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https://math.stackexchange.com/questions/3126698/is-it-true-that-given-a-set-e-every-point-in-the-set-of-boundary-points-of-e-i	[ "# Is it true that, given a set E, every point in the set of boundary points of E is an accumulation point of the set of boundary points?\n\nSuppose $$E\\subset \\mathbb{R}^n.$$ Is it true that for all $$x\\in\\partial E,$$ $$x$$ is an accumulation point of $$\\partial E$$?\n\nThe reason I think this is true (despite my feeling it is false) is that we have $$\\partial E=\\{ p \\in \\mathbb{R}^n, \\forall r>0 \\text{ one has } B(p,r)\\cap E\\neq\\emptyset \\text{ and } B(p,r)\\cap E^C \\neq \\emptyset \\}$$\n\nwhich implies, with some extra reasoning, that every ball about every point in $$\\partial E$$ Intersects $$S-\\{p\\}.$$\n\n• What is the boundary of the set $\\{0\\}$ in $\\mathbb R$? What are the accumulation points? – GEdgar Feb 25 at 22:15\n• @GEdgar the boundary is 0 and there are no accumulation points. Got it. Counter example. Thanks. – Jake Feb 25 at 22:17\n\n$$E=(0,1) \\subseteq \\mathbb R$$ only has two boundary points and these are not accumulation points of that finite boundary...." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8313551,"math_prob":0.9998779,"size":454,"snap":"2019-43-2019-47","text_gpt3_token_len":155,"char_repetition_ratio":0.12,"word_repetition_ratio":0.0,"special_character_ratio":0.31938326,"punctuation_ratio":0.091836736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T08:08:30Z\",\"WARC-Record-ID\":\"<urn:uuid:e1ad9351-4245-418b-b495-3262cf9c5998>\",\"Content-Length\":\"138071\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b41ed2a-c2fb-4808-b878-16d3ac971e1b>\",\"WARC-Concurrent-To\":\"<urn:uuid:cc04ce70-0593-41ed-88ec-0f9ccf2aa750>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3126698/is-it-true-that-given-a-set-e-every-point-in-the-set-of-boundary-points-of-e-i\",\"WARC-Payload-Digest\":\"sha1:NHO2YN2UPCK2K4XDURIPPFRRHM5BSAUS\",\"WARC-Block-Digest\":\"sha1:3CTIV37OGHG3B3QNDM45PO4L5MJGA2E6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668594.81_warc_CC-MAIN-20191115065903-20191115093903-00034.warc.gz\"}"}
https://developer.arm.com/docs/ddi0596/i/simd-and-floating-point-instructions-alphabetic-order/frint32x-scalar-floating-point-round-to-32-bit-integer-using-current-rounding-mode-scalar	[ "You copied the Doc URL to your clipboard.\n\n## FRINT32X (scalar)\n\nFloating-point Round to 32-bit Integer, using current rounding mode (scalar). This instruction rounds a floating-point value in the SIMD&FP source register to an integral floating-point value that fits into a 32-bit integer size using the rounding mode that is determined by the FPCR, and writes the result to the SIMD&FP destination register.\n\nA zero input returns a zero result with the same sign. When the result value is not numerically equal to the input value, an Inexact exception is raised. When the input is infinite, NaN or out-of-range, the instruction returns {for the corresponding result value} the most negative integer representable in the destination size, and an Invalid Operation floating-point exception is raised.\n\nA floating-point exception can be generated by this instruction. Depending on the settings in FPCR, the exception results in either a flag being set in FPSR, or a synchronous exception being generated. For more information, see Floating-point exception traps.\n\nDepending on the settings in the CPACR_EL1, CPTR_EL2, and CPTR_EL3 registers, and the current Security state and Exception level, an attempt to execute the instruction might be trapped.\n\n### Floating-point(Armv8.5)\n\n 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 0 1 1 1 1 0 0 x 1 0 1 0 0 0 1 1 0 0 0 0 Rn Rd ftype op\n\n#### Single-precision (ftype == 00)\n\nFRINT32X <Sd>, <Sn>\n\n#### Double-precision (ftype == 01)\n\nFRINT32X <Dd>, <Dn>\n\n```if !HaveFrintExt() then UNDEFINED;\ninteger d = UInt(Rd);\ninteger n = UInt(Rn);\n\ninteger esize;\ncase ftype of\nwhen '00' esize = 32;\nwhen '01' esize = 64;\nwhen '1x' UNDEFINED;\n\nFPRounding rounding = FPRoundingMode(FPCR[]);```\n\n### Assembler Symbols\n\n\nIs the 64-bit name of the SIMD&FP destination register, encoded in the \"Rd\" field.\n Is the 64-bit name of the SIMD&FP source register, encoded in the \"Rn\" field.\n Is the 32-bit name of the SIMD&FP destination register, encoded in the \"Rd\" field.\n Is the 32-bit name of the SIMD&FP source register, encoded in the \"Rn\" field.\n\n### Operation\n\n```CheckFPAdvSIMDEnabled64();\n\nFPCRType fpcr = FPCR[];\nboolean merge = IsMerging(fpcr);\nbits(128) result = if merge then V[d] else Zeros();\nbits(esize) operand = V[n];\n\nElem[result, 0, esize] = FPRoundIntN(operand, fpcr, rounding, 32);\n\nV[d] = result;```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66792864,"math_prob":0.9725795,"size":2272,"snap":"2021-04-2021-17","text_gpt3_token_len":650,"char_repetition_ratio":0.13580246,"word_repetition_ratio":0.13440861,"special_character_ratio":0.30853873,"punctuation_ratio":0.11659193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9614374,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-15T16:09:34Z\",\"WARC-Record-ID\":\"<urn:uuid:95798f23-5a29-48e5-9ce6-e867f870724a>\",\"Content-Length\":\"179186\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4cd9a3d7-cd93-46ff-9976-16c0905e69ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2f3fd80-e392-47d1-a98a-7e0cd98412ba>\",\"WARC-IP-Address\":\"104.111.77.174\",\"WARC-Target-URI\":\"https://developer.arm.com/docs/ddi0596/i/simd-and-floating-point-instructions-alphabetic-order/frint32x-scalar-floating-point-round-to-32-bit-integer-using-current-rounding-mode-scalar\",\"WARC-Payload-Digest\":\"sha1:Y42DDQVJPTMPXSCIIJH2MF7M4FQEVQU7\",\"WARC-Block-Digest\":\"sha1:UO2RZWKWIQWNJRF25QXBDSVWHFCN65V2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703495901.0_warc_CC-MAIN-20210115134101-20210115164101-00461.warc.gz\"}"}
https://rustcc.cn/article?id=13a7adbe-56fb-4459-9f99-ba31aa7138d7	[ "< 返回版块\n\n## Rust FFI 编程 - 手动绑定 C 库入门 05\n\nMike Tang 发表于 2020-06-27 19:26\n\nTags：rust,ffi\n\n1. 被调函数在 C 端，接收一个函数指针作为回调函数，并调用；\n2. 主函数在 Rust 中，在 Rust 中调用 C 端的这个函数；\n3. 在 Rust 中，传递一个 Rust 中定义的函数，到这个 C 端的被调函数中作为回调函数。\n\n1. 有可能想在底层事件（异步）框架中，注册一个函数，事件触发的时候，调用；\n2. 底层采用注册一个路由表的形式，在程序开始的时候，注册一堆函数操作进去；\n3. 其它。\n\n### 基础示例\n\n1. C 端，设计一个函数，sum_square_cb01，接收两个整型参数 a, b，和一个函数指针，计算 a2 + b2 的值，并且将值传递进第三个参数（函数中），进行打印；\n2. Rust 端，定义一个回调函数 cb_func，在这个回调函数中，打印上述平方和；\n3. Rust 端，引入 C 中定义的 sum_square_cb01；\n4. 在 Rust 的 main 中，调用 sum_square_cb01。\n\n``````// csrc/ccode01.c\n\n#include<stdio.h>\n\ntypedef void (SumSquareCB)(int result);\n\nvoid sum_square_cb01(int a, int b, SumSquareCB cb) {\nint result = aa + bb;\ncb(result);\n}\n``````\n\nRust 端：\n\n``````// src/r01.rs\n\nuse std::os::raw::c_int;\n\npub type SumSquareCB = unsafe extern fn(c_int);\n\nextern {\npub fn sum_square_cb01(a: c_int, b: c_int, cb: SumSquareCB);\n}\n\npub unsafe extern fn cb_func(result: c_int) {\nprintln!(\"The result in callback function is: {}\", result);\n}\n\nfn main() {\nunsafe {\nsum_square_cb01(3, 4, cb_func);\n}\n}\n``````\n\n• 两边都需要定义回调函数的类型（签名），而且定义要一致。\n\nC 中定义：\n\n``````typedef void (SumSquareCB)(int result);\n``````\n\nRust 中定义：\n\n``````pub type SumSquareCB = unsafe extern fn(c_int);\n``````\n\n`fn` 是 Rust 中的函数指针类型。具体可参见标准库文档 fn，解释得非常详尽。\n\n• Rust 中的回调函数定义\n``````pub unsafe extern fn cb_func(result: c_int) {\nprintln!(\"The result in callback function is: {}\", result);\n}\n``````\n\n• 代码的编译方式，见前一篇，此不赘述。\n\n``````RUSTFLAGS='-L .' LD_LIBRARY_PATH=\".\" cargo run --bin r01\n``````\n\n``````The result in callback function is: 25\n``````\n\n### 在回调函数中，更新外部数据\n\n``````// src/r01-1.rs\n\nuse std::os::raw::c_int;\n\npub type SumSquareCB = unsafe extern fn(c_int);\n\nextern {\npub fn sum_square_cb01(a: c_int, b: c_int, cb: SumSquareCB);\n}\n\nfn main() {\nlet mut sum = 0;\n\npub unsafe extern fn cb_func(result: c_int) {\nsum += result;\n}\n\nunsafe {\nsum_square_cb01(3, 4, cb_func);\n}\n\nprintln!(\"The result in callback function is: {}\", sum);\n}\n``````\n\n``````error[E0434]: can't capture dynamic environment in a fn item\n--> src/r01-1.rs:14:9\n\|\n14 \| sum += result;\n\| ^^^\n\|\n= help: use the `\|\| { ... }` closure form instead\n\nerror: aborting due to previous error\n``````\n\n``````// src/r01-2.rs\n\nuse std::os::raw::c_int;\n\npub type SumSquareCB = unsafe extern fn(c_int);\n\nextern {\npub fn sum_square_cb01(a: c_int, b: c_int, cb: SumSquareCB);\n}\n\nfn main() {\nlet mut sum = 0;\n\nunsafe {\nsum_square_cb01(3, 4, \|r\| sum += r );\n}\n\nprintln!(\"The result in callback function is: {}\", sum);\n}\n``````\n\n``````error[E0308]: mismatched types\n--> src/r01-1.rs:14:31\n\|\n14 \| sum_square_cb01(3, 4, \|r\| sum += r );\n\| ^^^^^^^^^^^^ expected fn pointer, found closure\n\|\n= note: expected fn pointer `unsafe extern \"C\" fn(i32)`\nfound closure `[closure@src/r01-1.rs:14:31: 14:43 sum:_]`\n\nerror: aborting due to previous error\n``````\n\n`void ` 是一种通用指针，意思是”指向某个东西的指针“，它的灵活和强大之处在于，可以强制转换到任何指针类型。这里，我们也可以使用 `void ` 来传递我们的”数据块“。\n\n1. 在 Rust 的 main 函数中，定义一个变量 sum；\n2. 在 Rust 中定义的回调函数中，更新这个变量 sum；\n3. 由于需要传递数据块地址，需要修改回调函数的签名定义；\n\nC端：\n\n``````// csrc/ccode02.c\n\n#include<stdio.h>\n\ntypedef void (SumSquareCB)(int result, void user_data);\n\nvoid sum_square_cb02(int a, int b, SumSquareCB cb, void user_data) {\nint result = aa + bb;\ncb(result, user_data);\n}\n``````\n\nRust 端：\n\n``````use std::os::raw::c_int;\nuse std::ffi::c_void;\n\npub type SumSquareCB = unsafe extern fn(c_int, mut c_void);\n\nextern {\npub fn sum_square_cb02(a: c_int, b: c_int, cb: SumSquareCB, user_data: mut c_void);\n}\n\npub unsafe extern fn cb_func(result: c_int, user_data: mut c_void) {\nlet data = &mut (user_data as mut c_int);\ndata += result;\n}\n\nfn main() {\nlet mut sum = 0;\n\nunsafe {\nsum_square_cb02(\n3,\n4,\ncb_func,\n&mut sum as mut c_int as mut c_void);\n}\n\nprintln!(\"The sum is {}\", sum);\n}\n``````\n\n``````RUSTFLAGS='-L .' LD_LIBRARY_PATH=\".\" cargo run --bin r01\n``````\n\n``````The sum is 25\n``````\n\nRust 端引入了 `std::ffi::c_void;`。这是 Rust 给我们提供的强大的基础设施，不然我们真要愁眉苦脸了。从标准库页面可以学习到，Rust 中的 `const c_void` 等于 C 的 `const void`，Rust 中的 `mut c_void` 等于 C 的 `void`。（C 中的 void 函数返回值本身，与 Rust 的空值类型 () 相等）\n\n• 回调函数中的类型强制转换\n``````pub unsafe extern fn cb_func(result: c_int, user_data: mut c_void) {\nlet data = &mut (user_data as mut c_int);\ndata += result;\n}\n``````\n\n``````let data = &mut (user_data as mut c_int);\n``````\n\n``````data += result;\n``````\n\n• 传参时的\n`````` unsafe {\nsum_square_cb02(\n3,\n4,\ncb_func,\n&mut sum as mut c_int as mut c_void);\n}\n``````\n\n``````&mut sum as mut c_int as mut c_void\n``````\n\n• 打印语句\n\n### 更新结构体\n\nC 端：\n\n``````#include<stdio.h>\n\ntypedef void (SumSquareCB)(int result, void user_data);\n\nvoid sum_square_cb03(int a, int b, SumSquareCB cb, void user_data) {\nint result = aa + bb;\ncb(result, user_data);\n}\n``````\n\nRust 端：\n\n``````use std::os::raw::c_int;\nuse std::ffi::c_void;\n\npub type SumSquareCB = unsafe extern fn(c_int, mut c_void);\n\nextern {\npub fn sum_square_cb03(a: c_int, b: c_int, cb: SumSquareCB, user_data: mut c_void);\n}\n\npub unsafe extern fn cb_func(result: c_int, user_data: mut c_void) {\nlet data = &mut (user_data as mut SumRecord);\ndata.sum += result;\ndata.elem_number += 1;\n}\n\n#[derive(Debug, Default, Clone, PartialEq)]\nstruct SumRecord {\nsum: c_int,\nelem_number: usize,\n}\n\nfn main() {\nlet mut sum = SumRecord::default();\n\nunsafe {\nsum_square_cb03(\n3,\n4,\ncb_func,\n&mut sum as mut SumRecord as mut c_void);\n}\n\nprintln!(\"The sum is {:?}\", sum);\n}\n``````\n\n``````RUSTFLAGS='-L .' LD_LIBRARY_PATH=\".\" cargo run --bin r03\n``````\n\n``````The sum is SumRecord { sum: 25, elem_number: 1 }\n``````\n\n• C 中代码没变\n\n• Rust 中加了结构体定义\n``````#[derive(Debug, Default, Clone, PartialEq)]\nstruct SumRecord {\nsum: c_int,\nelem_number: usize,\n}\n``````\n\n• 魔法在哪里？\n\nRust 中的回调函数签名都没有变化。变化在下面这里：\n\n`````` let data = &mut (user_data as mut SumRecord);\ndata.sum += result;\ndata.elem_number += 1;\n``````\n\n`````` unsafe {\nsum_square_cb03(\n3,\n4,\ncb_func,\n&mut sum as mut SumRecord as mut c_void);\n}\n``````\n\n1 共 0 条评论, 1 页" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.68516445,"math_prob":0.9911593,"size":7373,"snap":"2020-24-2020-29","text_gpt3_token_len":3777,"char_repetition_ratio":0.1613516,"word_repetition_ratio":0.4037461,"special_character_ratio":0.31669605,"punctuation_ratio":0.1912088,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941949,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T05:05:52Z\",\"WARC-Record-ID\":\"<urn:uuid:01495b20-d296-4a3c-b373-852a80e0f6b9>\",\"Content-Length\":\"19060\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:51f84395-9418-4f9a-8eab-1370ac61c344>\",\"WARC-Concurrent-To\":\"<urn:uuid:046ee52e-3f1a-4951-821d-fded3d2dc0f4>\",\"WARC-IP-Address\":\"39.100.77.211\",\"WARC-Target-URI\":\"https://rustcc.cn/article?id=13a7adbe-56fb-4459-9f99-ba31aa7138d7\",\"WARC-Payload-Digest\":\"sha1:U4VDS42HMXPQTZ6JHZOOMNZKX4H6DJFY\",\"WARC-Block-Digest\":\"sha1:UQ3VPABZQLT3ULJU7GDJ4K7357B5MOMQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655890105.39_warc_CC-MAIN-20200706042111-20200706072111-00430.warc.gz\"}"}
https://www.mycompiler.io/view/A51wt6Y	[ "# Write a python function to print the multiplication table of a given number.\n\n· September 23, 2021 ·", null, "Python\n```def reversed_Array(num):\nfor i in reversed(range(1,11)):\nprint(str(num)+\" X \",i,\"=\",num*i)\n\nreversed_Array(5)\n```" ]	[ null, "https://www.mycompiler.io/img/lang/python_small.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69250727,"math_prob":0.74237156,"size":359,"snap":"2021-43-2021-49","text_gpt3_token_len":90,"char_repetition_ratio":0.115492955,"word_repetition_ratio":0.0,"special_character_ratio":0.2534819,"punctuation_ratio":0.1388889,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95798004,"pos_list":[0,1,2],"im_url_duplicate_count":[null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T09:13:37Z\",\"WARC-Record-ID\":\"<urn:uuid:067c60df-f258-47b2-959a-93348b957f16>\",\"Content-Length\":\"12681\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc348b50-647b-446f-970c-f1eb9f4d5aad>\",\"WARC-Concurrent-To\":\"<urn:uuid:14ff3437-3f66-4515-b080-0332df18c8ff>\",\"WARC-IP-Address\":\"104.21.55.187\",\"WARC-Target-URI\":\"https://www.mycompiler.io/view/A51wt6Y\",\"WARC-Payload-Digest\":\"sha1:GTRPI4CJ3TBYHELMLXTKFAWI7BH7VL3M\",\"WARC-Block-Digest\":\"sha1:6SKFNJM3JIVXKI53AGSK72DTIFXMXGSS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363337.27_warc_CC-MAIN-20211207075308-20211207105308-00608.warc.gz\"}"}
https://www.colorhexa.com/1b462b	[ "# #1b462b Color Information\n\nIn a RGB color space, hex #1b462b is composed of 10.6% red, 27.5% green and 16.9% blue. Whereas in a CMYK color space, it is composed of 61.4% cyan, 0% magenta, 38.6% yellow and 72.5% black. It has a hue angle of 142.3 degrees, a saturation of 44.3% and a lightness of 19%. #1b462b color hex could be obtained by blending #368c56 with #000000. Closest websafe color is: #333333.\n\n• R 11\n• G 27\n• B 17\nRGB color chart\n• C 61\n• M 0\n• Y 39\n• K 73\nCMYK color chart\n\n#1b462b color description : Very dark cyan - lime green.\n\n# #1b462b Color Conversion\n\nThe hexadecimal color #1b462b has RGB values of R:27, G:70, B:43 and CMYK values of C:0.61, M:0, Y:0.39, K:0.73. Its decimal value is 1787435.\n\nHex triplet RGB Decimal 1b462b `#1b462b` 27, 70, 43 `rgb(27,70,43)` 10.6, 27.5, 16.9 `rgb(10.6%,27.5%,16.9%)` 61, 0, 39, 73 142.3°, 44.3, 19 `hsl(142.3,44.3%,19%)` 142.3°, 61.4, 27.5 333333 `#333333`\nCIE-LAB 26.121, -22.182, 11.9 3.078, 4.788, 3.047 0.282, 0.439, 4.788 26.121, 25.172, 151.788 26.121, -17.428, 15.081 21.88, -13.18, 7.059 00011011, 01000110, 00101011\n\n# Color Schemes with #1b462b\n\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #461b36\n``#461b36` `rgb(70,27,54)``\nComplementary Color\n• #21461b\n``#21461b` `rgb(33,70,27)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #1b4641\n``#1b4641` `rgb(27,70,65)``\nAnalogous Color\n• #461b21\n``#461b21` `rgb(70,27,33)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #411b46\n``#411b46` `rgb(65,27,70)``\nSplit Complementary Color\n• #462b1b\n``#462b1b` `rgb(70,43,27)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #2b1b46\n``#2b1b46` `rgb(43,27,70)``\n• #36461b\n``#36461b` `rgb(54,70,27)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #2b1b46\n``#2b1b46` `rgb(43,27,70)``\n• #461b36\n``#461b36` `rgb(70,27,54)``\n• #060f09\n``#060f09` `rgb(6,15,9)``\n• #0d2114\n``#0d2114` `rgb(13,33,20)``\n• #143420\n``#143420` `rgb(20,52,32)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #225836\n``#225836` `rgb(34,88,54)``\n• #296b42\n``#296b42` `rgb(41,107,66)``\n• #307d4d\n``#307d4d` `rgb(48,125,77)``\nMonochromatic Color\n\n# Alternatives to #1b462b\n\nBelow, you can see some colors close to #1b462b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #1b4620\n``#1b4620` `rgb(27,70,32)``\n• #1b4624\n``#1b4624` `rgb(27,70,36)``\n• #1b4627\n``#1b4627` `rgb(27,70,39)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #1b462f\n``#1b462f` `rgb(27,70,47)``\n• #1b4632\n``#1b4632` `rgb(27,70,50)``\n• #1b4636\n``#1b4636` `rgb(27,70,54)``\nSimilar Colors\n\n# #1b462b Preview\n\nThis text has a font color of #1b462b.\n\n``<span style=\"color:#1b462b;\">Text here</span>``\n#1b462b background color\n\nThis paragraph has a background color of #1b462b.\n\n``<p style=\"background-color:#1b462b;\">Content here</p>``\n#1b462b border color\n\nThis element has a border color of #1b462b.\n\n``<div style=\"border:1px solid #1b462b;\">Content here</div>``\nCSS codes\n``.text {color:#1b462b;}``\n``.background {background-color:#1b462b;}``\n``.border {border:1px solid #1b462b;}``\n\n# Shades and Tints of #1b462b\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, #050d08 is the darkest color, while #fefffe is the lightest one.\n\n• #050d08\n``#050d08` `rgb(5,13,8)``\n• #0b1c11\n``#0b1c11` `rgb(11,28,17)``\n• #102a1a\n``#102a1a` `rgb(16,42,26)``\n• #163822\n``#163822` `rgb(22,56,34)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #205434\n``#205434` `rgb(32,84,52)``\n• #26623c\n``#26623c` `rgb(38,98,60)``\n• #2b7045\n``#2b7045` `rgb(43,112,69)``\n• #317f4e\n``#317f4e` `rgb(49,127,78)``\n• #368d56\n``#368d56` `rgb(54,141,86)``\n• #3c9b5f\n``#3c9b5f` `rgb(60,155,95)``\n• #41a968\n``#41a968` `rgb(65,169,104)``\n• #47b771\n``#47b771` `rgb(71,183,113)``\n• #54bd7b\n``#54bd7b` `rgb(84,189,123)``\n• #63c386\n``#63c386` `rgb(99,195,134)``\n• #71c891\n``#71c891` `rgb(113,200,145)``\n• #7fce9c\n``#7fce9c` `rgb(127,206,156)``\n• #8dd3a7\n``#8dd3a7` `rgb(141,211,167)``\n• #9bd8b2\n``#9bd8b2` `rgb(155,216,178)``\n• #a9debd\n``#a9debd` `rgb(169,222,189)``\n• #b7e3c8\n``#b7e3c8` `rgb(183,227,200)``\n• #c6e9d3\n``#c6e9d3` `rgb(198,233,211)``\n• #d4eede\n``#d4eede` `rgb(212,238,222)``\n• #e2f4e9\n``#e2f4e9` `rgb(226,244,233)``\n• #f0f9f3\n``#f0f9f3` `rgb(240,249,243)``\n• #fefffe\n``#fefffe` `rgb(254,255,254)``\nTint Color Variation\n\n# Tones of #1b462b\n\nA tone is produced by adding gray to any pure hue. In this case, #2e3330 is the less saturated color, while #016024 is the most saturated one.\n\n• #2e3330\n``#2e3330` `rgb(46,51,48)``\n• #2a372f\n``#2a372f` `rgb(42,55,47)``\n• #263b2e\n``#263b2e` `rgb(38,59,46)``\n• #223f2d\n``#223f2d` `rgb(34,63,45)``\n• #1f422c\n``#1f422c` `rgb(31,66,44)``\n• #1b462b\n``#1b462b` `rgb(27,70,43)``\n• #174a2a\n``#174a2a` `rgb(23,74,42)``\n• #144d29\n``#144d29` `rgb(20,77,41)``\n• #105128\n``#105128` `rgb(16,81,40)``\n• #0c5527\n``#0c5527` `rgb(12,85,39)``\n• #085926\n``#085926` `rgb(8,89,38)``\n• #055c25\n``#055c25` `rgb(5,92,37)``\n• #016024\n``#016024` `rgb(1,96,36)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #1b462b 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":0.55881214,"math_prob":0.68883073,"size":3680,"snap":"2021-43-2021-49","text_gpt3_token_len":1674,"char_repetition_ratio":0.12350381,"word_repetition_ratio":0.011029412,"special_character_ratio":0.5597826,"punctuation_ratio":0.23555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99081784,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T21:24:56Z\",\"WARC-Record-ID\":\"<urn:uuid:ffd899f9-72e3-4779-850d-497c838d29bf>\",\"Content-Length\":\"36113\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b939275-a191-48d9-87aa-7a45d3d4262a>\",\"WARC-Concurrent-To\":\"<urn:uuid:06bb07fb-58a2-4c6a-8eef-2991ea536240>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/1b462b\",\"WARC-Payload-Digest\":\"sha1:ZBMJCAQ4U4LNWOAKTWNW72BG5SNRZQ3W\",\"WARC-Block-Digest\":\"sha1:BPY5T3DJAWSP64XRULTTHHSCTKT63F42\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363598.57_warc_CC-MAIN-20211208205849-20211208235849-00605.warc.gz\"}"}
https://tex.stackexchange.com/questions/54032/text-following-subequations-is-slightly-indented-if-a-label-is-used	[ "Text following subequations is slightly indented if a label is used\n\nThe following minimal code causes a small indentation at the beginning of the line of text, smaller than the usual indentation at the start of a paragraph. It disappears if the label in subequations is removed.\n\n\\documentclass{article}\n\\usepackage{mathtools}\n\n\\begin{document}\n\\begin{subequations}\n\\begin{gather}\na = b\\\\\nb = c\n\\end{gather}\n\\label{eqn:test}\n\\end{subequations}\nLorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua \\ref{eqn:test}.\n\\end{document}\n\nI haven't noticed this effect in any other maths environments - is this a bug in subequations?\n\nEDIT: replace gather* with gather\n\n• The * in gather was a mistake. It was in the minimal test code – mjr May 2 '12 at 0:16\n\nsubequations only plays around with the equation number and does nothing else. Therefore, placing it as the surrounding environment for gather actually sets it in text mode and leaves a spurious space if you don't end a line with %.\n\nSolution: Insert % at the end of the \\label: \\label{...}%\n\n• All the answers were essentially the same, except you called my attention to the fact about subequations and it working in text mode. Thanks – mjr May 2 '12 at 11:20\n\nPut a % after \\label{eqn:test}.\n\n\\begin{subequations}\n\\begin{gather}\na = b\\\\\nb = c\n\\end{gather}\n\\label{eqn:test}% %%<---here\n\\end{subequations}\nLorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua \\ref{eqn:test}.\n\nActually the label should be written before the begin{gather}. Then the Lorem... will start as a new para as in\n\n\\documentclass{article}\n\\usepackage{mathtools}\n\\begin{document}\n\\begin{subequations}\n\\label{eqn:test}%\n\\begin{gather}\na = b\\\\\nb = c\n\\end{gather}\n\\end{subequations}\nLorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua \\ref{eqn:test}.\n\\end{document}\n\nNote: Using \\label to a un-numbered equations doesn't mean anything (starred version does not produce equation number and hence referring to that equation!.....). Use gather instead.\n\nThe \\label is to subequations, so\n\n\\begin{subequations}\\label{eqn:test}\n\\begin{gather}\na = b\\\\\nb = c\n\\end{gather}\n\\end{subequations}\n\nwon't have spurious spaces. Notice that using subequations with gather is meaningless: there will be no equation number to reference. You probably mean gather, without *, so each equation in the alignment will receive a subnumber." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5793618,"math_prob":0.987002,"size":658,"snap":"2019-43-2019-47","text_gpt3_token_len":163,"char_repetition_ratio":0.12691131,"word_repetition_ratio":0.0,"special_character_ratio":0.21276596,"punctuation_ratio":0.08928572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99619204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T10:31:37Z\",\"WARC-Record-ID\":\"<urn:uuid:35181de6-f002-4944-9fae-03d6ec097a0b>\",\"Content-Length\":\"151940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c04a17b-abac-4079-ae09-55574786df3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:a10a53d7-72cd-4a9c-be5b-0ad5e01914f6>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/54032/text-following-subequations-is-slightly-indented-if-a-label-is-used\",\"WARC-Payload-Digest\":\"sha1:BDUYKHYMKTYIH4SEH3OIXGJWHQYWMENI\",\"WARC-Block-Digest\":\"sha1:SUCUTIIH7EJBKU2BN2TSTCVMFKI72DKT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986666959.47_warc_CC-MAIN-20191016090425-20191016113925-00320.warc.gz\"}"}
https://math.stackexchange.com/questions/1733746/faithful-character-of-degree-less-than-p-gives-abelian-p-sylow-groups	[ "# Faithful character of degree less than p gives abelian p-Sylow groups.\n\nI am trying to prove:\n\nSuppose $p$ divides $\|G\|$ and let $\\chi \\in Irr(G)$ if $\\chi$ is faithful and its degree is less than $p$ then any $p$-Sylow subgroup of $G$ is abelian.\n\nI have tried to decompose the restriction of $\\chi$ to $P$, a $p$-group, into irreducibles and I guess I should try to prove that this gives me all one dimensional constituents, but I have no idea how to proceed.\n\nAny hints, links to good sources or other help is much appreciated.\n\nThanks.\n\n• It's restriction to a Sylow $p$-subgroup $P$ is faithful and must be a sum of linear characters, so $P$ is abelian. (Remember that the degree of an irreducible character of $P$ must divide $\|P\|$.) – Derek Holt Apr 8 '16 at 19:18\n• That's a really nice quick little argument. Much obliged! (I'll accept if you choose to repost as an answer.) – Winston Apr 8 '16 at 20:02\n• @DerekHolt Sounds like an answer, not a comment. – zibadawa timmy Apr 8 '16 at 20:02\n\n## 1 Answer\n\nOK, I'll make it an answer. The restriction $\\chi_P$ of $\\chi$ to $P \\in {\\rm Sy}_p(G)$ decomposes into irreducible characters of $P$. Since the degree of any such character divides $\|P\|$, it must be power of $p$. But $\\deg(\\chi) < p$, so all of the irreducible constituents of $\\chi_P$ have degree $1$. Since $\\chi$ and hence $\\chi_P$ is faithful, so $P$ must be abelian.\n\nBy the way, there is no need to assume that $\\chi$ is irreducible, just that it is faithful." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93703943,"math_prob":0.9944254,"size":463,"snap":"2019-13-2019-22","text_gpt3_token_len":127,"char_repetition_ratio":0.100217864,"word_repetition_ratio":0.0,"special_character_ratio":0.26565874,"punctuation_ratio":0.0927835,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999554,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T12:58:55Z\",\"WARC-Record-ID\":\"<urn:uuid:cee4dc87-612f-4dd1-b6a3-9227e8e414e6>\",\"Content-Length\":\"133418\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2507e0b3-5b83-417b-b6ff-3a9a68643e93>\",\"WARC-Concurrent-To\":\"<urn:uuid:453f6f6d-a6d3-450c-9aee-4c2f953a4e5b>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1733746/faithful-character-of-degree-less-than-p-gives-abelian-p-sylow-groups\",\"WARC-Payload-Digest\":\"sha1:VRKHOM4W7FSRIZWND4DQ5ZDNHSPOX4V7\",\"WARC-Block-Digest\":\"sha1:MUJMDMXFPH46PQGFWEAHLS7XEGTFKPJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259177.93_warc_CC-MAIN-20190526125236-20190526151236-00379.warc.gz\"}"}
