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https://www.dailygk.in/2016/06/reasoning-quiz-46-for-sbi-po-clerk.html	[ "Daily GK - Free Current Affairs (GK Quiz)\n\nDaily GK Updates & Daily News in India with free current affairs. You will get free Current affairs on Daily Basis and GK questions which will help you in Banking and SSC Exams. Daily News in India. Check this Current affair page on daily basis for the improvement of General awareness section. Free GK Quiz on Daily Basis.\n\nTuesday, June 14, 2016\n\nREASONING QUIZ-46 FOR SBI PO & CLERK\n\nREASONING QUIZ-46 FOR SBI PO & CLERK\n\nDirections (Q. 1–2): Study the following information carefully and answer the questions given below:\nL is father of P who is son of S. M is grandmother of R who is son of O. N is paternal uncle of R who is brother of S.\n\n1. How is O related to L?\n1) Brother-in-law\n2) Brother\n3) Can’t be determine\n4) None of the above\n\n2. According to the above information how is S related to R?\n1) Paternal Aunt\n2) Maternal Aunt\n3) Niece\n4) Nephew\n\nDirections (Q. 3-5): Study the following information carefully and answer the questions given below:\nEight persons A, B, C, D, E, F, G and H are sitting in a straight line facing North, but not necessarily in the same order. B sits second to the left of A. Three persons are sitting between G and B. G sits on the extreme left end of the row. D is an immediate neighbor of G. F and B are not neighbours. C sits second to the right of D but is not the immediate neighbor of E. There are three persons between C and F.\n\n3. Who is the sitting 5th right of the 2nd to left of the above arrangement?\n(1)F\n(2)E\n(3)B\n(4)H\n(5)None of these\n\n4. If all the letters are arranged in alphabetical order from left end, then how many letters will obtain the same position after rearrangement?\n(1) Two\n(2) Three\n(3) Four\n(4) None\n\n5. How many meaningful words can be formed by using 2nd, 6th, 7th,8th letters of the above arrangement(using each letter once only)?\n(1) Three\n(2) Two\n(3) One\n(d) No meaningful words can be formed\n\nDirections (Q. 6– 10): In each question, relationship between different elements is shown in the statements. The statements are followed by conclusions. Study the conclusions based on the given statements and select the appropriate answer.\n\n6. Statements: B > E ≥ A < L; T > A > S\nConclusions:\nI. E ≥ T\nII. E<T\n\n(1) Both conclusion I and II are true.\n(2) Either conclusion I or II is true.\n(3) Conclusion II is true.\n(4) Neither conclusion I nor II is true\n(5) Only conclusion I is true.\n7. Statements: L = I ≥ N < E; N ≥ S\nConclusions:\nI. L >N\nII. L=N\n\n(1) Either conclusion I or II is true.\n(2) Neither conclusion I nor II is true.\n(3) Only conclusion II is true.\n(4) Both conclusion I and II are true.\n(5) Only conclusion I is true.\n\n8. Statements: V < E > B = H ≥ N; B ≤ T\nConclusions:\nI. B ≥N\nII. N ≤ T\n\n(1) Either conclusion I or II is true.\n(2) Only conclusion II is true.\n(3) Neither conclusion I nor II is true.\n(4) Only conclusion I is true.\n(5) Both conclusion I and II are true.\n\n9. Statements: F = K ≥ N < E; F < T\nConclusions:\nI. T>N\nII. F≥N\n\n(1) Only conclusion II is true.\n(2) Either conclusion I or II is true.\n(3) Both conclusion I and II are true.\n(4) Neither conclusion I nor II is true.\n(5) None follow\n\n10. Statements: J > E ≥ Z < L; Z ≥ A > S\nConclusions:\nI. J ≥ S\nII. E = L\n(1) Only conclusion II is true.\n(2) Both conclusion I and II is true.\n(3) Either conclusion I or II is true.\n(4) Neither conclusion I nor II is true.\n(5) Only conclusion I is true." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86659354,"math_prob":0.76022327,"size":3033,"snap":"2022-05-2022-21","text_gpt3_token_len":961,"char_repetition_ratio":0.22119512,"word_repetition_ratio":0.16202946,"special_character_ratio":0.31585887,"punctuation_ratio":0.16120219,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9628772,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-29T11:18:48Z\",\"WARC-Record-ID\":\"<urn:uuid:0b665cf3-4086-4108-b678-b234ceb63255>\",\"Content-Length\":\"150152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77a3b063-d4a3-4324-911d-41cb33e14f77>\",\"WARC-Concurrent-To\":\"<urn:uuid:4656ced0-a9c9-48b1-9252-ebc0f47e0c94>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://www.dailygk.in/2016/06/reasoning-quiz-46-for-sbi-po-clerk.html\",\"WARC-Payload-Digest\":\"sha1:45Z6IP4WAD4ISYRRTHGRLCHJTQIBB3HB\",\"WARC-Block-Digest\":\"sha1:L7DG6QMOR2GLPTBFJXVCCHPQH26SCF2W\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304883.8_warc_CC-MAIN-20220129092458-20220129122458-00074.warc.gz\"}"}
https://www.colorhexa.com/03729b	[ "# #03729b Color Information\n\nIn a RGB color space, hex #03729b is composed of 1.2% red, 44.7% green and 60.8% blue. Whereas in a CMYK color space, it is composed of 98.1% cyan, 26.5% magenta, 0% yellow and 39.2% black. It has a hue angle of 196.2 degrees, a saturation of 96.2% and a lightness of 31%. #03729b color hex could be obtained by blending #06e4ff with #000037. Closest websafe color is: #006699.\n\n• R 1\n• G 45\n• B 61\nRGB color chart\n• C 98\n• M 26\n• Y 0\n• K 39\nCMYK color chart\n\n#03729b color description : Dark blue.\n\n# #03729b Color Conversion\n\nThe hexadecimal color #03729b has RGB values of R:3, G:114, B:155 and CMYK values of C:0.98, M:0.26, Y:0, K:0.39. Its decimal value is 225947.\n\nHex triplet RGB Decimal 03729b `#03729b` 3, 114, 155 `rgb(3,114,155)` 1.2, 44.7, 60.8 `rgb(1.2%,44.7%,60.8%)` 98, 26, 0, 39 196.2°, 96.2, 31 `hsl(196.2,96.2%,31%)` 196.2°, 98.1, 60.8 006699 `#006699`\nCIE-LAB 44.828, -11.57, -29.684 11.97, 14.419, 33.161 0.201, 0.242, 14.419 44.828, 31.859, 248.706 44.828, -30.16, -42.177 37.973, -10.186, -25.196 00000011, 01110010, 10011011\n\n# Color Schemes with #03729b\n\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #9b2c03\n``#9b2c03` `rgb(155,44,3)``\nComplementary Color\n• #039b78\n``#039b78` `rgb(3,155,120)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #03269b\n``#03269b` `rgb(3,38,155)``\nAnalogous Color\n• #9b7803\n``#9b7803` `rgb(155,120,3)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #9b0326\n``#9b0326` `rgb(155,3,38)``\nSplit Complementary Color\n• #729b03\n``#729b03` `rgb(114,155,3)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #9b0372\n``#9b0372` `rgb(155,3,114)``\n• #039b2c\n``#039b2c` `rgb(3,155,44)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #9b0372\n``#9b0372` `rgb(155,3,114)``\n• #9b2c03\n``#9b2c03` `rgb(155,44,3)``\n• #023b50\n``#023b50` `rgb(2,59,80)``\n• #024d69\n``#024d69` `rgb(2,77,105)``\n• #036082\n``#036082` `rgb(3,96,130)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #0384b4\n``#0384b4` `rgb(3,132,180)``\n• #0497cd\n``#0497cd` `rgb(4,151,205)``\n• #04a9e6\n``#04a9e6` `rgb(4,169,230)``\nMonochromatic Color\n\n# Alternatives to #03729b\n\nBelow, you can see some colors close to #03729b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #03989b\n``#03989b` `rgb(3,152,155)``\n• #038b9b\n``#038b9b` `rgb(3,139,155)``\n• #037f9b\n``#037f9b` `rgb(3,127,155)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #03659b\n``#03659b` `rgb(3,101,155)``\n• #03599b\n``#03599b` `rgb(3,89,155)``\n• #034c9b\n``#034c9b` `rgb(3,76,155)``\nSimilar Colors\n\n# #03729b Preview\n\nThis text has a font color of #03729b.\n\n``<span style=\"color:#03729b;\">Text here</span>``\n#03729b background color\n\nThis paragraph has a background color of #03729b.\n\n``<p style=\"background-color:#03729b;\">Content here</p>``\n#03729b border color\n\nThis element has a border color of #03729b.\n\n``<div style=\"border:1px solid #03729b;\">Content here</div>``\nCSS codes\n``.text {color:#03729b;}``\n``.background {background-color:#03729b;}``\n``.border {border:1px solid #03729b;}``\n\n# Shades and Tints of #03729b\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000101 is the darkest color, while #edfaff is the lightest one.\n\n• #000101\n``#000101` `rgb(0,1,1)``\n• #000f14\n``#000f14` `rgb(0,15,20)``\n• #011d28\n``#011d28` `rgb(1,29,40)``\n• #012b3b\n``#012b3b` `rgb(1,43,59)``\n• #02394e\n``#02394e` `rgb(2,57,78)``\n• #024861\n``#024861` `rgb(2,72,97)``\n• #025675\n``#025675` `rgb(2,86,117)``\n• #036488\n``#036488` `rgb(3,100,136)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\n• #0380ae\n``#0380ae` `rgb(3,128,174)``\n• #048ec1\n``#048ec1` `rgb(4,142,193)``\n• #049cd5\n``#049cd5` `rgb(4,156,213)``\n• #04abe8\n``#04abe8` `rgb(4,171,232)``\n• #06b8fa\n``#06b8fa` `rgb(6,184,250)``\n• #19befb\n``#19befb` `rgb(25,190,251)``\n• #2cc3fb\n``#2cc3fb` `rgb(44,195,251)``\n• #40c9fb\n``#40c9fb` `rgb(64,201,251)``\n• #53cefc\n``#53cefc` `rgb(83,206,252)``\n• #66d4fc\n``#66d4fc` `rgb(102,212,252)``\n• #79d9fc\n``#79d9fc` `rgb(121,217,252)``\n• #8ddffd\n``#8ddffd` `rgb(141,223,253)``\n• #a0e4fd\n``#a0e4fd` `rgb(160,228,253)``\n• #b3e9fe\n``#b3e9fe` `rgb(179,233,254)``\n• #c6effe\n``#c6effe` `rgb(198,239,254)``\n• #daf4fe\n``#daf4fe` `rgb(218,244,254)``\n• #edfaff\n``#edfaff` `rgb(237,250,255)``\nTint Color Variation\n\n# Tones of #03729b\n\nA tone is produced by adding gray to any pure hue. In this case, #4c5052 is the less saturated color, while #03729b is the most saturated one.\n\n• #4c5052\n``#4c5052` `rgb(76,80,82)``\n• #465358\n``#465358` `rgb(70,83,88)``\n• #40565e\n``#40565e` `rgb(64,86,94)``\n• #3a5964\n``#3a5964` `rgb(58,89,100)``\n• #345c6a\n``#345c6a` `rgb(52,92,106)``\n• #2e5e70\n``#2e5e70` `rgb(46,94,112)``\n• #276177\n``#276177` `rgb(39,97,119)``\n• #21647d\n``#21647d` `rgb(33,100,125)``\n• #1b6783\n``#1b6783` `rgb(27,103,131)``\n• #156a89\n``#156a89` `rgb(21,106,137)``\n• #0f6c8f\n``#0f6c8f` `rgb(15,108,143)``\n• #096f95\n``#096f95` `rgb(9,111,149)``\n• #03729b\n``#03729b` `rgb(3,114,155)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #03729b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50901,"math_prob":0.6823155,"size":3674,"snap":"2019-43-2019-47","text_gpt3_token_len":1610,"char_repetition_ratio":0.12588556,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5647795,"punctuation_ratio":0.23634337,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9930873,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T02:25:51Z\",\"WARC-Record-ID\":\"<urn:uuid:ff5a51dd-5de5-4a3d-b0cb-41102865c61e>\",\"Content-Length\":\"36239\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a94da9f6-a860-4d4b-b791-c8b5a82493ca>\",\"WARC-Concurrent-To\":\"<urn:uuid:cde6df5d-948a-4933-9e98-458e44beca79>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/03729b\",\"WARC-Payload-Digest\":\"sha1:725AFBJI3H2IPK7YKFLGYHC7VRPRO2AN\",\"WARC-Block-Digest\":\"sha1:LE3NQ4KBZRKZOBKTMFUZ4BKEECTYUBBR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665976.26_warc_CC-MAIN-20191113012959-20191113040959-00353.warc.gz\"}"}
https://studyres.com/doc/8081256/general-relativity	[ "• Study Resource\n• Explore\n\nSurvey\n\n* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nAstronomical spectroscopy wikipedia, lookup\n\nOuter space wikipedia, lookup\n\nGravitational lens wikipedia, lookup\n\nFirst observation of gravitational waves wikipedia, lookup\n\nWormhole wikipedia, lookup\n\nGravitational microlensing wikipedia, lookup\n\nTranscript\n```General relativity\n• Special relativity applies only to observers moving with\nconstant velocity.\n• Can we generalize relativity to accelerated observers ?\n• No problem for Einstein, just a few more thought\nexperiments …\nCh. 11.1\nThe forces from acceleration and gravity\ncannot be distinguished\nGravity\nIn an accelerating spaceship\na stone drops just like Newton’s apple\nA stone thrown sideways follows an arc\n(again just like Newton)\nLight follows an arc, too\n(that’s Einstein)\nEinstein’s logic\nStar’s\nreal\nposition\nStar’s\napparent\nposition\n1) Acceleration and gravity\nStarlight deflected\nby Sun’s gravity\ncannot be distinguished.\n2) Acceleration deflects light.\n3) Therefore, gravity should\ndeflect light.\nConfirmed by experiment:\nStars can be detected close to\nthe Sun during a solar eclipse.\nTheir light is indeed deflected\nby the Sun’s gravity.\nMoon\nEddington’s Eclipse Expedition 1919\n• British astronomer Eddington\ntraveled to Principe Island in\nthe Gulf of Guinea to observe\ndeflection of starlight during\na solar eclipse.\n• After months of drought it was\npouring rain on the day of the\neclipse. The clouds parted just\nin time to see the eclipse.\n• The photographs revealed an\napparent deflection of stars\naway from the Sun, as predicted by Einstein.\n• Einstein was instantly famous.\nGravitational\nlensing\n• The Sun acts like\na lens deflecting\nthe light from\nstars behind it.\n• A cluster of galaxies can also\nact as lens by\ndeflecting light\nfrom galaxies\nbehind it.\n• The mass of the\ngalaxy cluster is\nobtained from\nfaint arcs produced by a lens.\nGravitational lensing simulation\nWhen a black hole with the mass of Saturn is placed in front\nof the two central towers , gravitational lensing bends them\ninto arcs. This explains the faint arcs in the Hubble image.\nCurved space-time\n• The bending of light by the gravity of the Sun\nbrings up the question: What is a straight line?\n• Laser beams are our best method to produce a\nstraight line.\n• If light does not go straight, what else would ?\n• We conclude that space itself must be curved.\nIt does not allow a straight line. The path of a\nlight beam is the straightest possible path.\nA light clock slows down in curved space\nMirror\nMirror\nLight\nbeam\nIn curved space, light takes a detour compared to flat space\n(dotted line). The detour increases the time to complete one\ncycle of a light clock (Lect. 14, Slide 3). The clock slows down.\nGravitational time dilation\n• Gravity warps both space and time!\n• Clocks run more slowly in a gravitational force field.\n• In a GPS satellite the clock speeds up during launch\ninto orbit , where gravity is reduced. The satellite\nclock needs to be set slightly slow before launch to\ncompensate (by a factor of 4.5 10-10).\n• This effect is much larger than the time dilation due\nto special relativity (Lect. 14, Slide 7), which acts in\nthe opposite direction.\nCurved space explains the planet’s orbits\n• In general relativity, gravity is a warp in space.\n• The orbits of the planets and comets can be explained\nby their movement around the dimple in space that is\ncreated by the mass of the Sun.\nEinstein’s equations of gravity\n• Einstein’s equations of gravity are rather complicated.\nEven Einstein needed help from expert mathematicians.\n• Einstein’s equations in words (by John Wheeler):\nMatter tells space how to curve,\nand space tells matter how to move.\n• General relativity connects space and matter, concepts\nthat seem to have nothing in common.\n• General relativity is reduced to special relativity in the\nabsence of gravity or acceleration. It includes the results\nof special relativity on length contraction, time dilation.\n```\nRelated documents" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9083226,"math_prob":0.8194045,"size":3755,"snap":"2019-51-2020-05","text_gpt3_token_len":786,"char_repetition_ratio":0.12796588,"word_repetition_ratio":0.0,"special_character_ratio":0.20639148,"punctuation_ratio":0.09224012,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96169466,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T01:14:58Z\",\"WARC-Record-ID\":\"<urn:uuid:66b824aa-f55e-4c44-8fb8-16221043f358>\",\"Content-Length\":\"87648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05a51839-558d-47a4-83c3-9aae3048073c>\",\"WARC-Concurrent-To\":\"<urn:uuid:89963719-8624-477d-845d-1e04900748bf>\",\"WARC-IP-Address\":\"104.28.20.25\",\"WARC-Target-URI\":\"https://studyres.com/doc/8081256/general-relativity\",\"WARC-Payload-Digest\":\"sha1:CVYZYTU2NUMNYAMJGEVPGY2WWLBMULNU\",\"WARC-Block-Digest\":\"sha1:GC3H7OPIMOZMKTES737ARIWT57T4SN7B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540529516.84_warc_CC-MAIN-20191210233444-20191211021444-00163.warc.gz\"}"}
https://www.guyuehome.com/40196	[ "## 双模型法离线神经网络辨识非线性动态系统\n\n_本文是我基于自己的理解实现的双模型法离线神经网络便是，可能不太正确，但仍记录下来。_", null, "### 辨识的具体流程\n\ny(k+1)=(y(k))/(1+y(k)*y(k))\ny(k+1)=u^3 (k)", null, "", null, "", null, "", null, "", null, "", null, "（需要设置采样频率，示波器中和unit delay中设置为0.001，采样时间为1ms）。将图1的非线性动态系统的输入u(k)、u(k+1)、y(k)、y(k+1)导出到工作区。（array数组形式）\n\nnnstart", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "" ]	[ null, "https://img-blog.csdnimg.cn/img_convert/7d40afe158368f99baf779fd2867cc38.png", null, "https://img-blog.csdnimg.cn/img_convert/c95ac695b4c8648cdac547431abc98c0.png", null, "https://img-blog.csdnimg.cn/img_convert/710b577bf1b351a55a615b85e5bebb29.png", null, "https://img-blog.csdnimg.cn/img_convert/d2897462fcb2fa678e29e3923ecac2be.png", null, "https://img-blog.csdnimg.cn/img_convert/7c3b770c4d746e4fbb8279d19a753cd9.png", null, "https://img-blog.csdnimg.cn/img_convert/269322478b6dd42998053e26ae4061ed.png", null, "https://img-blog.csdnimg.cn/img_convert/e1f37d5f103ab468930c9f654e09252d.png", null, "https://img-blog.csdnimg.cn/img_convert/f44a01cd98ef0e0345e446452ea5325a.png", null, "https://img-blog.csdnimg.cn/img_convert/98ee0e066f116a1af1e75b491a07abe4.png", null, "https://img-blog.csdnimg.cn/img_convert/88436e26bb30503bf0ff3515c7d6c2c4.png", null, "https://img-blog.csdnimg.cn/img_convert/64f165506f7c9f7af9c2f9f06eefa333.png", null, "https://img-blog.csdnimg.cn/img_convert/7ea957b9845561f68996aa94ae3bdb06.png", null, "https://img-blog.csdnimg.cn/img_convert/b7016c709474efa080399518fc66a939.png", null, "https://img-blog.csdnimg.cn/img_convert/51ef285381a8890472a30df3ccc0a0d3.png", null, "https://img-blog.csdnimg.cn/img_convert/508aec4c1ca7dae7892d200462ba04c8.png", null, "https://img-blog.csdnimg.cn/img_convert/bb3f941c611f56249b50f7d20f312d2f.png", null, "https://img-blog.csdnimg.cn/img_convert/5e05b1f572e988994c91589314e9754a.png", null, "https://img-blog.csdnimg.cn/img_convert/5bff39fa992261a173c5a2b049f1a3c6.png", null, "https://img-blog.csdnimg.cn/img_convert/61301499399e99f9de30f9627862d2bf.png", null, "https://img-blog.csdnimg.cn/img_convert/ca934612b3c5b0c76a48a0036819cc47.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.945046,"math_prob":0.9970105,"size":768,"snap":"2022-40-2023-06","text_gpt3_token_len":662,"char_repetition_ratio":0.087696336,"word_repetition_ratio":0.0,"special_character_ratio":0.17708333,"punctuation_ratio":0.011111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9582355,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-02T08:21:25Z\",\"WARC-Record-ID\":\"<urn:uuid:1546f809-82be-4c2a-b66a-e9f375e1ba30>\",\"Content-Length\":\"133824\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:300d9fc5-9378-4a36-8322-c5a619287c32>\",\"WARC-Concurrent-To\":\"<urn:uuid:91e0a75c-f838-4b04-bf27-8006b15e863f>\",\"WARC-IP-Address\":\"39.105.183.248\",\"WARC-Target-URI\":\"https://www.guyuehome.com/40196\",\"WARC-Payload-Digest\":\"sha1:TSK2ODRTJSPBW4EQFDPYECYUQCDJACQA\",\"WARC-Block-Digest\":\"sha1:TBQDZ67FJ3HUWCJORPA7X4GCJRRUWRIU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499967.46_warc_CC-MAIN-20230202070522-20230202100522-00617.warc.gz\"}"}
https://goprep.co/ex-3.e-q3-a-lady-has-only-25-paisa-and-50-paisa-coins-in-her-i-1njywy	[ "Q. 3\n\n# A lady has only 2\n\nLet the no. of 25 - paisa coins be x.", null, "the no of 50 - paisa coins = 50 - x\n\n[ the total no. of coins = 50]\n\nAccording to the question,\n\ntotal money = Rs. 19.50 = 1950 paise\n\n25x + 50(50 - x) = 1950\n\n2500 - 25x = 1950\n\n25x = 550\n\nx = 22\n\nThus, the no of 25 - paisa coins = x = 22 and,\n\nthe no of 50 - paisa coins = 50 - x = 50 - 22 = 28.\n\nRate this question :\n\nHow useful is this solution?\nWe strive to provide quality solutions. Please rate us to serve you better.\nTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts\nDedicated counsellor for each student\n24X7 Doubt Resolution\nDaily Report Card\nDetailed Performance Evaluation", null, "view all courses", null, "RELATED QUESTIONS :\n\nSolve each of theRS Aggarwal - Mathematics\n\nIf 2x – 3y = 7 anRD Sharma - Mathematics\n\nSolve the followiRD Sharma - Mathematics\n\nSolve each of theRD Sharma - Mathematics" ]	[ null, "https://gradeup-question-images.grdp.co/liveData/PROJ11416/1514360381661867.png", null, "https://grdp.co/cdn-cgi/image/height=128,quality=80,f=auto/https://gs-post-images.grdp.co/2020/8/group-7-3x-img1597928525711-15.png-rs-high-webp.png", null, "https://gs-post-images.grdp.co/2020/8/group-img1597139979159-33.png-rs-high-webp.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77197695,"math_prob":0.98175555,"size":914,"snap":"2021-43-2021-49","text_gpt3_token_len":301,"char_repetition_ratio":0.12307692,"word_repetition_ratio":0.086021505,"special_character_ratio":0.3358862,"punctuation_ratio":0.10465116,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9608814,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T03:42:37Z\",\"WARC-Record-ID\":\"<urn:uuid:9214c382-82d7-4bdd-bf5c-490950e43455>\",\"Content-Length\":\"309871\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4350d963-d501-44a6-ac42-ff579cedc6e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:48e750d8-0f7d-41e1-98dd-e4de9e6be082>\",\"WARC-IP-Address\":\"104.26.15.48\",\"WARC-Target-URI\":\"https://goprep.co/ex-3.e-q3-a-lady-has-only-25-paisa-and-50-paisa-coins-in-her-i-1njywy\",\"WARC-Payload-Digest\":\"sha1:M37X7IJTOUXMJDQUIM5ZKLTRNIMKIY4A\",\"WARC-Block-Digest\":\"sha1:QJRK6OWAMJYXUPJXEAYHJLR3FUFFNG4L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585196.73_warc_CC-MAIN-20211018031901-20211018061901-00598.warc.gz\"}"}
https://eqblog.com/python-disct-use.html	[ "```zidian={\n\"key1\":{\n\"keys1\":{\"a\",\"b\",\"c\"},\n\"keys2\":{\"d\",\"e\",\"f\"}\n},\n\"key2\":{\n\"keys3\":{\"g\",\"h\"}\n}\n}```\n\n`zidian[\"key1\"][\"keys1\"]`\n\n```del zidian[\"key2\"] #删除key2\nzidian[\"key1\"][\"keys1\"]=\"a\" #将key1键中的值修改为\"a\",其它内置函数下面介绍```\n\n```dict字典内置函数的使用\n1、dict()\n2、clear()字典清空\n3、keys()得到字典的键列表\n4、values()得到字典的值列表\n5、items()得到字典的项，即键值对\n6、D.get(key[,d]) 如果key在字典中在返回D[key],如果不存在在返回d，的默认为None\n7、pop(key[,d])如果key存在，则将对应的项删除，返回该项；如果key不存在，并且没有d参数则返回错误提示，如果传入d参数则显示d的内容\n8、popitem()删除字典中的一项，返回该删除项，如果字典为空则会报错\n9、D.update([E,]**F) E是可选参数，F是关键字参数。用可迭代对象或者一个字典的E去更新原字典;如个E存在并且有E.keys()方法，则k in E.keys() ,D[K]=E[K];\n\n10、D.copy()浅拷贝\n11、fromkeys(interable,value=none)\n\n'''\n# dict的创建方法\n# 直接通过花括号括起来，里面存放键值对，key:value，用逗号隔开\ndict1={'name':'lsm','age':30,'job':'teacher'} #dict是无序的\nprint(dict1)\n#通过dict函数创建一个字典\n#dict函数的参数有三种方式\n# 1、mapping,一个mapping对象（key,value）对\ndict2=dict(((1,'one'),(2,'tow'),(3,'three')))\nprint(dict2)\n# 2、可迭代对象 list,tuple\nlist1=[('name','cxf'),('age',28),('job','economic')]\ndict3=dict(list1)\nprint(dict3)\n#3、关键字参数\ndict4=dict(a='java',b='python',c='javascript')\nprint(dict4)\n# 4、得到字典的键列表\nprint(list(dict1.keys()))\n# 5、得到字典的项，一个键值对的元组列表\nprint(dict1.items())\n# 6、获取输入的key对应的value\nprint(dict1.get('name'))\nprint(dict1.get('city')) #不存在则返回None\nprint(dict1.get('interest','coding')) #如果有传入第二个参数则在key不存在的时候返回第二个参数的内容\n# 7、删除key对应的项，返回该项\ndict1.pop('name')\nprint(dict1)\nprint(dict1.pop('class','该项不存在')) #这种方式比较好，如果key不存在则会返回自定义提示内容\n# 8、删除一项\ndict1.popitem()\nprint(dict1)\n# 9、清空字典\ndict1.clear()\n\nanother_dict=dict2 #创建另外一个字典指向dict2所指向的字典\ndict2={} #采用这种方式也可以将字典元素置为空，但实际上只是将dict2重新指向了一个空的字典，原来字典的元素还存在内存中，这是一个不保险的方法\nprint(another_dict) #字典的元素还是存在的\nanother_dict.clear() #如果用clear方法则会将字典元素从内存清空\nprint(another_dict)\n\nprint(dict1)\n# 10、更新列表\nupdate_list=[('name','llx'),('age',20),('interest','play'),('city','amoi')] #可以是一个列表\nupdate_dict={'job':'engineer','school':'xiamen university'} #更可以是一个字典\ndict1.update(update_list)\nprint(dict1)\ndict1.update(update_dict)\nprint(dict1)\n\n#用关键字更新字典\ndict1.update(skill='python')\nprint(dict1)\n\n#11、字典的拷贝\ncopy_dict1=dict1.copy()\npoint_dict1=dict1 #用另外一个变量指向dict1\nprint('用另一个变量指向dict1的内容为：',point_dict1,'地址为:',id(point_dict1),'原dict的地址为：',id(dict1)) #可见用另一个变量所指向的地址跟原dict1所指向的地址一致\n\nprint('浅拷贝dict1的内容为：',copy_dict1,'地址为:',id(copy_dict1))\n\n# 12、fromkeys,第一个参数是一个序列，第二个参数是value值，会生成一个以序列为key，值全部为value的字典\ndict5=dict()\ndict5=dict5.fromkeys(range(10),'good')\nprint(dict5)```" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.6270754,"math_prob":0.9505401,"size":2531,"snap":"2019-51-2020-05","text_gpt3_token_len":1265,"char_repetition_ratio":0.15037593,"word_repetition_ratio":0.0,"special_character_ratio":0.3156855,"punctuation_ratio":0.2118451,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95342183,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T22:09:07Z\",\"WARC-Record-ID\":\"<urn:uuid:6ee181a3-c25d-49eb-8f46-35a0bbfaf7f7>\",\"Content-Length\":\"28463\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df987f42-01f5-44e8-b621-7c6da7baeeb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b5e34d15-b47f-45d4-8672-8ceae813e50a>\",\"WARC-IP-Address\":\"119.28.52.53\",\"WARC-Target-URI\":\"https://eqblog.com/python-disct-use.html\",\"WARC-Payload-Digest\":\"sha1:6VNBFSVTWEDPHTJWCNFZPJOADSEEJRLE\",\"WARC-Block-Digest\":\"sha1:MPXILEXU2GABLPDRKW375BSLG5DMY3EO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250599789.45_warc_CC-MAIN-20200120195035-20200120224035-00499.warc.gz\"}"}
https://www.geeksforgeeks.org/program-to-check-the-number-is-palindrome-or-not/?ref=rp	[ "Related Articles\n\n# Program to check the number is Palindrome or not\n\n• Difficulty Level : Basic\n• Last Updated : 14 Aug, 2021\n\nGiven an integer N, write a program that returns true if the given number is a palindrome, else return false.\nExamples:\n\n```Input: N = 2002\nOutput: true\n\nInput: N = 1234\nOutput: false```", null, "Approach:\nA simple method for this problem is to first reverse digits of n, then compare the reverse of n with n. If both are same, then return true, else false.\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C program to check whether a number``// is Palindrome or not.` `#include ` `/* Iterative function to reverse digits of num/``int` `reverseDigits(``int` `num)``{`` ``int` `rev_num = 0;`` ``while` `(num > 0) {`` ``rev_num = rev_num 10 + num % 10;`` ``num = num / 10;`` ``}`` ``return` `rev_num;``}` `/* Function to check if n is Palindrome/``int` `isPalindrome(``int` `n)``{` ` ``// get the reverse of n`` ``int` `rev_n = reverseDigits(n);` ` ``// Check if rev_n and n are same or not.`` ``if` `(rev_n == n)`` ``return` `1;`` ``else`` ``return` `0;``}` `/Driver program to test reversDigits/``int` `main()``{`` ``int` `n = 4562;`` ``printf``(``\"Is %d a Palindrome number? -> %s\\n\"``, n,`` ``isPalindrome(n) == 1 ? ``\"true\"` `: ``\"false\"``);` ` ``n = 2002;`` ``printf``(``\"Is %d a Palindrome number? -> %s\\n\"``, n,`` ``isPalindrome(n) == 1 ? ``\"true\"` `: ``\"false\"``);`` ``return` `0;``}`\n\n## Java\n\n `// Java program to check whether a number``// is Palindrome or not.` `class` `GFG``{`` ``/ Iterative function to reverse digits of num/`` ``static` `int` `reverseDigits(``int` `num)`` ``{`` ``int` `rev_num = ``0``;`` ``while` `(num > ``0``) {`` ``rev_num = rev_num ``10` `+ num % ``10``;`` ``num = num / ``10``;`` ``}`` ``return` `rev_num;`` ``}`` ` ` ``/* Function to check if n is Palindrome/`` ``static` `int` `isPalindrome(``int` `n)`` ``{`` ` ` ``// get the reverse of n`` ``int` `rev_n = reverseDigits(n);`` ` ` ``// Check if rev_n and n are same or not.`` ``if` `(rev_n == n)`` ``return` `1``;`` ``else`` ``return` `0``;`` ``}`` ` ` ``/Driver program to test reversDigits/`` ``public` `static` `void` `main(String []args)`` ``{`` ``int` `n = ``4562``;`` ``System.out.println(``\"Is\"` `+ n + ``\"a Palindrome number? -> \"` `+`` ``(isPalindrome(n) == ``1` `? ``\"true\"` `: ``\"false\"``));`` ` ` ``n = ``2002``;`` ` ` ``System.out.println(``\"Is\"` `+ n + ``\"a Palindrome number? -> \"` `+`` ``(isPalindrome(n) == ``1` `? ``\"true\"` `: ``\"false\"``));` ` ``}` `}` `// This code is contributed``// by Hritik Raj ( ihritik )`\n\n## Python3\n\n `# Python3 program to check whether a``# number is Palindrome or not.` `# Iterative function to reverse``# digits of num``def` `reverseDigits(num) :` ` ``rev_num ``=` `0``;`` ``while` `(num > ``0``) :`` ``rev_num ``=` `rev_num ``` `10` `+` `num ``%` `10`` ``num ``=` `num ``/``/` `10`` ` ` ``return` `rev_num` `# Function to check if n is Palindrome``def` `isPalindrome(n) :` ` ``# get the reverse of n`` ``rev_n ``=` `reverseDigits(n);` ` ``# Check if rev_n and n are same or not.`` ``if` `(rev_n ``=``=` `n) :`` ``return` `1`` ``else` `:`` ``return` `0` `# Driver Code``if` `__name__ ``=``=` `\"__main__\"` `:` ` ``n ``=` `4562`` ` ` ``if` `isPalindrome(n) ``=``=` `1` `:`` ``print``(``\"Is\"``, n, ``\"a Palindrome number? ->\"``, ``True``)`` ` ` ``else` `:`` ``print``(``\"Is\"``, n, ``\"a Palindrome number? ->\"``, ``False``)` ` ``n ``=` `2002`` ` ` ``if` `isPalindrome(n) ``=``=` `1` `:`` ``print``(``\"Is\"``, n, ``\"a Palindrome number? ->\"``, ``True``)`` ` ` ``else` `:`` ``print``(``\"Is\"``, n, ``\"a Palindrome number? ->\"``, ``False``)` `# This code is contributed by Ryuga`\n\n## C#\n\n `// C# program to check whether a number``// is Palindrome or not.` `using` `System;``class` `GFG``{`` ``/* Iterative function to reverse digits of num/`` ``static` `int` `reverseDigits(``int` `num)`` ``{`` ``int` `rev_num = 0;`` ``while` `(num > 0) {`` ``rev_num = rev_num 10 + num % 10;`` ``num = num / 10;`` ``}`` ``return` `rev_num;`` ``}`` ` ` ``/* Function to check if n is Palindrome/`` ``static` `int` `isPalindrome(``int` `n)`` ``{`` ` ` ``// get the reverse of n`` ``int` `rev_n = reverseDigits(n);`` ` ` ``// Check if rev_n and n are same or not.`` ``if` `(rev_n == n)`` ``return` `1;`` ``else`` ``return` `0;`` ``}`` ` ` ``/Driver program to test reversDigits/`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 4562;`` ``Console.WriteLine(``\"Is\"` `+ n + ``\"a Palindrome number? -> \"` `+`` ``(isPalindrome(n) == 1 ? ``\"true\"` `: ``\"false\"``));`` ` ` ``n = 2002;`` ` ` ``Console.WriteLine(``\"Is\"` `+ n + ``\"a Palindrome number? -> \"` `+`` ``(isPalindrome(n) == 1 ? ``\"true\"` `: ``\"false\"``));` ` ``}` `}` `// This code is contributed``// by Hritik Raj ( ihritik )`\n\n## PHP\n\n ` 0)`` ``{`` ``\\$rev_num` `= ``\\$rev_num` ` 10 +`` ``\\$num` `% 10;`` ``\\$num` `= ``\\$num` `/ 10;`` ``}`` ``return` `\\$rev_num``;``}` `// Function to check if n is Palindrome``function` `isPalindrome(``\\$n``)``{` ` ``// get the reverse of n`` ``\\$rev_n` `= reverseDigits(``\\$n``);` ` ``// Check if rev_n and n are same or not.`` ``if` `(``\\$rev_n` `== ``\\$n``)`` ``return` `1;`` ``else`` ``return` `0;``}` `// Driver Code``\\$n` `= 4562;``echo` `\"Is \"``, ``\\$n` `, ``\" a Palindrome number? ->\"``;` `if``(isPalindrome(``\\$n``) == 1)`` ``echo` `\"true\"` `;``else`` ``echo` `\"false\"``;``echo` `\"\\n\"``;` `\\$n` `= 2002;``echo` `\"Is \"``, ``\\$n` `, ``\" a Palindrome number? ->\"``;``if``(isPalindrome(!``\\$n``))`` ``echo` `\"true\"` `;``else`` ``echo` `\"false\"``;` `// This code is contributed by jit_t``?>`\n\n## Javascript\n\n ``\nOutput:\n```Is 4562 a Palindrome number? -> false\nIs 2002 a Palindrome number? -> true```\n\nTime Complexity: O(logN)\nAuxiliary Space: O(1)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up" ]	[ null, "https://media.geeksforgeeks.org/wp-content/cdn-uploads/program-to-check-if-a-number-is-palindrome-1024x512.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5469694,"math_prob":0.9085258,"size":5979,"snap":"2021-31-2021-39","text_gpt3_token_len":1887,"char_repetition_ratio":0.18292888,"word_repetition_ratio":0.5126263,"special_character_ratio":0.3602609,"punctuation_ratio":0.1617916,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985552,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T14:45:42Z\",\"WARC-Record-ID\":\"<urn:uuid:2e4c1ca8-c8fa-40a9-bd78-8efee8c1458c>\",\"Content-Length\":\"157942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3612a5a-47ee-4e24-99fa-d9918da6acff>\",\"WARC-Concurrent-To\":\"<urn:uuid:f39d6498-12d0-4d15-bbae-c5fe8e6499b1>\",\"WARC-IP-Address\":\"23.207.202.202\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/program-to-check-the-number-is-palindrome-or-not/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:5474DJ6NBT6JVP2JBXU6FPYHWGUYDLG3\",\"WARC-Block-Digest\":\"sha1:TPK5DOTZGYDB7QR5WXQPWD6DELGT2JRB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057366.40_warc_CC-MAIN-20210922132653-20210922162653-00647.warc.gz\"}"}
https://h3x.no/2009/08/16/display-clock-each-10-minutes-x-times-forward-in-php	[ "Categories\n\n# Display clock each 10 minutes x times forward in PHP\n\nHere is a simple code PHP snippet to display the clock each 10 minutes x times in the format of:\n\nphp5 test.php\n20:40\n20:50\n21:00\n21:10\n21:20\n21:30\n21:40\n\nEtc.\n\n```<?php\n\n\\$j = 0;\nfor(\\$i=0;\\$i<10;\\$i+=1) {\n\\$j += (1060);\n\\$hour = date(\"H\", (time()+\\$j));\n\\$min = ceil((date(\"i\", (time()+\\$j))/10))10;\nif(\\$min > 50) {\n\\$min = \"00\";\n\\$hour++;\n}\necho \\$hour.\":\".\\$min.\"n\";\n}\n\n?>\n```\n\n## 1 reply on “Display clock each 10 minutes x times forward in PHP”\n\n[…] to cancel reply. Name (required) Mail (will not be published) (required) Website. Recent Entries …Display clock each 10 minutes x times forward in PHP \| h3x.noHere is a simple code PHP snippet to display the clock each 10 minutes x times in the format of: […]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65366405,"math_prob":0.97353774,"size":687,"snap":"2020-24-2020-29","text_gpt3_token_len":230,"char_repetition_ratio":0.12445095,"word_repetition_ratio":0.3275862,"special_character_ratio":0.4250364,"punctuation_ratio":0.19018404,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9509719,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T10:07:13Z\",\"WARC-Record-ID\":\"<urn:uuid:4bf8e0df-1a2a-45a1-9424-2c4a867b4d80>\",\"Content-Length\":\"113655\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1b639c2c-efba-4ce2-8861-ba57f94feeb1>\",\"WARC-Concurrent-To\":\"<urn:uuid:e986c01b-0d11-4807-8849-0562212e2cae>\",\"WARC-IP-Address\":\"185.14.184.13\",\"WARC-Target-URI\":\"https://h3x.no/2009/08/16/display-clock-each-10-minutes-x-times-forward-in-php\",\"WARC-Payload-Digest\":\"sha1:CZYZRIIQAICDEWMGI7HLC22E4SH4L7UW\",\"WARC-Block-Digest\":\"sha1:AI565TFNLJ2AFDD3R4CDHIUXK3WO7GT4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657143365.88_warc_CC-MAIN-20200713100145-20200713130145-00404.warc.gz\"}"}
https://yyiki.org/wiki/Regression%20analysis/	[ "# Regression analysis\n\nA way to think and generalize the statistical models is breaking them into a stochastic and a systematic component.\n\n\\begin{aligned} Y_i \\sim f(\\theta_i, \\alpha) \\quad &\\text{(stochastic)} \\\\ \\theta_i = g(x_i, \\beta)\\quad &\\text{(systematic)} \\end{aligned}\n\nwhere there is always an estimation uncertainty – the lack of knowledge of $\\beta$ and $\\alpha$ – and fundamental uncertainty (stochastic component).\n\n## Marginal effect\n\nMarginal effect refers to “the slope of the regression surface with respect to a given covariate” (Leeper, 2017). In other words, it is about how much the dependent variable changes when we change the given independent variable. Of course this is not necessarily a constant. The slope can change based on the value of the independent variables. Thus we consider a representative marginal effect. A common method is calculating the “average marginal effects (AMEs)”, which is the mean of the marginal effect for every data point. Alternatively, the marginal effect can be calculated at the means of independent variables, “marginal effects at means (MEMs)”, or at a representative values (MERs). AMEs are most ‘data-driven’ and reflect the full data distribution." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85495746,"math_prob":0.9923313,"size":1438,"snap":"2021-04-2021-17","text_gpt3_token_len":287,"char_repetition_ratio":0.12831241,"word_repetition_ratio":0.0,"special_character_ratio":0.1891516,"punctuation_ratio":0.07860262,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987902,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-19T14:55:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b72c611d-38bf-41e9-9df2-8ef6f91d55ad>\",\"Content-Length\":\"13514\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:030da3ab-3f7f-4177-b88b-76cc32754c2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:03203644-0c6d-46e9-99e4-1cfc91eb2934>\",\"WARC-IP-Address\":\"45.56.105.151\",\"WARC-Target-URI\":\"https://yyiki.org/wiki/Regression%20analysis/\",\"WARC-Payload-Digest\":\"sha1:7ZFMBNBVMXIGIT4XMQOF4IKUH4ZQZBNP\",\"WARC-Block-Digest\":\"sha1:DG23L56IECKUUAAAH5RWKEIWPTD5NYWM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038887646.69_warc_CC-MAIN-20210419142428-20210419172428-00174.warc.gz\"}"}
http://greenteapress.com/thinkos/html/thinkos006.html	[ "", null, "", null, "", null, "Here is the PDF version of this book.\n\n# Chapter 5 More bits and bytes\n\n## 5.1 Representing integers\n\nYou probably know that computers represent numbers in base 2, also known as binary. For positive numbers, the binary representation is straightforward; for example, the representation for 510 is b101.\n\nFor negative numbers, the most obvious representation uses a sign bit to indicate whether a number is positive or negative. But there is another representation, called “two’s complement” that is much more common because it is easier to work with in hardware.\n\nTo find the two’s complement of a negative number, −x, find the binary representation of x, flip all the bits, and add 1. For example, to represent −510, start with the representation of 510, which is b0000 0101 if we write the 8-bit version. Flipping all the bits and adding 1 yields b1111 1011.\n\nIn two’s complement, the leftmost bit acts like a sign bit; it is 0 for positive numbers and 1 for negative numbers.\n\nTo convert from an 8-bit number to 16-bits, we have to add more 0’s for a positive number and add 1’s for a negative number. In effect, we have to copy the sign bit into the new bits. This process is called “sign extension”.\n\nIn C all integer types are signed (able to represent positive and negative numbers) unless you declare them unsigned. The difference, and the reason this declaration is important, is that operations on unsigned integers don’t use sign extension.\n\n## 5.2 Bitwise operators\n\nPeople learning C are sometimes confused about the bitwise operators `&` and `\|`. These operators treat integers as bit vectors and compute logical operations on corresponding bits.\n\nFor example, `&` computes the AND operation, which yields 1 if both operands are 1, and 0 otherwise. Here is an example of `&` applied to two 4-bit numbers:\n\n``` 1100\n& 1010\n----\n1000\n```\n\nIn C, this means that the expression `12 & 10` has the value 8.\n\nSimilarly, `\|` computes the OR operation, which yields 1 if either operand is 1, and 0 otherwise.\n\n``` 1100\n\| 1010\n----\n1110\n```\n\nSo the expression `12 \| 10` has the value 14.\n\nFinally, `^` computes the XOR operation, which yields 1 if either operand is 1, but not both.\n\n``` 1100\n^ 1010\n----\n0110\n```\n\nSo the expression `12 ^ 10` has the value 6.\n\nMost commonly, `&` is used to clear a set of bits from a bit vector, `\|` is used to set bits, and `^` is used to flip, or “toggle” bits. Here are the details:\n\nClearing bits: For any value x, x & 0 is 0, and x & 1 is x. So if you AND a vector with 3, it selects only the two rightmost bits, and sets the rest to 0.\n\n``` xxxx\n& 0011\n----\n00xx\n```\n\nIn this context, the value 3 is called a “mask” because it selects some bits and masks the rest.\n\nSetting bits: Similarly, for any x, x \| 0 is x, and x \| 1 is 1. So if you OR a vector with 3, it sets the rightmost bits, and leaves the rest alone:\n\n``` xxxx\n\| 0011\n----\nxx11\n```\n\nToggling bits: Finally, if you XOR a vector with 3, it flips the rightmost bits and leaves the rest alone. As an exercise, see if you can compute the two’s complement of 12 using `^`. Hint: what’s the two’s complement representation of -1?\n\nC also provides shift operators, << and >>, which shift bits left and right. Each left shift doubles a number, so 5 << 1 is 10, and 5 << 2 is 20. Each right shift divides by two (rounding down), so 5 >> 1 is 2 and 2 >> 1 is 1.\n\n## 5.3 Representing floating-point numbers\n\nFloating-point numbers are represented using the binary version of scientific notation. In decimal notation, large numbers are written as the product of a coefficient and 10 raised to an exponent. For example, the speed of light in m/s is approximately 2.998 · 108.\n\nMost computers use the IEEE standard for floating-point arithmetic. The C type float usually corresponds to the 32-bit IEEE standard; double usually corresponds to the 64-bit standard.\n\nIn the 32-bit standard, the leftmost bit is the sign bit, s. The next 8 bits are the exponent, q, and the last 23 bits are the coefficient, c. The value of a floating-point number is\n\n (−1)s c · 2q\n\nWell, that’s almost correct, but there’s one more wrinkle. Floating-point numbers are usually normalized so that there is one digit before the point. For example, in base 10, we prefer 2.998 · 108 rather than 2998 · 105 or any other equivalent expression. In base 2, a normalized number always has the digit 1 before the binary point. Since the digit in this location is always 1, we can save space by leaving it out of the representation.\n\nFor example, the integer representation of 1310 is b1101. In floating point, that’s 1.101 · 23, so the exponent is 3 and the part of the coefficient that would be stored is 101 (followed by 20 zeros).\n\nWell, that’s almost correct, but there’s one more wrinkle. The exponent is stored with a “bias”. In the 32-bit standard, the bias is 127, so the exponent 3 would be stored as 130.\n\nTo pack and unpack floating-point numbers in C, we can use a union and bitwise operations. Here’s an example:\n\n``` union {\nfloat f;\nunsigned int u;\n} p;\n\np.f = -13.0;\nunsigned int sign = (p.u >> 31) & 1;\nunsigned int exp = (p.u >> 23) & 0xff;\n\nunsigned int coef_mask = (1 << 23) - 1;\nunsigned int coef = p.u & coef_mask;\n\nprintf(\"%d\\n\", sign);\nprintf(\"%d\\n\", exp);\nprintf(\"0x%x\\n\", coef);\n```\n\nThis code is in float.c in the repository for this book (see Section 0.1).\n\nThe union allows us to store a floating-point value using p.f and then read it as an unsigned integer using p.u.\n\nTo get the sign bit, we shift the bits to the right 31 places and then use a 1-bit mask to select only the rightmost bit.\n\nTo get the exponent, we shift the bits 23 places, then select the rightmost 8 bits (the hexadecimal value 0xff has eight 1’s).\n\nTo get the coefficient, we need to extract the 23 rightmost bits and ignore the rest. We do that by making a mask with 1s in the 23 rightmost places and 0s on the left. The easiest way to do that is by shifting 1 to the left by 23 places and then subtracting 1.\n\nThe output of this program is:\n\n```1\n130\n0x500000\n```\n\nAs expected, the sign bit for a negative number is 1. The exponent is 130, including the bias. And the coefficient, which I printed in hexadecimal, is 101 followed by 20 zeros.\n\nAs an exercise, try assembling or disassembling a double, which uses the 64-bit standard. See http://en.wikipedia.org/wiki/IEEE_floating_point.\n\n## 5.4 Unions and memory errors\n\nThere are two common uses of C unions. One, which we saw in the previous section, is to access the binary representation of data. Another is to store heterogeneous data. For example, you could use a union to represent a number that might be an integer, float, complex, or rational number.\n\nHowever, unions are error-prone. It is up to you, as the programmer, to keep track of what type of data is in the union; if you write a floating-point value and then interpret it as an integer, the result is usually nonsense.\n\nActually, the same thing can happen if you read a location in memory incorrectly. One way that can happen is if you read past the end of an array.\n\nTo see what happens, I’ll start with a function that allocates an array on the stack and fills it with the numbers from 0 to 99.\n\n```void f1() {\nint i;\nint array;\n\nfor (i=0; i<100; i++) {\narray[i] = i;\n}\n}\n```\n\nNext I’ll define a function that creates a smaller array and deliberately accesses elements before the beginning and after the end:\n\n```void f2() {\nint x = 17;\nint array;\nint y = 123;\n\nprintf(\"%d\\n\", array[-2]);\nprintf(\"%d\\n\", array[-1]);\nprintf(\"%d\\n\", array);\nprintf(\"%d\\n\", array);\n}\n```\n\nIf I call f1 and then f2, I get these results:\n\n```17\n123\n98\n99\n```\n\nThe details here depend on the compiler, which arranges variables on the stack. From these results, we can infer that the compiler put x and y next to each other, “below” the array (at a lower address). And when we read past the array, it looks like we are getting values that were left on the stack by the previous function call.\n\nIn this example, all of the variables are integers, so it is relatively easy to figure out what is going on. But in general when you read beyond the bounds of an array, the values you read might have any type. For example, if I change f1 to make an array of floats, the results are:\n\n```17\n123\n1120141312\n1120272384\n```\n\nThe latter two values are what you get if you interpret a floating-point value as an integer. If you encountered this output while debugging, you would have a hard time figuring out what’s going on.\n\n## 5.5 Representing strings\n\nRelated issues sometimes come up with strings. First, remember that C strings are null-terminated. When you allocate space for a string, don’t forget the extra byte at the end.\n\nAlso, the letters and numbers in C strings are encoded in ASCII. The ASCII codes for the digits “0” through “9” are 48 through 57, not 0 through 9. The ASCII code 0 is the NUL character that marks the end of a string. And the ASCII codes 1 through 9 are special characters used in some communication protocols. ASCII code 7 is a bell; on some terminals, printing it makes a sound.\n\nThe ASCII code for the letter “A” is 65; the code for “a” is 97. Here are those codes in binary:\n\n```65 = b0100 0001\n97 = b0110 0001\n```\n\nA careful observer will notice that they differ by a single bit. And this pattern holds for the rest of the letters; the sixth bit (counting from the right) acts as a “case bit”, 0 for upper-case letters and 1 for lower case letters.\n\nAs an exercise, write a function that takes a string and converts from lower-case to upper-case by flipping the sixth bit. As a challenge, you can make a faster version by reading the string 32 or 64 bits at a time, rather than one character at a time. This optimization is made easier if the length of the string is a multiple of 4 or 8 bytes.\n\nIf you read past the end of a string, you are likely to see strange characters. Conversely, if you write a string and then accidentally read it as an int or float, the results will be hard to interpret.\n\nFor example, if you run:\n\n``` char array[] = \"allen\";\nfloat p = array;\nprintf(\"%f\\n\", p);\n```\n\nYou will find that the ASCII representation of the first 8 characters of my name, interpreted as a double-precision floating point number, is 69779713878800585457664.\n\n#### Are you using one of our books in a class?\n\nWe'd like to know about it. Please consider filling out this short survey.\n\nThink Bayes", null, "", null, "", null, "Think Python", null, "", null, "", null, "Think Stats", null, "", null, "", null, "Think Complexity", null, "", null, "", null, "", null, "", null, "", null, "" ]	[ null, "http://greenteapress.com/thinkos/html/back.png", null, "http://greenteapress.com/thinkos/html/up.png", null, "http://greenteapress.com/thinkos/html/next.png", null, "http://ir-na.amazon-adsystem.com/e/ir", null, "http://ws-na.amazon-adsystem.com/widgets/q", null, "http://ir-na.amazon-adsystem.com/e/ir", null, "http://www.assoc-amazon.com/e/ir", null, "http://ws-na.amazon-adsystem.com/widgets/q", null, "http://www.assoc-amazon.com/e/ir", null, "http://ir-na.amazon-adsystem.com/e/ir", null, "http://ws-na.amazon-adsystem.com/widgets/q", null, "http://ir-na.amazon-adsystem.com/e/ir", null, "http://www.assoc-amazon.com/e/ir", null, "http://ws-na.amazon-adsystem.com/widgets/q", null, "http://www.assoc-amazon.com/e/ir", null, "http://greenteapress.com/thinkos/html/back.png", null, "http://greenteapress.com/thinkos/html/up.png", null, "http://greenteapress.com/thinkos/html/next.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87734044,"math_prob":0.97775006,"size":9472,"snap":"2019-13-2019-22","text_gpt3_token_len":2410,"char_repetition_ratio":0.12019434,"word_repetition_ratio":0.010944701,"special_character_ratio":0.2783995,"punctuation_ratio":0.13141182,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956446,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,10,null,10,null,6,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,10,null,10,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T03:01:57Z\",\"WARC-Record-ID\":\"<urn:uuid:aec61952-8ce8-4fa7-a866-e6d1cb11abf8>\",\"Content-Length\":\"22970\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44449e30-cfc6-4d5f-a663-f0d91e55c2de>\",\"WARC-Concurrent-To\":\"<urn:uuid:e43fff06-8a60-4e8c-9805-8265886da186>\",\"WARC-IP-Address\":\"208.113.214.221\",\"WARC-Target-URI\":\"http://greenteapress.com/thinkos/html/thinkos006.html\",\"WARC-Payload-Digest\":\"sha1:VBEY76SQJSFKO2ZDJZV5ZA3TCAJ25GBN\",\"WARC-Block-Digest\":\"sha1:DMSKSO3QEOUGIVCYIBNLFUNLJHQMQQTC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232260658.98_warc_CC-MAIN-20190527025527-20190527051527-00549.warc.gz\"}"}
