Datasets:

Infi-MM
/

InfiMM-WebMath-40B

Dataset card Data Studio Files Files and versions Community

Split (1)

train · ~22.8M rows (showing the first 514k)

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.

URL stringlengths 15 1.68k	text_list sequencelengths 1 199	image_list sequencelengths 1 199	metadata stringlengths 1.19k 3.08k
https://physics.stackexchange.com/search?q=user:92654+%5Bgeneral-relativity%5D	[ "# Search Results\n\nResults tagged with Search options user 92654\n10 results\n\nA theory that describes how matter interacts dynamically with the geometry of space and time. It was first published by Einstein in 1915 and is currently used to study the structure and evolution of the universe, as well as having practical applications like GPS.\n\nThe rate of acceleration is independent of the rate of motion so this formula still applies even when the two objects are moving relative to each other. The only time you need to use a different form …\nanswered Feb 15 '16 by Anders Gustafson\nThe Schwarzchild metric is for the gravitational field of an object of mass $M$ with no electric charge and no angular momentum. The metric is $${ds}^{2} = \\frac{dr^2}{1 - \\frac{r_\\mathrm{s}}{r}} - … asked Aug 29 '18 by Anders Gustafson 4answers Time tends to slow down near objects with large amounts of positive mass relative to observers in micro gravity. Considering that negative mass is the opposite of normal mass and would time tend to s … asked Apr 12 '16 by Anders Gustafson 4answers In Newtonian Physics there is an equation that for the Gravitational Flux of an object known as Gauss's Law For Gravity. Gauss's Law for Gravity describes the number of Gravitational Field Lines comi … asked Dec 6 '15 by Anders Gustafson Repulsive Gravity would require Negative Energy and pure Negative Energy is not known to exist or at least as real particles. Negative Energy would also have Negative inertia and so it would tend to … answered Dec 28 '15 by Anders Gustafson If you travel faster than light you wouldn't be going back in time from all reference frames, just some reference frames. If you were going faster than light then from some reference frames you would … answered Sep 15 '15 by Anders Gustafson 0answers A spacetime with 2 indistinguishable dimensions and all spacetime directions equivalent would have the signature (++) meaning that there would be no difference between spacelike and timelike direction … asked 4 hours ago by Anders Gustafson 2answers If I understand correctly the equation for kinetic energy in relativity is$$ E_k= mc^2 \\left(\\frac{1}{\\sqrt{1-\\frac{v^2}{c^2}}}-1\\right) \\;, and the equation for escape velocity in General Relativi …\nIn the Schwarzschild Metric as the spacetime interval between two points in spacetime approaches $0$ for any ratio between the length of time and space the spacetime interval between the points in spa …" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90891767,"math_prob":0.9746984,"size":3335,"snap":"2019-35-2019-39","text_gpt3_token_len":850,"char_repetition_ratio":0.12668869,"word_repetition_ratio":0.0093632955,"special_character_ratio":0.23898052,"punctuation_ratio":0.062184874,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99075115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T02:12:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0696e043-a6c1-40e5-9a47-21979fae16bb>\",\"Content-Length\":\"117598\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c08c753-06ef-4126-aa9d-71eeff10a343>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f54e024-a2b8-4537-83ed-a5481f52a878>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/search?q=user:92654+%5Bgeneral-relativity%5D\",\"WARC-Payload-Digest\":\"sha1:WEY2TODROBR4CAHHTZ4Z2JCJXAFDWEVH\",\"WARC-Block-Digest\":\"sha1:Z2TBLJ26CW7IUYCGNA6E5HEGXALXOFKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573415.58_warc_CC-MAIN-20190919015534-20190919041534-00027.warc.gz\"}"}
http://www.archimedes-lab.org/monthly_puzzles_66.html	[ "Shortcuts", null, "Sitemap", null, "Contact", null, "Newsletter", null, "Store", null, "Books", null, "Features", null, "Gallery", null, "E-cards", null, "Games", null, "", null, "••• Smile! \"To do exactly the opposite is also a form of imitation\" - G. C. Lichtenberg ••• ••• Related links Puzzles workshops for schools & museums. Editorial content and syndication puzzles for the media, editors & publishers. Numbers, just numbers... Have a Math question? Ask Dr. Math! ••• Etymology Nostalgia f. Greek nostos 'return home' and -algos 'pain', translating German Heimweh 'homesickness', was coined in 1688 by the Alsatian physician Johannes Hofer. In his treatise 'Dissertatio Medica de Nostalgia, oder Heimweh', he described a new disease that affected some soldiers of the Swiss regiments serving the king of France as \"the pain a sick person feels because he wishes to return to his native land, and fears never to see it again\". •••", null, "Previous Puzzles of the Month + Solutions\n\nApril-May 2007, Puzzle nr 111", null, "Back to Puzzle-of-the-Month page \| Home", null, "Puzzle # 111 Italiano", null, "Français", null, "Difficulty level:", null, "", null, "", null, ", basic geometry knowledge.\n\n Soccer balls Two intersected semi-circles inscribed in a rectangle are tangent to 5 discs (soccer balls) as depicted in the image. What is the value of b/a?", null, "", null, "", null, "R = radius of the semi-circles; r = radius of the soccer balls. R - r = OA = radius of the circle (O, OA). SO(A') = A (point A' is symmetrical to A through point O) AA' = 2(R - r) [the diameter of the circle (O, OA)] = a AB = BA' = R = b/2 Triangle ABA' is inscribed in the circle (O, OA), then it is isosceles and right-angled.\n\nWe can therefore solve this puzzle with the help of the theorem of Pytagora:\nAA'2 = AB2 + BA'2\na2 = (b/2)2+ (b/2)2 = 2(b/2)2\nSimplifying: a = b", null, "2/2\n\nSo: b/a = b/(b", null, "2/2) =", null, "2\n\nHere below is a neat solution involving a general formula, suggested by Géry Huvent:", null, "a) Considering the basic diagram above; the circles centered in O with radius r1 , and in A with radius r3 are tangent, then:", null, "(n2 + r32) + r3 = r1\n\nb) The circles centered in B with radius r2 , and in A with radius r3 are tangent, then:", null, "[n2 + (m - r3)2] = AB = r2 + r3\n\nb) Simplifying both equations together, we obtain the following math relationship useful to solve other similar problems:\nr1(r1 - 2r3) = (m + r2)(r2+ 2r3 - m)\n\nc) Comparing this latter equation to the original puzzle diagram further above (), we can see that:\na = m ;\nb/2 = R = r1 = r2 ; r* = r3 ;\n\nand that:\nr* = R* - a/2 = b/2 - a/2 = (b - a*)/2\n\nd) Then, replacing all the values of the general formula in b), we obtain:\nb/2{b/2 - 2[(b - a)/2]} = (a + b/2) · {b/2 + 2[(b - a)/2] - a} ; and solving...\n\na2 = (b - a)(b + a) ; finally... b/a =", null, "2\n\nYou can find more interesting sangaku formulae at Géry's website.\n\nHere below another interesting solution posted by John Reidy:", null, "The radius of the larger semicircle on base b is b/2\nLet the radius of the projected soccer ball circular image = r\nAt the contact point of any of the circles, the radius line from the centre of each circle is at right angles to a tangent at that contact point. Therefore the lines AC and AD joining the centres of the contacting circles must pass through the contact point of these circles; these points being G and E respectively. Accordingly, as the sides of the rectangle are tangential at contact points B and F, the points B and F lie on the extension of the line joining circle centres D & C.\n\nFrom triangle ABC:\nAB2 = CA2 – BC2\nAnd from triangle ABD:\nAB2 = DA2 – BD2\nTherefore: CA2 – BC2 = DA2 – BD2\n\nFrom the diagram:\nCA = AG – GC = b/2 – r\nBC = r\nDA = AE + ED = b/2 + r\nBD = BF – DF = a – r\nTherefore:\n(b/2 – r)2 – r2 = (b/2 + r)2 – (a – r)2\nb2/4 – br + r2 – r2 = b2/4 + br + r2a2 + 2ar - r2\nRationalising:\n2br + 2ar = a2\nr = a2/2(b + a) -- Equation 1\n\nFrom inspection of AH in diagram:\na = AH = b/2 + (b/2 – 2r)\n\nTherefore: r = (ba)/2 -- Equation 2\n\nFrom Equation's 1 & 2:\na2/2(b + a) = (ba)/2\na2 = (ba)(b + a) = b2a2\n2a2 = b2\nb/a =", null, "2", null, "The Winners of the Puzzle of the Month are:\nJohn Reidy, Australia - Omar A. Alrefaie, Saudi Arabia - Amaresh G.S., India - Mohammed Ezz Abd El-Menaem El-Sayyed, Egypt - Geoffrey Harrison, Australia.\n\nCongratulations!\n\n More Math Facts behind the puzzle In geometry, a shape comprising two circular arcs, joined at their endpoints (fig. 1) is called a lens. The \"Vesica piscis\" (fish bladder) is one particular form of a symmetrical lens (fig. 2).", null, "", null, "A = r2[(", null, "", null, "/180) - sin", null, "] Vesica piscis\n\n© 2003 G. Sarcone, www.archimedes-lab.org\nYou can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.\n\nYou're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!\n\n Previous puzzles of the month...", null, "", null, "", null, "Back to Puzzle-of-the-Month page \| Home", null, "", null, "", null, "Follow us on Facebook \|", null, "Report any error, misspelling or dead link\nArchimedes' Laboratory™ \| How to contact us\n\|", null, "Come contattarci \|", null, "Comment nous contacter", null, "About Us \| Sponsorship \| Press-clippings \| [email protected] \| ©opyrights \| Link2us \| Sitemap\n© Archimedes' Lab \| Privacy & Terms \| The web's best resource for puzzling and mental activities", null, "", null, "", null, "" ]	[ null, "http://www.archimedes-lab.org/im_navpics2/icon_sitemap.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_contact.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_newsletter.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_store.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_book.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_syndication.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_gallery.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_ecards.gif", null, "http://www.archimedes-lab.org/im_navpics2/icon_game.gif", null, "http://www.archimedes-lab.org/images5/eureka_b_or.gif", null, "http://www.archimedes-lab.org/images/bottomtruc.gif", null, "http://www.archimedes-lab.org/images/corner2b.gif", null, "http://www.archimedes-lab.org/im_navpics2/arrow_leftSM.gif", null, "http://www.archimedes-lab.org/im_navpics2/arrow_upSM.gif", null, "http://www.archimedes-lab.org/images7/button-italy2Bl.gif", null, "http://www.archimedes-lab.org/images7/button-france2Bl.gif", null, "http://www.archimedes-lab.org/im_maths/Bulb2.gif", null, "http://www.archimedes-lab.org/im_maths/Bulb2.gif", null, "http://www.archimedes-lab.org/im_maths/Bulb2.gif", null, "http://www.archimedes-lab.org/im_maths/mon_p_apr_2k7_L2.gif", null, "http://www.archimedes-lab.org/images5/solution.gif", null, "http://www.archimedes-lab.org/im_maths/mon_p_apr_2k7_SOLU.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/im_maths/sangaku_huvent.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/im_maths/mon_p_apr_2k7_L3.gif", null, "http://www.archimedes-lab.org/numbers/images_num/radix.gif", null, "http://www.archimedes-lab.org/images8/Cup.gif", null, "http://www.archimedes-lab.org/im_maths/lens_math.gif", null, "http://www.archimedes-lab.org/im_maths/vesica_piscis.gif", null, "http://www.archimedes-lab.org/numbers/images_num/Alpha_small.gif", null, "http://www.archimedes-lab.org/numbers/images_num/Pi_small.gif", null, "http://www.archimedes-lab.org/numbers/images_num/Alpha_small.gif", null, "http://www.archimedes-lab.org/images5/content_solution.gif", null, "http://www.archimedes-lab.org/images3/puzzlesolver_petit.gif", null, "http://www.archimedes-lab.org/im_navpics2/arrow_leftSM.gif", null, "http://www.archimedes-lab.org/im_navpics2/arrow_upSM.gif", null, "http://www.archimedes-lab.org/images/transparent.gif", null, "http://www.archimedes-lab.org/im_navpics2/facebook2.gif", null, "http://www.archimedes-lab.org/im_navpics/discuss_comment.gif", null, "http://www.archimedes-lab.org/images5/italiano_flag.gif", null, "http://www.archimedes-lab.org/images5/francais_flagBIS.gif", null, "http://www.archimedes-lab.org/images/fondu.gif", null, "http://www.archimedes-lab.org/images/transparent.gif", null, "http://www.archimedes-lab.org/images/transparent.gif", null, "http://www.archimedes-lab.org/images/corner2.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8359361,"math_prob":0.9779523,"size":2903,"snap":"2019-43-2019-47","text_gpt3_token_len":1033,"char_repetition_ratio":0.114867195,"word_repetition_ratio":0.0192,"special_character_ratio":0.3689287,"punctuation_ratio":0.10810811,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9910673,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,7,null,7,null,7,null,10,null,10,null,2,null,2,null,null,null,null,null,null,null,1,null,5,null,1,null,null,null,null,null,null,null,1,null,null,null,null,null,null,null,1,null,null,null,5,null,1,null,1,null,2,null,1,null,2,null,5,null,5,null,10,null,10,null,null,null,6,null,6,null,7,null,7,null,8,null,null,null,null,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T02:06:13Z\",\"WARC-Record-ID\":\"<urn:uuid:428f001d-76bd-410f-83fb-fe184786842e>\",\"Content-Length\":\"40827\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77fa6883-7772-4b07-9b23-46bf7792ac58>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdfe4e14-5d3c-4819-93b3-2d4f4b45745d>\",\"WARC-IP-Address\":\"104.27.180.43\",\"WARC-Target-URI\":\"http://www.archimedes-lab.org/monthly_puzzles_66.html\",\"WARC-Payload-Digest\":\"sha1:AJ4N76UVX5YYFV4IGBHWBZYDNIGLDEQP\",\"WARC-Block-Digest\":\"sha1:JHTHAST7DJQJNEWML4UST57K34LYHQCO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987828425.99_warc_CC-MAIN-20191023015841-20191023043341-00083.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/3-56-plus-3-70	[ "Solutions by everydaycalculation.com\n\n3/56 + 3/70 is 27/280.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 56 and 70 is 280\n2. For the 1st fraction, since 56 × 5 = 280,\n3/56 = 3 × 5/56 × 5 = 15/280\n3. Likewise, for the 2nd fraction, since 70 × 4 = 280,\n3/70 = 3 × 4/70 × 4 = 12/280", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7564959,"math_prob":0.9992999,"size":300,"snap":"2020-10-2020-16","text_gpt3_token_len":116,"char_repetition_ratio":0.16891892,"word_repetition_ratio":0.0,"special_character_ratio":0.45333335,"punctuation_ratio":0.072463766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99656934,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T00:16:58Z\",\"WARC-Record-ID\":\"<urn:uuid:47fc5dd6-e633-451e-add5-2a5a5d7258fe>\",\"Content-Length\":\"7465\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69a6622b-f28e-437f-83fb-e7cb07013e31>\",\"WARC-Concurrent-To\":\"<urn:uuid:11b3fd4d-3a71-4373-8961-2a74b632464e>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/3-56-plus-3-70\",\"WARC-Payload-Digest\":\"sha1:O6EJT7ERF7KFFKWMXNUHWUDAARSIQT7R\",\"WARC-Block-Digest\":\"sha1:B4SB44VSBT2N5A4SRDBWYU6W4MEANFUL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875141460.64_warc_CC-MAIN-20200217000519-20200217030519-00155.warc.gz\"}"}
https://pos.sissa.it/375/016/	[ "", null, "Volume 375 - 14th International Symposium on Radiative Corrections (RADCOR2019) - Higher-Loop\nThe four-loop slope of the Dirac form factor\nS. Laporta\nFull text: pdf\nPre-published on: December 19, 2019\nPublished on: February 18, 2020\nAbstract\nWe have evaluated with 1100 digits of precision the 4-loop contribution to the slope of the Dirac form factor in QED. The value is\n$$m^2 F_1^{(4)'}(0) = 0.886545673946443145836821730610315359390424032660064745{\\ldots} \\left(\\frac{\\alpha}{\\pi}\\right)^4 \\ .$$\nWe have also obtained a semi-analytical fit to the numerical value. The expression contains harmonic polylogarithms of argument $e^{\\frac{i\\pi}{3}}$, $e^{\\frac{2i\\pi}{3}}$, $e^{\\frac{i\\pi}{2}}$, one-dimensional integrals of products of complete elliptic integrals and six finite parts of master integrals, evaluated up to 4800 digits. We show the correction to the energy levels of the hydrogen atom due to the slope.\nDOI: https://doi.org/10.22323/1.375.0016\nHow to cite\n\nMetadata are provided both in \"article\" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in \"proceeding\" format which is more detailed and complete.\n\nOpen Access" ]	[ null, "https://pos.sissa.it/images/headInternal.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6958683,"math_prob":0.8635445,"size":677,"snap":"2020-34-2020-40","text_gpt3_token_len":198,"char_repetition_ratio":0.121842496,"word_repetition_ratio":0.0,"special_character_ratio":0.35302806,"punctuation_ratio":0.08264463,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98286426,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T14:24:39Z\",\"WARC-Record-ID\":\"<urn:uuid:4f7d1cbd-231e-41d7-9f03-6b3321504f65>\",\"Content-Length\":\"8384\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:466ec75b-2821-4ee7-aedd-e9b33c500793>\",\"WARC-Concurrent-To\":\"<urn:uuid:9572fd86-fd5a-4c7e-911d-6ea379cddf1c>\",\"WARC-IP-Address\":\"93.189.185.133\",\"WARC-Target-URI\":\"https://pos.sissa.it/375/016/\",\"WARC-Payload-Digest\":\"sha1:MSKINP74PZAEMLE6I5GZ6WFL6WOSWDDW\",\"WARC-Block-Digest\":\"sha1:VUMKFQW52F22CVKXPGS63XJBXHAVL5QP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739328.66_warc_CC-MAIN-20200814130401-20200814160401-00447.warc.gz\"}"}
https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/quantization/fake_quant_with_min_max_vars_per_channel	[ "# tf.quantization.fake_quant_with_min_max_vars_per_channel\n\nFake-quantize the 'inputs' tensor of type float and one of the shapes: `[d]`,\n\n`[b, d]` `[b, h, w, d]` via per-channel floats `min` and `max` of shape `[d]` to 'outputs' tensor of same shape as `inputs`.\n\n`[min; max]` define the clamping range for the `inputs` data. `inputs` values are quantized into the quantization range (`[0; 2^num_bits - 1]` when `narrow_range` is false and `[1; 2^num_bits - 1]` when it is true) and then de-quantized and output as floats in `[min; max]` interval. `num_bits` is the bitwidth of the quantization; between 2 and 16, inclusive.\n\nBefore quantization, `min` and `max` values are adjusted with the following logic. It is suggested to have `min <= 0 <= max`. If `0` is not in the range of values, the behavior can be unexpected: If `0 < min < max`: `min_adj = 0` and `max_adj = max - min`. If `min < max < 0`: `min_adj = min - max` and `max_adj = 0`. If `min <= 0 <= max`: `scale = (max - min) / (2^num_bits - 1)`, `min_adj = scale * round(min / scale)` and `max_adj = max + min_adj - min`.\n\nThis operation has a gradient and thus allows for training `min` and `max` values.\n\n`inputs` A `Tensor` of type `float32`.\n`min` A `Tensor` of type `float32`.\n`max` A `Tensor` of type `float32`.\n`num_bits` An optional `int`. Defaults to `8`.\n`narrow_range` An optional `bool`. Defaults to `False`.\n`name` A name for the operation (optional).\n\nA `Tensor` of type `float32`." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5706963,"math_prob":0.9892006,"size":1490,"snap":"2020-45-2020-50","text_gpt3_token_len":458,"char_repetition_ratio":0.12853298,"word_repetition_ratio":0.05964912,"special_character_ratio":0.31543624,"punctuation_ratio":0.15463917,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99442816,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T01:30:56Z\",\"WARC-Record-ID\":\"<urn:uuid:ce082a3c-6fbb-4d37-bdf6-0ffef6b43cfe>\",\"Content-Length\":\"729038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a02ca95-19e5-40d0-9ab5-74f3564ea751>\",\"WARC-Concurrent-To\":\"<urn:uuid:2019c8c4-99c5-4a51-baad-4248adbccb5e>\",\"WARC-IP-Address\":\"172.217.13.238\",\"WARC-Target-URI\":\"https://www.tensorflow.org/versions/r2.1/api_docs/python/tf/quantization/fake_quant_with_min_max_vars_per_channel\",\"WARC-Payload-Digest\":\"sha1:TX75EQXF2NK7OV6YT6MWWOYAS6NIBVTM\",\"WARC-Block-Digest\":\"sha1:FNBIKFGIWGBJHPDLTSGDI3SS56AJXGMP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141717601.66_warc_CC-MAIN-20201203000447-20201203030447-00097.warc.gz\"}"}
https://www.chinaedu.com/article/1173728.html	[ "# 高一上学期物理曲线运动万有引力\n\n凡事预则立，不预则废。学习物理需要讲究方法和技巧，更要学会对知识点进行归纳整理。下面是学习啦小编为大家整理的高一上学期物理曲线运动万有引力，希望对大家有所帮助!", null, "质点的运动(2)——曲线运动万有引力\n\n1)平抛运动\n\n1、水平方向速度Vx=Vo2、竖直方向速度Vy=gt\n\n3、水平方向位移Sx=Vot4、竖直方向位移(Sy)=gt^2/2\n\n5、运动时间t=(2Sy/g)1/2(通常又表示为(2h/g)1/2)\n\n6、合速度Vt=(Vx^2+Vy^2)1/2=Vo^2+(gt)^21/2\n\n合速度方向与水平夹角β：tgβ=Vy/Vx=gt/Vo\n\n7、合位移S=(Sx^2+Sy^2)1/2，\n\n位移方向与水平夹角α：tgα=Sy/Sx=gt/2Vo\n\n注：(1)平抛运动是匀变速曲线运动，加速度为g，通常可看作是水平方向的匀速直线运动与竖直方向的自由落体运动的合成。(2)运动时间由下落高度h(Sy)决定与水平抛出速度无关。(3)θ与β的关系为tgβ=2tgα。(4)在平抛运动中时间t是解题关键。(5)曲线运动的物体必有加速度，当速度方向与所受合力(加速度)方向不在同一直线上时物体做曲线运动。\n\n2)匀速圆周运动\n\n1、线速度V=s/t=2πR/T2、角速度ω=Φ/t=2π/T=2πf\n\n3、向心加速度a=V^2/R=ω^2R=(2π/T)^2R4、向心力F心=Mv^2/R=mω^2R=m(2π/T)^2R\n\n5、周期与频率T=1/f6、角速度与线速度的关系V=ωR\n\n7、角速度与转速的关系ω=2πn(此处频率与转速意义相同)\n\n周期(T)：秒(s)转速(n)：r/s半径(R)：米(m)线速度(V)：m/s\n\n注：(1)向心力可以由具体某个力提供，也可以由合力提供，还可以由分力提供，方向始终与速度方向垂直。(2)做匀速度圆周运动的物体，其向心力等于合力，并且向心力只改变速度的方向，不改变速度的大小，因此物体的动能保持不变，但动量不断改变。\n\n3)万有引力\n\n1、开普勒第三定律T2/R3=K(=4π^2/GM)R：轨道半径T：周期K：常量(与行星质量无关)\n\n2、万有引力定律F=Gm1m2/r^2G=6、67×10^-11N·m^2/kg^2方向在它们的连线上\n\n3、天体上的重力和重力加速度GMm/R^2=mgg=GM/R^2R：天体半径(m)\n\n4、卫星绕行速度、角速度、周期V=(GM/R)1/2ω=(GM/R^3)1/2T=2π(R^3/GM)1/2\n\n5、第一(二、三)宇宙速度V1=(g地r地)1/2=7、9Km/sV2=11、2Km/sV3=16、7Km/s\n\n6、地球同步卫星GMm/(R+h)^2=m*4π^2(R+h)/T^2h≈3、6kmh：距地球表面的高度\n\n注：(1)天体运动所需的向心力由万有引力提供，F心=F万。(2)应用万有引力定律可估算天体的质量密度等。(3)地球同步卫星只能运行于赤道上空，运行周期和地球自转周期相同。(4)卫星轨道半径变小时，势能变小、动能变大、速度变大、周期变小。(5)地球卫星的最大环绕速度和最小发射速度均为7、9Km/S。\n\n以上就是本期整理的全部内容了，想要了解的更多内容，同学们请持续关注101教育。如果你觉得对你有所帮助，就请分享给你的小伙伴们吧!\n\n## 精品学习资料", null, "" ]	[ null, "https://www.chinaedu.com/uploads/2020/12/10/feb9b3b2de2dddfabe58c6abc7f9d461.jpg", null, "https://www.chinaedu.com/bundles/prceducms/frontend/images/downloadPopEwm.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.89765394,"math_prob":0.997746,"size":1416,"snap":"2021-31-2021-39","text_gpt3_token_len":1349,"char_repetition_ratio":0.05240793,"word_repetition_ratio":0.0,"special_character_ratio":0.34251413,"punctuation_ratio":0.006756757,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.980374,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T00:51:07Z\",\"WARC-Record-ID\":\"<urn:uuid:5b99cc05-145e-4bac-b593-9364673cec93>\",\"Content-Length\":\"51560\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75b2bd23-addc-4922-b738-e7f4c13e486e>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8ebbcbd-e0f9-4ac9-9404-1d7e2aad7dff>\",\"WARC-IP-Address\":\"116.55.250.239\",\"WARC-Target-URI\":\"https://www.chinaedu.com/article/1173728.html\",\"WARC-Payload-Digest\":\"sha1:DVJQ5YS4OBJNHKSJLOYS5YV65IO24HQL\",\"WARC-Block-Digest\":\"sha1:IFO2PELZ3TF7MT5U3ZSRTZJEKQM3ML5O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057584.91_warc_CC-MAIN-20210924231621-20210925021621-00343.warc.gz\"}"}
https://treasurytoday.com/treasury-practice/the-yield-curve-part-vi/	[ "## The yield curve part VI\n\n##### Published: Nov 2005\n\nContinuing our series on yield curves, this month we explain how to construct a zero-coupon yield curve. We consider two-year and three-year maturities.\n\nNext month we will conclude our series on yield curves by demonstrating how the resulting calculations are graphically represented.\n\nZero-coupon yield curves can actually be constructed from a series of coupon-paying bonds. This is a technique known as ‘bootstrapping’. For example, consider the following bonds which pay annual coupons and have a maturity of two and three years respectively.\n\n#### Two year maturity\n\nThe investor will pay 98.435 for the first bond, to receive 5 (the coupon) in one year and 105 (the coupon and repayment at par) at the end of two years.\n\nIn order to calculate the equivalent zero-coupon bond structure, we take the 98.435 that the investor pays to purchase the bond and assume the investor borrows 4.717 to offset the interim coupon payment – to make this calculation you need the one-year interest rate and we have assumed a oneyear rate of 6%. So the net cash outflow is 93.718.\n\nAt the end of the first year, the investor will use the interim coupon payment to repay the borrowing of 4.717, which with interest has become 5. At the end of the second year, the investor will receive 105.\n\nThis allows us to identify the two-year zero-coupon rate. This is the rate which discounts 105 to 93.718. To calculate this, we use the following formula:\n\n$$i\\:=\\:\\frac{FV}{PV}\\:1 / n\\:\\: – \\:1$$\n\nWhere,\n\n• $$i\\:= \\:interest\\: rate$$\n• $$FV\\:=\\: future \\:value$$\n• $$PV\\:= \\:present \\:value$$\n• $$n\\:= \\:number\\: of\\: years.$$\n\nIn this case,\n\n$$i\\:=\\:\\frac{105}{93.718} ^{1 / 2} \\:– \\:1$$\n\n$$i\\:=\\:0.05848$$\nSo the two-year zero-coupon rate is 5.848%\n\nthe present value must be entered as a negative value if using an HP12C or equivalent calculator because it is an expenditure.\n\n#### Three year maturity\n\nWe can also use this information to calculate a three-year zero-coupon rate. In this case, we need to construct an artificial zero-coupon bond, by identifying how much the investor would need to borrow to offset the payment of the two interim coupons of 4.\n\nBy using the one-year rate of 6%, we can calculate the investor would need to borrow 3.774 today to match the coupon of 4 in one year.\n\nWe also need to calculate how much the investor would need to borrow today, to match the anticipated two-year coupon receipt of another 4. For this, we use the two-year rate which will be the same as the zero-coupon rate we calculated earlier – 5.848%.\n\nThe two-year discount factor is $$\\frac{1}{1.058482}\\:=\\:0.8926\\:\\:$$ Using this, we can calculate that the investor would need to borrow 3.570 today to receive 4 in two years.\n\nIn order to calculate the zero-coupon rate, we assume the investor purchases the bond for 96.784 and borrows 7.344 to compensate for the two interim interest payments. The investor’s initial outlay is therefore 89.440.\n\nUsing the same formula as outlined above $$i\\:=\\:\\frac{FV}{PV}\\:1/n\\:–\\:1$$ , we can then identify the three-year zerocoupon rate. In other words, the rate which discounts 104 to 89.440.\n\nIn this case:\n\n$$i\\:=\\:\\frac{104}{89.440}\\:^{1 / 3} – 1$$\n\n$$i\\:=\\:0.05156$$\n\nSo the three-year zero-coupon rate is 5.156%\n\nUsing the different zero-coupon rates we have calculated, it is then possible to create a series of zero-coupon yields and then construct the yield curve." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9212681,"math_prob":0.994929,"size":3600,"snap":"2023-40-2023-50","text_gpt3_token_len":938,"char_repetition_ratio":0.15100111,"word_repetition_ratio":0.033955857,"special_character_ratio":0.28694445,"punctuation_ratio":0.14713542,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99985135,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T18:14:00Z\",\"WARC-Record-ID\":\"<urn:uuid:2a4f138e-3504-4efa-abbd-eacc561d588a>\",\"Content-Length\":\"96459\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29e6a827-90d2-46a8-aebf-efa8b3aa9b3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:835db2e3-14b9-4d69-b18c-d0c1d890ac7e>\",\"WARC-IP-Address\":\"160.153.137.128\",\"WARC-Target-URI\":\"https://treasurytoday.com/treasury-practice/the-yield-curve-part-vi/\",\"WARC-Payload-Digest\":\"sha1:KBVSPHRWTYJPJXHSERGXLJVBQ7TI2NVO\",\"WARC-Block-Digest\":\"sha1:T653FZJ6FCR22LYIEH5VAWUOFFGC3JRW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100942.92_warc_CC-MAIN-20231209170619-20231209200619-00464.warc.gz\"}"}
https://www.colorhexa.com/44badb	[ "In a RGB color space, hex #44badb is composed of 26.7% red, 72.9% green and 85.9% blue. Whereas in a CMYK color space, it is composed of 68.9% cyan, 15.1% magenta, 0% yellow and 14.1% black. It has a hue angle of 193.1 degrees, a saturation of 67.7% and a lightness of 56.3%. #44badb color hex could be obtained by blending #88ffff with #0075b7. Closest websafe color is: #33cccc.\n\n• R 27\n• G 73\n• B 86\nRGB color chart\n• C 69\n• M 15\n• Y 0\n• K 14\nCMYK color chart\n\n#44badb color description : Soft cyan.\n\nThe hexadecimal color #44badb has RGB values of R:68, G:186, B:219 and CMYK values of C:0.69, M:0.15, Y:0, K:0.14. Its decimal value is 4504283.\n\nHex triplet RGB Decimal 44badb `#44badb` 68, 186, 219 `rgb(68,186,219)` 26.7, 72.9, 85.9 `rgb(26.7%,72.9%,85.9%)` 69, 15, 0, 14 193.1°, 67.7, 56.3 `hsl(193.1,67.7%,56.3%)` 193.1°, 68.9, 85.9 33cccc `#33cccc`\nCIE-LAB 70.496, -22.383, -26.147 32.726, 41.458, 73.292 0.222, 0.281, 41.458 70.496, 34.419, 229.436 70.496, -44.125, -38.171 64.388, -21.956, -22.417 01000100, 10111010, 11011011\n\n``#44badb` `rgb(68,186,219)``\n• #db6544\n``#db6544` `rgb(219,101,68)``\nComplementary Color\n• #44dbb1\n``#44dbb1` `rgb(68,219,177)``\n``#44badb` `rgb(68,186,219)``\n• #446fdb\n``#446fdb` `rgb(68,111,219)``\nAnalogous Color\n• #dbb144\n``#dbb144` `rgb(219,177,68)``\n``#44badb` `rgb(68,186,219)``\n• #db446f\n``#db446f` `rgb(219,68,111)``\nSplit Complementary Color\n``#badb44` `rgb(186,219,68)``\n``#44badb` `rgb(68,186,219)``\n• #db44ba\n``#db44ba` `rgb(219,68,186)``\n• #44db65\n``#44db65` `rgb(68,219,101)``\n``#44badb` `rgb(68,186,219)``\n• #db44ba\n``#db44ba` `rgb(219,68,186)``\n• #db6544\n``#db6544` `rgb(219,101,68)``\n• #2291b1\n``#2291b1` `rgb(34,145,177)``\n• #26a3c6\n``#26a3c6` `rgb(38,163,198)``\n• #2fb2d7\n``#2fb2d7` `rgb(47,178,215)``\n``#44badb` `rgb(68,186,219)``\n• #59c2df\n``#59c2df` `rgb(89,194,223)``\n• #6fcae3\n``#6fcae3` `rgb(111,202,227)``\n• #84d2e7\n``#84d2e7` `rgb(132,210,231)``\nMonochromatic Color\n\nBelow, you can see some colors close to #44badb. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #44dbd6\n``#44dbd6` `rgb(68,219,214)``\n• #44d3db\n``#44d3db` `rgb(68,211,219)``\n• #44c7db\n``#44c7db` `rgb(68,199,219)``\n``#44badb` `rgb(68,186,219)``\n``#44addb` `rgb(68,173,219)``\n• #44a1db\n``#44a1db` `rgb(68,161,219)``\n• #4494db\n``#4494db` `rgb(68,148,219)``\nSimilar Colors\n\nThis text has a font color of #44badb.\n\n``<span style=\"color:#44badb;\">Text here</span>``\n\nThis paragraph has a background color of #44badb.\n\n``<p style=\"background-color:#44badb;\">Content here</p>``\n\nThis element has a border color of #44badb.\n\n``<div style=\"border:1px solid #44badb;\">Content here</div>``\nCSS codes\n``.text {color:#44badb;}``\n``.background {background-color:#44badb;}``\n``.border {border:1px solid #44badb;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #02090a is the darkest color, while #f9fdfe is the lightest one.\n\n• #02090a\n``#02090a` `rgb(2,9,10)``\n• #05161b\n``#05161b` `rgb(5,22,27)``\n• #08242b\n``#08242b` `rgb(8,36,43)``\n• #0b313c\n``#0b313c` `rgb(11,49,60)``\n• #0f3f4c\n``#0f3f4c` `rgb(15,63,76)``\n• #124c5d\n``#124c5d` `rgb(18,76,93)``\n• #155a6d\n``#155a6d` `rgb(21,90,109)``\n• #18677e\n``#18677e` `rgb(24,103,126)``\n• #1b758e\n``#1b758e` `rgb(27,117,142)``\n• #1e829e\n``#1e829e` `rgb(30,130,158)``\n• #2290af\n``#2290af` `rgb(34,144,175)``\n• #259ebf\n``#259ebf` `rgb(37,158,191)``\n• #28abd0\n``#28abd0` `rgb(40,171,208)``\n• #34b4d8\n``#34b4d8` `rgb(52,180,216)``\n``#44badb` `rgb(68,186,219)``\n• #54c0de\n``#54c0de` `rgb(84,192,222)``\n• #65c6e1\n``#65c6e1` `rgb(101,198,225)``\n• #75cce4\n``#75cce4` `rgb(117,204,228)``\n• #86d2e8\n``#86d2e8` `rgb(134,210,232)``\n• #96d8eb\n``#96d8eb` `rgb(150,216,235)``\n• #a7deee\n``#a7deee` `rgb(167,222,238)``\n• #b7e4f1\n``#b7e4f1` `rgb(183,228,241)``\n• #c8ebf4\n``#c8ebf4` `rgb(200,235,244)``\n• #d8f1f7\n``#d8f1f7` `rgb(216,241,247)``\n• #e8f7fb\n``#e8f7fb` `rgb(232,247,251)``\n• #f9fdfe\n``#f9fdfe` `rgb(249,253,254)``\nTint Color Variation\n\nA tone is produced by adding gray to any pure hue. In this case, #899396 is the less saturated color, while #22cdfd is the most saturated one.\n\n• #899396\n``#899396` `rgb(137,147,150)``\n• #80989f\n``#80989f` `rgb(128,152,159)``\n• #779da8\n``#779da8` `rgb(119,157,168)``\n• #6fa2b0\n``#6fa2b0` `rgb(111,162,176)``\n• #66a7b9\n``#66a7b9` `rgb(102,167,185)``\n• #5eacc1\n``#5eacc1` `rgb(94,172,193)``\n• #55b0ca\n``#55b0ca` `rgb(85,176,202)``\n• #4db5d2\n``#4db5d2` `rgb(77,181,210)``\n``#44badb` `rgb(68,186,219)``\n• #3bbfe4\n``#3bbfe4` `rgb(59,191,228)``\n• #33c4ec\n``#33c4ec` `rgb(51,196,236)``\n• #2ac8f5\n``#2ac8f5` `rgb(42,200,245)``\n• #22cdfd\n``#22cdfd` `rgb(34,205,253)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #44badb is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5374741,"math_prob":0.62599516,"size":3722,"snap":"2020-10-2020-16","text_gpt3_token_len":1667,"char_repetition_ratio":0.12076385,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5354648,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9899484,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-04T03:06:52Z\",\"WARC-Record-ID\":\"<urn:uuid:dcde6a91-a643-41da-b268-68ed128ae464>\",\"Content-Length\":\"36330\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78794d72-0900-42b0-8866-c69902e544f9>\",\"WARC-Concurrent-To\":\"<urn:uuid:27c1ea92-0e67-4e48-a622-a423b20842f9>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/44badb\",\"WARC-Payload-Digest\":\"sha1:UDNLWKSRI4OII5SJU35LTPDSCHALD2QK\",\"WARC-Block-Digest\":\"sha1:4LIVJ7DUC3UPZNNEPPDES2NQNPQTLXKE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370519111.47_warc_CC-MAIN-20200404011558-20200404041558-00151.warc.gz\"}"}
https://www.mrexcel.com/board/threads/converting-a-number-into-particular-format.1135911/	[ "# Converting a number into particular format\n\n#### Harry_1234\n\n##### New Member\nHello,\n\nWhat is the best way to convert following format into e.164 in excel i.e. i would like to convert numbers in the format 123 456-0001 to \\+11234560001.\nColumn A Column B\n\n123 456-0001 \\+11234560001\n123 456-0005 \\+11234560005\n123 456-1000 \\+11234561000\n\n### Excel Facts\n\nSpell Check in Excel\nPress F7 to start spell check in Excel. Be careful, by default, Excel does not check Capitalized Werds (whoops)\n\n#### kennypete\n\n##### Board Regular\nThe B1 formula here should do the trick:\nBook2\nAB\n1123 456-0001\n\\+11234560001\nSheet1\nCell Formulas\nRangeFormula\nB1B1=INT(SUBSTITUTE(SUBSTITUTE(A1,\"-\",\"\"),\" \",\"\"))\n\nYou can either prepend the \"\\+1\" or number format adding \"\\+1\" (what I did here).\nWhether this is the \"best\" way probably depends on factors that are not outlined here - e.g. I have presumed that the 123 456-0001 is text and not a representation presented by a formula, etc.\n\n#### Harry_1234\n\n##### New Member\nThe B1 formula here should do the trick:\nBook2\nAB\n1123 456-0001\n\\+11234560001\nSheet1\nCell Formulas\nRangeFormula\nB1B1=INT(SUBSTITUTE(SUBSTITUTE(A1,\"-\",\"\"),\" \",\"\"))\n\nYou can either prepend the \"\\+1\" or number format adding \"\\+1\" (what I did here).\nWhether this is the \"best\" way probably depends on factors that are not outlined here - e.g. I have presumed that the 123 456-0001 is text and not a representation presented by a formula, etc.\nThank you!! This was really helpful." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6340333,"math_prob":0.84211797,"size":568,"snap":"2020-24-2020-29","text_gpt3_token_len":199,"char_repetition_ratio":0.19503546,"word_repetition_ratio":0.8,"special_character_ratio":0.55985916,"punctuation_ratio":0.08928572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9668592,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T14:59:51Z\",\"WARC-Record-ID\":\"<urn:uuid:4ca79d39-d4dc-493b-b2a6-d4ffb117fe12>\",\"Content-Length\":\"63588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd36d950-dfd6-4968-bdbe-7bba5ea303cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:9a30ea61-9c66-4d68-897f-c07c61e88d91>\",\"WARC-IP-Address\":\"216.92.17.166\",\"WARC-Target-URI\":\"https://www.mrexcel.com/board/threads/converting-a-number-into-particular-format.1135911/\",\"WARC-Payload-Digest\":\"sha1:VZOCYJV5DA62CXZNRC2UFWABUVU3FRIX\",\"WARC-Block-Digest\":\"sha1:LX2DOGYBDFP65MEBBNNJC3JAQRH6TN2F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886178.40_warc_CC-MAIN-20200704135515-20200704165515-00098.warc.gz\"}"}
https://m.scirp.org/papers/70083	[ "Modified Diode Assisted Extended Boost Quasi Z-Source Inverter for PV Applications\nAuthor(s) N. Hemalatha1, R. Seyezhai2\nABSTRACT\nThe design, simulation and implementation of modified diode assisted extended boost q-ZSI (MDAEB q-ZSI) for photovoltaic application are proposed in this paper. It is the most efficient topology that provides a single stage conversion for PV systems by providing high input voltage gain, reduced number of components count, increased voltage boost property, reduced voltage ratings, reduced voltage stress across the switches and simplified control strategies. Its unique capability in single stage conversion with improved voltage gain is used for voltage buck and boost function. The operating modes and the steady state theoretical analysis of voltage boost, control methods and a system design guide for the proposed topology are investigated in this paper. A simulation model of the PV system based on MDAEB q-ZSI has been built in MATLAB/ SIMULINK. Performance parameters such as Total harmonic distortion (THD), voltage gain, voltage stress and boost factor are computed and compared with the conventional quasi z-source inverter. The prototype model for MDAEB q-ZSI is developed and the results are validated.\n\nReceived 7 April 2016; accepted 1 May 2016; published 25 August 2016", null, "1. Introduction\n\nIn standalone PV systems, the power electronic converters play a vital role in the conversion of DC current of PV panels into AC to supply the load, with maximum efficiency and superior performance. The two stages of DC-DC-AC power conversion may result in usage of more circuit components, lower efficiency, higher cost and larger size in comparison to the single stage one . The modified diode assisted extended boost quasi-Z-source inverter has a single power conversion stage which perfectly suits for interfacing of renewable energy sources.\n\nThe efficiency and the voltage gain of the conventional q-ZSI are limited and comparable with the traditional system of a voltage source inverter with the auxiliary step-up DC/DC converter in the input stage . The concept of extending the quasi ZSI gain without increasing the number of active switches has been reported in the literature . These new converter topologies are known as extended boost q-ZSI and can be generally classified as capacitor assisted, diode assisted topologies and hybrid topologies . In this paper, MDAEB q-ZSI with continuous input current is presented, analyzed and compared for the simple boost control technique. Simulation studies of the proposed inverter configuration are carried out in MATLAB/SIMULINK. The capacitor voltage in the impedance network, voltage gain, boost factor, voltage stress and THD are calculated and compared with the quasi z-source inverter. Hardware of the modified diode-assisted QZSI is developed and the simulation results are verified.\n\n2. Operating Modes and Steady State Analysis of MDAEB q-ZSI\n\nThe proposed topology of MDAEB q-ZSI is presented in Figure 1. The topology of MDAEB q-ZSI could be derived by the adding of one capacitor (C3), one inductor (L3) and two diodes (D2 and D3) to the conventional q-ZSI. The connection points of the capacitor C3 is interchanged to reduce its operating voltages. Figure 2(a) and Figure 2(b) show the equivalent circuits of the MDAEB q-ZSI for the shoot-through and the active states.\n\nThe extended boost q-ZSI has two operational modes at the dc side, non-shoot-through states and the shoot- through state . During the shoot through state the q-ZSI performs the voltage boost function. During the active state, the previously stored magnetic energy in turn provides the voltage boost at the load terminals. The unique LC impedance network provides the boosting function without disturbing the operation inverter - .\n\n2.1. Mode I (Shoot through Mode)\n\nLet T = Operating period of the q-ZSI,\n\nTa = Active state,\n\nTs = Shoot through state,\n\nDa = The duty cycle of an active state,\n\nDs = The duty cycles of shoot-through state,", null, "(1)", null, ". (2)\n\nFigure 1. Modified diode assisted extended boost q-ZSI based PV system.\n\n(a)(b)\n\nFigure 2. Operating modes of MDAEB q-ZSI: (a) Mode-I; (b) Mode-II.\n\nThe equivalent circuit of the MDAEB q-ZSI during the shoot-through state is shown in Figure 2(a). A unique LC impedance network is interfaced between the source and the inverter to achieve voltage boost and inversion in a single stage. During the shoot through state D3 is conducting and D1 and D2 diodes are in blocking state. All the inductors in the impedance network get charged up. Energy is transferred from the source to the inductor or the capacitor to the inductor when the capacitors are getting discharged.\n\nThe voltage across the inductors can be represented as", null, "(3)", null, "(4)", null, ". (5)\n\n2.2. Mode II (Non-Shoot through Mode)\n\nThe equivalent circuit of the MDAEB q-ZSI during the active state is shown in Figure 2(b). During the non shoot through state the diodes D1 and D2 are conducting and D3 is in blocking state. The inductors across the impedance network discharge and the capacitors get charged.\n\nThe voltage of the inductors can be represented as", null, "(6)", null, "(7)", null, ". (8)", null, ". (9)\n\nThe boost factor of the input voltage is", null, ". (10)\n\n3. Simulation Results\n\nFrom the design considerations and specifications, the component parameters are calculated and it is presented in Table 1. MDAEB q-ZSI can be used for grid connected and standalone applications - . The simulation model of PV energy generation system with MDAEB q-ZSI for the boost factor B = 2 is shown in Figure 3. This is constructed in the MATLAB/SIMULINK environment. Figure 4 shows the impedance network of MDAEB q-ZSI. The gating pattern of the pulse generation using the simple boost control technique is shown in Figure 5.\n\nFigure 3. Matlab/Simulink circuit of PV based MDAEB q-ZSI.\n\nTable 1. Simulation parameters.\n\nFigure 4. Impedance Network of MDAEB q-ZSI.\n\nFigure 5. Shoot through pulses.\n\n3.1. PV Array Characteristics\n\nFigure 6 shows the PV array characteristics. The characteristics are plotted for different insolation level at constant temperature.\n\n3.2. Continuous Input Current\n\nThe proposed topology has continuous input current as shown in Figure 7.\n\nThe proposed topology operates normally producing the demanded boost of the input voltage for the boost factor. The simulation results of DC link voltage of MDAEB q-ZSI is shown in Figure 8.\n\nFrom Figure 8, it is clear that for the input voltage of 21 V, MDAEB topology produce the boost voltage of 39 V.\n\n3.4. Operating Voltages of the Capacitors in the Cascaded QZS Network\n\nThe theoretical and simulated results of the capacitor voltages across the impedance network for the proposed topology are compared in Table 2.\n\nFigure 6. PV array characteristics.\n\nFigure 7. Input voltage & current waveforms of MDAEB q-ZSI.\n\nFigure 8. DC Link voltage of MDAEB q-ZSI.\n\nTable 2. Operating voltages of capacitors in the impedance network.\n\nFrom Table 2 it is clear that the operating voltage of the capacitor C3 of MDAEB q-ZSI was reduced by more than five times by changing the interconnection points of the capacitors C3 as in Figure 1.\n\nFigure 9 shows the operating voltages of the capacitors in the MDAEB q-ZSI.\n\nFigure 9. Capacitor voltages of MDAEB q-ZSI.\n\nFrom Figure 9 the average voltage across the capacitor C3 of MDAEB q-ZSI was reduced more than six times when compared to its basic diode assisted topology. Figure 10 shows that the effect of modulation index on capacitor voltages of MDAEB q-ZSI.\n\nFrom Figure 10 it is clear that the average value of voltage across the capacitor C3 of MDAEB q-ZSI was reduced with the increasing modulation index.\n\n3.5. Output Voltage and Current Waveforms\n\nThe simulation results of the output line voltage, load voltage and load current waveforms of MDAEB q-ZSI for the simple boost modulation technique without filter are shown in Figures 11-13.\n\nThe simulation results of line voltage, load voltage and load current waveforms of MDAEB q-ZSI for Simple Boost technique with filter are shown in Figures 14-16. Filtered three phase output line voltage is shown in Figure 17.\n\n4. Performance Characteristics of MDAEB Q-ZSI\n\nPerformance parameters of the MDAEB q-ZSI are analyzed for various modulation indices for the given boost factor. They are Total Harmonic Distortion, inductor current ripple of q-ZSI, capacitor voltage ripple, voltage gain and voltage stress, boost factor.\n\n4.1. Total Harmonic Distortion (THD)\n\nTotal Harmonic Distortion of MDAEB q-ZSI is analyzed and compared with the conventional quasi ZSI for the simple boost modulation technique.THD is calculated for various modulation index values and the effect of the modulation indices on THD and the comparison is shown in Figure 18.\n\nFrom Figure 18 THD increase with the decrease in the modulation index and the MDAEB q-ZSI has reduced THD, when compared to the quasi ZSI.\n\n4.2. Voltage Gain (G)\n\nVoltage Gain, G is calculated for the simple boost modulation technique. In Figure 19 the voltage gain G is compared with different values of modulation indices for the simple boost modulation technique.\n\nVoltage gain, G is given by,\n\n(13)\n\nwhere M = Modulation Index,\n\nB = Boost Factor.\n\nFrom Figure 19 it is clear that when compared with the conventional quasi ZSI, the proposed topology have increased voltage gain for the same value of the modulation index (Ma).\n\n4.3. Boost Factor (B)\n\nThe boost factor is calculated for the simple boost modulation technique. In Figure 20, the boost factor B is compared with different values of the shoot through duty cycle Ds. Boost factor is given by\n\nFigure 10. Effect of modulation index on capacitor voltage of MDAEB q-ZSI.\n\nFigure 11. Line voltage waveform for MDAEB q-ZSI.\n\nFigure 13. Load current waveform for MDAEB q-ZSI.\n\nFigure 14. Filtered output line voltage waveform.\n\nFigure 15. Filtered output load voltage waveform.\n\nFigure 16. Filtered output load current waveform.