https://imath.co.za/grade12e/nr1207graph.html	[ "#### Different kinds of graphs.\n\n1.\nThe diagrams below show the graphs of variables a and b. Variable a is plotted on the\nhorisontal axis, the x-axis, and b on the vertical axis, the y-axis.\nWhat is the relationship between a and b? Give reasons. Use the expressions directly\nproportional, inversely proportional or unknown to describe the relationship between a and b.\n1.1\n1.2\n1.3", null, "", null, "", null, "1.4\n1.5\n1.6", null, "", null, "", null, "1.7\n1.8\n1.9", null, "", null, "", null, "1.10\n1.11\n1.12", null, "", null, "", null, "2.\nThe table below shows the cost to manufacture n articles:\n Number of articles, n 1 2 5 6 8 Cost 4 8 20 24 32\n2.1\nWill the cost to manufacture n articles be directly proportional to n, the number of articles?\nExplain.\n2.2\nWhat shape will the graph have if number of articles is plotted on the x-axis and cost\n2.3\nDraw the graph of number of articles on the x-axis and cost on the y-axis.\n2.4\nDoes the graph have the shape which you predicted in 2.2?\n2.5\nRead from the graph the following values and clearly mark the point where you took the reading\nwith the letter in brackets:\n2.5.1\ncost to manufacture 4 articles (P).\n2.5.2\nthe number of articles that can be manufactured if the cost is R26. (Q)\n2.6\n3.\nThe table below shows the number of hours that a number of men must work to complete a task:\n Number of men 1 2 3 6 16 24 48 Number of hours 48 24 16 8 3 2 1\n3.1\nIs the number of hours directly proportional to the number of men? Explain.\n3.2\nWhat shape will the graph of the data above have if number of men is plotted on the x-axis and\nnumber of hours is plotted on the y-axis? Explain.\n3.3\nDraw the graph with number of men on the x-axis and the number of hours on the y-axis.\n3.4\nDoes the graph have the predicted shape?\n3.5\nRead the requested values from the graph and clearly mark the point where you made the reading.\n3.5.1\nHow many hours must 5 men work to complete the task? Is it a suitable answer? Explain.\n3.5.2\nHow many men must work for 20 hours to complete the task? Is it a suitable answer? Explain.\n3.6\nWhat is the greatest number of hours needed to complete the task if everybody works at the\nsame rate? Explain.\n3.7\nCan the work be completed in 15 minutes? How many men will be needed? Is it a suitable\n4.\nIn an experiment a student obtains the following values:\n a 0 1 2 3 4 8 b 0 3 6,05 7,9 12,5 23,9\n4.1\nCan the student claim that b is directly propotional to a? Explain.\n4.2\nWrite down the relationship between a and b in words and in symbols.\n4.3\nWhat shape will the graph of a, on the x-axis, and b, on the y-axis, have? Explain.\n4.4\nDraw the graph of a, on the x-axis, and b, on the y-axis.\n4.5\nRead from the graph the following values and mark the point where you take the reading with the\nletter in brackets:\n4.5.1\nthe value of b if a = 6 (P)\n4.5.2\nthe value of a if b = 10,5 (Q)\n4.6\nUse the formula that you wrote down in 4.2 to calculate the values of a and b in 4.5.1 and 4.5.2\nHow do the calculated values correspond with the values read from the graph?\n5.\nThe table below contains values for a and b.\n a 1 4 8 16 b 16 4 2 1\n5.1\n5.2\nDescribe the relationship between a and b in words and in symbols.\n5.3\nWhat shape will the graph of a against b have? Motivate your answer.\n5.4\nDraw the graph of a, on the x-axis, and b, on the y-axis.\n5.5\nRead from the graph the following values and mark the point where you take the reading with the\nletter in brackets:\n5.5.1\nthe value of b if a = 8 (P)\n5.5.2\nthe value of a if b = 10 (Q)\n5.6\nUse the formula that you wrote down in 5.2 to calculate the values of a and b in 5.5.1 and 5.5.2\nHow do the calculated values correspond with the values read from the graph?\n6.\nValues for a and b are given in the table below.\n a 0 1 2 4 b 0 1 4 16\n6.1\n6.2\nDescribe the relationship between a and b in words and in symbols.\n6.3\nWhat will the shape of the graph of a against b be? Motivate your answer.\n6.4\nDraw the graph of a, on the x-axis, and b, on the y-axis.\n6.5\nRead from the graph the following values and mark the point where you take the reading with the\nletter in brackets:\n6.5.1\nthe value of b if a = 1,5 (Q)\n6.5.2\nthe value of a if b = 9 (P)\n6.6\nUse the formula that you wrote down in 6.2 to calculate the values of a and b in 6.5.1 and 6.5.2\nHow do the calculated values correspond with the values read from the graph?\n7.\nA farmer wants to make a coop with an area of 36 m². Please help the farmer.\n7.1\nWrite down the relationship between the length and breadth of the coop in words and in symbols.\n7.2\nWhat shape will the graph of length, on the x-axis, and breadth, on the y-axis, have? Explain.\n7.3\nDraw up a table containing suitable possible values for the length and breadth of the coop.\n7.4\nDraw the graph, f, of length on the x-axis and breadth on the y-axis.\n7.5\nRead the following measurements from the graph and clearly mark the point where you take\nthe reading with the letter in brackets:\n7.5.1\nthe breadth if the length is 3 m (P).\n7.5.2\nthe length if the breadth is 20 m. (Q)\n7.6\nDraw the line g: y = x and let f and g intersect at R. Write down the co-ordinates of R.\n7.7\nWhat is special about point R?\n7.8\nThe farmer wants the coop to be a square. Read from the graph the length of the sides of the square\nand mark the point where the readings are taken.\n7.9\nWhat length of wire netting will be necessary to fence in the coop?\n7.10\nThe farmer has 24,4 m of wire netting to use. What will the measurements of the coop be?" ]	[ null, "https://imath.co.za/grafieke12/e12graf04.gif", null, "https://imath.co.za/grafieke12/e12graf05.gif", null, "https://imath.co.za/grafieke12/e12graf06.gif", null, "https://imath.co.za/grafieke12/e12graf07.gif", null, "https://imath.co.za/grafieke12/e12graf08.gif", null, "https://imath.co.za/grafieke12/e12graf09.gif", null, "https://imath.co.za/grafieke12/e12graf10.gif", null, "https://imath.co.za/grafieke12/e12graf11.gif", null, "https://imath.co.za/grafieke12/e12graf12.gif", null, "https://imath.co.za/grafieke12/e12graf20.gif", null, "https://imath.co.za/grafieke12/e12graf21.gif", null, "https://imath.co.za/grafieke12/e12graf22.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8707218,"math_prob":0.978705,"size":5607,"snap":"2021-43-2021-49","text_gpt3_token_len":1699,"char_repetition_ratio":0.14991969,"word_repetition_ratio":0.25065964,"special_character_ratio":0.30586767,"punctuation_ratio":0.15261324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99765307,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,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\":\"2021-10-20T16:46:12Z\",\"WARC-Record-ID\":\"<urn:uuid:0e8a2ee2-73b6-4fc4-9688-8cb3ee7a5821>\",\"Content-Length\":\"25564\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88e38a45-cdf7-4d23-96b7-e11477817515>\",\"WARC-Concurrent-To\":\"<urn:uuid:e4831e3b-dccc-42f4-a466-d2384f8bbeaf>\",\"WARC-IP-Address\":\"197.221.17.210\",\"WARC-Target-URI\":\"https://imath.co.za/grade12e/nr1207graph.html\",\"WARC-Payload-Digest\":\"sha1:XVYWQJFN2Y4AL3FVQUODLZ6JLHG46JS7\",\"WARC-Block-Digest\":\"sha1:6B7HQTAM6UTBD55A7YCAHXYBM67UUS6T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585322.63_warc_CC-MAIN-20211020152307-20211020182307-00006.warc.gz\"}"}
http://git.sourceforge.jp/view?p=pf3gnuchains/gcc-fork.git;a=commitdiff;h=ad2e42a870731f72f4813b6c41b10c8b7e768272	[ "author sandra Sat, 8 May 2010 15:53:59 +0000 (15:53 +0000) committer Masaki Muranaka Sun, 23 May 2010 05:03:46 +0000 (14:03 +0900)\nPR middle-end/28685\n\ngcc/\n* tree-ssa-reassoc.c (eliminate_redundant_comparison): New function.\n(optimize_ops_list): Call it.\n\ngcc/testsuite/\n* gcc.dg/pr28685-1.c: New file.\n\ngit-svn-id: svn+ssh://gcc.gnu.org/svn/gcc/trunk@159189 138bc75d-0d04-0410-961f-82ee72b054a4\n\n gcc/ChangeLog patch \| blob \| history gcc/testsuite/ChangeLog patch \| blob \| history gcc/tree-ssa-reassoc.c patch \| blob \| history\n\nindex cb1661c..28fd31f 100644 (file)\n@@ -1,3 +1,9 @@\n+2010-05-08 Sandra Loosemore <[email protected]>\n+\n+ PR middle-end/28685\n+ * tree-ssa-reassoc.c (eliminate_redundant_comparison): New function.\n+ (optimize_ops_list): Call it.\n+\n2010-05-08 Richard Guenther <[email protected]>\n\nPR tree-optimization/44030\nindex 66630e8..38501a7 100644 (file)\n@@ -1,3 +1,8 @@\n+2010-05-08 Sandra Loosemore <[email protected]>\n+\n+ PR middle-end/28685\n+ * gcc.dg/pr28685-1.c: New file.\n+\n2010-05-08 Richard Guenther <[email protected]>\n\nPR tree-optimization/44030\nindex aa08085..6bdb57e 100644 (file)\n@@ -1215,6 +1215,113 @@ undistribute_ops_list (enum tree_code opcode,\nreturn changed;\n}\n\n+/* If OPCODE is BIT_IOR_EXPR or BIT_AND_EXPR and CURR is a comparison\n+ expression, examine the other OPS to see if any of them are comparisons\n+ of the same values, which we may be able to combine or eliminate.\n+ For example, we can rewrite (a < b) \| (a == b) as (a <= b). /\n+\n+static bool\n+eliminate_redundant_comparison (enum tree_code opcode,\n+ VEC (operand_entry_t, heap) ops,\n+ unsigned int currindex,\n+ operand_entry_t curr)\n+{\n+ tree op1, op2;\n+ enum tree_code lcode, rcode;\n+ gimple def1, def2;\n+ int i;\n+ operand_entry_t oe;\n+\n+ if (opcode != BIT_IOR_EXPR && opcode != BIT_AND_EXPR)\n+ return false;\n+\n+ / Check that CURR is a comparison. /\n+ if (TREE_CODE (curr->op) != SSA_NAME)\n+ return false;\n+ def1 = SSA_NAME_DEF_STMT (curr->op);\n+ if (!is_gimple_assign (def1))\n+ return false;\n+ lcode = gimple_assign_rhs_code (def1);\n+ if (TREE_CODE_CLASS (lcode) != tcc_comparison)\n+ return false;\n+ op1 = gimple_assign_rhs1 (def1);\n+ op2 = gimple_assign_rhs2 (def1);\n+\n+ / Now look for a similar comparison in the remaining OPS. /\n+ for (i = currindex + 1;\n+ VEC_iterate (operand_entry_t, ops, i, oe);\n+ i++)\n+ {\n+ tree t;\n+\n+ if (TREE_CODE (oe->op) != SSA_NAME)\n+ continue;\n+ def2 = SSA_NAME_DEF_STMT (oe->op);\n+ if (!is_gimple_assign (def2))\n+ continue;\n+ rcode = gimple_assign_rhs_code (def2);\n+ if (TREE_CODE_CLASS (rcode) != tcc_comparison)\n+ continue;\n+ if (operand_equal_p (op1, gimple_assign_rhs1 (def2), 0)\n+ && operand_equal_p (op2, gimple_assign_rhs2 (def2), 0))\n+ ;\n+ else if (operand_equal_p (op1, gimple_assign_rhs2 (def2), 0)\n+ && operand_equal_p (op2, gimple_assign_rhs1 (def2), 0))\n+ rcode = swap_tree_comparison (rcode);\n+ else\n+ continue;\n+\n+ /* If we got here, we have a match. See if we can combine the\n+ two comparisons. /\n+ t = combine_comparisons (UNKNOWN_LOCATION,\n+ (opcode == BIT_IOR_EXPR\n+ ? TRUTH_OR_EXPR : TRUTH_AND_EXPR),\n+ lcode, rcode, TREE_TYPE (curr->op), op1, op2);\n+ if (!t)\n+ continue;\n+ if (dump_file && (dump_flags & TDF_DETAILS))\n+ {\n+ fprintf (dump_file, \"Equivalence: \");\n+ print_generic_expr (dump_file, curr->op, 0);\n+ fprintf (dump_file, \" %s \", op_symbol_code (opcode));\n+ print_generic_expr (dump_file, oe->op, 0);\n+ fprintf (dump_file, \" -> \");\n+ print_generic_expr (dump_file, t, 0);\n+ fprintf (dump_file, \"\\n\");\n+ }\n+\n+ / Now we can delete oe, as it has been subsumed by the new combined\n+ expression t. /\n+ VEC_ordered_remove (operand_entry_t, ops, i);\n+ reassociate_stats.ops_eliminated ++;\n+\n+ /* If t is the same as curr->op, we're done. Otherwise we must\n+ replace curr->op with t. Special case is if we got a constant\n+ back, in which case we add it to the end instead of in place of\n+ the current entry. /\n+ if (TREE_CODE (t) == INTEGER_CST)\n+ {\n+ VEC_ordered_remove (operand_entry_t, ops, currindex);\n+ add_to_ops_vec (ops, t);\n+ }\n+ else if (TREE_CODE (t) != lcode)\n+ {\n+ tree tmpvar;\n+ gimple sum;\n+ enum tree_code subcode;\n+ tree newop1;\n+ tree newop2;\n+ tmpvar = create_tmp_var (TREE_TYPE (t), NULL);\n+ add_referenced_var (tmpvar);\n+ extract_ops_from_tree (t, &subcode, &newop1, &newop2);\n+ sum = build_and_add_sum (tmpvar, newop1, newop2, subcode);\n+ curr->op = gimple_get_lhs (sum);\n+ }\n+ return true;\n+ }\n+\n+ return false;\n+}\n\n/* Perform various identities and other optimizations on the list of\noperand entries, stored in OPS. The tree code for the binary\n@@ -1276,7 +1383,8 @@ optimize_ops_list (enum tree_code opcode,\nif (eliminate_not_pairs (opcode, ops, i, oe))\nreturn;\nif (eliminate_duplicate_pair (opcode, ops, &done, i, oe, oelast)\n- \|\| (!done && eliminate_plus_minus_pair (opcode, ops, i, oe)))\n+ \|\| (!done && eliminate_plus_minus_pair (opcode, ops, i, oe))\n+ \|\| (!done && eliminate_redundant_comparison (opcode, ops, i, oe)))\n{\nif (done)\nreturn;\nAbout OSDN\nFind Software\nDevelop Software" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5905425,"math_prob":0.92562604,"size":3691,"snap":"2021-21-2021-25","text_gpt3_token_len":1146,"char_repetition_ratio":0.12584758,"word_repetition_ratio":0.020066889,"special_character_ratio":0.36060688,"punctuation_ratio":0.2585034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96233994,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T14:52:14Z\",\"WARC-Record-ID\":\"<urn:uuid:a94a901d-70b9-48af-86ae-d9422f4f573c>\",\"Content-Length\":\"32185\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:723c54ac-34be-4ef5-b590-9127c52e60ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c7e0761-e656-4a61-846c-5088389b5456>\",\"WARC-IP-Address\":\"44.240.209.230\",\"WARC-Target-URI\":\"http://git.sourceforge.jp/view?p=pf3gnuchains/gcc-fork.git;a=commitdiff;h=ad2e42a870731f72f4813b6c41b10c8b7e768272\",\"WARC-Payload-Digest\":\"sha1:QKMFZOAPH7CF6IPN2ZKJWQFYXHAA2DAY\",\"WARC-Block-Digest\":\"sha1:BD2TVKJVFGYRNTSD6AIQQWZVIYJYHRDK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487584018.1_warc_CC-MAIN-20210612132637-20210612162637-00134.warc.gz\"}"}
https://riptutorial.com/asp-net-mvc/example/2289/basic-syntax	[ "# asp.net-mvc Basic Syntax\n\n## Example\n\nRazor code can be inserted anywhere within HTML code. Razor code blocks are enclosed in `@{ ... }`. Inline variable and functions start with `@`. Code inside the Razor brackets follow the normal C# or VB rules.\n\nSingle line statement:\n\n``````@{ var firstNumber = 1; }\n``````\n\nMulti-line code block:\n\n``````@{\nvar secondNumber = 2;\nvar total = firstNumber + secondNumber;\n}\n``````\n\nUsing a variable inline:\n\n``````<h1>The total count is @total</h1>\n``````\n\nUsing a variable inline explicitly:\n\n``````<h2>Item@(item.Id)</h2>\n``````\n\nFor this particular example we will not be able to use the implicit syntax because `[email protected]` looks like an email and will be rendered as such by Razor.\n\nEnclose code inside control flow statements:\n\n``````<h1>Start with some HTML code</h1>\n\n@for (int i = 0; i < total; i++){\nConsole.Write(i);\n}\n\n<p>Mix in some HTML code for fun!</p>\n\n@if (total > 3)\n{\nConsole.Write(\"The total is greater than 3\");\n}\nelse\n{\nConsole.Write(\"The total is less than 3\");\n}\n``````\n\nThis same syntax would be used for all statements such as `for`, `foreach`, `while`, `if`, `switch`, etc.\n\n``````@if (total > 3)\n{\nif(total == 10)\n{\nConsole.Write(\"The total is 10\")\n}\n}\n``````\n\nNote that you don't need to type the `@` at the second `if`. After code you can just type other code behind the existing code.\n\nIf you want to add code after a HTML element you do need to type a `@`.", null, "PDF - Download asp.net-mvc for free" ]	[ null, "https://riptutorial.com/Images/icon-pdf-2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.75700283,"math_prob":0.9001495,"size":1340,"snap":"2020-24-2020-29","text_gpt3_token_len":353,"char_repetition_ratio":0.11377245,"word_repetition_ratio":0.008264462,"special_character_ratio":0.29328358,"punctuation_ratio":0.14338236,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97791135,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T15:37:39Z\",\"WARC-Record-ID\":\"<urn:uuid:7105a08a-08f2-464e-97d3-3f9bfd622675>\",\"Content-Length\":\"42890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f01dced4-57fa-4f91-92e5-0ee8d11086b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:12755813-78e3-4dfb-9ae2-8969967cffb2>\",\"WARC-IP-Address\":\"40.83.160.29\",\"WARC-Target-URI\":\"https://riptutorial.com/asp-net-mvc/example/2289/basic-syntax\",\"WARC-Payload-Digest\":\"sha1:OQYRSGQH3NFG2MOPBIYO5FEMCYLLKX5Z\",\"WARC-Block-Digest\":\"sha1:PAM7ALBQ4VNH6DUW3HX7OXNY2QPFBRLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655882634.5_warc_CC-MAIN-20200703153451-20200703183451-00554.warc.gz\"}"}
https://socratic.org/questions/57cb071c7c0149358fc7f631	[ "Question 7f631\n\nJan 13, 2017\n\n$- \\text{33.4g}$\n\nExplanation:\n\nStart by converting the speed of the car from kilometers per hour to meters per second\n\n85 color(red)(cancel(color(black)(\"km\")))/color(red)(cancel(color(black)(\"h\"))) * (10^3\"m\")/(1color(red)(cancel(color(black)(\"km\")))) * (1color(red)(cancel(color(black)(\"h\"))))/\"3600 s\" = \"23.61 m s\"^(-1)\n\nNow, notice that the car goes from a speed of ${\\text{23.61 m s}}^{- 1}$ to rest, i.e. to a speed of ${\\text{0 m s}}^{- 1}$, so right from the start, you should expect the average acceleration to come out negative.\n\nIn other words, the car is decelerating, which means that its acceleration is acting in the opposite direction to its direction of movement.\n\nAnother thing to notice here is that the speed of the car is decreasing significantly over a very short distance. From the moment of impact, it takes only\n\n$d = \\text{0.85 m}$\n\nfor the speed of the car to decrease by ${\\text{23.61 m s}}^{- 1}$, This tells you that the acceleration of the car will have a very high absolute value, i.e. its magnitude will be very high.\n\nYour tool of choice here will be this equation\n\n$\\textcolor{b l u e}{\\underline{\\textcolor{b l a c k}{{v}_{f}^{2} = {v}_{0}^{2} + 2 \\cdot a \\cdot d}}}$\n\nHere\n\n• ${v}_{f}$ is the final speed of the car\n• ${v}_{0}$ is its initial speed, i.e. its speed before the impact\n• $a$ is its acceleration\n• $d$ is the distance covered by the driver after the impact\n\nRearrange the above equation to solve for $a$\n\n$2 \\cdot a \\cdot d = {v}_{f}^{2} - {v}_{0}^{2} \\implies a = \\frac{{v}_{f}^{2} - {v}_{0}^{2}}{2 \\cdot d}$\n\nPlug in your values to find\n\na = ( 0^2 \"m\"^color(red)(cancel(color(black)(2))) \"s\"^(-2) - 23.61^2 \"m\"^color(red)(cancel(color(black)(2))) \"s\"^(-2))/(2 * 0.85 color(red)(cancel(color(black)(\"m\")))) = -\"327.9 m s\"^(-2)#\n\nAs you can see, the acceleration is indeed negative. Convert the acceleration to $\\text{g}$'s to get\n\n$- 327.9 \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{\\text{m s\"^(-2)))) * \"1.00g\"/(9.80color(red)(cancel(color(black)(\"m s\"^(-2))))) = color(darkgreen)(ul(color(black)(-\"33.4g}}}}$\n\nI'll leave the answer rounded to three sig figs." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82680744,"math_prob":0.999621,"size":1700,"snap":"2019-43-2019-47","text_gpt3_token_len":527,"char_repetition_ratio":0.19575472,"word_repetition_ratio":0.012295082,"special_character_ratio":0.35470587,"punctuation_ratio":0.068862274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994623,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T19:59:30Z\",\"WARC-Record-ID\":\"<urn:uuid:6c7b8ad5-b411-45e4-9506-16fce524c16f>\",\"Content-Length\":\"38620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:858408cd-db5a-4377-9eb4-d0a4a4c2afdc>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a68e971-5cc6-4904-b97c-eac457f1fc99>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/57cb071c7c0149358fc7f631\",\"WARC-Payload-Digest\":\"sha1:QSBZD5YDJODT6HU2Y76RAQGYAWOPNLV3\",\"WARC-Block-Digest\":\"sha1:ZTC3PJLD634XQCHZZLF5G5MYDZC6J5IE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986697760.44_warc_CC-MAIN-20191019191828-20191019215328-00033.warc.gz\"}"}
https://socratic.org/questions/how-do-you-find-the-definite-integral-of-int-x-2-2-x-1-from-0-2#361532	[ "# How do you find the definite integral of int (x^2-2)/(x+1) from [0,2]?\n\nJan 7, 2017\n\n$- \\ln 3$\n\n#### Explanation:\n\nStart by dividing the numerator by the denominator using long or synthetic division.", null, "Thus:\n\n${\\int}_{0}^{2} \\frac{{x}^{2} - 2}{x + 1} \\mathrm{dx} = {\\int}_{0}^{2} x - 1 + - \\frac{1}{x + 1} \\mathrm{dx}$\n\nWe can integrate using the rules $\\int \\left(\\frac{1}{x}\\right) \\mathrm{dx} = \\ln \| x \| + C$ and $\\int \\left({x}^{n}\\right) \\mathrm{dx} = {x}^{n + 1} / \\left(n + 1\\right) + C$.\n\n$= {\\left[\\frac{1}{2} {x}^{2} - x - \\ln \| x + 1 \|\\right]}_{0}^{2}$\n\nEvaluate using ${\\int}_{a}^{b} F \\left(x\\right) = f \\left(b\\right) - f \\left(a\\right)$, where $f ' \\left(x\\right) = F \\left(x\\right)$.\n\n$= \\frac{1}{2} {\\left(2\\right)}^{2} - 2 - \\ln \| 3 \| - \\left(\\frac{1}{2} {\\left(0\\right)}^{2} - 0 - \\ln \| 0 + 1 \|\\right)$\n\n$= \\frac{1}{2} \\left(4\\right) - 2 - \\ln \| 3 \| - 0$\n\n$= - \\ln 3$\n\nIn celebration of this being the 2000th answer I ever wrote for socratic, I've included a whole bunch of practice exercises for your improvement\n\nPractice exercises\n\n1. Evaluate each definite integral. Round answers to the nearest integer.\n\na) ${\\int}_{1}^{4} \\frac{{x}^{3} + 7 x + 14}{x + 2} \\mathrm{dx}$\n\nb) ${\\int}_{7}^{15} \\frac{2 {x}^{4} - 18 {x}^{3} + 2 {x}^{2} - 5 x + 1}{2 x + 1}$\n\nc) ${\\int}_{3}^{6} \\frac{5 {x}^{5} + 2 {x}^{2} - 8}{2 \\left(x + 4\\right)}$\n\nBonus\nHint: Use substitution\nb) ${\\int}_{e}^{e + 3} \\frac{{e}^{2 x} - {e}^{x}}{\\sqrt{{e}^{x}}} \\mathrm{dx}$\n\nHopefully this helps!\n\n$1. 33$\n$2. 2920$\n$3. 2102$\nbonus: $3474$", null, "" ]	[ null, "https://useruploads.socratic.org/8We6lFNhRgq50UkjQzXg_MSP641b593e8a0eb27gg70000396aefa89ci9f3i5.gif", null, "https://useruploads.socratic.org/JPI98gxpQG2smaf61FQP_2000.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7987694,"math_prob":1.0000094,"size":622,"snap":"2022-27-2022-33","text_gpt3_token_len":141,"char_repetition_ratio":0.116504855,"word_repetition_ratio":0.0,"special_character_ratio":0.22186495,"punctuation_ratio":0.10619469,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999106,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T02:42:03Z\",\"WARC-Record-ID\":\"<urn:uuid:8c7e7f42-d264-4727-9e4a-0bf1730f0730>\",\"Content-Length\":\"35464\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:34e9fb29-87c4-43e2-8d63-08ea14746886>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf0eb0b3-474a-49a3-b791-48a9a9189a0d>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-find-the-definite-integral-of-int-x-2-2-x-1-from-0-2#361532\",\"WARC-Payload-Digest\":\"sha1:HKRPIU26CAUC5SWFVGB2PPHJTJSF26F4\",\"WARC-Block-Digest\":\"sha1:3NFKKMCNYHVNJWJMKZNQMNRIXR74BBA6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103983398.56_warc_CC-MAIN-20220702010252-20220702040252-00155.warc.gz\"}"}
https://thebuckwheater.com/33-excel-chart-x-axis-text/	[ "By \| June 27, 2019\n\nLuxury 33 Illustration Excel Chart X Axis Text\nscatter chart with one text non numerical axis super user excel scatter plots cannot take names instead of values on their x axis they assume a number series for the x axis if you want to replicate the effect of a scatter plot but use named x axis values how to display text labels in the x axis of scatter chart 1 select the data you use and click insert insert line area chart line with markers to select a line chart see screenshot 2 then right click on the line in the chart to select format data s change axis labels in a chart office support your chart uses text from its source data for these axis labels dont confuse the horizontal axis labels qtr 1 qtr 2 qtr 3 and qtr 4 as shown below with the legend labels below them how to wrap x axis labels in an excel chart your best step 4 to show all parts of the names please double click on x axis then in the alignment change the custom angle to 0 degrees and text direction as horizontal many users may not use this steps graph how to create a text based y axis on excel chart the xy scatter series is on the secondary axis also not too useful i formatted the xy series and changed the axis to primary top left chart below the 0 5 values in column g make the points line adding text axis labels to excel scatter chart super user i have a spreadsheet with data like this label data pa 1 1 pa 2 2 pb 1 3 with hundreds of rows i want to create a scatter graph where the x axis contains the names in the label column how to wrap x axis labels in a chart in excel extendoffice how to wrap x axis labels in a chart in excel when the chart area is not wide enough to show its x axis labels in excel all the axis labels will be rotated and slanted in excel change the scale of the horizontal category axis in a chart the horizontal category axis also known as the x axis of a chart displays text labels instead of numeric intervals and provides fewer scaling options than are available for a vertical value axis text labels on a vertical column chart in excel peltier in excel 2003 go to the chart menu choose chart options and check the category x axis checkmark now the chart has four axes we want the rating labels at the left side of the chart and we how to position excel chart x axis text position c i am developing a chart programmatically using c and excel interop dll i have to develop this kind of char here is the screen shot just see the arrow sign i want to put that text at the right hand Luxury 33 Illustration Excel Chart X Axis Text\n\nExcel Chart X Axis Text Excel Adding Text to Legend Presenting Data with\n\nExcel Chart X Axis Text Adding Colored Regions to Excel Charts Duke Libraries\n\nExcel Chart X Axis Text How to Add Axis Label to Chart In Excel\n\nExcel Chart X Axis Text Excel Chart Tips Danjharrington\n\nExcel Chart X Axis Text Excel 2007 Line Graph Change X Axis Values Excel 2007\n\nexcel chart order, excel chart guide, excel chart exploded pie, excel chart variable range, excel chart tutorial, excel chart change order, excel," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80301464,"math_prob":0.7226464,"size":3039,"snap":"2019-51-2020-05","text_gpt3_token_len":646,"char_repetition_ratio":0.19110379,"word_repetition_ratio":0.06324787,"special_character_ratio":0.20631787,"punctuation_ratio":0.011744967,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96221983,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T20:08:04Z\",\"WARC-Record-ID\":\"<urn:uuid:00fe28c0-aa0b-4d3b-82c5-032d1d340222>\",\"Content-Length\":\"66195\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cee80099-8391-4261-95a2-29d2e3a103dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:fff9ac9d-558e-4299-b3e5-678ed8629889>\",\"WARC-IP-Address\":\"104.28.2.231\",\"WARC-Target-URI\":\"https://thebuckwheater.com/33-excel-chart-x-axis-text/\",\"WARC-Payload-Digest\":\"sha1:NI6Y44OE34JH72DI4JHLF646YRF56W3D\",\"WARC-Block-Digest\":\"sha1:WTXNSFJ2ISMNULVXJWQJSA7B4WJKKNNW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251802249.87_warc_CC-MAIN-20200129194333-20200129223333-00329.warc.gz\"}"}