https://mathemerize.com/tag/differentials/	[ "## Differentials Errors and Approximations\n\nHere you will learn what is differentials errors and approximations with examples. Let’s begin – Differentials Errors and Approximations In order to calculate the approximate value of a function, differentials may be used where the differential of a function is equal to its derivative multiplied by the differential of the independent variable, In general dy …" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90059143,"math_prob":0.98047495,"size":401,"snap":"2022-27-2022-33","text_gpt3_token_len":74,"char_repetition_ratio":0.19647355,"word_repetition_ratio":0.0,"special_character_ratio":0.16209476,"punctuation_ratio":0.048387095,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98452777,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T05:02:23Z\",\"WARC-Record-ID\":\"<urn:uuid:3b413ebf-7979-474d-96d5-5c42e99a0351>\",\"Content-Length\":\"189169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56ddf1da-4e22-4853-a404-3e512e55d4f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c436490-4394-46e9-8a86-15ecd5c5284d>\",\"WARC-IP-Address\":\"134.209.145.77\",\"WARC-Target-URI\":\"https://mathemerize.com/tag/differentials/\",\"WARC-Payload-Digest\":\"sha1:VWQNFFU7TQAIEV7YMAW2HQ34LOF3JOPA\",\"WARC-Block-Digest\":\"sha1:NT3OG6IWQFJQDSJL54APTJUGSPNPN3LK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215790.65_warc_CC-MAIN-20220703043548-20220703073548-00449.warc.gz\"}"}
https://mathvideoprofessor.com/courses/seventh-grade/lessons/unit-7-angles-triangles-and-prisms/topic/volume-of-right-prisms/	[ "# Volume of Right Prisms\n\nWarmup\n\nSelect all the prisms.\n\nActivity #1\n\nUse snap cubes to build a 3-D Shape.\n\nIn the applet below, you can keep dragging blocks out of the pile by their red points until you have enough to build what you want.", null, "• Use the snap cubes to build the shape you see on the paper. (You can keep dragging blocks out of the pile by their red points until you have enough to build what you want.)\n• Answer question 1 below. \n• Add another layer of cubes on top of the shape you have built.\n• Now answer the rest of the questions below.\n\n(1.) Using the face of a snap cube as your area unit, what is the area of the shape? Explain or show your reasoning using the applet below.\n\n(2.) Describe this three-dimensional object.\n\n(4.) Right now, your object has a height of 2. What would the volume be if it had a height of 5?\n\n(5.) Right now, your object has a height of 2. What would the volume be if it had a height of 8.5?\n\nActivity #2\n\nThe applet below has a set of three-dimensional figures. Note that each polyhedron has only one label per unique face. Where no measurements are shown, the faces are identical copies.", null, "• The first one has already been choosen.\n• Rotate the view using the Rotate 3D Graphics tool marked by two intersecting, curved arrows. If you do not see it click on the 3D Graphics window for the full toolbar to appear.\n• Use the distance tool to click on any segment to find the height or length.\n• Then fill in the required information in the table below.\n• Choose another figure using the slider and repeat all the steps above.\n\nActivity #3\n\nExplore Features of Polyhedra.\n\nThere are 4 different prisms that all have the same volume. Here below is what the base of each prism looks like.", null, "• Order the prisms from shortest to tallest.\n\n(2.) If the volume of each prism is 60 cubic units, what would be the height of each prism?\n\n(3.) For a volume other than 60 cubic units, what could be the height of each prism?\n\nChallenge #1\n\nFor each prism, shade one of its bases.\n\nChallenge #2\n\nThe volume of both of these trapezoidal prisms is 24 cubic units. Their heights are 6 and 8 units, as labeled.\n\nWhat is the area of a trapezoidal base of each prism?\n\nQuiz Time" ]	[ null, "https://i0.wp.com/mathvideoprofessor.com/wp-content/uploads/2021/08/Instructions-3.png", null, "https://i0.wp.com/mathvideoprofessor.com/wp-content/uploads/2021/08/Instructions-3.png", null, "https://i0.wp.com/mathvideoprofessor.com/wp-content/uploads/2021/08/Instructions-3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9210733,"math_prob":0.86973536,"size":1455,"snap":"2023-40-2023-50","text_gpt3_token_len":369,"char_repetition_ratio":0.14197105,"word_repetition_ratio":0.13284133,"special_character_ratio":0.25360826,"punctuation_ratio":0.13354038,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98705465,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T12:40:26Z\",\"WARC-Record-ID\":\"<urn:uuid:66fb7c44-0a13-4e2c-a355-f7fefae1d097>\",\"Content-Length\":\"210693\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1dd8926b-758d-457c-9a96-9f04001dbd9f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba92ed57-6472-454d-a8c9-ccdd3243c504>\",\"WARC-IP-Address\":\"35.208.47.198\",\"WARC-Target-URI\":\"https://mathvideoprofessor.com/courses/seventh-grade/lessons/unit-7-angles-triangles-and-prisms/topic/volume-of-right-prisms/\",\"WARC-Payload-Digest\":\"sha1:YWC4DKZZODATZXGE5FBVJ3JJKXLY23ZR\",\"WARC-Block-Digest\":\"sha1:L7TYQU5EEQ7GHSDIEAZZSDQSBOEDR3FM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100529.8_warc_CC-MAIN-20231204115419-20231204145419-00105.warc.gz\"}"}
https://yeahexp.com/i-can-not-understand-the-simplest-algorithm/	[ "# c++ – I can not understand the simplest algorithm\n\n## Question:\n\nIt is necessary to calculate the XORs of all numbers on a given segment. The xor operation is familiar to me, but I don't know how to calculate the xors of all numbers. Tried to XOR first the first number with all the others, and these XORs XOR-il among themselves, also until the last number and then these XOR-il results, but this is wrong.\n\nFor obvious reasons, the `xor` between an even number and the next odd number is `1` . Therefore, and also because `xor` is associative.\n\n``````0 ^ 1 ^ 2 ^ 3 ^ ... ^ 2n-2 ^ 2n-1 =\n(0 ^ 1) ^ (2 ^ 3) ^ ... ^ (2n-2 ^ 2n-1) =\n1 ^ 1 ^ ... ^ 1 ^ 1 (n раз)\n``````\n\nThose. if the sequence starts with an even number and ends with an odd number, then the result for it will be `1` if the total number of terms is not divisible by `4` , and `0` if it is divisible.\n\nThat is, for example\n\n``````40 ^ 41 ^ ... ^ 99999998 ^ 99999999 = 0\n``````\n\nbecause `40` is even, `99999999` is odd and `99999999 - 40 + 1` is divisible by 4\n\nIf the sequence starts with odd and/or ends with even, these numbers can be \"doxed\" to the result, because `xor` is also commutative.\n\n``````39 ^ 40 ^ 41 ^ ... ^ 99999998 ^ 99999999 ^ 100000000 =\n(40 ^ 41 ^ ... ^ 99999998 ^ 99999999) ^ 39 ^ 100000000 =\n0 ^ 39 ^ 100000000\n``````\n\nWell, perhaps the fact that the `xor` operation reverses itself will help\n\n``````a ^ b ^ a = b\n``````\n\nSo\n\n``````(0 ^ 1 ^ ... ^ n) ^ (0 ^ 1 ^ ... ^ m) = n+1 ^ n+2 ^ ... ^ m\n``````\n\nwhere n ≤ m, but this is up to you\n\nScroll to Top" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.847938,"math_prob":0.99678975,"size":1372,"snap":"2023-40-2023-50","text_gpt3_token_len":448,"char_repetition_ratio":0.16081871,"word_repetition_ratio":0.058631923,"special_character_ratio":0.43002915,"punctuation_ratio":0.16776316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997149,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T10:21:25Z\",\"WARC-Record-ID\":\"<urn:uuid:698ce0ef-aa28-4e63-a40b-23ac4ceb531c>\",\"Content-Length\":\"113867\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:143e1a87-844e-4626-ae21-6c78f2a75aa1>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd332d7b-baa6-4032-a702-9bb5df11c39f>\",\"WARC-IP-Address\":\"152.32.134.98\",\"WARC-Target-URI\":\"https://yeahexp.com/i-can-not-understand-the-simplest-algorithm/\",\"WARC-Payload-Digest\":\"sha1:252MIFKCM5YG2I6UWEONWO7FHBVZRAYO\",\"WARC-Block-Digest\":\"sha1:QZ7KKG2QO6WHIDSYIJIKIWEPF5IXY26L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510671.0_warc_CC-MAIN-20230930082033-20230930112033-00556.warc.gz\"}"}
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_18&diff=49895&oldid=49262	[ "# Difference between revisions of \"2008 AMC 8 Problems/Problem 18\"\n\n## Problem\n\nTwo circles that share the same center have radii", null, "$10$ meters and", null, "$20$ meters. An aardvark runs along the path shown, starting at", null, "$A$ and ending at", null, "$K$. How many meters does the aardvark run?", null, "$[asy] size((150)); draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle); draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle); draw((20,0)--(-20,0)); draw((0,20)--(0,-20)); draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow); draw((-18,1)--(-12, 1), EndArrow); draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow); draw((1,-9)--(1,9), EndArrow); draw((0,12)..(8.3, 8.3)..(12,0), EndArrow); draw((12,-1)--(18,-1), EndArrow); label(\"A\", (0,20), N); label(\"K\", (20,0), E); [/asy]$", null, "$\\textbf{(A)}\\ 10\\pi+20\\qquad\\textbf{(B)}\\ 10\\pi+30\\qquad\\textbf{(C)}\\ 10\\pi+40\\qquad\\textbf{(D)}\\ 20\\pi+20\\qquad \\\\ \\textbf{(E)}\\ 20\\pi+40$\n\n## See Also\n\n 2008 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions\nInvalid username\nLogin to AoPS" ]	[ null, "https://latex.artofproblemsolving.com/f/c/6/fc606f7f1e530731ab4f1cc364c01dc64a4455ee.png ", null, "https://latex.artofproblemsolving.com/f/8/3/f8366cd6196dd8a9da6d38a3e9eafb109e99d53e.png ", null, "https://latex.artofproblemsolving.com/0/1/9/019e9892786e493964e145e7c5cf7b700314e53b.png ", null, "https://latex.artofproblemsolving.com/d/f/b/dfb064112b6c94470339f6571f69d07afc1c024c.png ", null, "https://latex.artofproblemsolving.com/2/1/0/210387e06e9467a3ab329cbdf9e43ff63316fb45.png ", null, "https://latex.artofproblemsolving.com/2/5/7/257695d7e6bff3c0748a33ed6ad52f0344c8c5b5.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74577624,"math_prob":0.99870753,"size":1282,"snap":"2021-21-2021-25","text_gpt3_token_len":414,"char_repetition_ratio":0.17449139,"word_repetition_ratio":0.41484717,"special_character_ratio":0.3525741,"punctuation_ratio":0.085020244,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9920689,"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\":\"2021-06-18T06:38:34Z\",\"WARC-Record-ID\":\"<urn:uuid:8e83ee6d-efc6-4ad3-9da8-d122d1d7527e>\",\"Content-Length\":\"42077\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6ed5973-128c-4422-b03c-1c47d4ebb594>\",\"WARC-Concurrent-To\":\"<urn:uuid:07fef161-a4ab-44fb-8021-35ed2dbdfcc3>\",\"WARC-IP-Address\":\"172.67.69.208\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_8_Problems/Problem_18&diff=49895&oldid=49262\",\"WARC-Payload-Digest\":\"sha1:PXZUEOHVP7WTY7I5YEBJN24LNNAIZ6E4\",\"WARC-Block-Digest\":\"sha1:JCJBBS4W7KCCH3FFXP3COE35HJS73ZD2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487635724.52_warc_CC-MAIN-20210618043356-20210618073356-00377.warc.gz\"}"}
http://itp.jscottdutcher.com/?p=64	[ "# Infinite Pong, again\n\nhttp://itp.jscottdutcher.com/pong4/\n\nThis week, much like last week, was an exercise in not getting as much done as I would have liked. Adding functions to my previous code was a breeze, until I ran it and kept getting an uncaught reference error. It took me ages to figure out where I was missing a curly bracket and I felt like a right fool the whole time I was looking. Oh well! It remains remarkably satisfying when your code works, even when all it is is a meager refactor.\n\n```ar gameStart = false;\nx: 10,\ny: 100,\nw: 15,\nh: 100,\n};\nx: 770,\ny: 100,\nw: 15,\nh: 100,\n};\nvar ball = {\nx: 50,\ny: 20,\ndiam: 25,\nspeedX: 5,\nspeedY: 5,\n};\nvar speedX = 5;\nvar speedY = 5;\nvar s = \"Welcome to Infinte Pong! Player one controlls their paddle with the a and z keys. Player two controls their paddle with the /? key and the '\\\" key. Press the space bar to begin\";\n\n/function down(x){\nx = x + 5;\n}/\n\nfunction setup() {\ncreateCanvas(800, 600);\nsmooth();\n//background(0);\n//fill(255)\n//text(s, 10, 10, 70, 80);\n}\n\nfunction draw() {\n\n//if (keyPressed(32) === true) {\ngameStart === true;\n//}\n\nbackground(64, 161, 76);\nnoStroke();\n\ncreateBall();\nballBounceTopAndBottom();\nballBounceRight();\nballBounceLeft();\n\n}\nfunction createBall() {\n//Create ball\nfill(198, 237, 44);\nellipse(ball.x, ball.y, ball.diam, ball.diam);\n\nball.x = ball.x + speedX;\nball.y = ball.y + speedY;\n}\n\nfill(0, 50, 50);\nif (keyIsDown(90) === true) {\n}\n}\n}\n\nfill(50, 50, 0);\n\nif (keyIsDown(191) === true) { //move paddle down\n}\n}\n\n}\n\nfunction ballBounceTopAndBottom() {\n\n//If if the ball hits the top or bottom of the court it bounces\nspeedY = speedY * -1; //reverse the direction of the motion\nball.y = ball.y + speedY; //keeps things moving\nprint(\"bam top\");\n}\n\n//if the edge of the ball is higher than rect y plus height and\n//the x of the ball is greater than the x of the rect and less than the width\nelse if (ball.y + 12.5 < paddleR.y + paddleR.h && ball.y > paddleR.y && ball.x > paddleR.x && ball.x < paddleR.x + paddleR.h && ball.x > 0 && ball.x < width && ball.y > 0 && ball.y < height) {\nspeedY = speedY * -1; //reverse the direction of the motion\nball.y = ball.y + speedY; //keeps things moving\nprint(\"bam bottom\");\n}\n}\n\nfunction ballBounceLeft() {\n//if the ball hits the left wall\n/* if (ball.x < 0) {\nspeedX = speedX * -1; //This reverses the direction, I think\nball.x = ball.x + speedX; //This keeps the ball moving\nprint(\"pow\");/\n\n//if the ball hits the front of the left paddle\nif (ball.x - 12.5 < paddleL.x + paddleL.w && ball.y + 12.5 > paddleL.y && ball.y + 12.5 < paddleL.y + paddleL.h && ball.x > 0 && ball.x < width && ball.y > 0 && ball.y < height) {\nspeedX = speedX -1; //This reverses the direction, I think\nball.x = ball.x + speedX; //This keeps the ball moving\nprint(\"pow\");\n}\n}```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57246387,"math_prob":0.97983396,"size":4594,"snap":"2023-40-2023-50","text_gpt3_token_len":1433,"char_repetition_ratio":0.21938998,"word_repetition_ratio":0.3534994,"special_character_ratio":0.36525902,"punctuation_ratio":0.22789784,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97923076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T17:47:02Z\",\"WARC-Record-ID\":\"<urn:uuid:1912e777-162c-4f7b-9762-dddb10bd8400>\",\"Content-Length\":\"23460\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75cdbc65-5f39-44ff-9b3e-8ccc3e00bbc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f53eed85-8ac0-4f8b-8b5f-05b49206be34>\",\"WARC-IP-Address\":\"50.116.52.232\",\"WARC-Target-URI\":\"http://itp.jscottdutcher.com/?p=64\",\"WARC-Payload-Digest\":\"sha1:AULAOXNT2RYZKON3AGKIGNIOVETKVXEX\",\"WARC-Block-Digest\":\"sha1:RB5GMUIZPVSO6D3YX3P3L3PPCMZUPDJ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102612.80_warc_CC-MAIN-20231210155147-20231210185147-00752.warc.gz\"}"}
https://dbprohelp.fandom.com/wiki/MAKE_OBJECT_NEW	[ "## FANDOM\n\n540 Pages\n\n Syntax\n\n MAKE OBJECT NEW Object number, Vertex count, Index countMAKE OBJECT NEW Object number, Vertex count, Index count, FVFMAKE OBJECT NEW Object number, Vertex count, Index count, FVF, Initialise buffers\n\n Description\n\n Create an object with the number of vertices and indices specified. If the Index count is not 0, then it needs to be a multiple of 3, as 3 points are required for each polygon. If the Index count is 0, then the Vertex count needs to be a multiple of 3 for the same reason. The first form of this command will create an object using an FVF value of 274. The third form of this command allows you to skip initialising the buffers with zeros by using an Initialise value of 0 - use only when you are going to immediately set the values yourself using the VERTEXDATA commands. Constants for all allowable FVF settings:#constant FVF_XYZ = 0x002#constant FVF_XYZRHW = 0x004#constant FVF_XYZB1 = 0x006#constant FVF_XYZB2 = 0x008#constant FVF_XYZB3 = 0x00a#constant FVF_XYZB4 = 0x00c#constant FVF_XYZB5 = 0x00e#constant FVF_NORMAL = 0x010#constant FVF_PSIZE = 0x020#constant FVF_DIFFUSE = 0x040#constant FVF_SPECULAR = 0x080#constant FVF_TEX0 = 0x000#constant FVF_TEX1 = 0x100#constant FVF_TEX2 = 0x200#constant FVF_TEX3 = 0x300#constant FVF_TEX4 = 0x400#constant FVF_TEX5 = 0x500#constant FVF_TEX6 = 0x600#constant FVF_TEX7 = 0x700#constant FVF_TEX8 = 0x800\nCommunity content is available under CC-BY-SA unless otherwise noted." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7928467,"math_prob":0.9930232,"size":839,"snap":"2019-35-2019-39","text_gpt3_token_len":188,"char_repetition_ratio":0.14850299,"word_repetition_ratio":0.1292517,"special_character_ratio":0.2169249,"punctuation_ratio":0.10843374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9899589,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-18T16:57:53Z\",\"WARC-Record-ID\":\"<urn:uuid:52906d50-7672-4b42-9891-b650282777e3>\",\"Content-Length\":\"95016\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce0e4572-a5e2-49ab-812f-dab54d536fb1>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ac2430e-010b-4f2f-87fe-fe8d9508d8ce>\",\"WARC-IP-Address\":\"151.101.0.194\",\"WARC-Target-URI\":\"https://dbprohelp.fandom.com/wiki/MAKE_OBJECT_NEW\",\"WARC-Payload-Digest\":\"sha1:JXFEYQQJAKFTFAFRXLYSN7OYL7Q7KHMR\",\"WARC-Block-Digest\":\"sha1:XRPJFCFIW7JSOALZMBE47IDJFPOJZYLY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027313987.32_warc_CC-MAIN-20190818165510-20190818191510-00353.warc.gz\"}"}
https://volumeunits.com/convert/ml/dam3/109	[ "", null, "# Convert 109 ML To DAM3\n\nFind 109 Megalitre in Cubic Decametre :\n\n109 ML = 9.046676924E-10 DAM3\n\n### How to convert 109 Megalitres to Cubic Decametre\n\nBefore you start to calculate 109 Megalitres, you should know the conversion rate between Megalitres and Cubic Decametre.\n\nThe rate is 8.2997036E-12 and the math equation is 109 X 8.2997036E-12 = 9.046676924E-10 .\n\n## Quick conversion chart of Megalitres to Cubic Decametres\n\n1 Megalitre to Cubic Decametre = 8.2997036E-12 Cubic Decametre 5 Megalitre to Cubic Decametre = 4.1498518E-11 Cubic Decametre 10 Megalitre to Cubic Decametre = 8.2997036E-11 Cubic Decametre 20 Megalitre to Cubic Decametre = 1.65994072E-10 Cubic Decametre 30 Megalitre to Cubic Decametre = 2.48991108E-10 Cubic Decametre 40 Megalitre to Cubic Decametre = 3.31988144E-10 Cubic Decametre 50 Megalitre to Cubic Decametre = 4.1498518E-10 Cubic Decametre 75 Megalitre to Cubic Decametre = 6.2247777E-10 Cubic Decametre 100 Megalitre to Cubic Decametre = 8.2997036E-10 Cubic Decametre\n\n#### Minus 1 & Plus 1\n\n(-1)(+1)", null, "108 110", null, "" ]	[ null, "https://volumeunits.com/images/logo.png", null, "https://volumeunits.com/images/sol.png", null, "https://volumeunits.com/images/sag.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6839539,"math_prob":0.95296127,"size":926,"snap":"2023-14-2023-23","text_gpt3_token_len":350,"char_repetition_ratio":0.36984816,"word_repetition_ratio":0.06870229,"special_character_ratio":0.36825055,"punctuation_ratio":0.096969694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95685256,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T14:46:49Z\",\"WARC-Record-ID\":\"<urn:uuid:b9b355a4-c9ed-442a-a28f-f534eef36a1d>\",\"Content-Length\":\"9832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:497affd2-2cac-4504-bc90-31fa840491be>\",\"WARC-Concurrent-To\":\"<urn:uuid:01cf50ab-959e-459f-9fd4-bf4513b0b771>\",\"WARC-IP-Address\":\"176.9.111.7\",\"WARC-Target-URI\":\"https://volumeunits.com/convert/ml/dam3/109\",\"WARC-Payload-Digest\":\"sha1:GAUXFDOK377HMGKO2ARUFGFTJGLHKJ6I\",\"WARC-Block-Digest\":\"sha1:U6J45ERHLLMU6XZAEZ7IVSONMUG47ML6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649293.44_warc_CC-MAIN-20230603133129-20230603163129-00427.warc.gz\"}"}
https://nl.mathworks.com/matlabcentral/cody/problems/44282-minimum-possible-m-of-the-maximum-side-of-a-triangle-of-given-area-a/solutions/1322330	[ "Cody\n\n# Problem 44282. Minimum possible M of the maximum side of a triangle of given area A.\n\nSolution 1322330\n\nSubmitted on 1 Nov 2017 by H M Dipu Kabir\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\na = 0.4331; m = 1.0001; assert(tri(a)>m0.99)\n\n2 Pass\na = 43.31; m = 10.001; assert(tri(a)<m1.01)\n\n3 Pass\na = 4331; m = 100.01; assert(tri(a)>m0.99)\n\n4 Pass\na = 4331; m = 100.01; assert(tri(a)<m1.01)\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5840687,"math_prob":0.9457562,"size":538,"snap":"2020-45-2020-50","text_gpt3_token_len":199,"char_repetition_ratio":0.15168539,"word_repetition_ratio":0.06593407,"special_character_ratio":0.43680298,"punctuation_ratio":0.1627907,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95911914,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T09:27:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f33eaa45-9de9-437e-8cc8-ee53f0b2649a>\",\"Content-Length\":\"80837\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:399ad067-072e-42a3-a014-6fb85d5de716>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b1ce5ae-43c9-4b0c-8f9e-b84a5a6e1f0f>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/44282-minimum-possible-m-of-the-maximum-side-of-a-triangle-of-given-area-a/solutions/1322330\",\"WARC-Payload-Digest\":\"sha1:IILZUENQLK2O5ABA2BM34STT7Z3MZQIR\",\"WARC-Block-Digest\":\"sha1:UZ5NK6ZZZVI2GHTIF5VTWNWK2DBX6G4U\",\"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-20201205074417-20201205104417-00561.warc.gz\"}"}
https://owenduffy.net/blog/?p=18996	[ "# Working a common mode scenario – G3TXQ Radcom May 2015 – voltage balun solution\n\nAt (Hunt 2015) G3TXQ gave some measurements of his ‘balanced’ antenna system.", null, "Above is Hunt’s equivalent circuit of his antenna system and transmitter. It is along the lines of (Schmidt nd) with different notation.\n\nHe made measurements that led to calculation of these components for the dipole which he states is visually fairly symmetric.", null, "Above are the calculated values he gives at Leg a and Leg b. Lets assume that the measurements are made at the tuner interface with respect the the unbalanced tuner ground connection. Note that Za and Zb are not close to equal, so the antenna system is hardly symmetric or balanced.", null, "Above is a schematic of the antenna equivalent circuit driven by equal but opposite phase voltages (ie, an ideal voltage balun).\n\nFrom that we can write a set of mesh equations and solve them.\n\n$$1=(za+zc) \\cdot ia -zc \\cdot ib\\\\ 1=-zc \\cdot ia+(zb+zc) \\cdot ib\\\\$$\n\nThis is a system of 2 linear simultaneous equations in 2 unknowns. In matrix notation:\n\n$$\\begin{vmatrix}ia\\\\ib \\end{vmatrix}= \\begin{vmatrix} za+zc & -zc\\\\ -zc & zb+zc \\end{vmatrix}^{-1} \\times \\begin{vmatrix}1\\\\1\\end{vmatrix}\\\\$$\n\nLet’s do that in GNU Octave.\n\n#equations of mesh currents\n#1=(za+zc)ia -zcib\n#1=-zcia+(zb+zc)ib\nza=15.1+79i\nzb=1.6-109i\nzc=30.7+110i\n\nA=[za+zc,-zc;-zc,zb+zc]\nb=[1;1]\nx=A\\b\n\nia=x(1)\nib=x(2)\nic=(ia-ib)/2\nid=(ia+ib)/2\nica=abs(ic)\nida=abs(id)\nicrel=2abs(ic)/abs(id)\n#for calculator\nabs(ia)\nabs(ib)\n2abs(ic)\n\n\nThe console output is…\n\nza = 15.100 + 79.000i\nzb = 1.6000 - 109.0000i\nzc = 30.700 + 110.000i\nA =\n\n45.800 + 189.000i -30.700 - 110.000i\n-30.700 - 110.000i 32.300 + 1.000i\n\nb =\n\n1\n1\n\nx =\n\n0.0046179 + 0.0091411i\n0.0049694 + 0.0242611i\n\nia = 0.0046179 + 0.0091411i\nib = 0.0049694 + 0.0242611i\nic = -0.00017574 - 0.00756002i\nid = 0.0047937 + 0.0167011i\nica = 0.0075621\nida = 0.017375\nicrel = 0.87043\nans = 0.010241\nans = 0.024765\nans = 0.015124\n\n\nThe comparative statistic I will use is \| 2*ic\| (total common mode current) relative to \|id\|, it is given by the variable icrel above which has a calculated value of 0.870 for this scenario.\n\nBy way of comparison, the same scenario with the common mode choke (ie current balun) with Zcm=1500+j1500Ω, the same statistic is 0.0738 (Working a common mode scenario – G3TXQ Radcom May 2015). The modest current balun results in relative common mode current being 21dB lower than the ideal voltage balun.\n\nNote that differential current and common mode current will almost always each br standing waves and their phase velocity may differ.\n\nThe last three values calculated are those that would be measured by a clamp on RF ammeter around the a, b and a+b wires together. These values could be plugged into Resolve measurement of I1, I2 and I12 into Ic and Ic to resolve the measurements into common mode and differential mode components.", null, "Above, the calculator results reconcile with the results of the Octave script.\n\nYou don’t need complicated maths to asses an installed antenna system, measurements with a clamp on RF ammeter can be resolved into the common mode and differential mode components using the calculator.\n\nThis solution is based on Hunts measurements and equivalent circuit calculation, and the abject failure of a voltage balun on an antenna system that Hunt reported as apparently symmetric applies to this scenario, but it is not surprising as good current baluns will tend to be more effective in reduction of common mode current than voltage baluns." ]	[ null, "https://owenduffy.net/blog/wp-content/uploads/2020/10/Screenshot-201016095702-300x198.png", null, "https://owenduffy.net/blog/wp-content/uploads/2020/10/Screenshot-201016095742.png", null, "https://owenduffy.net/blog/wp-content/uploads/2020/10/balunmesh-300x138.jpg", null, "https://owenduffy.net/blog/wp-content/uploads/2020/10/Screenshot-201017141738-211x300.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8582265,"math_prob":0.99743223,"size":3690,"snap":"2020-45-2020-50","text_gpt3_token_len":1084,"char_repetition_ratio":0.11394466,"word_repetition_ratio":0.0068259384,"special_character_ratio":0.31761518,"punctuation_ratio":0.11578947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99795383,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,6,null,6,null,4,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T04:50:59Z\",\"WARC-Record-ID\":\"<urn:uuid:9ab190a8-fa95-41e9-8ffe-07faebb233a2>\",\"Content-Length\":\"43533\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5dbf7182-a3f1-4486-9cb9-c03e60e13f43>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e3c0482-000f-4f05-a3ed-7e55271fb4fb>\",\"WARC-IP-Address\":\"107.180.40.20\",\"WARC-Target-URI\":\"https://owenduffy.net/blog/?p=18996\",\"WARC-Payload-Digest\":\"sha1:7K6ZSJVXIAF4SE7WSFCT6CNFDXSWAYZQ\",\"WARC-Block-Digest\":\"sha1:PP2KYEOEHC7NMVAQUKMQ4LQXPS3M2PWD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107875980.5_warc_CC-MAIN-20201021035155-20201021065155-00631.warc.gz\"}"}
https://affairscloud.com/aptitude-questions-data-sufficiency-set-10/	[ "# 10% OFF \| Use Coupon Code “AffairsCloud10” \| Expire Anytime", null, "Aptitude Questions: Data Sufficiency Set 10\n\nHello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From DATA SUFFICIENCY topic, which is common for all the IBPS,SBI exam RBI,,SSC and other competitive exams. We have included Some questions that are repeatedly asked in exams !!!\n\nQuestions Penned by Yogit\n\nQ(1 – 10) Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the three statements and give answer: –\na) If the data in statement I alone is sufficient to answer the question.\nb) If the data in statement II alone is sufficient to answer the question.\nc) If the data either in statement I alone or statement II alone are sufficient to answer the question.\nd)If the data given in both I and II together are not sufficient to answer the question.\ne) If the data in both the statements I and II together are necessary to answer the question\n\n1. What is the profit earned by selling a bike for rupees 50000?\n1) The cost price of 4 such bikes is equal to selling price of 3 bikes\n2) 33.33% profit is earned by selling the bike.\nAnswer – c) If the data either in statement I alone or statement II alone are sufficient to answer the question\nExplanation :\nFrom 1st statement, 4cp = 3sp\n4cp = 350000, cp = 37500\nFrom second statement, 50000 = [(100 + 100/3)/100]c\nWe get cp = 37500\n\n2. What is the rate of interest p.c.p.a?\n1) The amount triples itself in 12 years\n2) The simple interest accrued in 4 years is Rs. 2000.\nAnswer – a) If the data in statement I alone is sufficient to answer the question\nExplanation :\nFrom first statement, 2p = p(r/100)12\nFrom second statement, 2000 = p(r/100)4\n\n3. The average salary of 10 teachers of a school is Rs.4500. A group of teachers whose average salary is Rs.3000 leave the school and 3 new teacher joins. What is the average salary of new group of teachers in the school?\n1) There are 2 teachers that leave the school.\n2) The salary of new teacher is 2500.\nAnswer – e) if the data in both the statements I and II together are necessary to answer the question\nExplanation :\nNew average salary = (450010 – 23000 + 32500)/11\n\n4. What is the perimeter of the rectangular field?\n1) Area of the field is 72m^2.\n2) Length and breadth are in the ratio of 2:1.\nAnswer – e) If the data in both the statements I and II together are necessary to answer the question\nExplanation :\nFrom both equations, we get l = 12 and b= 6 so perimeter = 36\n\n5. How much profit did Bharti made by selling a chair?\n1) She bought the chair with 30% discount on marked price\n2) She sold it with 10% profit on labelled price.\nAnswer – d)If the data given in both I and II together are not sufficient to answer the question.\nExplanation :\nCp = (70/100)Mp\nSp = (110/100)mp\nWe can find the profit/loss percent but we cannot the amount of profit.\n\n6. How long will machine B, working alone take to produce ‘y’ candles?\n1) Machine A produces ‘y’ candles in 10 minutes\n2) Machine A and Machine B working together produces ‘y’ candles in 4 minutes\nAnswer – e) If the data in both the statements I and II together are necessary to answer the question\nExplanation :\nFrom both statements,\n1/A + 1/B = 1/4\nSo, 1/B = 1/4 – 1/10 = 6/40\nB = 20/3 minutes\n\n7. A train crosses a pole in 20 seconds. What is the length of the train?\n1) The train crosses another train running in same direction with the speed of 80km/hr in 11 seconds.\n2) Speed of the train is 54 km/hr.\nAnswer – b) If the data in statement II alone is sufficient to answer the question.\nExplanation :\nFrom second statement, we get\nL = 20545/18 = 300 meter\n\n8. What is the speed of boat in still water?\n1) It takes 4 hours to cover the distance between P and Q downstream\n2) It takes 6 hours to cover the distance between P and Q upstream.\nAnswer – d) If the data given in both I and II together are not sufficient to answer the question.\nExplanation :\nD = 4(b+s) and D = 6(b-s)\nSo we can’t find the value of b\n\n9. What is the compound interest earned by Rahul at the end of 2 years?\n1) The amount invested is rupees 17000\n2) Simple interest at the same rate for 2 year is rupees 2400 and the rate of interest is 10 p.c.p.a\nAnswer – b) If the data in statement II alone is sufficient to answer the question\nExplanation :\nFrom second statement, we get\n2400 = p10/1002\nP = 12000\nSo CI = 12000*(1 + 10/100)^2 – 12000 = 2520\n\n10. How much time did P take to reach the destination?\n1) Q takes 24 minutes to reach the same destination\n2) Ratio of the speed of P and Q is 3:4\nAnswer – e) If the data in both the statements I and II together are necessary to answer the question\nExplanation :\n3x/4x = 24/t\nWe get t = 32 minutes\n\nNote: Dear Readers if you have any doubt in any chapter in Quants you can ask here. We will clear your doubts\n\nAffairsCloud Ebook - Support Us to Grow" ]	[ null, "https://www.affairscloud.com/assets/uploads/2020/04/AC-Current-Affairs.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90459704,"math_prob":0.8943857,"size":4872,"snap":"2020-24-2020-29","text_gpt3_token_len":1306,"char_repetition_ratio":0.18303205,"word_repetition_ratio":0.2601713,"special_character_ratio":0.29495075,"punctuation_ratio":0.07858546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974442,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T12:03:15Z\",\"WARC-Record-ID\":\"<urn:uuid:5f8185bd-d402-430e-a87c-3c5286484b0d>\",\"Content-Length\":\"186481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5bc10828-281d-46f8-bf2c-14c293acbfd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:20684dec-b07d-42d2-a1e0-5ab2323d37b7>\",\"WARC-IP-Address\":\"172.67.68.210\",\"WARC-Target-URI\":\"https://affairscloud.com/aptitude-questions-data-sufficiency-set-10/\",\"WARC-Payload-Digest\":\"sha1:37SZ7ISMZLPTR3S374IMW3OXHNDTSCHS\",\"WARC-Block-Digest\":\"sha1:4XFBACSJVQB2TPF6EXXJUOLLSOQVFUSU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348513230.90_warc_CC-MAIN-20200606093706-20200606123706-00151.warc.gz\"}"}
http://tjmm.edyropress.ro/past/2/	[ "", null, "", null, "ISSN: 2067-239X\nISSN(on-line): 2067-239X\n\nIndexed in:\nMathematical Reviews\nZentralblatt MATH\nEBSCO\n\nFront cover", null, "# Volume 2 (2010), Number 1\n\n## One Derivative of One Component Regularity Criterion for the Navier-Stokes Equations\n\n### Author(s): YUAN-SHAN ZHAO AND YUE HU\n\nAbstract: We study the incompressible Navier-Stokes equations in the entire three-dimensional space and we prove that if there exists one derivative of one component of the velocity ${\\partial }_{3}{u}_{3}\\in {L}_{t}^{s}{L}_{x}^{r},\\text{\\hspace{0.17em}}\\text{where}\\text{\\hspace{0.17em}}\\frac{2}{s}+\\frac{3}{r}\\le \\frac{1}{4},\\text{\\hspace{0.17em}}12\\le r\\le \\infty$ then the solution is regular. This extends one result of Patrick Penel, Toulon, Milan Pokorý, Praha [Appl.Math.,49,483-493(2004)]\n\n## On Some Multiplier Difference Sequence Spaces Defined over a 2-normed Linear Space\n\n### Author(s): B.SURENDER REDDY AND HEMEN DUTTA\n\nAbstract: In this paper, we introduce a new class of generalized difference sequences with base space, a real linear 2-normed space and by means of a fixed multiplier. We study the spaces of thus constructed classes of sequences for relevant linear topological structures. Further we investigate the spaces for solidity, monotonicity, symmetricity etc. We also obtain some relations between these spaces as well as prove some inclusion results.\n\n## On Some Inequalities of Simpson-type Via Quasi-Convex Functions and Applications\n\n### Author(s): MOHAMMAD ALOMARI AND MASLINA DARUS\n\nAbstract: Some inequalities of Simpson's type for quasi-convex functions are introduced. In the literature the error estimates for the midpoint rule is $\|{E}_{mid}\\left(f,d\\right)\|\\le \\frac{K}{24}\\sum _{i=0}^{n-1}{\\left({x}_{i+1}-{x}_{i}\\right)}^{3}$ , in this paper we restrict the conditions on f to get better error estimates than the original.\n\n## Solving Cauchy Problem for a Class of Sixth-order Hyperbolic Equations with Triple Characteristics\n\n### Author(s): LAZHAR BOUGOFFA AND HIND K. AL-JEAID\n\nAbstract: In this paper, the Cauchy problem for a class of the homogeneous hyperbolic equations for sixth-order with triple characteristics is considered and can be solved analytically by direct integration techniques. Also, an efficient modification of Adomian decomposition method is proposed for solving this type of problems. We then conduct a comparative study between the ADM and direct method with the help of several illustrative examples.\n\n## New Explicit and Implicit Solutions to Elliptic Equations with Two Space Variables\n\nAbstract: We present a direct method for finding a new explicit and implicit solutions of elliptic equations with hyperbolic, trigonometric and exponential nonlinearities.\n\n## BAICA-CARDU Paratrigonometry, a Generalization of the Classical and Some New Non-classical Trigonometries and Its Application in Mechanics and Wave Theory\n\n### Author(s): M. BAICA AND M.CARDU\n\nAbstract: In their previous papers the authors introduced some new Trigonometries as: 1. Quadratic Trigonometry (QT), 2. Polygonal Trigonometry (PT), 3. Trans Trigonometry (TT), 4. Infra Trigonometry (IT), 5. Ultra-Trigonometry (UT), 6. Extra Trigonometry (ET), 7. Para-Trigonometry (PRT). This time in this paper we perform a synthesis of all these Trigonometries and state some of their applications.\n\n## Oscillation of a Class of Two-variables Functional Equations with Variable Coefficients\n\n### Author(s): WENGUI YANG\n\nAbstract: In this paper we will establish some sufficient conditions of oscillation of a class of two-variables functional equations with variable coefficients. Our results extend Zhang and Zhou's results (B.G. Zhang and Y. Zhou, Comput. Math. Appl. 42 (3-5) (2001) 369-378).\n\n## On Algebraic Properties of the Generalized Chebyshev Polynomials\n\n### Author(s): AHMET İPEK\n\nAbstract: Chebyshev polynomials are of great importance in many areas of mathematics, particularly approximation theory. Numerous articles and books have been written about this topic. Analytical properties of Chebyshev polynomials are well understood, but algebraic properties less so. In this paper, new generalized Chebyshev polynomials of the first and second kinds have been introduced and studied. Many of the properties of these polynomials are proved.\n\n## A Note on Bounds for the Spectral Norms of Circulant-Cauchy-Toeplitz Matrices\n\n### Author(s): AHMET İPEK\n\nAbstract: In this paper, we established lower and upper bounds for the spectral norms of some Circulant-Cauchy-Toeplitz matrices.\n\n## A Nonlinear Mixed Type Volterra-Fredholm Functional Integral Equation Via Perov's Theorem\n\nAbstract: In this paper we study the following mixed type Volterra-Fredholm functional integral equation $x\\left(t\\right)=F\\left(t,x\\left(t\\right),{\\int }_{{a}_{1}}^{{t}_{1}}\\dots {\\int }_{{a}_{m}}^{{t}_{m}}K\\left(t,s,x\\left(s\\right)\\right)ds,{\\int }_{{a}_{1}}^{{b}_{1}}\\dots {\\int }_{{a}_{m}}^{{b}_{m}}H\\left(t,s,x\\left(s\\right)\\right)ds\\right)$. Using the Perov's Theorem and the Picard operator technique we establish existence, uniqueness, data dependence and Gronwall results for the solutions.\n\n## The Solution of a System of Nonlinear Integral Equations from Physics\n\n### Author(s): MARIA DOBRIŢOIU\n\nAbstract: Using the Perov's Theorem and the General data dependence Theorem, in this paper, we obtain some conditions concerning the existence and uniqueness of the solution in the Banach space $C\\left(\\left[a,b\\right],{ℝ}^{m}\\right)$ and the continuous data dependence of the solution of the following system of nonlinear integral equations from physics: $x\\left(t\\right)={\\int }_{a}^{b}K\\left(t,s,x\\left(s\\right),x\\left(a\\right),x\\left(b\\right)\\right)ds+f\\left(t\\right),t\\in \\left[a,b\\right]$. An example is also given here.\n\n## Double Inequalities of Boole's Quadrature Rule\n\n### Author(s): MARIUS HELJIU\n\nAbstract: In this paper double inequalities of Boole's type quadrature rule are given. These inequalities are sharp." ]	[ null, "http://tjmm.edyropress.ro/files/sigla.png", null, "http://tjmm.edyropress.ro/files/title.png", null, "http://tjmm.edyropress.ro/files/coperta.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86597353,"math_prob":0.9667154,"size":6083,"snap":"2023-40-2023-50","text_gpt3_token_len":1373,"char_repetition_ratio":0.13390361,"word_repetition_ratio":0.0093240095,"special_character_ratio":0.1857636,"punctuation_ratio":0.114423074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99671733,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T07:54:06Z\",\"WARC-Record-ID\":\"<urn:uuid:51e89938-02ae-401d-907a-dec18bed1f70>\",\"Content-Length\":\"14838\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc03dec0-1650-48f6-b687-355e1677a4cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:59e503d8-5032-4577-ae1b-ff6f78ebab7c>\",\"WARC-IP-Address\":\"37.187.16.183\",\"WARC-Target-URI\":\"http://tjmm.edyropress.ro/past/2/\",\"WARC-Payload-Digest\":\"sha1:IXNGDCY75NQAWAPFNFNQCOUGKQ2TZ7EJ\",\"WARC-Block-Digest\":\"sha1:YU2ZHN6JTBWZHIJEN6OHFQ5T2XNDQAJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510284.49_warc_CC-MAIN-20230927071345-20230927101345-00707.warc.gz\"}"}
https://letusprogram.com/2013/09/30/c-program-to-print-number-in-pyramid-shape/	[ "## C++ program to print number in pyramid shape\n\nThis is once of the basic program to start programming in C++.Here your main goal is to print the number’s in pyramid shape.The size of the shape depends on the input of the user.The output is similar to this picture:", null, "For this we need to write the control statements such as for loop.Here in the program we use the method print_pyramid to print the output.Now we shall write a program to this problem:\n\n```#include<iostream.h>\n#include<conio.h>\nclass pyramid\n{\npublic:\nint n;\nvoid print_pyramid();\n};\nvoid pyramid :: print_pyramid()\n{\nfor(int i=1;i<=n;i++)\n{\nfor(int j=1;j<=i;j++)\n{\ncout<<j<<\" \";\n}\ncout<<\"\\n\";\n}\n}\nvoid main()\n{\nclrscr();\ncout<<\"Enter number of lines to be printed in pyramid shape:\";\npyramid ob;\ncin>>ob.n;\nob.print_pyramid();\ngetch();\n}```\n\nOutput:", null, "", null, "" ]	[ null, "https://encrypted-tbn1.gstatic.com/images", null, "https://letusprogram.files.wordpress.com/2013/09/pyramid.png", null, "https://1.gravatar.com/avatar/181177e93002db86371e930b78d10cc4", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7317937,"math_prob":0.86023897,"size":794,"snap":"2020-34-2020-40","text_gpt3_token_len":197,"char_repetition_ratio":0.1493671,"word_repetition_ratio":0.0,"special_character_ratio":0.28463477,"punctuation_ratio":0.18390805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9570231,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,8,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T10:48:24Z\",\"WARC-Record-ID\":\"<urn:uuid:e1f4bb27-f5bb-4740-93c4-2ae307e86d6d>\",\"Content-Length\":\"78206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64526162-adb4-4728-afa7-44f913cb896e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff852d3d-3cf2-43d3-9bb2-3ae00f9fbbd2>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://letusprogram.com/2013/09/30/c-program-to-print-number-in-pyramid-shape/\",\"WARC-Payload-Digest\":\"sha1:ZMVZ22ZNSNNYIE5I4K67HTUWFQUTPXKQ\",\"WARC-Block-Digest\":\"sha1:ZBHKTUXFCAUQMNCXC6RWU5NNPZGXSI56\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401641638.83_warc_CC-MAIN-20200929091913-20200929121913-00724.warc.gz\"}"}
