{ // 获取包含Hugging Face文本的span元素 const spans = link.querySelectorAll('span.whitespace-nowrap, span.hidden.whitespace-nowrap'); spans.forEach(span => { if (span.textContent && span.textContent.trim().match(/Hugging\s*Face/i)) { span.textContent = 'AI快站'; } }); }); // 替换logo图片的alt属性 document.querySelectorAll('img[alt*="Hugging"], img[alt*="Face"]').forEach(img => { if (img.alt.match(/Hugging\s*Face/i)) { img.alt = 'AI快站 logo'; } }); } // 替换导航栏中的链接 function replaceNavigationLinks() { // 已替换标记，防止重复运行 if (window._navLinksReplaced) { return; } // 已经替换过的链接集合，防止重复替换 const replacedLinks = new Set(); // 只在导航栏区域查找和替换链接 const headerArea = document.querySelector('header') || document.querySelector('nav'); if (!headerArea) { return; } // 在导航区域内查找链接 const navLinks = headerArea.querySelectorAll('a'); navLinks.forEach(link => { // 如果已经替换过，跳过 if (replacedLinks.has(link)) return; const linkText = link.textContent.trim(); const linkHref = link.getAttribute('href') || ''; // 替换Spaces链接 - 仅替换一次 if ( (linkHref.includes('/spaces') || linkHref === '/spaces' || linkText === 'Spaces' || linkText.match(/^s*Spacess*$/i)) && linkText !== 'OCR模型免费转Markdown' && linkText !== 'OCR模型免费转Markdown' ) { link.textContent = 'OCR模型免费转Markdown'; link.href = 'https://fast360.xyz'; link.setAttribute('target', '_blank'); link.setAttribute('rel', 'noopener noreferrer'); replacedLinks.add(link); } // 删除Posts链接 else if ( (linkHref.includes('/posts') || linkHref === '/posts' || linkText === 'Posts' || linkText.match(/^s*Postss*$/i)) ) { if (link.parentNode) { link.parentNode.removeChild(link); } replacedLinks.add(link); } // 替换Docs链接 - 仅替换一次 else if ( (linkHref.includes('/docs') || linkHref === '/docs' || linkText === 'Docs' || linkText.match(/^s*Docss*$/i)) && linkText !== '模型下载攻略' ) { link.textContent = '模型下载攻略'; link.href = '/'; replacedLinks.add(link); } // 删除Enterprise链接 else if ( (linkHref.includes('/enterprise') || linkHref === '/enterprise' || linkText === 'Enterprise' || linkText.match(/^s*Enterprises*$/i)) ) { if (link.parentNode) { link.parentNode.removeChild(link); } replacedLinks.add(link); } }); // 查找可能嵌套的Spaces和Posts文本 const textNodes = []; function findTextNodes(element) { if (element.nodeType === Node.TEXT_NODE) { const text = element.textContent.trim(); if (text === 'Spaces' || text === 'Posts' || text === 'Enterprise') { textNodes.push(element); } } else { for (const child of element.childNodes) { findTextNodes(child); } } } // 只在导航区域内查找文本节点 findTextNodes(headerArea); // 替换找到的文本节点 textNodes.forEach(node => { const text = node.textContent.trim(); if (text === 'Spaces') { node.textContent = node.textContent.replace(/Spaces/g, 'OCR模型免费转Markdown'); } else if (text === 'Posts') { // 删除Posts文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } else if (text === 'Enterprise') { // 删除Enterprise文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } }); // 标记已替换完成 window._navLinksReplaced = true; } // 替换代码区域中的域名 function replaceCodeDomains() { // 特别处理span.hljs-string和span.njs-string元素 document.querySelectorAll('span.hljs-string, span.njs-string, span[class*="hljs-string"], span[class*="njs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换hljs-string类的span中的域名（移除多余的转义符号） document.querySelectorAll('span.hljs-string, span[class*="hljs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换pre和code标签中包含git clone命令的域名 document.querySelectorAll('pre, code').forEach(element => { if (element.textContent && element.textContent.includes('git clone')) { const text = element.innerHTML; if (text.includes('huggingface.co')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 处理特定的命令行示例 document.querySelectorAll('pre, code').forEach(element => { const text = element.innerHTML; if (text.includes('huggingface.co')) { // 针对git clone命令的专门处理 if (text.includes('git clone') || text.includes('GIT_LFS_SKIP_SMUDGE=1')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 特别处理模型下载页面上的代码片段 document.querySelectorAll('.flex.border-t, .svelte_hydrator, .inline-block').forEach(container => { const content = container.innerHTML; if (content && content.includes('huggingface.co')) { container.innerHTML = content.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 特别处理模型仓库克隆对话框中的代码片段 try { // 查找包含"Clone this model repository"标题的对话框 const cloneDialog = document.querySelector('.svelte_hydration_boundary, [data-target="MainHeader"]'); if (cloneDialog) { // 查找对话框中所有的代码片段和命令示例 const codeElements = cloneDialog.querySelectorAll('pre, code, span'); codeElements.forEach(element => { if (element.textContent && element.textContent.includes('huggingface.co')) { if (element.innerHTML.includes('huggingface.co')) { element.innerHTML = element.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { element.textContent = element.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); } // 更精确地定位克隆命令中的域名 document.querySelectorAll('[data-target]').forEach(container => { const codeBlocks = container.querySelectorAll('pre, code, span.hljs-string'); codeBlocks.forEach(block => { if (block.textContent && block.textContent.includes('huggingface.co')) { if (block.innerHTML.includes('huggingface.co')) { block.innerHTML = block.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { block.textContent = block.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); }); } catch (e) { // 错误处理但不打印日志 } } // 当DOM加载完成后执行替换 if (document.readyState === 'loading') { document.addEventListener('DOMContentLoaded', () => { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); }); } else { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); } // 增加一个MutationObserver来处理可能的动态元素加载 const observer = new MutationObserver(mutations => { // 检查是否导航区域有变化 const hasNavChanges = mutations.some(mutation => { // 检查是否存在header或nav元素变化 return Array.from(mutation.addedNodes).some(node => { if (node.nodeType === Node.ELEMENT_NODE) { // 检查是否是导航元素或其子元素 if (node.tagName === 'HEADER' || node.tagName === 'NAV' || node.querySelector('header, nav')) { return true; } // 检查是否在导航元素内部 let parent = node.parentElement; while (parent) { if (parent.tagName === 'HEADER' || parent.tagName === 'NAV') { return true; } parent = parent.parentElement; } } return false; }); }); // 只在导航区域有变化时执行替换 if (hasNavChanges) { // 重置替换状态，允许再次替换 window._navLinksReplaced = false; replaceHeaderBranding(); replaceNavigationLinks(); } }); // 开始观察document.body的变化，包括子节点 if (document.body) { observer.observe(document.body, { childList: true, subtree: true }); } else { document.addEventListener('DOMContentLoaded', () => { observer.observe(document.body, { childList: true, subtree: true }); }); } })(); \n```\n* **+20** if your code is still executable in both languages now (after the *charScore-changes*)\n* **+50** if running the code with the different interpreters does different things (eg order change, prints other things etc)\n* **+15** if your languages are Perl and Java\nThe answer with the highest score wins.\n**EDIT:** I don't think that [this](https://codegolf.stackexchange.com/questions/4481) is a duplicate (but very releated) because it has no \"different language\" limits and no score. On the other hand, I didn't know about so-called \"polyglot\"-code so maybe I am wrong.\n \n[Answer]\n## PHP + [Lenguage](http://esolangs.org/wiki/Lenguage), 165\nScoring calculation: 95 + 20 + 50.\n```\n !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~\n```\nIn PHP, this outputs the string verbatim (since there is no ` col=new ArrayList();BufferedImage i=ImageIO.read(new File(in.nextLine()));int h=i.getHeight(),w=i.getWidth(),k=in.nextInt(),x,y,z=0,r,g,b;while(col.size()1:i=[a for a in i if a];o+=[a[0]for a,b in combinations(i,2)if a[0]==b[0]];[j.pop(0)for j in i if j[0]==o[-1]]\n o+=i[0]\n return o\n```\n## Example:\n**Input:**\n```\ni = [['one', 'two', 'four'],\n ['two', 'three', 'five'],\n ['one', 'three', 'four']]\nprint f(i)\n```\n**Output:**\n```\n['one', 'two', 'three', 'four', 'five']\n```\n[Answer]\n# Mathematica, 115 bytes\nCould probably use some `Thread`s here...\n```\nSort[Union@##,Function[{a, b},Less@@Function[c,Mean[Flatten@DeleteCases[FirstPosition@c/@{##},_Missing]]]/@{a,b}]]&\n```\nWritten as a pure function, used via `*code*[, , ...]`. I don't see how this is dependent on the number of lists. If there is any ambiguity about the order, it chooses a pseudorandom working case.\n]"}}},{"rowIdx":13633,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/44451/edit).\nClosed 9 years ago.\n[Improve this question](/posts/44451/edit) \nYour task is to take the following simple C code and make a program which will interpret this code and produce a correct output, and be able to handle small modifications to the original code. The C code is\n```\nint x = 4;\nint y = 5;\nint z;\nz = x * y;\nif(z>10)\n{\n printf(\"%d is bigger than 10!\",z);\n}\n```\nOr as one line of text.\n```\nint x = 4; int y = 5; int z; z = x * y; if(z>10){ printf(\"%d is bigger than 10!\",z); }\n```\nThe program must satisfy the following conditions:\n* x and y can be changed to any integer.\n* \"is bigger than 10!\" must not be printed if z is smaller than 10.\n* The value of 10 can be changed to any integer (not including the 10 in the printed text).\n* The program must be able to read the c program in as one line of text.\n* Adding spaces shouldn't break the program.\nThe winning solution is the shortest code to satisfy the criteria. Libraries are not allowed.\n \n[Answer]\n# Perl 5, 76 bytes\n```\n$_=<>,s/ //g,s/[=>](-?\\d+)/$.*=$1/ge,($./=$1)>$1&&say\"$. is bigger than $1!\"\n```\nInput must be in one line. Negative integers are allowed. Tabs may break program (but spaces are fine, even between `-` and the number). \n**Ungolfed:**\n```\n$_ = <>; # read input\n# $. = 1; $. is the number of lines read on STDIN, which should be 1.\ns/ //g; # remove whitespace from $_\ns/[=>](-?\\d+)/ $. *= $1 /ge; # get product of all integers after = or >\n$. /= $1; # divide away the last integer found (inside the if)\nif($. > $1){ \n say \"$. is bigger than $1\";\n}\n```\nThis simply extracts the integers and does the product/comparison afterwards.\n**Example**:\n```\n$ perl -E '$_=<>,s/ //g,s/[=>](-?\\d+)/$.*=$1/ge,($./=$1)>$1&&say\"$. is bigger than $1!\"'\n> int x =4; int y = - 5 ; int f0_23; f0_23 = x* y ; if ( f0_23> - 30){ printf(\"%d is bigger than 10!\",f0_23) ; }\n-20 is bigger than -30!