http://frinklang.org/fsp/colorize.fsp?f=SolarCooker.frink	[ "# SolarCooker.frink\n\n``` // Solar cooker design program f = eval[input[\"Enter focal length [4 feet]: \", \"4 feet\"]] length = eval[input[\"Enter length of reflective material, long axis [100 in]: \", \"100 in\"]] width = eval[input[\"Enter width of reflective material, short axis [36 in]: \", \"36 in\"]] longspars = eval[input[\"Enter number of long spars : \", 3]] shortspars = eval[input[\"Enter number of short spars : \", 7]] inset = eval[input[\"Enter inset [2 in]: \", \"2 in\"]] longdiff = (width - (2inset)) / (longspars - 1) shortdiff = (length - (2inset)) / (shortspars - 1) println[] println[\"\\$longspars long spars will be \" + (longdiff -> \"in\") + \" apart.\"] println[\"\\$shortspars short spars will be \" + (shortdiff -> \"in\") + \" apart.\"] b = 100 in println[\"Arc length is \" + format[arclen[b,f], \"in\", 2]] a = 36 in println[\"Arc length is \" + format[arclen[a,f], \"in\", 2]] // Max height maxHeight = parabolaHeight[length/2, width/2, f] println[\"Maximum height is \" + format[maxHeight, \"in\", 2]] maxHeight = parabolaHeight[length/2, 0 in, f] println[\"Centerline height is \" + format[maxHeight, \"in\", 2]] println[\"long\\tshort\\theight\"] wmin = shortspars mod 2 == 1 ? 0 in : longdiff lmin = longspars mod 2 == 1 ? 0 in : shortdiff println[\"\\nCutting short spars:\"] for w = wmin to b/2 step shortdiff { if w == 0 in print[\"x 1\"] else print[\"x 2\"] println[\" (length \" + format[a, \"in\", 2] + \"), \" + format[w, \"in\", 1] + \" from center\"] for l = lmin to a/2 step longdiff { print[format[l, \"in\", 1] + \"\\t\"] print[format[parabolaHeight[l,w,f], \"in\", 1]] println[] } println[] } println[\"\\nCutting long spars:\"] for l = lmin to a/2 step longdiff { if l == 0 in print[\"x 1\"] else print[\"x 2\"] println[\" (length \" + format[b, \"in\", 2] + \"), \" + format[l, \"in\", 1] + \" from center\"] for w = wmin to b/2 step shortdiff { print[format[w, \"in\", 1] + \"\\t\"] print[format[parabolaHeight[l,w,f], \"in\", 1]] println[] } println[] } // Find arclength of the curve over a box length b: // (b sqrt[16 + (b/f)^2])/8 + 2 f arccsch[4 f/b] arclen[b, f] := (b sqrt[16 + (b/f)^2])/8 + 2 f arccsch[4 f/b] // Parabola height // Equation for a parabola with focal length f is z = d^2/(4 f) // where d is the distance from the center and z is the height. parabolaHeight[x, y, f] := { d = sqrt[x^2 + y^2] return d^2/(4 f) } ```\n\nThis is a program written in the programming language Frink.\nFor more information, view the Frink Documentation or see More Sample Frink Programs.\n\nAlan Eliasen was born 18354 days, 1 hours, 9 minutes ago." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6052155,"math_prob":0.9916371,"size":2582,"snap":"2019-43-2019-47","text_gpt3_token_len":876,"char_repetition_ratio":0.1625291,"word_repetition_ratio":0.22624435,"special_character_ratio":0.37955073,"punctuation_ratio":0.14426878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9894267,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T01:03:27Z\",\"WARC-Record-ID\":\"<urn:uuid:5d0a2bf7-69ff-4f13-aa0b-30d4afeb87f3>\",\"Content-Length\":\"7226\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7856809a-5737-4bac-8d4c-1452a9f0646d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d84b42a-daa1-4f6e-9ab6-2786f9816c32>\",\"WARC-IP-Address\":\"97.118.48.61\",\"WARC-Target-URI\":\"http://frinklang.org/fsp/colorize.fsp?f=SolarCooker.frink\",\"WARC-Payload-Digest\":\"sha1:6MRQE4TJGF7ASQ72HTCOGB4VSD7ST7BP\",\"WARC-Block-Digest\":\"sha1:224T2BNYSFSB7WGLBB5EFMYQPN4XQAAI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670389.25_warc_CC-MAIN-20191120010059-20191120034059-00288.warc.gz\"}"}
https://win-vector.com/2014/09/01/what-is-a-win-vector/	[ "# What is a win vector?\n\nFrom time to time we are asked “what is the company name Win-Vector LLC referring to?” It is a cryptic pun trying to be an encoding of “we deliver victory.”\n\nThe story is an inside joke referring to something really only funny to one of the founders. But a joke that amuses the teller is always enjoyed by at least one person. Win-Vector LLC’s John Mount had the honor of co-authoring a 1997 paper titled “The Polytope of Win Vectors.” The paper title is obviously mathematical terms in an odd combination. However the telegraphic grammar is coincidentally similar to deliberately ungrammatical gamer slang such as “full of win” and “so much win.”", null, "If we treat “win” as a concrete noun (say something you can put in a sack) and “vector” in its non-mathematical sense (as an entity of infectious transmission) we have “Win-Vector LLC is an infectious delivery of victory.” I.e.: we deliver success to our clients. Of course, we have now attempt to explain a weak joke. It is not as grand as “winged victory,” but it does encode a positive company value: Win-Vector LLC delivers successful data science projects and training to clients.", null, "Winged Victory: from Wikipedia\n\nLet’s take this as an opportunity to describe what a win vector is.\n\nWe take the phrase “win vector” from a technical article titled “The Polytope of Win Vectors” by J.E. Bartles, J. Mount, and D.J.A. Welsh (Annals of Combinatorics I, 1997, pp. 1-15. The topic of this paper concerns the possible outcomes of game tournaments (or other things that can be expressed as tournaments). For example: we could have four teams (A,B,C, and D) scheduled to play each other a number of times, as indicated in the diagram below.", null, "This graph is just saying in the tournament: A will play B 5 times, B will not play C, and so on. We assume each game can end in a win for one team (given them 1 point), or a loss or tie (giving them zero points). We can record a summary of the tournament outcomes as a vector (vector now back to its mathematical sense) that just records how often each team won. For example the vector [10,1,1,0] is a win vector compatible with the above diagram (it encodes A winning all matches and D losing all matches). The vector [0,0,0,5] is not a valid win vector for the digram as D did not play 5 games (so can not have 5 wins). (The Win-Vector LLC logo is itself a stylized single game tournament diagram, with the directed arrow representing both victory and reminiscent of vectors in the mathematical sense.)", null, "The idea is that a win vector might be treated as a sufficient statistic for the tournament. Or more accurately the win vector may be all that is known about a previously run tournament. Such censored observations may be all that is possible in field biology where wins represent territory or offspring. The question is then: given knowledge of the tournament structure (the graph) and the summary of outcomes (the win vector) is there evidence one team is dominant, or are the effects random? So we have well-formed statistical questions about effect strength and significance.\n\nThe question of significance is: when we introduce a notion of effect strength how likely are we to see an effect of that size assuming identical players. For example if we make our notion of effect strength the maximum ratio of wins to plays seen in the win vector should we consider this evidence of a strong player, or is it to be expected by random fluctuation? We need to estimate how strong a conditioning effect our tournament constraints impose on unobserved outcomes (to determine if irregularities in distribution are from player strengths our tournament mis-design).\n\nRelating distributions of unobserved details to observed totals (or margins) is one of the most fundamental problems in statistics. We have written on it many times (two examples: Google ad market reporting and checking scientific claims). In all cases you would be better off with direct detailed observations (i.e. without the censorship); but often you have to work with the data you have instead of the experiment you would design.\n\nThe math is a little easier to explain for a related problem: working out the number of ways to fill in a matrix with non-negative integers to meet given row and column totals. I’ll move on to discuss this contingency table problem a bit.\n\nThe statistical ideas largely come from “Testing for Independence in a Two-Way Table: New Interpretations of the Chi-Square Statistic”, Persi Diaconis and Bradley Efron, Ann. Statist., Vol. 13, No. 3, 1985 pp. 845-874. A contingency table is a matrix of non-negative integers, and the statistical problem is relating known row and column totals to possible fill-ins. In this paper the authors criticize some of the standard significance tests (chi-square, Fisher’s exact test) and propose a parameterized family of tests that at the extreme end considers a null-model of uniform fill-ins (each possible fill in equally likely). Obviously a uniform model is very different than the more standard distributions which tend to have cell counts more highly concentrated around their means. But the idea is: this proposed test takes more of the structure of the margin totals into account (or equivalently assumes away less of the margin mediated cell dependencies) and has its own merits.\n\nHowever, we are actually describing the work of mathematicians and theoretical computer scientists. In that style you only speak with “applied types” (such as theoretical statisticians) to justify working on a snappy math problem. In this case: counting the number of ways to fill in a contingency table or the number of detailed results compatible with a given win vector (the link between counting, and generation having been strongly established in “Randomised Algorithms for Counting and Generating Combinatorial Structures”, A.J. Sinclair, Ph.D. thesisUniversity of Edinburgh (1988) and related works).\n\nThe contingency table problem is partially solved in:\n\n• “Sampling contingency tables” Martin Dyer, Ravi Kannan, John Mount, Random Structures and Algorithms Vol. 10, no. 4, July 1997 pp. 487-506.\n• “Fast Unimodular Counting” John Mount, Combinatorics Probability and Computing, Vol. 9, No. 3, May 2000, pp 277-285.\n\nThe second paper (strengthening some results from my Ph.D. thesis) lets you calculate that the number of ways to fill in the following four by four contingency table with non-negative integers to meet the shown row and column totals is exactly `350854066054593772938684218633979710637454260` (about `3.508541e+44`).\n\n``` x(0,0) x(0,1) x(0,2) x(0,3) 154179 x(1,0) x(1,1) x(1,2) x(1,3) 255424 x(2,0) x(2,1) x(2,2) x(2,3) 277000 x(3,0) x(3,1) x(3,2) x(3,3) 160179 191780 288348 165221 201433 ```\n\nThe point being: the table could arise as the summary from a data set with `846782` (`=191780 + 288348 + 165221 + 201433`) items; to characterize probabilities over such a tables you need good methods to sample over the astronomical family of potential alternate fill-ins (and this is where you apply the link between counting and sampling for self-reducible problem families). We have example code, notes, improved runtime proof, and results here.\n\n“The Polytope of Win Vectors” introduced additional ideas from integral polymatroids to more strongly relate volume to number of integer vectors (and gets more complete theoretical results for its problem).\n\nAll the “big hammer” math is trying to extend some of the beauty of G.H. Hardy and J.E. Littlewood, “Some problems of Diophantine approximation: the lattice points of a right-angled triangle,” Hamburg. Math.Abh., 1 (1921) 212–249 to more general settings.\n\nOr more succinctly: we just like the word “win.”", null, "", null, "" ]	[ null, "https://i0.wp.com/win-vector.com/wp-content/uploads/2020/06/10215-2829551-full_of_win.jpg", null, "https://i0.wp.com/win-vector.com/wp-content/uploads/2020/06/0ae82-640px-nike_of_samothrake_louvre_ma2369_n4.jpg", null, "https://i0.wp.com/win-vector.com/wp-content/uploads/2020/06/af618-wv.png", null, "https://i0.wp.com/win-vector.com/wp-content/uploads/2020/06/1d142-page0_1.jpg", null, "https://i0.wp.com/win-vector.com/wp-content/uploads/2020/06/95083-somuchwin.jpg", null, "https://secure.gravatar.com/avatar/92a7340d55e6770e6659e961a10909e4", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9276849,"math_prob":0.901064,"size":7809,"snap":"2022-40-2023-06","text_gpt3_token_len":1795,"char_repetition_ratio":0.09352979,"word_repetition_ratio":0.014913658,"special_character_ratio":0.23805866,"punctuation_ratio":0.11205766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9672892,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,3,null,3,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T03:05:52Z\",\"WARC-Record-ID\":\"<urn:uuid:f492a468-8a45-4349-a24b-3095fab0188b>\",\"Content-Length\":\"54986\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ced5dac-12c1-4d10-ab01-ef0d3bfc2642>\",\"WARC-Concurrent-To\":\"<urn:uuid:52b7bc8e-8c5b-471f-83a7-c07dc4f549ba>\",\"WARC-IP-Address\":\"192.0.78.211\",\"WARC-Target-URI\":\"https://win-vector.com/2014/09/01/what-is-a-win-vector/\",\"WARC-Payload-Digest\":\"sha1:MZNJWBB37T3EOH7FH5JETZDX5S7GXW6J\",\"WARC-Block-Digest\":\"sha1:GY75KKB3MNWWCNO4EFZME3ZOGKBXBI6T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499697.75_warc_CC-MAIN-20230129012420-20230129042420-00199.warc.gz\"}"}
https://certainerror.com/portfolio_category/statistics/	[ "# Portfolio Category: Statistics\n\nThe use of curve-fitting or regression can have the additional result of the tolerance factor that defines confidence limits for the best-fit function. Often this is completed as a second effort following determination of the best-fit function. The traditional theory involves posing random errors about the data and minimizing deviations either by forming a linear…\nA ‘statistic’ refers to a single number measure of a sample. Statistics relevant to uncertainty are mean value, total covariance, total deviation, standard covariance and standard deviation. The formulas needed to calculate these with traditional arithmetic are known. However, the theory does not present error propagation. A comprehensive theory should be valid for choices…\nCurve fitting is the practical way of converting data to a best-fit function that can be carried forward into other calculations. The candidate function is a special case of datum that varies with an independent variable. What is ‘best’ is assessed by minimizing deviation of the data from the local datum. For example, the linear…" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9038139,"math_prob":0.98293084,"size":1147,"snap":"2020-34-2020-40","text_gpt3_token_len":209,"char_repetition_ratio":0.104986876,"word_repetition_ratio":0.0,"special_character_ratio":0.17436792,"punctuation_ratio":0.072164945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935516,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T17:16:12Z\",\"WARC-Record-ID\":\"<urn:uuid:80ac98eb-f20d-43ee-8d74-d519db2fb628>\",\"Content-Length\":\"29184\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73049de5-99e9-4ea1-8690-490274ea8b01>\",\"WARC-Concurrent-To\":\"<urn:uuid:fffefc99-c334-40e0-973f-4640dd9ee30d>\",\"WARC-IP-Address\":\"199.250.223.82\",\"WARC-Target-URI\":\"https://certainerror.com/portfolio_category/statistics/\",\"WARC-Payload-Digest\":\"sha1:2EKF3MMLLYJL6M55QYBNIJN3NJEKGLCG\",\"WARC-Block-Digest\":\"sha1:B4ICNXEF2D2YBLUU2L7US56PYS7X2X2W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738562.5_warc_CC-MAIN-20200809162458-20200809192458-00352.warc.gz\"}"}
http://www.loopandbreak.com/logic/	[ "### Logic\n\nLogic refers to the reasoning used by electronic machines. The term is also used in reference to the circuits that make up digital devices and systems.\n\n#### Boolean Algebra\n\nBoolean algebra is a system of mathematical logic using the numbers 0 and 1 with the operations AND (multiplication), OR (addition), and NOT (negation). Combinations of these operations are NAND (NOT AND) and NOR (NOT OR). This peculiar form of mathematical logic, which gets its name from the nineteenth century British mathematician George Boole, is used in the design of digital logic circuits.\n\n#### Symbology\n\nX Y -X XY X+Y\n0 0 1 0 0\n0 1 1 0 1\n1 0 0 0 1\n1 1 0 1 1\n\nIn Boolean algebra, the AND operation, also called logical conjunction, is written using an asterisk (), a multiplication symbol (×), or by running two characters together, for example, XY. The NOT operation, also called logical inversion, is denoted by placing a tilde (~) over the quantity, as a minus sign (−) or dash (-) followed by the quantity, as a “lazy inverted L” (¬) followed by the quantity, or as the quantity followed by an accent or “prime sign” (′). An example is −X. The OR operation, also called logical disjunction, is written using a plus sign (+), for example,\nX + Y.\nThe foregoing are the symbols used by engineers. Above table shows the values of these functions, where 0 indicates “falsity” and 1 indicates “truth.” In mathematics and philosophy courses involving logic, you may see other symbols used for conjunction and disjunction. The AND operation in some texts is denoted by a detached arrow head pointing up (\u0001) or by an ampersand (&), and the OR operation is denoted by a detached arrowhead pointing down (\u0002).\n\n#### Theorems\n\nTheorem (logic equation) What it’s called\nX+0 = X OR identity\nX1 = X AND identity\nX+1 = 1\nX0 = 0\nX+X = X\nXX = X\n−(−X) = X Double negation\nX+(−X) = 1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89585227,"math_prob":0.98279876,"size":2872,"snap":"2020-34-2020-40","text_gpt3_token_len":804,"char_repetition_ratio":0.10808926,"word_repetition_ratio":0.01622718,"special_character_ratio":0.27889973,"punctuation_ratio":0.08648649,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99751514,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-19T10:07:28Z\",\"WARC-Record-ID\":\"<urn:uuid:d8616b89-3576-4796-ab87-877540a39008>\",\"Content-Length\":\"85221\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cc90ba7a-4eda-4454-98f1-d262712c28c6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ce09e6e2-6038-4efe-9e2d-c565ea198a81>\",\"WARC-IP-Address\":\"104.238.95.146\",\"WARC-Target-URI\":\"http://www.loopandbreak.com/logic/\",\"WARC-Payload-Digest\":\"sha1:JNFDO2WEYD2MAWIOBCE7B6CGNQH7CRKM\",\"WARC-Block-Digest\":\"sha1:DSEFD652VKQBOGVBYKVCLNDG6HDTJBI3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400191160.14_warc_CC-MAIN-20200919075646-20200919105646-00377.warc.gz\"}"}
https://www.neuraldump.net/2016/05/neural-networks-the-node/	[ "# Neural Networks: The Node\n\nAs I covered previously in “Introduction to Neural Networks,” artificial neural networks (ANN) are simplified representations of biological neural networks in which the basic computational unit known as an artificial neuron, or node, represents its biological counterpart, the neuron. In order to understand how neural networks can be taught to identify and classify, it is first necessary to explore the characteristics and functionality of the basic building block itself, the node.\n\n### The Node\n\nFor a given node, there are n inputs labeled", null, ". Each input also has a weight associated with it:", null, ". Again, these inputs represent biological synapses connected to the dendrites of the neuron. The activation level of the node, a, is determined by calculating the sum of each input multiplied by it’s weight,", null, ".\n\nSo,", null, "or,", null, "This activation value is passed through a function known as the activation or transfer function,", null, ", to determine the output of the node, y.", null, "Substituting for a gives us:", null, "", null, "### Transfer Functions\n\nThe transfer function can either be linear or non-linear. Linear functions are seldom chosen for multi-layer neural networks since a multi-layer network made up of nodes with linear transfer functions has an equivalent single-layer network. In order to realize the advantages of a multi-layer network, non-linear transfer functions must be used.\n\nThe simplest transfer function is a discontinuous, step function also known as threshold function. The output is binary, usually a 1 or a 0. The outputs may also be -1 and 1 as long as the threshold value falls between the lower and upper value. If a node has a threshold value, θ, then when a >= θ, = 1, otherwise = 0.", null, "", null, "Non-linear transfer functions are almost always sigmoid functions. Sigmoid functions are a group of functions so named because of their “S” shape. The properties of a sigmoid function include:\n\n• is real-valued and differentiable\n• has a non-negative or non-positive first derivative\n• has one local minimum and one local maximum\n\nAll smooth, positive, bump-shaped functions have a sigmoidal integral so there are many functions to choose from, but the most common sigmoid transfer function is a special case of the logistic function:", null, "", null, "If an output range of -1 to 1 is desired instead, the hyperbolic tangent is often used:", null, "", null, "### McCulloch-Pitts Neurons\n\nThe first artificial neurons were proposed by Warren McCulloch and Walter Pitts in 1943. Their neuron’s primary function is as a Threshold Logic Unit (TLU), a computational unit that takes as it’s input an array of values and a corresponding array of weights for those values, sums them, compares that sum to a threshold value, and if that sum exceeds the threshold, outputs a value. In other words, a TLU is a single node with 2 or more inputs that uses a step function as it’s transfer function. These are also known as McCulloch-Pitts (MP) neurons, in deference to their creators, or threshold neurons.\n\nEven a single MP neuron can classify data. Imagine a node with two inputs both with a weight of 1. The threshold value, θ, is 1.5. For the inputs <0,0>, <0,1>, <1,0>, and <1,1> the respective outputs are: 0, 0, 0, and 1. Since the output is binary, the node is capable of identifying exactly two groups: in this case its the group where both inputs are 1s, designated by an output value of 1, and the group where at least one value is not a 1 (either two 0s or a 1 and 0), designated by an output of 0. This single TLU “knows” the Boolean AND function.\n\nMP neurons have been generalized and modified to serve as the building block of artificial neural networks. The most frequent modification is to use the sigmoid function as the transfer function instead of the original threshold operation.\n\nIn the next article, we will begin to put everything together. Although a single MP neuron is actually capable of classification as illustrated above, it is only when nodes are connected into single- and multi-layer networks does the true power of this AI method become apparent.\n\nThis site uses Akismet to reduce spam. 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https://groupprops.subwiki.org/wiki/Isomorph-normal_coprime_automorphism-invariant_implies_fusion_system-relatively_weakly_closed	[ "# Isomorph-normal coprime automorphism-invariant implies weakly closed for any fusion system\n\nSuppose", null, "$P$ is a group of prime power order, and", null, "$H$ is an Isomorph-normal coprime automorphism-invariant subgroup (?) of", null, "$P$, i.e.,", null, "$H$ is an Isomorph-normal subgroup (?) of", null, "$P$ and it is also coprime automorphism-invariant in", null, "$P$. In particular,", null, "$H$ is an Isomorph-normal coprime automorphism-invariant subgroup of group of prime power order (?).\nThen, for any fusion system", null, "$\\mathcal{F}$ on", null, "$P$,", null, "$H$ is a weakly closed subgroup for", null, "$\\mathcal{F}$." ]	[ null, "https://groupprops.subwiki.org/w/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png ", null, "https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ", null, "https://groupprops.subwiki.org/w/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png ", null, "https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ", null, "https://groupprops.subwiki.org/w/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png ", null, "https://groupprops.subwiki.org/w/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png ", null, "https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ", null, "https://groupprops.subwiki.org/w/images/math/2/6/a/26afd73f8c17f310707120691ccc4a35.png ", null, "https://groupprops.subwiki.org/w/images/math/4/4/c/44c29edb103a2872f519ad0c9a0fdaaa.png ", null, "https://groupprops.subwiki.org/w/images/math/c/1/d/c1d9f50f86825a1a2302ec2449c17196.png ", null, "https://groupprops.subwiki.org/w/images/math/2/6/a/26afd73f8c17f310707120691ccc4a35.png ", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8620544,"math_prob":0.9935695,"size":975,"snap":"2019-51-2020-05","text_gpt3_token_len":234,"char_repetition_ratio":0.17095777,"word_repetition_ratio":0.07575758,"special_character_ratio":0.18461539,"punctuation_ratio":0.11515152,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997516,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T18:33:04Z\",\"WARC-Record-ID\":\"<urn:uuid:e462ac72-8bbe-425d-8998-e3b200352748>\",\"Content-Length\":\"27552\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fac2208e-5fd7-4bc1-872f-058932ecf0fc>\",\"WARC-Concurrent-To\":\"<urn:uuid:67de7968-04c2-4493-adb9-daee270acbde>\",\"WARC-IP-Address\":\"96.126.114.7\",\"WARC-Target-URI\":\"https://groupprops.subwiki.org/wiki/Isomorph-normal_coprime_automorphism-invariant_implies_fusion_system-relatively_weakly_closed\",\"WARC-Payload-Digest\":\"sha1:QYTGDRPLECGFDVFZXUSJGNVHX3ILDEG5\",\"WARC-Block-Digest\":\"sha1:PM7EKFLENUJYQ7XYFOEN2FQCZ2JGAJKO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540490743.16_warc_CC-MAIN-20191206173152-20191206201152-00314.warc.gz\"}"}