https://tutorialcup.com/leetcode-solutions/binary-tree-zigzag-level-order-traversal-leetcode-solution.htm	[ "# Binary Tree Zigzag Level Order Traversal LeetCode Solution\n\n## Problem Statement\n\nBinary Tree Zigzag Level Order Traversal LeetCode Solution – Given the `root` of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).", null, "```Input: root = [3,9,20,null,null,15,7]\nOutput: [,[20,9],[15,7]]```\n\n## Explanation\n\nWe carry out simple level-order traversal but reverse alternate levels before adding them to the ans.\n\nIn this problem, we need to traverse the binary tree level by level. When we see levels in a binary tree, we need to think about BFS, because it is its logic: it first traverses all neighbors, before we go deeper. Here we also need to change direction on each level as well. So, the algorithm is the following:\n\n1. We create a queue, where we first put our root.\n2. `result` is to keep the final result and `direction`, equal to `1` or `-1` or we can keep it as a boolean in the direction of traverse.\n3. Then we start to traverse level by level: if we have `k` elements in queue currently, we remove them all and put their children instead. We continue to do this until our queue is empty. Meanwhile, we form `level` list and then add it to `result`, using correct direction and changing direction after.\n\n## Code\n\n### C++ Code for Binary Tree Zigzag Level Order Traversal\n\n```class Solution {\npublic:\nvector<vector<int> > zigzagLevelOrder(TreeNode* root) {\nif (root == NULL) {\nreturn vector<vector<int> > ();\n}\nvector<vector<int> > result;\n\nqueue<TreeNode> nodesQueue;\nnodesQueue.push(root);\nbool leftToRight = true;\n\nwhile ( !nodesQueue.empty()) {\nint size = nodesQueue.size();\nvector<int> row(size);\nfor (int i = 0; i < size; i++) {\nTreeNode node = nodesQueue.front();\nnodesQueue.pop();\n\n// find position to fill node's value\nint index = (leftToRight) ? i : (size - 1 - i);\n\nrow[index] = node->val;\nif (node->left) {\nnodesQueue.push(node->left);\n}\nif (node->right) {\nnodesQueue.push(node->right);\n}\n}\n// after this level\nleftToRight = !leftToRight;\nresult.push_back(row);\n}\nreturn result;\n}\n};```\n\n### Java Code for Binary Tree Zigzag Level Order Traversal\n\n```public class Solution {\npublic List<List<Integer>> zigzagLevelOrder(TreeNode root)\n{\nList<List<Integer>> sol = new ArrayList<>();\ntravel(root, sol, 0);\nreturn sol;\n}\n\nprivate void travel(TreeNode curr, List<List<Integer>> sol, int level)\n{\nif(curr == null) return;\n\nif(sol.size() <= level)\n{\n}\n\nList<Integer> collection = sol.get(level);\nif(level % 2 == 0) collection.add(curr.val);\n\ntravel(curr.left, sol, level + 1);\ntravel(curr.right, sol, level + 1);\n}\n}```\n\n### Python Code for Binary Tree Zigzag Level Order Traversal\n\n```class Solution:\ndef zigzagLevelOrder(self, root):\nif not root: return []\nqueue = deque([root])\nresult, direction = [], 1\n\nwhile queue:\nlevel = []\nfor i in range(len(queue)):\nnode = queue.popleft()\nlevel.append(node.val)\nif node.left: queue.append(node.left)\nif node.right: queue.append(node.right)\nresult.append(level[::direction])\ndirection *= (-1)\nreturn result```\n\nO(N)\n\nO(N)\n\nTranslate »" ]	[ null, "https://tutorialcup.com/wp-content/uploads/2022/04/tree11.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5996651,"math_prob":0.95502925,"size":3154,"snap":"2023-40-2023-50","text_gpt3_token_len":783,"char_repetition_ratio":0.12952381,"word_repetition_ratio":0.049356222,"special_character_ratio":0.27013317,"punctuation_ratio":0.19967794,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9968288,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T15:34:52Z\",\"WARC-Record-ID\":\"<urn:uuid:7319ce98-75eb-49ff-ae00-9512af2f9f9d>\",\"Content-Length\":\"153927\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3bc37361-d15b-4d06-a5c1-2b1d3c75a8e1>\",\"WARC-Concurrent-To\":\"<urn:uuid:718cea01-f375-444d-8c60-ffcd034889b8>\",\"WARC-IP-Address\":\"172.67.168.78\",\"WARC-Target-URI\":\"https://tutorialcup.com/leetcode-solutions/binary-tree-zigzag-level-order-traversal-leetcode-solution.htm\",\"WARC-Payload-Digest\":\"sha1:CE6LSQF2VG6UDSMHJ2ZXMXRULESSWBLX\",\"WARC-Block-Digest\":\"sha1:OJGIQOBAYPAN3BWGIEHTXIEMBHXFKHMS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100531.77_warc_CC-MAIN-20231204151108-20231204181108-00148.warc.gz\"}"}
https://solvedlib.com/n/fiil-in-each-blank-s0-that-the-resulling-stalement-is-true,10719433	[ "# FiIl In each blank s0 that the resulling stalement Is true.Consider (he following long division prablem.1OxBegin (he division process by\n\n###### Question:\n\nFiIl In each blank s0 that the resulling stalement Is true. Consider (he following long division prablem. 1Ox Begin (he division process by dividing In the dividend which oblalns_ Wrile Ihls result abovo Begin the division process by dividing which obtains Write this result above in the dividend:", null, "", null, "#### Similar Solved Questions\n\n##### . The most stable alkene is: Select one: a. 1-butene b. trans butenec. 2-methyl-2-butene d. cis butene\n. The most stable alkene is: Select one: a. 1-butene b. trans butenec. 2-methyl-2-butene d. cis butene...\n##### Usg random savala8; On 0 cormpukc _ +o estmate, Wth 7S7 20 nce range X1 -8x2+2X4 \| Vqume { Xe RH -2x,+4x+5x72 4x+Sx-2x2 43 ~Sx; +2Xs < 0\nUsg random savala8; On 0 cormpukc _ +o estmate, Wth 7S7 20 nce range X1 -8x2+2X4 \| Vqume { Xe RH -2x,+4x+5x72 4x+Sx-2x2 43 ~Sx; +2Xs < 0...\n##### TEICONOMEETRIC INTECRRALSEvaluate zach integral16.sin \" â‚¬ @16.gic? 0 co8? 0 a91 .sec2 rtal â‚¬ &\nTEICONOMEETRIC INTECRRALS Evaluate zach integral 16. sin \" â‚¬ @ 16. gic? 0 co8? 0 a9 1 . sec2 rtal â‚¬ &...\n##### [T] Find the minimum value of $N$ such that the remainder estimate $int_{N+1}^{infty} f<R_{N}<int_{N}^{infty} f$ guarantees that $sum_{n=1}^{N} a_{n}$ estimates $sum_{n=1}^{infty} a_{n}$, accurate to within the given error.$a_{n}=frac{1}{1+n^{2}}$, error $<10^{-3}$\n[T] Find the minimum value of $N$ such that the remainder estimate $int_{N+1}^{infty} f<R_{N}<int_{N}^{infty} f$ guarantees that $sum_{n=1}^{N} a_{n}$ estimates $sum_{n=1}^{infty} a_{n}$, accurate to within the given error.$a_{n}=frac{1}{1+n^{2}}$, error $<10^{-3}$...\n##### Question 2 On average; 2% of manufactured items from a factory are defective: The acccptance Or rejection ofa large batch fthe items is based on the following procedure: A random sample of 10 items is checked and the batch is accepted if the number of defective item is at most one, otherwise the batch is rejected.(a)Calculate the probability that the batch is accepted:.(6)Calculate the probability the batch has no defective item given that the batch is accepted.(c)A total of 35 batches is taken\nQuestion 2 On average; 2% of manufactured items from a factory are defective: The acccptance Or rejection ofa large batch fthe items is based on the following procedure: A random sample of 10 items is checked and the batch is accepted if the number of defective item is at most one, otherwise the bat...\n##### (4 points) Solve the initial value problem v' \| 2v f(t), %(0) Where: ( 2 if 0 <t<1. f(t) if 1 < +\n(4 points) Solve the initial value problem v' \| 2v f(t), %(0) Where: ( 2 if 0 <t<1. f(t) if 1 < +...\n##### Given millimeter-wave imaging system, with aperture diameter of 0.5 m; operating at wavelength of mm (300 GHz), what is the minimum feature size in the object that can be resolved;, if the object plane is at km distance?\nGiven millimeter-wave imaging system, with aperture diameter of 0.5 m; operating at wavelength of mm (300 GHz), what is the minimum feature size in the object that can be resolved;, if the object plane is at km distance?...\n##### Operating system 6) Banker's algorithm is used in deadlock avoidance, and the following table presents the...\noperating system 6) Banker's algorithm is used in deadlock avoidance, and the following table presents the current situation: hAvailable e Need Allocation 2 31vqas 5 2 3 1 10 PO 1 20 P1 2 0 1 4 20 P2 1 0 2 no0 1 1 to P3 43 1 3 3 2 P4 1 12 Questions: (1) is the current situation safe? (2) if ...\n##### (8) Lek Pax_Y be homeomorphism. ShJ thok Y fs {xdels} , BRe_ X fs indels} :\n(8) Lek Pax_Y be homeomorphism. ShJ thok Y fs {xdels} , BRe_ X fs indels} :...\n##### For the following exercises. use shells to lind the volumes of the given solids: Note that the rotated regions lie between the curve and the x-axis and are rotated around the y-axis\nFor the following exercises. use shells to lind the volumes of the given solids: Note that the rotated regions lie between the curve and the x-axis and are rotated around the y-axis...\n##### I need help with Part e. I have already constructed the histogram using R and Rstudio (picture below). I am just confused about the empirical rule and locating the data points on the intervals. If a...\nI need help with Part e. I have already constructed the histogram using R and Rstudio (picture below). I am just confused about the empirical rule and locating the data points on the intervals. If anyone could assist me, I'd appreciate it greatly! Thank you! e radon concentration (in pCi/liter)...\n##### The licorice plant, Glycyrrhiza glabra, grows in the wild in many Middle Eastern, European, and Western...\nThe licorice plant, Glycyrrhiza glabra, grows in the wild in many Middle Eastern, European, and Western Asian countries. The intensely sweet flavor of licorice root, which is thought to be about 150 times sweeter than table sugar, is due to the carbohydrate derivative, glycyrrhizic acid (1). How man...\n##### E E 10 00 8H1 8 2 J! 3Ez 0 a,b 0 0 2 8 H 3 2 1 = [ 5 7 { 1 1 ; [\nE E 10 00 8H1 8 2 J! 3 Ez 0 a,b 0 0 2 8 H 3 2 1 = [ 5 7 { 1 1 ; [...\n##### (Just need help with part F) File_size_(MB) Time_(sec) 77 31.1 93 35.1 85 35.3 94 35.7...\n(Just need help with part F) File_size_(MB) Time_(sec) 77 31.1 93 35.1 85 35.3 94 35.7 20 14.4 74 28.9 68 29.6 88 33.5 42 21.7 20 15.3 72 28.4 24 11.5 95 35.7 59 25.6 93 36.6 71 29.1 87 34.3 92 37.2 90 35.6 67 26.8 87 32.6 83 31.2 80 34.1 57 22.8 52 25.4 76 28.9 96 38.7 70 31.8 59 24.2 57 28.1 B...\n##### Discuss the importance of regular expressions in data analytics. Also, discuss the differences between the types...\nDiscuss the importance of regular expressions in data analytics. Also, discuss the differences between the types of regular expressions. Choose two types of regular expressions... For example: [brackets] (Matches the enclosed characters in any order anywhere in a string), and * wildcards (Matc...\n##### 15Find the solution set of the system of linear equations with the augmented matrix~3 52using the Gauss-Jordan elimnination method. (You must find the reduced row echelon form and determine the leadling variables and the free variables)\n15 Find the solution set of the system of linear equations with the augmented matrix ~3 52 using the Gauss-Jordan elimnination method. (You must find the reduced row echelon form and determine the leadling variables and the free variables)...\n##### FistonboonmKrndovHelp02 [ LeddanUreetol Cnloin (Goccmon attnn UmyorityCalgry)Cedn CClCNOTES W70 5 Booalc DoctFamelnCHEMAMenittFt19 CuckarHor; Atempt LQuestion 12 (0.25 points) Once you have the lotal numbor = molcs of H\" your lemcnace Fowdar sample und the number Moiee ascorb C acid in thu samplo you can determinc the moles citric acid in VouM tDsp sample.the total number mcles 0f Ht tDsp was 5900 mol and the number 0l moles of acorb C acid 0370 mol howmany males citric acid wcro this L\nFiston boonm Krndov Help 02 [ Leddan Ureetol Cnloin (Goccmon attnn Umyority Calgry) Cedn C ClC NOTES W70 5 Booalc Doct Fameln CHEMAM enitt Ft 19 CuckarHor; Atempt L Question 12 (0.25 points) Once you have the lotal numbor = molcs of H\" your lemcnace Fowdar sample und the number Moiee ascor...\n##### Question 3 (30 marks)Derive the Ordinary Least Squares (OLS) estimate for the simple linear regression model, i.e- Bo and 81 Be very specific.\nQuestion 3 (30 marks) Derive the Ordinary Least Squares (OLS) estimate for the simple linear regression model, i.e- Bo and 81 Be very specific....\n##### An auto repair shop has room for two cars: but there is only Ole repairman_ There are two types of cars that come to the shop_ Type and Type 2 and they arrive as independent Poisson processes with rates A and 42_ car arrives and there is room in the shop then it enters. otherwise it goes to the shop across the street. Type carsneed service times that are expon(/l;). the shop; Type cars have priority over Type 2 cars_ In other words _ if both types are present in the shop_ then the repairman work\nAn auto repair shop has room for two cars: but there is only Ole repairman_ There are two types of cars that come to the shop_ Type and Type 2 and they arrive as independent Poisson processes with rates A and 42_ car arrives and there is room in the shop then it enters. otherwise it goes to the shop..." ]	[ null, "https://cdn.numerade.com/ask_images/44ee3e7f66514ae6a259b60f675a6a66.jpg ", null, "https://cdn.numerade.com/previews/35a3ba2b-cd01-4e85-acf2-fb2883b19971_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8174657,"math_prob":0.97017294,"size":15884,"snap":"2023-14-2023-23","text_gpt3_token_len":4850,"char_repetition_ratio":0.096284635,"word_repetition_ratio":0.46300286,"special_character_ratio":0.3037648,"punctuation_ratio":0.15140133,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9687202,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T01:05:03Z\",\"WARC-Record-ID\":\"<urn:uuid:de91e867-917e-44e9-b97a-efc7a793fa4e>\",\"Content-Length\":\"87300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa28ee46-131e-45f9-88d1-f0fe21ece116>\",\"WARC-Concurrent-To\":\"<urn:uuid:cc80d86c-1163-4ca2-b5db-b93a647b96d2>\",\"WARC-IP-Address\":\"172.67.132.66\",\"WARC-Target-URI\":\"https://solvedlib.com/n/fiil-in-each-blank-s0-that-the-resulling-stalement-is-true,10719433\",\"WARC-Payload-Digest\":\"sha1:OFH3E3YOOHR5VFNBJ52E7HDYN72PUXE5\",\"WARC-Block-Digest\":\"sha1:F3BK6ETYMVKPMY5BHHZXDVQQ4EMSFZDK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649348.41_warc_CC-MAIN-20230603233121-20230604023121-00046.warc.gz\"}"}
https://proofwiki.org/wiki/Intersection_of_Power_Sets	[ "# Intersection of Power Sets\n\n## Theorem\n\nThe intersection of the power sets of two sets $S$ and $T$ is equal to the power set of their intersection:\n\n$\\powerset S \\cap \\powerset T = \\powerset {S \\cap T}$\n\n## Proof\n\n $\\displaystyle X$ $\\in$ $\\displaystyle \\powerset {S \\cap T}$ $\\displaystyle \\leadstoandfrom \\ \\$ $\\displaystyle X$ $\\subseteq$ $\\displaystyle S \\cap T$ Definition of Power Set $\\displaystyle \\leadstoandfrom \\ \\$ $\\displaystyle X$ $\\subseteq$ $\\displaystyle S \\land X \\subseteq T$ Definition of Set Intersection $\\displaystyle \\leadstoandfrom \\ \\$ $\\displaystyle X$ $\\in$ $\\displaystyle \\powerset S \\land X \\in \\powerset T$ Definition of Power Set $\\displaystyle \\leadstoandfrom \\ \\$ $\\displaystyle X$ $\\in$ $\\displaystyle \\powerset S \\cap \\powerset T$ Definition of Set Intersection\n\n$\\blacksquare$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.703402,"math_prob":0.9996979,"size":1300,"snap":"2020-34-2020-40","text_gpt3_token_len":405,"char_repetition_ratio":0.22762346,"word_repetition_ratio":0.24064171,"special_character_ratio":0.3423077,"punctuation_ratio":0.17961165,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999516,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-14T02:11:52Z\",\"WARC-Record-ID\":\"<urn:uuid:769d2cd0-e68a-4f53-bb7c-fc63d9e3aaf8>\",\"Content-Length\":\"36632\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29766c90-235f-4ad7-8e86-bbb30f504dba>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b9c40d7-18c0-4e21-b8f1-f1e4cd34a3f4>\",\"WARC-IP-Address\":\"104.27.169.113\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Intersection_of_Power_Sets\",\"WARC-Payload-Digest\":\"sha1:DSAYTFRSBBPQCUYPLNXAAQOP2JS45K2F\",\"WARC-Block-Digest\":\"sha1:FPHD2SKGG5AH56KNGUYZL4X6EZA2HEI4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439739134.49_warc_CC-MAIN-20200814011517-20200814041517-00231.warc.gz\"}"}
https://www.hepdata.net/search/?q=observables%3AASYM&author=Abazov%2C+Victor+Mukhamedovich	[ "Showing 25 of 43 results\n\n#### Measurement of the muon charge asymmetry in $p\\bar{p}$ $\\to$ W+X $\\to$ $\\mu$$\\nu$ + X events at $\\sqrt{s}$=1.96 TeV\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D88 (2013) 091102, 2013.\nInspire Record 1253555\n\nWe present a measurement of the muon charge asymmetry from the decay of the $W$ boson via W to mu nu using 7.3 fb^{-1} of integrated luminosity collected with the D0 detector at the Fermilab Tevatron Collider at sqrt{s} = 1.96 TeV. The muon charge asymmetry is presented in two kinematic regions in muon transverse momentum and event missing transverse energy: (p^{\\mu}_{T} > 25 GeV, \\met > 25 GeV) and (p^{\\mu}_{T} > 35 GeV, \\met > 35 GeV). The measured asymmetries are compared with theory predictions made using three parton distribution function sets. The predictions do not describe the data well for p^{\\mu}_{T} > 35 GeV, \\met > 35 GeV, and larger values of muon pseudorapidity.\n\n0 data tables match query\n\n#### Measurement of the forward-backward asymmetry in the distribution of leptons in $t\\bar{t}$ events in the lepton$+$jets channel\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D90 (2014) 072001, 2014.\nInspire Record 1283842\n0 data tables match query\n\n#### Measurement of the forward-backward asymmetry in top quark-antiquark production in ppbar collisions using the lepton+jets channel\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D90 (2014) 072011, 2014.\nInspire Record 1293918\n0 data tables match query\n\n#### Measurement of the electron charge asymmetry in $\\boldsymbol{p\\bar{p}\\rightarrow W+X \\rightarrow e\\nu +X}$ decays in $\\boldsymbol{p\\bar{p}}$ collisions at $\\boldsymbol{\\sqrt{s}=1.96}$ TeV\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D91 (2015) 032007, 2015.\nInspire Record 1333394\n\n<p>We present a measurement of the electron charge asymmetry in <inline-formula><mml:math display=\"inline\"><mml:mi>p</mml:mi><mml:mover accent=\"true\"><mml:mi>p</mml:mi><mml:mo accent=\"true\" stretchy=\"false\">¯</mml:mo></mml:mover><mml:mo stretchy=\"false\">→</mml:mo><mml:mi>W</mml:mi><mml:mo>+</mml:mo><mml:mi>X</mml:mi><mml:mo stretchy=\"false\">→</mml:mo><mml:mi>e</mml:mi><mml:mi>ν</mml:mi><mml:mo>+</mml:mo><mml:mi>X</mml:mi></mml:math></inline-formula> events at a center-of-mass energy of 1.96 TeV, using data corresponding to <inline-formula><mml:math display=\"inline\"><mml:mrow><mml:mn>9.7</mml:mn><mml:mtext> </mml:mtext><mml:mtext> </mml:mtext><mml:msup><mml:mrow><mml:mi>fb</mml:mi></mml:mrow><mml:mrow><mml:mo>−</mml:mo><mml:mn>1</mml:mn></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula> of integrated luminosity collected with the D0 detector at the Fermilab Tevatron Collider. The asymmetry is measured as a function of the electron pseudorapidity and is presented in five kinematic bins based on the electron transverse energy and the missing transverse energy in the event. The measured asymmetry is compared with next-to-leading-order predictions in perturbative quantum chromodynamics and provides accurate information for the determination of parton distribution functions of the proton. This is the most precise lepton charge asymmetry measurement to date.</p>\n\n0 data tables match query\n\n#### Measurement of the forward-backward asymmetry of $\\Lambda$ and $\\bar{\\Lambda}$ production in $p \\bar{p}$ collisions\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D93 (2016) 032002, 2016.\nInspire Record 1404885\n\nWe study $\\Lambda$ and $\\bar{\\Lambda}$ production asymmetries in $p \\bar{p} \\rightarrow \\Lambda (\\bar{\\Lambda}) X$, $p \\bar{p} \\rightarrow J/\\psi \\Lambda (\\bar{\\Lambda}) X$, and $p \\bar{p} \\rightarrow \\mu^\\pm \\Lambda (\\bar{\\Lambda}) X$ events recorded by the D0 detector at the Fermilab Tevatron collider at $\\sqrt{s} = 1.96$ TeV. We find an excess of $\\Lambda$'s ($\\bar{\\Lambda}$'s) produced in the proton (antiproton) direction. This forward-backward asymmetry is measured as a function of rapidity. We confirm that the $\\bar{\\Lambda}/\\Lambda$ production ratio, measured by several experiments with various targets and a wide range of energies, is a universal function of \"rapidity loss\", i.e., the rapidity difference of the beam proton and the lambda.\n\n0 data tables match query\n\n#### Measurement of the Forward-Backward Asymmetries in the Production of Ξ and Ω Baryons in $p\\overline{p}$ Collisions\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D93 (2016) 112001, 2016.\nInspire Record 1457606\n\nWe measure the forward-backward asymmetries AFB of charged Ξ and Ω baryons produced in pp¯ collisions recorded by the D0 detector at the Fermilab Tevatron collider at s=1.96 TeV as a function of the baryon rapidity y. We find that the asymmetries AFB for charged Ξ and Ω baryons are consistent with zero within statistical uncertainties.\n\n0 data tables match query\n\n#### Measurement of the W Boson Production Charge Asymmetry in $p\\bar{p}\\rightarrow W+X \\rightarrow e\\nu +X$ Events at $\\sqrt{s}=1.96$ TeV\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev.Lett. 112 (2014) 151803, 2014.\nInspire Record 1268647\n\n<p>We present a measurement of the <inline-formula><mml:math display=\"inline\"><mml:mi>W</mml:mi></mml:math></inline-formula> boson production charge asymmetry in <inline-formula><mml:math display=\"inline\"><mml:mrow><mml:mi>p</mml:mi><mml:mover accent=\"true\"><mml:mrow><mml:mi>p</mml:mi></mml:mrow><mml:mrow><mml:mo stretchy=\"false\">¯</mml:mo></mml:mrow></mml:mover><mml:mo stretchy=\"false\">→</mml:mo><mml:mi>W</mml:mi><mml:mo>+</mml:mo><mml:mi>X</mml:mi><mml:mo stretchy=\"false\">→</mml:mo><mml:mi>e</mml:mi><mml:mi>ν</mml:mi><mml:mo>+</mml:mo><mml:mi>X</mml:mi></mml:mrow></mml:math></inline-formula> events at a center of mass energy of 1.96 TeV, using <inline-formula><mml:math display=\"inline\"><mml:mrow><mml:mn>9.7</mml:mn><mml:mtext> </mml:mtext><mml:mtext> </mml:mtext><mml:msup><mml:mrow><mml:mi>fb</mml:mi></mml:mrow><mml:mrow><mml:mo>−</mml:mo><mml:mn>1</mml:mn></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula> of integrated luminosity collected with the D0 detector at the Fermilab Tevatron Collider. The neutrino longitudinal momentum is determined by using a neutrino weighting method, and the asymmetry is measured as a function of the <inline-formula><mml:math display=\"inline\"><mml:mi>W</mml:mi></mml:math></inline-formula> boson rapidity. The measurement extends over wider electron pseudorapidity region than previous results and is the most precise to date, allowing for precise determination of proton parton distribution functions in global fits.</p>\n\n0 data tables match query\n\n#### Measurement of the forward-backward asymmetry in $Λ_b^0$ and $\\bar{Λ}_b^0$ baryon production in $p\\bar{p}$ collisions at $\\sqrt{s}$=1.96 TeV\n\nThe collaboration Abazov, Victor Mukhamedovich ; Abbott, Braden Keim ; Acharya, Bannanje Sripath ; et al.\nPhys.Rev. D91 (2015) 072008, 2015.\nInspire Record 1352125\n\nWe measure the forward-backward asymmetry in the production of Λb0 and Λ¯b0 baryons as a function of rapidity in pp¯ collisions at s=1.96 TeV using 10.4 fb-1 of data collected with the D0 detector at the Fermilab Tevatron collider. The asymmetry is determined by the preference of Λb0 or Λ¯b0 particles to be produced in the direction of the beam protons or antiprotons, respectively. The measured asymmetry integrated over rapidity y in the range 0.1<\|y\|<2.0 is A=0.04±0.07(stat)±0.02(syst).\n\n0 data tables match query\n\n#### Measurement of the forward-backward charge asymmetry and extraction of sin*2 Theta(W)(eff) in p anti-p ---> Z/gamma + X ---> e+ e- + X events produced at s*(1/2) = 1.96$-TeV The collaboration Abazov, V.M. ; Abbott, B. ; Abolins, M. ; et al. Phys.Rev.Lett. 101 (2008) 191801, 2008. Inspire Record 783813 We present a measurement of the forward-backward charge asymmetry ($A_{FB}$) in$p\\bar{p} \\to Z/\\gamma^{}+X \\to e^+e^-+X$events at a center-of-mass energy of 1.96 TeV using 1.1 fb$^{-1}$of data collected with the D0 detector at the Fermilab Tevatron collider.$A_{FB}$is measured as a function of the invariant mass of the electron-positron pair, and found to be consistent with the standard model prediction. We use the$A_{FB}$measurement to extract the effective weak mixing angle sin$^2\\Theta^{eff}_W = 0.2327 \\pm 0.0018 (stat.) \\pm 0.0006 (syst.)$. 0 data tables match query #### rivet Analysis Measurement of the electron charge asymmetry in$p \\bar{p} \\to W + X \\to e \\nu + X$events at$\\sqrt{s}$= 1.96-TeV The collaboration Abazov, V.M. ; Abbott, B. ; Abolins, M. ; et al. Phys.Rev.Lett. 101 (2008) 211801, 2008. Inspire Record 791230 We present a measurement of the electron charge asymmetry in ppbar->W+X->enu+X events at a center of mass energy of 1.96 TeV using 0.75 fb-1 of data collected with the D0 detector at the Fermilab Tevatron Collider. The asymmetry is measured as a function of the electron transverse momentum and pseudorapidity in the interval (-3.2, 3.2) and is compared with expectations from next-to-leading order calculations in perturbative quantum chromodynamics. These measurements will allow more accurate determinations of the proton parton distribution functions. 0 data tables match query #### Inclusive and differential measurements of the$t\\overline{t}$charge asymmetry in pp collisions at$\\sqrt{s} =$8 TeV The collaboration Khachatryan, Vardan ; Sirunyan, Albert M ; Tumasyan, Armen ; et al. Phys.Lett. B757 (2016) 154-179, 2016. Inspire Record 1382590 0 data tables match query #### Measurement of the charge asymmetry in top quark pair production in pp collisions at$\\sqrt(s) =$8 TeV using a template method The collaboration Khachatryan, Vardan ; Sirunyan, Albert M ; Tumasyan, Armen ; et al. Phys.Rev. D93 (2016) 034014, 2016. Inspire Record 1388178 0 data tables match query #### Forward–backward asymmetry of Drell–Yan lepton pairs in pp collisions at$\\sqrt{s} = 8\\,\\mathrm{TeV}$The collaboration Khachatryan, Vardan ; Sirunyan, Albert M ; Tumasyan, Armen ; et al. Eur.Phys.J. C76 (2016) 325, 2016. Inspire Record 1415949 A measurement of the forward–backward asymmetry${A}_{\\mathrm{FB}}$of oppositely charged lepton pairs ($\\mu \\mu $and$\\mathrm{e}\\mathrm{e}$) produced via$\\mathrm{Z}/\\gamma ^$boson exchange in pp collisions at$\\sqrt{s} = 8\\,\\mathrm{TeV}$is presented. The data sample corresponds to an integrated luminosity of 19.7$\\,\\mathrm{fb}^{-1}$collected with the CMS detector at the LHC. The measurement of${A}_{\\mathrm{FB}}$is performed for dilepton masses between 40$\\,\\text {GeV}$and 2$\\,\\mathrm{TeV}$and for dilepton rapidity up to 5. The${A}_{\\mathrm{FB}}$measurements as a function of dilepton mass and rapidity are compared with the standard model predictions. 0 data tables match query #### rivet Analysis Measurement of the$W$charge asymmetry in the$W \\to \\mu \\nu$decay mode in$pp$collisions at$\\sqrt s=7$TeV with the ATLAS detector The collaboration Aad, Georges ; Abbott, Brad ; Abdallah, Jalal ; et al. Phys.Lett. B701 (2011) 31-49, 2011. Inspire Record 892704 This letter reports a measurement of the muon charge asymmetry from W Boson produced in proton-proton collisions at a centre-of-mass energy of 7 TeV with the ATLAS experiment at the LHC. The asymmetry is measured in the W Boson to muon decay mode as a function of the muon pseudorapidity using a data sample corresponding to a total integrated luminosity of 31 pb-1. The results are compared to predictions based on next-to-leading order calculations with various parton distribution functions. This measurement provides information on the u and d quark momentum fractions in the proton. 0 data tables match query #### Measurement of the electron charge asymmetry in inclusive$W$production in$pp$collisions at$\\sqrt{s}=7$TeV The collaboration Chatrchyan, Serguei ; Khachatryan, Vardan ; Sirunyan, Albert M ; et al. Phys.Rev.Lett. 109 (2012) 111806, 2012. Inspire Record 1118047 A measurement of the electron charge asymmetry in inclusive pp to W + X to e nu + X production at sqrt(s) = 7 TeV is presented based on data recorded by the CMS detector at the LHC and corresponding to an integrated luminosity of 840 inverse picobarns. The electron charge asymmetry reflects the unequal production of positive and negative W bosons in pp collisions. The electron charge asymmetry is measured in bins of absolute value of electron pseudorapidity in the range of abs(eta) < 2.4. The asymmetry rises from about 0.1 to 0.2 as a function of the pseudorapidity and is measured with a relative precision better than 7%. This measurement provides new stringent constraints for parton distribution functions. 0 data tables match query #### Measurement of the muon charge asymmetry in inclusive$pp \\to W+X$production at$\\sqrt s =$7 TeV and an improved determination of light parton distribution functions The collaboration Chatrchyan, Serguei ; Khachatryan, Vardan ; Sirunyan, Albert M ; et al. Phys.Rev. D90 (2014) 032004, 2014. Inspire Record 1273570 Measurements of the muon charge asymmetry in inclusive pp to WX production at sqrt(s) = 7 TeV are presented. The data sample corresponds to an integrated luminosity of 4.7 inverse femtobarns recorded with the CMS detector at the LHC. With a sample of more than twenty million W to mu nu events, the statistical precision is greatly improved in comparison to previous measurements. These new results provide additional constraints on the parton distribution functions of the proton in the range of the Bjorken scaling variable x from 10E-3 to 10E-1. These measurements and the recent CMS measurement of associated W + charm production are used together with the cross sections for inclusive deep inelastic ep scattering at HERA in a next-to-leading-order QCD analysis. The determination of the valence quark distributions is improved, and the strange-quark distribution is probed directly through the leading-order process g + s to W + c in proton-proton collisions at the LHC. 0 data tables match query #### Measurements of the$t\\bar{t}$charge asymmetry using the dilepton decay channel in pp collisions at$\\sqrt{s} =$7 TeV The collaboration Chatrchyan, Serguei ; Khachatryan, Vardan ; Sirunyan, Albert M ; et al. JHEP 1404 (2014) 191, 2014. Inspire Record 1281538 0 data tables match query #### rivet Analysis Forward-backward asymmetry of Drell-Yan lepton pairs in$pp$collisions at$\\sqrt{s} = 7$TeV The collaboration Chatrchyan, Serguei ; Khachatryan, Vardan ; Sirunyan, Albert M ; et al. Phys.Lett. B718 (2013) 752-772, 2013. Inspire Record 1122847 0 data tables match query #### rivet Analysis rivet Analysis rivet Analysis Measurement of the forward-backward asymmetry of electron and muon pair-production in$pp$collisions at$\\sqrt{s}$= 7 TeV with the ATLAS detector The collaboration Aad, Georges ; Abbott, Brad ; Abdallah, Jalal ; et al. JHEP 1509 (2015) 049, 2015. Inspire Record 1351916 This paper presents measurements from the ATLAS experiment of the forward-backward asymmetry in the reaction pp → Z/γ$^{}$→ l$^{+}$l$^{−}$, with l being electrons or muons, and the extraction of the effective weak mixing angle. The results are based on the full set of data collected in 2011 in pp collisions at the LHC at$ \\sqrt{s}=7 $TeV, corresponding to an integrated luminosity of 4.8 fb$^{−1}$. The measured asymmetry values are found to be in agreement with the corresponding Standard Model predictions. The combination of the muon and electron channels yields a value of the effective weak mixing angle of sin$^{2}$θ$_{eff}^{lept}$= 0.2308 ± 0.0005(stat.) ± 0.0006(syst.) ± 0.0009(PDF), where the first uncertainty corresponds to data statistics, the second to systematic effects and the third to knowledge of the parton density functions. This result agrees with the current world average from the Particle Data Group fit. 0 data tables match query #### Measurement of the forward-backward asymmetry in$Z/\\gamma^{\\ast} \\rightarrow \\mu^{+}\\mu^{-}$decays and determination of the effective weak mixing angle The collaboration Aaij, Roel ; Adeva, Bernardo ; Adinolfi, Marco ; et al. JHEP 1511 (2015) 190, 2015. Inspire Record 1394859 The forward-backward charge asymmetry for the process$ q\\overline{q}\\to Z/{\\gamma}^{\\ast}\\to {\\mu}^{+}{\\mu}^{-} $is measured as a function of the invariant mass of the dimuon system. Measurements are performed using proton proton collision data collected with the LHCb detector at$ \\sqrt{s}=7 $and 8 TeV, corresponding to integrated luminosities of 1 fb$^{−1}$and 2 fb$^{−1}$respectively. Within the Standard Model the results constrain the effective electroweak mixing angle to be$ { \\sin}^2{\\theta}_{\\mathrm{W}}^{\\mathrm{eff}}=0.23142\\pm 0.00073\\pm 0.00052\\pm 0.00056, $where the first uncertainty is statistical, the second systematic and the third theoretical. This result is in agreement with the current world average, and is one of the most precise determinations at hadron colliders to date. 0 data tables match query #### Measurement of the charge asymmetry in top-quark pair production in the lepton-plus-jets final state in pp collision data at$\\sqrt{s}=8\\,\\mathrm TeV{}$with the ATLAS detector The collaboration Aad, Georges ; Abbott, Brad ; Abdallah, Jalal ; et al. Eur.Phys.J. C76 (2016) 87, 2016. Inspire Record 1392455 This paper reports inclusive and differential measurements of the$t\\bar{t}$charge asymmetry$A_{\\text {C}}$in$20.3~{\\mathrm{fb}^{-1}}$of$\\sqrt{s} = 8~\\mathrm TeV{}pp$collisions recorded by the ATLAS experiment at the Large Hadron Collider at CERN. Three differential measurements are performed as a function of the invariant mass, transverse momentum and longitudinal boost of the$t\\bar{t}$system. The$t\\bar{t}$pairs are selected in the single-lepton channels (e or$\\mu $) with at least four jets, and a likelihood fit is used to reconstruct the$t\\bar{t}$event kinematics. A Bayesian unfolding procedure is performed to infer the asymmetry at parton level from the observed data distribution. The inclusive$t\\bar{t}$charge asymmetry is measured to be$A_{\\text {C}}{} = 0.009 \\pm 0.005$(stat.$+$syst.). The inclusive and differential measurements are compatible with the values predicted by the Standard Model. 0 data tables match query #### Measurement of the charge asymmetry in highly boosted top-quark pair production in$\\sqrt{s} =$8 TeV$pp$collision data collected by the ATLAS experiment The collaboration Aad, Georges ; Abbott, Brad ; Abdallah, Jalal ; et al. Phys.Lett. B756 (2016) 52-71, 2016. Inspire Record 1410588 In the pp→tt¯ process the angular distributions of top and anti-top quarks are expected to present a subtle difference, which could be enhanced by processes not included in the Standard Model. This Letter presents a measurement of the charge asymmetry in events where the top-quark pair is produced with a large invariant mass. The analysis is performed on 20.3 fb −1 of pp collision data at s=8TeV collected by the ATLAS experiment at the LHC, using reconstruction techniques specifically designed for the decay topology of highly boosted top quarks. The charge asymmetry in a fiducial region with large invariant mass of the top-quark pair ( mtt¯>0.75 TeV ) and an absolute rapidity difference of the top and anti-top quark candidates within −2<\|yt\|−\|yt¯\|<2 is measured to be 4.2±3.2% , in agreement with the Standard Model prediction at next-to-leading order. A differential measurement in three tt¯ mass bins is also presented. 0 data tables match query #### Measurement of the$\\Lambda_b$polarization and angular parameters in$\\Lambda_b\\to J/\\psi\\, \\Lambda$decays from pp collisions at$\\sqrt{s}=$7 and 8 TeV The collaboration Sirunyan, Albert M ; Tumasyan, Armen ; Adam, Wolfgang ; et al. Phys.Rev. D97 (2018) 072010, 2018. Inspire Record 1654926 An analysis of the bottom baryon decay Λb→J/ψ(→μ+μ-)Λ(→pπ-) is performed to measure the Λb polarization and three angular parameters in data from pp collisions at s=7 and 8 TeV, collected by the CMS experiment at the Large Hadron Collider. The Λb polarization is measured to be 0.00±0.06(stat)±0.06(syst) and the parity-violating asymmetry parameter is determined to be 0.14±0.14(stat)±0.10(syst). The measurements are compared to various theoretical predictions, including those from perturbative quantum chromodynamics. 0 data tables match query #### Study of the production of$\\Lambda_b^0$and$\\overline{B}^0$hadrons in$pp$collisions and first measurement of the$\\Lambda_b^0\\rightarrow J/\\psi pK^-$branching fraction The collaboration Aaij, R. ; Adeva, Bernardo ; Adinolfi, Marco ; et al. Chin.Phys. C40 (2016) 011001, 2016. Inspire Record 1391317 The product of the differential production cross-section and the branching fraction of the decay is measured as a function of the beauty hadron transverse momentum, p(T), and rapidity, y. The kinematic region of the measurements is p(T) < 20 GeV/c and 2.0 < y < 4.5. The measurements use a data sample corresponding to an integrated luminosity of 3fb(−)(1) collected by the LHCb detector in pp collisions at centre-of-mass energies in 2011 and in 2012. Based on previous LHCb results of the fragmentation fraction ratio the branching fraction of the decay is measured to bewhere the first uncertainty is statistical, the second is systematic, the third is due to the uncertainty on the branching fraction of the decay B̅(0) → J/ψK̅*(892)(0), and the fourth is due to the knowledge of . The sum of the asymmetries in the production and decay between and is also measured as a function of p(T) and y. The previously published branching fraction of , relative to that of , is updated. The branching fractions of are determined. 0 data tables match query #### Measurement of$D_s^{\\pm}$production asymmetry in$pp$collisions at$\\sqrt{s} =7$and 8 TeV The collaboration Aaij, Roel ; Adeva, Bernardo ; Adinolfi, Marco ; et al. JHEP 1808 (2018) 008, 2018. Inspire Record 1674916 The inclusive D$_{s}^{±}$production asymmetry is measured in pp collisions collected by the LHCb experiment at centre-of-mass energies of$ \\sqrt{s}=7 $and 8 TeV. Promptly produced D$_{s}^{±}$mesons are used, which decay as D$_{s}^{±}$→ ϕπ$^{±}$, with ϕ → K$^{+}$K$^{−}$. The measurement is performed in bins of transverse momentum, p$_{T}$, and rapidity, y, covering the range 2.5 < p$_{T}$< 25.0 GeV/c and 2.0 < y < 4.5. No kinematic dependence is observed. Evidence of nonzero D$_{s}^{±}\\$ production asymmetry is found with a significance of 3.3 standard deviations.\n\n0 data tables match query" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7596163,"math_prob":0.9515855,"size":14980,"snap":"2020-34-2020-40","text_gpt3_token_len":4115,"char_repetition_ratio":0.16700053,"word_repetition_ratio":0.1232023,"special_character_ratio":0.2528705,"punctuation_ratio":0.108052306,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98585755,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T21:49:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f7c66321-515f-4f66-a7a8-9b71aa603a10>\",\"Content-Length\":\"167886\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea025fa6-355e-470e-9440-4a6fb7cc8c08>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f381c19-999e-45d6-ad17-0961a3853f2f>\",\"WARC-IP-Address\":\"188.184.99.23\",\"WARC-Target-URI\":\"https://www.hepdata.net/search/?q=observables%3AASYM&author=Abazov%2C+Victor+Mukhamedovich\",\"WARC-Payload-Digest\":\"sha1:2HL76PKAZ6OJMAGKGJKKNTYL2T75JFAE\",\"WARC-Block-Digest\":\"sha1:4M7YTY7REU3RKS2MHP4RZHG44NNPLUPD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737225.57_warc_CC-MAIN-20200807202502-20200807232502-00484.warc.gz\"}"}
https://math.stackexchange.com/questions/263583/counting-no-of-ways-to-make-a-necklace	[ "Counting no. of ways to make a necklace\n\nClaudia wants to use 8 indistinguishable red beads and 32 indistinguishable blue beads to make a necklace such that there are at least 2 blue beads between any red beads. In how many ways can she do this? This is one of the unsolved problem in the book 'A Path to Combinatorics for Undergraduates' by Titu Andreescu and Zuming Feng. My approach: Denote blue by b, red by r. Then we create elements of the form 'brb'(8 elements), and 16'b's. We place the 16 'b's arounds a circle with a space in between them and we choose from those 16 available places 8 for the 'brb' and divide the whole be 2 to account for rotational symmetry.\n\nI am not sure whether I am fully accounting for the rotational symmetry and the fact that the beads are indistinguishable. Not Homework. Trying to learn. Thank you.\n\n• Look up Polyá Counting or Burnsides Lemma. – utdiscant Dec 22 '12 at 7:11\n• Okay I will but I was looking for something simple. I am only in high school. – Noel Dec 22 '12 at 13:32\n• Sorry - irrelevant - but I laughed a little. The book's called \"A Path to Combinatorics for Undergraduates,\" it's natural that you'll receive undergraduate tools here, regardless of what level of schooling you're actually at. :) – Gyu Eun Lee Dec 22 '12 at 23:31\n\nThis problem can indeed be solved by Polya counting. First place the eight red beads along a circle, leaving some kind of space between them for the blue beads. Now place two blue beads between each adjacent pair of red beads as well as an empty slot that will be filled later. That leaves $32-2*8 = 16$ blue beads. Now the empty slots may be filled by any number of blue beads as long as there are $16$ blue beads in total. This means that the blue beads going into the eight slots have generating function $$f(z) = \\frac{1}{1-z}.$$ What we have in fact is that the generating function is $$g(z) = 1 + z + z^2 + z^3 + \\cdots z^{15} + z^{16}$$ but we shall see that we can use $f(z)$ in place of $g(z)$ with no overcounting.\nNow the eight slots are being permuted by $C_8$, the cylic group of order $8.$ The cycle index of the cyclic group $C_n$ is $$Z(C_n) = \\frac{1}{n} \\sum_{d\|n} \\varphi(d) a_d^{n/d}$$ so that $Z(C_8)$ is $$Z(C_8) = \\frac{1}{8} \\left(a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8 \\right).$$ It follows that the number of orbits or necklace configurations is given by $$[z^{16}] Z(C_8)_{a_1=f(z); a_2=f(z^2); a_4=f(z^4); a_8=f(z^8)}$$ which is $$[z^{16}] \\frac{1}{8} \\left(\\frac{1}{(1-z)^8} + \\frac{1}{(1-z^2)^4} + 2 \\frac{1}{(1-z^4)^2} + 4 \\frac{1}{(1-z^8)} \\right).$$ Finally recall that $$[z^n] \\frac{1}{(1-z)^q} = \\binom{n+q-1}{q-1}$$ so that the value of the coefficient of $z^{16}$ is $$\\frac{1}{8} \\left( \\binom{16+7}{7} +\\binom{8+3}{3} +2\\binom{4+1}{1} +4\\binom{2+0}{0}\\right) = 30667.$$ In the case where instead of the cyclic group $C_n$ the dihedral group $D_n$ is acting on the slots on the necklace (i.e., the symmetry includes reflections) the cycle index is given by $$Z(D_n) = \\frac{1}{2} Z(C_n) + \\frac{1}{4} \\left( a_1^2 a_2^{(n-2)/2} + a_2^{n/2}\\right)$$ so that the answer becomes $$\\frac{1}{2} 30667 + \\frac{1}{4} [z^{16}] \\left( \\frac{1}{(1-z)^2}\\frac{1}{(1-z^2)^3} + \\frac{1}{(1-z^2)^4}\\right) = 15581.$$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92201686,"math_prob":0.9996542,"size":795,"snap":"2019-26-2019-30","text_gpt3_token_len":199,"char_repetition_ratio":0.11378002,"word_repetition_ratio":0.0,"special_character_ratio":0.22893082,"punctuation_ratio":0.08176101,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T19:53:56Z\",\"WARC-Record-ID\":\"<urn:uuid:eb7e0aa7-a123-4ffd-b4cd-665b46a1a937>\",\"Content-Length\":\"144829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a36329a-5e9e-44e0-b033-62f1f33fd1ed>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5afc007-b695-441b-9035-1fe72b336af6>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/263583/counting-no-of-ways-to-make-a-necklace\",\"WARC-Payload-Digest\":\"sha1:FXPGYCP7K24PXM2KYWDZL2CCDQU2NAS4\",\"WARC-Block-Digest\":\"sha1:SDFCKFGSNVH5U6EUWIFID2NP2F5BDFNB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999273.24_warc_CC-MAIN-20190620190041-20190620212041-00361.warc.gz\"}"}
https://www.colorhexa.com/242a27	[ "# #242a27 Color Information\n\nIn a RGB color space, hex #242a27 is composed of 14.1% red, 16.5% green and 15.3% blue. Whereas in a CMYK color space, it is composed of 14.3% cyan, 0% magenta, 7.1% yellow and 83.5% black. It has a hue angle of 150 degrees, a saturation of 7.7% and a lightness of 15.3%. #242a27 color hex could be obtained by blending #48544e with #000000. Closest websafe color is: #333333.\n\n• R 14\n• G 16\n• B 15\nRGB color chart\n• C 14\n• M 0\n• Y 7\n• K 84\nCMYK color chart\n\n#242a27 color description : Very dark (mostly black) cyan - lime green.\n\n# #242a27 Color Conversion\n\nThe hexadecimal color #242a27 has RGB values of R:36, G:42, B:39 and CMYK values of C:0.14, M:0, Y:0.07, K:0.84. Its decimal value is 2370087.\n\nHex triplet RGB Decimal 242a27 `#242a27` 36, 42, 39 `rgb(36,42,39)` 14.1, 16.5, 15.3 `rgb(14.1%,16.5%,15.3%)` 14, 0, 7, 84 150°, 7.7, 15.3 `hsl(150,7.7%,15.3%)` 150°, 14.3, 16.5 333333 `#333333`\nCIE-LAB 16.392, -3.41, 1.06 1.922, 2.177, 2.238 0.303, 0.344, 2.177 16.392, 3.571, 162.727 16.392, -2.497, 1.318 14.756, -2.578, 1.336 00100100, 00101010, 00100111\n\n# Color Schemes with #242a27\n\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #2a2427\n``#2a2427` `rgb(42,36,39)``\nComplementary Color\n• #242a24\n``#242a24` `rgb(36,42,36)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #242a2a\n``#242a2a` `rgb(36,42,42)``\nAnalogous Color\n• #2a2424\n``#2a2424` `rgb(42,36,36)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #2a242a\n``#2a242a` `rgb(42,36,42)``\nSplit Complementary Color\n• #2a2724\n``#2a2724` `rgb(42,39,36)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #27242a\n``#27242a` `rgb(39,36,42)``\n• #272a24\n``#272a24` `rgb(39,42,36)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #27242a\n``#27242a` `rgb(39,36,42)``\n• #2a2427\n``#2a2427` `rgb(42,36,39)``\n• #010101\n``#010101` `rgb(1,1,1)``\n• #0c0f0e\n``#0c0f0e` `rgb(12,15,14)``\n• #181c1a\n``#181c1a` `rgb(24,28,26)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #303834\n``#303834` `rgb(48,56,52)``\n• #3c4541\n``#3c4541` `rgb(60,69,65)``\n• #47534d\n``#47534d` `rgb(71,83,77)``\nMonochromatic Color\n\n# Alternatives to #242a27\n\nBelow, you can see some colors close to #242a27. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #242a26\n``#242a26` `rgb(36,42,38)``\n• #242a26\n``#242a26` `rgb(36,42,38)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #242a28\n``#242a28` `rgb(36,42,40)``\n• #242a28\n``#242a28` `rgb(36,42,40)``\n• #242a29\n``#242a29` `rgb(36,42,41)``\nSimilar Colors\n\n# #242a27 Preview\n\nThis text has a font color of #242a27.\n\n``<span style=\"color:#242a27;\">Text here</span>``\n#242a27 background color\n\nThis paragraph has a background color of #242a27.\n\n``<p style=\"background-color:#242a27;\">Content here</p>``\n#242a27 border color\n\nThis element has a border color of #242a27.\n\n``<div style=\"border:1px solid #242a27;\">Content here</div>``\nCSS codes\n``.text {color:#242a27;}``\n``.background {background-color:#242a27;}``\n``.border {border:1px solid #242a27;}``\n\n# Shades and Tints of #242a27\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #090a0a is the darkest color, while #ffffff is the lightest one.\n\n• #090a0a\n``#090a0a` `rgb(9,10,10)``\n• #121513\n``#121513` `rgb(18,21,19)``\n• #1b1f1d\n``#1b1f1d` `rgb(27,31,29)``\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #2d3531\n``#2d3531` `rgb(45,53,49)``\n• #363f3b\n``#363f3b` `rgb(54,63,59)``\n• #3f4a44\n``#3f4a44` `rgb(63,74,68)``\n• #48544e\n``#48544e` `rgb(72,84,78)``\n• #515f58\n``#515f58` `rgb(81,95,88)``\n• #5a6962\n``#5a6962` `rgb(90,105,98)``\n• #63746c\n``#63746c` `rgb(99,116,108)``\n• #6c7e75\n``#6c7e75` `rgb(108,126,117)``\n• #75897f\n``#75897f` `rgb(117,137,127)``\n• #809289\n``#809289` `rgb(128,146,137)``\n• #8b9b93\n``#8b9b93` `rgb(139,155,147)``\n• #95a49d\n``#95a49d` `rgb(149,164,157)``\n``#a0ada7` `rgb(160,173,167)``\n• #aab6b0\n``#aab6b0` `rgb(170,182,176)``\n• #b5bfba\n``#b5bfba` `rgb(181,191,186)``\n• #bfc8c4\n``#bfc8c4` `rgb(191,200,196)``\n``#cad2ce` `rgb(202,210,206)``\n• #d5dbd8\n``#d5dbd8` `rgb(213,219,216)``\n• #dfe4e1\n``#dfe4e1` `rgb(223,228,225)``\n• #eaedeb\n``#eaedeb` `rgb(234,237,235)``\n• #f4f6f5\n``#f4f6f5` `rgb(244,246,245)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #242a27\n\nA tone is produced by adding gray to any pure hue. In this case, #242a27 is the less saturated color, while #004e27 is the most saturated one.\n\n• #242a27\n``#242a27` `rgb(36,42,39)``\n• #212d27\n``#212d27` `rgb(33,45,39)``\n• #1e3027\n``#1e3027` `rgb(30,48,39)``\n• #1b3327\n``#1b3327` `rgb(27,51,39)``\n• #183627\n``#183627` `rgb(24,54,39)``\n• #153927\n``#153927` `rgb(21,57,39)``\n• #123c27\n``#123c27` `rgb(18,60,39)``\n• #0f3f27\n``#0f3f27` `rgb(15,63,39)``\n• #0c4227\n``#0c4227` `rgb(12,66,39)``\n• #094527\n``#094527` `rgb(9,69,39)``\n• #064827\n``#064827` `rgb(6,72,39)``\n• #034b27\n``#034b27` `rgb(3,75,39)``\n• #004e27\n``#004e27` `rgb(0,78,39)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #242a27 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.52884483,"math_prob":0.7148318,"size":3686,"snap":"2023-14-2023-23","text_gpt3_token_len":1662,"char_repetition_ratio":0.13986963,"word_repetition_ratio":0.010989011,"special_character_ratio":0.56294084,"punctuation_ratio":0.23444444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99190354,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-28T16:05:00Z\",\"WARC-Record-ID\":\"<urn:uuid:47098c26-427b-4a15-9b3c-bad38ec3c8a6>\",\"Content-Length\":\"36115\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:164f152e-90c7-4b75-8367-d473f6c47732>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9578878-9762-4cde-8e53-24f0e05c704e>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/242a27\",\"WARC-Payload-Digest\":\"sha1:H5JLPOMZQ5QWCJGNWR7XRY4277YVIXUN\",\"WARC-Block-Digest\":\"sha1:4TZPTIXIGIM56CKRKZIT4JTIX773TDGC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948867.32_warc_CC-MAIN-20230328135732-20230328165732-00048.warc.gz\"}"}
