system of exhaustive-gated service Tien Tzu Hsueh Pao/Acta Electronica Sinica 2014 42 4 774 778 10.3969/j.issn.0372-2112.2014.04.023 2-s2.0-84904046560 Guan Z. Zhao D. Zhao Y. A discrete time two-level mixed service parallel polling model Journal of Electronics 2012 29 1-2 103 110 10.1007/s11767-012-0781-3 2-s2.0-84861976713 Dorsman J. P. Boxma O. J. van der Mei R. D. On two-queue Markovian polling systems with exhaustive service Queueing Systems 2014 78 4 287 311 10.1007/s11134-014-9413-y MR3269260 2-s2.0-84910150691 Boon M. A. Adan I. J. Boxma O. J. A two-queue polling model with two priority levels in the first queue Discrete Event Dynamic Systems 2010 20 4 511 536 10.1007/s10626-009-0072-9 MR2721286 2-s2.0-77956790267 Dongfeng Z. Bihai L. sumin Z. Study of a polling systems with limited service Journal of Electronics 1997 19 1 44 49 Ibe O. C. Cheng X. Stability conditions for multi-queue systems with cyclic service IEEE Transactions on Automatic Control 1988 33 1 102 103 10.1109/9.370 2-s2.0-0023842924 Boxma O. J. Groenendijk W. P. Pseudoconservation laws in cyclic-service systems Journal of Applied Probability 1987 24 4 949 964 10.2307/3214218 MR913834 Zhao D. Zheng S. Analysis on a polling model with exhaustive service Acta Electronica Sinica 1994 22 5 102 107 2-s2.0-0028422934" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90160745,"math_prob":0.884889,"size":4030,"snap":"2019-43-2019-47","text_gpt3_token_len":850,"char_repetition_ratio":0.12468952,"word_repetition_ratio":0.07196402,"special_character_ratio":0.2191067,"punctuation_ratio":0.07032349,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9589004,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T13:51:14Z\",\"WARC-Record-ID\":\"<urn:uuid:9ed1eaa5-3f0b-403a-b7f4-52d8e5f0c072>\",\"Content-Length\":\"235499\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a528b2e7-c4c1-4867-8c5b-7e69d77a8557>\",\"WARC-Concurrent-To\":\"<urn:uuid:1e7e46b7-48d0-4d36-80b0-8d92fabcd87c>\",\"WARC-IP-Address\":\"52.216.130.3\",\"WARC-Target-URI\":\"http://downloads.hindawi.com/journals/mpe/2015/109325.xml\",\"WARC-Payload-Digest\":\"sha1:HYJOAI5SJL2L5RBIL6GRGNVRYJT7ZT7X\",\"WARC-Block-Digest\":\"sha1:QRPMM5636FJFUDJZYDKZEDBB5HWGL3P3\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669795.59_warc_CC-MAIN-20191118131311-20191118155311-00406.warc.gz\"}"}
https://communities.sas.com/t5/SAS-Procedures/Adjust-simulate-how-does-it-adjust/td-p/301201	[ "Hi, I am trying to understand what adjust=simulate does (within the lsmestimate option in proc mixed and other procs, SAS 9.4). I am comfortable with what it is doing at a high level, but I can not get my head around how it does it. I have read the reference below and the on line help in SAS (also below) but I am finding it hard to grasp the principle of how the the adjustment is done. I believe simulations are done and the contrasts are calculated for the specified contrasts for each simulation and a p-value is obtained from a mutivariate t but I'm unclear on how the simulations are generated and how the adjusted p-value is obtained. A simple explanation or even examples would be great. I have emailed SAS but they have no further info they can provide to me. thank you.\n\nhttp://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/viewer.htm#statug_glm_details2...\n\n1 ACCEPTED SOLUTION\n\nAccepted Solutions\n\nWhat is your statistical background and previous experience with simulation methods in statistics? From your questions, it sounds like you need to start with simpler situations to better unerstand the main ideas. Try reading about using simulation to estimate the power of a statistical test and Monte Carlo methods for contingency tables in SAS.\n\nAlso what is the application here? Are you merely intellectually curious about the details of the algorithm? Or do you need to implement a similar algoithm in a different situation?\n\n3 REPLIES 3\n\nIt's basically a multivariate generalization of using simulation to estimate the coverage probability of a 1-D confidence interval.\n\nSuppose you want to estimate ONE linear combination of the betas: c`beta. You also want a CI. The statistic c`beta is normally distributed, so the CI will be of the form\n\nc`beta +/- delta stderr(c`beta).\n\nThe challenge is to choose delta. For simple estimates (like c=(1 0 0 .. 0)), you can choose delta to be a critical value of the t distribution: delta=t(1-alpha/2, df). The t distribution is used instead of the normal distribution to adjust for the finite sample size.\n\nNow suppose that you have k linear combinations and you want SIMULTANEOUS CIs. The k estimates are jointly MVN with mean and covariance that can be computed because you are assuming a GLM. So simulate a bunch of values from the appropriate MVN distribution (actually multivariate t) and then use quantiles of the simulated distribution of estimates to find the critical value (delta) that works simultaneously for all the estimates. Make a bunch of draws of the form (t1, t2, ..., tk) and for each draw compute the statistic max(\|t1\|, ..., \|tk\|). The union of those max statistics has an empirical distribution. The Edwards/Berrry paper says that you can choose delta to be the (1-alpha)th quantile of the empirical distribution.\n\n When you say \"simulate a bunch of values from the appropriate MVN distribution (actually multivariate t)\". Do you mean, generate a lot of datasets, say 100 to keep things simple, from a mvt which has means which match the means of your k contrasts/estimates and covariance that matches the covariance of your k contrasts/means?\n\n* Then you say \"use quantiles of the simulated distribution of estimates to find the critical value (delta) that works simultaneously for all the estimates.\". My guess at what this means is, for the each datasets you have simulated, calculate the value of each of the k contrasts, so if k=2, and 100 datasets, you have 100 values for each contrast. Then I'm not sure how you find the critical value. You mention later to use (1-alpha)th quantile. So if alpha is 5%, is the 95th ordered value from the 100 values I have for each k my critical value?\n\n*Then \"Make a bunch of draws of the form (t1, t2, ..., tk) and for each draw compute the statistic max(\|t1\|, ..., \|tk\|).\". I'm lost here, am I randomly selecting a value from the 100 values I have for each k, so in my case I would have two values, then select the max of these and build up a distribution of these I guess. Sorry, I'm not sure what happens now.\n\nWhat is your statistical background and previous experience with simulation methods in statistics? From your questions, it sounds like you need to start with simpler situations to better unerstand the main ideas. Try reading about using simulation to estimate the power of a statistical test and Monte Carlo methods for contingency tables in SAS.\n\nAlso what is the application here? Are you merely intellectually curious about the details of the algorithm? Or do you need to implement a similar algoithm in a different situation?\n\nDiscussion stats\n• 3 replies\n• 1455 views\n• 1 like\n• 2 in conversation" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9092807,"math_prob":0.9375539,"size":5414,"snap":"2023-40-2023-50","text_gpt3_token_len":1183,"char_repetition_ratio":0.12606284,"word_repetition_ratio":0.27734807,"special_character_ratio":0.22072405,"punctuation_ratio":0.11797753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9873639,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T10:46:08Z\",\"WARC-Record-ID\":\"<urn:uuid:d3d6e280-4d55-49f3-b1d5-5893c896c69d>\",\"Content-Length\":\"208899\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d723028-06cd-409a-8b7f-e177f0d287ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2d7032b-9833-4a1d-aa62-3e6a39fc5cb3>\",\"WARC-IP-Address\":\"18.67.65.90\",\"WARC-Target-URI\":\"https://communities.sas.com/t5/SAS-Procedures/Adjust-simulate-how-does-it-adjust/td-p/301201\",\"WARC-Payload-Digest\":\"sha1:5CAUWI2BXQGXQY3NOXTXGNIPNBWCENVH\",\"WARC-Block-Digest\":\"sha1:RR7ZV4LK5TZVRBDDJSS3NRQXYNJPFQD6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100527.35_warc_CC-MAIN-20231204083733-20231204113733-00596.warc.gz\"}"}
https://www.jetpunk.com/user-quizzes/19203/intro-to-motion-physics-2/stats	[ "# Statistics for Introduction to Motion Physics Quiz #2\n\n### General Stats\n\n• This quiz has been taken 655 times\n• The average score is 9 of 15\n\nEnergy associated with movementKinetic Energy\n90%\nStored EnergyPotential Energy\n89%\nMeasure of how heavy you areMass\n82%\nEnergy and the above answer are measured in __________Joules (J)\n75%\nSpeed is to scalar as Velocity is to __________Vector\n73%\nMeasure of change in energy (force times distance)Work\n67%\nEnergy present in objects that are in a position from which they could fall as a result of the force of gravityGravitational Potential Energy\n66%\nThe force applied to an object by the Earth due to gravitational attractionWeight\n63%\nNegative AccelerationDeceleration\n62%\nHow useful a system as an energy converter is [(Useful energy output over energy input) times 100%]Efficiency\n57%\nProperty that makes objects resist changes in their motionInertia\n53%\nEnergy stored within the nucleus of all atomsNuclear Energy\n49%\nEnergy present in objects or groups of objects in which positively and negatively charged particles are separated ; also present when the charges are brought togetherElectrical Potential Energy\n32%\nEnergy can never be created nor destroyed - only transferred and storedLaw of Conservation of Energy\n29%\nEnergy present in all substances as a result of the electrical forces that hold atoms together ; A form of the above type of energyChemical Potential Energy\n26%" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9113481,"math_prob":0.86968017,"size":1388,"snap":"2021-21-2021-25","text_gpt3_token_len":308,"char_repetition_ratio":0.14378613,"word_repetition_ratio":0.017316017,"special_character_ratio":0.2615274,"punctuation_ratio":0.0091743115,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9618333,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T04:44:10Z\",\"WARC-Record-ID\":\"<urn:uuid:ac9a1561-5377-4bd2-8bb5-fcc0d2bf0bab>\",\"Content-Length\":\"29424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14e0724e-e9fc-474c-8b1c-b16d497e22cb>\",\"WARC-Concurrent-To\":\"<urn:uuid:91e0d72d-a2d1-4729-b227-5b089b9b63dc>\",\"WARC-IP-Address\":\"34.193.34.229\",\"WARC-Target-URI\":\"https://www.jetpunk.com/user-quizzes/19203/intro-to-motion-physics-2/stats\",\"WARC-Payload-Digest\":\"sha1:NG727TJXR7CUBPV4OBFVV2FK3PXCJ45F\",\"WARC-Block-Digest\":\"sha1:UB2FQ3WN3CZV5B5IC26AGNTOMRRWIEQ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988725.79_warc_CC-MAIN-20210506023918-20210506053918-00254.warc.gz\"}"}
https://rdrr.io/cran/rsimsum/man/nsim.html	[ "# nsim: Compute number of simulations required In rsimsum: Analysis of Simulation Studies Including Monte Carlo Error\n\n nsim R Documentation\n\n## Compute number of simulations required\n\n### Description\n\nThe function `nsim` computes the number of simulations B to perform based on the accuracy of an estimate of interest, using the following equation:\n\nB = [((Z(1 - α / 2) + Z(1 - θ)) σ) / δ] ^ 2\n\nwhere δ is the specified level of accuracy of the estimate of interest you are willing to accept (i.e. the permissible difference from the true value β), Z(1 - α / 2) is the (1 - α / 2) quantile of the standard normal distribution, Z(1 - θ) is the (1 - θ) quantile of the standard normal distribution with 1 - θ being the power to detect a specific difference from the true value as significant, and σ ^ 2] is the variance of the parameter of interest.\n\n### Usage\n\n```nsim(alpha, sigma, delta, power = 0.5)\n```\n\n### Arguments\n\n `alpha` Significance level. Must be a value between 0 and 1. `sigma` Variance for the parameter of interest. Must be greater than 0. `delta` Specified level of accuracy of the estimate of interest you are willing to accept. Must be greater than 0. `power` Power to detect a specific difference from the true value as significant. Must be a value between 0 and 1. Defaults to 0.5, e.g. a power of 50%.\n\n### Value\n\nA scalar value B representing the number of simulations to perform based on the accuracy required.\n\n### References\n\nBurton, A., Douglas G. Altman, P. Royston. et al. 2006. The design of simulation studies in medical statistics. Statistics in Medicine 25: 4279-4292 doi: 10.1002/sim.2673\n\n### Examples\n\n```# Number of simulations required to produce an estimate to within 5%\n# accuracy of the true coefficient of 0.349 with a 5% significance level,\n# assuming the variance of the estimate is 0.0166 and 50% power:\nnsim(alpha = 0.05, sigma = sqrt(0.0166), delta = 0.349 * 5 / 100, power = 0.5)\n\n# Number of simulations required to produce an estimate to within 1%\n# accuracy of the true coefficient of 0.349 with a 5% significance level,\n# assuming the variance of the estimate is 0.0166 and 50% power:\nnsim(alpha = 0.05, sigma = sqrt(0.0166), delta = 0.349 * 1 / 100, power = 0.5)\n```\n\nrsimsum documentation built on Aug. 17, 2022, 5:07 p.m." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79356915,"math_prob":0.99409664,"size":2074,"snap":"2023-40-2023-50","text_gpt3_token_len":585,"char_repetition_ratio":0.11932367,"word_repetition_ratio":0.3298153,"special_character_ratio":0.3090646,"punctuation_ratio":0.14942528,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99908984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T12:51:44Z\",\"WARC-Record-ID\":\"<urn:uuid:4f2922ea-3703-44e6-b47d-a31d6834d1cc>\",\"Content-Length\":\"30188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9507e2f-805a-4755-9419-731e91f8ac7b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9fa33f8-1071-42c9-b5f4-f14d1514cc23>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/rsimsum/man/nsim.html\",\"WARC-Payload-Digest\":\"sha1:X5QSESW6XDIAQQ7DDBFNGYHKTHJIPDOV\",\"WARC-Block-Digest\":\"sha1:3G5TNG7U6MQVH47CLI5FIN7PCDCBT4F5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510516.56_warc_CC-MAIN-20230929122500-20230929152500-00324.warc.gz\"}"}
https://www.frontiersin.org/articles/10.3389/fict.2016.00019/full	[ "World-class research. Ultimate impact.\nMore on impact ›\n\n# Frontiers in ICT", null, "## Original Research ARTICLE\n\nFront. ICT, 15 September 2016 \| https://doi.org/10.3389/fict.2016.00019\n\n# Energetic Cost of Superadiabatic Quantum Computation\n\n• 1Instituto de Física, Universidade Federal Fluminense, Niterói, Brazil\n• 2Information Sciences Institute, University of Southern California, Marina del Rey, CA, USA\n• 3Ming Hsieh Department of Electrical Engineering, Center for Quantum Information Science and Technology, University of Southern California, Los Angeles, CA, USA\n\nWe discuss the energetic cost of superadiabatic models of quantum computation. Specifically, we investigate the energy–time complementarity in general transitionless controlled evolutions and in shortcuts to the adiabatic quantum search over an unstructured list. We show that the additional energy resources required by superadiabaticity for arbitrary controlled evolutions can be minimized by using probabilistic dynamics, so that the optimal success probability is fixed by the choice of the evolution time. In the case of analog quantum search, we show that the superadiabatic approach induces a non-oracular counter-diabatic Hamiltonian, with the same energy–time complexity as equivalent adiabatic implementations.\n\n## 1. Introduction\n\nShortcuts to adiabatic passage (Demirplak and Rice, 2003, 2005; Berry, 2009; Torrontegui et al., 2013) provide a remarkable mechanism for speeding up quantum tasks, which can be achieved through the use of a counter-diabatic assistant driving. These techniques have been introduced to mimic the transitionless adiabatic dynamics, but with the usual constraint on the adiabatic runtime lifted. Transitionless quantum driving has been applied to a number of quantum information protocols, such as population transfer (Chen et al., 2014a; Lu et al., 2014a) and entanglement generation (Chen et al., 2014b, 2015a,b; Lu et al., 2014b). In the context of many-body systems, realizable settings have been investigated for assisted evolutions in quantum critical phenomena (del Campo et al., 2012; del Campo, 2013; Saberi et al., 2014). More recently, counter-diabatic approaches have been proposed for fast implementation of individual unitaries in quantum circuits, leading to universal superadiabatic schemes of quantum computing (QC) via local Hamiltonians (Santos and Sarandy, 2015; Santos et al., 2016). Such methods may be potentially relevant to accelerating the implementation of n-qubit controlled gates in digitized proposals of adiabatic quantum computing [see Kieferová and Wiebe (2014), Martinis and Geller (2014), Barends et al. (2015), and Hen (2015)].\n\nThe superadiabatic speedup is intrinsically connected with an increase of the energy resources demanded by the quantum computer (Santos and Sarandy, 2015; Santos et al., 2016), which in turn implies a rather versatile computational cost that is controlled by the energetic capacity available to the physical apparatus. Here, we show that this energy–time complementarity can be exploited in quantum information processing. First, we consider controlled evolutions (CE) as a mechanism to implement superadiabatic universal QC (Santos and Sarandy, 2015), which generalizes the original adiabatic approach introduced in Hen (2015). We then show that, within the superadiabatic scenario, the energetic cost can be minimized by replacing the deterministic realization of quantum gates for probabilistic implementations based on a probability distribution of a binary random variable described by an angle parameter. By doing so, the energy expense can be minimized by adjusting the probability distribution, provided the choice of the evolution time of the computational process. Second, we analyze the effects of the energy–time complementarity in analog quantum search (Grover, 1997), where the oracular approach designed by the local adiabatic Grover algorithm is known to be optimal (van Dam et al., 2001; Roland and Cerf, 2002). In this case, we show that the superadiabatic approach naturally requires an unphysical non-oracular counter-diabatic Hamiltonian, with the energy–time complexity equivalent to non-oracular adiabatic implementations.