\n```\n]"}}},{"rowIdx":13634,"cells":{"text":{"kind":"string","value":"[Question]\n [\nA little context:\n> \n> * Book has 411 Pages\n> * Read a random number of pages on the first day which is unknown\n> * Number of pages to read next day is the square of the sum of the digits of the page currently on\n> \n> \n> I.e. on Page 36: read 81 pages next day\n> * On day 6 The square answer directly adds to current page number to make 411\n> * Finish Book\n> \n> \n> How many pages did she read on the first day and how many pages did\n> she read on the 6 days?\n> \n> \n> \nThis is what I have so far:\n```\nfrom itertools import accumulate, count\ndef read_book( start ):\n for i in range( 6 ): yield start; start += sum( int( digit ) for digit in str( start ) ) ** 2\ndef find_start( pages ):\n for start in count( 1 ):\n for read in read_book( start ): pass\n if read == pages: return start\nprevious_answer, start = 0, find_start( 411 )\nfor i, read in enumerate( read_book( start ), 1 ):\n print( \"Pages read on day\", i ,\"was\", read-previous_answer ,\"\\t(Now on page\", str(read) + \")\\n\")\n previous_answer += read-previous_answer\n```\nOutput:\n> \n> Pages read on day 1 was 61 (Now on page 61)\n> \n> \n> Pages read on day 2 was 49 (Now on page 110)\n> \n> \n> Pages read on day 3 was 4 (Now on page 114)\n> \n> \n> Pages read on day 4 was 36 (Now on page 150)\n> \n> \n> Pages read on day 5 was 36 (Now on page 186)\n> \n> \n> Pages read on day 6 was 225 (Now on page 411)\n> \n> \n> \nIs there a more efficient way? Who can do it in the least no. chars\n \n[Answer]\n# gcc - 176 chars\nI don't grok Python (it is Python right?) so I don't understand how your example works. Is it actually treating the page number as a string?\nI'm not sure but (at least in C) string handling uses more characters than integers. So to sum the digits of an integer between 0 and 1000 I use:\n```\nfor(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;\n```\nWhere x is the input and s is the sum.\nIf you want the sum from some x for 7 days you could:\n```\nfor(c=2;c<7;x+=s*s,c++)\n for(s=0,i=1;i<1001;i*=10)s+=(x%i*10)/i;\n```\nWhere x is the input and the output.\nTo find the sum for 7 days which matches the number of pages in the book (p):\n```\nfor(y=1;y f(1,1e5)\n[1] 25201 52921 80641\n```\n]"}}},{"rowIdx":13636,"cells":{"text":{"kind":"string","value":"[Question]\n [\nAnyone who has spent any time around a deck of playing cards will recognize the riffle shuffle. The deck is cut in half and the two stacks are riffled simultaneously into the same pile before being tidied up in a sometimes visually appealing way.\nThis code-golf challenge is to write a function/method/(language equivalent) that performs the riffle shuffle on an array/stack/(language equivalent) of any size in the language of your choice. \nAs input, your function will except an array (language equivalent), plus how many times to run the shuffle. When done, it will return the shuffled array.\nYour riffle shuffle function must implement the following algorithm:\n1. Divide the `deck` into two `stacks`\nA `stack` must have a minimum of 40% of the total `cards` in the `deck` and a\nmaximum of 60%\n2. Choose a random `stack` to start with\n3. Drop 1 - 7 cards from the bottom of the `starting stack` to the top of the newly shuffled `deck`\n4. Drop 1 - 7 cards from the bottom of the `other stack` to the top of the newly shuffled `deck`\n5. Repeat #3 and #4 until only one `stack` remains\n6. Drop the remaining `cards` from the remaining `stack` on top of the newly shuffled `deck`\nSmallest character count will be chosen as the answer in 1 week (May 29th).\nThis is the sample deck of cards to use:\n```\n[\n'AH', '2H', '3H', '4H', '5H', '6H', '7H', '8H', '9H', '10H', 'JH', 'QH', 'KH',\n'AC', '2C', '3C', '4C', '5C', '6C', '7C', '8C', '9C', '10C', 'JC', 'QC', 'KC',\n'KD', 'QD', 'JD', '10D', '9D', '8D', '7D', '6D', '5D', '4D', '3D', '2D', 'AD',\n'KS', 'QS', 'JS', '10S', '9S', '8S', '7S', '6S', '5S', '4S', '3S', '2S', 'AS'\n]\n```\nYour answer should contain:\n* Code-golfed code\n* What the sample deck looks like after being shuffled twice\n* Un-golfed code\n \n[Answer]\n# Python (131)\nWaaaay too long, even for something that came to be after a \"careful\" reading of the rules: \n```\nDrop 1 - 7 cards from the bottom of the starting stack to the top of the newly shuffled deck\n```\nDoens't actually say this should be random, so I'll just always drop 1 card.\n```\nDivide the deck into two stacks\nA stack must have a minimum of 40% of the total cards in the deck and a maximum of 60%\n```\nSure. I'll just always do about 50%.\nAnyways, the code:\n```\ndef s(d):\n b=[d.pop()for _ in range(len(d)/2)];r=[];b,d=(b,d)if id(4)%2 else(d,b)\n while b and d:r+=[b.pop(),d.pop()]\n return r+b+d\n```\nSample output:\n```\nIn [17]: s(s([\n ...: 'AH', '2H', '3H', '4H', '5H', '6H', '7H', '8H', '9H', '10H', 'JH', 'QH', 'KH',\n ...: 'AC', '2C', '3C', '4C', '5C', '6C', '7C', '8C', '9C', '10C', 'JC', 'QC', 'KC',\n ...: 'KD', 'QD', 'JD', '10D', '9D', '8D', '7D', '6D', '5D', '4D', '3D', '2D', 'AD',\n ...: 'KS', 'QS', 'JS', '10S', '9S', '8S', '7S', '6S', '5S', '4S', '3S', '2S', 'AS'\n ...: ]))\nOut[17]: \n['AD',\n 'KH',\n 'AC',\n 'KS',\n '2D',\n 'QH',\n '2C',\n 'QS',\n '3D',\n 'JH',\n '3C',\n 'JS',\n '4D',\n '10H',\n '4C',\n '10S',\n '5D',\n '9H',\n '5C',\n '9S',\n '6D',\n '8H',\n '6C',\n '8S',\n '7D',\n '7H',\n '7C',\n '7S',\n '8D',\n '6H',\n '8C',\n '6S',\n '9D',\n '5H',\n '9C',\n '5S',\n '10D',\n '4H',\n '10C',\n '4S',\n 'JD',\n '3H',\n 'JC',\n '3S',\n 'QD',\n '2H',\n 'QC',\n '2S',\n 'KD',\n 'AH',\n 'KC',\n 'AS']\n```\nThis is actually the code as I wrote it, I didn't specifically golfed any code. If you just want an explanation, please say so.\n[Answer]\n# Cobra - 307\nDon't have access to a compiler atm, so this *should* work, but I'll be able to check later.\n```\ndef f(d,n) as List\n r=Random()\n x=d.count\n for m in n\n p=q=[]\n for c in x\n if c%2>0,p.add(d.pop)\n else,q.add(d.pop)\n if r.next(0,2)>0,a,b=p,q\n else,a,b=q,p\n for c in x\n for z in r.next(0,7),if a.count>0,d.add(a.pop)\n for z in r.next(0,7),if b.count>0,d.add(b.pop)\n return d\n```\n[Answer]\n# Python 84\n## Golfed:\n```\ndef s(d):\n e=len(d)/2;B=d[:e];C=d[e:];D=[]\n for x in range(e):D+=B[x],C[x]\n return D\n```\n## Output Deck:\n```\n['AH', 'AC', 'KD', 'KS', '2H', '2C', 'QD', 'QS', '3H', '3C', 'JD', 'JS', '4H', '4C', '10D', '10S', '5H', '5C', '9D', '9S', '6H', '6C', '8D', '8S', '7H', '7C', '7D', '7S', '8H', '8C', '6D', '6S', '9H', '9C', '5D', '5S', '10H', '10C', '4D', '4S', 'JH', 'JC', '3D', '3S', 'QH', 'QC', '2D', '2S', 'KH', 'KC', 'AD', 'AS']\n```\n## Un-golfed:\n```\ndef Shuffle(Deck):\n Stack1 = Deck[:len(Deck)/2] # 2nd Half of Deck\n Stack2 = Deck[len(Deck)/2:] # 1st Half of Deck\n Output = [] # Initialize Output List\n for index in range(len(Deck)/2):\n Output.append(Stack1[index]) # Append Output List with element from Stack1\n Output.append(Stack2[index]) # Append Output List with element from Stack2\n return Output # Return Output\n```\nUses the same loopholes as @synthetica. \n[Answer]\n# Python, 69 chars (actually takes number of times to shuffle as input)\nIf one follows the algorithm, always picking 1 card at each step of 1-7 cards (as it wasn't explicitly said to have to be random), then with an array like:\n```\n[1, 2, 3, 4, 5, 6]\n```\nWe have one of two results depending on which stack is picked first:\n```\n[1, 2, 3] [4, 5, 6] ==> [1, 4, 2, 5, 3, 6]\n[4, 5, 6] [1, 2, 3] ==> [4, 1, 5, 2, 6, 3]\n```\nThe following algorithm accomplishes this `n` times, using `id(b)%9>4` as a source of binary randomness[1]:\n```\ndef s(d,n):\n for i in range(n):b=d[:];F=id(b)%9>4;d[F::2]=b[26:];d[1-F::2]=b[:26]\n```\nSample output:\n```\n>>> o = ['AH', '2H', '3H', '4H', '5H', '6H', '7H', '8H', '9H', '10H', 'JH', \n 'QH', 'KH', 'AC', '2C', '3C', '4C', '5C', '6C', '7C', '8C', '9C', '10C', 'JC',\n 'QC', 'KC', 'KD', 'QD', 'JD', '10D', '9D', '8D', '7D', '6D', '5D', '4D', '3D', \n '2D', 'AD', 'KS', 'QS', 'JS', '10S', '9S', '8S', '7S', '6S', '5S', '4S', '3S',\n '2S', 'AS']\n>>> d = o[:]; s(d, 2); d\n['KD', 'KS', 'AH', 'AC', 'QD', 'QS', '2H', '2C', 'JD', 'JS', '3H', '3C', '10D', '10S', '4H', '4C', '9D', '9S', '5H', '5C', '8D', '8S', '6H', '6C', '7D', '7S', '7H', '7C', '6D', '6S', '8H', '8C', '5D', '5S', '9H', '9C', '4D', '4S', '10H', '10C', '3D', '3S', 'JH', 'JC', '2D', '2S', 'QH', 'QC', 'AD', 'AS', 'KH', 'KC']\n>>> d = o[:]; s(d, 2); d\n['AC', 'AH', 'KS', 'KD', '2C', '2H', 'QS', 'QD', '3C', '3H', 'JS', 'JD', '4C', '4H', '10S', '10D', '5C', '5H', '9S', '9D', '6C', '6H', '8S', '8D', '7C', '7H', '7S', '7D', '8C', '8H', '6S', '6D', '9C', '9H', '5S', '5D', '10C', '10H', '4S', '4D', 'JC', 'JH', '3S', '3D', 'QC', 'QH', '2S', '2D', 'KC', 'KH', 'AS', 'AD']\n```\n[1] This gives about a 44.4%/55.5% skew:\n```\n>>> r = {0: 0, 1: 0}\n>>> for i in range(1000000):\n r[int(id(i)%9>4)]+=1\n>>> r\n{0: 555517, 1: 444483}\n```\nFor two extra characters we can get a much better 50.5%/49.5% result:\n```\n>>> r = {0: 0, 1: 0}\n>>> for i in range(1000000):\n r[int(id(i)%99>49)]+=1\n>>> r\n{0: 505050, 1: 494950}\n```\n]"}}},{"rowIdx":13637,"cells":{"text":{"kind":"string","value":"[Question]\n [\nWrite the shortest POSIX-form regular expression that matches only non-negative integers divisible by 9. The non-negative integer should be given in base 10.\n \n[Answer]\n# 202,071 bytes\nObviously, [the solution](https://gist.github.com/nhahtdh/9405617#file-divisibleby9-re) is too large to be included in this post.\nI use [JFLAP](http://www.jflap.org/) to draw the [DFA](https://gist.github.com/nhahtdh/9405617#file-g23235-jff), then generate regex from it. The output is then processed in a text editor to replace `(0+9)` with `[09]` and `+` with `|`.\n[Here is the whole gist](https://gist.github.com/nhahtdh/9405617), with some testing code in Java.\n]"}}},{"rowIdx":13638,"cells":{"text":{"kind":"string","value":"[Question]\n [\nYour mission is to write 2 functions that accept a list of N integers (positive or negative).\nN can be higher than 2.\nThe first one returns the greater common divisor of the N numbers\n**Example**: [10, 15, 20] -> 5\nthe second one returns the smallest common multiple of the N numbers\n**Example**: [10, 15, 20] => 60\nSmallest code (in number of characters) wins!\n \n[Answer]\n### Ruby 38 characters\n```\ng=->a{a.inject:gcd}\nl=->a{a.inject:lcm}\n```\ndemo:\n```\nar = [10,15,20]\np g[ar] #=> 5\np l[ar] #=> 60\n```\n[Answer]\n## J, 12 characters\n```\ng=.