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https://www.mathewsopenaccess.com/full-text/finite-element-analysis-f-e-a-an-insight	[ "### Current Issue Volume 3, Issue 2 - 2018\n\nCommentary Article\n\n# Finite Element Analysis (F E A) - An Insight\n\nRohit Kulshrestha*\n\nSenior Lecturer, Department of Orthodontics and Dentofacial Orthopedics, Terna Dental College and Hospital, Navi Mumbai, Maharashtra, India.\n\nCorresponding Author: Rohit Kulshrestha, Senior Lecturer, Department of Orthodontics and Dentofacial Orthopedics, Terna Dental College and Hospital, Navi Mumbai, Maharashtra, India.\n\nE-Mail: [email protected]\n\nAccepted Date: 30 Oct 2018\nPublished Date: 31 Oct 2018\n\nCitation: Kulshrestha R. (2018). Finite Element Analysis (F E A) - An Insight. Mathews J Dentistry. 3(2): 022.\n\nINTRODUCTION\n\nFinite Element Analysis is a numerical method of structure analysis based on principle of dividing an infinite structure into a finite number of small elements connects each other at the corner points or nodes. It is a means of discretizing a continuous structure into sub-domains called Finite Elements. Essentially an attempt at simulating a physical object and analysing it's behaviour when subjected to various circumstances. This is a well-sophisticated engineering tool, which has been used extensively in design optimization and structural analysis. It is one of the most significant developments in the history of computational methods. This method is originated in aerospace industry to study stresses in complex airframe structures. Modern version of finite element method first used in engineering by Turner et al. The term finite element was coined by Argyris and Clough in 1960. First introduced to the dental arena in the 1970's and growth model was documented by MOSS in 1980.\n\n1. Accuracy\n2. Reproducibility\n3. No usage of materials\n4. Generation of intra-material results\n\nBasic steps involved in FEA\n\n1. Pre-processing\n2. Processing\n3. Post-processing\n\nPre-processing\nPre-Processing basically involves modelling of the structure being studied. It is the most crucial step in the finite element analysis. In Pre-processing, the structure being studied is discretised into smaller units termed the elements. Each element is free to get displaced in all the three planes of space.\n\nThe element co-ordinates (x,y,z) can be either\n\n1. Global Co-ordinate system or\n2. The Local Co-ordinate system\n\nVarious categories of elements exist. Examples are\n\n• Shell element\n• Beam element\n• Truss element etc.\n\nNewer possibilities of modelling of complex structures includes\n\n1. 3-D CT scanning\n2. 3-D Laser scanner\n3. Voxel modelling\n\nThese elements are connected at certain points termed 'Nodes'. The joining of elements into nodes and eliminating duplicate nodes is termed as 'Meshing'.\n\nThe mesh size is a crucial determinant of the accuracy of the result. However, it is inversely related to the time involved in the analysis. The meshed model is now a free-floating body. To simulate the exact structure, the material properties are assigned and boundary conditions enforced.\n\nMATERIAL PROPERTIES\n\nThe minimum properties to be assigned are\n\n1. The Modulus of elasticity\n2. Poisson's ratio\n\nModulus of elasticity (Young's modulus) refers to the stiffness of the material within its elastic range.\n\nE = Stress/ Strain\n\nModulus of elasticity of dental structures\n\n1. Enamel - 65 GPa\n2. Dentin - 15 GPa\n3. Alveolar bone - 10 GPa\n4. Periodontal ligament - 0.05 GPa\n\nPoisson's ratio denotes the strain imposed on the material relative to the axis of the load applied.\n\nP = Strain perpendicular to the force/ Strain parallel to the force\n\nPoisson's ratios for dental structures\n\n1. Enamel - 0.32\n2. Dentin - 0.28\n3. Alveolar bone - 0.33\n4. Periodontal ligament - 0.3\n\nAfter assigning the material properties, the material is constrained identical to the real situation. The freedom of the body to be displaced is termed as the \"degrees of freedom\". Each element has six degrees of freedom. The final step in Pre-processing is the application of loads. These can be either force or moments and be directed at any node in all the three planes of space.\n\nPROCESSING\n\n1. ESolving of differential equations\n2. Assemblage into matrices\n3. Summation of the matrix equations\n\nThe equation for the simplest linear static analysis is represented as\n\n[F] = {K} {u}\n\nThe non-linear analysis is solved usually by what is termed as the \"Newton-Raphson method\".\n\nPOST-PROCESSING\n\n1. Graphical output\n2. Numerical output\n3. Animated output", null, "" ]	[ null, "https://i.creativecommons.org/l/by/4.0/88x31.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9086543,"math_prob":0.80342305,"size":3593,"snap":"2020-45-2020-50","text_gpt3_token_len":767,"char_repetition_ratio":0.1128448,"word_repetition_ratio":0.05882353,"special_character_ratio":0.19510159,"punctuation_ratio":0.09822866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95351666,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-29T21:59:43Z\",\"WARC-Record-ID\":\"<urn:uuid:add9fbe5-4ebb-468d-9181-c1ec10782085>\",\"Content-Length\":\"31865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0eba9479-10e6-4904-8f62-9c73f087e439>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0ad58dd-11f9-4175-b9d7-2634a247199d>\",\"WARC-IP-Address\":\"165.22.76.182\",\"WARC-Target-URI\":\"https://www.mathewsopenaccess.com/full-text/finite-element-analysis-f-e-a-an-insight\",\"WARC-Payload-Digest\":\"sha1:VZD4SLZIRE4AD6TJMEPVXJB75GMLTJMM\",\"WARC-Block-Digest\":\"sha1:5LUPUFHO6HYAIF3EXAS3Y6Y3MG3U5Z7X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107905965.68_warc_CC-MAIN-20201029214439-20201030004439-00165.warc.gz\"}"}
https://www.studypug.com/ca/ib-chemistry/ka-and-kb-calculations	[ "# Ka and Kb calculations\n\n### Ka and Kb calculations\n\n#### Lessons\n\nIn this lesson, we will learn:\n\n• How to apply the Ka expression to find the pH of weak acid solutions.\n• The assumptions made in Ka calculations at equilibrium and how to justify them.\n• How to solve for Kb using Kw and find the pH of weak base solutions.\n\nNotes:\n\n• We know that weak acids and bases are any acid/base species that does not completely dissociate in water. Dissolving a weak acid in water then has two chemical effects:\n• Some of the weak acid, HX, will interact with water and dissociate into H3O+ and X- ions.\n• The rest of the HX will stay un-dissociated.\nWe can write an equilibrium for the dissociation of weak acid HX (with A concentration) in water, and show amounts in a table, in a ‘before and after’ format like below:\n\nHX + H2O $\\rightleftharpoons$ H3O+ + Cl-\n\n Start concentration (M) Equilibrium conc. (M) HX A A - B H3O+ $\\approx$ 0 B X- 0 B\n\nThe acid dissociation constant, Ka can be expressed in these terms:\n\nKa = $\\frac{[X^{-}][H_3O^+]}{[HX]}$ = $\\frac{[X^2]}{[A] -B}$\n\n• Taking an example with 0.1M methanoic acid HCOOH, we can write the following:\n Start concentration (M) Equilibrium conc. (M) HCOOH 0.1 0.1 - B H3O+ 0 B X- 0 B\n\nWe can apply the acid dissociation constant, Ka to this equilibrium. Methanoic acid1 has a Ka value of 1.8$$10-4 so the Ka equation can be written fully:\n\nKa = $\\frac{[HCOO^-][H_3O^+]}{[HCOOH]}$ = $\\frac{[H_3O^+]^2}{[0.1] -[H_3O^+]}$ = 1.8 $$ 10-4\n\nSome assumptions are made to complete this calculation:\n• The starting concentration of H3O+ ions, in neutral water is only 110-7 M (see Autoionization of water) and an equally tiny amount of hydroxide ions are also present. This is an incredibly small amount, so this H3O+ is not taken into the calculation; only H3O+ due to the weak acid is used in the calculation.\n• With weak acids, we assume that the acid is weak enough that the amount of dissociation doesn’t affect acid concentration. Using the table above, 0.1 M HCOOH added to neutral water will still have concentration of approximately 0.1M at equilibrium, and we can ignore the ‘– B’ in ‘A-B’. Where [HX]eq = equilibrium concentration, [HX]i = start concentration of HX:\n\n[HX]eq - B $\\cong$ [HX]i\n\nYOU MUST STATE THIS ASSUMPTION IN CALCULATIONS.\n\nWith the assumptions, we have a final expression:\n\nKa = $\\frac{[HCOO^-][H_3O^+]}{[HCOOH]}$ = $\\frac{[H_3O^+]^2}{[0.1] }$ = 1.8 $$ 10-4\n\n0.1 $$ 1.8 $$ 10-4 = [H3O+]2\n\n$\\small\\sqrt{1.8 * 10^{-5}}$ = [H3O+] = 4.24 $$ 10-3\n\npH = -log [H3O+] = 2.37\n\nNote that the assumption we made was that [HX] – B is approximately equal to [HX]. We now know that B = 4.2410-3, so our assumption in this example was to say that 0.1 – 4.2410-3 = 0.0958.\nThe assumption can be justified if percentage dissociation is less than 5% which we can work out:\n\n% dissociation = $\\frac{[H_3O^+]_{eq}}{[HX]_i}$ $$ 100\n\nTherefore:\n\n% dissociation = $\\frac{[4.24 10^{-3}]}{[0.1]}$ $$ 100 = 4.24%\n\nAs the calculation shows, the assumption was justified as only 4.24% dissociation occurs.\n\n• Kb calculations are similar to Ka calculations with some changes:\n• Because acidity strength tables give only Ka, Kb of a weak base will need to be found by the calculation in the autoionization of water expression. You will need to find the Ka of the conjugate acid in the acidity strength table to do this.\n• The equilibrium concentrations you obtain using Kb will give you [OH-], so pH will need to be found by solving: pH = 14 – pOH.\n\nTaking an example with 0.5M of the weak base ammonia, NH3, we can write the following:\n\nNH3 + H2O $\\rightleftharpoons$ NH4+ + OH-\n\n Start concentration (M) Equilibrium conc. (M) NH3 0.5 0.5 - B NH3+ 0 B OH- 0 B\n\nThe conjugate acid of ammonia is the ammonium ion, NH4+ which has a Ka value1 of 5.6 $$ 10-10. Solving the autoionization expression for Kb(NH3) gives:\n\nKb = $\\frac{K_w}{K_a}$ = $\\frac{10^{-14}}{5.6 10^{-10}}$ = 1.79 $$ 10-5\n\nUsing our value for Kb (now rounding to 1.8 $$ 10-5 or 2 significant figures) we can solve for the equilibrium concentration of hydroxide ions. Again, we make the assumption that the concentration of NH3 isn’t significantly affected by the dissociation into NH4+:\n\n[NH3]eq - B $\\cong$ [NH3]i\n\nNow we can find the hydroxide ion concentration:\n\nKb = $\\frac{[NH_{4}^{+}][OH^-]{}} {[NH_3]}$ = $\\frac{[OH^-]^2}{[0.5]}$ = 1.8 $$ 10-5\n\n[OH-] = $\\sqrt{((1.8 10^{-5}) * 0.5)}$ = 3 $$ 10-3\n\npOH = -log[OH-] = -log (3 $$ 10-3) = 2.52\n\npH = 14 - pOH = 14 - 2.52 = 11.48\n\nTesting the assumption can now be done:\n\n% dissociation = $\\frac{[OH^-]_{eq}}{[NH_3]_i}$ $$ 100 = $\\frac{3 10^{-3}}{0.5}$ $$ 100 = 0.6 %\n\nWith only 0.6% dissociation, the assumption is justified and the pH has been found.\n\n• Another calculation that shows the difference between strong and weak acids and bases is the effect of dilution on pH.\nThe effect on strong acids is straightforward, because we assume 100% dissociation:\n• If a solution of strong acid, e.g. HCl is 1M, then [H3O+ (aq)] = 1M. Taking the negative log of this:\npH = -log = 0\nDiluting this by a factor of 10 will give a concentration of 0.1M\npH = -log[0.1] = 1\nA further 10, or 100 fold from the original:\npH = -log[0.01] = 2\nIn short, diluting a strong acid or base has a direct logarithmic effect on pH.\nThe effect of dilution on weak acids and bases is different:\n• If a solution of weak acid, e.g. CH3COOH is 1M, then pH and [H3O+ (aq)] is worked out using the Ka expression:\nKa (CH3COOH) = 1.410-5\n\nKa = $\\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}$\n\nThe calculation to find pH using this expression has been explained above, so moving forward to an answer (using the assumptions needed)\n\n1.4 * 10-5 = $\\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{}$ where [H3O+] = [CH3COOH-]\n\n$\\sqrt{1.4 * 10^{-5}}$ = [H3O+] = 3.74 * 10-3\n\npH = -log[ 3.74 * 10-3] = 2.42\n\nA dilution of this weak acid solution to make it 0.1M would have the following effect on the calculation:\n\n1.4 * 10-5 = $\\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.1]}$ where [H3O+] = [CH3COOH-]\n\n$\\sqrt{1.4 * 10^{-6}}$ = [H3O+] = 1.18 * 10-3\n\npH = -log[ 1.18 * 10-3] = 2.92\n\nAnother dilution by a factor of ten:\n\n1.4 * 10-5 = $\\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.01]}$ where [H3O+] = [CH3COOH-]\n\n$\\sqrt{1.4 * 10^{-7}}$ = [H3O+] = 3.74 * 10-4\n\npH = -log[ 3.74 * 10-4] = 3.42\n\nIn short, diluting a weak acid has a lesser effect on pH than in strong acids.\n• Introduction\nApplying the Ka expression\na)\nWeak acids/bases at equilibrium.\n\nb)\n\nc)\nCalculating pH and percentage dissociation.\n\nd)\nCalculations using Kb.\n\n• 1.\nFind the pH of the weak acid solution and the percentage dissociation of the weak acid/base.\na)\nWhat is the pH of a solution of 0.5 M ethanoic acid, CH3COOH? Find the percentage dissociation of this ethanoic acid solution.\n\nb)\nWhat is the pH of a solution of 0.1 M ammonia, NH3? Find the percentage dissociation of this ammonia solution.\n\n• 2.\nFind the Ka of an unknown weak acid, given pH and concentration.\na)\nA solution of carbonic acid had a pH of 3.72. What was the initial concentration of this acid solution?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9006494,"math_prob":0.9996306,"size":6007,"snap":"2019-51-2020-05","text_gpt3_token_len":1850,"char_repetition_ratio":0.15808763,"word_repetition_ratio":0.0866426,"special_character_ratio":0.31030464,"punctuation_ratio":0.12509713,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99988115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T01:51:43Z\",\"WARC-Record-ID\":\"<urn:uuid:ec6cffd5-f27b-4a23-b53d-1b97dc7fc341>\",\"Content-Length\":\"291584\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2ea4622-f1bc-4046-b9cc-2d9b5844de81>\",\"WARC-Concurrent-To\":\"<urn:uuid:07dab1ec-1a30-4252-8096-1bd0ba3ca301>\",\"WARC-IP-Address\":\"34.200.169.6\",\"WARC-Target-URI\":\"https://www.studypug.com/ca/ib-chemistry/ka-and-kb-calculations\",\"WARC-Payload-Digest\":\"sha1:OEIYZTD3M4OCFSJPATO5TRNXEPEB75PR\",\"WARC-Block-Digest\":\"sha1:LNUVZRFAAMI2LA2PL6B3ZDERAFGJDWHX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594101.10_warc_CC-MAIN-20200119010920-20200119034920-00397.warc.gz\"}"}
https://percent.info/of/52/how-to-calculate-52-percent-of-278000.html	[ "52 percent of 278000", null, "", null, "Here we will show you how to calculate fifty-two percent of two hundred seventy-eight thousand. Before we continue, note that 52 percent of 278000 is the same as 52% of 278000. We will write it both ways throughout this tutorial to remind you that it is the same.\n\n52 percent means that for each 100, there are 52 of something. This page will teach you three different methods you can use to calculate 52 percent of 278000.\n\nWe think that illustrating multiple ways of calculating 52 percent of 278000 will give you a comprehensive understanding of what 52% of 278000 means, and provide you with percent knowledge that you can use to calculate any percentage in the future.\n\nTo solidify your understanding of 52 percent of 278000 even further, we have also created a pie chart showing 52% of 278000. On top of that, we will explain and calculate \"What is not 52 percent of 278000?\"\n\nCalculate 52 percent of 278000 using a formula\nThis is the most common method to calculate 52% of 278000. 278000 is the Whole, 52 is the Percent, and the Part is what we are calculating. Below is the math and answer to \"What is 52% of 278000?\" using the percent formula.\n\n(Whole × Percent)/100 = Part\n(278000 × 52)/100 = 144560\n52% of 278000 = 144560\n\nGet 52 percent of 278000 with a percent decimal number\nYou can convert any percent, such as 52.00%, to 52 percent as a decimal by dividing the percent by one hundred. Therefore, 52% as a decimal is 0.52. Here is how to calculate 52 percent of 278000 with percent as a decimal.\n\nWhole × Percent as a Decimal = Part\n278000 × 0.52 = 144560\n52% of 278000 = 144560\n\nGet 52 percent of 278000 with a fraction function\nThis is our favorite method of calculating 52% of 278000 because it best illustrates what 52 percent of 278000 really means. The facts are that it is 52 per 100 and we want to find parts per 278000. Here is how to illustrate and show you the answer using a function with fractions.\n\n Part 278000\n=\n 52 100\n\nPart = 144560\n\n52% of 278000 = 144560\n\nNote: To solve the equation above, we first multiplied both sides by 278000 and then divided the left side to get the answer.\n\n52 percent of 278000 illustrated\nBelow is a pie chart illustrating 52 percent of 278000. The pie contains 278000 parts, and the blue part of the pie is 144560 parts or 52 percent of 278000.", null, "Note that it does not matter what the parts are. It could be 52 percent of 278000 dollars, 52 percent of 278000 people, and so on. The pie chart of 52% of 278000 will look the same regardless what it is.\n\nWhat is not 52 percent of 278000?\nWhat is not 52 percent of 278000? In other words, what is the red part of our pie above? We know that the total is 100 percent, so to calculate \"What is not 52%?\" you deduct 52% from 100% and then take that percent from 278000:\n\n100% - 52% = 48%\n(278000 × 48)/100 = 133440\n\nAnother way of calculating the red part is to subtract 144560 from 278000.\n\n278000 - 144560 = 133440\n\nThat is the end of our tutorial folks. We hope we accomplished our goal of making you a percent expert - at least when it comes to calculating 52 percent of 278000.\n\nPercent of a Number\nGo here if you need to calculate the percent of a different number.\n\n52 percent of 279000\nHere is the next percent tutorial on our list that may be of interest.\n\nCopyright \| Privacy Policy \| Disclaimer \| Contact" ]	[ null, "https://percent.info/images/percent-of/52-percent.jpg", null, "https://percent.info/images/percent-of/of-278000.jpg", null, "https://percent.info/images/pie-charts/pie-chart-showing-52-percent.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91788507,"math_prob":0.9962296,"size":3234,"snap":"2021-04-2021-17","text_gpt3_token_len":856,"char_repetition_ratio":0.24458204,"word_repetition_ratio":0.05756579,"special_character_ratio":0.3521954,"punctuation_ratio":0.082822084,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994167,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T08:40:04Z\",\"WARC-Record-ID\":\"<urn:uuid:fa20cccc-cda7-41f9-b286-a5e626d97f41>\",\"Content-Length\":\"9434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c43ad319-2943-44ff-b262-6969039d1007>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe323988-5691-4c1d-a3dd-22590dfcdf6c>\",\"WARC-IP-Address\":\"13.32.200.19\",\"WARC-Target-URI\":\"https://percent.info/of/52/how-to-calculate-52-percent-of-278000.html\",\"WARC-Payload-Digest\":\"sha1:PQB4CX5TXAH5IMCFF4MFP322VVY5FRD4\",\"WARC-Block-Digest\":\"sha1:57M67SZSMX3ABVOVT5BQS5IZ5T6H2BII\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038469494.59_warc_CC-MAIN-20210418073623-20210418103623-00548.warc.gz\"}"}
http://msduncanchem.com/Unit_10/gas_stoich_std.htm	[ "Name: ID:\n\nEmail:\n\nGas Stoichiometry\n\nMultiple Choice\nIdentify the letter of the choice that best completes the statement or answers the question.\n\n1.\n\nGiven the reaction: 2 PbO --> 2 Pb + O2\nWhat is the total volume of O2, measured at STP, produced when 1.00 mole of PbO decomposes?\n a. 44.8 L b. 22.4 L c. 11.2 L d. 5.60 L\n\n2.\n\nGiven the balanced equation: Ca + 2 H2O --> Ca(OH)2 + H2\nWhen 80.2 grams of Ca react completely with the water, what is the total volume, at STP, of H2 produced?\n a. 44.8 L b. 22.4 L c. 2.00 L d. 1.00 L\n\n3.\n\nGiven the reaction at STP: 2 KClO3 (s) --> 2 KCl (s) + 3 O2 (g)\nWhat mass of KClO3 (s) is required to produce 32.0 liters of O2 (g)?\n a. 0.952 g b. 21.3 g c. 117 g d. 263 g\n\n4.\n\nGiven the balanced equation: C3H8 + 5 O2 --> 3 CO2 + 4 H2O\nWhat is the total number of liters of CO2 produced when 20.0 liters of O2 are completely consumed?\n a. 12.0 L b. 22.4 L c. 3.00 L d. 5.00 L" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7290856,"math_prob":0.99380517,"size":976,"snap":"2022-40-2023-06","text_gpt3_token_len":415,"char_repetition_ratio":0.113168724,"word_repetition_ratio":0.038793102,"special_character_ratio":0.42110655,"punctuation_ratio":0.18456376,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934794,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-31T02:58:10Z\",\"WARC-Record-ID\":\"<urn:uuid:88d82ec2-66d4-46dc-9c3f-4f29f4aacd50>\",\"Content-Length\":\"22062\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55b3d881-fb69-4c27-8c5e-a07deb755739>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9ee1d2c-98b1-4227-b0c9-b5c832c2a73a>\",\"WARC-IP-Address\":\"54.156.116.15\",\"WARC-Target-URI\":\"http://msduncanchem.com/Unit_10/gas_stoich_std.htm\",\"WARC-Payload-Digest\":\"sha1:JBDK5TVMOMQXDE64WBPI2E4SLIG6MJ4F\",\"WARC-Block-Digest\":\"sha1:NQDYTUOA5I2ZICZUXY3RMZYHKYVBETZK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499842.81_warc_CC-MAIN-20230131023947-20230131053947-00551.warc.gz\"}"}
https://www.themathcitadel.com/exploiting-chemistry-for-better-packet-flow-management-3-formal-analysis/	[ "\nExploiting Chemistry for Better Packet Flow Management 3: Formal Analysis\n\n# Exploiting Chemistry for Better Packet Flow Management 3: Formal Analysis\n\nThe previous two posts introduced the ideas of Meyer and Tschudin involving the application and exploitation of chemical kinetic theory to flow management in computer networking. The first part introduced the ideas and gave an overview of the entire work, and the second part took a deeper look into the formal model of a packet chemistry. This section discusses the analysis options available once a packet chemistry model has been created.\n\nThis section can also be skipped for those less interested in the formal mathematics. Suffice it to say that there are a multitude of already created methods now available for the elegant analysis of computer networks when modeled by an artificial packet chemistry.\n\n## Formal Analysis of Artificial Packet Chemistry\n\nBy representing packet flow in a computer network as an artificial chemistry, a multitude of analyses are available, from high to low granularity. The authors give a heavily brief survey (and a good bibliography) of works that can be utilized to analyze these networks pulled from the physics and chemistry literature. A particular advantage of this method is the ability to study the transient states of the network rather than just steady states. The authors also claim the ability to determine the stability of the network flow based only on topology, a heavy advantage in design.\n\n### Stochastic Analysis at the Microscopic Level\n\nThe stochastic behavior of chemical reaction networks is described by the chemical master equation which takes the form\n$$\\frac{\\text{d}\\mathbf{P}}{\\text{d}t} = \\mathbf{A}\\mathbf{P}$$\n\nwhich is a differential equation describing the evolution of state probabilities for a system. Here the states are discrete, and time is continuous. The matrix $\\mathbf{A}$ describes the transition rates (which can also be kinetic or reaction rates), and the stochastic process described is a Markov jump-process Since we’re on a network, the Markov jump process exists in an $\\mathcal{S}$-dimensional integer lattice. Some work has been done to analyze several classes of chemical reaction networks to find the steady-state probability distribution of the state space. For example, if the total number of packets in the network has a bound, and the network contains only first order (unimolecular to unimolecular) reactions, the steady state probability distribution for the lengths of the queues in the network is a multinomial distribution. On the other hand, if the network is open (we allow packets to exit the network completely), then the steady state probability distribution of the lengths of the queues follows a product of Poisson distributions (which is also Poisson). (This is an extremely desirable property, called a product-form.)\n\n### Deterministic Approximations\n\nThis is the most common approach utilized in computer network analysis today, simply because networks are so large and complex that stochastic modeling becomes too cumbersome. Here, the average trajectory is represented by a system of ordinary differential equations, building a fluid model. One downside to this in the networking space is that the analysis of protocols by this method requires manual extraction from source code and accuracy is uncertain.\n\nIn the chemistry sector (and now in the packet chemistry model), obtaining a fluid approximation is not only easier, but shown to be accurate. There are links between the stochastic master equation to several approximations[5,6] including a deterministic ODE model. Gillespie showed that the ODE model accurately predicts the network flow trajectory in many cases.\n\nOne thing the authors note here is that the ODE model can be directly and automatically generated from the network topology. For example, a single server with a single queue (M/M/1) is simply modeled as one chemical species $X$. The arrival rate (inflow) is $\\lambda$, and the service rate is proportional to the queue length, so $\\mu = kx$, where $x$ is the queue length. Then we get a simple differential equation\n$$\\dot{x} = \\lambda-kx$$ describing the change in queue length as the difference of inflow and outflow. In the steady state, $\\dot{x} = 0$, which lets us look for a fixed point $\\hat{x} = \\frac{\\lambda}{k}$. This is the steady-state queue length, which allows us to derive the expected waiting time $T = \\frac{1}{k}$, showing that the latency of a packet under this model is independent of the arrival rate and fill level. This model when implemented automatically adjusts the service rate such that in the steady state, every packet sees the same latency.\n\nIt’s also important to determine just how stable this steady state is by analyzing the sensitivity of the network and states to perturbations. The authors list several citations to show that no new approaches are needed to do this; one can look to signal and control theory literature. In particular, a network designer would desire to predict the stability of a complex network by studying the topology as opposed to an analysis of the system of ODEs. Fortunately, modeling a network this way allows for the use of the Deficiency Zero Theorem for complex chemical networks that gives conditions for stability of steady-state[2,7].", null, "The authors give a formal convergence proof that the example network above converges to a stable fixed point and is asymptotically stable, comparing it to the proof of a similar protocol Push-Sum (a gossip protocol in computer networks).\n\n## Continuation\n\nThe next post in this series will discuss Meyer and Tschudin’s implementation of a scheduler based on the principles discussed thus far.\n\n## References\n\n1. Dittrich, P., Ziegler, J., and Banzhaf, W. Artificial chemistries – a review. Artificial Life 7(2001), 225–275.\n1. Feinburg, M. Complex balancing in general kinetic systems. Archive for Rational Mechanics and Analysis 49 (1972).\n2. Gadgil, C., Lee, C., and Othmer, H. A stochastic analysis of first-order reaction networks. Bulletin of Mathematical Biology 67 (2005), 901–946.\n3. Gibson, M., and Bruck, J. Effcient stochastic simulation of chemical systems with many species and many channels. Journal of Physical Chemistry 104 (2000), 1876–1889.\n4. Gillespie, D. The chemical langevin equation. Journal of Chemical Physics 113 (2000).\n5. Gillespie, D. The chemical langevin and fokker-planck equations for the reversible isomerizationreaction. Journal of Physical Chemistry 106 (2002), 5063–5071.\n6. Horn, F. On a connexion between stability and graphs in chemical kinetics. Proceedings of the RoyalSociety of London 334 (1973), 299–330.\n7. Kamimura, K., Hoshino, H., and Shishikui, Y. Constant delay queuing for jitter-sensitive iptvdistribution on home network. IEEE Global Telecommunications Conference (2008).\n8. Laidler, K. Chemical Kinetics. McGraw-Hill, 1950.\n9. McQuarrie, D. Stochastic approach to chemical kinetics. Journal of Applied Probability 4 (1967), 413–478.\n10. Meyer, T., and Tschudin, C. Flow management in packet networks through interacting queues and law-of-mass-action-scheduling. Technical report, University of Basel.\n11. Pocher, H. L., Leung, V., and Gilles, D. An application- and management-based approach to atm scheduling. Telecommunication Systems 12 (1999), 103–122.\n12. Tschudin, C. Fraglets- a metabolistic execution model for communication protocols. Proceedings of the 2nd annual symposium on autonomous intelligent networks and systems (2003).\n\n### Attachments\n\n•", null, "chem-queuing-report" ]	[ null, "http://www.themathcitadel.com/wp-content/uploads/2019/01/Screen-Shot-2019-01-10-at-8.52.15-AM-1024x418.png", null, "https://www.themathcitadel.com/wp-content/plugins/download-attachments/images/ext/pdf.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8843673,"math_prob":0.90502816,"size":7519,"snap":"2021-43-2021-49","text_gpt3_token_len":1615,"char_repetition_ratio":0.12867598,"word_repetition_ratio":0.0070113936,"special_character_ratio":0.21239527,"punctuation_ratio":0.12091988,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98683065,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T15:18:27Z\",\"WARC-Record-ID\":\"<urn:uuid:5c3b22e8-0ede-4e01-8cc5-f8b90b319f4d>\",\"Content-Length\":\"51630\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e28c6b4a-cae3-4ed4-b8d2-eeb0d90f926e>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd0593e2-4125-4ab1-984b-16d2ae10d4c1>\",\"WARC-IP-Address\":\"165.227.218.96\",\"WARC-Target-URI\":\"https://www.themathcitadel.com/exploiting-chemistry-for-better-packet-flow-management-3-formal-analysis/\",\"WARC-Payload-Digest\":\"sha1:BHQ745RKASZCS74O72B675YRMPH53IQN\",\"WARC-Block-Digest\":\"sha1:QEKWKEC6HDWLU3BFHTOOJ37I5333O236\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358774.44_warc_CC-MAIN-20211129134323-20211129164323-00623.warc.gz\"}"}