\n\nThe paper is organized as follows. In Section 2, we describe the adiabatic implementation of quantum gates via CE and several adiabatic quantum search approaches. We then provide their superadiabatic versions and introduce the metric for energetic cost used in our work. In Section 3, we investigate the energy complexity of the superadiabatic realizations of both quantum gates via CE and analog quantum search. In particular, we consider the properties of the probabilistic model of QC through CE and the consequences of the energy–time complementarity for the search problem. In Section 4, we present our conclusions and future perspectives.\n\n## 2. Materials and Methods\n\nOur aim in this Section is to discuss adiabatic implementations of QC, their superadiabatic generalizations, and the energetic cost measure adopted in this work.\n\n### 2.1. Quantum Gates by Adiabatic Controlled Evolutions\n\nLet us begin by using adiabatic CE (Hen, 2015) to implement n-controlled gates (Santos and Sarandy, 2015). To this end, we will consider the adiabatic evolution of a composite system 𝒯 𝒜 associated with a Hilbert space ℋ𝒯 ⊗ ℋ𝒜, where 𝒯 denotes a target subsystem containing n + 1 qubits and 𝒜 denotes an auxiliary subsystem containing a single qubit. We will use the first n qubits of 𝒯 as the control register of the n-controlled gate, while the last qubit will play the role of its target register. Then, a rotation of the target qubit of an angle ϕ around a direction $n^$ in the Bloch sphere will be performed when the state of the control register is \|11⋯ 1⟩. We will adopt here the decimal representation \|11⋯ 1⟩ ≡ \|N − 1⟩, with N = 2n. An n-controlled rotation over a single qubit can be adibatically implemented by preparing the auxiliary qubit in the initial state \|0⟩, with the adiabatic Hamiltonian given by (Santos and Sarandy, 2015)\n\n$H(s)=𝟙−PN−1,n−⊗H0(s)+PN−1,n−⊗Hϕ(s),$\n\nwhere $Pk,n±=\|k⟩ ⟨k\|⊗\|n^±⟩ ⟨n^±\|$ is the set of all orthogonal projectors on the subspace 𝒯 and $\|n^±⟩ ⟨n^±\|=1∕2(𝟙±n^⋅σ→)$, with $σ→=σx,σy,σz$. The Hamiltonians H0(s) and Hϕ(s) are given by\n\n$Hξ(s)=−ℏωσzcosθ(s)+sinθ(s)[σxcosξ+σysinξ],$\n\nwhere θ (s) = θ0s, θ0 is a constant angle, ξ = {0, ϕ}, and s denotes the normalized time s = t/τ, with τ the total evolution time. The system is prepared in the initial state \|Ψ(0)⟩ = \|ψn⟩ ⊗ \|0⟩, where\n\n$\|ψn⟩=∑m=0N−1∑ϵ=±γm,ϵ\|m,n^ϵ⟩.$\n\nThen, by adiabatic evolution, the system will evolve to the final state \|Ψ(1)⟩ given by\n\n$\|Ψ(1)⟩=𝟙−PN−1,n^−ψn⊗\|E00(1)⟩+PN−1,n^−ψn⊗\|Eϕ0(1)⟩,$\n\nwhere $\|Eξ0(1)⟩=cosθ0∕20+eiξsinθ0∕21$ is the ground state of Hξ(1). Then, equivalently, we can write\n\n$\|Ψ(1)⟩=cosθ0∕2ψn⊗\|0⟩+sinθ0∕2\|ψnrot⟩⊗1,$\n\nwith\n\n$\|ψnrot⟩=∑k=0N−2∑ϵ=±γk,ϵk,n^ϵ+N−1⊗[γN−1,+n^++eiϕγN−1,−n^−].$\n\nThe rotated state $\|ψnrot⟩$ is the target of the n-controlled gate. However, note that \|Ψ(1)⟩ in equation (5) is an entangled state. Thus, a measurement must be performed on the auxiliary system, where the action of the gate will be considered successful if 𝒜 is measured in the state \|1⟩, which occurs with probability sin2 (θ0/2). On the other hand, if the outcome of a measurement on 𝒜 yields \|0⟩, the adiabatic evolution should be restarted through the Hamiltonian in equation (1), as the state of the system is projected onto the initial state \|Ψ(0)⟩. Naturally, by choosing θ0 = π, we deterministically ensure the success of the computation. However, as we will show, deterministic evolutions may demand more energy resources than probabilistic processes when transitionless drivings are considered. In particular, observe also that the scheme presented here allows for the implementation of arbitrary n-controlled gates, which lead to versatile sets of universal gates, e.g., single qubit rotations and controlled-NOT operations (Nielsen and Chuang, 2000).\n\nInstead of adiabatic implementations of quantum circuits, we can also consider the original approach of adiabatic QC (Farhi et al., 2001), where a single annealing process is performed using energy penalties attributed to quantum states that violate the solutions of an optimization problem. Here, we employ this method to analyze three possible adiabatic implementations of quantum search over an unstructured list. An adiabatic QC approach for the quantum search through Grover’s algorithm (Grover, 1997) was first proposed in Farhi et al. (2000) and improved by using local adiabaticity (van Dam et al., 2001; Roland and Cerf, 2002), where the adiabatic evolution is required for each local time interval, instead of being globally applied as in the original proposal. In both cases, the search for a marked element in an unstructured list of N = 2n elements (labeled by n qubits) can be achieved by employing a Hamiltonian of the form\n\n$H0(s)=f(s)(𝟙−\|+⟩⟨+\|)+g(s)(𝟙−\|m⟩⟨m\|),$\n\nwhere \|m⟩ is the marked state, s is the normalized time (0 ≤ s ≤ 1), $\|+⟩=1∕N∑i=0N−1\|i⟩$, and f (0) = g(1) = 1 and f (1) = g(0) = 0. The eigenspectrum of this Hamiltonian can be exactly derived [see Das et al. (2003) and Orus and Latorre (2004)]. In particular, the two lowest eigenstates can be written as\n\n$\|E±(s)⟩=N±(s)\|m⟩+b±(s)\|ϕ⟩,$\n\nwhere the normalization constant is $N±(s) = 1∕1+(N−1)b±(s)2$, $\|ϕ⟩=∑i≠m\|i⟩$, and\n\n$b±(s)=1−E±(s)f(s)N¯,$\n\nwith $N¯=1−1∕N$, and the corresponding energies E±(s) given by\n\n$E±(s)=f(s)+g(s)±[ f(s)+g(s)]2−4f(s)g(s)N¯2.$\n\nThe other higher-energy eigenstates form an (N − 2)-fold degenerate eigenspace, whose energy is given by\n\n$Edeg= f(s)+g(s).$\n\nIn order to explicitly provide the eigenstates $\|Edegk⟩$ (k = 1, ⋯ , N − 2) associated with the eigenenergy Edeg, we write\n\n$\|Edegk⟩=∑n=0N−1cnk\|n⟩.$\n\nThen, from the eigenvalue equation for H0(s), it directly follows that the set ${cnk}$ is just required to satisfy the constraints $∑n=0N−1cnk=0$ and $cmk=0$. As a consequence, the states $\|Edegk⟩$ can be suitably chosen as time-independent vectors.\n\nBy imposing a local adiabatic evolution (van Dam et al., 2001; Roland and Cerf, 2002), i.e., by requiring adiabaticity at each infinitesimal time interval, the runtime is minimized for the path [see also Kieferová and Wiebe (2014)]\n\n$f(s)=1−g(s),g(s)=N−1−tanarctanN−11−2s2N−1.$\n\nThis results in a quadratic speedup over the classical search, i.e., we obtain the time complexity $O(N)$ expected by the Grover quantum search (van Dam et al., 2001; Roland and Cerf, 2002).\n\nIt is possible to reduce the time complexity of the Grover quantum search by transferring the algorithmic cost to other physical resources. The second implementation of the adiabatic Grover search considered here has been introduced in Das et al. (2003) and Wen and Qiu (2008). It is also based on the Hamiltonian in equation (7) to perform the evolution, but requiring that the functions f (s) and g(s) satisfy\n\n$f(s)=1−s+N(1−s)s,$\n$g(s)=s+N(1−s)s.$\n\nThis implementation achieves the solution at constant time complexity O(1). As is apparent from equations (14) to (15), the original time resource has been transferred to the coupling strengths f (s) and g(s) and, as discussed in detail in the next Section, will be reflected in the energy scaling required by the system.\n\nThe two previous versions of the adiabatic Grover’s algorithm are based on oracular Hamiltonians, which we take here to be operators able to recognize the correct answer of a problem (Nielsen and Chuang, 2000). This is indeed the case if one chooses a Hamiltonian composed of an operator Om in the form Om = 𝟙 − \|m⟩ ⟨m\|. The action of Om in the computational basis {\|i⟩} is\n\n$Om\|i⟩=(𝟙−\|m⟩⟨m\|)\|i⟩=0(i=m),\|i⟩(i≠m),$\n\nso that this operator recognizes the marked state, providing no hint about its identity if acting upon any other state. Adiabatic versions of the quantum search have also been proposed via non-oracular Hamiltonians. Our third implementation of Grover’s algorithm is based on the non-linear non-oracular (NLNO) Hamiltonian proposed in Wen et al. (2009). In this work, the time-dependent Hamiltonian in equation (7) is replaced for\n\n$H0(s)=f(s)(𝟙−\|+⟩⟨+\|)+g(s)(𝟙−\|m⟩⟨m\|)+h(s)(\|+⟩⟨m\|+\|m⟩⟨+\|),$\n\nwhere h(0) = h(1) = 0. The Hamiltonian in equation (17) contains an operator $O¯m=\|+⟩⟨m\|+\|m⟩⟨+\|$. The action of $O¯m$ in the computational basis {\|i⟩} is\n\n$O¯m\|i⟩=(\|+⟩⟨m\|+\|m⟩⟨+\|)\|i⟩=1N\|m⟩+\|+⟩(i=m),1N\|m⟩(i≠m).$\n\nObserve that equation (18) implies that $O¯m$ cannot exactly recognize a marked element, even though it could effectively recover the marked state for N ≫ 1 with a single operation over the uniform superposition provided by the state \|+⟩. Naturally, the non-oracular form of the Hamiltonian involves all the individual computational states, requiring, therefore, much more than the capacity of the Hamiltonian to recognize the marked element. This is an obviously artificial approach, whose discussion here is kept just for comparison with the superadiabatic scenario. Assuming a restricted feasibility of such a Hamiltonian, we proceed by looking at its eigenspectrum. The ground and first excited states have the same structure as in equation (8), with\n\n$b±(s)=N¯f(s)+2h(s)N−E±(s)N¯ f(s)−h(s)N.$\n\nThe two lowest energy levels are given by\n\n$E±(s)=12f(s)+g(s)+2h(s)N±f(s)+g(s)2−4f(s)g(s)N¯+ 4h2(s)−4h(s)N f(s)+g(s) .$\n\nAs before, the higher-energy states form an (N − 2)-fold degenerate subspace, with energy given by f(s) + g(s). As shown in Wen et al. (2009), this formulation also shows constant time complexity O(1), which can be obtained by choosing a suitable interpolation, such as\n\n$f(s)=1−s,g(s)=s,h(s)=s(1−s).$\n\nThe performance of adiabatic QC is dictated by a long total evolution time compared to the inverse of a power of the energy gap (Messiah, 1962; Teufel, 2003; Sarandy et al., 2004; Jansen et al., 2007). However, the adiabatic evolution can be sped up through shortcuts to adiabaticity via counter-diabatic Hamiltonians (Demirplak and Rice, 2003, 2005; Berry, 2009). The fundamental idea underlying these shortcuts to adiabaticity is to add a new contribution HCD(t), called counter-diabatic Hamiltonian, to the original adiabatic Hamiltonian H(t). This term is constructed such that it allows the mimicking of the adiabatic evolution, however, without any constraint on the total time of evolution. The total composite Hamiltonian is\n\n$HSA(t)=H(t)+HCD(t),$\n\nwhich is called superadiabatic Hamiltonian. In particular, it is possible to show that the counter-diabatic term reads (Berry, 2009).\n\n$HCD(t)=iℏ∑nn˙(t)n(t)+n˙(t)\|n(t)n(t)n(t),$\n\nwhere \|n(t)⟩ is the eigenstate of H(t) associated with the energy En(t). The goal of the counter-diabatic term HCD(t) in the Hamiltonian HSA(t) is exactly to eliminate the diabatic contributions of H(t). Thus, if the system is initially prepared in the ground state of H(0), then the system will deterministically evolve to the instantaneous ground state of the Hamiltonian H(t) with no constraints over the evolution time. Note that, in general, one would need to be able to explicitly calculate all the eigenstates of H(t) to derive a shortcut to adiabaticity using the counter-diabatic driving. However, this may not be a hard requirement in the case of superadiabatic versions of circuit implementations, where one-qubit rotations and two-qubit entangling gates are enough to achieve QC universality (Nielsen and Chuang, 2000). In particular, as we shall see for this case, HCD(t) can be realized through a simple time-independent operator.\n\n### 2.4. Energetic Cost of Quantum Evolutions\n\nTo quantify the expense of energy in a quantum evolution driven by a Hamiltonian H(t), we adopt as the cost measure the average norm of H(t) computed for a total time of evolution τ. This yields (Kieferová and Wiebe, 2014; Santos and Sarandy, 2015; Zheng et al., 2015; Santos et al., 2016)\n\n$Σ(τ)=1τ∫0τH(t)dt=∫01H(s)ds,$\n\nwhere s = t/τ is the parameterized time and the norm here is defined by the Frobenius norm (Hilbert–Schmidt norm) $A=TrA†A$. Naturally, other norms can be adopted as, for instance, the spectral norm $A2=λmaxA†A$, where λmax denotes the maximum eigenvalue of [A A]. For the Hamiltonians investigated in this work, these norms will imply into a cost simply related by a constant D1/2, with D denoting the dimension of corresponding the Hilbert space. The Frobenius norm as well as arbitrary superadiabatic evolutions with total evolution time τ, the energetic cost can be written as\n\n$ΣSA(τ)=1τ∫0τTrHSA2(t)dt=1τ∫0τTrH2(t)+HCD2(t)dt,$\n\nwhere we have used that Tr({H(t), HCD(t)}) = 0 (Santos and Sarandy, 2015). This explicitly shows that a superadiabatic evolution has an energetic cost larger than its corresponding adiabatic evolution. By evaluating the trace in equation (25), we obtain\n\n$ΣSA(τ)=∫01∑mEm2(s)+ℏ2μm(s)τ2ds,$\n\nwhere μm(s) = ⟨∂sm(s)\|∂sm(s)⟩ − \|⟨m(s)\|∂sm(s)⟩\|2 and {Em(s)} is the energy spectrum of the adiabatic Hamiltonian H(t), with {\|m(s)⟩} denoting its eigenbasis. Notice that the adiabatic limit is recovered when taking τ → ∞. Thus, the speedup obtained by the superadiabatic dynamics is limited by the energetic cost of the evolution. Indeed, this energy–time complementarity can be formally discussed through the quantum speed limit (Deffner and Lutz, 2013), which suggests that the superadiabatic evolution time is compatible with arbitrarily short time intervals (implying into corresponding arbitrarily large energies) (Santos and Sarandy, 2015), while the adiabatic evolution time obeys the lower bound τAd ∝ 1/ωn, with ω associated with the energy gap and n ∈ ℕ+ (Messiah, 1962; Teufel, 2003; Sarandy et al., 2004; Jansen et al., 2007).\n\n## 3. Results\n\nWe now consider the performance of adiabatic and superadiabatic quantum computation, focusing on their time–energy complexity. This will be investigated both for the universal model of superadiabatic controlled gates and for the superadiabatic implementations of the Grover search.\n\n### 3.1. Quantum Gates by Superadiabatic Controlled Evolutions\n\nLet us begin by discussing the superadiabatic model of universal QC via CE implemented by shortcuts to adiabaticity (Santos and Sarandy, 2015). To this end, let us first write the complete set of eigenstates of H(t) as (Santos and Sarandy, 2015)\n\n$\|E0mϵk(s)⟩=\|m,n^ϵ⊗\|E0k(s)⟩,$\n$\|E0 (N−1)+k(s)⟩=\|N−1,n^+⊗\|E0k(s)⟩,$\n$\|Eϕ (N−1)−k(s)⟩=\|N−1,n^−⊗\|Eϕk(s)⟩,$\n\nwhere m ∈ {0, ⋯ , N − 2}, ϵ, k ∈ { ± }, and\n\n$\|Eξ+(s)⟩=−sinθ0s20+eiξcosθ0s21,$\n$\|Eξ−(s)⟩=cosθ0s20+eiξsinθ0s21,$\n\nwith ξ ∈ {0, ϕ} and ${\|Eξ±(s)⟩}$ denoting the set of eigenstates of each adiabatic Hamiltonian Hξ(s), as provided by equation (2). Thus, by using the equation (22), we can show that the superadiabatic Hamiltonian is given by (Santos and Sarandy, 2015)\n\n$HSA(s)=1−PN−1,n^−⊗H0SA(s)+PN−1,n^−⊗HϕSA(s),$\n\nwhere each term $HξSA(s)$ corresponds to the superadiabatic Hamiltonian associated with the adiabatic Hamiltonian Hξ(s), i.e., $HξSA(s)=Hξ(s)+HξCD$, with\n\n$HξCD=ℏθ02τσycosξ−σxsinξ$\n\nbeing the (time-independent) counter-diabatic contribution to achieve the evolution at total time τ (Santos and Sarandy, 2015).\n\n### 3.2. Energy–Time Complementarity in the CE Model of Quantum Gates\n\nLet us now consider equation (26) to investigate the time–energy complementarity relationship in both adiabatic and superadiabatic CE models of universal quantum gates. To this end, we need the set of eigenvalues and eigenstates of the adiabatic Hamiltonian in equation (1), which are given by Eqs 27–31. The spectrum of H(s) has (2N)-degenerate levels, with ${\|E0mϵ+(s)⟩,\|E0 (N−1)++(s)⟩,\|Eϕ (N−1)−+(s)⟩}$ and ${\|E0mϵ−(s)⟩,\|E0 (N−1)+−(s)⟩,\|Eϕ (N−1)−−(s)⟩}$ associated with the levels E+ = ℏω and E = −ℏω, respectively. So, by using Eqs 27–31, we can show that $μlm(s)=θ02∕4τ2$. In addition, the energetic cost to implement any gate controlled by n qubits is $ΣSA(τ,n)=2n∕2ΣSAsing(τ)$ (Santos and Sarandy, 2015), where $ΣSAsing$ is the energetic cost to implement any single qubit unitary transformation, with\n\n$ΣSAsingωτ,θ0=2ℏω1+θ024ωτ2.$\n\nA similar result can be obtained from the spectral norm, with energetic cost given by $ΣSAsingωτ,θ0\|2=(1∕2)ΣSAsingωτ,θ0$, since the Hilbert space has dimension D = 4 in this case. Note that the energetic cost is independent of the parameter θ0 in the adiabatic limit ωτ → ∞. Therefore, the best computational adiabatic strategy is to set θ = π, which deterministically ensures the implementation of the gate with probability one. On the other hand, probabilistic quantum computation can be energetically favored in the superadiabatic regime. Indeed, from equation (5), we can see that, by setting 0 < θ0 < π, the implementation of the quantum gate is achieved with a non-vanishing probability. Thus, we can investigate whether or not it is possible to find out a specific value of θ0 such that the energetic cost is better in average than the deterministic choice θ0 = π. To address this point, let us define the quantity\n\n$⟨N⟩=1sin2θ0∕2,$\n\nwhich is the average number of evolutions for a successful computation. So, the average energetic cost to implement a probabilistic evolution is\n\n$Σ¯=⟨N⟩Σ,$\n\nwhere Σ is the cost of a single evolution. Without loss of generality, we will consider the cost of single gates, since similar arguments apply for the cost of n-qubit controlled gates. So, by performing superadiabatic probabilistic quantum computing, the average energetic cost is given by\n\n$Σ¯SAsingωτ,θ0=NΣSAsingωτ,θ0=2ℏωcsc2θ021+θ024ωτ2.$\n\nThe function $Σ¯SAsingωτ,θ0 → ∞$ as θ0 → 0 and exhibits a minimum in the interval 0 < θ0 < π as a function of ωτ. Indeed, the angle $θ0min$ that minimizes $Σ¯SAsingωτ,θ0$ grows monotonically with ωτ, with $θ0min→π$ as ωτ → ∞ (adiabatic limit). Then, optimizing $Σ¯SAsingωτ,θ0$ for θ0, we obtain\n\n$∂∂θ0Σ¯SAsingωτ,θ0=ηθ0,ωτθ0−4ωτ2+θ02cotanθ02=0,$\n\nwhere we have defined the function\n\n$ηθ0,ωτ=csc2θ0∕22ωτ21+θ02∕4ωτ2.$\n\nNote that η(θ0, ωτ) is non-vanishing in the whole interval $Iθ0 ∈ 0,π$. Thus, to obtain the critical angle $θ0min$ in $Σ¯SAsingωτ,θ0$, we use equation (38) to note that ωτ satisfies\n\n$ωτ=θ0min2tanθ0min2−θ0min,$\n\nwhere we can see a dependence of $θ0min$ on the choice of ωτ. In addition, note that $θ0min$ is such that $tanθ0min2 ≥ θ0min$, since the quantity ωτ is required to be real and positive. The probabilistic advantage is plotted in Figure 1, where it is shown that the optimal value for θ0 is a continuous function of ωτ, being distinct of the deterministic implementation θ0 = π. In the inset, we show the global minimum of the average energy for ωτ = 0.01, which occurs for θ0 < π. In particular, $θ0min$ moves away from π as ωτ is lowered, i.e., in the strong superadiabatic regime. As ωτ shifts toward the adiabatic limit, we find that $θ0min→π$. The optimization of the energy cost is shown in the lower inset, where we define the fraction of energy required by the optimized probabilistic model as a function of ωτ as\n\n$Σrelωτ=Σ¯SAsing(ωτ,θ0min)ΣSAsingωτ,π.