+./\nl=.*./\n```\nA bit pointless giving them aliases like this really.\n[Answer]\n# Smalltalk (Smalltalk/X) (73 chars)\n```\na:=#(10 15 20). \ng:=a fold:[:i :e|i gcd:e]. \nl:=a fold:[:i :e|i lcm:e]. \n```\nwithout fold: (Ansi Smalltalk), use (102 chars):\n```\na := #(10 15 20).\ng := a inject:a first into:[:i :e | i gcd:e]. \nl := a inject:a first into:[:i :e | i lcm:e]. \n```\n]"}}},{"rowIdx":13639,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n Questions without an **objective primary winning criterion** are off-topic, as they make it impossible to indisputably decide which entry should win.\nClosed 6 years ago.\n[Improve this question](/posts/18710/edit) \nThe number 1 can be written as sum of fractions of the form 1/p (where p>1). There are different ways to write it. You will be given an integer k which denotes the number of fractions to be used to get the sum as 1. You need to find the largest possible value of p among all possible ways. See sample input for clarity\n**Input Format:**\nThe first line contains number of test cases t. Then t lines follow each having a single integer k.\n**Output:**\nYou need to print the largest value of p modulo 10^9 + 7.\n**Constraints:**\n```\n1<=t<=10^5\n2<=k<=10^6\n```\n**Sample Input** --------------- **Sample Output**\n```\n2 2\n2 6\n3\n```\n**Explanation**\nThere is only one way to write '1' as sum of two fractions. 1/2 + 1/2. hence the largest value is 2.\nIn the second test case, the only possible ways as sum of three fractions are:\n```\n1/3, 1/3, 1/3\n1/6, 1/2, 1/3\n1/4, 1/4, 1/2\n```\nAmong these the second representation has the largest denominator which is '6' and hence the answer.\n \n[Answer]\n# Java\nI solved this basic problem [on Stack Overflow](https://stackoverflow.com/a/18087894/752320) already, so I might as well do it here, too. Follow the link for a deeper explanation of the algorithm, but there's *no reason* to calculate all the possible fractions if you just want to know the largest denominator.\nThe only major difference between the code here and on SO is that I'm using modular multiplication, since each result should be `mod 10^9 + 7` and this speeds it up considerably.\n```\npublic class Egyptian {\n static int range = 1000000;\n static long[] mods;\n static long mod = 1000000007;\n public static void main(String[] args){\n mods = new long[range+1];\n mods[1] = 1;\n for(int k=2;k<=range;k++){\n mods[k] = (mods[k-1] * (mods[k-1]+1)) % mod;\n System.out.println(mods[k]);\n }\n }\n}\n```\n**Note:** This code simply calculates and outputs **every** value for inputs 2 to 10^6. As a popularity contest, I thought a simple proof of concept would be enough. Feel free to downvote if you disagree. \nIt runs in a few seconds, but most of that time is taken by the *million-line* output. If you just want to fill the array with correct values, it finishes in *well* under a second. You can then read/write the input/output at the speed of IO.\n**Output:** (only from 2-40, can't exactly paste a million lines here)\n```\n2\n6\n42\n1806\n3263442\n56876256\n530809436\n905136947\n994707547\n127363070\n613638513\n614624128\n690044800\n363952039\n128981958\n502041307\n671991252\n275513076\n791142674\n33665861\n222603452\n718053684\n214817910\n348558872\n748018254\n148507635\n647419241\n330373625\n661987133\n851958095\n407158983\n684304220\n916206786\n752612499\n438658510\n487129601\n996482080\n806772777\n958888343\n```\n[Answer]\n# APL (50)\n```\n⍪{⌈/∊v/⍨{g=+/(g←∧/⍵)÷⍵}¨v←,∘.,∘v⍣(n-1)⊢v←⍳!n←⎕}¨⍳⎕\n```\nThis code skips the 'p modulo 109 + 7' computation. But the algorithms is so slow (exponential) that you can barely get a result for k=5, let alone 109. Therefore by all practical means it can be considered equivalent to a piece of code that computed it :-)\n**Example**\n```\n ⍪{⌈/∊v/⍨{g=+/(g←∧/⍵)÷⍵}¨v←,∘.,∘v⍣(n-1)⊢v←⍳!n←⎕}¨⍳⎕\n⎕:\n 3\n⎕:\n 2\n⎕:\n 3\n⎕:\n 4\n 2\n 6\n24\n```\nThe `⎕:` is just the \"input prompt\" when used interactively.\n]"}}},{"rowIdx":13640,"cells":{"text":{"kind":"string","value":"[Question]\n [\nI am exploring a unique approach to the classic problem of [Reverse Polish Notation](https://en.wikipedia.org/wiki/Reverse_Polish_notation) on [code.golf](https://code.golf/). Unlike the conventional solutions, my focus lies on leveraging the `eval` function in GolfScript. This twist was inspired by a comment in a related [thread](https://codegolf.stackexchange.com/questions/221/reverse-polish-notation), where the absence of `eval` was deemed a limitation.\n> \n> \"It's a shame no eval is allowed, otherwise the GolfScript solution is 1 character: ~. :-P\" – C. K. Young\n> \n> \n> \nDespite being intrigued by the idea of solving the problem with minimal characters, I'm encountering challenges in implementing the solution. The specifications, as outlined on [code.golf](https://code.golf/reverse-polish-notation#golfscript), involve computing a list of arguments in polish notation stored in the stack `n`. The goal is to evaluate and print the result of each expression on a separate line. The expected results are non-negative integers not exceeding 32,767, with guaranteed exact integer results for division.\nI am currently attempting a concise GolfScript solution using the `n*` stack initialization for arguments. Following C. K. Young's hint, I have tried variations like `~n*` and `n*~` to no avail. I believe the solution is close but needs some expert insight to refine it further.\nIf you are well-versed in GolfScript or have experience with similar challenges, I would greatly appreciate your assistance. My goal is to gain a deeper understanding of this fascinating language and unlock the potential of the `eval` function within the constraints of the problem.\n \n[Answer]\n> \n> The specifications [...] involve computing a list of arguments in polish notation stored in the stack `n`.\n> \n> \n> \nI think you are slightly misunderstanding the code.golf specification. The input is not a variable `n`, but rather a list of strings already present on the stack when the program begins.\nIn GolfScript, there is a variable called `n`, but it is not the input to the program; it is a string containing a single newline.\nYour attempt `~n*` is pretty close. The problem is that `~` only does `eval` if its argument—the top of the stack—is a string. In this case, the top of the stack is actually a list of strings.\nTherefore, you have to use the map `%` function to loop over each element of the list:\n```\n{~}%\n```\nOnce we have this, we need to join the results on newlines. We can simply do:\n```\n{~}%n*\n```\nAs I think you figured out already, `n*` means join (`*`) on newlines (`n`).\nHowever, if we use the function each (`/`) instead of map (`%`), we can just push the newline to the stack during each iteration, saving a byte:\n```\n{~n}/\n```\nWhen a GolfScript program has finished execution, the entire stack is printed without a seperator. Here, each `eval`'d result is followed by a newline on the stack, which does the equivalent of joining the list on newlines.\nTry the final program on [TIO.run](https://tio.run/##RVbbjiw1DHyfr4jO46xa40viJN@DBEJCBwS8Ifj1PVXVmUWrzk7n4tjlcrl/@f23n//66c9f//j789vYLduzxZwzWnpbc@D1ah/N@57Zwmd7tdEx8dECu2ti3Zuvwk6Pvb2NWonVCwvWwoonPvg85s421554TRirHC3DZL4v2pxRRcOb06FzNca9I2DNeNJXLy1Z81p94Lfn3B3vT83St9cjI7Jl980h5M5skTFhOW3CZ59lrW@cv/CXHmPCeV@4yrGb007j07rpV@KIteGbd2IhemJ1Kjq8rlWrzUTAIQyXpqcJIdOuHLjyJe/hTneep3ccnU7o5qe2vB4EZfF0zllthSIVQvTMjalKn8DcN9J1YKenSMSI23s@yB0hi5ipdEaNTkjCO4B2oEGveDFdQRxnpJ9RfRcTzLEcKbxaB2pYUR5o1qZSxByDITkRdAZJIwcYUoAV42TwqQdBjTtnrqNIV2@Fa8656wE7aa4XbBrgyCZqmQf@7MmjA2EMpIQYLuUZaRr0aHsoB48sJgtBeiTCAEZTXsmP6FGI0nGzu0L2vdJaKtRhluIHmWeYRm30FIabeO@@HBctwkk/o6/qZ@TVdd/EEtllQiC2j/v2O3Lv@hHWM7WcsTVtU@S@N2WiJEHFwXNw5MYuMsVgP/G8BIBy@egG/gTuBXTbGfYYMQ@8wHLEu7aVFm/Ta8iOfrwOEwN8J5oxFk4wqigLIT9Ft7y3paGaRG8FPIcfMsOjh1fkauIGjviukkX8nTo3YuZUmxmGErQBSmNcSxsHD5OI6xQPn0DF5hmvN9HMOKfxkhSAHBMIQwxQJW9ZuW60Htb2SCkVEjB7kdGgyS5SRuPrNrL2Bj9y5J17H29L5DYWtkrewEBUMn1OW4UI3AffPE1klZKVAYbotpdSsbkAy9S5kfPUt4Fe8ACuM78d5is7Dow15WPo7pfSpGrB7dA647pTWi9oLbjBKncjPTtqEnyJ@9bRz2HvFKXqnSlZ7l2zIDv0cxon57C7L0BGVlv06UM1sUZJRjaoC9M@8lQBeF/hjxi24GnU3BTKoVSiMHbn/r7oUPGe0ZN1X1V3sUEcFlEdPk@qeGFuwnA9kAxoxSBJwZUttDo8U1PqSdITP50Lgy22lgWIyVnGgQEsRWmzC8RQTqLWyShK7QZmGlkncTNR1P7vL0doHDDGKrgJB9glPJl@Fbct3@/eAAfFNzB5dFGVRYdxdMpFUv6hxuieC/dphQxC80AKiFUuVUDed7A/9v42RFMIgLwnAQAACu2Q6zo1/vz6O4SRZ89TSXc6n1/74OiK6C0rQ9GnIYt0AK0QMUCKULsFVCExebert27rP1CoEI1cFLnRc2GX6DqpkSuOWmUnn@hIffa3jt7lCb0COpWkPM3VACHZDCacu6SDEpzZ1Qx5/3RVEDjiKlXwWq0InwqSEu7fZ6/2pz5nUAWkp4OT3A7NYD@Oue2rj93PcvE4xaqaRfptGJeuMeGP9DKQtyiAiIg9HErZ7RTsU45TQ8jmpHxpLEUOyoMJw9mXY45bxDyXgsI29WtQ1RHXPHmDiiL36JWdTXL2k0znHfgiIRCk40g/HzbPQwyoa7JnQ@9v3Ucvi1Bz3pswdMjwrVmbWpJlfE1oefHLQ5IFcaLzaK9DRhntNn0dfHx1BxVELMQ9zFlwfYSIbKEgVCKlbv3t@@vzn/@@//v6/PwB \"GolfScript – Try It Online\")!\n]"}}},{"rowIdx":13641,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/238370/edit).\nClosed 2 years ago.\n[Improve this question](/posts/238370/edit) \n# Challenge\nDraw lines in ASCII corresponding to given inputs:\nin the first line there are given size of rows and columns of the view and number of separators between each point\nin the second line is given list of lines separated by space\na line contains the position of head and tail coordinates (within the size of columns and rows) separated by space (x1 x2 y1 y2).\n> \n> In the below examples we used character '#' for filled spot, '.' for free spots and ' ' for the separator but you are free to use any other printable ASCII characters.\n> \n> \n> \nSample input 1:\n```\n33 15 1\n5 0 31 0 31 0 31 10 31 10 27 14 27 14 27 4 27 4 0 4 0 4 0 14 0 14 27 14 31 10 5 10 5 10 5 0 5 0 0 4 31 0 27 4 5 10 0 14\n```\nSample output 1:\n```\n. . . . . # # # # # # # # # # # # # # # # # # # # # # # # # # # .\n. . . . # # . . . . . . . . . . . . . . . . . . . . . . . . # # .\n. . # # . # . . . . . . . . . . . . . . . . . . . . . . . # . # .\n. # . . . # . . . . . . . . . . . . . . . . . . . . . . # . . # .\n# # # # # # # # # # # # # # # # # # # # # # # # # # # # . . . # .\n# . . . . # . . . . . . . . . . . . . . . . . . . . . # . . . # .\n# . . . . # . . . . . . . . . . . . . . . . . . . . . # . . . # .\n# . . . . # . . . . . . . . . . . . . . . . . . . . . # . . . # .\n# . . . . # . . . . . . . . . . . . . . . . . . . . . # . . . # .