https://gdl.graphisoft.com/forums/reply/2415	[ "#2415\n\nYour code would work with the following corrections:\n\n``````dim varia[]\nvaria=a1\nvaria=a2\nvaria=a3\nvaria=a4\nvaria=a5``````\n\nQuotation marks not needed.\n\n`if varia[LIT][i][/LIT]=1 then`\nparameters set with values{2} can be used as numbers.\n\nPéter Baksa\nLibrary Platform, Software Engineer\nGRAPHISOFT SE" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6445532,"math_prob":0.93586177,"size":460,"snap":"2023-40-2023-50","text_gpt3_token_len":150,"char_repetition_ratio":0.13815789,"word_repetition_ratio":0.0,"special_character_ratio":0.28478262,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9584256,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T18:38:12Z\",\"WARC-Record-ID\":\"<urn:uuid:4dd1047e-e11a-4441-a23d-f28a7c6c774d>\",\"Content-Length\":\"38804\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:915a30ea-81f6-40a3-98ae-830abff1374d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a332525-c672-48d2-b5ef-6a76180d9a68>\",\"WARC-IP-Address\":\"34.77.123.245\",\"WARC-Target-URI\":\"https://gdl.graphisoft.com/forums/reply/2415\",\"WARC-Payload-Digest\":\"sha1:IBQJP2VDKU4H5YIIJQJM2LSJBHEJIGSB\",\"WARC-Block-Digest\":\"sha1:SZZDCUKJJ2GJWDKV6JXYMXNIQEPAJ624\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100135.11_warc_CC-MAIN-20231129173017-20231129203017-00600.warc.gz\"}"}
https://arxiv.org/abs/1406.5372	[ "astro-ph.GA\n\nTitle:Two Conditions for Galaxy Quenching: Compact Centres and Massive Haloes\n\nAbstract: We investigate the roles of two classes of quenching mechanisms for central and satellite galaxies in the SDSS ($z<0.075$): those involving the halo and those involving the formation of a compact centre. For central galaxies with inner compactness $\\Sigma_{\\rm 1kpc} \\sim 10^{9-9.4}M_{\\odot} {\\rm kpc}^{-2}$, the quenched fraction $f_{q}$ is strongly correlated with $\\Sigma_{\\rm 1kpc}$ with only weak halo mass $M_{\\rm h}$ dependence. However, at higher and lower $\\Sigma_{\\rm 1kpc}$, sSFR is a strong function of $M_{\\rm h}$ and mostly independent of $\\Sigma_{\\rm 1kpc}$. In other words, $\\Sigma_{\\rm 1kpc} \\sim 10^{9-9.4} M_{\\odot} {\\rm kpc}^{-2}$ divides galaxies into those with high sSFR below and low sSFR above this range. In both the upper and lower regimes, increasing $M_{\\rm h}$ shifts the entire sSFR distribtuion to lower sSFR without a qualitative change in shape. This is true even at fixed $M_{}$, but varying $M_{}$ at fixed $M_{\\rm h}$ adds no quenching information. Most of the quenched centrals with $M_{\\rm h} > 10^{11.8}M_{\\odot}$ are dense ($\\Sigma_{\\rm 1kpc} > 10^{9}~ M_{\\odot} {\\rm kpc}^{-2}$), suggesting compaction-related quenching maintained by halo-related quenching. However, 21% are diffuse, indicating only halo quenching. For satellite galaxies in the outskirts of halos, quenching is a strong function of compactness and a weak function of host $M_{\\rm h}$. In the inner halo, $M_{\\rm h}$ dominates quenching, with $\\sim 90\\%$ of the satellites being quenched once $M_{\\rm h} > 10^{13}M_{\\odot}$. This regional effect is greatest for the least massive satellites. As demonstrated via semi-analytic modelling with simple prescriptions for quenching, the observed correlations can be explained if quenching due to central compactness is rapid while quenching due to halo mass is slow.\n Comments: 16 pages, 11 figures, MNRAS accepted Subjects: Astrophysics of Galaxies (astro-ph.GA) DOI: 10.1093/mnras/stu2755 Cite as: arXiv:1406.5372 [astro-ph.GA] (or arXiv:1406.5372v3 [astro-ph.GA] for this version)\n\nSubmission history\n\nFrom: Joanna Woo [view email]\n[v1] Fri, 20 Jun 2014 13:04:00 UTC (181 KB)\n[v2] Thu, 4 Dec 2014 22:49:21 UTC (200 KB)\n[v3] Fri, 16 Jan 2015 16:38:50 UTC (201 KB)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7553819,"math_prob":0.976279,"size":2322,"snap":"2019-43-2019-47","text_gpt3_token_len":701,"char_repetition_ratio":0.13373598,"word_repetition_ratio":0.005830904,"special_character_ratio":0.30534023,"punctuation_ratio":0.11261261,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98476064,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T14:45:47Z\",\"WARC-Record-ID\":\"<urn:uuid:e570c6a2-993e-47f8-8988-c4a85271a03b>\",\"Content-Length\":\"21541\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:493b9746-9cce-4391-beb4-80969d3e4187>\",\"WARC-Concurrent-To\":\"<urn:uuid:e5a4ffa9-7f71-463b-ac66-899dd5a84979>\",\"WARC-IP-Address\":\"128.84.21.199\",\"WARC-Target-URI\":\"https://arxiv.org/abs/1406.5372\",\"WARC-Payload-Digest\":\"sha1:5PYF5THS24YX57WNAWSDIM55N3FGJPYH\",\"WARC-Block-Digest\":\"sha1:526N3H6AZLVWSTBWNUDWHFMAALDYWX5J\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986696339.42_warc_CC-MAIN-20191019141654-20191019165154-00316.warc.gz\"}"}
http://lunlun.com/2021/06/	[ "## Solve x+3<=0 (first degree inequality)\n\n0 Solve x+3<=0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+2<=0 (first degree inequality)\n\n0 Solve x+2<=0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+2>=0 (first degree inequality)\n\n0 Solve x+2>=0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+4=0 (first degree equation)\n\n0 Solve x+4=0 (first degree equation) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+1<=0 (first degree inequality)\n\n0 Solve x+1<=0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+1>=0 (first degree inequality)\n\n0 Solve x+1>=0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+3<0 (first degree inequality)\n\n0 Solve x+3<0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry\n\n## Solve x+3>0 (first degree inequality)\n\n0 Solve x+3>0 (first degree inequality) Subcribe now to LUNLUN.com newsletter and you will get the cheatsheet “Top 10 Trigonometry" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8493061,"math_prob":0.97863054,"size":1830,"snap":"2021-31-2021-39","text_gpt3_token_len":505,"char_repetition_ratio":0.15607886,"word_repetition_ratio":0.62352943,"special_character_ratio":0.23825137,"punctuation_ratio":0.034591194,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9640619,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T03:56:25Z\",\"WARC-Record-ID\":\"<urn:uuid:8d41b97c-7bbc-40c6-8c3a-0b7553f05293>\",\"Content-Length\":\"36045\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:402a5456-32a4-47cf-b638-1fdb337b999e>\",\"WARC-Concurrent-To\":\"<urn:uuid:489aecd5-5263-4751-b0f3-f54bd00fc1b0>\",\"WARC-IP-Address\":\"166.62.119.124\",\"WARC-Target-URI\":\"http://lunlun.com/2021/06/\",\"WARC-Payload-Digest\":\"sha1:KZJBU2RYOCTJDFEQVHSERJ46KCG7JHWY\",\"WARC-Block-Digest\":\"sha1:GLLCG5CERGKRZCI2YWENUZCKRTHYZAYK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057796.87_warc_CC-MAIN-20210926022920-20210926052920-00706.warc.gz\"}"}
https://www.learndartprogramming.com/fundamentals/for-loop-in-dart/	[ "", null, "# For Loop in Dart\n\nPublished on 29 March 2020\nLast Updated on 29 March 2020\n\nFor loop is one of the control flow statements in Dart.\n\n## Use Of For Loop In Dart\n\nA For loop is used for performing repeated execution of instructions. Using for loops in Dart, we can iterate over same set of instructions over and over again.\n\n## Simple For Loop\n\n``````// for loop in dart\nmain(List<String> args) {\nfor (var i = 0; i < 4; i++) {\nprint('i is: \\${i}');\n}\n}\n``````\n\nAbove program produces following output:\n\n``````i is: 0\ni is: 1\ni is: 2\ni is: 3\n``````\n\n## For in loop in Dart\n\nFor in loop in Dart is used for iterating through object’s properties.\n\n``````// for in loop in dart\nmain(List<String> args) {\nvar collection = [1, 2, 3];\nfor (var obj in collection) {\nprint('obj is \\${obj}');\n}\n}\n``````\n\nAbove program produces following output:\n\n``````obj is 1\nobj is 2\nobj is 3\n``````\n\nAbove program demonstrates how we can iterate through elements in a list object through for in loop.\n\n## Nested For Loop in Dart\n\nWhen one or more for loops are placed into one another then they are called nested for loops.\n\n``````// nested for loop\nmain(List<String> args) {\nfor (var i = 0; i < 3; i++) {\nfor (var j = 0; j < 3; j++) {\nprint('i is: \\${i}, j is \\${j}');\n}\n}\n}\n``````\n\nAbove program produces following output:\n\n``````i is: 0, j is 0\ni is: 0, j is 1\ni is: 0, j is 2\ni is: 1, j is 0\ni is: 1, j is 1\ni is: 1, j is 2\ni is: 2, j is 0\ni is: 2, j is 1\ni is: 2, j is 2\n\nProcess finished with exit code 0\n``````\n\nWhile creating nested for loops one must make sure not to create an infinite loop.\n\n## Use of break statement in for loop\n\nOnce break is called inside a for loop, the program exits the loop. Subsequent statements placed after break statement are not executed.\n\n``````// use of break in for loop\nmain(List<String> args) {\nfor (var i = 0; i < 3; i++) {\nif (i == 3) {\n// for loop will exit when i becomes 3\nbreak;\n}\nprint('i is: \\${i}');\n}\n}\n``````\n\nAbove program produces following output:\n\n``````i is: 0\ni is: 1\ni is: 2\n``````\n\n## Use for continue statement in a for loop\n\nContinue statement is used for skipping statements that are placed after the continue statement in the current iteration and returning program execution pointer back to the beginning of the loop.\n\n``````// for loop continue\nmain(List<String> args) {\nfor (var i = 0; i < 4; i++) {\nif (i == 2) {\n// for loop will skip execution when i becomes 2\ncontinue;\n}\nprint('i is: \\${i}');\n}\n}\n``````\n\nAbove program produces following output:\n\n``````i is: 0\ni is: 1\ni is: 3\n``````" ]	[ null, "https://www.learndartprogramming.com/thumbnail/dart_3.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84290844,"math_prob":0.8446108,"size":2165,"snap":"2020-34-2020-40","text_gpt3_token_len":653,"char_repetition_ratio":0.1443776,"word_repetition_ratio":0.26741573,"special_character_ratio":0.32840645,"punctuation_ratio":0.14516129,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9838503,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T01:10:21Z\",\"WARC-Record-ID\":\"<urn:uuid:b2bd5475-e29a-4504-be57-6330c63c06c8>\",\"Content-Length\":\"28092\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2d59928f-0303-4cbe-b043-96362385ff1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:40bd31c5-af3d-41af-96e3-65d8bf4317cd>\",\"WARC-IP-Address\":\"151.101.1.195\",\"WARC-Target-URI\":\"https://www.learndartprogramming.com/fundamentals/for-loop-in-dart/\",\"WARC-Payload-Digest\":\"sha1:N5E6MOHWPIQOHIKATGMEQFQ6YJC2LDHN\",\"WARC-Block-Digest\":\"sha1:XG5RO3S7LFXZIDAMPHFCTTGNXHR3BD3W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401617641.86_warc_CC-MAIN-20200928234043-20200929024043-00299.warc.gz\"}"}
https://git.dynare.org/Dynare/dseries/commit/14478ca16c2cbdc5c6d922d131518be742d552b9	[ "### Added the ability to tag variables in dseries objects.\n\n``` - New class member tags\n\n- Member tags is a structure, initialized to empty.\n\n- Each user-defined field of this structure must be a vobs(o)1 cell\narray (each element is associated to a variable).\n\n- To add a new tag name:\n\n>> o.tag('type')\n\n- To associate a tag value to a variable:\n\n>> o.tag('type', 'Consumption', 'Flow')\n\nthe first argument is the tag name, the second argument is the name\nof the variable, the last argument is the value of the tag.```\nparent 03b223fa\n ... ... @@ -29,6 +29,7 @@ p.name = o.name; p.tex = o.tex; p.dates = o.dates; p.ops = o.ops; p.tags = o.tags; %@test:1 %\\$ % Define a dates object ... ...\n ... ... @@ -23,6 +23,7 @@ properties tex = {}; % TeX names of the variables. dates = dates(); % Dates associated to the observations. ops = {}; % History of operations on the variables. tags = struct(); % User defined tags on the variables. end methods ... ... @@ -54,6 +55,7 @@ methods o.tex = {}; o.dates = dates(); o.ops = {}; o.tags = struct(); return case 1 if isdates(varargin{1}) ... ... @@ -67,17 +69,19 @@ methods o.tex = {}; o.dates = varargin{1}; o.ops = {}; o.tags = struct(); otherwise error('dseries:WrongInputArguments', 'Input (identified as a dates object) must have a unique element!'); end return elseif ischar(varargin{1}) [init, data, varlist, tex, ops] = load_data(varargin{1}); [init, data, varlist, tex, ops, tags] = load_data(varargin{1}); o.data = data; o.name = varlist; o.dates = init:init+(nobs(o)-1); o.tex = tex; o.ops = ops; o.tags = tags; elseif ~isoctave() && istable(varargin{1}) % It is assumed that the dates are in the first column. thistable = varargin{1}; ... ... @@ -86,36 +90,40 @@ methods o.data = varargin{1}{:,2:end}; o.dates = dates(varargin{1}{1,1}{1})+(0:size(varargin{1}, 1)-1); o.ops = cell(length(o.name), 1); o.tags = struct(); elseif isnumeric(varargin{1}) && isequal(ndims(varargin{1}),2) o.data = varargin{1}; o.name = default_name(vobs(o)); o.tex = name2tex(o.name); o.dates = dates(1,1):dates(1,1)+(nobs(o)-1); o.ops = cell(length(o.name), 1); o.tags = struct(); end case {2,3,4} if isequal(nargin,2) && ischar(varargin{1}) && isdates(varargin{2}) % Instantiate dseries object with a data file and force the initial date to % be as given by the second input argument (initial period represented % with a dates object). [init, data, varlist, tex, ops] = load_data(varargin{1}); [init, data, varlist, tex, ops, tags] = load_data(varargin{1}); o.data = data; o.name = varlist; o.dates = varargin{2}:varargin{2}+(nobs(o)-1); o.tex = tex; o.ops = ops; o.tags = tags; return end if isequal(nargin,2) && ischar(varargin{1}) && ischar(varargin{2}) && isdate(varargin{2}) % Instantiate dseries object with a data file and force the initial date to % be as given by the second input argument (initial period represented with a % string). [init, data, varlist, tex, ops] = load_data(varargin{1}); [init, data, varlist, tex, ops, tags] = load_data(varargin{1}); o.data = data; o.name = varlist; o.dates = dates(varargin{2}):dates(varargin{2})+(nobs(o)-1); o.tex = tex; o.ops = ops; o.tags = tags; return end a = varargin{1}; ... ... @@ -177,6 +185,7 @@ methods o.name = default_name(vobs(o)); end o.ops = cell(length(o.name), 1); o.tags = struct(); if ~isempty(d) if vobs(o)==length(d) for i=1:vobs(o) ... ...\n ... ... @@ -78,6 +78,25 @@ end a.ops = vertcat(b.ops,c.ops); a.name = vertcat(b.name,c.name); a.tex = vertcat(b.tex,c.tex); btagnames = fieldnames(b.tags); ctagnames = fieldnames(c.tags); atagnames = union(btagnames, ctagnames); if isempty(atagnames) a.tags = struct(); else for i=1:length(atagnames) if ismember(atagnames{i}, btagnames) && ismember(atagnames{i}, ctagnames) a.tags.(atagnames{i}) = vertcat(b.tags.(atagnames{i}), b.tags.(atagnames{i})); elseif ismember(atagnames{i}, btagnames) a.tags.(atagnames{i}) = vertcat(b.tags.(atagnames{i}), cell(vobs(c), 1)); elseif ismember(atagnames{i}, ctagnames) a.tags.(atagnames{i}) = vertcat(cell(vobs(b), 1), c.tags.(atagnames{i})); else error('dseries::horzcat: This is a bug!') end end end if ~( d_nobs_flag(1) \|\| d_init_flag(1) ) a.data = [b.data,c.data]; a.dates = b.dates; ... ... @@ -331,3 +350,48 @@ end %\\$ %\\$ T = t; %@eof:7 %@test:8 %\\$ % Define a data set. %\\$ A = [transpose(1:10),2transpose(1:10)]; %\\$ B = [transpose(1:10),2transpose(1:10)]; %\\$ %\\$ % Define names %\\$ A_name = {'A1';'A2'}; %\\$ B_name = {'B1';'B2'}; %\\$ %\\$ % Define expected results. %\\$ e.init = dates(1,1); %\\$ e.freq = 1; %\\$ e.name = {'A1';'A2';'B1';'B2'}; %\\$ e.data = [A,B]; %\\$ %\\$ % Instantiate two time series objects. %\\$ ts1 = dseries(A,[],A_name,[]); %\\$ ts2 = dseries(B,[],B_name,[]); %\\$ ts1.tag('t1'); %\\$ ts1.tag('t1', 'A1', 'Stock'); %\\$ ts1.tag('t1', 'A2', 'Flow'); %\\$ ts2.tag('t2'); %\\$ ts2.tag('t2', 'B1', 0); %\\$ ts2.tag('t2', 'B2', 1); %\\$ %\\$ % Call the tested method. %\\$ try %\\$ ts3 = [ts1,ts2]; %\\$ t(1) = true; %\\$ catch %\\$ t(1) = false; %\\$ end %\\$ %\\$ % Check the results. %\\$ if t(1) %\\$ t(2) = dassert(ts3.init,e.init); %\\$ t(3) = dassert(ts3.freq,e.freq); %\\$ t(4) = dassert(ts3.data,e.data); %\\$ t(5) = dassert(ts3.name,e.name); %\\$ t(6) = dassert(ts3.tags.t1,{'Stock';'Flow';[];[]}); %\\$ t(7) = dassert(ts3.tags.t2,{[];[];0;1}); %\\$ end %\\$ T = all(t); %@eof:8\n ... ... @@ -48,23 +48,64 @@ end % Keep the second input argument constant. p = copy(p); % Add NaNs if necessary. [o, p] = align(o, p); n = length(id); % Get tag names in p ptagnames = fieldnames(p.tags); if n>1 [id, jd] = sort(id); p.data = p.data(:,jd); p.name = p.name(jd); p.tex = p.tex(jd); p.ops = p.ops(jd); if ~isempty(ptagnames) for i = 1:length(ptagnames) p.tags.(ptagnames{i}) = p.tags.(ptagnames{i})(jd); end end end % Get tag names in o otagnames = fieldnames(o.tags); % Merge tag names if isempty(otagnames) && isempty(ptagnames) notags = true; else notags = false; dtagnames_o = setdiff(ptagnames, otagnames); dtagnames_p = setdiff(otagnames, ptagnames); if ~isempty(dtagnames_o) % If p has tags that are not in o... for i=1:length(dtagnames_o) o.tags.(dtagnames_o{i}) = cell(vobs(o), 1); end end if ~isempty(dtagnames_p) % If o has tags that are not in p... for i=1:length(dtagnames_p) p.tags.(dtagnames_p{i}) = cell(vobs(p), 1); end end end % Update list of tag names in o. otagnames = fieldnames(o.tags); for i=1:n o.data = insert_column_vector_in_a_matrix(o.data, p.data(:,i),id(i)); o.name = insert_object_in_a_one_dimensional_cell_array(o.name, p.name{i}, id(i)); o.tex = insert_object_in_a_one_dimensional_cell_array(o.tex, p.tex{i}, id(i)); o.ops = insert_object_in_a_one_dimensional_cell_array(o.ops, p.ops{i}, id(i)); if ~notags for j=1:length(otagnames) o.tags.(otagnames{j}) = insert_object_in_a_one_dimensional_cell_array(o.tags.(otagnames{j}), p.tags.(otagnames{j}){i}, id(i)); end end id = id+1; end ... ... @@ -83,6 +124,16 @@ end %\\$ % Instantiate two dseries objects. %\\$ ts1 = dseries(A, A_init, A_name,[]); %\\$ ts2 = dseries(B, B_init, B_name,[]); %\\$ ts1.tag('t1'); %\\$ ts1.tag('t1','A1',1); %\\$ ts1.tag('t1','A2',1); %\\$ ts1.tag('t1','A3',0); %\\$ ts2.tag('t1'); %\\$ ts2.tag('t1','B1',1); %\\$ ts2.tag('t1','B2',1); %\\$ ts2.tag('t2'); %\\$ ts2.tag('t2','B1','toto'); %\\$ ts2.tag('t2','B2','titi'); %\\$ %\\$ try %\\$ ts1 = insert(ts1,ts2,[1,2]); ... ... @@ -91,11 +142,14 @@ end %\\$ t = 0; %\\$ end %\\$ %\\$ if length(t)>1 %\\$ t(2) = dassert(ts1.vobs,{'B1';'A1';'B2';'A3'}); %\\$ %\\$ if t(1) %\\$ t(2) = dassert(ts1.name,{'B1';'A1';'B2';'A2';'A3'}); %\\$ t(3) = dassert(ts1.nobs,10); %\\$ eB = [NaN(2,2); B; NaN(3,2)]; %\\$ t(4) = dassert(ts1.data,[eB(:,1), A(:,1), eB(:,2), A(:,2:3)], 1e-15); %\\$ t(5) = dassert(ts1.tags.t1,{1; 1; 1; 1; 0}); %\\$ t(6) = dassert(ts1.tags.t2,{'toto'; []; 'titi'; []; []}); %\\$ end %\\$ T = all(t); %@eof:1 ... ...\n ... ... @@ -73,6 +73,13 @@ if ~isequal(o.ops, p.ops) warning on backtrace end if ~isequal(o.tags, p.tags) warning off backtrace warning('dseries::isequal: Both input arguments have different tags!') warning on backtrace end if nargin<3 b = isequal(o.data, p.data); else ... ...\n ... ... @@ -42,12 +42,35 @@ if ~isequal(frequency(o), frequency(p)) end q = dseries(); [q.name, IBC, junk] = unique([o.name; p.name], 'last'); tex = [o.tex; p.tex]; q.tex = tex(IBC); ops = [o.ops; p.ops]; q.ops = ops(IBC); otagnames = fieldnames(o.tags); ptagnames = fieldnames(p.tags); qtagnames = union(otagnames, ptagnames); if isempty(qtagnames) q.tags = struct(); else for i=1:length(qtagnames) if ismember(qtagnames{i}, otagnames) && ismember(qtagnames{i}, ptagnames) q.tags.(qtagnames{i}) = vertcat(o.tags.(otagnames{i}), p.tags.(ptagnames{i})); elseif ismember(qtagnames{i}, otagnames) q.tags.(qtagnames{i}) = vertcat(o.tags.(qtagnames{i}), cell(vobs(p), 1)); elseif ismember(qtagnames{i}, ptagnames) q.tags.(qtagnames{i}) = vertcat(cell(vobs(o), 1), p.tags.(qtagnames{i})); else error('dseries::horzcat: This is a bug!') end q.tags.(qtagnames{i}) = q.tags.(qtagnames{i})(IBC); end end if nobs(o) == 0 q = copy(p); elseif nobs(p) == 0 ... ... @@ -93,22 +116,26 @@ q.dates = q_init:q_init+(nobs(q)-1); %\\$ % Define names %\\$ A_name = {'A1';'A2'}; B_name = {'A1'}; %\\$ %\\$ t = zeros(4,1); %\\$ %\\$ % Instantiate a time series object. %\\$ % Instantiate two time series objects and merge. %\\$ try %\\$ ts1 = dseries(A,[],A_name,[]); %\\$ ts1.tag('type'); %\\$ ts1.tag('type', 'A1', 'Stock'); %\\$ ts1.tag('type', 'A2', 'Flow'); %\\$ ts2 = dseries(B,[],B_name,[]); %\\$ ts2.tag('type'); %\\$ ts2.tag('type', 'A1', 'Flow'); %\\$ ts3 = merge(ts1,ts2); %\\$ t(1) = 1; %\\$ catch %\\$ t = 0; %\\$ end %\\$ %\\$ if length(t)>1 %\\$ if t(1) %\\$ t(2) = dassert(ts3.vobs,2); %\\$ t(3) = dassert(ts3.nobs,10); %\\$ t(4) = dassert(ts3.data,[B, A(:,2)],1e-15); %\\$ t(5) = dassert(ts3.tags.type, {'Flow';'Flow'}); %\\$ end %\\$ T = all(t); %@eof:1 ... ... @@ -120,12 +147,15 @@ q.dates = q_init:q_init+(nobs(q)-1); %\\$ % Define names %\\$ A_name = {'A1';'A2'}; B_name = {'B1'}; %\\$ %\\$ t = zeros(4,1); %\\$ %\\$ % Instantiate a time series object. %\\$ % Instantiate two time series objects and merge them. %\\$ try %\\$ ts1 = dseries(A,[],A_name,[]); %\\$ ts1.tag('t1'); %\\$ ts1.tag('t1', 'A1', 'Stock'); %\\$ ts1.tag('t1', 'A2', 'Flow'); %\\$ ts2 = dseries(B,[],B_name,[]); %\\$ ts2.tag('t2'); %\\$ ts2.tag('t2', 'B1', 1); %\\$ ts3 = merge(ts1,ts2); %\\$ t(1) = 1; %\\$ catch ... ... @@ -136,6 +166,8 @@ q.dates = q_init:q_init+(nobs(q)-1); %\\$ t(2) = dassert(ts3.vobs,3); %\\$ t(3) = dassert(ts3.nobs,10); %\\$ t(4) = dassert(ts3.data,[A, B],1e-15); %\\$ t(5) = dassert(ts3.tags.t1, {'Flow';'Flow';[]}); %\\$ t(6) = dassert(ts3.tags.t2, {[];[];1}); %\\$ end %\\$ T = all(t); %@eof:2\n ... ... @@ -69,6 +69,10 @@ if ~isequal(o.tex, p.tex) warning('dseries::ne: Both input arguments do not have the same tex names!') end if ~isequal(o.tags, p.tags) warning('dseries::ne: Both input arguments do not have the same tags!') end b = ne(o.data, p.data); %@test:1 ... ...\n ... ... @@ -42,6 +42,10 @@ o.data(:,id) = []; o.name(id) = []; o.tex(id) = []; o.ops(id) = []; otagnames = fieldnames(o.tags); for i=1:length(otagnames) o.tags.(otagnames{i})(id) = []; end %@test:1 %\\$ % Define a datasets. ... ... @@ -53,16 +57,21 @@ o.ops(id) = []; %\\$ % Instantiate a time series object. %\\$ try %\\$ ts1 = dseries(A,[],A_name,[]); %\\$ ts1.tag('type'); %\\$ ts1.tag('type', 'A1', 1); %\\$ ts1.tag('type', 'A2', 2); %\\$ ts1.tag('type', 'A3', 3); %\\$ ts1.pop_('A2'); %\\$ t(1) = 1; %\\$ catch %\\$ t(1) = 0; %\\$ end %\\$ %\\$ if length(t)>1 %\\$ if t(1) %\\$ t(2) = dassert(ts1.vobs,2); %\\$ t(3) = dassert(ts1.nobs,10); ts1 %\\$ t(3) = dassert(ts1.nobs,10); %\\$ t(4) = dassert(ts1.data,[A(:,1), A(:,3)],1e-15); %\\$ t(5) = dassert(ts1.tags.type, {1;3}); %\\$ end %\\$ T = all(t); %@eof:1 \\ No newline at end of file\n ... ... @@ -65,10 +65,31 @@ switch format fprintf(fid,[ '''' o.ops{i} '''']); end if i\n ... ... @@ -65,7 +65,7 @@ function B = subsref(A, S) % ---- Unitary tests ---- switch S(1).type case '.' switch S(1).subs case {'data','name','tex','dates','ops'} % Public members. case {'data','name','tex','dates','ops', 'tags'} % Public members. if length(S)>1 && isequal(S(2).type,'()') && isempty(S(2).subs) error(['dseries::subsref: ' S(1).subs ' is not a method but a member!']) end ... ... @@ -169,7 +169,7 @@ switch S(1).type else error('dseries::subsref: Call to size method must come in last position!') end case {'set_names','rename','rename_','tex_rename','tex_rename_'} case {'set_names','rename','rename_','tex_rename','tex_rename_', 'tag'} B = feval(S(1).subs,A,S(2).subs{:}); S = shiftS(S,1); case {'disp'} ... ...\n function o = tag(o, a, b, c) % ---- Unitary tests --*-- % Add tag to a dseries oject (in place modification). % INPUTS % - o [dseries] % - a [string] Name of the tag. % - b [string] Name of the variable. % - c [any] Value of the variable tag. % % OUTPUT % - o [dseries] Updated with tag % Copyright (C) 2017 Dynare Team % % This file is part of Dynare. % % Dynare is free software: you can redistribute it and/or modify % it under the terms of the GNU General Public License as published by % the Free Software Foundation, either version 3 of the License, or % (at your option) any later version. % % Dynare is distributed in the hope that it will be useful, % but WITHOUT ANY WARRANTY; without even the implied warranty of % MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the % GNU General Public License for more details. % % You should have received a copy of the GNU General Public License % along with Dynare. If not, see . if nargin<3 % Initialize a new tag name if ~ismember(a, fieldnames(o.tags)) o.tags.(a) = cell(vobs(o), 1); end else % Test if tag name (a) exists if ~ismember(a, fieldnames(o.tags)) error('dseries::tag: Tag name %s is unknown!', a) end % Test if variable (b) exists if ~ismember(b, o.name) error('dseries::tag: Variable %s is unknown!', b) else id = strmatch(b, o.name, 'exact'); end o.tags.(a)(id) = {c}; end %@test:1 %\\$ ts = dseries(randn(10, 3)); %\\$ try %\\$ tag(ts, 'name'); %\\$ tag(ts, 'name', 'Variable_1', 'Flow'); %\\$ tag(ts, 'name', 'Variable_2', 'Stock'); %\\$ tag(ts, 'name', 'Variable_3', 'Flow'); %\\$ t(1) = 1; %\\$ catch %\\$ t(1) = 0; %\\$ end %\\$ %\\$ if t(1) %\\$ t(2) = dassert(ts.tags.name, {'Flow'; 'Stock'; 'Flow'}); %\\$ end %\\$ %\\$ T = all(t); %@eof:1 %@test:2 %\\$ ts = dseries(randn(10, 3)); %\\$ try %\\$ tag(ts, 'name'); %\\$ tag(ts, 'name', 'Variable_1', 'Flow'); %\\$ tag(ts, 'name', 'Variable_3', 'Flow'); %\\$ t(1) = 1; %\\$ catch %\\$ t(1) = 0; %\\$ end %\\$ %\\$ if t(1) %\\$ t(2) = dassert(ts.tags.name, {'Flow'; []; 'Flow'}); %\\$ end %\\$ %\\$ T = all(t); %@eof:2 %@test:3 %\\$ ts = dseries(randn(10, 3)); %\\$ try %\\$ tag(ts, 'name'); %\\$ tag(ts, 'name', 'Variable_1', 'Flow'); %\\$ tag(ts, 'noname', 'Variable_3', 1); %\\$ t(1) = 0; %\\$ catch %\\$ t(1) = 1; %\\$ end %\\$ %\\$ if t(1) %\\$ t(2) = dassert(ts.tags.name, {'Flow'; []; []}); %\\$ end %\\$ %\\$ T = all(t); %@eof:3 %@test:4 %\\$ ts = dseries(randn(10, 3)); %\\$ try %\\$ ts.tag('name'); %\\$ ts.tag('name', 'Variable_1', 'Flow'); %\\$ ts.tag('name', 'Variable_2', 'Stock'); %\\$ ts.tag('name', 'Variable_3', 'Flow'); %\\$ t(1) = 1; %\\$ catch %\\$ t(1) = 0; %\\$ end %\\$ %\\$ if t(1) %\\$ t(2) = dassert(ts.tags.name, {'Flow'; 'Stock'; 'Flow'}); %\\$ end %\\$ %\\$ T = all(t); %@eof:4" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6630029,"math_prob":0.9988449,"size":555,"snap":"2020-10-2020-16","text_gpt3_token_len":140,"char_repetition_ratio":0.13793103,"word_repetition_ratio":0.0,"special_character_ratio":0.27747747,"punctuation_ratio":0.123893805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948908,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T09:53:28Z\",\"WARC-Record-ID\":\"<urn:uuid:e32161e0-c684-4c00-b765-571bb9fb4f63>\",\"Content-Length\":\"1049684\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a565dc1-694b-4ac6-9991-8235cfcece5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d1fb171-1b96-4ed6-aedb-d0f3125d49a3>\",\"WARC-IP-Address\":\"217.70.191.81\",\"WARC-Target-URI\":\"https://git.dynare.org/Dynare/dseries/commit/14478ca16c2cbdc5c6d922d131518be742d552b9\",\"WARC-Payload-Digest\":\"sha1:DGBJFLYYYLXHKYL5C4QFASTUEKRA3XV7\",\"WARC-Block-Digest\":\"sha1:5OKIXN27GHOAN7TABSJVCV5DBWNPMXH7\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875148850.96_warc_CC-MAIN-20200229083813-20200229113813-00323.warc.gz\"}"}