https://deepai.org/publication/diagnosing-bottlenecks-in-deep-q-learning-algorithms	[ "", null, "# Diagnosing Bottlenecks in Deep Q-learning Algorithms\n\nQ-learning methods represent a commonly used class of algorithms in reinforcement learning: they are generally efficient and simple, and can be combined readily with function approximators for deep reinforcement learning (RL). However, the behavior of Q-learning methods with function approximation is poorly understood, both theoretically and empirically. In this work, we aim to experimentally investigate potential issues in Q-learning, by means of a \"unit testing\" framework where we can utilize oracles to disentangle sources of error. Specifically, we investigate questions related to function approximation, sampling error and nonstationarity, and where available, verify if trends found in oracle settings hold true with modern deep RL methods. We find that large neural network architectures have many benefits with regards to learning stability; offer several practical compensations for overfitting; and develop a novel sampling method based on explicitly compensating for function approximation error that yields fair improvement on high-dimensional continuous control domains.\n\n## Authors\n\n##### This week in AI\n\nGet the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.\n\n## 1 Introduction\n\nQ-learning algorithms, which are based on approximating state-action value functions, are an efficient and commonly used class of RL methods. In recent years, such methods have been applied to great effect in domains such as playing video games from raw pixels (Mnih et al., 2015) and continuous control in robotics (Kalashnikov et al., 2018)\n\n. Methods based on approximate dynamic programming and Q-function estimation have several very appealing properties: they are generally moderately sample-efficient, when compared to policy gradient methods, they are simple to use, and they allow for off-policy learning. This makes them an appealing choice for a wide range of tasks, from robotic control\n\n(Kalashnikov et al., 2018) to off-policy learning from historical data for recommender (Shani et al., 2005) systems and other applications. However, although the basic tabular Q-learning algorithm is convergent and admits theoretical analysis (Sutton & Barto, 2018), its non-linear counterpart with function approximation (such as with deep neural networks) is poorly understood theoretically. In this paper, we aim to investigate the degree to which the theoretical issues with Q-learning actually manifest in practice. Thus, we empirically analyze aspects of the Q-learning method in a unit testing framework, where we can employ oracle solvers to obtain ground truth Q-functions and distributions for exact analysis. We investigate the following questions:\n\n1) What is the effect of function approximation on convergence? Most practical reinforcement learning problems, such as robotic control, require function approximation to handle large or continuous state spaces. However, the behavior of Q-learning methods under function approximation is not well understood. There are known counterexamples where the method diverges (Baird, 1995), and there are no known convergence guarantees (Sutton & Barto, 2018). To investigate these problems, we study the convergence behavior of Q-learning methods with function approximation, parametrically varying the function approximator power and analyzing the quality of the solution as compared to the optimal Q-function and the optimal projected Q-function under that function approximator. We find, somewhat surprisingly, that function approximation error is not a major problem in Q-learning algorithms, but only when the representational capacity of the function approximator is high. This makes sense in light of the theory: a high-capacity function approximator can perform a nearly perfect projection of the backed up Q-function, thus mitigating potentially convergence issues due to an imperfect norm projection. We also find that divergence rarely occurs, for example, we observed divergence in only 0.9% of our experiments. We discuss this further in Section 4.\n\n2) What is the effect of sampling error and overfitting? Q-learning is used to solve problems where we do not have access to the transition function of the MDP. Thus, Q-learning methods need to learn by collecting samples in the environment, and training on these samples incurs sampling error, potentially leading to overfitting. This causes errors in the computation of the Bellman backup, which degrades the quality of the solution. We experimentally show that overfitting exists in practice by performing ablation studies on the number of gradient steps, and by demonstrating that oracle based early stopping techniques can be used to improve performance of Q-learning algorithms. (Section 5). Thus, in our experiments we quantify the amount of overfitting which happens in practice, incorporating a variety of metrics, an performing a number of ablations and investigate methods to mitigate its effects.\n\n3) What is the effect of distribution shift and a moving target? The standard formulation of Q-learning prescribes an update rule, with no corresponding objective function (Sutton et al., 2009a). This results in a process which optimizes an objective that is non-stationary in two ways: the target values are updated during training, and the distribution under which the Bellman error is optimized changes, as samples are drawn from different policies. We refer to these problems as the moving target and distribution shift problems, respectively. These properties can make convergence behavior difficult to understand, and prior works have hypothesized that nonstationarity is a source of instability (Mnih et al., 2015; Lillicrap et al., 2015). In our experiments, we develop metrics to quantify the amount of distribution shift and performance change due to non-stationary targets. Surprisingly, we find that in a controlled experiment, distributional shift and non-stationary targets do not in fact correlate with reduction in performance. In fact, sampling strategies with large distributional shift often perform very well.\n\n4) What is the best sampling or weighting distribution? Deeply tied to the distribution shift problem is the choice of which distribution to sample from. Do moving distributions cause instability, as Q-values trained on one distribution are evaluated under another in subsequent iterations? Researchers have often noted that on-policy samples are typically superior to off-policy samples (Sutton & Barto, 2018), and there are several theoretical results that highlight favorable convergence properties under on-policy samples. However, there is little theoretical guidance on how to pick distributions so as to maximize learning rate. To this end, we investigate several choices for the sampling distribution. Surprisingly, we find that on-policy training distributions are not always preferable, and that a clear pattern in performance with respect to training distribution is that broader, higher-entropy distributions perform better, regardless of distributional shift. Motivated by our findings, we propose a novel weighting distribution, adversarial feature matching (AFM), which is explicitly compensates for function approximator error, while still producing high-entropy sampling distributions.\n\nOur contributions are as follows: We introduce a unit testing framework for Q-learning to disentangle issues related to function approximation, sampling, and distributional shift where approximate components are replaced by oracles. This allows for controlled analysis of different sources of error. We perform a detailed experimental analysis of many hypothesized sources of instability, error, and slow training in Q-learning algorithms on tabular domains, and show that many of these trends hold true in high dimensional domains. We propose novel choices of sampling distributions which lead to improved performance even on high-dimensional tasks. Our overall aim is to offer practical guidance for designing RL algorithms, as well as to identify important issues to solve in future research.\n\n## 2 Preliminaries\n\nQ-learning algorithms aim to solve a Markov decision process (MDP) by learning the optimal state-action value function, or Q-function. We define an MDP as a tuple\n\n. represent the state and action spaces, respectively. and represent the dynamics (transition distribution) and reward function, and represents the discount factor. The goal in RL is to find a policy that maximizes the expected cumulative discounted rewards, known as the returns:\n\n π∗=argmaxπ Est+1∼T(⋅\|st,at),at∼π(⋅\|st)[∞∑t=0γtR(st,at)]\n\nThe quantity of interest in Q-learning methods are state-action value functions, which give the expected future return starting from a particular state-action tuple, denoted . The state value function can also be denoted as . Q-learning algorithms are based on iterating the Bellman backup operator , defined as\n\n (TQ)(s,a)=R(s,a)+γEs′∼T[V(s′)]\n V(s)=maxa′Q(s,a′)\n\nThe (tabular) Q-iteration algorithm is a dynamic programming algorithm that iterates the Bellman backup . Because the Bellman backup is a -contraction in the L- norm, and (the Q-values of ) is its fixed point, Q-iteration can be shown to converge to (Sutton & Barto, 2018). A deterministic optimal policy can then be obtained as .\n\nWhen state spaces cannot be enumerated in a tabular format, function approximators can be used to represent the Q-values. An important class of such Q-learning methods are fitted Q-iteration (FQI) (Ernst et al., 2005), or approximate dynamic programming (ADP) methods, which form the basis of modern deep RL methods such as DQN (Mnih et al., 2015). FQI projects the values of the Bellman backup onto a family of Q-function approximators :\n\n Qt+1←Πμ(TQt).\n\nHere, denotes a -weighted L2 projection, which minimizes the Bellman error\n\n Πμ(Q)\\makebox[0.0pt]\\tiny def=argminQ′∈Q Es,a∼μ[(Q′(s,a)−Q(s,a))2]. (1)\n\nThe values produced by the Bellman backup, are commonly referred to as target values, and when neural networks are used for function approximation, the previous Q-function is referred to as the target network. In this work, we distinguish between the cases when the Bellman error is estimated with Monte-Carlo sampling or computed exactly (see Section 3.1). The sampled variant corresponds to FQI as described in the literature (Ernst et al., 2005; Riedmiller, 2005), while the exact variant is analogous to conventional ADP (Bertsekas & Tsitsiklis, 1996).\n\nConvergence guarantees for Q-iteration do not cleanly translate to FQI. is an projection, but is a contraction in the norm – this norm mistmatch means the composition of the backup and projection is no longer guaranteed to be a contraction under any norm (Bertsekas & Tsitsiklis, 1996), and hence the convergence is not guaranteed.\n\nA related branch of Q-learning methods are online Q-learning methods, in which Q-values are updated while samples are being collected in the MDP. This includes classic algorithms such as Watkin’s Q-learning (Watkins & Dayan, 1992). Online Q-learning methods can be viewed as a form of stochastic approximation (such as Robbins-Monro) applied to Q-iteration and FQI (Bertsekas & Tsitsiklis, 1996), and share many of its theoretical properties (Szepesvári, 1998). Modern deep RL algorithms such as DQN (Mnih et al., 2015) have characteristics of both online Q-learning and FQI – using replay buffers means the sampling distribution changes very little between target updates (see Section 6.3), and target networks are justified from the viewpoint of FQI. Because FQI corresponds to the case when the sampling distribution is static between target updates, the behavior of modern deep RL methods more closely resembles FQI than a true online method without target networks.\n\n## 3 Experimental Setup\n\nOur experimental setup is centered around unit-testing. We first introduce a spectrum of Q-learning algorithms, starting with exact approximate dynamic programming and gradually replacing oracle components, such as knowledge of dynamics, until the algorithm resembles modern deep Q-learning methods. We then introduce a suite of tabular environments where oracle solutions can be computed and compared against, to aid in diagnosis, as well as testing in high-dimensional environments to verify our hypotheses.\n\nIn order to provide consistent metrics across domains, we normalize returns and errors involving Q-functions (such as Bellman error) by the returns of the expert policy on each environment.\n\n### 3.1 Algorithms\n\nIn the analysis presented in Section 4, 5, 6 and 7, we will use three different Q-learning variants, each of which remove some of the approximations in the standard Q-learning method used in the literature – Exact-FQI, Sampling-FQI, and Replay-FQI. Although FQI is not exactly identical to commonly used deep RL methods, such as DQN (Mnih et al., 2015), DDPG (Lillicrap et al., 2015), and SAC (Haarnoja et al., 2017), it is structurally similar and, when the replay buffer for the commonly used methods becomes large, the difference becomes negligible, since the sampling distribution changes very little between target network updates. However, FQI methods are much more amenable for controlled analysis, since we can separately isolate target values, update rates, and the number of samples used for each iteration. We therefore use variants of FQI as the basis for our analysis, but we also confirm that similar trends hold with more commonly used algorithms on standard benchmark problems.\n\nExact-FQI (Algorithm 1): Exact-FQI computes the backup and projection on all state-action tuples without any sampling error. It also assumes knowledge of dynamics and reward function to compute Bellman backups exactly. We use Exact-FQI to study convergence, distribution shift (by varying weighting distributions on transitions), and function approximation in the absence of sampling error. Exact-FQI eliminates errors due to sampling states, and computing inexact, sampled backups.\n\nSampled-FQI (Algorithm 2): Sampled-FQI is a special case of Exact-FQI, where the Bellman error is approximated with Monte-Carlo estimates from a sampling distribution , and the Bellman backup is approximated with samples from the dynamics as . We use Sampled-FQI to study effects of overfitting. Sampled-FQI incorporates all sources of error – arising from function approximation, sampling and also distribution shift.\n\nReplay-FQI (Algorithm 3): Replay-FQI is a special case of Sampled-FQI that uses a replay buffer (Lin, 1992), that saves past transition samples , which are used for computing Bellman error. Replay-FQI strongle resembles DQN (Mnih et al., 2015), lacking the online updates that allow to change within an FQI iteration. With large replay buffers, we expect the difference between Replay-FQI and DQN to be minimal as changes slowly.\n\nWe additionally investigate the following choices of weighting distributions () for the Bellman error. When sampling the Bellman error, these can be implemented by sampling directly from the distribution, or via importance sampling.\n\nUnif: Uniform weights over state-action space. This is the weighting distribution typically used by dynamic programming algorithms, such as FQI.\n\n: The on-policy state-action marginal induced by .\n\n: The state-action marginal induced by .\n\nRandom: State-action marginal induced by executing uniformly random actions.\n\nPrioritized(s,a): Weights Bellman errors proportional to . This is similar to prioritized replay (Schaul et al., 2015) without importance sampling.\n\nReplay and Replay10: Averaged state-action marginal of all policies (or the previous 10) produced during training. This simulates sampling uniformly from a replay buffer where infinite samples are collected from each policy.\n\n### 3.2 Domains\n\nWe evaluate our methods on suite of tabular environments where we can compute oracle values. This will help us compare, analyze and fix various sources of error by means of comparing the learned Q-functions to the true, oracle-compute Q-functions. We selected 8 tabular domains, each with different qualitative attributes, including: gridworlds of varying sizes and observations, blind Cliffwalk (Schaul et al., 2015), discretized Pendulum and Mountain Car based on implementations in OpenAI Gym (Plappert et al., 2018), and a random sparsely connected graph. We give full details of these environments in Appendix A, as well as their motivation for inclusion.\n\n### 3.3 Function Approximators\n\nThroughout our experiments, we use 2-layer ReLU networks, denoted by a tuple\n\nwhere N represents the number of units in a layer. The “Tabular” architecture refers to the case when no function approximation is used.\n\n### 3.4 High-Dimensional Testing\n\nIn addition to diagnostic experiments on tabular domains, we also wish to see if the observed trends hold true on high-dimensional environments. To this end, we include experiments on continuous control tasks in the OpenAI Gym benchmark (Plappert et al., 2018) (HalfCheetah-v2, Hopper-v2, Ant-v2, Walker2d-v2). In continuous domains, computing the maximum over actions of the Q-value is difficult (). A common choice in this case is to use a second “actor” neural network to approximate (Lillicrap et al., 2015; Fujimoto et al., 2018; Haarnoja et al., 2018). This approach most closely resembles Replay-FQI, but using the actor network in place of the max.\n\n## 4 Function Approximation and Convergence\n\nThe first issue we investigate is the connection between function approximation and convergence properties.\n\n### 4.1 Technical Background\n\nAs discussed in Section 2, when function approximation is introduced to Q-learning, convergence guarantees are lost. This interaction between approximation and convergence has been a long-studied topic in reinforcement learning. In the control literature, it is closely related to the problems of state-aliasing or interference (Farrell & Berger, 1995). Baird (1995) introduces a simple counterexample in which Watkin’s Q-learning with linear approximators can cause unbounded divergence. In the policy evaluation scenario, Tsitsiklis & Van Roy (1997) prove that on-policy TD-learning with linear function approximators converge, and methods such as GTD (Sutton et al., 2009a) and ETD (Sutton et al., 2016) have extended results to off-policy cases. In the control scenario, convergent algorithms such as SBEED (Dai et al., 2018) and Greedy-GQ (Maei et al., 2010) have been developed. However, several works have noted that divergence need not occur. Munos (2005) theoretically addresses the norm-mismatch problem, which show that unbounded divergence is impossible provided has adequate support and projections are non-expansive in p-norms. Concurrently to us, Van Hasselt et al. (2018) experimentally find that unbounded divergence rarely occurs with DQN variants on Atari games.\n\n### 4.2 How does function approximation affect convergence properties and suboptimality of solutions?\n\nThe crucial quantities we wish to measure are a trend between function approximation and performance, and a measure for the bias in the learning procedure introduced by function approximation. Thus, using Exact-FQI with uniform weighting (to remove sampling error), we measure the returns of the learning policy, and the error between and the solution found by Exact-FQI () or the projection of the optimal solution (). represents the best solution inside the model class, in absence of error from the bootstrapping process of FQI. Thus, the difference between FQI error and projection error represents the bias introduced by the bootstrapping procedure, while controlling for bias that is simply due to function approximation – this quantity is roughly the inherent Bellman error of the function class (Munos & Szepesvári, 2008). This is the gap which can possibly be improved upon via better Q-learning algorithm design. We plot our results in Fig. 1.", null, "Figure 1: Normalized returns and normalized Q-function error with function approximation, averaged across domains and seeds. We see that for small architectures, there is a significant gap between the solution found by FQI (FQI Error) and the best solution within the model class (Project Error).\n\nWe first note the obvious trend that smaller architectures produce lower returns, and converge to more suboptimal solutions. However, we also find that smaller architectures introduce significant bias in the learning process, and there is often a significant gap between the solution found by Exact-FQI and the best solution within the model class. This gap may be due to the fact that when the target is bootstrapped, we must be able to represent all Q-function along the path to the solution, and not just the final result (Bertsekas & Tsitsiklis, 1996). This observation implies that using large architectures is crucial not only because they have capacity to represent a better solution, but also because they are significantly easier to train using bootstrapping, and suffer less from nonconvergence issues. We also note that divergence rarely happens in practice. We observed divergence in 0.9% of our experiments using function approximation, measured by the largest Q-value growing larger than 10 times that of .\n\nFor high-dimensional problems, we present experiments on varying the architecture of the Q-network in SAC (Haarnoja et al., 2018) in Appendix Fig. 13. We still observe that large networks have the best performance, and that divergence rarely happens even in high-dimensional continuous spaces. We briefly discuss theoretical intuitions on apparent discrepancy between the lack of unbounded divergence in relation known counterexamples in Appendix B.\n\n## 5 Sampling Error and Overfitting\n\nA second source of error in minimizing the Bellman error, orthogonal to function approximation, is that of sampling or generalization error. The next issue we investigate is the effect of sampling error on Q-learning methods.\n\n### 5.1 Technical Background", null, "Figure 2: Samples plotted with returns for a 256x256 network. More samples yields better performance.\n\nApproximate dynamic programming assumes that the projection of the Bellman backup (Eqn. 1) is computed exactly, but in reinforcement learning we can normally only compute the empirical Bellman error\n\nover a finite set of samples. In the PAC framework, overfitting can be quantified by a bounded error in between the empirical and expected loss with high probability, which decays with sample size\n\n(Shalev-Shwartz & Ben-David, 2014). Munos & Szepesvári (2008); Maillard et al. (2010); Tosatto et al. (2017) provide such PAC-bounds which account for sampling error in the context of Q-learning and value-based methods, and quantify the quality of the final solution in terms of sample complexity.\n\nWe analyze several key points that relate to sampling error. First, we show that Q-learning is prone to overfitting, and that this overfitting has a real impact on performance, in both tabular and high-dimensional settings. We also show that the replay buffer is in fact a very effective technique in addressing this issue, and discuss several methods to migitate the effects of overfitting in practice.", null, "Figure 3: On-policy validation losses for varying amounts of on-policy data (or replay buffer), averaged across environments and seeds. Note that sampling from the replay buffer has lower on-policy validation loss, despite bias from distribution shift.\n\n### 5.2 Quantifying Overfitting\n\nWe first quantify the amount of overfitting that happens during training, by varying the number of samples. In order provide comparable validation errors across different experiments, we fix a reference sequence of Q-functions, , obtained during a normal training run. We then retrace the training sequence, and minimize the projection error for each training iteration, using varying amounts of on-policy data or sampling from a replay buffer. We measure the exact validation error (the expected Bellman error) at each iteration under the on-policy distribution, plotted in Fig. 3. We note the obvious trend that more samples leads to lower validation loss, confirming that overfitting can in fact occur. A more interesting observation is that sampling from the replay buffer results in the lowest on-policy validation loss, despite bias due to distribution mismatch from sampling off-policy data. As we discuss in Section 6, we believe that replay buffers are mainly effective because they greatly reduce the effect of overfitting and create relatively good coverage over the state space, not necessarily due to reducing the effects of distribution shift.\n\nNext, Fig. 2 shows the relationship between number of samples and returns. We see a clear trend that higher sample count leads to improved learning speed and a better final solution, confirming our hypothesis that overfitting has a significant effect on the performance of Q-learning. A full sweep including architectures is presented in Appendix Fig. 14. We observe that despite overfitting being an issue, larger architectures still perform better because the bias introduced by smaller architectures dominates.", null, "Figure 4: Normalized returns plotted over training iterations (32 samples are taken per iteration), for different ratios of gradient steps taken per sample during projection using Replay-FQI. We observe that intermediate values of gradient steps work best, and too many gradient steps hinders performance.", null, "Figure 5: Normalized returns plotted over training iterations (32 samples are taken per iteration), for different early stopping methods using Replay-FQI. We observe that using proper early stopping can result in a modest performance increase.\n\n### 5.3 What methods can be used to compensate for overfitting?\n\nFinally, we discuss methods to compensate for overfitting. One common method for reducing overfitting is to regularize the function approximator to reduce its capacity. However, as we have seen before that weaker architectures can give rise to suboptimal convergence, we instead study early stopping\n\nmethods to mitigate overfitting without reducing model size. First, we observe that the number of gradient steps taken per sample in the projection step has an important effect on performance – too few steps and the algorithm learns slowly, but too many steps and the algorithm may initially learn quickly but overfit. To show this, we run a hyperparameter sweep over the number of gradient steps taken per environment step in Replay-FQI and TD3 (TD3 uses 1 by default). Results for FQI are shown in Fig.\n\n4, and for TD3 in Appendix Fig. 15.\n\nIn order to understand whether better early stopping criteria can possibly help with overfitting, we employ oracle\n\nearly stopping rules. While neither of these rules can be used to solve overfitting in practice, these experiments can provide guidance for future methods and an “upper bound” on the best improvement that can be obtained from optimal stopping. We investigate two oracle early stopping criteria for setting the number of gradient steps: using the expected Bellman error and the expected returns of the greedy policy w.r.t. the current Q-function (oracle returns). We implement both methods by running the projection step of Replay-FQI to convergence using gradient descent, and afterwards selecting the intermediate Q-function which is judged best by the evaluation metric (lowest Bellman error or highest returns). Using such oracle stopping metrics results in a modest boost in performance in tabular domains (Fig.\n\n5). Thus, we believe that there is promise in further improving such early-stopping methods for reducing overfitting in deep RL algorithms.\n\nWe might draw a few actionable conclusions from these experiments. First, overfitting is indeed a serious issue with Q-learning, and too many gradient steps or too few samples can lead to poor performance. Second, replay buffers and early stopping can be used to mitigate the effects of overfitting. Third, although overfitting is a problem, large architectures are still preferred, because the harm from function approximation bias outweighs the harm from increased overfitting with large models.\n\n## 6 Non-Stationarity\n\nIn this section, we discuss issues related to the non-stationarity of the Q-learning process (relating to the Bellman backup and Bellman error minimization).\n\n### 6.1 Technical Background\n\nInstability in Q-learning methods is often attributed to the nonstationarity of the regression objective (Lillicrap et al., 2015; Mnih et al., 2015). Nonstationarity occurs in two places: in the changing target values , and in a changing weighting distribution (“distribution shift”) (i.e., due to samples being taken from different policies). Note that a non-stationary objective, by itself, is not indicative of instability. For example, gradient descent can be viewed as successively minimizing linear approximations to a function: for gradient descent on with parameter and learning rate , we have the “moving” objective . However, the fact that the Q-learning algorithm prescribes an update rule and not a stationary objective complicates analysis. Indeed, the motivation behind algorithms such as GTD (Sutton et al., 2009b, a) and residual methods (Baird, 1995; Scherrer, 2010) can be seen as introducing a stationary objective that can be optimized with standard procedures such as gradient descent for increased stability. Therefore, a key question to investigate is whether these non-stationarities are detrimental to the learning process.\n\n### 6.2 Does a moving target cause instability in the absence of a moving distribution?\n\nTo study the moving target problem, we must first isolate the effects of a moving target, and study how the rate at which the target changes impacts performance. To control the rate at which the target changes, we introduce an additional smoothing parameter to Q-iteration, where the target values are now computed as an -moving average over previous targets. We define the -smoothed Bellman backup, , as follows:\n\n TαQ=αTQ+(1−α)Q\n\nThis scheme is inspired by the soft target update used in algorithms such as DDPG (Lillicrap et al., 2015) and SAC (Haarnoja et al., 2017) to improve the stability of learning. Standard Q-iteration uses a “hard” update where . A soft target update weakens the contraction of Q-iteration from to (See Appendix C), so we expect slower convergence, but perhaps it is more stable under heavy function approximation error. We performed experiments with this modified backup using Exact-FQI under the weighting distribution.\n\nOur results are presented in Appendix Fig. 12. We find that the most cases, the hard update with results in the fastest convergence and highest asymptotic performance. However, for the smallest two architectures we used, and , lower values of (such as 0.1) achieve slightly higher asymptotic performance. Thus, while more expressive architectures are still stable under fast-changing targets, we believe that a slowly moving target may have benefits under heavy approximation error. This evidence points to either using large function approximators, in line with the conclusions drawn in the previous sections, or adaptively slowing the target updates when the architecture is weak (relative to the problem difficulty) and the projected Bellman error is therefore high.\n\n### 6.3 Does distribution shift impact performance?", null, "Figure 6: Distribution shift and loss shift plotted against time. Prioritized and on-policy distributions induce the greatest shift, whereas replay buffers greatly reduce the amount of shift.\n\nTo study the distribution shift problem, we exactly compute the amount of distribution shift between iterations in total-variation distance, and the “loss shift”:\n\n Eμt+1[(Qt−TQt)2]−Eμt[(Qt−TQt)2].\n\nThe loss shift quantifies the Bellman error objective when evaluated under a new distribution - if the distribution shifts to previously unseen or low support states, we would expect a highly inaccurate Q-value in such states, and a correspondingly high loss shift.", null, "Figure 7: Average distribution shift across time for different weighting distributions, plotted against returns for a 256x256 model. We find that distribution shift does not have strong correlation with returns.\n\nWe run our experiments using Exact-FQI with a 256x256 layer architecture, and plot the distribution discrepancy and the loss discrepancy in Fig. 6. We find that Prioritized has the greatest shift, followed by on-policy variants. Replay buffers greatly reduce distribution shift compared to on-policy learning, which is similar to the de-correlation argument cited for its use by Mnih et al. (2015). However, we find that this metric correlates very little with the actual performance of the algorithm (Fig. 7). For example, prioritized weighting performs well yet has high distribution shift.\n\nOverall, our experiments indicate that nonstationarities in both distributions and target values, when isolated, do not cause significant stability issues. Instead, other factors such as sampling error and function approximation appear to have more significant effects on performance. In the light of these findings, we might therefore ask: can we design a better sampling distribution, without regard for distributional shift and with regard for high-entropy, that results in better final performance, and is realizable in practice? We investigate this in the following section.\n\n## 7 Sampling Distributions\n\nAs alluded to in Section 6, the choice of sampling distribution is an important design decision can have a large impact on performance. Indeed, it is not immediately clear which distribution is ideal for Q-learning. In this section, we hope to shed some light on this issue.\n\n### 7.1 Technical Background\n\nOff-policy data has been cited as one of the “deadly triads” for Q-learning (Sutton & Barto, 2018), which has potential to cause instabilities in learning. On-policy distributions (Tsitsiklis & Van Roy, 1997) and fixed behavior distributions (Sutton et al., 2009b; Maei et al., 2010) have often been targeted for theoretical convergence analysis, and many works use importance sampling to correct for off-policyness (Precup et al., 2001; Munos et al., 2016) However, to our knowledge, there is relatively little guidance which compares how different weighting distributions compare in terms of convergence rate and final solutions.\n\nNevertheless, several works give hypotheses on good choices for weighting distributions. (Munos, 2005) provides an error bound which suggests that “more uniform” weighting distributions can guarantee better worst-case performance. (Geist et al., 2017) suggests that when the state-distribution is fixed, the action distribution should be weighted by the optimal policy for residual Bellman errors. In deep RL, several methods have been developed to prevent instabilities in Q-Learning, such as prioritized replay (Schaul et al., 2015), and mixing replay buffer with on-policy data (Hausknecht & Stone, 2016; Zhang & Sutton, 2017) have been found to be beneficial. In our experiments, we aim to empirically analyze multiple choices for weighting distributions to determine which are the most effective.\n\n### 7.2 What Are the Best Weighting Distributions in Absence of Sampling Error?", null, "Figure 8: Weighting distribution versus architecture in Exact-FQI. Replay(s, a) consistently provides the highest performance. Note that Adversarial Feature Matching is comparable to Replay(s, a), but surprisingly better for small networks.", null, "Figure 9: Normalized returns plotted against normalized entropy for different weighting distributions. All experiments use Exact-FQI with a 256x256 network. We see a general trend that high-entropy distributions lead to greater performance.\n\nWe begin by studying the effect of weighting distributions when disentangled from sampling error. We run Exact-FQI with varying choices of architectures and weighting distributions and report our results in Fig. 8. , and consistently result in the highest returns across all architectures. We believe that these results are in favor of the uniformity hypothesis: the top performing distributions spread weight across larger support of the state-action space. For example, a replay buffer contains state-action tuples from many policies, and therefore would be expected to have wider support than the state-action distribution of a single policy. We can see this general trend in Fig. 9. These distributions generally result in the tightest contraction rates, and allow the Q-function to focus on locations where the error is high. In the sampled setting, this observation motivates exploration algorithms that maximize state coverage (for example, Hazan et al. (2018) solve an exploration objective which maximizes state-space entropy). However, note that in this particular experiment, there is no sampling. All states are observed, just with different weights, thus isolating the issue of distributions from the issue of sampling.\n\n### 7.3 Designing a Better Off-Policy Distribution: Adversarial Feature Matching\n\nIn our final study, we attempt to design a better weighting distribution using insights from previous sections that can be easily integrated into deep RL methods. We refer to this method as adversarial feature-matching (AFM). We draw upon three specific insights outlined in previous analysis. First, the function approximator should be incentivized to maximize its ability to distinguish states to minimize function approximation bias (Section 4). Second, the weighting distribution should emphasize areas where the Q-function incurs high Bellman error, in order to minimize the discrepancy between norm error and norm error. Third, more-uniform weighting distributions tend to be higher performant. The first insight was also demonstrated in (Liu et al., 2018) where enforcing sparsity in the Q-function was found to provide locality in the Q-function which prevented catastrophic interference and provided better values for bootstrapping.\n\nWe propose to model our problem as a minimax game, where the weighting distribution is a parameterized adversary which tries to maximize the Bellman error, while the Q-function () tries to minimize it. Note that in the unconstrained setting, this game is equivalent to minimizing the norm error in its dual-norm representation. However, in practical settings where minimizing stochastic approximations of the norm can be difficult for neural networks (also noticed when using PER (Van Hasselt et al., 2018)\n\n), it is crucial to introduce constraints to limit the power of the adversary. These constraints also make the adversary closer to the uniform distribution while still allowing it to be sufficiently different at specific state-action pairs.\n\nWe elect to use a feature matching constraint which enforces the expected feature vectors,\n\n, under to roughly match the expected feature vector under uniform sampling from the replay buffer. We can express the output of a neural network Q-function as or, in the continuous case, as , where the feature vector represent the the output of all but the final layer. Intuitively, this constraint restricts the adversarial sampler to distributing probability mass among states (or state-action pairs) that are perceptually similar to the Q-function, which in turn forces the Q-function to reduce state-aliasing by learning features that are more separable. Note that, in our case, . This also provides a natural extension of our method by performing expected gradient matching over all parameters (), instead of matching only (we leave it to future work to explore this direction). Formally, this objective is given as follows:\n\n minθ,wmaxϕEpϕ(s,a)[(Qw,θ(s,a)−y(s,a))2]s.t. \|\|Epϕ(s,a)[Φ(s)]−∑iΦ(si)N\|\|≤ε\n\nNote that is a function of but, while solving the maximization, is assumed to be a constant. This is equivalent to solving only the inner maximization with a constraint, and empirically provides better stability. Implementation details for AFM are provided in Appendix D. The denotes an estimator for the true expectation under some sampling distribution, such as a uniform distribution over all states and actions (in exact FQI) or the replay buffer distribution. So, holds when using a replay buffer.\n\nWhile both AFM and PER tend to upweight samples in the buffer with a high Bellman error, PER explicitly attempts to reduce distribution shift via importance sampling. As we observed in Section 7, distributional shift is not actually harmful in practice, and AFM dispenses with this goal, instead explicitly aiming to rebalance the buffer to attain better coverage via adversarial optimization. In our experiments, this results in substantially better performance, consistent with the hypothesis that coverage, rather than reduction of distributional shift, is the most important property in a sampling distribution.\n\nIn tabular domains with Exact-FQI, we find that AFM performs at par with the top performing weighting distributions, such as and better than (Fig. 8). This confirms that adaptive prioritization works better than Prioritized(). Another benefit of AFM is its robustness to function approximation and the performance gains in the case of small architectures (say, ) are particularly noticeable. (Fig. 8)\n\nIn tabular domains with Replay-FQI (Table 1), we also compare AFM to prioritized replay (PER) (Schaul et al., 2015), where AFM and PER perform similarly in terms of normalized returns. Note that AFM reweights samples drawn uniformly from the buffer, whereas PER changes which samples are actually drawn. We also evaluate a variant of AFM (AFM+Sampling in Table 1) which changes which samples instead of reweighting. Essentially, in this version we sample from the replay buffer using probabilities determined by the AFM optimization, rather than using importance sampling while making bellman updates. We note that, in Table 1, AFM+Sampling performs strictly better than AFM and PER.\n\nWe further evaluate AFM on MuJoCo tasks with the TD3 algorithm (Fujimoto et al., 2018) and the entropy constrained SAC algorithm (Haarnoja et al., 2018). We find that in all 3 tested domains (Half-Cheetah, Hopper and Ant), AFM yields substantial empirical improvement in the case of TD3 (Fig. 10) and performs slightly better than entropy constrained SAC (Fig. 11). Surprisingly, we found PER to not work very well in these domains. In light of these results, we conclude that: (1) the choice of sampling distribution is very important for performance, and (2) considerations such as incorporating knowledge about the function approximator (for example, through ) into the choice of (the sampling/weighting distribution) can be very effective.\n\n## 8 Conclusions and Discussion\n\nFrom our analysis, we have several broad takeaways for the design of deep Q-learning algorithms.\n\nPotential convergence issues with Q-learning do not seem to be endemic empirically, but function approximation still has a strong impact on the solution to which these methods converge. This impact goes beyond just approximation error, suggesting that Q-learning methods do find suboptimal solutions (within the given function class) with smaller function approximators. However, expressive architectures largely mitigate this problem, suffer less from bootstrapping error, converge faster, and more stable with moving targets.\n\nSampling error can cause substantial overfitting problems with Q-learning. However, replay buffers and early stopping can mitigate this problem, and the biases incurred from small function approximators outweigh any benefits they may have in terms of overfitting. We believe the best strategy is to keep large architectures but carefully select the number of gradient steps used per sample. We showed that employing oracle early stopping techniques can provide huge benefits in the performance in Q-learning. This motivates the future research direction of devising early stopping techniques to dynamically control the number of gradient steps in Q-learning, rather than setting it as a hyperparameter as this can give rise to big difference in performance.\n\nThe choice of sampling or weighting distribution has significant effect on solution quality, even in the absence of sampling error. Surprisingly, we do not find on-policy distributions to be the most performant, but rather methods which have high state-entropy and spread mass uniformly among state-action pairs, seem to be highly effective for training. Based on these insights, we propose a new weighting distribution which balances high-entropy and state aliasing, AFM, that yields fair improvements in both tabular and continuous domains with state-of-the-art off-policy RL algorithms.\n\nFinally, we note that there are many other topics in Q-learning that we did not investigate, such as overestimation bias and multi-step returns. We believe that these issues too could be studied in future work with our oracle-based analysis framework.\n\n## Acknowledgements\n\nWe thank Vitchyr Pong and Kristian Hartikainen for providing us with implementations of RL algorithms. We thank Chelsea Finn for comments on an earlier draft of this paper. SL thanks George Tucker for helpful discussion. We thank Google, NVIDIA, and Amazon for providing computational resources. This research was supported by Berkeley DeepDrive, NSF IIS-1651843 and IIS-1614653, the DARPA Assured Autonomy program, and ARL DCIST CRA W911NF-17-2-0181.