\n# . . . . # . . . . . . . . . . . . . . . . . . . . . # . . . # .\n# . . . . # # # # # # # # # # # # # # # # # # # # # # # # # # # .\n# . . . # . . . . . . . . . . . . . . . . . . . . . . # . . # . .\n# . # # . . . . . . . . . . . . . . . . . . . . . . . # . # . . .\n# # . . . . . . . . . . . . . . . . . . . . . . . . . # # . . . .\n# # # # # # # # # # # # # # # # # # # # # # # # # # # # . . . . .\n```\nSample input 2:\n```\n31 15 0\n0 0 30 14 0 14 30 0 15 0 15 14 0 7 30 7\n```\nSample output 2:\n```\n##.............#.............##\n..##...........#...........##..\n....##.........#.........##....\n.....###.......#......###......\n........##.....#.....##........\n..........###..#..###..........\n.............#####.............\n###############################\n.............#####.............\n..........###..#..###..........\n.........##....#.....##........\n......###......#......###......\n....##.........#.........##....\n..##...........#...........##..\n##.............#.............##\n```\n**Bonus Points**\nUse line position format like so `(y1,x1)-(y2,x2)`.\ne.g.\n```\n31 15 0\n(0,0)-(14,30) (14,0)-(0,30) (0,15)-(14,15) (7,0)-(7,30)\n```\n**Scoring**\nThis is `code-golf`, so the shortest solution wins.\n---\nJust to finish this up, simple python(3) answer would be:\n[Try it online!](https://tio.run/##fVTRbtowFH3nK67oA3YxKClsrYiotI6t0/ZSbXtL8@CQEDwZJ42htVX129m1IdCuYQgcfM7xOfZ1dCu7XpZqdFXV22oq@SrNOGiWT3u9SVULtSY4Udk0p527N/S9eifQe8GkIt17E6aXXdqp/8GuEJs3Ro9sifjCE/HzYz98iZ6XOH47Q5VoVGIihV6TFa8IxjEx1JUUa0Ip7ZyBKuFJqKx80sArAZXUw06mm1RQTe5H5YI3m9dEE6xePnkyaydvPLloJz97Mm0nZ45Usp384sjqBPnVkUveTt46MtPt5HdHyhPkDySf9Iki/PLkiSL8RnL5WJ24OXdvLwt/oPp/kp8o0UXdHrFyr0txKGXNCpa@EtToUOAv9cIzuKlznaslX/U0SKFy4LIoa7FerjpZvoCUmIDZgJmQ2ZBOoAP4yQxMwYQwABPsAIuAdYDdAUaLQiEWglg4@TUEkEudwyD0vH3D23e8T@CpJpmhxwQP2B3Q@GZ24uc@1TAwloHFp3UL/DYYBP7rM73WJR1XZaj29hkuzcxpt8aDHXy9dobUBZw7j0Gz3sv9v0VZgwGhoOaqyPE80Acs5CHFilxmWEeEDZoYx1v841MPqG1Qe1iIBZjBNcYcvXbJfSzrG2gGg/0Oj4ebOdlu1/6eOdnvqZBlyiUIVrE7DzxMBTaMauNaRcRfT/Y9Yd6lUW7yOeneEdIb9obyzwYbzUN8kfRDev4QBwmNJ4MwoVEXZ@7JpzGPxWA8EYmrjzjWZ8xkrgin/TEb0yQqp3ESxeUwN2vsjsR3sJRItGQSnXC4cMMowTbmnKRz4kmkCY3iORFOI1B93myHHvLKJKoJ7Wzx5NvtKITwA95YgO/hKIBwDH4Yubkj3ODBS4dd/gU \"Python 3.8 (pre-release) – Try It Online\")\n```\na()\n```\nor python(1789) \\*without escape characters:\n[Try it online!](https://tio.run/##hVTLbtswELznK7buQWQiGVLiIIUA5@RjURS9uoKgSLTNQKYESW7F/ry7S1KvpEENJzR3hrO75JC17k6VevhSN9fVagXfKhUcq/IgCvglmlZWCjB8c/NDtNml7GALNLspxAFeGtEKdcrOrA990PjXRzhGPL4B/BQ9kvsIAuhDG9AY0BTQNtC38qgwFoE8EP0ZQhBlKyCIDK4XuH6HmwzZS8uKnk8ZTEDbwKBbaFuTydpjoRoLxVHTAlOGD6H5mpyGS5mmVQWyjXyBS4v@Y7VBwx91DXeH0D3ckkYwrDd08@tQNdCDVNBk6iiwH7iDYR8NU4qywH3EcI8iPeEaf5isY1QPUT0uxA3YwTOmmbRs5jvc1kVoB4GrcGpuRzRbNZ55XmZtC98rqTp3wmiCNJVKdmnKWlEesOcYu@iwKDPOOiB4bRyxDBlPjOJfpRIfap9EVsQ2vw9dJks3eZuFeKhKwxKgNQjQ4CxcNNnvEnOyVv4RMXSXuhQ@UKSNcWgxUX5q2i3z1p4P3mf6Bx7nEDyDuxE2@bGsXrJyiJlQXmGufTKeL6nSEVv1sbK8Wou@E6pglI9Nl2pxOrTINLbu/dlEuwm1NCJmojnnY@q8Ki9nNfmLut1HyWzjiNVUv99QwjnF2YkhzXeKnPh5teTQZ3gt0D60f5hrQVlerY8Whcnb3FThpy242oLo/xr3ya1h3yfzK@H2Y5CK3knNZFY/1cpgjegujRqgqyR/quws0hS2SEvTcyZVmq6sEgmT18hR7JzVzLhWqvrSMb5u61Li6I4or6qmkCrrREuWkfs4DqLEnImkHd4bZ4wir26556MT@Yw2tTgwAm@GD8kbUZdZLpjHyc7eFFhsgccc6qTQ9rakV6OVJP8qfTbdy2ATy1kX1lcbNKlQbEbkdxjbcKtnLsfy3uTWZCN/OilDXmd1TbeHng5m3gOWoznwiy6lH@hzHwYgGoDIAO4A6obQxWPgXgHOr9eHCKJHfKhZ6Ic8YNHGfwg50EjT0M5CP3q0II7Angz2RNhf \"Python 3.8 (pre-release) – Try It Online\")\n```\n\"\"\" Non-golfed version \"\"\"\nResault = \"\"\ndef bresenham(x0, y0, x1, y1):\n dx = x1 - x0\n dy = y1 - y0\n xsign = 1 if dx > 0 else -1\n ysign = 1 if dy > 0 else -1\n dx = abs(dx)\n dy = abs(dy)\n if dx > dy:\n xx, xy, yx, yy = xsign, 0, 0, ysign\n else:\n dx, dy = dy, dx\n xx, xy, yx, yy = 0, ysign, xsign, 0\n D = 2 * dy - dx\n y = 0\n for x in range(dx + 1):\n yield x0 + x * xx + y * yx, y0 + x * xy + y * yy\n if D >= 0:\n y += 1\n D -= 2 * dx\n D += 2 * dy\nclass Point:\n def __init__(self, x: int, y: int):\n self.x = x\n self.y = y\nclass Line:\n def __init__(self, head: Point, tail: Point):\n self.head = head\n self.tail = tail\ndef drawline(size: tuple, lines: list, chrs=('.', '#', ' ')) -> Resault:\n global Resault\n co = []\n for line in lines:\n co.extend(list(bresenham(\n line.head.x, line.head.y, line.tail.x, line.tail.y)))\n for column in range(size[1]):\n for row in range(size[0]):\n if (row, column) in co:\n Resault += chrs[1]\n else:\n Resault += chrs[0]\n if row != size[0]-1:\n Resault += chrs[2]*size[2]\n if column != size[1]-1:\n Resault += \"\\n\"\n return Resault\nif __name__ == \"__main__\":\n size = tuple(map(int, input().split()))\n coordinates = [i[::-1] for i in [list(map(int, j.split(','))) for i in [\n i.split('-') for i in input().replace(')', '').replace(\n '(', '').split(' ')] for j in i]]\n coordinates = [coordinates[i-4:i] for i in range(4, len(coordinates)+4, 4)]\n lines = []\n for c in coordinates:\n lines.append(Line(Point(c[0][0], c[0][1]), Point(c[1][0], c[1][1])))\n print(drawline(size, lines))\n```\n \n[Answer]\n# [Charcoal](https://github.com/somebody1234/Charcoal), 84 bytes, score unknown\n```\n≔⮌Ｉ⪪Ｓ θＵＯ⊟θ⊟θ.ＦＩＥ⪪ＳＥ⪪ι-⪪✂λ¹±¹¦¹,«≔Ｅ²↨Ｅι§μκ±¹ηＦ⊕⌈↔η«≔⁺·⁵Ｅ§ι⁰⁺λ∕×κ§ημ⌈↔ηκＪ⊟κ⊟κ#»»ＵＥ⊟θ\n```\n[Try it online!](https://tio.run/##dZFNboMwEIXXcAqLbsaSE5mmURZZpT@LVGobNbkAATdY2IaAiSJVOTudgdBs2gUazzz85uOR5kmdlonpulXT6IODT3VSdaPgKWk8bCujPaxd1fqtr7U7ABcsYhHnWI98GX7sTYnTTVnBEUdjjaYRil9lzQaft6T630uwm6xxNKHR0G6NThUYwWLB3tUh8QpiFOmJBFFwzr7D4IpONveCPSaIT2c0W/m1y9QZrGAFMf@a4DlHxKBnXLu0VlY5rzK8eNa2tbDaN6Vp8d183DKu2Zi2ATmdD9zjBlwmKQESEfhZn3SmYKetaqC4geSCWd5/8l9rSCkIK3htbbUr@2CLa7CDsMHoPER3FHBwCS/hy9krl11/AV923Sxm8ZzJEKSQfALxg5hJzqhSK4dOing@iFgZLHptQVo3OZkf \"Charcoal – Try It Online\") Link is to verbose version of code. Explanation:\n```\n≔⮌Ｉ⪪Ｓ θ\n```\nInput the first line, split on spaces, cast to integer and reverse.\n```\nＵＯ⊟θ⊟θ.\n```\nDraw the background of the desired size.\n```\nＦＩＥ⪪ＳＥ⪪ι-⪪✂λ¹±¹¦¹,«\n```\nInput the second line, splice on spaces, split each piece on `-`, chop off the `()`s, split again on `,`, cast to integer, and loop over the sets of coordinates.\n```\n≔Ｅ²↨Ｅι§μκ±¹η\n```\nCalculate the difference between the start and end points.\n```\nＦ⊕⌈↔η«\n```\nLoop over the number of steps needed to draw the line.\n```\n≔⁺·⁵Ｅ§ι⁰⁺λ∕×κ§ημ⌈↔ηκ\n```\nInterpolate between the start and end point.\n```\nＪ⊟κ⊟κ#\n```\nJump to the calculated position and output a `#`.\n```\n»»ＵＥ⊟θ\n```\nSpace the output horizontally as desired.\n]"}}},{"rowIdx":13642,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Is this even or odd?](/questions/113448/is-this-even-or-odd)\n (261 answers)\n \nClosed 4 years ago.\n### Task\nYour task is to write as short as possible program preferably in Brainf\\*\\*k which determines the last digit of given number after it's conversion from decimal to binary system.\n### Input\nAn integer in decimal system consisting of up to 200 digits. There is a newline character (ASCII 10) after the given number\n### Output\nOne digit (1 or 0) which is the last digit of given number after it's conversion to binary system.\n### Example\n`4372667135165131576213 -> 1`\n \n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), 13 bytes\n```\nBoole@OddQ@#&\n```\n[Try it online!](https://tio.run/##y00syUjNTSzJTE78n2b73yk/PyfVwT8lJdBBWe1/QFFmXomCg0JatImxuZGZmbmhsamhmamhsaGpuZmRoXHsfwA \"Wolfram Language (Mathematica) – Try It Online\")\n]"}}},{"rowIdx":13643,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Yo boy, must it sum](/questions/146084/yo-boy-must-it-sum)\n (26 answers)\n \nClosed 5 years ago.\nLet `n` be a positive integer then n = a + b + c for some a, b, and c that are palindrome integers. What is the largest possible integer `a` for `k` = 1 to 1\\_000\\_000?\nGolf this or have the fastest running time.\nNOTE: it's NOT the same as this [question](https://codegolf.stackexchange.com/questions/146084/yo-boy-must-it-sum) as I am asking for the largest palindrome component. The question just asks for ANY combination.\n \n[Answer]\nMy Julia solution\n```\nusing Base.Iterators\nis_palindrome(n) = begin\n dn = digits(n)\n all(dn .== reverse(dn))\nend\nlim = Int64(1e6)\nres = filter(is_palindrome,lim:-1:0)\nusing DataFrames\nlp = DataFrame(k = 1:1_000_000, largest_palindrome = Array{Int64,1}(lim))\n# found stores the integers backwardds\nok(lim, res, lp) = begin\n found = BitArray(lim)\n found .= false\n tot = Int128(0)\n cap = lim\n r = res[1]\n for r in res[1:end-1]\n bs = res[res .<= (cap - r)]\n cs = res[res .<= ceil((cap - r)/2)]\n res1 = vec([r + b + c for (b, c) in product(bs,cs)])\n res1 = res1[res1 .<= cap]\n for rr in res1\n if !found[rr]\n lp[rr,:largest_palindrome] = r\n tot = tot + r\n found[rr] = true\n end\n end\n for i in lim:-1:1\n if !found[i]\n cap = i\n break\n end\n end\n end\n (tot, found, lp)\nend\n```\n]"}}},{"rowIdx":13644,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Reverse stdin and place on stdout](/questions/242/reverse-stdin-and-place-on-stdout)\n (140 answers)\n \nClosed 6 years ago.\nJust take an input in any way and then print it backward (trailing newlines allowed). For example:\n```\nHello World -> dlroW olleH\nHi -> iH\n```\nYes, it is case sensitive. Shortest code wins\n \n[Answer]\n# Mathematica, 13 bytes\n```\nStringReverse\n```\n]"}}},{"rowIdx":13645,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Approximation of e](/questions/82604/approximation-of-e)\n (55 answers)\n \nClosed 7 years ago.\nFrom Wikipedia:\n> \n> The number e is an important mathematical constant that is the base of the natural logarithm. It is approximately equal to 2.71828, and is the limit of (1 + 1/n)n as n approaches infinity.\n> \n> \n> \n## Challenge\nCalculate the number e to 15 digits after the decimal point. \nThis means that your output must start with `2.718281828459045` or equivalent.\nYour program can output digits after the required output - these may be accurate or inaccurate, inaccuracy after 15 digits does not disqualify you from the challenge. Your program may also output nothing after the required 15 digits.\nThis is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"), shortest implementation in bytes wins.\n## Constraints\n* You may not use built in constants or hardcode the number.\n* Standard loopholes and rules apply.