https://git.sesse.net/?p=pitch;a=blobdiff;f=pitchdetector.cpp;h=cf0abe0d01fc313dd596aa44cdc956053c6b77cf;hp=71e5875c7b328a31a38ba45540d967c861af316f;hb=af5720a7e5ece0711550fa86f30a59f30b819bbd;hpb=61ad39f32700534f5750d9405efd05d33e12ad14;ds=sidebyside	[ "index 71e5875..cf0abe0 100644 (file)\n@@ -38,7 +38,7 @@ PitchDetector::~PitchDetector()\nstd::pair<double, double> PitchDetector::detect_pitch(short buf)\n{\nunsigned buf_len = fft_length / pad_factor / overlap;\nstd::pair<double, double> PitchDetector::detect_pitch(short buf)\n{\nunsigned buf_len = fft_length / pad_factor / overlap;\n- memmove(in, in + buf_len, (fft_length - buf_len) * sizeof(double));\n+ memmove(in, in + buf_len, (fft_length / pad_factor - buf_len) * sizeof(double));\n\nfor (unsigned i = 0; i < buf_len; ++i)\nin[i + (fft_length / pad_factor - buf_len)] = double(buf[i]);\n\nfor (unsigned i = 0; i < buf_len; ++i)\nin[i + (fft_length / pad_factor - buf_len)] = double(buf[i]);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66055435,"math_prob":0.93393964,"size":1080,"snap":"2020-10-2020-16","text_gpt3_token_len":325,"char_repetition_ratio":0.1598513,"word_repetition_ratio":0.31578946,"special_character_ratio":0.32407406,"punctuation_ratio":0.21787709,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97698975,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-23T21:18:23Z\",\"WARC-Record-ID\":\"<urn:uuid:59e61b27-0aeb-4854-af3f-8fca97efee1b>\",\"Content-Length\":\"8876\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:280cdf37-f09a-43d1-940e-9a531041437f>\",\"WARC-Concurrent-To\":\"<urn:uuid:be36f0ed-0569-4b2d-96ea-2bf4811d06bd>\",\"WARC-IP-Address\":\"193.35.52.50\",\"WARC-Target-URI\":\"https://git.sesse.net/?p=pitch;a=blobdiff;f=pitchdetector.cpp;h=cf0abe0d01fc313dd596aa44cdc956053c6b77cf;hp=71e5875c7b328a31a38ba45540d967c861af316f;hb=af5720a7e5ece0711550fa86f30a59f30b819bbd;hpb=61ad39f32700534f5750d9405efd05d33e12ad14;ds=sidebyside\",\"WARC-Payload-Digest\":\"sha1:INUUXCSK2L4E4PNZCPG7DTR65SLHYZLF\",\"WARC-Block-Digest\":\"sha1:5X2FAJVYUZKPSKDZAONB3IPJIKEDBIP7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145839.51_warc_CC-MAIN-20200223185153-20200223215153-00495.warc.gz\"}"}
https://stats.stackexchange.com/questions/64562/compare-two-sets-of-data	[ "# Compare two sets of data\n\nI have three sets of data, measured by three different devices: A,B and C of air balloon whose fall is influenced by wind. Each data sheet looks like:\n\nA:\n\nLongitude Latitude Altitude Weight Rotation(about main axis) Time\n... ... ... ... ... ...\n\nB:\n\nLongitude Latitude Altitude Weight Rotation(about main axis) Time\n... ... ... ... ... ...\n\n\nand similarly for C.\n\nAre there standard techniques to compare the error measurement of B and C with respect to A (considered standard)?.\n\nNote that except time, all other parameter can both increase and decrease, and longitude/latitude and rotation are the only parameters that can be positive and negative.\n\nThe problem with regression is that I do not have independent/dependent variables. The method of error analysis I generally use is to calculate ${\\rm Err}_{x}$, ${\\rm Err}_{y}$, and ${\\rm Err}_{z}$, and combine them suitably when I have a function explaining something. I do not have 'the function'. (I mean I do not have a function to use 'propagation of error'.)\n\nPrecisely, I am trying to measure accuracy of devices B and C\n\nSide note: I could not find an appropriate tag.\n\n• And where is the question? Jul 17, 2013 at 7:41\n• @sashkello The was small thinking error. I intended what are? but wrote these are. Sorry. Jul 17, 2013 at 7:44\n• Can you give examples of results you expect or even hypotheses? This will influence the answer. For example, I suppose combined change over time in longitude and latitude (i.e. speed) might be interesting. Or only change in altitude over time? On the other hand, change in weight over time maybe not? Jul 17, 2013 at 8:56\n• @robert, thank you for your comment. I expect change in all quantities. Longitude and latitude and altitude definitely change. Weight also change slightly when you consider significant change in altitude. I am merely measuring accuracy of two devices B and C with respect to A(consider standard). Hypothesis would be accurate measurement. I don't know what else? Jul 17, 2013 at 10:19\n• For clarification: (1) You want to know how much measurement error there is in the data that come from each device; & (2) A is considered the 'gold standard' / measured w/o error. Is that correct? Jul 17, 2013 at 12:49\n\n1. Compute relative differences, e.g. $(longitude_{B}-longitude_{A})/longitude_{A}$ for all variables at all time points (assuming the devices measured all variables at the same time points). Then look at the distribution and change over time of the differences.\n2. If the differences increase over time (which might or might not be plausible) you could use linear regression with time as predictor and relative difference in e.g., longitude, as variable you want to explain. This could be done for each variable and pair of devices.\n• If you only care about total error then you don't need linear regression. However, if (and only if) the error changes over time I could imagine it would make a lot of sense to include that aspect in your analysis. For example, sensor error could increase over time because the accuracy of the sensors deteriorates over time or because measurements at $t_2$ depend on measurements at $t_1$ (and error accumulates). Whether such a scenario makes sense or not is of course up to you to decide. Jul 17, 2013 at 14:33" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85301155,"math_prob":0.86023283,"size":1204,"snap":"2022-27-2022-33","text_gpt3_token_len":281,"char_repetition_ratio":0.12166667,"word_repetition_ratio":0.10152284,"special_character_ratio":0.25415283,"punctuation_ratio":0.23371647,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9910997,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T03:34:14Z\",\"WARC-Record-ID\":\"<urn:uuid:80bccdef-6b7c-416b-886f-21ec04050546>\",\"Content-Length\":\"230795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:813f661b-b238-43d5-9b70-c0899cbe6605>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa82d024-a730-4893-81dc-7c48cc6422d2>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stats.stackexchange.com/questions/64562/compare-two-sets-of-data\",\"WARC-Payload-Digest\":\"sha1:4Q75IHWRSLIE6P2OLV7AARF6PNKWEWV2\",\"WARC-Block-Digest\":\"sha1:QKRTKHY65WHDLEVEIDZT7GL6LY34YWHA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103661137.41_warc_CC-MAIN-20220630031950-20220630061950-00059.warc.gz\"}"}
https://wiki.haskell.org/index.php?title=Examples/Random_list&diff=15445&printable=yes	[ "# Difference between revisions of \"Examples/Random list\"\n\n## Create a random list\n\nGenerate a random list of numbers, without using the System.Random.randoms method:\n\n```import System.Random\nimport Data.List\n\nmain = do\nseed <- newStdGen\nlet rs = randomlist 10 seed\nprint rs\n\nrandomlist :: Int -> StdGen -> [Int]\nrandomlist n = take n . unfoldr (Just . random)\n```\n\n## Delete an element at random\n\n``` unpick and unpick' are by osfameron and are from http://osfameron.vox.com/library/post/more-random-fun.html (no explicit license)\nremoveOne is by Chris Kuklewicz (BSD3 licence, 2007)\n\n> import System.Random\n> import Debug.Trace -- for removeOne' demonstration\n\nThe unpick function and its helper unpick' are strict in the entire\nlist being operated on (forcing it all into memory at once). And IO\n[a] cannot lazily return any initial values.\n\n> unpick :: [a] -> IO [a]\n> unpick [] = undefined\n> unpick [x] = do return []\n> unpick (x:xs) = do zs <- unpick' [] [x] xs 2\n> return (reverse zs)\n>\n> unpick' :: (Num p, Random p) => [t] -> [t] -> [t] -> p -> IO [t]\n> unpick' curr orig [] _\n> = do return curr\n> unpick' curr orig (next:rest) prob\n> = do r <- getStdRandom (randomR (1,prob))\n> let curr' = if r == 1 then orig else (next:curr)\n> unpick' curr' (next:orig) rest (prob+1)\n\nTo run in the IO Monad just use (getStdRandom . removeOne) :: [a] -> IO [a].\n\nremoveOne returns the output list lazily as soon as it has decided\nnot to delete any element in a prefix of the input list.\nThe resulting list is constructed efficiently, with no wasted\nintermediate list construction. removeOne allows any output it\ngenerates to be garbage collected, it holds no references to it.\n\n\"removeOne\" is presented in curried form, without a binding for the\nRandomGen g. The StdGen is hidden inside a State\nmonad. removeOne is designed for use with Strict.Lazy. It may not be\noptimal to use with Strict.Strict.\n\nLike \"tail\" this function is partial and will produce an error if\ngiven the empty list.\n\n> removeOne :: (RandomGen g) => [a] -> g -> ([a],g)\n> removeOne [] = error \"Cannot removeOne from empty list\"\n> removeOne whole@(_:xs) = runState (helper whole xs 0 1) where\n\nThe laziness is needed in helper to make \"rest\" a lazy thunk. The\n\"start\" list parameter to helper is a suffix of \"whole\" that has the\ncurrent candidate for deletion as its head. \"oldIndex\" is the index\nof the current candidate for deletion in the \"whole\" list. \"here\" is a\nsuffix of \"whole\" with the \"index\" element of whole as its head. The\nrandomR decides if the head of \"here\" replaces the head of \"start\" as\nthe candidate to remove. If it does replace the old candidate then\na prefix of \"start\" of length \"(index-oldIndex)\" is immediately\noutput, counted off by prependSome.\n\nAssert \"start\" is never [].\nAssert 0 <= oldIndex < index.\n\n> helper start [] oldIndex index = return (tail start)\n> helper start here@(_:ys) oldIndex index = do\n> r <- State (randomR (0,index))\n> if r==0 then do rest <- helper here ys index \\$! succ index\n> return (prependSome (index-oldIndex) start rest)\n> else helper start ys oldIndex \\$! succ index\n\nI assert that \"prependSome n xs ys == take n xs ++ ys\" but slightly\noptimized (without depending on the compiler). Assert n >= length xs.\n\n> prependSome :: Int -> [a] -> [a] -> [a]\n> prependSome 0 _ rest = rest\n> prependSome n (x:xs) rest = x : prependSome (pred n) xs rest\n> prependSome _ [] _ = error \"impossible error in removeOne.prependSome\"\n\n\"removeOne'\" is a tracing version for demonstration below:\n\n> removeOne' :: (Show a,RandomGen g) => [a] -> g -> ([a],g)\n> removeOne' [] _ = error \"Cannot removeOne from empty list\"\n> removeOne' whole@(x:xs) g = runState (helper whole xs 0 1) g where\n> helper start [] oldIndex index = return (tail start)\n> helper start here@(_:ys) oldIndex index = do\n> r <- State (randomR (0,index))\n> if r==0 then do rest <- helper here ys index \\$! succ index\n> let rest' = trace \".\" rest\n> return (prependSome (index-oldIndex) start rest')\n> else do let ys' = trace \"_\" ys\n> helper start ys' oldIndex \\$! succ index\n\nUse \"removeOne'\" to demonstrate when random decisions to drop\nelements are made. This also demonstrates that removeOne is lazy,\nreturning elements as soon as the removal decision has moved on to a\nlater element (the \".\" is output instead of \"_\").\n\nThe element after the last \".\" is the one actually removed,\ndefaulting to the first element.\n\nSince the probability of \".\" decreases, the average length of the\nrun of output produced by appendSome increases as the list is\nprocessed.\n\nMain> getStdRandom (removeOne' [1..10])\n[1.\n_\n_\n,2,3,4.\n_\n_\n_\n_\n,5,6,7,8,9.\n]\nMain> getStdRandom (removeOne' [1..10])\n_\n_\n[1,2,3.\n_\n_\n_\n_\n_\n_\n,5,6,7,8,9,10]\nMain> getStdRandom (removeOne' [1..10])\n_\n_\n_\n_\n_\n_\n_\n_\n_\n[2,3,4,5,6,7,8,9,10]\nMain> getStdRandom (removeOne' [1..10])\n[1.\n_\n_\n_\n_\n_\n_\n_\n_\n,3,4,5,6,7,8,9,10]\nMain> getStdRandom (removeOne' [1..10])\n[1.\n,2.\n_\n_\n_\n_\n_\n_\n_\n,4,5,6,7,8,9,10]\nMain> getStdRandom (removeOne' [1..10])\n[1.\n_\n_\n_\n_\n,2,3,4,5,6.\n_\n_\n,7,8,9.\n]\n\nIf I use \":m + Data.List\" then I can demonstrate how fair the removal is:\n\n*Main Data.List> sequence (replicate 1000 \\$ getStdRandom (removeOne [1..4])) >>= return . map length . group . sort\n[241,255,239,265]\n\nwhere a perfect balance would be [250,250,250,250]\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73738384,"math_prob":0.945395,"size":11585,"snap":"2020-34-2020-40","text_gpt3_token_len":3723,"char_repetition_ratio":0.17787756,"word_repetition_ratio":0.5686801,"special_character_ratio":0.380492,"punctuation_ratio":0.16689466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97339994,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-23T02:41:23Z\",\"WARC-Record-ID\":\"<urn:uuid:48bdd15d-cf29-407a-a6db-4ff52dcdab4a>\",\"Content-Length\":\"81833\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0dffb103-2445-40d5-b769-017ca33685d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:883f7221-98ae-4030-9083-f2274b8e158a>\",\"WARC-IP-Address\":\"199.232.65.175\",\"WARC-Target-URI\":\"https://wiki.haskell.org/index.php?title=Examples/Random_list&diff=15445&printable=yes\",\"WARC-Payload-Digest\":\"sha1:HMYWPITGIP7IG6VEDAW2TXC2JAJYKG2G\",\"WARC-Block-Digest\":\"sha1:YJE234N67DZQF3G46HHLDMV3FIPHDKDC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400209665.4_warc_CC-MAIN-20200923015227-20200923045227-00187.warc.gz\"}"}
http://dreamhosting.xyz/hvds-case-study-63/	[ "# HVDS CASE STUDY\n\nThe feeder voltage and feeder current are two constraints which should be within the standard range. The power factor assumed to be used as 0. Now voltage drop can be estimated at various power factors and also at various temperatures of conductors. The proposed case study includes the conversion of existing LVDS into HVDS in order to minimise the distribution losses and pilferage thereby, improving the voltage profile and quality of supply to the consumers in existing 11kV feeder. Log In Sign Up. This paper demonstrates the capability of load factor and load loss factor to calculate the power losses of the network. Nuclear Power 20 nuclear reactors in operation, generating 5.", null, "Total losses per annum The total installed capacity in India is Thus, this sector especially, the distribution sectors require economical system to provide electrical energy at a suitable prize and at a minimum voltage drop to reduce the voltage regulation. Estimation of Current at different power factor On the basis of the current estimation at reference power factor of 0. Thus, the net power loss per annum can be calculated as considering working for hours of a day with days per annum. The performance of the feeder may be improved by the HVDS system installation, which results in saving of total losses per annum and an annual savings of In 1st stage, the power losses including the theft and losses in the line and transformer losses for both LT and HT systems are determined and then in 2 nd stage, determination of annual savings and payback period is carried out alongwith the complete comparison of LVDS and HVDS system of 11kV feeder.\n\nThus, this sector especially, the distribution sectors require economical system to provide electrical energy at a suitable prize and at a minimum voltage drop to reduce the voltage regulation. The results obtained can be used for financial loss calculation and can be presented to regulate wtudy tariff determination process. The results are satisfactorily obtained in the above case study, theoretically.\n\n# Case Study: High Voltage Distribution System (HVDS) Implementation in BESCOM and MGVCL\n\nThe present paper emphasis on the re-designing for the existing distribution network for fulfilment of the above objectives of distribution feeder. Estimation of Current at different power factor On the basis of the current estimation at reference power factor of 0. Calculative Analysis of 11kV Distribution Feeder. Thus core and copper losses occurred in the transformer also contributed to the total power losses per annum in LT system So, for kVA rating transformer, the fixed value of no-load losses and full load losses are W and W respectively.\n\nDISSERTATION SUR LE MARIAGE PUTATIF\n\nHelp Center Find new research papers in: Therefore, loss minimization in power system has assumed greater significance HVDS scheme has led to the formulation of new strategy of energy conservation and minimization of transmission and distribution losses by reducing the power theft. The installation of HVDS system in considerable area of the sub-division, is the main technique which is also applied in the present work to evaluate the proposed re-designing of existing distribution network and its future planning.\n\nGeneration Total capacity till Calculation of voltage drop at various power factors and temperature As the values of current at various power factors had been determined as per above table. Installed capacity generation in Punjab IV.\n\nThis results in improving the stability as well as energy handling capacity atudy the system at minimum cost. Thus, the net power loss per annum can be calculated as considering working for hours stydy a day with days per annum.\n\nFrom the result, it is also realised that the causes of voltage drop on the feeder was mainly due to high impedance level as compared to the permissible value and this high impedance is caused by poor jointing and terminations, use of undersized conductors and different types of conductor materials etc.\n\nNuclear Power 20 nuclear reactors in operation, generating 5. Ritula Thakur et al in presented a paper analysing and designing with the observation that the existing feeder is to be operated on 0. The subdivision is working on the installation of HVDS system in this feeder and calculations made above is an attempt for success of the above work made by State Electricity Board. Sub- transmission and distribution systems constitute the link between electricity utilities and consumers.\n\nLOKTAK LAKE ESSAY\n\nIn fact, it has become essential ingredient for improving the quality of life and its absence is associated with poverty and poor quality of life. Pilferage on HT system is assumed to negligible. Total load losses per annum units hvvds 5. Thus required calculation of HT transformer losses are as under in table no. Total power losses Sub-transmission and distribution systems constitute the link between electricity utilities and consumers.\n\nThis work was mainly focused on low voltage distribution system LVDS. As the distribution network is located far away from the sources of power generation and the other infrastructure of electrical power system.\n\n## Case Study: High Voltage Distribution System (HVDS) Implementation in BESCOM and MGVCL\n\nBut the weight of 80mm 2 conductor is The power factor assumed to be used as 0. Total iron losses per annum units units 4. Due to this unplanned expansion in the system, the supply conditions were sacrificed to meet the required targets.", null, "The performance of the feeder may be hvrs by the HVDS system installation, which results in saving of total losses per annum and an annual savings of LT Transformer losses As per the information derived from the 33kV State Electricity board of subdivision, on LT side, a large number of transformer of capacity of kVA is used to supply the power to the consumers at the end point of each section.\n\nThe technical losses are the losses occurred in the electrical elements during of transmission of energy from source to consumer and mainly comprises of ohmic losses. The details are as under in table 1 .", null, "Efficient functioning of these segments of the electricity utility is essential to sustain the growth of the power sector and the economy of the country." ]	[ null, "http://dreamhosting.xyz/essay.png", null, "https://cdn.slidesharecdn.com/ss_thumbnails/tpddlcasestudy-141230062517-conversion-gate02-thumbnail-4.jpg", null, "https://imgv2-1-f.scribdassets.com/img/document/49451269/original/364fc74a27/1545306266", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9359864,"math_prob":0.9164949,"size":6195,"snap":"2020-10-2020-16","text_gpt3_token_len":1180,"char_repetition_ratio":0.14165725,"word_repetition_ratio":0.1998002,"special_character_ratio":0.1803067,"punctuation_ratio":0.06481481,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9579147,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,4,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-06T15:57:01Z\",\"WARC-Record-ID\":\"<urn:uuid:18f8881b-6595-41ba-8d09-3562b6af0792>\",\"Content-Length\":\"29069\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:16271137-3bd5-4c7b-9415-5bd4e0707845>\",\"WARC-Concurrent-To\":\"<urn:uuid:490e5c03-c440-4c21-8a18-831d2c5eaa7e>\",\"WARC-IP-Address\":\"104.31.70.247\",\"WARC-Target-URI\":\"http://dreamhosting.xyz/hvds-case-study-63/\",\"WARC-Payload-Digest\":\"sha1:7T5K6LWPIOMNBACJMRD7UD5SZCW3OK4D\",\"WARC-Block-Digest\":\"sha1:J65U42M7XMKLU6RSWZLUNLMI6BQ5WW7O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371637684.76_warc_CC-MAIN-20200406133533-20200406164033-00245.warc.gz\"}"}
https://www.univ-smb.fr/listic/en/production-scientifique/revue-busefal/version-electronique/ebusefal-86/	[ "# eBUSEFAL #86\n\n Amiya Kumar Shyamal and Madhumangal Pal Distances Between Interval-Valued Intuitionistic Fuzzy Sets Liu Huawen Decompositions of Intuitionistic Fuzzy Sets Ma Baoguo On Fuzzy weakly semi-precontinuous multifunctions Liu Jinlu Pseudo-fuzzy Linear Inequation and Equation Wang Hongxu and Li Guishan Notes on the decision theorems that the equation type II of a fuzzy matrix has the solutions when the index is one Witold Pedrycz and George Vukovich An algorithmic Framework of Collaborative Clustering Witold Pedrycz and George Vukovich Relevance and consistency in rule-based systems Zou Li (u,v)-Implication of IOFL Jacek M. Leski An e-insensitive fuzzy c-means clustering Wang Xin and Liu Xiadong The Base of Finite EI Algebra 1 Wen-Xiang Gu and Su-yun Li The necessary and sufficient condition that a fuzzy set is a pointwise fuzzy group Bai Shi-Zhong Fuzzy Pre-Urysohn Spaces Bai Shi-Zhong Countably Strong Lowen's Compactness in L-fts Yang Jie The decision theorems that the equation type III of a fuzzy matrix has a solution when the index is one Zhang Chengyi and Su Jinlin Fuzzy Congruence on Rings Liu Jinlu On A Kind of Fuzzy Relational Equation" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7848195,"math_prob":0.8213113,"size":1185,"snap":"2023-14-2023-23","text_gpt3_token_len":359,"char_repetition_ratio":0.11346316,"word_repetition_ratio":0.057142857,"special_character_ratio":0.19324894,"punctuation_ratio":0.010256411,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T18:02:28Z\",\"WARC-Record-ID\":\"<urn:uuid:1632a9c8-cad0-44c5-992f-e0a6df6301c8>\",\"Content-Length\":\"693724\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9a06353-bbfb-42d3-964d-9a1f0d5d3c21>\",\"WARC-Concurrent-To\":\"<urn:uuid:9803d22f-28d8-4db2-825f-d7f09a5e2b0c>\",\"WARC-IP-Address\":\"145.239.14.239\",\"WARC-Target-URI\":\"https://www.univ-smb.fr/listic/en/production-scientifique/revue-busefal/version-electronique/ebusefal-86/\",\"WARC-Payload-Digest\":\"sha1:4V4QO6L44OWRAUWE4SVMAHCQ3VZMFACY\",\"WARC-Block-Digest\":\"sha1:JLPO2AZDYULMHXUBNTQRW5GH2NZQGRNH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654012.67_warc_CC-MAIN-20230607175304-20230607205304-00430.warc.gz\"}"}