$\nFIGURE 1", null, "Figure 1. Optimal value $θ0min$ for the angle parameter θ0 as a function of ωτ, with ωτ in logarithmic scale. The points are obtained from equation (40), with the curve denoting the numerical fit. Upper inset: average energy in units of ℏω as a function of θ0 for ωτ = 0.01. The results are obtained from equation (37). Lower inset: fraction Σrel(ωτ) of energy required by the optimized probabilistic model as a function of ωτ, with data in logarithmic scale. The points are obtained from equation (41), with the curve denoting the numerical fit.\n\nNotice that Σrel(ωτ) decreases in the superadiabatic regime, implying into a large reduction of the energetic cost for small values of ωτ. On the other hand, Σrel(ωτ) →1 in the adiabatic limit, since $θ0min→π$.\n\nHere, we derive a superadiabatic Hamiltonian HSA(s) for the oracular quantum search governed by the adiabatic Hamiltonian H0(s) in equation (7). We will adopt linear interpolation, with f (s) = 1 − s and g(s) = s, as in Farhi et al. (2000) and write HSA(s) = H0(s) + HCD(s). In order to determine the counter-diabatic Hamiltonian HCD(s), we observe that, since H0(s) has real eigenstates, we use that $⟨n˙(s)\|n(s)⟩=0$ in equation (23), which implies that\n\n$HCD(s)=iℏτ∑ξ=±\|E˙ξ(s)⟩⟨Eξ(s)\|,$\n\nwhere the energies \|E±(s)⟩ are given by equation (8) and\n\n$\|E˙±(s)⟩=−(N−1)b±b˙±(1+(N−1)b±2)3∕2\|m⟩+b˙±(1+(N−1)b±2)3∕2\|ϕ⟩.$\n\nNote that only the ground and first excited states contribute to HCD(s), since the higher-energy degenerate sector ${\|Edegk⟩}$ is composed by time-independent eigenvectors [see equation (12)]. Note also that the counter-diabatic Hamiltonian will naturally be non-oracular [see equation (18)], with contributions from operators, such as \|ϕ⟩ ⟨m\| and \|m⟩ ⟨ϕ\|. This is the reason behind the time complexity O(1) for the superadiabatic Hamiltonian. Naturally, such a result leads to an artificial approach. In a more physical scenario, superadiabaticity could be applied to the quantum search via the direct implementation of the Grover quantum circuit, through the controlled evolution approach discussed in Section 3.1.\n\n### 3.4. Energy–Time Complementarity in the Quantum Search\n\nLet us now analyze the time–energy complementarity relationship in the adiabatic and superadiabatic versions of the Grover search. In the adiabatic regime, the energetic cost can be computed from equation (26) and using τ → ∞. Therefore, the adiabatic cost can be written as\n\n$Σad=∫01ds∑mEm2(s)=∫01dsE+(s)2+E−(s)2+(N−2)Edeg(s)2.$\n\nLet us initially consider the oracular Hamiltonian H0(s) in equation (7), whose eigenvalues are given by equations (10) and (11). By considering the case of local adiabatic evolution provided by the interpolation in equation (13) and by taking N ≫ 1, we obtain E±(s) ∼ Edeg(s) ∼ O(1), which implies from equation (44) into an energetic cost $ΣadLA$ that scales as $O(N)$. On the other hand, in the superenergetic version of the quantum search, we adopt the interpolation in Eqs 14 and 15. Then, by taking N ≫ 1, we obtain now $E±(s)∼Edeg(s)∼O(N)$, which implies into an energetic cost $ΣadSE$ that scales as O(N). This higher energetic cost is a consequence of the complementarity between energy and time, which arises to compensate the constant time complexity O(1) of the superenergetic version. Naturally, the composite energy–time complexity is kept constant for both cases. This overall complexity is reduced by taking non-oracular artificial Hamiltonians. In the case of the adiabatic NLNO model, we use the Hamiltonian in equation (17), whose ground state and first excited state energies are given now by equation (20), with the higher energies kept as in equation (11). Its energetic cost $ΣadNO$ can also be computed from equation (44) by considering the interpolation in equation (21) and by taking N ≫ 1. Then, we obtain E±(s) ∼ Edeg(s) ∼ O(1), which yields $ΣadNO$ scaling as $O(N)$.\n\nFor the superadiabatic algorithm, equation (26) must be used. Without loss of generality, we set energy units such that /τ = 1. We find that for N ≫ 1, the value of μ±(s) in equation (26) evaluate to\n\n$μ±(s)=⟨E˙±(s)\|E˙±(s)⟩=(N−1)b˙±(s)2(1+(N−1)b±(s)2)2,$\n\nwhich in turn gives the superadiabatic search energetic cost ΣSA of order $O(N)$, just reproducing the scaling of the NLNO adiabatic search. Similar results can be obtained if one chooses the spectral norm in the energetic cost, up to a common scaling factor $D1∕2=N$ related to the dimension of the Hilbert space. These results are summarized in Table 1.\n\nTABLE 1", null, "Table 1. Energy–time complexity for several versions of oracular and non-oracular Hamiltonians for the Grover quantum search.\n\n## 4. Discussion\n\nImplications of probabilistic superadiabatic QC under decoherence is a further challenge of immediate interest. In a quantum open-systems scenario, there is a compromise between the time required by adiabaticity and the decoherence time of the quantum device. Therefore, a superadiabatic implementation may provide a direction to obtain an optimal running time for the quantum algorithm while keeping an inherent protection against decoherence. In this context, it is our interest to understand to what extent decoherence can affect the optimal angle $θ0min$, investigating in particular if it can be robust against classes of decohering processes. Concerning specifically the Grover search, it would be interesting to understand whether superadiabatic implementations are equivalent to arbitrary non-oracular adiabatic Hamiltonians, as suggested in our present discussion. Moreover, the behavior of correlations, such as entanglement and the investigation of experimental proposals in the superadiabatic scenario are also topics under investigation.\n\n## Author Contributions\n\nAll the authors equally contributed to the conception of the work, development of the main results, and writing of the manuscript.\n\n## Conflict of Interest Statement\n\nThe authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.\n\n## Funding\n\nMS thanks Daniel Lidar for his hospitality at the University of Southern California. IC, AS, and MS acknowledge support from CNPq/Brazil and the Brazilian National Institute for Science and Technology of Quantum Information (INCT-IQ).\n\n## References\n\nBarends, R., Shabani, A., Lamata, L., Kelly, J., Mezzacapo, A., Las Heras, U., et al. (2016). Digitized adiabatic quantum computing with a superconducting circuit. Nature 534, 222–226. doi: 10.1038/nature17658\n\nBerry, M. V. (2009). Transitionless quantum driving. J. Phys. A: Math. Theor. 42, 365303. doi:10.1088/1751-8113/42/36/365303\n\nBocharov, A., Roetteler, M., and Svore, K. M. (2015). Efficient synthesis of universal repeat-until-success quantum circuits. Phys. Rev. Lett. 114, 080502. doi:10.1103/PhysRevLett.114.080502\n\nChen, Y. H., Xia, Y., Chen, Q. Q., and Song, J. (2014a). Efficient shortcuts to adiabatic passage for fast population transfer in multiparticle systems. Phys. Rev. A 89, 033856. doi:10.1103/PhysRevA.89.033856\n\nChen, Y.-H., Xia, Y., Chen, Q. Q., and Song, J. (2014b). Shortcuts to adiabatic passage for multiparticles in distant cavities: applications to fast and noise-resistant quantum population transfer, entangled states preparation and transition. Laser Phys. Lett. 11, 115201. doi:10.1088/1612-2011/11/11/115201\n\nChen, Y.-H., Xia, Y., Chen, Q. Q., and Song, J. (2015a). Fast and noise-resistant implementation of quantum phase gates and creation of quantum entangled states. Phys. Rev. A 91, 012325. doi:10.1103/PhysRevA.91.012325\n\nChen, Y.-H., Xia, Y., Song, J., and Chen, Q.-Q. (2015b). Shortcuts to adiabatic passage for fast generation of Greenberger-Horne-Zeilinger states by transitionless quantum driving. Sci. Rep. 5, 15616. doi:10.1038/srep15616\n\nDas, S., Kobes, R., and Kunstatter, G. (2003). Energy and efficiency of adiabatic quantum search algorithms. J. Phys. A: Math. Gen. 36, 2839. doi:10.1088/0305-4470/36/11/313\n\nDeffner, S., and Lutz, E. (2013). Energy time uncertainty relation for driven quantum systems. J. Phys. A: Math. Theor. 46, 335302. doi:10.1088/1751-8113/46/33/335302\n\ndel Campo, A. (2013). Shortcuts to adiabaticity by counter-diabatic driving. Phys. Rev. Lett. 111, 100502. doi:10.1103/PhysRevLett.111.100502\n\ndel Campo, A., Rams, M. M., and Zurek, W. H. (2012). Assisted finite-rate adiabatic passage across a quantum critical point: exact solution for the quantum Ising model. Phys. Rev. Lett. 109, 115703. doi:10.1103/PhysRevLett.109.115703\n\nDemirplak, M., and Rice, S. A. (2003). Adiabatic population transfer with control fields. J. Phys. Chem. A 107, 9937. doi:10.1021/jp030708a\n\nDemirplak, M., and Rice, S. A. (2005). Assisted adiabatic passage revisited. J. Phys. Chem. B 109, 6838. doi:10.1021/jp040647w\n\nFarhi, E., Goldstone, J., Gutmann, S., Lapan, J., Lundgren, A., and Preda, D. (2001). A quantum adiabatic evolution algorithm applied to random instances of an NP-complete problem. Science 292, 472. doi:10.1126/science.1057726\n\nFarhi, E., Goldstone, J., Gutmann, S., and Sipser, M. (2000). Quantum computation by adiabatic evolution. MIT-CTP-2936 (arXiv:quant-ph/0001106).\n\nGrover, L. K. (1997). Quantum mechanics helps in searching for a needle in a haystack. Phys. Rev. Lett. 79, 325. doi:10.1103/PhysRevLett.79.325\n\nHen, I. (2015). Quantum gates with controlled adiabatic evolutions. Phys. Rev. A 91, 022309. doi:10.1103/PhysRevA.91.022309\n\nJansen, S., Ruskai, M.-B., and Seiler, R. (2007). Bounds for the adiabatic approximation with applications to quantum computation. J. Math. Phys. 48, 102111. doi:10.1063/1.2798382\n\nKieferová, M., and Wiebe, N. (2014). On the power of coherently controlled quantum adiabatic evolutions. New J. Phys. 16, 123034. doi:10.1088/1367-2630/16/12/123034\n\nLu, M., Xia, Y., Shen, L. T., and Song, J. (2014a). An effective shortcut to adiabatic passage for fast quantum state transfer in a cavity quantum electronic dynamics system. Laser Phys. 24, 105201. doi:10.1088/1054-660X/24/10/105201\n\nLu, M., Xia, Y., Shen, L. T., Song, J., and An, N. B. (2014b). Shortcuts to adiabatic passage for population transfer and maximum entanglement creation between two atoms in a cavity. Phys. Rev. A 89, 012326. doi:10.1103/PhysRevA.89.012326\n\nMartinis, J. M., and Geller, M. R. (2014). Fast adiabatic qubit gates using only σz control. Phys. Rev. A 90, 022307. doi:10.1103/PhysRevA.90.022307\n\nMessiah, A. (1962). Quantum Mechanics. North-Holland, Amsterdam: North-Holland Pub. Co.\n\nNielsen, M. A., and Chuang, I. L. (2000). Quantum Computation and Quantum Information. Cambridge: Cambridge University Press.\n\nOrus, R., and Latorre, J. I. (2004). Universality of entanglement and quantum computation complexity. Phys. Rev. A 69, 052308. doi:10.1103/PhysRevA.69.052308\n\nPaetznick, A., and Svore, K. M. (2014). Repeat-until-success: non-deterministic decomposition of single-qubit unitaries. Quantum Inf. Comput. 14, 1277.\n\nRoland, J., and Cerf, N. J. (2002). Quantum search by local adiabatic evolution. Phys. Rev. A 65, 042308. doi:10.1103/PhysRevA.65.042308\n\nSaberi, H., Opatrny, T., Molmer, K., and del Campo, A. (2014). Adiabatic tracking of quantum many-body dynamics. Phys. Rev. A 90, 060301(R). doi:10.1103/PhysRevA.90.060301\n\nSantos, A. C., and Sarandy, M. S. (2015). Superadiabatic controlled evolutions and universal quantum computation. Sci. Rep. 5, 15775. doi:10.1038/srep15775\n\nSantos, A. C., Silva, R. D., and Sarandy, M. S. (2016). Shortcut to adiabatic gate teleportation. Phys. Rev. A 93, 012311. doi:10.1103/PhysRevA.93.012311\n\nSarandy, M. S., Wu, L.-A., and Lidar, D. (2004). Consistency of the adiabatic theorem. Quantum Inf. Process. 3, 331. doi:10.1007/s11128-004-7712-7\n\nTeufel, S. (2003). “Adiabatic perturbation theory in quantum dynamics,” in Lecture Notes in Mathematics, Vol. 1821 (Berlin Heidelberg: Springer-Verlag).\n\nTorrontegui, E., Ibáñez, S., Martínez-Garaot, S., Modugno, M., del Campo, A., Guéry-Odelin, D., et al. (2013). Shortcuts to adiabaticity. Adv. Atom. Mol. Opt. Phys. 62, 117. doi:10.1016/B978-0-12-408090-4.00002-5\n\nvan Dam, W., Mosca, M., and Vazirani, U. (2001). “How powerful is adiabatic quantum computation?,” in Proceedings of the 42nd Annual Symposium on Foundations of Computer Science, Las Vegas, 279.\n\nWen, J., Huang, Y., and Qiu, D. (2009). Entanglement proprieties of adiabatic quantum algorithms. Int. J. Quantum Inform. 7, 1531. doi:10.1142/S0219749909006000\n\nWen, J., and Qiu, D. (2008). Entanglement in adiabatic quantum searching algorithms. Int. J. Quantum Inform. 6, 997. doi:10.1142/S0219749908004249\n\nZheng, Y., Campbell, S., Chiara, G., and Poletti, D. (2015). Cost of transitionless driving and work output. e-print arXiv:1509.01882\n\nKeywords: quantum computing, quantum information, shortcuts to adiabaticity, superadiabaticity, quantum gates, quantum search\n\nCitation: Coulamy IB, Santos AC, Hen I and Sarandy MS (2016) Energetic Cost of Superadiabatic Quantum Computation. Front. ICT 3:19. doi: 10.3389/fict.2016.00019\n\nReceived: 29 March 2016; Accepted: 23 August 2016;\nPublished: 15 September 2016\n\nEdited by:" ]	[ null, "https://crossmark-cdn.crossref.org/widget/v2.0/logos/CROSSMARK_Color_square.svg", null, "https://www.frontiersin.org/files/Articles/202223/fict-03-00019-HTML/image_t/fict-03-00019-g001.gif", null, "https://www.frontiersin.org/files/Articles/202223/fict-03-00019-HTML/image_t/fict-03-00019-t001.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8398711,"math_prob":0.9868201,"size":35434,"snap":"2019-26-2019-30","text_gpt3_token_len":9244,"char_repetition_ratio":0.1732148,"word_repetition_ratio":0.044207036,"special_character_ratio":0.25193316,"punctuation_ratio":0.18330465,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949713,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-19T14:25:29Z\",\"WARC-Record-ID\":\"<urn:uuid:920a7f45-1837-4578-a822-c6ec617912f6>\",\"Content-Length\":\"258532\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5473216-9ebe-429d-9fa2-c7e108b2dca1>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d0b7e4c-3af7-4d66-979d-8086254ea8fc>\",\"WARC-IP-Address\":\"134.213.70.247\",\"WARC-Target-URI\":\"https://www.frontiersin.org/articles/10.3389/fict.2016.00019/full\",\"WARC-Payload-Digest\":\"sha1:JIWLQU4LTDK74HWSPIHWNVVB6XUV47DF\",\"WARC-Block-Digest\":\"sha1:IFJFJX5JMNOGS5C65HB6WX2RT7XCOABC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195526254.26_warc_CC-MAIN-20190719140355-20190719162355-00547.warc.gz\"}"}
https://stablemarkets.wordpress.com/2019/03/	[ "# Efficient MCMC with Caching\n\nThis post is part of a running series on Bayesian MCMC tutorials. For updates, follow @StableMarkets. Metropolis Review Metropolis-Hastings is an MCMC algorithm for drawing samples from a distribution known up to a constant of proportionality, $latex p(\\theta \| y) \\propto p(y\|\\theta)p(\\theta)$. Very briefly, the algorithm works by starting with some initial draw $latex \\theta^{(0)}$ then running … Continue reading Efficient MCMC with Caching" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86679035,"math_prob":0.83698344,"size":445,"snap":"2019-43-2019-47","text_gpt3_token_len":107,"char_repetition_ratio":0.09977324,"word_repetition_ratio":0.0,"special_character_ratio":0.21348314,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811637,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T00:06:14Z\",\"WARC-Record-ID\":\"<urn:uuid:619e0aa9-11f1-49f4-a319-ab1612673d14>\",\"Content-Length\":\"78093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ddb1b20-4000-47be-a957-eaa9560ee549>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d2d211e-34ad-45fa-ad59-eaf01fcedf6a>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://stablemarkets.wordpress.com/2019/03/\",\"WARC-Payload-Digest\":\"sha1:DKKC6755X3XJTT6CDA5CTWZS4KUUEAWV\",\"WARC-Block-Digest\":\"sha1:W5STQMIUOQNJQTJX2HLREAJLEEDCEEKI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668544.32_warc_CC-MAIN-20191114232502-20191115020502-00184.warc.gz\"}"}
https://timsong-cpp.github.io/cppwp/n4659/valarray.cons	[ "# 29 Numerics library [numerics]\n\n## 29.7 Numeric arrays [numarray]\n\n### 29.7.2 Class template valarray[template.valarray]\n\n#### 29.7.2.2valarray constructors [valarray.cons]\n\n```valarray(); ```\n\nEffects: Constructs a valarray that has zero length.277\n\n```explicit valarray(size_t n); ```\n\nEffects: Constructs a valarray that has length n. Each element of the array is value-initialized.\n\n```valarray(const T& v, size_t n); ```\n\nEffects: Constructs a valarray that has length n. Each element of the array is initialized with v.\n\n```valarray(const T* p, size_t n); ```\n\nRequires: p points to an array ([dcl.array]) of at least n elements.\n\nEffects: Constructs a valarray that has length n. The values of the elements of the array are initialized with the first n values pointed to by the first argument.278\n\n```valarray(const valarray& v); ```\n\nEffects: Constructs a valarray that has the same length as v. The elements are initialized with the values of the corresponding elements of v.279\n\n```valarray(valarray&& v) noexcept; ```\n\nEffects: Constructs a valarray that has the same length as v. The elements are initialized with the values of the corresponding elements of v.\n\nComplexity: Constant.\n\n```valarray(initializer_list<T> il); ```\n\nEffects: Equivalent to valarray(il.begin(), il.size()).\n\n```valarray(const slice_array<T>&); valarray(const gslice_array<T>&); valarray(const mask_array<T>&); valarray(const indirect_array<T>&); ```\n\nThese conversion constructors convert one of the four reference templates to a valarray.\n\n```~valarray(); ```\n\nEffects: The destructor is applied to every element of *this; an implementation may return all allocated memory.\n\nThis default constructor is essential, since arrays of valarray may be useful. After initialization, the length of an empty array can be increased with the resize member function.\n\nThis constructor is the preferred method for converting a C array to a valarray object.\n\nThis copy constructor creates a distinct array rather than an alias. Implementations in which arrays share storage are permitted, but they shall implement a copy-on-reference mechanism to ensure that arrays are conceptually distinct." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6846369,"math_prob":0.95534515,"size":1539,"snap":"2019-13-2019-22","text_gpt3_token_len":328,"char_repetition_ratio":0.19413681,"word_repetition_ratio":0.12903225,"special_character_ratio":0.20987654,"punctuation_ratio":0.14234875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9907688,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T00:26:25Z\",\"WARC-Record-ID\":\"<urn:uuid:86bc15a6-cc70-4714-99ea-2dc45c1183e6>\",\"Content-Length\":\"9784\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:494ad421-a400-44ff-839d-11d68816fcfa>\",\"WARC-Concurrent-To\":\"<urn:uuid:5776ab81-b948-4a64-9e81-651449db63b2>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://timsong-cpp.github.io/cppwp/n4659/valarray.cons\",\"WARC-Payload-Digest\":\"sha1:IVE3GJUEH7IY5K6VXJOCS4RLC6E65X5J\",\"WARC-Block-Digest\":\"sha1:3KGQT4KO6JONCSTYNMQ6WGPQGCCJ66BM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202704.58_warc_CC-MAIN-20190323000443-20190323022443-00480.warc.gz\"}"}