\n\nFigure 17. Filtered output line voltage waveform.\n\nFigure 18. Effect of modulation index on THD.\n\nFigure 19. Comparison of voltage gain.\n\nFigure 20. Boost factor comparison.\n\n(14)\n\nFrom Figure 20, the proposed topology have increased boost factor of the input voltage for the same value of the shoot through duty cycle Ds when compared with the conventional q-ZSI.\n\n4.4. Voltage Stress\n\nVoltage stress is compared with voltage gain for the simple boost modulation technique. Voltage stress is calculated from the voltage gain G. Figure 21 shows the variation of voltage stress/DC voltage with voltage gain (G) for MDAEB q-ZSI and the traditional quasi ZSI.\n\nVoltage stress across the devices is given by\n\n(15)\n\nFrom Figure 21 it is clear that the simple boost PWM control technique gives better voltage gain and reduced voltage stress for the MDAEB q-ZSI than the conventional quasi ZSI.\n\nFrom the simulation results it is observed that MDAEB q-ZSI gives higher RMS value of the output voltage, higher voltage gain, increased boost factor, reduced voltage stress and reduced THD when compared with the traditional quasi ZSI for the simple boost modulation technique. It provides reduced operating voltages of the capacitor C3. MDAEB q-ZSI is the preferred topology for the photovoltaic applications when compared to the conventional q-ZSI.\n\n5. Experimental Results\n\nIn order to verify the theoretical assumptions the laboratory setup for MDAEB q-ZSI was assembled. The experimental setup for the PV connected MDAEB q-ZSI was shown in Figure 22.\n\nThe shoot through pulses and the gate pulses during the switching sequence of the three phase inverter are shown in Figure 23 and Figure 24. The shoot through duty cycle is 0.178 and the modulation index is 0.822.\n\nThe boost voltage and the load voltage of the MDAEB q-ZSI are shown in Figure 25 and Figure 26. For the input voltage of 21 V, the proposed topology produces the boost voltage of 42 V for the boost factor B = 2.\n\nFrom the hardware results it is shown that modified diode assisted extended boost q-ZSI have continuous input current, high dc link voltage ,high demanded boost ,reduced voltage stress, increased voltage gain and increased load voltage.\n\nFigure 21. Effect of voltage gain on voltage stress.\n\nFigure 22. Experimental setup.\n\nFigure 23. The shoot through pulses.\n\nFigure 24. Gate pulses.\n\nFigure 25. Boost voltage.\n\nFigure 26. Output line-line voltage.\n\n6. Conclusion\n\nIn this paper, the topology of modified diode assisted extended boost quasi ZSI for PV applications has been discussed and compared with the conventional q-ZSI. The detailed steady state operation of the proposed topology is analyzed and simulated. Simulation results validate the theoretical analysis. From the results, it is observed that the voltage stress across the impedance network is reduced and the operating voltages of the capacitors are reduced by five times; reduced THD, higher boost factor and better spectral quality of the output are obtained compared to QZSI configuration. Therefore, the proposed topology of extended QZSI is suited for PV applications.\n\nAcknowledgements\n\nThe authors wish to thank the management of SSN institutions for providing the computational facilities to carry out this work.\n\nNOTES\n\nCorresponding author.\n\nCite this paper\nHemalatha, N. and Seyezhai, R. (2016) Modified Diode Assisted Extended Boost Quasi Z-Source Inverter for PV Applications. Circuits and Systems, 7, 3271-3284. doi: 10.4236/cs.2016.710279.\nReferences\n Yang, L.-S., Liang, T.-J. and Chen, J.-F. (2009) Transformer Less DC-DC Converters with High Step-Up Voltage Gain. IEEE Transactions on Industrial Electronics, 56, 3144-3152.\nhttp://dx.doi.org/10.1109/TIE.2009.2022512\n\n Badin, R., Huang, Y., Peng, F.Z. and Kim, H.G. (2007) Grid Interconnected Z Source PV System. IEEE Power Electronics Specialists Conference (PESC’07), Orlando, 17-21 June 2007, 2328-2333.\nhttp://dx.doi.org/10.1109/pesc.2007.4342373\n\n Huang, Y., Shen, M.S., Peng, F.Z. and Wang, J. (2006) Z-Source Inverter for Residential Photovoltaic Systems. IEEE Transactions on Power Electronics, 21, 1776-1782.\nhttp://dx.doi.org/10.1109/TPEL.2006.882913\n\n Gajanayake, J., Luo, F.L., Gooi, H.B., So, P.L. and Siow, L.K. (2009) Extended Boost Z-Source Inverters. IEEE Energy Conversion Congress and Exposition (ECCE’09), San Jose, San Jose, CA, 20-24 September 2009, 3845-3852.\n\n Gajanayake, C.J., et al. (2010) Extended-Boost Z-Source Inverters. IEEE Transactions on Power Electronics, 25, 2642-2652.\nhttp://dx.doi.org/10.1109/TPEL.2010.2050908\n\n Vinnikov, D., Roasto, I., Strzelecki, R. and Adamowicz, M. (2010) Performance Improvement Method for the Voltage- Fed q-ZSI with Continuous Input Current. IEEE Mediterranean Electrotechnical. Conference (MELECON’10), Valletta, 26-28 April 2010, 1459-1464.\n\n Vinnikov, D., Roasto, I. and Jalakas, T. (2010) Comparative Study of Capacitor-Assisted Extended Boost qZSIs Operating in Continuous Conduction Mode. 12th Biennial Baltic Electronics Conference (BEC), Tallinn, 4-6 October 2010, 297-300.\n\n Li, Y., Jiang, S., Cintron-Rivera, J.G. and Peng, F.Z. (2013) Modeling and Control of Quasi-Z-Source Inverter for Distributed Generation Applications. IEEE Transactions on Industrial Electronics, 60, 1532-1541.\nhttp://dx.doi.org/10.1109/TIE.2012.2213551\n\n Peng, F.Z. (2003) Z-Source Inverter. IEEE Transactions on Industry Applications, 39, 504-510.\nhttp://dx.doi.org/10.1109/TIA.2003.808920\n\n Anderson, J. and Peng, F.Z. (2008) Four Quasi-Z-Source Inverters. IEEE Power Electronics Specialists Conference (PESC’08), Rhodes, 15-19 June 2008, 2743-2749.\nhttp://dx.doi.org/10.1109/pesc.2008.4592360\n\n Li, Y., Anderson, J., Peng, F.Z. and Liu, F.Z. (2009) Quasi-Z-Source Inverter for Photovoltaic Power Generation Systems. Applied Power Electronics Conference and Exposition (APEC 2009), Washington, 15-19 February 2009, 918- 924.\nhttp://dx.doi.org/10.1109/APEC.2009.4802772\n\n Cintron-Rivera, J.G., Li, Y., Jiang, S. and Peng, F.Z. (2011) Quasi-Z-Source Inverter with Energy Storage for Photovoltaic Power Generation Systems. Applied Power Electronics Conference and Exposition (APEC 2009), Washington, 15-19 February 2009, 401-406.\nhttp://dx.doi.org/10.1109/apec.2011.5744628\n\n Park, J.-H., Kim, H.-G., Nho, E.-C., Chun, T.-W. and Choi, J. (2009) Grid-Connected PV System Using a Quasi-Z- Source Inverter. Applied Power Electronics Conference and Exposition (APEC 2009), Washington, 15-19 February 2009, 925-929.\nhttp://dx.doi.org/10.1109/APEC.2009.4802773\n\n Carrasco, J.M., Franquelo, L.G., Bialasiewicz, J.T., Galvan, E., PortilloGuisado, R.C., Prats, M.A.M., Leon, J.I. and Moreno-Alfonso, N. (2006) Power-Electronic Systems for the Grid Integration of Renewable Energy Sources. IEEE Transactions on Industrial Electronics, 53, 1002-1016.\nhttp://dx.doi.org/10.1109/TIE.2006.878356\n\n Liu, J.F., Jiang, S., Cao, D. and Peng, F.Z. (2013) A Digital Current Control of Quasi-Z-Source Inverter With Battery. IEEE Transactions on Industrial Informatics, 9, 928-937.\nhttp://dx.doi.org/10.1109/TII.2012.2222653\n\n Zakis, J., Vinnikov, D., Roasto, I. and Ribickis, L. (2011) Quasi-Z-Source Inverter Based Bi-Directional DC/DC Converter: Analysis of Experimental Results. International Conference Workshop Compatibility and Power Electronices (CPE), Tallin, 1-3 June 2011, 394-399.\nhttp://dx.doi.org/10.1109/cpe.2011.5942267\n\nTop" ]	[ null, "https://html.scirp.org/file/40-7600744x5.png", null, "https://html.scirp.org/file/40-7600744x7.png", null, "https://html.scirp.org/file/40-7600744x8.png", null, "https://html.scirp.org/file/40-7600744x12.png", null, "https://html.scirp.org/file/40-7600744x13.png", null, "https://html.scirp.org/file/40-7600744x14.png", null, "https://html.scirp.org/file/40-7600744x15.png", null, "https://html.scirp.org/file/40-7600744x16.png", null, "https://html.scirp.org/file/40-7600744x17.png", null, "https://html.scirp.org/file/40-7600744x18.png", null, "https://html.scirp.org/file/40-7600744x19.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8686272,"math_prob":0.8510209,"size":12013,"snap":"2019-51-2020-05","text_gpt3_token_len":2711,"char_repetition_ratio":0.18719293,"word_repetition_ratio":0.07395994,"special_character_ratio":0.21376842,"punctuation_ratio":0.09937611,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.97751266,"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],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T16:27:31Z\",\"WARC-Record-ID\":\"<urn:uuid:1df33caf-536b-44bd-b27d-3d12fc1baa9d>\",\"Content-Length\":\"57571\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5e93e3d-14d3-4d85-bd5b-1b2a699af4ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:eba1869c-6bb3-4d74-bb2f-f68ed852baab>\",\"WARC-IP-Address\":\"198.204.224.91\",\"WARC-Target-URI\":\"https://m.scirp.org/papers/70083\",\"WARC-Payload-Digest\":\"sha1:7VKCLDKHBWJB2UVK5G4XRCN3PQ4H2TDB\",\"WARC-Block-Digest\":\"sha1:INUZPC4PGOKDVKCSBQCPEGSVSM4ZRKGQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250604849.31_warc_CC-MAIN-20200121162615-20200121191615-00029.warc.gz\"}"}
https://discuss.codecademy.com/t/9-more-with-for---what-was-your-mistake-here/8339	[ "", null, "# 9. More with \"for\" _ what was your mistake here?\n\nI’m writing this to track back what I was thinking while I was solving this part. Some steps are easy to follow, but some are not, while instructions are written simply and easy. I am curious why I can’t understand certain instruction, and trying to share the course of my thoughts.\n\n## I got stuck at 9. More with “for”\n\n``````start_list = [5, 3, 1, 2, 4]\nsquare_list = []\n\nfor number in start_list:\nsquare_list.append(number2)\nsquare_list.sort()\n\nprint square_list\n``````\n\nbecause I thought I cannot mention name of another list in one list.\nI tried to figure out how to combine two lists at the same time without like above.\nI saw the right answer by others’ Q&A.\n\nIs it just me? If you stuck here, why were you? ( I think reviewing like this helps me to learning it.)\n\nI repeatedly practiced it and it didn’t work.\nMy code was like below.\n\n``````A_list = [ \"1\", \"4\", \"5\", \"6\", \"7\"]\nB_list = [ ]\n\nfor number in A_list:\nB_list.append(number2)\nB_list.sort()\n\nprint B_list\n``````\n\nCan you see why it didn’t work?\n\nBut the next code worked.\n\n``````A_list = [ 1, 4, 5, 6, 7]\nB_list = [ ]\n\nfor number in A_list:\nB_list.append(number2)\nB_list.sort()\n\nprint B_list\n``````\n\n(It may seem a very easy and simple to more advanced ppl) My mistake here was that I didn’t write the right type of values in the list. To make number2 possible, it should be in the form of number, but I put “”, and made it as strings.\n\n3 Likes\n\nHi medouxa,\n\nI notice, on the code that didn’t work, your A_list = [“1” , “2”, “3” etc]\n\n1. The quotation marks, which surround the list indices, thereby turns this list into a *‘string’.\n2. You cannot then do a numerical calculation (i.e X.append(i2)) on an alphabetical value.\n\nThis is my understanding of your problem in the code here. I hope this helps.\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor number in start_list:\nsquare_list.append(number2)\nsquare_list.sort()\n\nprint square_list\n\nlike this bro/sis\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor number in start_list:\nsquare_list.append(number2)\nsquare_list.sort()\nprint square_list\n\nwhy m i till not able to get my result?\nmy code i shown above\n\nyour “(number2)” is meant to be \"(number 2) @aku\n\nNo, the problem is that it should be (numbers 2) not (number 2)\n\nGood luck =)\n\nUsing the “for” command was confusing to me. I didn’t (and still do not all the way) understand it’s direct function. I tried doing it this way first:\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor number in start_list:\nsquare_list = start_list.append(number2)\n\nsquare_list.sort()\nprint square_list\n\nI thought you would have to assign (square_list) to the action of appending (start_list). Now I realize that the “for” command would not be useful in that sense.\n\nI also wondered if you could achieve the same result with this code:\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nsquare_list = start_list.append(number2)\nsquare_list.sort()\n\nprint square_list\n\nIf the above code works, wouldn’t that be easier and more straight forward?\n\nI ended up using this code to get through the lesson:\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor numbers in start_list:\nsquare_list.append(number2)\n\nsquare_list.sort()\nprint square_list\n\nStill confused as to the main purpose of using “for ___ in___:”.\n\nedit 11/15/2015\n\nI have learned that whatever variable you type after “for ___” can be anything you want it to (without using reserved terms that are special in python). You are basically assigning a variable name to the values of the list or dictionary that comes after \" in ___\". So you could put “for taco in dictionary” and use an indented code afterwards to manipulate the content in the dictionary. Hope this makes sense to someone else too!\n\n2 Likes\n\nsquare_list.sort() should be outside loop for\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor number in start_list:\nsquare_list.append(number2)\n\nsquare_list.sort()\nprint square_list\n\nIf square_list placed outside the loop, then it will append only the last value to square list i.e 42 which results in 16 to the square_list \nsquare_list.append(number\n2) and square_list.sort() both should be inside the loop.\n\nfor x in start_list:\nprint x2\nsquare_list.append(x\n2)\nsquare_list.sort()\nprint square_list\n\nstart_list = [5, 3, 1, 2, 4]\nsquare_list =\n\nfor number in start_list:\nsquare_list.append(number**2)\nsquare_list.sort()\nprint square_list" ]	[ null, "https://aws1.discourse-cdn.com/codecademy/original/5X/8/3/1/1/83117c5c7564a9b51c2981be72ef9873faf7acdd.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9314362,"math_prob":0.9417856,"size":1371,"snap":"2019-43-2019-47","text_gpt3_token_len":367,"char_repetition_ratio":0.12289686,"word_repetition_ratio":0.06374502,"special_character_ratio":0.29248723,"punctuation_ratio":0.14851485,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941241,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T15:29:32Z\",\"WARC-Record-ID\":\"<urn:uuid:ad7e8cd2-7293-48e9-ae67-e557a5c9b7bd>\",\"Content-Length\":\"32458\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:061d1dc8-829e-46f2-b389-7ab317d73954>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c0ea1e4-514b-4191-8b63-0ee59455974e>\",\"WARC-IP-Address\":\"216.218.159.21\",\"WARC-Target-URI\":\"https://discuss.codecademy.com/t/9-more-with-for---what-was-your-mistake-here/8339\",\"WARC-Payload-Digest\":\"sha1:4SZI5YKIL2VDQJAVB54W3BHXFHSWRRQO\",\"WARC-Block-Digest\":\"sha1:3LMHE35SNCQNGC62YTDSDV3JZQA5NMFH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670156.86_warc_CC-MAIN-20191119144618-20191119172618-00060.warc.gz\"}"}
https://lessonplanet.com/teachers/mixed-practice-word-problems-5	[ "", null, "Worksheet\n\n# Mixed Practice Word Problems #5\n\n##### This Mixed Practice Word Problems #5 worksheet also includes:\n\nSix word problems make up this mixed practice worksheet. Pupils work with money, adding, subtracting, multiplying, and dividing using numbers up to 250.\n\n##### Instructional Ideas\n• Assign the worksheet as homework\n• Direct learners to complete problems in small group, math rotations\n• Review problems step-by-step with small groups, or as a class\n##### Classroom Considerations\n• The worksheet is the fifth of 12 created by K5 Learning to support third grade math skills\n• Copies are required\n##### Pros\n• Answers are available in number and word form\n• Space is provided to show work\n• The objective is clear and to the point\n• None" ]	[ null, "data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90429336,"math_prob":0.44013578,"size":911,"snap":"2022-40-2023-06","text_gpt3_token_len":199,"char_repetition_ratio":0.099228226,"word_repetition_ratio":0.0,"special_character_ratio":0.20417124,"punctuation_ratio":0.08387097,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.961205,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-07T11:08:59Z\",\"WARC-Record-ID\":\"<urn:uuid:a4d07d05-a0ec-4086-9595-5d7ff5dd7091>\",\"Content-Length\":\"152264\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4941a183-ea9b-4f1b-b1d4-b3eca3e6bbcd>\",\"WARC-Concurrent-To\":\"<urn:uuid:b75a7b14-ce26-42c8-b5d9-bc7feb267a9e>\",\"WARC-IP-Address\":\"3.129.199.180\",\"WARC-Target-URI\":\"https://lessonplanet.com/teachers/mixed-practice-word-problems-5\",\"WARC-Payload-Digest\":\"sha1:S56TKADQPEIQO3WU52RV7GFW77TYP6AN\",\"WARC-Block-Digest\":\"sha1:MITFI7SVJXGZCUY3TXCWHQPNO3VF3G5W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500456.61_warc_CC-MAIN-20230207102930-20230207132930-00657.warc.gz\"}"}
http://www.gicgac.com/FirstSet/KIDS/Practice_Free_Multiply_3_numbers_Questions_online.html	[ "# Learn and Practice Free Multiply 3 numbers questions - Kids Mathematics\n\n## Practice Free Multiply 3 numbers Questions online\n\nBy Practicing the below question set :\nThe student will get the basic and advanced knowledge on Multiply 3 numbers.\nThe student will learned the priciples and methedology of Multiply 3 numbers.\nThe student will evaluate the knowledge level on Multiply 3 numbers.\nStudent can practice number of times the question till he is clear.\nStudent can provide the review / feedback / suggestion on the questions which will help to others.\nThe questions can be practiced by All the standards / Grades.\nThe question contains the simple to complex level difficulties so that student can improve the knowledge.\n\n## You can get answer of the below sample questions by going through the above Question sets.\n\nsolve for K: 1 × 5 × K = 20 What is the value of K ?\nMultiply: 3 × 1 × 0 = ?\nMultiply: 3 × 1 × 9 = ?\nsolve for K: K × 9 × 1 = 54 What is the value of K ?\nsolve for K: 1 × 4 × K = 20 What is the value of K ?\nMultiply: 9 × 1 × 0 = ?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80839163,"math_prob":0.9891877,"size":1651,"snap":"2020-34-2020-40","text_gpt3_token_len":402,"char_repetition_ratio":0.15421979,"word_repetition_ratio":0.1402214,"special_character_ratio":0.2337977,"punctuation_ratio":0.15625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997084,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T10:35:45Z\",\"WARC-Record-ID\":\"<urn:uuid:d7c4c11d-99da-427b-a81b-12c87197691a>\",\"Content-Length\":\"19113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b99498a-f176-4a85-b135-2ea04dbe186c>\",\"WARC-Concurrent-To\":\"<urn:uuid:75501593-b7ab-4879-a6d9-b725f895b5ef>\",\"WARC-IP-Address\":\"95.217.197.119\",\"WARC-Target-URI\":\"http://www.gicgac.com/FirstSet/KIDS/Practice_Free_Multiply_3_numbers_Questions_online.html\",\"WARC-Payload-Digest\":\"sha1:UB6EQW7IG5WX4DVDKA6XN3Y7UDNHAC4S\",\"WARC-Block-Digest\":\"sha1:P5WC4PE2M4VUGBLSSO5MT7AA3EFISIC7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400274441.60_warc_CC-MAIN-20200927085848-20200927115848-00024.warc.gz\"}"}
https://it.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/1529466	[ "Cody\n\n# Problem 94. Target sorting\n\nSolution 1529466\n\nSubmitted on 15 May 2018 by Suraj Mankulangara\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\na = [1 2 3 4]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))\n\n2 Pass\na = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))\n\n3 Pass\na = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))\n\n4 Pass\na = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.56367934,"math_prob":0.9955248,"size":714,"snap":"2019-51-2020-05","text_gpt3_token_len":261,"char_repetition_ratio":0.15211268,"word_repetition_ratio":0.0,"special_character_ratio":0.44257703,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9800574,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T10:00:14Z\",\"WARC-Record-ID\":\"<urn:uuid:31775b83-0ec5-45a3-bd87-a9f43e878270>\",\"Content-Length\":\"73883\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8177ba32-b673-4a73-acc0-fca6f6486802>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0ca1711-d526-4b6a-8cda-a35e11ca73f7>\",\"WARC-IP-Address\":\"23.32.68.178\",\"WARC-Target-URI\":\"https://it.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/1529466\",\"WARC-Payload-Digest\":\"sha1:DRKXUJQPWLDPKD7X6TPYOCNRWTUUABEC\",\"WARC-Block-Digest\":\"sha1:N22ZHUSBJ4SPAHYOOF5WLQEV4PBAYI52\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251687958.71_warc_CC-MAIN-20200126074227-20200126104227-00358.warc.gz\"}"}
http://stephengoeddel.com/blog/post/10/sorting-algorithms/	[ "# Sorting Algorithms\n\n### Some basic sorting algorithms implemented in Java and timed\n\n#### May 18, 2014, 12:56 p.m.\n\nI wanted some more Java practice before I start my new job as a Software Engineer working with Java, so I decided to implement some basic sorting algorithms and time them. For each algorithm I chose to start by coping the ArrayList into a new one as to preserve the original input. I started with bubble sort, as I remembered it was simple but incredibly slow:\n\n```public static ArrayList<Integer> bubble(ArrayList<Integer> input ) {\nArrayList<Integer> result = new ArrayList<Integer>();\nwhile(true) {\nboolean done = true;\nfor(int i = 1; i < result.size(); i++) {\nif (result.get(i) < result.get(i-1)) {\nint temp = result.get(i - 1);\nresult.set(i - 1, result.get(i));\nresult.set(i, temp);\ndone = false;\n}\n}\nif(done) {\nbreak;\n}\n}\nreturn result;\n}\n```\n\nWith my test of 100,000 randomly generated Integers between 0 and 99,999, bubble sort was able to complete the sort in 64.82 seconds.\nNext I moved on to quick sort as I knew it would be much better:\n\n```public static ArrayList<Integer> quick(ArrayList<Integer> input) {\nArrayList<Integer> result = new ArrayList<Integer>();\nif (result.size() <= 1) {\nreturn result;\n}\nInteger pivot = result.get(0);\nresult.remove(0);\nArrayList<Integer> less = new ArrayList<Integer>();\nArrayList<Integer> more = new ArrayList<Integer>();\nfor (int i = 0; i < result.size(); i++) {\nif (result.get(i) <= pivot) {\n} else {\n}\n}\nArrayList<Integer> output = new ArrayList<Integer>();\nreturn output;\n}\n```\n\nUsing the same ArrayList of Integers quick sort was able to finish in merely 0.71 seconds, obviously a huge speed up! I then decided to implement insertion sort:\n\n```public static ArrayList<Integer> insertion(ArrayList<Integer> input) {\nArrayList<Integer> result = new ArrayList<Integer>(); // make new ArrayList from input in order to preserve input\nfor (int i = 0; i < result.size(); i++) {\nint x = result.get(i);\nint j = i;\nwhile (j > 0 && result.get(j-1) > x) {\nresult.set(j, result.get(j-1));\nj--;\n}\nresult.set(j, x);\n}\nreturn result;\n}\n```\n\nWith insertion sort, our ArrayList was sorted in 4.319 seconds. Next, I chose to implement selection sort:\n\n```public static ArrayList<Integer> selection(ArrayList<Integer> input) {\nArrayList<Integer> result = new ArrayList<Integer>(); // make new ArrayList from input in order to preserve input\nint min;\nfor (int i = 0; i < result.size() - 1; i++) {\nmin = i;\nfor (int j = i + 1; j < result.size(); j++) {\nif(result.get(j) < result.get(min)) {\nmin = j;\n}\n}\nif (min != i) {\nint temp = result.get(i);\nresult.set(i, result.get(min));\nresult.set(min, temp);\n}\n}\nreturn result;\n}\n```\n\nSelection sort was able to sort the same ArrayList in 14.187 seconds. These results all make sense with the complexities of each respective algorithm. For more information on each algorithm and some pseudo code to work off of check out wikipedia." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59712285,"math_prob":0.9181068,"size":2981,"snap":"2023-40-2023-50","text_gpt3_token_len":719,"char_repetition_ratio":0.24084648,"word_repetition_ratio":0.135255,"special_character_ratio":0.2955384,"punctuation_ratio":0.19504133,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967509,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T04:17:01Z\",\"WARC-Record-ID\":\"<urn:uuid:07d02ad3-2a78-4aa9-8573-c15b9b8d0408>\",\"Content-Length\":\"24416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c4b7552-a234-47e8-a834-580257169ab3>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e88406a-b79b-41aa-821a-0209a18e6268>\",\"WARC-IP-Address\":\"104.236.236.151\",\"WARC-Target-URI\":\"http://stephengoeddel.com/blog/post/10/sorting-algorithms/\",\"WARC-Payload-Digest\":\"sha1:FGCVL7KQ2ECQXLSZV7LQKV4PI2RROHHT\",\"WARC-Block-Digest\":\"sha1:FJFEZLAVLOEK7OA4LQEPXAL644TEZ6GE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101195.85_warc_CC-MAIN-20231210025335-20231210055335-00444.warc.gz\"}"}
https://www.arxiv-vanity.com/papers/0711.0815/	[ "# A local–global problem for linear differential equations\n\nMarius van der Put and Marc Reversat\nDepartment of Mathematics, University of Groningen, P.O.Box 800,\n9700 AV Groningen, The Netherlands, , and\nInstitut de Mathématiques de Toulouse, UMR 5219, Université Paul Sabatier\n31602 Toulouse cedex 9, France, email:\n\nAbstract. An inhomogeneous linear differential equation over a global differential field can have a formal solution for each place without having a global solution. The vector space measures this phenomenon. This space is interpreted in terms of cohomology of linear algebraic groups and is computed for abelian differential equations and for regular singular equations. An analogue of Artin reciprocity for abelian differential equations is given. Malgrange’s work on irregularity is reproved in terms cohomology of linear algebraic groups.\n\n## 1 Introduction\n\nThe topics: Elliptic curves over a number field ; Drinfeld modules over a field like ; linear differential equations over a differential field , e.g., a finite extension of the differential field , have many common features.\nFor every place of one considers the completion . An example of a local–global problem is the following. Consider an elliptic curve over and an integer . Suppose that with has a solution in every . Does there exists a solution ? By ‘folklore’ the answer is positive and the analogous problem for Drinfeld modules, where the integer is replaced by a non zero element of , has a negative answer (see [H] for both statements).\n\nHere we consider a differential operator , where is the derivation on extending on and , acting upon . Suppose that the equation with has a solution in every completion . Then, in general, there is no solution in . One defines the -vector space as the kernel of the obvious -linear map . This vector space measures this global–local problem. The theme of this paper is the interpretation and the computation of . The main results are:\n* A formula expressing in terms of the cohomology for differential Galois groups acting on the solution space of and a proof of ,\n* Computation of for certain affine group schemes ,\n* Classification of abelian differential equations and Artin reciprocity,\n* Explicit computations of for abelian differential operators and for regular singular operators ,\n* A new proof of B. Malgrange’s results on irregularity.\n\nThe last item requires a precise knowledge of the universal differential Galois group for the differential field of the convergent Laurent series. The multisummation theory of J.-P. Ramis et al. provides this knowledge.\n\n## 2 K/l(k) and M/∂M as cohomology groups\n\ndenotes a differential field and let or denote the derivative of . We suppose that its field of constants is algebraically closed, different from and has characteristic 0. A linear differential equation over can be written in an operator form\n\n (an∂n+⋯+a1∂+a0)y=f with all ai,f∈K.\n\nAn equivalent formulation is given by a differential module where is a finite dimensional vector space over and the additive operator satisfies . (The meaning of the symbols and will be clear from the context).\n\nWe recall (see [vdP-S] for details), that for every linear differential equation (or module) over there is a differential ring , called the Picard–Vessiot ring of over , such that ‘all solutions’ of live in this differential ring and this ring has only trivial differential ideals. The (differential) Galois group of the module is the linear algebraic group over consisting of all -linear automorphisms of , commuting with the differentiation on . The direct limit of all is the universal differential extension of . Its Galois group is the affine group scheme, which is the projective limit of the Galois groups of all differential modules over .\n\nFor a differential operator , the solution space of is the -vector space is the solution space of . The action of on leaves invariant and the restriction of the action to is the (differential) Galois group of . Similarly, for a differential module over , the -vector space is the solution space of and the restriction to of the natural action of on is the (differential) Galois group of .\n\n###### Proposition 2.1\n\nand for all .\n\n• Using that is a direct limit of Picard–Vessiot rings, one concludes that it suffices to consider a differential module over with Picard–Vessiot ring and Galois group . In this case one has to prove and for all .\n\nThe first statement is well known. The affine variety corresponding to is known to be a -torsor for the linear algebraic group over . In other words, there exists a finite Galois extension such that . Further the action of the Galois group on this object commutes with the action of .\n\nWe recall that the cohomology groups , where is any -module, are the cohomology groups of the Hochschild complex . Moreover, one has for all , since is an injective module (see [Jan] for these statements). It follows that also for . Let and let be an element in with . Then for some . Then belongs to and satisfies .\n\n###### Lemma 2.2\n\nLet be a differential operator and be as before. The following sequences are exact.\n\n 0→ker(L,UK)→UKL→UK→0 and\n 0→ker(L,K)→KL→K→H1(GK,∂,ker(L,U))→0.\n\nIn particular, is canonically isomorphic to and moreover, for .\n\n• For the exactness of the first sequence we have to show that with has a solution . For any there exists a such that . Indeed, lies in some and apply now [vdP-S], Corollary 1.38. The equation has all its solutions in . Hence, for a suitable solution of this equation one has .\n\nTaking in the first exact sequence, invariants under , one obtains, by using Proposition 2.1, the exact sequence\n\n 0→ker(L,K)→KL→K→H1(G,ker(L,UK))→0.\n\nFor a differential module over one has a similar result, namely:\nThere is a canonical isomorphism where again provided with its structure of -module.\n\nA direct comparison between modules and differential operators is given by the theorem of the cyclic element which states that any differential module has the form for some operator . Let denote the adjoint of . Then one can verify that can be identified with and with .\n\n## 3 On cohomology of linear algebraic groups\n\nThe base field is supposed to be algebraically closed and to have characteristic 0. Let be a linear algebraic group, or, more generally, an affine group scheme, over . A -module is a finite dimensional vector space over , provided with an algebraic action of , i.e., a morphism of affine group schemes . The cohomology groups are defined as the derived functors of . The -vector space can, as in the case of ordinary group cohomology, be described as the space of all -cocycles , divided out by the subspace of the trivial -cocycles. The only difference is that the -cocycles are supposed to be morphisms of algebraic varieties (or more generally, of affine schemes). We will also allow -modules of infinite dimension, namely direct limits of finite dimensional -modules. Now we collect here (with some comments) the facts and results that we will need in the sequel and refer to [Jan] for the general theory.\n\n###### Fact 3.1\n\nLet be a reductive affine group scheme (not necessarily connected), then for every -module one has for all .\n\nIndeed, since the characteristic of is 0, the functor is exact.\n\n###### Fact 3.2\n\nLet be a closed, normal subgroup of the affine group scheme . Suppose that is reductive. Let be a -module. Then is canonically isomorphic to .\n\nIndeed, the functor factors as and the functor is exact and maps injective objects to acyclic objects. The special case of 3.2, where is the unipotent radical , reduces the computations to the case of (connected) unipotent groups.\n\n###### Fact 3.3 (the five terms exact sequence)\n\nLet be a closed normal subgroup of the affine group and let be a -module. Then the exact sequence of five terms reads\n\n 0→H1(G/N,VN)→H1(G,V)→H1(N,V)G/N→H2(G/N,VN)→H2(G,V).\n\nThis exact sequence is derived from the spectral sequence converging to .\n\n###### Remark 3.4\n\nExplicit action of on .\nLet be a -module and a normal closed subgroup of . The action of on can be made explicit on the level of 1-cocycles. Let be a 1-cocycle and then is the 1-cocycle defined by . The trivial 1-cocycle (for a fixed ) is mapped to the trivial 1-cocycle . Thus acts on . For and a 1-cocycle one observes that is the trivial 1-cocycle . Thus acts trivially on .\n\n###### Fact 3.5\n\nLet be the Lie algebra of a connected affine group scheme . Any -module has the structure of a -module. The cohomology groups are canonically isomorphic to the cohomology groups .\n\nSketch of the proof. Using Fact 3.2, we may suppose that is a connected unipotent linear algebraic group over . In particular, is simply connected. Therefore the category of the finite dimensional representations of and those of its Lie algebra are equivalent. This equivalence extends to an equivalence between the representations of which are direct limits of finite dimensional representations and those of . The latter categories contain enough injective objects and the cohomology groups can be obtained from injective resolutions.\n\nWe note that the group can also be described by -cocycles modulo trivial -cocycles (see [Jac]). In some cases the computations of the are easier than those for .\n\n###### Lemma 3.6\n\nLet be a generator of the Lie algebra of . The -module , given by , induces a nilpotent map . Then ,\n\nand for .\n\n• By 3.5, it suffices to compute the cohomology of as module over the Lie algebra of . An obvious calculation of these cohomology groups in terms of cocycles gives the required answer. The same computation can be done in terms of cocycles for the group .\n\n###### Corollary 3.7\n\nLet the group scheme be topologically generated by one element . Suppose that the unipotent factor of the Jordan decomposition is non trivial. Then , where the first factor is topologically generated by and the second by .\n\nLet the -module be given by . Then for and .\n\n• and this equals since commutes with . Apply now 3.6. The final statement follows from the decomposition of into eigenspaces for .\n\n###### 3.8\n\nGroup schemes with free unipotent generators.\nLet be some non empty set. One considers tuples where is a finite dimensional -vector space and maps every to a unipotent . This defines an abelian category which is in an obvious way a Tannakian category. The fibre functor is given by . Thus is isomorphic to the category of the finite dimensional representations of a certain affine group scheme over .\n\nConsider an object and the Tannakian subcategory generated by it. The affine group scheme corresponding to this subcategory can be seen to be the smallest algebraic subgroup of containing all . The representation , corresponding to , has the property . In particular, if is a finite group then . Thus is a connected affine group scheme. Moreover, is the projective limit of the , taken over all objects . For a fixed , each contains an element . The projective limit of the elements can be considered as an element, again called , in . Thus for every object and corresponding representation . Therefore, we will call the group with free unipotent generators . Indeed, the elements of , seen as elements of , are topological generators, they are unipotent and they have no relations.\n\nWe want to show that for any -module , the are the cohomology groups of the complex , where the non trivial map is given by . A direct proof is maybe possible, however we will prove this using the (pro) Lie algebra of .\n\n###### 3.9\n\nLie algebras with free nilpotent generators.\nLet be again a non empty set. The free Lie algebra over with generators has the universal property that the representations on vector spaces over are in bijection with the maps . One considers now representation such that is finite dimensional and every is nilpotent. The image is an algebraic Lie algebra, because it is generated by nilpotent maps. The corresponding (connected) algebraic subgroup of is the smallest algebraic subgroup containing all . Define by . Then is an object of , the Tannakian group of this object is the smallest algebraic group containing all and its Lie algebra is .\n\nThe projective limit , taken over all these will be called the (pro)-Lie algebra with free nilpotent generators . It is clear from the above that this pro Lie algebra is the Lie algebra of the affine group scheme of 3.8. The bijection between the representations of and those of its (pro)-Lie algebra can be interpreted as being simply connected.\n\n###### Proposition 3.10\n\nLet be the affine group scheme with free unipotent generators and let be a -module. The are the cohomology groups of the complex , where the non trivial map is given by .\n\n• Let denote the pro-Lie algebra of . According to 3.5 and the constructions in 3.8, 3.9, the proposition is equivalent to the statement that the are the cohomology groups of the complex , where the non trivial map is given by .\n\nFor , this is obvious. For we use the explicit definition of a 1-cocycle (see [Jac]) and conclude that the elements are arbitrary and that they determine completely. The trivial 1-cocycles are of the form for a fixed and where is induced by . This proves the statement for . The verification of for is easy.\n\n###### 3.11\n\nFiniteness conditions.\nLet the set be a disjoint union of (non empty) subsets with (and an infinite set). Let denote the category for which the objects are the pairs with a finite dimensional -vector space and such that is unipotent for every and there is a finite subset of such that for . As in 3.8 and 3.9, this defines an affine group scheme and a pro-Lie algebra . The analogue of Proposition 3.10 is:\n\nLet be an -module. The are the cohomology groups of the complex , where denotes the -vector space of the maps such that there exists a finite subset of with for . The non trivial map is again .\n\nThe above situation occurs in connection with the Stokes phenomenon (see [vdP-S]), where is the set of alien derivations. The set is the disjoint union, over , of the sets . The corresponding affine group scheme is the kernel of the surjective morphism of affine schemes . We will return to this in Section 5. We note that the finiteness condition stated in [vdP-S] is slightly wrong.\n\n## 4 Formal differential equations\n\nLet be the differential field with derivation . A differential equation or module over will be called formal. We recall the explicit descriptions of and , slightly extending the one given in [vdP-S]. The aim of this section is to make both (where is the solution space of ) and the canonical isomorphism explicit.\n\nDescription of .\nWrite with a -vector space. First one introduces the universal Picard–Vessiot ring for the regular singular differential modules over . This ring is where is the algebraic closure of and where the symbols and satisfy only the identities . The differentiation is given by and . Then one introduces the set and symbols for satisfying only the identities . Further . Now where . This is the universal Picard–Vessiot ring for . Put and write for . Then .\n\nDescription of .\nFor convenience, we will identify the affine group scheme with its set of -valued points which is the group of the differential automorphisms of . A special (and very natural) element in this group is the formal monodromy defined by:\n(i) acts on by for all ,\n(ii) for all ,\n(iii) for all ,\n(iv) .\n\nThe exponential torus (in the terminology of J.-P. Ramis) is the group . An element acts on by is the identity on , on and on the elements . Further for all . The group together with the element generate topologically (for the Zariski topology). So far we have followed [vdP-S]. The Zariski closure of the group generated by is rather big and we prefer to split this group into smaller pieces. For this purpose we introduce more special elements in .\n\nOne decomposes as a product of commuting automorphisms , where has the same definition as except for (iv) which is replaced by . Further is the identity for the elements in , the elements , and . We note that is the Jordan decomposition of as a product of a semi-simple element and a unipotent element.\n\nWe still want to decompose the semi-simple as a product of commuting elements and . The direct sum yields, using , a direct product decomposition . Now and are the unique semi-simple elements in with eigenvalues in and such that . One verifies that and can also be defined by\n(i) , for all ;\n(ii) , for all ,\n(iii) , for all ,\n(iv) .\n\nFor every element there is an integer such that . It follows that the algebraic subgroup of , generated (topologically) by is , the projective limit of the groups . The algebraic subgroup generated (topologically) by can be identified with the torus . The algebraic subgroup generated by can be identified with .\n\nThus , the Zariski closure of the group generated by , can be identified with . Moreover, is the topological generator of , i.e., the group of the differential automorphisms of , in other words the universal differential Galois group for the regular singular equations over .\n\nWe extend the exponential torus to a larger torus . An element acts on by is the identity on and . Further for .\n\nWe conclude that has the form . The subgroup is reductive and the subgroup is the unipotent radical of . Further with topological generator .\n\nDescription of differential modules over .\nOne associates to a differential module over its solution space:\nwith the following data and relations:\n(i) a direct sum decomposition where ,\n(ii) an element which is the restriction of on to the subspace .\n(iii) for all .\n\nThe functor defines an equivalence of categories. We note that has a decomposition as a product of commuting elements induced by .\n\nComputation of .\nBy Section 3, with . From the above one sees that . The subspace of , consisting of the elements invariant under , is . Hence . The action of on is induced by its action on , given by . The group , generated for the Zariski topology by , acts on by . Hence can be identified with for and with for . In particular, the two cohomology groups have the same dimension. This proves Corollary 4.1.\n\nThe explicit canonical isomorphism .\nAny module is, after a finite extension of , a direct sum of isotypical submodules modules with . This is just a translation of the decomposition . Now is bijective on for . Thus for , the summand gives no contribution for the above map . The direct summand , which is the regular singular part of , can be represented by a matrix differential operator with a constant matrix such that the eigenvalues of satisfy . Only the generalized eigenspace for the eigenvalue of can give a contribution to . This generalized eigenspace is the submodule of corresponding to . After restricting to this submodule, is nilpotent. The kernel and cokernel of on coincide with the kernel and cokernel of on . The fundamental matrix for the equation is . The columns of this matrix form a basis for . The differential Galois group is generated by the action of which multiplies the fundamental matrix by . Thus on can be identified with on .\n\n###### Corollary 4.1 (B. Malgrange)\n\n[M]. For any differential module over one has .\n\n## 5 Analytic differential equations\n\ndenotes the differential field , consisting of the convergent Laurent series, with derivation . A differential equation or module over this field will be called analytic. We recall and extend results of [vdP-S].\n\nDescription of and .\nAs in Section 4, denotes the field of the formal Laurent series. Let denote the -subalgebra consising of the elements that satisfy a linear differential equation with coefficients in . Then\n\n ¯¯¯¯F[{e(q)}q∈Q+,ℓ]⊂UF=D[{e(q)}q∈Q+,ℓ]⊂UˆF=¯¯¯¯¯ˆF[{e(q)}q∈Q+.ℓ].\n\nThe precise structure of the differential algebra is unknown. However the multisummations in the directions lead to ‘locally unipotent’ elements of , the Stokes maps (multipliers) for every direction . One considers the ‘locally nilpotent’ logarithms of the Stokes maps. These are elements of the pro-Lie algebra of . The element is an -linear derivation of , commuting with , and trivial on . Thus is determined by its restriction . The resulting maps are important ingredients for the structure of . Put and let and denote the pro-Lie algebra and the affine group scheme corresponding to defined in Section 3. Then . We recall that is topologically generated by and the formal monodromy . The action, by conjugation, of on induces an action on . One can verify that and that, for any , one has .\n\nThe group is reductive and therefore is the unipotent radical of . The affine group scheme is topologically generated by . The relations between these generators are unknown. The same holds for their logarithms .\nFor the group with topological generators one observes that the only relations are . A good translation in terms of generators of Lie algebras does not seem to exists.\n\nDescription of the analytic differential modules.\nAs in [vdP-S] one can describe a differential module over by a structure on its space of solutions , namely . The maps can be replaced by and can be replaced by its components (with is a singular direction for ). These elements in map each to etc. Thus is represented by a tuple .\n\nComputation of . The computation uses the formula . First an example.\n\n###### Example 5.1\n\nwhere and with and . Then has dimension (see below for the definition of the irregularity of ).\n\n• Let denote the action of on . Then is trivial for and . Further, for the map is multiplication by .\n\nNow . The cohomology group identifies with the complex vector space of the algebraic homomorphism , such that there are only finitely many for which there is a with .\n\nLet be such a map and suppose that is invariant under . For and one has . Since the latter is one has . This applied with and or with yields and . Now we apply this with and . Then . Using that and one has .\n\nWe conclude that the invariant algebraic homomorphism are described by: for and for all . The dimension of the under consideration is therefore equal to the number of the singular directions modulo of . This number is easily seen to be . We note that this result coincides with the explicit calculations in [vdP-S], Section 7.3.\n\nB. Malgrange has introduced the irregularity of a differential module over as follows. Put . The action of the operator on the exact sequence induces the long exact sequence" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91550535,"math_prob":0.97769517,"size":21627,"snap":"2022-27-2022-33","text_gpt3_token_len":4696,"char_repetition_ratio":0.17060538,"word_repetition_ratio":0.054607507,"special_character_ratio":0.21255837,"punctuation_ratio":0.11927504,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99662584,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T18:35:53Z\",\"WARC-Record-ID\":\"<urn:uuid:9e2fe80a-2618-42cd-8829-1debe42255ac>\",\"Content-Length\":\"1049512\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81bc005a-ec40-4fec-af6d-c5c6ae714595>\",\"WARC-Concurrent-To\":\"<urn:uuid:e62e3990-d295-402e-a0d9-36f5a58fbed6>\",\"WARC-IP-Address\":\"104.21.14.110\",\"WARC-Target-URI\":\"https://www.arxiv-vanity.com/papers/0711.0815/\",\"WARC-Payload-Digest\":\"sha1:VYA4BU3N535SMVLSZC2Q6BTAN2W3OT5L\",\"WARC-Block-Digest\":\"sha1:7VWD6CFRA4A7XKTBV2ZUQTGPWLDU33OR\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570871.10_warc_CC-MAIN-20220808183040-20220808213040-00051.warc.gz\"}"}