http://www.machineintellegence.com/knn-k-nearest-neighbour/	[ "# KNN: K-Nearest Neighbour\n\nThis is the story of a person found at JORDAN border, the country is surrounded by 5 neighbors, so how can we find the person belong to which country. KNN plays a vital role here,\n\nKNN is a algorithm which stores the whole model, i.e. it will never make any function. It will check the person with N numbers of countries who are neighbor to it. This algorithm has following names as it consider the whole sample set.\n\n1-Instance-Based Learning\n\n2-Lazy Learner\n\n3-Non-Parametric\n\nIt is used for classification as well as regression problems. It works good for small data sets but hold problem when the dimensions increases and suffer from “Curse of Dimensionality\n\nLets Take the problem of Iris Data set,\n\n`import numpy as npimport pandas as pdfrom sklearn.datasets import load_irisimport seaborn as sns`\n\niris.data.shape–>150,4–>means 150 rows and 4 column\n\nWe can create a data frame like this,\n\n`df=pd.DataFrame(iris.data,columns=iris.feature_names)df['Target']=iris.target`\n\nOutput:\n\n`from sklearn.neighbors import NearestNeighborsnn = NearestNeighbors(5) #The arguements specify to return the Fast 5 most among the #dataset nn.fit(iris.data)ornn.fit(df.iloc[:,:-1])`\n\nOutput:\n\n```NearestNeighbors(algorithm='auto', leaf_size=30, metric='minkowski',\nmetric_params=None, n_jobs=None, n_neighbors=5, p=2,\n\nThe distance is of many types , please refer: here\n\nIn Order to predict a sample:\n\n`sample = np.array([[5.4,2,2,2.3]])`\n`nn.kneighbors(sample)`\n\nOutput:\n\n```(array([[1.6673332 , 1.90525589, 1.94679223, 2.02484567, 2.09523268]]),\narray([[98, 93, 57, 60, 79]]))```\n\nprint(df.ix[[98, 93, 23, 60, 79],])\n\nSo, 4 neighbors show it belongs to class1 and sample 23 states it belongs to class 0.\n\nThe probability of class is defines as:\n\n•", null, "oprolevorter\n•", null, "vurtil opmer" ]	[ null, "http://2.gravatar.com/avatar/8c6933fb74ee741e550ba14201b7e1af", null, "http://0.gravatar.com/avatar/3f2544ec73f133a919247770d76b21d9", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8156357,"math_prob":0.976463,"size":2217,"snap":"2023-40-2023-50","text_gpt3_token_len":590,"char_repetition_ratio":0.08359693,"word_repetition_ratio":0.0,"special_character_ratio":0.27289128,"punctuation_ratio":0.17118998,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99060035,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T14:15:38Z\",\"WARC-Record-ID\":\"<urn:uuid:41cc9d57-a9a2-4c4e-b416-5c744cae160a>\",\"Content-Length\":\"53876\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:551fbf92-36d7-4f51-8840-e94d31bef586>\",\"WARC-Concurrent-To\":\"<urn:uuid:aa7b4f74-4b1e-43ae-986f-55b15001eb60>\",\"WARC-IP-Address\":\"43.255.154.8\",\"WARC-Target-URI\":\"http://www.machineintellegence.com/knn-k-nearest-neighbour/\",\"WARC-Payload-Digest\":\"sha1:YQORFPP33KIMLHJDOWIN4UL6KZVCU6V2\",\"WARC-Block-Digest\":\"sha1:KZNKJ7O7YKOK2LIC3ZWZZIHYXC3JQ2SI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511106.1_warc_CC-MAIN-20231003124522-20231003154522-00611.warc.gz\"}"}
https://www.math.fsu.edu/e-prints/archive/paper21.abs.html	[ "Jacobian of the Picard Curve\n\nJack Quine\n\nWe compute the Jacobian of the genus $3$ curve with $48$ automorphisms and show that the minimum non-zero norm is $1 + \\sqrt3/3$. This show that Klein's surface with with minimum non-zero norm $4/\\sqrt7$ is not extremal for genus 3. We show that the lattice obtained is the isodual lattice $M(E_6)$ discovered by Conway and Sloane. Our computation gives a cyclotomic construction of this lattice." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87971437,"math_prob":0.992236,"size":436,"snap":"2022-27-2022-33","text_gpt3_token_len":113,"char_repetition_ratio":0.12962963,"word_repetition_ratio":0.0,"special_character_ratio":0.24311927,"punctuation_ratio":0.05,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98037124,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-02T20:57:42Z\",\"WARC-Record-ID\":\"<urn:uuid:8b71f703-d842-4549-8e01-8ea144f0d9ea>\",\"Content-Length\":\"768\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ee65388-81d5-4766-b2bc-e19c6c3f30f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e827003-4e64-4e13-9842-aa6913322e81>\",\"WARC-IP-Address\":\"128.186.104.71\",\"WARC-Target-URI\":\"https://www.math.fsu.edu/e-prints/archive/paper21.abs.html\",\"WARC-Payload-Digest\":\"sha1:COS7YSYQAR3MLIVTBZPTRSIPIVFCCEZX\",\"WARC-Block-Digest\":\"sha1:5S3FNEUI7VF4RPU262LLAY35OWC5J2ZI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104204514.62_warc_CC-MAIN-20220702192528-20220702222528-00730.warc.gz\"}"}
https://www.statisticsviews.com/details/book/8679561/Linear-Algebra-Ideas-and-Applications-4th-Edition.html	[ "# Linear Algebra: Ideas and Applications, 4th Edition\n\n## Books", null, "Praise for the Third Edition\n\n“This volume is ground-breaking in terms of mathematical texts in that it does not teach from a detached perspective, but instead, looks to show students that competent mathematicians bring an intuitive understanding to the subject rather than just a master of applications.”\n\n– Electric Review\n\nA comprehensive introduction, Linear Algebra: Ideas and Applications, Fourth Edition provides a discussion of the theory and applications of linear algebra that blends abstract and computational concepts. With a focus on the development of mathematical intuition, the book emphasizes the need to understand both the applications of a particular technique and the mathematical ideas underlying the technique.\n\nThe book introduces each new concept in the context of an explicit numerical example, which allows the abstract concepts to grow organically out of the necessity to solve specific problems. The intuitive discussions are consistently followed by rigorous statements of results and proofs.\n\nLinear Algebra: Ideas and Applications, Fourth Edition also features:\n\n• Two new and independent sections on the rapidly developing subject of wavelets\n• A thoroughly updated section on electrical circuit theory\n• Illuminating applications of linear algebra with self-study questions for additional study\n• End-of-chapter summaries and sections with true-false questions to aid readers with further comprehension of the presented material\n• Numerous computer exercises throughout using MATLAB code\n\nLinear Algebra: Ideas and Applications, Fourth Edition is an excellent undergraduate-level textbook for one or two semester courses for students majoring in mathematics, science, computer science, and engineering. With an emphasis on intuition development, the book is also an ideal self-study reference.\n\nPreface xi\n\nFeatures of the Text xiii\n\nAcknowledgments xvii\n\n1 Systems of Linear Equations 1\n\n1.1 The Vector Space of m × n Matrices 1\n\nThe Space Rn 4\n\nLinear Combinations and Linear Dependence 6\n\nWhat is a Vector Space? 11\n\nExercises 17\n\n1.1.1 Computer Projects 22\n\n1.1.2 Applications to Graph Theory I 25\n\nExercises 27\n\n1.2 Systems 28\n\nRank: The Maximum Number of Linearly Independent Equations 35\n\nExercises 38\n\n1.2.1 Computer Projects 41\n\n1.2.2 Applications to Circuit Theory 41\n\nExercises 46\n\n1.3 Gaussian Elimination 47\n\nSpanning in Polynomial Spaces 58\n\nComputational Issues: Pivoting 61\n\nExercises 63\n\nComputational Issues: Counting Flops 68\n\n1.3.1 Computer Projects 69\n\n1.3.2 Applications to Traffic Flow 72\n\n1.4 Column Space and Nullspace 74\n\nSubspaces 77\n\nExercises 86\n\n1.4.1 Computer Projects 94\n\nChapter Summary 95\n\n2 Linear Independence and Dimension 97\n\n2.1 The Test for Linear Independence 97\n\nBases for the Column Space 104\n\nTesting Functions for Independence 106\n\nExercises 108\n\n2.1.1 Computer Projects 113\n\n2.2 Dimension 114\n\nExercises 123\n\n2.2.1 Computer Projects 127\n\n2.2.2 Applications to Differential Equations 128\n\nExercises 131\n\n2.3 Row Space and the rank-nullity theorem 132\n\nBases for the Row Space 134\n\nComputational Issues: Computing Rank 142\n\nExercises 143\n\n2.3.1 Computer Projects 146\n\nChapter Summary 147\n\n3 Linear Transformations 149\n\n3.1 The Linearity Properties 149\n\nExercises 157\n\n3.1.1 Computer Projects 162\n\n3.2 Matrix Multiplication (Composition) 164\n\nPartitioned Matrices 171\n\nComputational Issues: Parallel Computing 172\n\nExercises 173\n\n3.2.1 Computer Projects 178\n\n3.2.2 Applications to Graph Theory II 180\n\nExercises 181\n\n3.3 Inverses 182\n\nComputational Issues: Reduction versus Inverses 188\n\nExercises 190\n\n3.3.1 Computer Projects 195\n\n3.3.2 Applications to Economics 197\n\nExercises 202\n\n3.4 The LU Factorization 203\n\nExercises 212\n\n3.4.1 Computer Projects 214\n\n3.5 The Matrix of a Linear Transformation 215\n\nCoordinates 215\n\nIsomorphism 228\n\nInvertible Linear Transformations 229\n\nExercises 230\n\n3.5.1 Computer Projects 235\n\nChapter Summary 236\n\n4 Determinants 238\n\n4.1 Definition of the Determinant 238\n\n4.1.1 The Rest of the Proofs 246\n\nExercises 249\n\n4.1.2 Computer Projects 251\n\n4.2 Reduction and Determinants 252\n\nUniqueness of the Determinant 256\n\nExercises 258\n\n4.2.1 Volume 261\n\nExercises 263\n\n4.3 A Formula for Inverses 264\n\nExercises 268\n\nChapter Summary 269\n\n5 Eigenvectors and Eigenvalues 271\n\n5.1 Eigenvectors 271\n\nExercises 279\n\n5.1.1 Computer Projects 282\n\n5.1.2 Application to Markov Processes 283\n\nExercises 285\n\n5.2 Diagonalization 287\n\nPowers of Matrices 288\n\nExercises 290\n\n5.2.1 Computer Projects 292\n\n5.2.2 Application to Systems of Differential Equations 293\n\nExercises 295\n\n5.3 Complex Eigenvectors 296\n\nComplex Vector Spaces 303\n\nExercises 304\n\n5.3.1 Computer Projects 305\n\nChapter Summary 306\n\n6 Orthogonality 308\n\n6.1 The Scalar Product in RN 308\n\nOrthogonal/Orthonormal Bases and Coordinates 312\n\nExercises 316\n\n6.2 Projections: The Gram-Schmidt Process 318\n\nThe QR Decomposition 325\n\nUniqueness of the QR Factorization 327\n\nExercises 328\n\n6.2.1 Computer Projects 331\n\n6.3 Fourier Series: Scalar Product Spaces 333\n\nExercises 341\n\n6.3.1 Application to Data Compression: Wavelets 344\n\nExercises 352\n\n6.3.2 Computer Projects 353\n\n6.4 Orthogonal Matrices 355\n\nHouseholder Matrices 361\n\nExercises 364\n\nDiscrete Wavelet Transform 367\n\n6.4.1 Computer Projects 369\n\n6.5 Least Squares 370\n\nExercises 377\n\n6.5.1 Computer Projects 380\n\n6.6 Quadratic Forms: Orthogonal Diagonalization 381\n\nThe Spectral Theorem 385\n\nThe Principal Axis Theorem 386\n\nExercises 392\n\n6.6.1 Computer Projects 395\n\n6.7 The Singular Value Decomposition (SVD) 396\n\nApplication of the SVD to Least-Squares Problems 402\n\nExercises 404\n\nComputing the SVD Using Householder Matrices 406\n\nDiagonalizing Matrices Using Householder Matrices 408\n\n6.8 Hermitian Symmetric and Unitary Matrices 410\n\nExercises 417\n\nChapter Summary 419\n\n7 Generalized Eigenvectors 421\n\n7.1 Generalized Eigenvectors 421\n\nExercises 429\n\n7.2 Chain Bases 431\n\nJordan Form 438\n\nExercises 443\n\nThe Cayley-Hamilton Theorem 445\n\nChapter Summary 445\n\n8 Numerical Techniques 446\n\n8.1 Condition Number 446\n\nNorms 446\n\nCondition Number 448\n\nLeast Squares 451\n\nExercises 451\n\n8.2 Computing Eigenvalues 452\n\nIteration 453\n\nThe QR Method 457\n\nExercises 462\n\nChapter Summary 464\n\nIndex 487\n\n## Books & Journals\n\n### Books", null, "#### Simulation Modeling and Arena, 2nd Edition", null, "#### Weight-of-Evidence for Forensic DNA Profiles, 2nd Edition", null, "View all\n\n### Journals", null, "#### Mathematical Logic Quarterly", null, "#### Statistica Neerlandica", null, "View all" ]	[ null, "https://media.wiley.com/product_data/coverImage/85/11189095/1118909585.jpg", null, "https://media.wiley.com/product_data/coverImage/10/11186079/1118607910.jpg", null, "https://media.wiley.com/product_data/coverImage/5X/11188145/111881455X.jpg", null, "https://media.wiley.com/product_data/coverImage/54/11190619/1119061954.jpg", null, "https://www.statisticsviews.com/common/images/thumbnails/small/13ab0de8af1.gif", null, "https://www.statisticsviews.com/common/images/thumbnails/small//1389e6d7a6f.gif", null, "https://www.statisticsviews.com/common/images/thumbnails/small///1389ad71abd.gif", null 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https://slideplayer.com/slide/5329040/	[ "", null, "# Summarizing Quantitative Data Frequency Distribution Relative Frequency and Percent Frequency Distributions Histogram Cumulative Distributions Ogive.\n\n## Presentation on theme: \"Summarizing Quantitative Data Frequency Distribution Relative Frequency and Percent Frequency Distributions Histogram Cumulative Distributions Ogive.\"— Presentation transcript:\n\nSummarizing Quantitative Data Frequency Distribution Relative Frequency and Percent Frequency Distributions Histogram Cumulative Distributions Ogive\n\nConstructing a Frequency Distribution for Quantitative Data 3 initial steps 1.Determine the number of nonoverlapping classes. 2.Determine the width of each class. 3.Determine the class limits\n\nExample: Hudson Auto Repair The manager of Hudson Auto would like to have a better understanding of the cost of parts used in the engine tune-ups performed in the shop. She examines 50 customer invoices for tune-ups. The costs of parts, rounded to the nearest dollar, are listed on the next slide.\n\nExample: Hudson Auto Repair n Sample of Parts Cost for 50 Tune-ups\n\nFrequency Distribution Guidelines for Selecting Number of Classes Use between 5 and 20 classes. Use between 5 and 20 classes. Data sets with a larger number of elements Data sets with a larger number of elements usually require a larger number of classes. usually require a larger number of classes. Smaller data sets usually require fewer classes Smaller data sets usually require fewer classes\n\nFrequency Distribution Guidelines for Selecting Width of Classes Use classes of equal width. Use classes of equal width. Approximate Class Width = Approximate Class Width =\n\nFrequency Distribution We decide that SIX (6) classes are appropriate for our purposes Thus the approximate class width is given by: Approximate Class Width = (109 - 52)/6 = 9.5  10\n\nThus we have: Parts Cost(\\$)Frequency 50-592 60-6913 70-7916 80-897 90-997 100-1095 Total50\n\nUsing Excel’s Frequency Function to Construct a Frequency Distribution Note: Rows 9-51 are not shown. Formula Worksheet (showing data entered)\n\nUsing Excel’s Frequency Function to Construct a Frequency Distribution Note: Rows 9-51 are not shown. Formula Worksheet\n\nRelative Frequency and Percent Frequency Parts Cost (\\$) Relative Frequency Percent Frequency 50-59.044 60-69.2626 70-79.3232 80-89.1414 90-99.1414 100-109.1010 Total1.00100 2/50.04(100 )\n\nUsing Excel’s Frequency Function to Construct a Frequency Distribution Note: Rows 9-51 are not shown. Value Worksheet\n\nAudit Time Example We want to construct a frequency distribution to describe the days required to complete year end audits for our clients. Here we have data for 20 of our clients Year End Audit Times (in Days) 12141918 15 1817 20272223 22213328 14181630.5\n\n5 classes should be sufficient, given the size of our data set. Approximate class width is given by: Audit example We round up to the next highest integer, so our class interval is 5.\n\nUsing Excel’s Frequency Function to Construct a Frequency Distribution 1.Step 1: Select cells D2:D6, the cells in which we want the frequencies to appear 2.Step 2: Type the following formula in the formula bar: =FREQUENCY(A2:A21,{14,19,24,29,34}) 3.Step 3: Press Ctrl-Shift-Enter\n\nThe Audit Example Audit TimeFrequencyRelative FrequencyPercent Frequency 10-1440.220 15-1980.440 20-2450.2525 25-2920.110 30-3410.055 Total201100\n\nThe Histogram This is a bar graph of a frequency distribution for quantitative data. The variable of interest is measured on the horizontal axis. Frequency, relative frequency, or percent frequency for each class is measured on the vertical axis. There are no spaces between the bars on a histogram—the rectangles are adjacent.\n\nHistogram Parts Cost (\\$) Parts Cost (\\$) 2 2 4 4 6 6 8 8 10 12 14 16 18 Frequency 50 60 70 80 90 100 110 120 Tune-up Parts Cost\n\nUsing Excel’s Chart Wizard to Construct a Histogram: Audit Example 1.Step 1: Select cells C1:D6. 2.Click the Chart Wizard button. 3.When Chart Type dialog box appears, select Clustered Column from the Chart sub-type display. Click Next>. 4.At Step 2 —Chart Data Source —Click Next>. 5.At Step 3 – Chart Options – select Titles Tab and then type Histogram for Audit Time Data in the Chart title box. Type Audit Time in Days in the Category (X) axis box and type Frequency in the Value (Y) axis box. Select the Legend tab and then remove the check in the Show Legend box. Click Next>. 6.Step 6: Click Finish\n\nRemoving the Gaps Between Rectangles (bars) Step 1: Click right on any rectangle (bar) in the chart to produce a list of options. Step 2: Select the Format Data Series option Step 3: Select the Options tab and then enter 0 in the Gap width box. Click OK\n\nIt looks like this:\n\nCumulative Distributions These show the number of data items with values less than or equal to the class limit of each class. Cumulative relative frequency distributions show the proportion of data items with values less than or equal to the class limit of each class. Cumulative frequency distributions show the proportion of data items with values less than or equal to the class limit of each class.\n\nCumulative Distributions Hudson Auto Repair <59 <69 <79 <89 <99 <109 Cost (\\$) Cumulative CumulativeFrequency RelativeFrequencyCumulativePercent Frequency Frequency 2 15 31 38 45 50.04.30.62.76.90 1.00 4 30 62 76 90 100 2 + 13 15/50.30(100)\n\nParts Parts Cost (\\$) Parts Parts Cost (\\$) 20 40 60 80 100 Cumulative Percent Frequency 50 60 70 80 90 100 110 (89.5, 76) Ogive with Cumulative Percent Frequencies Cumulative Percent Frequencies Tune-up Parts Cost\n\nOgive for Audit Example\n\nDownload ppt \"Summarizing Quantitative Data Frequency Distribution Relative Frequency and Percent Frequency Distributions Histogram Cumulative Distributions Ogive.\"\n\nSimilar presentations" ]	[ null, "https://slideplayer.com/static/blue_design/img/slide-loader4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7427646,"math_prob":0.8726592,"size":5393,"snap":"2021-43-2021-49","text_gpt3_token_len":1348,"char_repetition_ratio":0.15976991,"word_repetition_ratio":0.15028901,"special_character_ratio":0.28258854,"punctuation_ratio":0.10581506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95473313,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T21:25:17Z\",\"WARC-Record-ID\":\"<urn:uuid:528ef288-927c-45fa-ac6b-5e9391696bbe>\",\"Content-Length\":\"179583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0d9163f-401f-4403-926f-4d7cbae849dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:0f329d84-ac8a-4410-acd7-98f3fde0d846>\",\"WARC-IP-Address\":\"138.201.54.25\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/5329040/\",\"WARC-Payload-Digest\":\"sha1:BPTBNQPO2ODZKH4IJCGBKVJX6UVJ4YOQ\",\"WARC-Block-Digest\":\"sha1:OBALWO5MUCOKRN77PLPCU7MVFI2ZQJLW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585209.43_warc_CC-MAIN-20211018190451-20211018220451-00384.warc.gz\"}"}
https://physics.stackexchange.com/questions/122741/capillary-action-meniscus-height-in-a-tube-fitted-inside-another-tube	[ "# Capillary action meniscus height in a tube fitted inside another tube?\n\nI was thinking about how would capillary action change in a tube (classic example) and in a tube fitted inside another tube (considering water as the liquid involved).\n\nHeight of liquid column:", null, "where:\n\n$\\gamma$ = liquid-air surface tension\n\n$\\theta$ = contact angle\n\n$\\rho$ = density of liquid\n\n$g$ = gravity acceleration\n\n$r$ = radius\n\nI tried my best to draw the examples I'm interested in order to help my explanation.\n\nI didn't consider the capillarity inside the smaller tube in both example #2 and #3 because I'd like to assume that $a/2$ in example #1 is close to $c$ in example #2 and #3 (drawings not to scale).", null, "Since from what I understand the column height is given, among other things (most of which can't be changed, like liquid-air surface tension, contact angle, density of liquid and gravity acceleration), by the tube radius, I'd like to know if $c$ in example #2 can be considered as $a/2$ in example #1 to calculate column height using above formula.\n\nAlso I'd like to know how having beads of slightly smaller diameter than $c$ between the two tubes (example #3) would affect the column height.\n\nIf said beads were less dense than water, could they still improve column height or would they just form a floating mat on top of 1 unit thickness?\n\nWhat'd be the column height of example #2 and #3 assuming $c$ as 1mm?\n\nI'm quite sure that given the same reached height $h$ in example #2 and #3, $c$ of #2 has to be smaller than $c$ in #3.\n\nIf $c\\ll R$ the radius of your outer tube, the total curvature is approximately $\\cos \\theta/c$, so you will get $$h = \\frac{\\gamma}{\\rho g} \\frac{\\cos \\theta}{c}.$$" ]	[ null, "https://i.stack.imgur.com/P5s6S.png", null, "https://i.stack.imgur.com/Au1cp.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9558098,"math_prob":0.9819818,"size":1443,"snap":"2019-35-2019-39","text_gpt3_token_len":353,"char_repetition_ratio":0.14246005,"word_repetition_ratio":0.0,"special_character_ratio":0.25433126,"punctuation_ratio":0.06498195,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997491,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T08:41:30Z\",\"WARC-Record-ID\":\"<urn:uuid:e8bbd332-5b38-4cdc-a6c4-cfd47ed86cb9>\",\"Content-Length\":\"138832\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4675bece-ba59-4682-a54a-a17c426c5de9>\",\"WARC-Concurrent-To\":\"<urn:uuid:b839871e-2234-4876-bfff-d7d6c69dd6e6>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/122741/capillary-action-meniscus-height-in-a-tube-fitted-inside-another-tube\",\"WARC-Payload-Digest\":\"sha1:ZZ4SYY35K4K2DHZO334JALNCCVZ2DIDT\",\"WARC-Block-Digest\":\"sha1:WSMF7VGP4D2FXDT5745KW7ROLC3CCFL4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573065.17_warc_CC-MAIN-20190917081137-20190917103137-00044.warc.gz\"}"}