\n## Reference Implementation (Ruby)\n```\ndef e(i) \n e = 1.0 # make it floating point\n i.times { |n| # loop repeats n times\n e += 1.0/(1..n+1).reduce(:*) # factorial - no factorial function\n } # in the Ruby standard library\n return e\nend\ne(200) # Almost certainly will return e\n```\n \n[Answer]\n# Haskell, ~~33~~ 31 Bytes\n*-2 Bytes thanks to @xnor*\n```\nsum[1/product[1..i]|i<-[0..99]]\n```\nStraightforward implementation of the series definition. (I have posted a version of this to Reddit before.)\n[Answer]\n## Haskell, 25 bytes\n```\nf 99=0;f n=1+f(n+1)/n;f 0\n```\n26 bytes:\n```\nfoldr(\\x y->1+y/x)0[1..99]\n```\n]"}}},{"rowIdx":13646,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/79138/edit).\nClosed 7 years ago.\n[Improve this question](/posts/79138/edit) \n## Encrypting Challenge\nYour encrypting program must be able to do following things:\n1. Input password\n2. Generate random key\n3. Encrypt random key with set key \n \nThen, the decrypting program must be able to these things:\n4. Input encrypted password\n5. Input encrypted random key\n6. Decrypt random key\n7. Decrypt password with decrypted random key\n8. Print password\nThe encryption must be setted with the key. It must be a table encryption or a encryption which makes the unencrypted values to a part of the key.\nThis is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"). The shortest program wins!\n \n[Answer]\n# [Pyth](https://github.com/isaacg1/pyth), 16+16 = 32 bytes\nEncryption:\n```\ns@LG+VxLGQxLG*zl\n```\n[Try it online!](http://pyth.herokuapp.com/?code=s%40LG%2BVxLG*zlQxLG&input=%22encryption%22%0Akey&debug=0)\nDecryption:\n```\ns@LG-VxLGQxLG*zl\n```\n[Try it online!](http://pyth.herokuapp.com/?code=s%40LG-VxLGQxLG*zl&input=%22orabcndmmx%22%0Akey&debug=0)\nUses [Vigenère cipher](https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher).\n[![Table of Vigenère cipher](https://i.stack.imgur.com/uzXOD.png)](https://i.stack.imgur.com/uzXOD.png)\n]"}}},{"rowIdx":13647,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Mathematical Combination](/questions/1744/mathematical-combination)\n (36 answers)\n \nClosed 10 years ago.\nThe problem is to compute nCr = n!/(r!) \\* (n-r)! in the fewest characters possible. The input will be in the form: 1st line will have number of test cases and second line will have n and r for each testcase. You need to print ncr\ne.g.\n```\nInput\n1\n100 10\nOutput:\n17310309456440\n```\nHere's what I have in python. Is there a way to reduce the number of characters further ?[143 chars presently]\n```\nr=raw_input\nf=lambda x:0**x or x*f(x-1)\nC=lambda n,r:f(n)/f(r)/f(n-r)\nfor i in range(int(r())):print C(*map(int, r().split()))\n```\nThanks!\n \n[Answer]\n# Python 3 (100 chars)\n```\nI=input\nC=lambda n,r:r and C(n-1,r-1)*n//r or 1\nfor _ in' '*int(I()):print(C(*map(int,I().split())))\n```\n]"}}},{"rowIdx":13648,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/127/edit).\nClosed 6 years ago.\n[Improve this question](/posts/127/edit) \nWhat is the shortest function (implemented in C) that will return a random draw from a Poisson distribution (given the mean or `λ` parameter)?\n```\nint random_poisson(double lambda);\n```\n \n[Answer]\nTaken [from the wikipedia article which credits Knuth](http://en.wikipedia.org/wiki/Poisson_distribution#Generating_Poisson-distributed_random_variables) (and notes that there are implementations with better time performance) this code is a complete program implementing a minimal test scaffold. The code that does the actual work is 169 characters still formatted:\n```\n#include \n#include \n#include \n#include \n#include \n/* For serious work you need a high quality random number generator \n *\n * And these macros may need to be adjusted for your system and choice\n * of random number engine.\n */\n#define GOODRAND random()\n#define GOODRANDMAX INT_MAX\n#define RANDTYPE long\n/* poisson.c\n *\n * Implementation straight from \n * http://en.wikipedia.org/wiki/Poisson_distribution#Generating_Poisson-distributed_random_variables\n * which credits Knuth.\n *\n * Time complexity is O(lambda), which is not optimal.\n*/\nRANDTYPE poisson(double lambda){\n RANDTYPE k=0;\n double L=exp(-lambda), p=1;\n do {\n ++k;\n p *= GOODRAND/(double)GOODRANDMAX;\n } while (p > L);\n return --k;\n}\n/* minimal testing scaffold \n*\n* Note that with no initialzation the sequece will repeat on every test.\n*/\nint main(int argc, char**argv){\n while (argc > 1) {\n RANDTYPE in = atoi(argv[--argc]);\n printf(\"lambda=%d: %d\\n\",in,poisson(in));\n }\n}\n```\n]"}}},{"rowIdx":13649,"cells":{"text":{"kind":"string","value":"[Question]\n [\nYou need to print A-Z like this:\n```\n1. A\n2. B\n3. C\n4. D\n5. E\n6. F\n7. G\n8. H\n9. I\n10. J\n11. K\n12. L\n13. M\n14. N\n15. O\n16. P\n17. Q\n18. R\n19. S\n20. T\n21. U\n22. V\n23. W\n24. X\n25. Y\n26. Z\n```\nBut your source code cannot use `1,2,4,6`.\nTrailing newlines in output allowed,, but leading ones not allowed.\nStandard loopholes apply, shortest code wins.\n \n[Answer]\n# [Zsh](https://www.zsh.org/), 27 bytes\n```\neval ';echo $[++i]. '{A..Z}\n```\n[Try it online!](https://tio.run/##qyrO@P8/tSwxR0HdOjU5I19BJVpbOzNWT0G92lFPL6r2/38A \"Zsh – Try It Online\")\n[Answer]\n# [JavaScript (Node.js)](https://nodejs.org), 48 bytes\n```\nf=(n=033)=>--n?f(n)+n+`. ${Buffer([8*8|n])}\n`:''\n```\n[Try it online!](https://tio.run/##BcHRCoIwFADQd79iSOC9ybQyQgIL@g0RHLpbi3EX27SH6tvXOU@1qjB584qS3axTog642zUNdhcp@UrAWHI5VmLzuS1E2kPfbtsvD/jLxnNRJENQ9/vD8TTUVdQhAiGK@PDuLXLDq7JmFsEtftJ5NjkOzurKujsQIKY/ \"JavaScript (Node.js) – Try It Online\")\n[Answer]\n# [Python 2](https://docs.python.org/2/), 44 bytes\n```\ni=8*8\nwhile i<90:i=-~i;print`i-8*8`+'. %c'%i\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWQsuCqzwjMydVIdPG0sAq01a3LtO6oCgzryQhUxcomaCtrqegmqyumvn/PwA \"Python 2 – Try It Online\")\n**46 bytes**\n```\ni=0\nwhile i<33-7:i=-~i;print'%d. %c'%(i,i+8*8)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWgKs8IzMnVSHTxthY19wq01a3LtO6oCgzr0RdNUVPQTVZXVUjUydT20LLQvP/fwA \"Python 2 – Try It Online\")\nThe loop condition can also be `while-~i/9-3:` or `while~-i/5-5` or `while-i/5-~5:` or `while~-i-5*5:` or `while~-033>i`.\n**46 bytes**\n```\ni=0\nexec 78/3*\"i=-~i;print'%d. %c'%(i,i+8*8);\"\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWgCu1IjVZwdxC31hLKdNWty7TuqAoM69EXTVFT0E1WV1VI1MnU9tCy0LTWun/fwA \"Python 2 – Try It Online\")\n**46 bytes**\n```\ni=8*8\nwhile i<90:i=-~i;print'%d. %c'%(i-8*8,i)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWQsuCqzwjMydVIdPG0sAq01a3LtO6oCgzr0RdNUVPQTVZXVUjUxeoSidT8/9/AA \"Python 2 – Try It Online\")\n**46 bytes**\n```\ni=0\nwhile~i-5*5:i=~-i;print'%d. %c'%(-i,8*8-i)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWgKs8IzMntS5T11TL1CrTtk4307qgKDOvRF01RU9BNVldVUM3U8dCy0I3U/P/fwA \"Python 2 – Try It Online\")\n**46 bytes**\n```\ni=8*8\nwhile i<90:i=-~i;print'%d. %c'%(i&799,i)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/P9PWQsuCqzwjMydVIdPG0sAq01a3LtO6oCgzr0RdNUVPQTVZXVUjU83c0lInU/P/fwA \"Python 2 – Try It Online\")\n**47 bytes**\n```\nfor i in range(33-7):print\"%d. %c\"%(-~i,8*8-~i)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/r/Py2/SCFTITNPoSgxLz1Vw9hY11zTqqAoM69ESTVFT0E1WUlVQ7cuU8dCywJIaf7/DwA \"Python 2 – Try It Online\")\n[Answer]\n# [APL (Dyalog Extended)](https://github.com/abrudz/dyalog-apl-extended), 19 [bytes](https://codegolf.meta.stackexchange.com/a/9429/43319 \"When can APL characters be counted as 1 byte each?\") ([SBCS](https://github.com/abrudz/SBCS \".dyalog files using a single byte character set\"))\n```\n↑(⍳∘≢,⍥⍕¨'. '∘,¨)⎕A\n```\n[Try it online!](https://tio.run/##SyzI0U2pTMzJT9dNrShJzUtJTfn/qKO94P@jtokaj3o3P@qY8ahzkc6j3qWPeqceWqGup6AOFNI5tELzUd9UR5DS/wpgUAAA \"APL (Dyalog Extended) – Try It Online\")\n`⎕A` the uppercase **A**lphabet\n`(`…`)` apply the following tacit function:\n `'. '∘,¨` prepend \". \" to each letter.\n `,⍥⍕¨` concatenate the string representation of each to the string representation of:\n `⍳∘≢` the **ɩ**ndices of the length of the alphabet\n`↑` \"mix\" the list of strings into a character matrix\n[Answer]\n# [PHP](https://php.net/), ~~45~~ ~~40~~ ~~39~~ 38 bytes\n```\nfor($c=A;$c<>AA;$c++)echo++$i,\". $c\n\";\n```\n[Try it online!](https://tio.run/##K8go@G9jXwAk0/KLNFSSbR2tVZJt7BxBlLa2ZmpyRr62tkqmjpKegkoyl5L1//8A \"PHP – Try It Online\")\n-1 thanks to [Kaddath](https://codegolf.stackexchange.com/users/90841/kaddath)\n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E), 9 bytes\n```\nAuā'.«sø»\n```\n[Try it online!](https://tio.run/##yy9OTMpM/f/fsfRIo7reodXFh3cc2v3/PwA \"05AB1E – Try It Online\")\n```\nA # push the lowercase alphabet\n u # convert to uppercase\n ā # push the range [1 .. len(alphabet)]\n '.« '# append '.' to each number\n s # swap to the alphabet\n ø # zip both lists => [['1.', 'A'], ..., ['26.', 'Z']]\n » # join each inner list by spaces and the outer list by newlines\n```\n[Answer]\n# [Charcoal](https://github.com/somebody1234/Charcoal), 9 bytes\n```\nＥα⁺⁺⊕κ. ι\n```\n[Try it online!](https://tio.run/##S85ILErOT8z5/z@gKDOvRMM3sUAjUUchIKe0WANMeOYlF6XmpuaVpKZoZGvqKCjpKSgBqUxNTU3r////65blAAA \"Charcoal – Try It Online\") Link is to verbose version of code. Explanation:\n```\n α Predefined variable uppercase alphabet\nＥ Map over characters\n κ Current index (0-indexed)\n ⊕ Incremented\n . Literal string `. `\n ι Current character\n ⁺⁺ All concatenated\n Implicitly print on separate lines\n```\n[Answer]\n# [Perl 5.10](https://www.perl.org/), 22 bytes\n```\nsay++$..\". $_\"for A..Z\n```\n[Try it online!](https://tio.run/##K0gtyjH9/784sVJbW0VPT0lPQSVeKS2/SMFRTy/q//9/@QUlmfl5xf91fU31DA30DAA \"Perl 5 – Try It Online\")\n[Answer]\n# [R](https://www.r-project.org/), 33 bytes\n```\nwrite.table(LETTERS,,,F,\". \",c=F)\n```\n[Try it online!](https://tio.run/##K/r/v7wosyRVryQxKSdVw8c1JMQ1KFhHR8dNR0lPQUkn2dZN8/9/AA \"R – Try It Online\")\n[Answer]\n# [Canvas](https://github.com/dzaima/Canvas), 8 [bytes](https://github.com/dzaima/Canvas/blob/master/files/chartable.md)\n```\nＺ｛²Ｏ. ｏｏ\n```\n[Try it here!](https://dzaima.github.io/Canvas/?u=JXVGRjNBJXVGRjVCJUIyJXVGRjJGLiUyMCV1RkY0RiV1RkY0Rg__,v=8), [alternate version](https://dzaima.github.io/Canvas/?u=JXVGRjNBJXVGRjVCJUIyLiUyMCV1MjUxOCV1RkYwOSV1RkYzRA__,v=8)\nThe only reason this is shorter than charcoal is because it's got a 1-indexed iteration count.