https://pygraphviz.github.io/documentation/stable/_modules/pygraphviz/agraph.html	[ "# Source code for pygraphviz.agraph\n\n```\"\"\"\nA Python interface to Graphviz.\n\"\"\"\nimport os\nimport re\nimport shlex\nimport subprocess\nimport sys\nimport warnings\nfrom collections.abc import MutableMapping\nimport tempfile\nimport io\n\nfrom . import graphviz as gv\n\n_DEFAULT_ENCODING = \"UTF-8\"\n\ndef __init__(self, result, pipe):\nself.result = result\nself.pipe = pipe\n\ndef run(self):\ntry:\nwhile True:\nif not chunk:\nbreak\nself.result.append(chunk)\nfinally:\nself.pipe.close()\n\nclass _Action:\nfind, create = 0, 1\n\nclass DotError(ValueError):\n\"\"\"Dot data parsing error\"\"\"\n\n[docs]class AGraph:\n\"\"\"Class for Graphviz agraph type.\n\nExample use\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G = pgv.AGraph(directed=True)\n>>> G = pgv.AGraph(\"file.dot\") # doctest: +SKIP\n\nGraphviz graph keyword parameters are processed so you may add\nthem like\n\n>>> G = pgv.AGraph(landscape=\"true\", ranksep=\"0.1\")\n\nor alternatively\n\n>>> G = pgv.AGraph()\n>>> G.graph_attr.update(landscape=\"true\", ranksep=\"0.1\")\n\nand\n\n>>> G.node_attr.update(color=\"red\")\n>>> G.edge_attr.update(len=\"2.0\", color=\"blue\")\n\nSee http://www.graphviz.org/doc/info/attrs.html\nfor a list of attributes.\n\nKeyword parameters:\n\nthing is a generic input type (filename, string, handle to pointer,\ndictionary of dictionaries). An attempt is made to automaticaly\ndetect the type so you may write for example:\n\n>>> d = {\"1\": {\"2\": None}, \"2\": {\"1\": None, \"3\": None}, \"3\": {\"2\": None}}\n>>> A = pgv.AGraph(d)\n>>> s = A.to_string()\n>>> B = pgv.AGraph(s)\n>>> h = B.handle\n>>> C = pgv.AGraph(h)\n\nParameters::\n\nname: Name for the graph\n\nstrict: True\|False (True for simple graphs)\n\ndirected: True\|False\n\ndata: Dictionary of dictionaries or dictionary of lists\nrepresenting nodes or edges to load into initial graph\n\nstring: String containing a dot format graph\n\nhandle: Swig pointer to an agraph_t data structure\n\n\"\"\"\n\ndef __init__(\nself,\nthing=None,\nfilename=None,\ndata=None,\nstring=None,\nhandle=None,\nname=\"\",\nstrict=True,\ndirected=False,\n*attr,\n):\nself.handle = None # assign first in case the __init__ bombs\nself._owns_handle = True\n# initialization can take no arguments (gives empty graph) or\n# a file name\n# a string of graphviz dot language\n# a swig pointer (handle) to a graph\n# a dict of dicts (or dict of lists) data structure\n\nself.has_layout = False # avoid creating members outside of init\n\n# backward compability\nfilename = attr.pop(\"file\", filename)\n# guess input type if specified as first (nonkeyword) argument\nif thing is not None:\n# can't specify first argument and also file,data,string,handle\nfilename = None\ndata = None\nstring = None\nhandle = None\nif isinstance(thing, dict):\ndata = thing # a dictionary of dictionaries (or lists)\nelif hasattr(thing, \"own\"): # a Swig pointer - graph handle\nhandle = thing\nelif isinstance(thing, str):\npattern = re.compile(r\"(strict)?\\s(graph\|digraph).{.}\\s\", re.DOTALL)\nif pattern.match(thing):\nstring = thing # this is a dot format graph in a string\nelse:\nfilename = thing # assume this is a file name\nelif hasattr(thing, \"open\"):\nfilename = thing # assume this is a file name (in a path obj)\nelse:\nraise TypeError(f\"Unrecognized input {thing}\")\n\nif handle is not None:\n# if handle was specified, reference it\nself.handle = handle\nself._owns_handle = False\nelif filename is not None:\n# load new graph from file (creates self.handle)\nelif string is not None:\n# load new graph from string (creates self.handle)\n# get the charset from the string to properly encode it for\n# writing to the temporary file in from_string()\nmatch = re.search(r'charset\\s=\\s\"([^\"]+)\"', string)\nif match is not None:\nself.encoding = match.group(1)\nelse:\nself.encoding = _DEFAULT_ENCODING\nself.from_string(string)\nelse:\n# no handle, need to\nself.handle = None\n\nif self.handle is not None:\n# the handle was specified or created\n# get the encoding from the \"charset\" graph attribute\nitem = gv.agget(self.handle, b\"charset\")\nif item is not None:\nself.encoding = (\nitem if type(item) is not bytes else item.decode(\"utf-8\")\n)\nelse:\nself.encoding = _DEFAULT_ENCODING\nelse:\n# no handle was specified or created\n# get encoding from the \"charset\" kwarg\nself.encoding = attr.get(\"charset\", _DEFAULT_ENCODING)\ntry:\nif name is None:\nname = \"\"\n# instantiate a new, empty graph\nself.handle = gv.agraphnew(name.encode(self.encoding), strict, directed)\nexcept TypeError:\nraise TypeError(f\"Graph name must be a string: {name}\")\n\n# encoding is already set but if it was specified explicitly\n# as an attr, then set it explicitly for the graph\nif \"charset\" in attr:\ngv.agattr_label(self.handle, 0, \"charset\", self.encoding)\n\n# if data is specified, populate the newly created graph\nif data is not None:\n# load from dict of dicts or dict of lists\nfor node in data:\nfor nbr in data[node]:\n\n# throw away the charset attribute, if one exists,\n# since we've already set it, and now it should not be changed\nif \"charset\" in attr:\ndel attr[\"charset\"]\n\n# assign any attributes specified through keywords\nself.graph_attr = Attribute(self.handle, 0) # graph attributes\nself.graph_attr.update(attr) # apply attributes passed to init\nself.node_attr = Attribute(self.handle, 1) # default node attributes\nself.edge_attr = Attribute(self.handle, 2) # default edge attribtes\n\ndef __enter__(self):\nreturn self\n\ndef __exit__(self, ext_type, exc_value, traceback):\npass\n\ndef __str__(self):\nreturn self.string()\n\ndef __repr__(self):\nname = gv.agnameof(self.handle)\nif name is None:\nreturn f\"<AGraph {self.handle}>\"\nreturn f\"<AGraph {name} {self.handle}>\"\n\ndef _svg_repr(self):\nreturn self.draw(format=\"svg\").decode(self.encoding)\n\ndef _repr_mimebundle_(self, include=None, exclude=None):\nif self.has_layout:\nrepr_dict = {\"image/svg+xml\": self._svg_repr()}\nelse:\nrepr_dict = {\"text/plain\": self.__repr__()}\nreturn repr_dict\n\ndef __eq__(self, other):\n# two graphs are equal if they have exact same nodes and edges\n# and attributes. This is not graph isomorphism.\nif sorted(self.nodes()) != sorted(other.nodes()):\nreturn False\nif sorted(self.edges()) != sorted(other.edges()):\nreturn False\n# check attributes\nself_all_nodes_attr = {n: n.attr.to_dict() for n in sorted(self.nodes_iter())}\nother_all_nodes_attr = {n: n.attr.to_dict() for n in sorted(other.nodes_iter())}\nif self_all_nodes_attr != other_all_nodes_attr:\nreturn False\nself_all_edges_attr = {e: e.attr.to_dict() for e in sorted(self.edges_iter())}\nother_all_edges_attr = {e: e.attr.to_dict() for e in sorted(other.edges_iter())}\nif self_all_edges_attr != other_all_edges_attr:\nreturn False\n\n# All checks pass. They are equal\nreturn True\n\ndef __hash__(self):\n# include nodes and edges in hash\n# Could do attributes too, but hash should be fast\nreturn hash(\n(\ntuple(sorted(self.nodes_iter())),\ntuple(sorted(self.edges_iter())),\n)\n)\n\ndef __iter__(self):\n# provide \"for n in G\"\nreturn self.nodes_iter()\n\ndef __contains__(self, n):\n# provide \"n in G\"\nreturn self.has_node(n)\n\ndef __len__(self):\nreturn self.number_of_nodes()\n\ndef __getitem__(self, n):\n# \"G[n]\" returns nodes attached to n\nreturn self.neighbors(n)\n\n# not implemented, but could be...\n# def __setitem__(self,u,v):\n\ndef __del__(self):\nself._close_handle()\n\n[docs] def get_name(self):\nname = gv.agnameof(self.handle)\nif name is not None:\nname = name.decode(self.encoding)\nreturn name\n\nname = property(get_name)\n\nIf n is not a string, conversion to a string will be attempted.\nString conversion will work if n has valid string representation\n(try str(n) if you are unsure).\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.nodes()\n['a']\n>>> G.add_node(1) # will be converted to a string\n>>> G.nodes()\n['a', '1']\n\nAttributes can be added to nodes on creation or updated after creation\n(attribute values must be strings)\n\nSee http://www.graphviz.org/doc/info/attrs.html\nfor a list of attributes.\n\nAnonymous Graphviz nodes are currently not implemented.\n\"\"\"\nif not isinstance(n, str):\nn = str(n)\nn = n.encode(self.encoding)\ntry:\nnh = gv.agnode(self.handle, n, _Action.find)\nexcept KeyError:\nnh = gv.agnode(self.handle, n, _Action.create)\nnode = Node(self, nh=nh)\nnode.attr.update(attr)\n\n\"\"\"Add nodes from a container nbunch.\n\nnbunch can be any iterable container such as a list or dictionary\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> nlist = [\"a\", \"b\", 1, \"spam\"]\n>>> sorted(G.nodes())\n['1', 'a', 'b', 'spam']\n\nAttributes can be added to nodes on creation or updated after creation\n\n>>> G.add_nodes_from(nlist, color=\"red\") # set all nodes in nlist red\n\"\"\"\nfor n in nbunch:\n\n[docs] def remove_node(self, n):\n\"\"\"Remove the single node n.\n\nAttempting to remove a node that isn't in the graph will produce\nan error.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.remove_node(\"a\")\n\"\"\"\nif not isinstance(n, str):\nn = str(n)\nn = n.encode(self.encoding)\ntry:\nnh = gv.agnode(self.handle, n, _Action.find)\ngv.agdelnode(self.handle, nh)\nexcept KeyError:\nraise KeyError(f\"Node {n.decode(self.encoding)} not in graph.\")\n\ndelete_node = remove_node\n\n[docs] def remove_nodes_from(self, nbunch):\n\"\"\"Remove nodes from a container nbunch.\n\nnbunch can be any iterable container such as a list or dictionary\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> nlist = [\"a\", \"b\", 1, \"spam\"]\n>>> G.remove_nodes_from(nlist)\n\"\"\"\nfor n in nbunch:\nself.remove_node(n)\n\ndelete_nodes_from = remove_nodes_from\n\n[docs] def nodes_iter(self):\n\"\"\"Return an iterator over all the nodes in the graph.\n\nNote: modifying the graph structure while iterating over\nthe nodes may produce unpredictable results. Use nodes()\nas an alternative.\n\"\"\"\nnh = gv.agfstnode(self.handle)\nwhile nh is not None:\nyield Node(self, nh=nh)\ntry:\nnh = gv.agnxtnode(self.handle, nh)\nexcept StopIteration:\nreturn\n\niternodes = nodes_iter\n\n[docs] def nodes(self):\n\"\"\"Return a list of all nodes in the graph.\"\"\"\nreturn list(self.nodes_iter())\n\n[docs] def number_of_nodes(self):\n\"\"\"Return the number of nodes in the graph.\"\"\"\nreturn gv.agnnodes(self.handle)\n\n[docs] def order(self):\n\"\"\"Return the number of nodes in the graph.\"\"\"\nreturn self.number_of_nodes()\n\n[docs] def has_node(self, n):\n\"\"\"Return True if n is in the graph or False if not.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.has_node(\"a\")\nTrue\n>>> \"a\" in G # same as G.has_node('a')\nTrue\n\n\"\"\"\ntry:\nnode = Node(self, n)\nreturn True\nexcept KeyError:\nreturn False\n\n[docs] def get_node(self, n):\n\"\"\"Return a node object (Node) corresponding to node n.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> node = G.get_node(\"a\")\n>>> print(node)\na\n\"\"\"\nreturn Node(self, n)\n\n[docs] def add_edge(self, u, v=None, key=None, attr):\n\"\"\"Add a single edge between nodes u and v.\n\nIf the nodes u and v are not in the graph they will added.\n\nIf u and v are not strings, conversion to a string will be attempted.\nString conversion will work if u and v have valid string representation\n(try str(u) if you are unsure).\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.edges()\n[('a', 'b')]\n\nThe optional key argument allows assignment of a key to the\nedge. This is especially useful to distinguish between\nparallel edges in multi-edge graphs (strict=False).\n\n>>> G = pgv.AGraph(strict=False)\n>>> sorted(G.edges(keys=True))\n[('a', 'b', 'first'), ('a', 'b', 'second')]\n\nAttributes can be added when edges are created or updated after creation\n\nAttributes must be valid strings.\n\nSee http://www.graphviz.org/doc/info/attrs.html\nfor a list of attributes.\n\n\"\"\"\nif v is None:\n(u, v) = u # no v given, assume u is an edge tuple\ntry:\nuh = Node(self, u).handle\nexcept:\nuh = Node(self, u).handle\ntry:\nvh = Node(self, v).handle\nexcept:\nvh = Node(self, v).handle\nif key is not None:\nif not isinstance(key, str):\nkey = str(key)\nkey = key.encode(self.encoding)\ntry:\n# new\neh = gv.agedge(self.handle, uh, vh, key, _Action.create)\nexcept KeyError:\neh = gv.agedge(self.handle, uh, vh, key, _Action.find)\ne = Edge(self, eh=eh)\ne.attr.update(attr)\n\n\"\"\"Add nodes to graph from a container ebunch.\n\nebunch is a container of edges such as a list or dictionary.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> elist = [(\"a\", \"b\"), (\"b\", \"c\")]\n\nAttributes can be added when edges are created or updated after creation\n\n\"\"\"\nfor e in ebunch:\n\n[docs] def get_edge(self, u, v, key=None):\n\"\"\"Return an edge object (Edge) corresponding to edge (u,v).\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> edge = G.get_edge(\"a\", \"b\")\n>>> print(edge)\n('a', 'b')\n\nWith optional key argument will only get edge matching (u,v,key).\n\n\"\"\"\nreturn Edge(self, u, v, key)\n\n[docs] def remove_edge(self, u, v=None, key=None):\n\"\"\"Remove edge between nodes u and v from the graph.\n\nWith optional key argument will only remove an edge\nmatching (u,v,key).\n\n\"\"\"\nif v is None:\n(u, v) = u # no v given, assume u is an edge tuple\ne = Edge(self, u, v, key)\ntry:\ngv.agdeledge(self.handle, e.handle)\nexcept KeyError:\nraise KeyError(f\"Edge {u}-{v} not in graph.\")\n\ndelete_edge = remove_edge\n\n[docs] def remove_edges_from(self, ebunch):\n\"\"\"Remove edges from ebunch (a container of edges).\"\"\"\nfor e in ebunch:\nself.remove_edge(e)\n\ndelete_edges_from = remove_edges_from\n\n[docs] def has_edge(self, u, v=None, key=None):\n\"\"\"Return True an edge u-v is in the graph or False if not.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.has_edge(\"a\", \"b\")\nTrue\n\nOptional key argument will restrict match to edges (u,v,key).\n\n\"\"\"\n\nif v is None:\n(u, v) = u # no v given, assume u is an edge tuple\ntry:\nEdge(self, u, v, key)\nreturn True\nexcept KeyError:\nreturn False\n\n[docs] def edges(self, nbunch=None, keys=False):\n\"\"\"Return list of edges in the graph.\n\nIf the optional nbunch (container of nodes) only edges\nadjacent to nodes in nbunch will be returned.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> print(sorted(G.edges()))\n[('a', 'b'), ('c', 'd')]\n>>> print(G.edges(\"a\"))\n[('a', 'b')]\n\"\"\"\nreturn list(self.edges_iter(nbunch=nbunch, keys=keys))\n\n[docs] def has_neighbor(self, u, v, key=None):\n\"\"\"Return True if u has an edge to v or False if not.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.has_neighbor(\"a\", \"b\")\nTrue\n\nOptional key argument will only find edges (u,v,key).\n\"\"\"\nreturn self.has_edge(u, v)\n\n[docs] def neighbors_iter(self, n):\n\"\"\"Return iterator over the nodes attached to n.\n\nNote: modifying the graph structure while iterating over\nnode neighbors may produce unpredictable results. Use neighbors()\nas an alternative.\n\"\"\"\nn = Node(self, n)\nnh = n.handle\neh = gv.agfstedge(self.handle, nh)\nwhile eh is not None:\n(s, t) = Edge(self, eh=eh)\nif s == n:\nyield Node(self, t)\nelse:\nyield Node(self, s)\ntry:\neh = gv.agnxtedge(self.handle, eh, nh)\nexcept StopIteration:\nreturn\n\n[docs] def neighbors(self, n):\n\"\"\"Return a list of the nodes attached to n.\"\"\"\nreturn list(self.neighbors_iter(n))\n\niterneighbors = neighbors_iter\n\n[docs] def out_edges_iter(self, nbunch=None, keys=False):\n\"\"\"Return iterator over out edges in the graph.\n\nIf the optional nbunch (container of nodes) only out edges\nadjacent to nodes in nbunch will be returned.\n\nNote: modifying the graph structure while iterating over\nedges may produce unpredictable results. Use out_edges()\nas an alternative.\n\"\"\"\n\nif nbunch is None: # all nodes\nnh = gv.agfstnode(self.handle)\nwhile nh is not None:\neh = gv.agfstout(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtout(self.handle, eh)\nexcept StopIteration:\nbreak\ntry:\nnh = gv.agnxtnode(self.handle, nh)\nexcept StopIteration:\nreturn\nelif nbunch in self: # if nbunch is a single node\nn = Node(self, nbunch)\nnh = n.handle\neh = gv.agfstout(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtout(self.handle, eh)\nexcept StopIteration:\nreturn\nelse: # if nbunch is a sequence of nodes\ntry:\nbunch = [n for n in nbunch if n in self]\nexcept TypeError:\nraise TypeError(\"nbunch is not a node or a sequence of nodes.\")\nfor n in nbunch:\ntry:\nnh = Node(self, n).handle\nexcept KeyError:\ncontinue\neh = gv.agfstout(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtout(self.handle, eh)\nexcept StopIteration:\nbreak\n\niteroutedges = out_edges_iter\n\n[docs] def in_edges_iter(self, nbunch=None, keys=False):\n\"\"\"Return iterator over out edges in the graph.\n\nIf the optional nbunch (container of nodes) only out edges\nadjacent to nodes in nbunch will be returned.\n\nNote: modifying the graph structure while iterating over\nedges may produce unpredictable results. Use in_edges()\nas an alternative.\n\"\"\"\nif nbunch is None: # all nodes\nnh = gv.agfstnode(self.handle)\nwhile nh is not None:\neh = gv.agfstin(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtin(self.handle, eh)\nexcept StopIteration:\nbreak\ntry:\nnh = gv.agnxtnode(self.handle, nh)\nexcept StopIteration:\nreturn\nelif nbunch in self: # if nbunch is a single node\nn = Node(self, nbunch)\nnh = n.handle\neh = gv.agfstin(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtin(self.handle, eh)\nexcept StopIteration:\nbreak\nelse: # if nbunch is a sequence of nodes\ntry:\nbunch = [n for n in nbunch if n in self]\nexcept TypeError:\nraise TypeError(\"nbunch is not a node or a sequence of nodes.\")\nfor n in nbunch:\ntry:\nnh = Node(self, n).handle\nexcept KeyError:\ncontinue\neh = gv.agfstin(self.handle, nh)\nwhile eh is not None:\ne = Edge(self, eh=eh)\nif keys:\nyield (e, e, e.name)\nelse:\nyield e\ntry:\neh = gv.agnxtin(self.handle, eh)\nexcept StopIteration:\nbreak\n\n[docs] def edges_iter(self, nbunch=None, keys=False):\n\"\"\"Return iterator over edges in the graph.\n\nIf the optional nbunch (container of nodes) only edges\nadjacent to nodes in nbunch will be returned.\n\nNote: modifying the graph structure while iterating over\nedges may produce unpredictable results. Use edges()\nas an alternative.\n\"\"\"\nif nbunch is None: # all nodes\nfor e in self.out_edges_iter(keys=keys):\nyield e\nelif nbunch in self: # only one node\nfor e in self.out_edges_iter(nbunch, keys=keys):\nyield e\nfor e in self.in_edges_iter(nbunch, keys=keys):\nif e != (nbunch, nbunch):\nyield e\nelse: # a group of nodes\nused = set()\nfor e in self.out_edges_iter(nbunch, keys=keys):\nyield e\nfor e in self.in_edges_iter(nbunch, keys=keys):\nif e not in used:\nyield e\n\niterinedges = in_edges_iter\n\niteredges = edges_iter\n\n[docs] def out_edges(self, nbunch=None, keys=False):\n\"\"\"Return list of out edges in the graph.\n\nIf the optional nbunch (container of nodes) only out edges\nadjacent to nodes in nbunch will be returned.\n\"\"\"\nreturn list(self.out_edges_iter(nbunch=nbunch, keys=keys))\n\n[docs] def in_edges(self, nbunch=None, keys=False):\n\"\"\"Return list of in edges in the graph.\nIf the optional nbunch (container of nodes) only in edges\nadjacent to nodes in nbunch will be returned.\n\"\"\"\nreturn list(self.in_edges_iter(nbunch=nbunch, keys=keys))\n\n[docs] def predecessors_iter(self, n):\n\"\"\"Return iterator over predecessor nodes of n.\n\nNote: modifying the graph structure while iterating over\nnode predecessors may produce unpredictable results. Use\npredecessors() as an alternative.\n\"\"\"\nn = Node(self, n)\nnh = n.handle\neh = gv.agfstin(self.handle, nh)\nwhile eh is not None:\n(s, t) = Edge(self, eh=eh)\nif s == n:\nyield Node(self, t)\nelse:\nyield Node(self, s)\ntry:\neh = gv.agnxtin(self.handle, eh)\nexcept StopIteration:\nreturn\n\niterpred = predecessors_iter\n\n[docs] def successors_iter(self, n):\n\"\"\"Return iterator over successor nodes of n.\n\nNote: modifying the graph structure while iterating over\nnode successors may produce unpredictable results. Use\nsuccessors() as an alternative.\n\"\"\"\nn = Node(self, n)\nnh = n.handle\neh = gv.agfstout(self.handle, nh)\nwhile eh is not None:\n(s, t) = Edge(self, eh=eh)\nif s == n:\nyield Node(self, t)\nelse:\nyield Node(self, s)\ntry:\neh = gv.agnxtout(self.handle, eh)\nexcept StopIteration:\nreturn\n\nitersucc = successors_iter\n\n[docs] def successors(self, n):\n\"\"\"Return list of successor nodes of n.\"\"\"\nreturn list(self.successors_iter(n))\n\n[docs] def predecessors(self, n):\n\"\"\"Return list of predecessor nodes of n.\"\"\"\nreturn list(self.predecessors_iter(n))\n\n# digraph definitions\nout_neighbors = successors\nin_neighbors = predecessors\n\n[docs] def degree_iter(self, nbunch=None, indeg=True, outdeg=True):\n\"\"\"Return an iterator over the degree of the nodes given in\nnbunch container.\n\nReturns pairs of (node,degree).\n\"\"\"\nfor n in self._prepare_nbunch(nbunch):\nyield (Node(self, n), gv.agdegree(self.handle, n.handle, indeg, outdeg))\n\n[docs] def in_degree_iter(self, nbunch=None):\n\"\"\"Return an iterator over the in-degree of the nodes given in\nnbunch container.\n\nReturns pairs of (node,degree).\n\"\"\"\nreturn self.degree_iter(nbunch, indeg=True, outdeg=False)\n\n[docs] def out_degree_iter(self, nbunch=None):\n\"\"\"Return an iterator over the out-degree of the nodes given in\nnbunch container.\n\nReturns pairs of (node,degree).\n\n\"\"\"\nreturn self.degree_iter(nbunch, indeg=False, outdeg=True)\n\niteroutdegree = out_degree_iter\niterindegree = in_degree_iter\n\n[docs] def out_degree(self, nbunch=None, with_labels=False):\n\"\"\"Return the out-degree of nodes given in nbunch container.\n\nUsing optional with_labels=True returns a dictionary\nkeyed by node with value set to the degree.\n\"\"\"\nif with_labels:\nreturn dict(self.out_degree_iter(nbunch))\nelse:\ndlist = list(d for n, d in self.out_degree_iter(nbunch))\nif nbunch in self:\nreturn dlist\nelse:\nreturn dlist\n\n[docs] def in_degree(self, nbunch=None, with_labels=False):\n\"\"\"Return the in-degree of nodes given in nbunch container.\n\nUsing optional with_labels=True returns a dictionary\nkeyed by node with value set to the degree.\n\"\"\"\n\nif with_labels:\nreturn dict(self.in_degree_iter(nbunch))\nelse:\ndlist = list(d for n, d in self.in_degree_iter(nbunch))\nif nbunch in self:\nreturn dlist\nelse:\nreturn dlist\n\n[docs] def reverse(self):\n\"\"\"Return copy of directed graph with edge directions reversed.\"\"\"\nif self.directed:\n# new empty DiGraph\nH = self.__class__(strict=self.strict, directed=True, name=self.name)\nH.graph_attr.update(self.graph_attr)\nH.node_attr.update(self.node_attr)\nH.edge_attr.update(self.edge_attr)\nfor n in self.nodes():\nnew_n = Node(H, n)\nnew_n.attr.update(n.attr)\nfor e in self.edges():\n(u, v) = e\nuv = H.get_edge(v, u)\nuv.attr.update(e.attr)\nreturn H\nelse:\nreturn self\n\n[docs] def degree(self, nbunch=None, with_labels=False):\n\"\"\"Return the degree of nodes given in nbunch container.\n\nUsing optional with_labels=True returns a dictionary\nkeyed by node with value set to the degree.\n\n\"\"\"\nif with_labels:\nreturn dict(self.degree_iter(nbunch))\nelse:\ndlist = list(d for n, d in self.degree_iter(nbunch))\nif nbunch in self:\nreturn dlist\nelse:\nreturn dlist\n\niterdegree = degree_iter\n\n[docs] def number_of_edges(self):\n\"\"\"Return the number of edges in the graph.\"\"\"\nreturn gv.agnedges(self.handle)\n\n[docs] def clear(self):\n\"\"\"Remove all nodes, edges, and attributes from the graph.\"\"\"\nself.remove_edges_from(self.edges())\nself.remove_nodes_from(self.nodes())\n# now \"close\" existing graph and create a new graph\nname = gv.agnameof(self.handle)\nstrict = self.strict\ndirected = self.directed\nself._close_handle()\nself.handle = gv.agraphnew(name, strict, directed)\nself._owns_handle = True\nself._update_handle_references()\n\n[docs] def close(self):\nself._close_handle()\n\ndef _close_handle(self):\n# may be useful to clean up graphviz data\n# this should completely remove all of the existing graphviz data\nif self._owns_handle:\nif self.handle is not None:\ngv.agclose(self.handle)\nself.handle = None\nself._owns_handle = False\nelse:\nself.handle = None\n\n[docs] def copy(self):\n\"\"\"Return a copy of the graph.\n\nNotes\n=====\nVersions <=1.6 made a copy by writing and the reading a dot string.\nThis version loads a new graph with nodes, edges and attributes.\n\"\"\"\nG = self.__class__(directed=self.is_directed())\nfor node in self.nodes():\nG.get_node(node).attr.update(self.get_node(node).attr)\nfor edge in self.edges():\nG.get_edge(edge).attr.update(self.get_edge(edge).attr)\nG.graph_attr.update(self.graph_attr)\nG.node_attr.update(self.node_attr)\nG.edge_attr.update(self.edge_attr)\nreturn G\n\n\"\"\"Add the path of nodes given in nlist.\"\"\"\nfromv = nlist.pop(0)\nwhile len(nlist) > 0:\ntov = nlist.pop(0)\nfromv = tov\n\n\"\"\"Add the cycle of nodes given in nlist.\"\"\"\n\ndef _prepare_nbunch(self, nbunch=None):\n# private function to build bunch from nbunch\nif nbunch is None: # include all nodes via iterator\nbunch = self.nodes_iter()\nelif nbunch in self: # if nbunch is a single node\nbunch = [Node(self, nbunch)]\nelse: # if nbunch is a sequence of nodes\ntry: # capture error for nonsequence/iterator entries.\nbunch = [Node(self, n) for n in nbunch if n in self]\n# bunch=(n for n in nbunch if n in self) # need python 2.4\nexcept TypeError:\nraise TypeError(\"nbunch is not a node or a sequence of nodes.\")\nreturn bunch\n\n[docs] def add_subgraph(self, nbunch=None, name=None, attr):\n\"\"\"Return subgraph induced by nodes in nbunch.\"\"\"\nif name is not None:\nname = name.encode(self.encoding)\ntry:\nhandle = gv.agsubg(self.handle, name, _Action.create)\nexcept TypeError:\nraise TypeError(\nf\"Subgraph name must be a string: {name.decode(self.encoding)}\"\n)\n\nH = self.__class__(\nstrict=self.strict, directed=self.directed, handle=handle, name=name, attr\n)\nif nbunch is None:\nreturn H\n# add induced subgraph on nodes in nbunch\nbunch = self._prepare_nbunch(nbunch)\nfor n in bunch:\nnode = Node(self, n)\nnh = gv.agsubnode(handle, node.handle, _Action.create)\nfor (u, v, k) in self.edges(keys=True):\nif u in H and v in H:\nedge = Edge(self, u, v, k)\neh = gv.agsubedge(handle, edge.handle, _Action.create)\n\nreturn H\n\n[docs] def remove_subgraph(self, name):\n\"\"\"Remove subgraph with given name.\"\"\"\ntry:\nhandle = gv.agsubg(self.handle, name.encode(self.encoding), _Action.find)\nexcept TypeError:\nraise TypeError(f\"Subgraph name must be a string: {name}\")\nif handle is None:\nraise KeyError(f\"Subgraph {name} not in graph.\")\ngv.agdelsubg(self.handle, handle)\n\ndelete_subgraph = remove_subgraph\n\n[docs] def subgraph_parent(self, nbunch=None, name=None):\n\"\"\"Return parent graph of subgraph or None if graph is root graph.\"\"\"\nhandle = gv.agparent(self.handle)\nif handle is None:\nreturn None\nH = self.__class__(\nstrict=self.strict, directed=self.directed, handle=handle, name=name\n)\nreturn H\n\n[docs] def subgraph_root(self, nbunch=None, name=None):\n\"\"\"Return root graph of subgraph or None if graph is root graph.\"\"\"\nhandle = gv.agroot(self.handle)\nif handle is None:\nreturn None\nH = self.__class__(\nstrict=self.strict, directed=self.directed, handle=handle, name=name\n)\nreturn H\n\n[docs] def get_subgraph(self, name):\n\"\"\"Return existing subgraph with specified name or None if it\ndoesn't exist.\n\"\"\"\ntry:\nhandle = gv.agsubg(self.handle, name.encode(self.encoding), _Action.find)\nexcept TypeError:\nraise TypeError(f\"Subgraph name must be a string: {name}\")\n\nif handle is None:\nreturn None\nH = self.__class__(strict=self.strict, directed=self.directed, handle=handle)\nreturn H\n\n[docs] def subgraphs_iter(self):\n\"\"\"Iterator over subgraphs.\"\"\"\nhandle = gv.agfstsubg(self.handle)\nwhile handle is not None:\nyield self.__class__(\nstrict=self.strict, directed=self.directed, handle=handle\n)\ntry:\nhandle = gv.agnxtsubg(handle)\nexcept StopIteration:\nreturn\n\n[docs] def subgraphs(self):\n\"\"\"Return a list of all subgraphs in the graph.\"\"\"\nreturn list(self.subgraphs_iter())\n\n# directed, undirected tests and conversions\n\n[docs] def is_strict(self):\n\"\"\"Return True if graph is strict or False if not.\n\nStrict graphs do not allow parallel edges or self loops.\n\"\"\"\nif gv.agisstrict(self.handle) == 1:\nreturn True\nelse:\nreturn False\n\nstrict = property(is_strict)\n\n[docs] def is_directed(self):\n\"\"\"Return True if graph is directed or False if not.\"\"\"\nif gv.agisdirected(self.handle) == 1:\nreturn True\nelse:\nreturn False\n\ndirected = property(is_directed)\n\n[docs] def is_undirected(self):\n\"\"\"Return True if graph is undirected or False if not.\"\"\"\nif gv.agisundirected(self.handle) == 1:\nreturn True\nelse:\nreturn False\n\n[docs] def to_undirected(self):\n\"\"\"Return undirected copy of graph.\"\"\"\nif not self.directed:\nreturn self.copy()\nelse:\nU = AGraph(strict=self.strict)\nU.graph_attr.update(self.graph_attr)\nU.node_attr.update(self.node_attr)\nU.edge_attr.update(self.edge_attr)\nfor n in self.nodes():\nnew_n = Node(U, n)\nnew_n.attr.update(n.attr)\nfor e in self.edges():\n(u, v) = e\nuv = U.get_edge(u, v)\nuv.attr.update(e.attr)\nreturn U\n\n[docs] def to_directed(self, *kwds):\n\"\"\"Return directed copy of graph.\n\nEach undirected edge u-v is represented as two directed\nedges u->v and v->u.\n\"\"\"\nif not self.directed:\nD = AGraph(strict=self.strict, directed=True)\nD.graph_attr.update(self.graph_attr)\nD.node_attr.update(self.node_attr)\nD.edge_attr.update(self.edge_attr)\nfor n in self.nodes():\nnew_n = Node(D, n)\nnew_n.attr.update(n.attr)\nfor e in self.edges():\n(u, v) = e\nuv = D.get_edge(u, v)\nvu = D.get_edge(v, u)\nuv.attr.update(e.attr)\nuv.attr.update(e.attr)\nvu.attr.update(e.attr)\nreturn D\nelse:\nreturn self.copy()\n\n# io\n\"\"\"Read graph from dot format file on path.\n\npath can be a file name or file handle\n\nuse::\n\n\"\"\"\nfh = self._get_fh(path)\ntry:\nself._close_handle()\ntry:\nexcept ValueError:\nraise DotError(\"Invalid Input\")\nelse:\nself._owns_handle = True\nself._update_handle_references()\nexcept OSError:\n\n[docs] def write(self, path=None):\n\"\"\"Write graph in dot format to file on path.\n\npath can be a file name or file handle\n\nuse::\n\nG.write('file.dot')\n\"\"\"\nif path is None:\npath = sys.stdout\nfh = self._get_fh(path, \"w\")\n# NOTE: TemporaryFile objects are not instances of IOBase on windows.\nif not isinstance(fh, (io.IOBase, tempfile._TemporaryFileWrapper)):\nraise TypeError(f\"{fh} is not a file handle\")\ntry:\ngv.agwrite(self.handle, fh)\nexcept OSError:\nprint(\"IO error writing file\")\nfinally:\nif hasattr(fh, \"close\") and not hasattr(path, \"write\"):\nfh.close()\n\n[docs] def string_nop(self):\n\"\"\"Return a string (unicode) representation of graph in dot format.\"\"\"\n# this will fail for graphviz-2.8 because of a broken nop\n# so use tempfile version below\nreturn self.draw(format=\"dot\", prog=\"nop\").decode(self.encoding)\n\n[docs] def to_string(self):\n\"\"\"Return a string representation of graph in dot format.\n\n`to_string()` uses \"agwrite\" to produce \"dot\" format w/o rendering.\nThe function `string_nop()` layouts with \"nop\" and renders to \"dot\".\n\"\"\"\nfh = tempfile.TemporaryFile()\nself.write(fh)\nfh.seek(0)\nfh.close()\nreturn data.decode(self.encoding)\n\n[docs] def string(self):\n\"\"\"Return a string (unicode) representation of graph in dot format.\"\"\"\nreturn self.to_string()\n# return self.string_nop()\n\n[docs] def from_string(self, string):\n\"\"\"Load a graph from a string in dot format.\n\nOverwrites any existing graph.\n\nTo make a new graph from a string use\n\n>>> import pygraphviz as pgv\n>>> s = \"digraph {1 -> 2}\"\n>>> A = pgv.AGraph()\n>>> t = A.from_string(s)\n>>> A = pgv.AGraph(string=s) # specify s is a string\n>>> A = pgv.AGraph(s) # s assumed to be a string during initialization\n\"\"\"\n# allow either unicode or encoded string\ntry:\nstring = string.decode(self.encoding)\nexcept (UnicodeEncodeError, AttributeError):\npass\nfrom tempfile import TemporaryFile\n\nfh = TemporaryFile()\nfh.write(string.encode(self.encoding))\nfh.seek(0)\nfh.close()\nreturn self\n\ndef _get_prog(self, prog):\n# private: get path of graphviz program\nprogs = {\n\"neato\",\n\"dot\",\n\"twopi\",\n\"circo\",\n\"fdp\",\n\"nop\",\n\"osage\",\n\"patchwork\",\n\"gc\",\n\"acyclic\",\n\"gvpr\",\n\"gvcolor\",\n\"ccomps\",\n\"sccmap\",\n\"tred\",\n\"sfdp\",\n\"unflatten\",\n}\nif prog not in progs:\nraise ValueError(f\"Program {prog} is not one of: {', '.join(progs)}.\")\n\ntry: # user must pick one of the graphviz programs...\nrunprog = self._which(prog)\nexcept:\n\nreturn runprog\n\ndef _run_prog(self, prog=\"nop\", args=\"\"):\n\"\"\"Apply graphviz program to graph and return the result as a string.\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph()\n>>> s = A._run_prog() # doctest: +SKIP\n>>> s = A._run_prog(prog=\"acyclic\") # doctest: +SKIP\n\n\"\"\"\nrunprog = r'\"%s\"' % self._get_prog(prog)\ncmd = \" \".join([runprog, args])\ndotargs = shlex.split(cmd)\np = subprocess.Popen(\ndotargs,\nshell=False,\nstdin=subprocess.PIPE,\nstdout=subprocess.PIPE,\nstderr=subprocess.PIPE,\nclose_fds=False,\n)\n(child_stdin, child_stdout, child_stderr) = (p.stdin, p.stdout, p.stderr)\n# Use threading to avoid blocking\ndata = []\nerrors = []\nt.start()\n\nself.write(child_stdin)\nchild_stdin.close()\n\nt.join()\np.wait()\n\nif not data:\nraise OSError(b\"\".join(errors).decode(self.encoding))\n\nif len(errors) > 0:\nwarnings.warn(b\"\".join(errors).decode(self.encoding), RuntimeWarning)\nreturn b\"\".join(data)\n\n[docs] def unflatten(self, args=\"\"):\n\"\"\"Adjust directed graphs to improve layout aspect ratio.\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph()\n>>> A_unflattened = A.unflatten(\"-f -l 3\")\n>>> A.unflatten(\"-f -l 1\").layout()\n\n\"\"\"\ndata = self._run_prog(\"unflatten\", args)\nself.from_string(data)\nreturn self\n\n[docs] def tred(self, args=\"\", copy=False):\n\"\"\"Transitive reduction of graph. Modifies existing graph.\n\nTo create a new graph use\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph(directed=True)\n>>> B = A.tred(copy=True) # doctest: +SKIP\n\nSee the graphviz \"tred\" program for details of the algorithm.\n\"\"\"\nif not self.directed:\nraise TypeError(\"tred requires a directed graph\")\n\ndata = self._run_prog(\"tred\", args)\nif copy:\nreturn self.__class__(string=data.decode(self.encoding))\nelse:\nreturn self.from_string(data)\n\n[docs] def acyclic(self, args=\"\", copy=False):\n\"\"\"Reverse sufficient edges in digraph to make graph acyclic.\nModifies existing graph.\n\nTo create a new graph use\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph(directed=True)\n>>> B = A.acyclic(copy=True) # doctest: +SKIP\n\nSee the graphviz \"acyclic\" program for details of the algorithm.\n\"\"\"\nif not self.directed:\nraise TypeError(\"acyclic requires a directed graph\")\n\ndata = self._run_prog(\"acyclic\", args)\nif copy:\nreturn self.__class__(string=data.decode(self.encoding))\nelse:\nreturn self.from_string(data)\n\n[docs] def layout(self, prog=\"neato\", args=\"\"):\n\"\"\"Assign positions to nodes in graph.\n\nOptional prog=['neato'\|'dot'\|'twopi'\|'circo'\|'fdp'\|'nop']\nwill use specified graphviz layout method.\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph()\n>>> A.layout()\n>>> A.layout(prog=\"neato\", args=\"-Nshape=box -Efontsize=8\")\n\nThe layout might take a long time on large graphs.\n\n\"\"\"\noutput_fmt = \"dot\"\ndata = self._run_prog(prog, \" \".join([args, \"-T\", output_fmt]))\nself.from_string(data)\nself.has_layout = True\nreturn\n\ndef _layout(self, prog=\"neato\", args=\"\"):\n\"\"\"Assign positions to nodes in graph.\n\n.. caution:: EXPERIMENTAL\n\nThis version of the layout command uses libgvc for layout instead\nof command line GraphViz tools like in versions <1.6 and the default.\n\nOptional prog=['neato'\|'dot'\|'twopi'\|'circo'\|'fdp'\|'nop']\nwill use specified graphviz layout method.\n\n>>> import pygraphviz as pgv\n>>> A = pgv.AGraph()\n>>> A.layout()\n>>> A.layout(prog=\"neato\", args=\"-Nshape=box -Efontsize=8\")\n\nThe layout might take a long time on large graphs.\n\nNote: attaching positions in the AGraph usually doesn't affect the\nnext rendering. The positions are recomputed. But if you use prog=\"nop\"\nwhen rendering, it will take node positions from the AGraph attributes.\nIf you use prog=\"nop2\" it will take node and edge positions from the\nAGraph when rendering.\n\"\"\"\n_, prog = self._manually_parse_args(args, None, prog)\n\n# convert input strings to type bytes (encode it)\nif isinstance(prog, str):\nprog = prog.encode(self.encoding)\n\ngvc = gv.gvContext()\ngv.gvLayout(gvc, self.handle, prog)\ngv.gvRender(gvc, self.handle, format=b\"dot\", out=None)\n\ngv.gvFreeLayout(gvc, self.handle)\ngv.gvFreeContext(gvc)\n\nself.has_layout = True\nreturn\n\n[docs] def draw(self, path=None, format=None, prog=None, args=\"\"):\n\"\"\"Output graph to path in specified format.\n\nAn attempt will be made to guess the output format based on the file\nextension of `path`. If that fails, then the `format` parameter will\nbe used.\n\nNote, if `path` is a file object returned by a call to os.fdopen(),\nthen the method for discovering the format will not work. In such\ncases, one should explicitly set the `format` parameter; otherwise, it\nwill default to 'dot'.\n\nIf path is None, the result is returned as a Bytes object.\n\nFormats (not all may be available on every system depending on\nhow Graphviz was built)\n\n'canon', 'cmap', 'cmapx', 'cmapx_np', 'dia', 'dot',\n'fig', 'gd', 'gd2', 'gif', 'hpgl', 'imap', 'imap_np',\n'ismap', 'jpe', 'jpeg', 'jpg', 'mif', 'mp', 'pcl', 'pdf',\n'pic', 'plain', 'plain-ext', 'png', 'ps', 'ps2', 'svg',\n'svgz', 'vml', 'vmlz', 'vrml', 'vtx', 'wbmp', 'xdot', 'xlib'\n\nIf prog is not specified and the graph has positions\n(see layout()) then no additional graph positioning will\nbe performed.\n\nOptional prog=['neato'\|'dot'\|'twopi'\|'circo'\|'fdp'\|'nop']\nwill use specified graphviz layout method.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.add_edges_from([(0, 1), (1, 2), (2, 0), (2, 3)])\n>>> G.layout()\n\n# use current node positions, output pdf in 'file.pdf'\n>>> G.draw(\"file.pdf\")\n\n# use dot to position, output png in 'file'\n>>> G.draw(\"file\", format=\"png\", prog=\"dot\")\n\n# use keyword 'args' to pass additional arguments to graphviz\n>>> G.draw(\"test.pdf\", prog=\"twopi\", args=\"-Gepsilon=1\")\n>>> G.draw(\"test2.pdf\", args=\"-Nshape=box -Edir=forward -Ecolor=red \")\n\nThe layout might take a long time on large graphs.\n\n\"\"\"\n# try to guess format from extension\nif format is None and path is not None:\np = path\n# in case we got a file handle get its name instead\nif not isinstance(p, str):\np = path.name\nformat = os.path.splitext(p)[-1].lower()[1:]\n\nif format is None or format == \"\":\nformat = \"dot\"\n\nif prog is None:\nif self.has_layout:\nprog = \"neato\"\nargs += \"-n2\"\nelse:\nraise AttributeError(\n\"Graph has no layout information, see layout() or specify prog=%s.\"\n% (\"\|\".join([\"neato\", \"dot\", \"twopi\", \"circo\", \"fdp\", \"nop\"]))\n)\n\nelse:\nif self.number_of_nodes() > 1000:\nsys.stderr.write(\n\"Warning: graph has %s nodes...layout may take a long time.\\n\"\n% self.number_of_nodes()\n)\n\nif prog == \"nop\": # nop takes no switches\nargs = \"\"\nelse:\nargs = \" \".join([args, \"-T\" + format])\n\ndata = self._run_prog(prog, args)\n\nif path is not None:\nfh = self._get_fh(path, \"w+b\")\nfh.write(data)\nif isinstance(path, str):\nfh.close()\nd = None\nelse:\nd = data\nreturn d\n\ndef _draw(self, path=None, format=None, prog=None, args=\"\"):\n\"\"\"Output graph to path in specified format.\n\n.. caution:: EXPERIMENTAL\n\nThis version of the draw command uses libgvc for drawing instead\nof command line GraphViz tools like in versions <1.6 and the default.\n\nAn attempt will be made to guess the output format based on the file\nextension of `path`. If that fails, then the `format` parameter will\nbe used.\n\nNote, if `path` is a file object returned by a call to os.fdopen(),\nthen the method for discovering the format will not work. In such\ncases, one should explicitly set the `format` parameter; otherwise, it\nwill default to 'dot'.\n\nIf path is None, the result is returned as a Bytes object.\n\nFormats (not all may be available on every system depending on\nhow Graphviz was built)\n\n'canon', 'cmap', 'cmapx', 'cmapx_np', 'dia', 'dot',\n'fig', 'gd', 'gd2', 'gif', 'hpgl', 'imap', 'imap_np',\n'ismap', 'jpe', 'jpeg', 'jpg', 'mif', 'mp', 'pcl', 'pdf',\n'pic', 'plain', 'plain-ext', 'png', 'ps', 'ps2', 'svg',\n'svgz', 'vml', 'vmlz', 'vrml', 'vtx', 'wbmp', 'xdot', 'xlib'\n\nIf prog is not specified and the graph has positions\n(see layout()) then no additional graph positioning will\nbe performed.\n\nOptional prog=['neato'\|'dot'\|'twopi'\|'circo'\|'fdp'\|'nop']\nwill use specified graphviz layout method.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> G.add_edges_from([(0, 1), (1, 2), (2, 0), (2, 3)])\n>>> G.layout()\n\n# use current node positions, output pdf in 'file.pdf'\n>>> G.draw(\"file.pdf\")\n\n# use dot to position, output png in 'file'\n>>> G.draw(\"file\", format=\"png\", prog=\"dot\")\n\n# use keyword 'args' to pass additional arguments to graphviz\n>>> G.draw(\"test.pdf\", prog=\"twopi\", args=\"-Gepsilon=1\")\n>>> G.draw(\"test2.pdf\", args=\"-Nshape=box -Edir=forward -Ecolor=red \")\n\nThe layout might take a long time on large graphs.\n\n\"\"\"\n# try to guess format from extension\nif format is None and path is not None:\np = path\n# in case we got a file handle get its name instead\nif not isinstance(p, str):\np = path.name\nformat = os.path.splitext(p)[-1].lower()[1:]\n\nif format is None or format == \"\":\nformat = \"dot\"\n\nif prog is None:\nif self.has_layout:\nprog = \"neato\"\nargs += \" -n2\"\nelse:\nraise AttributeError(\n\"\"\"Graph has no layout information, see layout() or specify prog=%s.\"\"\"\n% (\"\|\".join([\"neato\", \"dot\", \"twopi\", \"circo\", \"fdp\", \"nop\"]))\n)\n\nelse:\nif self.number_of_nodes() > 1000:\nsys.stderr.write(\n\"Warning: graph has %s nodes...layout may take a long time.\\n\"\n% self.number_of_nodes()\n)\n\n# process args\nformat, prog = self._manually_parse_args(args, format, prog)\n\n# convert input strings to type bytes (encode it)\nif isinstance(format, str):\nformat = format.encode(self.encoding)\nif isinstance(prog, str):\nprog = prog.encode(self.encoding)\n\n# Start the drawing\ngvc = gv.gvContext()\nG = self.handle\n\n# Layout\nerr = gv.gvLayout(gvc, G, prog)\nif err:\nif err != -1:\nraise ValueError(\"Graphviz raised a layout error.\")\nprog = prog.decode(self.encoding)\nraise ValueError(f\"Can't find prog={prog} in this graphviz installation\")\n\n# Render\nif path is None:\nout = gv.gvRenderData(gvc, G, format)\nif out:\nraise ValueError(f\"Graphviz Error creating dot representation:{out}\")\nerr, dot_string, length = out\nassert len(dot_string) == length\ngv.gvFreeLayout(gvc, G)\ngv.gvFreeContext(gvc)\nreturn dot_string\n\n# path is string holding the filename, a file handle, or pathlib.Path\nfh = self._get_fh(path, \"wb\")\nerr = gv.gvRender(gvc, G, format, fh)\nif err:\nraise ValueError(\"Graphviz raised a render error. Maybe bad format?