https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_13&diff=30348&oldid=28428	[ "# Difference between revisions of \"1994 AIME Problems/Problem 13\"\n\n## Problem\n\nThe equation", null, "$x^{10}+(13x-1)^{10}=0\\,$\n\nhas 10 complex roots", null, "$r_1, \\overline{r_1}, r_2, \\overline{r_2}, r_3, \\overline{r_3}, r_4, \\overline{r_4}, r_5, \\overline{r_5},\\,$ where the bar denotes complex conjugation. Find the value of", null, "$\\frac 1{r_1\\overline{r_1}}+\\frac 1{r_2\\overline{r_2}}+\\frac 1{r_3\\overline{r_3}}+\\frac 1{r_4\\overline{r_4}}+\\frac 1{r_5\\overline{r_5}}.$\n\n## Solution\n\nLet", null, "$t = 1/x$. After multiplying the equation by", null, "$t^{10}$,", null, "$1 + (13 - t)^{10} = 0\\Rightarrow (13 - t)^{10} = - 1$.\n\nUsing DeMoivre,", null, "$13 - t = \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right)$ where", null, "$k$ is an integer between", null, "$0$ and", null, "$9$.", null, "$t = 13 - \\text{cis}\\left(\\frac {(2k + 1)\\pi}{10}\\right) \\Rightarrow \\bar{t} = 13 - \\text{cis}\\left(-\\frac {(2k + 1)\\pi}{10}\\right)$.\n\nSince", null, "$\\text{cis}(\\theta) + \\text{cis}(-\\theta) = 2\\cos(\\theta)$,", null, "$t\\bar{t} = 170 - 26\\cos \\left(\\frac {(2k + 1)\\pi}{10}\\right)$ after expanding. Here", null, "$k$ ranges from 0 to 4 because two angles which sum to", null, "$2\\pi$ are involved in the product.\n\nThe expression to find is", null, "$\\sum t\\bar{t} = 850 - 26\\sum_{k = 0}^4 \\cos \\frac {(2k + 1)\\pi}{10}$.\n\nBut", null, "$\\cos \\frac {\\pi}{10} + \\cos \\frac {9\\pi}{10} = \\cos \\frac {3\\pi}{10} + \\cos \\frac {7\\pi}{10} = \\cos \\frac {\\pi}{2} = 0$ so the sum is", null, "$\\boxed{850}$." ]	[ null, "https://latex.artofproblemsolving.com/7/3/e/73ef5b72d76840f828b8eded9fe5a63bba1d2958.png ", null, "https://latex.artofproblemsolving.com/9/6/0/96085a8136e254533acb15da0fa3a66dd66f6675.png ", null, "https://latex.artofproblemsolving.com/3/d/f/3df94cc625c07e4055d82e8eef058b98b14fde99.png ", null, "https://latex.artofproblemsolving.com/6/3/9/6394005e2cce77fd6160d821d368074decc504dc.png ", null, "https://latex.artofproblemsolving.com/9/0/8/908eb84285dc4794d5bf9cd93bfd22cadcf9e0d8.png ", null, "https://latex.artofproblemsolving.com/1/2/2/122d20f8fafc6474f3f3d3c7317e812e83610418.png ", null, "https://latex.artofproblemsolving.com/4/e/b/4ebe7135919c2ec9a555dc3d5346061cdaa953f6.png ", null, "https://latex.artofproblemsolving.com/8/c/3/8c325612684d41304b9751c175df7bcc0f61f64f.png ", null, "https://latex.artofproblemsolving.com/b/c/1/bc1f9d9bf8a1b606a4188b5ce9a2af1809e27a89.png ", null, "https://latex.artofproblemsolving.com/b/f/2/bf2c9074b396e3af0dea52d792660eea1c77f10f.png ", null, "https://latex.artofproblemsolving.com/9/d/b/9db63ea6ee7ed40f3da11331fefc4c5f3820363c.png ", null, "https://latex.artofproblemsolving.com/4/c/5/4c5cc1e448556a53bc38c7963c8700d30b026dd5.png ", null, "https://latex.artofproblemsolving.com/9/6/2/96252c856625cb6370a1022de069b478ce1edf61.png ", null, "https://latex.artofproblemsolving.com/8/c/3/8c325612684d41304b9751c175df7bcc0f61f64f.png ", null, "https://latex.artofproblemsolving.com/7/0/9/709acef3551b886e6754de72d59a88f00660f0d3.png ", null, "https://latex.artofproblemsolving.com/2/2/f/22f53ad4b1e4cfe03be2ce6eff08e645d24f1dae.png ", null, "https://latex.artofproblemsolving.com/4/1/f/41f599490089a11f562b33f0eabe91120e03029d.png ", null, "https://latex.artofproblemsolving.com/9/3/1/93171c7d4fcdd25e8c3690afeb801531b871947a.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6129785,"math_prob":1.0000085,"size":2246,"snap":"2021-31-2021-39","text_gpt3_token_len":833,"char_repetition_ratio":0.18108831,"word_repetition_ratio":0.32152587,"special_character_ratio":0.44033837,"punctuation_ratio":0.073903,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000072,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T01:14:54Z\",\"WARC-Record-ID\":\"<urn:uuid:59c2a334-46ea-4343-b649-76128fa5212f>\",\"Content-Length\":\"48161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9bf5e59f-0aa1-4aa7-933f-15437d1b3fba>\",\"WARC-Concurrent-To\":\"<urn:uuid:f25ea4ee-0198-4c16-9010-8135628be829>\",\"WARC-IP-Address\":\"104.26.11.229\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_13&diff=30348&oldid=28428\",\"WARC-Payload-Digest\":\"sha1:47ZNIF46NQ6SPG4DFAL3CKEMFCMLOHI5\",\"WARC-Block-Digest\":\"sha1:H4QVY6H25NI54EPKDFU4CPCWWV5UNUKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057479.26_warc_CC-MAIN-20210923225758-20210924015758-00243.warc.gz\"}"}
https://runestone.academy/ns/books/published//TeacherCSP/CSPNameNumbers/expressionTable.html	[ "Summary of Expression Types¶\n\n Expression Arithmetic meaning 1 + 2 Addition, the result is 3 3 * 4 Multiplication, the result is 12 1 / 3 Integer division, the result is 0 in older Python environments, but 0.333333333333 in Python 3 2.0 / 4.0 Division, the result is 0.5, since you are using decimal numbers in the calculation 2 % 3 Modulo (remainder), the result is 2 -1 Negation, the result is -1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82041097,"math_prob":0.9975239,"size":470,"snap":"2022-05-2022-21","text_gpt3_token_len":150,"char_repetition_ratio":0.22103004,"word_repetition_ratio":0.0,"special_character_ratio":0.34893617,"punctuation_ratio":0.13043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9927589,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-20T10:13:33Z\",\"WARC-Record-ID\":\"<urn:uuid:df8a49c2-31aa-408d-9990-f0adeeded3d8>\",\"Content-Length\":\"26285\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b58c2655-b54f-4e76-9732-b7dd539488b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5f5fb6e-d875-4105-838d-45f24797aefd>\",\"WARC-IP-Address\":\"134.122.31.165\",\"WARC-Target-URI\":\"https://runestone.academy/ns/books/published//TeacherCSP/CSPNameNumbers/expressionTable.html\",\"WARC-Payload-Digest\":\"sha1:XK3UFHHCARCUQHE7QUFHUEVMDQ62VGWX\",\"WARC-Block-Digest\":\"sha1:U7HZWHQV3ARQPUR2Z6VBTEXNXLKC4A6I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301737.47_warc_CC-MAIN-20220120100127-20220120130127-00679.warc.gz\"}"}
https://www.coursehero.com/file/p5kgta7/pts-Consider-the-linear-transformation-T-R-3-P-3-R-given-byT-a-b-c-a-bx-a-b-x-3/	[ "# Pts consider the linear transformation t r 3 p 3 r\n\n• Test Prep\n• 8\n• 100% (2) 2 out of 2 people found this document helpful\n\nThis preview shows page 3 - 8 out of 8 pages.\n\nProblem 1 (20 pts).Consider the linear transformationT:R3P3(R) given byTabc=a+bx+ (a+b)x3.(a) Find the matrix [T]γβofTrelative to the standard basesβandγofR3andP3(R).(b) Explain whyTis neither 1-1 nor onto.3\nProblem 2 (30 pts).LetVbe the set of matrices defined as follows:V=a-bbaa, bR.(a) Show thatVis a vector space overR.(b) Give a basis forVand find dimV.4\n(c) ConsiderC, the set of complex numbers, as a vector space overR.Give a basis forCand find its dimension.(d) Letf:VCbe a function defined byfa-bba=a+ibfor alla, bR.Prove thatfis an isomorphism of vectors spaces overR.5\nProblem 3 (20 pts).LetVbe a vector space and letL(V, V) be the vector space of all lineartransformations fromVtoV.(a) State what it means for several linear transformationsT1, T2, . . . , TkfromVtoVto belinearly independent inL(V, V).(b) LetV=P5(R). LetTandUbe linear transformations fromVtoVdefined byT(f(x)) =f(2) +f0(x) andU(f(x)) =f00(x) for all polynomialsf(x)V.Prove thatTandUare linearly independent inL(V, V).6\nProblem 4 (10 pts).LetVbe a vector space of dimension 2017. LetT:VVbe a lineartransformation and let~vbe some vector inV. Prove that for a large enough positive integermthevectors~v, T(~v), T2(~v), T3(~v), . . . , Tm(~v)will be linearlydependentinV.7\n•", null, "•", null, "•", null, "" ]	[ null, "https://www.coursehero.com/assets/img/doc-landing/start-quote.svg", null, "https://www.coursehero.com/assets/img/doc-landing/start-quote.svg", null, "https://www.coursehero.com/assets/img/doc-landing/start-quote.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6594304,"math_prob":0.99656105,"size":1257,"snap":"2021-43-2021-49","text_gpt3_token_len":434,"char_repetition_ratio":0.12769353,"word_repetition_ratio":0.0,"special_character_ratio":0.2728719,"punctuation_ratio":0.17343174,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996373,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-21T05:37:55Z\",\"WARC-Record-ID\":\"<urn:uuid:295e7259-dd49-4d9e-ae0a-54397c103310>\",\"Content-Length\":\"278626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c77ffdd0-6550-487e-a717-4d562ea42f40>\",\"WARC-Concurrent-To\":\"<urn:uuid:05d68b14-8800-49b9-87a2-235967ebe7be>\",\"WARC-IP-Address\":\"104.17.93.47\",\"WARC-Target-URI\":\"https://www.coursehero.com/file/p5kgta7/pts-Consider-the-linear-transformation-T-R-3-P-3-R-given-byT-a-b-c-a-bx-a-b-x-3/\",\"WARC-Payload-Digest\":\"sha1:OKTZVZ3CIRGQJ5G47AGVNHAN7T4FRJFF\",\"WARC-Block-Digest\":\"sha1:FCSZDFF7HOGQZCBLQQ3J674HZCRSDVIJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585381.88_warc_CC-MAIN-20211021040342-20211021070342-00157.warc.gz\"}"}
https://everything.explained.today/Closure_(topology)/	[ "Closure (topology) explained\n\nIn mathematics, the closure of a subset S of points in a topological space consists of all points in S together with all limit points of S. The closure of S may equivalently be defined as the union of S and its boundary, and also as the intersection of all closed sets containing S. Intuitively, the closure can be thought of as all the points that are either in S or \"near\" S. A point which is in the closure of S is a point of closure of S. The notion of closure is in many ways dual to the notion of interior.\n\nDefinitions\n\nPoint of closure\n\nFor\n\nS\n\na subset of a Euclidean space,\n\nx\n\nis a point of closure of\n\nS\n\nif every open ball centered at\n\nx\n\ncontains a point of\n\nS\n\n(this point may be\n\nx\n\nitself).\n\nThis definition generalizes to any subset\n\nS\n\nof a metric space\n\nX.\n\nFully expressed, for\n\nX\n\na metric space with metric\n\nd,\n\nx\n\nis a point of closure of\n\nS\n\nif for every\n\nr>0\n\nthere exists some\n\ns\\inS\n\nsuch that the distance\n\nd(x,s)<r\n\n(again,\n\nx=s\n\nis allowed). Another way to express this is to say that\n\nx\n\nis a point of closure of\n\nS\n\nif the distance\n\nd(x,S):=infsd(x,s)=0.\n\nThis definition generalizes to topological spaces by replacing \"open ball\" or \"ball\" with \"neighbourhood\". Let\n\nS\n\nbe a subset of a topological space\n\nX.\n\nThen\n\nx\n\nis a or of\n\nS\n\nif every neighbourhood of\n\nx\n\ncontains a point of\n\nS.\n\nNote that this definition does not depend upon whether neighbourhoods are required to be open.\n\nLimit point\n\nSee main article: Limit point.\n\nThe definition of a point of closure is closely related to the definition of a limit point. The difference between the two definitions is subtle but important – namely, in the definition of limit point, every neighbourhood of the point\n\nx\n\nin question must contain a point of the set . The set of all limit points of a set\n\nS\n\nis called the\n\nThus, every limit point is a point of closure, but not every point of closure is a limit point. A point of closure which is not a limit point is an isolated point. In other words, a point\n\nx\n\nis an isolated point of\n\nS\n\nif it is an element of\n\nS\n\nand if there is a neighbourhood of\n\nx\n\nwhich contains no other points of\n\nS\n\nother than\n\nx\n\nitself.\n\nFor a given set\n\nS\n\nand point\n\nx,\n\nx\n\nis a point of closure of\n\nS\n\nif and only if\n\nx\n\nis an element of\n\nS\n\nor\n\nx\n\nis a limit point of\n\nS\n\n(or both).\n\nClosure of a set\n\nThe of a subset\n\nS\n\nof a topological space\n\n(X,\\tau),\n\ndenoted by\n\n\\operatorname{cl}(X,S\n\nor possibly by\n\n\\operatorname{cl}XS\n\n(if\n\n\\tau\n\nis understood), where if both\n\nX\n\nand\n\n\\tau\n\nare clear from context then it may also be denoted by\n\n\\operatorname{cl}S,\n\n\\overline{S},\n\nor\n\nS{}-\n\n(moreover,\n\n\\operatorname{cl}\n\nis sometimes capitalized to\n\n\\operatorname{Cl}\n\n) can be defined using any of the following equivalent definitions:\n1. \\operatorname{cl}S\n\nis the set of all points of closure of\n\nS.\n\n2. \\operatorname{cl}S\n\nis the set\n\nS\n\ntogether with all of its limit points.\n3. \\operatorname{cl}S\n\nis the intersection of all closed sets containing\n\nS.\n\n4. \\operatorname{cl}S\n\nis the smallest closed set containing\n\nS.\n\n5. \\operatorname{cl}S\n\nis the union of\n\nS\n\nand its boundary\n\n\\partial(S).\n\n6. \\operatorname{cl}S\n\nis the set of all\n\nx\\inX\n\nfor which there exists a net (valued) in\n\nS\n\nthat converges to\n\nx\n\nin\n\n(X,\\tau).\n\nThe closure of a set has the following properties.\n\n\\operatorname{cl}S\n\nis a closed superset of\n\nS\n\n• The set\n\nS\n\nis closed if and only if\n\nS=\\operatorname{cl}S\n\n• If\n\nS\\subseteqT\n\nthen\n\n\\operatorname{cl}S\n\nis a subset of\n\n\\operatorname{cl}T.\n\n• If\n\nA\n\nis a closed set, then\n\nA\n\ncontains\n\nS\n\nif and only if\n\nA\n\ncontains\n\n\\operatorname{cl}S.\n\nSometimes the second or third property above is taken as the of the topological closure, which still make sense when applied to other types of closures (see below).\n\nIn a first-countable space (such as a metric space),\n\n\\operatorname{cl}S\n\nis the set of all limits of all convergent sequences of points in\n\nS.\n\nFor a general topological space, this statement remains true if one replaces \"sequence\" by \"net\" or \"filter\".\n\nNote that these properties are also satisfied if \"closure\", \"superset\", \"intersection\", \"contains/containing\", \"smallest\" and \"closed\" are replaced by \"interior\", \"subset\", \"union\", \"contained in\", \"largest\", and \"open\". For more on this matter, see closure operator below.\n\nExamples\n\nConsider a sphere in 3 dimensions. Implicitly there are two regions of interest created by this sphere; the sphere itself and its interior (which is called an open 3-ball). It is useful to be able to distinguish between the interior of 3-ball and the surface, so we distinguish between the open 3-ball, and the closed 3-ball – the closure of the 3-ball. The closure of the open 3-ball is the open 3-ball plus the surface.\n\n• In any space,\n\n\\varnothing=\\operatorname{cl}\\varnothing.\n\n• In any space\n\nX,\n\nX=\\operatorname{cl}X.\n\nGiving\n\nR\n\nand\n\nC\n\nthe standard (metric) topology:\n• If\n\nX\n\nis the Euclidean space\n\nR\n\nof real numbers, then\n\n\\operatorname{cl}X((0,1))=[0,1].\n\n• If\n\nX\n\nis the Euclidean space\n\nR\n\nthen the closure of the set\n\nQ\n\nof rational numbers is the whole space\n\nR.\n\nWe say that\n\nQ\n\nis dense in\n\nR.\n\n• If\n\nX\n\nis the complex plane\n\nC=R2,\n\nthen\n\n\\operatorname{cl}X\\left(\\{z\\inC:\|z\|>1\\}\\right)=\\{z\\inC:\|z\|\\geq1\\}.\n\n• If\n\nS\n\nis a finite subset of a Euclidean space\n\nX,\n\nthen\n\n\\operatorname{cl}XS=S.\n\n(For a general topological space, this property is equivalent to the T1 axiom.)\n\nOn the set of real numbers one can put other topologies rather than the standard one.\n\n• If\n\nX=R\n\nis endowed with the lower limit topology, then\n\n\\operatorname{cl}X((0,1))=[0,1).\n\n• If one considers on\n\nX=R\n\nthe discrete topology in which every set is closed (open), then\n\n\\operatorname{cl}X((0,1))=(0,1).\n\n• If one considers on\n\nX=R\n\nthe trivial topology in which the only closed (open) sets are the empty set and\n\nR\n\nitself, then\n\n\\operatorname{cl}X((0,1))=R.\n\nThese examples show that the closure of a set depends upon the topology of the underlying space. The last two examples are special cases of the following.\n\n• In any discrete space, since every set is closed (and also open), every set is equal to its closure.\n\nX,\n\nsince the only closed sets are the empty set and\n\nX\n\nitself, we have that the closure of the empty set is the empty set, and for every non-empty subset\n\nA\n\nof\n\nX,\n\n\\operatorname{cl}XA=X.\n\nIn other words, every non-empty subset of an indiscrete space is dense.\n\nThe closure of a set also depends upon in which space we are taking the closure. For example, if\n\nX\n\nis the set of rational numbers, with the usual relative topology induced by the Euclidean space\n\nR,\n\nand if\n\nS=\\{q\\inQ:q2>2,q>0\\},\n\nthen\n\nS\n\nis both closed and open in\n\nQ\n\nbecause neither\n\nS\n\nnor its complement can contain\n\n\\sqrt2\n\n, which would be the lower bound of\n\nS\n\n, but cannot be in\n\nS\n\nbecause\n\n\\sqrt2\n\nis irrational. So,\n\nS\n\nhas no well defined closure due to boundary elements not being in\n\nQ\n\n. However, if we instead define\n\nX\n\nto be the set of real numbers and define the interval in the same way then the closure of that interval is well defined and would be the set of all greater than\n\n\\sqrt2\n\n.\n\nClosure operator\n\nA on a set\n\nX\n\nis a mapping of the power set of\n\nX,\n\nl{P}(X)\n\n, into itself which satisfies the Kuratowski closure axioms. Given a topological space\n\n(X,\\tau)\n\n, the topological closure induces a function\n\n\\operatorname{cl}X:\\wp(X)\\to\\wp(X)\n\nthat is defined by sending a subset\n\nS\\subseteqX\n\nto\n\n\\operatorname{cl}XS,\n\nwhere the notation\n\n\\overline{S}\n\nor\n\nS-\n\nmay be used instead. Conversely, if\n\nc\n\nis a closure operator on a set\n\nX,\n\nthen a topological space is obtained by defining the closed sets as being exactly those subsets\n\nS\\subseteqX\n\nthat satisfy\n\nc(S)=S\n\n(so complements in\n\nX\n\nof these subsets form the open sets of the topology).\n\nThe closure operator\n\n\\operatorname{cl}X\n\nis dual to the interior operator, which is denoted by\n\n\\operatorname{int}X,\n\nin the sense that\n\n\\operatorname{cl}XS=X\\setminus\\operatorname{int}X(X\\setminusS),\n\nand also\n\n\\operatorname{int}XS=X\\setminus\\operatorname{cl}X(X\\setminusS).\n\nTherefore, the abstract theory of closure operators and the Kuratowski closure axioms can be readily translated into the language of interior operators by replacing sets with their complements in\n\nX.\n\nIn general, the closure operator does not commute with intersections. However, in a complete metric space the following result does hold:\n\nA subset\n\nS\n\nis closed in\n\nX\n\nif and only if\n\n\\operatorname{cl}XS=S.