https://stuffsure.com/what-is-4-out-of-16-as-a-percentage/	[ "# What is 4 out of 16 as a Percentage?\n\nIf you’re not sure what 4 out of 16 as a percentage, don’t worry – we’ll show you how to calculate it. Just follow these simple steps and you’ll have your answer in no time.\n\nCheckout this video:\n\n## Introduction\n\n4 out of 16 as a percentage is equal to 25 percent. To find the percentage, divide 4 by 16 and then multiply the answer by 100. The answer will be 25 percent.\n\n## What is 4 out of 16 as a Percentage?\n\n4 out of 16 as a percentage is 25%.\n\n## Conclusion\n\n4 out of 16 is equivalent to 25 percent." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9275967,"math_prob":0.9681767,"size":491,"snap":"2023-40-2023-50","text_gpt3_token_len":132,"char_repetition_ratio":0.20123203,"word_repetition_ratio":0.125,"special_character_ratio":0.29531568,"punctuation_ratio":0.096491225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.988052,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T16:04:28Z\",\"WARC-Record-ID\":\"<urn:uuid:a537b59c-2530-4ef5-86bd-db34378a9a69>\",\"Content-Length\":\"55751\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:929384c3-c6c8-4122-94e3-d0ac282512dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:90f5b7ad-d600-4933-ae3f-2196e5a5befc>\",\"WARC-IP-Address\":\"50.16.223.119\",\"WARC-Target-URI\":\"https://stuffsure.com/what-is-4-out-of-16-as-a-percentage/\",\"WARC-Payload-Digest\":\"sha1:4OPFDDODLRHJA73OGMGUDSOXT6VBMMNC\",\"WARC-Block-Digest\":\"sha1:3IP4DWHH2YA3ZNEW6WYSNOXVJ4EVWY64\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100762.64_warc_CC-MAIN-20231208144732-20231208174732-00721.warc.gz\"}"}
https://www.mathworksheets4kids.com/volume-mixed-pyramids.php	[ "1. Worksheets>\n2. Math>\n3. Geometry>\n4. Volume>\n5. Mixed Pyramids\n\n# Volume of a Pyramid Worksheets\n\nWork out the volume of the pyramids with rectangular, triangular and polygonal base faces. Calculate the volume by plugging in the measures expressed as integers and decimals in the appropriate formulas. The 8th grade and high school printable worksheets are classified into two levels. Level 1 comprises polygons with 3 or 4-sided base faces, while level 2 includes polygonal base faces. Practice finding the missing measures as well. Explore some of these worksheets for free!\n\nSelect the Measurement Units\n\nFinding Volume using Base Area \| Integers\n\nJump-start your learning with this batch of PDF worksheets for 8th grade and 9th grade students on mixed pyramids featuring rectangular and triangular pyramids. Apply apt formulas to find the volume using the base area measure expressed as integers.\n\nFinding Volume using Base Area \| Decimals\n\nAdvance to the next level consisting of pyramids whose base faces are rectangles or triangles, and the base area is indicated as decimals. Multiply the base area with the height and divide by 3 to compute the volume.\n\nVolume of Rectangular Pyramids\n\nExclusively dealing with rectangular pyramids, these PDF worksheets for 9th grade and 10th grade students are a must-have for thorough knowledge and practice in finding the volume of rectangular pyramids offering varied levels of difficulty.\n\n(24 Worksheets)\n\nVolume of Triangular Pyramids\n\nMake strides in your practice with these printable worksheets on finding the volume of triangular pyramids or tetrahedrons. Try the easy, moderate and challenging levels with integer and decimal dimensions.\n\n(24 Worksheets)\n\nFinding Volume of Polygonal Pyramids using Apothem \| Level 1\n\nAnother essential step is to determine the volume of pyramids with polygonal base faces. Plug in the apothem(a), number of sides(n), side length(s) and height(h) in the formula V = 1/6 * ansh and compute the volume.\n\nVolume of Polygonal Pyramids using Side length or Perimeter \| Level 2\n\nThe side length or perimeter and height are provided. First, find the apothem (apothem = side / 2 tan (180/n)), and then assign the known values in the volume formula to solve for volume of polygonal pyramids.\n\nVolume of Pyramids \| Level 1 - Integers - Easy\n\nThe easy level contains pyramids with integer dimensions ≤ 20. One-third of the base area of the square, rectangle or a triangle multiplied by the height results in the volume of the pyramid. Recommended for grade 8 and grade 9 children.\n\nVolume of Pyramids \| Level 1 - Integers - Moderate\n\nPolish up your skills in finding the volume of pyramids with this section of high school geometry worksheets presenting pyramids as 3D shapes and in the word format with 3 or 4-sided base faces involving integer dimensions ≥ 20.\n\nVolume of Pyramids \| Level 1 - Decimals\n\nScale new heights as you practice solving for volume with these pdf worksheets posing a challenge with decimal dimensions. Apply relevant formula, substitute and calculate the volume of each pyramid.\n\nVolume of Pyramids \| Level 2\n\nWith the sides of the base face, increases the level of difficulty. Figure out the volume of the triangular, rectangular and polygonal pyramids applying suitable formulas and bolster skills.\n\nVolume of Pyramids \| Missing Measure\n\nMotivate learners to give these printable worksheets a quick try. All they have to do is simply rearrange the formula, making the missing measure (x) the subject, plug in the known values and solve!\n\nRelated Worksheets" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86409026,"math_prob":0.9495832,"size":3441,"snap":"2023-40-2023-50","text_gpt3_token_len":722,"char_repetition_ratio":0.1754437,"word_repetition_ratio":0.052158274,"special_character_ratio":0.20023249,"punctuation_ratio":0.0713073,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98783106,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T19:40:00Z\",\"WARC-Record-ID\":\"<urn:uuid:c551efc3-336a-4570-86d7-39322f5aca2b>\",\"Content-Length\":\"40665\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76804cb3-33fb-417a-9d69-97ed2a2ea4e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:cc27697f-6928-429c-9607-eb556322afb8>\",\"WARC-IP-Address\":\"67.225.178.45\",\"WARC-Target-URI\":\"https://www.mathworksheets4kids.com/volume-mixed-pyramids.php\",\"WARC-Payload-Digest\":\"sha1:ZI2DN2JT4IX666H5UP72XF54TVKHIZKQ\",\"WARC-Block-Digest\":\"sha1:4BSQOA2UPDLRGZSRIIDN6CO245NZLWMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100686.78_warc_CC-MAIN-20231207185656-20231207215656-00511.warc.gz\"}"}
https://www.mediagust.com/2019/11/artificial-intelligence-ideal-mapping.html	[ "Once we realize that an agent's behavior depends only on its percept sequence to date, then we can describe any particular agent by making a table of the action it takes in response to each possible percept sequence. (For most agents, this would be a very long list—infinite, in fact, unless we place a bound on the length of percept sequences we want to consider.) Such a list is called a mapping from percept sequences to actions. We can, in principle, find out which mapping correctly describes an agent by trying out all possible percept sequences and recording which actions the agent does in response. (If the agent uses some randomization in its computations, then we would have to try some percept sequences several times to get a good idea of the agent's average behavior.) And if mappings describe agents, then ideal mappings describe ideal agents.\n\nSpecifying which action an agent ought to take in response to any given percept sequence provides a design for an ideal agent. This does not mean, of course, that we have to create an explicit table with an entry for every possible percept sequence. It is possible to define a specification of the mapping without exhaustively enumerating it. Consider a very simple agent: the square-root function on a calculator. The percept sequence for this agent is a sequence of keystrokes representing a number, and the action is to display a number on the display screen. The ideal mapping is that when the percept is a positive number x, the right action is to display a positive number z such that z 2 « x, accurate to, say, 15 decimal places.\n\nThis specification of the ideal mapping does not require the designer to actually construct a table of square roots. Nor does the square-root function have to use a table to behave correctly: Figure 2.2 shows part of the ideal mapping and a simple program that implements the mapping using Newton's method. The square-root example illustrates the relationship between the ideal mapping and an ideal agent design, for a very restricted task. Whereas the table is very large, the agent is a nice,; compact program. It turns out that it is possible to design nice, compact agents that implement part of the ideal mapping for the square-root problem (accurate to 1 5 digits), and a corresponding program that implements the ideal mapping." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.895473,"math_prob":0.9700656,"size":2420,"snap":"2019-51-2020-05","text_gpt3_token_len":508,"char_repetition_ratio":0.15811259,"word_repetition_ratio":0.009876544,"special_character_ratio":0.19545455,"punctuation_ratio":0.09409191,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9635108,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T10:14:54Z\",\"WARC-Record-ID\":\"<urn:uuid:928962a7-db19-4176-b374-66db91cf4a33>\",\"Content-Length\":\"364457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a62c9eba-0ad9-44e5-b7f8-03ee1bfdf0b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2a1590f-63bd-43f0-8139-1253675b5f2b>\",\"WARC-IP-Address\":\"172.217.164.179\",\"WARC-Target-URI\":\"https://www.mediagust.com/2019/11/artificial-intelligence-ideal-mapping.html\",\"WARC-Payload-Digest\":\"sha1:HETYZW5HJJQ7SLUVAO2ITRIXMBE4COKC\",\"WARC-Block-Digest\":\"sha1:YEH4CHEJJ3DCRGN64KPAYKSHOB2YZF6E\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540518627.72_warc_CC-MAIN-20191209093227-20191209121227-00052.warc.gz\"}"}
https://nl.mathworks.com/help/deeplearning/ref/dlarray.dlconv.html	[ "# dlconv\n\nDeep learning convolution\n\n## Syntax\n\n``dlY = dlconv(dlX,weights,bias)``\n``dlY = dlconv(dlX,weights,bias,'DataFormat',FMT)``\n``dlY = dlconv(___Name,Value)``\n\n## Description\n\nThe convolution operation applies sliding filters to the input data. Use 1-D and 2-D filters with ungrouped or grouped convolutions and 3-D filters with ungrouped convolutions.\n\nUse grouped convolution for channel-wise separable (also known as depth-wise separable) convolution. For each group, the operation convolves the input by moving filters along spatial dimensions of the input data, computing the dot product of the weights and the data and adding a bias. If the number of groups is equal to the number of channels, then this function performs channel-wise convolution. If the number of groups is equal to `1`, this function performs ungrouped convolution.\n\nNote\n\nThis function applies the deep learning convolution operation to `dlarray` data. If you want to apply convolution within a `layerGraph` object or `Layer` array, use one of the following layers:\n\nexample\n\n````dlY = dlconv(dlX,weights,bias)` computes the deep learning convolution of the input `dlX` using sliding convolutional filters defined by `weights`, and adds a constant `bias`. The input `dlX` is a formatted `dlarray` with dimension labels. Convolution acts on dimensions that you specify as `'S'` dimensions. The output `dlY` is a formatted `dlarray` with the same dimension labels as `dlX`.```\n\nexample\n\n````dlY = dlconv(dlX,weights,bias,'DataFormat',FMT)` also specifies dimension format `FMT` when `dlX` is not a formatted `dlarray`. The output `dlY` is an unformatted `dlarray` with the same dimension order as `dlX`. ```\n\nexample\n\n````dlY = dlconv(___Name,Value)` specifies options using one or more name-value pair arguments in addition to the input arguments in previous syntaxes. For example, `'Stride',3` sets the stride of the convolution operation. ```\n\n## Examples\n\ncollapse all\n\nConvolve all channels of an image input using a single filter.\n\nImport the image data and convert it to a `dlarray`.\n\n```X = imread('sherlock.jpg'); dlX = dlarray(single(X),'SSC');```\n\nDisplay the image.\n\n`imshow(X,'DisplayRange',[])`", null, "Initialize the convolutional filters. Specify an ungrouped convolution that applies a single filter to all three channels of the input data.\n\n```filterHeight = 10; filterWidth = 10; numChannelsPerGroup = 3; numFiltersPerGroup = 1; numGroups = 1; weights = rand(filterHeight,filterWidth,numChannelsPerGroup,numFiltersPerGroup,numGroups);```\n\nInitialize the bias term.\n\n`bias = rand(numFiltersPerGroupnumGroups,1);`\n\nPerform the convolution. Use a `'Stride'` value of `2` and a `'DilationFactor'` value of `2`.\n\n`dlY = dlconv(dlX,weights,bias,'Stride',2,'DilationFactor',2);`\n\nDisplay the convolved image.\n\n```Y = extractdata(dlY); imshow(Y,'DisplayRange',[])```", null, "Convolve the input data in three groups of two channels each. Apply four filters per group.\n\nCreate the input data as 10 observations of size 100-by-100 with six channels.\n\n```height = 100; width = 100; channels = 6; numObservations = 10; X = rand(height,width,channels,numObservations); dlX = dlarray(X,'SSCB');```\n\nInitialize the convolutional filters. Specify three groups of convolutions that each apply four convolution filters to two channels of the input data.\n\n```filterHeight = 8; filterWidth = 8; numChannelsPerGroup = 2; numFiltersPerGroup = 4; numGroups = 3; weights = rand(filterHeight,filterWidth,numChannelsPerGroup,numFiltersPerGroup,numGroups);```\n\nInitialize the bias term.\n\n`bias = rand(numFiltersPerGroupnumGroups,1);`\n\nPerform the convolution.\n\n```dlY = dlconv(dlX,weights,bias); size(dlY)```\n```ans = 1×4 93 93 12 10 ```\n`dims(dlY)`\n```ans = 'SSCB' ```\n\nThe 12 channels of the convolution output represent the three groups of convolutions with four filters per group.\n\nSeparate the input data into channels and perform convolution on each channel separately.\n\nCreate the input data as a single observation with a size of 64-by-64 and 10 channels. Create the data as an unformatted `dlarray`.\n\n```height = 64; width = 64; channels = 10; X = rand(height,width,channels); dlX = dlarray(X);```\n\nInitialize the convolutional filters. Specify an ungrouped convolution that applies a single convolution to all three channels of the input data.\n\n```filterHeight = 8; filterWidth = 8; numChannelsPerGroup = 1; numFiltersPerGroup = 1; numGroups = channels; weights = rand(filterHeight,filterWidth,numChannelsPerGroup,numFiltersPerGroup,numGroups);```\n\nInitialize the bias term.\n\n`bias = rand(numFiltersPerGroupnumGroups,1);`\n\nPerform the convolution. Specify the dimension labels of the input data using the `'DataFormat'` option.\n\n```dlY = dlconv(dlX,weights,bias,'DataFormat','SSC'); size(dlY)```\n```ans = 1×3 57 57 10 ```\n\nEach channel is convolved separately, so there are 10 channels in the output.\n\n## Input Arguments\n\ncollapse all\n\nInput data, specified as a `dlarray` with or without dimension labels or a numeric array. When `dlX` is not a formatted `dlarray`, you must specify the dimension label format using `'DataFormat',FMT`. If `dlX` is a numeric array, at least one of `weights` or `bias` must be a `dlarray`.\n\nConvolution acts on dimensions that you specify as spatial dimensions using the `'S'` dimension label. You can specify up to three dimensions in `dlX` as `'S'` dimensions.\n\nData Types: `single` \| `double`\n\nConvolutional filters, specified as a `dlarray` with or without labels or a numeric array. The `weights` argument specifies the size and values of the filters, as well as the number of filters and the number of groups for grouped convolutions.\n\nSpecify weights as a `filterSize`-by-`numChannelsPerGroup`-by-`numFiltersPerGroup`-by-`numGroups` array.\n\n• `filterSize` — Size of the convolutional filters. `filterSize` can have up to three dimensions, depending on the number of spatial dimensions in the input data.\n\nInput Data `'S'` Dimensions`filterSize`\n1-Dh, where h corresponds to the height of the filter\n2-D h-by-w, where h and w correspond to the height and width of the filter, respectively\n3-Dh-by-w-by-d, where h, w, and d correspond to the height, width, and depth of the filter, respectively\n\n• `numChannelsPerGroup` — Number of channels to convolve within each group. `numChannelsPerGroup` must equal the number of channels in the input data divided by `numGroups`, the number of groups. For ungrouped convolutions, where `numGroups = 1`, `numChannelsPerGroup` must equal the number of channels in the input data.\n\n• `numFiltersPerGroup` — Number of filters to apply within each group.\n\n• `numGroups` — Number of groups (optional). When `numGroups > 1`, the function performs grouped convolutions. Grouped convolutions are not supported for input data with more than two `'S'` dimensions. When `numGroups = 1`, the function performs ungrouped convolutions; in this case, this dimension is singleton and can be omitted.\n\nIf `weights` is a formatted `dlarray`, it can have multiple spatial dimensions labeled `'S'`, one channel dimension labeled `'C'`, and up to two other dimensions labeled `'U'`. The number of `'S'` dimensions must match the number of `'S'` dimensions of the input data. The labeled dimensions correspond to the filter specifications as follows.\n\nFilter SpecificationDimension Labels\n`filterSize`Up to three `'S'` dimensions\n`numChannelsPerGroup``'C'` dimension\n`numFiltersPerGroup`First `'U'` dimension\n`numGroups` (optional)Second `'U'` dimension\n\nData Types: `single` \| `double`\n\nBias constant, specified as a `dlarray` vector or `dlarray` scalar with or without labels, a numeric vector, or a numeric scalar.\n\n• If `bias` is a scalar or has only singleton dimensions, the same bias is applied to each output.\n\n• If `bias` has a nonsingleton dimension, each element of `bias` is the bias applied to the corresponding convolutional filter specified by `weights`. The number of elements of `bias` must match the number of filters specified by `weights`.\n\n• If `bias` is a scalar numeric array with value `0`, the bias term is disabled and no bias is added during the convolution operation.\n\nIf `bias` is a formatted `dlarray`, the nonsingleton dimension must be a channel dimension labeled `'C'`.\n\nData Types: `single` \| `double`\n\n### Name-Value Pair Arguments\n\nSpecify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.\n\nExample: `'DilationFactor',2` sets the dilation factor for each convolutional filter to `2`.\n\nDimension order of unformatted input data, specified as the comma-separated pair consisting of `'DataFormat'` and a character array or string `FMT` that provides a label for each dimension of the data. Each character in `FMT` must be one of the following:\n\n• `'S'` — Spatial\n\n• `'C'` — Channel\n\n• `'B'` — Batch (for example, samples and observations)\n\n• `'T'` — Time (for example, sequences)\n\n• `'U'` — Unspecified\n\nYou can specify multiple dimensions labeled `'S'` or `'U'`. You can use the labels `'C'`, `'B'`, and `'T'` at most once.\n\nYou must specify `'DataFormat'` when the input data `dlX` is not a formatted `dlarray`.\n\nExample: `'DataFormat','SSCB'`\n\nData Types: `char` \| `string`\n\nStep size for traversing the input data, specified as the comma-separated pair consisting of `'Stride'` and a numeric scalar or numeric vector. If you specify `'Stride'` as a scalar, the same value is used for all spatial dimensions. If you specify `'Stride'` as a vector of the same size as the number of spatial dimensions of the input data, the vector values are used for the corresponding spatial dimensions.\n\nThe default value of `'Stride'` is `1`.\n\nExample: `'Stride',3`\n\nData Types: `single` \| `double`\n\nFilter dilation factor, specified as the comma-separated pair consisting of `'DilationFactor'` and one of the following.\n\n• Numeric scalar — The same dilation factor value is applied for all spatial dimensions.\n\n• Numeric vector — A different dilation factor value is applied along each spatial dimension. Use a vector of size `d`, where `d` is the number of spatial dimensions of the input data. The `i`th element of the vector specifies the dilation factor applied to the `i`th spatial dimension.\n\nUse the dilation factor to increase the receptive field of the filter (the area of the input that the filter can see) on the input data. Using a dilation factor corresponds to an effective filter size of `filterSize + (filterSize-1)(dilationFactor-1)`.\n\nExample: `'DilationFactor',2`\n\nData Types: `single` \| `double`\n\nSize of padding applied to edges of data, specified as the comma-separated pair consisting of `'Padding'` and one of the following:\n\n• `'same'` — Padding size is set so that the output size is the same as the input size when the stride is `1`. More generally, the output size of each spatial dimension is `ceil(inputSize/stride)`, where `inputSize` is the size of the input along a spatial dimension.\n\n• Numeric scalar — The same amount of padding is applied to both ends of all spatial dimensions.\n\n• Numeric vector — A different amount of padding is applied along each spatial dimension. Use a vector of size `d`, where `d` is the number of spatial dimensions of the input data. The `i`th element of the vector specifies the size of padding applied to the start and the end along the `i`th spatial dimension.\n\n• Numeric matrix — A different amount of padding is applied to the start and end of each spatial dimension. Use a matrix of size 2-by-`d`, where `d` is the number of spatial dimensions of the input data. The element `(1,d)` specifies the size of padding applied to the start of spatial dimension `d`. The element `(2,d)` specifies the size of padding applied to the end of spatial dimension `d`. For example, in 2-D, the format is ```[top, left; bottom, right]```.\n\nIn each case, the input data is padded with zeros.\n\nExample: `'Padding','same'`\n\nData Types: `single` \| `double`\n\n## Output Arguments\n\ncollapse all\n\nConvolved feature map, returned as a `dlarray`. The output `dlY` has the same underlying data type as the input `dlX`.\n\nIf the input data `dlX` is a formatted `dlarray`, `dlY` has the same dimension labels as `dlX`. If the input data is not a formatted `dlarray`, `dlY` is an unformatted `dlarray` with the same dimension order as the input data.\n\nThe size of the `'C'` channel dimension of `dlY` depends on the size of the `weights` input. The size of the `'C'` dimension of output `Y` is the product of the size of the dimensions `numFiltersPerGroup` and `numGroups` in the `weights` argument. If `weights` is a formatted `dlarray`, this product is the same as the product of the size of the `'U'` dimensions.\n\ncollapse all\n\n### Deep Learning Convolution\n\nThe `dlconv` function applies sliding convolution filters to the spatial dimensions of the input data. The `dlconv` function supports convolution in one, two, or three spatial dimensions. For more information, see the definition of convolutional layer on the `convolution2dLayer` reference page." ]	[ null, "https://nl.mathworks.com/help/examples/nnet/win64/PerformUngroupedConvolutionExample_01.png", null, "https://nl.mathworks.com/help/examples/nnet/win64/PerformUngroupedConvolutionExample_02.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7025217,"math_prob":0.9862056,"size":1551,"snap":"2020-34-2020-40","text_gpt3_token_len":368,"char_repetition_ratio":0.16548158,"word_repetition_ratio":0.041666668,"special_character_ratio":0.19406834,"punctuation_ratio":0.09328358,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995226,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T18:05:15Z\",\"WARC-Record-ID\":\"<urn:uuid:d2acf812-ee25-4595-ab27-68b086520983>\",\"Content-Length\":\"123004\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0a3fc98-3681-406f-8908-547eb6e172ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:6a3d086c-ef39-4de4-9041-a57b8156484a>\",\"WARC-IP-Address\":\"96.7.70.236\",\"WARC-Target-URI\":\"https://nl.mathworks.com/help/deeplearning/ref/dlarray.dlconv.html\",\"WARC-Payload-Digest\":\"sha1:3GJ4LDS2ZIJEHUFADDVXUJJGKPKFNY57\",\"WARC-Block-Digest\":\"sha1:PI6665DS2TXN7UEHZC3ASGJTKYNLSEB3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400188049.8_warc_CC-MAIN-20200918155203-20200918185203-00395.warc.gz\"}"}
https://socratic.org/questions/58cbf8e77c01492623633e19	[ "# Question 33e19\n\nMar 17, 2017\n\nThe divisor is\n\n${x}^{2} - 5 x + 6$\n\n$= {x}^{2} - 3 x - 2 x + 6$\n\n$= x \\left(x - 3\\right) - 2 \\left(x - 3\\right)$\n\n$= \\left(x - 3\\right) \\left(x - 2\\right)$\nso it has two linear factors $\\left(x - 3\\right) \\mathmr{and} \\left(x - 2\\right)$\n\nThe polynomial of $x$ to be divided is\n\n$P \\left(x\\right) = a {x}^{3} - 9 {x}^{2} + b x + 3 a$\nThis is to be exactly divisible by the divisor containing two linear factors $\\left(x - 3\\right) \\mathmr{and} \\left(x - 2\\right)$.\n\nSo\n\n$P \\left(3\\right) = 0$\n\n$\\implies a \\cdot {3}^{3} - 9 \\cdot {3}^{2} + b \\cdot 3 + 3 a = 0$\n\n$\\implies 30 a + 3 b = 729$\n\n$\\implies 10 a + b = 243. \\ldots \\ldots . \\left(1\\right)$\n\nAgain\n\n$P \\left(2\\right) = 0$\n\n$\\implies a \\cdot {2}^{3} - 9 \\cdot {2}^{2} + b \\cdot 2 + 3 a = 0$\n\n$\\implies 11 a + 2 b = 36. \\ldots \\ldots . \\left(2\\right)$\n\nMultiplying (1) by 2 and subtracting (2) from the product we get\n\n$20 a - 11 a = 486 - 36$\n\n$\\implies 9 a = 450$\n\n$\\implies a = \\frac{450}{9} = 50$\n\nInserting the value of a in equation (1)\n\n=>10xx50+b=243)#\n\n$\\implies b = 243 - 500 = - 257$\n\nMar 17, 2017\n\n$a = 2 , b = 7$\n\n#### Explanation:\n\nIf $a {x}^{3} - 9 {x}^{2} + b x + 3 a$ is exactly divisible by ${x}^{2} - 5 x + 6$ then\n\n$a {x}^{3} - 9 {x}^{2} + b x + 3 a \\equiv 0 \\mod {x}^{2} - 5 x + 6$\n\nthen $\\exists \\left(c x + d\\right) \| a {x}^{3} - 9 {x}^{2} + b x + 3 a = \\left(c x + d\\right) \\left({x}^{2} - 5 x + 6\\right)$\n\nor\n\n$\\left(a - c\\right) {x}^{3} + \\left(5 c - 9 - d\\right) {x}^{2} + \\left(5 d - 6 c + b\\right) x + 3 a - 6 d = 0$\n\nSolving\n\n$\\left\\{\\begin{matrix}a - c = 0 \\\\ 5 c - 9 - d = 0 \\\\ 5 d - 6 c + b = 0 \\\\ 3 a - 6 d = 0\\end{matrix}\\right.$\n\nwe have\n\n$a = 2 , b = 7 , c = 2 , d = 1$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85162044,"math_prob":1.00001,"size":368,"snap":"2020-45-2020-50","text_gpt3_token_len":93,"char_repetition_ratio":0.10714286,"word_repetition_ratio":0.0,"special_character_ratio":0.2201087,"punctuation_ratio":0.015151516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000033,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T09:27:28Z\",\"WARC-Record-ID\":\"<urn:uuid:6f401be7-0f46-41a6-b173-3c097a940287>\",\"Content-Length\":\"35972\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ceea8afe-e6bf-49fc-8887-379b9dda2807>\",\"WARC-Concurrent-To\":\"<urn:uuid:fae76dd8-aac2-4f09-b729-df1db552a0d1>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/58cbf8e77c01492623633e19\",\"WARC-Payload-Digest\":\"sha1:2WLPY7NYXUGV7Q37LLQE26CUE6AAD2Q2\",\"WARC-Block-Digest\":\"sha1:GHU7HWNM6YWCCVXFUIZEQETDXTEQYCUZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141747323.98_warc_CC-MAIN-20201205074417-20201205104417-00583.warc.gz\"}"}