\")\nif isinstance(path, str):\nfh.close()\ngv.gvFreeLayout(gvc, G)\ngv.gvFreeContext(gvc)\n\n# some private helper functions\n\ndef _manually_parse_args(self, args, format=None, prog=None):\n\"\"\"Experimental code to parse args relevant for libgvc drawing and layout\"\"\"\narg_list = shlex.split(args)\nfor arg in arg_list:\nvalue = arg[2:]\nif arg[:2] == \"-T\":\nif format and format != value:\nraise ValueError(\"format doesnt match in args and format inputs\")\nformat = value\nif arg[:2] == \"-K\":\nif prog and prog != value:\nprog = value\n# raise ValueError(\"prog doesnt match in args and prog inputs\")\nprog = value\nif arg[:2] == \"-G\":\nkey, val = value.split(\"=\")\nself.graph_attr[key] = val\nif arg[:2] == \"-N\":\nkey, val = value.split(\"=\")\nself.node_attr[key] = val\nif arg[:2] == \"-E\":\nkey, val = value.split(\"=\")\nself.edge_attr[key] = val\nreturn format, prog\n\ndef _get_fh(self, path, mode=\"r\"):\n\"\"\"Return a file handle for given path.\n\nPath can be a string, pathlib.Path, or a file handle.\nAttempt to uncompress/compress files ending in '.gz' and '.bz2'.\n\"\"\"\nimport os\n\nif isinstance(path, str):\nif path.endswith(\".gz\"):\n# import gzip\n# fh = gzip.open(path,mode=mode) # doesn't return real fh\nfh = os.popen(\"gzcat \" + path) # probably not portable\nelif path.endswith(\".bz2\"):\n# import bz2\n# fh = bz2.BZ2File(path,mode=mode) # doesn't return real fh\nfh = os.popen(\"bzcat \" + path) # probably not portable\nelse:\nfh = open(path, mode=mode)\nelif hasattr(path, \"write\"):\n# Note, mode of file handle is unchanged.\nfh = path\nelif hasattr(path, \"open\"):\nfh = path.open(mode=mode)\nelse:\nraise TypeError(\"path must be a string, path, or file handle.\")\nreturn fh\n\ndef _which(self, name):\n\"\"\"Searches for name in exec path and returns full path\"\"\"\nimport glob\nimport platform\n\nif platform.system() == \"Windows\":\nname += \".exe\"\n\npaths = os.environ[\"PATH\"]\nfor path in paths.split(os.pathsep):\nmatch = glob.glob(os.path.join(path, name))\nif match:\nreturn match\nraise ValueError(f\"No prog {name} in path.\")\n\ndef _update_handle_references(self):\ntry:\nself.graph_attr.handle = self.handle\nself.node_attr.handle = self.handle\nself.edge_attr.handle = self.handle\nexcept AttributeError:\npass # ignore as likely still in __init__()\n\nclass Node(str):\n\"\"\"Node object based on unicode.\n\nIf G is a graph\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n\nthen\n\nwill create a node object labeled by the string \"1\".\n\nTo get the object use\n\n>>> node = pgv.Node(G, 1)\n\nor\n>>> node = G.get_node(1)\n\nThe node object is derived from a string and can be manipulated as such.\n\nEach node has attributes that can be directly accessed through\nthe attr dictionary:\n\n>>> node.attr[\"color\"] = \"red\"\n\n\"\"\"\n\ndef __new__(self, graph, name=None, nh=None):\nif nh is not None:\nn = super().__new__(self, gv.agnameof(nh), graph.encoding)\nelse:\nn = super().__new__(self, name)\ntry:\nnh = gv.agnode(graph.handle, n.encode(graph.encoding), _Action.find)\nexcept KeyError:\nraise KeyError(f\"Node {n} not in graph.\")\n\nn.ghandle = graph.handle\nn.attr = ItemAttribute(nh, 1)\nn.handle = nh\nn.encoding = graph.encoding\nreturn n\n\ndef get_handle(self):\n\"\"\"Return pointer to graphviz node object.\"\"\"\nreturn gv.agnode(self.ghandle, self.encode(self.encoding), _Action.find)\n\n# handle=property(get_handle)\n\ndef get_name(self):\nname = gv.agnameof(self.handle)\nif name is not None:\nname = name.decode(self.encoding)\nreturn name\n\nname = property(get_name)\n\nclass Edge(tuple):\n\"\"\"Edge object based on tuple.\n\nIf G is a graph\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n\nthen\n\nwill add the edge 1-2 to the graph.\n\n>>> edge = pgv.Edge(G, 1, 2)\n\nor\n>>> edge = G.get_edge(1, 2)\n\nwill get the edge object.\n\nAn optional key can be used\n\n>>> edge = pgv.Edge(G, 2, 3, \"spam\")\n\nThe edge is represented as a tuple (u,v) or (u,v,key)\nand can be manipulated as such.\n\nEach edge has attributes that can be directly accessed through\nthe attr dictionary:\n\n>>> edge.attr[\"color\"] = \"red\"\n\n\"\"\"\n\ndef __new__(self, graph, source=None, target=None, key=None, eh=None):\n# edge handle given, reconstruct node object\nif eh is not None:\ns = Node(graph, nh=source)\nt = Node(graph, nh=target)\n# no edge handle, search for edge and construct object\nelse:\ns = Node(graph, source)\nt = Node(graph, target)\nif key is not None:\nif not isinstance(key, str):\nkey = str(key)\nkey = key.encode(graph.encoding)\ntry:\neh = gv.agedge(graph.handle, s.handle, t.handle, key, _Action.find)\nexcept KeyError:\nraise KeyError(f\"Edge {source}-{target} not in graph.\")\n\ntp = tuple.__new__(self, (s, t))\ntp.ghandle = graph.handle\ntp.handle = eh\ntp.attr = ItemAttribute(eh, 3)\ntp.encoding = graph.encoding\nreturn tp\n\ndef get_name(self):\nname = gv.agnameof(self.handle)\nif name is not None:\nname = name.decode(self.encoding)\nreturn name\n\nname = property(get_name)\nkey = property(get_name)\n\nclass Attribute(MutableMapping):\n\"\"\"Default attributes for graphs.\n\nAssigned on initialization of AGraph class.\nand manipulated through the class data.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph() # initialize, G.graph_attr, G.node_attr, G.edge_attr\n>>> G.graph_attr[\"splines\"] = \"true\"\n>>> G.node_attr[\"shape\"] = \"circle\"\n>>> G.edge_attr[\"color\"] = \"red\"\n\nSee\nhttp://graphviz.org/doc/info/attrs.html\nfor a list of all attributes.\n\n\"\"\"\n\n# use for graph, node, and edge default attributes\n# atype:graph=0, node=1,edge=3\ndef __init__(self, handle, atype):\nself.handle = handle\nself.type = atype\n# get the encoding\nghandle = gv.agraphof(handle)\nroot_handle = gv.agroot(ghandle) # get root graph\ntry:\nitem = gv.agattrdefval(gv.agattr(root_handle, 0, b\"charset\", None))\nself.encoding = item if type(item) is not bytes else item.decode(\"utf-8\")\nexcept KeyError:\nself.encoding = _DEFAULT_ENCODING\n\ndef __setitem__(self, name, value):\nif name == \"charset\" and self.type == 0:\nraise ValueError(\"Graph charset is immutable!\")\nif not isinstance(value, str):\nvalue = str(value)\nghandle = gv.agroot(self.handle) # get root graph\nif ghandle == self.handle:\ngv.agattr_label(\nself.handle,\nself.type,\nname.encode(self.encoding),\nvalue.encode(self.encoding),\n)\nelse:\ngv.agsafeset_label(\nghandle,\nself.handle,\nname.encode(self.encoding),\nvalue.encode(self.encoding),\nb\"\",\n)\n\ndef __getitem__(self, name):\nitem = gv.agget(self.handle, name.encode(self.encoding))\nif item is None:\nah = gv.agattr(self.handle, self.type, name.encode(self.encoding), None)\nitem = gv.agattrdefval(ah)\nreturn item.decode(self.encoding)\n\ndef __delitem__(self, name):\ngv.agattr(self.handle, self.type, name.encode(self.encoding), b\"\")\n\ndef __contains__(self, name):\ntry:\nself.__getitem__(name)\nreturn True\nexcept:\nreturn False\n\ndef __len__(self):\nreturn len(list(self.__iter__()))\n\ndef has_key(self, name):\nreturn self.__contains__(name)\n\ndef keys(self):\nreturn list(self.__iter__())\n\ndef __iter__(self):\nfor (k, v) in self.iteritems():\nyield k\n\ndef iteritems(self):\nah = None\nwhile True:\ntry:\nah = gv.agnxtattr(self.handle, self.type, ah)\nyield (\ngv.agattrname(ah).decode(self.encoding),\ngv.agattrdefval(ah).decode(self.encoding),\n)\nexcept KeyError: # gv.agattrdefval returned KeyError, skip\ncontinue\nexcept StopIteration: # gv.agnxtattr is done, as are we\nreturn\n\nclass ItemAttribute(Attribute):\n\"\"\"Attributes for individual nodes and edges.\n\nAssigned on initialization of Node or Edge classes\nand manipulated through the class data.\n\n>>> import pygraphviz as pgv\n>>> G = pgv.AGraph()\n>>> n = pgv.Node(G, \"a\")\n>>> n.attr[\"shape\"] = \"circle\"\n>>> e = pgv.Edge(G, \"a\", \"b\")\n>>> e.attr[\"color\"] = \"red\"\n\nSee\nhttp://graphviz.org/doc/info/attrs.html\nfor a list of all attributes.\n\"\"\"\n\n# use for individual item attributes - either a node or an edge\n# graphs and default node and edge attributes use Attribute\ndef __init__(self, handle, atype):\nself.handle = handle\nself.type = atype\nself.ghandle = gv.agraphof(handle)\n# get the encoding\nroot_handle = gv.agroot(self.ghandle) # get root graph\ntry:\nitem = gv.agattrdefval(gv.agattr(root_handle, 0, b\"charset\", None))\nself.encoding = item if type(item) is not bytes else item.decode(\"utf-8\")\nexcept KeyError:\nself.encoding = _DEFAULT_ENCODING\n\ndef __setitem__(self, name, value):\nif not isinstance(value, str):\nvalue = str(value)\nif self.type == 1 and name == \"label\":\ndefault = \"\\\\N\"\nelse:\ndefault = \"\"\ngv.agsafeset_label(\nself.ghandle,\nself.handle,\nname.encode(self.encoding),\nvalue.encode(self.encoding),\ndefault.encode(self.encoding),\n)\n\ndef __getitem__(self, name):\nval = gv.agget(self.handle, name.encode(self.encoding))\nif val is not None:\nval = val.decode(self.encoding)\nreturn val\n\ndef __delitem__(self, name):\ngv.agset(self.handle, name.encode(self.encoding), b\"\")\n\ndef iteritems(self):\nah = None\nwhile 1:\ntry:\nah = gv.agnxtattr(self.ghandle, self.type, ah)\nvalue = gv.agxget(self.handle, ah)\ntry:\ndefval = gv.agattrdefval(ah) # default value\nif defval == value:\ncontinue # don't report default\nexcept: # no default, gv.getattrdefval raised error\npass\n# unique value for this edge\nyield (\ngv.agattrname(ah).decode(self.encoding),\nvalue.decode(self.encoding),\n)\nexcept KeyError: # gv.agxget returned KeyError, skip\ncontinue\nexcept StopIteration: # gv.agnxtattr is done, as are we\nreturn\n\ndef to_dict(self):\nah = None\nattrdict = {}\nwhile 1:\ntry:\nah = gv.agnxtattr(self.ghandle, self.type, ah)\nexcept StopIteration: # gv.agnxtattr is done, as are we\nbreak\nkey = gv.agattrname(ah).decode(self.encoding)\nvalue = gv.agxget(self.handle, ah).decode(self.encoding)\nattrdict[key] = value\nreturn attrdict\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5177441,"math_prob":0.74180037,"size":52790,"snap":"2022-05-2022-21","text_gpt3_token_len":14388,"char_repetition_ratio":0.17432652,"word_repetition_ratio":0.35065293,"special_character_ratio":0.29316157,"punctuation_ratio":0.22255193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9781181,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T10:20:24Z\",\"WARC-Record-ID\":\"<urn:uuid:6f8129eb-c59d-4d88-92a6-5c0a0a6ff88e>\",\"Content-Length\":\"289743\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b5baf0e-32f7-4304-8212-4402f4a6c0e6>\",\"WARC-Concurrent-To\":\"<urn:uuid:566e71b2-b29f-4e21-950b-30ca905d3bb6>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://pygraphviz.github.io/documentation/stable/_modules/pygraphviz/agraph.html\",\"WARC-Payload-Digest\":\"sha1:FYESSE6ZVAXJBV5OSSGOSBWB7SD4MPUT\",\"WARC-Block-Digest\":\"sha1:5BJS3IPWU7HTP5N3NWWUXKTEDOGHDCC2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662604794.68_warc_CC-MAIN-20220526100301-20220526130301-00001.warc.gz\"}"}
https://www.algebra-class.com/linear-inequalities.html	[ "# Graphing Linear InequalitiesPractice Problems\n\nHow are you doing with graphing inequalities? Can you remember how to draw the boundary line and which side of the half plane to shade?\n\nOn this page, you will find two practice problems for graphing inequalities. Try them on your own and see how you do.\n\nYou may need to go back and review the examples on graphing linear inequalities.\n\nBefore you begin, let's review a few things!\n\n## Steps for Graphing Linear Inequalities\n\n• Rewrite the inequality in slope intercept form if needed.\n\n• Graph a dotted line if the inequality is greater than or less than.\n\n• Graph a solid line if the inequality is greater than or equal to or less than or equal to.\n\n• Substitute an ordered pair to determine which side of the line to shade. (0,0) is the easiest ordered pair to choose, if it is not on the line.\n\n• Shade the side of the line that contains all of the solutions to the inequality.\n\n• Check your work by substituting a point to determine whether or not it is a solution.\n\n## Practice Problems\n\n1. y < -x - 4\n\n2. y > 3x + 3\n\nNow you must check your answers. If you had difficulty, check the steps below to help find where you are making your mistake. Remember that making mistakes is OK, as long as you learn from those mistakes. This is the reason why I feel that step by step answer keys are very important.\n\n## Solutions\n\nThe inequality in Example 1 is already written in slope intercept form. Therefore, you can graph the y-intercept at y = -4 and use a slope of -1 to find the next point.\n\nNotice that we used a solid boundary line since the inequality symbol is \"less than or equal to\" in this problem.", null, "All of the points in the yellow shaded area are solutions to this inequality. You can choose any point in this area, substitute those values for x and y into the original inequality, and end up with a true math statement.\n\n### Problem 2\n\nProblem 2 is already written in slope intercept form as well, so we can begin graphing the y-intercept at y = 3 and using a slope of 3 to find the next point.\n\nNotice that we have a dotted boundary line this time because the inequality symbol is \"greater than\" in this problem. This means that solutions to the inequality are NOT included on the boundary line.", null, "Since (0,0) is not a solution, we shaded the left hand side of the dotted line. This half plane contains all of the points that are solutions to this inequality.\n\nTry it! Pick any point in the shaded area and substitute the values for x and y into the original inequality. Your math sentence should be a true statement with any point in the shaded area.\n\nGreat Job! You are doing great with graphing linear inequalities. Next we will study Systems of Inequalities.", null, "" ]	[ null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 421 506'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 452 498'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 208 122.451612903226'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9408894,"math_prob":0.9716502,"size":2498,"snap":"2023-40-2023-50","text_gpt3_token_len":539,"char_repetition_ratio":0.12951082,"word_repetition_ratio":0.057777777,"special_character_ratio":0.21417134,"punctuation_ratio":0.0996016,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9873533,"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\":\"2023-12-06T00:20:00Z\",\"WARC-Record-ID\":\"<urn:uuid:0eb2ef35-030d-4681-a424-f43b40ded6cb>\",\"Content-Length\":\"26950\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6dea085b-7616-4d59-9503-e59723d77849>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f221a4a-d76e-4559-9dbc-df2e77072c74>\",\"WARC-IP-Address\":\"173.247.219.209\",\"WARC-Target-URI\":\"https://www.algebra-class.com/linear-inequalities.html\",\"WARC-Payload-Digest\":\"sha1:MYFXHH6TLNZDN5QF6ZXEJEDC7V4YTIHE\",\"WARC-Block-Digest\":\"sha1:FL542D5DRULEBXUWOK4CA6JV3VYKC4WI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00288.warc.gz\"}"}
http://samplingplans.com/tpzerousermanual.htm	[ "\| Home \| TPZero \|", null, "TPZero User Manual\n\nTool for Ac=0 Acceptance Sampling Plans", null, "Related Methods Example Input: Consumer's Point: Example Output: Decision Rule: Example Output: Probability Statement of the Consumer's Point: File Menu Design Menu Help Menu The Right mouse button pop-up menu Entering the Consumer's Point, RQL & Beta Entering the acceptance number for AC>0 Calculated Sample Size, n Points on the OC Curve Producer's Point (AQL,Alpha) Possible Fractions Defective Nomenclature Assumptions and Limitations of TPZero Range of Input variables Binomial Distribution Getting Help Buy and Register TPZero How to Contact H & H Servicco Corp. Glossary Credits: TPZero uses the following products:\n\n# What does this tool for Ac = 0 do?\n\nThis software tool calculates the sample size for Ac=0 acceptance sampling plans. The user specifies the sampling requirement by entering a Consumer's Point of the oc curve that the sampling plan should have. That is the only input required.\n\n## Related Methods\n\nThese zero acceptance number plans are usually called C=0 plans when used for manufacturing and quality control applications, and to \"Discovery Sampling\" plans when used for auditing financial records.\n\n### Example Input: a Consumer's Point:\n\nRQL = 0.10 fraction defective, Beta = 0.05\n\nn = 28, Ac = 0\n\n### Example Output: of Probability Statement describing the Consumer's Point:\n\nIf the true lot fraction defective is 0.10 or more, then the probability that a sample of n =28 will contain 0 defectives is 0.05 or less.\n\nTPZero is a simple tool with only three main menus.\n\nThe File menu will print a copy of the report on the default printer, copy it to the Windows clipboard, or exit the program.", null, "The Design menu has only one option -- to design a C=0 sampling plan.", null, "The Help menu provides help in using TPZero.", null, "## The Right mouse button pop-up menu", null, "# User Interface\n\n## Entering the Consumer's Point, RQL & Beta\n\nYou start this form from the Design Menu. [Design][Ac=0].\n\nTo calculate the sample size required, you only need to enter two values, RQL and Beta.\n\nTo determine a sample size for your application, change the example values of RQL and Beta to your values. Then push the Calculate Button.", null, "## Entering the acceptance number for AC>0\n\nYou start this form from the Design Menu. [Design][Ac>0].\n\nAs you increase the Acceptance number ( Ac) and do nothing else, the oc curve moves to the right (towards higher fraction defective)\n\nAs you move Ac up and RQL down the oc curve becomes steeper, making the plan more discriminating.\n\n## Calculated Sample Size, n\n\nWhen you push the Calculate Button, TPZero tool calculates the sample size required by the consumer's point ( RQL, Beta).", null, "### Interpretation of the output:\n\nThe Decision Rule", null, "· The sample size, n, is rounded to the nearest whole number.", null, "· The exact-n displays the value of n that exactly satisfies your consumer's point -- before rounding to the whole number, n. (Shown only for sample size less than 1,000.)", null, "· The Ac = 0 decision rule is stated.\n\n### Performance of the Sampling Plan", null, "· The exact meaning of your consumer's point is shown as a probability statement.", null, "· The producer's point is displayed and interpreted for Alpha = 0.05.\n\n# Using the OC Curve, Plot of Pa versus p'", null, "This oc curve describes the performance of the decision rule n=28, Ac=0. This plan was designed by specifying the consumer's point: RQL=0.10, Beta=0.05\n\nEvery decision rule ( n , Ac ) has an oc curve, which describes the sampling plans' probability of accepting lots having various fractions defective.\n\nThe program TPZero displays two points on the oc curve, the Consumer's Point and the Producer's Point.\n\n## Points on the OC Curve\n\n( RQL, Beta) and ( AQL, Alpha) define the consumer's and producer's points on an oc curve. Actually, any point on the oc curve, regardless of what you call it, can be entered for (RQL, Beta). TPZero will calculate the sample size for the one Ac=0 sampling plan that has an oc curve going through that point..\n\n## Producer's Point (AQL,Alpha)\n\nFor your information, TPZero shows the producers point, ( AQL and Alpha ) at Alpha = 0.05 for all sampling plans. This second point is not necessary in designing an Ac=0 sampling plan because for a given Ac, you only need to specify one point.\n\nHowever, knowing the producer's point can be useful if you need to know the lot quality that will have a high probability of acceptance with the plan.\n\nFor Alpha = 0.05, Pa = (1-Alpha) = 0.95 (95%)\n\n# Appendix\n\n## Possible Fractions Defective\n\nTPZero does not use lot size, so it can not automatically limit RQL to the possible series of values of lot p'. For any lot containing N items, the possible fractions defective are:\n\np': 0, 1/N, 2/N, 3/N ....\n\nFor example, if N=100, the possible numbers of defective items are:\n\ndefectives = 0, 1, 2, 3 ...\n\nSo the values of p' are limited to:\n\np' = 0.00, 0.01, 0.02, 0.03 ...\n\nConsequently, the probability statements will be inexact when RQL is smaller than1/N.\n\n## Nomenclature\n\nTerms and Symbols used by TPZero\n\n SYMBOL NAME Pa probability of acceptance of a lot. p' fraction defective of a lot. AQL Acceptable Quality Level. RQL Rejectable Quality Level. Alpha risk of rejecting a lot if p'=AQL. Beta risk of accepting a lot if p'=RQL. n Sample Size Exact-n Sample Size before rounding to whole number N Lot Size Ac Acceptance Number, Same as C C Acceptance number, same as Ac X The number of defective items in a sample.\n\n## Assumptions and Limitations of TPZero\n\n### Range of Input variables\n\nRQL: 0.000001 to 0.999999 (decimal fractions)\n\nBeta: 0.000001 to 0.999999 (decimal fractions)\n\nFor Ac>0 option. Ac: 0 to 700\n\n### Binomial Distribution\n\nTPZero calculates the sample size using the binomial distribution. The binomial does not use the lot size in calculating n. As such, the binomial is an approximation to the true distribution, the hypergeometric. It is a very accurate approximation and is commonly used when small samples are taken from large lots. As a Rule of Thumb, the binomial distribution is very accurate when N / n =>10. That is, when the lot is 10 times larger than the sample.\n\n## Getting Help\n\nContext sensitive help: help F1\n\nSampling plan help: www.samplingplans.com\n\nThe menu [Help][About] displays the version of TPZero, whether it is registered, and if not, how many days of free usage are left.\n\n## How to Contact H & H Servicco Corp.\n\n Website: www.samplingplans.com Discussion forum: www.samplingplans.com/forum/forum.htm Email service@samplingplans.com Phone: 651-777-0152 Mail: PO Box 9340 North St Paul, MN 55109-0340 USA\n\n## Glossary\n\nAc\nThe Acceptance Number is the maximum allowable number of defective items in the sample and still accept the lot. If the number of defectives in the sample exceeds Ac, then the lot is rejected. The symbols \"Ac\" and \"C\" are commonly used to represent the acceptance number.\n\nacceptance number\nThe acceptance number, symbol=Ac,or C, is the maximum allowable number of defectives in the sample for a lot to be accepted. For the acceptance number of zero, the lot is rejected when a sample contains one or more defective items.\n\nAlpha\nThe Producer's Risk, Alpha, is the risk of rejecting a good lot, that is, a lot that contains AQL fraction defective.\n\nAQL\nThe Acceptable Quality Level, AQL, is the fraction defective of a lot that would have a high probability of acceptance. AQL is often associated with 0.95 probability of acceptance ( Pa). AQL and Alpha define the Producer's Point.\n\nBeta\nThe Consumer's Risk, Beta, is the risk of accepting a rejectable lot, that is, a lot that contains RQL fraction defective. A typical value for the Beta risk is 0.05.\n\nC=0\nA C=0 sampling plan is one in which the lot is only accepted if zero defectives occur in the sample. Same as Ac=0.\n\nConsumer's Point\nThe Consumer's Point is the point on the oc curve defined by RQL on the X-axis and Beta on the Y-axis. Example: beta=0.05 probability of accepting a lot if its fraction defective is RQL=0.10. (10%)\n\ndiscriminating\nThe ability of a sampling plan to discriminate (tell the difference) between an AQL lot and RQL lot. A sampling plan can discriminate well when its AQL and RQL are sufficiently close together. If one sampling plan is more discriminating than another, its oc curve is steeper (more vertical).\n\nexact-n\nThe exact sample size shows the value of n -- before rounding to a whole number -- that exactly satisfies the consumer's point that you entered. The exact n can be useful to know when the sample size is small and the direction of rounding can have economic consequences.\n\nn\nn: Lower case n is the symbol for the sample size to be inspected.\n\nN: Upper case N is the symbol for the lot size (Number of items in the lot.)\n\noc curve\nThe oc curve (Operating Characteristic Curve) of a sampling plan is a graph of Pa versus p', where Pa = probability of acceptance, p' = true lot fraction defective. The oc curve tells you how the sampling plan will perform in making accept/reject decisions.\n\np'\np' is the symbol for the fraction of items in a lot that are defective.\n\nPa\nPa is the probability that a sampling plan will accept a lot. Pa depends on the decision rule (n, C) and the lot's fraction defective.( p)'\n\nprobability statement\nThese two probability statements are produced by TPZero:\n\nThe probability of a type II error:\n\nThe probability of accepting a lot if it contains RQL fraction defective\n\nThe probability of a type I error:\nThe probability of rejecting a lot if it contains AQL fraction defective.\n\nProducer's Point\nThe producer's point of the OC Curve is defined by AQL and Alpha.\n\nRQL\nThe Rejectable Quality Level, RQL, is the fraction defective of a lot that would have a low probability of acceptance. RQL and Beta define the consumer's Point of the oc curve.\n\nSample Size\nThe sample size is number of items drawn from the lot for inspection. The symbol for sample size is n, whereas the symbol for the lot size is N.\n\nTPZero\nThis program is a component of the more comprehensive program \"Sampling Plans for Attributes (TP105).\n\nTP105 allows input of BOTH consumer's point and producer's point, or (n, Ac) to calculate the Alpha and Beta risk, or to calculate AQL and RQL.\n\nTP105 also calculates, in addition to the complete oc curve, the AOQ, ARL, ASN, and AFI curves.\n\nTP105 also calculates, for any (n, Ac), the matched sequential sampling plan. Or, from two points on the oc curve, both fixed-n and sequential sampling plans.\n\n## Credits: TPZero uses the following products:\n\nThe author of TPZero\n\nStan Hilliard, H & H Servicco corp.\n\nVisual Basic V6 - Microsoft\n\nTo design and compile the program code.\n\nVB Helpwriter\n\nTo design the Help system.\n\nFabLock - Teletech Systems\n\nTo design the copy protection system\n\n\| Home \| TPZero \|" ]	[ null, "http://samplingplans.com/_vti_bin/fpcount.exe/", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/tpzerousermanual/image28.gif", null, "http://samplingplans.com/tpzerousermanual/image29.gif", null, "http://samplingplans.com/tpzerousermanual/image30.gif", null, "http://samplingplans.com/tpzerousermanual/image31.gif", null, "http://samplingplans.com/tpzerousermanual/image32.gif", null, "http://samplingplans.com/tpzerousermanual/image33.gif", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/_themes/technorg/tecbul1d.gif", null, "http://samplingplans.com/tpzerousermanual/image34.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83458704,"math_prob":0.9655208,"size":6950,"snap":"2021-21-2021-25","text_gpt3_token_len":1706,"char_repetition_ratio":0.13374604,"word_repetition_ratio":0.018456375,"special_character_ratio":0.24460432,"punctuation_ratio":0.13890858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9919036,"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,2,null,2,null,2,null,2,null,2,null,2,null,null,null,null,null,null,null,null,null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T11:53:24Z\",\"WARC-Record-ID\":\"<urn:uuid:4d521f95-97a3-4b50-96c1-42b36525dccd>\",\"Content-Length\":\"39218\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e6ac45f-ac9f-44ea-ab9d-28a8a1f17041>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c5c31e4-de44-40bd-a9cf-9779bd7063d1>\",\"WARC-IP-Address\":\"192.99.161.21\",\"WARC-Target-URI\":\"http://samplingplans.com/tpzerousermanual.htm\",\"WARC-Payload-Digest\":\"sha1:JEHO4K34OAUVSDRF4OXCNKKQLFLK6KUH\",\"WARC-Block-Digest\":\"sha1:OMOIHNMBYS3XG3GYXGRL3NFNVYJE3TWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989819.92_warc_CC-MAIN-20210518094809-20210518124809-00530.warc.gz\"}"}
https://artflexmobiliario.com.br/k2oq/qtjq	[ "# Travelling Salesman Problem Vba Code\n\nThe procedure finds the shortest tour of 20 locations randomly oriented on a two-dimensional -plane, where and. com offers free software downloads for Windows, Mac, iOS and Android computers and mobile devices. I also think there's opportunity to make the code more readable by using arrays or lists instead of so many. 5 • Nearest Neighbor Algorithm (Heuristic) Class 8. Travelling Salesman Problem example in Operation Research. It should accept data as triples in the format: (city #1, city #2, distance) , where distance is the distance in a standard unit (like miles or kilometers) between. The traveling salesman problem is important because it is NP complete. I need a distance matrix and a cost matrix. The Traveling Salesman Problem (TSP) Given a set ofcitiesalong with the cost of travel between them, ﬁnd the cheapest route visiting all cities and returning to your starting point. A General VNS heuristic for the traveling salesman problem with time windows. This is the collection of benchmark instances used in our papers Beam-ACO for the travelling salesman problem with time windows and The Travelling Salesman Problem with Time Windows: Adapting Algorithms from Travel-time to Makespan Optimization. Code for the paper 'An Efficient Graph Convolutional Network Technique for the Travelling Salesman Problem' (arXiv Pr… deep-learning combinatorial-optimization travelling-salesman-problem pytorch geometric-deep-learning graph-neural-networks. In this application we solve a large Sudoku problem using a MIP formulation. The package provides some simple algorithms and an interface to the Concorde TSP solver and its implementation of the Chained-Lin-Kernighan heuristic. TSPSG is intended to generate and solve Travelling Salesman Problem (TSP) tasks. before and the distance (a,b) is the. If we want to check a tour for credibility, we check that the tour contains each vertex once. I've a base point and 235 clients. JavaScript Console +1. It is an NP complete problem. The traveling salesman problem asks: Given a collection of cities connected by highways, what is the shortest route that visits every city and returns to the starting place? The answer has. The Travelling Salesman Challenge is returning for 2018, and this time you’re finding the cheapest route between whole areas. The cost of the transportation among the cities (whichever combination possible) is given. Travelling Salesman is a hard sci-fi story that adds one new discovery to our world and imagines the consequences for the discoverers. It is focused on optimization. It is just probably close. The exact application involved finding the shortest distance to fly between eight cities without. Can the Travelling Salesman Problem be solved in a recursive algorithm? I'm trying to learn recursion and a friend has suggested to solve this problem using recursion. Transition: The problem we just explored is actually a very famous one in computer science, technically known as the vertex cover problem. The source codes for Travelling Salesman Problem (TSP) Travelling Salesman Problem (TSP) Source Codes. The problem is to find a tour of all the cities so that the salesman ends up at the same place he started so that he visits each city exactly once and travels the shortest possible distance. If I were given a 2D array or a 2D vector with distances between the cities how can I use the data to calculate the shortest possible distance and shortest possible route?. pdf from ISEN 230 at Texas A&M University. distance) between any two cities does not. An alldifferent constraint can be used to model problems involving ordering or sequencing of choices, such as the Traveling Salesman Problem. In this case study we shall: show how to compactly define a model with Fusion;. Secondly changing the dimensions of the underlying problem (like adopting the objective function, adding new decision variables or adding new constraints) usually requires changes in the worksheet structures and the VBA code. How to formulate Traveling Salesman Problem (TSP) as Integer Linear Program (ILP)? I'm afraid that debugging your ILP code isn't on-topic for this site. In what order should she visit the cities? Let there be n cities, numbered from 1 up to n. The objective of this senior design project is to implement a divide and conquer algorithm to code a more efficient search algorithm than conventional exhaustive methods. A man who travels in a given territory to solicit business orders or sell merchandise. The code for this is available here. This is the third problem in a series of traveling salesman problems. The Traveling Salesman Problem is a well known problem which has become a comparison benchmark test for different computational methods. The Travelling Salesman Problem (TSP) is probably the most known and studied problem in Operations Research. VBA code help for looping the macro. the maximal clique problem [10, 13]. You will also learn how to handle constraints in optimization problems. The formulation uses a subtour elimination based on logic to find all subtours first, and then add appropriate eliminations constraints. io Find an R package R language docs Run R in your browser R Notebooks. Typically, these improved algorithms have been tested again on the TSP. I'm looking for a solution to the salesman problem where the algo is minimum of each row. Research has shown that this can result in significant savings, which has led to the formulation of various truck and drone routing and scheduling optimization problems. Fixed Start End Point Multiple Traveling Salesmen Problem Genetic Algorithm - Top4Download. The objective of this senior design project is to implement a divide and conquer algorithm to code a more efficient search algorithm than conventional exhaustive methods. As already mentioned in the introduction, the TSP is still very popular in Operations Research. There are two primary forms of the problem. Here is the definition of a TSP from Wikipedia: “Given a list of cities and their pairwise distances, the task is to find a shortest possible tour that visits each city exactly once. Problem StatementGiven a map with cities locations, what is the least-cost round-trip route that visits each city exactly once and then returns to the starting city. Dynamic Programming Travelling Salesman Problem - Dynamic Programming Travelling Salesman Problem - Analysis of Algorithm Video Tutorial - Analysis of Algorithm video tutorials for GATE, IES and other PSUs exams preparation and to help Mechanical Engineering Students covering Introduction, Definition of Algorithm, Space and Time Complexity, Time Complexity Big-Oh Notation, Classification, Back. It is important in theory of computations. Description Basic infrastructure and some algorithms for the traveling salesperson problem (also traveling salesman problem; TSP). ###Problem Description### The traveling salesman problem was first formulated in 1930 . Ci j =1, if i = j. Prüfer code. Transition: The problem we just explored is actually a very famous one in computer science, technically known as the vertex cover problem. The Traveling Salesman Problem for Cubic Graphs David Eppstein Computer Science Department University of California, Irvine Irvine, CA 92697-3435, USA eppstein@ics. The traveling salesman problem The traveling salesman problem, or TSP for short, is this: given a finite number of ``cities'' along with the cost of travel between each pair of them, find the cheapest way of visiting all the cities and returning to your starting point. The Traveling Salesman Problem is a well known problem which has become a comparison benchmark test for different computational methods. The origins of the travelling salesman problem are unclear. Keywords: Traveling salesman problem, Heuristic algorithm, Excel VBA 1. Traveling Salesman Problem The traveling salesman problem (TSP) is a famous difficult problem. Ahmed zhahmed@gmail. It can thus be seen that TSP's can be viewed as special cases of VRP's. The traveling salesman problem is a good example: the salesman is looking to visit a set of cities in the order that minimizes the total number of miles he travels. to-traveling-salesman/ machine of blockchain in few lines of code. This is of course done in the subtour constraint handler ConshdlrSubtour. What are the decisions to be made? For this problem, we need Excel to find out if an arc is on the shortest path or not (Yes=1, No=0). Third, it is generally believed that the with gate-based quantum computation one cannot solve in polinomial time NP complete problems. When neither constraint applies, the VRP reduces to a traveling salesman problem: if the objective is simply to minimize total distance, it is a 1-TSP [from (6. I've been testing out some ideas around the Travelling Salesman Problem using TSP and ggmap. The issue with the travelling salesman problem is that it is an NP-Hard problem. The term Branch and Bound refers to all state space search methods in which all the children of E-node are generated before any other live node can become the E-node. Problem Statement. This is outside the scope of the routing functionality supported in TatukGIS products, though a DK customer can implement their own multi-point routing algorithm into a DK developed application. When I run the program, the cost calculation is correct. TSP as integer linear programs (FULL encoding) TSP encoded using Random Keys (RANDOMKEYS encoding). Where's the Traveling Salesman for Google I wrote the code, and gave it a \"standard\" traveling salesman problem from a book. In the movie, four mathematicians confront the US government official who has just overseen their successful breakthrough in math that will enable code breaking of every communication code. In this particular case, the salesman moves from city to city by selecting the longest route possible between two points. Al final de esta descripción esta el paso a paso que encuentra en el paper. If not, cell F5 equals 0. An input is a number of cities and a matrix of city-to-city travel prices. Algorithm of tsp based on genetic algorithm (traveling salesman problem) tsp (travelling salesman problem-Traveling SalesmanProblem), is a classical NP-complete problem, namely, the worst-case time complexity as the problem grows exponentially, up to now cannot find a polynomial-time algorithm. 5 synonyms for travelling salesman: bagman, commercial traveler, commercial traveller, roadman, traveling salesman. This page contains the useful online traveling salesman problem calculator which helps you to determine the shortest path using the nearest neighbour algorithm. This is the third problem in a series of traveling salesman problems. Find helpful customer reviews and review ratings for The Traveling Salesman Problem: A Computational Study (Princeton Series in Applied Mathematics) at Amazon. ingsalesmanproblem. Third, it is generally believed that the with gate-based quantum computation one cannot solve in polinomial time NP complete problems. After using all the formulas, i get a new resultant matrix. Short description of this problem is to find the shortest path by visiting all cities only once. A salesperson wishes to visit a number of cities before returning home using the shortest possible route, whilst only visiting each city once. INTRODUCTION Travelling Salesman Problem (TSP) is a well-known problem in computer science. She knows the distance between each of the cities and wishes to minimize the total distance traveled while visiting all of the cities. A man who travels in a given territory to solicit business orders or sell merchandise. For those not. length code to the most frequent character. 10)]; if the number of vehicles to be used is specified, it is a m-TSP. Hi, Does someone have code to solve the Traveling Salesman Problem (TSP) algorithm. 260n and linear space. Traveling Salesman Problems with PyGMO¶. Why choose simulated annealing?. VBA code help for looping the macro. As in the classical travelling salesman problem (TSP), the travelling salesman problem with time windows (TSPTW) requires the specification of a Hamiltonian cycle through a graph of nodes, but adds the requirement that each node must be visited within a predefined time window. Shop Lousy T-Shirt - Traveling Salesman created by DiscreteOptimization. 1 Introduction The traveling salesman problem consists of a salesman and a set of cities. A salesperson must visit n cities, passing through each city only once, beginning from one of the city that is considered as a base or starting city and returns to it. problem more quickly when classic methods are too slow (from Wikipedia). Read honest and unbiased product reviews from our users. This is the collection of benchmark instances used in our papers Beam-ACO for the travelling salesman problem with time windows and The Travelling Salesman Problem with Time Windows: Adapting Algorithms from Travel-time to Makespan Optimization. Travelling Salesman Problem example in Operation Research. This paper explains and analyzes a new approach to the Drone Traveling Salesman Problem (DTSP) based on ant colony optimization (ACO). Source code for travelling salesman problem using hamiltonian cycle in c? But, to find the shortest path, i. It is used by the warehouse to pick orders, and we'd like the software to present the items to be picked in the most efficient. 1 Introduction The Traveling Salesman Problem (TSP) is a problem in combinatorial optimization studied in operations research and theoretical computer science. I did exactly this for a course project in 2012. The following Matlab project contains the source code and Matlab examples used for open traveling salesman problem genetic algorithm. ) This example illustrates the use of the GA procedure to solve a traveling salesman problem (TSP); it combines a genetic algorithm with a local optimization strategy. This problem is known to be NP-hard, and cannot be solved exactly in polynomial time. Problem Description. 10 This is the demonstration version (and at present the only version) of \"The Travelling Salesman\" program, written by Peter Meyer. Our problem is given a matrix costs in which costs[i][j] is the traveling cost going from node i to node j, we are asked to return the minimum cost for traveling from node 0 and visit every other node exactly once and return back to node 0. I also think there's opportunity to make the code more readable by using arrays or lists instead of so many. A Classical Traveling Salesman Problem (TSP) can be defined as a problem where starting from a node is required to visit every other node only once in a way that the total distance covered is minimized. I am hoping that i can do a good job with at least a rough version of the project in the 10 days i have[its not really limited to 10 days, could probably get a whole month to do this. It is significant because, just like the Traveling Salesman Problem, it is computationally hard to solve perfectly. prolog travelling salesman problem, Search on prolog travelling salesman problem VBA Help With Chart Object Problem: Jan 18: Problems in writing VBA code for an. To simplify parameters setting, we present a list-based simulated annealing (LBSA) algorithm to solve traveling salesman problem (TSP). For example, if the terrain from A to B was uphill, the energy required to travel from A. ) This example illustrates the use of the GA procedure to solve a traveling salesman problem (TSP); it combines a genetic algorithm with a local optimization strategy. As the number of cities gets large, it becomes too computationally intensive to check every possible itinerary. the maximal clique problem [10, 13]. Description: traveling salesman problem (dynamic regulation of) a salesman to a number of cities to sell commodities, known the distance between cities (or travel). 4D code for solving Travelling Salesman Problem. The matrix can be populated with random values in a given range (useful for generating tasks). I am trying to develop a program in C++ from Travelling Salesman Problem Algorithm. Perhaps the most famous combinatorial optimization problem is the Traveling Salesman Problem (TSP). Gendreau et al. Dynamic Programming Travelling Salesman Problem - Dynamic Programming Travelling Salesman Problem - Analysis of Algorithm Video Tutorial - Analysis of Algorithm video tutorials for GATE, IES and other PSUs exams preparation and to help Mechanical Engineering Students covering Introduction, Definition of Algorithm, Space and Time Complexity, Time Complexity Big-Oh Notation, Classification, Back. Given a distance matrix, the optimal path for TSP is found using simplex solver module. Travelling Salesman is a hard sci-fi story that adds one new discovery to our world and imagines the consequences for the discoverers. Programming Team Lecture: DP Algorithm for Traveling Salesman Problem One version of the traveling salesman problem is as follows: Given a graph of n vertices, determine the minimum cost path to start at a given vertex and travel to each other vertex exactly once, returning to the starting vertex. I did exactly this for a course project in 2012. How to formulate Traveling Salesman Problem (TSP) as Integer Linear Program (ILP)? I'm afraid that debugging your ILP code isn't on-topic for this site. The experts on this site are more than happy to help you with your problems but they cannot do your assignment/program for you. the hometown) and returning to the same city. Integer constraints have many important applications, but the presence of even one such constraint in a Solver model makes the problem an integer programming problem, which may be much more difficult to. Quiz & Worksheet - Computation's Traveling Salesman their understanding of the traveling salesman problem can use this quiz/worksheet to help. Travelling Salesman is a hard sci-fi story that adds one new discovery to our world and imagines the consequences for the discoverers. For eachsubset a lowerbound onthe length ofthe tourstherein. I need matlab code for bitonic euclidean Learn more about travelling salesman problem. Secondly changing the dimensions of the underlying problem (like adopting the objective function, adding new decision variables or adding new constraints) usually requires changes in the worksheet structures and the VBA code. The goal of the problem is, given a list of cities and their locations, to find the shortest possible path that passes through each city exactly once and ends at the starting city. Description Basic infrastructure and some algorithms for the traveling salesperson problem (also traveling salesman problem; TSP). Putting it All Together. I have done a little project in the past involving Genetic Algorithms (GAs), written in VB6 (really poorly). This is outside the scope of the routing functionality supported in TatukGIS products, though a DK customer can implement their own multi-point routing algorithm into a DK developed application. Applying the 2-opt algorithm to travelling salesman problems in C# / WPF the user interface is a simple window with buttons for opening the test problem (. Este paper contiene todas las formulas y el paso a paso que se requiere para desarrollar el algoritmo de Colonia de Hormigas. QUANTUM INSPIRED PARTICLE SWARM COMBINED WITH LIN-KERNIGHAN-HELSGAUN METHOD TO THE TRAVELING SALESMAN PROBLEM. At that time, he was a grad student working on his thesis. We introduce a new model that incorporates some ideas from existing robust optimization models - most importantly, the ability to control the model’s level of conservatism - but does so. Solving The Traveling Salesman Problem (TSP) in JAVA w/ Brute Force, Genetic Algorithms, Hill Climbing, Random Restart Hill Climbing, Nearest Neighbor, Simulated Annealing, Ant Colony Optimization,. Traveling Salesman Problems with PyGMO¶. 