https://dsp.stackexchange.com/questions/52738/ofdm-rf-signal-waveform/52745#52745	[ "# OFDM RF signal waveform\n\nFrom the Wikipedia article: \"An inverse FFT is computed on each set of symbols, giving a set of complex time-domain samples. These samples are then quadrature-mixed to passband in the standard way. The real and imaginary components are first converted to the analogue domain using digital-to-analogue converters (DACs); the analogue signals are then used to modulate cosine and sine waves at the carrier frequency, f{c}, respectively. These signals are then summed to give the transmission signal, s(t).\"\n\nSo, information is added to a pure sine (or cosine) wave. This causes the wave to have not just one frequency, but a range of frequencies within a bandwidth that can be seen in the frequency domain of the signal. Having found out about this only yesterday, there are a number of questions. I've added an image of a regular time domain OFDM signal found somewhere on the internet.\n\n1. Does the modulated signal really pass through the very narrow (ideally, 20 MHz in regular LTE) bandpass filter at the receiver? At a center frequency of 2 GHz that would mean to bandpass only what is within 1.99 GHz and 2.01 GHz, which sounds impressive to me.\n2. Wouldn't even the slightest distortion from the emitter move the frequency of the signal out of the passband?\n3. At 2 GHz operating frequency (analog time domain signal, at the LNA), there are 210^9 cycles in a second. If the number of carriers is 2048, and a symbol is a string of 2048 cycles or slightly more including the cyclic prefix, doesn't it mean that the theoretical maximum symbol rate is (210^-9)/(2048+prefix length)? This would mean that the signal is transmitted continuously. This is not true, however, since the symbol rate is a function of the sampling time, Ts, which is 32.552 ns. So, I don't really understand what FFT samples has to do with the 2 GHz analog signal that is present at the receiving LNA. My reasoning is that I could digitize the signal with any number of samples, get a plot for each 2048-cycle sequence then apply FFT.\n\nIf the graph below is the LNA signal of 2 GHz LTE, what would the period of a cycle be in nanoseconds? There are about 20 cycles from what I understand. How many cycles of this type would a symbol take?\n\nLater edit: Regarding the third question, I've found an answer. Digital processing of the signal is done in the baseband, which is independent of carrier frequency. It needs to be converted into the passband. These links have more information: IFFT and OFDM upconversion\n\nbaseband and passband modulation\n\nLater edit 2: it seems that there are many ways to convert the baseband OFDM into the passband. Constant Envelope OFDM is one interesting way to do it, the baseband OFDM signal is phase modulating the carrier. This is in contrast to regular OFDM where the baseband is somesort of amplitude modulating the carrier. More info at \"Constant Envelope OFDM Phase Modulation\", for example.", null, "(1) Yes, if the technology permits it for the power handling requirements of the design, the strategy is to filter as much as possible immediately at the front-end of the receiver prior to the LNA. The primary reason is jamming from stronger signals. Depending on receiver topology used we would be concerned with 2nd and 3rd order intermodulation products and receiver phase noise that can be induced on out of band signals if significantly stronger all creating noise in band that would limit our receiver sensitivity.\n\nFBAR technology is an excellent choice for front-end filters combining low loss and exceptional rejection. I have used FBAR filters in the design of femtocell cellular base-stations where the constraint was to receive the entire band (in this case it was a duplexer so would be used to receive 1850-1910 MHz and transmit 1930-1990 MHz --- notice the 20 MHz spacing between these two bands that this technology was able to provide exceptional isolation up at 2 GHz). We use the low loss band-selecting filter due to linearity/intermodulation concerns with the LNA, however with FBAR one must also be very careful with inter-modulation effects especially when using these filters in multi-carrier designs. Here is a good online article with more details about FBAR and other filtering considerations.\n\nhttp://www.linleygroup.com/mpr/article.php?id=11440\n\n(2) The distortion you are referring to is concern with frequency offset. Frequency offset will always exist since the typical transmitter/receiver set do not use a shared clock. The two dominant sources of frequency offset are the offset of the clocks themselves and additional offsets under dynamic conditions due to Doppler. The relevant standards will specify the maximum allowable clock offset for the design, but < 5 KHz is a reasonable maximum possible frequency offset at 2 GHz for lower cost technologies (this is driven by the ppm specification for the master oscillator used). If I calculated correctly the Doppler at this frequency when the transmitter and receiver are moving apart from each other at 60 mph would only be 179 Hz. The combination of both of these numbers will be well within the passband of whatever front-end filter is used, as no such filter would have a brick-wall response.\n\n(3) I think you are confusing the carrier (2 GHz) with the bandwidth at that carrier (20 MHz). The actual carrier frequency is irrelevant to how much data we can transmit (without consideration to path loss effects on signal power for the SNR component of capacity). The theoretical maximum symbol rate is given by the Shannon-Hartley equation for maximum channel capacity which accounts for both SNR and BW:\n\n$$C = B \\log_2(1+ SNR)$$\n\nWhere:\n\nC is the maximum capacity in B/sec B is the bandwidth of the channel is Hz SNR is the signal to noise power ratio (not in dB but ratio of power)\n\nNote MIMO technologies exceed this by utilizing multipath to send multiple \"channels\" simultaneously." ]	[ null, "https://i.stack.imgur.com/n8L4I.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9513316,"math_prob":0.9212299,"size":2898,"snap":"2021-43-2021-49","text_gpt3_token_len":658,"char_repetition_ratio":0.119212165,"word_repetition_ratio":0.0,"special_character_ratio":0.22084196,"punctuation_ratio":0.10350877,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96537817,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T15:01:27Z\",\"WARC-Record-ID\":\"<urn:uuid:b196b7e4-5a53-4fac-8732-89b5a3f5897e>\",\"Content-Length\":\"172256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:14cee990-689b-4267-96a2-e49eddc90dee>\",\"WARC-Concurrent-To\":\"<urn:uuid:6eb77729-bc91-479d-be3d-dbc8c2693d9d>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/52738/ofdm-rf-signal-waveform/52745#52745\",\"WARC-Payload-Digest\":\"sha1:CTXO3UEHXSUERQ5RXA5APHAOQSFYMZC5\",\"WARC-Block-Digest\":\"sha1:QBRL3YVL32F2KRLPSDGMSEC4YUZ53565\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585178.60_warc_CC-MAIN-20211017144318-20211017174318-00314.warc.gz\"}"}
https://www.daniweb.com/programming/software-development/threads/482482/search-in-file-handling-in-c	[ "Hello Friends ! I am working on a project of CD CAFE. In which I have arranged movies like this :\nMovie Name::::Genre::::Hero of Movie::::Heroine of Movie\n\nand I have entered data in this. Now I want to add search option in my program that I can search by Movie name or genre or hero or heroine it searches all records from texgt file and give me all matches searched results.\nCan Anyone help me ??\n\nI would suggest that when the program starts you read in all of the data in the file and load it into an collection of \"Movie Records\". Then you can search that collection for the records you want.\n\nYou may say more details are not\n\nCreate a class\n\n``````class Movie\n\npublic name\npublic genre\n``````\n\nThen you search by those class attributes, or you could create an associative array same thing.\n\nGoogle classes or assoc arrays in c++.\n\nYou may like to file your data using commas to delimit the data record fields.\n\nThat way you can easily handle missing fields:\n\nMovieName1,Genre1,HeroOfMovie1,HeroineOfMovie1\nMovieName2,Genre2,,HeroineOfMovie2\nMovieName3,Genre3,HeroOfMovie3,\n\nAnd display as:\n\nMovieName1::::Genre1::::HeroOfMovie1::::HeroineOfMovie1\nMovieName2::::Genre2::::NA_HERO::::HeroineOfMovie2\nMovieName3::::Genre3::::HeroOfMovie3::::NA_HEROINE\n\nThis example of sorting data records and finding data records using the C++ library functions may help you to get started.\n\nThis example uses an array of Student records. It demo's the idea of sorting on various fields and finding records that have particular field values.\n\n``````// sortStudentsById_with_find2.cpp //\n\n#include <iostream>\n#include <iomanip>\n#include <string>\n#include <cctype> // re. tolower ...\n\n#include <algorithm> // re array sort, find, etc...\n#include <vector>\n\nusing namespace std;\n\nconst int SIZE_STUDENT_ARY = 5;\n\nclass Student\n{\npublic:\n// ctors...\nStudent() : age(0), grade(0), id(0) {}\nStudent( int age, int grade, int id, const string& name )\n\nvoid setAge( int age ) { this->age = age; }\nvoid setId( int id ) { this->id = id ; }\nvoid setName( const string& name ) { this->name = name ; }\n\nint getAge() const { return age; }\nint getId() const { return id; }\nstring getName() const { return name; }\n\nvoid print( ostream& os ) const\n{\nos << left << setw(14) << name << ' '\n<< setw(7) << id << ' '\n<< setw(7) << grade << ' '\n<< age << right;\n}\nprivate:\nint age;\nint id;\nstring name;\n\n// to permit sorting by id's ... //\nfriend bool cmpById( const Student& a, const Student& b );\n\n// to permit finding by id and thus if id was used already ... //\nfriend bool operator == ( const Student& a, const Student& b );\n\n// to permit sorting and binary_search by name ... //\nfriend bool cmpByName( const Student& a, const Student& b );\n} ;\n\n// def'n of friend functions ...\nbool cmpById( const Student& a, const Student& b )\n{\nreturn a.id < b.id;\n}\nbool operator == ( const Student& a, const Student& b )\n{\nreturn a.id == b.id;\n}\nbool cmpByName( const Student& a, const Student& b )\n{\nreturn a.name < b.name;\n}\n\n// def'n of overloaded << for Student objects ...\nostream& operator << ( ostream& os, const Student& s )\n{\ns.print( os );\nreturn os;\n}\n\n// some utilities used here ..\n\n{\ncout << left << setw(14) << \"NAME\" << ' '\n<< setw(7) << \"ID\" << ' '\n<< setw(7) << \"GRADE\" << ' '\n<< \"AGE\" << right << endl;\n}\n\nstring toCapsOnAllFirstLetters( const string& strIn )\n{\nstring str = strIn;\nint prev_was_space = 1;\nint len = str.size();\nfor( int i = 0; i < len; ++ i )\n{\nif( prev_was_space )\nstr[i] = std::toupper( str[i] );\nprev_was_space = isspace( str[i] );\n}\nreturn str;\n}\nstring takeInString( const string& msg = \"\" )\n{\ncout << msg << flush;\nstring val;\ngetline( cin, val );\nreturn val;\n}\nchar takeInChr( const std::string& msg = \"\" )\n{\nstring reply = takeInString( msg );\n// else ...\nreturn 0;\n}\nbool more()\n{\nif( tolower( takeInChr( \"More (y/n) ? \" )) == 'n' )\nreturn false;\n// else ...\nreturn true;\n}\nint takeInValidInt( const char* msg )\n{\nint val;\nwhile( true )\n{\ncout << msg << flush;\nif( cin >> val && cin.get() == '\\n' )\nbreak;\nelse\n{\ncout << \"Invalid input ... numbers only!\\n\";\ncin.clear(); // clear error flags\ncin.sync(); // 'flush' cin stream ...\n}\n}\nreturn val;\n}\n\nint main()\n{\nStudent students[SIZE_STUDENT_ARY];\n\n// Ok ... take in an array of Student records ...\n\nint size = 0;\ndo\n{\nstudents[size].setName(\ntoCapsOnAllFirstLetters( takeInString(\"Enter Name : \" )));\nstudents[size].setAge( takeInValidInt( \"Enter Age : \" ));\nstudents[size].setId( takeInValidInt( \"Enter Id : \" ));\n\nif( find( students, students+size, students[size] )\n!= students+size ) // then found\n{\ncout << \"Sorry ... that ID is taken ... \\n\";\ncontinue;\n}\n\ncout << students[size];\nif( tolower(takeInChr( \" ... Ok (y/n) ? \" )) == 'y' )\n++size;\nelse\ncout << \"Ok ... that student was NOT entered ...\\n\";\n\nif( size == SIZE_STUDENT_ARY )\n{\ncout << \"You have reached \"\n<< SIZE_STUDENT_ARY\n<< \", the max space here!\\n\";\nbreak;\n}\n}\nwhile( more() );\n\n// Ok ... output results ...\n\ncout<< \"\\nStudents in entered order ... \\n\";\nfor( int x = 0; x < size; ++x )\n{\nstudents[x].print( cout );\ncout << endl;\n}\n\ncout<< \"Students ordered by ID ... \\n\";\nsort( students, students+size, cmpById );\n\nfor( int x = 0; x < size; ++x )\n{\ncout << students[x] << endl; // using overloaded <<\n}\n\ncout<< \"Students ordered by name ... \\n\";\nstable_sort( students, students+size, cmpByName );\nfor( int x = 0; x < size; ++x )\n{\ncout << students[x] << endl; // using overloaded <<\n}\n\ncout << \"\\nLoop to test binary_search by name ...\\n\";\ndo\n{\nStudent tmp;\ntmp.setName( toCapsOnAllFirstLetters(takeInString(\n\"Enter name to find: \" )) );\nif( binary_search( students, students+size, tmp, cmpByName ) )\n{\ncout << \"'\"<< tmp.getName() << \"' was found ...\\n\";\n}\n}\nwhile( more() );\n\ncout << \"\\nloop to test find by name using 'lower bound' ...\\n\";\ndo\n{\nStudent tmp;\ntmp.setName( toCapsOnAllFirstLetters(takeInString(\n\"Enter name to find: \" )) );\n\nStudent* low = lower_bound( students, students+size, tmp,\ncmpByName );\nbool found = false;\nwhile( low->getName() == tmp.getName() )\n{\ncout << \"'\"<< tmp.getName() << \"' was found: \"\n<< *low << endl;\nfound = true;\nlow = lower_bound( low+1, students+size, tmp,\ncmpByName );\n}\nif( !found )" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86296594,"math_prob":0.87436926,"size":397,"snap":"2020-10-2020-16","text_gpt3_token_len":89,"char_repetition_ratio":0.117048346,"word_repetition_ratio":0.0,"special_character_ratio":0.23677582,"punctuation_ratio":0.2,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9849239,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-18T04:28:22Z\",\"WARC-Record-ID\":\"<urn:uuid:16289a3c-3c2a-4731-bcee-1e8fcb72e615>\",\"Content-Length\":\"61809\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dccf2b41-c91a-4609-a310-f3fd5e11605e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9d3d92e3-c318-41dd-b055-e98a46eadcee>\",\"WARC-IP-Address\":\"169.55.25.107\",\"WARC-Target-URI\":\"https://www.daniweb.com/programming/software-development/threads/482482/search-in-file-handling-in-c\",\"WARC-Payload-Digest\":\"sha1:UYMMU3BBAPXOQP2E7L3XZWZ4GKZPCKH7\",\"WARC-Block-Digest\":\"sha1:YZJXNBAMMBQLN7T6OYSMSOT555OAOUOU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875143505.60_warc_CC-MAIN-20200218025323-20200218055323-00017.warc.gz\"}"}
https://dsp.stackexchange.com/questions/tagged/2d?tab=Votes	[ "# Questions tagged [2d]\n\nThe tag has no usage guidance.\n\n60 questions\nFilter by\nSorted by\nTagged with\n4k views\n\n### Fast/efficient way to decompose separable integer 2D filter coefficients\n\nI would like to be able to quickly determine whether a given 2D kernel of integer coefficients is separable into two 1D kernels with integer coefficients. E.g. ...\n1k views\n\n### Determining Type and Bandwidth of a filter\n\nGiven a filter, if it is given as an equation such as: $$f(x,y) = \\left(\\frac{\\partial^2}{\\partial x^2} + \\frac{\\partial^2}{\\partial y^2}\\right) \\exp\\left(-\\frac{x^2 + y^2}{\\sigma^2}\\right)$$ Or in ...\n628 views\n\n### Data fusion using 2d discrete wavelet transform (DWT)\n\nI am working on a project that employs a linear array sensor that provides data from the same object at two different energies. Collected in time, I end up with two images (16-bit sensor values, MxM ...\n4k views\n\n### Standard Deviation in Gaussian Blur\n\nI have a function that performs gaussian blur on image for some specific $\\sigma$ (the standard deviation). It first computes kernel of size $\\lceil 3\\sigma \\rceil$ and then performs convolution with ...\n295 views\n\n### Efficient implementation of 2-d circularly symmetric low-pass filter\n\nHow can a two-dimensional ideal circularly symmetric low-pass filter or its approximation be efficiently implemented on data sampled on a square grid? I'm referring to an ideal filter with a spatial ...\n1k views\n\nDoes anyone know about different adaptive filtering implementations (LMS, RLS ...) in 2D or even 3D ? I have sequences of 2D images and 3D volumes with repeating patterns but small differences. I was ...\n308 views\n\n### How to separate the upwards-propagating from the downwards-propagating waves?\n\nI've got a real 2D image of propagating waves in both directions. How can I separate this image into one with the upward-propagating waves and one with the downward-propagating ones? This example has ...\n1k views\n\n### Image Processing and applicability of 2D Fourier Transform\n\nAs a newbie in the world of signal processing, I am having a hard time in appreciating image 2-D fourier transforms. I am fully able to appreciate the concept of 1-D Fourier transform. Essentially, ...\n453 views\n\n### 2D Convolution in MATLAB Causes Artifacts (Boundary Issues)\n\nIm trying to do a 2D fast convolution in Matlab of large matrices. If I use FFT version based on convolution theorem ( https://en.wikipedia.org/wiki/Convolution_theorem ), there are some artefacts in ...\n4k views\n\n### How do you interpolate between points in an image (2D), e.g. using splines?\n\nI can understand just fine how to use 1-dimensional interpolation on data points where one coordinate is a function of the other: y = f(x). However, when we have an ...\n759 views\n\n### 2D FFT of Vector\n\nI have a bunch of 8x8 images which have been vectorized and formed into a 64xN matrix, X. I'd like to take the 2D-FFT of each of these images without reshaping each column of X into an 8x8 image and ...\n1k views\n\n### Laplacian Operator with and without Diagonal Direction Elements in the Kernel\n\nThis is a general question on the laplacian operator, which has two different versions. The first version is : \\begin{matrix} 0 & 1 & 0 \\\\ 1 & -4 & 1 \\\\ 0 & 1 & 0 \\end{...\n788 views\n\n### Image 2D Real Cepstrum with DFT, Is ifftshift` Needed?\n\nThis is my testing image,it is taken from this paper: I tried to transform it into its real cepstrum domain with this simple MATLAB code: ...\n261 views\n\n### Easy way to get rid of noise in a hand drawing\n\nI would like to take a user's hand drawn input, and filter out \"noise\", aka small sketches or blobs, where the user might have screwed up and accidentally touched his hand to the drawing device, or ...\n2k views\n\n### Harmonics in 2-D FFT Signal\n\nIn case of 1-D FFT it is far more easier to find number of harmonics and their Respective frequencies and amplitudes. Lets say if i have 8 points signal the harmonics of it then would be 7 with one DC ...\n8k views\n\n### 2D Deconvolution in matlab\n\nI am trying to solve the following equation for h (an [MxM matrix]): $$k[\\tau_1,\\tau_2]=\\sum_{i_1=0}^M \\sum_{i_2=0}^M h[i_1,i_2]x[\\tau_1-i_1]x[\\tau_2-i_2]$$ I have k, which is a 2D [MxM] symmetric ...\n1k views\n\n### Finding an Image's Orientation With DFT Frequency Amplitudes\n\nI'm looking to get some general idea of an image's orientation so that I can then rotate it to the nearest 90 degree angle. My idea on how to do this is to take the DFT on the image, then do a \"radar ...\n418 views\n\n### Understanding 2D FFT result of an image having a pattern\n\nI have created Fast Fourier Transform (FFT 2D) of the following image (without pattern angle markers) using ejectamenta online tool. Input (Original image is here) Output I got is the following As ...\n276 views\n\nI've read there are two admissibility criteria for wavelets, both of which are designed to preserve total power of the signal (source: http://en.wikipedia.org/wiki/Wavelet#Mother_wavelet, as well as ...\n4k views\n\n### Calculate 1D Power Spectrum from 2D Images\n\nImagine satellite images, these are irregular sampled in X and Y direction and the shapes are of course are oddly off. We now want to estimate a 1D power spectrum from the whole image to estimate the ...\n133 views\n\n### Upsampling Methods for Computed-Tomography\n\nI have two sets of data of given Field of view, one of them only covers a subset of the FOV of the other. I therefore want to upsample the one with the larger FOV to combine it with the other one. So ...\n1k views\n\n### How does MATLAB recover picture from magnitude spectrum alone?\n\nThis is the transformation I did. The code fft2() the Lena picture than ifft2() it back to the original. Add some ...\n43 views\n\n### Why does this plot correspond to this function?\n\nI have a function: $$u_1(\\mathbf{x})= \\begin{cases} 1, & \\text{if} \\;\\;\|x_1\|\\le r_1, \|x_2\| \\le r_2 \\\\ 0, & \\text{otherwise} \\end{cases} ,\\mathbf{x}=[x_1,x_2]^T \\in \\mathbb{R}^2$$ And I'm ...\n87 views\n\n32 views\n\n### how to interpret the 2D FFT\n\nI know how to compute the 1D FFT (and interpret values from 0 to Nyq). When computing the 2D FFT, do we compute the FFT of row then the FFT of row then the FFT of row up to the last row. ...\n1k views\n\n### MATLAB phase of 2D rectangular pulse's Fourier transform\n\nI'm trying to plot the graph of the phase of the Fourier transform of a 2D rectangular pulse. I've been able to evaluate the FFT but I'm not sure if the phase is correct because there are some tilts I ...\n356 views\n\n### Linearity and shift-invariance of 2-D system on lattice\n\nI know how to check a 1-D system for these conditions, but am confused about translating this to a 2-D system over a lattice. The system $H$ is define as: w[\\mathbf x] = H\\left\\{u[\\mathbf x]\\right\\}...\n329 views\n\n### Fast/efficient way to compute Laplacian edge enhancement filter\n\nI would like to implement a somewhat smarter Laplacian edge enhancement convolution. Right now it is implemented as (generic 3x3 convolution): ...\n3k views\n\n### partial derivative of image\n\nI have to find the partial derivative of an image with respect to its x dimension. I am using central difference method i.e $\\partial_x F(x) = \\frac{F(x+1,y)-F(x-1,y)}{2}$ Here $F(x,y)$ ...\n178 views\n\n### How do I implement the CDF wavelet inverse?\n\nI understand and have implemented a 2D discrete wavelet transform that uses a Haar wavelet to process images. Now I would like to implement a 2D Cohen-Daubechies-Feauveau wavelet transform, but I am ...\n96 views\n\n### Two-dimensional wavelet analysis\n\nCould you tell me why wavelets without scaling functions can't be used for two-dimensional analysis? What is the role of scaling function?\n35 views\n\n### Require explanation for Matlab freqz2\n\nQuestion I browsed the code of the Matlab freqz2 function (to find out how it is different to simple fft2.) First of all it ..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8894385,"math_prob":0.9182748,"size":13310,"snap":"2020-10-2020-16","text_gpt3_token_len":3606,"char_repetition_ratio":0.12971592,"word_repetition_ratio":0.014175258,"special_character_ratio":0.26235914,"punctuation_ratio":0.11706422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99668825,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-27T05:27:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a890704a-b792-4400-b3ba-bee275020523>\",\"Content-Length\":\"253278\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4580a8b-f0c7-4663-9f18-edb32e9b3052>\",\"WARC-Concurrent-To\":\"<urn:uuid:347aa03a-d126-497e-a491-dc929daab335>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/tagged/2d?tab=Votes\",\"WARC-Payload-Digest\":\"sha1:AIODV5GJKLSWLKJNENGPG346I46A7B5W\",\"WARC-Block-Digest\":\"sha1:7V4JODZR43BCRFBXDUQ2ANUGMWD6GSBT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146647.82_warc_CC-MAIN-20200227033058-20200227063058-00039.warc.gz\"}"}
https://www.alibabacloud.com/help/doc-detail/63451.htm	[ "Log Service supports time functions, date functions, interval functions, and a time series padding function. You can use the functions described in this topic in analysis statements.\n\n## Date and time data types\n\n1. Unix timestamp: the number of seconds that have elapsed since 00:00:00 (UTC) on January 1, 1970. The value is of the integer type. For example, `1512374067` indicates the time ```Mon Dec 4 15:54:27 CST 2017```. The built-in time `__time__` in each log of Log Service is of this type.\n2. timestamp: the time in the format of string, for example, `2017-11-01 13:30:00`.\n\n## Date functions\n\nThe following table describes the common date functions supported by Log Service.\n\nFunction Description Example\n`current_date` Returns the current date. `latency>100\| select current_date`\n`current_time` Returns the current time. `latency>100\| select current_time`\n`current_timestamp` Returns the current timestamp by combining the results of current_date and current_time. `latency>100\| select current_timestamp`\n`current_timezone()` Returns the current time zone. `latency>100\| select current_timezone()`\n`from_iso8601_timestamp(string)` Parses an ISO 8601-formatted time into a timestamp that contains the time zone. ```latency>100\| select from_iso8601_timestamp(iso8601)```\n`from_iso8601_date(string)` Parses an ISO 8601-formatted time into a date. ```latency>100\| select from_iso8601_date(iso8601)```\n`from_unixtime(unixtime)` Returns a Unix timestamp as a timestamp. ```latency>100\| select from_unixtime(1494985275) ```\n`from_unixtime(unixtime,string)` Returns a Unix timestamp as a timestamp that uses the specified string as the time zone. `latency>100\| select from_unixtime (1494985275,'Asia/Shanghai')`\n`localtime` Returns the current time. `latency>100\| select localtime `\n`localtimestamp` Returns the current timestamp. `latency>100\| select localtimestamp `\n`now()` Functions the same as `current_timestamp`. N/A\n`to_unixtime(timestamp)` Returns a timestamp as a Unix timestamp. `\| select to_unixtime('2017-05-17 09:45:00.848 Asia/Shanghai')`\n\n## Time functions\n\nMySQL time formats\n\nLog Service supports the MySQL time formats, such as %a, %b, and %y.\n\nFunction Description Example\n`date_format(timestamp, format)` Formats a timestamp as a string by using the specified format. `latency>100`\| `select date_format (date_parse('2017-05-17 09:45:00','%Y-%m-%d %H:%i:%S'), '%Y-%m-%d')`\n`date_parse(string, format)` Parses a string into a timestamp by using the specified format. `latency>100`\|`select date_format (date_parse(time,'%Y-%m-%d %H:%i:%S'), '%Y-%m-%d')`\nTable 1. Format description\nFormat Description\n%a The abbreviation of a day in a week, such as Sun and Sat.\n%b The abbreviation of a month, such as Jan and Dec.\n%c The month, of the numeric type. Valid values: [1, 12].\n%D The day in a month with a suffix, such as 0th, 1st, 2nd, and 3rd.\n%d The day in a month, in decimal format. Valid values: [01, 31].\n%e The day in a month, in decimal format. Valid values: [1, 31].\n%H The hour in 24-hour format.\n%h The hour in 12-hour format.\n%I The hour in 12-hour format.\n%i The minutes, of the numeric type. Valid values: [00, 59].\n%j The day in a year. Valid values: [001, 366].\n%k The hour. Valid values: [0, 23].\n%l The hour. Valid values: [1, 12].\n%M The month. Valid values: January, February, March, April, May, June, July, August, September, October, November, and December.\n%m The month, of the numeric type. Valid values: [01, 12].\n%p The abbreviation of a period in 12-hour format. Valid values: AM and PM.\n%r The time in 12-hour format: `hh:mm:ss AM/PM`.\n%S The seconds. Valid values: [00, 59].\n%s The seconds. Valid values: [00, 59].\n%T The time in 24-hour format: `hh:mm:ss`.\n%U The week in a year, where Sunday is the first day of each week. Valid values: [00, 53].\n%u The week in a year, where Monday is the first day of each week. Valid values: [00, 53].\n%V The week in a year, where Sunday is the first day of each week. Valid values: [01, 53]. %V is used with %X.