https://socratic.org/questions/how-do-you-write-the-equation-of-a-circle-with-a-center-at-6-7-and-a-diameter-of	[ "# How do you write the equation of a circle with a center at (6, 7) and a diameter of 4?\n\n${\\left(x - 6\\right)}^{2} + {\\left(y - 7\\right)}^{2} = {2}^{2}$\n${\\left(x - h\\right)}^{2} + {\\left(y - k\\right)}^{2} = {r}^{2}$ is the general equation where $\\left(h , k\\right)$ is the center. $r$ is radius thus it equals to $\\frac{1}{2}$ of diameter, $r$= 2." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6922903,"math_prob":1.00001,"size":249,"snap":"2020-24-2020-29","text_gpt3_token_len":65,"char_repetition_ratio":0.114285715,"word_repetition_ratio":0.0,"special_character_ratio":0.26104417,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000012,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-08T01:46:36Z\",\"WARC-Record-ID\":\"<urn:uuid:a59ec779-ab6b-4b60-9a2e-281e79d8f561>\",\"Content-Length\":\"32813\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7e1ea4f9-2bb1-4245-891d-078b808b4717>\",\"WARC-Concurrent-To\":\"<urn:uuid:d654405f-b07d-4c0d-9594-6d7f9882c112>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-write-the-equation-of-a-circle-with-a-center-at-6-7-and-a-diameter-of\",\"WARC-Payload-Digest\":\"sha1:BULHBMT733RSO4LQR4XBGLC7GKSUCM2U\",\"WARC-Block-Digest\":\"sha1:YWSSF7AVKQSFYKZGWD2E4YFCWUHETUBF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655896169.35_warc_CC-MAIN-20200708000016-20200708030016-00493.warc.gz\"}"}
https://homework.cpm.org/category/CC/textbook/cca2/chapter/8/lesson/8.3.2/problem/8-158	[ "", null, "", null, "### Home > CCA2 > Chapter 8 > Lesson 8.3.2 > Problem8-158\n\n8-158.\n\nSolve each equation.\n\n1. $\\log_3(2x - 1) = -2$\n\nConvert the equation into exponential form.\n$3^{-2} = 2x - 1$\n\nSolve for $x$.\n\n$x = \\frac{5}{9}$\n\n1. $5^{\\log_5(x)} = 3$\n\nRemember the Inverse Relationship of Logarithms:\n\n$b^{ \\log_b (N)}$\n\n$x = 3$\n\n1. $\\log_2(x) − \\log_2(3) = 4$\n\nRemember the Quotient Property of Logarithms:\n\n1. $\\log_3(5) = x$" ]	[ null, "https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png", null, 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https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-area/v/area-of-a-kite	[ "Main content\n\n# Area of kites\n\nCCSS Math: 6.G.A.1\n\n## Video transcript\n\nWhat is the area of this figure? And this figure right over here is sometimes called a kite for obvious reasons. If you tied some string here, you might want to fly it at the beach. And another way to think about what a kite is, it's a quadrilateral that is symmetric around a diagonal. So this right over here is the diagonal of this quadrilateral. And it's symmetric around it. This top part and this bottom part are mirror images. And to think about how we might find the area of it given that we've been given essentially the width of this kite, and we've also been given the height of this kite, or if you view this as a sideways kite, you could view this is the height and that the eight centimeters as the width. Given that we've got those dimensions, how can we actually figure out its area? So to do that, let me actually copy and paste half of the kite. So this is the bottom half of the kite. And then let's take the top half of the kite and split it up into sections. So I have this little red section here. I have this red section here. And actually, I'm going to try to color the actual lines here so that we can keep track of those as well. So I'll make this line green and I'll make this line purple. So imagine taking this little triangle right over here-- and actually, let me do this one too in blue. So this one over here is blue. You get the picture. Let me try to color it in at least reasonably. So I'll color it in. And then I could make this segment right over here, I'm going to make orange. So let's start focusing on this red triangle here. Imagine flipping it over and then moving it down here. So what would it look like? Well then the green side is going to now be over here. This kind of mauve colored side is still on the bottom. And my red triangle is going to look something like this. My red triangle is going to look like that. Now let's do the same thing with this bigger blue triangle. Let's flip it over and then move it down here. So this green side, since we've flipped it, is now over here. And this orange side is now over here. And we have this blue right over here. And the reason that we know that it definitely fits is the fact that it is symmetric around this diagonal, that this length right over here is equivalent to this length right over here. That's why it fits perfectly like this. Now, what we just constructed is clearly a rectangle, a rectangle that is 14 centimeters wide and not 8 centimeters high, it's half of 8 centimeters high. So it's 8 centimeters times 1/2 or 4 centimeters high. And we know how to find the area of this. This is 4 centimeters times 14 centimeters. So the area is equal to 4 centimeters times 14 centimeters which is equal to-- let's see, that's 40 plus 16-- 56 square centimeters. So if you're taking the area of a kite, you're really just taking 1/2 the width times the height, or 1/2 the width times the height, any way you want to think about it." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9281157,"math_prob":0.94198525,"size":3012,"snap":"2019-35-2019-39","text_gpt3_token_len":775,"char_repetition_ratio":0.15026596,"word_repetition_ratio":0.017421603,"special_character_ratio":0.23373175,"punctuation_ratio":0.10138249,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946609,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-17T15:43:27Z\",\"WARC-Record-ID\":\"<urn:uuid:86de4927-879f-4389-989c-0357eef7f11f>\",\"Content-Length\":\"194938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23495942-b383-4e8a-8c59-4eca8833b0ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:ce6e6ae6-bba7-478f-8252-ebd3a1e2e681>\",\"WARC-IP-Address\":\"151.101.249.42\",\"WARC-Target-URI\":\"https://www.khanacademy.org/math/geometry/hs-geo-foundations/hs-geo-area/v/area-of-a-kite\",\"WARC-Payload-Digest\":\"sha1:MSDAHTHXCJO5AR4ULM5HFTQGZ4P4TEDR\",\"WARC-Block-Digest\":\"sha1:6E4ZNTQEMOGONRLDP57CY4MHT424WMBM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313428.28_warc_CC-MAIN-20190817143039-20190817165039-00087.warc.gz\"}"}
https://math.stackexchange.com/questions/tagged/closed-form	[ "Questions tagged [closed-form]\n\nA \"closed form expression\" is any representation of a mathematical expression in terms of \"known\" functions, \"known\" usually being replaced with \"elementary\".\n\n2,397 questions\n26 views\n\n24 views\n\nSolve and asymptotic expansion of $\\sum_{a=1}^{H} \\sum_{b=a+1}^{H} \\left\\lfloor{\\frac{H}{a\\, b}}\\right\\rfloor$\n\nI am solving constrained polynomial systems resulting in constrained sums. I am looking to see if $$\\sum_{a=1}^{H} \\sum_{b=a+1}^{H} \\left\\lfloor{\\frac{H}{a\\, b}}\\right\\rfloor$$ is expressible in ...\n63 views\n\n152 views\n\n114 views\n\n431 views\n\nIntegral $\\int_0^1 \\frac{\\ln(1+x+x^2)\\ln(1-x+x^2)}{x}dx$\n\nProve $$\\sf I=\\int_0^1 \\frac{\\ln(1+x+x^2)\\ln(1-x+x^2)}{x}dx=\\frac{\\pi}{6\\sqrt{3}}\\psi_1\\left(\\frac{1}{3}\\right)-\\frac{\\pi^3}{9\\sqrt{3}}-\\frac{19}{18}\\zeta(3).$$ I have thought about the integral ...\n96 views\n\nClosed form for $\\int_0^t(x+c)^p(1-2x)^{N-1}\\text{d}x$\n\nI am trying to find a closed form for the integral $$I\\equiv\\int_0^t(x+c)^p(1-2x)^{N-1}\\text{d}x,$$ where $N\\in\\mathbb{N}$, $p>0$, $c\\ge 0$ and $t\\in\\left(0,\\frac{1}{2}\\right)$. I thought to ...\n41 views\n\nFinding closed form of exponential generating function involving identity permutation\n\nFix a prime number $p > 1$ and for a positive integer $n$, let $a_n$ be the number of permutations $π ∈ S_n$ such that $π^p = id$, where $id$ is the identity permutation. Find a closed form for the ...\n101 views" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.710646,"math_prob":0.99993,"size":14916,"snap":"2019-26-2019-30","text_gpt3_token_len":5674,"char_repetition_ratio":0.14511803,"word_repetition_ratio":0.011789925,"special_character_ratio":0.3795924,"punctuation_ratio":0.106796116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000008,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T01:42:16Z\",\"WARC-Record-ID\":\"<urn:uuid:3d0d71bb-79e1-4b65-a69c-644a69acf81a>\",\"Content-Length\":\"240588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df57a462-944f-4b7d-b75a-b2c334ea6d96>\",\"WARC-Concurrent-To\":\"<urn:uuid:a465a3fe-0d44-423a-9250-4e3d125c28ee>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/tagged/closed-form\",\"WARC-Payload-Digest\":\"sha1:QBC2XO5UHD7QQH7TEBEFZFCXTWN2SX6K\",\"WARC-Block-Digest\":\"sha1:EQIZYIZOCKXQGNWSUOZIHWK5TZ5PHMUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000044.37_warc_CC-MAIN-20190626013357-20190626035357-00507.warc.gz\"}"}
http://topcoder.bgcoder.com/print.php?id=858	[ "### Problem Statement\n\nDetermine the number of ways to cut a convex polygon with n sides if the only cuts allowed are from vertex to vertex, each cut divides exactly one polygon into exactly two polygons, and you must end up with exactly k polygons. Consider each vertex distinct. For example, there are three ways to cut a square - the two diagonals and not cutting at all - but only two ways to cut it to form 2 polygons, and only one way to cut it to form 1 polygon. The order of cuts does not matter. Since the number of ways is very large, you should return the number taken modulo 1000000000 (one billion). In other words, if the answer would have at least 10 digits, return only the 9 least significant. If there is no way to cut the polygon into k pieces, return -1.\n\n### Definition\n\n Class: PolygonDecomposition Method: howMany Parameters: int, int Returns: int Method signature: int howMany(int n, int k) (be sure your method is public)\n\n### Notes\n\n-The vertices are distinct - there are 5 ways to cut a pentagon into 3 triangles, not just one way.\n-Only one polygon at a time may be cut - you cannot cut two triangles into four triangles with one cut.\n\n### Constraints\n\n-n is between 3 and 100, inclusive.\n-k is between 1 and 100, inclusive.\n\n### Examples\n\n0)\n\n `4` `2`\n`Returns: 2`\n A quadrilateral can be cut into two triangles in two different ways, one for each diagonal.\n1)\n\n `100` `1`\n`Returns: 1`\n Any polygon can be cut into one polygon by not cutting at all, but no other way.\n2)\n\n `6` `4`\n`Returns: 14`\n3)\n\n `31` `20`\n`Returns: 956146480`\n The actual number of ways is about 6.5 x 10^18, but we return only the final 9 digits.\n4)\n\n `3` `4`\n`Returns: -1`\n\n#### Problem url:\n\nhttp://www.topcoder.com/stat?c=problem_statement&pm=4445\n\n#### Problem stats url:\n\nhttp://www.topcoder.com/tc?module=ProblemDetail&rd=7998&pm=4445\n\nEnogipe\n\n#### Testers:\n\nPabloGilberto , brett1479 , Olexiy\n\n#### Problem categories:\n\nDynamic Programming" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87897855,"math_prob":0.95873,"size":1609,"snap":"2020-24-2020-29","text_gpt3_token_len":425,"char_repetition_ratio":0.12523365,"word_repetition_ratio":0.0066225166,"special_character_ratio":0.28216285,"punctuation_ratio":0.11890244,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98866546,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-14T14:13:28Z\",\"WARC-Record-ID\":\"<urn:uuid:a01897ac-b266-4444-bf3f-8efecda2831d>\",\"Content-Length\":\"7081\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e949b2ca-038f-49be-ac5a-a193171a12f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:bda1e739-a57c-46f9-b83f-78800199f241>\",\"WARC-IP-Address\":\"78.128.6.10\",\"WARC-Target-URI\":\"http://topcoder.bgcoder.com/print.php?id=858\",\"WARC-Payload-Digest\":\"sha1:IXVNJDB5BP3G2I224CP3IMOK3HYESEUC\",\"WARC-Block-Digest\":\"sha1:VWXWNOEFKZA3ULEADMZMIYGG5LDZACOQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655880665.3_warc_CC-MAIN-20200714114524-20200714144524-00274.warc.gz\"}"}
https://silently9527.cn/?p=52	[ "# 图算法系列之无向图的数据结构", null, "2021-08-14 / 0 评论 / 410 阅读\n\n## 前言\n\n1. 基于散列函数将被查找键转换为数组的下标\n2. 处理散列值冲突的情况，有两种方式来处理冲突：拉链式和线性探测\n\n## 散列函数\n\njava中的约定：如果两个对象的equals相等，那么hashCode一定相同；如果hashCode相同，equals不一定相同。对于自定义类型的键我们通常需要自定义实现hashCode和equals；默认的hashCode返回的是对象的内存地址，这种散列值不会太好。\n\n#### Integer\n\n``````@Override\npublic int hashCode() {\nreturn Integer.hashCode(value);\n}\npublic static int hashCode(int value) {\nreturn value;\n}\n``````\n\n#### Long\n\nJava中Long类型的hashCode计算是先把值无符号右移32位，之后再与值相异或，保证每一位都用用到，最后强制转换成int值\n\n``````@Override\npublic int hashCode() {\nreturn Long.hashCode(value);\n}\n\npublic static int hashCode(long value) {\nreturn (int)(value ^ (value >>> 32));\n}\n``````\n\n#### Double、Float\n\n``````public static int hashCode(float value) {\nreturn floatToIntBits(value);\n}\npublic static int floatToIntBits(float value) {\nint result = floatToRawIntBits(value);\n// Check for NaN based on values of bit fields, maximum\n// exponent and nonzero significand.\nif ( ((result & FloatConsts.EXP_BIT_MASK) ==\nresult = 0x7fc00000;\nreturn result;\n}\n``````\n\n#### String\n\njava中每个char都可以表示成一个int值，所以字符串转换成一个int值\n\n``````public int hashCode() {\nint h = hash;\nif (h == 0 && value.length > 0) {\nchar val[] = value;\n\nfor (int i = 0; i < value.length; i++) {\nh = 31 * h + val[i];\n}\nhash = h;\n}\nreturn h;\n}\n\n``````\n\n## 拉链式的散列表", null, "``````public class SeparateChainingHashMap<K, V> implements Map<K, V> {\n\nprivate int size;\n\npublic SeparateChainingHashMap(int capacity) {\nfor (int i = 0; i < capacity; i++) {\n}\n}\n\n@Override\npublic void put(K key, V value) {\nthis.table[hash(key)].put(key, value);\nsize++;\n}\n\nprivate int hash(K key) {\nreturn (key.hashCode() & 0x7fffffff) % table.length;\n}\n\n@Override\npublic V get(K key) {\nreturn this.table[hash(key)].get(key);\n}\n\n@Override\npublic void delete(K key) {\nthis.table[hash(key)].delete(key);\nsize--;\n}\n\n@Override\npublic int size() {\nreturn size;\n}\n\n}\n\n``````\n\n## 线性探测式散列表\n\n1. 下一个位置和待插入的键相等，那么值就修改值\n2. 下一个位置和待插入的键不相等，那么索引加一继续查找\n3. 如果下一个位置还是一个空位，那么直接把待插入对象放入到这个空位\n\n#### 初始化\n\n``````private int size;\nprivate int capacity;\nprivate K[] keys;\nprivate V[] values;\n\npublic LinearProbingHashMap(int capacity) {\nthis.capacity = capacity;\nthis.keys = (K[]) new Object[capacity];\nthis.values = (V[]) new Object[capacity];\n}\n\n``````\n\n#### 插入\n\n1. 当插入键的位置超过了数组的大小，就需要回到数组的开始位置继续查找，直到找到一个位置为null的才结束；`index = (index + 1) % capacity`\n2. 当数组已存放的容量超过了数组总容量的一半，就需要扩容到原来的2倍\n``````@Override\npublic void put(K key, V value) {\nif (Objects.isNull(key)) {\nthrow new IllegalArgumentException(\"Key can not null\");\n}\nif (this.size > this.capacity / 2) {\nresize(2 * this.capacity);\n}\nint index;\nfor (index = hash(key); this.keys[index] != null; index = (index + 1) % capacity) {\nif (this.keys[index].equals(key)) {\nthis.values[index] = value;\nreturn;\n}\n}\nthis.keys[index] = key;\nthis.values[index] = value;\nsize++;\n}\n``````\n\n#### 动态调整数组的大小\n\n``````private void resize(int cap) {\nLinearProbingHashMap<K, V> linearProbingHashMap = new LinearProbingHashMap<>(cap);\nfor (int i = 0; i < capacity; i++) {\nlinearProbingHashMap.put(keys[i], values[i]);\n}\nthis.keys = linearProbingHashMap.keys;\nthis.values = linearProbingHashMap.values;\nthis.capacity = linearProbingHashMap.capacity;\n}\n``````\n\n#### 查询\n\n``````@Override\npublic V get(K key) {\nif (Objects.isNull(key)) {\nthrow new IllegalArgumentException(\"Key can not null\");\n}\nint index;\nfor (index = hash(key); this.keys[index] != null; index = (index + 1) % capacity) {\nif (this.keys[index].equals(key)) {\nreturn this.values[index];\n}\n}\nreturn null;\n}\n``````\n\n#### 删除元素\n\n``````@Override\npublic void delete(K key) {\nif (Objects.isNull(key)) {\nthrow new IllegalArgumentException(\"Key can not null\");\n}\nint index;\nfor (index = hash(key); this.keys[index] != null; index = (index + 1) % capacity) {\nif (this.keys[index].equals(key)) {\nthis.keys[index] = null;\nthis.values[index] = null;\nbreak;\n}\n}\n\nfor (index = (index + 1) % capacity; this.keys[index] != null; index = (index + 1) % capacity) {\nthis.size--;\nthis.put(this.keys[index], this.values[index]);\nthis.keys[index] = null;\nthis.values[index] = null;\n}\nthis.size--;\nif (size > 0 && size < capacity / 4) {\nresize(capacity / 2);\n}\n\n}\n``````\n\nhttps://github.com/silently9527/JavaCore" ]	[ null, "data:image/png;base64,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", null, "https://tva1.sinaimg.cn/large/008eGmZEgy1gpg7i9y3qoj30ct06ra9w.jpg", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.64887166,"math_prob":0.9786487,"size":5912,"snap":"2023-14-2023-23","text_gpt3_token_len":2953,"char_repetition_ratio":0.14759648,"word_repetition_ratio":0.23076923,"special_character_ratio":0.23393099,"punctuation_ratio":0.19278607,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9898664,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-10T11:35:01Z\",\"WARC-Record-ID\":\"<urn:uuid:e0715a53-9dbc-46d2-9e48-ef868ae6ae3f>\",\"Content-Length\":\"112994\",\"Content-Type\":\"application/http; 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https://statkat.com/stattest.php?t=7	[ "# Paired sample t test - overview\n\nThis page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table\n\nPaired sample $t$ test\nIndependent variable\n2 paired groups\nDependent variable\nOne quantitative of interval or ratio level\nNull hypothesis\nH0: $\\mu = \\mu_0$\n\nHere $\\mu$ is the population mean of the difference scores, and $\\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.\nAlternative hypothesis\nH1 two sided: $\\mu \\neq \\mu_0$\nH1 right sided: $\\mu > \\mu_0$\nH1 left sided: $\\mu < \\mu_0$\nAssumptions\n• Difference scores are normally distributed in the population\n• Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another\nTest statistic\n$t = \\dfrac{\\bar{y} - \\mu_0}{s / \\sqrt{N}}$\nHere $\\bar{y}$ is the sample mean of the difference scores, $\\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).\n\nThe denominator $s / \\sqrt{N}$ is the standard error of the sampling distribution of $\\bar{y}$. The $t$ value indicates how many standard errors $\\bar{y}$ is removed from $\\mu_0$.\nSampling distribution of $t$ if H0 were true\n$t$ distribution with $N - 1$ degrees of freedom\nSignificant?\nTwo sided:\nRight sided:\nLeft sided:\n$C\\%$ confidence interval for $\\mu$\n$\\bar{y} \\pm t^* \\times \\dfrac{s}{\\sqrt{N}}$\nwhere the critical value $t^$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^$ and $t^$ (e.g. $t^$ = 2.086 for a 95% confidence interval when df = 20).\n\nThe confidence interval for $\\mu$ can also be used as significance test.\nEffect size\nCohen's $d$:\nStandardized difference between the sample mean of the difference scores and $\\mu_0$: $$d = \\frac{\\bar{y} - \\mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean of the difference scores $\\bar{y}$ is removed from $\\mu_0.$\nVisual representation\nEquivalent to\n• One sample $t$ test on the difference scores.\n• Repeated measures ANOVA with one dichotomous within subjects factor.\nExample context\nIs the average difference between the mental health scores before and after an intervention different from $\\mu_0 = 0$?\nSPSS\nAnalyze > Compare Means > Paired-Samples T Test...\n• Put the two paired variables in the boxes below Variable 1 and Variable 2\nJamovi\nT-Tests > Paired Samples T-Test\n• Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line\n• Under Hypothesis, select your alternative hypothesis\nPractice questions" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81178,"math_prob":0.99873954,"size":2411,"snap":"2022-27-2022-33","text_gpt3_token_len":664,"char_repetition_ratio":0.16701289,"word_repetition_ratio":0.09186352,"special_character_ratio":0.28204066,"punctuation_ratio":0.071599044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999843,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T13:27:55Z\",\"WARC-Record-ID\":\"<urn:uuid:d08c6f66-a729-410b-81b6-db04e45b8861>\",\"Content-Length\":\"17978\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fc752211-39d7-4e4f-b6c8-927628244f98>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba2b13e1-029c-4d36-8595-f5594ab9605c>\",\"WARC-IP-Address\":\"141.138.168.125\",\"WARC-Target-URI\":\"https://statkat.com/stattest.php?t=7\",\"WARC-Payload-Digest\":\"sha1:AMLH56C4UUANZEZTENS3I3N4UEB7O6ZT\",\"WARC-Block-Digest\":\"sha1:2XZGQITFUSVRUZRDQWBPTKOKT4BVNPZX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104692018.96_warc_CC-MAIN-20220707124050-20220707154050-00737.warc.gz\"}"}
https://zbmath.org/?q=an:0873.45006	[ "## About the non cutoff Kac equation: Uniqueness and asymptotic behaviour.(English)Zbl 0873.45006\n\nThe authors investigate the asymptotic behaviour of the solution of Kac’s equation by showing uniform boundedness in time of various convex functionals of the solution. The uniqueness of the solution of the non cutoff variant of Kac equation has been proved by means of Tanaka’s functional.\n\n### MSC:\n\n 45K05 Integro-partial differential equations 45M05 Asymptotics of solutions to integral equations\n\n### Keywords:\n\nnon cutoff Kac equation; asymptotic behaviour" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.788694,"math_prob":0.98703015,"size":821,"snap":"2022-27-2022-33","text_gpt3_token_len":237,"char_repetition_ratio":0.12974297,"word_repetition_ratio":0.052173913,"special_character_ratio":0.26187575,"punctuation_ratio":0.19875777,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9736755,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T15:01:11Z\",\"WARC-Record-ID\":\"<urn:uuid:ec62a5cb-7215-4d1b-b01f-da10ca38b853>\",\"Content-Length\":\"49535\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:abc249ae-70d8-407c-a5ad-4186ced73cfa>\",\"WARC-Concurrent-To\":\"<urn:uuid:ae20ce77-e19b-485a-b583-3c49279a5128>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0873.45006\",\"WARC-Payload-Digest\":\"sha1:RIMN2Y5PQTGCDRF5YRUHSWGUKSJ4YA22\",\"WARC-Block-Digest\":\"sha1:FLGFKIP4QWUA3B6VLYXHCQYBOB4KCLW7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103334753.21_warc_CC-MAIN-20220627134424-20220627164424-00550.warc.gz\"}"}
https://matplotlib.org/3.0.2/api/_as_gen/matplotlib.pyplot.fill.html	[ "# matplotlib.pyplot.fill¶\n\nmatplotlib.pyplot.fill(args, data=None, kwargs)[source]\n\nPlot filled polygons.\n\nParameters: args : sequence of x, y, [color] Each polygon is defined by the lists of x and y positions of its nodes, optionally followed by a color specifier. See matplotlib.colors for supported color specifiers. The standard color cycle is used for polygons without a color specifier. You can plot multiple polygons by providing multiple x, y, [color] groups. For example, each of the following is legal: ax.fill(x, y) # a polygon with default color ax.fill(x, y, \"b\") # a blue polygon ax.fill(x, y, x2, y2) # two polygons ax.fill(x, y, \"b\", x2, y2, \"r\") # a blue and a red polygon a list of :class:~matplotlib.patches.Polygon *kwargs : Polygon properties\n\nNotes\n\nUse fill_between() if you would like to fill the region between two curves.\n\nNote\n\nIn addition to the above described arguments, this function can take a data keyword argument. If such a data argument is given, the following arguments are replaced by data[<arg>]:\n\n• All arguments with the following names: 'x', 'y'.\n\nObjects passed as data must support item access (data[<arg>]) and membership test (<arg> in data).\n\n## Examples using matplotlib.pyplot.fill¶", null, "Interactive functions", null, "Fill Spiral" ]	[ null, "https://matplotlib.org/3.0.2/_images/sphx_glr_ginput_manual_clabel_sgskip_thumb.png", null, "https://matplotlib.org/3.0.2/_images/sphx_glr_fill_spiral_thumb.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55996287,"math_prob":0.58768785,"size":1270,"snap":"2021-21-2021-25","text_gpt3_token_len":323,"char_repetition_ratio":0.13112165,"word_repetition_ratio":0.0,"special_character_ratio":0.25826773,"punctuation_ratio":0.20689656,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98239744,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T04:30:50Z\",\"WARC-Record-ID\":\"<urn:uuid:0c67513c-69e7-40a8-aad5-f314461dd638>\",\"Content-Length\":\"14172\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5bf4c12-e38c-4ad4-95f9-803bf352156a>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c7bacb5-91b9-45c0-a331-0f7e8e8d4f9b>\",\"WARC-IP-Address\":\"104.26.0.8\",\"WARC-Target-URI\":\"https://matplotlib.org/3.0.2/api/_as_gen/matplotlib.pyplot.fill.html\",\"WARC-Payload-Digest\":\"sha1:XDVP2A2B37EQA3KFQ5YTN7EFRI4DQ7KR\",\"WARC-Block-Digest\":\"sha1:V7OIFGQRPF4U7BYEUPOQBMRJDO7XWK64\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991737.39_warc_CC-MAIN-20210514025740-20210514055740-00351.warc.gz\"}"}