\n\nIn particular:\n• The closure of the empty set is the empty set;\n• The closure of\n\nX\n\nitself is\n\nX.\n\n• The closure of an intersection of sets is always a subset of (but need not be equal to) the intersection of the closures of the sets.\n• In a union of finitely many sets, the closure of the union and the union of the closures are equal; the union of zero sets is the empty set, and so this statement contains the earlier statement about the closure of the empty set as a special case.\n• The closure of the union of infinitely many sets need not equal the union of the closures, but it is always a superset of the union of the closures.\n\nIf\n\nS\\subseteqT\\subseteqX\n\nand if\n\nT\n\nis a subspace of\n\nX\n\n(meaning that\n\nT\n\nis endowed with the subspace topology that\n\nX\n\ninduces on it), then\n\n\\operatorname{cl}TS\\subseteq\\operatorname{cl}XS\n\nand the closure of\n\nS\n\ncomputed in\n\nT\n\nis equal to the intersection of\n\nT\n\nand the closure of\n\nS\n\ncomputed in\n\nX\n\n:\n\n\\operatorname{cl}TS~=~T\\cap\\operatorname{cl}XS.\n\n\n\nIn particular,\n\nS\n\nis dense in\n\nT\n\nif and only if\n\nT\n\nis a subset of\n\n\\operatorname{cl}XS.\n\nIf\n\nS,T\\subseteqX\n\nbut\n\nS\n\nis not necessarily a subset of\n\nT\n\nthen only\n\n\\operatorname{cl}T(S\\capT)~\\subseteq~T\\cap\\operatorname{cl}XS\n\nis guaranteed in general, where this containment could be strict (consider for instance\n\nX=\\R\n\nwith the usual topology,\n\nT=(-infty,0],\n\nand\n\nS=(0,infty)\n\n) although if\n\nT\n\nis an open subset of\n\nX\n\nthen the equality\n\n\\operatorname{cl}T(S\\capT)=T\\cap\\operatorname{cl}XS\n\nwill hold (no matter the relationship between\n\nS\n\nand\n\nT\n\n). Consequently, if\n\nl{U}\n\nis any open cover of\n\nX\n\nand if\n\nS\\subseteqX\n\nis any subset then:\n\n\\operatorname{cl}XS=cupU\n\n} \\operatorname_U (U \\cap S)\n\nbecause\n\n\\operatorname{cl}U(S\\capU)=U\\cap\\operatorname{cl}XS\n\nfor every\n\nU\\inl{U}\n\n(where every\n\nU\\inl{U}\n\nis endowed with the subspace topology induced on it by\n\nX\n\n). This equality is particularly useful when\n\nX\n\nis a manifold and the sets in the open cover\n\nl{U}\n\nare domains of coordinate charts. In words, this result shows that the closure in\n\nX\n\nof any subset\n\nS\\subseteqX\n\ncan be computed \"locally\" in the sets of any open cover of\n\nX\n\nand then unioned together.In this way, this result can be viewed as the analogue of the well-known fact that a subset\n\nS\\subseteqX\n\nis closed in\n\nX\n\nif and only if it is \"locally closed in\n\nX\n\n\", meaning that if\n\nl{U}\n\nis any open cover of\n\nX\n\nthen\n\nS\n\nis closed in\n\nX\n\nif and only if\n\nS\\capU\n\nis closed in\n\nU\n\nfor every\n\nU\\inl{U}.\n\nCategorical interpretation\n\nOne may elegantly define the closure operator in terms of universal arrows, as follows.\n\nThe powerset of a set\n\nX\n\nmay be realized as a partial order category\n\nP\n\nin which the objects are subsets and the morphisms are inclusion maps\n\nA\\toB\n\nwhenever\n\nA\n\nis a subset of\n\nB.\n\nFurthermore, a topology\n\nT\n\non\n\nX\n\nis a subcategory of\n\nP\n\nwith inclusion functor\n\nI:T\\toP.\n\nThe set of closed subsets containing a fixed subset\n\nA\\subseteqX\n\ncan be identified with the comma category\n\n(A\\downarrowI).\n\nThis category — also a partial order — then has initial object\n\n\\operatorname{cl}A.\n\nThus there is a universal arrow from\n\nA\n\nto\n\nI,\n\ngiven by the inclusion\n\nA\\to\\operatorname{cl}A.\n\nSimilarly, since every closed set containing\n\nX\\setminusA\n\ncorresponds with an open set contained in\n\nA\n\nwe can interpret the category\n\n(I\\downarrowX\\setminusA)\n\nas the set of open subsets contained in\n\nA,\n\nwith terminal object\n\n\\operatorname{int}(A),\n\nthe interior of\n\nA.\n\nAll properties of the closure can be derived from this definition and a few properties of the above categories. Moreover, this definition makes precise the analogy between the topological closure and other types of closures (for example algebraic closure), since all are examples of universal arrows.\n\nNotes and References\n\n1. , and use the second property as the definition.\n2. Because\n\n\\operatorname{cl}XS\n\nis a closed subset of\n\nX,\n\nthe intersection\n\nT\\cap\\operatorname{cl}XS\n\nis a closed subset of\n\nT\n\n(by definition of the subspace topology), which implies that\n\n\\operatorname{cl}TS\\subseteqT\\cap\\operatorname{cl}XS\n\n(because\n\n\\operatorname{cl}TS\n\nis the closed subset of\n\nT\n\ncontaining\n\nS\n\n). Because\n\nT\\cap\\operatorname{cl}XS\n\nis a closed subset of\n\nT,\n\nfrom the definition of the subspace topology, there must exist some set\n\nC\\subseteqX\n\nsuch that\n\nC\n\nis closed in\n\nX\n\nand\n\n\\operatorname{cl}TS=T\\capC.\n\nBecause\n\nS\\subseteq\\operatorname{cl}TS\\subseteqC\n\nand\n\nC\n\nis closed in\n\nX,\n\nthe minimality of\n\n\\operatorname{cl}XS\n\nimplies that\n\n\\operatorname{cl}XS\\subseteqC.\n\nIntersecting both sides with\n\nT\n\nshows that\n\nT\\cap\\operatorname{cl}XS\\subseteqT\\capC=\\operatorname{cl}TS.\n\n\\blacksquare\n\n3. From\n\nT:=(-infty,0\n\n4. Let\n\nS,T\\subseteqX\n\nand assume that\n\nT\n\nis open in\n\nX.\n\nLet\n\nC:=\\operatorname{cl}T(T\\capS),\n\nwhich is equal to\n\nT\\cap\\operatorname{cl}X(T\\capS)\n\n(because\n\nT\\capS\\subseteqT\\subseteqX\n\n). The complement\n\nT\\setminusC\n\nis open in\n\nT,\n\nwhere\n\nT\n\nbeing open in\n\nX\n\nnow implies that\n\nT\\setminusC\n\nis also open in\n\nX.\n\nConsequently\n\nX\\setminus(T\\setminusC)=(X\\setminusT)\\cupC\n\nis a closed subset of\n\nX\n\nwhere\n\n(X\\setminusT)\\cupC\n\ncontains\n\nS\n\nas a subset (because if\n\ns\\inS\n\nis in\n\nT\n\nthen\n\ns\\inT\\capS\\subseteq\\operatorname{cl}T(T\\capS)=C\n\n), which implies that\n\n\\operatorname{cl}XS\\subseteq(X\\setminusT)\\cupC.\n\nIntersecting both sides with\n\nT\n\nproves that\n\nT\\cap\\operatorname{cl}XS\\subseteqT\\capC=C.\n\nThe reverse inclusion follows from\n\nC\\subseteq\\operatorname{cl}X(T\\capS)\\subseteq\\operatorname{cl}XS.\n\n\\blacksquare" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7067444,"math_prob":0.93707913,"size":6400,"snap":"2022-05-2022-21","text_gpt3_token_len":2191,"char_repetition_ratio":0.26626018,"word_repetition_ratio":0.0068965517,"special_character_ratio":0.25140625,"punctuation_ratio":0.11904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9979797,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T20:24:03Z\",\"WARC-Record-ID\":\"<urn:uuid:3cc730bd-b2d7-4283-929e-856e76db9273>\",\"Content-Length\":\"59802\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7fa74a8a-03fb-4fbb-a4dc-f64cc67fcd11>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa3106cb-ff60-4f87-8ef7-d192e801ea64>\",\"WARC-IP-Address\":\"85.25.210.18\",\"WARC-Target-URI\":\"https://everything.explained.today/Closure_(topology)/\",\"WARC-Payload-Digest\":\"sha1:454NWRM4JPCKUIKNTCFALI7ZVZMGAIES\",\"WARC-Block-Digest\":\"sha1:DUYV2BBEL5YEE3XEMG5INH5AIDRTEQBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305288.57_warc_CC-MAIN-20220127193303-20220127223303-00040.warc.gz\"}"}
https://www.informatik.uni-kiel.de/~curry/listarchive/0142.html	[ "# Re: Puzzle\n\nFrom: Michael Hanus <hanus_at_informatik.rwth-aachen.de>\nDate: Tue, 1 Dec 1998 15:35:37 +0100\n\nDear Wolfgang,\n\nthanks for your interesting example. It shows that one has\nto be careful if there is too much syntactic sugar.\nMy initial motivation for the remark \"a defining equation f = g\nbetween functions will be interpreted in Curry as syntactic sugar\nfor the corresponding defining equation f x = g x on base types\"\nwas to express that the semantics of Curry is not based on\nhigher-order rewriting which might complicate things.\nThus, I assumed that this kind of desugaring is done\nafter lambda lifting. It might be interesting to note\nthat Haskell has also surprisingly restrictions.\nFor instance, the program\n\nf x = 1\nf x = 2\n\nis a valid Haskell program, whereas the version\n\nf = \\x->1\nf = \\x->2\n\nseems to be invalid (although Hugs accepts currently both).\n\nYour remark reminds me that I wanted to suggest a further\nrestriction for local pattern definitions in Curry which solves\nthese problems:\n\nAll variables occurring in lhs patterns in some let/where\nshould occur at most once in a lhs of a local pattern declaration.\n\nThis restriction would forbid programs like\n\nf x = coin where coin = 1\ncoin = 2" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91382134,"math_prob":0.91390336,"size":1304,"snap":"2023-40-2023-50","text_gpt3_token_len":319,"char_repetition_ratio":0.08846154,"word_repetition_ratio":0.0,"special_character_ratio":0.24156442,"punctuation_ratio":0.08949416,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9692447,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T02:54:08Z\",\"WARC-Record-ID\":\"<urn:uuid:32e57ad9-ef92-4b8f-a41e-e8dc672df9e0>\",\"Content-Length\":\"7237\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f63dd7f4-4c5d-43ff-88e8-2e8c116f5426>\",\"WARC-Concurrent-To\":\"<urn:uuid:ca0133c8-30ac-42d9-aa25-e0048bff6730>\",\"WARC-IP-Address\":\"134.245.248.185\",\"WARC-Target-URI\":\"https://www.informatik.uni-kiel.de/~curry/listarchive/0142.html\",\"WARC-Payload-Digest\":\"sha1:JY52QEVOSKTLYZFDTO2YBZFDTB3NB7EM\",\"WARC-Block-Digest\":\"sha1:ONWY6F2ZGD6SIKLPEYI42R7DMEFQICST\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100264.9_warc_CC-MAIN-20231201021234-20231201051234-00444.warc.gz\"}"}
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/10%3A_Theories_of_Electronic_Molecular_Structure/10.0S%3A_10.S%3A_Theories_of_Electronic_Molecular_Structure_(Summary)	[ "# 10.S: Theories of Electronic Molecular Structure (Summary)\n\nOther terms account for the interactions between all the magnetic dipole moments and the interactions with any external electric or magnetic fields. The charge distribution of an atomic nucleus is not always spherical and, when appropriate, this asymmetry must be taken into account as well as the relativistic effect that a moving electron experiences as a change in mass. This complete Hamiltonian is too complicated and is not needed for many situations. In practice, only the terms that are essential for the purpose at hand are included. Consequently in the absence of external fields, interest in spin-spin and spin-orbit interactions, and in electron and nuclear magnetic resonance spectroscopy (ESR and NMR), the molecular Hamiltonian usually is considered to consist only of the kinetic and potential energy terms, and the Born-Oppenheimer approximation is made in order to separate the nuclear and electronic motion.\n\nIn general, electronic wavefunctions for molecules are constructed from approximate one-electron wavefunctions. These one-electron functions are called molecular orbitals. The expectation value expression for the energy is used to optimize these functions, i.e. make them as good as possible. The criterion for quality is the energy of the ground state. According to the Variational Principle, an approximate ground state energy always is higher than the exact energy, so the best energy in a series of approximations is the lowest energy. In this chapter we describe how the variational method, perturbation theory, the self-consistent field method, and configuration interaction all are used to describe the electronic states of molecules. The ultimate goal is a mathematical description of electrons in molecules that enables chemists and other scientists to develop a deep understanding of chemical bonding, to calculate properties of molecules, and to make predictions based on these calculations. For example, an active area of research in industry involves calculating changes in chemical properties of pharmaceutical drugs as a result of changes in their chemical structure.\n\nStudy Guide\n\n• What is meant by the expression ab initio calculation?\n• List all the terms in a complete molecular Hamiltonian.\n• Why are calculations on closed-shell systems more easily done than on open-shell systems?\n• How is it possible to reduce a multi-electron Hamiltonian operator to a single-electron Fock operator?\n• Why is the calculation with the Fock operator called a self-consistent field calculation?\n• What is the physical meaning of a SCF one-electron energy?\n• Why is the nonlinear variational method not used in every case to optimize basis functions, and what usually is done instead?\n• Why is it faster for a computer to use the variational principle to determine the coefficients in a linear combination of functions than to determine the parameters in the functions?\n• Identify the characteristics of hydrogenic, Slater, and Gaussian basis sets.\n• What is meant by the Hartree-Fock wavefunction and energy?\n• What is the difference between a restricted and unrestricted Hartree-Fock calculation?\n• What is neglected that makes the Hartree-Fock energy necessarily greater than the exact energy?\n• What is meant by correlation energy?\n• What purpose is served by including configuration interaction in a calculation?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9239626,"math_prob":0.96149075,"size":3566,"snap":"2019-51-2020-05","text_gpt3_token_len":679,"char_repetition_ratio":0.13054463,"word_repetition_ratio":0.0073937154,"special_character_ratio":0.1805945,"punctuation_ratio":0.09166667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9748172,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T03:14:49Z\",\"WARC-Record-ID\":\"<urn:uuid:e42da2b0-0f1f-4350-8cec-f52aec91434e>\",\"Content-Length\":\"86019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:faf34fa5-9af5-4d85-9600-bbc0df9ca645>\",\"WARC-Concurrent-To\":\"<urn:uuid:24c61003-89c5-4b7e-9856-37c789f5f155>\",\"WARC-IP-Address\":\"99.84.181.53\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/10%3A_Theories_of_Electronic_Molecular_Structure/10.0S%3A_10.S%3A_Theories_of_Electronic_Molecular_Structure_(Summary)\",\"WARC-Payload-Digest\":\"sha1:WPKMW32JHGRTHTQWSESV2RTSP6RMVYVF\",\"WARC-Block-Digest\":\"sha1:TPVBO5NDHPWH4GMBYVBNQCZLBCNXPGQH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251684146.65_warc_CC-MAIN-20200126013015-20200126043015-00152.warc.gz\"}"}
https://en.wikipedia.org/wiki/Riemann_surface	[ "# Riemann surface", null, "Riemann surface for the function f(z) = z. The two horizontal axes represent the real and imaginary parts of z, while the vertical axis represents the real part of z. The imaginary part of z is represented by the coloration of the points. For this function, it is also the height after rotating the plot 180° around the vertical axis.\n\nIn mathematics, particularly in complex analysis, a Riemann surface is a connected one-dimensional complex manifold. These surfaces were first studied by and are named after Bernhard Riemann. Riemann surfaces can be thought of as deformed versions of the complex plane: locally near every point they look like patches of the complex plane, but the global topology can be quite different. For example, they can look like a sphere or a torus or several sheets glued together.\n\nThe main interest in Riemann surfaces is that holomorphic functions may be defined between them. Riemann surfaces are nowadays considered the natural setting for studying the global behavior of these functions, especially multi-valued functions such as the square root and other algebraic functions, or the logarithm.\n\nEvery Riemann surface is a two-dimensional real analytic manifold (i.e., a surface), but it contains more structure (specifically a complex structure) which is needed for the unambiguous definition of holomorphic functions. A two-dimensional real manifold can be turned into a Riemann surface (usually in several inequivalent ways) if and only if it is orientable and metrizable. So the sphere and torus admit complex structures, but the Möbius strip, Klein bottle and real projective plane do not.\n\nGeometrical facts about Riemann surfaces are as \"nice\" as possible, and they often provide the intuition and motivation for generalizations to other curves, manifolds or varieties. The Riemann–Roch theorem is a prime example of this influence.\n\n## Definitions\n\nThere are several equivalent definitions of a Riemann surface.\n\n1. A Riemann surface X is a connected complex manifold of complex dimension one. This means that X is a connected Hausdorff space that is endowed with an atlas of charts to the open unit disk of the complex plane: for every point xX there is a neighbourhood of x that is homeomorphic to the open unit disk of the complex plane, and the transition maps between two overlapping charts are required to be holomorphic.\n2. A Riemann surface is an oriented manifold of (real) dimension two – a two-sided surface – together with a conformal structure. Again, manifold means that locally at any point x of X, the space is homeomorphic to a subset of the real plane. The supplement \"Riemann\" signifies that X is endowed with an additional structure which allows angle measurement on the manifold, namely an equivalence class of so-called Riemannian metrics. Two such metrics are considered equivalent if the angles they measure are the same. Choosing an equivalence class of metrics on X is the additional datum of the conformal structure.\n\nA complex structure gives rise to a conformal structure by choosing the standard Euclidean metric given on the complex plane and transporting it to X by means of the charts. Showing that a conformal structure determines a complex structure is more difficult.\n\n## Examples\n\n• The complex plane C is the most basic Riemann surface. The map f(z) = z (the identity map) defines a chart for C, and {f} is an atlas for C. The map g(z) = z* (the conjugate map) also defines a chart on C and {g} is an atlas for C. The charts f and g are not compatible, so this endows C with two distinct Riemann surface structures. In fact, given a Riemann surface X and its atlas A, the conjugate atlas B = {f* : f ∈ A} is never compatible with A, and endows X with a distinct, incompatible Riemann structure.\n• In an analogous fashion, every non-empty open subset of the complex plane can be viewed as a Riemann surface in a natural way. More generally, every non-empty open subset of a Riemann surface is a Riemann surface.\n• Let S = C ∪ {∞} and let f(z) = z where z is in S \\ {∞} and g(z) = 1 / z where z is in S \\ {0} and 1/∞ is defined to be 0. Then f and g are charts, they are compatible, and { fg } is an atlas for S, making S into a Riemann surface. This particular surface is called the Riemann sphere because it can be interpreted as wrapping the complex plane around the sphere. Unlike the complex plane, it is compact.