5 • Nearest Neighbor Algorithm (Heuristic) Class 8. The goal of the problem is, given a list of cities and their locations, to find the shortest possible path that passes through each city exactly once and ends at the starting city. Since I am new to VBA I could use some help. TSP can be described as follows: Given a number of cities to visit. And, for the case of trips regularly planned, e. It has long been known to be NP-hard and hence research on developing algorithms for the TSP has focused on approximate methods. In this question, you will write some code that would be used in solving this problem. The traveling salesman problem is NP-complete. The exact application involved finding the shortest distance to fly between eight cities without. In Computer Science, The Problem Can Be Applied To The Most Efficient Calculation. The spreadsheet uses VBA code to call GAMS and to read and write GDX files to exchange data with GAMS. In this example, we consider a salesman traveling in the US. com offers free software downloads for Windows, Mac, iOS and Android computers and mobile devices. Traveling Salesperson Problem Consider a traveling salesperson who must visit each of 20 cities before returning home. Hence a different order is a different value of the variable (e. Join ResearchGate to discover and stay up-to-date with the latest research from leading experts in Travelling Salesman Problem and many other. Personalize it with photos & text or purchase as is! One of the best solution to the Traveling Salesman problem 6, courtesy of mkal (a. In this paper, we examine a variant of the symmetric traveling salesman problem in which travel time uncertainty is modeled by interval ranges. A magazine contest in 1964 challenged its readers to find the optimum route for a traveling salesman who was to visit 33 specific cities in the United States. These are all greedy algorithms that give an approximate result. It is Orienteering Problem in which visiting all of the nodes are not necessary. It uses Branch and Bound method for solving. Find node r such that c ir is minimal and form sub-tour i-r-i. The TSP problem is probably the most studied problem in the OR field. Adopting declaration statements would take a few seconds in C# or C++ source code. I followed instructions, copy-typed from books, and made minor adjustments to. In Computer Science, The Problem Can Be Applied To The Most Efficient Calculation. If you want to see working code for a travelling salesman problem, look at this example. In this article, we propose a new crossover operator for traveling salesman problem to minimize the total distance. Approximation Algorithms for the Traveling Salesman Problem. He knows the distance between each pair of cities, and wishes to minimize the total distance he is to travel. Using RouteXL is very easy. Travelling Salesman Problem using Branch and Bound Approach in PHP - July 5, 2016 Javascript: Print content of particular div - March 4, 2015 Group by with Order Desc in MySQL - June 3, 2014. It is described as follows: a salesman, in order to sell his goods out, needs to arrive to n cities in proper order to sell products, and the distance between each city is known. mTSP: The mTSP is defined as: In a given set of nodes, let there are m salesmen located at a single depot node. Second, the TSP is solved. As these Japanese researchers demonstrated, a certain type of amoeba can be used to calculate nearly optimal. edu Abstract We show how to ﬁnd a Hamiltonian cycle in a graph of degree at most three with n vertices, in time O(2n/3) ≈ 1. Secondly changing the dimensions of the underlying problem (like adopting the objective function, adding new decision variables or adding new constraints) usually requires changes in the worksheet structures and the VBA code. Free traveling salesman problem download - traveling salesman problem script - Top 4 Download - Top4Download. 1s) and use this script on your gpx-files before osm-upload. The code for Concorde. The origins of the travelling salesman problem are unclear. As already mentioned in the introduction, the TSP is still very popular in Operations Research. Genetic algorithm is a kind of evolutionary. When I run the program, the cost calculation is correct. The salesman has to visit each one of the cities starting from a certain one (e. The Traveling Salesman Problem (TSP) The travelling salesman problem, which was first formulated in 1930, asks the following question: Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city?\". More info can be found here. He selected from a resident, after each city again, the last resident to return to the line, making the total distance (or travel) is the smallest. Approximation Algorithms for the Traveling Salesman Problem. A simple process: convert an image to black and white, sample the black points, and then solve the Traveling Salesman Problem for those points. The parameters for toll k are given as x_k and y_k. Find node r such that c ir is minimal and form sub-tour i-r-i. Multiple Turbo Code; multiple twin quad. before and the distance (a,b) is the. io Find an R package R language docs Run R in your browser R Notebooks. The routes are calculated using cartesian coordinates and Euclidean distances. First, the expression data is converted to a Traveling Salesman Problem (TSP), based on the desired number of clusters. Start with a sub-graph consisting of node i only. Find more Mathematics widgets in Wolfram\|Alpha. In this particular case, the salesman moves from city to city by selecting the longest route possible between two points. A JAVA IMPLEMENTATION OF THE BRANCH AND BOUND ALGORITHM: THE ASYMETRIC TRAVELING SALESMAN PROBLEM 156 JOURNAL OF OBJECT TECHNOLOGY VOL. Description Basic infrastructure and some algorithms for the traveling salesperson problem (also traveling salesman problem; TSP). among all. The Travelling Salesman Problem. After using all the formulas, i get a new resultant matrix. Introduction. Recently, Narayanan and Zorbalas proposed a method to represent weights in DNA codes to solve traveling salesman problems. Third, it is generally believed that the with gate-based quantum computation one cannot solve in polinomial time NP complete problems. The main task of a constraint handler is to decide whether a given solution is feasible for all constraints of the constraint handler's type (i. There are K toll booths. Here we solve a small instance as a MIP. The Traveling salesman problem is the problem that demands the shortest possible route to visit and come back from one point to another. The origins of the travelling salesman problem are unclear. Rational numbers class. The Simulated Annealing Method for the Traveling Salesman Model demonstrates the use of the \"simulated annealing algorithm\" to attempt to solve the \"travelling salesman\" problem. A traveler needs to visit all the cities from a list, where distances between all the cities are known and each city should be visited just once. Read honest and unbiased product reviews from our users. He wishes to visit all cities exactly once and return back to city 0. JavaScript Console +1. edu Abstract We show how to ﬁnd a Hamiltonian cycle in a graph of degree at most three with n vertices, in time O(2n/3) ≈ 1. And, for the case of trips regularly planned, e. I followed instructions, copy-typed from books, and made minor adjustments to. The traveling-salesman problem is that of finding a permutation P = (1 i2 i3 … in) of the integers from 1 through n that minimizes the quantity \\documentclass{aastex} \\usepackage{amsbsy} \\usepackag. Description Basic infrastructure and some algorithms for the traveling salesperson problem (also traveling salesman problem; TSP). Travelling Salesman Problem using Branch and Bound Approach in PHP - July 5, 2016 Javascript: Print content of particular div - March 4, 2015 Group by with Order Desc in MySQL - June 3, 2014. In Computer Science, The Problem Can Be Applied To The Most Efficient Calculation. I am trying to develop a program in C++ from Travelling Salesman Problem Algorithm. Travelling Salesman Problem. pdf from ISEN 230 at Texas A&M University. It is focused on optimization. RouteXL is a Google Maps route planner that can help you solve the 'travelling salesman problem' of finding the optimum route for multiple stops. Tropofy is an innovative web deployment platform for problem solvers. Third, it is generally believed that the with gate-based quantum computation one cannot solve in polinomial time NP complete problems. 4D code for solving Travelling Salesman Problem. I want to show the resulted path (after solving) in a scatter-plot contains all nodes and connects some of nodes (not all of them) and not just with lines but with arrows. Operations Research, 43, 330 - 335. The TSP problem is. The traveling salesman problem asks: Given a collection of cities connected by highways, what is the shortest route that visits every city and returns to the starting place? The answer has. Problem Statement. The Simulated Annealing Method for the Traveling Salesman Model demonstrates the use of the \"simulated annealing algorithm\" to attempt to solve the \"travelling salesman\" problem. Permutation Variables and Traveling Salesman Problem • Permutation- an ordered list of the numbers 1 to N. \" The problem seems very interesting. The goal is to minimize the total distance traveled. In this post, Travelling Salesman Problem using Branch and Bound is discussed. These instances are available from different sources, sometimes along with instructions on how to. In the ACS, a set of cooperating agents called ants cooperate to find good solutions to TSP’s. TSP is an extension of the Hamiltonian circuit problem. Given a list of cities and their pair wise distances, the task is to find a shortest possible tour that visits each city exactly once. Win a trip around the world based on an algorithm you write! This is your chance to get your hands on something that’s normally purely theoretical and have some fun with it. Start with a sub-graph consisting of node i only. This post shows how to attack Traveling Salesman Problem using Darwin approach. He knows the distance between each pair of cities, and wishes to minimize the total distance he is to travel. You will learn how to code the TSP and VRP in Python programming. This is an implementation of TSP using backtracking in C. The Traveling Salesman Problem (TSP) is one of the. travelling salesman problem, met heuristics, ant colony optimization 1. The traveling salesman problem is a good example: the salesman is looking to visit a set of cities in the order that minimizes the total number of miles he travels. Asymmetric TSP allows for distances between nodes to be unequal. traveling salesman problem (TSP). DMCA Notice ADA Compliance Honor Code For. Have you used a traveling salesman algorithm to solve a problem? I studied TSP in college in the context of NP Completeness. A little bit of research shows that it has been use…. DMCA Notice ADA Compliance Honor Code For. 1 Introduction The Traveling Salesman Problem (TSP) is a problem in combinatorial optimization studied in operations research and theoretical computer science. This generates a solution only, and doesn't do any input/output. The Travelling Salesman Problem (TSP) is probably the most known and studied problem in Operations Research. Continuous Line Drawings via the Traveling Salesman Problem Robert Bosch and Adrianne Herman Dept. The traveling salesman problem asks: Given a collection of cities connected by highways, what is the shortest route that visits every city and returns to the starting place? The answer has. optimization of Traveling Salesman problem is NP hard…. • Wrote a MATLAB code that could solve a simplified traveling salesman problem, and integrated the code into a graphical user interface that animated the results. There is a wealth of information on genetic algorithms available online. It calculates the shortest path between cities. The Christofides Approximation Algorithm, therefore, helps to achieve 3/2 approximation to the Travelling Salesperson Problem Algorithm. (View the complete code for this example. Traveling salesman problem (TSP) is a classical algorithm problem. Wikipedia defines the \"Traveling Salesman Problem\" this way:. Short description of this problem is to find the shortest path by visiting all cities only once. In this case, the SUBMIT block calls the SGPLOT procedure to display intermediate results during the solution of an instance of the traveling salesman problem (TSP). The algorithm we will use is called the Nearest Neighbor Algorithm. There are many applications, like finding the fastest way of moving a robot arm between. JavaScript Console +1. (View the complete code for this example. Hi, Does someone have code to solve the Traveling Salesman Problem (TSP) algorithm. Using commercial IP software, and a short (60line long) MATLAB code, students can optimally solve instances with up to 70cities in a few minutes by adding cuts from the stronger formulation to the weaker. I'm also given the time ( in minutes ) to reach from various clients and the base. The exact application involved finding the shortest distance to fly between eight cities without. The African Buffalo Optimization builds a mathematical model from the behavior of this animal and uses the model to solve 33 benchmark symmetric Traveling Salesman's Problem and six difficult asymmetric instances from the TSPLIB. A traveler needs to visit all the cities from a list, where distances between all the cities are known and each city should be visited just once. It is just probably close. The Traveling Salesman Problem (TSP) and its allied problems like Vehicle Routing Problem (VRP) are one of the most widely studied problems in combinatorial optimization. Asymmetric TSP allows for distances between nodes to be unequal. It is significant because, just like the Traveling Salesman Problem, it is computationally hard to solve perfectly. The Traveling Salesman Problem is a complex multiple point routing problem requiring a specialized, and typically expensive, algorithm to resolve. 1 edition of The traveling salesman found in the catalog. The algorithm we will use is called the Nearest Neighbor Algorithm. Charitable threads donate to threads that are out of work. Putting it All Together. The formulation uses a subtour elimination based on logic to find all subtours first, and then add appropriate eliminations constraints. All these improved versions of AS have in common a stronger exploita-. Visual Basic 4 / 5 / 6 Forums on Bytes. Java Program to Implement Traveling. The experts on this site are more than happy to help you with your problems but they cannot do your assignment/program for you. in OPL model and script for Symmetric Travelling Salesman Problem, there is code: tuple Subtour {int size; int subtour[Cities travelling salesman. Al final de esta descripción esta el paso a paso que encuentra en el paper. 60 CPU-seconds later, I got the. Travelling salesman problem: genetic algorithm (with demo) - Algorithms and Data Structures Algorithms and Data Structures. Algorithms and data structures source codes on Java and C++. VBA code help for looping the macro. This is the third problem in a series of traveling salesman problems." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90401983,"math_prob":0.8937213,"size":35396,"snap":"2019-35-2019-39","text_gpt3_token_len":7249,"char_repetition_ratio":0.20586573,"word_repetition_ratio":0.32761973,"special_character_ratio":0.19253588,"punctuation_ratio":0.09440836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9709727,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T01:11:51Z\",\"WARC-Record-ID\":\"<urn:uuid:2044c4ec-b221-47eb-b627-5c4f328dae13>\",\"Content-Length\":\"46609\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e8f8fd1-b44f-4c6e-93ac-8e5faf544a08>\",\"WARC-Concurrent-To\":\"<urn:uuid:0cdd512c-5637-401c-98a1-9ffb6895da68>\",\"WARC-IP-Address\":\"177.234.153.17\",\"WARC-Target-URI\":\"https://artflexmobiliario.com.br/k2oq/qtjq\",\"WARC-Payload-Digest\":\"sha1:PKOZZOA2EYWMFMMCYHE3DMIZRN3G7LEJ\",\"WARC-Block-Digest\":\"sha1:JOUXNOFGGMNYWSOBMFE6EL4UM7I34MDK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027322160.92_warc_CC-MAIN-20190825000550-20190825022550-00419.warc.gz\"}"}
https://abakbot.com/en/algebra-en/polynom-tangent-en	[ "Polynomial elements Variable X at which we find values of a derivative\n Given function\n\nConsider one of the simple and undeservedly forgotten on the Internet Internet methods for determining the derivative of a polynomial, an arbitrary (positive) degree.\n\nUntil recently, I was sure that if a polynomial of the form\n\nand it is necessary to find out the value of a derivative, for example, of the 5th order at some point, you must first calculate this derivative (of the fifth order), and then substitute the value, calculate the derivative.\n\nIt turns out there is a simpler and algorithmically easier way to find the derivative at a point.\n\nTo do this, we need the technique described in the materials: Expand the polynomial in degrees and Horner method online calculator. Division of a polynomial.\n\nYes, yes, it turns out the Horner method successfully solves the problem.\n\nConsider an example:\n\nCalculate the third-order derivative for x = 3 of the next polynomial\n\n1. Divide the given polynomial by\n\nGet and the remainder of 19.\n\nThe number 19 is the value of the function if we substitute x = 3 there\n\n2. Divide again on\n\nGet and the remainder of 25.\n\nSince this is the first result, we multiply the result by 1! (One factorial) = 1. Got the same number 25\n\nThe number 25 is the value of the first derivative of the given function for x = 3. That is, if we calculate the first derivative\n\nand substitute the value 3 there, we get the same answer = 25.\n\n3. Divide again on\n\nwe get and residue 13.\n\nMultiply this number by 2! (two factorial) = 2 and we get the value of the derivative of the second order function for x = 3\n\nThis number = 26\n\n4. The third-order derivative is calculated in this case simply, since it’s impossible to divide further, this is the remainder. It must be multiplied by 3! (Three factorial) = 6\n\nAnd we get that the third-order derivative for a given polynomial for x = 3 is 12.\n\nIn such a straightforward way, we can find the values of any derivative of any polynomial.\n\nThe algorithm is simple, but with polynomials with degrees above 10, we are faced with the need to calculate factorials above 10, which is very laborious, since the factorial from 10 is 3628800 and the factorial from 16 is already 20922789888000\n\nBut we benefit from one of the properties of Horner's methodology, which states: If we multiply a function by a number, then the remainder of the branch will increase by the same amount.\n\nTherefore, it is enough for us to multiply the obtained coefficients of the polynomial by dividing by the numbers 1,2,3,4,5, etc. depending on which derivative we’ll calculate at the moment and calculate the remainder.\n\nThe calculator also works in the field of complex numbers, so let's solve this example.\n\nThere is a function\n\nIt is necessary to find out all possible derivatives of this function for x = i\n\nIt is easy to make sure that solving it manually, you can make a mistake and go the wrong way.\n\nIt is much easier to use the bot and write through XMPP client\n\npropol 2 1-5i 0 -7 i 2 -9 -1; i\n\nand we will get all the results\n\nThe polynomial derivative values are found\n0 derivative. Function Value -10-6i\n1 derivative. Function Value 7 + 35i\n2 derivative. Function Value 112-66i\n3 derivative. Function Value -180-282i\n4 derivative. Function Value -528 + 120i\n5 derivative. Function Value -1440 + 720i\n6 derivative. Function Value 720 + 6480i\n7 derivative. Function Value 10080\n\nThe logical question is what is the zero derivative?\nAnswer - this is the original function. And the value -10-6i is obtained if we substitute -i into the original function\n\nLet's try to solve another equation\nwe know what the fourth derivative of the function is equal to\nfor x = 2 + i\n\nA polynomial of the 17th degree .. this is serious as well as computation with a complex argument.\nWell try\nPreset function\n Derivative The value of the derivative at X = 2 + i 0 707043 + 6123674i one 25630678 + 39273242i 2 289802562 + 169486216i 3 2247959580 + 147950190i 4 13006113720-5465417040i 5 53432793120-62240220840i 6 107126132400-427018989600i 7 -468058852800-2114656795440i 8 -6101588908800-7522728998400i nine -35506871769600-16099283692800i 10 -1.393813225728E + 14 + 5293047513600i eleven -3.828579156864E + 14 + 2.0995438464E + 14i 12 -6.6691392768E + 14 + 9.6332011776E + 14i thirteen -3.705077376E + 14 + 6.1024803840002E + 14i 14 1.4820309504E + 15 + 7.8460462080004E + 14i fifteen 5.2306974720004E + 14 + 5.230697472E + 14i sixteen 3.1384184832005E + 14 + 1.0461394944E + 14i 17 24.89811996672https://abak.pozitiv-r.ru\n\nwhen x = 2 + i, the value of the function when taking the fourth derivative will be\n 4 13006113720-5465417040i\nWhat else can you notice?\nWhat you need to carefully look at the calculations.\nIn our example, when taking 17 winding, the number 24.898 is obtained\nalthough it should certainly be where is 17! this is the factorial from 17 = 355687428096000\nThis small flaw (an error in calculating large derivatives) will be eliminated soon. But the calculation of derivatives is not higher than 10 orders of magnitude, the bot performs correctly.\n\nGood luck!\n\nCopyright © 2021 AbakBot-online calculators. All Right Reserved. Author by Dmitry Varlamov" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89641625,"math_prob":0.99386275,"size":3089,"snap":"2021-31-2021-39","text_gpt3_token_len":730,"char_repetition_ratio":0.15948136,"word_repetition_ratio":0.010830325,"special_character_ratio":0.24538685,"punctuation_ratio":0.09886548,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9995598,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-26T05:19:48Z\",\"WARC-Record-ID\":\"<urn:uuid:e6201378-cba4-4f9c-89ca-99e7d7c9d3d3>\",\"Content-Length\":\"36960\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2285fa81-d429-470c-a8bb-f4c4b29102f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd4f91b0-7a0b-482d-ae00-ee86b1c33d09>\",\"WARC-IP-Address\":\"31.28.24.244\",\"WARC-Target-URI\":\"https://abakbot.com/en/algebra-en/polynom-tangent-en\",\"WARC-Payload-Digest\":\"sha1:P7Z3TMQHJEKJ3XHFUVYRJIFMBF5UYPNH\",\"WARC-Block-Digest\":\"sha1:BYVZQ44VHNIE3T7GJGK3DDI6BMROW2VY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152000.25_warc_CC-MAIN-20210726031942-20210726061942-00057.warc.gz\"}"}
https://ask.learncbse.in/t/relative-density-of-gold-is-19-3-if-the-density-of-water-is-1000-kg-m3-what-is-the-density-of-gold-in-si-units/64514	[ "", null, "# Relative density of gold is 19.3. If the density of water is 1000 kg/m3. What is the density of gold in SI units?\n\nRelative density of gold is 19.3. If the density of water is 1000 kg/m3. What is the density of gold in SI units ?\n\nRelative density = Density of Substance ÷ Density of Water\n\nRelative density = 19.3\n\nDensity of Water = 1000 kg/m³\n\nDensity of Gold = Density of Water ×Relative density\n\n`````` = 1000 × 19.3\n\n= 19300 kg/m³``````" ]	[ null, "https://ask.learncbse.in/images/discourse-logo-sketch.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84195393,"math_prob":0.9989109,"size":422,"snap":"2021-21-2021-25","text_gpt3_token_len":125,"char_repetition_ratio":0.28947368,"word_repetition_ratio":0.4556962,"special_character_ratio":0.3364929,"punctuation_ratio":0.10869565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99925125,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T16:15:31Z\",\"WARC-Record-ID\":\"<urn:uuid:4ae7c503-7d96-4cc6-bca7-d4a449f52ca9>\",\"Content-Length\":\"12553\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a78aac33-36bc-4f62-8d37-f02c4389644f>\",\"WARC-Concurrent-To\":\"<urn:uuid:e4650a12-101c-4b7b-8296-a97de674651e>\",\"WARC-IP-Address\":\"46.165.252.193\",\"WARC-Target-URI\":\"https://ask.learncbse.in/t/relative-density-of-gold-is-19-3-if-the-density-of-water-is-1000-kg-m3-what-is-the-density-of-gold-in-si-units/64514\",\"WARC-Payload-Digest\":\"sha1:F3BX4R3557QCBCOSA57PI6NOJE7LSXHC\",\"WARC-Block-Digest\":\"sha1:SYD2IRZQLFC3R4EM3DGLM236S2ILJSAS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988882.94_warc_CC-MAIN-20210508151721-20210508181721-00171.warc.gz\"}"}
https://puckon.net/articles/preweighting-and-playoffs.php	[ "### Preweighting and Playoffs\n\nMar 20, 2017\n\nAs per my last Corsi improvement, the adjusted Corsi formula has been getting rather unwieldy. It's not likely to get much more predictive without some major revisions, and even then it's a bit of a pain to work with. There is, however, an alternate form that's slightly less predictive but a little more malleable. To show this I'm going to take a step back to the Corsi SA formula, which is the following:\n\n$$C_{SA} = {\\sum \\limits_{n=down3}^{up3} {({F_{n} + A_{n}})({{F_{n}} \\over {F_{n} + A_{n}}} - {{F_{avg_n}} \\over {F_{avg_n} + A_{avg_n}}})} \\over \\sum \\limits_{n=down3}^{up3} F_{n} + A_{n}} + 50\\%$$\n\nOn a basic level, what this formula is doing is calculating the difference between a team's performance and average league performance in each score state, and then weighting that difference by the number of the team's events in that state. An alternative approach is to just weight each event by its rarity in that state:\n\n$$C_{SA} = {\\sum \\limits_{n=down3}^{up3} {F_{n} {{F_{avg_n} + A_{avg_n}} \\over {F_{avg_n}}}} \\over \\sum \\limits_{n=down3}^{up3} F_{n} {{F_{avg_n} + A_{avg_n}} \\over {F_{avg_n}}} + A_{n} {{F_{avg_n} + A_{avg_n}} \\over {A_{avg_n}}}}$$\n\nThese formulas aren't exactly equal, but they're pretty close. The main advantage to the latter formula is that the weight terms, $$W_{F_n} = {{F_{avg_n} + A_{avg_n}} \\over {F_{avg_n}}}, W_{A_n} = {{F_{avg_n} + A_{avg_n}} \\over {A_{avg_n}}}$$ are terms that we can calculate using past seasons, essentially making them constants, and once we do it really simplifies the formula down to\n\n$$C_{SA} = {\\sum \\limits_{n=down3}^{up3} {F_{n} W_{F_n}} \\over \\sum \\limits_{n=down3}^{up3} F_{n} W_{F_n} + A_{n} W_{A_n}}$$\n\nThis form is also applicable to venue and event adjustment by splitting the data accross those vectors and similarly calculating a weight for them. Score adjustment creates 7 weights, venue adjustment creates 2 weights, and event adjustment creates 4 weights, totalling a multiplicative 56 weights if we use all three adjustments. I've contemplated using this form previously, but it does have a small correlational hit to it when compared to the previous formula.\n\nGenerally speaking, the calculated versions of these metrics take a little longer to mature; before about 15 games played per team (or 225 games league wide), the preweighted versions outperform the calculated ones. After that, however, there's a slight benefit to using the calculated form. My guess on that behavior is this: before 15 games, not only has the data not converged for individual teams, but it hasn't converged for the league as a whole. After that point, preweighting introduces a slight lopsidedness to the formula as it slightly overrepresents teams that shoot more and underrepresents teams that shoot less, which introduces some noise to the weights.\n\nThe difference between these two methods is so small - an averaged 0.18 percentage points in correlational R2 - that it can effectively be ignored. I'm not particularly interested in arguing which method is better as they both perform similarly. That we employ these adjustments far outperforms the specifics as to how they are adjusted. Preweighting isn't new either - I think most other sites that use adjusted numbers use the preweighted method. That being said, I've modified the site to employ the preweighted formulas for any dataset that includes less than 225 games total, or about 15 games per team. This decision is more aesthetic than anything as the preweighted formulas tend to vary a little less over smaller data sets.\n\nWhat preweighting does allow me is a little more confidence in posting playoff numbers. Historically I've considered that playoffs are too small a data set to adjust - As Corsi doesn't reach it's most predictive state until after around 225 games, playoffs would never reach that number given that even if all series went to a game 7, there would only be 105 games.\n\nThe site now has an additional filter that lets you select if you want to chose regular season games, playoff games, or both. Note that if you chose any option that includes playoff games, the \"Points\" column will be replaced by a Win-Loss-OTL \"Record\" column, as points are not awarded in the playoffs. I personally haven't looked much at the playoff numbers, but they've been requested by quite a few of my viewers, so I'm hoping that making them available will allow others to find some insight." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9538288,"math_prob":0.9884093,"size":4422,"snap":"2023-40-2023-50","text_gpt3_token_len":1117,"char_repetition_ratio":0.12494341,"word_repetition_ratio":0.0056737587,"special_character_ratio":0.26028946,"punctuation_ratio":0.07777778,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811701,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T22:27:54Z\",\"WARC-Record-ID\":\"<urn:uuid:191e7097-af5c-4a37-9d99-141a032a2f9a>\",\"Content-Length\":\"9276\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f107650-185f-4090-b07a-0ea58df93d8b>\",\"WARC-Concurrent-To\":\"<urn:uuid:28018c3c-daf2-4b4b-b7eb-48dbaa8e6abe>\",\"WARC-IP-Address\":\"52.36.42.173\",\"WARC-Target-URI\":\"https://puckon.net/articles/preweighting-and-playoffs.php\",\"WARC-Payload-Digest\":\"sha1:AEMENKORWNOHW6T3A3ZFPKI3GUBOPHW4\",\"WARC-Block-Digest\":\"sha1:4SCDWAR45AG75TVPIWAHP7SO267HPXNV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510326.82_warc_CC-MAIN-20230927203115-20230927233115-00220.warc.gz\"}"}
https://www.jpost.com/conferences/article-725380	[ "(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] \|\| {}, a[c].trigger = a[c].trigger \|\| function () { (a[c].trigger.arg = a[c].trigger.arg \|\| []).push(arguments)}, a[c].on = a[c].on \|\| function () {(a[c].on.arg = a[c].on.arg \|\| []).push(arguments)}, a[c].off = a[c].off \|\| function () {(a[c].off.arg = a[c].off.arg \|\| []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96655923,"math_prob":0.96919554,"size":273,"snap":"2022-40-2023-06","text_gpt3_token_len":62,"char_repetition_ratio":0.10408922,"word_repetition_ratio":0.9047619,"special_character_ratio":0.1941392,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789482,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T13:51:59Z\",\"WARC-Record-ID\":\"<urn:uuid:f4f7ae36-5f5c-466d-9115-7ab17072a6ba>\",\"Content-Length\":\"53370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85ad6840-2769-4166-b7c6-21180a92d408>\",\"WARC-Concurrent-To\":\"<urn:uuid:075a1a49-f9f2-45b8-a283-93e7f308b274>\",\"WARC-IP-Address\":\"104.22.43.245\",\"WARC-Target-URI\":\"https://www.jpost.com/conferences/article-725380\",\"WARC-Payload-Digest\":\"sha1:AG3UCNA46UWHDUQVRJIHYY53HNY3F5XE\",\"WARC-Block-Digest\":\"sha1:I75SVMQBBJV4YIEUYSRRTW2EPKTIILWM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500028.12_warc_CC-MAIN-20230202133541-20230202163541-00185.warc.gz\"}"}
https://computational-communication.com/forum/ergm-python/	[ "Exponential random graph models (ERGMs) are statistical models for explaining the structure of social and other networks (also called graphs). If we have a network, and some hypotheses about the factors that make it looks the way it does, an ERGM is meant to tell us how big a role (if any) these factors actually play.\n\nI think of it as a regression model: we have several independent variables we think determine the dependent variable, and the model estimates the size and significance of the effect of each.\n\nFor someone like me who works a lot with networks, ERGMs can be a very powerful tool. There is even an excellent R package called, appropriately enough, ergm, that makes estimating ERGMs just about as easy as specifying a regression formula.\n\nI’ve always felt a bit nervous about using them, though, because I didn’t feel confident I really understood how they worked, and how they were being estimated.\n\n# Toy ERGM from Scratch\n\nTo help with that, I decided I needed to implement a simple toy ERGM from scratch. It didn’t need to be fast, or really applicable at all. The most important thing was to take the methodology apart and put it back together again.\n\nIn this post, I’m going to walk through my implementation of a highly simplified ERGM estimation, using Python and the NetworkX package. I’m hoping that it will help others with similar questions, and that people who know more than I do will point out anything I’m getting wrong.\n\nimport random\nimport numpy as np\nimport networkx as nx\n%matplotlib inline\nimport matplotlib.pyplot as plt\nimport matplotlib\n\n\n# Graph probabilities\n\nThe basic idea of ERGMs is to define a probability distribution over all possible graphs of a given number of nodes, where the probability of each graph is proportional to some of its network statistics. In this post, the statistics I’ll use throughout are the edge count and the number of triangles, but any network statistics will do. If edge count and triangles have coefficients a and b, then the probability of observing a particular graph G is:\n\n$Pr(G)∝a∗edges(G)+b∗triangles(G)$\n\nor more specifically:\n\n$Pr(G)∝e^{a∗edges(G)+b∗triangles(G)}$\n\n(hence the exponential).\n\nThink of this as the ‘weight’ we put on a particular network. In Python, this looks like:\n\ndef compute_weight(G, edge_coeff, tri_coeff):\n'''\nCompute the probability weight on graph G\n'''\nedge_count = len(G.edges())\ntriangles = sum(nx.triangles(G).values())\nreturn np.exp(edge_count * edge_coeff + triangles * tri_coeff)\n\n\nTo turn that weight into a probability, we need to normalize it by the weights of all other possible networks. Mathematically:\n\n$Pr(G)=\\frac{exp(a∗edges(G)+b∗triangles(G))}{∑_{g∈G}exp(a∗edges(g)+b∗triangles(g))}$\n\nNow we come to the first problem: calculating the denominator of that equation. There are many possible graphs. To take the simplest example, an undirected graph with 3 nodes has 8 possible configurations:", null, "There are 64 possible configurations for a 4-node graph, and the number of possible configurations grows mind-bogglingly quickly after that. Most of the time, we simply can’t calculate the weights for all possible graphs. Instead, we need some sort of approximation.\n\nWe could just generate some large number of random graphs, and use that – but we have no way of knowing how representative they are of the actual distribution. We need a way to generate a sample that we think covers the most probable area of the distribution. In terms of the equation above, we want to sample heavily from networks where the weights are large, and less where the weights are small, so that the sum of the weights of the sample gets as close as possible to the actual total weight. Which brings us to the next step:\n\n# Markov Chain Monte Carlo\n\nMarkov Chain Monte Carlo (MCMC) actually refers a broad class of techniques for generating samples from complicated distributions. For our purposes, it means exploring the space of possible networks, moving toward the ones which are more likely given the distribution coefficients.\n\nWe’ll implement a simple version of the Metropolis-Hastings algorithm. We start with some network, and randomly add or subtract an edge.\n\ndef permute_graph(G):\n'''\nReturn a new graph with an edge randomly added or subtracted from G\n'''\nG1 = nx.copy.deepcopy(G)\nd = nx.density(G1)\nr = random.random()\nif (r < 0.5 or d == 0) and d != 1:\nnodes = G.nodes()\nn1 = random.choice(nodes)\nn2 = random.choice(nodes)\nelse:\n# Remove an edge\nn1, n2 = random.choice(G1.edges())\nG1.remove_edge(n1, n2)\nreturn G1\n\n\n\nAfter the change, we check the weight on this new network.\n\n• If it’s greater than the previous network’s weight (that is, it’s more probable under this distribution), we accept it and add it to our sample; it is also our new ‘current’ network.\n• Otherwise, we might still accept it, deciding randomly based on the ratio between the new and old weights.\n\nWe repeat this many times, until we decide we have enough samples. The starting network can be any graph – I found that a random network worked well for my relatively small sample sizes, but we can also use the observed network itself, or a completely empty one. In theory, it shouldn’t matter after enough iterations.\n\ndef mcmc(G, edge_coeff, triangle_coeff, n):\n'''\nUse MCMC to generate a sample of networks from an ERG distribution.\n\nArgs:\nG: The observed network, to seed the graph with\nedge_coeff: The coefficient on the number of edges\ntriangle_coeff: The coefficient on number of triangles\nn: The number of samples to generate\nReturns:\nA list of graph objects\n'''\n\nv = len(G) # number of nodes in G\np = nx.density(G) # Probability of a random edge existing\ncurrent_graph = nx.erdos_renyi_graph(v, p) # Random graph\ncurrent_w = compute_weight(G, edge_coeff, triangle_coeff)\ngraphs = []\nwhile len(graphs) < n:\nnew_graph = permute_graph(current_graph)\nnew_w = compute_weight(new_graph, edge_coeff, triangle_coeff)\nif new_w > current_w or random.random() < (new_w/current_w):\ngraphs.append(new_graph)\ncurrent_w = new_w\nreturn graphs\n\n\nSo this code lets us generate a sample of networks from an exponential random graph distribution when we know the coefficients.\n\nHowever, what we really want to do is find the coefficients. To do this, we want to look for the coefficients that describe an ERG distribution that makes the observed graph as likely as possible.\n\n# Fitting the model\n\nThe first thing we need to do is calculate the probability of observing any given graph, which means summing the weights of all the graphs in a sample:\n\ndef sum_weights(graphs, edge_coeff, tri_coeff):\n'''\nSum the probability weights on every graph in graphs\n'''\ntotal = 0.0\nfor g in graphs:\ntotal += compute_weight(g, edge_coeff, tri_coeff)\n\n\nWith this, we can calculate the probability of a given graph by dividing its weight by the sum of all weights from the sample.\n\nNext, we can use our favorite optimization technique to hunt through the space of possible edge and triangle coefficients.\n\nInstead of changing the network, we’ll randomly change the parameters, and use our MCMC function to estimate the probability of getting the observed network at those parameter values.\n\nOne simple way to do it uses a similar Metropolis-Hastings approach to the one we used above, with one tweak: early on, we want to try larger jumps around the parameter space; as we get further along, and hopefully closer to the ‘best’ value, the jumps should get smaller and smaller. There’s not really a right way to tweak the jump size\n\nHere’s the code I came up with to do it:\n\ndef fit_ergm(G, coeff_samples=100, graph_samples=1000, return_all=False):\n'''\nUse MCMC to sample possible coefficients, and return the best fits.\n\nArgs:\nG: The observed graph to fit\ncoeff_samples: The number of coefficient combinations to sample\ngraph_samples: The number of graphs to sample for each set of coeffs\nreturn_all: If True, return all sampled values. Otherwise, only best.\nReturns:\nIf return_all=False, returns a tuple of values,\n(best_edge_coeff, best_triangle_coeff, best_p)\nwhere p is the estimated probability of observing the graph G with\nthe fitted parameters.\n\nOtherwise, return a tuple of lists:\n(edge_coeffs, triangle_coeffs, probs)\n'''\nedge_coeffs = \ntriangle_coeffs = \nprobs = [None]\n\nwhile len(probs) < coeff_samples:\n# Make the jump size larger early on, and smaller toward the end\nw = coeff_samples/50.0\ns = np.sqrt(w/len(probs))\n# Pick new coefficients to try:\nedge_coeff = edge_coeffs[-1] + random.normalvariate(0, s)\ntriangle_coeff = triangle_coeffs[-1] + random.normalvariate(0, s)\n# Check how likely the observed graph is under this distribution:\ngraphs = mcmc(G, edge_coeff, triangle_coeff, graph_samples)\nsum_weight = sum_weights(graphs, edge_coeff, triangle_coeff)\np = compute_weight(G, edge_coeff, triangle_coeff) / sum_weight\n# Decide whether to accept the jump:\nif p > probs[-1] or random.random() < (p / probs[-1]):\nedge_coeffs.append(edge_coeff)\ntriangle_coeffs.append(triangle_coeff)\nprobs.append(p)\nelse:\nedge_coeffs.append(edge_coeffs[-1])\ntriangle_coeffs.append(triangle_coeffs[-1])\nprobs.append(probs)\n# Return either the best values, or all of them:\nif not return_all:\ni = np.argmax(probs)\nbest_p = probs[i]\nbest_edge_coeff = edge_coeffs[i]\nbest_triangle_coeff = triangle_coeffs[i]\nreturn (best_edge_coeff, best_triangle_coeff, best_p)\nelse:\nreturn (edge_coeffs, triangle_coeffs, probs)\n\n\n# Example application\n\nNow the real test – fitting the ERGM to an actual network. The canonical example is the Florentine marriage network, which is included in NetworkX.\n\nG = nx.florentine_families_graph()\n\nnx.draw(G)", null, "%%timeit\ngraphs = mcmc(G, -1.25, 0.15, 10000)\n\n1 loop, best of 3: 6.79 s per loop\n\n\nI fit the ERGM with 100 coefficient iterations and 10,000 random networks for each candidate distribution. I also have it return the full series of steps. Note: this part takes a while.\n\n%%timeit\nedge_coeffs, triangle_coeffs, probs = fit_ergm(G, 10, 10, True)\n\nThe slowest run took 25.50 times longer than the fastest. This could mean that an intermediate result is being cached.\n10 loops, best of 3: 199 ms per loop\n\nedge_coeffs, triangle_coeffs, probs = fit_ergm(G, 100, 10, True)\n\ni = np.argmax(probs)\nprint max(probs)\nprint edge_coeffs[i]\nprint triangle_coeffs[i]\n\n0.59541550664\n0.445701809562\n-0.492293584485\n\n\nOnce it’s done (about 20 minutes later, for me), we can find the values it converged to. For me, they were:\n\nEdge coefficient: -1.667\n\nTriangle coefficient: -0.258\n\n==========================\nSummary of model fit\n==========================\n\nFormula: flomarriage ~ edges + triangles\n\nIterations: 20\n\nMonte Carlo MLE Results:\nEstimate Std. Error MCMC % p-value\nedges -1.6773 0.3499 0 <1e-04 *\ntriangle 0.1574 0.5831 0 0.788\n---\nSignif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1\nNot too far off! The edge coefficient is very close; the triangle coefficient is not too far off, and in any case isn't statistically significant.\n\n\nOne problem I discovered is that I obtained some estimated probabilities for the observed network that were greater than 1. This means that the network MCMC was sampling far away from the observed network, so that the weight on the observed network was greater than the sum of weights on all the sampled networks. This isn’t a great outcome, though it still does tell us something about how likely the observed network is under that distribution. Increasing the network sample size might help make this problem go away, as would improving the Markov chain procedure (e.g. with more permutation between networks).\n\nWe can do a few more diagnostics on the results. For example, the trace of the value of the coefficients at each iteration is:\n\nfig = plt.figure(figsize=(8,4))\np1, = ax.plot(edge_coeffs)\np2, = ax.plot(triangle_coeffs)\nax.set_ylabel(\"Coefficient value\")\nax.set_xlabel(\"Iteration\")\nl = ax.legend([p1, p2], [\"Edge coefficient\", \"Triangle coefficient\"])", null, "weighted_edge_coeffs = np.array(edge_coeffs[1:]) * np.array(probs[1:])\nprint np.sum(weighted_edge_coeffs)/np.sum(probs[1:])\n\n\n0.297715535504\n\nweighted_tri_coeffs = np.array(triangle_coeffs[1:]) * np.array(probs[1:])\nprint np.sum(weighted_tri_coeffs)/np.sum(probs[1:])\n\n-0.481915973737\n\n\nWe can also see the distribution of possible coefficients, by making a histogram weighted by the likelihood we observed for each value. (To deal with the likelihoods estimated at greater than 1, I actually weight each coefficient value by the log of the estimated likelihood).\n\nfig = plt.figure(figsize=(12,4))\nax1.hist(edge_coeffs[1:], weights=-np.log(probs[1:]))\nax1.set_title(\"Edge coefficient\")\nax1.set_xlabel(\"Coefficient value\")\n\nax2.hist(triangle_coeffs[1:], weights=-np.log(probs[1:]))\nax2.set_title(\"Triangle coefficient\")\nax2.set_xlabel(\"Coefficient value\")\nplt.show()", null, "This shows that the bulk of the weight on the edge coefficient is between -1 and -2, while the triangle coefficient is much more evenly distributed on both sides of 0. Using a distribution like this, we can statistically test whether the coefficients are significantly different from 0, estimate the standard error, and do various other statistical tests and operations I won’t cover here.\n\nAnd that’s it! We have a slow, simplistic but apparently somewhat-working ERGM model fitter, built up from scratch.\n\nIf you want to learn how to actually do ERGM analysis in R , Benjamin Lind has a great hands-on tutorial over at Bad Hessian on using ERGMs to model the hookup network on the TV show Grey’s Anatomy.\n\nThe paper introducing the ergm package by Hunter et al., the package developers, is a great guide\n\nas is the (paywalled) An introduction to exponential random graph (p*) models for social networks by Robins et al.\n\nMarkov Chain Monte Carlo Estimation of Exponential Random Graph Models by Tom Snijders goes into more detail on estimation methodology, and includes an appendix with a much better estimation algorithm, which I may try to implement in the future.\n\nTags:\n\nUpdated:" ]	[ null, "https://socrateslab.github.io/forum/assets/img2017/ergm0.png", null, "https://socrateslab.github.io/forum/assets//img2017/ergm1.png", null, "https://socrateslab.github.io/forum/assets//img2017/ergm2.png", null, "https://socrateslab.github.io/forum/assets/img2017/ergm3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86825955,"math_prob":0.9890786,"size":14058,"snap":"2021-43-2021-49","text_gpt3_token_len":3336,"char_repetition_ratio":0.15703714,"word_repetition_ratio":0.0037771482,"special_character_ratio":0.2474036,"punctuation_ratio":0.1411985,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995865,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T05:09:18Z\",\"WARC-Record-ID\":\"<urn:uuid:f727bc62-3659-4b52-a874-5154fe0f4996>\",\"Content-Length\":\"59322\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db2a6952-0fad-42b6-83f8-86e46c150098>\",\"WARC-Concurrent-To\":\"<urn:uuid:450a858f-d4d1-4fd2-8449-24ebc30a97ce>\",\"WARC-IP-Address\":\"172.67.171.251\",\"WARC-Target-URI\":\"https://computational-communication.com/forum/ergm-python/\",\"WARC-Payload-Digest\":\"sha1:NIQLUVAT5TZX432GF54OHCEL27DMDM7Y\",\"WARC-Block-Digest\":\"sha1:JJK5OD6IKTHXY43KHJYLMPM3UCJMRDO3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587623.1_warc_CC-MAIN-20211025030510-20211025060510-00136.warc.gz\"}"}