\n\n## References\n\n• Arjovsky et al. (2017) Arjovsky, M., Chintala, S., and Bottou, L. Wasserstein generative adversarial networks. In\n\nInternational Conference on Machine Learning (ICML)\n\n, pp. 214–223, 2017.\n• Baird (1995) Baird, L. Residual Algorithms : Reinforcement Learning with Function Approximation. 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R., and Szepesvári, C. A convergent o(n) temporal-difference algorithm for off-policy learning with linear function approximation. In Advances in Neural Information Processing Systems (NeurIPS), 2009b.\n• Sutton et al. (2016) Sutton, R. S., Mahmood, A. R., and White, M. An emphatic approach to the problem of off-policy temporal-difference learning. The Journal of Machine Learning Research, 17(1):2603–2631, 2016.\n• Szepesvári (1998) Szepesvári, C. The asymptotic convergence-rate of q-learning. In Advances in Neural Information Processing Systems, pp. 1064–1070, 1998.\n• Tosatto et al. (2017) Tosatto, S., Pirotta, M., D’Eramo, C., and Restelli, M. Boosted fitted q-iteration. In International Conference on Machine Learning (ICML), pp. 3434–3443. JMLR. org, 2017.\n• Tsitsiklis & Van Roy (1997) Tsitsiklis, J. N. and Van Roy, B. Analysis of temporal-diffference learning with function approximation. In Advances in Neural Information Processing Systems (NeurIPS), pp. 1075–1081, 1997.\n• Tuomas Haarnoja & Levine (2018) Tuomas Haarnoja, Aurick Zhou, K. H. G. T. S. H. J. T. V. K. H. Z. A. G. P. A. and Levine, S. Soft actor-critic algorithms and applications. Technical report, 2018.\n• Van Hasselt et al. (2018) Van Hasselt, H., Doron, Y., Strub, F., Hessel, M., Sonnerat, N., and Modayil, J. Deep reinforcement learning and the deadly triad. arXiv preprint arXiv:1812.02648, 2018.\n• Watkins & Dayan (1992) Watkins, C. J. and Dayan, P. Q-learning. Machine learning, 8(3-4):279–292, 1992.\n• Yazıcı et al. (2019) Yazıcı, Y., Foo, C.-S., Winkler, S., Yap, K.-H., Piliouras, G., and Chandrasekhar, V. The unusual effectiveness of averaging in GAN training. In International Conference on Learning Representations, 2019.\n• Zhang & Sutton (2017) Zhang, S. and Sutton, R. S. A deeper look at experience replay. CoRR, abs/1712.01275, 2017. URL http://arxiv.org/abs/1712.01275.\n\n## Appendix A Benchmark Tabular Domains\n\nWe evaluate on a benchmark of 8 tabular domains, selected for qualitative differences.\n\n4 Gridworlds. The Gridworld environment is an NxN grid with randomly placed walls. The reward is proportional to Manhattan distance to a goal state (1 at the goal, 0 at the initial position), and there is a 5% chance the agent travels in a different direction than commanded. We vary two parameters: the size ( and ), and the state representations. We use a “one-hot” representation, an (X, Y) coordinate tuple (represented as two one-hot vectors), and a “random” representation, a vector drawn from , where N is the width or height of the Gridworld. The random observation significantly increases the challenge of function approximation, as significant state aliasing occurs.\n\nCliffwalk: Cliffwalk is a toy example from Schaul et al. (2015). It consists of a sequence of states, where each state has two allowed actions: advance to the next state or return to the initial state. A reward of 1.0 is obtained when the agent reaches the final state. Observations consist of vectors drawn from .\n\nInvertedPendulum and MountainCar: InvertedPendulum and MountainCar are discretized versions of continuous control tasks found in OpenAI gym (Plappert et al., 2018), and are based on problems from classical RL literature. In the InvertedPendulum task, an agent must swing up an pendulum and hold it in its upright position. The state consists of the angle and angular velocity of the pendulum. Maximum reward is given when the pendulum is upright. The observation consists of the and of the pendulum angle, and the angular velocity. In the MountainCar task, the agent must push a vehicle up a hill, but the hill is steep enough that the agent must gather momentum by swinging back and forth within a valley in order to reach the top. The state consists of the position and velocity of the vehicle.\n\nSparseGraph: The SparseGraph environment is a 256-state graph with randomly drawn edges. Each state has two edges, each corresponding to an action. One state is chosen as the goal state, where the agent receives a reward of one.\n\n## Appendix B Fitted Q-iteration with Bounded Projection Error\n\nWhen function approximation is introduced to Q-iteration, we lose guarantees that our solution will converge to the optimal solution , because the composition of projection and backup is no longer guaranteed to be a contraction under any norm. However, this does not imply divergence, and in most cases it merely degrades the quality of solution found.\n\nThis can be seen by recalling the following result from (Bertsekas & Tsitsiklis, 1996), that describes the quality of the solution obtained by fitted Q-iteration (FQI) when the projection error at each step is bounded. The conclusion is that FQI converges to an ball around the optimal solution which scales proportionally with the projection error. While this statement does not claim that divergence cannot occur in general (this theorem can only be applied in retrospect, since we cannot always uniformly bound the projection error at each iteration), it nevertheless offers important intuitions on the behavior of FQI under approximation error. For similar results concerning -weighted norms, see (Munos, 2005).\n\n###### Theorem B.1 (Bounded error in fitted Q-iteration).\n\nLet the projection or Bellman error at each iteration of FQI be uniformly bounded by , i.e. . Then, the error in the final solution is bounded as\n\n limi→∞∥^Qi−Q∗∥∞≤δ1−γ\n###### Proof.\n\nSee of Chapter 6 of Bertsekas & Tsitsiklis (1996). ∎\n\nWe can use this statement to provide a bound on the performance of the final policy.\n\n###### Corollary B.1.1.\n\nSuppose we run fitted Q-iteration, and let the projection error at each iteration be uniformly bounded by , i.e. . Letting denote the returns of a policy , the the performance of the final policy is bounded as:\n\n limi→∞\|η(πi)−η(π∗)\|≤2γδ(1−γ)2\n###### Proof.\n\nThis result is obtained by substituting Thm. B.1 into Propositon 6.1 of Bertsekas & Tsitsiklis (1996). ∎\n\n### b.1 Unbounded divergence in FQI\n\nBecause norms are bounded by the norm, Thm. B.1 implies that unbounded divergence is impossible when weighting distribution has positive support at all states and actions (i.e. ), and the projection is non-expansive in the norm (such as when using linear approximators).\n\nWe can bound the -weighted in terms of the as follows: . Thus, we can apply Thm. B.1 with to show that unbounded divergence is impossible. Note that because this bound scales with the size of the state and action spaces, it is fairly loose in many practical cases, and practitioners may nevertheless see Q-values grow to large values (tighter bounds concerning L2 norms can be found in (Munos, 2005), which depend on the transition distribution). It also suggests that distributions which are fairly uniform (so as to maximize the denominator) can perform well.\n\nWhen the weighting distribution does not have support over all states and actions, divergence can still occur, as noted in the counterexamples such as Section 11.2 of Sutton & Barto (2018). In this case, we consider two states (state 1 and 2) with feature vectors 1 and 2, respectively, and a linear approximator with parameter . There exists a single action with a deterministic transition from state 1 to state 2, and we only sample the transition from state 1 to state 2 (i.e. is 1 for state 1 and 0 for state 2). All rewards are 0. In this case, the projected Bellman backup takes the form:\n\n wt+1=argminw (w−2γwt)2\n\nWhich will cause unbounded growth when iterated, provided . However, if we add a transition from state 2 back to itself or to state 1, and place nonzero probability on sampling these transitions, divergence can be avoided.\n\n## Appendix C α-smoothed Q-iteration\n\nIn this section we show that the -smoothed Bellman backup introduced in Section 6 is still a valid Q-iteration method, in that it is a contraction (for ) and thus converges to .\n\nWe define the -smoothed Bellman backup as:\n\n TαQ=αTQ+(1−α)Q\n###### Theorem C.1 (Contraction rate of the α-smoothed Bellman backup).\n\nis a -contraction:\n\n ∥TαQ1−TαQ2∥∞≤(1−α+γα)∥Q1−Q2∥∞\n###### Proof.\n\nThis statement follows from straightforward application of the triangle rule and the fact that is a -contraction:\n\n ∥TαQ1−TαQ2∥∞ =∥(αTQ1−(1−α)Q1)−(αTQ2−(1−α)Q2∥∞ =∥α(TQ1−TQ2)+(1−α)(Q1−Q2)∥∞ ≤α∥TQ1−TQ2∥∞+(1−α)∥Q1−Q2∥∞ ≤αγ∥Q1−Q2∥∞+(1−α)∥Q1−Q2∥∞ =(1−α+αγ)∥Q1−Q2∥∞", null, "Figure 12: Results for the α-smoothed Bellman backup experiment. Normalized ℓ∞ norm error to Q∗ and normalized returns plotted for different values of α and architectures. Values are averaged over all domains and 5 seeds. For large architectures, higher values of α result in faster convergence and higher asymptotic returns. However, for smaller architectures, low values of α slightly outperform higher values.\n\n## Appendix D Adversarial Feature Matching (AFM): Detailed Explanation and Practical Implementation\n\nAs described in section 7.3, we devise a novel weighting scheme for the Bellman error objective based on an adversarial minimax game. The adversary computes weights (representing the weighting distribution ), for the Bellman error: . Recalling from Section 7.3, the optimization problem is given by:\n\n minθ,wmaxϕEpϕ(s,a)[(Qw,θ(s,a)−y(s,a))2]s.t. ∣∣∣∣Epϕ(s,a)[Φ(s)]−∑iΦ(si)N\|\|≤ε\n\nwhere are the state features learned by the Q-function approximator. is easy to extract out of the multiheaded () model typically used for discrete action control, as one choice is to let be the output of the penultimate layer of the Q-network. For continuous control tasks, however, we model (which is a function of the actions as well) as state-only features are unavailable, unless separately modeled. This can also be interpreted as modelling a feature matching constraint on the gradient of with respect to the last linear parameters . A possible extension is to take into account the entire gradient as the features in the feature matching constraint, that is, .\n\nThis choice of the constraint is suitable and can be interpreted in two ways. First, an adversary constrained in this manner has enough power to exploit the Q-network at states which get aliased under the chosen function class, thereby promoting more separable feature learning and reducing some negative aspects of function approximation that can arise in Q-learning. This is also similar in motivation to (Liu et al., 2018). Second, this feature constraint also bears a similarity the Maximum Mean Discrepancy (MMD) distance between two distributions and that can be written as , where the set of functions is the canonical feature map, (from real space to the RKHS). In our context, this is analogous to optimizing a distance between the adversarial distribution and the replay buffer distribution (as the average is a Monte-Carlo estimator of the expected under the replay buffer distribution ). In the light of these arguments, AFM, and other associated methods that take into account the properties of the function approximator into account (for example, here), can greatly reduce the bias incurred due to function approximation in the due course of Q-learning/FQI, as depicted in 1.\n\n#### Solving the optimization\n\nWe solve this saddle point problem using alternating dual gradient descent. We first solve the inner maximization problem, and then use its solution to then solve the outer minimization problem. We first compute the Lagrangian for the maximization, by introducing a dual variable ,\n\n Linner(ϕ;λ,θ)=−Epϕ(s,a)[(Qθ(s,a)−y(s,a))2]+λ(\|\|Epϕ(s,a)[Φ(s)]−∑Φ(s)N\|\|−ε)\n\n(Note that this Lagrangian is flipped in sign because we first convert the maximization problem to standard minimization form.) We now solve the inner problem using dual gradient descent. We then plug in the solutions (approximate solutions obtained after gradient descent), into the Lagrangian, to then solve the outside minimization over . Note that while depends on\n\n(as it is the feature layer of the Q-network), we don not backpropagate through\n\nwhile solving the minimization. This improves stability of the Q-network training in practice and to makes sure that Q-function is only affected by FQI updates. In practice, we take up to 10 gradient steps for the inner problem every 1 gradient step of the outer problem. The algorithm is summarized in Algorithm 4. Our results provided in the main paper and here don’t particularly assume any other tricks like Optimistic Gradient (Daskalakis et al., 2018), using exponential moving average of the parameters (Yazıcı et al., 2019). Our tabular experiments seemed to benefit some what using these tricks.\n\n#### Practical implementation with replay buffers\n\nWe incorporate this weighting/sampling distribution into Q-learning in the setting with replay buffers and with state-action sampling. We evaluate the weighting version of our method, AFM, where, we usually sample a large batch of state-action pairs from a usual replay buffer used in Q-learning, but use importance weights to then match in expectation. Thus, we use a parametric function approximator to model – that is, the importance weights of the adversarial distribution with respect to the replay buffer distribution . Mathematically, we estimate: , where\n\n. The latter expectation is then approximated using a set of finite samples. It has been noted in literature that importance sampling (IS) suffers from high variance especially if the number of samples is small. Hence, we use the self-normalized importance sampling estimator, which averages the importance weights in a set of samples or a large number of samples. That is, let\n\n, then instead of using as the importance weights, we use (where and represent state-action tuples; concisely mentioned for visual clarity) as the importance weights. We also regularize the second-order Renyi Divergence between and for stability. Mathematically, it can be shown that this is a lower bound on the true expectation of under , which is being estimated using importance sampling. This result has also been shown in (Metelli et al., 2018) (Theorem 4.1), where the authors use this lower bound in policy optimization via importance sampling. We state the theorem below for completeness.\n\n###### Theorem D.1.\n\n(Metelli et al., 2018) Let and be two probability measures on the measurable space such that and . Let\n\nbe i.i.d. random variables sampled from\n\n, and be a bounded function. Then, for any and with probability at least it holds that:\n\n Ex∼Q[f(x)]≥1N∑wP/Q(xi)f(xi)−\|\|f\|\|∞√(1−δ)d2(P\|\|Q)Nδ\n\nwhere is the exponentiated second-order Renyi Divergence between and .\n\nHence, our objective for the inner loop now becomes: is now computed using samples with an additional renyi regularisation term. Since, we end up modeling this ratio, through out parameteric model, we can hence easily compute an estimator for the Renyi divergence term. The overall lower bound inner maximization problem is:\n\n maxϕ1N∑(s,a)∼prb[fϕ(s,a)(Qθ,w(s,a)−y(s,a))2]−C ⎷(1−δ)(∑fϕ(s,a))2N)Nδs.t.~{}~{}\|\|∑s,a∈prb[fϕ(s,a)Φ(s)]N−∑s,a∈prbΦ(s)N\|\|≤ε\n\nWe found that this Renyi penalty helped stabilize training. In practice, we model the importance weights:\n\nas a parametric model with an identical architecture to the Q-network. We use parameter clipping for\n\n, where the parameter are clipped to , analogous to Wasserstein GANs (Arjovsky et al., 2017). We also found that self-normalization during importance sampling has a huge practical benefit. Note that as the true norm of the Bellman error is not known, for computing in the Renyi Divergence term, and hence we either replace it by constant, or compute a stochastic approximation to the norm over the current batch. We found the former to be more stable, and hence, used that in all our experiments. This coefficient of the Renyi divergence penalty is tuned uniformly between . The learning rate for the adversary was chosen to be 1e-4 for the tabular environments, and 5e-4 for TD3. The batch size for our algorithm was chosen to be 128 for the tabular environments and 500 for TD3/SAC. Note that a larger batch size ensures smoothness in the minmax optimization problem. We also found that instead of having a Lagrange multiplier for the feature matching constraint, having Lagrange multipliers for constraining each of the individual dimensions of the features also helps very much. This is to ensure that the hyperparameters remain the same across different architectures regardless of the dimension of the penultimate layer of the Q-network. The algorithm in this case is exactly the same as the algorithm before with a vector valued dual variable . We used TD3 and SAC implementations from rlkit (https://github.com/vitchyr/rlkit/tree/master/rlkit)\n\n## Appendix E Function approximation analysis on Mujoco Tasks\n\nAs discussed in Section 4, we validate our findings on the effect of function approximation on 3 MuJoCo tasks from OpenAI Gym with the SAC algorithm from the author’s implementation at (Tuomas Haarnoja & Levine, 2018). 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https://curriculum.illustrativemathematics.org/HS/teachers/1/4/11/index.html	[ "# Lesson 11\n\nDomain and Range (Part 2)\n\n## 11.1: Which One Doesn't Belong: Unlabeled Graphs (5 minutes)\n\n### Warm-up\n\nThis warm-up prompts students to carefully analyze and compare the properties of four graphs. Each graph represents a function, but no labels or scales are shown on the coordinate axes, so students need to look for and make use of the structure of the graphs in determining how each one is like or unlike the others (MP7).\n\nIn making comparisons, students have a reason to use language precisely (MP6), especially mathematical terms that describe features of graphs or properties of functions.\n\n### Student Facing\n\nWhich one doesn't belong?\n\n### Student Response\n\nFor access, consult one of our IM Certified Partners.\n\n### Activity Synthesis\n\nAsk each group to share one reason why a particular item does not belong. Record and display the responses for all to see. Encourage students to use relevant mathematical vocabulary in their explanations, and ask students to explain the meaning of any terminology that they do use, such as \"intercepts,\" \"minimum,\" or \"linear functions.\"\n\nAfter each response, ask the class if they agree or disagree. Since there is no single correct answer to the question of which one does not belong, attend to students’ explanations and ensure the reasons given are correct.\n\nBy now students are well aware that certain features of a graph have special significance in that they tell us something about the quantities or relationships in the situation. Tell students that features of graphs can also help us understand the domain and range of a function. We will explore this in the lesson.\n\n## 11.2: Time on the Swing (20 minutes)\n\n### Activity\n\nIn this activity, students are given the same four graphs they saw in the warm-up and four descriptions of functions and are asked to match them. All of the functions share the same context. Students then use these features to reason about likely domain and range of each function.\n\nTo make the matches, students analyze and interpret features of the graphs, looking for and making use of structure in the situation and in the graphs (MP7). Here are some possible ways students may reason about each match:\n\n• The swing goes up and down while the child is swinging, so D could be a graph for function $$h$$.\n• The time left on the swing decreases as the time on the swing increases, so A is a possible graph for function $$r$$.\n• The distance of the child from the top beam of the swing doesn't change as long as the child is on the swing, so C is a possible graph for function $$d$$.\n• The total number of times the swing is pushed must be a counting number and cannot be fractional. Graph B has multiple pieces and each one could represent the total number of push for certain intervals of time, so B is a possible graph of $$s$$.\n\nStudents reason quantitatively and abstractly as they connect verbal and graphical representations of functions and as they think about the domain and range of each function (MP2).\n\n### Launch\n\nAsk students to imagine a child getting on a swing, swinging for 30 seconds, and then getting off the swing. Explain that they will look at four functions that can be found in this situation. Their job is to match verbal descriptions and graphs that define the same functions, and then to think about reasonable domain and range for each function. Tell students they will need additional information for the last question.\n\nArrange students in groups of 2. Give students a few minutes of quiet time to think about the first two questions, and then time to discuss their thinking with their partner. Follow with a whole-class discussion.\n\nInvite students to share their matching decisions and explanations as to how they know each pair of representations belong together. Make sure that students can offer an explanation for each match, including for their last pair (other than because the description and the graph are the only pair left). See some possible explanations in the Activity Narrative.\n\nNext, ask students to share the points that they think would be helpful for determining the domain and range of each function. If students gesture to the intercepts, a maximum, or a minimum on a graph but do not use those terms to refer to points, ask them to use mathematical terms to clarify what they mean.\n\nHere is the information students will need for the last question. Display it for all to see, or provide it as requested. If the requested information is not shown or cannot be reasoned from what is available, ask students to request a different piece of information.\n\n• The child is given 30 seconds on the swing.\n• While the child is on the swing, an adult pushes the swing a total of 5 times.\n• The swing is 1.5 feet (18 inches) above ground.\n• The chains that hold the seat and suspend it from the top beam are 7 feet long.\n• The highest point that the child swings up to is 4 feet above the ground.\n\nIf time is limited, ask each partner to choose two functions (different than their partner's) and write the domain and range only for those functions.\n\nSpeaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion as students explain how they matched the function description to the graph. Display the following sentence frames for all to see: “_____ matches the graph _____ because . . .” and “I noticed _____, so . . . .” Encourage students to challenge each other when they disagree. The student frames will help students consider the domain and range of functions as they connect written descriptions with graphical representations.\nDesign Principle(s): Support sense-making\nRepresentation: Internalize Comprehension. Use color coding and annotations to highlight connections between representations in a problem. For example, invite students to highlight the input in each description using one color, label the horizontal axis of each graph, and highlight the label in the same color. Then they can repeat the process for the outputs and the vertical axes.\nSupports accessibility for: Visual-spatial processing\n\n### Student Facing\n\nA child gets on a swing in a playground, swings for 30 seconds, and then gets off the swing.\n\n1. Here are descriptions of four functions in the situation and four graphs representing them.\n\nThe independent variable in each function is time, measured in seconds.\n\nMatch each function with a graph that could represent it. Then, label the axes with the appropriate variables. Be prepared to explain how you make your matches.\n\n• Function $$h$$: The height of the swing, in feet, as a function of time since the child gets on the swing\n• Function $$r$$: The amount of time left on the swing as a function of time since the child gets on the swing\n• Function $$d$$: The distance, in feet, of the swing from the top beam (from which the swing is suspended) as a function of time since the child gets on the swing\n• Function $$s$$: The total number of times an adult pushes the swing as a function of time since the child gets on the swing\n\n\n\n2. On each graph, mark one or two points that—if you have the coordinates—could help you determine the domain and range of the function. Be prepared to explain why you chose those points.\n3. Once you receive the information you need from your teacher, describe the domain and range that would be reasonable for each function in this situation.\n\n### Student Response\n\nFor access, consult one of our IM Certified Partners.\n\n### Anticipated Misconceptions\n\nSome students may struggle to match the descriptions and the graphs because they confuse the independent and dependent variables and think that, in each situation, time is represented by the vertical axis. Encourage them to re-read the activity statement, clarify the input and output in each situation, label the horizontal graph with the input, and then try interpreting the graphs again.\n\n### Activity Synthesis\n\nSelect students to share the domain and range for each function and their reasoning. Record and display their responses for all to see.\n\nOne key point to highlight is that the range of a function could be a single value (say 7, as shown in graph C), a bunch of isolated values (say, only some whole numbers, as shown in graph B), all values in an interval (say, all values from 1.5 to 4, as shown in graph D, or all values between 0 and 30, as in graph A), or a combination of these.\n\nThe domain of a function may also be limited in similar ways. Tell students that, in upcoming lessons, we will look at functions in which the rules relating their input and output are a bit more complex, so their domain and range are also a bit more so.\n\n## 11.3: Back to the Bouncing Ball (10 minutes)\n\n### Activity\n\nIn this activity, students continue to interpret a graph of a function in terms of a situation and relate the features of the graph to the domain and range of the function. The context is a familiar one, allowing students to focus their reasoning on domain and range.\n\nUnlike in the activity about a child on a swing, the graph includes a scale on each axis and the coordinate pairs of some points, allowing students to identify the range more definitively. On the other hand, the graph is a partial representation of the function, as it does not show what happens after a dropped ball hits the ground the fifth time. When describing the domain, students need to attend to what is reasonable in this situation (that is, noting that the ball likely does not just stop after the fifth bounce).\n\n### Student Facing\n\nA tennis ball was dropped from a certain height. It bounced several times, rolled along for a short period, and then stopped. Function $$H$$ gives its height over time.\n\nHere is a partial graph of $$H$$. Height is measured in feet. Time is measured in seconds.\n\nBe prepared to explain what each value or set of values means in this situation.\n\n1. Find $$H(0)$$.\n2. Solve $$H(x) = 0$$.\n3. Describe the domain of the function.\n4. Describe the range of the function.\n\n### Student Response\n\nFor access, consult one of our IM Certified Partners.\n\n### Student Facing\n\n#### Are you ready for more?\n\nIn function $$H$$, the input was time in seconds and the output was height in feet.\n\nThink about some other quantities that could be inputs or outputs in this situation.\n\n1. Describe a function whose domain includes only integers. Be sure to specify the units.\n2. Describe a function whose range includes only integers. Be sure to specify the units.\n3. Sketch a graph of each function.\n\n### Student Response\n\nFor access, consult one of our IM Certified Partners.\n\n### Activity Synthesis\n\nFocus the discussion on how students reasoned about the domain and the range for the function. Highlight explanations that account for what is realistic in the context.\n\n• The domain tells us all the possible amounts of time that passed since the moment the tennis ball was dropped until it stopped rolling.\n• The range includes all the possible heights of the tennis ball from the time it was dropped until the time it stopped rolling.\n\nAsk students if there are points on the graph whose coordinates are particularly useful for identifying the domain and range of the function. (The heights of the bounces? The points where the ball hits the ground? Points between the two? Others?)\n\nEmphasize that, just like in the activity about the swing, some points and features on a graph can give us more information than others about possible input-output values of a function.\n\nSpeaking, Listening, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion of the meaning of domain and range in the context of a tennis ball bouncing. As students share explanations of their reasoning about the domain and the range for the function, encourage students to press for details and precision in language in peers’ responses by asking questions such as, “Can you explain this using points on the graph?” or “Are there values for the domain and range not shown on the graph? How do you know?”\nDesign Principle(s): Optimize output (for explanation); Cultivate conversation\n\n## Lesson Synthesis\n\n### Lesson Synthesis\n\nDisplay several familiar graphs of functions from this unit. Here are some examples.\n\nAsk students to examine the graphs and use them to help summarize what graphs can tell us about the domain and range of functions. Ask students to complete the following prompts as thoroughly as they can.\n\n• I can learn about the domain and range of a function from a graph by looking for . . .\n• A graph may not always show all that is needed to fully describe the domain and range, however. For example, it may not show . . .\n\n## 11.4: Cool-down - A Pot of Water (5 minutes)\n\n### Cool-Down\n\nFor access, consult one of our IM Certified Partners.\n\n## Student Lesson Summary\n\n### Student Facing\n\nThe graph of a function can sometimes give us information about its domain and range.\n\nHere are graphs of two functions we saw earlier in the unit. The first graph represents the best price of bagels as a function of the number of bagels bought. The second graph represents the height of a bungee jumper as a function of seconds since the jump began.\n\nWhat are the domain and range of each function?\n\nThe number of bagels cannot be negative but could include 0 (no bagels bought). The domain of the function therefore includes 0 and positive whole numbers, or $$n \\geq 0$$.\n\nThe best price can be \\\\$0 (for buying 0 bagels), certain multiples of 1.25, certain multiples of 6, and so on. The range includes 0 and certain positive values.\n\nThe domain of the height function would include any amount of time since the jump began, up until the jump is complete. From the graph, we can tell that this happened more than 70 seconds after the jump began, but we don't know the exact value of $$t$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93561643,"math_prob":0.93029946,"size":12077,"snap":"2022-27-2022-33","text_gpt3_token_len":2531,"char_repetition_ratio":0.17667523,"word_repetition_ratio":0.050400376,"special_character_ratio":0.21354641,"punctuation_ratio":0.09383033,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97605366,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T02:12:29Z\",\"WARC-Record-ID\":\"<urn:uuid:a8709f54-efb4-4df3-9266-c178b206cb99>\",\"Content-Length\":\"130648\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a343feb-f0ea-4820-9c7d-29e58187c3ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:f5c22c01-9b5f-4a2b-8d44-28eba1e7172e>\",\"WARC-IP-Address\":\"34.201.80.84\",\"WARC-Target-URI\":\"https://curriculum.illustrativemathematics.org/HS/teachers/1/4/11/index.html\",\"WARC-Payload-Digest\":\"sha1:E3AFIPY74UH3ZF22A3LKMFJRX7Z4DQJX\",\"WARC-Block-Digest\":\"sha1:BWVPIQV5S4FE2JQLB7F7IJXGOIGM5C75\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103033925.2_warc_CC-MAIN-20220625004242-20220625034242-00157.warc.gz\"}"}
https://git.proxmox.com/?p=mirror_edk2.git;a=blob;f=MdeModulePkg/Universal/Network/SnpDxe/Callback.c;h=c4789ba11dc81be576666a40eb7e1bca2e180f9e;hb=4cda7726e5fd30aaf3e05c80207ae1b264bfa123	[ "1 / @file\\r\n2 This file contains two sets of callback routines for undi3.0 and undi3.1.\\r\n3 the callback routines for Undi3.1 have an extra parameter UniqueId which\\r\n4 stores the interface context for the NIC that snp is trying to talk.\\r\n5 \\r\n6 Copyright (c) 2006 - 2008, Intel Corporation. <BR>\\r\n7 All rights reserved. This program and the accompanying materials\\r\n9 which accompanies this distribution. The full text of the license may be found at\\r\n11 \\r\n12 THE PROGRAM IS DISTRIBUTED UNDER THE BSD LICENSE ON AN \"AS IS\" BASIS,\\r\n13 WITHOUT WARRANTIES OR REPRESENTATIONS OF ANY KIND, EITHER EXPRESS OR IMPLIED.\\r\n14 \\r\n15 /\\r\n16 \\r\n17 #include \"Snp.h\"\\r\n18 \\r\n19 //\\r\n20 // Global variables\\r\n21 // these 2 global variables are used only for 3.0 undi. we could not place\\r\n22 // them in the snp structure because we will not know which snp structure\\r\n23 // in the callback context!\\r\n24 //\\r\n25 BOOLEAN mInitializeLock = TRUE;\\r\n26 EFI_LOCK mLock;\\r\n27 \\r\n28 //\\r\n29 // End Global variables\\r\n30 //\\r\n31 extern EFI_PCI_IO_PROTOCOL mPciIo;\\r\n32 \\r\n33 /\\r\n34 This is a callback routine supplied to UNDI at undi_start time.\\r\n35 UNDI call this routine with a virtual or CPU address that SNP provided to \\r\n36 convert it to a physical or device address. Since EFI uses the identical \\r\n37 mapping, this routine returns the physical address same as the virtual address\\r\n38 for most of the addresses. an address above 4GB cannot generally be used as a \\r\n39 device address, it needs to be mapped to a lower physical address. This \\r\n40 routine does not call the map routine itself, but it assumes that the mapping\\r\n41 was done at the time of providing the address to UNDI. This routine just \\r\n42 looks up the address in a map table (which is the v2p structure chain). \\r\n43 \\r\n46 The DeviceAddrPtr will contain 0 in case of any error.\\r\n47 \\r\n48 /\\r\n49 VOID\\r\n50 SnpUndi32CallbackV2p30 (\\r\n53 )\\r\n54 {\\r\n55 V2P V2p;\\r\n56 //\\r\n57 // Do nothing if virtual address is zero or physical pointer is NULL.\\r\n58 // No need to map if the virtual address is within 4GB limit since\\r\n59 // EFI uses identical mapping\\r\n60 //\\r\n62 DEBUG ((EFI_D_NET, \"\\nv2p: Null virtual address or physical pointer.\\n\"));\\r\n63 return ;\\r\n64 }\\r\n65 \\r\n66 if (CpuAddr < FOUR_GIGABYTES) {\\r\n68 return ;\\r\n69 }\\r\n70 //\\r\n71 // SNP creates a vaddr tp paddr mapping at the time of calling undi with any\\r\n72 // big address, this callback routine just looks up in the v2p list and\\r\n74 //\\r\n75 if (FindV2p (&V2p, (VOID ) (UINTN) CpuAddr) != EFI_SUCCESS) {\\r\n77 } else {\\r\n79 }\\r\n80 }\\r\n81 \\r\n82 /\\r\n83 This is a callback routine supplied to UNDI at undi_start time.\\r\n84 UNDI call this routine when it wants to have exclusive access to a critical\\r\n85 section of the code/data.\\r\n86 \\r\n87 @param Enable non-zero indicates acquire\\r\n88 zero indicates release\\r\n89 \\r\n90 /\\r\n91 VOID\\r\n92 SnpUndi32CallbackBlock30 (\\r\n93 IN UINT32 Enable\\r\n94 )\\r\n95 {\\r\n96 //\\r\n97 // tcpip was calling snp at tpl_notify and if we acquire a lock that was\\r\n98 // created at a lower level (TPL_CALLBACK) it gives an assert!\\r\n99 //\\r\n100 if (mInitializeLock) {\\r\n101 EfiInitializeLock (&mLock, TPL_NOTIFY);\\r\n102 mInitializeLock = FALSE;\\r\n103 }\\r\n104 \\r\n105 if (Enable != 0) {\\r\n106 EfiAcquireLock (&mLock);\\r\n107 } else {\\r\n108 EfiReleaseLock (&mLock);\\r\n109 }\\r\n110 }\\r\n111 \\r\n112 /\\r\n113 This is a callback routine supplied to UNDI at undi_start time.\\r\n114 UNDI call this routine with the number of micro seconds when it wants to\\r\n115 pause.\\r\n116 \\r\n117 @param MicroSeconds number of micro seconds to pause, ususlly multiple of 10.\\r\n118 \\r\n119 /\\r\n120 VOID\\r\n121 SnpUndi32CallbackDelay30 (\\r\n122 IN UINT64 MicroSeconds\\r\n123 )\\r\n124 {\\r\n125 if (MicroSeconds != 0) {\\r\n126 gBS->Stall ((UINTN) MicroSeconds);\\r\n127 }\\r\n128 }\\r\n129 \\r\n130 /\\r\n131 This is a callback routine supplied to UNDI at undi_start time.\\r\n132 This is the IO routine for UNDI. This is not currently being used by UNDI3.0\\r\n133 because Undi3.0 uses io/mem offsets relative to the beginning of the device\\r\n134 io/mem address and so it needs to use the PCI_IO_FUNCTION that abstracts the\\r\n135 start of the device's io/mem addresses. Since SNP cannot retrive the context\\r\n136 of the undi3.0 interface it cannot use the PCI_IO_FUNCTION that specific for\\r\n137 that NIC and uses one global IO functions structure, this does not work.\\r\n138 This however works fine for EFI1.0 Undis because they use absolute addresses\\r\n139 for io/mem access.\\r\n140 \\r\n142 @param NumBytes number of bytes to read or write\\r\n144 @param BufferAddr memory location to read into or that contains the bytes to \\r\n145 write\\r\n146 \\r\n147 /\\r\n148 VOID\\r\n149 SnpUndi32CallbackMemio30 (\\r\n151 IN UINT8 NumBytes,\\r\n154 )\\r\n155 {\\r\n156 EFI_PCI_IO_PROTOCOL_WIDTH Width;\\r\n157 \\r\n158 switch (NumBytes) {\\r\n159 case 2:\\r\n160 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 1;\\r\n161 break;\\r\n162 \\r\n163 case 4:\\r\n164 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 2;\\r\n165 break;\\r\n166 \\r\n167 case 8:\\r\n168 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 3;\\r\n169 break;\\r\n170 \\r\n171 default:\\r\n172 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 0;\\r\n173 }\\r\n174 \\r\n178 mPciIo,\\r\n179 Width,\\r\n180 1, // BAR 1, IO base address\\r\n182 1, // count\\r\n184 );\\r\n185 break;\\r\n186 \\r\n187 case PXE_IO_WRITE:\\r\n188 mPciIo->Io.Write (\\r\n189 mPciIo,\\r\n190 Width,\\r\n191 1, // BAR 1, IO base address\\r\n193 1, // count\\r\n195 );\\r\n196 break;\\r\n197 \\r\n200 mPciIo,\\r\n201 Width,\\r\n202 0, // BAR 0, Memory base address\\r\n204 1, // count\\r\n206 );\\r\n207 break;\\r\n208 \\r\n209 case PXE_MEM_WRITE:\\r\n210 mPciIo->Mem.Write (\\r\n211 mPciIo,\\r\n212 Width,\\r\n213 0, // BAR 0, Memory base address\\r\n215 1, // count\\r\n217 );\\r\n218 break;\\r\n219 }\\r\n220 \\r\n221 return ;\\r\n222 }\\r\n223 \\r\n224 /\\r\n225 This is a callback routine supplied to UNDI3.1 at undi_start time.\\r\n226 UNDI call this routine when it wants to have exclusive access to a critical\\r\n227 section of the code/data.\\r\n228 New callbacks for 3.1:\\r\n229 there won't be a virtual2physical callback for UNDI 3.1 because undi3.1 uses\\r\n230 the MemMap call to map the required address by itself!\\r\n231 \\r\n232 @param UniqueId This was supplied to UNDI at Undi_Start, SNP uses this to \\r\n233 store Undi interface context (Undi does not read or write\\r\n234 this variable)\\r\n235 @param Enable non-zero indicates acquire\\r\n236 zero indicates release\\r\n237 /\\r\n238 VOID\\r\n239 SnpUndi32CallbackBlock (\\r\n240 IN UINT64 UniqueId,\\r\n241 IN UINT32 Enable\\r\n242 )\\r\n243 {\\r\n244 SNP_DRIVER Snp;\\r\n245 \\r\n246 Snp = (SNP_DRIVER ) (UINTN) UniqueId;\\r\n247 //\\r\n248 // tcpip was calling snp at tpl_notify and when we acquire a lock that was\\r\n249 // created at a lower level (TPL_CALLBACK) it gives an assert!\\r\n250 //\\r\n251 if (Enable != 0) {\\r\n252 EfiAcquireLock (&Snp->Lock);\\r\n253 } else {\\r\n254 EfiReleaseLock (&Snp->Lock);\\r\n255 }\\r\n256 }\\r\n257 \\r\n258 /\\r\n259 This is a callback routine supplied to UNDI at undi_start time.\\r\n260 UNDI call this routine with the number of micro seconds when it wants to\\r\n261 pause.\\r\n262 \\r\n263 @param UniqueId This was supplied to UNDI at Undi_Start, SNP uses this to\\r\n264 store Undi interface context (Undi does not read or write\\r\n265 this variable)\\r\n266 @param MicroSeconds number of micro seconds to pause, ususlly multiple of 10.\\r\n267 /\\r\n268 VOID\\r\n269 SnpUndi32CallbackDelay (\\r\n270 IN UINT64 UniqueId,\\r\n271 IN UINT64 MicroSeconds\\r\n272 )\\r\n273 {\\r\n274 if (MicroSeconds != 0) {\\r\n275 gBS->Stall ((UINTN) MicroSeconds);\\r\n276 }\\r\n277 }\\r\n278 \\r\n279 /\\r\n280 This is a callback routine supplied to UNDI at undi_start time.\\r\n281 This is the IO routine for UNDI3.1 to start CPB.\\r\n282 \\r\n283 @param UniqueId This was supplied to UNDI at Undi_Start, SNP uses this \\r\n284 to store Undi interface context (Undi does not read or\\r\n285 write this variable)\\r\n287 @param NumBytes number of bytes to read or write.\\r\n289 @param BufferPtr memory location to read into or that contains the bytes\\r\n290 to write.\\r\n291 /\\r\n292 VOID\\r\n293 SnpUndi32CallbackMemio (\\r\n294 IN UINT64 UniqueId,\\r\n296 IN UINT8 NumBytes,\\r\n298 IN OUT UINT64 BufferPtr\\r\n299 )\\r\n300 {\\r\n301 SNP_DRIVER Snp;\\r\n302 EFI_PCI_IO_PROTOCOL_WIDTH Width;\\r\n303 \\r\n304 Snp = (SNP_DRIVER ) (UINTN) UniqueId;\\r\n305 \\r\n306 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 0;\\r\n307 switch (NumBytes) {\\r\n308 case 2:\\r\n309 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 1;\\r\n310 break;\\r\n311 \\r\n312 case 4:\\r\n313 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 2;\\r\n314 break;\\r\n315 \\r\n316 case 8:\\r\n317 Width = (EFI_PCI_IO_PROTOCOL_WIDTH) 3;\\r\n318 break;\\r\n319 }\\r\n320 \\r\n324 Snp->PciIo,\\r\n325 Width,\\r\n326 Snp->IoBarIndex, // BAR 1 (for 32bit regs), IO base address\\r\n328 1, // count\\r\n329 (VOID ) (UINTN) BufferPtr\\r\n330 );\\r\n331 break;\\r\n332 \\r\n333 case PXE_IO_WRITE:\\r\n334 Snp->PciIo->Io.Write (\\r\n335 Snp->PciIo,\\r\n336 Width,\\r\n337 Snp->IoBarIndex, // BAR 1 (for 32bit regs), IO base address\\r\n339 1, // count\\r\n340 (VOID ) (UINTN) BufferPtr\\r\n341 );\\r\n342 break;\\r\n343 \\r\n346 Snp->PciIo,\\r\n347 Width,\\r\n348 Snp->MemoryBarIndex, // BAR 0, Memory base address\\r\n350 1, // count\\r\n351 (VOID ) (UINTN) BufferPtr\\r\n352 );\\r\n353 break;\\r\n354 \\r\n355 case PXE_MEM_WRITE:\\r\n356 Snp->PciIo->Mem.Write (\\r\n357 Snp->PciIo,\\r\n358 Width,\\r\n359 Snp->MemoryBarIndex, // BAR 0, Memory base address\\r\n361 1, // count\\r\n362 (VOID ) (UINTN) BufferPtr\\r\n363 );\\r\n364 break;\\r\n365 }\\r\n366 \\r\n367 return ;\\r\n368 }\\r\n369 \\r\n370 /\\r\n371 This is a callback routine supplied to UNDI at undi_start time.\\r\n372 UNDI call this routine when it has to map a CPU address to a device\\r\n374 \\r\n375 @param UniqueId - This was supplied to UNDI at Undi_Start, SNP uses this to store\\r\n376 Undi interface context (Undi does not read or write this variable)\\r\n378 @param NumBytes - size of memory to be mapped\\r\n379 @param Direction - direction of data flow for this memory's usage:\\r\n380 cpu->device, device->cpu or both ways\\r\n382 \\r\n383 /\\r\n384 VOID\\r\n385 SnpUndi32CallbackMap (\\r\n386 IN UINT64 UniqueId,\\r\n388 IN UINT32 NumBytes,\\r\n389 IN UINT32 Direction,\\r\n391 )\\r\n392 {\\r\n394 EFI_PCI_IO_PROTOCOL_OPERATION DirectionFlag;\\r\n395 UINTN BuffSize;\\r\n396 SNP_DRIVER Snp;\\r\n397 UINTN Index;\\r\n398 EFI_STATUS Status;\\r\n399 \\r\n400 BuffSize = (UINTN) NumBytes;\\r\n401 Snp = (SNP_DRIVER ) (UINTN) UniqueId;\\r\n403 \\r\n404 if (CpuAddr == 0) {\\r\n406 return ;\\r\n407 }\\r\n408 \\r\n409 switch (Direction) {\\r\n410 case TO_AND_FROM_DEVICE:\\r\n411 DirectionFlag = EfiPciIoOperationBusMasterCommonBuffer;\\r\n412 break;\\r\n413 \\r\n414 case FROM_DEVICE:\\r\n415 DirectionFlag = EfiPciIoOperationBusMasterWrite;\\r\n416 break;\\r\n417 \\r\n418 case TO_DEVICE:\\r\n420 break;\\r\n421 \\r\n422 default:\\r\n424 //\\r\n425 // any non zero indicates error!\\r\n426 //\\r\n427 return ;\\r\n428 }\\r\n429 //\\r\n430 // find an unused map_list entry\\r\n431 //\\r\n432 for (Index = 0; Index < MAX_MAP_LENGTH; Index++) {\\r\n433 if (Snp->MapList[Index].VirtualAddress == 0) {\\r\n434 break;\\r\n435 }\\r\n436 }\\r\n437 \\r\n438 if (Index >= MAX_MAP_LENGTH) {\\r\n439 DEBUG ((EFI_D_INFO, \"SNP maplist is FULL\\n\"));\\r\n441 return ;\\r\n442 }\\r\n443 \\r\n445 \\r\n446 Status = Snp->PciIo->Map (\\r\n447 Snp->PciIo,\\r\n448 DirectionFlag,\\r\n450 &BuffSize,\\r\n453 );\\r\n454 if (Status != EFI_SUCCESS) {\\r\n457 }\\r\n458 \\r\n459 return ;\\r\n460 }\\r\n461 \\r\n462 /\\r\n463 This is a callback routine supplied to UNDI at undi_start time.\\r\n464 UNDI call this routine when it wants to unmap an address that was previously\\r\n465 mapped using map callback.\\r\n466 \\r\n467 @param UniqueId This was supplied to UNDI at Undi_Start, SNP uses this to store.\\r\n468 Undi interface context (Undi does not read or write this variable)\\r\n470 @param NumBytes size of memory mapped\\r\n471 @param Direction direction of data flow for this memory's usage:\\r\n472 cpu->device, device->cpu or both ways\\r\n474 \\r\n475 /\\r\n476 VOID\\r\n477 SnpUndi32CallbackUnmap (\\r\n478 IN UINT64 UniqueId,\\r\n480 IN UINT32 NumBytes,\\r\n481 IN UINT32 Direction,\\r\n483 )\\r\n484 {\\r\n485 SNP_DRIVER Snp;\\r\n486 UINT16 Index;\\r\n487 \\r\n488 Snp = (SNP_DRIVER ) (UINTN) UniqueId;\\r\n489 \\r\n490 for (Index = 0; Index < MAX_MAP_LENGTH; Index++) {\\r\n492 break;\\r\n493 }\\r\n494 }\\r\n495 \\r\n496 if (Index >= MAX_MAP_LENGTH)\\r\n497 {\\r\n498 DEBUG ((EFI_D_ERROR, \"SNP could not find a mapping, failed to unmap.\\n\"));\\r\n499 return ;\\r\n500 }\\r\n501 \\r\n505 return ;\\r\n506 }\\r\n507 \\r\n508 /\\r\n509 This is a callback routine supplied to UNDI at undi_start time.\\r\n510 UNDI call this routine when it wants synchronize the virtual buffer contents\\r\n511 with the mapped buffer contents. The virtual and mapped buffers need not\\r\n512 correspond to the same physical memory (especially if the virtual address is\\r\n513 > 4GB). Depending on the direction for which the buffer is mapped, undi will\\r\n514 need to synchronize their contents whenever it writes to/reads from the buffer\\r\n516 \\r\n517 EFI does not provide a sync call, since virt=physical, we sould just do\\r\n518 the synchronization ourself here!\\r\n519 \\r\n520 @param UniqueId This was supplied to UNDI at Undi_Start, SNP uses this to store\\r\n521 Undi interface context (Undi does not read or write this variable)\\r\n523 @param NumBytes size of memory mapped.\\r\n524 @param Direction direction of data flow for this memory's usage:\\r\n525 cpu->device, device->cpu or both ways.\\r\n527 \\r\n528 */\\r\n529 VOID\\r\n530 SnpUndi32CallbackSync (\\r\n531 IN UINT64 UniqueId,\\r\n533 IN UINT32 NumBytes,\\r\n534 IN UINT32 Direction,\\r\n536 )\\r\n537 {\\r\n538 if ((CpuAddr == 0) \|\| (DeviceAddr == 0) \|\| (NumBytes == 0)) {\\r\n539 return ;\\r\n540 \\r\n541 }\\r\n542 \\r\n543 switch (Direction) {\\r\n544 case FROM_DEVICE:\\r" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.683985,"math_prob":0.6770538,"size":13026,"snap":"2020-45-2020-50","text_gpt3_token_len":3781,"char_repetition_ratio":0.1452158,"word_repetition_ratio":0.206507,"special_character_ratio":0.32135728,"punctuation_ratio":0.08945821,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9620816,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T18:42:25Z\",\"WARC-Record-ID\":\"<urn:uuid:18df5f03-0905-49c1-89a1-13f7e8cd2159>\",\"Content-Length\":\"194463\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:13ba4960-7de4-466f-aa95-aadb91ae1fea>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec20110d-adf6-4efe-92dc-a803c32d61fa>\",\"WARC-IP-Address\":\"79.133.36.253\",\"WARC-Target-URI\":\"https://git.proxmox.com/?p=mirror_edk2.git;a=blob;f=MdeModulePkg/Universal/Network/SnpDxe/Callback.c;h=c4789ba11dc81be576666a40eb7e1bca2e180f9e;hb=4cda7726e5fd30aaf3e05c80207ae1b264bfa123\",\"WARC-Payload-Digest\":\"sha1:KST7P55M266FLQ73C7LA4AZEYN6CDNWX\",\"WARC-Block-Digest\":\"sha1:3NJTTV62GPOGNHMOLG3ZMYUUT7KBQKBS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141188899.42_warc_CC-MAIN-20201126171830-20201126201830-00500.warc.gz\"}"}