\n• The theory of compact Riemann surfaces can be shown to be equivalent to that of projective algebraic curves that are defined over the complex numbers and non-singular. For example, the torus C/(Z + τ Z), where τ is a complex non-real number, corresponds, via the Weierstrass elliptic function associated to the lattice Z + τ Z, to an elliptic curve given by an equation\ny2 = x3 + a x + b.\n\nTori are the only Riemann surfaces of genus one; surfaces of higher genera g are provided by the hyperelliptic surfaces\n\ny2 = P(x),\nwhere P is a complex polynomial of degree 2g + 1.\n• All compact Riemann surfaces are algebraic curves since they can be embedded into some $\\mathbb {CP} ^{n}$", null, ". This follows from the Kodaira embedding theorem and the fact there exists a positive line bundle on any complex curve.\n• Important examples of non-compact Riemann surfaces are provided by analytic continuation.\n\n## Further definitions and properties\n\nAs with any map between complex manifolds, a function f: MN between two Riemann surfaces M and N is called holomorphic if for every chart g in the atlas of M and every chart h in the atlas of N, the map hfg−1 is holomorphic (as a function from C to C) wherever it is defined. The composition of two holomorphic maps is holomorphic. The two Riemann surfaces M and N are called biholomorphic (or conformally equivalent to emphasize the conformal point of view) if there exists a bijective holomorphic function from M to N whose inverse is also holomorphic (it turns out that the latter condition is automatic and can therefore be omitted). Two conformally equivalent Riemann surfaces are for all practical purposes identical.\n\n### Orientability\n\nEach Riemann surface, being a complex manifold, is orientable as a real manifold. For complex charts f and g with transition function h = f(g−1(z)), h can be considered as a map from an open set of R2 to R2 whose Jacobian in a point z is just the real linear map given by multiplication by the complex number h'(z). However, the real determinant of multiplication by a complex number α equals \|α\|2, so the Jacobian of h has positive determinant. Consequently, the complex atlas is an oriented atlas.\n\n### Functions\n\nEvery non-compact Riemann surface admits non-constant holomorphic functions (with values in C). In fact, every non-compact Riemann surface is a Stein manifold.\n\nIn contrast, on a compact Riemann surface X every holomorphic function with values in C is constant due to the maximum principle. However, there always exist non-constant meromorphic functions (holomorphic functions with values in the Riemann sphere C ∪ {∞}). More precisely, the function field of X is a finite extension of C(t), the function field in one variable, i.e. any two meromorphic functions are algebraically dependent. This statement generalizes to higher dimensions, see Siegel (1955). Meromorphic functions can be given fairly explicitly, in terms of Riemann theta functions and the Abel–Jacobi map of the surface.\n\n## Analytic vs. algebraic\n\nThe existence of non-constant meromorphic functions can be used to show that any compact Riemann surface is a projective variety, i.e. can be given by polynomial equations inside a projective space. Actually, it can be shown that every compact Riemann surface can be embedded into complex projective 3-space. This is a surprising theorem: Riemann surfaces are given by locally patching charts. If one global condition, namely compactness, is added, the surface is necessarily algebraic. This feature of Riemann surfaces allows one to study them with either the means of analytic or algebraic geometry. The corresponding statement for higher-dimensional objects is false, i.e. there are compact complex 2-manifolds which are not algebraic. On the other hand, every projective complex manifold is necessarily algebraic, see Chow's theorem.\n\nAs an example, consider the torus T := C/(Z + τ Z). The Weierstrass function $\\wp _{\\tau }(z)$", null, "belonging to the lattice Z + τ Z is a meromorphic function on T. This function and its derivative $\\wp _{\\tau }'(z)$", null, "generate the function field of T. There is an equation\n\n$[\\wp '(z)]^{2}=4[\\wp (z)]^{3}-g_{2}\\wp (z)-g_{3},$", null, "where the coefficients g2 and g3 depend on τ, thus giving an elliptic curve Eτ in the sense of algebraic geometry. Reversing this is accomplished by the j-invariant j(E), which can be used to determine τ and hence a torus.\n\n## Classification of Riemann surfaces\n\nThe set of all Riemann surfaces can be divided into three subsets: hyperbolic, parabolic and elliptic Riemann surfaces. Geometrically, these correspond to surfaces with negative, vanishing or positive constant sectional curvature. That is, every connected Riemann surface $X$", null, "admits a unique complete 2-dimensional real Riemann metric with constant curvature equal to $-1,0$", null, "or $1$", null, "which belongs to the conformal class of Riemannian metrics determined by its structure as a Riemann surface. This can be seen as a consequence of the existence of isothermal coordinates.\n\nIn complex analytic terms, the Poincaré–Koebe uniformization theorem (a generalization of the Riemann mapping theorem) states that every simply connected Riemann surface is conformally equivalent to one of the following:\n\n• The Riemann sphere ${\\widehat {\\mathbf {C} }}:=\\mathbf {C} \\cup \\{\\infty \\}$", null, ", which is isomorphic to the $\\mathbf {P} ^{1}(\\mathbf {C} )$", null, ";\n• The complex plane $\\mathbf {C}$", null, ";\n• The open disk $\\mathbf {D} :=\\{z\\in \\mathbf {C} :\|z\|<1\\}$", null, "which is isomorphic to the upper half-plane $\\mathbf {H} :=\\{z\\in \\mathbf {C} :\\mathrm {Im} (z)>0\\}$", null, ".\n\nA Riemann surface is elliptic, parabolic or hyperbolic according to whether its universal cover is isomorphic to $\\mathbf {P} ^{1}(\\mathbf {C} )$", null, ", $\\mathbf {C}$", null, "or $\\mathbf {D}$", null, ". The elements in each class admit a more precise description.\n\n### Elliptic Riemann surfaces\n\nThe Riemann sphere $\\mathbf {P} ^{1}(\\mathbf {C} )$", null, "is the only example, as there is no group acting on it by biholomorphic transformations freely and properly discontinuously and so any Riemann surface whose universal cover is isomorphic to $\\mathbf {P} ^{1}(\\mathbf {C} )$", null, "must itself be isomorphic to it.\n\n### Parabolic Riemann surfaces\n\nIf $X$", null, "is a Riemann surface whose universal cover is isomorphic to the complex plane $\\mathbf {C}$", null, "then it is isomorphic to one of the following surfaces:\n\n• $\\mathbf {C}$", null, "itself;\n• The quotient $\\mathbf {C} /\\mathbf {Z}$", null, ";\n• A quotient $\\mathbf {C} /(\\mathbf {Z} +\\mathbf {Z} \\tau )$", null, "where $\\tau \\in \\mathbf {C}$", null, "with $\\mathrm {Im} (\\tau )>0$", null, ".\n\nTopologically there are only three types: the plane, the cylinder and the torus. But while in the two former case the (parabolic) Riemann surface structure is unique, varying the parameter $\\tau$", null, "in the third case gives non-isomorphic Riemann surfaces. The description by the parameter $\\tau$", null, "gives the Teichmüller space of \"marked\" Riemann surfaces (in addition to the Riemann surface structure one adds the topological data of a \"marking\", which can be seen as a fixed homeomorphism to the torus). To obtain the analytic moduli space (forgetting the marking) one takes the quotient of Teichmüller space by the mapping class group. In this case it is the modular curve.\n\n### Hyperbolic Riemann surfaces\n\nIn the remaining cases $X$", null, "is a hyperbolic Riemann surface, that is isomorphic to a quotient of the upper half-plane by a Fuchsian group (this is sometimes called a Fuchsian model for the surface). The topological type of $X$", null, "can be any orientable surface save the torus and sphere.\n\nA case of particular interest is when $X$", null, "is compact. Then its topological type is described by its genus $g\\geq 2$", null, ". Its Teichmüller space and moduli space are $6g-6$", null, "-dimensional. A similar classification of Riemann surfaces of finite type (that is homeomorphic to a closed surface minus a finite number of points) can be given. However in general the moduli space of Riemann surfaces of infinite topological type is too large to admit such a description.\n\n## Maps between Riemann surfaces\n\nThe geometric classification is reflected in maps between Riemann surfaces, as detailed in Liouville's theorem and the Little Picard theorem: maps from hyperbolic to parabolic to elliptic are easy, but maps from elliptic to parabolic or parabolic to hyperbolic are very constrained (indeed, generally constant!). There are inclusions of the disc in the plane in the sphere: $\\Delta \\subset \\mathbf {C} \\subset {\\widehat {\\mathbf {C} }},$", null, "but any holomorphic map from the sphere to the plane is constant, any holomorphic map from the plane into the unit disk is constant (Liouville's theorem), and in fact any holomorphic map from the plane into the plane minus two points is constant (Little Picard theorem)!\n\n### Punctured spheres\n\nThese statements are clarified by considering the type of a Riemann sphere ${\\widehat {\\mathbf {C} }}$", null, "with a number of punctures. With no punctures, it is the Riemann sphere, which is elliptic. With one puncture, which can be placed at infinity, it is the complex plane, which is parabolic. With two punctures, it is the punctured plane or alternatively annulus or cylinder, which is parabolic. With three or more punctures, it is hyperbolic – compare pair of pants. One can map from one puncture to two, via the exponential map (which is entire and has an essential singularity at infinity, so not defined at infinity, and misses zero and infinity), but all maps from zero punctures to one or more, or one or two punctures to three or more are constant.\n\n### Ramified covering spaces\n\nContinuing in this vein, compact Riemann surfaces can map to surfaces of lower genus, but not to higher genus, except as constant maps. This is because holomorphic and meromorphic maps behave locally like $z\\mapsto z^{n},$", null, "so non-constant maps are ramified covering maps, and for compact Riemann surfaces these are constrained by the Riemann–Hurwitz formula in algebraic topology, which relates the Euler characteristic of a space and a ramified cover.\n\nFor example, hyperbolic Riemann surfaces are ramified covering spaces of the sphere (they have non-constant meromorphic functions), but the sphere does not cover or otherwise map to higher genus surfaces, except as a constant.\n\n## Isometries of Riemann surfaces\n\nThe isometry group of a uniformized Riemann surface (equivalently, the conformal automorphism group) reflects its geometry:\n\n• genus 0 – the isometry group of the sphere is the Möbius group of projective transforms of the complex line,\n• the isometry group of the plane is the subgroup fixing infinity, and of the punctured plane is the subgroup leaving invariant the set containing only infinity and zero: either fixing them both, or interchanging them (1/z).\n• the isometry group of the upper half-plane is the real Möbius group; this is conjugate to the automorphism group of the disk.\n• genus 1 – the isometry group of a torus is in general translations (as an Abelian variety), though the square lattice and hexagonal lattice have addition symmetries from rotation by 90° and 60°.\n• For genus g ≥ 2, the isometry group is finite, and has order at most 84(g − 1), by Hurwitz's automorphisms theorem; surfaces that realize this bound are called Hurwitz surfaces.\n• It is known that every finite group can be realized as the full group of isometries of some Riemann surface.\n• For genus 2 the order is maximized by the Bolza surface, with order 48.\n• For genus 3 the order is maximized by the Klein quartic, with order 168; this is the first Hurwitz surface, and its automorphism group is isomorphic to the unique simple group of order 168, which is the second-smallest non-abelian simple group. This group is isomorphic to both PSL(2,7) and PSL(3,2).\n• For genus 4, Bring's surface is a highly symmetric surface.\n• For genus 7 the order is maximized by the Macbeath surface, with order 504; this is the second Hurwitz surface, and its automorphism group is isomorphic to PSL(2,8), the fourth-smallest non-abelian simple group.\n\n## Function-theoretic classification\n\nThe classification scheme above is typically used by geometers. There is a different classification for Riemann surfaces which is typically used by complex analysts. It employs a different definition for \"parabolic\" and \"hyperbolic\". In this alternative classification scheme, a Riemann surface is called parabolic if there are no non-constant negative subharmonic functions on the surface and is otherwise called hyperbolic. This class of hyperbolic surfaces is further subdivided into subclasses according to whether function spaces other than the negative subharmonic functions are degenerate, e.g. Riemann surfaces on which all bounded holomorphic functions are constant, or on which all bounded harmonic functions are constant, or on which all positive harmonic functions are constant, etc.\n\nTo avoid confusion, call the classification based on metrics of constant curvature the geometric classification, and the one based on degeneracy of function spaces the function-theoretic classification. For example, the Riemann surface consisting of \"all complex numbers but 0 and 1\" is parabolic in the function-theoretic classification but it is hyperbolic in the geometric classification." ]	[ null, "https://upload.wikimedia.org/wikipedia/commons/thumb/9/9c/Riemann_sqrt.svg/310px-Riemann_sqrt.svg.png", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/2f02cfe21a83e074d5589b15f2e5f0947cb39c4d", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/0b7906cd754770b5fdbcdf47696755df0cd4bcd4", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/b13aa27846469ee752b04ccd1e5907256cf8b1a7", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/d094adcb0b3a44c6e5cbdc747eef133ab81f4d67", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab", null, "https://wikimedia.org/api/rest_v1/media/math/render/svg/2ecc0cf14129eadc29034c37c3514979e9a22f62", null, 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https://eng.kakprosto.ru/how-242383-how-to-calculate-average-wage	[ "You will need\n• settlement statements for the three months of the employee;\n• - production calendar;\n• calculator;\n• - HR documents employee.\nInstruction\n1\nUse the settlement statements for preceding three months of dismissal of the employee. Fold his wages for the period. Enable this amount of salary, bonuses, allowances and other payments that are remuneration for the performance of his official duties, spelled out in the agreement (contract). Do not include in the calculation of the cash issued to an employee as financial assistance or a lump sum.\n2\nNow the sum of the number of working days in the last three months of employment of the employee. To do this, use the production calendar. Then calculate the number of days worked in the period.\n3\nFind the value of the average daily wage of a specialist. To do this, divide the amount of payments on the actual number of days worked.\n4\nThen determine the average number of working days. The actual number of days worked, divide by three.\n5\nAfter that the average daily earnings of an employee, multiply by the average number of days worked in a month. Thus, to get the average wage of an employee. This value will be used for the calculation and accrual of unemployment benefits.\n6\nIf you need to calculate the average wage for the calculation of sickness benefit or vacation pay, the calculation will be made in the same way. The difference lies in the following. The billing period is twelve calendar months. Payments for performance of duties for the year are summarized. Then calculated the average daily wage by dividing total earnings by the number of calendar days in the year. The obtained value is multiplied by 29.5 days.\n7\nWhen filling out the reference to the employment center please note that the document contains graphs in which you need to specify the number of days of sick leave, vacation, if any." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9312381,"math_prob":0.97260046,"size":1875,"snap":"2023-14-2023-23","text_gpt3_token_len":378,"char_repetition_ratio":0.15499733,"word_repetition_ratio":0.0062111802,"special_character_ratio":0.20266667,"punctuation_ratio":0.10335196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9601221,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T21:19:56Z\",\"WARC-Record-ID\":\"<urn:uuid:8802ee12-70b3-478d-acad-c17163f5d10b>\",\"Content-Length\":\"33108\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e020816-b00f-4e1a-b407-1027936be37b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1828741c-6a91-481d-b314-cdbe8fa928bd>\",\"WARC-IP-Address\":\"178.154.246.3\",\"WARC-Target-URI\":\"https://eng.kakprosto.ru/how-242383-how-to-calculate-average-wage\",\"WARC-Payload-Digest\":\"sha1:UPDETR5JWYOFFNKCLEVLBD3MHE6BRLD5\",\"WARC-Block-Digest\":\"sha1:4437Z5LTIKML7UDOGFSGE3RVDKYTQI5L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656833.99_warc_CC-MAIN-20230609201549-20230609231549-00534.warc.gz\"}"}
https://bloginonline.com/diagram-of-a-typical-bone/diagram-of-a-typical-bone-diagram-of-a-typical-long-bone-anatomical-diagram-of-typical-long/	[ "# Diagram Of A Typical Bone Diagram Of A Typical Long Bone Anatomical Diagram Of Typical Long\n\nDiagram Of A Typical Bone Diagram Of A Typical Long Bone Anatomical Diagram Of Typical Long", null, "Tagged: a well labelled diagram of a typical long bone, diagram of a typical bone, diagram of a typical long bone, draw a labelled diagram of a typical long bone, draw and label a diagram of a typical bone, labeled diagram of the structure of a typical long bone, structure of a typical bone diagram" ]	[ null, "https://bloginonline.com/wp-content/uploads/2018/07/diagram-of-a-typical-bone-diagram-of-a-typical-long-bone-anatomical-diagram-of-typical-long.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62901187,"math_prob":0.9147955,"size":1508,"snap":"2019-51-2020-05","text_gpt3_token_len":312,"char_repetition_ratio":0.37367022,"word_repetition_ratio":0.37269372,"special_character_ratio":0.19363396,"punctuation_ratio":0.028571429,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97503364,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T07:20:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b7ce2b63-76e5-4f5a-b257-e4a1c5294575>\",\"Content-Length\":\"58433\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a0a2a5a-f9ce-4aae-ac31-a51863eceb8b>\",\"WARC-Concurrent-To\":\"<urn:uuid:4bc1fe6a-b977-4bd8-961b-2f188c9c794f>\",\"WARC-IP-Address\":\"145.239.7.237\",\"WARC-Target-URI\":\"https://bloginonline.com/diagram-of-a-typical-bone/diagram-of-a-typical-bone-diagram-of-a-typical-long-bone-anatomical-diagram-of-typical-long/\",\"WARC-Payload-Digest\":\"sha1:4Q7ZG3VMXYF5JO24FCKDYAVYRLG5ALCH\",\"WARC-Block-Digest\":\"sha1:2WUSBIQ5QEKUEPI2TNKIL6N73KQPL5HF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541307797.77_warc_CC-MAIN-20191215070636-20191215094636-00355.warc.gz\"}"}