https://docs.sqlalchemy.org/en/14/orm/self_referential.html	[ "# Adjacency List Relationships¶\n\nThe adjacency list pattern is a common relational pattern whereby a table contains a foreign key reference to itself, in other words is a self referential relationship. This is the most common way to represent hierarchical data in flat tables. Other methods include nested sets, sometimes called “modified preorder”, as well as materialized path. Despite the appeal that modified preorder has when evaluated for its fluency within SQL queries, the adjacency list model is probably the most appropriate pattern for the large majority of hierarchical storage needs, for reasons of concurrency, reduced complexity, and that modified preorder has little advantage over an application which can fully load subtrees into the application space.\n\nThis section details the single-table version of a self-referential relationship. For a self-referential relationship that uses a second table as an association table, see the section Self-Referential Many-to-Many Relationship.\n\nIn this example, we’ll work with a single mapped class called `Node`, representing a tree structure:\n\n```class Node(Base):\n__tablename__ = 'node'\nid = Column(Integer, primary_key=True)\nparent_id = Column(Integer, ForeignKey('node.id'))\ndata = Column(String(50))\nchildren = relationship(\"Node\")```\n\nWith this structure, a graph such as the following:\n\n```root --+---> child1\n+---> child2 --+--> subchild1\n\| +--> subchild2\n+---> child3```\n\nWould be represented with data such as:\n\n```id parent_id data\n--- ------- ----\n1 NULL root\n2 1 child1\n3 1 child2\n4 3 subchild1\n5 3 subchild2\n6 1 child3```\n\nThe `relationship()` configuration here works in the same way as a “normal” one-to-many relationship, with the exception that the “direction”, i.e. whether the relationship is one-to-many or many-to-one, is assumed by default to be one-to-many. To establish the relationship as many-to-one, an extra directive is added known as `relationship.remote_side`, which is a `Column` or collection of `Column` objects that indicate those which should be considered to be “remote”:\n\n```class Node(Base):\n__tablename__ = 'node'\nid = Column(Integer, primary_key=True)\nparent_id = Column(Integer, ForeignKey('node.id'))\ndata = Column(String(50))\nparent = relationship(\"Node\", remote_side=[id])```\n\nWhere above, the `id` column is applied as the `relationship.remote_side` of the `parent` `relationship()`, thus establishing `parent_id` as the “local” side, and the relationship then behaves as a many-to-one.\n\nAs always, both directions can be combined into a bidirectional relationship using the `backref()` function:\n\n```class Node(Base):\n__tablename__ = 'node'\nid = Column(Integer, primary_key=True)\nparent_id = Column(Integer, ForeignKey('node.id'))\ndata = Column(String(50))\nchildren = relationship(\"Node\",\nbackref=backref('parent', remote_side=[id])\n)```\n\nThere are several examples included with SQLAlchemy illustrating self-referential strategies; these include Adjacency List and XML Persistence.\n\n## Composite Adjacency Lists¶\n\nA sub-category of the adjacency list relationship is the rare case where a particular column is present on both the “local” and “remote” side of the join condition. An example is the `Folder` class below; using a composite primary key, the `account_id` column refers to itself, to indicate sub folders which are within the same account as that of the parent; while `folder_id` refers to a specific folder within that account:\n\n```class Folder(Base):\n__tablename__ = 'folder'\n__table_args__ = (\nForeignKeyConstraint(\n['account_id', 'parent_id'],\n['folder.account_id', 'folder.folder_id']),\n)\n\naccount_id = Column(Integer, primary_key=True)\nfolder_id = Column(Integer, primary_key=True)\nparent_id = Column(Integer)\nname = Column(String)\n\nparent_folder = relationship(\"Folder\",\nbackref=\"child_folders\",\nremote_side=[account_id, folder_id]\n)```\n\nAbove, we pass `account_id` into the `relationship.remote_side` list. `relationship()` recognizes that the `account_id` column here is on both sides, and aligns the “remote” column along with the `folder_id` column, which it recognizes as uniquely present on the “remote” side.\n\n## Self-Referential Query Strategies¶\n\nQuerying of self-referential structures works like any other query:\n\n```# get all nodes named 'child2'\nsession.query(Node).filter(Node.data=='child2')```\n\nHowever extra care is needed when attempting to join along the foreign key from one level of the tree to the next. In SQL, a join from a table to itself requires that at least one side of the expression be “aliased” so that it can be unambiguously referred to.\n\nRecall from Selecting ORM Aliases in the ORM tutorial that the `aliased()` construct is normally used to provide an “alias” of an ORM entity. Joining from `Node` to itself using this technique looks like:\n\n```from sqlalchemy.orm import aliased\n\nnodealias = aliased(Node)\nsession.query(Node).filter(Node.data=='subchild1').\\\njoin(Node.parent.of_type(nodealias)).\\\nfilter(nodealias.data==\"child2\").\\\nall()\nSELECT node.id AS node_id,\nnode.parent_id AS node_parent_id,\nnode.data AS node_data\nFROM node JOIN node AS node_1\nON node.parent_id = node_1.id\nWHERE node.data = ?\nAND node_1.data = ?\n['subchild1', 'child2']\n```\n\nFor an example of using `aliased()` to join across an arbitrarily long chain of self-referential nodes, see XML Persistence.\n\nEager loading of relationships occurs using joins or outerjoins from parent to child table during a normal query operation, such that the parent and its immediate child collection or reference can be populated from a single SQL statement, or a second statement for all immediate child collections. SQLAlchemy’s joined and subquery eager loading use aliased tables in all cases when joining to related items, so are compatible with self-referential joining. However, to use eager loading with a self-referential relationship, SQLAlchemy needs to be told how many levels deep it should join and/or query; otherwise the eager load will not take place at all. This depth setting is configured via `relationships.join_depth`:\n\n```class Node(Base):\n__tablename__ = 'node'\nid = Column(Integer, primary_key=True)\nparent_id = Column(Integer, ForeignKey('node.id'))\ndata = Column(String(50))\nchildren = relationship(\"Node\",\nlazy=\"joined\",\njoin_depth=2)\n\nsession.query(Node).all()\nSELECT node_1.id AS node_1_id,\nnode_1.parent_id AS node_1_parent_id,\nnode_1.data AS node_1_data,\nnode_2.id AS node_2_id,\nnode_2.parent_id AS node_2_parent_id,\nnode_2.data AS node_2_data,\nnode.id AS node_id,\nnode.parent_id AS node_parent_id,\nnode.data AS node_data\nFROM node\nLEFT OUTER JOIN node AS node_2\nON node.id = node_2.parent_id\nLEFT OUTER JOIN node AS node_1\nON node_2.id = node_1.parent_id\n[]\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7403133,"math_prob":0.8112532,"size":7479,"snap":"2022-40-2023-06","text_gpt3_token_len":1719,"char_repetition_ratio":0.14622073,"word_repetition_ratio":0.073746316,"special_character_ratio":0.22890761,"punctuation_ratio":0.12967382,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95671463,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T13:54:23Z\",\"WARC-Record-ID\":\"<urn:uuid:b7152566-f5c0-4a02-8c71-361eb5d53458>\",\"Content-Length\":\"49568\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f21ccd1e-5454-44a1-a9ab-fca2a2cf2498>\",\"WARC-Concurrent-To\":\"<urn:uuid:c8728c0d-ab7d-44d8-8382-82c72378f1bf>\",\"WARC-IP-Address\":\"45.33.79.102\",\"WARC-Target-URI\":\"https://docs.sqlalchemy.org/en/14/orm/self_referential.html\",\"WARC-Payload-Digest\":\"sha1:STIJRUMHSPO5YBXYCGAFCTPDYLVP5NY2\",\"WARC-Block-Digest\":\"sha1:4TQNBLK3AZM7ZLHCY4FUSYPVUYS5ZNSZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337421.33_warc_CC-MAIN-20221003133425-20221003163425-00053.warc.gz\"}"}
http://bbcbasic.co.uk/wiki/doku.php?id=queueing_20event_20interrupts	[ "", null, "BBC BASIC Programmers' Reference\n\nSite Tools\n\nqueueing_20event_20interrupts\n\nQueueing event interrupts\n\nby Richard Russell, June 2006; amended June 2009 and December 2011\n\nThe ON CLOSE, ON MOUSE, ON MOVE, ON SYS and ON TIME statements interrupt your program when one of the specified events occurs. Perhaps the most obvious way to use these statements is to cause a procedure (an interrupt service routine) to be called when the interrupt happens, as follows:\n\nON SYS PROCsys(@wparam%,@lparam%) : RETURN\n\nIt is important that any event parameters in which you are interested (@msg%, @wparam% and/or @lparam%) are read in the statement immediately following the ON statement otherwise they might have changed as a result of a subsequent interrupt. For example the following code will not work reliably:\n\nON SYS wp%=@wparam%:lp%=@lparam%:PROCsys(wp%,lp%):RETURN : REM Don't do this!\n\nAlthough wp% will contain the @wparam% parameter relevant to the ON SYS call, lp% may not contain the relevant value of @lparam%, because another interrupt may have occurred in between.\n\nSo with care it is possible to ensure that every event is processed by your program, with no chance of any being missed, but not necessarily in the order in which they occurred! To see why, consider the following code:\n\nON SYS PROCsys(@wparam%) : RETURN\n\nDEF PROCsys(W%)\nPRINT W%\nENDPROC\n\nSuppose two ON SYS interrupts happen in quick succession, with @wparam% values of 1 and 2 in that order. The first will result in PROCsys being called, but before the PRINT statement gets a chance to be executed the second interrupt will occur. So the actual sequence of events will be:\n\n1. PROCsys(1)\n2. PROCsys(2)\n3. PRINT 2\n4. ENDPROC\n5. PRINT 1\n6. ENDPROC“\n\nresulting in the following output:\n\n2\n1\n\nSo the events were processed in the opposite order to that in which they occurred!\n\nOften this won't matter, and indeed in many circumstances it will be irrelevant because there is no likelihood of two interrupts happening sufficiently close together. For example this will be the situation if you are using ON SYS only to respond to menu selections, since you should only get one interrupt for each selection. In such a case you can safely use an alternative approach where the interrupt simply sets a global variable that you can poll elsewhere:\n\nON SYS Click% = @wparam% : RETURN\n\nClick% = -1\nREPEAT\ntemp% = INKEY(1)\nIF temp%=-1 SWAP temp%,Click%\nCASE temp% OF\nWHEN ....\nWHEN ....\nREM. etc.\nENDCASE\nUNTIL FALSE\n\nThis routine conveniently uses INKEY as both a delay (to avoid using too much CPU time) and to monitor for keypresses, which is handy for implementing keyboard shortcuts for menu items.\n\nHowever life isn't always this easy! Occasionally it may be necessary to ensure that every event is registered, even if two or more occur in very quick succession, and to ensure that the events are processed in the order in which they happen. An example might be handling events from a dialogue box. One way of dealing with this is to implement a First In First Out queue of events which can be written as the events occur (however fast) and read as they are processed, even if relatively slowly.\n\nAt first thought this doesn't sound too difficult - creating a FIFO queue in software is straightforward - but there is a major difficulty: we must transfer the event into the queue in just one statement. If we don't it won't work: either data could be lost or the events could be processed in the wrong order! Achieving this sounds like a tall order, but it can be done.\n\nMethod 1: Using an array\n\nFor a queue with six entries the code for writing into the queue would be as follows:\n\nDIM Q%(6)\n\nON SYS Q%()=Q%(0)+1,@wparam%,Q%(1),Q%(2),Q%(3),Q%(4),Q%(5) : RETURN\n\nThis may need a little explanation. To do it all in a single statement we use the ability to load an entire array from a comma-separated list of values. The clever bit is that some of the values we load into the array depend on elements of that same array, as follows:\n\n• Q%(0) = Q%(0)+1\n• Q%(1) = @wparam%\n• Q%(2) = Q%(1)\n• Q%(3) = Q%(2)\n• Q%(4) = Q%(3)\n• Q%(5) = Q%(4)\n• Q%(6) = Q%(5)\n\nHopefully you can see what is happening here. Elements Q(1) to Q(6) act as a shift register: each time an event occurs the previously-stored data is shifted one place along the queue (the data in element Q%(6) is discarded) and the new data is stored in Q%(1). Element Q%(0) is loaded with the value Q%(0)+1, in other words it is incremented. This zeroth element of the array acts a pointer to the oldest event stored in the queue.\n\nSo by using this cunning method we manage to store each event into the queue, and increment a pointer, in just one statement! How, then, do we read the data out? This is the code:\n\nWHILE Q%(0)\nevent% = Q%(0)<=DIM(Q%(),1) AND Q%(Q%(0) AND Q%(0)<=DIM(Q%(),1))\nQ%(0) -= 1\nREM Do something with event%\nENDWHILE\n\nFirstly we examine Q%(0), which is the pointer. If this is zero the queue is empty and we need take no further action. If it is non-zero there is at least one event in the queue. The next line reads the oldest event in the queue, by using Q%(0) as the subscript; the comparisons ensure that a 'Bad subscript' error doesn't occur if the queue overflows.\n\nNote that as Q%(0) is accessed and the queue's contents retrieved in the same statement, there is no possibility of reading the wrong event. Even if another interrupt has occurred since Q%(0) was tested in the previous line it makes no difference: the oldest event will always be read.\n\nFinally the next line simply decrements the pointer, so it points to the next event to be read (if any), and leaves the other elements in the array unchanged. Again, it doesn't matter if another interrupt occurs between the previous statement and this one.\n\nThe array can, of course, be any length (within reason). If events occur more quickly than they can be processed the queue will eventually fill and the oldest event(s) will be discarded; in that case event% will be set to zero.\n\nIf you need to know the value of @msg% and @lparam% as well as @wparam% you can extend the technique as follows. To write into the queue:\n\nDIM Q%(18)\n\nON SYS Q%()=Q%(0)+3,@msg%,@wparam%,@lparam%,Q%(1),Q%(2),Q%(3),Q%(4),Q%(5),...,Q%(15) : RETURN\n\nWHILE Q%(0)\nlpar% = Q%(0)<=DIM(Q%(),1) AND Q%(Q%(0)-0 AND Q%(0)<=DIM(Q%(),1))\nwpar% = Q%(0)<=DIM(Q%(),1) AND Q%(Q%(0)-1 AND Q%(0)<=DIM(Q%(),1))\nmsg% = Q%(0)<=DIM(Q%(),1) AND Q%(Q%(0)-2 AND Q%(0)<=DIM(Q%(),1))\nQ%(0) -= 3\nREM Do something with msg%, wpar% and lpar%\nENDWHILE\n\nThe maximum length of queue that can be fitted into one line is 11 events (33 values):\n\nDIM Q%(33)\n\nON SYS Q%()=Q%(0)+3,@msg%,@wparam%,@lparam%,Q%(1),Q%(2),Q%(3),Q%(4),Q%(5),...,Q%(30) : RETURN\n\nTo create a longer queue split the line using line-continuation characters (BBC BASIC for Windows version 5.91a or later only):\n\nDIM Q%(99)\n\nON SYS Q%() = Q%(0)+3,@msg%,@wparam%,@lparam%,Q%(1),Q%(2),Q%(3),Q%(4),Q%(5),Q%(6),Q%(7),Q%(8),Q%(9),Q%(10),Q%(11),Q%(12),Q%(13),Q%(14),Q%(15),Q%(16),Q%(17),Q%(18),Q%(19),Q%(20),Q%(21),Q%(22),Q%(23),Q%(24),Q%(25),Q%(26),Q%(27),Q%(28),Q%(29),\\\n\\ Q%(30),Q%(31),Q%(32),Q%(33),Q%(34),Q%(35),Q%(36),Q%(37),Q%(38),Q%(39),Q%(40),Q%(41),Q%(42),Q%(43),Q%(44),Q%(45),Q%(46),Q%(47),Q%(48),Q%(49),Q%(50),Q%(51),Q%(52),Q%(53),Q%(54),Q%(55),Q%(56),Q%(57),Q%(58),Q%(59),Q%(60),Q%(61),Q%(62),Q%(63),\\\n\\ Q%(64),Q%(65),Q%(66),Q%(67),Q%(68),Q%(69),Q%(70),Q%(71),Q%(72),Q%(73),Q%(74),Q%(75),Q%(76),Q%(77),Q%(78),Q%(79),Q%(80),Q%(81),Q%(82),Q%(83),Q%(84),Q%(85),Q%(86),Q%(87),Q%(88),Q%(89),Q%(90),Q%(91),Q%(92),Q%(93),Q%(94),Q%(95),Q%(96) : RETURN\n\nUsing this technique you can make the queue as long as you like, within reason. However, requiring a very long queue is suggestive that you may be able to find a better solution by restructuring your program. For example you may be able to increase the frequency with which you poll the queue, or use an interrupt approach rather than polling.\n\nMethod 2: Using a string\n\nThis method is easier to understand than the foregoing one, and the maximum practical queue length is greater (over 5000 events), but it is more expensive of CPU time and memory.\n\nThe code for writing into the queue is as follows:\n\nQueue\\$ = \"\"\n!^wParam\\$ = ^@wparam% : ?(^wParam\\$+4) = 4\nON SYS Queue\\$ += wParam\\$ : RETURN\n\nHere wParam\\$ is a global string variable containing four characters, corresponding to the 4-byte (32-bit) value of @wparam%.\n\nThis is the code for reading the data out:\n\nWHILE Queue\\$<>\"\"\nevent% = !!^Queue\\$\nQueue\\$ = MID\\$(Queue\\$,5)\nREM Do something with event%\nENDWHILE\n\nIf events occur more quickly than they can be processed the queue will eventually fill and a String too long error will result.\n\nIf you need to know the value of @msg% and @lparam% as well as @wparam% you can extend the technique as follows. To write into the queue:\n\nQueue\\$ = \"\"\n!^iMsg\\$ = ^@msg% : ?(^iMsg\\$+4) = 4\n!^wParam\\$ = ^@wparam% : ?(^wParam\\$+4) = 4\n!^lParam\\$ = ^@lparam% : ?(^lParam\\$+4) = 4\nON SYS Queue\\$ += iMsg\\$ + wParam\\$ + lParam\\$ : RETURN\n\nHere iMsg\\$, wParam\\$ and lParam\\$ are global string variables containing the values of @msg%, @wparam% and @lparam% respectively.\n\nWHILE Queue\\$<>\"\"\nevent\\$ = LEFT\\$(Queue\\$,12)\nQueue\\$ = MID\\$(Queue\\$,13)\n\nevent% = !^event\\$\nmsg% = event%!0\nwpar% = event%!4\nlpar% = event%!8\nREM Do something with msg%, wpar% and lpar%\nENDWHILE\n\nNote that the use of the temporary string event\\$ is important because the memory address of the string Queue\\$ alters as it changes length, and so could change between reading the msg%, wpar% and lpar% values.", null, "" ]	[ null, "http://bbcbasic.co.uk/wiki/lib/exe/fetch.php", null, "http://bbcbasic.co.uk/wiki/lib/exe/indexer.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86675197,"math_prob":0.98600423,"size":9235,"snap":"2022-05-2022-21","text_gpt3_token_len":2589,"char_repetition_ratio":0.13714658,"word_repetition_ratio":0.07713499,"special_character_ratio":0.3122902,"punctuation_ratio":0.14836505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9802211,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-19T17:36:41Z\",\"WARC-Record-ID\":\"<urn:uuid:90e38d66-0058-4e78-a03d-2033720451e1>\",\"Content-Length\":\"27890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12ff275a-d437-4596-8a5b-74c369818b54>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a9433d6-162e-472f-b5e4-eb3379b1a1b6>\",\"WARC-IP-Address\":\"94.229.68.18\",\"WARC-Target-URI\":\"http://bbcbasic.co.uk/wiki/doku.php?id=queueing_20event_20interrupts\",\"WARC-Payload-Digest\":\"sha1:64I2GOXBO46REHS2ZFAHFJUQ5MCFZ3ZU\",\"WARC-Block-Digest\":\"sha1:BCUV4PPXCLCEHLTSNKVT5TYB6COA6XJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320301475.82_warc_CC-MAIN-20220119155216-20220119185216-00263.warc.gz\"}"}
http://www.softmath.com/parabola-in-math/point-slope/write-equation-for-function.html	[ "English \| Español\n\nTry our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nwrite equation for function for variable\nRelated topics:\nPrintable Linear Equation Quiz \| how to write 65 divided into a number algebracally \| hard math problems for kids \| 3rd order quadratic program for ti-83 \| algebraic expressions class7 \| review guide for math 117 test 2 \| solving by substiution grade 9, practice \| maths formulaes \| greatest commn factor between terms calculator \| binomial fraction subtraction\n\nAuthor Message Author Message\n-=ET=-", null, "Reg.: 25.05.2003", null, "Posted: Thursday 04th of Jan 07:05 DoubniDoom", null, "Reg.: 26.06.2002", null, "Posted: Monday 08th of Jan 07:29\n\nI have this home assignment due and I would really I would like to try this thing if it can really help me learn\nappreciate if anyone can help to solve write equation math efficiently . I like maths, but after the job, I don’t\nfor function for variable on which I’m stuck and have any energy left in my body to solve equations .\ndon’t know where to start from. Can you give me\nassistance with quadratic equations, dividing fractions\nand multiplying fractions. I would rather get assistance\nfrom you than hire a math tutor who are very costly .\nAny pointer will be really treasured very much.\nJrobhic", null, "Reg.: 09.08.2002", null, "Posted: Wednesday 10th of Jan 08:57\nkfir", null, "Reg.: 07.05.2006", null, "Posted: Saturday 06th of Jan 08:04\nAlgebrator is the program that I have used through\nI could help you if you can be more specific and provide several algebra classes - Pre Algebra, Remedial\ndetails about write equation for function for variable. A Algebra and Algebra 2. It is a truly a great piece of\ngood program would be ideal rather than a costly math software. I remember of going through problems\nalgebra tutor. After trying a number of program I found with linear equations, simplifying expressions and\nthe Algebrator to be the best I have so far found . It trinomials. I would simply type in a problem from the\nsolves any math problem that you may want solved. It workbook , click on Solve – and step by step\nalso shows all the steps (of the solution). You can just solution to my algebra homework. I highly recommend\ncopy it as your homework . However, you should use it the program.\nto learn math , and simply not use it to copy answers.\nerx", null, "Reg.: 26.10.2001", null, "Posted: Thursday 11th of Jan 19:27\nDolknankey", null, "Reg.: 24.10.2003", null, "Posted: Saturday 06th of Jan 12:17\nTry this link :https://softmath.com/news.html. I found mine\nAlgebrator indeed is a very good software to help you there. You will see, algebra won't seem such a hard\nlearn math, sitting at home . You won’t just get the thing after you get this software ! Have a good time\nproblem solved but the entire solution as well, that’s with it!\nhow concepts are built . And to do well in math, it’s\nimportant to have strong concepts. I would highly\nrecommend using this software if you want to finish\nyour project on time." ]	[ null, "http://www.softmath.com/images/video-pages/solver-top.png", null, "http://www.softmath.com/images/avatars/328.gif", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null, "http://www.softmath.com/images/avatars/none.png", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null, "http://www.softmath.com/images/avatars/115.gif", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null, "http://www.softmath.com/images/avatars/363.jpg", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null, "http://www.softmath.com/images/avatars/163.gif", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null, "http://www.softmath.com/images/avatars/none.png", null, "http://www.softmath.com/images/forum/icon_minipost.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8965615,"math_prob":0.7215145,"size":2499,"snap":"2019-43-2019-47","text_gpt3_token_len":657,"char_repetition_ratio":0.0993988,"word_repetition_ratio":0.009237875,"special_character_ratio":0.2705082,"punctuation_ratio":0.15047619,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9749709,"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],"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],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T21:12:16Z\",\"WARC-Record-ID\":\"<urn:uuid:cb3d2f44-4b50-4c19-924a-a5b91c21089b>\",\"Content-Length\":\"103294\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c08b1c5f-b2a7-4e16-9cfb-76b2db2fb9c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdfc0780-16e6-494b-9170-6997c9d3d4cc>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"http://www.softmath.com/parabola-in-math/point-slope/write-equation-for-function.html\",\"WARC-Payload-Digest\":\"sha1:ZCURB6AYV3KNBUWKALAVJMKNXNMLBW3A\",\"WARC-Block-Digest\":\"sha1:ROYQEJ6XDUJFCQKGRAVC3X4QF4DCES4B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987787444.85_warc_CC-MAIN-20191021194506-20191021222006-00275.warc.gz\"}"}
https://drostlab.github.io/myTAI/reference/PlotCategoryExpr.html	[ "This function visualizes the expression level distribution of each phylostratum during each time point or experiment as boxplot, dot plot, or violin plot enabling users to quantify the age (PS) or divergence (DS) category specific contribution to the corresponding transcriptome.\n\nPlotCategoryExpr(\nExpressionSet,\nlegendName,\ntest.stat = TRUE,\ntype = \"category-centered\",\ndistr.type = \"boxplot\",\ny.ticks = 10,\nlog.expr = FALSE,\ngene.set = NULL\n)\n\n## Arguments\n\nExpressionSet a standard PhyloExpressionSet or DivergenceExpressionSet object. a character string specifying whether \"PS\" or \"DS\" are used to compute relative expression profiles. a logical value indicating whether a Benjamini-Hochberg adjusted kruskal.test should be applied to determine significant differences in age or divergence category specific expression. type of age or divergence category comparison. Specifications can be type = \"category-centered\" or type = \"stage-centered\". format of visualizing age or divergence category specific expression distributions. Either distr.type = \"boxplot\", distr.type = \"dotplot\", or distr.type = \"violin\". number of y-axis ticks (default y.ticks = 10). a logical value specifying whether or not expression levels should internally be log2-transformed before visualization. a character vector storing the gene ids for which gene expression levels shall be visualized.\n\n## Value\n\nA boxplot, violin plot, or dot plot visualizing the gene expression levels of different PS or DS categories.\n\nFurthermore, the statistical test results returned from the kruskal.test are printed to the console.\n\n(1) '' = P-Value <= 0.05\n\n(2) '' = P-Value <= 0.005\n\n(3) '**' = P-Value <= 0.0005\n\n(4) 'n.s.' = not significant = P-Value > 0.05\n\n## Details\n\nThis way of visualizing the gene expression distribution of each age (PS) or divergence (DS) category during all developmental stages or experiments allows users to detect specific age or divergence categories contributing significant levels of gene expression to the underlying biological process (transcriptome).\n\nThis quantification allows users to conclude that genes originating in specific PS or DS contribute significantly more to the overall transcriptome than other genes originating from different PS or DS categories. More specialized analyses such as PlotMeans, PlotRE, PlotBarRE, etc. will then allow to study the exact mean expression patterns of these age or divergence categories.\n\nThe statistical quantification of differences between expression levels of different age or divergence categories is done by performing a kruskal.test with Benjamini & Hochberg p-value adjustment for multiple comparisons.\n\n• type = \"category-centered\" Here, the kruskal.test quantifies the differences of gene expression between all combinations of age or divergence categories for each stage or experiment separately. Here, a significant p-value quantifies that there is at least one pairwise comparison for which age or divergence categories significantly differ in their gene expression distribution. This type of analysis allows users to detect stages or experiments that show high diviation between age or divergence category contributions to the overall transcriptome or no significant deviations of age or divergence categories, suggesting equal age or divergence category contributions to the overall transcriptome.\n\n• type = \"stage-centered\" Here, the kruskal.test quantifies the differences of gene expression between all stages or experiments for each age or divergence category separately. Hence, the test quantifies whether or not the gene expression distribution of a single age or divergence category significantly changes throughout development or experiments. This type of analysis allows users to detect specific age or divergence categories that significantly change their expression levels throughout development or experiments.\n\nArgument Specifications:\n\nArgument: type\n\n• type = \"category-centered\" This specification allows users to compare the differences between all age or divergence categories during the same stage or experiment.\n\n• type = \"stage-centered\" This specification allows users to compare the differences between all age or divergence categories between stages or experiments.\n\nArgument: distr.type\n\n• distr.type = \"boxplot\" This specification allows users to visualize the expression distribution of all PS or DS as boxplot.\n\n• distr.type = \"violin\" This specification allows users to visualize the expression distribution of all PS or DS as violin plot.\n\n• distr.type = \"dotplot\" This specification allows users to visualize the expression distribution of all PS or DS as dot plot.\n\nFinally, users can specify a gene.set (a subset of genes included in the input ExpressioSet) for which expression levels should be visualized as boxplot, dotplot, or violinplot.\n\nPlotMeans, PlotRE, PlotBarRE, age.apply, pTAI, pTDI, pStrata, pMatrix, TAI, TDI\n\nHajk-Georg Drost\n\n## Examples\n\n\ndata(PhyloExpressionSetExample)\ndata(DivergenceExpressionSetExample)\n\nif (FALSE) {\n\n# category-centered visualization of PS specific expression level distributions (log-scale)\nPlotCategoryExpr(ExpressionSet = PhyloExpressionSetExample,\nlegendName = \"PS\",\ntest.stat = TRUE,\ntype = \"category-centered\",\ndistr.type = \"boxplot\",\nlog.expr = TRUE)\n\n# stage-centered visualization of PS specific expression level distributions (log-scale)\nPlotCategoryExpr(ExpressionSet = PhyloExpressionSetExample,\nlegendName = \"PS\",\ntest.stat = TRUE,\ndistr.type = \"boxplot\",\ntype = \"stage-centered\",\nlog.expr = TRUE)\n\n# category-centered visualization of PS specific expression level distributions (log-scale)\n# as violoin plot\nPlotCategoryExpr(ExpressionSet = PhyloExpressionSetExample,\nlegendName = \"PS\",\ntest.stat = TRUE,\ndistr.type = \"violin\",\ntype = \"stage-centered\",\nlog.expr = TRUE)\n\n# analogous for DivergenceExpressionSets\nPlotCategoryExpr(ExpressionSet = DivergenceExpressionSetExample,\nlegendName = \"DS\",\ntest.stat = TRUE,\ntype = \"category-centered\",\ndistr.type = \"boxplot\",\nlog.expr = TRUE)\n\n# visualize the expression levels of 500 example genes\nset.seed(234)\nexample.gene.set <- PhyloExpressionSetExample[sample(1:25260,500) , 2]\n\nPlotCategoryExpr(ExpressionSet = PhyloExpressionSetExample,\nlegendName = \"PS\",\ntest.stat = TRUE,\ntype = \"category-centered\",\ndistr.type = \"boxplot\",\nlog.expr = TRUE,\ngene.set = example.gene.set)\n\n}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65097773,"math_prob":0.90122175,"size":6066,"snap":"2022-40-2023-06","text_gpt3_token_len":1273,"char_repetition_ratio":0.17238535,"word_repetition_ratio":0.23200993,"special_character_ratio":0.1983185,"punctuation_ratio":0.13493724,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9511658,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T12:08:00Z\",\"WARC-Record-ID\":\"<urn:uuid:ce8644bb-3b9f-4e46-b9bb-593fa01666d7>\",\"Content-Length\":\"22148\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9087297e-b3d3-4506-b22e-20dcc6c15ec7>\",\"WARC-Concurrent-To\":\"<urn:uuid:47bb2792-af1d-4d32-b0ca-341dc61ac1b9>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://drostlab.github.io/myTAI/reference/PlotCategoryExpr.html\",\"WARC-Payload-Digest\":\"sha1:5IQBZXAIMXK5BU4GHJRVYK3BAHSTXWBZ\",\"WARC-Block-Digest\":\"sha1:GWK2DO6KBTPKMKVXE7HSXGHWLPT3T7PT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764494976.72_warc_CC-MAIN-20230127101040-20230127131040-00129.warc.gz\"}"}
https://www.pluralsight.com/guides/creating-nested-and-special-purpose-functions-in-r	[ "Important Update\nThe Guide Feature will be discontinued after December 15th, 2023. Until then, you can continue to access and refer to the existing guides.", null, "Deepika Singh\n\n# Creating Nested and Special-Purpose Functions in R\n\n• Jan 17, 2020\n• 13,308 Views\n• Jan 17, 2020\n• 13,308 Views\nData\nR\n\n## Introduction\n\nA function is a set of scripts organized together to carry out a specific task. Writing efficient functions is an important skill that can significantly improve the productivity of data scientists and data science solutions. In this guide, you will learn the basics of writing a function and the types of functions, which will enable you perform analytical tasks more efficiently.\n\n### Components of Functions\n\nThe main components of a function can be expressed as below.\n\n``````1name_function <- function(argument 1, argument 2) {\n2 # Function body which executes what is to be done\n3}``````\n{r}\n\nIn the above expression, the component `name_function` defines the name of the function, which gets stored as an object in R. It is possible to write a function without the name, but for code readability and re-usability, naming a function is strongly recommended.\n\nThe second component is `argument`—argument 1, argument 2, and so on. An argument is a placeholder where the values are passed when a function gets invoked. Again, arguments are optional, but it is recommended to specify them for code reproducibility.\n\nThe third component is the function body, which is a collection of scripts defining what is to be done by the function.\n\nThe R programming library has a rich set of functions, but before we explore them it's important to see an example of the usefulness of writing functions.\n\n## Importance of Functions\n\nFunctions are important because they substantially reduce repetition in code. This decreases the chance of errors in the code, thereby reducing the workload. The other advantage is that if written properly, functions can be reproduced by other users. We will understand this better with an illustration as discussed below.\n\n### Data\n\nTo understand the importance of functions, and also for the other sections in this guide, we will use a fictitious dataset of loan applicants containing 600 observations and ten variables, as described below:\n\n1. `Marital_status`: Whether the applicant is married (\"Yes\") or not (\"No\")\n\n2. `Is_graduate`: Whether the applicant is a graduate (\"Yes\") or not (\"No\")\n\n3. `Income`: Annual Income of the applicant (in USD)\n\n4. `Loan_amount`: Loan amount (in USD) for which the application was submitted\n\n5. `Credit_score`: Whether the applicant's credit score is good (\"Good\") or not (\"Bad\")\n\n6. `Approval_status`: Whether the loan application was approved (\"Yes\") or not (\"No\")\n\n7. `Age`: The applicant's age in years\n\n8. `Sex`: Whether the applicant is a male (\"M\") or a female (\"F\")\n\n9. `Investment`: Total investment in stocks and mutual funds (in USD) as declared by the applicant\n\n10. `Purpose`: Purpose of applying for the loan\n\n``````1library(plyr)\n3library(dplyr)\n4library(ggplot2)\n5library(repr)\n6\n8df2 = df1\n9glimpse(df1)``````\n{r}\n\nOutput:\n\n``````1Observations: 600\n2Variables: 10\n3\\$ Marital_status <chr> \"Yes\", \"No\", \"Yes\", \"No\", \"Yes\", \"Yes\", \"Yes\", \"Yes\", ...\n4\\$ Is_graduate <chr> \"No\", \"Yes\", \"Yes\", \"Yes\", \"Yes\", \"Yes\", \"Yes\", \"Yes\",...\n5\\$ Income <int> 30000, 30000, 30000, 30000, 89900, 133300, 136700, 136...\n6\\$ Loan_amount <int> 60000, 90000, 90000, 90000, 80910, 119970, 123030, 123...\n7\\$ Credit_score <chr> \"Satisfactory\", \"Satisfactory\", \"Satisfactory\", \"Satis...\n8\\$ approval_status <chr> \"Yes\", \"Yes\", \"No\", \"No\", \"Yes\", \"No\", \"Yes\", \"Yes\", \"...\n9\\$ Age <int> 25, 29, 27, 33, 29, 25, 29, 27, 33, 29, 25, 29, 27, 33...\n10\\$ Sex <chr> \"F\", \"F\", \"M\", \"F\", \"M\", \"M\", \"M\", \"F\", \"F\", \"F\", \"M\",...\n11\\$ Investment <int> 21000, 21000, 21000, 21000, 62930, 93310, 95690, 95690...\n12\\$ Purpose <chr> \"Education\", \"Travel\", \"Others\", \"Others\", \"Travel\", \"...``````\n\nThe output shows that the dataset has five numerical (labeled as `int`) and five character variables (labeled as `chr`). For building machine learning algorithms, we will convert the `chr` into factor variables using the lines of code below.\n\n``````1df1\\$Marital_status = as.factor(df1\\$Marital_status)\n2\n4\n5df1\\$Credit_score = as.factor(df1\\$Credit_score)\n6\n7df1\\$approval_status = as.factor(df1\\$approval_status)\n8\n9df1\\$Sex = as.factor(df1\\$Sex)\n10\n11df1\\$Purpose = as.factor(df1\\$Purpose)\n12\n13glimpse(df1)``````\n{r}\n\nOutput:\n\n``````1Observations: 600\n2Variables: 10\n3\\$ Marital_status <fct> Yes, No, Yes, No, Yes, Yes, Yes, Yes, Yes, Yes, No, No...\n4\\$ Is_graduate <fct> No, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, No, Y...\n5\\$ Income <int> 30000, 30000, 30000, 30000, 89900, 133300, 136700, 136...\n6\\$ Loan_amount <int> 60000, 90000, 90000, 90000, 80910, 119970, 123030, 123...\n7\\$ Credit_score <fct> Satisfactory, Satisfactory, Satisfactory, Satisfactory...\n8\\$ approval_status <fct> Yes, Yes, No, No, Yes, No, Yes, Yes, Yes, No, No, No, ...\n9\\$ Age <int> 25, 29, 27, 33, 29, 25, 29, 27, 33, 29, 25, 29, 27, 33...\n10\\$ Sex <fct> F, F, M, F, M, M, M, F, F, F, M, F, F, M, M, M, M, M, ...\n11\\$ Investment <int> 21000, 21000, 21000, 21000, 62930, 93310, 95690, 95690...\n12\\$ Purpose <fct> Education, Travel, Others, Others, Travel, Travel, Tra...``````\n\nThe output shows that the variables have been converted into factor variables. It's interesting to see that it took six lines of code to perform the conversion. We can do the same operation with just two lines of code, as shown below. The first line specifies the position of the columns to be changed to factor variables, while the second line does the conversion using the `lapply()` function.\n\n``````1names <- c(1,2,5,6,8,10)\n2df2[,names] <- lapply(df2[,names] , factor)\n3glimpse(df2) ``````\n{r}\n\nOutput:\n\n``````1Observations: 600\n2Variables: 10\n3\\$ Marital_status <fct> Yes, No, Yes, No, Yes, Yes, Yes, Yes, Yes, Yes, No, No...\n4\\$ Is_graduate <fct> No, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, Yes, No, Y...\n5\\$ Income <int> 30000, 30000, 30000, 30000, 89900, 133300, 136700, 136...\n6\\$ Loan_amount <int> 60000, 90000, 90000, 90000, 80910, 119970, 123030, 123...\n7\\$ Credit_score <fct> Satisfactory, Satisfactory, Satisfactory, Satisfactory...\n8\\$ approval_status <fct> Yes, Yes, No, No, Yes, No, Yes, Yes, Yes, No, No, No, ...\n9\\$ Age <int> 25, 29, 27, 33, 29, 25, 29, 27, 33, 29, 25, 29, 27, 33...\n10\\$ Sex <fct> F, F, M, F, M, M, M, F, F, F, M, F, F, M, M, M, M, M, ...\n11\\$ Investment <int> 21000, 21000, 21000, 21000, 62930, 93310, 95690, 95690...\n12\\$ Purpose <fct> Education, Travel, Others, Others, Travel, Travel, Tra...``````\n\nThe above output confirms that with only two lines of code and using the `lapply()` function, we can perform an operation that initially required six lines of code. It also removed chances of manual error and made the code more robust. This was for a small dataset containing ten variables. For larger datasets with millions of observations and thousands of variables, the use of functions becomes absolutely vital.\n\nHaving understood the importance of functions, we'll now dive deeper into the various options available in R.\n\n## Custom Functions\n\nThere are several in-built functions in R that can be used to perform analytical tasks, some of which are discussed below.\n\n### Mean\n\nThe arithmetic mean of a numeric vector or variable can be calculated using the `mean()` function. In its simplest form, it takes the syntax `mean(x)`, where x stands for the numeric vector. The line of code below calculates the mean of the `Income` variable.\n\n``1mean(df1\\$Income)``\n{r}\n\nOutput:\n\n``````1 658614.7\n2 ``````\n\nWe can add more arguments to the function as shown below. The argument `na.rm` ignores the missing values, while the argument `trim` removes the proportion of outliers from each end before performing the calculation.\n\n``1mean(df1\\$Income, na.rm = TRUE, trim = 0.05)``\n{r}\n\nOutput:\n\n``````1 592630.7\n2 ``````\n\nThe output shows that the mean income is now reduced from \\$658,615.00 to \\$592,631.00. Similarly, we can calculate the other summary statistics using the functions below.\n\n``````1min(df1\\$Income)\n2\n3max(df1\\$Income)\n4\n5quantile(df1\\$Income)``````\n{r}\n\nOutput:\n\n``````1 30000\n2\n3 3173700\n4\n5 0% 25% 50% 75% 100%\n6 30000 381750 500800 760400 3173700 ``````\n\nAlternatively, we can use the `summary()` function to get these values, which shows there is more than one function to get a particular computation in R.\n\n``1summary(df1\\$Income)``\n{r}\n\nOutput:\n\n``````1 Min. 1st Qu. Median Mean 3rd Qu. Max.\n2 30000 381750 500800 658615 760400 3173700 ``````\n\n## User-Defined Functions\n\nThe in-built functions in R are powerful, but often in data science we have to create our own functions. Such user-defined functions have a name, argument and a body. For example, the summary function above does not compute the standard deviation. To do this, we can create a user-defined function using the code below.\n\n``````1compute_sd <- function(x) {\n2 sqrt(sum((x-mean(x))^2/(length(x)-1)))\n3}``````\n{r}\n\nOnce we have created the function, we can call the function to get the desired computation.\n\n``1compute_sd(df1\\$Income)``\n{r}\n\nOutput:\n\n``````1 486281.1\n2 ``````\n\n## Nested Functions\n\nIn the previous sections, we have learned about in-built and user-defined functions. In complex data science use cases, we may have to work on nested functions, which contain functions within a function.\n\nFor example, if we want to visualize class separation by numeric features for our dataset, we can do that using a box plot. There are four numeric features in the data, namely `Income`, `Loan_amount`, `Age`, and `Investment`. If we want to draw a box plot for each of these features with respect to the target variable, `approval_status`, it will require several lines of code.\n\nInstead, we'll try to create a nested function to achieve this objective as shown below.\n\n``````1create_boxplot = function(df, num_cols, target_col = 'approval_status'){\n2 for(col in num_cols){\n3 bp = ggplot(df, aes_string(target_col, col)) +\n4 geom_boxplot() +\n5 ggtitle(paste('Box plot of', col, '\\n vs.', target_col))\n6 print(bp)\n7 }\n8}``````\n{r}\n\nHaving created the function, the next step is to use the above function to generate the plots. The first line of code specifies the list to be used as an argument in the function. The second line calls the function we created above and plots the four box plots.\n\n``````1num_cols = c('Income', 'Loan_amount', 'Age', 'Investment')\n2\n3create_boxplot(df1, num_cols)``````\n{r}\n\nOutput:", null, "Output:", null, "Output:", null, "Output:", null, "The above plots show how we can use the `ggplot()`, `geom_boxplot()`, and `ggtitle` functions nested within the `create_boxplot` function to perform complex data science tasks in fewer lines of code.\n\n## Conclusion\n\nIn this guide, you have learned the basics of writing functions in R. You were introduced to in-built custom functions, and went on to learn how to create user-built and more complex nested functions in R. This knowledge of functions will help you make your analyses more robust, scalable, and reproducible, in addition to reducing errors in the code." ]	[ null, "https://pluralsight.imgix.net/author/lg/c83827e7-f216-4494-a2d3-5c84aa7047e8.png", null, "https://imgur.com/oJVWXDe.png", null, "https://imgur.com/mgwRash.png", null, "https://imgur.com/7gQM6ZI.png", null, "https://imgur.com/YBvBKh1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8333195,"math_prob":0.9674029,"size":10452,"snap":"2023-40-2023-50","text_gpt3_token_len":2982,"char_repetition_ratio":0.14088821,"word_repetition_ratio":0.20205687,"special_character_ratio":0.32902795,"punctuation_ratio":0.238326,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97673553,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T07:07:36Z\",\"WARC-Record-ID\":\"<urn:uuid:61382d9d-b122-44d8-9298-3f12c3ad040c>\",\"Content-Length\":\"143188\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75d75f00-eff6-4bef-83bd-ec675cd03a27>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9e49c4d-33be-4cfb-816a-80b10f6e91ed>\",\"WARC-IP-Address\":\"104.17.35.85\",\"WARC-Target-URI\":\"https://www.pluralsight.com/guides/creating-nested-and-special-purpose-functions-in-r\",\"WARC-Payload-Digest\":\"sha1:4JAJOLJUW64TXEMTWNWRNTUGBQLYTECJ\",\"WARC-Block-Digest\":\"sha1:J3FB7RYEEOSEP5OCWZTCZ63E655LZQ6U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100724.48_warc_CC-MAIN-20231208045320-20231208075320-00673.warc.gz\"}"}