https://math.stackexchange.com/questions/2574815/find-an-angle-of-a-triangle-on-a-larger-triangle-which-cut-through-its-midpoint	[ "# Find an angle of a triangle on a larger triangle which cut through its midpoint\n\nIn triangle $$\\triangle BAC$$ with $$\\angle ABC = 30\\deg$$. $$D$$ is the midpoint of $$BC$$. We join $$A$$ and $$D$$ and $$\\angle CDA = 45 \\deg$$. Find $$\\angle BAC$$.", null, "On applying Sine rule, $$\\frac{2x}{\\sin {(15+\\theta)}}=\\frac{AC}{\\sin 30}$$ and also $$\\frac{x}{\\sin \\theta}=\\frac{AC}{\\sin 45}$$\nWhere $$x$$ is $$CD$$ or $$DB$$ and $$\\theta$$ is $$\\angle CAD$$.\n\nBut solving this gives $$\\frac{\\sin {(15+\\theta)}}{\\sin \\theta}=\\sqrt 2$$\n\nIs this correct?\n\n• I think that this can be solved suing just F and Z angles. Draw a line parallel to $AB$ that goes through $C$. – stuart stevenson Dec 20 '17 at 18:57\n• @stuartstevenson Ok....but....can you please go through my method?....i want to know why it is not working. – ami_ba Dec 20 '17 at 19:01\n• i think your second equation is wrong! – Dr. Sonnhard Graubner Dec 20 '17 at 19:08\n• I don't think there's a problem with it. – stuart stevenson Dec 20 '17 at 19:10\n• but i think so, you can not use $$45^{\\circ}$$ AND $$\\theta$$ in the triangle $$\\Delta ADC$$ – Dr. Sonnhard Graubner Dec 20 '17 at 19:14\n\nYour reasoning looks good to me. Using your second equation, $$AC=\\frac{x}{\\sqrt{2}\\sin{\\theta}}$$ Now substituting $AC$ in the first equation, $$\\frac{2x}{\\sin{(15+\\theta)}}=\\frac{2x}{\\sqrt{2}\\sin{\\theta}}$$ or $$\\sin{(15+\\theta)}=\\sqrt{2}\\sin{\\theta}$$ Using trig identity, $$\\cos15\\sin\\theta+\\sin15\\cos\\theta=\\sqrt{2}\\sin{\\theta}$$ Dividing by $\\sin\\theta$ we get $$\\cot\\theta=\\frac{\\sqrt{2}-\\cos15}{\\sin15}$$ Knowing that $\\sin15=\\frac{\\sqrt{3}-1}{2\\sqrt{2}}$, $\\cos15=\\frac{\\sqrt{3}+1}{2\\sqrt{2}}$ we get $\\cot\\theta=\\sqrt{3}$, $\\theta=30°$\n\n• Perfect....thank you☺️☺️ – ami_ba Dec 21 '17 at 3:35\n\nyou will Need three equations $$a^2=b^2+c^2-2bc\\cos(\\alpha)$$ $$a=\\frac{b\\sin(\\alpha)}{\\sin(30^{\\circ})}$$ $$c=\\frac{b\\sin(150^{\\circ})}{\\sin(30^{\\circ})}$$ then you Can divide by $b^2$ and you will get only an equation for $$\\alpha$$\n\nNever Underestimate the Power of Euclidean Geometry", null, "• By Exterior Angle Theorem $$\\angle DAB = 15°$$.\n• Draw from $$C$$ the line perpendicular to $$AB$$, intersecting $$AB$$ in $$E$$. We then have $$\\angle ECB = 60°$$ and thus $$\\triangle EBC$$ is half of an equilater triangle and $$CE \\cong \\frac{BC}{2} \\cong CD \\cong BD$$.\n• Now connect $$E$$ with $$D$$. $$\\triangle CED$$ is isosceles, but having $$\\angle ECB = 60°$$, it is also equilateral and thus $$ED \\cong CE$$, and $$\\angle EDA = \\angle DAB = 15°$$.\n• So $$\\triangle AED$$ is isosceles and we get also $$AE \\cong CE$$.\n• We conclude that $$\\triangle ACE$$ is isosceles and right-angled. Therefore $$\\angle BAC = 45°$$.\n\n$$\\blacksquare$$" ]	[ null, "https://i.stack.imgur.com/obpU1.jpg", null, "https://i.stack.imgur.com/R1DDO.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7145418,"math_prob":1.0000091,"size":1415,"snap":"2021-31-2021-39","text_gpt3_token_len":509,"char_repetition_ratio":0.1438696,"word_repetition_ratio":0.0,"special_character_ratio":0.37738517,"punctuation_ratio":0.06741573,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T14:09:38Z\",\"WARC-Record-ID\":\"<urn:uuid:cfa9b035-3aa2-4124-a6a2-78c052c66c61>\",\"Content-Length\":\"190545\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cfabcd2e-04b4-431b-8002-1800873129f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:f48b5ea5-1806-43a5-83f3-098f55fa8930>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2574815/find-an-angle-of-a-triangle-on-a-larger-triangle-which-cut-through-its-midpoint\",\"WARC-Payload-Digest\":\"sha1:IO5G2BGO7RNDZYLLMDY5FDGN3UVBZD6H\",\"WARC-Block-Digest\":\"sha1:74SCL2JDBYKSUJXLFFE2KCBGMJWF5XRN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153392.43_warc_CC-MAIN-20210727135323-20210727165323-00644.warc.gz\"}"}
https://scicomp.stackexchange.com/questions/503/can-diagonal-plus-fixed-symmetric-linear-systems-be-solved-in-quadratic-time-aft	[ "# Can diagonal plus fixed symmetric linear systems be solved in quadratic time after precomputation?\n\nIs there an $O(n^3+n^2 k)$ method to solve $k$ linear systems of the form $(D_i + A) x_i = b_i$ where $A$ is a fixed SPD matrix and $D_i$ are positive diagonal matrices?\n\nFor example, if each $D_i$ is scalar, it suffices to compute the SVD of $A$. However, this breaks down for general $D$ due to lack of commutativity.\n\nUpdate: The answers so far are \"no\". Does anyone have any interesting intuition as to why? A no answer means that there's no nontrivial way to compress the information between two noncommuting operators. It isn't terribly surprisingly, but it'd be great to understand it better.\n\n• SPD = semi-positive definite? – rcollyer Jan 2 '12 at 15:50\n• Yes, though the problem is essentially the same without SPD. I added that constraint only to ensure that the systems are never singular. – Geoffrey Irving Jan 2 '12 at 19:40\n\n## 3 Answers\n\nThe closest positive answers to your question that I could find is for sparse diagonal perturbations (see below).\n\nWith that said, I do not know of any algorithms for the general case, though there is a generalization of the technique you mentioned for scalar shifts from SPD matrices to all square matrices:\n\nGiven any square matrix $A$, there exists a Schur decomposition $A=U T U^H$, where $U$ is unitary and $T$ is upper triangular, and $A+\\sigma I = U (T + \\sigma I) U^H$ provides a Schur decomposition of $A + \\sigma I$. Thus, your precomputation idea extends to all square matrices through the algorithm:\n\n• Compute $[U,T]=\\mathrm{schur}(A)$ in at most $\\mathcal{O}(n^3)$ work.\n• Solve each $(A+\\sigma I) x = b$ via $x := U (T +\\sigma I)^{-1} U^H b$ in $\\mathcal{O}(n^2)$ work (the middle inversion is simply back substitution).\n\nThis line of reasoning reduces to the approach you mentioned when $A$ is SPD since the Schur decomposition reduces to an EVD for normal matrices, and the EVD coincides with the SVD for Hermitian positive-definite matrices.\n\nResponse to update: Until I have a proof, which I do not, I refuse to claim that the answer is \"no\". However, I can give some insights as to why it's hard, as well as another subcase where the answer is yes.\n\nThe essential difficulty is that, even though the update is diagonal, it is still in general full rank, so the primary tool for updating an inverse, the Sherman-Morrison-Woodbury formula, does not appear to help. Even though the scalar shift case is also full rank, it is an extremely special case since it commutes with every matrix, as you mentioned.\n\nWith that said, if each $D$ was sparse, i.e., they each had $\\mathcal{O}(1)$ nonzeros, then the Sherman-Morrison-Woodbury formula yields an $\\mathcal{O}(n^2)$ solve with each pair $\\{D,b\\}$. For example, with a single nonzero at the $j$th diagonal entry, so that $D=\\delta e_j e_j^H$:\n\n$$[A^{-1}+\\delta e_j e_j^H]^{-1} = A^{-1} - \\frac{\\delta A^{-1} e_j e_j^H A^{-1}}{1+\\delta (e_j^H A^{-1} e_j)},$$\n\nwhere $e_j$ is the $j$th standard basis vector.\n\nAnother update: I should mention that I tried the $A^{-1}$ preconditioner that @GeoffOxberry suggested on a few random SPD $1000 \\times 1000$ matrices using PCG and, perhaps not surprisingly, it seems to greatly reduce the number of iterations when $\|\|D\|\|_2/\|\|A\|\|_2$ is small, but not when it is $\\mathcal{O}(1)$ or greater.\n\nIf $(D_{i} + A)$ is diagonally dominant for each $i$, then recent work by Koutis, Miller, and Peng (see Koutis' website for work on symmetric diagonally dominant matrices) could be used to solve each system in $\\mathcal{O}(n^2 \\log(n))$ time (actually $\\mathcal{O}(m\\log(n))$ time, where $m$ is the maximum number of nonzero entries in $(D_{i} + A)$ over all $i$, so you could take advantage of sparsity as well). Then, the total running time would be $\\mathcal{O}(n^2 \\log(n) k)$, which is better than the $\\mathcal{O}(n^3 k)$ approach of solving each system naïvely using dense linear algebra, but slightly worse than the quadratic run time you're asking for.\n\nSignificant sparsity in $(D_{i} + A)$ for all $i$ could be exploited by sparse solvers to yield an $\\mathcal{O}(n^2 k)$ algorithm, but I'm guessing that if you had significant sparsity, then you would have mentioned it.\n\nYou could also use $A^{-1}$ as a preconditioner to solve each system using iterative methods, and see how that works out.\n\nResponse to update: @JackPaulson makes a great point from the standpoint of numerical linear algebra and algorithms. I will focus on computational complexity arguments instead.\n\nThe computational complexity of the solution of linear systems and the computational complexity of matrix multiplication are essentially equal. (See Algebraic Complexity Theory.) If you could find an algorithm that could compress the information between two non-commuting operators (ignoring the positive semidefinite part) and directly solve the collection of systems you're proposing in quadratic time in $n$, then it's likely that you could use such an algorithm to make inferences about faster matrix multiplication. It's difficult to see how positive semidefinite structure could be used in a dense, direct method for linear systems to decrease its computational complexity.\n\nLike @JackPaulson, I'm unwilling to say that the answer is \"no\" without a proof, but given the connections above, the problem is very difficult and of current research interest. The best you could do from an asymptotic standpoint without leveraging special structure is an improvement on the Coppersmith and Winograd algorithm, yielding a $\\mathcal{O}(n^{\\alpha}k)$ algorithm, where $\\alpha \\approx 2.375$. That algorithm would be difficult to code, and would likely be slow for small matrices, because the constant factor preceding the asymptotic estimate is probably huge relative to Gaussian elimination.\n\n• I have yet to see a concrete statement of where the crossover might be, but several reputable sources have stated that (implementation issues aside), Coppersmith-Winograd cannot beat standard methods for matrix sizes that will be able to fit in memory in the near future (a few decades). Given that the Linpack benchmark takes more than a day to run on the current top machines, it does not seem likely that Coppersmith-Winograd will ever be used in practice. Strassen is actually practical for large problems, though it is somewhat less numerical stable. – Jed Brown Dec 25 '11 at 5:16\n• That does not surprise me. +1 for the implementation details. – Geoff Oxberry Dec 25 '11 at 13:42\n\nA first order Taylor expansion can be used to improve convergence over simple lagging. Suppose we have a preconditioner (or factors for a direct solve) available for $A+D$, and we want to use it for preconditioning $A$. We can compute\n\n\\begin{align} A^{-1} &= (A+D-D)^{-1} (A+D) (A+D)^{-1} \\\\ &= [(A+D)^{-1} (A+D-D)]^{-1} (A+D)^{-1} \\\\ &= [I - (A+D)^{-1} D]^{-1} (A+D)^{-1} \\\\ &\\approx [I + (A+D)^{-1} D] (A+D)^{-1} \\end{align}\n\nwhere the Taylor expansion was used to write the last line. Application of this preconditioner requires two solves with $A+D$.\n\nIt works fairly well when the preconditioner is shifted away from 0 by a similar or larger amount than the operator we are trying to solve with (e.g. $D\\gtrsim 0$). If the shift in the preconditioner is smaller ($D \\lesssim \\min \\sigma(A)$), the preconditioned operator becomes indefinite.\n\nIf the shift in the preconditioner is much larger than in the operator, this method tends to produce a condition number about half that of preconditioning by the lagged operator (in the random tests I ran, it could be better or worse for a specific class of matrices). That factor of 2 in condition number gives a factor of $\\sqrt 2$ in iteration count. If the iteration cost is dominated by the solves with $A+D$, then this is not a sufficient factor to justify the first order Taylor expansion. If matrix application is proportionately expensive (e.g. you only have an inexpensive-to-apply preconditioner for $A+D$), then this first order method may make sense." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8979222,"math_prob":0.99537927,"size":2390,"snap":"2019-35-2019-39","text_gpt3_token_len":664,"char_repetition_ratio":0.09974853,"word_repetition_ratio":0.0,"special_character_ratio":0.28075314,"punctuation_ratio":0.10330579,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99948853,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T14:35:48Z\",\"WARC-Record-ID\":\"<urn:uuid:9684601c-6e26-442e-a42b-a54c6ae0036a>\",\"Content-Length\":\"158224\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9fc87c1-e90c-4337-a9b5-dfb5fa3543c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:86f9ba78-d2d7-4244-a8d8-5d0e112cd0d9>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://scicomp.stackexchange.com/questions/503/can-diagonal-plus-fixed-symmetric-linear-systems-be-solved-in-quadratic-time-aft\",\"WARC-Payload-Digest\":\"sha1:RLOLMTINX7BYA6L2WF7AK4NWK2DXLYB7\",\"WARC-Block-Digest\":\"sha1:3Y3IEYGYH6JVSQBJY7ZOFEPL65JPBY6E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330233.1_warc_CC-MAIN-20190825130849-20190825152849-00404.warc.gz\"}"}
https://johnmuschelli.com/intro_to_r/HW/homework3.html	[ "### Instructions\n\n1. You must submit both the RMD and “knitted” HTML files as one compressed .zip to the Homework 3 Drop Box on CoursePlus.\n2. All assignments are due by the end of the grading period for this term (26 June 2020).\n\n### Getting Started\n\nIn this assignment, we will be working with the infant mortality data set, found here: http://johnmuschelli.com/intro_to_r/data/indicatordeadkids35.csv.\n\nThe packages listed below are simply suggestions, but please edit this list as you see fit.\n\n## you can add more, or change...these are suggestions\nlibrary(tidyverse)\nlibrary(dplyr)\nlibrary(ggplot2)\nlibrary(tidyr)\n\n### Problem Set\n\n1. Read the data using read_csv() and name it mort. Rename the first column to country using the rename() command in dplyr. Create an object year variable by extracting column names (using colnames()) and make it to an integer as.integer() ), excluding the first column either with string manipulations or bracket subsetting or subsetting with is.na().\n\n2. Reshape the data so that there is a variable named year corresponding to year (key) and a column of the mortalities named mortality (value), using the tidyr package and its gather() function. Name the output long and make year a numeric variable.\nHint: remember that -COLUMN_NAME removes that column, gather all the columns but country.\n\n3. Read in this the tab-delim file and call it pop: http://johnmuschelli.com/intro_to_r/data/country_pop.txt. The file contains population information on each country. Rename the second column to \"Country\" and the column \"% of world population\", to percent.\nHint: use read_tsv()\n\n4. Determine the population of each country in pop using arrange(). Get the order of the countries based on this (first is the highest population), and extract that column and call it pop_levels. Make a variable in the long data set named sorted that is the country variable coded as a factor based on pop_levels.\n\n5. Parts a, b, and c below are only broken up here for clarity, but all three components can be addressed in one chunk of code/as one function, using %>% as necessary.\n\na. Subset long based on years 1975-2010, including 1975 and 2010 and call this long_sub using & or the between() function.\nb. Further subset long_sub for the following countries using dplyr::filter() and the %in% operator on the sorted country factor (sorted):c(\"Venezuela\", \"Bahrain\", \"Estonia\", \"Iran\", \"Thailand\", \"Chile\", \"Western Sahara\", \"Azerbaijan\", \"Argentina\", \"Haiti\").\nc. Lastly, remove missing rows for mortality using filter() and is.na().\n\nHint: Be sure to assign your final object created from a through c as long_sub so you can use it in questions 6 and 7.\n\n6. Plotting: create “spaghetti”/line plots for the countries in long_sub, using different colors for different countries, using sorted. The x-axis should be year, and the y-axis should be mortality. Make the plot using a.qplot and b. ggplot.\n\n7. Bonus: load the plotly package (library(plotly)) and assign the plot from question 6 to g and run ggplotly(g)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79947716,"math_prob":0.5554885,"size":2762,"snap":"2021-43-2021-49","text_gpt3_token_len":681,"char_repetition_ratio":0.113850616,"word_repetition_ratio":0.0,"special_character_ratio":0.23750906,"punctuation_ratio":0.14694656,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96610963,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T01:42:49Z\",\"WARC-Record-ID\":\"<urn:uuid:5bdba052-f4d7-4803-a706-7515388874c7>\",\"Content-Length\":\"635024\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15e29e19-dd9f-41d0-8276-83eb252c49a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ad5f232-ebbe-43b3-ad2f-fcc2c1dcbaa5>\",\"WARC-IP-Address\":\"172.67.144.169\",\"WARC-Target-URI\":\"https://johnmuschelli.com/intro_to_r/HW/homework3.html\",\"WARC-Payload-Digest\":\"sha1:ZZ6CDRRKUKFFZVLH54ZINDGE6T666BH5\",\"WARC-Block-Digest\":\"sha1:D3TUBPYZQ6KHJRLZSAS2DDTCD6NCHRQJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587608.86_warc_CC-MAIN-20211024235512-20211025025512-00155.warc.gz\"}"}
https://apboardsolutions.in/ap-board-6th-class-maths-solutions-chapter-8-intext-questions/	[ "# AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions\n\nAP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions and Answers.\n\n## AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts InText Questions", null, "(Page No. 111)\n\ni) How many rays are there ?\nSolution:\nFour", null, "ii) How many coins are close to player Q ?\nSolution:\nB, C and D coins are close to player Q.\n\niii) While striking with striker there is a possibility of a coin touches with any other. Draw all such possibilities in the given picture by means of the line segments.\nSolution:\n$$\\overline{\\mathrm{CB}}$$ and $$\\overline{\\mathrm{DE}}$$ .\n\niv) How many such line segments can be drawn in the picture ?\nSolution:\n$$\\overline{\\mathrm{AB}}, \\overline{\\mathrm{BC}}, \\overline{\\mathrm{ED}}, \\overline{\\mathrm{AC}}, \\overline{\\mathrm{AE}}, \\overline{\\mathrm{CD}}, \\overline{\\mathrm{CE}}, \\overline{\\mathrm{AD}}, \\overline{\\mathrm{BD}} \\text { and } \\overline{\\mathrm{BE}}$$", null, "Let’s Do (Page No. 112)\n\nQuestion 1.\nObserve the table and their notations and fill the gaps.", null, "Solution:", null, "Lets Explore (Page No. 114)\n\nQuestion 1.\nMeasure the lengths of all line segments in the given figures by using divider and scale. Then compare the sides of the given figures.", null, "Solution:\ni) In ΔABC; $$\\overline{\\mathrm{AB}}$$ = 2.2 cm; $$\\overline{\\mathrm{BC}}$$ = 2 cm and $$\\overline{\\mathrm{AC}}$$ = 2.2 cm ‘\n2.2 cm = 2.2 cm > 2 cm i.e., $$\\overline{\\mathrm{AB}}=\\overline{\\mathrm{AC}}>\\overline{\\mathrm{BC}}$$\nTwo sides are equal and one side is different in length.\n\nii) In PQRS rectangle $$\\overline{\\mathrm{PS}}=\\overline{\\mathrm{QR}}$$ = 2.7 cm; $$\\overline{\\mathrm{P Q}}=\\overline{\\mathrm{RS}}$$ = 1.8 cm and $$\\overline{\\mathrm{PR}}=\\overline{\\mathrm{QS}}$$ = 3.2 cm.\nOpposite sides are equal and diagonals are equal in length,\n\niii) In KLMN square $$\\overline{\\mathrm{KL}}=\\overline{\\mathrm{LM}}=\\overline{\\mathrm{MN}}=\\overline{\\mathrm{KN}}$$ = 1.8 cm\nAll sides are equal in length.", null, "Let’s Do (Page No. 115)\n\nQuestion 1.\n(i) Find the parallel lines in the below figure. Name, write and read them.", null, "Solution:\na) l, m are parallel lines.\nWe denote this by writing l \|\| m and can be read as l is parallel to m.\nb) n, o are parallel lines.\nWe denote this by writing n\|\|o and can be read as n is parallel to o.\nc) p, q are parallel lines.\nWe denote this by writing p\|\|q and can be read as p is parallel to q.\n\nii) Find the intersecting lines in the above figure. Name, write and read them.\nSolution:\nIntersecting lines are (l, q); (m,q); (m, r); (n,q); (p, r); (o, r); (o, q); (q, r);\n\niii) Find the concurrent lines in the above figure. Name, write and read them.\nSolution:\nThree or more lines passing through the same point are called concurrent lines. Concurrent lines are (l, o, p) & (m, n, p).\n\niv) Find the perpendicular lines in the above figure. Name, write and read them.\nSolution:\na) p, o are perpendicular lines.\nWe denote this by writing p ⊥ o and can be read as p is perpendicular to o.\nb) p, n are perpendicular lines.\nWe denote this by writing p ⊥ n and can be read as p is perpendicular to n.\nc) n, q are perpendicular lines.\nWe denote this by writing n ⊥ q and can be read as n is perpendicular to q.\nd) q, o are perpendicular lines.\nWe denote this by writing q⊥ o and can be read as q is perpendicular to o.", null, "(Page No. 119)\n\nQuestion 1.\nMeasure the angles at the vertices.", null, "Solution:", null, "In triangle ABC,\nm∠BAC = 60°\nm∠ABC = 60°\nm∠ACB = 60°", null, "In triangle XYZ,\nm∠YXZ = 40°\nm∠XYZ = 70°\nm∠XZY = 70°", null, "", null, "In triangle PQR,\nm∠QPR = 35°\nm∠PQR = 38°\nm∠PRQ = 107°" ]	[ null, "https://apboardsolutions.in/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-1.png", null, "https://apboardsolutions.in/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-2.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-3.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-4.png", null, "https://apboardsolutions.in/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-5.png", null, "https://apboardsolutions.in/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-6.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-7.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-8.png", null, "https://apboardsolutions.in/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.in/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Solutions-Chapter-8-Basic-Geometric-Concepts-InText-Questions-9.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7627685,"math_prob":0.9989384,"size":3493,"snap":"2022-27-2022-33","text_gpt3_token_len":1076,"char_repetition_ratio":0.1994841,"word_repetition_ratio":0.1520979,"special_character_ratio":0.30375037,"punctuation_ratio":0.15579228,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99963653,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,null,null,1,null,null,null,1,null,1,null,1,null,null,null,1,null,null,null,1,null,1,null,1,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T06:48:59Z\",\"WARC-Record-ID\":\"<urn:uuid:64c2faed-43a9-43c5-b99b-8cf06218f8e5>\",\"Content-Length\":\"41746\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f51092e9-1794-48e7-89b1-fbc43b16dc83>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e1cf7ac-99bf-474f-81d9-3de22fbb4497>\",\"WARC-IP-Address\":\"143.244.140.135\",\"WARC-Target-URI\":\"https://apboardsolutions.in/ap-board-6th-class-maths-solutions-chapter-8-intext-questions/\",\"WARC-Payload-Digest\":\"sha1:4KRXLNAADKWGX6M6MTDNPYBTFMTTMHVH\",\"WARC-Block-Digest\":\"sha1:KUNJIWG4NLY2V43TDEKNP64CRO4JT57J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104514861.81_warc_CC-MAIN-20220705053147-20220705083147-00613.warc.gz\"}"}
https://www.markedbyteachers.com/as-and-a-level/science/rates-of-halogenoalkanes.html	[ "• Join over 1.2 million students every month\n• Accelerate your learning by 29%\n• Unlimited access from just £6.99 per month\n\n# Rates of Halogenoalkanes\n\nExtracts from this document...\n\nIntroduction\n\nRates of Reaction of Halogenoalkanes I have been asked to design an experiment to examine how the rate of reaction of halogenoalkanes varies with respect to the C-X bond. I will be varying the C-X bond using the halogens chlorine, bromine and iodine (C-Cl, C-Br or C-I) to see which of these bonds gives the fastest rate of hydrolysis. Other factors that affect the rate of reaction are temperature, concentration of reagents, pressure, surface area and the use of a catalyst, however I will make sure that these remain constant, in order to ensure that only the chosen variable (the halogenoalkanes) is causing any difference to the rate of hydrolysis. The hydrolysis of the halogenoalkanes involves a nucleophilic substitution reaction. The mechanism for the reaction is as follows. Firstly as the halogen atom is more electronegative than the carbon atom, the C-X bond is polarised to become, C?+ - X?- . The positive charge of the carbon atom promotes attack by a nucleophile (a chemical that donates a pair of electrons to form a new covalent bond) this causes heterolytic fission of the carbon-halogen bond to occur and the halogen atom is displaced as a halide ion. A new covalent bond is then formed between the nucleophile and the carbon. ...read more.\n\nMiddle\n\nis added to the halogenoalkanes. Calculations: From my preliminary experiment I know that a reasonable mass of 1-bromobutane to be hydrolysed is 0.20g. Using this I can calculate the mass of 1-chlorobutane and 1-iodobutane needed. For the experiment to be a fair test, the amount (in moles) of halogenoalkane being reacted must remain constant. I will use the equation: To work out the amount (in moles) that represents 0.20g of 1-bromobutane: Mr of C4H9Br (4 x 12.0) + (9x1.0) + (1x79.9) = 0.00146092 mol. of C4H9Br If the amount of halogenoalkane is to be constant then I must use this number of moles of the other two halogenoalkane compounds. Therefore for each reagent I will use the same equation as above, but rearrange it to make mass the subject and use the number just calculated for the moles value: For 1-chlorobutane (C4H9Cl): Mass = moles x Mr Mass = 0.00146092 x (4x12.0) + (9x1.0) + (35.5) Mass = 0.14g For 1-iodobutane (C4H9I): Mass = moles x Mr Mass = 0.00146092 x (4x12.0) + (9x1.0) + (127.0) Mass = 0.27g I also need to calculate the amount of AgNO3 that reacts with each hydrogen halide compound. Knowing the amount required by the reactions I can then ensure there is excess silver nitrate available, so as not to limit any of the reactions. ...read more.\n\nConclusion\n\n10) Put the bungs back in and shake each test tube twice. Shaking ensures complete mixing. Each test tube must be shaken the same number of times so that shaking does not become a variable (it must not be shaking that has any effect on the rate of hydrolysis). 11) Arrange the test tubes so that the sides with the crosses drawn on are furthest away from you. Look through each tube. The moment a cross becomes invisible, stop the stopwatch corresponding to the test tube number. Continue with this until all the crosses have become invisible, or 20 minutes has elapsed since the last silver nitrate was added. The halogenoalkanes should react well before 20 minutes, so this extra time is given to demonstrate that the control will never react, even when left for much longer than the other test tubes. A time of 20 minutes is sensible as it allows the experiment to be completed in a practical amount of time, whilst still giving adequate time to the control. 12) Record any observations of the test tube contents. Detailed and well presented observations will allow proper analysis of the results. 13) Record the time shown on each stopwatch in the relevant column of a results table. 14) Repeat the above procedure a further 2 times. To increase the reliability of the results. ...read more.\n\nThe above preview is unformatted text\n\nThis student written piece of work is one of many that can be found in our AS and A Level Organic Chemistry section.\n\n## Found what you're looking for?\n\n• Start learning 29% faster today\n• 150,000+ documents available\n• Just £6.99 a month\n\nNot the one? Search for your essay title...\n• Join over 1.2 million students every month\n• Accelerate your learning by 29%\n• Unlimited access from just £6.99 per month\n\n# Related AS and A Level Organic Chemistry essays\n\n1.", null, "## Find the enthalpy change of combustion of a number of alcohol's' so that you ...\n\nFrom my table you can see that the enthalpy change of combustion form methanol to ethanol increases by '134.34 Kj mol-1' and from ethanol to propan-1-ol is '158.96 Kj mol-1'. This shows the increase in enthalpy is roughly the same meaning that as the number of carbon atoms increase so does the enthalpy change of combustion.\n\n2.", null, "## The aim of this experiment is to produce Aspirin. This is an estrification in ...\n\n* Once the crystals were dry, they were weighed and the theoretical yield and percentage yield of aspirin was calculated. Results Appearance of crystals = white crystals Amount of salicylic acid used = 4.0029g Amount of re-crystallised product produced = 2.6586g Determine the melting point of Aspirin Apparatus * Melting\n\n1.", null, "## Comparing the enthalpy changes of combustion of different alcohols\n\nThis means that when taking the reading from a thermometer, it was possible to gain a recording within 0.05 of a degree. Therefore the percentage uncertainty will be found by the formula (0.05/temperature change) x 100. Percentage Errors Methanol 1st Recording Mass of fuel burned: 1.96g; Percentage uncertainty: (0.005/1.96)\n\n2.", null, "## Investigating the Enthalpy Changes of Combustion of Alcohols.\n\nIt is known that branching an alcohol reduced the enthalpy of combustion (2), but why? Another possible source of error is the fact that some of the heat energy evaporated the water in the calorimeter, rather than just heating up the water.\n\n1.", null, "## The aim of this experiment is to investigate the enthalpy change of combustion for ...\n\n* Light the spirit burner and replace the base of the calorimeter. * Wait for the temperature to rise several degrees while continuously stirring the water. * Extinguish the flame and switch off the suction pump. * Keep stirring for a further minute to ensure that the temperature of the water increases to its true value.\n\n2.", null, "## The Preparation of 1-bromobutane\n\nIt is used in the preparation of alcohols e.g. hydrolysis of 1-bromobutane produces 1-butan-1-ol 2. William synthesis: it is used in the preparation of ether. When a halogeno alkane is refluxed with an alcoholic solution of sodium alkoxide, ether is produced. E.g. CH3CH2CH2CH2Br + CH3COO-Na+ CH3CH2CH2CH2-O-CH2CH3 + NaBr 3)\n\n1.", null, "## The aim of this experiment is to obtain the rate equation for the reaction ...\n\nThe colorimeter was set zero again by putting the tube with distilled water. 7. Step 6 was repeated 4 times with solutions in test tube 3-6 and all of the absorbance was recorded. The colorimeter was set zero after using the tube of distilled water before every reading. 8.\n\n2.", null, "## Hydrolysing Organic Halogen Compounds. The purpose of this experiment is to find out ...\n\nThe leaving group is the halide ion which varies from chlorides, bromides and iodides depending on the substrates. Therefore the bonding energy / bonding length also varies depending on the halide group.", null, "• Over 160,000 pieces\nof student written work\n• Annotated by\nexperienced teachers\n• Ideas and feedback to", null, "" ]	[ null, "https://static1.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/61656/images/preview/img_138_1.jpg", null, "https://static1.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/907541/images/preview/img_138_1.jpg", null, "https://static2.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/836662/images/preview/img_138_1.jpg", null, "https://static2.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/60455/images/preview/img_138_1.jpg", null, "https://static3.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/20748/images/preview/img_138_1.jpg", null, "https://static3.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/946474/images/preview/img_138_1.jpg", null, "https://static3.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/935638/images/preview/img_138_1.jpg", null, "https://static2.mbtfiles.co.uk/media/docs/newdocs/as_and_a_level/science/chemistry/organic_chemistry/933447/images/preview/img_138_1.jpg", null, "https://static2.mbtfiles.co.uk/skin/frontend/mbt/mbt/images/cms/study-guides/what-is-mbt-text.png", null, "https://static2.mbtfiles.co.uk/skin/frontend/mbt/mbt/images/cms/study-guides/what-is-mbt.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9048117,"math_prob":0.86489344,"size":4156,"snap":"2021-31-2021-39","text_gpt3_token_len":1054,"char_repetition_ratio":0.116329476,"word_repetition_ratio":0.025352113,"special_character_ratio":0.23869105,"punctuation_ratio":0.11015912,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95586103,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,6,null,4,null,5,null,4,null,6,null,1,null,1,null,1,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T00:03:57Z\",\"WARC-Record-ID\":\"<urn:uuid:3268ac5a-18a6-4e0b-a7f1-2d032c68383a>\",\"Content-Length\":\"78580\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ebabbfc6-da2d-46aa-b2b5-ba53188f8a0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a4b6465f-60fd-43a2-885c-dadb5366ef48>\",\"WARC-IP-Address\":\"149.86.98.195\",\"WARC-Target-URI\":\"https://www.markedbyteachers.com/as-and-a-level/science/rates-of-halogenoalkanes.html\",\"WARC-Payload-Digest\":\"sha1:COBO5UBLXSGJ6IQEH3NWYR722QBPFY5W\",\"WARC-Block-Digest\":\"sha1:TD5GZ5MVMSOQKVTGBXCUAGUW2ABUBBBH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057403.84_warc_CC-MAIN-20210922223752-20210923013752-00339.warc.gz\"}"}
https://www.scirp.org/journal/paperinformation.aspx?paperid=110796	[ "Mutual Influence of the Atmosphere and the Ocean under Wave Processes\n\nAbstract\n\nThe article solves the problem of surface gravitational waves using the theory of tangential discontinuity between media: air-water. Using the improved equation of mass continuity and taking into account the atmosphere inhomogeneity in the gravitational field of the Earth, it is shown that during wave processes, these two media mutually influence each other, which explains the reason for the formation of a stormy condition over the ocean and the drop in atmospheric pressure before the storm. The mechanism of the formation of the “killer wave” has been established and thus the “greatest mystery of nature” has been solved. The scale of wind and tsunami wavelengths has been established.\n\nKeywords\n\nShare and Cite:\n\nKirtskhalia, V. and Ninidze, K. (2021) Mutual Influence of the Atmosphere and the Ocean under Wave Processes. Journal of Modern Physics, 12, 1346-1365. doi: 10.4236/jmp.2021.129081.\n\n1. Introduction\n\nIn early works (see e.g. ), we argued that linear theory describes only capillary waves. As it turned out, this statement was erroneous, but we came to this conclusion thanks to the existing theory of capillary waves, which is based on the assumptions of the potentiality motion of a liquid in the gravitational field of the Earth and its incompressibility. In these assumptions, the linear theory of tangential discontinuity is used in solving the problem, and in the dynamic boundary condition on the liquid surface, the surface tension force is used as the only stabilizing factor of the pressure perturbation ( , §62). Nevertheless, in the dispersion equation of capillary waves, a gravitational acceleration is present along with the surface tension coefficient and therefore, these waves are called capillary-gravitational. After we showed that the liquid cannot be incompressible, i.e. $\\nabla \\stackrel{¯}{V}\\ne 0$ , where $\\stackrel{\\to }{V}$ -is the velocity of the liquid particle, the gravitational acceleration from the dispersion equation disappeared . This fact led us to believe that gravitational waves should be described by a nonlinear theory. However, a simple analysis shows that in most cases gravitational waves are linear. Indeed, the maximum perturbation of pressure on the water surface in a gravitational wave is equal to ${{P}^{\\prime }}_{\\mathrm{max}}={\\rho }_{0}ga$, where a is the amplitude of the wave and ${\\rho }_{0}$ is the undisturbed density of water. At $a=3\\text{\\hspace{0.17em}}\\text{m}$, which is approximately equal to the maximum value of the amplitude of the wind wave away from the coast and is greater than the amplitude of the tsunami wave, this disturbance is equal to 2.7 × 104 Pa, while the equilibrium pressure is P0 = 1 atm. = 1.013 × 105 Pa. Thus, the pressure on the surface of water can be represented in the following form $P={P}_{0}+{P}^{\\prime }$, where ${P}^{\\prime }/{P}_{0}<1$, and therefore, linear theory can be used.\n\nA surface gravitational wave is generated and propagated at the interface between two media, i.e. it is a typical problem of tangential discontinuity. Therefore, the dispersion equation must contain the thermodynamic parameters of both media. Despite this, when solving this problem, many authors do not take into account the influence of atmospheric parameters on the phenomenon under study (see for example. ), but nevertheless, they get results that match the results of observations. For example, in work the two-dimensional problem of the surface gravitational wave was solved, in which the following are used:\n\n1) Equation of motion of fluid— $\\rho \\frac{\\text{d}\\stackrel{\\to }{V}}{\\text{d}t}=-\\nabla P+\\rho \\stackrel{\\to }{g}$ (1)\n\n2) Equation of incompressibility of liquid— $\\nabla \\stackrel{\\to }{V}=0$ (2)\n\nHere: $\\stackrel{\\to }{V}$ is velocity of a liquid particle, P—pressure, $\\rho$ —density, $\\stackrel{\\to }{g}$ —gravitational acceleration. It is assumed that the velocity is small and after linearization, the system of Equations (1), (2) takes the form:\n\n$\\frac{\\partial u}{\\partial t}=-\\frac{1}{\\rho }\\frac{\\partial P}{\\partial x},$ (3)\n\n$\\frac{\\partial w}{\\partial t}=-\\frac{1}{\\rho }\\frac{\\partial P}{\\partial z}-g,$ (4)\n\n$\\frac{\\partial u}{\\partial x}+\\frac{\\partial w}{\\partial z}=0.$ (5)\n\nBy denoting $\\xi \\left(x,t\\right)={\\xi }_{0}\\mathrm{cos}\\left(\\omega t-kx\\right)$, the displacement of the liquid surface along the Z axis and writing the boundary conditions on the free surface and on the bottom as\n\n${w\|}_{z=0}=\\frac{\\partial \\xi }{\\partial t}$ and ${w\|}_{z=-H}=0$, (6)\n\nsolutions of the system of Equations (3), (4), (5) will be:\n\n$u={\\xi }_{0}\\omega \\frac{\\mathrm{cosh}\\left[k\\left(z+H\\right)\\right]}{\\mathrm{sinh}\\left(kH\\right)}\\mathrm{cos}\\left(\\omega t-kx\\right),$ (7)\n\n$w=-{\\xi }_{0}\\omega \\frac{\\mathrm{sinh}\\left[k\\left(z+H\\right)\\right]}{\\mathrm{sinh}\\left(kH\\right)}\\mathrm{sin}\\left(\\omega t-kx\\right),$ (8)\n\n$P=-\\rho gz+\\rho g{\\xi }_{0}\\frac{\\mathrm{cosh}\\left[k\\left(z+H\\right)\\right]}{\\mathrm{cosh}\\left(kH\\right)}\\mathrm{cos}\\left(\\omega t-kx\\right).$ (9)\n\nSubstituting solutions (7) and (9) in (3), or (8) and (9) in (4) at, the author obtains an expression for the phase velocity in the form:\n\n$\|{U}_{p}\|=\\sqrt{\\frac{g}{k}\\mathrm{tanh}\\left(kH\\right)}.$ (10)\n\nFor deep water, when $kH=2\\pi H/\\lambda >1⇒\\mathrm{tanh}\\left(kH\\right)\\cong 1$, where $\\lambda$ is the wavelength, from (10) follows\n\n$\|{U}_{p}\|=\\sqrt{\\frac{g}{k}}$ (11)\n\nand for shallow water, when $kH=2\\pi H/\\lambda <1⇒\\mathrm{tanh}\\left(kH\\right)\\cong kH$,\n\n$\|{U}_{p}\|=\\sqrt{gH}.$ (12)\n\nSimilar results were obtained in using the hydrostatic approximation method, in which Equation (4) is integrated under the following assumption $\\partial w/\\partial t=0$. In the work it is noted that the validity of this assumption is not proven and this doubt is well-founded, because firstly- $\\partial w/\\partial t=0$ means that $w=const=0$, and therefore, no vertical movement of the liquid particle occurs, and second-at $\\partial w/\\partial t=0$, from Equation (4) follows $\\nabla P=\\rho \\stackrel{\\to }{g}$ which is a condition for the equilibrium of the liquid and therefore, vibrations are impossible.\n\nAs noted above, while analyzing the problem of surface gravity waves, both media should be taken into account—air and water, i.e. should be obtained in a linear approximation of the equations of gravitational waves in these media, and then, tied to their solutions on the boundary surface using boundary conditions.\n\nWhen performing these procedures, the perturbed values of density and pressure in each medium are linked by the equation of the state of the medium $\\rho ={P}^{\\prime }/{C}^{2}$, where C is the speed of sound in the medium. We called attention to the absurdity of the fact that the speed of sound in the entire atmosphere is calculated by the formula $C=\\sqrt{\\gamma {k}_{B}T/{m}_{0}}$ , where $\\gamma ={c}_{p}/{c}_{\\upsilon }=1.4$ is an adiabatic index of air and is equal to the ratio of heat capacities at constant pressure and volume, ${k}_{B}=1.38×{10}^{-23}\\text{J}/\\text{K}$ —the Boltzmann constant, ${m}_{0}=4.81×{10}^{-26}\\text{ }\\text{ }\\text{kg}$ —mass of one air molecule, T—absolute temperature. This means that on the altitude of 60 km and on the North Pole, where the temperatures are the same and equal to −40˚C, the speeds of sounds should have the same values.\n\nWe have found the answer to this paradox and it is that the abovementioned formula is just only for a homogeneous medium where the density depends only on the pressure $\\rho =\\rho \\left(P\\right)$. In the inhomogeneous medium, which is the earth’s atmosphere, due to the influence of the gravitational field on it, the density also depends on entropy $\\rho =\\rho \\left(P,S\\right)$. In that case the density perturbation equals to the following:\n\n${\\rho }^{\\prime }={\\left(\\frac{\\partial {\\rho }_{0}}{\\partial {P}_{0}}\\right)}_{s}{P}^{\\prime }+{\\left(\\frac{\\partial {\\rho }_{0}}{\\partial {S}_{0}}\\right)}_{p}{S}^{\\prime },$ (13)\n\nwhere ${P}_{0},{\\rho }_{0},{S}_{0}$ -represent unperturbed values of pressure, density and entropy accordingly. Considering that ${S}^{\\prime }={\\left(\\partial {S}_{0}/\\partial {P}_{0}\\right)}_{T}{P}^{\\prime }$, from (13) we easily receive :\n\n${\\rho }^{\\prime }=\\left(\\frac{1}{{C}_{s}^{2}}+\\frac{1}{{C}_{p}^{2}}\\right){P}^{\\prime }=\\frac{1}{{C}^{2}}{P}^{\\prime }.$ (14)\n\nHere:\n\n${C}_{S}^{2}=\\left(\\frac{\\partial {P}_{0}}{\\partial {\\rho }_{0}}\\right)=\\gamma \\frac{{k}_{B}T}{{m}_{0}}$ is the speed of adiabatic sound, (15)\n\n${C}_{P}^{2}={\\left[{\\left(\\frac{\\partial {\\rho }_{0}}{\\partial {S}_{0}}\\right)}_{P}{\\left(\\frac{\\partial {S}_{0}}{\\partial {P}_{0}}\\right)}_{T}\\right]}^{-1}=\\frac{{c}_{p}{\\rho }_{0}^{2}}{T{\\left(\\partial {\\rho }_{0}/\\partial T\\right)}_{s}^{2}}$ is the speed of isobaric sound, (16)\n\n${C}^{2}=\\frac{{C}_{s}^{2}{C}_{p}^{2}}{{C}_{s}^{2}+{C}_{p}^{2}}$ is the true value of the speed of sound. (17)\n\nThus, the square of the true value of the speed of sound is reduced from the squares of the velocities of adiabatic and isobaric sounds. Substituting into (16) the value ${\\rho }_{0}$ from the Laplace equation\n\n${\\rho }_{0}={\\rho }_{0}^{0}\\mathrm{exp}\\left(-{m}_{0}gz/{k}_{B}T\\right),$ (18)\n\nwhere ${\\rho }_{0}^{0}$ -is unperturbed density value at sea level, for the speed of isobaric sound we receive the following:\n\n${C}_{p}=\\sqrt{\\frac{{c}_{p}{k}_{B}^{2}{T}^{3}}{{m}_{0}^{2}{g}^{2}{z}^{2}}}.$ (19)\n\nUsing (15) and (19), from (17) we find:\n\n$C\\left(z,T\\right)=\\sqrt{\\frac{\\gamma {k}_{B}T}{{m}_{0}\\left(1+\\frac{\\gamma m{g}^{2}{z}^{2}}{{c}_{p}{k}_{B}{T}^{2}}\\right)}}.$ (20)\n\nWe can see that the true value of the speed of sound in the Earth’s atmosphere truly depends on the altitude (density) and condition $z=0$ is equivalent to the condition $g=0$. Thus, at the sea level, air is a homogeneous medium and the speed of sound should be calculated by the Equation (15).\n\nFigure 1 shows the temperature distribution over height in the Earth’s atmosphere . We can see that up to the altitude of approximately 11 km., the temperature drops according to strictly linear law $T=\\alpha +\\beta z$, where $\\alpha =288.15\\text{\\hspace{0.17em}}\\text{K}$\n\nFigure 1. Temperature as a function of geometrical altitude.\n\nand $\\beta =-6.52×{10}^{-3}\\text{K}/\\text{m}$. This part of the atmosphere is called the troposphere, which has no heat source, hence the entropy S is constant, i.e. can be used the adiabatic equation\n\n$\\frac{\\text{d}S}{\\text{d}t}=\\frac{\\partial S}{\\partial t}+\\left(\\stackrel{\\to }{V}\\nabla \\right)S=0,$ (21)\n\nOn the top boundary of the troposphere, the drop in temperature abruptly stops and remains constant till the altitude of approximately 20 km (tropopause), and then increases in the stratosphere. In this regard, the law of entropy constancy, which was used in our calculations, ceases to be true. Therefore, our theory is valid up to the height $z\\cong 11\\text{\\hspace{0.17em}}\\text{km}$.\n\nSubstituting in (15), (19) and (20) $T=288.15-6.53×{10}^{-3}z$ and taking into account that ${c}_{p}={10}^{3}\\text{J}/\\text{kg}\\cdot \\text{K}$, we get:\n\n${C}_{s}\\left(z\\right)=\\sqrt{401.66\\left(288.15-6.53×{10}^{-3}z\\right)},$ (22)\n\n${C}_{p}\\left(z\\right)=925.70\\frac{\\sqrt{{\\left(288.15-6.53×{10}^{-3}z\\right)}^{3}}}{z},$ (23)\n\n$C\\left(z\\right)=20.05\\sqrt{\\frac{{\\left(288.15-6.53×{10}^{-3}z\\right)}^{3}}{{\\left(288.15-6.53×{10}^{-3}z\\right)}^{2}+4.69×{10}^{-4}{z}^{2}}}.