\n## Explanation\n```\nZ{²O. oo\nZ{ loop over the alphabet\n ²O output the iteration number\n . o output \". \"\n o output the current letter \n```\n[Answer]\n# [PowerShell](https://github.com/TryItOnline/TioSetup/wiki/Powershell), 25 bytes\n```\n'A'..'Z'|%{++$i;\"$i. $_\"}\n```\n[Try it online!](https://tio.run/##K8gvTy0qzkjNyfn/X91RXU9PPUq9RrVaW1sl01pJJVNPQSVeqfb/fwA \"PowerShell – Try It Online\")\n[Answer]\n# [BASIC](http://www.yabasic.de), 41 bytes\n```\nfor i=!0 to 33-7:?i,\". \",chr$(8*8+i):next\n```\n[Try it online!](https://tio.run/##q0xMSizOTP7/Py2/SCHTVtFAoSRfwdhY19zKPlNHSU9BSSc5o0hFw0LLQjtT0yovtaLk/38A \"Yabasic – Try It Online\")\n[Answer]\n# [Retina 0.8.2](https://github.com/m-ender/retina/wiki/The-Language/a950ad7d925ec9316e3e2fb2cf5d49fd15d23e3d), ~~49~~ ~~29~~ 28 bytes\n```\nT`@L`_o\n}`$\nZ\n.\n$.`. $&¶\nA`@\n```\n[Try it online!](https://tio.run/##K0otycxL/P8/JMHBJyE@n6s2QYUrikuPS0UvQU9BRe3QNi7HBIf//wE \"Retina 0.8.2 – Try It Online\") Explanation:\n```\nT`@L`_o\n}`$\nZ\n```\nGenerate the uppercase alphabet prefixed by `@`. This works by feeding in `Z`s and transliterating a step backwards each time until the transformation becomes idempotent.\n```\n.\n$.`. $&¶\n```\nPrefix each character with its 0-based index and `.` , and split each character on to its own line.\n```\nA`@\n```\nDelete the `@` entry.\n[Answer]\n# [PowerShell](https://github.com/TryItOnline/TioSetup/wiki/Powershell), 41 bytes\n```\n(8-7)..(33-7)|%{\"$($_). \"+[char]($_+8*8)}\n```\n[Try it online!](https://tio.run/##K8gvTy0qzkjNyfn/X8NC11xTT0/D2BhI16hWK6loqMRr6ikoaUcnZyQWxQJ52hZaFpq1//8DAA \"PowerShell – Try It Online\")\n[Answer]\n# [Python 3](https://docs.python.org/3/), ~~52~~ 50 bytes\n```\nfor x in range(8-7,33-6):print(f\"{x}.\",chr(x+8*8))\n```\n[Try it online!](https://tio.run/##K6gsycjPM/7/Py2/SKFCITNPoSgxLz1Vw0LXXMfYWNdM06qgKDOvRCNNqbqiVk9JJzmjSKNC20LLQlPz/38A \"Python 3 – Try It Online\")\n[Answer]\n# [Retina](https://github.com/m-ender/retina/wiki/The-Language), 27 bytes\n```\n3*9*@\nY`@`L_\nL$`.\n$.>`. $&\n```\n[Try it online!](https://tio.run/##K0otycxLNPz/n8tYy1LLgSsywSHBJ57LRyVBj0tFzy5BT0FF7f9/AA \"Retina – Try It Online\") Explanation:\n```\n3*9*@\n```\nInsert 27 `@`s. (Best I can do since I can't use `2` or `6`.)\n```\nY`@`L_\n```\nReplace each `@` with subsequent uppercase letters, but delete the last one.\n```\nL$`.\n```\nMatch each letter.\n```\n$.>`. $&\n```\nFor each letter, output its 1-indexed index, `.` , then the letter.\n[Answer]\n# [Vyxal](https://github.com/Lyxal/Vyxal), `j`, 10 bytes\n```\nαɾkAZƛı. j\n```\n[Try it Online!](http://lyxal.pythonanywhere.com?flags=j&code=%CE%B1%C9%BEkAZ%C6%9B%C4%B1.%20j&inputs=&header=&footer=)\n[Answer]\n# [Japt](https://github.com/ETHproductions/japt), 12 bytes\n```\n;B£[YÄLSXR]q\n```\n[Try it online!](https://tio.run/##y0osKPn/39rp0OLoyMMtPsERQbGF//8DAA \"Japt – Try It Online\")\nStraightforward, though I feel like there's still room for improvement.\nExplanation:\n```\n;B£[YÄLSXR]q #\n; # Store the uppercase alphabet in B\n B£ # Replace each character in B with the following:\n [ ] # Make an array containing:\n YÄ # The 0-based index of the character + 1 (i.e. 1-based index)\n L # Period\n S # Space\n X # The original character\n R # A newline\n q # Join the array into a string with no separator\n # Implicitly output the resulting string\n```\n[Answer]\n# [Rust](https://rust-lang.org/), 56 bytes\n```\n||for(i,x)in(9..).zip('A'..'['){print!(\"{}. {}\n\",i-8,x)}\n```\n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), ~~10~~ 9 bytes\n```\nØAĖj€⁾. Y\n```\n[Try it online!](https://tio.run/##y0rNyan8///wDMcj07IeNa151LhPTyHy/38A \"Jelly – Try It Online\")\n## How it works\n```\nØAĖj€⁾. Y - Main link. Takes no arguments\nØA - Yield the uppercase alphabet and set as the left and right arguments\n Ė - Enumerate\n j€⁾. - Join each pair with \". \"\n Y - Join with newlines and output\n```\n[Answer]\n# Java, 62 bytes\n```\n$->{for(char c='@';++c<'[';)System.out.println(c-'@'+\". \"+c);}\n```\n[Try it online!](https://tio.run/##LYxBCoMwFETX5hSfUlAr5gKppdB1V0I30kX6qzY2JpL8CEU8u83C2QzzZphBzrIc3t9NjZN1BEPMPJDSvAsGSVnDT4JN4aUVAmrpPdylMgtLduZJUrTZqjeMsclqcsr0zROk632@MNiV3KzxYWzd@RG3F@iq7Vhels66DD/SAVbpNRVFgee0SUVe/zy1I7eB@BQPSZsMy7goDhwOBeZi3QRLko5LxHaizASt80hWtm5/)\n# Java, 95 bytes\n```\n$->java.util.stream.IntStream.range('A','[').forEach(i->System.out.println(i-'@'+\". \"+(char)i))\n```\n[Try it online!](https://tio.run/##RY3LasMwEEXX1leIEJDUxPqBNKaldJFFVoZuQhdTRXbG1cNIo0AJ@XZX0EBnc@eeGTgTXKGdzt8L@jkm4lPtuhA6PZRgCGPQTzs2ly@HhhsHOfMjYLix5sEyAdW4RjxzXy@yp4RhPH1ySGNWN8Yf07zFkIu36fmj/nZ82C/rtvv3ZUoWvD4E6v@2BGG0UryKrTgJpYeY3sFcJLZd/5PJeh0L6bnKyIVKxYvYrDRfbaS5QFKo1LJjTTNoMMbOJENxTlVyZ/flFw)\n[Answer]\n# Lua (55 bytes)\n```\nfor i=7%3,33-7 do print(i..\". \"..(8*8+i..\"\"):char())end\n```\n[Answer]\n# Javascript (Browser console), 58 bytes\n```\nfor(i=0;i<35-9;)alert(++i+\". \"+String.fromCharCode(i+8*8))\n```\n-2 bytes thanks to @expressjs123\n-3 bytes thanks to @tsh\n[Answer]\n# [Bash](https://www.gnu.org/software/bash/), 41 bytes\n```\necho {A..Z}|tr ' ' '\n'|nl -s\". \" -w$[8-7]\n```\n[Try it online!](https://tio.run/##S0oszvj/PzU5I1@h2lFPL6q2pqRIQR0EudRr8nIUdIuV9BSUFHTLVaItdM1j//8HAA \"Bash – Try It Online\")\n[Answer]\n# [Julia 1.0](http://julialang.org/), ~~37~~ 35 bytes\n```\n'A':'Z'.|>i->print(\"$(i-'@'). $i\n\")\n```\n[Try it online!](https://tio.run/##yyrNyUw0rPj/X91R3Uo9Sl2vxi5T166gKDOvRENJRSNTV91BXVNPQSWTS0nz/38A \"Julia 1.0 – Try It Online\")\n[Answer]\n# [Japt](https://github.com/ETHproductions/japt) [`-R`](https://codegolf.meta.stackexchange.com/a/14339/), 11 [bytes](https://en.wikipedia.org/wiki/ISO/IEC_8859-1)\n```\n;B¬ËiSiLiEÄ\n```\n[Test it](https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&flags=LVI&code=O0Ksy2lTaUxpRcQ)\n[Answer]\n# C\n```\nmain() { for(int i='A',j=0;i<='Z';j++,printf(\"%d. %c\\n\",j,i),i++);}\n```\nIf for some, this wouldn't work:\n```\n#include \nvoid main() { for(int i='A',j=0;i<='Z';j++,printf(\"%d. %c\\n\",j,i),i++);}\n```\n[Answer]\n# [C (gcc)](https://gcc.gnu.org/), 50 bytes\n```\nmain(i){for(;i<3*9;printf(\"%d. %c\\n\",i++,i+8*8));}\n```\n[Try it online!](https://tio.run/##S9ZNT07@/z83MTNPI1OzOi2/SMM608ZYy9K6oCgzryRNQ0k1RU9BNTkmT0knU1sbiC20LDQ1rWv//wcA \"C (gcc) – Try It Online\")\n```\nf(i){for(i=0;++i<3*9;printf(\"%d. %c\\n\",i,i+8*8));}\n```\n[Try it online!](https://tio.run/##S9ZNT07@/z9NI1OzOi2/SCPT1sBaWzvTxljL0rqgKDOvJE1DSTVFT0E1OSZPSSdTJ1PbQstCU9O69j9QTiE3MTNPQ1OhmotLAQjSNDStuWr/AwA \"C (gcc) – Try It Online\")\n[Answer]\n# [C (gcc)](https://gcc.gnu.org/), 50 bytes\n```\nf(i){for(i=0;++i<033;)printf(\"%d. %c\\n\",i,i+'@');}\n```\n[Try it online!](https://tio.run/##DcHBCkBAEADQu6/YlMy0VspxKR/iotFsUyyhHDa/bniPXCBSZRBMvB0gfeOtla5pW4/7IfFiyIu5NgWNMa@kElsOJfpH10kioEmZ@TGgzx59iZcpnOruDw \"C (gcc) – Try It Online\")\n[Answer]\n# [Stax](https://github.com/tomtheisen/stax), ~~9~~ 8 [bytes](https://github.com/tomtheisen/stax/blob/master/docs/packed.md#packed-stax)\n```\nτ┌Z*µ╝╘Q\n```\n[Run and debug it](https://staxlang.xyz/#p=e7da5a2ae6bcd451&i=)\n]"}}},{"rowIdx":13650,"cells":{"text":{"kind":"string","value":"[Question]\n [\n**This question already has answers here**:\n \n \n[Sing Happy Birthday to your favourite programming language](/questions/39752/sing-happy-birthday-to-your-favourite-programming-language)\n (207 answers)\n \nClosed 4 years ago.\nYour task is to write a program or function that takes a string as input, and prints `Hello (input), I'm (language name)`.\n## Examples\n**Python:** \nInput: `Oliver` \nOutput: `Hello Oliver, I'm Python`\n**C++:** \nInput: `Jacob` \nOutput: `Hello Jacob, I'm C++`\netc.\nRemember, this is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"), so the code with the lowest number of bytes wins.\n## Leaderboards\nHere is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.\nTo make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:\n```\n# Language Name, N bytes\n```\nwhere `N` is the size of your submission. If you improve your score, you *can* keep old scores in the headline, by striking them through. For instance:\n```\n# Ruby, ~~104~~ ~~101~~ 96 bytes\n```\nIf there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the *last* number in the header:\n```\n# Perl, 43 + 2 (-p flag) = 45 bytes\n```\nYou can also make the language name a link which will then show up in the leaderboard snippet:\n```\n# [><>](http://esolangs.org/wiki/Fish), 121 bytes\n```\n```\nvar QUESTION_ID=94485,OVERRIDE_USER=12537;function answersUrl(e){return\"https://api.stackexchange.com/2.2/questions/\"+QUESTION_ID+\"/answers?page=\"+e+\"&pagesize=100&order=desc&sort=creation&site=codegolf&filter=\"+ANSWER_FILTER}function commentUrl(e,s){return\"https://api.stackexchange.com/2.2/answers/\"+s.join(\";\")+\"/comments?page=\"+e+\"&pagesize=100&order=desc&sort=creation&site=codegolf&filter=\"+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:\"get\",dataType:\"jsonp\",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:\"get\",dataType:\"jsonp\",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r=\"

\"+e.body.replace(OVERRIDE_REG,\"\")+\"

\")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery(\"#answer-template\").html();t=t.replace(\"{{PLACE}}\",n+\".\").replace(\"{{NAME}}\",e.user).replace(\"{{LANGUAGE}}\",e.language).replace(\"{{SIZE}}\",e.size).replace(\"{{LINK}}\",e.link),t=jQuery(t),jQuery(\"#answers\").append(t);var o=e.language;/s.lang?1:e.lang\\s*([^\\n,]*[^\\s,]),.*?(\\d+)(?=[^\\n\\d<>]*(?:<(?:s>[^\\n<>]*<\\/s>|[^\\n<>]+>)[^\\n\\d<>]*)*<\\/h\\d>)/,OVERRIDE_REG=/^Override\\s*header:\\s*/i;\n```\n```\nbody{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}\n```\n```\n

Leaderboard

	Author	Language	Size

Link

\n```\n[Answer]\n## JavaScript (ES7), 239 bytes\nEncodes the shape and the line breaks separately.\n```\nlet f =\n_=>'h7hbigkfigjdjdldl9lbkfihhhz6fkwbeoi2ab9vf36d812937991qa1b2db193d472189731i7671d9451h927721486212f10e10fvfvftfrfqgphlhkihjhjgkfldldldn9n7o5'.replace(/1?./g,n=>' X'[i&1].repeat(parseInt(n,36)-1)+((++i|716000693/2**(i/2))&1?'':`\n`),i=-26)\no.innerHTML = f()\n```\n```\n