https://physics.stackexchange.com/questions/220638/time-independent-kerr-metric	[ "# Time independent Kerr metric\n\nThe Kerr metric expressed in terms of polar coordinates $r,\\theta,\\phi$, such that $x = r\\sin(\\theta)\\cos(\\phi)$, $y = r\\sin(\\theta)\\sin(\\phi)$, $z = r\\cos(\\theta)$. Then the Kerr metric is given as \\begin{align} ds^2 = &-\\left(1 - \\frac{2GMr}{r^2+a^2\\cos^2(\\theta)}\\right) dt^2 + \\left(\\frac{r^2+a^2\\cos^2(\\theta)}{r^2-2GMr+a^2}\\right) dr^2 + \\left(r^2+a^2\\cos(\\theta)\\right) d\\theta^2\\\\ &+ \\left(r^2+a^2+\\frac{2GMra^2}{r^2+a^2\\cos^2(\\theta)}\\right)\\sin^2(\\theta) d\\phi^2 - \\left(\\frac{4GMra\\sin^2(\\theta)}{r^2+a^2\\cos^2(\\theta)}\\right) d\\phi\\, dt \\end{align} where $a \\equiv S/M$ is the object's angular momentum per unit mass, and $G$ is the gravitational constant. This is an exact solution for the empty-space Einstein equation.\n\nSay, If we are to consider the metric for a constant time, $t_0$. Is it then possible to define the Kerr metric on a submanifold of spacetime, say only in space? If so how can I accomlish this? Is it as simple as dropping the time dependent terms, i.e \\begin{align} ds^2 = & \\left(\\frac{r^2+a^2\\cos^2(\\theta)}{r^2-2GMr+a^2}\\right) dr^2 + \\left(r^2+a^2\\cos(\\theta)\\right) d\\theta^2\\\\ &+ \\left(r^2+a^2+\\frac{2GMra^2}{r^2+a^2\\cos^2(\\theta)}\\right)\\sin^2(\\theta) d\\phi^2 \\end{align} or do I need to use the induced metric to describe the metric on the submanifold?\n\nEdit : I solved the geodesic differential equations using a \"time independent\" Kerr metric, with a = 0 (i.e this reduces Kerr metric to the Schwarzschild metric), and the Schwarzschild radius to define the other parameters :", null, "Most plots I got spiraled around a singularity at the origo.\n\nHere is a plot where I set $\\phi$ to a constant, the z-axis becomes the \"time\" :", null, "Update : I have found the following figure which seem to verify my first figure.", null, "Strategies for Direct Visualization of Second-Rank Tensor Fields by Werner Benger and Hans-Christian Hege\n\n• At heart, something like this has to be properly understood as a 3+1 decomposition of the spacetime, which can be understood using the ADM decomposition: en.wikipedia.org/wiki/ADM_formalism Note that different choices for the time coordinate will give you radically different 3-geometries. – Jerry Schirmer Aug 13 '18 at 19:11\n\n## 1 Answer\n\nThe metric is telling you how to calculate the proper time along a path of your choosing. If you select a path where the time is everywhere constant then as you integrate along that path $dt = 0$ and any terms involving $dt$ disappear. It is as simple as that.\n\n• Its nice to know that. That simplifies things a lot for me. I intend to visualize the metric by solving the geodesic differential equations. And it makes things much easier if I can simply consider the problem in 3D. – imranal Nov 26 '15 at 9:51\n• @imranal: I don't think you can actually do that. Geodesics in space are not the same as geodesics in spacetime. – Javier Nov 26 '15 at 13:53\n• @imranal: I think what you're describing is a bit different to what I thought you meant. The hypersurface of constant time is a Riemannian manifold, with a metric obtained by setting $ds=0$. You could solve the geodesic equation for this manifold and maybe this is a good way to understand the shape of the manifold. However the curves you get have no physical relevance in the sense that they are not physically meaningful trajectories. – John Rennie Nov 26 '15 at 16:40\n• @javier: Can I drop one of the space components instead, say $\\phi$ ? – imranal Nov 27 '15 at 23:32" ]	[ null, "https://i.stack.imgur.com/0I7yw.png", null, "https://i.stack.imgur.com/CpcZc.png", null, "https://i.stack.imgur.com/DKaYt.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6911236,"math_prob":0.9976437,"size":1852,"snap":"2021-21-2021-25","text_gpt3_token_len":620,"char_repetition_ratio":0.16666667,"word_repetition_ratio":0.03265306,"special_character_ratio":0.3201944,"punctuation_ratio":0.072864324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988675,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-17T09:38:51Z\",\"WARC-Record-ID\":\"<urn:uuid:1464f91f-3fb6-4858-9ccb-dac0caf51173>\",\"Content-Length\":\"168572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f847ad0-ea44-495b-9a97-1866af530d56>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a17445e-9684-4ee4-aad7-b6938270a022>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/220638/time-independent-kerr-metric\",\"WARC-Payload-Digest\":\"sha1:AM5NKNHOXQ3Y4KLR647DL5WAXCTSR6YZ\",\"WARC-Block-Digest\":\"sha1:5TJFTTQCE4OKNSJKZE4DSGIZGKF4F6DZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487629632.54_warc_CC-MAIN-20210617072023-20210617102023-00010.warc.gz\"}"}
https://mpm.spbstu.ru/article/2014.32.5/	[ "# Thermoelectric figure-of-merit effects on fluid flow\n\nАвторы:\nАннотация:\n\nThis work is related to flow of an electro conducting fluid presented thermoelectric figure-of-merit effect, in the presence of magnetic field. The electro conducting thermofluid equation heat transfer with one relaxation time is derived. The flow of electro conducting fluid over a suddenly moved plate is considered. The governing coupled equations in the frame of the boundary layer model are applied to Stokes' first problem with heat sources. Laplace transforms and Fourier transforms techniques are used to get the solution. The inverses of Fourier transforms are obtained analytically. Laplace transforms are obtained using the complex inversion formula of the transform together with Fourier expansion techniques. Numerical results for the temperature distribution and the velocity component are represented graphically. Thermoelectric figure-of-merit effect on the fluid flow is studied." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91735125,"math_prob":0.94878787,"size":949,"snap":"2021-43-2021-49","text_gpt3_token_len":174,"char_repetition_ratio":0.123809524,"word_repetition_ratio":0.0,"special_character_ratio":0.15700738,"punctuation_ratio":0.06711409,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9818638,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T14:27:40Z\",\"WARC-Record-ID\":\"<urn:uuid:0fb52c92-4521-4a21-b2d4-3148c089affd>\",\"Content-Length\":\"35714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:60b8696e-d82a-4b15-a2be-463eb8d13479>\",\"WARC-Concurrent-To\":\"<urn:uuid:a746f38e-cd88-46b6-b2d0-fde1e1e57bf0>\",\"WARC-IP-Address\":\"194.190.225.184\",\"WARC-Target-URI\":\"https://mpm.spbstu.ru/article/2014.32.5/\",\"WARC-Payload-Digest\":\"sha1:DD2P7BK3AFDTTF4LTVR7IXL7TJKX5GMI\",\"WARC-Block-Digest\":\"sha1:P3XJJBRXOZDGBPY52URS7QL6KZMBU5CO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363189.92_warc_CC-MAIN-20211205130619-20211205160619-00406.warc.gz\"}"}
https://jsciences.ut.ac.ir/article_31257.html	[ "Abstract\n\nLet D be a division ring with centre K and dim, D< ? a valuation on K and v a\nnoninvariant extension of ? to D. We define the initial ramfication index of v over\n?, ?(v/ ?) .Let A be a valuation ring of o with maximal ideal m, and v , v ,…, v noninvariant extensions of w to D with valuation rings A , A ,…, A .\nIf B= A , it is shown that the following conditions are equivalent: (i) B is a finite A-module, (ii) B is a free A-module, (iii) [B/mB: A/m] = [D: k], (iv) e(v / ?) f(v / ?)= [D: K] and ? (v / ?)= e(v / ?). It is also proved that if ? (v/ ?) = e(v/ ?), and any of (i) - (iv) holds, then v is invariant" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7300024,"math_prob":0.99808705,"size":622,"snap":"2020-34-2020-40","text_gpt3_token_len":210,"char_repetition_ratio":0.11812298,"word_repetition_ratio":0.0,"special_character_ratio":0.3697749,"punctuation_ratio":0.22222222,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988682,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T02:53:33Z\",\"WARC-Record-ID\":\"<urn:uuid:712160d8-5c28-42e1-9039-a2bfd56014c1>\",\"Content-Length\":\"43271\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee643999-a608-4cf6-9ccc-e19120c26782>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d7902b0-a0db-4b2f-9ca1-9ec64dab7b62>\",\"WARC-IP-Address\":\"80.66.179.110\",\"WARC-Target-URI\":\"https://jsciences.ut.ac.ir/article_31257.html\",\"WARC-Payload-Digest\":\"sha1:QIU6MPGJDMVFVOX4IQ2DIMLN3GUA62LE\",\"WARC-Block-Digest\":\"sha1:FQQWCK2OQG45XIAVUEGIER34AJQRJHV5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739134.49_warc_CC-MAIN-20200814011517-20200814041517-00168.warc.gz\"}"}
https://www.nag.com/numeric/nl/nagdoc_28.4/examples/source/c06pqfe.f90.html	[ "NAG Library Manual, Mark 28.4\n``` Program c06pqfe\n\n! C06PQF Example Program Text\n\n! Mark 28.4 Release. NAG Copyright 2022.\n\n! .. Use Statements ..\nUse nag_library, Only: c06pqf, nag_wp\n! .. Implicit None Statement ..\nImplicit None\n! .. Parameters ..\nInteger, Parameter :: nin = 5, nout = 6\n! .. Local Scalars ..\nInteger :: i, ieof, ifail, j, m, n\n! .. Local Arrays ..\nReal (Kind=nag_wp), Allocatable :: work(:), x(:)\n! .. Executable Statements ..\nWrite (nout,) 'C06PQF Example Program Results'\n! Skip heading in data file\nloop: Do\nIf (ieof<0) Then\nExit loop\nEnd If\n\nAllocate (work((m+2)n+15),x(m(n+2)))\nDo j = 1, m(n+2), n + 2\nEnd Do\nWrite (nout,)\nWrite (nout,) 'Original data values'\nWrite (nout,)\nDo j = 1, m(n+2), n + 2\nWrite (nout,99999) ' ', (x(j+i),i=0,n-1)\nEnd Do\n\n! ifail: behaviour on error exit\n! =0 for hard exit, =1 for quiet-soft, =-1 for noisy-soft\nifail = 0\nCall c06pqf('F',n,m,x,work,ifail)\n\nWrite (nout,)\nWrite (nout,) &\n'Discrete Fourier transforms in complex Hermitian format'\nDo j = 1, m(n+2), n + 2\nWrite (nout,)\nWrite (nout,99999) 'Real ', (x(j+2i),i=0,n/2)\nWrite (nout,99999) 'Imag ', (x(j+2i+1),i=0,n/2)\nEnd Do\nWrite (nout,)\nWrite (nout,) 'Fourier transforms in full complex form'\n\nDo j = 1, m(n+2), n + 2\nWrite (nout,)\nWrite (nout,99999) 'Real ', (x(j+2i),i=0,n/2), &\n(x(j+2(n-i)),i=n/2+1,n-1)\nWrite (nout,99999) 'Imag ', (x(j+2i+1),i=0,n/2), &\n(-x(j+2(n-i)+1),i=n/2+1,n-1)\nEnd Do\n\nCall c06pqf('B',n,m,x,work,ifail)\n\nWrite (nout,)\nWrite (nout,) 'Original data as restored by inverse transform'\nWrite (nout,)\nDo j = 1, m(n+2), n + 2\nWrite (nout,99999) ' ', (x(j+i),i=0,n-1)\nEnd Do\nDeallocate (x,work)\nEnd Do loop\n\n99999 Format (1X,A,9(:,1X,F10.4))\nEnd Program c06pqfe\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6084833,"math_prob":0.9852582,"size":1745,"snap":"2023-14-2023-23","text_gpt3_token_len":728,"char_repetition_ratio":0.190695,"word_repetition_ratio":0.21641791,"special_character_ratio":0.44584528,"punctuation_ratio":0.28629857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99449605,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T12:16:36Z\",\"WARC-Record-ID\":\"<urn:uuid:c5514a74-e4f7-4711-9426-b4d6639a9425>\",\"Content-Length\":\"17351\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9bd46dc2-6571-4cda-b0f1-52f18c617e26>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a0aa72c-d8c0-4aab-96e3-7e2456a7da09>\",\"WARC-IP-Address\":\"78.129.168.4\",\"WARC-Target-URI\":\"https://www.nag.com/numeric/nl/nagdoc_28.4/examples/source/c06pqfe.f90.html\",\"WARC-Payload-Digest\":\"sha1:J5IAMPVKU3H244IYD5HMG2ALJFGV7VAN\",\"WARC-Block-Digest\":\"sha1:Q7BL7B4HCKBOYTMIZKUIHZB2HNQ6TTDM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648635.78_warc_CC-MAIN-20230602104352-20230602134352-00695.warc.gz\"}"}
https://ncatlab.org/nlab/show/Faa%20di%20Bruno%20formula	[ "# nLab Faa di Bruno formula\n\nFaà di Bruno formula is a remarkable combinatorial formula for higher derivatives of a composition of functions. There are various modern approaches to the related mathematics, using Joyal’s theory of species, operads, graphs/trees, combinatorial Hopf algebras and so on.\n\n### Literature\n\nWe prove a Faà di Bruno formula for the Green function in the bialgebra of P-trees, for any polynomial endofunctor P. The formula appears as relative homotopy cardinality of an equivalence of groupoids. For suitable choices of P, the result implies also formulae for Green functions in bialgebras of graphs.\n\n• Doron Zeilberger, Toward a combinatorial proof of the Jacobian conjecture? in Combinatoire énumérative (Montreal, Que., 1985/Quebec, Que., 1985), 370–380, Lecture Notes in Math. 1234, Springer 1986. MR89c:05009\n• Eliahu Levy, Why do partitions occur in Faa di Bruno’s chain rule for higher derivatives?, math.GM/0602183.\n• E. Di Nardo, G. Guarino, D. Senato, A new algorithm for computing the multivariate Faà di Bruno’s formula, arxiv/1012.6008\n• Miguel A. Mendez, Combinatorial differential operators in: Faà di Bruno formula, enumeration of ballot paths, enriched rooted trees and increasing rooted trees, arXiv:1610.03602\n\nIn works of T. J. Robinson the formula is treated in the context of vertex algebras, calculus with formal power series and in logarithmic calculus, as well as in a connection to the umbral calculus:\n\n• Thomas J. Robinson, New perspectives on exponentiated derivations, the formal Taylor theorem, and Faà di Bruno’s formula, Proc.Conf.Vert.Op.Alg., Cont.Math. 497 (2009) 185-198 arxiv/0903.3391; Formal calculus and umbral calculus, Electronic Journal of Combinatorics, 17(1) (2010) R95 arxiv/0912.0961\n\n### Faà di Bruno Hopf algebra\n\n• Christian Brouder, Alessandra Frabetti, Christian Krattenthaler, Non-commutative Hopf algebra of formal diffeomorphisms, Adv. Math. 200:2 (2006) 479-524 pdf\n\n• Kurusch Ebrahimi-Fard, Frederic Patras, Exponential renormalization, Annales Henri Poincare 11:943-971,2010, arxiv/1003.1679 doi\n\nUsing Dyson’s identity for Green’s functions as well as the link between the Faà di Bruno Hopf algebra and the Hopf algebras of Feynman graphs, its relation to the composition of formal power series is analyzed.\n\n• Hector Figueroa, Jose M. Gracia-Bondia, Joseph C. Varilly, Faà di Bruno Hopf algebras, article at Springer eom, math.CO/0508337\n\n• Jean-Paul Bultel, Combinatorial properties of the noncommutative Faà di Bruno algebra, J. of Algebraic Combinatorics 38:243–273 (2013) MR3081645\n\nWe give a new combinatorial interpretation of the noncommutative Lagrange inversion formula, more precisely, of the formula of Brouder–Frabetti–Krattenthaler for the antipode of the noncommutative Faà di Bruno algebra.\n\nLast revised on October 13, 2016 at 08:46:57. See the history of this page for a list of all contributions to it." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82196134,"math_prob":0.87468964,"size":1882,"snap":"2021-43-2021-49","text_gpt3_token_len":498,"char_repetition_ratio":0.1315229,"word_repetition_ratio":0.014652015,"special_character_ratio":0.22316684,"punctuation_ratio":0.14730878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9764139,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-15T20:46:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a4bc8cd6-d601-4f11-8d7a-8eee87c8058e>\",\"Content-Length\":\"17986\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:07ff80b8-dfec-4e8f-b6f2-7dcedefb331c>\",\"WARC-Concurrent-To\":\"<urn:uuid:10802e90-8fe9-49b1-a056-452c09171bbe>\",\"WARC-IP-Address\":\"104.21.81.15\",\"WARC-Target-URI\":\"https://ncatlab.org/nlab/show/Faa%20di%20Bruno%20formula\",\"WARC-Payload-Digest\":\"sha1:QO4DON6DPZSSKPDTJGZYFFUBLWL46S6Y\",\"WARC-Block-Digest\":\"sha1:M4HTC2XS7D7I3VYYDQWG2MG2HJ2YONRQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323583083.92_warc_CC-MAIN-20211015192439-20211015222439-00062.warc.gz\"}"}
https://codegolf.stackexchange.com/questions/133754/one-oeis-after-another/152800	[ "# One OEIS after another\n\nAs of 13/03/2018 16:45 UTC, the winner is answer #345, by Khuldraeseth na'Barya. This means the contest is officially over, but feel free to continue posting answers, just so long as they follow the rules.\n\nAs well, just a quick shout out to the top three answerers in terms of numbers of answers:\n\n3. Hyper Neutrino - 26 answers\n\nThis is an answer chaining question that uses sequences from OEIS, and the length of the previous submission.\n\nThis answer chaining question will work in the following way:\n\n• I will post the first answer. All other solutions must stem from that.\n• The next user (let's call them userA) will find the OEIS sequence in which its index number (see below) is the same as the length of my code.\n• Using the sequence, they must then code, in an unused language, a program that takes an integer as input, n, and outputs the nth number in that sequence.\n• Next, they post their solution after mine, and a new user (userB) must repeat the same thing.\n\nThe nth term of a sequence is the term n times after the first, working with the first value being the first value given on its OEIS page. In this question, we will use 0-indexing for these sequences. For example, with A000242 and n = 3, the correct result would be 25.\n\n## However!\n\nThis is not a , so shortest code doesn't matter. But the length of your code does still have an impact. To prevent the duplication of sequences, your bytecount must be unique. This means that no other program submitted here can be the same length in bytes as yours.\n\nIf there isn't a sequence for then length of the last post, then the sequence for your post is the lowest unused sequence. This means that the sequences used also have to be unique, and that the sequence cannot be the same as your bytecount.\n\nAfter an answer has been posted and no new answers have been posted for more than a week, the answer before the last posted (the one who didn't break the chain) will win.\n\n## Input and Output\n\nGeneric input and output rules apply. Input must be an integer or a string representation of an integer and output must be the correct value in the sequence.\n\n## Formatting\n\n# N. language, length, [sequence](link)\n\ncode\n\nanything else\n\n\n## Rules\n\n• You must wait for at least 1 hour before posting an answer, after having posted.\n• You may not post twice (or more) in a row.\n• The index number of a sequence is the number after the A part, and with leading zeros removed (e.g. for A000040 the index number is 40)\n• You can assume that neither the input nor the required output will be outside your languages numerical range, but please don't abuse this by choosing a language that can only use the number 1, for example.\n• If the length of your submission is greater than 65536 characters long, please provide a link to a way to access the code (pastebin for example).\n• n will never be larger than 1000, or be out of bounds for the sequence, simply to prevent accuracy discrepancies from stopping a language from competing.\n• Every 150 (valid) answers, the number of times a language may be used increases. So after 150 solutions have been posted, every language may be used twice (with all previous answers counting towards this). For instance, when 150 answers have been posted, Python 3 may be used twice, but due to the fact that it has already been used once, this means it can only be used once more until 300 answers have been posted.\n• Please be helpful and post a link to the next sequence to be used. This isn't required, but is a recommendation.\n• Different versions of languages, e.g. Python 2 and Python 3 are different languages. As a general rule, if the different versions are both available on Try It Online, they are different languages, but keep in mind that this is a general rule and not a rigid answer.\n• It is not banned, but please try not to copy the code from the OEIS page, and actually try to solve it.\n• Hardcoding is only allowed if the sequence is finite. Please note that the answer that prompted this (#40) is the exception to the rule. A few answers early in the chain hardcode, but these can be ignored, as there is no good in deleting the chain up to, say, #100.\n\nvar QUESTION_ID=133754,OVERRIDE_USER=66833;function shareUrl(i){return\"https://codegolf.stackexchange.com/a/\"+i}function answersUrl(e){return\"https://api.stackexchange.com/2.2/questions/\"+QUESTION_ID+\"/answers?page=\"+e+\"&pagesize=100&order=desc&sort=creation&site=codegolf&filter=\"+ANSWER_FILTER}function commentUrl(e,s){return\"https://api.stackexchange.com/2.2/answers/\"+s.join(\";\")+\"/comments?page=\"+e+\"&pagesize=100&order=desc&sort=creation&site=codegolf&filter=\"+COMMENT_FILTER}function getTemplate(s){return jQuery(jQuery(\"#answer-template\").html().replace(\"{{PLACE}}\",s.index+\".\").replace(\"{{NAME}}\",s.user).replace(\"{{LANGUAGE}}\",s.language).replace(\"{{SEQUENCE}}\",s.sequence).replace(\"{{SIZE}}\",s.size).replace(\"{{LINK}}\",s.link))}function search(l,q){m=jQuery(\"<tbody id='answers'></tbody>\");e.forEach(function(s){if(!q\|\|(l==0&&RegExp('^'+q,'i').exec(s.lang_name))\|\|(l==1&&q===''+s.size)){m.append(jQuery(getTemplate(s)))}});jQuery(\"#answers\").remove();jQuery(\".answer-list\").append(m)}function sortby(ix){t=document.querySelector('#answers');_els=t.querySelectorAll('tr');els=[];for(var i=0;i<_els.length;i++){els.push(_els[i]);}els.sortBy(function(a){a=a.cells[ix].innerText;return ix==0\|\|ix==4?Number(a):a.toLowerCase()});for(var i=0;i<els.length;i++)t.appendChild(els[i]);}function checkSize(x){if(!x)return jQuery(\"#size-used\").text(\"\");var i=b.indexOf(+x);if(i<0)return jQuery(\"#size-used\").text(\"Available!\");var low=+x,high=+x;while(~b.indexOf(low))low--;while(~b.indexOf(high))high++;jQuery(\"#size-used\").text((\"Not available. The nearest are \"+low+\" and \"+high).replace(\"are 0 and\",\"is\"))}function checkLang(x){}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:\"get\",dataType:\"jsonp\",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.answer_id;answer_ids.push(s),answers_hash[s]=e}),e.has_more\|\|(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:\"get\",dataType:\"jsonp\",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return (e.owner.user_id==OVERRIDE_USER?\"<span id='question-author'>\"+e.owner.display_name+\"</span>\":e.owner.display_name)}function process(){b=[];c=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r=\"<h1>\"+e.body.replace(OVERRIDE_REG,\"\")+\"</h1>\")});var a=r.match(SCORE_REG);if(a){e.push({user:getAuthorName(s),size:+a,language:a,lang_name:a,index:+a,sequence:a,link:shareUrl(s.answer_id)});if(b.indexOf(+a)>=0&&c.indexOf(+a)<0){c.push(+a)};b.push(+a)}else{jQuery('#weird-answers').append('<a href=\"'+shareUrl(s.answer_id)+'\">This answer</a> is not formatted correctly. <b>Do not trust the information provided by this snippet until this message disappears.</b><br />')}}),e.sortBy(function(e){return e.index});e.forEach(function(e){jQuery(\"#answers\").append(getTemplate(e))});var q=\"A\"+(\"000000\"+e.slice(-1).size).slice(-6);jQuery(\"#next\").html(\"<a href='http://oeis.org/\"+q+\"'>\"+q+\"</a>\");c.forEach(function(n){jQuery('#weird-answers').append('The bytecount '+n+' was used more than once!<br />')})}Array.prototype.sortBy=function(f){return this.sort(function(a,b){if(f)a=f(a),b=f(b);return(a>b)-(a<b)})};var ANSWER_FILTER=\"!RB.h_bK(IAWbmRBLe\",COMMENT_FILTER=\"!owfmI7e3fd9oB\",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page,e=[];getAnswers();var SCORE_REG=/<h\\d>\\s(\\d+)\\.\\s((?:<a [^>]+>\\s)?((?:[^\\n,](?!<\\/a>))[^\\s,])(?:<\\/a>)?),.?(\\d+)(?=[^\\n\\d<>](?:<(?:s>[^\\n<>]<\\/s>\|[^\\n<>]+>)[^\\n\\d<>]), ((?:<a[^>]+>\\s)?A\\d+(?:\\s<\\/a>)?)\\s<\\/h\\d>)/,OVERRIDE_REG=/^Override\\sheader:\\s/i;\nbody{text-align:left!important;font-family:Roboto,sans-serif}#answer-list,#language-list{padding:10px;/width:290px/;float:left;display:flex;flex-wrap:wrap;list-style:none;}table thead{font-weight:700}table td{padding:5px}ul{margin:0px}#board{display:flex;flex-direction:column;}#language-list li{padding:2px 5px;}#langs-tit{margin-bottom:5px}#byte-counts{display:block;margin-left:15px;}#question-author{color:purple;text-shadow: 0 0 15px rgba(128,0,128,0.1);}#label-info{font-weight: normal;font-size: 14px;font-style: italic;color: dimgray;padding-left: 10px;vertical-align: middle; }\n<script src=\"https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js\"></script><link rel=\"stylesheet\" type=\"text/css\" href=\"//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b\"><p id=\"weird-answers\"></p><p>Currently waiting on <span id=\"next\"></span></p><span>Search by Byte Count: <input id=\"search\" type=\"number\" min=1 oninput=\"checkSize(this.value);search(1,this.value)\" onclick=\"document.getElementById('search2').value='';!this.value&&search(0,'')\"/> <span id=\"size-used\"></span></span><br><span>Search by Language: <input id=\"search2\" oninput=\"checkLang(this.value);search(0,this.value)\" onclick=\"document.getElementById('search').value='';!this.value&&search(0,'')\"/> <span id=\"language-used\"></span></span><h2>Answer chain <span id=\"label-info\">click a label to sort by column</span></h2><table class=\"answer-list\"><thead><tr><td onclick=\"sortby(0)\">#</td><td onclick=\"sortby(1)\">Author</td><td onclick=\"sortby(2)\">Language</td><td onclick=\"sortby(3)\">Sequence</td><td onclick=\"sortby(4)\">Size</td></tr></thead><tbody id=\"answers\"></tbody></table><table style=\"display: none\"><tbody id=\"answer-template\"><tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SEQUENCE}}</td><td>{{SIZE}}</td><td><a href=\"{{LINK}}\">Link</a></td></tr></tbody></table><table style=\"display: none\"><tbody id=\"language-template\"><tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SEQUENCE}}</td><td>{{SIZE}}</td><td><a href=\"{{LINK}}\">Link</a></td></tr></tbody></table>\n\n• Comments are not for extended discussion; this conversation has been moved to chat. Oct 31, 2017 at 2:49\n• Is it OK if a program would need a better floating-point accuracy for the builtin float/double type in order to produce values for larger n?\n– Maya\nNov 21, 2017 at 15:15\n• @Giuseppe No, as you're generating the numbers by doing the maths, rather than just placing them into an array/string Dec 15, 2017 at 22:14\n• @cairdcoinheringaahing In my opinion that's hardcoding the gamma constant. It doesn't work \"in theory\" for larger numbers. Dec 22, 2017 at 12:44\n• Chat room Dec 22, 2017 at 12:45\n\n# 22. FiM++, 982 bytes, A000024\n\nNote: if you are reading this, you might want to sort by \"oldest\".\n\nDear PPCG: I solved A000024!\n\nI learned how to party to get a number using the number x and the number y.\nDid you know that the number beers was x?\nFor every number chug from 1 to y,\nbeers became beers times x!\nThat's what I did.\nThen you get beers!\nThat's all about how to party.\n\nToday I learned how to do math to get a number using the number n.\nDid you know that the number answer was 0?\nFor every number x from 1 to n,\nFor every number y from 1 to n,\nDid you know that the number tmp1 was how to party using x and 2?\nDid you know that the number tmp2 was how to party using y and 2?\nDid you know that the number max was how to party using 2 and n?\ntmp2 became tmp2 times 10!\ntmp1 became tmp1 plus tmp2!\nIf tmp1 is more than max then: answer got one more.\nThat's what I did.\nThat's what I did.\nThat's all about how to do math.\n\nPS: This is the best answer\nPPS: This really is the best answer\n\n\nNext sequence\n\n• Hahaha, laughed so hard through the whole thing. +1 for choice of language :-) Jul 22, 2017 at 13:49\n• Amazing, take my upvote Jul 23, 2017 at 11:11\n\n# 1. Triangular, 10 bytes, A000217\n\n$\\:_%i/2< Try it online! Next Sequence ## How it works The code formats into this triangle $\n\\ :\n_ % i\n/ 2 <\n\n\nwith the IP starting at the $ and moving South East (SE), works like this: $ Take a numerical input (n); STACK = [n]\n: Duplicate it; STACK = [n, n]\ni Increment the ToS; STACK = [n, n+1]\n< Set IP to W; STACK = [n, n+1]\n* Multiply ToS and 2ndTos; STACK = [n(n+1)]\n2 Push 2; STACK = [n(n+1), 2]\n/ Set IP to NE; STACK = [n(n+1), 2]\n_ Divide ToS by 2ndToS; STACK = [n(n+1)/2]\n\\ Set IP to SE; STACK = [n(n+1)/2]\n% Output ToS as number; STACK = [n(n+1)/2]\n* Multiply ToS by 2ndToS (no op); STACK = [n(n+1)/2]\n\n• 1. Triangular, 10 bytes, A000217. follows link A000217 Triangular numbers ... Jul 23, 2017 at 3:39\n\n# 73. Starry, 363 bytes, A000252\n\n, + + * '. \n+ + + + * * * +\n+ +* + \n+ + + + + + * '\n+ ' #### + + +\n+ + #### +* + \n' ##### + + '\n ######+ + + +\n+ + + ######### '\n+ + + #####+ + +\n* + + * + * * +\n+ * + + + + * \n+ + + + + +\n+ + + + ' + +.\n\n\nTry it online!\n\nNext sequence\n\nUses the formula \"a(n) = n^4 product p^(-3)(p^2 - 1)(p - 1) where the product is over all the primes p that divide n\" from OEIS.\n\nThe moon's a no-op, but hey, this isn't code-golf.\n\n• stars in moon? hmmm Jan 1, 2018 at 14:45\n\n# 97. Python 3 (PyPy), 1772 bytes, A000236\n\nFirst of all, many thanks to Dr. Max Alekseyev for being patient with me. I'm very fortunate that I was able to contact him by email to understand this challenge. His Math.SE answer here helped me out a lot. Thanks to Wheat Wizard for helping me as well. :)\n\nplist = []\n\ndef primes(maximal = -1): # Semi-efficient prime number generator with caching up to a certain max.\nindex = plist and plist[-1] or 2\nfor prime in plist:\nif prime <= maximal or maximal == -1: yield prime\nelse: break\nwhile index <= maximal or maximal == -1:\ncomposite = False\nfor prime in plist:\nif index % prime == 0:\ncomposite = True\nbreak\nif not composite:\nyield index\nplist.append(index)\nindex += 1\n\ndef modinv(num, mod): # Multiplicative inverse with a modulus\nindex = 1\nwhile num index % mod != 1: index += 1\nreturn index\n\ndef moddiv(num, dnm, mod):\nreturn num * modinv(dnm, mod) % mod\n\ndef isPowerResidue(num, exp, mod):\nfor base in range(mod):\nif pow(base, exp, mod) == num:\nreturn base\nreturn False\n\ndef compute(power, prime):\nfor num in range(2, prime):\nif isPowerResidue(moddiv(num - 1, num, prime), power, prime):\nreturn num - 1\nreturn -1\n\n# file = open('output.txt', 'w')\n\ndef output(string):\nprint(string)\n# file.write(str(string) + '\\n')\n\ndef compPrimes(power, count):\nmaximum = 0\nindex = 0\nfor prime in getValidPrimes(power, count):\nresult = compute(power, prime)\nif result > maximum: maximum = result\nindex += 1\n# output('Computed %d / %d = %d%% [result = %d, prime = %d]' % (index, count, (100 * index) // count, result, prime))\nreturn maximum\n\ndef isValidPrime(power, prime):\nreturn (prime - 1) % power == 0\n\ndef getValidPrimes(power, count):\ncollected = []\nfor prime in primes():\nif isValidPrime(power, prime):\ncollected.append(prime)\nif len(collected) >= count:\nreturn collected\n# output('Collected %d / %d = %d%% [%d]' % (len(collected), count, (100 * len(collected)) // count, prime))\n\npower = int(input()) + 2\n\noutput(compPrimes(power, 100))\n\n# file.close()\n\n\nTry it online!\n\nIf it gives the wrong result, just increase the 100 to something larger. I think 10000 will work for 4 but I'll leave my computer running overnight to confirm that; it may take a couple of hours to finish.\n\nNote that the (PyPy) part is just so that I can use Python again. I really don't know many other languages and I'm not going to try to port this to Java and risk not finishing in time.\n\nNext Sequence (Also please don't do any more crazy math stuff; I don't have any Python versions left so someone else will have to save this challenge D:)\n\n• well there's always pypy3 Aug 6, 2017 at 4:29\n\n# 107. TrumpScript, 1589 bytes, A000047\n\nMy cat hears everything really well\nbecause with me every cat is a safe cat\nEverybody knows that one is 1000001 minus 1000000\nbut only most of you that two is, one plus one;\nAs always nothing is, one minus one;\nMy dog is one year old.\nI promise you that as long as you vote on me, nothing will be less cool than a cat;:\nMuch dog is, dog times two;\nDead cat is, cat minus one;!\nI will make dog feel good, food for dog plus one;\nRoads can be made using different things. Asphalt is one of them.\nAs long as Hillary jailed, I love asphalt less than my dog;:\nI promise my roadways are, two times asphalt than you want;\nVladimir is nothing more than my friend.\nAs long as, Putin eat less roadways;:\nChina is nothing interesting.\nWe all know people speaking Chinese are from China.\nAs long as, Chinese makes less roads;:\nI will make economy, for Putin - Chinese will love me;\nIf it will mean, economy is asphalt in Russia?;:\nI will make cat feel good, cat plus one dollar on food;\nI show you how great China is, China plus one; You can add numbers to China.\nLike Chinese is, China times China makes sense;\nLike Chinese is, two times Chinese letter;!\nI also show you how great Putin is, Vladimir times Vladimir; You can do number stuff to Putin too!\nI will make asphalt roads a lot!\nEverybody say cat. You did it? America is great.\n\n\nTry it online!\n\nFirst time programming in TrumpScript, it is possible that I reinvented the wheel a few times - 4 lines are dedicated to calculating 2 ^ n. I tried to make it look like something that (drunk) Trump could say. As a bonus, here is a Python script I wrote to verify that I'm doing everything right. There are some differences to the above program, but much of it is directly equivalent.\n\ncat = int(input())\ndog = 2 ** cat + 1\nasphalt = 1\ncat = 0\nwhile asphalt < dog:\nroadways = 2 * asphalt + 1\nchina = 0\nchinese = china\nchair = putin - chinese\nif chair == asphalt:\ncat += 1\nchina += 1\nchinese = 2 * china * china\nprint(cat)\n\n\nNext sequence!\n\n• I will make cat feel good O_O Aug 15, 2017 at 16:20\n• Sadly I will make Business Cat feel good won't work...\n– Maya\nAug 15, 2017 at 16:26\n\n# 2. Haskell, 44 bytes, A000010\n\nf k\|n<-k+1=length.filter(==1)$gcd n<$>[1..n]\n\n\nTry it online!\n\nNext Sequence\n\n• The name of the next sequence though... Jul 21, 2017 at 15:16\n• @totallyhuman poor rabbits... Jul 21, 2017 at 15:17\n• Should we link to the previous post? Jul 21, 2017 at 15:17\n• It pains me that I cannot golf it now. I had to be first you see Jul 21, 2017 at 15:20\n• What is that next sequence? I don't understand the three ones :P Jul 21, 2017 at 15:20\n\n# 30. Python 1, 1112 bytes, A000046\n\ndef rotations(array):\nrotations = []\nfor divider_index in range(len(array)):\nrotations.append(array[divider_index:] + array[:divider_index])\nreturn rotations\n\ndef next(array):\nfor index in range(len(array) - 1, -1, -1):\narray[index] = 1 - array[index]\nif array[index]: break\nreturn array\n\ndef reverse(array):\nreversed = []\nfor index in range(len(array) - 1, -1, -1):\nreversed.append(array[index])\nreturn reversed\n\ndef primitive(array):\nfor index in range(1, len(array)):\nif array == array[:index] * (len(array) / index): return 1\nreturn 0\n\ndef necklaces(size):\nprevious_necklaces = []\narray = * size\nnecklaces = 0\nfor iteration in range(2 ** size):\nif not primitive(array) and array not in previous_necklaces:\nnecklaces = necklaces + 1\nfor rotation in rotations(array):\ncomplement = []\nfor element in rotation:\ncomplement.append(1 - element)\nprevious_necklaces.append(rotation)\nprevious_necklaces.append(complement)\nprevious_necklaces.append(reverse(rotation))\nprevious_necklaces.append(reverse(complement))\narray = next(array)\nreturn necklaces\n\n\nTry it online!\n\nNot even going to bother to golf this. Hey, it's not my longest Python answer on this site!\n\nNext sequence\n\n• Congratulations on decoding the maths :D Jul 22, 2017 at 4:27\n• 313 bytes, lol Jul 22, 2017 at 4:57\n• @LeakyNun As I was saying, I didn't bother to golf this lol. Besides, it's not my longest Python answer on this site so idc :P but nice Jul 22, 2017 at 15:16\n• @LeakyNun And thanks :D It took me a while to understand all of it lol Jul 22, 2017 at 16:22\n• @LeakyNun 309 bytes because the actual value of _ is irrelevant; we just need to repeat that many times Sep 24, 2017 at 1:36\n\n# 9. Pyth, 19 bytes, A000025\n\n?>Q0sm@_B1-edld./Q1\n\n\nNext sequence\n\na(n) = number of partitions of n with even rank minus number with odd rank. The rank of a partition is its largest part minus the number of parts.\n\n• For those who know Pyth, I deliberately used >Q0 instead of Q in order to, you know, have the next sequence to be A000019. Jul 21, 2017 at 16:41\n• From the OEIS page Keywords: easy,nice Jul 21, 2017 at 16:46\n• @LeakyNun Yeah since otherwise I'd have to solve A000017...gross. Jul 21, 2017 at 16:56\n\n# 308. ENIAC (simulator), 3025 bytes, A006060\n\nPseudocode:\n\nrepeat{\nM←input\nN←-M\nA←1\nB←253\nwhile(N<0){\nC←60\nC←C-A\nrepeat(194){\nC←C+B\n}\nA←B\nB←C\nN←N+1\n}\noutput←A\n}\n\n\nNo online simulator, execution result:", null, "", null, "Registers and constants:\n\nA: 1-2\nB: 3-4\nC: 5-6\nM: 7\nN: 8\n\ninput: const. A\n253: const. J\n60: const. K\n194: Master programmer decade limit 1B\n\n\nProgram signal flow and data flow:", null, "Full \"code\" on pastebin or in HTML comments in the markup of this answer, to prevent linkrot and a quite long answer to scroll through at the same time. This is fun!\n\nNext sequence\n\n• @Zacharý The link is in the post. I'll move it to the end of the post so it's easier to find. Jan 8, 2018 at 17:02\n\n# 8. Mathematica (10.1), 25 bytes, A000070\n\nTr@PartitionsP@Range@#+1&\n\n\nNext sequence\n\n• The perfect sequence to use Mathematica for. Jul 21, 2017 at 15:52\n• A000025 is an incredibly difficult one. You should add a byte to get A000026 instead. :P Jul 21, 2017 at 16:29\n\n# 206. Proton, 3275 bytes, A000109\n\n# This took me quite a while to write; if it's wrong, please tell me and I'll try to fix it without changing the byte count..\n\npermutations = x => {\nif len(x) == 0 return [ ]\nif len(x) == 1 return [x]\nresult = []\nfor index : range(len(x)) {\nfor permutation : permutations(x[to index] + x[index + 1 to]) {\nresult.append([x[index]] + permutation)\n}\n}\nreturn result\n}\n\ncycles = cycles[to]\nsize = cycles.pop()\nmatrix = [ * size for i : range(size)]\nfor cycle : cycles {\ni, j, k = cycle, cycle, cycle\nmatrix[i][j] = matrix[i][k] = matrix[j][i] = matrix[j][k] = matrix[k][i] = matrix[k][j] = 1\n}\nreturn matrix\n}\n\ntransform = a => [[a[j][i] for j : range(len(a[i]))] for i : range(len(a))]\n\nisomorphic = (a, b) => {\nreturn any(sorted(b) == sorted(transform(A)) for A : permutations(transform(a)))\n}\n\nintersection = (a, b) => [x for x : a if x in b]\n\nunion = (a, b) => [x for x : a if x not in b] + list(b)\n\nvalidate = graph => {\nrowsums = map(sum, matrix)\nr = 0\nfor s : rowsums if s + 1 < graph[-1] r++\nreturn 2 \|\| r\n}\n\ngraphs = nodes => {\nif nodes <= 2 return []\nif nodes == 3 return [[(0, 1, 2), 3]]\nresult = []\nexisting = []\nfor graph : graphs(nodes - 1) {\ngraph = graph[to]\nnext = graph.pop()\nfor index : range(len(graph)) {\ng = graph[to]\ncycle = g.pop(index)\nn = g + [(cycle, cycle, next), (cycle, cycle, next), (cycle, cycle, next), next + 1]\nif N not in existing {\nexisting += [sorted(transform(a)) for a : permutations(transform(adjacency(n)))]\nresult.append(n)\n}\nfor secondary : index .. len(graph) - 1 {\ng = graph[to]\nc1 = g.pop(index)\nc2 = g.pop(secondary)\nq = union(c1, c2)\ng = [k for k : g if len(intersection(k, intersection(c1, c2))) <= 1]\nif len(intersection(c1, c2)) == 2 {\nfor i : range(3) {\nfor j : i + 1 .. 4 {\nif len(intersection(q[i, j], intersection(c1, c2))) <= 1 {\ng.append((q[i], q[j], next))\n}\n}\n}\n}\ng.append(next + 1)\nif N not in existing {\nexisting += [sorted(transform(a)) for a : permutations(transform(adjacency(g)))]\nresult.append(g)\n}\nfor tertiary : secondary .. len(graph) - 2 {\ng = graph[to]\nc1 = g.pop(index)\nc2 = g.pop(secondary)\nc3 = g.pop(tertiary)\nq = union(union(c1, c2), c3)\ng = [k for k : g if len(intersection(k, intersection(c1, c2))) <= 1 and len(intersection(k, intersection(c2, c3))) <= 1]\nif len(q) == 5 and len(intersection((q1 = intersection(c1, c2)), (q2 = intersection(c2, c3)))) <= 1 and len(q1) == 2 and len(q2) == 2 {\nfor i : range(4) {\nfor j : i + 1 .. 5 {\nif len(intersection(q[i, j], q1)) <= 1 and len(intersection(q[i, j], q2)) <= 1 {\ng.append((q[i], q[j], next))\n}\n}\n}\ng.append(next + 1)\nif N not in existing {\nexisting += [sorted(transform(a)) for a : permutations(transform(adjacency(g)))]\nresult.append(g)\n}\n}\n}\n}\n}\n}\nreturn [k for k : result if max(sum(k[to -1], tuple([]))) + 1 == k[-1] and validate(k)]\n}\n\nx = graphs(int(input()) + 3)\nprint(len(x))\n\n\nTry it online!\n\nNext Sequence\n\n• Wait, you actually did it? If you don't write a paper with these freaking programs and go talk to some professor, you're passing up on something cool :P Oct 10, 2017 at 2:54\n• @Stephen Currently bugfixing lol Oct 10, 2017 at 2:58\n• Is this the approach of splitting triangles, squares, and pentagons as per plantri? Looks like it might be, but some of the syntax is unfamiliar. Oct 10, 2017 at 6:50\n• @PeterTaylor Assuming I understand the approach you're describing, yes, it looks for triangles and places a vertex adjacent to all 3 vertices, or two adjacent cycles and deletes the common edge and places a vertex adjacent to all 4, same for 3 triangles on a pentagon. I think that's one you're describing. Oct 10, 2017 at 12:01\n• @ChristianSievers math.stackexchange.com/a/2463430/457091 Oct 10, 2017 at 15:06\n\n# 15. CJam, 85 bytes, A000060\n\n{ee\\f{\\~\\0a@+f}:.+}:C;2,qi:Q,2f+{_ee1>{~2\\,:!XX<a~}%{CX<}W=+}fX_0a1$_C.- .+Q)= Online demo Next sequence ### Dissection OEIS gives G.f.: S(x)+S(x^2)-S(x)^2, where S(x) is the generating function for A000151. - Pab Ter, Oct 12 2005 where $$\\begin{eqnarray}S(x) & = & x \\prod_{i \\ge 1} \\frac{1}{(1 - x^i)^{2s(i)}} \\\\ & = & x \\prod_{i \\ge 1} (1 + x^i + x^{2i} + \\ldots)^{2s(i)}\\end{eqnarray}$$ { e# Define a block to convolve two sequences (multiply two polynomials) ee\\f{ e# Index one and use the other as an extra parameter for a map \\~\\0a e# Stack manipulations; create a sequence of index 0s @+f* e# Shift the extra parameter poly and multiply by the coefficient } :.+ e# Fold pointwise add to sum the polys }:C; e# Assign the block to C (for \"convolve\") 2, e# Initial values of S: S(0) = 0, S(1) = 1 qi:Q e# Read integer and assign it to Q ,2f+{ e# For X = 2 to Q+1 _ee1> e# Duplicate accumulator of S, index, and ditch 0th term { e# Map (over notional variable i) ~2\\ e# Double S(i) and flip i to top of stack ,:! e# Create an array with a 1 and i-1 0s XX< e# Replicate X times and truncate to X values e# This gives g.f. 1/(1-x^i) to the first X terms a~ e# Create 2S(i) copies of this polynomial }% {CX<} e# Fold convolution and truncation to X terms W=+ e# Append the final coefficient, which is S(X), to the accumulator }fX _0a* e# Pad a copy to get S(X^2) 1$_C e# Convolve two copies to get S(X)^2\n.- e# Pointwise subtraction\n.+ e# Pointwise addition. Note the leading space because the parser thinks\ne# -. is an invalid number\nQ)= e# Take the term at index Q+1 (where the +1 adjusts for OEIS offset)\n\n• 1 minute and 33 seconds ahead of me... while I was typing the explanation Jul 21, 2017 at 19:05\n\n# 67. LOLCODE, 837 bytes, A000043\n\nHAI 1.2\nCAN HAS STDIO?\n\nI HAS A CONT ITZ 0\nI HAS A ITRZ ITZ 1\nI HAS A NUMBAH\nGIMMEH NUMBAH\nNUMBAH R SUM OF NUMBAH AN 1\n\nIM IN YR GF\nITRZ R SUM OF ITRZ AN 1\n\nI HAS A PROD ITZ 1\nIM IN YR MOM UPPIN YR ASS WILE DIFFRINT ITRZ AN SMALLR OF ITRZ AN ASS\nPROD R PRODUKT OF PROD AN 2\nIM OUTTA YR MOM\nPROD R DIFF OF PROD AN 1\n\nI HAS A PRAIME ITZ WIN\nI HAS A VAR ITZ 1\nIM IN YR MOM\nVAR R SUM OF VAR AN 1\nBOTH SAEM VAR AN PROD, O RLY?\nYA RLY, GTFO\nOIC\nBOTH SAEM 0 AN MOD OF PROD AN VAR, O RLY?\nYA RLY\nPRAIME R FAIL\nGTFO\nOIC\nIM OUTTA YR MOM\n\nBOTH SAEM PRAIME AN WIN, O RLY?\nYA RLY, CONT R SUM OF CONT AN 1\nOIC\n\nBOTH SAEM NUMBAH AN CONT, O RLY?\nYA RLY, GTFO\nOIC\nIM OUTTA YR GF\n\nVISIBLE ITRZ\nKTHXBYE\n\n\nMy capslock key is bound to escape, so I wrote this entire thing while holding shift ..\n\nTry it online!\n\nNext sequence\n\n• +1 for using PRAIME Jul 23, 2017 at 9:07\n• You're a programmer, you could have written this and then run it through a Python script that upper'd it -.- Jul 23, 2017 at 12:25\n• @StepHen Or simply gggUG in vim where I wrote it, but I am not that clever Jul 23, 2017 at 12:36\n\n# 10. Magma, 65 bytes, A000019\n\nf:=function(n);return NumberOfPrimitiveGroups(n+1);end function;\n\n\nTry it here\n\nlol builtin\n\nNext sequence\n\n• @ETHproductions :) no problem, thank the OEIS page though cuz it has the exact builtin there lol Jul 21, 2017 at 17:20\n• ;_; I solved A000064 and you changed it. Downvoted. Jul 21, 2017 at 17:26\n• My gosh, so many partition sequences Jul 21, 2017 at 17:27\n• I accidentally solved A007317 while trying to do this in Python (TIO) :P Jul 21, 2017 at 17:36\n• Re-upvoted! \\o/ Jul 21, 2017 at 17:38\n\n# 281. Java 5, 11628 bytes, A000947\n\n// package oeis_challenge;\n\nimport java.util.;\nimport java.lang.;\n\nclass Main {\n\n// static void assert(boolean cond) {\n// if (!cond)\n// throw new Error(\"Assertion failed!\");\n// }\n\n/* Use the formula a(n) = A000063(n + 2) - A000936(n).\nIt's unfair that I use the formula of \"number of free polyenoid with n\nnodes and symmetry point group C_{2v}\" (formula listed in A000063)\nwithout understanding why it's true...