https://thetextchemistry.org/qa/question-how-much-is-1-cm-on-a-ruler.html	[ "", null, "# Question: How Much Is 1 Cm On A Ruler?\n\n## Is centimeter a metric unit?\n\nIn addition to the difference in the basic units, the metric system is based on 10s, and different measures for length include kilometer, meter, decimeter, centimeter, and millimeter.\n\nThis means that a meter is 100 times larger than a centimeter, and a kilogram is 1,000 times heavier than a gram..\n\n## How can I measure cm without a ruler?\n\nOne inch (2.5 cm) is roughly the measurement from the top knuckle on your thumb to your thumb tip. Measure yours to see how close it is to 1 inch. After all, you should always have a thumb handy for a guide for measuring items under 6 inches (15cm)!\n\n## Is 1 cm the same as 1 inch?\n\nOne inch is equal to 2.54 centimeters, so reverse it. For example, if you have 50 cm. and the problem is to convert into inches. … One inch equals 2.54 centimeters, so 62 inches would be 157.5 centimeters.\n\n## What stage is a 2 cm tumor?\n\nIn general, stage IIB describes invasive breast cancer in which: the tumor is larger than 2 cm but no larger than 5 centimeters; small groups of breast cancer cells — larger than 0.2 mm but not larger than 2 mm — are found in the lymph nodes or.\n\n## What is the centimeter symbol?\n\nA centimetre (international spelling) or centimeter (American spelling) (SI symbol cm) is a unit of length in the metric system, equal to one hundredth of a metre, centi being the SI prefix for a factor of 1100.\n\n## How many cm exactly is an inch?\n\n2.54cmHow many cm in an inch? 1 inch = 2.54cm. To convert inches to centimeters multiply your figure by 2.54.\n\n## How big is a pea in CM?\n\nLNCtips.com: Wound SizingCMInchesObject0.1 cm0.04 inchesGrain of sugar0.5 cm0.2 inchesPea0.6 cm0.2 inchesPencil eraser0.9 cm0.4 inchesLadybug20 more rows\n\n## How thick is a centimeter?\n\nEquals: 0.39 inches thickness (in) in dimension. Converting centimeter thickness to inches thickness value in the thickness units scale.\n\n## What size is 2.5 cm in inches?\n\nTo convert 2.5 centimeters to inches you have to divide the value in cm by 2.54. 2.5 cm in inches: 2.5 cm are equal to 2.5/2.54 = 0.98425 inches.\n\n## What is a sentence for centimeter?\n\nCentimeter sentence examples. It’s a 38 centimeter chain with the typical Chanel logo at the end, studded in pave cubic zirconias.\n\n## How long is an inch on your finger?\n\n* Use your own body for fast, approximate measuring. The first joint of an index finger is about 1 inch long. When a hand is spread wide, the span from the tip of the thumb to the tip of the pinkie is about 9 inches; from the tip of the thumb to the tip of the index finger, around 6 inches.\n\n## Do you count the zero on a ruler?\n\nStart Measuring With Your Ruler Place the flat end of the ruler against whatever it is you’re measuring, and line the zero mark on the ruler up with one end of the object to be measured. … Remember that however many marks you’ve counted along the ruler equal the number of millimeters you’ve measured.\n\n## What objects are 1 cm long?\n\nA centimeter (cm) is about:about as long as a staple.the width of a highlighter.the diameter of a belly button.the width of 5 CD’s stacked on top of each other.the thickness of a notepad.the radius (half the diameter) of a US penny.\n\n## Do you start measuring at 0 or 1?\n\nPlace the starting end of the measuring tool where it says “0” against the closest edge of the object or distance you’re trying to measure. Make sure the starting edge of the measuring tool and the edge of the object are perfectly aligned in order to get an accurate measurement.\n\n## What’s a centimeter look like?\n\nA centimeter is a metric unit of length. … 1 centimeter is equal to 0.3937 inches, or 1 inch is equal to 2.54 centimeters. In other words, 1 centimeter is less than half as big as an inch, so you need about two-and-a-half centimeters to make one inch.\n\n## What length is represented by 1 centimeter?\n\n0.39370 inchesThe centimetre is a unit of length in the metric system, equal to one-hundredth of a metre. 1cm is equivalent to 0.39370 inches.\n\n## Is 1 cm half an inch?\n\nInches to centimeters conversion tableInches (“)Centimeters (cm)1/2 in1.27 cm1 in2.54 cm2 in5.08 cm3 in7.62 cm23 more rows\n\n## What does 1 cm look like on a ruler?\n\nThe longest line represents the biggest unit on the ruler: 1 cm. Each centimeter is labeled on the ruler (1-30). Example: You take out a ruler to measure the width of your fingernail. The ruler stops at 1 cm, meaning that your nail is precisely 1 cm wide.\n\n## How big is a 2 cm tumor?\n\nPrimary breast tumors vary in shape and size. The smallest lesion that can be felt by hand is typically 1.5 to 2 centimeters (about 1/2 to 3/4 inch) in diameter. Sometimes tumors that are 5 centimeters (about 2 inches) — or even larger — can be found in the breast.\n\n## Does tumor size determine stage?\n\nIn general, the smaller the tumor, the better the prognosis tends to be . Tumor size is part of breast cancer staging. In the TNM staging system, a “T” followed by a number shows the size of the tumor. In some cases, the size of the tumor cannot be determined (TX) or a tumor cannot be found (T0).\n\n## What is something that is 1 cm?\n\nNote a few objects that are roughly 1 cm wide. The easiest objects to use are a standard pencil, pen, or highlighter. The width of a pencil is close to 1 cm. Other options include the length of a staple, the width of five CDs or DVDs stacked together, the thickness of a standard notepad, and the radius of a U.S. penny." ]	[ null, "https://mc.yandex.ru/watch/68554720", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89741576,"math_prob":0.9659364,"size":6022,"snap":"2020-45-2020-50","text_gpt3_token_len":1601,"char_repetition_ratio":0.1540379,"word_repetition_ratio":0.12982456,"special_character_ratio":0.264364,"punctuation_ratio":0.13101406,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95611846,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T02:28:08Z\",\"WARC-Record-ID\":\"<urn:uuid:72b8443d-33f8-4386-8b2c-fdd1e78c68bb>\",\"Content-Length\":\"36225\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1e3e7623-b4de-4ce7-ae5d-89da851d611e>\",\"WARC-Concurrent-To\":\"<urn:uuid:c89c2713-b526-44e7-aa6f-434ef093d8d7>\",\"WARC-IP-Address\":\"87.236.16.33\",\"WARC-Target-URI\":\"https://thetextchemistry.org/qa/question-how-much-is-1-cm-on-a-ruler.html\",\"WARC-Payload-Digest\":\"sha1:FJKB27RKA6DLOE7265MF2XSMSDPQI7N4\",\"WARC-Block-Digest\":\"sha1:YHJW3TAQGMKGYH2NF26O7R723YQX7XGW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141733120.84_warc_CC-MAIN-20201204010410-20201204040410-00177.warc.gz\"}"}
http://cpr-mathph.blogspot.com/2013/07/10094062-jesper-l-jacobsen-et-al.html	[ "## Is the five-flow conjecture almost false? [PDF]\n\nJesper L. Jacobsen, Jesus Salas\nThe number of nowhere zero Z_Q flows on a graph G can be shown to be a polynomial in Q, defining the flow polynomial \\Phi_G(Q). According to Tutte's five-flow conjecture, \\Phi_G(5) > 0 for any bridgeless G.A conjecture by Welsh that \\Phi_G(Q) has no real roots for Q \\in (4,\\infty) was recently disproved by Haggard, Pearce and Royle. These authors conjectured the absence of roots for Q \\in [5,\\infty). We study the real roots of \\Phi_G(Q) for a family of non-planar cubic graphs known as generalised Petersen graphs G(m,k). We show that the modified conjecture on real flow roots is also false, by exhibiting infinitely many real flow roots Q>5 within the class G(nk,k). In particular, we compute explicitly the flow polynomial of G(119,7), showing that it has real roots at Q\\approx 5.0000197675 and Q\\approx 5.1653424423. We moreover prove that the graph families G(6n,6) and G(7n,7) possess real flow roots that accumulate at Q=5 as n\\to\\infty (in the latter case from above and below); and that Q_c(7)\\approx 5.2352605291 is an accumulation point of real zeros of the flow polynomials for G(7n,7) as n\\to\\infty.\nView original: http://arxiv.org/abs/1009.4062" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8842048,"math_prob":0.9576258,"size":1223,"snap":"2020-24-2020-29","text_gpt3_token_len":347,"char_repetition_ratio":0.12797375,"word_repetition_ratio":0.0,"special_character_ratio":0.28209323,"punctuation_ratio":0.123595506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.984818,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-28T15:34:26Z\",\"WARC-Record-ID\":\"<urn:uuid:3f29deb9-a63e-42ef-be48-d7330de5f471>\",\"Content-Length\":\"202683\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8c2b78e5-01fd-4082-9c24-37d36eecfd15>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf8845e2-80c5-4023-9c2f-c8456a379bed>\",\"WARC-IP-Address\":\"172.217.13.225\",\"WARC-Target-URI\":\"http://cpr-mathph.blogspot.com/2013/07/10094062-jesper-l-jacobsen-et-al.html\",\"WARC-Payload-Digest\":\"sha1:6P4PKJN4ZWKOVVC2XSSYWKQVNSFAYIWH\",\"WARC-Block-Digest\":\"sha1:BYPPRPKDDJTGWUVIDOAG23FK5WFZSGFP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347399820.9_warc_CC-MAIN-20200528135528-20200528165528-00088.warc.gz\"}"}
https://socratic.org/questions/how-do-you-determine-the-theoretical-probability-of-rolling-a-five-on-a-die	[ "# How do you determine the theoretical probability of rolling a five on a die?\n\nFeb 13, 2015\n\nThe probability of any event is based on given probabilities of elementary events and a composition of our event as a set of elementary events.\n\nIf we know the probabilities of elementary events and a composition of our event, the probability of our event is a sum of probabilities of all elementary events that comprise it.\n\nIn case of a die, the elementary events are numbers rolled on this die. The so-called \"fair\" die has all its 6 numbers equally probable. Since the total probability always equals to 1, the probability of each elementary event (rolling each number from 1 to 6) equals to\n$P \\left\\{1\\right\\} = P \\left\\{2\\right\\} = P \\left\\{3\\right\\} = P \\left\\{4\\right\\} = P \\left\\{5\\right\\} = P \\left\\{6\\right\\} = \\frac{1}{6}$.\n\nAn event offered in the problem, \"rolling a five on a die\", contains only one elementary event - number 5, whose probability we know as being equal to $\\frac{1}{6}$. Therefore, the probability of rolling five on a die equals to\n$P \\left\\{5\\right\\} = \\frac{1}{6}$.\n\nJust as an example, an event \"Rolling a number that is less than five\" contains 4 different elementary events - rolling 1, 2, 3 or 4. Each of them has a probability $\\frac{1}{6}$. The sum of all elementary events comprising our event, that is the probability of our event, is\n$P \\left\\{< 5\\right\\} = P \\left\\{1 \\mathmr{and} 2 \\mathmr{and} 3 \\mathmr{and} 4\\right\\} = \\frac{4}{6} = \\frac{2}{3}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9430478,"math_prob":0.99980134,"size":1044,"snap":"2020-34-2020-40","text_gpt3_token_len":226,"char_repetition_ratio":0.21538462,"word_repetition_ratio":0.05464481,"special_character_ratio":0.21360153,"punctuation_ratio":0.07462686,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T21:19:44Z\",\"WARC-Record-ID\":\"<urn:uuid:818697bd-edcb-4125-b60e-042392ed7b04>\",\"Content-Length\":\"35820\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:939870bc-684b-41eb-9d7a-fd40667bf436>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f965b34-88fb-4d34-988e-f6b3e31e9150>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-determine-the-theoretical-probability-of-rolling-a-five-on-a-die\",\"WARC-Payload-Digest\":\"sha1:4O3R6NGJAKZHN4POUDNUV3EN7LNKAAVR\",\"WARC-Block-Digest\":\"sha1:2F4NPXPOWHWICC3CZYXG3UXRS5A4KASJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738573.99_warc_CC-MAIN-20200809192123-20200809222123-00402.warc.gz\"}"}
http://currency7.com/MWK-to-BOB-exchange-rate-converter?amount=300	[ "# 300 Malawian Kwacha (MWK) to Bolivian Boliviano (BOB)\n\nThe currency calculator will convert exchange rate of Malawian kwacha (MWK) to Bolivian boliviano (BOB).\n\n• Malawian kwacha\nThe Malawian kwacha (MWK) is the currency of Malawi. The currency code is MWK and currency symbol is MK. The Malawian kwacha is subdivided into 100 tambala. Plural of kwacha is kwacha. Frequently used Malawian kwacha coins are in denominations of MK 1, MK 5, MK 10. Frequently used Malawian kwacha banknotes are in denominations of MK 20, MK 50, MK 100, MK 200, MK 500, MK 1000.\n• Bolivian boliviano\nThe Bolivian boliviano (BOB) is the currency of Bolivia. The currency code is BOB and currency symbol is Bs. The Bolivian boliviano is subdivided into 100 centavos (singular: centavo; symbol: Cvs.). Frequently used Bolivian boliviano coins are in denominations of Bs.1, Bs.2, Bs.5, 10 centavos, 20 centavos, 50 centavos. Frequently used Bolivian boliviano banknotes are in denominations of Bs.10, Bs.20, Bs.50, Bs.100, Bs.200.\n• 100 MWK = 0.41 BOB\n• 200 MWK = 0.82 BOB\n• 500 MWK = 2.06 BOB\n• 1,000 MWK = 4.11 BOB\n• 2,000 MWK = 8.22 BOB\n• 5,000 MWK = 20.55 BOB\n• 6,000 MWK = 24.66 BOB\n• 10,000 MWK = 41.10 BOB\n• 20,000 MWK = 82.20 BOB\n• 30,000 MWK = 123.31 BOB\n• 50,000 MWK = 205.51 BOB\n• 100,000 MWK = 411.02 BOB\n• 500,000 MWK = 2,055.11 BOB\n• 1,000,000 MWK = 4,110.22 BOB\n• 5,000,000 MWK = 20,551.11 BOB\n• 1 BOB = 243.30 MWK\n• 5 BOB = 1,216.48 MWK\n• 10 BOB = 2,432.96 MWK\n• 20 BOB = 4,865.92 MWK\n• 25 BOB = 6,082.40 MWK\n• 50 BOB = 12,164.80 MWK\n• 100 BOB = 24,329.59 MWK\n• 200 BOB = 48,659.18 MWK\n• 250 BOB = 60,823.98 MWK\n• 500 BOB = 121,647.95 MWK\n• 1,000 BOB = 243,295.91 MWK\n• 2,000 BOB = 486,591.81 MWK\n• 2,500 BOB = 608,239.77 MWK\n• 3,500 BOB = 851,535.67 MWK\n• 5,000 BOB = 1,216,479.53 MWK\n\n## Popular MWK pairing\n\n` <a href=\"http://currency7.com/MWK-to-BOB-exchange-rate-converter?amount=300\">300 MWK in BOB</a> `" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.529612,"math_prob":0.99880385,"size":2933,"snap":"2023-40-2023-50","text_gpt3_token_len":1117,"char_repetition_ratio":0.26254696,"word_repetition_ratio":0.029354207,"special_character_ratio":0.37572452,"punctuation_ratio":0.18032786,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9559932,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T03:43:54Z\",\"WARC-Record-ID\":\"<urn:uuid:65fd71b0-ab09-40fc-93a3-6eab25d14318>\",\"Content-Length\":\"29275\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9d6beb3-cd69-430b-acf3-2eb6a0477f81>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc660320-5545-4fa5-b827-e584865cdc77>\",\"WARC-IP-Address\":\"70.35.206.41\",\"WARC-Target-URI\":\"http://currency7.com/MWK-to-BOB-exchange-rate-converter?amount=300\",\"WARC-Payload-Digest\":\"sha1:AR6SONG2R7DKUQJJID4BQKPPJW6LN6KE\",\"WARC-Block-Digest\":\"sha1:VZFUTKVQIEFJLMDVIZLM4SUMYUZ4CYLJ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100632.0_warc_CC-MAIN-20231207022257-20231207052257-00307.warc.gz\"}"}
http://gpl4you.com/discussion.php?q=mcq&company=HCL&type=Mathematical%20Skills&subtype=Number%20System	[ "## Company\n\n•", null, "### which is the 4 digit number whose second digit is thrice the first digit and 3'rd digit is sum of 1’st and 2'nd and last digit is twice the second digit. 1. 2674 2. 1349 3. 3343 4. 3678 (HCL Question)\n\nlast reply by sandeep tiwari • 4 years ago • asked by Saikat\n\n•", null, "### How many integers n greater than and less than 100 are there such that,if the digits of n are reversed, the resulting integer is n+9 ? (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 (HCL Question)\n\nlast reply by saranya • 5 years ago • asked by Saikat\n\n•", null, "### What does the hex number E78 correspond to in radix 7 ? a) 12455 b) 14153 c) 14256 d) 13541 e) 13112 (HCL Question)\n\nlast reply by rakhi • 7 years ago • asked by Saikat\n\n•", null, "### How many integers n greater than and less than 100 are there such that,if the digits of n are reversed, the resulting integer is n+9 ? (A)5 (B)6 (C)7 (D)8 (E)9 (HCL Question)\n\nlast reply by rakhi • 7 years ago • asked by Bhushan\n\n•", null, "### How many integers n greater than and less than 100 are there such that, if the digits of n are reversed, the resulting integer is n+9 ? (A)5 (B)6 (C)7 (D)8 (E)9 (HCL Question)\n\nlast reply by Anita • 8 years ago • asked by Bhushan\n\n•", null, "" ]	[ null, "http://gpl4you.com/images/noavatar_small.gif", null, "http://gpl4you.com/images/noavatar_small.gif", null, "http://gpl4you.com/images/noavatar_small.gif", null, "http://gpl4you.com/images/noavatar_small.gif", null, "http://gpl4you.com/images/noavatar_small.gif", null, "http://gpl4you.com/images/noavatar_small.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80843616,"math_prob":0.944885,"size":1115,"snap":"2019-13-2019-22","text_gpt3_token_len":305,"char_repetition_ratio":0.115211524,"word_repetition_ratio":0.02970297,"special_character_ratio":0.26816145,"punctuation_ratio":0.043956045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95179063,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-20T05:00:55Z\",\"WARC-Record-ID\":\"<urn:uuid:f2ff5336-cb4a-4e4c-805f-cce8114a0124>\",\"Content-Length\":\"96307\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:223cf0c1-b393-4ba8-9b39-7ad5079353b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:ecba1d25-cc7c-49cc-b9fe-e2dd8c822d2b>\",\"WARC-IP-Address\":\"13.127.215.39\",\"WARC-Target-URI\":\"http://gpl4you.com/discussion.php?q=mcq&company=HCL&type=Mathematical%20Skills&subtype=Number%20System\",\"WARC-Payload-Digest\":\"sha1:SOHFABTLU5BDAG3DW6F3IYND72D5W4JC\",\"WARC-Block-Digest\":\"sha1:XONJ6T7QEGQD43P3QB5H6BGTG7TMBT5R\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202299.16_warc_CC-MAIN-20190320044358-20190320070358-00340.warc.gz\"}"}
https://crypto.stackexchange.com/questions/67592/el-gamal-cryptosystem-proof-step-by-step-explanation-needed-i-feel-like-i-don	[ "# El Gamal Cryptosystem Proof, Step-by-Step Explanation Needed. I feel like I don't understand what I have on this proof\n\nCan someone provide a good and thorough explanation of the El Gamal proof? Basically, I need a step-by-step breakdown of what is happening at each important part in the algorithm.\n\nI haven't been able to find any good notes on El Gamala that explain the proof in an easy to understand format.\n\nHere is the proof in this image:", null, "• Generally when people talk about the \"proof of ElGamal encryption\" they refer to the proof of security. What you have there is just the derivation that shows that it is correct, i.e. that decryption actually works. Are you aware of that or is that part of the confusion? – Maeher Feb 25 '19 at 7:36\n• Oh dear, I must be so confused then. This is what I took from my professor's notes and I assumed it was the proof because it is labeled as such. – John Doe X Feb 25 '19 at 8:10\n• I think I mean that I need an explanation of the derivation, which is the image posted in the body of my message. – John Doe X Feb 25 '19 at 8:10\n\nFirst, we should make clear what we're proving here. The derivation you're showing is part of a proof of correctness of ElGamal encryption, not security.\n\nPerfect correctness (also referred to as completeness) of a public key encryption scheme is defined as follows.\n\nLet $$(\\mathsf{Gen},\\mathsf{Enc},\\mathsf{Dec})$$ be a public key encryption scheme. The scheme is said to be perfectly correct if it holds that for any security parameter $$n\\in\\mathbb{N}$$, any key pair $$(\\mathsf{ek},\\mathsf{dk})\\gets\\mathsf{Gen}(1^n)$$, any message $$m$$ from the message space, and any ciphertext $$c \\gets \\mathsf{Enc}(\\mathsf{ek},m)$$ it holds that $$\\mathsf{Dec}(\\mathsf{dk},c)=m$$.\n\nTo see whether ElGamal encryption is correct, we first recall the definition of ElGamal encryption. I will try to follow the notation in your notes as close as possible. From the notes it's not clear what kind of group is being used. But $$e_1$$ is a generator of some subgroup of $$\\mathbb{Z}_p^$$. Let $$q$$ be the order of that subgroup. (If $$e_1$$ is a generator of $$\\mathbb{Z}_p^$$, then $$q=p-1$$, if we are in a safe prime setting, then $$q=(p-1)/2$$ and prime.\n\n\\begin{align} &\\mathsf{Gen}(1^n) && \\mathsf{Enc}(e_2,P) && \\mathsf{Dec}(d,(c_1,c_2))\\\\ &d\\gets\\mathbb{Z}_q&&r \\gets \\mathbb{Z}_q&&P' := c_2\\cdot (c_1^d)^{-1} \\bmod p\\\\ &e_2 := e_1^d \\bmod p&&c_1:= e_1^r\\bmod p&&\\text{return }P'\\\\ &\\text{return } (e_2,d)&&c_2 := P\\cdot e_2^r\\bmod p&\\\\ &&&\\text{return } (c_1,c_2)&&\\\\ \\end{align}\n\nTo verify that the scheme is correct, we need to verify that for any choice of key pair, and message it always holds that $$\\mathsf{Dec}(d,\\mathsf{Enc}(e_2,P)) = P$$. This is where the derivation you're looking at comes in.\n\n\\begin{align} P' =& c_2\\cdot (c_1^d)^{-1} \\bmod p \\tag{1}\\\\ =& c_2\\cdot (e_1^{rd})^{-1} \\bmod p\\tag{2}\\\\ =& P\\cdot e_2^r\\cdot (e_1^{rd})^{-1} \\bmod p\\tag{3}\\\\ =& P\\cdot e_1^{rd}\\cdot (e_1^{rd})^{-1} \\bmod p\\tag{4}\\\\ =& P\\cdot e_1^{rd}\\cdot e_1^{-rd} \\bmod p\\tag{5}\\\\ =& P\\cdot e_1^{rd-rd} \\bmod p\\tag{6}\\\\ =& P \\bmod p\\tag{7}\\\\ \\end{align}\n\nNow let's go through this line by line.\n\n1. In line (1) we simply have the definition of the decryption algorithm.\n2. In line (2) we use the definition of $$c_1 = e_1^r\\bmod p$$ from the encryption algorithm and replace $$c_1$$ by its definition.\n3. In line (3) we do exactly the same with $$c_2$$ and replace it with its definition $$c_2 := P\\cdot e_2^r\\bmod p$$ from the encryption algorithm.\n4. In line (4) we now look at the key generation algorithm and replace $$e_2$$ by its definition from there $$e_2 := e_1^d \\bmod p$$. Note that we have not changed anything. We have simply replaced the variables $$c_1,c_2,e_2$$ by their respective definitions.\n5. In line (5) we use a basic rule of exponentiation, namely that $$(x^a)^b = x^{a\\cdot b}$$, therefore $$(e_1^{rd})^{-1} = e_1^{-rd}$$.\n6. In line (6) we again use a basic rule of exponentiation, namely that $$x^a\\cdot x^b = x^{a+ b}$$, therefore $$e_1^{rd}\\cdot e_1^{-rd} = e_1^{rd-rd}$$.\n7. Since $$rd-rd=0$$ and anything raised to the $$0$$th power equals the identity, we are left with $$P\\cdot 1 \\bmod p$$ and thereby, since $$1$$ is the identity of the group with $$P$$ in line (7).\n\nSince this derivation makes no assumptions whatsoever about $$P,d,e_2,c_1$$ and $$c_2$$, besides that they were generated according to the above definition of ElGamal encryption, it holds for any choice of key-pair and any plaintext $$P$$.\n\nThe derivation thus shows that decryption always works and results in the same plaintext that was encrypted, thus showing that the encryption scheme is in fact correct.\n\nWhat it does not show is anything about the security of the scheme.\n\n• This is amazing and exactly what I needed. However, I think my professor is using this as a proof for El Gamal's encryption and decryption process and why it works, and I don't think she is using it as a proof of the security of El Gamal. She didn't really specify, but thanks for clearing this up. – John Doe X Mar 1 '19 at 8:59" ]	[ null, "https://i.stack.imgur.com/Po1m7.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8156,"math_prob":0.9990081,"size":3587,"snap":"2021-04-2021-17","text_gpt3_token_len":1197,"char_repetition_ratio":0.1373151,"word_repetition_ratio":0.018348623,"special_character_ratio":0.33398384,"punctuation_ratio":0.09375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998696,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T00:25:57Z\",\"WARC-Record-ID\":\"<urn:uuid:4fdc1ee9-106e-4377-9601-f4c27de437ad>\",\"Content-Length\":\"156246\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b09e708-b95e-4658-ab5f-61abae7c510b>\",\"WARC-Concurrent-To\":\"<urn:uuid:8056e076-ddb5-41db-bd01-65a774060412>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://crypto.stackexchange.com/questions/67592/el-gamal-cryptosystem-proof-step-by-step-explanation-needed-i-feel-like-i-don\",\"WARC-Payload-Digest\":\"sha1:C4DIARUWXHV6LILLFV3MVT3RWZSOJRTU\",\"WARC-Block-Digest\":\"sha1:CC77KRZXE5UF6U4NJSUY4DZ3OFL4XSER\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703538741.56_warc_CC-MAIN-20210123222657-20210124012657-00401.warc.gz\"}"}
http://ffmpeg.org/pipermail/ffmpeg-cvslog/2006-June/003120.html	[ "# [Ffmpeg-cvslog] r5493 - in trunk/libavcodec/ppc: dsputil_altivec.c dsputil_h264_template_altivec.c\n\nlu_zero subversion\nSat Jun 17 20:46:07 CEST 2006\n\n```Author: lu_zero\nDate: Sat Jun 17 20:46:06 2006\nNew Revision: 5493\n\nModified:\ntrunk/libavcodec/ppc/dsputil_altivec.c\ntrunk/libavcodec/ppc/dsputil_h264_template_altivec.c\n\nLog:\nCosmetics: should not hurt performance, scream if are\n\nModified: trunk/libavcodec/ppc/dsputil_altivec.c\n==============================================================================\n--- trunk/libavcodec/ppc/dsputil_altivec.c\t(original)\n+++ trunk/libavcodec/ppc/dsputil_altivec.c\tSat Jun 17 20:46:06 2006\n@@ -1311,9 +1311,9 @@\nint hadamard8_diff8x8_altivec(/MpegEncContext/ void s, uint8_t dst, uint8_t src, int stride, int h){\nint sum;\nregister const_vector unsigned char vzero = (const_vector unsigned char)vec_splat_u8(0);\nregister vector signed short temp0, temp1, temp2, temp3, temp4, temp5, temp6, temp7;\n{\nregister const_vector signed short vprod1 = (const_vector signed short)AVV( 1,-1, 1,-1, 1,-1, 1,-1);\nregister const_vector signed short vprod2 = (const_vector signed short)AVV( 1, 1,-1,-1, 1, 1,-1,-1);\n@@ -1338,6 +1338,8 @@\n{ \\\nregister vector unsigned char src1, src2, srcO; \\\nregister vector unsigned char dst1, dst2, dstO; \\\n+ register vector signed short srcV, dstV; \\\n+ register vector signed short but0, but1, but2, op1, op2, op3; \\\nsrc1 = vec_ld(stride i, src); \\\nif ((((stride * i) + (unsigned long)src) & 0x0000000F) > 8) \\\nsrc2 = vec_ld((stride * i) + 16, src); \\\n@@ -1348,17 +1350,19 @@\ndstO = vec_perm(dst1, dst2, vec_lvsl(stride * i, dst)); \\\n/* promote the unsigned chars to signed shorts / \\\n/ we're in the 8x8 function, we only care for the first 8 / \\\n- register vector signed short srcV = \\\n- (vector signed short)vec_mergeh((vector signed char)vzero, (vector signed char)srcO); \\\n- register vector signed short dstV = \\\n- (vector signed short)vec_mergeh((vector signed char)vzero, (vector signed char)dstO); \\\n+ srcV = \\\n+ (vector signed short)vec_mergeh((vector signed char)vzero, \\\n+ (vector signed char)srcO); \\\n+ dstV = \\\n+ (vector signed short)vec_mergeh((vector signed char)vzero, \\\n+ (vector signed char)dstO); \\\n/ substractions inside the first butterfly / \\\n- register vector signed short but0 = vec_sub(srcV, dstV); \\\n- register vector signed short op1 = vec_perm(but0, but0, perm1); \\\n- register vector signed short but1 = vec_mladd(but0, vprod1, op1); \\\n- register vector signed short op2 = vec_perm(but1, but1, perm2); \\\n- register vector signed short but2 = vec_mladd(but1, vprod2, op2); \\\n- register vector signed short op3 = vec_perm(but2, but2, perm3); \\\n+ but0 = vec_sub(srcV, dstV); \\\n+ op1 = vec_perm(but0, but0, perm1); \\\n+ but1 = vec_mladd(but0, vprod1, op1); \\\n+ op2 = vec_perm(but1, but1, perm2); \\\n+ but2 = vec_mladd(but1, vprod2, op2); \\\n+ op3 = vec_perm(but2, but2, perm3); \\\nres = vec_mladd(but2, vprod3, op3); \\\n}\nONEITERBUTTERFLY(0, temp0);\n@@ -1481,37 +1485,63 @@\n\n#define ONEITERBUTTERFLY(i, res1, res2) \\\n{ \\\n- register vector unsigned char src1 REG_v(v22), src2 REG_v(v23); \\\n- register vector unsigned char dst1 REG_v(v24), dst2 REG_v(v25); \\\n+ register vector unsigned char src1 REG_v(v22), \\\n+ src2 REG_v(v23), \\\n+ dst1 REG_v(v24), \\\n+ dst2 REG_v(v25), \\\n+ srcO REG_v(v22), \\\n+ dstO REG_v(v23); \\\n+ \\\n+ register vector signed short srcV REG_v(v24), \\\n+ dstV REG_v(v25), \\\n+ srcW REG_v(v26), \\\n+ dstW REG_v(v27), \\\n+ but0 REG_v(v28), \\\n+ but0S REG_v(v29), \\\n+ op1 REG_v(v30), \\\n+ but1 REG_v(v22), \\\n+ op1S REG_v(v23), \\\n+ but1S REG_v(v24), \\\n+ op2 REG_v(v25), \\\n+ but2 REG_v(v26), \\\n+ op2S REG_v(v27), \\\n+ but2S REG_v(v28), \\\n+ op3 REG_v(v29), \\\n+ op3S REG_v(v30); \\\n+ \\\nsrc1 = vec_ld(stride i, src); \\\nsrc2 = vec_ld((stride * i) + 16, src); \\\n- register vector unsigned char srcO REG_v(v22) = vec_perm(src1, src2, vec_lvsl(stride * i, src)); \\\n+ srcO = vec_perm(src1, src2, vec_lvsl(stride * i, src)); \\\ndst1 = vec_ld(stride * i, dst); \\\ndst2 = vec_ld((stride * i) + 16, dst); \\\n- register vector unsigned char dstO REG_v(v23) = vec_perm(dst1, dst2, vec_lvsl(stride * i, dst)); \\\n+ dstO = vec_perm(dst1, dst2, vec_lvsl(stride * i, dst)); \\\n/* promote the unsigned chars to signed shorts / \\\n- register vector signed short srcV REG_v(v24) = \\\n- (vector signed short)vec_mergeh((vector signed char)vzero, (vector signed char)srcO); \\\n- register vector signed short dstV REG_v(v25) = \\\n- (vector signed short)vec_mergeh((vector signed char)vzero, (vector signed char)dstO); \\\n- register vector signed short srcW REG_v(v26) = \\\n- (vector signed short)vec_mergel((vector signed char)vzero, (vector signed char)srcO); \\\n- register vector signed short dstW REG_v(v27) = \\\n- (vector signed short)vec_mergel((vector signed char)vzero, (vector signed char)dstO); \\\n+ srcV = \\\n+ (vector signed short)vec_mergeh((vector signed char)vzero, \\\n+ (vector signed char)srcO); \\\n+ dstV = \\\n+ (vector signed short)vec_mergeh((vector signed char)vzero, \\\n+ (vector signed char)dstO); \\\n+ srcW = \\\n+ (vector signed short)vec_mergel((vector signed char)vzero, \\\n+ (vector signed char)srcO); \\\n+ dstW = \\\n+ (vector signed short)vec_mergel((vector signed char)vzero, \\\n+ (vector signed char)dstO); \\\n/ substractions inside the first butterfly / \\\n- register vector signed short but0 REG_v(v28) = vec_sub(srcV, dstV); \\\n- register