\n```\n[Answer]\n# [C (gcc)](https://gcc.gnu.org/), ~~255 244~~ 239 bytes\n```\nb,c;main(a){for(;a=\"TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq TNn*RPn/QPbEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBLOFAkHFOuFETpHCStHAUFAgcEAelclcn^r^r\\\\tZvYxXyT|S~Pn SPm SOn TNn ULo0ULo#ULo-WHq!WFs XDt!\"[b++];)for(;a-->64;putchar(++c%80?33-b%2:10));}\n```\nSlightly bugfixed and golfed version of OP's reference implementation.\n[Try it online!](https://tio.run/##JY7RToMwAEV/pcwsAWsjOmOMjZqiNB0ho6xFpk4MNHMaWRkMzJY5fx0hPpych/twrkJLpdo2O1F4lX5qM7X270Vl4vRmIOnOCL0GTIMCyIk2O1vCL0FP5BcwYhsQeWW/HU@5Pg155sbizRNxSMZekEwc1@cL5lAmD1IT/zF3/ICSL0aDhrpyze5FzUhEyVK5ZJGrXOmkSqr5vH7@ftrOdvJH/HINBF8BEeg@02ftjqMOFLPSiOkGzB5qY/CSQfiKrf/rCN1eXuB1U6uPtDIhVMMr@240Qtnw/PrMtix8aNs/ \"C (gcc) – Try It Online\")\nMinimally less golfed:\n```\nb,c;\nmain(a){\n  for(;a=\"TFy!QJu ROo TNn(ROo)SLq SLq ULo+UHs UJq TNn*RPn/QPbEWS_JSWQAIJO^NBELPeHBFHT}TnALVlBLOFAkHFOuFETpHCStHAUFAgcEAelclcn^r^r\\\\tZvYxXyT|S~Pn SPm SOn TNn ULo0ULo#ULo-WHq!WFs XDt!\"[b++];)\n    for(;a-->64;)\n      putchar(++c%80?33-b%2:10);\n}\n```\n[Answer]\n# [C (gcc)](https://gcc.gnu.org/), 205 bytes\nMy solution quickly converged on something extremely close to [ceilingcat](https://codegolf.stackexchange.com/users/52904/ceilingcat)'s [golfed version of OP's implementation](https://codegolf.stackexchange.com/a/195252/75886), but output does differ slightly when it comes to leading spaces on each line, since I worked from the text found in OP's post, which does not exactly match the output of the reference implementation, most likely due to OP missing the leading spaces required for code text, but there *are* answers using that text as reference, so it's somewhat ambiguous.\n```\ni,j;main(c){for(;c=\"B8wq>s>q:u=W:48:AoA`3>C^4><83]:8q>q>s:u8x6b\"[i++];)for(;c---50;j++)printf(\"\\n%c\"+(!j||j%75),33-i%2);}\n```\n[Try it online!](https://tio.run/##Jc4LasIwAADQs1goNMSALO0SkhiSOAfTMSooUtT5CVbSrR@rojB39lrwBO9ZdLC2aVw34/nWFYEFf2lZB9z2PUOv4qwrletSHuVJHtlF1Co3v2ZK3oYToT4GmAsRq5DI/pyFlOlSb7AcfIdSULxiVBwo0TsWqZRhQ3ESEzxPZslsEa/j9df20473o/TdDX9MYVpFVa3TSuxCb687b@EgXHHw7CCEoh7PIARV7YpzGnjLwrceDDrZ/Z75JAJdjJHzXwD/b5oH \"C (gcc) – Try It Online\")\n[Answer]\n# [Vyxal](https://github.com/Lyxal/Vyxal) `j`, 442 bytes\n```\n»⌐ṫżṫṗƛ&%≠ɽ_~ẇ*⁼3L ≈;n~U7⋏₇ḭǔ°₁>Aµ1I(□]:ȯ≤KWƒʁY•₇⟩-RCa⟨&8ẋżṫẏL_ġN⇧ṗḞΠṄẇ%¨ḃ₆Q> 1\n  p1, q1, t1 = piks(a, m)\n  p2, q2, t2 = piks(m+1, b)\n  return (p1*p2, q1*q2, q2*t1 + p1*t2)\nn = int(sys.argv[1])-1\nm = n*20//3\nz = mpz(10)**n\n# n / log(777924, 10)\npi_terms = mpz(n*0.16975227728583067)\npp, pq, pt = piks(0, pi_terms)\npq *= 3528\npi2m = (pq << m) // pt\na, b = 2 << m, 8\nwhile a != b:\n  a, b = (a + b) >> 1, isqrt(a*b)\nmlog2_pi = (z << m) // a\na, b = 2*pi2m, 8\nwhile a != b:\n  a, b = (a + b) >> 1, isqrt(a*b)\nlogpi_pi = z * pi2m // a - mlog2_pi\nmlog2 = mlog2_pi * pq // pt\nd = e = (15044673 << m) // 10450451\npt //= z\nwhile d:\n  a, b = 2*e, 8\n  while a != b:\n    a, b = (a + b) >> 1, isqrt(a*b)\n  lnx = (pq * e) // (pt * a) - mlog2\n  d = e * (lnx - logpi_pi) // z\n  e -= d\nprint(e * z >> m)\n```\n[Try it online!](https://tio.run/##nVLbjpswFHz3V0zVF@zAYhsIpNrsj1TViigkixqIIe4l/Hx6jgNK1JdK5SLhc8Yzc8a4q/8499nt1nbuPHpcrhdxGM8djp27WszV9jKMPkbnJiH2zQGu/X6J6hg7@UUA7QE1tlvsePFY6vsSGBv/Y@wRmRj8GJvJ0HHYolYRPZlVdZJXcmWtTLLQHEKTb5tXm6xYV6HsqWxNvtaqXjGTeBZwMZIhhlOeBTqCRvVqJ/H2BkMFR@oDvd5QZxmhY6iz1KHX26XTrQyPJ57IjQowo4aAVkSzIk7lrRSip41t7yPK76Uejz@/mm8yMYI99MrqNM3ERN@UYGS0VKoX4jN6pDidj1FZlhubUzRaCte@@2bsLjO4V/rFrDdlYW1Z2qqoMr0uSc/RrI5n9YtjTYt5L7EMUFtkha0I2toQBdVeX2lepCltE4LPj9MM1RiV@PXRnho6u0/zUc6AqKY55xTj@69AZ0bZiI7M23fXMmh6kNcPbsXi/8NNzDRNoJ6gEGZgaiRYVGd9DmrxQcBhGW9PjYYlTKHzfF1mD4NG51QrjKD00pQUZnv7J2dWNewb@Nv5v70Dp/73nLhCEyQjklKo5eKfQHeDChGjEywTB/hE/QbJFns6v5F/LEZOrNXJ2@1mNV1/AA \"Python 3 – Try It Online\")\nInput *n = 20000* finishes in less than one second.\n\\$\\pi\\$ can be computed in subquadratic time by use of Karatsuba splitting a.k.a. [Fast E-function Evaluation](https://en.wikipedia.org/wiki/FEE_method), which reduces a summation to *n* terms to a single rational value *p/q*, splitting in binary descent. I've chosen to use [Ramanujan #39](https://books.google.com/books?id=oSioAM4wORMC&pg=PA38&hl=en&redir_esc=y#v=onepage&q&f=false), which is the fastest converging series of its kind that doesn't require an arbitrary precision square root, to my knowledge. Explicitly, this is computed as:\n\\$\\large{\\left.{3528}\\middle/{\\sum\\limits\\_{n=0}^{\\infty}\\frac{(-1)^n (4n)! (1123+21460n)}{(n!)^4 14112^{2n}}}\\right.}\\$\nFor \\$\\ln(x)\\$ and \\$e^x\\$, using the same technique wouldn't provide any benefit, because both are computed as a power series of an arbitrary precision variable. Fortunately, Gauss has gifted us with an elegant quadratically converging formula for \\$\\ln(x)\\$, based on the [Arithmetic-Geometric Mean](https://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean):\n\\$\\DeclareMathOperator{\\AGM}{AGM}\\large\\ln(x)=\\lim\\limits\\_{n\\rightarrow\\infty}\\frac{\\pi x^n}{2n\\AGM(x^n,4)}\\$\nor, in cases when large powers of *x* are inconvenient, this can also be computed as:\n\\$\\large\\ln(x)\\approx\\frac{\\pi x 2^m}{2\\AGM(x 2^m,4)}-m\\log(2)\\$\nConveniently, division by \\$\\pi\\$ can be acheived simply by not multiplying through by \\$\\pi\\$.\nWith the natural logarithm thusly defined, \\$e^\\frac{\\ln(\\pi)}{\\pi}\\$ can be computed via Newton's method on \\$x\\_{n+1}=x\\_n-x\\_n(\\ln(x\\_n)-\\frac{\\ln(\\pi)}{\\pi})\\$. The overall complexity is then \\$\\mathcal{O}(n^\\*\\log^3(n))\\$, where \\$n^\\*\\$ will vary with the complexity of the multiplication algorithm GMP is using for any given bit length.\n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), 62 bytes\nIn my first try I used Zeta function but this one uses the imaginary part of ln(-1) for pi. (@someone)  \nPrints more than 4000 digits in the first 10 seconds, \n```\nq=Im@Log@-1;Do[Print@RealDigits[N[q^(1/q),2k]][[1,k]],{k,∞}]\n```\n[Try it online!](https://tio.run/##y00syUjNTSzJTE78/7/Q1jPXwSc/3UHX0NolPzqgKDOvxCEoNTHHJTM9s6Q42i@6ME7DUL9QU8coOzY2OtpQB0jpVGfrPOqYVxv7//9/AA \"Wolfram Language (Mathematica) – Try It Online\") \n*9 bytes saved from @someone* \n[Answer]\n# AXIOM, 221 bytes\n```\nm(k)==(v:=8.*k;(4/(v+1)-2/(v+4)-1/(v+5)-1/(v+6))*16^-k)\np(n:PI):String==(d:=digits(n+9);e:=10.^-digits();i:=s:=0;repeat(k:=m(i);kbreak;s:=s+k;i:=i+1);r:=concat split((s^(1/s))::String,char \" \");digits(d);r.(1..(n+1)))\n```\ntest and ungolf:\n```\n(3) -> p 1000\n   (3)\n  \"1.43961949584759068833649080497375567869829647445664098223316064189024343948\n  91758478197750465984130420344294359334315186918367329519847221194330793016711\n  10102682697604070399426193641233251599869541114696602206159806187886346672286\n  57856367519525119750661263299495113759810214853630906903937015021965897319237\n  10165099328745976281768843995002409093956556773589445623638530076425378312931\n  87261467105359527145042168086291313192180728142873804095866545892671050160887\n  16124784115980120121434100886405695749644308230493847450694342636218668197581\n  84867550375317450302818249455173467218279517584627294354040035671684811472191\n  01725582782513814164998627344159575816112484930170874421446660340640765307896\n  24071974858079626467839175875265777541603392496832537982189139796664242012704\n  72417497759453219873255463704766434211076771925114046204043479455332360344963\n  38423479342775816430095564073314320146986193356277171415551195734877053835347\n  93046480854891219035971014474191677302358635371602655346261468634197528218364\n  3\"\n                                                             Type: String\n           Time: 0.03 (IN) + 0.53 (EV) + 0.25 (OT) + 0.15 (GC) = 0.97 sec\n-- Bailey-Borwein-Plouffe formula for pi\n-- https://en.m.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula   \n--       +oo\n--      -----          --                                      --\n--      \\          1   |     4         2         1         1    |\n--   pi= |      -------|  ------- - ------- - ------- - ------- |\n--      /            k |   8*k+1     8*k+4     8*k+5     8*k+6  |\n--      ----- k   16   --                                      --\n--        0\nm(k)==(v:=8.*k;(4/(v+1)-2/(v+4)-1/(v+5)-1/(v+6))*16^-k)\np(n:PI):String==\n      d:=digits(n+9); e:=10.^-digits(); i:=s:=0\n      repeat\n          k:=m(i)\n          kbreak\n          s:=s+k;i:=i+1\n      r:=concat split((s^(1/s))::String,char \" \")\n      digits(d)\n      r.(1..(n+1))\n```\nthe function m calculate the term of the sum; the p() function loop, sum them in the variable s until the term is < than min float value (one can see as epsilon);\np() function return one string of that number pi^(1/pi); it seems the required digits are returned in less than 1 second for input to p() function 1000.\n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), ~~23~~ 22 bytes\n```\n(p=Log@-1/I)^p^-1~N~#&\n```\n[Try it online!](https://tio.run/##y00syUjNTSzJTE78n6ZgqxDzX6PA1ic/3UHXUN9TM64gTtewzq9OWe1/QFFmXkl0SH5wCZCRHp0WbWhgEBsb@x8A \"Wolfram Language (Mathematica) – Try It Online\")\n-1 byte by @someone — the use and reuse of `p` beats the pure function\nTaking a lot of inspiration from @J42161217 and displaying n digits instead of an infinite stream.\nNote that I'm using `ToString` in Tio to suppress the trailing precision specifier in the output, which does not appear when this code is executed in Mathematica.\nMaybe using `180°` for pi would work; but maybe that's too close to being trigonometric and is disallowed.\n[Answer]\n# APL(NARS), 304 chars\n```\nr←a P w;⎕ct;s;y;k;b\n   ⎕ct←0⋄s←''⋄→B×⍳∼w<0⋄s←'¯'⋄w←-w\nB: y←⌊w×10x*a⋄b←''\nC: b,←10∣y⋄→C×⍳0\n\n\n\n```\n      \n[Answer]\n# perl 17055\n```\nprint '\n\n\n'\n```\nkolmogorov who?\n[Answer]\n# HTML/JavaScript 14184 Bytes\nUsing [Obfuscatweet] we can get 7146 chars! 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Original HTML/JS\n[Demo](http://c99.nl/f/115183.html)\nOriginal HTML is ~~15039~~ 14182 ~~and pushes me over the size limit~~. The entire SVG has been reduced in complexity using parsing and substring searches replaced with the letters D and so on.\n```\n\n```\n[Answer]\n# PHP - 8559 bytes\n```\n because 5's smoothness factor is greater than 10's) multiplies your value by x0.7. Command-line flags don't count for anything.  \nHave fun! \nThis challenge **ends** the 19th of March.  \nCurrent placeholders: The functional language holder, with 45 chars, Nahuel Fouilleul, and the code golf language holder, with 7 chars, Luis Mendo.