\n/\n\nstatic int catalan(int x) {\nint ans = 1;\nfor (int i = 1; i <= x; ++i)\nans = ans (2x+1-i) / i;\nreturn ans / -~x;\n}\n\nstatic int A63(int n) {\nint ans = catalan(n/2 - 1);\nif (n%4 == 0) ans -= catalan(n/4 - 1);\nif (n%6 == 0) ans -= catalan(n/6 - 1);\nreturn ans;\n}\n\nstatic class Point implements Comparable<Point> {\nfinal int x, y;\nPoint(int _x, int _y) {\nx = _x; y = _y;\n}\n\n/// @return true if this is a point, false otherwise (this is a vector)\npublic boolean isPoint() {\nreturn (x + y) % 3 != 0;\n}\n\n/// Translate this point by a vector.\nassert(this.isPoint() && ! p.isPoint());\nreturn new Point(x + p.x, y + p.y);\n}\n\n/// Reflect this point along x-axis.\npublic Point reflectX() {\nreturn new Point(x - y, -y);\n}\n\n/// Rotate this point 60 degrees counter-clockwise.\npublic Point rot60() {\nreturn new Point(x - y, x);\n}\n\n@Override\npublic boolean equals(Object o) {\nif (!(o instanceof Point)) return false;\nPoint p = (Point) o;\nreturn x == p.x && y == p.y;\n}\n\n@Override\npublic int hashCode() {\nreturn 21521 (3491 + x) + y;\n}\n\npublic String toString() {\n// return String.format(\"(%d, %d)\", x, y);\nreturn String.format(\"setxy %d %d\", x * 50 - y * 25, y * 40);\n}\n\npublic int compareTo(Point p) {\nint a = Integer.valueOf(x).compareTo(p.x);\nif (a != 0) return a;\nreturn Integer.valueOf(y).compareTo(p.y);\n}\n\n/// Helper class.\nstatic interface Predicate {\nabstract boolean test(Point p);\n}\n\nstatic abstract class UnaryFunction {\nabstract Point apply(Point p);\n}\n\n}\n\nstatic class Edge implements Comparable<Edge> {\nfinal Point a, b; // guarantee a < b\nEdge(Point x, Point y) {\nassert x != y;\nif (x.compareTo(y) > 0) { // y < x\na = y; b = x;\n} else {\na = x; b = y;\n}\n}\n\npublic int compareTo(Edge e) {\nint x = a.compareTo(e.a);\nif (x != 0) return x;\nreturn b.compareTo(e.b);\n}\n}\n\n/// A graph consists of multiple {@code Point}s.\nstatic class Graph {\nprivate HashMap<Point, Point> points;\n\npublic Graph() {\npoints = new HashMap<Point, Point>();\n}\n\npublic Graph(Graph g) {\npoints = new HashMap<Point, Point>(g.points);\n}\n\npublic void add(Point p, Point root) {\nassert(p.isPoint());\nassert(root.isPoint());\nassert(p == root \|\| points.containsKey(root));\npoints.put(p, root);\n}\n\npublic Graph map(Point.UnaryFunction fn) {\nGraph result = new Graph();\nfor (Map.Entry<Point, Point> pq : points.entrySet()) {\nPoint p = pq.getKey(), q = pq.getValue();\nassert(p.isPoint()) : p;\nassert(q.isPoint()) : q;\np = fn.apply(p); assert(p.isPoint()) : p;\nq = fn.apply(q); assert(q.isPoint()) : q;\nresult.points.put(p, q);\n}\nreturn result;\n}\n\npublic Graph reflectX() {\nreturn this.map(new Point.UnaryFunction() {\npublic Point apply(Point p) {\nreturn p.reflectX();\n}\n});\n}\n\npublic Graph rot60() {\nreturn this.map(new Point.UnaryFunction() {\npublic Point apply(Point p) {\nreturn p.rot60();\n}\n});\n}\n\n@Override\npublic boolean equals(Object o) {\nif (o == null) return false;\nif (o.getClass() != getClass()) return false;\nGraph g = (Graph) o;\nreturn points.equals(g.points);\n}\n\n@Override\npublic int hashCode() {\nreturn points.hashCode();\n}\n\nGraph[] expand(Point.Predicate fn) {\nList<Graph> result = new ArrayList<Graph>();\n\nfor (Point p : points.keySet()) {\nint[] deltaX = new int[] { -1, 0, 1, 1, 0, -1};\nint[] deltaY = new int[] { 0, 1, 1, 0, -1, -1};\nfor (int i = 6; i --> 0;) {\nPoint p1 = new Point(p.x + deltaX[i], p.y + deltaY[i]);\nif (points.containsKey(p1) \|\| !fn.test(p1)\n\|\| !p1.isPoint()) continue;\n\nGraph g = new Graph(this);\n}\n}\n\nreturn result.toArray(new Graph);\n}\n\npublic static Graph[] expand(Graph[] graphs, Point.Predicate fn) {\nSet<Graph> result = new HashSet<Graph>();\n\nfor (Graph g0 : graphs) {\nGraph[] g = g0.expand(fn);\nfor (Graph g1 : g) {\nif (result.contains(g1)) continue;\n}\n}\n\nreturn result.toArray(new Graph);\n}\n\nprivate Edge[] edges() {\nList<Edge> result = new ArrayList<Edge>();\nfor (Map.Entry<Point, Point> pq : points.entrySet()) {\nPoint p = pq.getKey(), q = pq.getValue();\nif (p.equals(q)) continue;\n}\nreturn result.toArray(new Edge);\n}\n\n/*\n Check if two graphs are isomorphic... under translation.\n* @return {@code true} if {@code this} is isomorphic\n* under translation, {@code false} otherwise.\n/\npublic boolean isomorphic(Graph g) {\nif (points.size() != g.points.size()) return false;\nEdge[] a = this.edges();\nEdge[] b = g.edges();\nArrays.sort(a);\nArrays.sort(b);\n\n// for (Edge e : b)\n// System.err.println(e.a + \" - \" + e.b);\n// System.err.println(\"------- >><< \");\n\nassert (a.length > 0);\nassert (a.length == b.length);\nint a_bx = a.a.x - b.a.x, a_by = a.a.y - b.a.y;\nfor (int i = 0; i < a.length; ++i) {\nif (a_bx != a[i].a.x - b[i].a.x \|\|\na_by != a[i].a.y - b[i].a.y) return false;\nif (a_bx != a[i].b.x - b[i].b.x \|\|\na_by != a[i].b.y - b[i].b.y) return false;\n}\n\nreturn true;\n}\n\n// C_{2v}.\npublic boolean correctSymmetry() {\n\nGraph[] graphs = new Graph;\ngraphs = this.reflectX();\nfor (int i = 1; i < 6; ++i) graphs[i] = graphs[i-1].rot60();\nassert(graphs.rot60().isomorphic(graphs));\nint count = 0;\nfor (Graph g : graphs) {\nif (this.isomorphic(g)) ++count;\n// if (count >= 2) {\n// return false;\n// }\n}\n// if (count > 1) System.err.format(\"too much: %d%n\", count);\nassert(count > 0);\nreturn count == 1; // which is, basically, true\n}\n\npublic void reflectSelfType2() {\nGraph g = this.map(new Point.UnaryFunction() {\npublic Point apply(Point p) {\nreturn new Point(p.y - p.x, p.y);\n}\n});\n\nPoint p = new Point(1, 1);\nassert (p.equals(points.get(p)));\n\npoints.putAll(g.points);\n\nassert (p.equals(points.get(p)));\nPoint q = new Point(0, 1);\nassert (q.equals(points.get(q)));\npoints.put(p, q);\n}\n\npublic void reflectSelfX() {\nGraph g = this.reflectX();\npoints.putAll(g.points); // duplicates doesn't matter\n}\n\n}\n\nstatic int A936(int n) {\n// if (true) return (new int[]{0, 0, 0, 1, 1, 2, 4, 4, 12, 10, 29, 27, 88, 76, 247, 217, 722, 638, 2134, 1901, 6413})[n];\n\n// some unreachable codes here for testing.\nint ans = 0;\n\nif (n % 2 == 0) { // reflection type 2. (through line 2x == y)\nGraph[] graphs = new Graph;\ngraphs = new Graph();\n\nPoint p = new Point(1, 1);\n\nfor (int i = n / 2 - 1; i --> 0;)\ngraphs = Graph.expand(graphs, new Point.Predicate() {\npublic boolean test(Point p) {\nreturn 2p.x > p.y;\n}\n});\n\nint count = 0;\nfor (Graph g : graphs) {\ng.reflectSelfType2();\nif (g.correctSymmetry()) {\n++count;\n\n// for (Edge e : g.edges())\n// System.err.println(e.a + \" - \" + e.b);\n// System.err.println(\"------\");\n\n}\n// else System.err.println(\"Failed\");\n}\n\nassert (count%2 == 0);\n\n// System.err.println(\"A936(\" + n + \") count = \" + count + \" -> \" + (count/2));\n\nans += count / 2;\n\n}\n\n// Reflection type 1. (reflectX)\n\nGraph[] graphs = new Graph;\ngraphs = new Graph();\n\nPoint p = new Point(1, 0);\n\nif (n % 2 == 0) graphs.add(new Point(2, 0), p);\n\nfor (int i = (n-1) / 2; i --> 0;)\ngraphs = Graph.expand(graphs, new Point.Predicate() {\npublic boolean test(Point p) {\nreturn p.y > 0;\n}\n});\n\nint count = 0;\nfor (Graph g : graphs) {\ng.reflectSelfX();\nif (g.correctSymmetry()) {\n++count;\n// for (Edge e : g.edges())\n\n// System.err.printf(\n\n// \"pu %s pd %s\\n\"\n// // \"%s - %s%n\"\n\n// , e.a, e.b);\n// System.err.println(\"-------/\");\n\n}\n// else System.err.println(\"Failed\");\n}\n\nif(n % 2 == 0) {\nassert(count % 2 == 0);\ncount /= 2;\n}\nans += count;\n\n// System.err.println(\"A936(\" + n + \") = \" + ans);\n\nreturn ans;\n}\n\npublic static void main(String[] args) {\n\n// Probably\nif (! \"1.5.0_22\".equals(System.getProperty(\"java.version\"))) {\nSystem.err.println(\"Warning: Java version is not 1.5.0_22\");\n}\n\n// A936(6);\n\nfor (int i = 0; i < 20; ++i)\nSystem.out.println(i + \" \| \" + (A63(i+9) - A936(i+7)));\n//A936(i+2);\n}\n}\n\n\nTry it online!\n\nSide note:\n\n1. Tested locally with Java 5. (such that the warning is not printed - see TIO debug tab)\n2. Don't. Ever. Use. Java. 1. It's more verbose than Java in general.\n3. This may break the chain.\n4. The gap (7 days and 48 minutes) is no more than the gap created by this answer, which is 7 days and 1 hours 25 minutes later than the previous one.\n5. New record on large bytecount! Because I (mistakenly?) use spaces instead of tabs, the bytecount is larger than necessary. On my machine it's 9550 bytes. (at the time of writing this revision)\n6. Next sequence.\n7. The code, in its current form, only prints the first 20 terms of the sequence. However it's easy to change so that it will prints first 1000 items (by change the 20 in for (int i = 0; i < 20; ++i) to 1000)\n\nYay! This can compute more terms than listed on the OEIS page! (for the first time, for a challenge I need to use Java) unless OEIS has more terms somewhere...\n\n# Quick explanation\n\n### Explanation of the sequence description.\n\nThe sequence ask for the number of free nonplanar polyenoid with symmetry group C2v, where:\n\n• polyenoid: (mathematical model of polyene hydrocarbons) trees (or in degenerate case, single vertex) with can be embedded in hexagonal lattice.\n\nFor example, consider the trees\n\n O O O O (3)\n\| \\ / \\\n\| \\ / \\\nO --- O --- O O --- O O --- O\n\| \\\n\| (2) \\\n(1) O O\n\n\nThe first one cannot be embedded in the hexagonal lattice, while the second one can. That particular embedding is considered different from the third tree.\n\n• nonplanar polyenoid: embedding of trees such that there exists two overlapping vertices.\n\n(2) and (3) tree above are planar. This one, however, is nonplanar:\n\n O---O O\n/ \\\n/ \\\nO O\n\\ /\n\\ /\nO --- O\n\n\n(there are 7 vertices and 6 edges)\n\n• free polyenoid: Variants of one polyenoid, which can be obtained by rotation and reflection, is counted as one.", null, "• C2v group: The polyenoid are only counted if they have 2 perpendicular planes of reflection, and no more.\n\nFor example, the only polyenoid with 2 vertices\n\nO --- O\n\n\nhas 3 planes of reflection: The horizontal one -, the vertical one \|, and the one parallel to the computer screen ■. That's too much.\n\nOn the other hand, this one\n\nO --- O\n\\\n\\\nO\n\n\nhas 2 planes of reflection: / and ■.\n\n### Explanation of the method\n\nAnd now, the approach on how to actually count the number.\n\nFirst, I take the formula a(n) = A000063(n + 2) - A000936(n) (listed on the OEIS page) for granted. I didn't read the explanation in the paper.\n\n[TODO fix this part]\n\nOf course, counting planar is easier than counting nonplanar. That's what the paper does, too.\n\nGeometrically planar polyenoids (without overlapping vertices) are enumerated by computer programming. Thus the numbers of geometrically nonplanar polyenoids become accessible.\n\nSo... the program counts the number of planar polyenoid, and subtract it from the total.\n\nBecause the tree is planar anyway, it obviously has the ■ plane of reflection. So the condition boils down to \"count number of tree with an axis of reflection in its 2D representation\".\n\nThe naive way would be generate all trees with n nodes, and check for correct symmetry. However, because we only want to find the number of trees with an axis of reflection, we can just generate all possible half-tree on one half, mirror them through the axis, and then check for correct symmetry. Moreover, because the polyenoids generated are (planar) trees, it must touch the axis of reflection exactly once.\n\nThe function public static Graph[] expand(Graph[] graphs, Point.Predicate fn) takes an array of graphs, each have n nodes, and output an array of graph, each has n+1 nodes, not equal to each other (under translation) - such that the added node must satisfy the predicate fn.\n\nConsider 2 possible axes of reflection: One that goes through an vertex and coincide with edges (x = 0), and one that is the perpendicular bisector of an edge (2x = y). We can take only one of them because the generated graphs are isomorphic, anyway.", null, "So, for the first axis x = 0, we start from the base graph consists of a single node (1, 0) (in case n is odd) or two nodes with an edge between (1, 0) - (2, 0) (in case n is even), and then expand nodes such that y > 0. That's done by the \"Reflection type 1\" section of the program, and then for each generated graph, reflect (mirror) itself through the X axis x = 0 (g.reflectSelfX()), and then check if it has the correct symmetry.\n\nHowever, note that if n is divisible by 2, by this way we counted each graph twice, because we also generate its mirror image by the axis 2x = y + 3.", null, "(note the 2 orange ones)\n\nSimilar for the axis 2x = y, if (and only if) n is even, we start from the point (1, 1), generate graphs such that 2x > y, and reflect each of them over the 2x = y axis (g.reflectSelfType2()), connect (1, 0) with (1, 1), and check if they have correct symmetry. Remember to divide by 2, too.\n\n• Given that I was asleep when this (and the other one) were posted, I'll give you the benefit of the doubt and not accept an answer yet. Dec 24, 2017 at 6:21\n• @cairdcoinheringaahing You were online 3 minutes before the deadline... Dec 24, 2017 at 6:23\n• Uh oh, the next sequence can be hard-coded... (although it's infinite) if I read it correctly. The calculation itself is ---pretty--- very easy, so don't do it. Dec 24, 2017 at 12:39\n\n# 24. Julia 0.5, 33 bytes, A000023\n\nExpansion of e.g.f. exp(−2x)/(1−x).\n\n!x=foldl((a,b)->ab+(-2)^b,1,1:x)\n\n\nTry it online!\n\nNext sequence.\n\n# 156. C# (Mono), 2466 bytes, A000083\n\nNote: the score is 2439 bytes for the code and 27 for the compiler flag -reference:System.Numerics.\n\nusing Num = System.Numerics.BigInteger;\nnamespace PPCG\n{\nclass A000083\n{\nstatic void Main(string[] a)\n{\nint N = int.Parse(a) + 1;\n\nvar phi = new int[N + 1];\nfor (int i = 1; i <= N; i++)\nphi[i] = 1;\nfor (int p = 2; p <= N; p++)\n{\nif (phi[p] > 1) continue;\nfor (int i = p; i <= N; i += p)\nphi[i] = p - 1;\nint pa = p p;\nwhile (pa <= N)\n{\nfor (int i = pa; i <= N; i += pa)\nphi[i] = p;\npa = p;\n}\n}\n\nvar aik = new Num[N + 1, N + 1];\nvar a035350 = new Num[N + 1];\nvar a035349 = new Num[N + 1];\naik[0, 0] = aik[1, 1] = a035350 = a035350 = a035349 = a035349 = 1;\nfor (int n = 2; n <= N; n++)\n{\n// A000237 = EULER(A035350)\nNum nbn = 0;\nfor (int k = 1; k < n; k++)\nfor (int d = 1; d <= k; d++)\nif (k % d == 0) nbn += d * a035350[d] * aik[1, n - k];\naik[1, n] = nbn / (n - 1);\n\n// Powers of A000237 are used a lot\nfor (int k = 2; k <= N; k++)\nfor (int i = 0; i <= n; i++)\naik[k, n] += aik[k - 1, i] * aik[1, n - i];\n\n// A035350 = BIK(A000237)\nNum bn = 0;\nfor (int k = 1; k <= n; k++)\n{\nbn += aik[k, n];\nif (k % 2 == 1)\nfor (int i = n & 1; i <= n; i += 2)\nbn += aik[1, i] * aik[k / 2, (n - i) / 2];\nelse if (n % 2 == 0)\nbn += aik[k / 2, n / 2];\n}\na035350[n] = bn / 2;\n\n// A035349 = DIK(A000237)\nNum dn = 0;\nfor (int k = 1; k <= n; k++)\n{\n// DIK_k is Polyà enumeration with the cyclic group D_k\n// The cycle index for D_k has two parts: C_k and what Bower calls CPAL_k\n// C_k\nNum cikk = 0;\nfor (int d = 1; d <= k; d++)\nif (k % d == 0 && n % d == 0)\ncikk += phi[d] * aik[k / d, n / d];\ndn += cikk / k;\n\n// CPAL_k\nif (k % 2 == 1)\nfor (int i = 0; i <= n; i += 2)\ndn += aik[1, n - i] * aik[k / 2, i / 2];\nelse\n{\nNum cpalk = 0;\nfor (int i = 0; i <= n; i += 2)\ncpalk += aik[2, n - i] * aik[k / 2 - 1, i / 2];\nif (n % 2 == 0)\ncpalk += aik[k / 2, n / 2];\ndn += cpalk / 2;\n}\n}\na035349[n] = dn / 2;\n}\n\n// A000083 = A000237 + A035350 - A000237 * A035349\nvar a000083 = new Num[N + 1];\nfor (int i = 0; i <= N; i++)\n{\na000083[i] = aik[1, i] + a035349[i];\nfor (int j = 0; j <= i; j++) a000083[i] -= aik[1, j] * a035350[i - j];\n}\n\nSystem.Console.WriteLine(a000083[N - 1]);\n}\n}\n}\n\n\nOnline demo. This is a full program which takes input from the command line.\n\nNext sequence\n\n### Dissection\n\nI follow Bowen's comment in OEIS that the generating function A000083(x) = A000237(x) + A035349(x) - A000237(x) * A035350(x) where the component generating functions are related by transforms as\n\n• A000237(x) = x EULER(A035350(x))\n• A035350(x) = BIK(A000237(x))\n• A035349(x) = DIK(A000237(x))\n\nI use the definitions of BIK and DIK from https://oeis.org/transforms2.html but the formulae seem to have a number of typos. I corrected LPAL without much difficulty, and independently derived a formula for DIK based on applying Pólya enumeration to the cycle index of the dihedral group. Between #121 and #156 I'm learning a lot about Pólya enumeration. I have submitted some errata, which may prove useful to other people if these transforms come up again in the chain.\n\n# 3. JavaScript (ES6), 38 bytes, A000044\n\nf=n=>n<0?0:n<3?1:f(n-1)+f(n-2)-f(n-13)\n\n\nTry it online!\n\nNext sequence (should be an easy one :P)\n\n• \"(should be an easy one :P)\" it was Jul 21, 2017 at 15:24\n\n# 13. VB.NET (.NET 4.5), 1246 bytes, A000131\n\nPublic Class A000131\nPublic Shared Function Catalan(n As Long) As Long\nDim ans As Decimal = 1\nFor k As Integer = 2 To n\nans = (n + k) / k\nNext\nReturn ans\nEnd Function\nShared Function Answer(n As Long) As Long\n\nn += 7\n\nDim a As Long = Catalan(n - 2)\n\nDim b As Long = Catalan(n / 2 - 1)\nIf n Mod 2 = 0 Then\nb = Catalan(n / 2 - 1)\nElse\nb = 0\nEnd If\n\nDim c As Long = Catalan(n \\ 2 - 1) ' integer division (floor)\n\nDim d As Long\nIf n Mod 3 = 0 Then\nd = Catalan(n / 3 - 1)\nElse\nd = 0\nEnd If\n\nDim e As Long = Catalan(n / 4 - 1)\nIf n Mod 4 = 0 Then\ne = Catalan(n / 4 - 1)\nElse\ne = 0\nEnd If\n\nDim f As Long = Catalan(n / 6 - 1)\nIf n Mod 6 = 0 Then\nf = Catalan(n / 6 - 1)\nElse\nf = 0\nEnd If\n\nReturn (\na -\n(n / 2) b -\nn * c -\n(n / 3) * d +\nn * e +\nn * f\n) /\n(2 * n)\nEnd Function\nEnd Class\n\n\nA001246\n\nTry it Online!\n\n# 26. TI-BASIC, 274 bytes, A000183\n\n.5(1+√(5→θ\n\"int(.5+θ^X/√(5→Y₁\n\"2+Y₁(X-1)+Y₁(X+1→Y₂\n{0,0,0,1,2,20→L₁\nPrompt A\nLbl A\nIf A≤dim(L₁\nThen\nDisp L₁(A\nElse\n1+dim(L₁\n(~1)^Ans(4Ans+Y₂(Ans))+(Ans/(Ans-1))((Ans+1))-(2Ans/(Ans-2))((Ans-3)L₁(Ans-2)+(~1)^AnsY₂(Ans-2))+(Ans/(Ans-3))((Ans-5)L₁(Ans-3)+2(~1)^(Ans-1)Y₂(Ans-3))+(Ans/(Ans-4))(L₁(Ans-4)+(~1)^(Ans-1)Y₂(Ans-4→L₁(Ans\nGoto A\nEnd\n\n\nEvaluates the recursive formula found on the OEIS link.\n\nNext Sequence\n\n• Agh I knew when the site went down that it would be a mad rush when it came back up. Barely beat me. Jul 22, 2017 at 0:33\n• I didn't realize the site went down... Jul 22, 2017 at 0:33\n• twitter.com/Nick_Craver/status/888553194313449472 Jul 22, 2017 at 0:34\n\n# 39. Add++, 1 byte, A000004\n\nO\n\n\nTry it online!\n\nNext sequence\n\n# 91. Python 2 (PyPy), 1733 bytes, A000066\n\nimport itertools\n\ngirth = int(input()) + 3\n\nv = 4\n\nr = range\n\ndef p(v):\na = [0 for i in r(v)]\nk = int((v * 2) ** .5)\na[k - 1] = a[k - 2] = a[k - 3] = 1\nj = len(a) - 1\nfor i in r(1, 3):\na[j] = 1\nj -= i\nyield [x for x in a]\nwhile not all(a):\nfor index in r(len(a) - 1, -1, -1):\na[index] ^= 1\nif a[index]: break\nyield [x for x in a]\n\ndef wrap_(p, v):\nm = [[0 for j in r(v)] for i in r(v)]\nk = 0\nfor i in r(0, v - 1):\nfor j in r(i + 1, v):\nm[i][j] = m[j][i] = p[k]\nk += 1\nreturn m\n\ndef completes_cycle(edgelist):\nif not edgelist or not edgelist[1:]: return False\nstart = edgelist\nedge = edgelist\ne = [x for x in edgelist]\nedgelist = edgelist[1:]\nwhile edgelist:\n_edges = [_edge for _edge in edgelist if _edge in edge or _edge in edge]\nif _edges:\nedgelist.remove(_edges)\nif _edges in edge: _edges = (_edges, _edges)\nedge = _edges\nelse:\nreturn False\nflat = sum(e, ())\nfor i in flat:\nif flat.count(i) != 2: return False\nreturn edge in start\n\ndef powerset(a):\nreturn sum([list(itertools.combinations(a, t)) for t in r(len(a))], [])\n\nwhile True:\nps = (v * (v - 1)) // 2\nskip = False\nfor Q in p(ps):\nm = wrap_(Q, v)\noutput = [row + for row in m]\noutput.append([0 for i in r(len(m))])\nfor i in r(len(m)):\noutput[i][-1] = sum(m[i])\noutput[-1][i] = sum(row[i] for row in m)\nif all(map(lambda x: x == 3, map(sum, m))):\nedges = []\nfor i in r(v):\nfor j in r(i, v):\nif m[i][j]: edges.append((i, j))\nfor edgegroup in powerset(edges):\nif completes_cycle(list(edgegroup)):\nif len(edgegroup) == girth:\nprint(v)\nexit(0)\nelse:\nskip = True\nbreak\nif skip: break\nv += 1\n\n\nTry it online!\n\nI hope using Python 2 PyPy counts as another major version. If someone could get me a Python 0 interpreter, I could use that too, but I hope this is valid.\n\nThis starts at 1 vertex and works up, creating the adjacency matrix representation of every possible undirected graph with that many vertices. If it is trivalent, then it will look through the powerset of the edges, which will be sorted by length. If the first cycle it finds is too short, then it will move on. If the first cycle it finds matches the input (offset by 3) then it will output the correct vertex count and terminate.\n\nNext Sequence <-- have an easy one as a break from all this math nonsense :D\n\nEDIT: I added some optimizations to make it a bit faster (still can't compute the third term within TIO's 60 second limit though) without changing the bytecount.\n\n• ... and I was seriously thinking the chain would end with answer 90 Jul 31, 2017 at 0:04\n• @ppperry :) I like doing hard challenges because most people can't even make a solution so I don't have to worry about getting outgolfed :) (e.g. the carbon chain namer problem) Jul 31, 2017 at 0:25\n• Unless someone takes your solution and converts it into a terser language Jul 31, 2017 at 0:26\n• @ppperry that too o_O :P Jul 31, 2017 at 0:26\n• @HyperNeutrino Congrats on solving that! I was worried I had broken the chain, and was considering padding the byte count to point to a different sequence. Good job! Jul 31, 2017 at 22:27\n\n# 250. Coconut, 711 bytes, A000171\n\nfrom collections import Counter\nfrom math import gcd, factorial\n\ndef sparts(n, m=4):\nif n%2==1:\nreturn [ + p for p in sparts(n-1) ]\nelif n==0:\nreturn [ [] ]\nelif n<m:\nreturn []\nelse:\nreturn [ [m] + p for p in sparts(n-m,m) ] + sparts(n,m+4)\n\ndef ccSize(l) =\ncentSize = [ val*mult factorial(mult)\nfor val, mult in Counter(l).items() ] \|> reduce\\$()\nfactorial(sum(l)) // centSize\n\ndef edgeorbits(l) =\nsamecyc = sum(l) // 2\ndiffcyc = sum ([ gcd(l[i],l[j])\nfor i in range(len(l)) for j in range(i) ])\nsamecyc + diffcyc\n\ndef a(n) =\nsum ([ ccSize(l)2*edgeorbits(l)\nfor l in sparts(n) ]) // factorial(n)\n\ndef f(n) = a(n+1)\n\n\nTry it online!\n\nNext sequence\n\nThis can easily compute more values than the 31 that OEIS has.\n\nOnce again we have a symmetric group acting on nodes, inducing an action on the set of possible edges and on graphs. It acts on the set of self-complementary graphs, but I don't see how to count how many of these are fixed by a given permutation to apply the lemma that is often named after Burnside. So instead I use the formula in the comment of the OEIS entry that says that the number of self-complementary graphs with n nodes is equal to the difference of the number of graphs with n nodes with an even number of edges and with an odd number of edges. The symmetric group acts on these sets of graphs as well. Instead of computing both numbers independently, we will consider the cases of an even number and an odd number of edges in parallel, which will turn out to be simpler and allow optimizations. We'll see that \"evaluate the graph polynomial at -1\" will lead to the same result.\n\nAs usual, we take the sum over the group elements in Burnside's lemma conjugacy class wise, and represent the conjugacy classes by partitions of n. For each conjugacy class, we want to know the difference between the numbers of graphs with an even and odd number of edges that is fixed by a permutation from that class.\n\nAssume we are given a concrete permutation g. The group generated by g acts on the set of possible edges and partitions them into orbits. A graph is fixed by g if for each orbit, it contains either all or none of its edges. Let's look at the orbits. Each edge in one orbit has its endpoints in the same two (possibly same) cycles of g.\n\nLet's first assume the endpoints are from the same cycle of length k>=2. If k is odd, then there are (k-1)/2 orbits of length k. If k is even, then there are k/2-1 orbits of length k and one of length k/2 (for example, if g contains the cycle (123456), then the there is an edge orbit of size 3 containing {1,4}). Note that we get all orbits of even size exactly when k is a multiple of four.\n\nNext consider the case that the endpoints are from different cycles of lengths k and l. Then the orbit has size lcm(k,l), and there are gcd(k,l) such orbits.\n\nIn order to combine all the possibilities, consider the product, for each edge orbit, of 1+x^r, where r is the size of the corresponding orbit. The coefficient of x^s of the product gives the number of graphs that are fixed by g and have s edges. We want the sum of the coefficients at even exponents minus the sum of them at odd coefficients, which we get by evaluating the polynomial at x=-1.\n\nIf we kept the polynomials and added them together, we'd get the graph polynomial. Instead, we don't even create these polynomials, but do the substitution immediately. For an edge orbit of even size, we get a factor of 2, for an odd size we get 0. (There is also a direct combinatorial argument that if there is an edge orbit of odd size then there are as many fixed graphs with an even number of edges as there are with an odd number of edges. And the factor 2 corresponds to the choice of including the edges of one orbit to the fixed graph or not).\n\nSo, by the considerations about edges with points from only one cycle, we get a difference of 0 whenever g contains a cycle of odd length greater than one, or a cycle of even length not divisible by four. We also get a difference of 0 whenever g contains at least two fixed points (cycles of length one).\n\nThis means we only need to sum over conjugacy classes with at most one fixed point and all other cycle lengths multiples of four. sparts computes the corresponding partitions of n. There are none unless n is of the form 4k or 4k+1, so in the other cases there are no self-complementary graphs, which can also easily seen by noting that in these cases the total number of possible edges is odd. But this observations alone doesn't lead to the optimization of only considering special partitions, of which there are only as much as there are partitions of k.\n\nccSize computes the size of a conjugacy class with given cycle lengths. edgeorbits computes the number of edge orbits of the action of the group generated by a permutation with the given cycle lengths, assuming they are of the special form. In the general case, samecyc, the number of edge orbits with edges that have both nodes from the same cycle, could not be calculated like this. (When I wrote the code, I didn't notice that it only depends on n in the relevant cases.)\n\na puts everything together, and f fixes the different indexing.\n\n• This looks like Python with a touch of JavaScript added :P Nov 12, 2017 at 19:07\n• @ChristianSievers I'm stupid. Anyway, don't forget that some PPCGers have OEIS accounts and could add more values to A000171 :) Nov 12, 2017 at 21:38\n• @Stephen I consider doing that Nov 12, 2017 at 22:11\n• Can you explain what algorithm are you using and/or what does the abbreviated function names (sparts, ccSize) means? Nov 13, 2017 at 14:17\n• @user202729 I finally added some explanation. Nov 15, 2017 at 18:18\n\n# 11. Pari/GP, 64 bytes, A000065\n\n{a(n) = if( n<0, 0, polcoeff ( 1 / eta(x + xO(x^n) ), n) - 1)};\n\n\nTry it online!\n\nNext sequence\n\n• Is that valid input? Jul 21, 2017 at 17:35\n• Didya have to get 64 bytes? :P Jul 21, 2017 at 17:40\n• @totallyhuman yes: ;_; I solved A000064 and you changed it. Downvoted. Jul 21, 2017 at 17:41\n• @totallyhuman compromises lol. see chat Jul 21, 2017 at 17:41\n• Dang Jul 21, 2017 at 17:42\n\n# 318. Funciton, 1688 bytes, A000652\n\n╔═══╗ ┌─────────╖ ┌─┐ ┌───╖ ╔═══╗\n║ ╟──┤ str→int ╟──┤ └─┤ ↑ ╟──╢ 2 ║\n╚═══╝ ╘═════════╝ │ ╘═╤═╝ ╚═══╝\n┌─────────────────┘ │\n┌─┴┐ │\n│┌─┴─╖ ┌───┴───┐\n││ ♭ ║ ┌─┴─╖ ┌─┴─╖\n│╘═╤═╝┌───╖ ╔═══╗│ ♭ ║ │ ! ║\n│ └──┤ ↑ ╟────╢ 2 ║╘═╤═╝ ╘═╤═╝\n│ ╘═╤═╝ ╚═╤═╝ │ │\n│ ┌──┴─┐ │ └───┐ │\n│ ┌─┴─╖ │┌───╖ │ ┌───╖ │ │\n│ │ ! ║ └┤ ↑ ╟─┴──┤ ↑ ╟─┘ │\n│ ╘═╤═╝ ╘═╤═╝ ╘═╤═╝ │\n│ │ ┌───╖ │ ┌───╖│ │\n│ └─┤ × ╟─┘ ┌─┤ × ╟┘ │\n│ ╘═╤═╝ │ ╘═╤═╝ ┌───╖ │\n│ ╔═══╗ └─────┘ └───┤ + ╟──┘\n│ ║ 2 ╟┐ ╘═╤═╝\n│ ╚═╤═╝│ ┌───╖ │\n│ │ └──┐ ┌─┤ ÷ ╟───┘\n│ └──┐┌─┴─╖ │ ╘═╤═╝┌─────────╖\n│ ┌───╖ ││ ↑ ╟─┘ └──┤ int→str ╟─\n└─┤ × ╟─┘╘═╤═╝ ╘═════════╝\n╘═╤═╝ │\n└──────┘\n\n\nNext Sequence\n\nCalculates the formula from the OEIS page. I'm sure there's a more compact way to do this, without multiple 2 constants, but this took me long enough.\n\nTry it online!\n\n# 121. Pip, 525 bytes, A000022\n\nn:(a+1)//2\nt:[RL(3a+3)PE1]\nFh,n{\nm:(RL(3a+3))\nFi,(a+1){\nFj,(a+1){\nFk,(a+1)m@(i+j+k)+:(t@h@i)(t@h@j)(t@h@k)\nm@(i+2j)+:3(t@h@i)(t@h@j)\n}\nm@(3i)+:2(t@h@i)\n}\nt:(tAE(m//6PE1))\n}\nk:t@n\no:0\nFh,aFi,aFj,aI(h+i+j<a)o+:(k@h)(k@i)(k@j)k@(a-1-h-i-j)\nFh,((a+1)//2){\nFi,aI(2h+i<a){o+:6(k@h)(k@i)(k@(a-1-2h-i))}\nI(a%2=1)o+:3(k@h)(k@((a-1-2h)//2))\n}\nFh,((a+2)//3)o+:8(k@h)(k@(a-1-3h))\nI(a%4=1)o+:6k@(a//4)\no//:24\nIa(o+:t@n@a)\nFh,nFj,(a+1)o-:(t@(h+1)@j-t@h@j)(t@(h+1)@(a-j))\no\n\n\nOnline demo\n\nNext sequence\n\nFun fact: when the challenge was first posted, I drew up a list of small nasty sequence numbers that I wanted to aim for with CJam, and A000022 was at the top of the list.\n\nThis implements the generating function described in E. M. Rains and N. J. A. Sloane, On Cayley's Enumeration of Alkanes (or 4-Valent Trees), Journal of Integer Sequences, Vol. 2 (1999), taking the sum for Ck to as many terms as a necessary for the nth coefficient to be fixed and then telescoping three quarters of the sum. In particular, telescoping the first half means that the cycle index of S4 only has to be applied to one of the Th rather than to all of them.\n\nThe code breaks down as\n\n; Calculate the relevant T_h\nt:[RL(3a+3)PE1]\nFh,n{\nm:(RL(3a+3))\nFi,(a+1){\nFj,(a+1){\nFk,(a+1)m@(i+j+k)+:(t@h@i)(t@h@j)(t@h@k)\nm@(i+2j)+:3(t@h@i)(t@h@j)\n}\nm@(3i)+:2(t@h@i)\n}\nt:(tAE(m//6PE1))\n}\n\n; Calculate the cycle index of S_4 applied to the last one\nk:t@n\no:0\nFh,aFi,aFj,aI(h+i+j<a)o+:(k@h)(k@i)(k@j)k@(a-1-h-i-j)\nFh,((a+1)//2){\nFi,aI(2h+i<a){o+:6(k@h)(k@i)(k@(a-1-2h-i))}\nI(a%2=1)o+:3(k@h)(k@((a-1-2h)//2))\n}\nFh,((a+2)//3)o+:8(k@h)(k@(a-1-3h))\nI(a%4=1)o+:6k@(a//4)\no//:24\n\n; Handle the remaining convolution,\n; pulling out the special case which involves T_{-2}\nIa(o+:t@n@a)\nFh,nFj,(a+1)o-:(t@(h+1)@j-t@h@j)(t@(h+1)@(a-j))\n\n\nNote that this is my first ever Pip program, so is probably not very idiomatic.\n\n• Comments are not for extended discussion; this conversation has been moved to chat. Aug 31, 2017 at 12:28\n\n# 6. R, 71 bytes, A000072\n\nfunction(n)length(unique((t<-outer(r<-(0:2^n)^2,r4,\"+\"))[t<=2^n&t>0]))\n\n\nTry it online!\n\nNext sequence\n\n• For the love of God, I didn't check the next sequence before I posted this answer. Jul 21, 2017 at 15:38\n• Isn't an easy next sequence a strategic advantage? Jul 21, 2017 at 15:39\n• @BlackCap They can't answer twice in a row or less than 1 hour after they last answered. Jul 21, 2017 at 15:41\n• @EriktheOutgolfer the answer before the last posted (the one who didn't break the chain) will win Jul 21, 2017 at 15:41\n• @BlackCap at this point that isn't going to happen Jul 21, 2017 at 15:42\n\n# 14. Python 2, 60 bytes, A001246\n\nf=lambda x:x<1or xf(x-1)\nc=lambda n:(f(2n)/f(n)/f(n+1))2\n\n\nTry it online!\n\nNext sequence.\n\n• Wow you ninja'd by 2 seconds Jul 21, 2017 at 18:13\n\n# 34. Prolog (SWI), 168 bytes, A000073\n\ntribonacci(0,0).\ntribonacci(1,0).\ntribonacci(2,1).\ntribonacci(A,B):-\nC is A-1,\nD is A-2,\nE is A-3,\ntribonacci(C,F),\ntribonacci(D,G),\ntribonacci(E,H),\nB is F+G+H.\n\n\nTry it online!\n\nNext sequence\n\n# 49. SageMath, 74 bytes, A000003\n\nlambda n: len(BinaryQF_reduced_representatives(-4n, primitive_only=True))\n\n\nTry it online!\n\nNext sequence\n\n• And I just spent an hour trying to work this sequence out using JavaScript... oh well, I'll just have to move on to the next one... Jul 22, 2017 at 15:20" ]	[ null, "https://i.stack.imgur.com/v2DNh.png", null, "https://i.stack.imgur.com/wJBcY.png", null, "https://i.stack.imgur.com/rtNMc.png", null, "https://i.stack.imgur.com/Jr65M.png", null, "https://i.stack.imgur.com/1USCZm.png", null, "https://i.stack.imgur.com/SGf7Um.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6062263,"math_prob":0.98059654,"size":10373,"snap":"2023-40-2023-50","text_gpt3_token_len":2830,"char_repetition_ratio":0.11167904,"word_repetition_ratio":0.0021413276,"special_character_ratio":0.29326135,"punctuation_ratio":0.2060164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99742264,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,7,null,7,null,7,null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T11:55:28Z\",\"WARC-Record-ID\":\"<urn:uuid:1de183e8-dfd4-4756-a4d2-9174a0e12c2b>\",\"Content-Length\":\"655414\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b2f55606-4245-4a24-887f-0a407b0c9a3e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e6de97e3-a888-42b7-81b9-1b0a272a6d59>\",\"WARC-IP-Address\":\"104.18.11.86\",\"WARC-Target-URI\":\"https://codegolf.stackexchange.com/questions/133754/one-oeis-after-another/152800\",\"WARC-Payload-Digest\":\"sha1:D5APDILFXA5FGQRUHDSWJRDF5XMZRZM3\",\"WARC-Block-Digest\":\"sha1:REVHUX4QMPRERGIRBUJKNA4IDKON2GRH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511075.63_warc_CC-MAIN-20231003092549-20231003122549-00181.warc.gz\"}"}
https://groupprops.subwiki.org/wiki/Minimal_splitting_field_need_not_be_unique	[ "# Minimal splitting field need not be unique\n\n## Statement\n\n### In characteristic zero\n\nLet", null, "$G$ be a finite group. It is possible for", null, "$G$ to have two distinct non-isomorphic minimal splitting fields", null, "$K$ and", null, "$L$ in characteristic zero. In other words, both", null, "$K$ and", null, "$L$ are splitting fields, no proper subfield of either is a splitting field, and", null, "$K$ is not isomorphic to", null, "$L$.\n\n### In prime characteristic\n\nNot sure whether there are examples here.\n\n## Proof\n\n### Example of the quaternion group\n\nThe quaternion group of order eight has many different minimal splitting fields in characteristic zero. Specifically the following are true:\n\n•", null, "$\\mathbb{Q}$ is not a splitting field.\n• Any field of the form", null, "$\\mathbb{Q}(\\alpha,\\beta)$ where", null, "$\\alpha^2 + \\beta^2 = -1$ is a splitting field.\n\nThus, any field of the form", null, "$\\mathbb{Q}(\\sqrt{-m^2 - 1}) = \\mathbb{Q}[t]/(t^2 + m^2 + 1)$, where", null, "$m \\in \\mathbb{Q}$, is a quadratic extension of", null, "$\\mathbb{Q}$ satisfying the condition for being a splitting field, and hence is a minimal splitting field. There are multiple non-isomorphic fields of this type, such as", null, "$\\mathbb{Q}(i) = \\mathbb{Q}[t]/(t^2 + 1)$ and", null, "$\\mathbb{Q}(\\sqrt{-2}) = \\mathbb{Q}[t]/(t^2 + 2)$." ]	[ null, "https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ", null, "https://groupprops.subwiki.org/w/images/math/d/f/c/dfcf28d0734569a6a693bc8194de62bf.png ", null, "https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ", null, "https://groupprops.subwiki.org/w/images/math/d/2/0/d20caec3b48a1eef164cb4ca81ba2587.png ", null, "https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ", null, "https://groupprops.subwiki.org/w/images/math/d/2/0/d20caec3b48a1eef164cb4ca81ba2587.png ", null, "https://groupprops.subwiki.org/w/images/math/a/5/f/a5f3c6a11b03839d46af9fb43c97c188.png ", null, "https://groupprops.subwiki.org/w/images/math/d/2/0/d20caec3b48a1eef164cb4ca81ba2587.png ", null, "https://groupprops.subwiki.org/w/images/math/d/4/5/d45a4aa156a8ac07ab80e7d9cf5fa79f.png ", null, "https://groupprops.subwiki.org/w/images/math/a/2/9/a292f6614e2b102f6a06142b95c833d1.png ", null, "https://groupprops.subwiki.org/w/images/math/0/2/3/023af01b243a17ffa191d06673b46722.png ", null, "https://groupprops.subwiki.org/w/images/math/c/6/e/c6e197d4adef8e7fee9a18a4f9ba7a09.png ", null, "https://groupprops.subwiki.org/w/images/math/b/d/c/bdc13cfd5747267a53eb5837f8cf92ab.png ", null, "https://groupprops.subwiki.org/w/images/math/d/4/5/d45a4aa156a8ac07ab80e7d9cf5fa79f.png ", null, "https://groupprops.subwiki.org/w/images/math/3/9/2/3926b27c8ca6938bb61c5b75f728c58f.png ", null, "https://groupprops.subwiki.org/w/images/math/a/7/4/a74e9acb254a4ce8d83d3b2a52cd089f.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89056444,"math_prob":0.9999032,"size":1367,"snap":"2022-05-2022-21","text_gpt3_token_len":263,"char_repetition_ratio":0.1995598,"word_repetition_ratio":0.037558686,"special_character_ratio":0.17922458,"punctuation_ratio":0.09130435,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998534,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,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,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T15:10:19Z\",\"WARC-Record-ID\":\"<urn:uuid:364d0ae8-d562-4b7f-81ce-49a32e97c6e2>\",\"Content-Length\":\"27003\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bed64b93-bc7b-4775-9a34-9c1230e4f5b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ac21cf4-5273-4668-a25b-ae251a5f0ca9>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/wiki/Minimal_splitting_field_need_not_be_unique\",\"WARC-Payload-Digest\":\"sha1:XTBZ7FUXRRF33AX65F443FMEWA6NKGQT\",\"WARC-Block-Digest\":\"sha1:HOU7KWMMUP3H7UWR2BK35HKKAVLID25C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662606992.69_warc_CC-MAIN-20220526131456-20220526161456-00315.warc.gz\"}"}
https://www.learncram.com/dav-solutions/dav-class-8-maths-chapter-14-worksheet-3/	[ "# DAV Class 8 Maths Chapter 14 Worksheet 3 Solutions\n\nThe DAV Maths Book Class 8 Solutions Pdf and DAV Class 8 Maths Chapter 14 Worksheet 3 Solutions of Mensuration offer comprehensive answers to textbook questions.\n\n## DAV Class 8 Maths Ch 14 Worksheet 3 Solutions\n\nQuestion 1.\nFind the length of the side of a cube whose total surface area measures 600 cm2.\nSolution:\nLet the length of the cube be l cm\n∴ The total surface area of the cube = 6l2\n⇒ 600 = 6l2\n⇒ l2 = 100\n⇒ l = 10 cm\nHence, the length of the side of the cube is 10 cm.", null, "Question 2.\nThe length, breadth, and height of a room are 5 m, 4 m, and 3 m respectively. Find the cost of whitewashing the inner of the room and ceiling at the rate of ₹ 50 per square meter.\nSolution:\nHere l = 5 m, 6 = 4 m and h = 3 m\n∴ Area of the walls and the ceiling to be whitewashed = area of the 4 walls + area of the ceiling\n= 2[bh + lh] + lb\n= 2[4 × 3 + 5 × 3] + 5 × 4\n= 2[12 + 15] + 20\n= 54 + 20\n= 74 m2\nThe cost of whitewashing = 74 × 50 = ₹ 3700.\n\nQuestion 3.\nFind the total surface area of a closed cardboard box of length 0.5 m, breadth 25 cm, and height 15 cm.\nSolution:\nHere l = 0.5 m = 50 cm, b = 25 cm and h = 15 cm\n∴ T.S.A. = 2[lb + bh + lh]\n= 2[50 × 25 + 25 × 15 + 50 × 15]\n= 2[1250 + 375 + 750]\n= 2\n= 4750 cm2\nHence, T.S.A. = 4750 cm2.", null, "Question 4.\nYou are given two boxes. Which box will need more paper to cover the whole box?", null, "", null, "Solution:\n(i) Total surface of area of the cuboid = 2[lb + bh + lh]\n= 2[50 × 40 + 40 × 30 + 50 × 30]\n= 2[2000 + 1200 + 1500]\n= 2\n= 9400 cm2\n\n(ii) Total surface area of the cube = 6l2\n= 6 × (40)2\n= 6 × 40 × 40\n= 9600 cm2\nHence, (ii) box will need more paper to cover.\n\nQuestion 5.\nThe dimensions of an oil tin are 26 cm × 26 cm × 45 cm. Find the area of tin sheet required to make 20 such tins.\nSolution:\nHere l = 26 cm, b = 26 cm and h = 45 cm\n∴ T.S.A. of the tin sheet = 2[lb + bh + lh]\n= 2126 × 26 + 26 × 45 + 26 × 45]\n= 2[676 + 1170 + 1170]\n= 2\n= 6032 cm2\n∴ The area of the tin sheet to make 20 such tins = 6032 × 20 = 120640 cm2 = 12.064 m2.\n\nQuestion 6.\nA swimming pool is 20 m in length, 15 m in breadth, and 4 m in depth. Find the cost of cementing its floor and walls at ₹ 35 per m2.\nSolution:\nArea of the walls and floor = 2 [lh + bh] + lb\n= 2[20 × 4 + 15 × 4] + 20 × 15\n= 2[80 + 60] + 300\n= 2 × 140 + 300\n= 280 + 300\n= 580 m2\n∴ The cost of cementing = 580 × 35 = ₹ 20300.", null, "Question 7.\nA cubical box with a lid has a length of 30 cm. Find the cost of painting the inside and outside of the box at ₹ 5.50 of per m2.\nSolution:\nThe total surface area of the cubical box outside and inner side = 2 × 6l2\n= 2 × 6 × 30 × 30\n= 10800 cm2\n= 1.08 m2\n∴ The cost of painting the box on both sides = ₹ 5.50 × 1.08 = ₹ 5.94.\n\nQuestion 8.\nTwo cubes of side 4 cm are fixed together. Find the total surface area of the new solid formed.\nSolution:\nWhen the two cubes are fixed together the new solid formed will be a cuboid whose length = 4 + 4 = 8 cm" ]	[ null, "https://i1.wp.com/www.learncram.com/wp-content/uploads/2023/07/Learn-Cram.jpg", null, "https://i1.wp.com/www.learncram.com/wp-content/uploads/2023/07/Learn-Cram.jpg", null, "https://i2.wp.com/www.learncram.com/wp-content/uploads/2023/08/DAV-Class-8-Maths-Chapter-14-Worksheet-3-Solutions-Q4.png", null, "https://i0.wp.com/www.learncram.com/wp-content/uploads/2023/08/DAV-Class-8-Maths-Chapter-14-Worksheet-3-Solutions-Q4.1.png", null, "https://i1.wp.com/www.learncram.com/wp-content/uploads/2023/07/Learn-Cram.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7530399,"math_prob":0.9996444,"size":3014,"snap":"2023-40-2023-50","text_gpt3_token_len":1171,"char_repetition_ratio":0.14385381,"word_repetition_ratio":0.06901218,"special_character_ratio":0.45686796,"punctuation_ratio":0.10419682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998221,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T18:48:19Z\",\"WARC-Record-ID\":\"<urn:uuid:7faa0474-ad3f-4695-9508-57fbed0cfbe9>\",\"Content-Length\":\"46111\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4e49fee-ebdc-47c5-9df3-6e5d851e4105>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e22215e-aa73-4af1-a100-b504298c8aac>\",\"WARC-IP-Address\":\"68.183.85.35\",\"WARC-Target-URI\":\"https://www.learncram.com/dav-solutions/dav-class-8-maths-chapter-14-worksheet-3/\",\"WARC-Payload-Digest\":\"sha1:7LBPPMRLNHTC2LF7CN4Y5HXMAKSAZ3TZ\",\"WARC-Block-Digest\":\"sha1:4HOEUNSHRZHUTS7AWR6BKOPZCWRPKMSW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510219.5_warc_CC-MAIN-20230926175325-20230926205325-00244.warc.gz\"}"}
https://www.numbersaplenty.com/524863	[ "Search a number\nBaseRepresentation\nbin10000000001000111111\n3222122222101\n42000020333\n5113243423\n615125531\n74314133\noct2001077\n9878871\n10524863\n11329379\n122138a7\n13154b91\n14d93c3\nhex8023f\n\n524863 has 2 divisors, whose sum is σ = 524864. Its totient is φ = 524862.\n\nThe previous prime is 524857. The next prime is 524869. The reversal of 524863 is 368425.\n\nIt is a balanced prime because it is at equal distance from previous prime (524857) and next prime (524869).\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 524863 - 25 = 524831 is a prime.\n\nIt is a congruent number.\n\nIt is not a weakly prime, because it can be changed into another prime (524869) by changing a digit.\n\nIt is a good prime.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 262431 + 262432.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (262432).\n\n2524863 is an apocalyptic number.\n\n524863 is a deficient number, since it is larger than the sum of its proper divisors (1).\n\n524863 is an equidigital number, since it uses as much as digits as its factorization.\n\n524863 is an evil number, because the sum of its binary digits is even.\n\nThe product of its digits is 5760, while the sum is 28.\n\nThe square root of 524863 is about 724.4742921595. The cubic root of 524863 is about 80.6644145566.\n\nThe spelling of 524863 in words is \"five hundred twenty-four thousand, eight hundred sixty-three\"." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8755578,"math_prob":0.99538887,"size":1574,"snap":"2023-14-2023-23","text_gpt3_token_len":498,"char_repetition_ratio":0.16242038,"word_repetition_ratio":0.014545455,"special_character_ratio":0.42947903,"punctuation_ratio":0.123809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99259835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T07:12:46Z\",\"WARC-Record-ID\":\"<urn:uuid:db15a9b7-c977-4950-aa25-e6bda4bdcd4a>\",\"Content-Length\":\"7861\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2179faf2-bf13-4f17-84c9-674a8ac2c759>\",\"WARC-Concurrent-To\":\"<urn:uuid:350ee5de-adfb-42d0-8b31-cb3bb98492bf>\",\"WARC-IP-Address\":\"89.46.108.74\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/524863\",\"WARC-Payload-Digest\":\"sha1:XIUKBAHK64ABOEQLZHGGJEB5PT7M5WBX\",\"WARC-Block-Digest\":\"sha1:WVWPR3ZZ5RZC6KUKAZABLSNNIPACOOEZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655446.86_warc_CC-MAIN-20230609064417-20230609094417-00274.warc.gz\"}"}
http://safecurves.cr.yp.to/proof/81371.html	[ "Primality proof for n = 81371:\n\nTake b = 2.\n\nb^(n-1) mod n = 1.\n\n103 is prime.\nb^((n-1)/103)-1 mod n = 27717, which is a unit, inverse 320.\n\n79 is prime.\nb^((n-1)/79)-1 mod n = 76290, which is a unit, inverse 4340.\n\n(79 * 103) divides n-1.\n\n(79 * 103)^2 > n.\n\nn is prime by Pocklington's theorem." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72311294,"math_prob":0.9997826,"size":289,"snap":"2019-13-2019-22","text_gpt3_token_len":120,"char_repetition_ratio":0.14385965,"word_repetition_ratio":0.036363635,"special_character_ratio":0.5467128,"punctuation_ratio":0.18181819,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993274,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T10:54:15Z\",\"WARC-Record-ID\":\"<urn:uuid:492fcf99-098e-4bd6-ab71-ac941b669154>\",\"Content-Length\":\"546\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a12b8369-7168-4d6e-9a8e-bc1a9a556e9f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6a9f033-235e-4db1-9d59-5355745dd92a>\",\"WARC-IP-Address\":\"131.193.32.108\",\"WARC-Target-URI\":\"http://safecurves.cr.yp.to/proof/81371.html\",\"WARC-Payload-Digest\":\"sha1:APPLE6KAOYSQZBNSBLMH76D4OK53XGHU\",\"WARC-Block-Digest\":\"sha1:S3A7JQ2BEU2TRXAPXIEDT4IUEG2CQ7E6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201953.19_warc_CC-MAIN-20190319093341-20190319115341-00176.warc.gz\"}"}