$ (24)\n\nIn Figure 2, the graphs of expressions (22) and (23) are shown. Calculations show that the relative inaccuracy, between values ${C}_{s}\\left(z\\right)$ and $C\\left(z\\right)$ at heights $z=1\\text{\\hspace{0.17em}}\\text{km}$ and $z=10\\text{\\hspace{0.17em}}\\text{km}$ is equal to 0.3% and 33%, respectively, i.e. increases 110 times. Obviously, such an error cannot be ignored when calculating the Mach number .\n\nIn the work , the author suggests that at the upper boundary of the troposphere ( $z\\cong \\left(10\\text{\\hspace{0.17em}}\\text{-}\\text{\\hspace{0.17em}}11\\right)\\text{km}$ which is often called the ozone layer), there is an exothermic reaction of ozone synthesis ( ${\\text{O}}_{2}+\\text{O}\\to {\\text{O}}_{3}+24$ k.cal/mol), which is the reason for such dynamics of the temperature distribution. Let us show that our theory fully confirms this hypothesis.\n\nFigure 2. Dependence of isobaric ( ${C}_{s}\\left(z\\right)$ -the blue curve) and true ( $C\\left(z\\right)$ -the green curve) speeds of sounds on the altitude in the troposphere.\n\nIn Figure 3, the graphs of the height distribution of the adiabatic ${C}_{s}\\left(z\\right)$ and isobaric ${C}_{p}\\left(z\\right)$ velocities of sounds are shown. It is seen that these graphs intersect at a height of $z\\cong 10200\\text{\\hspace{0.17em}}\\text{m}$. Equating ${C}_{s}^{2}\\left(T\\right)$ and ${C}_{p}^{2}\\left(z,T\\right)$ from Formulas (15) and (19), we obtain\n\n$\\frac{\\gamma kT}{{m}_{0}}=\\frac{{c}_{p}{k}^{2}{T}^{3}}{{m}_{0}^{2}{g}^{2}{z}^{2}}⇒z=\\sqrt{\\frac{{c}_{p}k}{\\gamma {m}_{0}}}\\frac{T}{g}.$ (25)\n\nSubstituting in (25) $T=\\alpha +\\beta z$ we find:\n\n$z=\\frac{\\sqrt{{c}_{p}k{\\alpha }^{2}/\\gamma {m}_{0}{g}^{2}}}{1+\\sqrt{{c}_{p}k{\\beta }^{2}/\\gamma {m}_{0}{g}^{2}}}=10230\\text{\\hspace{0.17em}}\\text{m}.$ (26)\n\nThis altitude almost exactly coincides with the upper boundary of the troposphere, presented in Figure 1, which proves the high reliability of our theory. Thus, it can be said with high confidence, that expressions (25) and/or (26) are equations of the upper border of the troposphere, where adiabatic and isobaric speeds of sound are equated, i.e. a resonance ${\\omega }_{s}={\\omega }_{p}$ occurs. As it is known, the abrupt change in the dynamics of the process, which is observed in Figure 1, is as a general rule in connection with resonance. Thus, it can be assumed that the resonance of frequencies of the adiabatic and isobaric sounds is a trigger mechanism for the exothermic reaction of ozone synthesis and, as a consequence, the release of a large amount of heat.\n\nThe discovery of an isobaric sound leads to another important result. Let us transform the following expression:\n\n${\\left[\\frac{{\\rho }_{0}}{{\\left(\\partial {\\rho }_{0}/\\partial T\\right)}_{p}}\\right]}^{2}={\\left[\\frac{m}{\\upsilon }{\\left(\\frac{\\partial \\left(m/\\upsilon \\right)}{\\partial T}\\right)}_{p}^{-1}\\right]}_{m=const}^{2}={\\left[\\frac{1}{\\upsilon }{\\left(\\frac{\\partial \\upsilon }{\\partial T}\\right)}_{p}\\right]}^{-2}=\\frac{1}{{\\beta }_{p}^{2}},$ (27)\n\nFigure 3. The dependences of adiabatic ( ${C}_{s}\\left(T\\right)$ -green curve) and isobaric ( ${C}_{p}\\left(z,T\\right)$ -red curve) sound velocities from the altitude in the troposphere.\n\nwhere $\\upsilon$ -is gas volume and ${\\beta }_{p}$ -is the coefficient of thermal expansion. Then from (16) we shall find that:\n\n${C}_{p}\\left(z,T\\right)=\\frac{1}{{\\beta }_{p}}{\\left(\\frac{{c}_{p}}{T}\\right)}^{1/2}.$ (28)\n\nFrom (28) it is derived, that ${\\beta }_{p}={\\beta }_{p}\\left(z,T\\right)$, i.e. the coefficient of the thermal expansion depends not only on the temperature, as is usual to modern thermodynamics, but also on the altitude of the atmosphere. This fact invalidates the universality of the laws of the ideal gas. More detailed account of this please see the work .\n\nSince the sound wave carries the density perturbation, an idea of generalization of the equation of mass continuity for an inhomogeneous medium arose. This equation was also received for the homogeneous medium and it determines the change of density as a result of substance mass change in the constant volume. However, the change in density is possibly also due to the change of volume of the constant mass of the substance, i.e.:\n\n$\\frac{\\text{d}\\rho }{\\text{d}t}=\\frac{\\text{d}}{\\text{d}t}\\left(\\frac{m}{\\upsilon }\\right)=\\frac{\\upsilon \\frac{\\text{d}m}{\\text{d}t}-m\\frac{\\text{d}\\upsilon }{\\text{d}t}}{{\\upsilon }^{2}}={\\left(\\frac{\\text{d}\\rho }{\\text{d}t}\\right)}_{\\upsilon }-{\\left(\\rho \\frac{\\text{d}}{\\text{d}t}\\mathrm{ln}\\upsilon \\right)}_{m},$ (29)\n\nEquation (29) is reduced to the following :\n\n$\\frac{\\text{d}\\rho }{\\text{d}t}=-\\rho \\nabla \\stackrel{\\to }{V}-\\frac{\\stackrel{\\to }{V}\\nabla P}{{C}_{p}^{2}}.$ (30)\n\nThe first summand in the right side of the Equation (30) determines the change of mass in the constant volume while the second summand is the isobaric change of the volume of the substance constant mass as a result of the change in temperature, which is caused by the change in entropy in the inhomogeneous medium. This work has radically altered the existing concepts of compressibility and incompressibility of medium. It is considered, that these concepts have a mechanical meaning. In reality, they have thermodynamic meaning and characterize the homogeneity or non homogeneity of the medium. The homogeneous medium is always compressible. The incompressibility is a consequence of its non homogeneity and it manifests as strongly as inhomogeneous the medium is, as it occurs in the Earth’s atmosphere with increasing altitude. The speed of sound in a homogeneous medium is adiabatic, and in an inhomogeneous medium it is a combination of the speeds of adiabatic and isobaric sounds.\n\nThe work considers the problem of internal waves from the monograph . The authors consider the water to be incompressible and contemplate, that the density perturbation is isobaric, i.e. only the second summand is considered in the Equation (13). As a result, they receive a so-called internal wave, the frequency of which depends only on the direction of the wave vector, the magnitude of which can be any. It is clear that such wave does not exist in nature and the reason for this paradox is the use of incompressibility condition towards the water. Hence, water is a compressible (homogeneous) medium. Though it may seem paradoxical, water is a much more compressible medium (in the thermodynamic sense), than the atmosphere in the higher layers. Work has been dedicated to this problem.\n\nAnother paradox that we noticed is that the Euler equation in its current form contradicts the basic principle of physics. Indeed, let us consider the Euler equation in its generally accepted form: $\\rho \\text{d}\\stackrel{\\to }{V}/\\text{d}t=\\rho \\left[\\partial \\stackrel{\\to }{V}/\\partial t+\\left(\\stackrel{\\to }{V}\\nabla \\right)\\stackrel{\\to }{V}\\right]=-\\nabla P+\\rho \\stackrel{\\to }{g}$. If ${\\stackrel{\\to }{V}}_{0}$ there is a stationary velocity of motion of the liquid and ${\\stackrel{\\to }{V}}^{\\prime }$ is its small perturbation, then after linearization its left side, which determines the acceleration of the liquid particle, will be equal to $\\left[\\partial {\\stackrel{\\to }{V}}^{\\prime }/\\partial t+\\left({\\stackrel{\\to }{V}}_{0}\\nabla \\right){\\stackrel{\\to }{V}}^{\\prime }\\right]$. Thus, the acceleration of a liquid particle depends on the stationary velocity of the medium, which contradicts the principle of relativity. This contradiction is caused by the assumption $\\text{d}\\rho /\\text{d}t=0⇒\\rho =const$, i.e. the assumption of incompressibility of the liquid $\\nabla \\stackrel{\\to }{V}=0$ and neglect of the second term in Equation (30). In fact, if the first term is equal to zero (incompressible medium), remains the second term, and if the second term is equal to zero (compressible medium), the first remains. Thus, $\\text{d}\\rho /\\text{d}t\\ne 0$ and density $\\rho$ must be entered under the derivative\n\n$\\frac{\\text{d}\\left(\\rho \\stackrel{\\to }{V}\\right)}{\\text{d}t}=\\stackrel{\\to }{V}\\frac{\\text{d}\\rho }{\\text{d}t}+\\rho \\frac{\\text{d}\\stackrel{\\to }{V}}{\\text{d}t}=-\\nabla P+\\rho \\stackrel{\\to }{g}.$ (31)\n\nLet’s substitute in the right side of Equation (31) the value $\\text{d}\\rho /\\text{d}t$ from Equation (30)\n\n$\\begin{array}{l}-\\stackrel{\\to }{V}\\rho \\nabla \\stackrel{\\to }{V}-\\frac{{V}^{2}\\nabla P}{{C}_{p}^{2}}+\\rho \\frac{\\text{d}\\stackrel{\\to }{V}}{\\text{d}t}=-\\nabla P+\\rho \\stackrel{\\to }{g}\\\\ ⇒\\rho \\left[\\frac{\\text{d}\\stackrel{\\to }{V}}{\\text{d}t}-\\left(V\\nabla \\right)\\stackrel{←}{V}\\right]-\\frac{{V}^{2}\\nabla P}{{C}_{p}^{2}}=-\\nabla P+\\rho \\stackrel{\\to }{g}\\end{array}$ (32)\n\nNeglect the term ${V}^{2}\\nabla P/{C}_{p}^{2}$, the Euler equation takes the form:\n\n$\\rho \\frac{\\partial \\stackrel{\\to }{V}}{\\partial t}=-\\nabla P+\\rho \\stackrel{\\to }{g}.$ (33)\n\nAs can be seen the nonlinear term dropped out of the Euler equation and, therefore, there is no contradiction. The Equation (30) along with the Equation (27) was used by us in the problem of the capillary waves , after which all existing contradictions were removed.\n\nIn our opinion, plasma is an incompressible medium, because it is a substantially inhomogeneous, due to the fact that each electron and ion are in a force field created by neighboring particles. It can be assumed that this is precisely the reason for the difficulties in implementing a controlled thermonuclear reaction, since it is impossible to compress the plasma to the required state and keep it in this state for a long enough period of time.\n\n2. A New Approach to the Theory of Surface Gravitational Waves\n\nLet us now apply these equations to the problem of surface gravitational waves:\n\n$\\left\\{\\begin{array}{l}\\rho \\frac{\\text{d}\\stackrel{\\to }{V}}{\\text{d}t}=\\rho \\left[\\frac{\\partial \\stackrel{\\to }{V}}{\\partial t}+\\left(\\stackrel{\\to }{V}\\nabla \\right)\\stackrel{\\to }{V}\\right]=-\\nabla P+\\rho \\stackrel{\\to }{g}\\\\ \\frac{\\text{d}\\rho }{\\text{d}t}=\\frac{\\partial \\rho }{\\partial t}+\\left(\\stackrel{\\to }{V}\\nabla \\right)\\rho =-\\rho \\nabla \\stackrel{\\to }{V}-\\frac{\\stackrel{\\to }{V}\\nabla P}{{C}_{p}^{2}}\\end{array}$ (34)\n\nSuppose that ${V}_{0}=0$ and let’s represent all the variables as the sum of their stationary values and small perturbations:\n\n$P={P}_{0}+{P}^{\\prime };\\rho ={\\rho }_{0}+{\\rho }^{\\prime };\\stackrel{\\to }{V}={\\stackrel{\\to }{V}}^{\\prime }.$\n\nUsing under the linearization of system (34) the condition of equilibrium of the medium in the gravitational field of the Earth $\\nabla {P}_{0}={\\rho }_{0}\\stackrel{\\to }{g}$, which is true for both air and water, we get:\n\n$\\left\\{\\begin{array}{l}{\\rho }_{0}\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}=-\\nabla {P}^{\\prime }+{\\rho }^{\\prime }\\stackrel{\\to }{g}\\\\ \\frac{\\partial {\\rho }^{\\prime }}{\\partial t}+\\left({\\stackrel{\\to }{V}}^{″}\\nabla \\right){\\rho }_{0}=-{\\rho }_{0}\\nabla {\\stackrel{\\to }{V}}^{\\prime }-\\frac{{\\rho }_{0}\\stackrel{\\to }{g}{\\stackrel{\\to }{V}}^{\\prime }}{{C}_{p}^{2}}\\end{array}$ (35)\n\nLet’s apply the operator $\\nabla$ to the first equation of the system (32) and the operator $\\partial /\\partial t$ to the second, after which we have\n\n$\\left\\{\\begin{array}{l}\\nabla {\\rho }_{0}\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}+{\\rho }_{0}\\nabla \\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}=-\\Delta {P}^{\\prime }+\\nabla {\\rho }^{\\prime }\\stackrel{\\to }{g}\\\\ \\frac{{\\partial }^{2}{\\rho }^{\\prime }}{\\partial {t}^{2}}+\\left(\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}\\nabla \\right){\\rho }_{0}=-{\\rho }_{0}\\nabla \\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}-\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}\\frac{\\nabla {P}_{0}}{{C}_{p}^{2}}\\end{array}$ (36)\n\nGiven that ${P}_{0}$ and ${\\rho }_{0}$ in air and in water depend only on the vertical z coordinate, the following relations are valid:\n\n$z\\nabla {\\rho }_{0}\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}=\\left(\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}\\nabla \\right){\\rho }_{0}=\\frac{\\text{d}{\\rho }_{0}}{\\text{d}z}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t};\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}\\frac{\\nabla {P}_{0}}{{C}_{p}^{2}}=-\\frac{{\\rho }_{0}g}{{C}_{p}^{2}}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t}\\nabla {\\rho }^{\\prime }\\stackrel{\\to }{g}=-g\\frac{\\partial {\\rho }^{\\prime }}{\\partial z},$\n\nafter that, the system (36) can be rewritten as\n\n$\\left\\{\\begin{array}{l}\\frac{\\text{d}{\\rho }_{0}}{\\text{d}z}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t}+{\\rho }_{0}\\nabla \\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}=-\\Delta {P}^{\\prime }-g\\frac{\\partial {\\rho }^{\\prime }}{\\partial z}\\\\ \\frac{\\text{d}{\\rho }_{0}}{\\text{d}z}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t}+{\\rho }_{0}\\nabla \\frac{\\partial {\\stackrel{\\to }{V}}^{\\prime }}{\\partial t}=-\\frac{{\\partial }^{2}{\\rho }^{\\prime }}{\\partial {t}^{2}}+\\frac{{\\rho }_{0}g}{{C}_{p}^{2}}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t}\\end{array}$ (37)\n\nEquating the right parts of the system (37) we get\n\n$\\Delta {P}^{\\prime }+g\\frac{\\partial {\\rho }^{\\prime }}{\\partial z}+\\frac{{\\rho }_{0}g}{{C}_{p}^{2}}\\frac{\\partial {{V}^{\\prime }}_{z}}{\\partial t}-\\frac{{\\partial }^{2}{\\rho }^{\\prime }}{\\partial {t}^{2}}=0.$ (38)\n\nUsing the equation of state of the medium ${\\rho }^{\\prime }=\\left(1/{C}^{2}\\right){P}^{\\prime }$ and expressing $\\partial {{V}^{\\prime }}_{z}/\\partial t$ from the first equation of the system (35) $\\partial {{V}^{\\prime }}_{z}/\\partial t=-\\left(1/{\\rho }_{0}\\right)\\left(\\partial {P}^{\\prime }/\\partial z+g{P}^{\\prime }/{C}^{2}\\right)$, Equation (38) takes the form:\n\n$\\Delta {P}^{\\prime }+g\\frac{\\partial }{\\partial z}\\left(\\frac{{P}^{\\prime }}{{C}^{2}}\\right)-\\frac{g}{{C}_{p}^{2}}\\left(\\frac{\\partial {P}^{\\prime }}{\\partial z}+g\\frac{{P}^{\\prime }}{{C}^{2}}\\right)-\\frac{1}{{C}^{2}}\\frac{{\\partial }^{2}{P}^{\\prime }}{\\partial {t}^{2}}=0.$ (39)\n\nC-is the speed of sound in the gravitational field of the Earth and is an important thermodynamic parameter of the medium, which, like other parameters, must depend on the vertical z coordinate. It is expressed in terms of the adiabatic ${C}_{s}$ and isobaric ${C}_{p}$ speeds of sounds by the following ratio :\n\n$\\frac{1}{{C}^{2}\\left(z\\right)}=\\frac{1}{{C}_{s}^{2}\\left(z\\right)}+\\frac{1}{{C}_{p}^{2}\\left(z\\right)}⇒{C}^{2}\\left(z\\right)=\\frac{{C}_{s}^{2}\\left(z\\right){C}_{p}^{2}\\left(z\\right)}{{C}_{s}^{2}\\left(z\\right)+{C}_{p}^{2}\\left(z\\right)}.$ (40)\n\nTaking into account (40), Equation (39) can be written as:\n\n$\\Delta {P}^{\\prime }+\\frac{g}{{C}_{s}^{2}\\left(z\\right)}\\frac{\\partial {P}^{\\prime }}{\\partial z}-\\frac{g}{{C}^{2}\\left(z\\right)}\\left(\\frac{1}{{C}^{2}\\left(z\\right)}\\frac{\\text{d}{C}^{2}\\left(z\\right)}{\\text{d}z}+\\frac{g}{{C}_{p}^{2}\\left(z\\right)}\\right){P}^{\\prime }-\\frac{1}{{C}^{2}\\left(z\\right)}\\frac{{\\partial }^{2}{P}^{\\prime }}{\\partial {t}^{2}}=0.$ (41)\n\nWe introduce the notation:\n\n$\\begin{array}{l}\\Gamma \\left(z\\right)=\\frac{1}{{C}^{4}\\left(z\\right)}\\frac{\\text{d}{C}^{2}\\left(z\\right)}{\\text{d}z}=\\frac{1}{{C}_{s}^{4}\\left(z\\right)}\\frac{\\text{d}{C}_{s}^{2}\\left(z\\right)}{\\text{d}z}+\\frac{1}{{C}_{p}^{4}\\left(z\\right)}\\frac{\\text{d}{C}_{p}^{2}\\left(z\\right)}{\\text{d}z};\\\\ \\Sigma \\left(z\\right)=\\frac{1}{{C}^{2}\\left(z\\right){C}_{p}^{2}\\left(z\\right)}\\end{array}$ (42)\n\nand then, Equation (39) finally takes the form:\n\n$\\Delta {P}^{\\prime }+\\frac{g}{{C}_{s}^{2}\\left(z\\right)}\\frac{\\partial {P}^{\\prime }}{\\partial z}-g\\left[\\Gamma \\left(z\\right)+g\\Sigma \\left(z\\right)\\right]{P}^{\\prime }-\\frac{1}{{C}^{2}\\left(z\\right)}\\frac{{\\partial }^{2}{P}^{\\prime }}{\\partial {t}^{2}}=0.$ (43)\n\nEquation (43) is the equation of mechanical waves in any medium in the earth’s gravitational field, or the generalized equation of gravitational waves. We will search ${P}^{\\prime }$ in the form:\n\n${P}^{\\prime }\\left(x,z,t\\right)={P}_{a}\\left(z\\right)\\mathrm{exp}\\left[i\\left(kx-\\omega t\\right)\\right]$ (44)\n\nafter then from (43) we get:\n\n$\\frac{{\\text{d}}^{2}{P}_{a}\\left(z\\right)}{\\text{d}{z}^{2}}+\\frac{g}{{C}_{s}^{2}\\left(z\\right)}\\frac{\\text{d}{P}_{a}\\left(z\\right)}{\\text{d}z}-\\left\\{{k}^{2}+g\\left[\\Gamma \\left(z\\right)+g\\Sigma \\left(z\\right)\\right]-\\frac{{\\omega }^{2}}{{C}^{2}\\left(z\\right)}\\right\\}{{P}^{\\prime }}_{a}\\left(z\\right)=0.$ (45)\n\nLet’s denote the values related to air $\\left(z>0\\right)$ by index 1, and to water $\\left(z<0\\right)$ -by index 2 and consider this equation for air, where:\n\n${C}_{s1}^{2}=\\gamma \\frac{{k}_{B}T}{{m}_{0}}$ and ${C}_{p1}^{2}=\\frac{{c}_{p}{k}_{B}^{2}{T}^{3}}{{m}_{0}^{2}{g}^{2}{z}^{2}}.$ (46)\n\nSubstituting in (46) $T=\\alpha +\\beta z$, where $\\alpha =288.15\\text{\\hspace{0.17em}}\\text{K}$ and $\\beta =-6.52×{10}^{-3}\\text{K}/\\text{m}$, we get:\n\n${C}_{s1}^{2}\\left(z\\right)=\\gamma \\frac{{k}_{B}\\left(\\alpha +\\beta z\\right)}{{m}_{0}}$ and ${C}_{p1}^{2}\\left(z\\right)=\\frac{{c}_{p}{k}_{B}^{2}{\\left(\\alpha +\\beta z\\right)}^{3}}{{m}_{0}^{2}{g}^{2}{z}^{2}}.$ (47)\n\nThus, Equation (45) is a second-order differential equation with variable coefficients. Considering that ${c}_{p}={10}^{3}\\text{J}/\\text{kg}\\cdot \\text{K}$, to simplify the problem, we will average these coefficients in the range of heights from $z={z}_{0}=10000\\text{\\hspace{0.17em}}\\text{m}$ after which the following relations are valid\n\n$\\frac{1}{{\\stackrel{¯}{C}}_{s1}^{2}}=\\frac{1}{{z}_{0}}\\underset{0}{\\overset{{Z}_{0}}{\\int }}\\frac{1}{{C}_{s1}^{2}\\left(z\\right)}\\text{d}z=9.80×{10}^{-6}\\frac{{\\mathrm{sec}}^{2}}{{\\text{m}}^{2}},$\n\n$\\frac{1}{{\\stackrel{¯}{C}}_{p1}^{2}}=\\frac{1}{{z}_{0}}\\underset{0}{\\overset{{Z}_{0}}{\\int }}\\frac{1}{{C}_{p1}^{2}\\left(z\\right)}\\text{d}z=2.90×{10}^{-6}\\frac{{\\mathrm{sec}}^{2}}{{\\text{m}}^{\\text{2}}},$\n\n$\\frac{1}{{\\stackrel{¯}{C}}_{1}^{2}}=\\frac{1}{{z}_{0}}\\underset{0}{\\overset{{Z}_{0}}{\\int }}\\frac{1}{{C}_{1}^{2}\\left(z\\right)}\\text{d}z=12.70×{10}^{-6}\\frac{{\\mathrm{sec}}^{2}}{{\\text{m}}^{2}},$ (48)\n\n${\\stackrel{¯}{\\Gamma }}_{1}=\\frac{1}{{z}_{0}}\\underset{0}{\\overset{{Z}_{0}}{\\int }}\\Gamma \\left(z\\right)\\text{d}z=-1.30×{10}^{-9}\\frac{{\\mathrm{sec}}^{2}}{{\\text{m}}^{3}},$\n\n${\\stackrel{¯}{\\Sigma }}_{1}=\\frac{1}{{z}_{0}}\\underset{0}{\\overset{{Z}_{0}}{\\int }}{\\Sigma }_{1}\\left(z\\right)\\text{d}z=4.75×{10}^{-11}\\frac{{\\mathrm{sec}}^{4}}{{\\text{m}}^{4}}.$\n\nDenoting $g\\left({\\stackrel{¯}{\\Gamma }}_{1}+g{\\stackrel{¯}{\\Sigma }}_{1}\\right)={\\stackrel{¯}{\\Omega }}_{1}=-8.19×{10}^{-9}\\text{ }\\text{ }{\\text{m}}^{-2}$, Equation (45) for air takes the form:\n\n$\\frac{{\\text{d}}^{2}{P}_{a1}\\left(z\\right)}{\\text{d}{z}^{2}}+\\frac{g}{{\\stackrel{¯}{C}}_{s1}^{2}}\\frac{\\text{d}{P}_{a1}\\left(z\\right)}{\\text{d}z}-\\left({k}^{2}+{\\stackrel{¯}{\\Omega }}_{1}-\\frac{{\\omega }^{2}}{{\\stackrel{¯}{C}}_{1}^{2}}\\right){P}_{a1}\\left(z\\right)=0.$ (49)\n\nWe will seek a solution to Equation (49) in the form\n\n${P}_{a1}\\left(z\\right)=A\\mathrm{exp}\\left(\\gamma z\\right),$ (50)\n\nwhich gives\n\n${P}_{a1}\\left(z\\right)={A}_{1}\\mathrm{exp}\\left({\\gamma }_{1}z\\right)+{A}_{2}\\mathrm{exp}\\left({\\gamma }_{2}z\\right),$ (51)\n\nwhere:\n\n${\\gamma }_{1}=-\\frac{k}{{\\theta }_{s1}}\\left[1+\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)}\\right],$ (52)\n\n${\\gamma }_{2}=-\\frac{k}{{\\theta }_{s1}}\\left[1-\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1`}}{{k}^{2}}-{x}^{2}\\right)}\\right].$ (53)\n\nHere $x={U}_{p}/{\\stackrel{¯}{C}}_{1}$ and ${U}_{p}=\\omega /k$ -is the phase velocity of the wave. ${\\theta }_{s1}$ -dimensionless quantity that is equal to\n\n${\\theta }_{s1}=\\frac{2k{\\stackrel{¯}{C}}_{s1}^{2}}{g}.$ (54)\n\nIt is easy to see that if the condition is met\n\n$1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)>1,$ (55)\n\nwe have: ${\\gamma }_{1}<0$, ${\\gamma }_{2}>0$ and then, based on the wave surface condition ( ${P}_{a1}\\left(z\\right)\\to 0$ when $z\\to \\infty$ ), in (51) must be put ${A}_{2}=0$. On condition\n\n$0\\le 1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)<1,$ (56)\n\nwe have: ${\\gamma }_{1}<0$, ${\\gamma }_{2}<0$ a and then in the right part (51), both terms should be taken into account. Substituting the value ${\\stackrel{¯}{\\Omega }}_{1}=-8.19×{10}^{-9}\\text{ }\\text{ }{\\text{m}}^{-2}$, the solutions of inequalities (56) and (56) are:\n\n$k>9.05×{10}^{-5}/\\sqrt{1-{x}^{2}}{\\text{m}}^{-1}⇒\\lambda <6.90×{10}^{4}\\sqrt{1-{x}^{2}}\\text{ }\\text{ }\\text{m},$ (57)\n\n$k<9.05×{10}^{-5}/\\sqrt{1-{x}^{2}}{\\text{m}}^{-1}⇒\\lambda >6.90×{10}^{4}\\sqrt{1-{x}^{2}}\\text{ }\\text{ }\\text{m}.$ (58)\n\nlet’s call the waves satisfying condition (57) wind waves, and condition (58) tsunami waves.\n\nFor water $\\left(z<0\\right)$ dependencies ${C}_{s2}$ and ${C}_{p2}$ on z are unknown, but we can assume with a high probability that they are very weak. As for their numerical values, they can be determined from the experimental data. Substituting in (28) the values of the coefficient of thermal expansion and specific heat capacity at constant pressure for water ${\\beta }_{p}=1.50×{10}^{-4}\\text{ }\\text{ }{\\text{K}}^{-1}$, ${c}_{p}=4.19×{10}^{3}\\text{J}/\\text{kg}\\cdot \\text{K}$ at a temperature of $T=288\\text{\\hspace{0.17em}}\\text{K}$ we obtain ${C}_{p2}=25210\\text{\\hspace{0.17em}}\\text{m}/\\text{sec}$. On the other hand, the speed of sound in water, measured experimentally with great accuracy ${C}_{2}=1480\\text{\\hspace{0.17em}}\\text{m}/\\text{sec}$, and then, from formula (17), we have ${C}_{s2}={C}_{2}{C}_{p2}/\\sqrt{{C}_{p2}^{2}-{C}_{2}^{2}}=1482.60\\text{\\hspace{0.17em}}\\text{m}/\\text{sec}$. As we can see, the speed of sound in water is practically equal to the adiabatic speed of sound, i.e. ${C}_{2}={C}_{s2}$. This result irrefutably proves that the mechanism of sound propagation in water is adiabatic $\\left({C}_{p2}=\\infty \\right)$ and on the right-hand side of Equation (30) only the first term remains. Thus, Equation (45) for water has the form:\n\n$\\frac{{\\text{d}}^{2}{P}_{a2}\\left(z\\right)}{\\text{d}{z}^{2}}+\\frac{g}{{C}_{2}^{2}}\\frac{\\text{d}{P}_{a2}\\left(z\\right)}{\\text{d}z}-\\left({k}^{2}-\\frac{{\\omega }^{2}}{{C}_{2}^{2}}\\right){P}_{a2}\\left(z\\right)=0$ (59)\n\nRepresenting the amplitude of the pressure perturbation in the form ${P}_{a2}\\left(z\\right)=B\\mathrm{exp}\\left(\\delta z\\right)$ from (59), we obtain\n\n${P}_{a2}\\left(z\\right)={B}_{1}\\mathrm{exp}\\left({\\delta }_{1}z\\right)+{B}_{2}\\mathrm{exp}\\left({\\delta }_{2}z\\right)$, (60)\n\nwhere:\n\n${\\delta }_{1}=-\\frac{k}{{\\theta }_{s2}}\\left[1+\\sqrt{1+{\\theta }_{s2}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{2}}{{k}^{2}}-\\frac{{U}_{p2}^{2}}{{C}_{2}^{2}}\\right)}\\right]<0,$ (61)\n\n${\\delta }_{2}=-\\frac{k}{{\\theta }_{s2}}\\left[1-\\sqrt{1+{\\theta }_{s2}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{2}}{{k}^{2}}-\\frac{{U}_{p}^{2}}{{C}_{2}^{2}}\\right)}\\right]>0,$ (62)\n\n${\\theta }_{s2}=\\frac{2k{C}_{s2}^{2}}{g}=\\frac{2k{C}_{2}^{2}}{g}.$ (63)\n\nObviously, since the region of water is bounded along the vertical coordinate, both terms in (60) must be preserved for it.\n\n3. Waves of Wind\n\nLet us first consider the wind waves, when the amplitudes of pressure perturbations in air and water are determined by a system of expressions:\n\n$\\left\\{\\begin{array}{l}{P}_{a1}\\left(z\\right)=A\\mathrm{exp}\\left({\\gamma }_{1}z\\right)\\\\ {P}_{a2}\\left(z\\right)={B}_{1}\\mathrm{exp}\\left({\\delta }_{1}z\\right)+{B}_{2}\\mathrm{exp}\\left({\\delta }_{2}z\\right)\\end{array}$ (64)\n\nNow it is necessary to determine the boundary condition connecting the perturbed pressures of two media on the water surface. Obviously, this condition must have the form\n\n${{P}_{2}\|}_{z=0}={{P}_{1}\|}_{z=0}+{\\rho }_{02}g\\varsigma \\left(x,t\\right),$ (65)\n\nwhere\n\n$\\xi \\left(x,t\\right)=a\\mathrm{exp}\\left[i\\left(kx-\\omega t\\right)\\right]$ (66)\n\nis a displacement of the free surface of the liquid along the Z axis, a is the amplitude of the surface wave. Conditions for the continuity of the z components of perturbed air and water velocities at the air-water interface will give:\n\n${{V}_{z1}\|}_{z=0}={{V}_{z2}\|}_{z=0}=\\frac{\\partial \\xi }{\\partial t}$ (67)\n\nand at the bottom of the water $\\left(z=-H\\right)$ we will have:\n\n${{V}_{z2}\|}_{z=-H}=0.$ (68)\n\nBy representing ${V}_{z}\\left(x,z,t\\right)$ from the first equation of the system (35) in the form ${V}_{z}\\left(x,z,t\\right)={V}_{a}\\left(z\\right)\\mathrm{exp}\\left[i\\left(kx-\\omega t\\right)\\right]$, we obtain:\n\n${V}_{z}\\left(x,z,t\\right)=-\\frac{i}{{\\rho }_{0}\\omega }\\left[\\frac{\\text{d}{P}_{a}\\left(z\\right)}{\\text{d}z}+\\frac{g}{{C}^{2}}{P}_{a}\\left(z\\right)\\right]\\mathrm{exp}\\left[i\\left(kx-\\omega t\\right)\\right].$ (69)\n\nTaking into account expressions (64) and (69), the boundary conditions (47), (49) and (50) will have the form\n\n$\\left\\{\\begin{array}{l}{A}_{1}-{B}_{1}-{B}_{2}+{\\rho }_{02}ga=0\\hfill \\\\ \\frac{1}{{\\rho }_{01}\\omega }\\left({\\gamma }_{1}+\\frac{g}{{\\stackrel{¯}{C}}_{1}^{2}}\\right)A-\\omega a=0\\hfill \\\\ \\frac{1}{{\\rho }_{02}\\omega }\\left({\\delta }_{1}+\\frac{g}{{C}_{2}^{2}}\\right){B}_{1}+\\frac{1}{{\\rho }_{02}\\omega }\\left({\\delta }_{2}+\\frac{g}{{C}_{2}^{2}}\\right){B}_{2}-\\omega a=0\\hfill \\\\ \\left({\\delta }_{1}+\\frac{g}{{C}_{2}^{2}}\\right)\\mathrm{exp}\\left(-{\\delta }_{1}H\\right){B}_{1}+\\left({\\delta }_{2}+\\frac{g}{{C}_{2}^{2}}\\right)\\mathrm{exp}\\left(-{\\delta }_{2}H\\right){B}_{2}=0\\hfill \\end{array}$ (70)\n\nBy equating the determinant of the system (70) to zero, we obtain the dispersion equation for surface gravitational waves in the form\n\n${\\stackrel{˜}{\\delta }}_{1}\\mathrm{exp}\\left(-{\\delta }_{1}H\\right)\\left[\\frac{{\\stackrel{˜}{\\gamma }}_{1}{\\stackrel{˜}{\\delta }}_{2}g}{{\\rho }_{01}{\\omega }^{2}}+\\frac{{\\stackrel{˜}{\\delta }}_{2}}{{\\rho }_{02}}-\\frac{{\\stackrel{˜}{\\gamma }}_{1}}{{\\rho }_{01}}\\right]-{\\stackrel{˜}{\\delta }}_{2}\\mathrm{exp}\\left(-{\\delta }_{2}H\\right)\\left[\\frac{{\\stackrel{˜}{\\gamma }}_{1}{\\stackrel{˜}{\\delta }}_{1}g}{{\\rho }_{01}{\\omega }^{2}}+\\frac{{\\stackrel{˜}{\\delta }}_{1}}{{\\rho }_{02}}-\\frac{{\\stackrel{˜}{\\gamma }}_{1}}{{\\rho }_{01}}\\right]=0,$ (71)\n\nwhere:\n\n${\\stackrel{˜}{\\gamma }}_{1}={\\gamma }_{1}+\\frac{g}{{\\stackrel{¯}{C}}_{1}^{2}}=\\frac{2g}{{\\stackrel{¯}{C}}_{p1}^{2}}+\\frac{k}{{\\theta }_{s1}}\\left[1-\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)}\\right],$ (72)\n\n${\\stackrel{˜}{\\delta }}_{1}={\\delta }_{1}+\\frac{g}{{\\stackrel{¯}{C}}_{2}^{2}}=\\frac{k}{{\\theta }_{s2}}\\left[1-\\sqrt{1+{\\theta }_{s2}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{2}}{{k}^{2}}-\\frac{{\\stackrel{¯}{C}}_{1}^{2}}{{C}_{2}^{2}}{x}^{2}\\right)}\\right],$ (73)\n\n${\\stackrel{˜}{\\delta }}_{2}={\\delta }_{2}+\\frac{g}{{\\stackrel{¯}{C}}_{2}^{2}}=\\frac{k}{{\\theta }_{s2}}\\left[1+\\sqrt{1+{\\theta }_{s2}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{2}}{k}-\\frac{{\\stackrel{¯}{C}}_{1}^{2}}{{C}_{2}^{2}}{x}^{2}\\right)}\\right].$ (74)\n\nConsidering that $x\\le 1$, in formula (72) ${x}^{2}$ must be saved. As for the quantity ${U}_{p}^{2}/{C}_{2}^{2}=\\left({\\stackrel{¯}{C}}_{1}^{2}/{C}_{2}^{2}\\right){x}^{2}\\cong 0.03{x}^{2}$ it is of the second or greater order of smallness and in the linear approximation should be discarded in formulas (73) and (74). For the minimum value ${k}_{\\mathrm{min}}=9.05×{10}^{-5}\\text{ }\\text{ }{\\text{m}}^{-1}$ from (63) we have ${\\left({\\theta }_{s2}\\right)}_{\\mathrm{min}}\\cong 40$. Thus, we can neglect the unit in comparison with ${\\theta }_{s2}$ and from (61), (62), (73) and (74) we have: ${\\delta }_{1}={\\stackrel{˜}{\\delta }}_{1}=-k$, ${\\delta }_{2}={\\stackrel{˜}{\\delta }}_{2}=k$. Then from (71) we easily obtain:\n\n$\\begin{array}{l}\\left\\{\\frac{2}{{\\theta }_{p1}}+\\frac{1}{{\\vartheta }_{s1}}\\left[1-\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-x\\right)}\\right]\\right\\}×\\left[\\frac{kg}{{\\omega }^{2}}\\mathrm{tanh}\\left(kH\\right)-1\\right]\\\\ \\text{ }-k\\frac{{\\rho }_{01}}{{\\rho }_{02}}\\mathrm{tanh}\\left(kH\\right)=0,\\end{array}$ (75)\n\nwhere ${\\theta }_{p1}=2k{\\stackrel{¯}{C}}_{p1}^{2}/g$. Equation (75) is a dispersion equation for wind waves and, as we see, it contains the thermodynamic parameters of both water and air. The last summand in (75) can be neglected due to its smallness, and then, it splits into two equations:\n\n$\\frac{2}{{\\theta }_{p1}}+\\frac{1}{{\\theta }_{s1}}\\left[1-\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)}\\right]=0,$ (76)\n\n$\\frac{kg}{{\\omega }^{2}}\\mathrm{tanh}\\left(kH\\right)-1=0.$ (77)\n\nEquation (76) describes longitudinal waves in the air excited by perturbations on the surface of water and propagating along this surface. Let’s consider this equation, ignoring the dependence of the sounds velocities on the coordinate z, i.e. take their values at sea level: ${\\stackrel{¯}{C}}_{1}={C}_{1}\\left(0\\right)={C}_{s1}\\left(0\\right)=340\\text{\\hspace{0.17em}}\\text{m}/\\text{sec}$, ${C}_{p1}=\\infty$. Then ${\\stackrel{¯}{\\Omega }}_{1}\\left(z\\right)=0$ and it will take the for\n\n$1-\\sqrt{1+{\\theta }_{s1}^{2}\\left(1-{x}^{2}\\right)}=0⇒\|x\|=1⇒\|{U}_{p}\|={C}_{1}\\left(0\\right).$ (78)\n\nObviously, in this case, the condition (57) is optional and the wave vector k can take any value. We see that in this assumption, the speed of the wave in the air does not depend on the wavelength and is equal to the speed of sound at sea level. There is no doubt that perturbations whose wave lengths are comparable to the lengths of surface gravitational waves cannot propagate in the air at the speed of sound, and thus it is impossible to ignore the dependence of the speed of sound on the z coordinate.\n\nThe solution to Equation (76) is:\n\n$\|x\|=\\sqrt{1+\\frac{{\\stackrel{\\to }{\\Omega }}_{1}}{{k}^{2}}-\\frac{4}{{\\theta }_{p1}{\\theta }_{s1}}\\left(1+\\frac{{\\theta }_{s1}}{{\\theta }_{p1}}\\right)}=\\sqrt{1+\\frac{1}{{k}^{2}}\\left[{\\stackrel{¯}{\\Omega }}_{1}-\\frac{{g}^{2}}{{\\stackrel{¯}{C}}_{p1}^{2}{\\stackrel{¯}{C}}_{s1}^{2}}\\left(1+\\frac{{\\stackrel{¯}{C}}_{s1}^{2}}{{\\stackrel{¯}{C}}_{p1}^{2}}\\right)\\right]}.$ (79)\n\nSubstituting in (79) the numerical values of the parameters from (48), we get\n\n$\|x\|=\\sqrt{1-\\frac{1.17×{10}^{-8}}{{k}^{2}}}⇒\|{U}_{p}\|=\\sqrt{1-\\frac{1.17×{10}^{-8}}{{k}^{2}}}{\\stackrel{¯}{C}}_{1}.$ (80)\n\nFrom (80) we find\n\n$k=\\frac{1.08×{10}^{-4}}{\\sqrt{1-{x}^{2}}}$ (81)\n\nwhich is consistent with the condition (57) and therefore the roots (79) are not extraneous for any valid valuesk. Table 1 shows the roots of Equation (62) for those values k $\\left(\\lambda \\right)$, that satisfy condition (57)\n\nWe see that for $k>{10}^{-3}\\text{ }\\text{ }{\\text{m}}^{-1}⇒\\lambda <6.28×{10}^{3}\\text{ }\\text{ }\\text{m}$, the phase velocity of the wave in the air is constant and equal to ${U}_{p}={\\stackrel{¯}{C}}_{1}\\cong 281.72\\text{\\hspace{0.17em}}\\text{m}/\\text{sec}$, and then, with a decrease of k, it falls and at $k=1.08×{10}^{-4}\\text{ }\\text{ }{\\text{m}}^{-1}⇒\\lambda =5.80×{10}^{4}\\text{ }\\text{ }\\text{m}$, we have $x=0$, i.e., the wave stops. This means that at this wavelength, two regions of about 30 km long, with high and low pressures are formed in the atmosphere above water. In the area of low atmospheric pressure, the amplitude of the surface wave will increase and the pressure difference between the two areas will lead to the appearance of wind. When is further reduced k to its minimum value, which is defined from (57) ${k}_{\\mathrm{min}}\\cong 9×{10}^{-5}\\text{ }\\text{ }{\\text{m}}^{-1}⇒{\\lambda }_{\\mathrm{max}}\\cong 7×{10}^{4}\\text{ }\\text{ }\\text{m}$, the roots of Equation (76) and the corresponding frequencies of standing waves in the atmosphere become imaginary, which leads to a sharp increase in the pressure difference and, consequently, the amplitude of the surface wave and the wind force. This result clearly explains the reason for the drop in atmospheric pressure over the sea and ocean before the storm, as well as the reason for its strengthening.\n\nTable 1. The dependence of the roots of Equation (76) on $k\\ge 9.05×{10}^{-5}\\text{ }\\text{ }{\\text{m}}^{-1}⇒\\lambda \\le 6.94×{10}^{4}\\text{ }\\text{ }\\text{m}$.\n\nThe solution to Equation (77), which describes surface gravitational waves on water, is:\n\n$\|{U}_{p}\|=\\sqrt{\\frac{g}{k}\\mathrm{tanh}\\left(kH\\right)}=\\sqrt{\\frac{g\\lambda }{2\\pi }\\mathrm{tanh}\\left(\\frac{2\\pi H}{\\lambda }\\right)}$ (82)\n\nwhich coincides with (10) and thus, the existing theory for wind waves gives the correct result.\n\nThe phase velocity of a wave in the atmosphere does not depend on the depth of the ocean and decreases from the speed of sound to zero with increasing wavelength. In the ocean, on the contrary, the phase velocity increases from zero with increasing wavelength and depth. It is evident, that at certain wavelengths, which will depend on the depth of the ocean, these speeds will coincide and resonance of the frequencies will occur of waves in the atmosphere and the ocean. We can assume that this resonance is the cause of the “killer wave”, especially since there is no other explanation yet. It should be noted, that the resonance alone is not enough for the appearance of a “killer wave”—it is essential that the oscillations in the air and in the water are occurring in antiphase.\n\nThe resonant wavelengths are obtained by equating the right sides of the Equations (80) and (82), i.e.\n\n$\\sqrt{1-\\frac{1.17×{10}^{-8}{\\lambda }^{2}}{4{\\pi }^{2}}}{\\stackrel{¯}{C}}_{1}=\\sqrt{\\frac{g\\lambda }{2\\pi }\\mathrm{tanh}\\left(\\frac{2\\pi H}{\\lambda }\\right)}$ (83)\n\nFor deep water $\\left(2\\pi H/\\lambda >1⇒\\mathrm{tanh}\\left(2\\pi H/\\lambda \\right)=1\\right)$, the Equation (83) obtains\n\n$1-\\frac{1.17×{10}^{-8}}{4\\pi {}^{2}}{\\lambda }^{2}=\\frac{g}{2\\pi {\\stackrel{¯}{C}}_{1}^{2}}\\lambda$ (84)\n\nThe Equation (84) does not depend on H and its solution is $\\lambda \\cong 6.4×{10}^{4}\\text{ }\\text{ }\\text{m}$ so therefore $H>{10}^{4}\\text{ }\\text{ }\\text{m}$. Since there are practically no such depths with a flat relief, it can be said with great confidence that “killer waves” do not arise in deep water.\n\nFor the shallow water $\\left(2\\pi H/\\lambda <1⇒\\mathrm{tanh}\\left(2\\pi H/\\lambda \\right)=2\\pi H/\\lambda \\right)$ we have\n\n$\\lambda =2\\pi ×{10}^{4}\\sqrt{\\frac{1}{1.17}\\left(1-\\frac{gH}{{\\stackrel{¯}{C}}_{1}^{2}}\\right)}$ (85)\n\nIn order for the values $\\lambda$ from (85) to satisfy the shallow water condition, the following condition must be fulfilled\n\n$\\left(1-\\frac{gH}{{\\stackrel{¯}{C}}_{1}^{2}}\\right)>1.17×{10}^{-8}{H}^{2}$ (86)\n\nwhich obtains the following $H<5.3×{10}^{3}\\text{ }\\text{ }\\text{m}$. Therefore, the “killer waves” arise in shallow water, the depth of which does not exceed 5.3 km.\n\nThe values of the killer wavelength for different depths and the corresponding values of the phase velocities, calculated by the formulas (85) and ${U}_{p}=\\sqrt{gH}$ as well as the values of the periods and the ratio $2\\pi H/\\lambda$ are given in Table 2 below.\n\nTable 2. Values of lengths, phase velocities and periods of the “killer wave” for various depths.\n\nFigure 4 presents the graphs of the dependences of the phase velocities of longitudinal waves in the air (80) and on the water surface (82) for the same depths. We can see that the values of the resonant wavelengths coincide with the tabulated values.\n\nFigure 4. Plots of dependences of the phase velocities of longitudinal waves on the air (80) (blue curve) and on the water surface (82) (green curves) for different depths.\n\n4. Tsunami Waves\n\nLet us now consider tsunami waves, i.e. surface gravitational waves for those values k that satisfy condition (58): $k<9.05×{10}^{-5}\\text{ }\\text{ }{\\text{m}}^{-1}⇒\\lambda >6.90×{10}^{4}\\text{ }\\text{ }\\text{m}$. In this case, the amplitude of longitudinal waves in the atmosphere is determined by the expression (51), where ${A}_{1}\\cong {A}_{2}<1$ and according to (52), (53) and (56), ${\\gamma }_{1}<0$ and ${\\gamma }_{2}<0$, i.e. both terms give the wave attenuation at $z\\to \\infty$. It can be seen that the wave corresponding to the first term decays faster than the wave corresponding to the second term, and therefore, to fulfill the condition of the wave’s superficiality, it is sufficient to leave only the second term and thus we have:\n\n$\\left\\{\\begin{array}{l}{P}_{a1}\\left(z\\right)=A\\mathrm{exp}\\left({\\gamma }_{2}z\\right)\\\\ {P}_{a2}\\left(z\\right)={B}_{1}\\mathrm{exp}\\left({\\delta }_{1}z\\right)+{B}_{2}\\mathrm{exp}\\left({\\delta }_{2}z\\right)\\end{array}$ (87)\n\nNote that boundary conditions (65), (67), and (68) are also valid for tsunami waves, and then, comparing (61) and (83), it is easy to verify that for tsunami waves we obtain an equation similar to Equation (71)\n\n${\\stackrel{˜}{\\delta }}_{1}\\mathrm{exp}\\left(-{\\delta }_{1}H\\right)\\left[\\frac{{\\stackrel{˜}{\\gamma }}_{2}{\\stackrel{˜}{\\delta }}_{2}g}{{\\rho }_{01}{\\omega }^{2}}+\\frac{{\\stackrel{˜}{\\delta }}_{2}}{{\\rho }_{02}}-\\frac{{\\stackrel{˜}{\\gamma }}_{2}}{{\\rho }_{01}}\\right]-{\\stackrel{˜}{\\delta }}_{2}\\mathrm{exp}\\left(-{\\delta }_{2}H\\right)\\left[\\frac{{\\stackrel{˜}{\\gamma }}_{2}{\\stackrel{˜}{\\delta }}_{1}g}{{\\rho }_{01}{\\omega }^{2}}+\\frac{{\\stackrel{˜}{\\delta }}_{1}}{{\\rho }_{02}}-\\frac{{\\stackrel{˜}{\\gamma }}_{2}}{{\\rho }_{01}}\\right]=0.$ (88)\n\nFor waves, the length of which is $\\lambda \\ge 500\\text{\\hspace{0.17em}}\\text{km}$, the value ${\\theta }_{s2}$ changes in the interval ${\\theta }_{s2}\\le 5.5$ and, therefore, it cannot be neglected in comparison with the unit, in contrast to ${\\theta }_{s2}^{2}$. Then we will have:\n\n$\\left\\{\\begin{array}{l}{\\stackrel{˜}{\\gamma }}_{2}=\\frac{2}{{\\theta }_{p1}}+\\frac{1}{{\\theta }_{s1}}\\left[1+\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)}\\right]\\\\ {\\delta }_{1}=-{\\stackrel{˜}{\\delta }}_{2}=-k\\left(1+\\frac{1}{{\\theta }_{s2}}\\right),{\\delta }_{2}=-{\\stackrel{˜}{\\delta }}_{1}=k\\left(1-\\frac{1}{{\\theta }_{s2}}\\right)\\end{array}$ (89)\n\nafter which, Equation (88) takes the form\n\n$\\begin{array}{l}\\left\\{\\frac{2}{{\\theta }_{p1}}+\\frac{1}{{\\vartheta }_{s1}}\\left[1+\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-x\\right)}\\right]\\right\\}\\\\ \\text{ }×\\left[\\left(\\frac{kg}{{\\omega }^{2}}+\\frac{1}{{\\theta }_{s2}}\\right)\\mathrm{tanh}\\left(kH\\right)-1\\right]-k\\frac{{\\rho }_{01}}{{\\rho }_{02}}\\mathrm{tanh}\\left(kH\\right)=0\\end{array}$ (90)\n\nIgnoring the last term, Equation (90) again splits into two equations:\n\n$\\frac{2}{{\\theta }_{p1}}+\\frac{1}{{\\theta }_{s1}}\\left[1+\\sqrt{1+{\\theta }_{s1}^{2}\\left(1+\\frac{{\\stackrel{¯}{\\Omega }}_{1}}{{k}^{2}}-{x}^{2}\\right)}\\right]=0,$ (91)\n\n$\\left(\\frac{kg}{{\\omega }^{2}}+\\frac{1}{{\\theta }_{s2}}\\right)\\mathrm{tanh}\\left(kH\\right)-1=0.$ (92)\n\nFrom condition (56) it follows that Equation (91) has no solution and this means that during a tsunami, wave processes in the atmosphere are not generated. As for Equation (92), its solution is\n\n${U}_{p}=\\sqrt{\\frac{\\left(g/k\\right)\\mathrm{tanh}\\left(kH\\right)}{1-\\mathrm{tanh}\\left(kH\\right)/{\\theta }_{s2}}}.$ (93)\n\nConsidering that $kH<1⇒\\mathrm{tanh}\\left(kH\\right)\\cong kH$, we will have:\n\n${U}_{p}=\\sqrt{\\frac{gH}{1-\\frac{gH}{2{C}_{2}^{2}}}}.$ (94)\n\nFor great depths, for example, at $H={10}^{4}\\text{ }\\text{ }\\text{m}$, the value $gH/2{C}_{2}^{2}=0.02$ and it can be ignored. Thus, the phase velocity of the tsunami wave is ${U}_{p}=\\sqrt{gH}$.\n\n5. Conclusions\n\nThis article does not claim to be highly accurate or to be the ultimate truth. The upper boundary of the troposphere changes depending on geographic parameters, and this concludes, that the average values of the problem parameters calculated here and, therefore, all the numerical data given in the article are rather conditional. However, undoubtedly the proposed method for solving the problem is new and makes it possible to trace the correlation between the ocean and the atmosphere during wave processes. In particular, it became clear why the atmospheric pressure in the ocean drops before the storm, as well as differentiating between the wavelengths of wind and tsunami became possible.\n\nIts apparent advantage is also that at the level of a highly plausible hypothesis, it reveals the greatest mystery of nature called the “Killer Wave”. Now it is clear why this wave is solitary. This is due to the fact that the flat topography of the ocean floor is disturbed at the distances of the order of the wavelengths that we calculated. It is also clear why a cavity, is formed before the wave. This is attributed to the fact that there is a region in front of the wave, where the pressure in the water sharply drops and in the atmosphere sharply increases.\n\nAll this became possible after the discovery of the isobaric speed of sound and the dependence of the true value of the speed of sound in the atmosphere on the vertical coordinate. This led to a radical change in many established dogmas and ideas in aero and hydrodynamics, which are recognized by the international scientific community (work is posted on several sites on the Internet (see, for example, ) and work is posted in the NASA database ). Despite this, the Internet to this day gives the values of the speed of sound at different heights of the atmosphere, calculated by the formula (15). Obviously, this can be explained by the fact that our theoretical results have not been experimentally confirmed. We hope that this article will be able to help popularize this problem and in the relevant scientific community will show desire in conducting the necessary experiments. The importance of such experiments also lies in the fact that if it is possible in laboratories to create conditions corresponding to the upper boundary of the troposphere and to confirm the fact of heat release, we will get an alternative source of energy.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest regarding the publication of this paper.\n\n Kirtskhalia, V.G. (2016) Journal of Fluids, 2016, Article ID: 4519201. https://doi.org/10.1155/2016/4519201 Landau, L.D. and Lifchitz, E.N. (1988) Theoretical Physics, Hydrodynamics. Vol. 6, Nauka, Moscow. Kowalik, Z. (2012) Introduction to Numerical Modeling of Tsunami Waves. Institute of Marine Science University of Alaska, Fairbank. https://www.sfos.uaf.edu/directory/faculty/kowalik/Tsunami_Book Stoker, J.J. (1957) Water Waves. Interscience, New York. Whitham, G. (1974) Linear and Nonlinear Waves. John Wiley & Sons (Wiley-Interscience), New York. Gossard, E.E. and Hooke, W.H. (1975) Waves in the Atmosphere. Wang, G.S.K. (1986) The Journal of the Acoustical Society of America, 79, 1359-1366. Kirtskhalia, V.G. (2012) Open Journal of Acoustics, 2, 80-85. https://doi.org/10.4236/oja.2012.22009 Kirtskhalia, V. (2012) Open Journal of Acoustics, 2, 115-120. https://doi.org/10.4236/oja.2012.23013 National Aeronautics and Space Administration (1976) U.S. Standard Atmosphere. Kirtskhalia, V.G. (2021) IOP Conference Series: Materials Science and Engineering, 1024, Article ID: 012037. https://doi.org/10.1088/1757-899X/1024/1/012037 Sorokhtin, O.G. (2009) The Process of Absorption of Ultra-Violet Radiation of the Sun Terrestrial Atmosphere. Bulletin of the Russian Academy of Natural Sciences, 2009/3. Kirtskhalia, V.G. (2015) Journal of Modern Physics, 7, 948-954. https://doi.org/10.4236/jmp.2015.67099 Kirtskhalia, V.G. (2013) Journal of Modern Physics, 4, 1075-1079. https://doi.org/10.4236/jmp.2013.48144 Kirtskhalia, V.G. (2019) Journal of Modern Physics, 10, 452-458. https://doi.org/10.4236/jmp.2019.104030 Kirtskhalia, V.G. 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https://blog.ednalyn.com/tag/python/	[ "## Programming Environment Setup\n\nI was bored over the weekend so I decided to restore my Macbook Pro to factory settings so that I can set up my programming environment the proper way. After all, what’s a data scientist without her toys?\n\nLet’s start with a replacement to the default terminal and pyenv installation to manage different Python versions.\n\nLet’s move on to managing different Python interpreters and virtual environments using pyenv-virtualenv.\n\n## Into the Heart of Darkness - Pt. 2\n\nExploring the Trump Twitter Archive with spaCy.\n\nIn a previous post, we set out to explore the dataset provided by the Trump Twitter Archive. My initial goal was to do something fun by using a very interesting dataset. However, it didn’t quite turn out that way.\n\nOn this post, we’ll continue our journey but this time we’ll be using spaCy.\n\nFor this project, we’ll be using pandas for data manipulation, spaCy for natural language processing, and joblib to speed things up.\n\nLet’s get started by firing up a Jupyter notebook!\n\n### Housekeeping\n\nLet’s import pandas and also set the display options so Jupyter won’t truncate our columns and rows. Let’s also set a random seed for reproducibility.\n\n`# for manipulating dataimport pandas as pd`\n`# setting the random seed for reproducibilityimport randomrandom.seed(493)`\n`# to print out all the outputsfrom IPython.core.interactiveshell import InteractiveShellInteractiveShell.ast_node_interactivity = \"all\"`\n`# set display optionspd.set_option('display.max_columns', None)pd.set_option('display.max_rows', None)pd.set_option('display.max_colwidth', -1)`\n\n### Getting the Data\n\nLet’s read the data into a dataframe. If you want to follow along, you can download the cleaned dataset here along with the file for stop words¹. This dataset contains Trump’s tweets from the moment he took office on January 20, 2017 to May 30, 2020.\n\n`df = pd.read_csv('trump_20200530_clean.csv', parse_dates=True, index_col='datetime')`\n\nLet’s take a quick look at the data.\n\n`df.head()df.info()`\n\n### Using spaCy\n\nNow let’s import spaCy and begin natural language processing.