https://betterlesson.com/lesson/425905/working-with-expressions-and-equations-part-3?from=breadcrumb_lesson	[ "# Working with Expressions and Equations Part 3\n\n15 teachers like this lesson\nPrint Lesson\n\n## Objective\n\nSWBAT: â¢ Create algebraic expressions and equations based on word problems. â¢ Use substitution to evaluate algebraic expressions.\n\n#### Big Idea\n\nBenâs mom gives him the option of grooming 3 horses or mowing 2 fields. Which job should Ben choose? Why? Students use their knowledge of algebraic expressions and equations to write expressions and equations.\n\n## Do Now\n\n7 minutes\n\nPart of my class routine is a do now at the beginning of every class. Students walk into class and pick up the packet for the day. They get to work quickly on the problems. Often, I create do nows that have problems that connect to the task that students will be working on that day. For this lesson I want students to start thinking about expressions that involve more than one operation.\n\nI ask for a volunteer to read number 1. I assert that I think that the answer is a. 8h + 2. I tell students to discuss my answer with their neighbor and decide whether they agree or disagree. I can on a few students to explain their thinking. I want students to articulate that a. would be correct if it cost \\$2 to enter and \\$8 for every extra hour. For number 3 I ask students to identify one answer choice that is incorrect and explain why it does not work. Why might a student choose that answer choice?\n\n## Guided Practice\n\n7 minutes\n\nAfter the Do Now, I have a student read the objectives for the day. I tell students that they will again be creating expressions and equations to model situations. These expressions and equations will be more complicated than the ones that we created previously.\n\nBefore students move into groups, we work through page 2 together. I ask students who has ever ridden a horse before. Students share out what they know about horses and we look at the picture about the tools that you need in order to groom, or clean, a horse.\n\nWe read #1 and I call on a student to share his/her answer. If the student gives me 90 minutes, I ask for another way to write that answer (1 ½ hours or 1.5 hours). How long would it take him to group 2 horses? 4? 9? Using that data, I ask students to write an expression that shows how many minutes it will take Ben to groom h horses and to create an expression that is equivalent. I want students to articulate that since it takes 30 minutes to groom each horse, you need to multiply the number of horses he grooms by 30 minutes and the order does not matter.\n\nStudents will be working in groups of 3-4 students on the rest of the packet. I will review group expectations and using the Group Work Rubric. See the Using the Group Work Rubric video in my Strategy Folder.\n\n## Group Work\n\n31 minutes\n\nI walk around and monitor student progress and fill out the group work rubric for each group. I am also looking for student work that has different strategies for solving #4 on page 5 to use during the closure.\n\nSome common mistakes are students incorrectly using substitution with the expression they have created. Some students also get confused with number 6 on page 4, since both chores will take the same amount of time. If students are struggling with pages 3-4, I may ask them the following questions:\n\n• What is going on in this situation?\n• How long would it take if he/she needed to groom 10 horses? What operation did you use to figure that out?\n• How many unknowns are in the situation?\n• What does your answer mean? What units belong with your answer? Could you write the answer in another way?\n\nSome students may struggle when they get to the “Making Money” questions on page 5 and 6. Here are some questions I may ask:\n\n• How many minutes are in an hour?\n• How much does he make for 1 hour of work? How can you use that information to help you?\n• How many hours did it take Ben to complete that chore?\n\nFor students who correctly work through problems I may ask them these questions and then have them work on the substitution practice on page 7:\n\n• Would Ben make more money grooming 11 horses or mowing 8 fields? How do you know?\n• How much money would Ben earn if he mowed 3 ½ fields?\n• Ben has 4 ½ hours, how many fields can he mow?\n\n## Closure and Ticket to Go\n\n15 minutes\n\nI ask students to share and compare their answers to question 6 on page 6. I want students to articulate that it doesn’t matter which chore he chooses, each of them will take 2 hours to complete. Then we’ll move on to talk about #4 on page 5. I will call up the students I identified during the group work time to put their work under the document camera and explain their thinking. Some students may use equivalent rates to show that if he earns \\$6/1 hour than he’ll earn \\$15/2 ½ hours of work. Other students may start with the rate \\$6/60 minutes and go from there. Other student may create a unit rate that \\$1/10 minutes and use that to find the amount of money Ben earns. Other students may use go through a similar process but not use rates. Some students will use decimals. If I did not see students using some of these strategies I will model them under the document camera.\n\nI will ask whether I could change 150 minutes --> 2 hours 30 minutes --> 2.3 hours to use in my calculations. I want students to be able to articulate that 2 hours and 30 minutes is not equivalent to 2.3 hours, since an hour is out of 60 minutes, so 30 minutes represents 30/60 or ½ or .5.\n\nIf I have time, I ask students to share struggles they had and how they overcame them. I also ask students if there are still struggles they are having. I ask other students to give advice. In my classroom I try to consistently show students that they will struggle with different problems but that they need to use their creative problem solving minds to try a strategy. If that strategy doesn’t work, try another one! The main question in this lesson was a difficult one that required students to problem solve and persevere through challenges and set-backs. I want to acknowledge my students hard work and their persistence.\n\nWith 5 minutes remaining, pass out the ticket to go. Students work independently to complete it." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96570885,"math_prob":0.6255729,"size":4864,"snap":"2019-35-2019-39","text_gpt3_token_len":1063,"char_repetition_ratio":0.15884773,"word_repetition_ratio":0.008908686,"special_character_ratio":0.22327302,"punctuation_ratio":0.07947686,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9535845,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-26T05:55:00Z\",\"WARC-Record-ID\":\"<urn:uuid:3194cbd6-abea-439a-b0b9-aa3b4f3e60dc>\",\"Content-Length\":\"143842\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2fb2849-ffff-4cea-9b10-70424314fd60>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5d36c5b-8c04-44a6-8789-953d81a269a1>\",\"WARC-IP-Address\":\"107.21.10.124\",\"WARC-Target-URI\":\"https://betterlesson.com/lesson/425905/working-with-expressions-and-equations-part-3?from=breadcrumb_lesson\",\"WARC-Payload-Digest\":\"sha1:LYKPFN6QQWYNM5DCLFZZRIY3ASUR7IC2\",\"WARC-Block-Digest\":\"sha1:ZIT5RCU7OX2AVYTWMTZYMKAZHVAK367R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330968.54_warc_CC-MAIN-20190826042816-20190826064816-00547.warc.gz\"}"}
https://refractiveindex.info/?shelf=main&book=AgGaSe2&page=Boyd-e	[ "# RefractiveIndex.INFO\n\nRefractive index database\n\nShelf\n\nBook\n\nPage\n\n## Optical constants of AgGaSe2 (Silver gallium selenide, AGSe)Boyd et al. 1972: n(e) 0.725–13.5 µm\n\nWavelength: µm\n(0.725 – 13.5)\n\n### Complex refractive index (n+ik)[ i ]\n\nn k LogX LogY eV\n\n### Dispersion formula [ i ]\n\n$$n^2-1=4.2912+\\frac{1.3970λ^2}{λ^2-0.2845}+\\frac{1.9282λ^2}{λ^2-1600}$$\n\nExtraordinary ray(e)\n\n### References\n\n1) G. D. Boyd, H. M. Kasper, J. H McFee, F. G Storz. Linear and nonlinear optical properties of some ternary selenides, IEEE J. Quant. Electron., 8, 900-908 (1972)\n2) G. C. Bhar, Refractive index interpolation in phase-matching, Appl. Opt. 15, 305-307 (1976)\n*Ref. 2 provides a dispersion formula based on data from Ref. 1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5535143,"math_prob":0.91295236,"size":657,"snap":"2022-40-2023-06","text_gpt3_token_len":251,"char_repetition_ratio":0.07810107,"word_repetition_ratio":0.0,"special_character_ratio":0.3850837,"punctuation_ratio":0.23125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9831254,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T11:46:46Z\",\"WARC-Record-ID\":\"<urn:uuid:3de00029-b4e0-4114-b140-061267af1587>\",\"Content-Length\":\"37660\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:292c9800-734e-49b8-8055-511f35e2eb25>\",\"WARC-Concurrent-To\":\"<urn:uuid:9019a275-f745-4cb3-81e5-0b835bcdf67f>\",\"WARC-IP-Address\":\"50.63.7.197\",\"WARC-Target-URI\":\"https://refractiveindex.info/?shelf=main&book=AgGaSe2&page=Boyd-e\",\"WARC-Payload-Digest\":\"sha1:RWFIVTVKCAJ6RXFZOOCXZMTWY4BCJTAE\",\"WARC-Block-Digest\":\"sha1:JTZ7BMXKC6QMHT6Z32H6FKXPSUCDGJ55\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335254.72_warc_CC-MAIN-20220928113848-20220928143848-00783.warc.gz\"}"}
https://carafilms.com/mastering-linear-regression-a-step-by-step-guide/	[ "# Mastering Linear Regression: A Step-By-Step Guide\n\nAre you ready to become a master of linear regression? If you’re looking to gain a deep understanding of this powerful statistical technique, then you’ve come to the right place.\n\nIn this step-by-step guide, we will walk you through the process of mastering linear regression, helping you to build a solid foundation and develop the skills needed to confidently apply it to your own data analysis projects.\n\nLinear regression is a fundamental tool in statistics and machine learning, allowing you to model the relationship between a dependent variable and one or more independent variables. It provides a way to make predictions, understand the strength of relationships, and identify the most important factors influencing the outcome.\n\nBy mastering linear regression, you will unlock the ability to uncover valuable insights from your data and make informed decisions based on evidence. Whether you’re a beginner or have some experience with linear regression, this guide will equip you with the knowledge and practical skills needed to excel in this field.\n\nSo, let’s dive in and take your understanding of linear regression to new heights!\n\n## Understanding the Basics of Linear Regression\n\nSo, you’re ready to dive into the world of linear regression? Let’s start by understanding the basics and wrap our heads around this popular statistical technique.\n\nLinear regression is a powerful tool used to analyze the relationship between two variables. It helps you predict the value of a dependent variable based on the values of one or more independent variables.\n\nThe main idea behind linear regression is to find the best-fitting line that represents the relationship between these variables. This line is determined by minimizing the sum of the squared differences between the predicted values and the actual values of the dependent variable. By doing so, linear regression allows you to make predictions and understand the impact of the independent variables on the dependent variable.\n\nTo get started with linear regression, you need to have a clear understanding of the key components involved. The dependent variable, also known as the response variable, is the variable you want to predict or explain using the independent variables.\n\nOn the other hand, the independent variables, also known as predictor variables, are the variables used to predict the value of the dependent variable. It’s important to note that linear regression assumes a linear relationship between the independent variables and the dependent variable. This means that the relationship can be represented by a straight line on a scatter plot.\n\nBy understanding these basics, you can begin to grasp the foundations of linear regression and move towards mastering this essential statistical technique.\n\n## Defining Dependent and Independent Variables\n\nTo define your dependent and independent variables, picture yourself as a researcher identifying the factors that influence a particular outcome. The dependent variable is the outcome or the variable that you’re trying to predict or explain. It’s the variable that you believe is influenced by one or more independent variables.\n\nFor example, if you’re studying the effect of temperature on plant growth, the dependent variable would be the growth of the plants. In this case, temperature would be the independent variable as it’s believed to have an effect on the growth of the plants.\n\nThe independent variables, on the other hand, are the variables that you believe have an effect on the dependent variable. These variables are usually manipulated or controlled by the researcher to observe their impact on the dependent variable.\n\nIn the plant growth example, the independent variable would be the temperature. Other independent variables that could be considered include the amount of sunlight, the type of soil, or the amount of water provided to the plants.\n\nBy defining your dependent and independent variables clearly, you can set up your study in a way that allows you to analyze the relationship between them and make predictions or explanations based on your findings.\n\n## Assumptions and Types of Linear Regression Models\n\nLinear regression models make certain assumptions about the relationship between the dependent and independent variables, and understanding these assumptions is crucial for accurate analysis and interpretation of the data.\n\nThe first assumption is linearity, which means that there is a linear relationship between the dependent variable and the independent variable(s). This assumption implies that as the independent variable(s) change, the dependent variable changes in a constant and predictable manner. If this assumption is violated, the results of the linear regression model may not be reliable.\n\nThe second assumption is independence, which means that the observations in the dataset are independent of each other. This assumption is important because if the observations are not independent, it can lead to biased and inefficient estimates. To check for independence, it is important to ensure that there is no systematic pattern or correlation in the residuals, which are the differences between the observed and predicted values of the dependent variable.\n\nAnother assumption is homoscedasticity, which means that the variance of the residuals is constant across all levels of the independent variable(s). If this assumption is violated, it indicates that the spread of the residuals is not consistent, and it can lead to unreliable estimates and incorrect inferences. To check for homoscedasticity, one can plot the residuals against the predicted values and look for any patterns or trends.\n\nLastly, the assumption of normality states that the residuals follow a normal distribution. This assumption is important because many statistical tests and confidence intervals rely on the assumption of normality. Violation of this assumption can lead to incorrect p-values and confidence intervals.\n\nOverall, understanding the assumptions and types of linear regression models is crucial for accurate analysis and interpretation of data. Violation of these assumptions can lead to biased and unreliable results, so it’s important to check for them and consider alternative models if necessary.\n\n## Step 1: Data Collection and Cleaning\n\nCollecting and cleaning data is the first exciting step in building a successful linear regression model. It involves gathering all the necessary data points and ensuring that they’re accurate and reliable.\n\nYou start by identifying the variables that are relevant to your model and collecting data for each of them. This may involve conducting surveys, performing experiments, or extracting data from existing sources.\n\nOnce you’ve collected the data, you need to clean it by removing any errors, outliers, or missing values. This is crucial because using flawed data can lead to inaccurate and unreliable results. Cleaning the data involves techniques such as imputation, where missing values are estimated and replaced, and outlier detection, where extreme values are identified and either corrected or removed.\n\nThe process of data cleaning also includes checking for inconsistencies, such as duplicate entries or contradictory information. It’s important to ensure that all the data is in the same format and follows the same conventions. This may require converting data types, standardizing units of measurement, or merging datasets.\n\nAdditionally, it’s crucial to validate the data by cross-checking it with known values or independent sources. This helps to identify any potential errors or discrepancies.\n\nBy meticulously collecting and cleaning your data, you lay the foundation for a robust and reliable linear regression model. It ensures that your analysis is based on accurate and trustworthy information, leading to more meaningful insights and better predictions.\n\n## Step 2: Model Building and Evaluation\n\nOnce you’ve gathered and cleaned your data, it’s time to start building and evaluating your model to uncover valuable insights and make accurate predictions. The first step in model building is to select the appropriate variables for your regression analysis. This involves identifying the independent variables that are most likely to have an impact on the dependent variable you’re trying to predict.\n\nIt’s important to consider the theoretical and practical significance of each variable and ensure they’re relevant to your research question.\n\nAfter selecting the variables, you can start building your regression model. There are various techniques and algorithms available, but a common approach is to use the ordinary least squares (OLS) method. This method minimizes the sum of squared residuals to find the best-fitting line that represents the relationship between the independent and dependent variables.