\n%v The week in a year, where Monday is the first day of each week. Valid values: [01, 53]. %v is used with %x.\n%W The day in a week. Valid values: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.\n%w The day in a week, where Sunday is the first day of each week. Valid values: [0, 6].\n%Y The year in four-digit format.\n%y The year in two-digit format.\n%% The escape character %.\n\n## Truncation function\n\nLog Service supports a truncation function, which can return a time truncated by the second, minute, hour, day, month, and year. The truncation function is applicable to time-based statistics.\n\n• Function syntax:\n``date_trunc(unit, x)``\n• Parameters:\n\nThe value of the x parameter can be a timestamp or Unix timestamp.\n\nThe following table lists the values of the unit parameter and the output results when the x parameter is set to `2001-08-22 03:04:05.000`.\nValue Output result\nsecond 2001-08-22 03:04:05.000\nminute 2001-08-22 03:04:00.000\nhour 2001-08-22 03:00:00.000\nday 2001-08-22 00:00:00.000\nweek 2001-08-20 00:00:00.000\nmonth 2001-08-01 00:00:00.000\nquarter 2001-07-01 00:00:00.000\nyear 2001-01-01 00:00:00.000\n• Example:\n\nThe date_trunc function is applicable only to statistics at a fixed time interval. For statistics based on flexible time dimensions, for example, every 5 minutes, you need to use a GROUP BY clause according to the mathematical modulus method.\n\n`` \| SELECT count(1) as pv, __time__ - __time__% 300 as minute5 group by minute5 limit 100``\n\nIn the preceding statement, `%300` indicates that modulus and truncation are performed every 5 minutes.\n\nThe following example shows how to use the date_trunc function:\n``````\|select date_trunc('minute' , __time__) as t,\ntruncate (avg(latency) ) ,\ncurrent_date\ngroup by t\norder by t desc\nlimit 60``````\n\n## Interval functions\n\nInterval functions are used to perform interval-related calculation. For example, add or subtract an interval based on a date, or calculate the interval between two dates.\n\nFunction Description Example\n`date_add(unit, value, timestamp)` Adds an interval `value` of the `unit` type to a `timestamp`. To subtract an interval, use a negative `value`. `date_add('day', -7, '2018-08-09 00:00:00')`: indicates seven days before August 9.\n`date_diff(unit, timestamp1, timestamp2)` Returns the time difference between `timestamp1` and `timestamp2` expressed in terms of `unit`. `date_diff('day', '2018-08-02 00:00:00', '2018-08-09 00:00:00') = 7`\nThe following table lists the values of the unit parameter supported by the interval functions.\nValue Description\nmillisecond The millisecond.\nsecond The second.\nminute The minute.\nhour The hour.\nday The day.\nweek The week.\nmonth The month.\nquarter The quarter, namely, three months.\nyear The year.\n\n## Time series padding function\n\nThe time series padding function time_series is used to pad time series data with the missing time.\n\n• Function format:\n``time_series(time_column, window, format, padding_data)``\n• The following table describes the parameters.\nParameter Description\ntime_column The sequence of time, such as the default time field `__time__` provided by Log Service. The value is of the long or timestamp type.\nwindow The size of the time window, which is composed of a number and a unit. Unit: s (seconds), m (minutes), H (hours), or d (days). For example, 2h, 5m, or 3d.\nformat The MySQL time format, which is the final output format.\npadding_data The content to be added. Valid values:\n• 0: adds 0.\n• null: adds null.\n• last: adds the last value.\n• next: adds the next value.\n• avg: adds the average value of the last and next values.\n• Example:\n\nThe following statement is used to format data every 2 hours:\n\n`````` \| select time_series(__time__, '2h', '%Y-%m-%d %H:%i:%s', '0') as stamp, count(*) as num from log group by stamp order by stamp\n``````\nThe following figure shows the output result.", null, "" ]	[ null, "http://static-aliyun-doc.oss-cn-hangzhou.aliyuncs.com/assets/img/13111/156205726337530_en-US.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6646607,"math_prob":0.9225431,"size":7692,"snap":"2019-35-2019-39","text_gpt3_token_len":2179,"char_repetition_ratio":0.15816857,"word_repetition_ratio":0.1015953,"special_character_ratio":0.32163286,"punctuation_ratio":0.19358894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9575681,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T17:57:13Z\",\"WARC-Record-ID\":\"<urn:uuid:09d7d67a-cc11-402d-b3b1-963d2d293f31>\",\"Content-Length\":\"208617\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c102c19e-cc9b-423f-b742-fef8a3f77f87>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7f218c4-f981-4568-85bb-a8cebcd0e1bb>\",\"WARC-IP-Address\":\"47.88.73.19\",\"WARC-Target-URI\":\"https://www.alibabacloud.com/help/doc-detail/63451.htm\",\"WARC-Payload-Digest\":\"sha1:WBKBV6WJWDDP7KAIJVXMKJ5QXPZ5GRRW\",\"WARC-Block-Digest\":\"sha1:W662LWEGKUUY5GEEBTKGR6W4EMIM4LWQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317339.12_warc_CC-MAIN-20190822172901-20190822194901-00527.warc.gz\"}"}
https://www.thecrazyprogrammer.com/2021/01/c-map-check-if-key-exists.html	[ "# C++ Map Check if Key Exists – 2 Ways to Check\n\nHello everyone, In this article, we will see different ways by which we can check if a particular key exists in a map or not. Let’s first go through what a map is:\n\nC++ map stores information in <key, value> pairs and we can access the value field from the key in O(logn) time. Now the problem is to check if there exists a particular key in the map or not.\n\n## Method 1: Using map::find\n\nWe can take the help of the standard library function find for map. map::find returns an iterator to the <key, value> pair if the key exists or it points to the standard iterator end in the map. So if for any particular key if find is returning an iterator pointing to the standard operator end then we can say that map doesn’t have that particular key. Here is the code:\n\nThe time complexity to check is the same as map::find function which is O(logn).\n\n## Method 2: Using map::count\n\nWe can also make use of the c + + count method which returns 1 if the particular key is present otherwise returns 0. Here is the code illustrating the same:\n\nThe time complexity for the above code is O(logn) as well. This is widely used to check if a key is present or not.\n\nNote: If the particular key is not present and if somewhere we use it then the key will be created in the map and initiated default value (0 for int. “” for string) will be created. So never check a key exists or not with the below condition:\n\nCheck the below code for better understanding:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68480825,"math_prob":0.94250894,"size":2494,"snap":"2022-27-2022-33","text_gpt3_token_len":793,"char_repetition_ratio":0.1124498,"word_repetition_ratio":0.32572615,"special_character_ratio":0.38171613,"punctuation_ratio":0.15833333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9560228,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T16:46:00Z\",\"WARC-Record-ID\":\"<urn:uuid:72f0c61e-fbb1-4794-a040-debeb7ffe684>\",\"Content-Length\":\"161475\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:399804dc-5226-41e4-bda9-a7c2f1f715e7>\",\"WARC-Concurrent-To\":\"<urn:uuid:783edfee-eb24-40f4-9862-8ebbe247620a>\",\"WARC-IP-Address\":\"104.21.59.152\",\"WARC-Target-URI\":\"https://www.thecrazyprogrammer.com/2021/01/c-map-check-if-key-exists.html\",\"WARC-Payload-Digest\":\"sha1:AHGFWHXFQYIADZBYUZV6E4GYIO65DLMV\",\"WARC-Block-Digest\":\"sha1:KI4PSN4PJNBRO3HWYTJEM24PUNK4XL2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104495692.77_warc_CC-MAIN-20220707154329-20220707184329-00570.warc.gz\"}"}
https://intellinuts.com/java/java-pow-method	[ "", null, "", null, "# Java Pow() Method\n\nShow\n\n## Description\n\nThe process returns the price of the first disagreement raised to the supremacy of the second argument. The java.lang.Math.pow() is utilized to calculate an integer raised to the power of several other numbers. This meaning accepts two parameters and returns the charge of the first stricture raised to the second parameter.\n\n### Syntax\n\n`double pow(double base, double exponent)`\n\n### Parameters\n\nHere is the detail of parameters −\n\n• base − Any primitive data type.\n• exponent − Any primitive data type.\n\n### Return Value\n\n• This method returns the value of the first argument raised to the power of the second argument.\n\n#### Example\n\n```public class Test {\n\npublic static void main(String args[]) {\ndouble x = 11.635;\ndouble y = 2.76;\n\nSystem.out.printf(\"The value of e is %.4f%n\", Math.E);\nSystem.out.printf(\"pow(%.3f, %.3f) is %.3f%n\", x, y, Math.pow(x, y));\n}\n}```\n\nThis will produce the following result:\n\n#### Output\n\n```The value of e is 2.7183\npow(11.635, 2.760) is 874.008```\n\nHere at Intellinuts, we have created a complete Java tutorial for Beginners to get started in Java." ]	[ null, "https://intellinuts.com/image/logo.png", null, "http://sdimg.intellinuts.com/img/java.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7201405,"math_prob":0.87456673,"size":821,"snap":"2022-05-2022-21","text_gpt3_token_len":208,"char_repetition_ratio":0.11138311,"word_repetition_ratio":0.0,"special_character_ratio":0.28501827,"punctuation_ratio":0.2,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98917174,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T19:13:06Z\",\"WARC-Record-ID\":\"<urn:uuid:f035bed8-80c6-475d-9b02-ac0cc5be3b4c>\",\"Content-Length\":\"17925\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c9a89fe-4d37-434a-8b64-8c8a0cb5c394>\",\"WARC-Concurrent-To\":\"<urn:uuid:1639e90f-5b83-4269-876b-34f0318b1fc4>\",\"WARC-IP-Address\":\"68.168.220.124\",\"WARC-Target-URI\":\"https://intellinuts.com/java/java-pow-method\",\"WARC-Payload-Digest\":\"sha1:J6EZX6HH2TE6F4D5VTP4C3JJ3F3ZV7EF\",\"WARC-Block-Digest\":\"sha1:KTMIG75TRSK3M2RDSNK4DWJYIYHZGIUK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522309.14_warc_CC-MAIN-20220518183254-20220518213254-00647.warc.gz\"}"}
https://in.mathworks.com/examples/matlab-hdl-coder/mw/hdlcoder_product-mlhdlc_tutorial_image_heq-image-enhancement-by-histogram-equalization	[ "MATLAB Examples\n\n# Image Enhancement by Histogram Equalization\n\nThis example shows how to generate HDL code from a MATLAB® design that does image enhancement using histogram equalization.\n\n## Algorithm\n\nThe Histogram Equalization algorithm enhances the contrast of images by transforming the values in an intensity image so that the histogram of the output image is approximately flat.\n\n```I = imread('pout.tif');\nJ = histeq(I);\nsubplot(2,2,1);\nimshow( I );\nsubplot(2,2,2);\nimhist(I)\nsubplot(2,2,3);\nimshow( J );\nsubplot(2,2,4);\nimhist(J)\n```", null, "## MATLAB Design\n\n```design_name = 'mlhdlc_heq'; testbench_name = 'mlhdlc_heq_tb'; ```\n\nLet us take a look at the MATLAB design\n\n```type(design_name); ```\n```%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % heq.m % Histogram Equalization Algorithm %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function [x_out, y_out, pixel_out] = ... mlhdlc_heq(x_in, y_in, pixel_in, width, height) % Copyright 2011-2015 The MathWorks, Inc. persistent histogram persistent transferFunc persistent histInd persistent cumSum if isempty(histogram) histogram = zeros(1, 2^14); transferFunc = zeros(1, 2^14); histInd = 0; cumSum = 0; end % Figure out indexes based on where we are in the frame if y_in < height && x_in < width % valid pixel data histInd = pixel_in + 1; elseif y_in == height && x_in == 0 % first column of height+1 histInd = 1; elseif y_in >= height % vertical blanking period histInd = min(histInd + 1, 2^14); elseif y_in < height % horizontal blanking - do nothing histInd = 1; end %Read histogram (must be outside conditional logic) histValRead = histogram(histInd); %Read transfer function (must be outside conditional logic) transValRead = transferFunc(histInd); %If valid part of frame add one to pixel bin and keep transfer func val if y_in < height && x_in < width histValWrite = histValRead + 1; %Add pixel to bin transValWrite = transValRead; %Write back same value cumSum = 0; elseif y_in >= height %In blanking time index through all bins and reset to zero histValWrite = 0; transValWrite = cumSum + histValRead; cumSum = transValWrite; else histValWrite = histValRead; transValWrite = transValRead; end %Write histogram (must be outside conditional logic) histogram(histInd) = histValWrite; %Write transfer function (must be outside conditional logic) transferFunc(histInd) = transValWrite; pixel_out = transValRead; x_out = x_in; y_out = y_in; ```\n```type(testbench_name); ```\n```%Test bench for Histogram Equalization % Copyright 2011-2015 The MathWorks, Inc. testFile = 'mlhdlc_img_peppers.png'; imgOrig = imread(testFile); [height, width] = size(imgOrig); imgOut = zeros(height,width); hBlank = 20; % make sure we have enough vertical blanking to filter the histogram vBlank = ceil(2^14/(width+hBlank)); for frame = 1:2 disp(['working on frame: ', num2str(frame)]); for y_in = 0:height+vBlank-1 %disp(['frame: ', num2str(frame), ' of 2, row: ', num2str(y_in)]); for x_in = 0:width+hBlank-1 if x_in < width && y_in < height pixel_in = double(imgOrig(y_in+1, x_in+1)); else pixel_in = 0; end [x_out, y_out, pixel_out] = ... mlhdlc_heq(x_in, y_in, pixel_in, width, height); if x_out < width && y_out < height imgOut(y_out+1,x_out+1) = pixel_out; end end end % normalize image to 255 imgOut = round(255imgOut/max(max(imgOut))); figure(1) subplot(2,2,1); imshow(imgOrig, [0,255]); title('Original Image'); subplot(2,2,2); imshow(imgOut, [0,255]); title('Equalized Image'); subplot(2,2,3); hist(double(imgOrig(:)),2^14-1); axis([0, 255, 0, 1500]) title('Histogram of original Image'); subplot(2,2,4); hist(double(imgOut(:)),2^14-1); axis([0, 255, 0, 1500]) title('Histogram of equalized Image'); end ```\n\n## Simulate the Design\n\nIt is always a good practice to simulate the design with the testbench prior to code generation to make sure there are no runtime errors.\n\n```mlhdlc_heq_tb ```\n```working on frame: 1 working on frame: 2 ```", null, "## Setup for the Example\n\nExecuting the following lines copies the necessary files into a temporary folder\n\n```mlhdlc_demo_dir = fullfile(matlabroot, 'toolbox', 'hdlcoder', 'hdlcoderdemos', 'matlabhdlcoderdemos'); mlhdlc_temp_dir = [tempdir 'mlhdlc_heq']; % create a temporary folder and copy the MATLAB files cd(tempdir); [~, ~, ~] = rmdir(mlhdlc_temp_dir, 's'); mkdir(mlhdlc_temp_dir); cd(mlhdlc_temp_dir); % copy files to the temp dir copyfile(fullfile(mlhdlc_demo_dir, [design_name,'.m']), mlhdlc_temp_dir); copyfile(fullfile(mlhdlc_demo_dir, [testbench_name,'.m*']), mlhdlc_temp_dir); copyfile(fullfile(mlhdlc_demo_dir, 'mlhdlc_img_peppers.png'), mlhdlc_temp_dir); ```\n\n## Create a New HDL Coder™ Project\n\n```coder -hdlcoder -new mlhdlc_heq_prj\n```\n\nNext, add the file 'mlhdlc_heq.m' to the project as the MATLAB Function and 'mlhdlc_heq_tb.m' as the MATLAB Test Bench.\n\nYou can refer to Getting Started with MATLAB to HDL Workflow tutorial for a more complete tutorial on creating and populating MATLAB HDL Coder projects.\n\n## Run Fixed-Point Conversion and HDL Code Generation\n\nLaunch HDL Advisor and right click on the 'Code Generation' step and choose the option 'Run to selected task' to run all the steps from the beginning through the HDL code generation.\n\nExamine the generated HDL code by clicking on the hyperlinks in the Code Generation Log window.\n\n## Clean up the Generated Files\n\nYou can run the following commands to clean up the temporary project folder.\n\n```mlhdlc_demo_dir = fullfile(matlabroot, 'toolbox', 'hdlcoder', 'hdlcoderdemos', 'matlabhdlcoderdemos');\nmlhdlc_temp_dir = [tempdir 'mlhdlc_heq'];\nclear mex;\ncd (mlhdlc_demo_dir);\nrmdir(mlhdlc_temp_dir, 's');\n```" ]	[ null, "https://www.mathworks.com/help/releases/R2018b/examples/hdlcoder_product/win64/mlhdlc_histeq_io.png", null, "https://www.mathworks.com/help/releases/R2018b/examples/hdlcoder_product/win64/mlhdlc_tutorial_image_heq_01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5864854,"math_prob":0.97753406,"size":5507,"snap":"2019-13-2019-22","text_gpt3_token_len":1503,"char_repetition_ratio":0.14228603,"word_repetition_ratio":0.0685871,"special_character_ratio":0.30107138,"punctuation_ratio":0.19263157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99125123,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-23T07:31:21Z\",\"WARC-Record-ID\":\"<urn:uuid:d13e120d-67fd-4d65-9218-59b8a4230a58>\",\"Content-Length\":\"67602\",\"Content-Type\":\"application/http; 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https://doc.cgal.org/Manual/3.3/doc_html/cgal_manual/Kernel_23_ref/Class_Segment_2.html	[ "", null, "## CGAL::Segment_2<Kernel>\n\n### Definition\n\nAn object s of the data type Segment_2<Kernel> is a directed straight line segment in the two-dimensional Euclidean plane", null, "2, i.e. a straight line segment [p,q] connecting two points p,q in", null, "2. The segment is topologically closed, i.e. the end points belong to it. Point p is called the source and q is called the target of s. The length of s is the Euclidean distance between p and q. Note that there is only a function to compute the square of the length, because otherwise we had to perform a square root operation which is not defined for all number types, which is expensive, and may not be exact.\n\n### Creation\n\n Segment_2 s ( Point_2 p, Point_2 q); introduces a segment s with source p and target q. The segment is directed from the source towards the target.\n\n### Operations\n\n bool s.operator== ( q) Test for equality: Two segments are equal, iff their sources and targets are equal. bool s.operator!= ( q) Test for inequality. Point_2 s.source () returns the source of s. Point_2 s.target () returns the target of s. Point_2 s.min () returns the point of s with lexicographically smallest coordinate. Point_2 s.max () returns the point of s with lexicographically largest coordinate. Point_2 s.vertex ( int i) returns source or target of s: vertex(0) returns the source of s, vertex(1) returns the target of s. The parameter i is taken modulo 2, which gives easy access to the other vertex. Point_2 s.point ( int i) returns vertex(i). Point_2 s.operator[] ( int i) returns vertex(i). Kernel::FT s.squared_length () returns the squared length of s. Direction_2 s.direction () returns the direction from source to target of s. Vector_2 s.to_vector () returns the vector s.target() - s.source(). Segment_2 s.opposite () returns a segment with source and target point interchanged. Line_2 s.supporting_line () returns the line l passing through s. Line l has the same orientation as segment s.\n\n### Predicates\n\nbool s.is_degenerate () segment s is degenerate, if source and target are equal.\n\nbool s.is_horizontal ()\nbool s.is_vertical ()\n\nbool s.has_on ( Point_2<Kernel> p) A point is on s, iff it is equal to the source or target of s, or if it is in the interior of s.\n\nbool s.collinear_has_on ( Point_2<Kernel> p)\nchecks if point p is on segment s. This function is faster than function has_on().\n Precondition: p is on the supporting line of s.\n\n### Miscellaneous\n\n Bbox_2 s.bbox () returns a bounding box containing s. Segment_2 s.transform ( Aff_transformation_2 t) returns the segment obtained by applying t on the source and the target of s." ]	[ null, "https://doc.cgal.org/Manual/3.3/doc_html/cgal_manual/Kernel_23_ref/cc_Class.gif", null, "https://doc.cgal.org/Manual/3.3/doc_html/cgal_manual/Kernel_23_ref/cc_mathbb_E.gif", null, "https://doc.cgal.org/Manual/3.3/doc_html/cgal_manual/Kernel_23_ref/cc_mathbb_R.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7058557,"math_prob":0.98666686,"size":1338,"snap":"2023-14-2023-23","text_gpt3_token_len":332,"char_repetition_ratio":0.26536733,"word_repetition_ratio":0.11372549,"special_character_ratio":0.27952167,"punctuation_ratio":0.13492064,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9875635,"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\":\"2023-06-01T02:55:39Z\",\"WARC-Record-ID\":\"<urn:uuid:8f4f65ee-2184-44ff-b854-e89155d7f443>\",\"Content-Length\":\"17116\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:169e8c36-2f60-4b26-98ad-265a2a230bdb>\",\"WARC-Concurrent-To\":\"<urn:uuid:99c92751-164b-4202-9a3e-a6e6ae5f2df9>\",\"WARC-IP-Address\":\"213.186.33.40\",\"WARC-Target-URI\":\"https://doc.cgal.org/Manual/3.3/doc_html/cgal_manual/Kernel_23_ref/Class_Segment_2.html\",\"WARC-Payload-Digest\":\"sha1:MO436FWO2RKP4UZZHIFFNJ5E7OGR44ZP\",\"WARC-Block-Digest\":\"sha1:DZWCS4C4XW5A7NZH7NTN66OXE5HER77C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647525.11_warc_CC-MAIN-20230601010402-20230601040402-00556.warc.gz\"}"}
https://itvoyagers.in/5-best-java-program-to-convert-decimal-to-binary/	[ "# 5 – Best Java program to convert decimal to binary\n\n## Java program to convert decimal to binary\n\nDecimal number have base 10 and they ranges from 0 to 9.\n\nBinary number have base 2 and they ranges from 0 to 1.\n\nFollowing is the java program to convert decimal to binary.\n\n### Logic to convert decimal to binary in java\n\n`//always check name of classimport java.util.*;//itvoyagers.inclass DecToBin{ public static void main(String agr[]) { int[] bin = new int; Scanner in = new Scanner(System.in); System.out.println(\"Enter any number : \"); int dnum = in.nextInt(); int i; for(i = 0; dnum>0; i++) { bin[i] = dnum % 2; dnum = dnum / 2; } for(int j=i-1; j>=0; j--) { System.out.print(bin[j]); } }}`\n\nOutput :\n\nEnter any number :\n54\n110110\n\n### Connect with us on following platforms\n\nWe are aiming to explain all concepts of Java in easiest terms as possible.", null, "ITVoyagers\nAuthor" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6672828,"math_prob":0.9074246,"size":876,"snap":"2023-40-2023-50","text_gpt3_token_len":239,"char_repetition_ratio":0.10091743,"word_repetition_ratio":0.054054055,"special_character_ratio":0.29680365,"punctuation_ratio":0.15846995,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96083707,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T09:39:01Z\",\"WARC-Record-ID\":\"<urn:uuid:d5623aa4-7d98-468c-9a57-bf0f41640cea>\",\"Content-Length\":\"180968\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bdc3df32-08ed-46a5-b999-77b976480c3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6948c2d-c741-4cb4-bbef-55875c54856f>\",\"WARC-IP-Address\":\"104.21.22.36\",\"WARC-Target-URI\":\"https://itvoyagers.in/5-best-java-program-to-convert-decimal-to-binary/\",\"WARC-Payload-Digest\":\"sha1:HV4P2TUORRCHZV53QROWJGISL3BNC6EU\",\"WARC-Block-Digest\":\"sha1:XTMAWQXGD53VKHPSL3RKERULOQHUCZZA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506632.31_warc_CC-MAIN-20230924091344-20230924121344-00258.warc.gz\"}"}
https://www.mcsweeneys.net/articles/more-academic-euphemisms-for-masturbating	[ "Integrating the derivative\nC=2∏r\nSterilizing the test tube\nExperiment with a scientific control\nDosing out the placebo\nForce = Mass x Acceleration\nNormalizing the units\nGraphing the cosine\nConverting from Cartesian to Polar coordinates\nMeasuring plate tectonics\nPosing a hypothesis\nConcluding with the empirical method\nProving a theorem" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70901686,"math_prob":0.87143165,"size":451,"snap":"2023-40-2023-50","text_gpt3_token_len":109,"char_repetition_ratio":0.13199106,"word_repetition_ratio":0.0,"special_character_ratio":0.14412417,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872757,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T01:02:33Z\",\"WARC-Record-ID\":\"<urn:uuid:91c39bf5-3745-479c-a3e0-f0802b7205a8>\",\"Content-Length\":\"78491\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f07563e8-4761-47ba-ac6f-ea5223cb4642>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0848a8f-04f9-49f2-800f-ae8039c879ed>\",\"WARC-IP-Address\":\"54.162.128.250\",\"WARC-Target-URI\":\"https://www.mcsweeneys.net/articles/more-academic-euphemisms-for-masturbating\",\"WARC-Payload-Digest\":\"sha1:QGNXST6I7D76OOGKEIFSMJV5QQVPHHFU\",\"WARC-Block-Digest\":\"sha1:LKGEVEBBM7GP3L7Q2LO7LJP5AXZILXOP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100258.29_warc_CC-MAIN-20231130225634-20231201015634-00134.warc.gz\"}"}