vector signed short but0S REG_v(v29) = vec_sub(srcW, dstW); \\\n- register vector signed short op1 REG_v(v30) = vec_perm(but0, but0, perm1); \\\n- register vector signed short but1 REG_v(v22) = vec_mladd(but0, vprod1, op1); \\\n- register vector signed short op1S REG_v(v23) = vec_perm(but0S, but0S, perm1); \\\n- register vector signed short but1S REG_v(v24) = vec_mladd(but0S, vprod1, op1S); \\\n- register vector signed short op2 REG_v(v25) = vec_perm(but1, but1, perm2); \\\n- register vector signed short but2 REG_v(v26) = vec_mladd(but1, vprod2, op2); \\\n- register vector signed short op2S REG_v(v27) = vec_perm(but1S, but1S, perm2); \\\n- register vector signed short but2S REG_v(v28) = vec_mladd(but1S, vprod2, op2S); \\\n- register vector signed short op3 REG_v(v29) = vec_perm(but2, but2, perm3); \\\n+ but0 = vec_sub(srcV, dstV); \\\n+ but0S = vec_sub(srcW, dstW); \\\n+ op1 = vec_perm(but0, but0, perm1); \\\n+ but1 = vec_mladd(but0, vprod1, op1); \\\n+ op1S = vec_perm(but0S, but0S, perm1); \\\n+ but1S = vec_mladd(but0S, vprod1, op1S); \\\n+ op2 = vec_perm(but1, but1, perm2); \\\n+ but2 = vec_mladd(but1, vprod2, op2); \\\n+ op2S = vec_perm(but1S, but1S, perm2); \\\n+ but2S = vec_mladd(but1S, vprod2, op2S); \\\n+ op3 = vec_perm(but2, but2, perm3); \\\nres1 = vec_mladd(but2, vprod3, op3); \\\n- register vector signed short op3S REG_v(v30) = vec_perm(but2S, but2S, perm3); \\\n+ op3S = vec_perm(but2S, but2S, perm3); \\\nres2 = vec_mladd(but2S, vprod3, op3S); \\\n}\nONEITERBUTTERFLY(0, temp0, temp0S);\n@@ -1526,6 +1556,12 @@\n#undef ONEITERBUTTERFLY\n{\nregister vector signed int vsum;\n+ register vector signed short line0S, line1S, line2S, line3S, line4S,\n+ line5S, line6S, line7S, line0BS,line2BS,\n+ line1BS,line3BS,line4BS,line6BS,line5BS,\n+ line7BS,line0CS,line4CS,line1CS,line5CS,\n+ line2CS,line6CS,line3CS,line7CS;\n+\nregister vector signed short line0 = vec_add(temp0, temp1);\nregister vector signed short line1 = vec_sub(temp0, temp1);\nregister vector signed short line2 = vec_add(temp2, temp3);\n@@ -1562,32 +1598,32 @@\nvsum = vec_sum4s(vec_abs(line6C), vsum);\nvsum = vec_sum4s(vec_abs(line7C), vsum);\n\n- register vector signed short line0S = vec_add(temp0S, temp1S);\n- register vector signed short line1S = vec_sub(temp0S, temp1S);\n- register vector signed short line2S = vec_add(temp2S, temp3S);\n- register vector signed short line3S = vec_sub(temp2S, temp3S);\n- register vector signed short line4S = vec_add(temp4S, temp5S);\n- register vector signed short line5S = vec_sub(temp4S, temp5S);\n- register vector signed short line6S = vec_add(temp6S, temp7S);\n- register vector signed short line7S = vec_sub(temp6S, temp7S);\n-\n- register vector signed short line0BS = vec_add(line0S, line2S);\n- register vector signed short line2BS = vec_sub(line0S, line2S);\n- register vector signed short line1BS = vec_add(line1S, line3S);\n- register vector signed short line3BS = vec_sub(line1S, line3S);\n- register vector signed short line4BS = vec_add(line4S, line6S);\n- register vector signed short line6BS = vec_sub(line4S, line6S);\n- register vector signed short line5BS = vec_add(line5S, line7S);\n- register vector signed short line7BS = vec_sub(line5S, line7S);\n-\n- register vector signed short line0CS = vec_add(line0BS, line4BS);\n- register vector signed short line4CS = vec_sub(line0BS, line4BS);\n- register vector signed short line1CS = vec_add(line1BS, line5BS);\n- register vector signed short line5CS = vec_sub(line1BS, line5BS);\n- register vector signed short line2CS = vec_add(line2BS, line6BS);\n- register vector signed short line6CS = vec_sub(line2BS, line6BS);\n- register vector signed short line3CS = vec_add(line3BS, line7BS);\n- register vector signed short line7CS = vec_sub(line3BS, line7BS);\n+ line1S = vec_sub(temp0S, temp1S);\n+ line3S = vec_sub(temp2S, temp3S);\n+ line5S = vec_sub(temp4S, temp5S);\n+ line7S = vec_sub(temp6S, temp7S);\n+\n+ line2BS = vec_sub(line0S, line2S);\n+ line3BS = vec_sub(line1S, line3S);\n+ line6BS = vec_sub(line4S, line6S);\n+ line7BS = vec_sub(line5S, line7S);\n+\n+ line4CS = vec_sub(line0BS, line4BS);\n+ line5CS = vec_sub(line1BS, line5BS);\n+ line6CS = vec_sub(line2BS, line6BS);\n+ line7CS = vec_sub(line3BS, line7BS);\n\nvsum = vec_sum4s(vec_abs(line0CS), vsum);\nvsum = vec_sum4s(vec_abs(line1CS), vsum);\n\nModified: trunk/libavcodec/ppc/dsputil_h264_template_altivec.c\n==============================================================================\n--- trunk/libavcodec/ppc/dsputil_h264_template_altivec.c\t(original)\n+++ trunk/libavcodec/ppc/dsputil_h264_template_altivec.c\tSat Jun 17 20:46:06 2006\n@@ -19,13 +19,13 @@\n/ this code assume that stride % 16 == 0 /\nvoid PREFIX_h264_chroma_mc8_altivec(uint8_t dst, uint8_t * src, int stride, int h, int x, int y) {\nPOWERPC_PERF_DECLARE(PREFIX_h264_chroma_mc8_num, 1);\n- POWERPC_PERF_START_COUNT(PREFIX_h264_chroma_mc8_num, 1);\n- signed int ABCD __attribute__((aligned(16)));\n+ signed int ABCD __attribute__((aligned(16))) =\n+ {((8 - x) * (8 - y)),\n+ ((x) * (8 - y)),\n+ ((8 - x) * (y)),\n+ ((x) * (y))};\nregister int i;\n- ABCD = ((8 - x) * (8 - y));\n- ABCD = ((x) * (8 - y));\n- ABCD = ((8 - x) * (y));\n- ABCD = ((x) * (y));\n+ vector unsigned char fperm;\nconst vector signed int vABCD = vec_ld(0, ABCD);\nconst vector signed short vA = vec_splat((vector signed short)vABCD, 1);\nconst vector signed short vB = vec_splat((vector signed short)vABCD, 3);\n@@ -34,55 +34,61 @@\nconst vector signed int vzero = vec_splat_s32(0);\nconst vector signed short v32ss = vec_sl(vec_splat_s16(1),vec_splat_u16(5));\nconst vector unsigned short v6us = vec_splat_u16(6);\n+ register int loadSecond = (((unsigned long)src) % 16) <= 7 ? 0 : 1;\n+ register int reallyBadAlign = (((unsigned long)src) % 16) == 15 ? 1 : 0;\n\n- vector unsigned char fperm;\n+ vector unsigned char vsrcAuc, vsrcBuc, vsrcperm0, vsrcperm1;\n+ vector unsigned char vsrc0uc, vsrc1uc;\n+ vector signed short vsrc0ssH, vsrc1ssH;\n+ vector unsigned char vsrcCuc, vsrc2uc, vsrc3uc;\n+ vector signed short vsrc2ssH, vsrc3ssH, psum;\n+ vector unsigned char vdst, ppsum, vfdst, fsum;\n+\n+ POWERPC_PERF_START_COUNT(PREFIX_h264_chroma_mc8_num, 1);\n\nif (((unsigned long)dst) % 16 == 0) {\n- fperm = (vector unsigned char)AVV(0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17,\n- 0x08, 0x09, 0x0A, 0x0B, 0x0C, 0x0D, 0x0E, 0x0F);\n+ fperm = (vector unsigned char)AVV(0x10, 0x11, 0x12, 0x13,\n+ 0x14, 0x15, 0x16, 0x17,\n+ 0x08, 0x09, 0x0A, 0x0B,\n+ 0x0C, 0x0D, 0x0E, 0x0F);\n} else {\n- fperm = (vector unsigned char)AVV(0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07,\n- 0x18, 0x19, 0x1A, 0x1B, 0x1C, 0x1D, 0x1E, 0x1F);\n+ fperm = (vector unsigned char)AVV(0x00, 0x01, 0x02, 0x03,\n+ 0x04, 0x05, 0x06, 0x07,\n+ 0x18, 0x19, 0x1A, 0x1B,\n+ 0x1C, 0x1D, 0x1E, 0x1F);\n}\n\n- register int loadSecond = (((unsigned long)src) % 16) <= 7 ? 0 : 1;\n- register int reallyBadAlign = (((unsigned long)src) % 16) == 15 ? 1 : 0;\n-\n- vector unsigned char vsrcAuc;\n- vector unsigned char vsrcBuc;\n- vector unsigned char vsrcperm0;\n- vector unsigned char vsrcperm1;\nvsrcAuc = vec_ld(0, src);\n+\nvsrcBuc = vec_ld(16, src);\nvsrcperm0 = vec_lvsl(0, src);\nvsrcperm1 = vec_lvsl(1, src);\n\n- vector unsigned char vsrc0uc;\n- vector unsigned char vsrc1uc;\nvsrc0uc = vec_perm(vsrcAuc, vsrcBuc, vsrcperm0);\nvsrc1uc = vsrcBuc;\nelse\nvsrc1uc = vec_perm(vsrcAuc, vsrcBuc, vsrcperm1);\n\n- vector signed short vsrc0ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc0uc);\n- vector signed short vsrc1ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc1uc);\n+ vsrc0ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc0uc);\n+ vsrc1ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc1uc);\n\nfor (i = 0 ; i < h ; i++) {\n- vector unsigned char vsrcCuc;\n+\n+\nvsrcCuc = vec_ld(stride + 0, src);\n\n- vector unsigned char vsrc2uc;\n- vector unsigned char vsrc3uc;\nvsrc2uc = vec_perm(vsrcCuc, vsrcCuc, vsrcperm0);\nvsrc3uc = vec_perm(vsrcCuc, vsrcCuc, vsrcperm1);\n\n- vector signed short vsrc2ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc2uc);\n- vector signed short vsrc3ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc3uc);\n-\n- vector signed short psum;\n+ vsrc2ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc2uc);\n+ vsrc3ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc3uc);\n\n@@ -91,11 +97,9 @@\npsum = vec_sra(psum, v6us);\n\n- vector unsigned char vdst = vec_ld(0, dst);\n- vector unsigned char ppsum = (vector unsigned char)vec_packsu(psum, psum);\n-\n- vector unsigned char vfdst = vec_perm(vdst, ppsum, fperm);\n- vector unsigned char fsum;\n+ vdst = vec_ld(0, dst);\n+ ppsum = (vector unsigned char)vec_packsu(psum, psum);\n+ vfdst = vec_perm(vdst, ppsum, fperm);\n\nOP_U8_ALTIVEC(fsum, vfdst, vdst);\n\n@@ -108,24 +112,21 @@\nsrc += stride;\n}\n} else {\n- for (i = 0 ; i < h ; i++) {\n- vector unsigned char vsrcCuc;\nvector unsigned char vsrcDuc;\n+ for (i = 0 ; i < h ; i++) {\nvsrcCuc = vec_ld(stride + 0, src);\nvsrcDuc = vec_ld(stride + 16, src);\n\n- vector unsigned char vsrc2uc;\n- vector unsigned char vsrc3uc;\nvsrc2uc = vec_perm(vsrcCuc, vsrcDuc, vsrcperm0);\nvsrc3uc = vsrcDuc;\nelse\nvsrc3uc = vec_perm(vsrcCuc, vsrcDuc, vsrcperm1);\n\n- vector signed short vsrc2ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc2uc);\n- vector signed short vsrc3ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero, (vector unsigned char)vsrc3uc);\n-\n- vector signed short psum;\n+ vsrc2ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc2uc);\n+ vsrc3ssH = (vector signed short)vec_mergeh((vector unsigned char)vzero,\n+ (vector unsigned char)vsrc3uc);\n\n@@ -134,11 +135,9 @@\npsum = vec_sr(psum, v6us);\n\n- vector unsigned char vdst = vec_ld(0, dst);\n- vector unsigned char ppsum = (vector unsigned char)vec_pack(psum, psum);\n-\n- vector unsigned char vfdst = vec_perm(vdst, ppsum, fperm);\n- vector unsigned char fsum;\n+ vdst = vec_ld(0, dst);\n+ ppsum = (vector unsigned char)vec_pack(psum, psum);\n+ vfdst = vec_perm(vdst, ppsum, fperm);\n\nOP_U8_ALTIVEC(fsum, vfdst, vdst);\n\n@@ -157,7 +156,6 @@\n/* this code assume stride % 16 == 0 /\nstatic void PREFIX_h264_qpel16_h_lowpass_altivec(uint8_t dst, uint8_t * src, int dstStride, int srcStride) {\nPOWERPC_PERF_DECLARE(PREFIX_h264_qpel16_h_lowpass_num, 1);\n- POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_h_lowpass_num, 1);\nregister int i;\n\nconst vector signed int vzero = vec_splat_s32(0);\n@@ -172,13 +170,30 @@\nconst vector signed short v20ss = vec_sl(vec_splat_s16(5),vec_splat_u16(2));\nconst vector signed short v16ss = vec_sl(vec_splat_s16(1),vec_splat_u16(4));\nconst vector unsigned char dstperm = vec_lvsr(0, dst);\n- const vector unsigned char neg1 = (const vector unsigned char)vec_splat_s8(-1);\n- const vector unsigned char dstmask = vec_perm((const vector unsigned char)vzero, neg1, dstperm);\n+ const vector unsigned char neg1 =\n+ (const vector unsigned char) vec_splat_s8(-1);\n+\n+ const vector unsigned char dstmask =\n+ vec_perm((const vector unsigned char)vzero,\n+ neg1, dstperm);\n+\n+ vector unsigned char srcM2, srcM1, srcP0, srcP1, srcP2, srcP3;\n\nregister int align = ((((unsigned long)src) - 2) % 16);\n\n+ vector signed short srcP0A, srcP0B, srcP1A, srcP1B,\n+ srcP2A, srcP2B, srcP3A, srcP3B,\n+ srcM1A, srcM1B, srcM2A, srcM2B,\n+ sum1A, sum1B, sum2A, sum2B, sum3A, sum3B,\n+ pp1A, pp1B, pp2A, pp2B, pp3A, pp3B,\n+ psumA, psumB, sumA, sumB;\n+\n+ vector unsigned char sum, dst1, dst2, vdst, fsum,\n+ rsum, fdst1, fdst2;\n+\n+ POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_h_lowpass_num, 1);\n+\nfor (i = 0 ; i < 16 ; i ++) {\n- vector unsigned char srcM2, srcM1, srcP0, srcP1, srcP2, srcP3;\nvector unsigned char srcR1 = vec_ld(-2, src);\nvector unsigned char srcR2 = vec_ld(14, src);\n\n@@ -237,55 +252,54 @@\n} break;\n}\n\n- const vector signed short srcP0A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP0);\n- const vector signed short srcP0B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP0);\n- const vector signed short srcP1A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP1);\n- const vector signed short srcP1B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP1);\n-\n- const vector signed short srcP2A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP2);\n- const vector signed short srcP2B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP2);\n- const vector signed short srcP3A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP3);\n- const vector signed short srcP3B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP3);\n-\n- const vector signed short srcM1A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM1);\n- const vector signed short srcM1B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM1);\n- const vector signed short srcM2A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM2);\n- const vector signed short srcM2B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM2);\n-\n- const vector signed short sum1A = vec_adds(srcP0A, srcP1A);\n- const vector signed short sum1B = vec_adds(srcP0B, srcP1B);\n- const vector signed short sum2A = vec_adds(srcM1A, srcP2A);\n- const vector signed short sum2B = vec_adds(srcM1B, srcP2B);\n- const vector signed short sum3A = vec_adds(srcM2A, srcP3A);\n- const vector signed short sum3B = vec_adds(srcM2B, srcP3B);\n-\n- const vector signed short pp1A = vec_mladd(sum1A, v20ss, v16ss);\n- const vector signed short pp1B = vec_mladd(sum1B, v20ss, v16ss);\n-\n- const vector signed short pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n- const vector signed short pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n-\n- const vector signed short pp3A = vec_add(sum3A, pp1A);\n- const vector signed short pp3B = vec_add(sum3B, pp1B);\n-\n- const vector signed short psumA = vec_sub(pp3A, pp2A);\n- const vector signed short psumB = vec_sub(pp3B, pp2B);\n-\n- const vector signed short sumA = vec_sra(psumA, v5us);\n- const vector signed short sumB = vec_sra(psumB, v5us);\n-\n- const vector unsigned char sum = vec_packsu(sumA, sumB);\n-\n- const vector unsigned char dst1 = vec_ld(0, dst);\n- const vector unsigned char dst2 = vec_ld(16, dst);\n- const vector unsigned char vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n+ srcP0A = vec_mergeh((vector unsigned char)vzero, srcP0);\n+ srcP0B = vec_mergel((vector unsigned char)vzero, srcP0);\n+ srcP1A = vec_mergeh((vector unsigned char)vzero, srcP1);\n+ srcP1B = vec_mergel((vector unsigned char)vzero, srcP1);\n+\n+ srcP2A = vec_mergeh((vector unsigned char)vzero, srcP2);\n+ srcP2B = vec_mergel((vector unsigned char)vzero, srcP2);\n+ srcP3A = vec_mergeh((vector unsigned char)vzero, srcP3);\n+ srcP3B = vec_mergel((vector unsigned char)vzero, srcP3);\n+\n+ srcM1A = vec_mergeh((vector unsigned char)vzero, srcM1);\n+ srcM1B = vec_mergel((vector unsigned char)vzero, srcM1);\n+ srcM2A = vec_mergeh((vector unsigned char)vzero, srcM2);\n+ srcM2B = vec_mergel((vector unsigned char)vzero, srcM2);\n+\n+\n+ pp1A = vec_mladd(sum1A, v20ss, v16ss);\n+ pp1B = vec_mladd(sum1B, v20ss, v16ss);\n+\n+ pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n+ pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n+\n+\n+ psumA = vec_sub(pp3A, pp2A);\n+ psumB = vec_sub(pp3B, pp2B);\n+\n+ sumA = vec_sra(psumA, v5us);\n+ sumB = vec_sra(psumB, v5us);\n+\n+ sum = vec_packsu(sumA, sumB);\n+\n+ dst1 = vec_ld(0, dst);\n+ dst2 = vec_ld(16, dst);\n+ vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n\n- vector unsigned char fsum;\nOP_U8_ALTIVEC(fsum, sum, vdst);\n\n- const vector unsigned char rsum = vec_perm(fsum, fsum, dstperm);\n- const vector unsigned char fdst1 = vec_sel(dst1, rsum, dstmask);\n- const vector unsigned char fdst2 = vec_sel(rsum, dst2, dstmask);\n+ rsum = vec_perm(fsum, fsum, dstperm);\n+ fdst1 = vec_sel(dst1, rsum, dstmask);\n+ fdst2 = vec_sel(rsum, dst2, dstmask);\n\nvec_st(fdst1, 0, dst);\nvec_st(fdst2, 16, dst);\n@@ -299,7 +313,6 @@\n/* this code assume stride % 16 == 0 /\nstatic void PREFIX_h264_qpel16_v_lowpass_altivec(uint8_t dst, uint8_t * src, int dstStride, int srcStride) {\nPOWERPC_PERF_DECLARE(PREFIX_h264_qpel16_v_lowpass_num, 1);\n- POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_v_lowpass_num, 1);\n\nregister int i;\n\n@@ -318,49 +331,71 @@\nconst vector unsigned char srcM2a = vec_ld(0, srcbis);\nconst vector unsigned char srcM2b = vec_ld(16, srcbis);\nconst vector unsigned char srcM2 = vec_perm(srcM2a, srcM2b, perm);\n- srcbis += srcStride;\n- const vector unsigned char srcM1a = vec_ld(0, srcbis);\n+// srcbis += srcStride;\n+ const vector unsigned char srcM1a = vec_ld(0, srcbis += srcStride);\nconst vector unsigned char srcM1b = vec_ld(16, srcbis);\nconst vector unsigned char srcM1 = vec_perm(srcM1a, srcM1b, perm);\n- srcbis += srcStride;\n- const vector unsigned char srcP0a = vec_ld(0, srcbis);\n+// srcbis += srcStride;\n+ const vector unsigned char srcP0a = vec_ld(0, srcbis += srcStride);\nconst vector unsigned char srcP0b = vec_ld(16, srcbis);\nconst vector unsigned char srcP0 = vec_perm(srcP0a, srcP0b, perm);\n- srcbis += srcStride;\n- const vector unsigned char srcP1a = vec_ld(0, srcbis);\n+// srcbis += srcStride;\n+ const vector unsigned char srcP1a = vec_ld(0, srcbis += srcStride);\nconst vector unsigned char srcP1b = vec_ld(16, srcbis);\nconst vector unsigned char srcP1 = vec_perm(srcP1a, srcP1b, perm);\n- srcbis += srcStride;\n- const vector unsigned char srcP2a = vec_ld(0, srcbis);\n+// srcbis += srcStride;\n+ const vector unsigned char srcP2a = vec_ld(0, srcbis += srcStride);\nconst vector unsigned char srcP2b = vec_ld(16, srcbis);\nconst vector unsigned char srcP2 = vec_perm(srcP2a, srcP2b, perm);\n- srcbis += srcStride;\n+// srcbis += srcStride;\n+\n+ vector signed short srcM2ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcM2);\n+ vector signed short srcM2ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcM2);\n+ vector signed short srcM1ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcM1);\n+ vector signed short srcM1ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcM1);\n+ vector signed short srcP0ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP0);\n+ vector signed short srcP0ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP0);\n+ vector signed short srcP1ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP1);\n+ vector signed short srcP1ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP1);\n+ vector signed short srcP2ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP2);\n+ vector signed short srcP2ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP2);\n+\n+ vector signed short pp1A, pp1B, pp2A, pp2B, pp3A, pp3B,\n+ psumA, psumB, sumA, sumB,\n+ srcP3ssA, srcP3ssB,\n+ sum1A, sum1B, sum2A, sum2B, sum3A, sum3B;\n\n- vector signed short srcM2ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM2);\n- vector signed short srcM2ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM2);\n- vector signed short srcM1ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM1);\n- vector signed short srcM1ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM1);\n- vector signed short srcP0ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP0);\n- vector signed short srcP0ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP0);\n- vector signed short srcP1ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP1);\n- vector signed short srcP1ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP1);\n- vector signed short srcP2ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP2);\n- vector signed short srcP2ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP2);\n+ vector unsigned char sum, dst1, dst2, vdst, fsum, rsum, fdst1, fdst2,\n+ srcP3a, srcP3b, srcP3;\n+\n+ POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_v_lowpass_num, 1);\n\nfor (i = 0 ; i < 16 ; i++) {\n- const vector unsigned char srcP3a = vec_ld(0, srcbis);\n- const vector unsigned char srcP3b = vec_ld(16, srcbis);\n- const vector unsigned char srcP3 = vec_perm(srcP3a, srcP3b, perm);\n- const vector signed short srcP3ssA = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP3);\n- const vector signed short srcP3ssB = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP3);\n- srcbis += srcStride;\n-\n- const vector signed short sum1A = vec_adds(srcP0ssA, srcP1ssA);\n- const vector signed short sum1B = vec_adds(srcP0ssB, srcP1ssB);\n- const vector signed short sum2A = vec_adds(srcM1ssA, srcP2ssA);\n- const vector signed short sum2B = vec_adds(srcM1ssB, srcP2ssB);\n- const vector signed short sum3A = vec_adds(srcM2ssA, srcP3ssA);\n- const vector signed short sum3B = vec_adds(srcM2ssB, srcP3ssB);\n+ srcP3a = vec_ld(0, srcbis += srcStride);\n+ srcP3b = vec_ld(16, srcbis);\n+ srcP3 = vec_perm(srcP3a, srcP3b, perm);\n+ srcP3ssA = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP3);\n+ srcP3ssB = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP3);\n+// srcbis += srcStride;\n+\n\nsrcM2ssA = srcM1ssA;\nsrcM2ssB = srcM1ssB;\n@@ -373,33 +408,32 @@\nsrcP2ssA = srcP3ssA;\nsrcP2ssB = srcP3ssB;\n\n- const vector signed short pp1A = vec_mladd(sum1A, v20ss, v16ss);\n- const vector signed short pp1B = vec_mladd(sum1B, v20ss, v16ss);\n+ pp1A = vec_mladd(sum1A, v20ss, v16ss);\n+ pp1B = vec_mladd(sum1B, v20ss, v16ss);\n\n- const vector signed short pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n- const vector signed short pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n+ pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n+ pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n\n- const vector signed short pp3A = vec_add(sum3A, pp1A);\n- const vector signed short pp3B = vec_add(sum3B, pp1B);\n\n- const vector signed short psumA = vec_sub(pp3A, pp2A);\n- const vector signed short psumB = vec_sub(pp3B, pp2B);\n+ psumA = vec_sub(pp3A, pp2A);\n+ psumB = vec_sub(pp3B, pp2B);\n\n- const vector signed short sumA = vec_sra(psumA, v5us);\n- const vector signed short sumB = vec_sra(psumB, v5us);\n+ sumA = vec_sra(psumA, v5us);\n+ sumB = vec_sra(psumB, v5us);\n\n- const vector unsigned char sum = vec_packsu(sumA, sumB);\n+ sum = vec_packsu(sumA, sumB);\n\n- const vector unsigned char dst1 = vec_ld(0, dst);\n- const vector unsigned char dst2 = vec_ld(16, dst);\n- const vector unsigned char vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n+ dst1 = vec_ld(0, dst);\n+ dst2 = vec_ld(16, dst);\n+ vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n\n- vector unsigned char fsum;\nOP_U8_ALTIVEC(fsum, sum, vdst);\n\n- const vector unsigned char rsum = vec_perm(fsum, fsum, dstperm);\n- const vector unsigned char fdst1 = vec_sel(dst1, rsum, dstmask);\n- const vector unsigned char fdst2 = vec_sel(rsum, dst2, dstmask);\n+ rsum = vec_perm(fsum, fsum, dstperm);\n+ fdst1 = vec_sel(dst1, rsum, dstmask);\n+ fdst2 = vec_sel(rsum, dst2, dstmask);\n\nvec_st(fdst1, 0, dst);\nvec_st(fdst2, 16, dst);\n@@ -412,7 +446,6 @@\n/* this code assume stride % 16 == 0 and tmp is properly aligned /\nstatic void PREFIX_h264_qpel16_hv_lowpass_altivec(uint8_t dst, int16_t * tmp, uint8_t * src, int dstStride, int tmpStride, int srcStride) {\nPOWERPC_PERF_DECLARE(PREFIX_h264_qpel16_hv_lowpass_num, 1);\n- POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_hv_lowpass_num, 1);\nregister int i;\nconst vector signed int vzero = vec_splat_s32(0);\nconst vector unsigned char permM2 = vec_lvsl(-2, src);\n@@ -430,8 +463,38 @@\n\nregister int align = ((((unsigned long)src) - 2) % 16);\n\n- src -= (2 * srcStride);\n+ const vector unsigned char neg1 = (const vector unsigned char)\n+ vec_splat_s8(-1);\n+\n+ vector signed short srcP0A, srcP0B, srcP1A, srcP1B,\n+ srcP2A, srcP2B, srcP3A, srcP3B,\n+ srcM1A, srcM1B, srcM2A, srcM2B,\n+ sum1A, sum1B, sum2A, sum2B, sum3A, sum3B,\n+ pp1A, pp1B, pp2A, pp2B, psumA, psumB;\n+\n+ const vector unsigned char dstperm = vec_lvsr(0, dst);\n+\n+ const vector unsigned char dstmask = vec_perm((const vector unsigned char)vzero, neg1, dstperm);\n+\n+ const vector unsigned char mperm = (const vector unsigned char)\n+ AVV(0x00, 0x08, 0x01, 0x09, 0x02, 0x0A, 0x03, 0x0B,\n+ 0x04, 0x0C, 0x05, 0x0D, 0x06, 0x0E, 0x07, 0x0F);\n+ int16_t tmpbis = tmp;\n\n+ vector signed short tmpM1ssA, tmpM1ssB, tmpM2ssA, tmpM2ssB,\n+ tmpP0ssA, tmpP0ssB, tmpP1ssA, tmpP1ssB,\n+ tmpP2ssA, tmpP2ssB;\n+\n+ vector signed int pp1Ae, pp1Ao, pp1Be, pp1Bo, pp2Ae, pp2Ao, pp2Be, pp2Bo,\n+ pp3Ae, pp3Ao, pp3Be, pp3Bo, pp1cAe, pp1cAo, pp1cBe, pp1cBo,\n+ pp32Ae, pp32Ao, pp32Be, pp32Bo, sumAe, sumAo, sumBe, sumBo,\n+ ssumAe, ssumAo, ssumBe, ssumBo;\n+ vector unsigned char fsum, sumv, sum, dst1, dst2, vdst,\n+ rsum, fdst1, fdst2;\n+ vector signed short ssume, ssumo;\n+\n+ POWERPC_PERF_START_COUNT(PREFIX_h264_qpel16_hv_lowpass_num, 1);\n+ src -= (2 srcStride);\nfor (i = 0 ; i < 21 ; i ++) {\nvector unsigned char srcM2, srcM1, srcP0, srcP1, srcP2, srcP3;\nvector unsigned char srcR1 = vec_ld(-2, src);\n@@ -492,36 +555,48 @@\n} break;\n}\n\n- const vector signed short srcP0A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP0);\n- const vector signed short srcP0B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP0);\n- const vector signed short srcP1A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP1);\n- const vector signed short srcP1B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP1);\n-\n- const vector signed short srcP2A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP2);\n- const vector signed short srcP2B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP2);\n- const vector signed short srcP3A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcP3);\n- const vector signed short srcP3B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcP3);\n-\n- const vector signed short srcM1A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM1);\n- const vector signed short srcM1B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM1);\n- const vector signed short srcM2A = (vector signed short)vec_mergeh((vector unsigned char)vzero, srcM2);\n- const vector signed short srcM2B = (vector signed short)vec_mergel((vector unsigned char)vzero, srcM2);\n-\n- const vector signed short sum1A = vec_adds(srcP0A, srcP1A);\n- const vector signed short sum1B = vec_adds(srcP0B, srcP1B);\n- const vector signed short sum2A = vec_adds(srcM1A, srcP2A);\n- const vector signed short sum2B = vec_adds(srcM1B, srcP2B);\n- const vector signed short sum3A = vec_adds(srcM2A, srcP3A);\n- const vector signed short sum3B = vec_adds(srcM2B, srcP3B);\n+ srcP0A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP0);\n+ srcP0B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP0);\n+ srcP1A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP1);\n+ srcP1B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP1);\n+\n+ srcP2A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP2);\n+ srcP2B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP2);\n+ srcP3A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcP3);\n+ srcP3B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcP3);\n+\n+ srcM1A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcM1);\n+ srcM1B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcM1);\n+ srcM2A = (vector signed short)\n+ vec_mergeh((vector unsigned char)vzero, srcM2);\n+ srcM2B = (vector signed short)\n+ vec_mergel((vector unsigned char)vzero, srcM2);\n+\n\n- const vector signed short pp1A = vec_mladd(sum1A, v20ss, sum3A);\n- const vector signed short pp1B = vec_mladd(sum1B, v20ss, sum3B);\n+ pp1A = vec_mladd(sum1A, v20ss, sum3A);\n+ pp1B = vec_mladd(sum1B, v20ss, sum3B);\n\n- const vector signed short pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n- const vector signed short pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n+ pp2A = vec_mladd(sum2A, v5ss, (vector signed short)vzero);\n+ pp2B = vec_mladd(sum2B, v5ss, (vector signed short)vzero);\n\n- const vector signed short psumA = vec_sub(pp1A, pp2A);\n- const vector signed short psumB = vec_sub(pp1B, pp2B);\n+ psumA = vec_sub(pp1A, pp2A);\n+ psumB = vec_sub(pp1B, pp2B);\n\nvec_st(psumA, 0, tmp);\nvec_st(psumB, 16, tmp);\n@@ -530,35 +605,25 @@\ntmp += tmpStride; /* int16_t, and stride is 16, so it's OK here /\n}\n\n- const vector unsigned char dstperm = vec_lvsr(0, dst);\n- const vector unsigned char neg1 = (const vector unsigned char)vec_splat_s8(-1);\n- const vector unsigned char dstmask = vec_perm((const vector unsigned char)vzero, neg1, dstperm);\n- const vector unsigned char mperm = (const vector unsigned char)\n- AVV(0x00, 0x08, 0x01, 0x09, 0x02, 0x0A, 0x03, 0x0B,\n- 0x04, 0x0C, 0x05, 0x0D, 0x06, 0x0E, 0x07, 0x0F);\n-\n- int16_t tmpbis = tmp - (tmpStride 21);\n-\n- vector signed short tmpM2ssA = vec_ld(0, tmpbis);\n- vector signed short tmpM2ssB = vec_ld(16, tmpbis);\n+ tmpM2ssA = vec_ld(0, tmpbis);\n+ tmpM2ssB = vec_ld(16, tmpbis);\ntmpbis += tmpStride;\n- vector signed short tmpM1ssA = vec_ld(0, tmpbis);\n- vector signed short tmpM1ssB = vec_ld(16, tmpbis);\n+ tmpM1ssA = vec_ld(0, tmpbis);\n+ tmpM1ssB = vec_ld(16, tmpbis);\ntmpbis += tmpStride;\n- vector signed short tmpP0ssA = vec_ld(0, tmpbis);\n- vector signed short tmpP0ssB = vec_ld(16, tmpbis);\n+ tmpP0ssA = vec_ld(0, tmpbis);\n+ tmpP0ssB = vec_ld(16, tmpbis);\ntmpbis += tmpStride;\n- vector signed short tmpP1ssA = vec_ld(0, tmpbis);\n- vector signed short tmpP1ssB = vec_ld(16, tmpbis);\n+ tmpP1ssA = vec_ld(0, tmpbis);\n+ tmpP1ssB = vec_ld(16, tmpbis);\ntmpbis += tmpStride;\n- vector signed short tmpP2ssA = vec_ld(0, tmpbis);\n- vector signed short tmpP2ssB = vec_ld(16, tmpbis);\n+ tmpP2ssA = vec_ld(0, tmpbis);\n+ tmpP2ssB = vec_ld(16, tmpbis);\ntmpbis += tmpStride;\n\nfor (i = 0 ; i < 16 ; i++) {\nconst vector signed short tmpP3ssA = vec_ld(0, tmpbis);\nconst vector signed short tmpP3ssB = vec_ld(16, tmpbis);\n- tmpbis += tmpStride;\n\nconst vector signed short sum1A = vec_adds(tmpP0ssA, tmpP1ssA);\nconst vector signed short sum1B = vec_adds(tmpP0ssB, tmpP1ssB);\n@@ -567,6 +632,8 @@\nconst vector signed short sum3A = vec_adds(tmpM2ssA, tmpP3ssA);\nconst vector signed short sum3B = vec_adds(tmpM2ssB, tmpP3ssB);\n\n+ tmpbis += tmpStride;\n+\ntmpM2ssA = tmpM1ssA;\ntmpM2ssB = tmpM1ssB;\ntmpM1ssA = tmpP0ssA;\n@@ -578,57 +645,56 @@\ntmpP2ssA = tmpP3ssA;\ntmpP2ssB = tmpP3ssB;\n\n- const vector signed int pp1Ae = vec_mule(sum1A, v20ss);\n- const vector signed int pp1Ao = vec_mulo(sum1A, v20ss);\n- const vector signed int pp1Be = vec_mule(sum1B, v20ss);\n- const vector signed int pp1Bo = vec_mulo(sum1B, v20ss);\n-\n- const vector signed int pp2Ae = vec_mule(sum2A, v5ss);\n- const vector signed int pp2Ao = vec_mulo(sum2A, v5ss);\n- const vector signed int pp2Be = vec_mule(sum2B, v5ss);\n- const vector signed int pp2Bo = vec_mulo(sum2B, v5ss);\n-\n- const vector signed int pp3Ae = vec_sra((vector signed int)sum3A, v16ui);\n- const vector signed int pp3Ao = vec_mulo(sum3A, v1ss);\n- const vector signed int pp3Be = vec_sra((vector signed int)sum3B, v16ui);\n- const vector signed int pp3Bo = vec_mulo(sum3B, v1ss);\n-\n- const vector signed int pp1cAe = vec_add(pp1Ae, v512si);\n- const vector signed int pp1cAo = vec_add(pp1Ao, v512si);\n- const vector signed int pp1cBe = vec_add(pp1Be, v512si);\n- const vector signed int pp1cBo = vec_add(pp1Bo, v512si);\n-\n- const vector signed int pp32Ae = vec_sub(pp3Ae, pp2Ae);\n- const vector signed int pp32Ao = vec_sub(pp3Ao, pp2Ao);\n- const vector signed int pp32Be = vec_sub(pp3Be, pp2Be);\n- const vector signed int pp32Bo = vec_sub(pp3Bo, pp2Bo);\n-\n- const vector signed int sumAe = vec_add(pp1cAe, pp32Ae);\n- const vector signed int sumAo = vec_add(pp1cAo, pp32Ao);\n- const vector signed int sumBe = vec_add(pp1cBe, pp32Be);\n- const vector signed int sumBo = vec_add(pp1cBo, pp32Bo);\n-\n- const vector signed int ssumAe = vec_sra(sumAe, v10ui);\n- const vector signed int ssumAo = vec_sra(sumAo, v10ui);\n- const vector signed int ssumBe = vec_sra(sumBe, v10ui);\n- const vector signed int ssumBo = vec_sra(sumBo, v10ui);\n-\n- const vector signed short ssume = vec_packs(ssumAe, ssumBe);\n- const vector signed short ssumo = vec_packs(ssumAo, ssumBo);\n-\n- const vector unsigned char sumv = vec_packsu(ssume, ssumo);\n- const vector unsigned char sum = vec_perm(sumv, sumv, mperm);\n-\n- const vector unsigned char dst1 = vec_ld(0, dst);\n- const vector unsigned char dst2 = vec_ld(16, dst);\n- const vector unsigned char vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n+ pp1Ae = vec_mule(sum1A, v20ss);\n+ pp1Ao = vec_mulo(sum1A, v20ss);\n+ pp1Be = vec_mule(sum1B, v20ss);\n+ pp1Bo = vec_mulo(sum1B, v20ss);\n+\n+ pp2Ae = vec_mule(sum2A, v5ss);\n+ pp2Ao = vec_mulo(sum2A, v5ss);\n+ pp2Be = vec_mule(sum2B, v5ss);\n+ pp2Bo = vec_mulo(sum2B, v5ss);\n+\n+ pp3Ae = vec_sra((vector signed int)sum3A, v16ui);\n+ pp3Ao = vec_mulo(sum3A, v1ss);\n+ pp3Be = vec_sra((vector signed int)sum3B, v16ui);\n+ pp3Bo = vec_mulo(sum3B, v1ss);\n+\n+\n+ pp32Ae = vec_sub(pp3Ae, pp2Ae);\n+ pp32Ao = vec_sub(pp3Ao, pp2Ao);\n+ pp32Be = vec_sub(pp3Be, pp2Be);\n+ pp32Bo = vec_sub(pp3Bo, pp2Bo);\n+\n+\n+ ssumAe = vec_sra(sumAe, v10ui);\n+ ssumAo = vec_sra(sumAo, v10ui);\n+ ssumBe = vec_sra(sumBe, v10ui);\n+ ssumBo = vec_sra(sumBo, v10ui);\n+\n+ ssume = vec_packs(ssumAe, ssumBe);\n+ ssumo = vec_packs(ssumAo, ssumBo);\n+\n+ sumv = vec_packsu(ssume, ssumo);\n+ sum = vec_perm(sumv, sumv, mperm);\n+\n+ dst1 = vec_ld(0, dst);\n+ dst2 = vec_ld(16, dst);\n+ vdst = vec_perm(dst1, dst2, vec_lvsl(0, dst));\n\n- vector unsigned char fsum;\nOP_U8_ALTIVEC(fsum, sum, vdst);\n\n- const vector unsigned char rsum = vec_perm(fsum, fsum, dstperm);\n- const vector unsigned char fdst1 = vec_sel(dst1, rsum, dstmask);\n- const vector unsigned char fdst2 = vec_sel(rsum, dst2, dstmask);\n+ rsum = vec_perm(fsum, fsum, dstperm);\n+ fdst1 = vec_sel(dst1, rsum, dstmask);\n+ fdst2 = vec_sel(rsum, dst2, dstmask);\n\nvec_st(fdst1, 0, dst);\nvec_st(fdst2, 16, dst);\n\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5322732,"math_prob":0.99971193,"size":41620,"snap":"2022-27-2022-33","text_gpt3_token_len":14818,"char_repetition_ratio":0.37218857,"word_repetition_ratio":0.50117135,"special_character_ratio":0.35504565,"punctuation_ratio":0.24784878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9951869,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T21:46:25Z\",\"WARC-Record-ID\":\"<urn:uuid:96b53827-bc5e-425c-9ada-11de5b8ffc35>\",\"Content-Length\":\"52875\",\"Content-Type\":\"application/http; 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https://encyclopediaofmath.org/wiki/Subrepresentation_of_a_representation	[ "# Subrepresentation of a representation\n\nA linear representation $\\rho$ in an invariant subspace $F \\subset E$ of a representation $\\pi$ of a group (algebra, ring or semi-group) $X$ in a (topological) vector space $E$ defined by the formula $\\rho ( x) \\xi = \\pi ( x) \\xi$ for all $\\xi \\in F$, $x \\in X$. If $\\pi$ is a continuous representation (of a topological group, algebra, ring, or semi-group), then any subrepresentation of it is also continuous." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7200645,"math_prob":0.99802554,"size":1027,"snap":"2021-21-2021-25","text_gpt3_token_len":251,"char_repetition_ratio":0.2170088,"word_repetition_ratio":0.07092199,"special_character_ratio":0.24732229,"punctuation_ratio":0.1597633,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99975365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-10T04:15:38Z\",\"WARC-Record-ID\":\"<urn:uuid:418e6c4d-87a0-4a17-a554-a84b35f289c0>\",\"Content-Length\":\"14700\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d9ff18d-1bba-4a58-a982-17a19d908cb3>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ccdf969-3162-4df6-b920-bc958936343d>\",\"WARC-IP-Address\":\"34.96.94.55\",\"WARC-Target-URI\":\"https://encyclopediaofmath.org/wiki/Subrepresentation_of_a_representation\",\"WARC-Payload-Digest\":\"sha1:JSIFCL7T3FZ6JARO7GDBZ66OYTD4IG3J\",\"WARC-Block-Digest\":\"sha1:SW2YPTZJN7ALYHVTL2FZ35VJLVKDTWGK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989030.87_warc_CC-MAIN-20210510033850-20210510063850-00172.warc.gz\"}"}
https://www.learnatnoon.com/s/a-pass-on-is-numbered-so-that-its-faces-show/6036/	[ "A pass on is numbered so that its faces show the numbers 1, 2, 2, 3, 3, 6. It is tossed multiple times and the complete score in two tosses is noted. Complete the accompanying table which gives a couple of upsides of the absolute score on the two tosses:\nA pass on is numbered so that its faces show the numbers 1, 2, 2, 3, 3, 6. It is tossed multiple times and the complete score in two tosses is noted. Complete the accompanying table which gives a couple of upsides of the absolute score on the two tosses:\n\nWhat is the likelihood that the absolute score is\n\n(I) even?\n\n(ii) 6?\n\n(iii) something like 6?\n\nSolution:\n\nThe table will be as per the following:\n\n• 1 2 2 3 3 6\n\n1 2 3 3 4 4 7\n\n2 3 4 4 5 5 8\n\n2 3 4 4 5 5 8\n\n3 4 5 5 6 6 9\n\n3 4 5 5 6 6 9\n\n6 7 8 8 9 9 12\n\nAlong these lines, the all out number of results = 6×6 = 36\n\n(I) E (Even) = 18\n\nP (Even) = 18/36 = ½\n\n(ii) E (aggregate is 6) = 4\n\nP (aggregate is 6) = 4/36 = 1/9\n\n(iii) E (aggregate is atleast 6) = 15\n\nP (aggregate is atleast 6) = 15/36 = 5/12" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86242723,"math_prob":0.99917173,"size":479,"snap":"2022-40-2023-06","text_gpt3_token_len":242,"char_repetition_ratio":0.15157895,"word_repetition_ratio":0.11023622,"special_character_ratio":0.5407098,"punctuation_ratio":0.045801528,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9848254,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-05T04:47:52Z\",\"WARC-Record-ID\":\"<urn:uuid:44cdacc4-4049-4e96-ad2e-8891a678ff54>\",\"Content-Length\":\"110916\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c0b02362-6a60-40b6-8f9b-124173347461>\",\"WARC-Concurrent-To\":\"<urn:uuid:f669b5df-a91c-464e-951c-238186a996dd>\",\"WARC-IP-Address\":\"18.158.64.80\",\"WARC-Target-URI\":\"https://www.learnatnoon.com/s/a-pass-on-is-numbered-so-that-its-faces-show/6036/\",\"WARC-Payload-Digest\":\"sha1:KXZLIVADGHVCU4L42DV6SVMP45676GWE\",\"WARC-Block-Digest\":\"sha1:4MA22ZNN7XRLY4TGBL3HMAEE2SHSXOLB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500215.91_warc_CC-MAIN-20230205032040-20230205062040-00537.warc.gz\"}"}
https://javascript.tutorialink.com/opencv-js-perspective-transform/	[ "# opencv.js perspective transform\n\nI’m trying to use opencv.js to find a document in a provided image (detect edges, apply perspective transform, etc.\n\nI’ve got a reasonable set of code that (occasionally) detects edges of a document and grabs the bounding box for that. However, I’m struggling to do the perspective transform steps. There are some helpers for this (not in JS) here and here.\n\nUnfortunately I’m getting stuck on something simple. I can find the matching `Mat` that has 4 edges. Displaying that shows it to be accurate. However, I have no idea how to get some simple X/Y info out of that `Mat`. I thought `minMaxLoc()` would be a good option, but I keep getting an error passing in my matching `Mat`. Any idea why I can draw `foundContour` and get bounding box info from it, but I can’t call `minMaxLoc` on it?\n\nCode:\n\n```//<Get Image>\n//<Convert to Gray, do GaussianBlur, and do Canny edge detection>\nlet contours = new cv.MatVector();\ncv.findContours(matDestEdged, contours, hierarchy, cv.RETR_LIST, cv.CHAIN_APPROX_SIMPLE);\n\n//<Sort resulting contours by area to get largest>\n\nlet foundContour = null;\nfor (let sortableContour of sortableContours) {\nlet peri = cv.arcLength(sortableContour.contour, true);\nlet approx = new cv.Mat();\ncv.approxPolyDP(sortableContour.contour, approx, 0.1 * peri, true);\n\nif (approx.rows == 4) {\nconsole.log('found it');\nfoundContour = approx\nbreak;\n}\nelse {\napprox.delete();\n}\n}\n\n//<Draw foundContour and a bounding box to ensure it's accurate>\n\n//TODO: Do a perspective transform\nlet result = cv.minMaxLoc(foundContour);\n```\n\nThe last line above results in a runtime error (`Uncaught (in promise): 6402256 - Exception catching is disabled`). I can run `minMaxLoc()` on other `Mat` objects.\n\nFor anyone else looking to do this in OpenCV.JS, what I commented above seems to still be accurate. The contour found can’t be used with `minMaxLoc`, but the X/Y data can be pulled out of `data32S[]`. That should be all that’s needed to do this perspective transform. Some code is below.\n\n```//Find all contours\nlet contours = new cv.MatVector();\nlet hierarchy = new cv.Mat();\ncv.findContours(matDest, contours, hierarchy, cv.RETR_LIST, cv.CHAIN_APPROX_SIMPLE);\n\n//Get area for all contours so we can find the biggest\nlet sortableContours: SortableContour[] = [];\nfor (let i = 0; i < contours.size(); i++) {\nlet cnt = contours.get(i);\nlet area = cv.contourArea(cnt, false);\nlet perim = cv.arcLength(cnt, false);\n\nsortableContours.push(new SortableContour({ areaSize: area, perimiterSize: perim, contour: cnt }));\n}\n\n//Sort 'em\nsortableContours = sortableContours.sort((item1, item2) => { return (item1.areaSize > item2.areaSize) ? -1 : (item1.areaSize < item2.areaSize) ? 1 : 0; }).slice(0, 5);\n\n//Ensure the top area contour has 4 corners (NOTE: This is not a perfect science and likely needs more attention)\nlet approx = new cv.Mat();\ncv.approxPolyDP(sortableContours.contour, approx, .05 * sortableContours.perimiterSize, true);\n\nif (approx.rows == 4) {\nconsole.log('Found a 4-corner approx');\nfoundContour = approx;\n}\nelse{\nconsole.log('No 4-corner large contour!');\nreturn;\n}\n\n//Find the corners\n//foundCountour has 2 channels (seemingly x/y), has a depth of 4, and a type of 12. Seems to show it's a CV_32S \"type\", so the valid data is in data32S??\nlet corner1 = new cv.Point(foundContour.data32S, foundContour.data32S);\nlet corner2 = new cv.Point(foundContour.data32S, foundContour.data32S);\nlet corner3 = new cv.Point(foundContour.data32S, foundContour.data32S);\nlet corner4 = new cv.Point(foundContour.data32S, foundContour.data32S);\n\n//Order the corners\nlet cornerArray = [{ corner: corner1 }, { corner: corner2 }, { corner: corner3 }, { corner: corner4 }];\n//Sort by Y position (to get top-down)\ncornerArray.sort((item1, item2) => { return (item1.corner.y < item2.corner.y) ? -1 : (item1.corner.y > item2.corner.y) ? 1 : 0; }).slice(0, 5);\n\n//Determine left/right based on x position of top and bottom 2\nlet tl = cornerArray.corner.x < cornerArray.corner.x ? cornerArray : cornerArray;\nlet tr = cornerArray.corner.x > cornerArray.corner.x ? cornerArray : cornerArray;\nlet bl = cornerArray.corner.x < cornerArray.corner.x ? cornerArray : cornerArray;\nlet br = cornerArray.corner.x > cornerArray.corner.x ? cornerArray : cornerArray;\n\n//Calculate the max width/height\nlet widthBottom = Math.hypot(br.corner.x - bl.corner.x, br.corner.y - bl.corner.y);\nlet widthTop = Math.hypot(tr.corner.x - tl.corner.x, tr.corner.y - tl.corner.y);\nlet theWidth = (widthBottom > widthTop) ? widthBottom : widthTop;\nlet heightRight = Math.hypot(tr.corner.x - br.corner.x, tr.corner.y - br.corner.y);\nlet heightLeft = Math.hypot(tl.corner.x - bl.corner.x, tr.corner.y - bl.corner.y);\nlet theHeight = (heightRight > heightLeft) ? heightRight : heightLeft;\n\n//Transform!\nlet finalDestCoords = cv.matFromArray(4, 1, cv.CV_32FC2, [0, 0, theWidth - 1, 0, theWidth - 1, theHeight - 1, 0, theHeight - 1]); //\nlet srcCoords = cv.matFromArray(4, 1, cv.CV_32FC2, [tl.corner.x, tl.corner.y, tr.corner.x, tr.corner.y, br.corner.x, br.corner.y, bl.corner.x, bl.corner.y]);\nlet dsize = new cv.Size(theWidth, theHeight);\nlet M = cv.getPerspectiveTransform(srcCoords, finalDestCoords)\ncv.warpPerspective(matDestTransformed, finalDest, M, dsize, cv.INTER_LINEAR, cv.BORDER_CONSTANT, new cv.Scalar());\n```\n\nFor reference, here is the class definition I was using for `SortableContour`. The code above is meant as a guide, not as something that can run on its own, however.\n\n```export class SortableContour {\nperimiterSize: number;\nareaSize: number;\ncontour: any;\n\nconstructor(fields: Partial<SortableContour>) {\nObject.assign(this, fields);\n}\n}\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6003427,"math_prob":0.7680381,"size":5644,"snap":"2022-05-2022-21","text_gpt3_token_len":1602,"char_repetition_ratio":0.18457447,"word_repetition_ratio":0.0232859,"special_character_ratio":0.2834869,"punctuation_ratio":0.27483442,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9602185,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T01:55:01Z\",\"WARC-Record-ID\":\"<urn:uuid:4c612283-60c5-4d03-b4b5-596cda62f4da>\",\"Content-Length\":\"41563\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06d7d279-0231-4c97-8c29-601cc6fd858c>\",\"WARC-Concurrent-To\":\"<urn:uuid:1d880aef-37d9-4d3a-ac4c-910581be7084>\",\"WARC-IP-Address\":\"104.21.20.67\",\"WARC-Target-URI\":\"https://javascript.tutorialink.com/opencv-js-perspective-transform/\",\"WARC-Payload-Digest\":\"sha1:SS5542SDCSSFBPTIKIUBYYQUTZPC2EGZ\",\"WARC-Block-Digest\":\"sha1:OPRFMQPRE5WYZP2MBVVJJLGFWLGPNI2D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662595559.80_warc_CC-MAIN-20220526004200-20220526034200-00461.warc.gz\"}"}
http://crux.ms/commands/spectral-counts.html	[ "# spectral-counts\n\n## Usage:\n\ncrux spectral-counts [options] <input PSMs>\n\n## Description:\n\nGiven a collection of scored PSMs, produce a list of proteins or peptides ranked by a quantification score. Spectral-counts supports four types of quantification: Normalized Spectral Abundance Factor (NSAF), Distributed Normalized Spectral Abundance (dNSAF), Normalized Spectral Index (SIN) and Exponentially Modified Protein Abundance Index (emPAI). The NSAF method is from Paoletti et al. (2006). The SIN method is from Griffin et al. (2010). The emPAI method was first described in Ishihama et al (2005). The quantification methods are defined below and in the following paper:\n\nS McIlwain, M Mathews, M Bereman, EW Rubel, MJ MacCoss, and WS Noble. \"Estimating relative abundances of proteins from shotgun proteomics data.\" BMC Bioinformatics. 13:308, 2012.\n\n### Protein Quantification\n\n1. For each protein in a given database, the NSAF score is:\n$$NSAF_N=\\frac{S_N/L_N}{\\sum_{i=1}^ns_i/L_i}$$\nwhere:\n• N is protein index\n• SN is the number of peptide spectra matched to the protein\n• LN is the length of protein N\n• n is the total number of proteins in the input database\n2. For each protein in a given database, the dNSAF score is:\n$$NSAF_N=\\frac{\\frac{uSpc_N+(d)sSpc_N}{uL_N+sL_N}}{\\frac{uSpc_i+(d)sSpc_i}{uL_i+sL_i}}$$\nwhere:\n• N is the protein index\n• uSpcN is the unique number spectra matched to the protein index\n• sSpcN is the shared number peptide spectra matched to the protein index\n• LN is the length of protein N\n• n is the total number of proteins in the input database\n• d is the distribution factor of peptide K to protein N, given by\n$$d=\\frac{uSpc_N}{\\sum_{i=1}^nuSpc_i}$$\n3. For each protein in a given database, the SIN score is:\n$$SI_N=\\frac{\\sum_{j=1}^{p_N}(\\sum_{k=1}^{s_j}i_k)}{L_N(\\sum_{j=1}^nSI_j)}$$\nwhere:\n• N is protein index\n• pn is the number of unique peptides in protein N\n• sj is the number of spectra assigned to peptide j\n• ik is the total fragment ion intensity of spectrum k\n• LN is the length of protein N\n4. For each protein in a given database, the emPAI score is:\n$$emPAI=10^{\\frac{N_{observed}}{N_{observable}}}-1$$\nwhere:\n• Nobserved is the number of experimentally observed peptides with scores above a specified threshold.\n• Nobservable is the calculated number of observable peptides for the protein given the search constraints.\n\n### Peptide Quantification\n\n1. For each peptide in a given database, the NSAF score is:\n$$NSAF_N=\\frac{S_N/L_N}{\\sum_{i=1}^ns_i/L_i}$$\nwhere:\n• N is the peptide index\n• SN is the number spectra matched to peptide N\n• LN is the length of peptide N\n• n is the total number of peptides in the input database\n2. For each peptide in a given database, the SIN score is:\n$$SI_N=\\frac{(\\sum_{k=1}^{S_N}i_k)}{L_N(\\sum_{j=1}^nSI_J)}$$\nwhere:\n• N is the peptide index\n• SN is the number of spectra assigned to peptide N\n• ik is the total fragment ion intensity of spectrum k\n• LN is the length of peptide N\n\n## Input:\n\n• input PSMs – A PSM file in either tab delimited text format (as produced by percolator), or pepXML format.\n\n## Output:\n\nThe program writes files to the folder crux-output by default. The name of the output folder can be set by the user using the --output-dir option. The following files will be created:\n\n• spectral-counts.target.txt – a tab-delimited text file containing the protein IDs and their corresponding scores, in sorted order.\n• spectral-counts.params.txt – a file containing the name and value of all parameters/options for the current operation. Not all parameters in the file may have been used in the operation. The resulting file can be used with the --parameter-file option for other Crux programs.\n• spectral-counts.log.txt – All messages written to standard error.\n\n## Options:\n\n• ### spectral-counts options\n\n• --parsimony none\|simple\|greedy – Perform a parsimony analysis on the proteins, and report a \"parsimony rank\" column in the output file. This column contains integers indicating the protein's rank in a list sorted by spectral counts. If the parsimony analysis results in two proteins being merged, then their parsimony rank is the same. In such a case, the rank is assigned based on the largest spectral count of any protein in the merged meta-protein. The \"simple\" parsimony algorithm only merges two proteins A and B if the peptides identified in protein A are the same as or a subset of the peptides identified in protein B. The \"greedy\" parsimony algorithm does additional merging, by identifying the longest protein (i.e., the protein with the most peptides) that contains one or more shared peptides. The shared peptides are assigned to the identified protein and removed from any other proteins that contain them, and the process is then repeated. Note that, with this option, some proteins end up being assigned no peptides at all; these orphan proteins are not reported in the output. Default = none.\n• --threshold <float> – Only consider PSMs with a threshold value. By default, q-values are thresholded using a specified threshold value. This behavior can be changed using the --custom-threshold and --threshold-min parameters. Default = 0.01.\n• --threshold-type none\|qvalue\|custom – Determines what type of threshold to use when filtering matches. none : read all matches, qvalue : use calculated q-value from percolator, custom : use --custom-threshold-name and --custom-threshold-min parameters. Default = qvalue.\n• --input-ms2 <string> – MS2 file corresponding to the psm file. Required to measure the SIN. Ignored for NSAF, dNSAF and EMPAI. Default = <empty>.\n• --unique-mapping T\|F – Ignore peptides that map to multiple proteins. Default = false.\n• --quant-level protein\|peptide – Quantification at protein or peptide level. Default = protein.\n• --measure RAW\|NSAF\|dNSAF\|SIN\|EMPAI – Type of analysis to make on the match results: (RAW\|NSAF\|dNSAF\|SIN\|EMPAI). With exception of the RAW metric, the database of sequences need to be provided using --protein-database. Default = NSAF.\n• --custom-threshold-name <string> – Specify which field to apply the threshold to. The direction of the threshold (<= or >=) is governed by --custom-threshold-min. By default, the threshold applies to the q-value, specified by \"percolator q-value\", \"decoy q-value (xcorr)\". Default = <empty>.\n• --custom-threshold-min T\|F – When selecting matches with a custom threshold, custom-threshold-min determines whether to filter matches with custom-threshold-name values that are greater-than or equal (F) or less-than or equal (T) than the threshold. Default = true.\n• --mzid-use-pass-threshold T\|F – Use mzid's passThreshold attribute to filter matches. Default = false.\n• --protein-database <string> – The name of the file in FASTA format. Default = <empty>.\n• ### Input and output\n\n• --verbosity <integer> – Specify the verbosity of the current processes. Each level prints the following messages, including all those at lower verbosity levels: 0-fatal errors, 10-non-fatal errors, 20-warnings, 30-information on the progress of execution, 40-more progress information, 50-debug info, 60-detailed debug info. Default = 30.\n• --parameter-file <string> – A file containing parameters. See the parameter documentation page for details. Default = <empty>.\n• --spectrum-parser pwiz\|mstoolkit – Specify the parser to use for reading in MS/MS spectra. The default, ProteoWizard parser can read the MS/MS file formats listed here. The alternative is MSToolkit parser. If the ProteoWizard parser fails to read your files properly, you may want to try the MSToolkit parser instead. Default = pwiz.\n• --fileroot <string> – The fileroot string will be added as a prefix to all output file names. Default = <empty>.\n• --output-dir <string> – The name of the directory where output files will be created. Default = crux-output.\n• --overwrite T\|F – Replace existing files if true or fail when trying to overwrite a file if false. Default = false." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76553607,"math_prob":0.81404495,"size":7817,"snap":"2021-21-2021-25","text_gpt3_token_len":1975,"char_repetition_ratio":0.15295021,"word_repetition_ratio":0.14652957,"special_character_ratio":0.2362799,"punctuation_ratio":0.10846746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9665535,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T08:43:55Z\",\"WARC-Record-ID\":\"<urn:uuid:e8edddd6-458e-495e-8379-fe7ea3d68600>\",\"Content-Length\":\"15536\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8eaf5992-986a-40e1-a307-968f0ba1fa7c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e789043-2a9a-4e02-8ee6-31ee1ee3e6e0>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://crux.ms/commands/spectral-counts.html\",\"WARC-Payload-Digest\":\"sha1:WNSTCJL4YVGX752VHTQOJ5QX6ZLOH6NU\",\"WARC-Block-Digest\":\"sha1:HKFBBETMQHOMEMJO5GYQHYJWPNQTF5E3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988753.91_warc_CC-MAIN-20210506083716-20210506113716-00245.warc.gz\"}"}