\n\nOnce the model is built, you can evaluate its performance by assessing the goodness of fit. This can be done by analyzing the R-squared value, which indicates the proportion of the variance in the dependent variable that’s explained by the independent variables. Additionally, you can examine the p-values of the coefficients to determine if they’re statistically significant. A low p-value suggests that the coefficient has a significant impact on the dependent variable.\n\nBy carefully evaluating your model, you can ensure its accuracy and reliability, allowing you to make informed decisions and predictions based on the insights gained from the regression analysis.\n\n### What are some common challenges faced during the data collection and cleaning process in linear regression?\n\nSome common challenges you may face during the data collection and cleaning process in linear regression include missing values, outliers, data inconsistencies, and dealing with large datasets.\n\n### How can we handle missing data in linear regression analysis?\n\nTo handle missing data in linear regression analysis, you can use techniques like imputation, where missing values are filled in based on other variables, or you can remove the incomplete cases from the analysis.\n\n### Are there any specific techniques or methods to deal with outliers in linear regression?\n\nTo handle outliers in linear regression, you can use techniques like removing outliers based on statistical tests, transforming variables, or using robust regression methods. These methods help improve the accuracy of the regression analysis.\n\n### What are some common evaluation metrics used to assess the performance of a linear regression model?\n\nCommon evaluation metrics used to assess linear regression models include mean squared error, mean absolute error, root mean squared error, and R-squared. These metrics help you measure how well the model fits the data and make comparisons between different models.\n\n### How can multicollinearity affect the results of a linear regression analysis and how can it be addressed?\n\nMulticollinearity can distort the results of a linear regression analysis by making it difficult to determine the independent effects of predictor variables. It can be addressed by removing correlated variables or using techniques like principal component analysis.\n\n## Conclusion\n\nSo there you have it, a step-by-step guide to mastering linear regression. By understanding the basics of linear regression and the importance of defining dependent and independent variables, you’re well on your way to becoming a pro in this field.\n\nAdditionally, by familiarizing yourself with the assumptions and types of linear regression models, you can choose the most suitable approach for your data.\n\nThe key to success lies in thorough data collection and cleaning, followed by model building and evaluation. This process ensures that your model is accurate and reliable.\n\nWith practice and dedication, you’ll become proficient in mastering linear regression and be able to confidently apply it to various real-world scenarios.\n\nSo go ahead, take the plunge, and start your journey towards becoming a linear regression expert!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9060462,"math_prob":0.9049163,"size":11985,"snap":"2023-40-2023-50","text_gpt3_token_len":2086,"char_repetition_ratio":0.19288874,"word_repetition_ratio":0.0413679,"special_character_ratio":0.17046309,"punctuation_ratio":0.08787432,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98242915,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T22:37:19Z\",\"WARC-Record-ID\":\"<urn:uuid:1141d720-358a-4a1f-9b6c-fa4cffb59ecd>\",\"Content-Length\":\"66407\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49655865-de66-473c-99ea-a21c363948da>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7f0f73f-5e9c-463f-87a2-d333a1044fed>\",\"WARC-IP-Address\":\"172.67.150.147\",\"WARC-Target-URI\":\"https://carafilms.com/mastering-linear-regression-a-step-by-step-guide/\",\"WARC-Payload-Digest\":\"sha1:BDKHYBUOHF35UG7MORJHKN3ZOVDZKFXG\",\"WARC-Block-Digest\":\"sha1:MQG4UPREB2KUTZ4EWBCRTQSZTY7SJM6V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100705.19_warc_CC-MAIN-20231207221604-20231208011604-00128.warc.gz\"}"}
https://meangreenmath.com/category/popular-culture/page/2/	[ "# Predicate Logic and Popular Culture (Part 237): Psycho\n\nLet", null, "$P$ be the set of all people, let", null, "$T$ be the set of all times, and let", null, "$M(x,t)$ be the statement “", null, "$x$ goes mad at time", null, "$t$.” Translate the logical statement", null, "$\\forall x \\in P \\exists t \\in T (\\sim M(x,t))$.\n\nThis matches a line from the movie Psycho.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 236): Dirty Dancing\n\nLet", null, "$P$ be the set of all people, and let", null, "$C(x)$ be the statement “", null, "$x$ puts Baby in a corner.” Translate the logical statement", null, "$\\forall x \\in P (\\sim C(x))$.\n\nThis matches a line from the movie Dirty Dancing.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 235): Suits\n\nLet", null, "$p$ be the statement “Winners make excuses,” and let", null, "$q$ be the statement “The other side plays the game.” Translate the logical statement", null, "$q Rightarrow \\sim p$.\n\nThis matches a line from the TV show Suits.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 234): Linkin Park\n\nLet", null, "$p$ be the statement “They turn down the lights,” and let", null, "$q$ be the statement “I hear my battle symphony.” Translate the logical statement", null, "$p \\Rightarrow q$.\n\nThis matches part of the chorus of “Battle Symphony” by Linkin Park.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 233): Panic! At The Disco\n\nLet", null, "$F(x)$ be the statement “", null, "$x$ feels good,” let", null, "$H(x)$ be the statement “", null, "$x$ tastes good,” let", null, "$M(x)$ be the statement “", null, "$x$ is mine,” and let", null, "$H$ be the set of all things. Translate the logical statement", null, "$\\forall x \\in H( (F(x) \\land H(x)) \\Rightarrow M(x))$.\n\nThis matches a line from “Emperor’s New Clothes” by Panic! At The Disco.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 232): Limp Bizkit\n\nLet", null, "$B(x)$ be the statement “", null, "$x$ knows what it’s like to be the bad man,” let", null, "$H(x)$ be the statement “", null, "$x$ knows what it’s like to be hated,” and let", null, "$P$ be the set of all people. Translate the logical statement", null, "$\\forall x \\in P(\\lnot B(x) \\land \\lnot H(x))$.\n\nThis matches the opening lines of “Behind Blue Eyes” by Limp Bizkit.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 231): Aristocats\n\nLet", null, "$C(x)$ be the statement “", null, "$x$ wants to be a cat,” and let", null, "$P$ be the set of all people. Translate the logical statement", null, "$\\forall x in P(C(x))$.\n\nThis matches the opening line of “Everyone Wants to be a Cat” from the movie “The Aristocats.”\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 230): Dean Lewis\n\nLet", null, "$W(t)$ be the statement “It is easy to walk away at time", null, "$t$,” and let", null, "$T$ be the set of all times. Translate the logical statement", null, "$\\forall t \\in T(\\lnot W(t))$.\n\nThis matches part of the chorus of “Be Alright” by Dean Lewis.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 229): Mean Girls\n\nLet", null, "$W(t)$ be the statement “", null, "$t$ is a Wednesday,” let", null, "$P(t)$ be the statement “We wear pink at time", null, "$t$,” and let", null, "$T$ be the set of all times. Translate the logical statement", null, "$\\forall t \\in T(W(t) \\Rightarrow P(t))$.\n\nThis matches a line from the movie “Mean Girls.”\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. However, for the purposes of this series, I’ll just present the statement in predicate logic first.\n\n# Predicate Logic and Popular Culture (Part 228): Hannah Montana\n\nLet", null, "$M(x)$ be the statement “", null, "$x$ makes mistakes,” let", null, "$D(x)$ be the statement “", null, "$x$ has those days,” and let", null, "$P$ be the set of all people. Translate the logical statement", null, "$\\forall x \\in P(M(x) \\land D(x))$.\n\nThis matches the opening lines of “Nobody’s Perfect” by Hannah Montana.\n\nContext: Part of the discrete mathematics course includes an introduction to predicate and propositional logic for our math majors. As you can probably guess from their names, students tend to think these concepts are dry and uninteresting even though they’re very important for their development as math majors.\n\nIn an effort to making these topics more appealing, I spent a few days mining the depths of popular culture in a (likely futile) attempt to make these ideas more interesting to my students. In this series, I’d like to share what I found. Naturally, the sources that I found have varying levels of complexity, which is appropriate for students who are first learning prepositional and predicate logic.\n\nWhen I actually presented these in class, I either presented the logical statement and had my class guess the statement in actual English, or I gave my students the famous quote and them translate it into predicate logic. 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https://www.nagwa.com/en/videos/748137847924/	[ "# Video: Determining the Definite Integral of a Trigonometric Function\n\nDetermine ∫_(−𝜋) ^(−𝜋/4) 8 cos 5𝜃 d𝜃.\n\n04:50\n\n### Video Transcript\n\nDetermine the integral of eight cos five 𝜃 d𝜃 between negative 𝜋 by four and negative 𝜋.\n\nOur first step here is to take out the constant eight. This leaves us with eight multiplied by the integral of cos five 𝜃 d𝜃. The integral of cos 𝜃 is a standard one that we should know. It is equal to sin 𝜃, as integrating is the opposite, or inverse, of differentiating.\n\nWe want to integrate cos five 𝜃. This means we need to use another general rule. This states that the integral of cos 𝑛𝜃 is equal to one over 𝑛 multiplied by sin 𝜃. We differentiate the 𝑛𝜃 to give us 𝑛. The integral of cos five 𝜃 is equal to one-fifth multiplied by sin five 𝜃. We need to multiply this by eight and have limits of negative 𝜋 by four and negative 𝜋. As with the eight at the start, we can take the constant one-fifth outside of the bracket. This gives us eight-fifths multiplied by sin five 𝜃.\n\nOur next step is to substitute in our two limits and subtract the answers. Substituting in the upper limit gives us sin of five multiplied by negative 𝜋 over four. This can be rewritten as sin of negative five 𝜋 over four. Substituting in our lower limit gives us sin of five multiplied by negative 𝜋. Once again, this can be rewritten as sin of negative five 𝜋.\n\nAt this point, it is worth drawing the sine curve to see if negative five 𝜋 by four and negative five 𝜋 correspond with any our known angles. The sine curve has a maximum value of one and a minimum value of negative one. It has key values on the 𝜃-, or 𝑥-axis, of 𝜋 by two and 𝜋, negative 𝜋 by two and negative 𝜋. If you prefer to think of these angles in degrees. It’s worth remembering that 𝜋 radians is equal to 180 degrees.\n\nThe sine curve looks as shown in the diagram. However, at the moment, we have a slight problem as our two angles negative five 𝜋 by four and negative five 𝜋 don’t fit in the range. As the sine curve has a period of two 𝜋, it repeats every two 𝜋 radians, we can continue the graph as shown.\n\nWe can see clearly from the graph that the sin of negative five 𝜋 is equal to zero. Negative five 𝜋 by four is shown in the diagram. By going vertically upwards to the sine curve and then horizontally along to the 𝑦-axis, we can see what this value will take. Due to the symmetry of the sine curse, sin of negative five 𝜋 by four is equal to sin of 𝜋 by four.\n\n𝜋 by four is equal to 45 degrees and is one of our known angles. This is equal to root two over two. The sin of 45 degrees equals root two over two. Therefore, the sin of 𝜋 by four radians must also equal root two over two. The sin of negative five 𝜋 by four is equal to root two over two. And the sin of negative five 𝜋 is equal to zero.\n\nRoot two over two minus zero is root two over two. So, we need to multiply this by eight-fifths. Multiplying the numerators gives us eight root two. And multiplying the denominators, gives us 10. We have eight root two over 10. Eight and 10 have a common factor of two. So, we can divide the numerator and denominator by two. This gives us four root two over five. We can, therefore, say that the integral of eight cos five 𝜃 d𝜃 between negative 𝜋 by four and negative 𝜋 is equal to four root two over five." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9327611,"math_prob":0.9968589,"size":3176,"snap":"2020-45-2020-50","text_gpt3_token_len":894,"char_repetition_ratio":0.1793821,"word_repetition_ratio":0.16242038,"special_character_ratio":0.22638538,"punctuation_ratio":0.09348442,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99968016,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T11:26:27Z\",\"WARC-Record-ID\":\"<urn:uuid:c54b4209-38e1-4ee9-8f68-bb5864c06cb2>\",\"Content-Length\":\"29399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44f32aaf-3362-4552-89f7-64fc02af3f3e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3275d99d-2306-4c3e-adba-c86a87116a6f>\",\"WARC-IP-Address\":\"23.23.60.1\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/videos/748137847924/\",\"WARC-Payload-Digest\":\"sha1:VAT25YWXSMW5SMCTLBAJUAUCMC4ZWA47\",\"WARC-Block-Digest\":\"sha1:FBEHRGKZBDAPLBGDPI2RRUK47CUZ4P6A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879537.28_warc_CC-MAIN-20201022111909-20201022141909-00319.warc.gz\"}"}
https://www.npmjs.com/package/geometric-pack	[ "# npm\n\n## geometric-pack", null, "0.7.1 • Public • Published\n\n# Geometric pack\n\nGeometric calculator created with TypeScript\n\nThe package supports accuracy up to 12 decimal places.\n\n## General info\n\nGeometric pack with lots of available calculations for 2D and 3D geometry\n\n## Example\n\n```import { Triangle } from \"geometric-pack\";\n\nconst firstTriangle = new Triangle(5, 3, 4);\nconst secondTriangle = new Triangle(10, 6, 8);\n\nconsole.log(firstTriangle.isCongruent(secondTriangle)); // False\nconsole.log(firstTriangle.isSimilar(secondTriangle)); // True```\n\nYou can also check more examples by running `.getDefinition()` method on other classes or by running example file\n\n``````\\$ node node_modules/geometric-pack/dist/utils/example\n``````\n\n## Launch\n\nTo use available calculations in this project you will have to create object of class you want to use and start available methods\n\n```import { Triangle } from \"geometric-pack\";\n\nconst firstTriangle = new Triangle(5, 3, 4);\n\nconsole.log(firstTriangle.getDefinition());\n\n// returns:\n// {\n// sideLengthA: 3,\n// sideLengthB: 4,\n// sideLengthC: 5,\n// circumference: 12,\n// area: 6,\n// rightAngle: true,\n// heights: { heightOfBaseA: 4, heightOfBaseB: 3, heightOfBaseC: 2.4 },\n// angles: { gamma: 90, beta: 53.13010235415598, alpha: 36.86989764584402 },\n// }```\n\nAvailable classes:\n\n• Two-dimensional\n\n• `Triangle`, You have to use three arguments and program will take them as a sides of triangle 'a', 'b', 'c', you can write them in any order you want because program will sort them a < b < c\n\n• `Rectangle`, You have to use two arguments and program will take them as a sides of rectangle 'a', 'b', program will not sort them in any order\n\n• `Square`, You have to use one argument and program will take it as a side of square 'a'\n\n• `Circle`, You have to use one argument and program will take it as a radius of circle 'r'\n\n• `Rhombus`, You have to use two arguments and program will take them as side 'a' and height 'h' in exact order\n\n• `Parallelogram`, You have to use three arguments and program will take them as side 'a', side 'b' and height 'h' in exact order\n\n• `Polygon`, It's a regular polygon so you have to use two arguments and program will take them as side 'a' and number of sides 'n' where 3 <= n <= 14 in exact order\n\n• `Stadium`, You have to use two arguments and program will take them as side length 'a' and radius 'r' in exact order\n\n• `Distance2D`, You have to use four arguments and program will take them as coordinates 'x1', 'y1', 'x2', 'y2' in exact order\n\n• `Annulus`, You have to use two arguments and program will take them as outerRadius 'r1' and innerRadius 'r2' in exact order\n\n• `QuadraticFunction`, You have two use three arguments and program will take them as 'a', 'b' and 'c' of function in exact order\n\n• `LinearFunction`, You have to use two arguments and program will take them as 'a' and 'b' of function in exact order\n\n• Three-dimensional\n\n• `Cone`, You have to use two arguments and program will take them as radius 'r' and height 'h' in exact order\n\n• `Cylinder`, You have to use two arguments and program will take them as radius 'r' and height 'h' in exact order\n\n• `Sphere`, You have to use one argument and program will take it as radius 'r'\n\n• `Cube`, You have to use one argument and program will take it as length side 'a'\n\n• `ConicalFrustum`, You have to use three arguments and program will take them as radius1 'r1', radius2 'r2' and height 'h' in exact order\n\n• `Capsule`, You have to use two arguments and program will take them as sideLength 'a' and radius 'r' in exact order\n\n• `Hemisphere`, You have to use two one argument and program will take it as radius 'r'\n\n• `Pyramid`, You have to use two arguments and program will take them as side length 'a' and height 'h' in exact order\n\n• `Cuboid`, You have to use three arguments and program will take them as length 'l', width 'w' and height 'h' in exact order\n\n• `Distance3D`, You have to use six arguments and program will take them as coordinates 'x1', 'y1', 'z1', 'x2', 'y2', 'z2' in exact order\n\n• `Tube`, You have to use three arguments and program will take them as outerRadius 'r1', innerRadius 'r2' and height 'h' in exact order\n\n## Setup\n\nTo run this project, install it locally using npm\n\n``````\\$ cd ../lorem\n\\$ npm init\n\\$ npm i geometric-pack\n``````\n\n## Technologies\n\nProject is created with:\n\n• TypeScript: 4.3.5\n\n### Install\n\n`npm i geometric-pack`\n\n### Repository\n\ngithub.com/MateuszCaputa/geometric-pack\n\n### Homepage\n\ngithub.com/MateuszCaputa/geometric-pack\n\n8\n\n0.7.1" ]	[ null, "https://static.npmjs.com/255a118f56f5346b97e56325a1217a16.svg", null 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