\n\n`# for natural language processing: named entity recognitionimport spacyimport en_core_web_sm`\n\nWe’re only going to use spaCy’s ner functionality or named-entity recognition so we’ll disable the rest of the functionalities. This will save us a lot of loading time later.\n\n`nlp = spacy.load(‘en_core_web_sm’, disable=[‘tagger’, ‘parser’, ‘textcat’])`\n\nNow let’s load the contents stopwords file into the variable `stopswords`. Note that we converted the list into a set to also save some processing time later.\n\n`with open(‘twitter-stopwords — TA — Less.txt’) as f: contents = f.read().split(‘,’)stopwords = set(contents)`\n\nNext, we’ll import joblib and define a few functions to help with parallel processing.\n\n```from joblib import Parallel, delayed\n\ndef chunker(iterable, total_length, chunksize):\nreturn (iterable[pos: pos + chunksize] for pos in range(0, total_length, chunksize))\n\ndef flatten(list_of_lists):\n\"Flatten a list of lists to a combined list\"\nreturn [item for sublist in list_of_lists for item in sublist]\n\ndef process_chunk(texts):\npreproc_pipe = []\nfor doc in nlp.pipe(texts, batch_size=20):\npreproc_pipe.append([(ent.text) for ent in doc.ents if ent.label_ in ['NORP', 'PERSON', 'FAC', 'ORG', 'GPE', 'LOC', 'PRODUCT', 'EVENT']])\nreturn preproc_pipe\n\ndef preprocess_parallel(texts, chunksize=100):\nexecutor = Parallel(n_jobs=7, backend='multiprocessing', prefer=\"processes\")\ndo = delayed(process_chunk)\ntasks = (do(chunk) for chunk in chunker(texts, len(df), chunksize=chunksize))\nreturn flatten(result)```\n\nIn the code above², the function `preprocess_parallel` executes the other function `process_chunks` in parallel to help with speed. The function `process_chunks` iterates through a series of texts — in our case, the column `'tweet'` of our the `df` dataframe — and inspects the entity if it belongs to either NORP, PERSON, FAC, ORG, GPE, LOC, PRODUCT, or EVENT. If it is, the entity is then appended to `'preproc_pipe'` and subsequently returned to its caller. Prashanth Rao has a very good article on making spaCy super fast.\n\nLet’s call the main driver for the functions now.\n\n`df['entities'] = preprocess_parallel(df['tweet'], chunksize=1000)`\n\nDoing a quick `df.head()` will reveal the new column `'entities'` that we added earlier to hold the entities found in the `'tweet'` column.\n\n### Prettifying the Results\n\nIn the code below, we’re making a list of lists called `'entities'` and then flattening it for easier processing. We’re also converting it into a set called `'entities_set'`.\n\n`entities = [entity for entity in df.entities if entity != []]entities = [item for sublist in entities for item in sublist]`\n`entities_set = set(entities)`\n\nNext, let’s count the frequency of the entities and append it to the list of tuples `entities_counts`. Then let’s convert the results into a dataframe `df_counts`.\n\n`df_counts = pd.Series(entities).value_counts()[:20].to_frame().reset_index()df_counts.columns=['entity', 'count']df_counts`\n\nFor this step, we’re going to reinitialize an empty list `entity_counts` and manually construct a list of tuples with a combined set of entities and the sum of their frequencies or count.\n\n```entity_counts = []\n\nentity_counts.append(('Democrats', df_counts.loc[df_counts.entity.isin(['Democrats', 'Dems', 'Democrat'])]['count'].sum()))\nentity_counts.append(('Americans', df_counts.loc[df_counts.entity.isin(['American', 'Americans'])]['count'].sum()))\nentity_counts.append(('Congress', df_counts.loc[df_counts.entity.isin(['House', 'Senate', 'Congress'])]['count'].sum()))\nentity_counts.append(('America', df_counts.loc[df_counts.entity.isin(['U.S.', 'the United States', 'America'])]['count'].sum()))\nentity_counts.append(('Republicans', df_counts.loc[df_counts.entity.isin(['Republican', 'Republicans'])]['count'].sum()))\n\nentity_counts.append(('China', 533))\nentity_counts.append(('FBI', 316))\nentity_counts.append(('Russia', 313))\nentity_counts.append(('Fake News', 248))\nentity_counts.append(('Mexico', 213))\nentity_counts.append(('Obama', 176))```\n\nLet’s take a quick look before continuing.\n\nFinally, let’s convert the list of tuples into a dataframe.\n\n`df_ner = pd.DataFrame(entity_counts, columns=[\"entity\", \"count\"]).sort_values('count', ascending=False).reset_index(drop=True)`\n\nAnd that’s it!\n\nWe’ve successfully created a ranking of the named entities that President Trump most frequently talked about in his tweets since taking office.\n\nThank you for reading! Exploratory data analysis uses a lot of techniques and we’ve only explored a few on this post. I encourage you to keep practicing and employ other techniques to derive insights from data.\n\nIn the next post, we shall continue our journey into the heart of darkness and do some topic-modeling using LDA.\n\nStay tuned!\n\n GONG Wei’s Homepage. (May 30, 2020). Stop words for tweets. https://sites.google.com/site/iamgongwei/home/sw\n\n Towards Data Science. (May 30, 2020). Turbo-charge your spaCy NLP pipeline. https://towardsdatascience.com/turbo-charge-your-spacy-nlp-pipeline-551435b664ad\n\n## Into the Heart of Darkness - Pt. 1\n\nExploring the Trump Twitter Archive with Python. For beginners.\n\nIn this post, we’ll explore the dataset provided by the Trump Twitter Archive. My goal was to do something fun by using a very interesting dataset. However, as it turned out, exposure to Trump’s narcissism and shenanigans were quite depressing — if not traumatic.\n\nYou’d been warned!\n\nFor this project, we’ll be using pandas and numpy for data manipulation, matplotlib for visualizations, datetime for working with timestamps, unicodedata and regex for processing strings, and finally, nltk for natural language processing.\n\nLet’s get started by firing up a Jupyter notebook!\n\n### Environment\n\nWe’re going to import pandas and matplotlib, and also set the display options for Jupyter so that the rows and columns are not truncated.\n\n`# for manipulating dataimport pandas as pdimport numpy as np`\n`# for visualizations%matplotlib inlineimport matplotlib.pyplot as plt`\n`# to print out all the outputsfrom IPython.core.interactiveshell import InteractiveShellInteractiveShell.ast_node_interactivity = \"all\"`\n`# set display optionspd.set_option('display.max_columns', None)pd.set_option('display.max_rows', None)pd.set_option('display.max_colwidth', -1)`\n\n### Getting the Data\n\nLet’s read the data into a dataframe. If you want to follow along, you can download the dataset here. This dataset contains Trump’s tweets from the moment he took office on January 20, 2017 to May 30, 2020.\n\n`df = pd.read_csv('trump_20200530.csv')`\n\nLet’s look at the first five rows and see the number of records (rows) and fields (columns).\n\n`df.head()df.shape`\n\nLet’s do a quick renaming of the columns to make it easier for us later.\n\n`df.columns=['source', 'tweet', 'date_time', 'retweets', 'favorites', 'is_retweet', 'id']`\n\nLet’s drop the id column since it’s not really relevant right now.\n\n`df = df.drop(columns=['id'])`\n\nLet’s do a quick sanity check, this time let’s also check the dtypes of the columns.\n\n`df.head()df.info()`\n\n### Working with Timestamps\n\nWe can see from the previous screenshot that the ‘date_time’ column is a string. Let’s parse it to a timestamp.\n\n`# for working with timestampsfrom datetime import datetimefrom dateutil.parser import parse`\n`dt = []for ts in df.date_time: dt.append(parse(ts))dt[:5]`\n\nLet’s add a column with ‘datetime’ that contains the timestamp information.\n\n`df['datetime'] = df.apply(lambda row: parse(row.date_time), axis=1)`\n\nLet’s double-check the data range of our dataset.\n\n`df.datetime.min()df.datetime.max()`\n\n### Trimming the Data\n\nLet’s see how many sources there are for the tweets.\n\n`df.source.value_counts()`\n\nLet’s only keep the ones that were made using the ‘Twitter for iPhone’ app.\n\n`df = df.loc[df.source == 'Twitter for iPhone']`\n\nWe should drop the old ‘date_time’ column and the ‘source’ column as well.\n\n`df = df.drop(columns=['date_time', 'source'])`\n\n### Separating the Retweets\n\nLet’s see how many are retweets.\n\n`df.is_retweet.value_counts()`\n\nLet’s make another dataframe that contains only retweets and drop the ‘is_retweet’ column.\n\n`df_retweets = df.loc[df.is_retweet == True]df_retweets = df_retweets.drop(columns=['is_retweet'])`\n\nSanity check:\n\n`df_retweets.head()df_retweets.shape`\n\nBack on the original dataframe, let’s remove the retweets from the dataset and drop the ‘is_retweet’ column altogether.\n\n`df = df.loc[df.is_retweet == False]df = df.drop(columns=['is_retweet'])`\n\nAnother sanity check:\n\n`df.head()df.shape`\n\n### Exploring the Data\n\nLet’s explore both of the dataframes and answer a few questions.\n\n#### What time does the President tweet the most? What time does he tweet the least?\n\nThe graph below shows that the President most frequently tweets around 12pm. He also tweets the least around 8am. Maybe he’s not a morning person?\n\n`title = 'Number of Tweets by Hour'df.tweet.groupby(df.datetime.dt.hour).count().plot(figsize=(12,8), fontsize=14, kind='bar', rot=0, title=title)plt.xlabel('Hour')plt.ylabel('Number of Tweets')`\n\n#### What day does the President tweet the most? What day does he tweet the least?\n\nThe graph below shows that the President most frequently tweets on Wednesday. He also tweets the least on Thursday.\n\n`title = 'Number of Tweets by Day of the Week'df.tweet.groupby(df.datetime.dt.dayofweek).count().plot(figsize=(12,8), fontsize=14, kind='bar', rot=0, title=title)plt.xlabel('Day of the Week')plt.ylabel('Number of Tweets')plt.xticks(np.arange(7),['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'])`\n\n#### Isolating Twitter Handles from the Retweets\n\nLet’s import regex so we can use it to parse the text and isolate the Twitter handles of the original tweets. In the code below, we add another column that contains the Twitter handle.\n\n`import re`\n`pattern = re.compile('(?<=RT @).*?(?=:)')df_retweets['original'] = [re.search(pattern, tweet).group(0) for tweet in df_retweets.tweet]`\n\nLet’s create another dataframe that will contain only the original Twitter handles and their associated number of retweets.\n\n`df_originals = df_retweets.groupby(['original']).sum().sort_values('retweets').reset_index().sort_values('retweets', ascending=False)`\n\nLet’s check the data real quick:\n\n`df_originals.head()df_originals.shape`\n\nLet’s visualize the results real quick so we can get an idea if the data is disproportionate or not.\n\n`df_originals = df_retweets.groupby(['original']).sum().sort_values('retweets').reset_index().sort_values('retweets', ascending=False)[:10].sort_values('retweets')df_originals.plot.barh(x='original', y='retweets', figsize=(16,10), fontsize=16)plt.xlabel(\"Originating Tweet's Username\")plt.xticks([])`\n\n#### Which Twitter user does the President like to retweet the most?\n\nThe graph below shows that the President likes to retweet the tweets from ‘@realDonaldTrump’. Does this mean the president likes to retweet himself? You don’t say!\n\nThe interesting handle on this one is ‘@charliekirk11’. Charlie Kirk is the founder of Turning Point USA. CBS News described the organization as a far-right organization that is “shunned or at least ignored by more established conservative groups in Washington, but embraced by many Trump supporters”.¹\n\n#### The Top 5 Retweets\n\nLet’s look at the top 5 tweets that were retweeted the most by others based on the original Twitter handle.\n\n`df_retweets.loc[df_retweets.original == 'realDonaldTrump'].sort_values('retweets', ascending=False)[:5]`\n\nAnd another one with ‘@charliekirk11’.\n\n`df_retweets.loc[df_retweets.original == 'charliekirk11'].sort_values('retweets', ascending=False)[:5]`\n\n### Examining Retweets’ Favorites count\n\nLet’s find out how many of the retweets are favorited by others.\n\n`df_retweets.favorites.value_counts()`\n\nSurprisingly, none of the retweets seemed to have been favorited by anybody. Weird.\n\nWe should drop it.\n\n### Counting N-Grams\n\nTo do some n-gram ranking, we need to import unicodedata and nltk. We also need to specify additional stopwords that we may need to exclude from our analysis.\n\n`# for cleaning and natural language processingimport unicodedataimport nltk`\n`# add appropriate words that will be ignored in the analysisADDITIONAL_STOPWORDS = ['rt']`\n\nHere are a few of my favorite functions for natural language processing:\n\n```def clean(text):\n\"\"\"\nA simple function to clean up the data. All the words that\nare not designated as a stop word is then lemmatized after\nencoding and basic regex parsing are performed.\n\"\"\"\nwnl = nltk.stem.WordNetLemmatizer()\ntext = (unicodedata.normalize('NFKD', text)\n.encode('ascii', 'ignore')\n.decode('utf-8', 'ignore')\n.lower())\nwords = re.sub(r'[^\\w\\s]', '', text).split()\nreturn [wnl.lemmatize(word) for word in words if word not in stopwords]\n\ndef get_words(df, column):\n\"\"\"\nTakes in a dataframe and columns and returns a list of\nwords from the values in the specified column.\n\"\"\"\nreturn clean(''.join(str(df[column].tolist())))\n\ndef get_bigrams(df, column):\n\"\"\"\nTakes in a list of words and returns a series of\nbigrams with value counts.\n\"\"\"\nreturn (pd.Series(nltk.ngrams(get_words(df, column), 2)).value_counts())[:10]\n\ndef get_trigrams(df, column):\n\"\"\"\nTakes in a list of words and returns a series of\ntrigrams with value counts.\n\"\"\"\nreturn (pd.Series(nltk.ngrams(get_words(df, column), 3)).value_counts())[:10]\n\ndef viz_bigrams(df ,column):\nget_bigrams(df, column).sort_values().plot.barh(color='blue', width=.9, figsize=(12, 8))\n\nplt.title('20 Most Frequently Occuring Bigrams')\nplt.ylabel('Bigram')\nplt.xlabel('# Occurances')\n\ndef viz_trigrams(df, column):\nget_trigrams(df, column).sort_values().plot.barh(color='blue', width=.9, figsize=(12, 8))\n\nplt.title('20 Most Frequently Occuring Trigrams')\nplt.ylabel('Trigram')\nplt.xlabel('# Occurances')\n```\n\nLet’s look at the top 10 bigrams in the `df` dataframe using the ‘tweet’ column.\n\n`get_bigrams(df, 'tweet')`\n\nAnd now, for the top 10 trigrams:\n\nLet’s use the `viz_bigrams()` function and visualize the bigrams.\n\n`viz_bigrams(df, ‘tweet’)`\n\nSimilarly, let’s use the `viz_trigrams()` function and visualize the trigrams.\n\n`viz_trigrams(df, 'tweet')`\n\nAnd there we have it!\n\nFrom the moment that Trump took office, we can confidently say that the “fake news media” has been on top of the president’s mind.\n\n### Conclusion\n\nUsing basic Python and the nltk library, we’ve explored the dataset from the Trump Twitter Archive and did some n-gram ranking out of it.\n\nThank you for reading! Exploratory data analysis uses a lot of techniques and we’ve only explored a few on this post. I encourage you to keep practicing and employ other techniques to derive insights from data.\n\nIn the next post, we shall continue our journey into the heart of darkness and use spaCy to extract named-entities from the same dataset.\n\nStay tuned!\n\n CBS News. “Trump speaks to conservative group Turning Point USA”. www.cbsnews.com. Archived from the original on July 31, 2019. Retrieved August 5, 2019.\n\n## Populating a Network Graph with Named-Entities\n\nAn early attempt of using networkx to visualize the results of natural language processing.\n\nI do a lot of natural language processing and usually, the results are pretty boring to the eye. When I learned about network graphs, it got me thinking, why not use keywords as nodes and connect them together to create a network graph?\n\nYupp, why not!\n\nIn this post, we’ll do exactly that. We’re going to extract named-entities from news articles about coronavirus and then use their relationships to connect them together in a network graph.\n\n# A Brief Introduction\n\nNetwork graphs are a cool visual that contains nodes (vertices) and edges (lines). It’s often used in social network analysis and network analysis but data scientists also use it for natural language processing.\n\nNatural Language Processing or NLP is a branch of artificial intelligence that deals with programming computers to process and analyze large volumes of text and derive meaning out of them.¹ In other words, it’s all about teaching computers how to understand human language… like a boss!\n\nEnough introduction, let’s get to coding!\n\nTo get started, let’s make sure to take care of all dependencies. Open up a terminal and execute the following commands:\n\n`pip install -U spacypython -m spacy download enpip install networkxpip install fuzzywuzzy`\n\nThis will install spaCy and download the trained model for English. The third command installs networkx. This should work for most systems. If it doesn’t work for you, check out the documentation for spaCy and networkx. Also, we’re using fuzzywuzzy for some text preprocessing.\n\nWith that out of the way, let’s fire up a Jupyter notebook and get started!\n\n# Imports\n\nRun the following code block into a cell to get all the necessary imports into our Python environment.\n\n`import pandas as pdimport numpy as npimport picklefrom operator import itemgetterfrom fuzzywuzzy import process, fuzz# for natural language processingimport spacyimport en_core_web_sm# for visualizations%matplotlib inlinefrom matplotlib.pyplot import figureimport networkx as nx`\n\n# Getting the Data\n\nIf you want to follow along, you can download the sample dataset here. The file was created using newspaper to import news articles from the npr.org. If you’re feeling adventurous, use the code snippet below to build your own dataset.\n\n```import requests\nimport json\nimport time\nimport newspaper\nimport pickle\n\ncorpus = []\ncount = 0\nfor article in npr.articles:\ntime.sleep(1)\narticle.parse()\ntext = article.text\ncorpus.append(text)\nif count % 10 == 0 and count != 0:\nprint('Obtained {} articles'.format(count))\ncount += 1\n\ncorpus300 = corpus[:300]\n\nwith open(\"npr_coronavirus.txt\", \"wb\") as fp: # Pickling\npickle.dump(corpus300, fp)\n\n# with open(\"npr_coronavirus.txt\", \"rb\") as fp: # Unpickling\n\nLet’s get our data.\n\n`with open('npr_coronavirus.txt', 'rb') as fp: # Unpickling corpus = pickle.load(fp)`\n\n# Extract Entities\n\n`nlp = en_core_web_sm.load()`\n\nThen, we’ll extract the entities:\n\n`entities = []for article in corpus[:50]: tokens = nlp(''.join(article)) gpe_list = [] for ent in tokens.ents: if ent.label_ == 'GPE': gpe_list.append(ent.text) entities.append(gpe_list)`\n\nIn the above code block, we created an empty list called `entities` to store a list of lists that contains the extracted entities from each of the articles. In the for-loop, we looped through the first 50 articles of the corpus. For each iteration, we converted each articles into tokens (words) and then we looped through all those words to get the entities that are labeled as `GPE` for countries, states, and cities. We used `ent.text` to extract the actual entity and appended them one by one to `entities`.\n\nHere’s the result:\n\nNote that North Carolina has several variations of its name and some have “the” prefixed in their names. Let’s get rid of them.\n\n`articles = []for entity_list in entities: cleaned_entity_list = [] for entity in entity_list: cleaned_entity_list.append(entity.lstrip('the ').replace(\"'s\", \"\").replace(\"’s\",\"\")) articles.append(cleaned_entity_list)`\n\nIn the code block above, we’re simply traversing the list of lists `articles` and cleaning the entities one by one. With each iteration, we’re stripping the prefix “the” and getting rid of `'s.`\n\n# Optional: FuzzyWuzzy\n\nLooking at the entities, I’ve noticed that there are also variations in the “United States” is represented. There exists “United States of America” while some are just “United States”. We can trim these down into a more standard naming convention.\n\nFuzzyWuzzy can help with this.\n\nDescribed by pypi.org as “string matching like a boss,” FiuzzyWuzzy uses Levenshtein distance to calculate the similarities between words.¹ For a really good tutorial on how to use FuzzyWuzzy, check out Thanh Huynh’s article.FuzzyWuzzy: Find Similar Strings within one column in PythonToken Sort Ratio vs. Token Set Ratiotowardsdatascience.com\n\nHere’s the optional code for using FuzzyWuzzy:\n\n```choices = set([item for sublist in articles for item in sublist])\n\ncleaned_articles = []\nfor article in articles:\narticle_entities = []\nfor entity in set(article):\narticle_entities.append(process.extractOne(entity, choices))\ncleaned_articles.append(article_entities)```\n\nFor the final step before creating the network graph, let’s get rid of the empty lists within our list of list that were generated by articles who didn’t have any `GPE` entity types.\n\n`articles = [article for article in articles if article != []]`\n\n# Create the Network Graph\n\nFor the next step, we’ll create the world into which the graph will exist.\n\n`G = nx.Graph()`\n\nThen, we’ll manually add the nodes with `G.add_nodes_from()`.\n\n`for entities in articles: G.add_nodes_from(entities)`\n\nLet’s see what the graph looks like with:\n\n`figure(figsize=(10, 8))nx.draw(G, node_size=15)`\n\nNext, let’s add the edges that will connect the nodes.\n\n`for entities in articles: if len(entities) > 1: for i in range(len(entities)-1): G.add_edges_from([(str(entities[i]),str(entities[i+1]))])`\n\nFor each iteration of the code above, we used a conditional that will only entertain a list of entities that has two or more entities. Then, we manually connect each of the entities with `G.add_edges_from()`.\n\nLet’s see what the graph looks like now:\n\n`figure(figsize=(10, 8))nx.draw(G, node_size=10)`\n\nThis graph reminds me of spiders! LOL.\n\nTo organize it a bit, I decided to use the shell version of the network graph:\n\n`figure(figsize=(10, 8))nx.draw_shell(G, node_size=15)`\n\nWe can tell that some nodes are heavier on connections than others. To see which nodes have the most connections, let’s use `G.degree()`.\n\n`G.degree()`\n\nThis gives the following degree view:\n\nLet’s find out which node or entity has the most number of connections.\n\n`max(dict(G.degree()).items(), key = lambda x : x)`\n\nTo find out which other nodes have the most number of connections, let’s check the top 5:\n\n`degree_dict = dict(G.degree(G.nodes()))nx.set_node_attributes(G, degree_dict, 'degree')sorted_degree = sorted(degree_dict.items(), key=itemgetter(1), reverse=True)`\n\nAbove, `sorted_degrees` is a list that contains all the nodes and their degree values. We only wanted the top 5 like so:\n\n`print(\"Top 5 nodes by degree:\")for d in sorted_degree[:5]: print(d)`\n\n# Bonus Round: Gephi\n\nGephi is an open-source and free desktop application that lets us visualize, explore, and analyze all kinds of graphs and networks.²\n\nLet’s export our graph data into a file so we can import it into Gephi.\n\n`nx.write_gexf(G, \"npr_coronavirus_GPE_50.gexf\")`\n\nCool beans!\n\n# Next Steps\n\nThis time, we only processed 50 articles from npr.org. What would happen if we processed all 300 articles from our dataset? What will we see if we change the entity type from `GPE` to `PERSON`? How else can we use network graphs to visualize natural language processing results?\n\nThere’s always more to do. The possibilities are endless!\n\nI hope you enjoyed today’s post. The code is not perfect and we have a long way to go towards realizing insights from the data. I encourage you to dive deeper and learn more about spaCynetworkxfuzzywuzzy, and even Gephi.\n\nStay tuned!\n\n: Wikipedia. (May 25, 2020). Natural language processing https://en.wikipedia.org/wiki/Natural_language_processing\n\n: Gephi. (May 25, 2020). The Open Graph Viz Platform https://gephi.org/\n\n## From DataFrame to Named-Entities\n\nA quick-start guide to extracting named-entities from a Pandas dataframe using spaCy.\n\nA long time ago in a galaxy far away, I was analyzing comments left by customers and I noticed that they seemed to mention specific companies much more than others. This gave me an idea. Maybe there is a way to extract the names of companies from the comments and I could quantify them and conduct further analysis.\n\nThere is! Enter: named-entity-recognition.\n\n# Named-Entity Recognition\n\nAccording to Wikipedia, named-entity recognition or NER “is a subtask of information extraction that seeks to locate and classify named entity mentioned in unstructured text into pre-defined categories such as person names, organizations, locations, medical codes, time expressions, quantities, monetary values, percentages, etc.”¹ In other words, NER attempts to extract words that categorized into proper names and even numerical entities.\n\nIn this post, I’ll share the code that will let us extract named-entities from a Pandas dataframe using spaCy, an open-source library provides industrial-strength natural language processing in Python and is designed for production use.²\n\nTo get started, let’s install spaCy with the following pip command:\n\n`pip install -U spacy`\n\n`python -m spacy download en`\n\nWith that out of the way, let’s open up a Jupyter notebook and get started!\n\n# Imports\n\nRun the following code block into a cell to get all the necessary imports into our Python environment.\n\n`# for manipulating dataframesimport pandas as pd# for natural language processing: named entity recognitionimport spacyfrom collections import Counterimport en_core_web_smnlp = en_core_web_sm.load()# for visualizations%matplotlib inline`\n\nThe important line in this block is `nlp = en_core_web_sm.load()` because this is what we’ll be using later to extract the entities from the text.\n\n# Getting the Data\n\n`df = pd.read_csv('ever_trump.csv')`\n\nRunning `df.head()` in a cell will get us acquainted with the data set quickly.\n\n# Getting the Tokens\n\nSecond, let’s create tokens that will serve as input for spaCy. In the line below, we create a variable `tokens` that contains all the words in the `'text'` column of the `df` dataframe.\n\n`tokens = nlp(''.join(str(df.text.tolist())))`\n\nThird, we’re going to extract entities. We can just extract the most common entities for now:\n\n`items = [x.text for x in tokens.ents]Counter(items).most_common(20)`\n\n# Extracting Named-Entities\n\nNext, we’ll extract the entities based on their categories. We have a few to choose from people to events and even organizations. For a complete list of all that spaCy has to offer, check out their documentation on named-entities.\n\nTo start, we’ll extract people (real and fictional) using the `PERSON` type.\n\n`person_list = []for ent in tokens.ents: if ent.label_ == 'PERSON': person_list.append(ent.text) person_counts = Counter(person_list).most_common(20)df_person = pd.DataFrame(person_counts, columns =['text', 'count'])`\n\nIn the code above, we started by making an empty list with `person_list = []`.\n\nThen, we utilized a for-loop to loop through the entities found in tokens with `tokens.ents`. After that, we made a conditional that will append to the previously created list if the entity label is equal to `PERSON` type.\n\nWe’ll want to know how many times a certain entity of `PERSON` type appears in the tokens so we did with `person_counts = Counter(person_list).most_common(20)`. This line will give us the top 20 most common entities for this type.\n\nFinally, we created the `df_person` dataframe to store the results and this is what we get:\n\nWe’ll repeat the same pattern for the `NORP` type which recognizes nationalities, religious and political groups.\n\n`norp_list = []for ent in tokens.ents: if ent.label_ == 'NORP': norp_list.append(ent.text) norp_counts = Counter(norp_list).most_common(20)df_norp = pd.DataFrame(norp_counts, columns =['text', 'count'])`\n\nAnd this is what we get:\n\n# Bonus Round: Visualization\n\nLet’s create a horizontal bar graph of the `df_norp` dataframe.\n\n`df_norp.plot.barh(x='text', y='count', title=\"Nationalities, Religious, and Political Groups\", figsize=(10,8)).invert_yaxis()`\n\nVoilà, that’s it!\n\nI hope you enjoyed this one. Natural language processing is a huge topic but I hope that this gentle introduction will encourage you to explore more and expand your repertoire.\n\nStay tuned!\n\n: Wikipedia. (May 22, 2020). Named-entity recognition https://en.wikipedia.org/wiki/Named-entity_recognition\n\n: spaCy. (May 22, 2020). Industrial-Strength Natural Language Processing in Python https://spacy.io/\n\n## Create an N-Gram Ranking in Power BI\n\nA quick start guide on building a Python visual with a few simple clicks of the mouse and a dash of code.\n\nIn a previous article, I wrote a quick start guide on creating and visualizing n-gram ranking using nltk for natural language processing. However, I needed a way to share my findings with others who don’t have Python or Jupyter Notebook installed in their machines. I needed to use our organization’s BI reporting tool: Power BI.\n\nEnter Python Visual.\n\nThe Python visual allows you to create a visualization generated by running Python code. In this post, we’ll walk through the steps needed to visualize the results of our n-gram ranking using this visual.\n\nFirst, let’s get our data. You can download the sample dataset here. Then, we could load the data into Power BI Desktop as shown below:\n\nSelect Text/CSV and click on “Connect”.\n\nSelect the file in the Windows Explorer folder and click open:\n\nNext, find the Py icon on the “Visualizations” panel.\n\nThen, click on “Enable” at the prompt that appears to enable script visuals.\n\nYou’ll see a placeholder appear in the main area and a Python script editor panel at the bottom of the dashboard.\n\nSelect the ‘text’ column on the “Fields” panel.\n\nYou’ll see a predefined script that serves as a preamble for the script that we’re going to write.\n\nIn the Python script editor panel, place your cursor at the end of line #6 and hit enter twice.\n\nThen, copy and paste the following code:\n\n`import reimport unicodedataimport nltkfrom nltk.corpus import stopwordsADDITIONAL_STOPWORDS = ['covfefe']import matplotlib.pyplot as pltdef basic_clean(text): wnl = nltk.stem.WordNetLemmatizer() stopwords = nltk.corpus.stopwords.words('english') + ADDITIONAL_STOPWORDS text = (unicodedata.normalize('NFKD', text) .encode('ascii', 'ignore') .decode('utf-8', 'ignore') .lower()) words = re.sub(r'[^\\w\\s]', '', text).split() return [wnl.lemmatize(word) for word in words if word not in stopwords]words = basic_clean(''.join(str(dataset['text'].tolist())))bigrams_series = (pandas.Series(nltk.ngrams(words, 2)).value_counts())[:12]bigrams_series.sort_values().plot.barh(color='blue', width=.9, figsize=(12, 8))plt.show()`\n\nIn a nutshell, the code above transforms extracts n-grams from the `'text'` column of the`dataset` dataframe and creates a horizontal bar graph out of it using matplotlib. The result of `plt.show()` is what Power BI displays on the Python visual.\n\nFor more information on this code, please visit my previous tutorial.From DataFrame to N-GramsA quick-start guide to creating and visualizing n-gram ranking using nltk for natural language processing.towardsdatascience.com\n\nAfter you’re done pasting the code, click on the “play” icon at the upper right corner of the Python script editor panel.\n\nAfter a few moments, you should now be able to see the horizontal bar graph like the one below:\n\nAnd that’s it!\n\nWith a few simple clicks of the mouse, along with some help from our Python script, we’re able to visualize the results of our n-gram ranking.\n\nI hope you enjoyed today’s post on one of Power BI’s strongest features. Power BI already has some useful and beautiful built-in visuals but sometimes, you just need a little bit more flexibility. Running Python code helps with this. I hope this gentle introduction will encourage you to explore more and expand your repertoire.\n\nIn the next article, I’ll share a quick-start guide to extracting named-entities from a Pandas dataframe using spaCy.\n\nStay tuned!\n\n## From DataFrame to N-Grams\n\nA quick-start guide to creating and visualizing n-gram ranking using nltk for natural language processing.\n\nWhen I was first starting to learn NLP, I remember getting frustrated or intimidated by information overload so I’ve decided to write a post that covers the bare minimum. You know what they say, “Walk before you run!”\n\nThis is a very gentle introduction so we won’t be using any fancy code here.\n\nIn a nutshell, natural language processing or NLP simply refers to the process of reading and understanding written or spoken language using a computer. At its simplest use case, we can use a computer to read a book, for example, and count how many times each word was used instead of us manually doing it.\n\nNLP is a big topic and there’s already been a ton of articles written on the subject so we won’t be covering that here. Instead, we’ll focus on how to quickly do one of the simplest but useful techniques in NLP: N-gram ranking.\n\n# N-Gram Ranking\n\nSimply put, an n-gram is a sequence of n words where n is a discrete number that can range from 1 to infinity! For example, the word “cheese” is a 1-gram (unigram). The combination of the words “cheese flavored” is a 2-gram (bigram). Similarly, “cheese flavored snack” is a 3-gram (trigram). And “ultimate cheese flavored snack” is a 4-gram (qualgram). So on and so forth.\n\nIn n-gram ranking, we simply rank the n-grams according to how many times they appear in a body of text — be it a book, a collection of tweets, or reviews left by customers of your company.\n\nLet’s get started!\n\n# Getting the Data\n\n`import pandas as pddf = pd.read_csv('tweets.csv')`\n\nUsing `df.head()` we can quickly get acquainted with the dataset.\n\n# Importing Packages\n\nNext, we’ll import packages so we can properly set up our Jupyter notebook:\n\n`# natural language processing: n-gram rankingimport reimport unicodedataimport nltkfrom nltk.corpus import stopwords# add appropriate words that will be ignored in the analysisADDITIONAL_STOPWORDS = ['covfefe']import matplotlib.pyplot as plt`\n\nIn the code block above, we imported pandas so that we can shape and manipulate our data in all sorts of different and wonderful ways! Next, we imported `re` for regex, `unicodedata` for Unicode data, and `nltk` to help with parsing the text and cleaning them up a bit. And then, we specified additional stop words that we want to ignore. This is helpful in trimming down the noise. Lastly, we imported `matplotlib` matplotlib so we can visualize the result of our n-gram ranking later.\n\nNext, let’s create a function that will perform basic cleaning of the data.\n\n# Basic Cleaning\n\n`def basic_clean(text): \"\"\" A simple function to clean up the data. All the words that are not designated as a stop word is then lemmatized after encoding and basic regex parsing are performed. \"\"\" wnl = nltk.stem.WordNetLemmatizer() stopwords = nltk.corpus.stopwords.words('english') + ADDITIONAL_STOPWORDS text = (unicodedata.normalize('NFKD', text) .encode('ascii', 'ignore') .decode('utf-8', 'ignore') .lower()) words = re.sub(r'[^\\w\\s]', '', text).split() return [wnl.lemmatize(word) for word in words if word not in stopwords]`\n\nThe function above takes in a list of words or text as input and returns a cleaner set of words. The function does normalization, encoding/decoding, lower casing, and lemmatization.\n\nLet’s use it!\n\n`words = basic_clean(''.join(str(df['text'].tolist())))`\n\nAbove, we’re simply calling the function `basic_lean()` to process the `'text'` column of our dataframe `df` and making it a simple list with `tolist()`. We then assign the results to `words`.\n\n# N-grams\n\nHere comes the fun part! In one line of code, we can find out which bigrams occur the most in this particular sample of tweets.\n\n`(pd.Series(nltk.ngrams(words, 2)).value_counts())[:10]`\n\nWe can easily replace the number 2 with 3 so we can get the top 10 trigrams instead.\n\n`(pd.Series(nltk.ngrams(words, 3)).value_counts())[:10]`\n\nVoilà! We got ourselves a great start. But why stop now? Let’s try it and make a little eye candy.\n\n# Bonus Round: Visualization\n\nTo make things a little easier for ourselves, let’s assign the result of n-grams to variables with meaningful names:\n\n`bigrams_series = (pd.Series(nltk.ngrams(words, 2)).value_counts())[:12]trigrams_series = (pd.Series(nltk.ngrams(words, 3)).value_counts())[:12]`\n\nI’ve replaced `[:10]` with `[:12]` because I wanted more n-grams in the results. This is an arbitrary value so you can choose whatever makes the most sense to you according to your situation.\n\nLet’s create a horizontal bar graph:\n\n`bigrams_series.sort_values().plot.barh(color='blue', width=.9, figsize=(12, 8))`\n\nAnd let’s spiffy it up a bit by adding titles and axis labels:\n\n`bigrams_series.sort_values().plot.barh(color='blue', width=.9, figsize=(12, 8))plt.title('20 Most Frequently Occuring Bigrams')plt.ylabel('Bigram')plt.xlabel('# of Occurances')`\n\nAnd that’s it! With a few simple lines of code, we quickly made a ranking of n-grams from a Pandas dataframe and even made a horizontal bar graph out of it.\n\nI hope you enjoyed this one. Natural Language Processing is a big topic but I hope that this gentle introduction will encourage you to explore more and expand your repertoire.\n\nIn the next article, we’ll visualize an n-gram ranking in Power BI with a few simple clicks of the mouse and a dash of Python!\n\nStay tuned!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79824954,"math_prob":0.7847999,"size":28702,"snap":"2020-34-2020-40","text_gpt3_token_len":6547,"char_repetition_ratio":0.11596627,"word_repetition_ratio":0.15630755,"special_character_ratio":0.22792837,"punctuation_ratio":0.13564214,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95765543,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-20T17:32:31Z\",\"WARC-Record-ID\":\"<urn:uuid:818196ae-072c-44e2-89d2-e23408b9ad6f>\",\"Content-Length\":\"363166\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1448404b-29a5-4249-a8c2-5b35d437aaad>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7c0c845-f5bb-4383-a58a-1900dd917e52>\",\"WARC-IP-Address\":\"74.208.236.174\",\"WARC-Target-URI\":\"https://blog.ednalyn.com/tag/python/\",\"WARC-Payload-Digest\":\"sha1:O45AKSZ6FXTHU3SKPI6WFCX37D47MMJU\",\"WARC-Block-Digest\":\"sha1:CKKMW3WAYBWHHTT7G7TOK4ZWIIQYDZ7I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400198287.23_warc_CC-MAIN-20200920161009-20200920191009-00213.warc.gz\"}"}
https://bootstrapworld.org/materials/latest/en-us/courses/data-science/lessons/defining-functions/pages/match-examples-definitions-math.html	[ "Referenced from lesson Defining Functions (Spring, 2021)\n\nLook at each set of examples on the left and circle what is changing from one example to the next.\n\nThen,match the examples on the left to the definitions on the right.\n\nExamples: Functions:\nx$\\displaystyle x$ fx$\\displaystyle f(x)$\n\n1\n\n2 × 1$\\displaystyle 2 \\times 1$\n\n2\n\n2 × 2$\\displaystyle 2 \\times 2$\n\n3\n\n2 × 3$\\displaystyle 2 \\times 3$\n\n1\n\nA\n\nfx=x-3$\\displaystyle f(x)=x-3$\n\nx$\\displaystyle x$ fx$\\displaystyle f(x)$\n\n15\n\n15 - 3$\\displaystyle 15 - 3$\n\n25\n\n25 - 3$\\displaystyle 25 - 3$\n\n35\n\n35 - 3$\\displaystyle 35 - 3$\n\n2\n\nB\n\nfx=2x$\\displaystyle f(x)=2x$\n\nx$\\displaystyle x$ fx$\\displaystyle f(x)$\n\n10\n\n10 + 2$\\displaystyle 10 + 2$\n\n15\n\n15 + 2$\\displaystyle 15 + 2$\n\n20\n\n20 + 2$\\displaystyle 20 + 2$\n\n3\n\nC\n\nfx=2x+1$\\displaystyle f(x)=2x+1$\n\nx$\\displaystyle x$ fx$\\displaystyle f(x)$\n\n0\n\n30 -2$\\displaystyle 3(0) -2$\n\n1\n\n31 -2$\\displaystyle 3(1) -2$\n\n2\n\n32 -2$\\displaystyle 3(2) -2$\n\n4\n\nD\n\nfx=3x-2$\\displaystyle f(x)=3x-2$\n\nx$\\displaystyle x$ fx$\\displaystyle f(x)$\n\n10\n\n210 + 1$\\displaystyle 2(10) + 1$\n\n20\n\n220 + 1$\\displaystyle 2(20) + 1$\n\n30\n\n230 + 1$\\displaystyle 2(30) + 1$\n\n5\n\nE\n\nfx=x+2$\\displaystyle f(x)=x+2$\n\nThese materials were developed partly through support of the National Science Foundation, (awards 1042210, 1535276, 1648684, and 1738598).", null, "Bootstrap:Data Science by the Bootstrap Community is licensed under a Creative Commons 4.0 Unported License. This license does not grant permission to run training or professional development. Offering training or professional development with materials substantially derived from Bootstrap must be approved in writing by a Bootstrap Director. Permissions beyond the scope of this license, such as to run training, may be available by contacting contact@BootstrapWorld.org." ]	[ null, "https://bootstrapworld.org/materials/latest/en-us/lib/CCbadge.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86929584,"math_prob":0.99757904,"size":956,"snap":"2021-43-2021-49","text_gpt3_token_len":243,"char_repetition_ratio":0.1144958,"word_repetition_ratio":0.0,"special_character_ratio":0.27196652,"punctuation_ratio":0.115606934,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949968,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T08:43:27Z\",\"WARC-Record-ID\":\"<urn:uuid:5f5f9af8-7092-43bf-ac45-457b4db53a2e>\",\"Content-Length\":\"14646\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:944dabbe-1932-4290-a51f-e4422df9d4ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:c108b1ce-eb3e-414d-bb63-26c62643c42e>\",\"WARC-IP-Address\":\"128.148.32.110\",\"WARC-Target-URI\":\"https://bootstrapworld.org/materials/latest/en-us/courses/data-science/lessons/defining-functions/pages/match-examples-definitions-math.html\",\"WARC-Payload-Digest\":\"sha1:ULHC6OICUWLP6J72MK5YRD5I5VZU5IIG\",\"WARC-Block-Digest\":\"sha1:3QGVWMWU6P35IX6YLR5OSNPLDNRSZXNG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363689.56_warc_CC-MAIN-20211209061259-20211209091259-00414.warc.gz\"}"}
https://physics.stackexchange.com/questions/238180/why-do-non-hydrogen-atomic-orbitals-have-the-same-degeneracy-structure-as-hydrog/238227	[ "# Why do non-hydrogen atomic orbitals have the same degeneracy structure as hydrogen orbitals?\n\nThe solutions of the Schrödinger equation for hydrogen are the \"electronic orbitals\", shown in this picture:\n\nThey have the following degeneracy structure:\n\nIt is often said that atoms in other elements simply have more electrons \"filling up\" the orbitals in increasing order of energy, as required by the Pauli exclusion principle. This is somewhat confirmed via xray spectroscopy:\n\nBut why would non-hydrogen atoms have the same orbital degeneracy structure as hydrogen, if the actual Hamiltonian is different from one atom to another?\n\n• I suspect that the shell theorem plays a part in why this is a working approximation. – dmckee --- ex-moderator kitten Feb 18 '16 at 17:15\n\nPolyelectronic atoms don't have atomic orbitals - though they are a very useful approximation for describing the properties of polyelectronic atoms.\n\nThe 1s, 2s, etc orbitals are solutions for a central potential, and for any smooth monotonic central potential we'll get solutions of this form. The radial part of the orbitals will be different for different central potentials but the angular part is dictated by the spherical symmetry and is the same for all (smooth monotonic) central potentials.\n\nBut for any atom with more than two electrons the potential is not centrally symmetric because it includes terms like $1/r_{ij}$ for the interaction between the $i$th and $j$th electrons. This means the hydrogenic orbitals are not solutions.\n\nHowever because the electrons are delocalised over the whole atom the potential is approximately central. By this I mean that if we take a time average potential the $1/r_{ij}$ terms tend to average out to a central force. In that case we do get hydrogenic type orbital as solutions, but we have to bear in mind that they are approximate solutions. They should be regarded as a useful way of building up the complete electronic structure but they are not themselves real. For example in a lithium atom there is not actually two electrons in a 1s orbital and one in a 2s orbital. There is a single three electron wavefunction that can be approximately decomposed into hydrogenic 1s and 2s orbitals for convenience.\n\nHaving said this, for most purposes the hydrogenic orbitals are very good approximations. For example when we are considering atomic spectra we usually describe them as transitions between the hydrogenic orbitals and this works pretty well.\n\n• why with two electrons the potential is still centrally symmetric? – Sparkler Feb 18 '16 at 17:07\n• @Sparkler: suppose you're sitting on electron $i$ looking at electron $j$. From your perspective electron $j$ will spend as much time to the left of the nucleus as to the right of the nucleus, and as much time above the nucleus as below the nucleus. So even though the force due to electron $j$ is in lots of different directions if you take a time average the force vector will point towards the nucleus i.e. on average it is a central force. – John Rennie Feb 18 '16 at 17:12\n• so Li is not actually $\\mathrm{1s^2 2s^1}$ but rather $\\mathrm{1\\square^3}$..? – Sparkler Feb 18 '16 at 17:12\n• @Sparkler: yes, though obviously your notation is not standard. We would normally describe the polyelectron state using term symbols. – John Rennie Feb 18 '16 at 17:13\n• @Sparkler: yes! Basically, as you increase the atomic number the 1s orbital shrinks inwards. – John Rennie Feb 18 '16 at 17:28\n\nAlthough John Rennie is right to remind us that when talking about orbitals for a multi-electron atom, it is only an approximation, I would like to point out that the source of the exact degeneracy of the (many-electron) atomic states, it is the isotropy of the atom in the absence of external polarizing fields. This makes the (total) angular momentum J and its projection m, exact quantum numbers, applicable to classification of atomic terms." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.938609,"math_prob":0.9638571,"size":1695,"snap":"2019-51-2020-05","text_gpt3_token_len":350,"char_repetition_ratio":0.14488469,"word_repetition_ratio":0.0,"special_character_ratio":0.18997051,"punctuation_ratio":0.06020067,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9768158,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T02:50:33Z\",\"WARC-Record-ID\":\"<urn:uuid:d060d909-cf92-475e-ae67-1543d1b5ff96>\",\"Content-Length\":\"155386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed65efa6-df7c-44e0-8231-24a10cedcbd8>\",\"WARC-Concurrent-To\":\"<urn:uuid:fbdd5242-ca24-4853-a1a8-231dff5c942d>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/238180/why-do-non-hydrogen-atomic-orbitals-have-the-same-degeneracy-structure-as-hydrog/238227\",\"WARC-Payload-Digest\":\"sha1:YZAZOG62BVWCTUE3EO5SRZESBOK4ZVRM\",\"WARC-Block-Digest\":\"sha1:2LBHAUS6UN5IK5VOJZ2GUVN2FYKQX5N4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251684146.65_warc_CC-MAIN-20200126013015-20200126043015-00339.warc.gz\"}"}