https://swlaschin.gitbooks.io/fsharpforfunandprofit/content/posts/fvsc-quicksort.html	[ "# Comparing F# with C#: Sorting\n\nIn this next example, we will implement a quicksort-like algorithm for sorting lists and compare an F# implementation to a C# implementation.\n\nHere is the logic for a simplified quicksort-like algorithm:\n\n```If the list is empty, there is nothing to do.\nOtherwise:\n1. Take the first element of the list\n2. Find all elements in the rest of the list that\nare less than the first element, and sort them.\n3. Find all elements in the rest of the list that\nare >= than the first element, and sort them\n4. Combine the three parts together to get the final result:\n(sorted smaller elements + firstElement +\nsorted larger elements)\n```\n\nNote that this is a simplified algorithm and is not optimized (and it does not sort in place, like a true quicksort); we want to focus on clarity for now.\n\nHere is the code in F#:\n\n``````let rec quicksort list =\nmatch list with\n\| [] -> // If the list is empty\n[] // return an empty list\n\| firstElem::otherElements -> // If the list is not empty\nlet smallerElements = // extract the smaller ones\notherElements\n\|> List.filter (fun e -> e < firstElem)\n\|> quicksort // and sort them\nlet largerElements = // extract the large ones\notherElements\n\|> List.filter (fun e -> e >= firstElem)\n\|> quicksort // and sort them\n// Combine the 3 parts into a new list and return it\nList.concat [smallerElements; [firstElem]; largerElements]\n\n//test\nprintfn \"%A\" (quicksort [1;5;23;18;9;1;3])\n``````\n\nAgain note that this is not an optimized implementation, but is designed to mirror the algorithm closely.\n\nLet's go through this code:\n\n• There are no type declarations anywhere. This function will work on any list that has comparable items (which is almost all F# types, because they automatically have a default comparison function).\n• The whole function is recursive -- this is signaled to the compiler using the `rec` keyword in \"`let rec quicksort list =`\".\n• The `match..with` is sort of like a switch/case statement. Each branch to test is signaled with a vertical bar, like so:\n``````match x with\n\| caseA -> something\n\| caseB -> somethingElse\n``````\n• The \"`match`\" with `[]` matches an empty list, and returns an empty list.\n• The \"`match`\" with `firstElem::otherElements` does two things.\n• First, it only matches a non-empty list.\n• Second, it creates two new values automatically. One for the first element called \"`firstElem`\", and one for the rest of the list, called \"`otherElements`\". In C# terms, this is like having a \"switch\" statement that not only branches, but does variable declaration and assignment at the same time.\n• The `->` is sort of like a lambda (`=>`) in C#. The equivalent C# lambda would look something like `(firstElem, otherElements) => do something`.\n• The \"`smallerElements`\" section takes the rest of the list, filters it against the first element using an inline lambda expression with the \"`<`\" operator and then pipes the result into the quicksort function recursively.\n• The \"`largerElements`\" line does the same thing, except using the \"`>=`\" operator\n• Finally the resulting list is constructed using the list concatenation function \"`List.concat`\". For this to work, the first element needs to be put into a list, which is what the square brackets are for.\n• Again note there is no \"return\" keyword; the last value will be returned. In the \"`[]`\" branch the return value is the empty list, and in the main branch, it is the newly constructed list.\n\nFor comparison here is an old-style C# implementation (without using LINQ).\n\n``````public class QuickSortHelper\n{\npublic static List<T> QuickSort<T>(List<T> values)\nwhere T : IComparable\n{\nif (values.Count == 0)\n{\nreturn new List<T>();\n}\n\n//get the first element\nT firstElement = values;\n\n//get the smaller and larger elements\nvar smallerElements = new List<T>();\nvar largerElements = new List<T>();\nfor (int i = 1; i < values.Count; i++) // i starts at 1\n{ // not 0!\nvar elem = values[i];\nif (elem.CompareTo(firstElement) < 0)\n{\n}\nelse\n{\n}\n}\n\n//return the result\nvar result = new List<T>();\nreturn result;\n}\n}\n``````\n\nComparing the two sets of code, again we can see that the F# code is much more compact, with less noise and no need for type declarations.\n\nFurthermore, the F# code reads almost exactly like the actual algorithm, unlike the C# code. This is another key advantage of F# -- the code is generally more declarative (\"what to do\") and less imperative (\"how to do it\") than C#, and is therefore much more self-documenting.\n\n## A functional implementation in C#\n\nHere's a more modern \"functional-style\" implementation using LINQ and an extension method:\n\n``````public static class QuickSortExtension\n{\n/// <summary>\n/// Implement as an extension method for IEnumerable\n/// </summary>\npublic static IEnumerable<T> QuickSort<T>(\nthis IEnumerable<T> values) where T : IComparable\n{\nif (values == null \|\| !values.Any())\n{\nreturn new List<T>();\n}\n\n//split the list into the first element and the rest\nvar firstElement = values.First();\nvar rest = values.Skip(1);\n\n//get the smaller and larger elements\nvar smallerElements = rest\n.Where(i => i.CompareTo(firstElement) < 0)\n.QuickSort();\n\nvar largerElements = rest\n.Where(i => i.CompareTo(firstElement) >= 0)\n.QuickSort();\n\n//return the result\nreturn smallerElements\n.Concat(new List<T>{firstElement})\n.Concat(largerElements);\n}\n}\n``````\n\nThis is much cleaner, and reads almost the same as the F# version. But unfortunately there is no way of avoiding the extra noise in the function signature.\n\n## Correctness\n\nFinally, a beneficial side-effect of this compactness is that F# code often works the first time, while the C# code may require more debugging.\n\nIndeed, when coding these samples, the old-style C# code was incorrect initially, and required some debugging to get it right. Particularly tricky areas were the `for` loop (starting at 1 not zero) and the `CompareTo` comparison (which I got the wrong way round), and it would also be very easy to accidentally modify the inbound list. The functional style in the second C# example is not only cleaner but was easier to code correctly.\n\nBut even the functional C# version has drawbacks compared to the F# version. For example, because F# uses pattern matching, it is not possible to branch to the \"non-empty list\" case with an empty list. On the other hand, in the C# code, if we forgot the test:\n\n``````if (values == null \|\| !values.Any()) ...\n``````\n\nthen the extraction of the first element:\n\n``````var firstElement = values.First();\n``````\n\nwould fail with an exception. The compiler cannot enforce this for you. In your own code, how often have you used `FirstOrDefault` rather than `First` because you are writing \"defensive\" code. Here is an example of a code pattern that is very common in C# but is rare in F#:\n\n``````var item = values.FirstOrDefault(); // instead of .First()\nif (item != null)\n{\n// do something if item is valid\n}\n``````\n\nThe one-step \"pattern match and branch\" in F# allows you to avoid this in many cases.\n\n## Postscript\n\nThe example implementation in F# above is actually pretty verbose by F# standards!\n\nFor fun, here is what a more typically concise version would look like:\n\n``````let rec quicksort2 = function\n\| [] -> []\n\| first::rest ->\nlet smaller,larger = List.partition ((>=) first) rest\nList.concat [quicksort2 smaller; [first]; quicksort2 larger]\n\n// test code\nprintfn \"%A\" (quicksort2 [1;5;23;18;9;1;3])\n``````\n\nNot bad for 4 lines of code, and when you get used to the syntax, still quite readable." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.77154505,"math_prob":0.83464366,"size":7480,"snap":"2022-27-2022-33","text_gpt3_token_len":1767,"char_repetition_ratio":0.14245586,"word_repetition_ratio":0.051779937,"special_character_ratio":0.2528075,"punctuation_ratio":0.13531114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9638849,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-15T09:25:47Z\",\"WARC-Record-ID\":\"<urn:uuid:0df4bc9b-6718-4fc4-80a8-2583718f9178>\",\"Content-Length\":\"87916\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:531a1673-4082-4163-aca5-9587d5415776>\",\"WARC-Concurrent-To\":\"<urn:uuid:9fe2aa80-6691-45b1-a9c6-58c3c6e6ba49>\",\"WARC-IP-Address\":\"104.18.3.206\",\"WARC-Target-URI\":\"https://swlaschin.gitbooks.io/fsharpforfunandprofit/content/posts/fvsc-quicksort.html\",\"WARC-Payload-Digest\":\"sha1:AIQEGQ3EQVMVKD4W7TZTUNUBEIUUAXEO\",\"WARC-Block-Digest\":\"sha1:K76P4NGRI57RRU4AZXN33TAR47MF2ORP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572163.61_warc_CC-MAIN-20220815085006-20220815115006-00142.warc.gz\"}"}
http://www.webpgomez.com/docencia/discrete-mathematics-2011/287-solution-to-project-1	[ "Menú\n\n### 1 Description of the Magic Effect\n\n1. The magic effect consists of the following steps:\n2. Take a 3-digit number and reverse it.\n3. Subtract the smallest number from the greatest number.\n4. Take the result and reverse it again.\n5. Sum the result and the reversed number.\n\n### 2 Mathematical Principle Behind the Magic Effect\n\nLet n be the chosen number and write it in positional notation as n = xyz. This means that n = 100x + 10y + z. When the volunteer is asked to reverse the number and then subtract the smallest from the greatest, the following operations are carried out:", null, "At this point it is clear that the effect will not work if the number is palindromic, that is, if we obtain the same number when read backwards. In this case the subtraction will be zero. Without loss of generality, we can assume that x > z. We re-write the result as 99(x - z) = 11 ⋅ 9 ⋅ (x - z). Since x - z < 9, the product 9 ⋅ (x - z) is a multiple of 9 between 9 and 81. This multiple can be written as 9 ⋅ (x - z) = a10 + b, for certain a,b in ℕ. By substituting this expression in the product 99(x - z), we will obtain:", null, "We reverse the number and sum the original number:", null, "Since a10 + b is a multiple of 9, the sum of its digits is 9, and, therefore, (a + b)121 = 9 ⋅ 121 = 1089. This shows that the magic effect always works as long as the chosen number is not a palindrome.\n\nNote: Some people have put forward a simpler, not so deep explanation. I have taken it as valid as this one.\n\n### 3 The 2-digit Version\n\nLet n be a 2-digit number. As before, we write it in positional notation, n = xy = 10x + y. Then, the number is reversed and subtracted from each other.", null, "Again, we must require x is not equal to y, that is, the number cannot be a palindrome. Let us assume, for example, that x > y. Let us write number 9(x - y) as 10a + b, as done before.", null, "### 4 The 4-digit Version\n\nFirst of all, the same situation arises when a palindromic number is chosen. Let us take a non-palindromic number. Even in that case, the final result is not unique. Consider the following numbers as a point in case:\n\nn = 1,000: 1,000 - 0001 = 999", null, "999 + 999 = 1,998.\nn = 1,003: 3,001 - 1,003 = 1,998", null, "1,998 + 8,991 = 10,989.\nn = 1,011 : 1,101 - 1,011 = 90", null, "90 + 09 = 99.\nn = 7,451 : 7,451 - 1,547 = 5,904", null, "5,904 + 4,095 = 9,999.\n\n### Docencia\n\n• ##### Proyectos de fin de carrera", null, "" ]	[ null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution0x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution1x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution2x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution3x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution5x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution6x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution6x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution6x.png", null, "http://www.webpgomez.com/images/stories/Docencia/Discrete-Mathematics/Project-1/Class-Project-1-solution6x.png", null, "http://www.webpgomez.com/images/jsn_air_pro/backgrounds/hr-bg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9036305,"math_prob":0.9986843,"size":2262,"snap":"2020-34-2020-40","text_gpt3_token_len":677,"char_repetition_ratio":0.12533215,"word_repetition_ratio":0.004310345,"special_character_ratio":0.34305924,"punctuation_ratio":0.15779468,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997161,"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,3,null,3,null,3,null,3,null,3,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-19T02:16:41Z\",\"WARC-Record-ID\":\"<urn:uuid:9d196b85-ff79-4279-9b14-60c514dff9b9>\",\"Content-Length\":\"31137\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47727b4f-8232-49f2-9263-65166cba2bec>\",\"WARC-Concurrent-To\":\"<urn:uuid:a33c653f-55a2-4f74-a046-ce039c528b2c>\",\"WARC-IP-Address\":\"5.135.40.7\",\"WARC-Target-URI\":\"http://www.webpgomez.com/docencia/discrete-mathematics-2011/287-solution-to-project-1\",\"WARC-Payload-Digest\":\"sha1:XLO5QBZ3HHHLB4PYLQ3ELQD77ARE2E5J\",\"WARC-Block-Digest\":\"sha1:WGBCZBA7XKA23RYNJL4THIHBPPZNLQPJ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400189928.2_warc_CC-MAIN-20200919013135-20200919043135-00462.warc.gz\"}"}
https://davidlovezoe.club/wordpress/archives/419	[ "# Go语言单元测试入门\n\n#### 单元测试的定义\n\nThe difference between unit and integration tests is that unit tests usually isolate dependencies that communicate with network, disk etc. Unit tests normally test only one thing such as a function.\n\n#### 官方自带的测试姿势\n\nGo语言之所以令人感到欣喜，很大的一个原因是那些设计者们已经为广大的开发者做好了很多轮子，包括测试框架，试问自带测试框架的编程语言有几个。\n\n```func twoSum(nums []int, target int) []int {\nmapv := make(map[int]int)\nfor i, v := range nums {\nif val, ok := mapv[target-v]; ok {\nreturn []int{i, val}\n}\nmapv[v] = i\n}\nreturn nil\n}```\n\n```import (\n\"testing\"\n\"reflect\"\n)\n\nfunc TestTwoSum(t testing.T) {\ntc := []int{1, 2, 3, 4}\ntarget := 5\nexpected := []int{2, 1}\ngot := TwoSum(tc, target)\n\n// because slice type can only be compared to nil via ==\n// use DeepEqual instead, it does check the length and each elements of two slice array\n// you can find details via source code or here(https://gist.github.com/nevermosby/d373dfc07335c51477087f0f544e3b7c).\nif !reflect.DeepEqual(got, expected) {\nt.Errorf(\"TwoSum(%v, %v) = %v, expect %v\", tc, target,got, expected)\n}\n}```\n\n• -v，老司机肯定懂，通过添加verbose参数，输出更多信息，这里如果你的单元测试里使用Log或Logf，就会全量输出相关信息\n• -cover，大家也能猜出一二，顾名思义，就是开启测试覆盖率分析，即在测试执行完毕后同时输出该单元测试可以覆盖源代码行数的比例\n\n1. 单元测试文件是以_test.go命名\n2. 测试函数是以Test作为命名前缀\n3. 测试函数的入参是testing.T\n4. 单元测试文件没有main函数\n5. 对单个测试文件执行测试\n\n• 普通测试函数（Test）\n• 基准测试函数（Benchmark）\n• 示例函数（Example）\n\n#### 个人推荐的测试姿势——表格驱动\n\n```func TestFoo(t testing.T) {\n// setup code\nt.Run(\"A=1\", func(t testing.T) { … })\nt.Run(\"A=2\", func(t testing.T) { … })\nt.Run(\"B=1\", func(t testing.T) { … })\n// teardown code\n}```\n\n```func TestFoo(t testing.T) {\n// define test data by table\ntestData := []struct {\ncaseName string\ninput string\nwant bool\n}{\n{\"if-A\",\"A\",true},\n{\"if-B\",\"B\",true},\n{\"if-A&B\",\"AB\",false},\n{\"if-C\",\"C\",false},\n}\nfor _, test := range testData {\nt.Run(test.name, func(t testing.T){\nif got := Foo(test.input); got != test.want {\nt.Errorf(\"Foo(%v) = %v, expect %v\", test.input ,got, want)\n}\n})\n}\n}```\n\n`gotests` makes writing Go tests easy. It’s a Golang commandline tool that generates table driven tests based on its target source files’ function and method signatures. Any new dependencies in the test files are automatically imported.\n\n```# Firstly, create the target test file\n# Otherwise \"gotests\" will throw errors for \"no such file or directory\"\n> touch twosum_test.go\n# \"-all\" means to generate go tests for all functions and methods in target file\n# \"-w\" means to write output to files instead of stdout\n> gotests -all -w twosum_test.go twosum.go\n> Generated TestTwoSum```\n\n```import (\n\"reflect\"\n\"testing\"\n)\n\nfunc TestTwoSum(t testing.T) {\ntype args struct {\nnums []int\ntarget int\n}\ntests := []struct {\nname string\nargs args\nwant []int\n}{\n}\nfor _, tt := range tests {\nt.Run(tt.name, func(t *testing.T) {\nif got := TwoSum(tt.args.nums, tt.args.target); !reflect.DeepEqual(got, tt.want) {\nt.Errorf(\"TwoSum() = %v, want %v\", got, tt.want)\n}\n})\n}\n}```\n\nAlt+Insert 可以选择生成函数的类型，但内容上只是一个模板，比gotests生成的要简单很多，参见以下操作视频：\n\n1.", null, "雪儿?说道：" ]	[ null, "https://secure.gravatar.com/avatar/c7293066193afab9b205629571374688", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.64749986,"math_prob":0.87800866,"size":4454,"snap":"2021-43-2021-49","text_gpt3_token_len":2285,"char_repetition_ratio":0.09820225,"word_repetition_ratio":0.015521064,"special_character_ratio":0.2429277,"punctuation_ratio":0.1704871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96005267,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T23:48:54Z\",\"WARC-Record-ID\":\"<urn:uuid:902c3bac-3760-41ad-8aeb-57f5deeaa0fa>\",\"Content-Length\":\"63120\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a599bff-ca57-4ea6-ab0b-453c5a266aeb>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea9b3d81-e70d-4464-b005-c08481877cab>\",\"WARC-IP-Address\":\"35.189.182.55\",\"WARC-Target-URI\":\"https://davidlovezoe.club/wordpress/archives/419\",\"WARC-Payload-Digest\":\"sha1:3U7CY3X25YVZQ2X46HK7OO6NWRTCAHLE\",\"WARC-Block-Digest\":\"sha1:HFXZNI4QBYNOQXQOKWWGO3NRFJHL5C4H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363226.68_warc_CC-MAIN-20211205221915-20211206011915-00025.warc.gz\"}"}
http://bonneshonneur.pw/8159592748/4702/one-tailed-two-tailed-test-difference.html	[ " One tailed two tailed test difference - so, depending on\nHome\n\n# One tailed two tailed test difference\n\n### Difference Between One-tailed and Two-tailed Test (with\n\nBasis of Comparison One-tailed Test Two-tailed Test; Meaning: A statistical hypothesis test in which alternative hypothesis has only one end, is known as one tailed test A two-tailed test allows you to determine if two means are different from one another. A direction does not have to be specified prior to testing. A direction does not have to be specified prior to testing One-tailed tests have more statistical power to detect an effect in one direction than a two-tailed test with the same design and significance level. One-tailed tests occur most frequently for studies where one of the following is true statisticslectures.com - where you can find free lectures, videos, and exercises, as well as get your questions answered on our forums In statistical significance testing, a one-tailed test and a two-tailed test are alternative ways of computing the statistical significance of a parameter inferred from a data set, in terms of a test statistic. A two-tailed test is appropriate if the estimated value may be more than or less than the reference value, for example, whether a test taker may score above or below the historical.\n\n### The Difference Between One-Tailed & Two-Tailed Testing\n\nThis project was created with Explain Everything ™ Interactive Whiteboard for iPad In essence, one-tailed tests allow for the possibility of an effect in just one direction where with two-tailed tests, you are testing for the possibility of an effect in two directions - both positive and negative\n\n### One-Tailed and Two-Tailed Hypothesis Tests Explained\n\nA statistical test is based on two competing hypotheses: the null hypothesis H0 and the alternative hypothesis Ha. The type of alternative hypothesis Ha defines if a. The choice of one-tailed or two-tailed tests follows the logic of the hypothesis that is being tested! The one-tailed test, if appropriate, will be more powerful I'm not sure that the difference between one and two tails would particularly matter for sample size. In general, the difference between a one-tailed test and a two. This lesson explores the difference between the one-tailed and two-tailed tests. We will look at what they mean in statistical testing, as well as when you should and should not use them\n\n### One-Tailed and Two-Tailed Tests - YouTub\n\n• I am currently working on my dissertation and one of my committee members suggested that I should have used a one-tailed test as I have a directional hypothesis, but I think that a two-tailed test is just as appropriate based on several of the reasons listed on the blog\n• A two tailed test tests for a difference in either direction. Thus the P value would be the area under the t distribution to the right of t=1.92 PLUS the area under.\n\n### One- and two-tailed tests - Wikipedi\n\n1. This entry was posted in A/B Testing, Conversion Optimization, Statistical Significance, Statistics and tagged composite hypothesis, null hypothesis, one-sided, one-tailed, statistical significance, t test, two-sided, two-tailed, z test\n2. As you can see, the one-tailed test requires a less extreme t-value (1.725) to produce a statistically significant result in the right tail than the two-tailed test (2.086). In other words, a smaller effect is statistically significant in the one-tailed test\n3. The difference between a one-tailed and two-tailed test of significance is that a one-tailed test looks for an increase or decrease in the parameter and a two-tailed.\n4. · One-tailed test looks at the probability that the sample mean was either greater than, or less than or equal to · Two-tailed test, sees if two means are different from each other (ie from.\n\n### choosing between a one tailed and a two tailed test - YouTub\n\nIf you were able to reject the null in a two-tailed test you will be able to reject it in a one-tailed too (with the right side of the association chosen). Conduct a one-tailed test if your 2-tailed test was not enough powered While studying for my stats course, I was trying to understand the difference between one-tailed and two-tailed hypothesis tests. Specifically, why does the one-tailed test reject the null while th.. A couple of assumptions were made: that the distribution is normal, A two-tailed test was used at a confidence level of 95%, the critical value = 1.96. canadacouncil.ca Les hypothèses suivantes ont été posées : la distribution est normal e, un te st bilatéral a été fait à un niveau de confiance de 95 %, et la valeur critique est de 1,96 The test of such a hypothesis is nondirectional or two‐tailed because an extreme test statistic in either tail of the distribution (positive or negative) will lead to the rejection of the null hypothesis of no difference Two-tailed tests are more widely used in clinical trials than the one-tailed test, in view of the fact that the two-tailed test is more stringent and more conservative (4646Norman GR, Streiner DL. From: Clinical Trials (Second Edition) , 201", null, "You will often see a statistical test or a hypothesis referred to as one-tailed or two-tailed. A one-tailed hypothesis is simply one that specifies the direction of a difference or correlation, while a two-tailed hypothesis is one that does not In the 2003 report, Table 2, the one-tailed Fisher exact test p-value for the 2003 nest occupancy rate (Nesting Activity) is reported as 0.109: it appears that the wrong tail was used in the test - the correct value is 0.957 Should you use a one-tailed test or a two-tailed test for your data analysis? Posted March 14, 2017 When creating your data analysis plan or working on your results, you may have to decide if your statistical test should be a one-tailed test or a two-tailed test (also known as directional and non-directional tests respectively)\n\nTwo-tailed test (Image Source) One-tailed Test (Image Source) Chris Stucchio does a great job explaining the difference between the two tests in context Assess individual situations to determine whether a one-tailed or two-tailed test is necessary. Define acceptance sampling. Skill Level Intermediate. 1h 56m Duration. 733,992 Views. Show More Show Less. Related Courses. Preview course. Excel Statistics E.\n\nA one-tailed test is a statistical test in which the critical area of a distribution is one-sided so that it is either greater than or less than a certain value, but not both Figure showing one-tailed and two-tailed tests with the corresponding values of the z statistic As can be seen from the above figure, the value of the z statistic in a one-tailed test is 1.65. This means that a test statistic greater than 1.65 will be considered statistically significant, resulting in rejection of the null hypothesis A two-tailed test splits your significance level and applies it in both directions, thus each direction is only half as strong as a one-tailed test (which puts all the significance in one direction) and thus requires more subjects to reach significance A two-tailed test would be used to test these null hypotheses: There will be no significant difference in IQ scores between males and females. There will be no significant difference between blue collar and white collar workers. There is no significant difference in strength between Superman and the average person. A one-tailed probability is exactly half the value of a two-tailed probability A one-tailed test looks for an increase or decrease in the parameter whereas a two-tailed test looks for any change in the parameter (which can be any change- increase or decrease). We can perform the test at any level (usually 1%, 5% or 10%)\n\n### One-Tailed vs Two-Tailed Tests (Does It Matter?) - CX\n\n• One-tailed and two-tailed tests. More significance testing videos. Hypothesis testing and p-values. One-tailed and two-tailed tests . This is the currently selected item. Z-statistics vs. T-statistics. Small sample hypothesis test. Large sample proportion.\n• Note that, if we wanted to use the same 95% confidence level we employed in the one-tailed test, we would need the white areas to each correspond to 2.5% of the possible values; that is, we would need to use critical values corrseponding to those for a one-tailed test at the 97.5% confidence level\n• g a hypothesis test, its time to learn about the difference between two-tailed and one tail tests. To help you understand this let us take two different scenarios\n• One of the first steps is to look up a z-score, and in order to do that, you need to know if it's a one tailed test or two. You can figure this out in just a couple of steps. You can figure this out in just a couple of steps\n\n### What is the difference between a two-tailed and a one-tailed\n\n• By way of example, if you used a two-tailed test (which you should not), you would reject the null if the test statistic is large, regardless of whether it is negative or positive\n• It's not always true that the significance of a two-tailed test is twice the significance of a one-tailed test, that's a feature of this particular example. But the two-tailed significance is always the same or higher as the significance of a one-tailed test with the same data. In this case, we are testing the hypothesis that the program makes no difference against the alternative that it.\n• The null hypothesis for the two-tailed test is π = 0.5. By contrast, the null hypothesis for the one-tailed test is π ≤ 0.5. Accordingly, we reject the two-tailed hypothesis if the sample proportion deviates greatly from 0.5 in either direction. The one-tailed hypothesis is rejected only if the sample proportion is much greater than 0.5. The alternative hypothesis in the two-tailed test is.\n• One-Tailed and Two-Tailed Tests. One important concept in significance testing is whether to use a one-tailed or two-tailed test of significance\n\nBest Answer: That depends on your data. Look at the grades of people taking sports and the grades of people that are not taking sports A two-tailed test is a statistical test in which the critical area of a distribution is two-sided and tests whether a sample is greater than or less than a certain. Two-tailed tests are more widely used in clinical trials than the one-tailed test, in view of the fact that the two-tailed test is more stringent and more conservative (46, 47). Read full chapter Biostatistics—Part Yes, a two-tailed test is correct because the maximum amount consumed by men can be more or less than that of women. Experimentally, you should establish the rejection region before you collect the data, and as it is possible in theory at least that men could drink less than women, you should include that The two tailed test is called a two tailed test because the rejection region can be in either tail. Here's what the right tailed test looks like on a graph: As you can see, the rejection region (shaded in yellow) is to the right of the graph\n\n### Difference between one and two tailed tests - answers\n\nVariations of the t-Test: 2 Sample 1 tail 1 2 Sample t-Test (1 tailed, equal variance) Suppose we have two samples of ceramic sherd thickness collected from a One- and two-tailed tests explained In statistical significance testing , a one-tailed test and a two-tailed test are alternative ways of computing the statistical significance of a parameter inferred from a data set, in terms of a test statistic One-tailed and two-tailed Test: When comparing two groups of continuous data, the null hypothesis is that there is no real difference between the groups (A and B)\n\nTwo-tailed tests are much more common than one-tailed tests in scientific research because an outcome signifying that something other than chance is operating is usually worth noting. One-tailed tests are appropriate when it is not important to distinguish between no effect and an effect in the unexpected direction. For example, consider an experiment designed to test the efficacy of a. Now, if you are doing a one-tailed test, this means that you predicted that the difference will be in one side of the normal curve but not the other, so if you had a .05 significant level then all of the .05 is in the tail that you predicted the difference would be in The difference from the one‐tailed situation discussed earlier occurs because the alternative hypothesis for a two‐tailed test is non‐contiguous, being divided into two separate regions by the null hypothesis When I am doing two-tailed t-test to compare two means (Null Hypothesis: µ1 =µ2, Alternative: µ1≠ µ2) the result no significant difference at alpha = 0.05 and. A two-tailed test is a statistical test in which the critical area of a distribution is two-sided and tests whether a sample is greater than or less than a certain range of values\n\nTraductions en contexte de one tailed test en anglais-français avec Reverso Context : Treatment groups should be compared to a control group using techniques such as ANOVA followed by pairwise comparisons (e.g. Dunnett's one tailed test) and the criterion for statistical difference, for example, p <= 0,05 The significance test is always defined as either one-tailed or two-tailed. A two-tailed test is a test that will be interpreted if the effect meets the criterion for significance and falls in either direction, and is appropriate for the vast majority of research studies. A one-tailed test is a test that will be interpreted only if the effect meets the criterion for significance and falls in. Testing Hypotheses • Overview • 5 Steps for testing hypotheses Chapter 13 - 1 • Research and null hypotheses • One and two-tailed tests • Type 1 and Type. Here we should apply a two-tailed test because we are interested in a difference in either direction. Our null hypothesis is that the two population means are equal, H 0 : μ 1 = μ 2 while the alternative hypothesis is that the two population means are not equal, H 1 : μ 1 ≠ μ 2\n\n### What is the difference between one-tailed test or two-tailed\n\nTwo of these correspond to one-tailed tests and one corresponds to a two-tailed test. What is Two Tail Test : A two-tailed test is a statistical test in which the critical area of a distribution is two-sided and tests whether a sample is greater than or less than a certain range of values One-Sided or One-Tailed Hypothesis TestsIn most applications, a two-sided or two-tailed hypothesis testis the most appropriate approach. This approach is based onthe expression of the null and alternative hypotheses asfollows: H0: = 170 vs H1: ≠ 170To test the above hypothesis, we set up the rejection andacceptance regions as shown on the next slide, where we areusing = 0.05\n\n### One-Tailed Vs. Two-Tailed Tests: Differences & Examples ..\n\n• CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you.\n• In other words, a single tailed t-test at 10% significance level have the rejection area either in left or right side of the mean, while for two tailed t-test at 10% significance level have 5% rejection area on the left side & remaining 5% rejection area on the right side of the mean\n• ed. Hypothesis states related for studies are provided\n• Comparing one- and two-tailed tests (with a constant alpha level and sample size), the probability of rejection will be higher for a. the two-tailed test b. the one-tailed test, if you have correctly predicted the direction of the difference c. neither, the probability of rejection does not change d. the test with the most conservative test statisti\n• If the test is performed using the actual population mean and variance, rather than an estimate from a sample, it would be called a one- or two-tailed Z test The statistical tables for Z and for t provide critical values for both one- and two-tailed tests\n\nA two-sided hypothesis and a two-tailed test should be used only when we would act the same way, or draw the same conclusions if we discover a statistically significant difference in any direction. But this is clearly not how one would act in making an actual business decision with a two-tailed test One-Tailed vs. Two-Tailed Tests Remember that, in looking at the differences between the two groups, if the researcher is trying to determine if there is an improvement or an increase in the dependent variable being measured (or a deterioration or a decrease), then this is a one-tailed test H 1: μ ≠ μ 0, where a difference is hypothesized and this is called a two-tailed test. The exact form of the research hypothesis depends on the investigator's belief about the parameter of interest and whether it has possibly increased, decreased or is different from the null value\n\n### One-tailed and Two-tailed Tests - The Analysis Facto\n\nYep one-tailed means that you are predicting a certain thing will happen (e.g. X will be more than Y, X will increase Y etc) and two-tailed means you're just predicting some relationship (e.g. X will affect Y) but you're not predicting exactly how it will affect it This means that given p and t values from a two-tailed test, you would reject the null hypothesis of a greater-than test when p/2 < alpha and t > 0, and of a less-than test when p/2 < alpha and t < 0 Questions. When to use one-tailed and when two-tailed test? One-tailed probability. In the James Bond case study, Mr. Bond was given 16 trials on which he judged. If a population parameter is hypothesized to be greater than or less than some value, a one-tailed test is used. When no direction is indicated in the research hypothesis, a two-tailed test is used. A two-tailed test will show whether or not there is a difference in the values of the variables involved\n\nMyth #4: One should not do a one-sided test after a two-tailed test or after a one-tailed test in the opposite direction This claim is encountered in much of the literature and is enshrined in statistical guidelines and requirements in the form of requiring pre-specification One-Tailed Test: Definition & Examples . Chapter 14 for two separate one-tailed tests of significance. A one-tailed test is a type of significance test used when the hypothesis being tested is. one tailed or two tailed hypothesis One tailed or two tailed hypothesis 8.9 out of 10 based on 786 ratings. A one-tailed writing a college level paper test has one tailed or two tailed hypothesis the entire 5% of the alpha level in one tail god is one essay (in either the left, thesis statement for a research paper or the right tail) If the test is performed using the actual population mean and variance, rather than an estimate from a sample, it would be called a one- or two-tailed Z test The statistical tables for Z and for t provide critical values for both one- and two-tailed tests The following simulation is like one on the previous page, but a 1-tailed test is used to compare the population means. Samples of size 20 are again selected from two populations, both of which are normal with mean 75 and standard deviation 8\n\nA one-tailed test is appropriate if the estimated value may depart from the reference value in only one direction, for example, whether a machine produces more than one-percent defective products. Alternative names are one-sided and two-sided tests; the t. Chi-square tests were preplanned to test the difference between positive and negative attitudes for each questionnaire, with the test being one-tailed for the. one-tail test is applied. If words such as ^change, the same, different/difference and so on _ are used in the claim of the question (≠ is used in H 1), a two-tailed test is applied. In a one-tailed test, the critical region has just one part (the gree. There are really two different one-tailed t-tests, one for each tail. In a one-tailed t-test, all the area associated with a is placed in either one tail or the other When doing a two-tailed t-test, the null hypothesis is a particular value, and there are two alternative hypotheses, one positive and one negative. When the null hypothesis is rejected, one of the alternative hypotheses is accepted and the other rejected, depending upon the direction of the results\n\nThe two ways of carrying out statistical significance test of a A one-tailed test is used to ascertain if there is any relationship between. When you conduct a test. Calculates the T-test on TWO RELATED samples of scores, a and b. This is a two-sided test for the null hypothesis that 2 related or repeated samples have identical average (expected) values. Two-sided means two-tailed Attachments. One-tailed Vs.Two-tailed test.docx What is the difference between a one-tailed and a two-tailed test of significanc... read more by clicking on the.\n\nThe right tailed test and the left tailed test are examples of one-tailed tests. They are called one tailed tests because the rejection region (the area where you would reject the null hypothesis) is only in one tail. The two tailed test is called a two tailed test because the rejection region can be in either tail Most of the tests that you would use to compare two samples in R have been set by default to run a two-tailed test. This means that, when running these tests, you.\n\n### Difference between one-tailed and two-tailed testing? - Cross\n\n1. Two tailed test is a hypothesis test performed if there is no reason to think the parameter is higher or lower than the assumed parameter of the null hypothesis. The null hypothesis should be rejected when the test value is in either of the two critical regions. Two tailed test simply show a difference\n2. The U test is used to find the difference of distribution between two small groups and also for finding the difference of means between two small groups. It make me confused because my data is.\n3. e if there is a difference between the statistic and a value, then we have a two-tail test\n4. Explain the difference between a left-tailed, two-tailed, and right-tailed test and how do wetell the direction of the test by looking at the pair of hypothesis. read more Sk1ll\n5. The analysis is done using formal test of hypothesis for the difference between two means of financial performances of corporations with three different supervisory board characteristics using both directional one-tailed and nondirectional two-tailed tests\n\nOn the other hand, A one-sided hypothesis and a one-tailed test should be used when we would act a certain way, or draw certain conclusions, if we discover a statistically significant difference in a particular direction, but not in the other direction The use of a one or two tailed test is an important issue. What every analysis I've read fails to understand is that the issue cannot be decided at the level of statistics alone - it needs to be decided from within the context of a decision procedure Two-Tailed Test of Population Proportion The null hypothesis of the two-tailed test about population proportion can be expressed as follows: where p 0 is a hypothesized value of the true population proportion p E) One-tailed tests and two-tailed tests - A two-tailed test is a significance test in which the alternative hypothesis specifies no direction of difference from the population parameter value such as m. Equal rejection regions are in both tails of the distribution. In a one-tailed test w", null, "A two-tailed test is used when the two samples being compared are related in some way. For example, you would use a two-tailed test if one random sample was 15 quarter horses and the second sample was 15 sires or dams of those same horses A two-tailed test requires us to consider both sides of the Ho distribution, so we split and place half in each tail. A directional hypothesis only considers one tail (the other tail is ignored as irrelevant to H1), thus all of can be placed in that one tail\n\n### The Lonely Journey: What are the differences between one\n\n1. We didn't say one tailed two tailed hypothesis whether the drug would lower the response time one tailed two tailed hypothesis or raise the response time. in case the research hypothesis entails the direction of interrelation or difference, then one-tailed test is applied, but if the research hypothesis does not signifies the direction of interaction or difference, we use two-tailed test.\n2. It depends on what the experimenter is interested in. Suppose a new drug is tested to see if it reduces blood pressure. The experimenter would use a one-tailed test\n3. e whether there is a difference at all, no directionality implied. They test for differences in.\n4. Many translated example sentences containing two-tailed test - French-English dictionary and search engine for French translations\n5. Chapter 9: One-Tailed Tests, Two-Tailed Tests, and Logarithms Chapter 9 Outline • A One-Tailed Hypothesis Test: The Downward Sloping Demand Curv\n\n### One-tailed vs Two-tailed Tests of Significance in A/B Testing\n\n1. Two-tailed test is used when the alternate hypothesis does not have a single direction of deviation from the null rather, it only says that the null is not true\n2. A test of a statistical hypothesis , where the region of rejection is on only one side of the sampling distribution , is called a one-tailed test. For example, suppose the null hypothesis states that the mean is less than or equal to 10\n3. Meilleure réponse: If you are testing whether two quantities are different, use a two-tailed test, as the first quantity can be either less than or.\n4. A two tailed test looks at whether the sample is is greater than and/or less than the distribution. Two tailed means that you are looking at both sides (known as tails) of a distribution and seeing their relationship to the sample\n5. One and Two Tail Tests We use the Analyze->Compare Means->Independent Samples T Test SPSS command sequence to test difference between means hypotheses. In the Output for this procedure, the cell headed Sig. (2-tailed) provides the basis.", null, "### When Can I Use One-Tailed Hypothesis Tests? - Statistics By Ji\n\n1. Answer: One-tailed test happens if we want to test the difference between two groups in one direction (ie. either greater or lesser) Two-tailed test happens if we want to test the difference between two groups in two.\n2. However, if the alternate hypothesis was stated so that a one-tailed test could be used (that is, male lawyers make more money than female lawyers), it is entirely conceivable that the one-tailed test could find that a \\$2,000 difference is statistically significant at a P value less than\n3. A two-tailed test will test both if the mean is significantly greater than x and if the mean significantly less than x. The mean is considered significantly different from x if the test statistic is in the top 2.5% or bottom 2.5% of its probability distribution, resulting in a p-value less than 0.05\n4. I am doing hypothesis testing using t-test and p-value. I calculate my p-value based on the degrees of freedom and the t-test value. The p-value is a two-tailed.\n5. We calculated the volume ratio in 52 donors and tested the difference between the segmental volume ratios with a one-tailed t-test. 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https://math.stackexchange.com/questions/3784325/can-we-apply-reidemeister-moves-to-self-crossings-of-one-component-only-of-a-tr	[ "# Can we apply Reidemeister moves to self crossings of one component only of a “trivial” link and keep the others unchanged?\n\nSuppose $$L=L_1 \\cup L_2 \\cup L_3$$ be a classical link of three components. Suppose $$L$$ is an unlink, that is $$L$$ can be splitted into three simple closed curves. Assume that $$L$$ has a diagram in 2-plane such that\n\n• There are some crossings between $$L_1$$ and $$L_2$$ And between $$L_1$$ and $$L_3$$.\n• There are some crossings between $$L_2$$ and $$L_3$$.\n• There are some self intersections of each $$L_i, i=1,2,3$$.\n\nI have two questions:\n\nQ1: Can we apply a finite sequence of Reidemeister moves to the self crossings of $$L_1$$ only so that all self crossings of $$L_1$$ are eliminated without affecting the other crossings of $$L$$. If the answer is yes, then this can be generalized to $$L_2$$ and $$L_3$$. This means we can first eliminate self crossings of $$L$$ and then we may deal with the crossings between the components. Is this possible always?\n\nQ2: Can we first apply a finite sequence of Reidemeister moves to split $$L_1$$ from the link $$L$$ without affecting the crossings which are not between $$L_1$$ and $$L_i, i=1,2,3$$. If so, then we may split the components of the link first and then deal with the self crossings of each component.\n\nI know that the answer of both of my questions is no if the link $$L$$ is not trivial link. The whitehead link with five crossings is a counterexample. How about the case of an unlink?\n\nSo, I believe the answer is no for each question. Here is not so much as a proof, but a couple candidate links for you to consider. I think it is easy to see these are indeed unlinks. But I got away with two components in each. Making similar links with 3 components should not be hard, but I leave that to you.", null, "The first link has twists in each component that cannot be removed without some type II moves or type III moves first, between the shared crossings.\n\nThe second question is harder. I think this link does it though. You cannot unlink the red, keeping the blue fixed without a type II first. If you fix the red, then you will need a few moves between red and blue to get to a type I move with the blue before you can go any further.\n\nNow, this is in now way a proof. I am not sure how you might have to go about showing these are in fact, counter-examples. Perhaps looking at the Reidemeister graph and somehow showing that every path to the trivial diagram must go though a \"bad\" point, or something. Cooking up counter-examples to conjectures like this is a good way to build intuition into how crappy diagrams really are in knot theory. But its what we got. Good luck.\n\nEdit\n\nSo upon further reflection, I am now sure that my link for Question 2 does not hold up. You can slide the blue link down the red and off of it without any blue only Reidemeister move.\n\nOne definition of the unlink is a link where each component simultaneously bounds an embedded disk and these disks are all disjoint. This allows us to pick any component $$L_1$$ and isotope it along this disk to a small unknot component which has no crossings with any other components. This lets us keep the other components fixed the whole time we are running this isotopy. But I don't see how to guarantee that the crossings of $$L_1$$ are not subjected to Reidemeister moves or not in this.\n\nQuick summary: We can fix any components of an unlink we like and isotope the rest away.\n\n• I think the answer is yes for the second question. Since each of the three components bounds an embedded disk in 3-space, we may move the closed disk of the component $L_1$ freely in 3-space so that all crossings between $L_1$ and the other components are eliminated, that is $L_1$ is splitted from $L$. In general, we may split the components first and then eliminate the self crossings of each component. Am I right? Aug 17, 2020 at 23:53\n• @user113715 This is where it gets sticky. Maybe it doesn't matter for your situation, but the isotopy we have from the disk might produce Reidemeister moves between the crossing of just $L_1$. You can definitely split the components like you say, but I can't see a guarantee that the diagram of $L_1$ alone will be unmodified after you split it. But if you only care about getting them apart, leaving the other components unchanged, you are good. 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http://global-sci.com/intro/article_detail/aamm/12108.html	[ "Volume 8, Issue 4\nSpectral-Collocation Method for Volterra Delay Integro-Differential Equations with Weakly Singular Kernels\n\nAdv. Appl. Math. Mech., 8 (2016), pp. 648-669.\n\nPublished online: 2018-05\n\nPreview Full PDF 3 694\nExport citation\n\nCited by\n\n• Abstract\n\nA spectral Jacobi-collocation approximation is proposed for Volterra delay integro-differential equations with weakly singular kernels. In this paper, we consider the special case that the underlying solutions of equations are sufficiently smooth. We provide a rigorous error analysis for the proposed method, which shows that both the errors of approximate solutions and the errors of approximate derivatives decay exponentially in L ∞ norm and weighted L 2 norm. Finally, two numerical examples are presented to demonstrate our error analysis.\n\n• Keywords\n\nVolterra integro-differential equations, spectral Jacobi-collocation method, pantograph delay, weakly singular kernel.\n\n65R20, 65M70, 45D05\n\n• BibTex\n• RIS\n• TXT\n@Article{AAMM-8-648, author = {}, title = {Spectral-Collocation Method for Volterra Delay Integro-Differential Equations with Weakly Singular Kernels}, journal = {Advances in Applied Mathematics and Mechanics}, year = {2018}, volume = {8}, number = {4}, pages = {648--669}, abstract = {\n\nA spectral Jacobi-collocation approximation is proposed for Volterra delay integro-differential equations with weakly singular kernels. In this paper, we consider the special case that the underlying solutions of equations are sufficiently smooth. We provide a rigorous error analysis for the proposed method, which shows that both the errors of approximate solutions and the errors of approximate derivatives decay exponentially in L ∞ norm and weighted L 2 norm. Finally, two numerical examples are presented to demonstrate our error analysis.\n\n}, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.2015.m1088}, url = {http://global-sci.org/intro/article_detail/aamm/12108.html} }\nTY - JOUR T1 - Spectral-Collocation Method for Volterra Delay Integro-Differential Equations with Weakly Singular Kernels JO - Advances in Applied Mathematics and Mechanics VL - 4 SP - 648 EP - 669 PY - 2018 DA - 2018/05 SN - 8 DO - http://dor.org/10.4208/aamm.2015.m1088 UR - https://global-sci.org/intro/aamm/12108.html KW - Volterra integro-differential equations, spectral Jacobi-collocation method, pantograph delay, weakly singular kernel. AB -\n\nA spectral Jacobi-collocation approximation is proposed for Volterra delay integro-differential equations with weakly singular kernels. In this paper, we consider the special case that the underlying solutions of equations are sufficiently smooth. We provide a rigorous error analysis for the proposed method, which shows that both the errors of approximate solutions and the errors of approximate derivatives decay exponentially in L ∞ norm and weighted L 2 norm. Finally, two numerical examples are presented to demonstrate our error analysis.\n\nXiulian Shi & Yanping Chen. (2020). Spectral-Collocation Method for Volterra Delay Integro-Differential Equations with Weakly Singular Kernels. Advances in Applied Mathematics and Mechanics. 8 (4). 648-669. doi:10.4208/aamm.2015.m1088\nCopy to clipboard\nThe citation has been copied to your clipboard" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87683845,"math_prob":0.75821954,"size":1334,"snap":"2020-24-2020-29","text_gpt3_token_len":291,"char_repetition_ratio":0.10977443,"word_repetition_ratio":0.8087432,"special_character_ratio":0.22713643,"punctuation_ratio":0.14529915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.962568,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-11T20:03:01Z\",\"WARC-Record-ID\":\"<urn:uuid:b18ca145-2234-4cc9-910c-20175d0d4b24>\",\"Content-Length\":\"41171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:97ea5842-a126-4620-8310-353fb4a89da8>\",\"WARC-Concurrent-To\":\"<urn:uuid:39b48913-cfd1-46a6-a33e-6fbc52e485ba>\",\"WARC-IP-Address\":\"47.52.67.10\",\"WARC-Target-URI\":\"http://global-sci.com/intro/article_detail/aamm/12108.html\",\"WARC-Payload-Digest\":\"sha1:WRUFYU6A7R4MMG3NRYWAAMAAYAMAYO7K\",\"WARC-Block-Digest\":\"sha1:DNFMNB4FPGKRHIB7EBSY3ZJXJ6QIYE4Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655937797.57_warc_CC-MAIN-20200711192914-20200711222914-00202.warc.gz\"}"}