\n      \n[Answer]\n# [MATL](https://github.com/lmendo/MATL), 7 characters\n```\nBtnwdz/\n```\n[Try it online!](https://tio.run/##y00syfn/36kkrzylSv//f0MDAA \"MATL – Try It Online\")\n### Explanation\n```\nB    % Convert to binary\nt    % Duplicate\nn    % Number of elements\nw    % Swap\nd    % Consecutive differences\nz    % Number of nonzeros\n/    % Divide\n```\n[Answer]\n# [Japt](https://github.com/ETHproductions/japt), 8 bytes\n*NULL for infinity*\n```\n¤Ê/¢ä¦ x\n```\n[Try it online!](https://ethproductions.github.io/japt/?v=1.4.6&code=pMovouSmIHg=&input=MTA=)\n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), 8 [bytes](https://github.com/DennisMitchell/jelly/Code-page), score 8\nMaybe there is a terser way... edit: I don't think there is.\n```\nBL÷BITLƲ\n```\nInfinity is given as `inf`.\n**[Try it online!](https://tio.run/##y0rNyan8/9/J5/B2J88Qn2Ob/v//bwwA \"Jelly – Try It Online\")**\n...other 8's are possible too, for example `BµITL÷@L` or `BL÷BnƝSƊ`.\n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), 8 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)/characters\n```\nbgIbγg(g=s=>n&&1+g(x=s-(n^(n>>=1))%2))``/~x\n```\n[Try it online!](https://tio.run/##ZclLCoAgEADQffcoZog@Ri7Ho4RSJkWMkRGuurq1jbbvreYyYTyW/azYTzbNlJgUOAqkuChE6SBSqIAHYKVIIOYdotbNHdPoOfjN1pt3MMNb2VfkT0T7o072iOkB \"JavaScript (Node.js) – Try It Online\")\n### Commented\n```\nn => (                  // n = input integer\n  g = s =>              // g = recursive function taking the number s of bit switches\n    n &&                //   stop if n is equal to 0\n    1 +                 //   otherwise, add 1 to the final returned value\n    g(                  //   and do a recursive call to g:\n      x =               //     update s and save the result in x:\n        s -             //       subtract 1 from s if ...\n        (n ^ (n >>= 1)) //       ... there is a bit switch; and shift n to the right\n        % 2             //       NB: an extra bit switch is counted on the last bit\n    )                   //   end of recursive call\n)``                     // initial call to g with s = [''], which is coerced to 0\n                        // as soon as something is subtracted from it\n/ ~x                    // divide the result of g by -(x + 1), which compensates for\n                        // the extra switch\n```\n[Answer]\n# Perl 5 (`-p -Mbignum`), 45 bytes\n```\n$_=sprintf\"%b\",$_;$_=y///c/(s/(.)(?!\\1)//g-1)\n```\n[TIO](https://tio.run/##K0gtyjH9/18l3ra4oCgzryRNSTVJSUcl3hooUqmvr5@sr1Gsr6GnqWGvGGOoqa@frmuo@f@/KZehAZchlzGX@b/8gpLM/Lzi/7oFOf91fZMy0/NKcwE)\n[Answer]\n# [Kotlin](https://kotlinlang.org), ~~83~~ 82 bytes\n```\n{s->s.toString(2).run{length.toFloat()/(0..length-2).count{this[it]!=this[it+1]}}}\n```\n[Try it online!](https://tio.run/##LY4xC8IwEIX3/orrdkEaraNQR0Fwc5QOodY2GC/SXF1Cfnu8Vm86Hh/ve0/PzlL@GAePmboD4JlYQXWEk/OGockxVMeg2V95sjTgXulppuh6GniUeMVQbXGn9S@sBOn8TBx5tOFmuS2b/7ep25RSFhO8jCVUEAuQW/TcB9HB1Jv7xVKPqiylXtagWpm36NkRLjNxgZUqUq6/)\n[Answer]\n# [R](https://www.r-project.org/), 56 bytes\n```\nlength(y<-(x=scan())%/%2^(0:log2(x))%%2)/sum(diff(y)!=0)\n```\n[Try it online!](https://tio.run/##K/r/Pyc1L70kQ6PSRlejwrY4OTFPQ1NTVV/VKE7DwConP91IowLIVzXS1C8uzdVIyUxL06jUVLQ10Pxv@h8A \"R – Try It Online\")\n[Answer]\n# [Charcoal](https://github.com/somebody1234/Charcoal), 24 characters\n```\n≔⍘Ｎ²θ≔⁺№θ10№θ01η¿ηＩ∕Ｌθη∞\n```\n[Try it online!](https://tio.run/##S85ILErOT8z5/9@xuDgzPU/DKbE4NbikKDMvXcMzr6C0xK80Nym1SENTR8EIiAs1rbmgCgNySos1nPNL80o0CnUUlAwNlIDyCL6BoZImUCADqCEzTUEjQ1MhAGhoiYZzYnGJhktmWWZKqoZPal56SYZGIVidJlBlak5xKlSd0qOOeUqa1v//m//XLcsBAA \"Charcoal – Try It Online\") Link is to verbose version of code. Explanation:\n```\n≔⍘Ｎ²θ\n```\nInput the number and convert it to base 2 as a string.\n```\n≔⁺№θ10№θ01η\n```\nCalculate the number of of switches by counting the occurrences of `10` or `01` in the string.\n```\n¿ηＩ∕Ｌθη∞\n```\nIf the total is nonzero then output the smoothness otherwise print Infinity.\n[Answer]\n# 1. Python 3, 131 bytes (154 with file header)\nHi. I know that my code is way longer than others, but I will try it ;) Indentation by tabs.\nScript takes sequence of numbers in aguments and prints \"smoothness\" for each argument on own line. If number of changes is 0, prints \"inf\".\n```\n$ ./script.py 12\n4.0\n$ ./script.py 12 5 6\n4.0\n1.5\n3.0\n$ ./script.py `seq 5`\nNone\n2.0\nNone\n3.0\n1.5\n```\nfile header\n```\n#!/usr/bin/env python3\n```\ncode\n```\nimport sys\nfor n in sys.argv[1:]:\n    b=bin(int(n))[2:];c=0;l=b[0]\n    for o in b:\n        if o!=l:c+=1\n        l=o\n    print(len(b)/c if c>0 else\"inf\")\n```\n## Input single number from STDIN: 102+23 bytes (code + file header)\n```\nn=input();b=bin(int(n))[2:];c=0;l=b[0]\nfor o in b:\n    if o!=l:c+=1\n    l=o\nprint(len(b)/c if c>0 else\"inf\")\n```\n[Answer]\n# [Ruby](https://www.ruby-lang.org/), 42 bytes\n```\n->n{1.0*(w=(n^n/2).digits 2).size/~-w.sum}\n```\n[Try it online!](https://tio.run/##KypNqvyfZvtf1y6v2lDPQEuj3FYjLy5P30hTLyUzPbOkWAHIKs6sStWv0y3XKy7Nrf0fbapjaKBjqGOsYx6rl5tYUF1TUVOgkBZdEVv7HwA \"Ruby – Try It Online\")\n### How?\nFirst step: bitwise XOR of x and x/2. The result will have a bit set to 1 for every switch in the input number plus 1, and so we just need to get the number of digits in base 2, and their sum. Then add some parentheses, and make it a float.\n[Answer]\n# Python 3, 105 chars\n```\ndef s(n):b=bin(n)[2:];l=len(b);c=sum([0if b[i]==b[i+1]else 1for i in range(l-1)]);return l/c if c>0else-1\n```\nReturns -1 in the case of an infinity.\n[Answer]\n# [Python 3.8 (pre-release)](https://docs.python.org/3.8/), 60 bytes\nPort of [G B](https://codegolf.stackexchange.com/users/18535/g-b)'s Ruby [answer](https://codegolf.stackexchange.com/a/180999/64121).\n```\nlambda l:(c:=(b:=bin(l^l>>1)).count('1'))>1and(len(b)-2)/~-c\n```\n[Try it online!](https://tio.run/##DctLDsIgEADQ/ZyCXWcSq07rp5LASYwJ0GJJcNqQunDj1dG3f@tnmxfph7XUaO41u5cfncoagzbotfFJMD@ytUy0D8tbNmy4IbLsZMQ8CXpqOzp821DjUlRSSVRx8pyQd8ykQam1pH@LmIgqQwc9nOAMF7jCADfg4w8 \"Python 3.8 (pre-release) – Try It Online\")\n---\n# [Python 2](https://docs.python.org/2/), 68 bytes\nReturns false if the value is infinite.\n```\nk=input()\nn=s=0\nwhile k:l=k%2;n+=1.;k/=2;s+=l^k%2\nprint s>1and n/~-s\n```\n[Try it online!](https://tio.run/##LY1PC4JAFMTv71M8hKCQ0t3@K6@bXbt0DkwXlJV1ebtSXvrqm1KnGYbfzNjRN72RoeprhYRRFAVNrbGDX67AkKMUXk3bKdRZR3ohcxOT2OQ6IZm7mLrHlIHl1nh0F1GaGk3yWbswDcG/eedBZYDoeZwFUb1VhfMhzL5S1mNxuxbMPf@AJ6tSBwEStrCDPRzgCCc4g0i/ \"Python 2 – Try It Online\")\n[Answer]\n# [J-uby](https://github.com/cyoce/J-uby), 42 bytes\n```\n~:digits&2|:/%[:+@|Q,:chunk+I|A|:+@|:pred]\n```\n[Attempt This Online!](https://ato.pxeger.com/run?1=m700K740qXLBIjfbpaUlaboWN7XqrFIy0zNLitWMaqz0VaOttB1qAnWskjNK87K1PWsca0ACVgVFqSmxEB07o011DA10DHWMdcxj9XITC6oLFNyi4w1jayHyCxZAaAA)\n[Answer]\n# [q](https://code.kx.com/home/), 30 bytes\nOutputs infinity `0w` for 1, 3, 7 etc.\n```\n{count[b]%-1+/differ b:2 vs x}\n```\n# k, 21 bytes\n```\n{(#x)%-1+/~~':x:2\\:x}\n```\n]"}}},{"rowIdx":13660,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n# Connecting Dots\n## We define a type of question on the test, connecting the dots\n### Question parameters\nThere are two parameters. Suppose they are `5` and `4`. The second one **must be less than or equal to the first one**.\nThus, the question will look like this:\n```\n*\n         *\n*\n         *\n*\n         *\n*\n         *\n*\n```\n### Possible answers\nAn answer is termed `logically possible` if and only if:\n* Each dot on the left corresponds to one and only one dot on the right\n* Each dot on the right corresponds to at least one dot on the left (there is no maximum)\n### We describe an answer using a `matrix`, or `list of lists`.\nFor instance, `[[0,0],[1,0],[2,1]]` will link the dot indexed 0 on the left to the dot indexed 0 on the right, et cetera. You may choose to use 0-indexed or 1-indexed.\n### We define an answer's `complexity`...\n...as the number of intersections there are. For instance, the complexity of the answer `[[0,2],[1,1],[2,0]]` to be `1` as they intersect at the same point (assuming they are **evenly spaced out**).\n# Purpose\n* Calculate the complexity of a given solution\n# Grading\nThis is **`code golf`**.\n      \n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), 17 bytes\n```\n#2!StirlingS2@##&\n```\n[Try it online!](https://tio.run/##y00syUjNTSzJTE78n6Zg@1/ZSDG4JLMoJzMvPdjIQVlZ7X9AUWZeiYK@g4JjUVFiZXSajkK1hY6CRa2OgkHsfwA \"Wolfram Language (Mathematica) – Try It Online\")\nThis is [OEIS A019538](http://oeis.org/A019538).\n[Answer]\n# [C (gcc)](https://gcc.gnu.org/), 42 bytes\n```\nf(a,b){return a--?b*(f(a,b)+f(a,b-1)):!b;}\n```\n[Try it online!](https://tio.run/##hY5BDoIwEEXX7SkGDEkr1GhiYiKiF3EDBXSIFlNhBZwdW0HFjc7mp5N5fV@Kk5R9n7M4SHijs6rWCmIhDsmcDUv/GWLF@dZJwq6foZKXOs1gd69SLBfnPUVVwTVGxThtKLlp886ZC63LQ0ryUgOzFwVEsAxNOBFsTPo@pzCZN@et0Q2gsPBrd1TCF//nW4iDEEchWiExBSe/etgaFVrsd1HyocZ@OcOg4JbsaNc/AA \"C (gcc) – Try It Online\")\nThis uses a recurrence relation. Let \\$f(a,b)\\$ be the number of answers for \\$a\\$ left dots and \\$b\\$ right dots.\nConsider the first left dot; it is joined to one right dot, with \\$b\\$ possibilities.\n* If that right dot is joined to at least one other left dot, then the remaining \\$a-1\\$ left dots cover the \\$b\\$ right dots at least once each, for \\$f(a-1,b)\\$ possibilities.\n* If that right dot is not joined to any other left dot, then the remaining \\$a-1\\$ left dots cover the \\$b-1\\$ other right dots at least once each, for \\$f(a-1,b-1)\\$ possibilities.\nTherefore, \\$ f(a,b) = b \\times (f(a-1,b) + f(a-1,b-1)) \\$.\nThe base case is \\$ f(0,0) = 1\\$ and \\$ f(0,b) = 0\\$ for \\$ b > 0 \\$.\n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), 10 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)\n```\nLIãεÙg¹Q}O\n```\nTwo loose inputs in reversed order.\nPort of [*@MatteoC.*'s Python answer](https://codegolf.stackexchange.com/a/247126/52210), so make sure to upvote him as well!\n[Try it online](https://tio.run/##yy9OTMpM/f/fx/Pw4nNbD89MP7QzsNb//39jLhMA) or [verify all test cases below 6](https://tio.run/##yy9OTMpM/W96bJLfoZUgMvTQuohHDbPslRQetU1SULIHcv/7HFp3ePG5rYdnpkcE1vr/1/kPAA).\n**Explanation:**\n```\n          #  E.g. inputs: b=2,a=3\nL         # Push a list in the range [1, first (implicit) input `b`]\n          #  → [1,2]\n Iã       # Create all possible `a`-sized combinations from this list, using the\n          # cartesian product\n          #  → [[1,1,1],[1,1,2],[1,2,1],[1,2,2],[2,1,1],[2,1,2],[2,2,1],[2,2,2]]\n   ε      # Map over each inner list:\n    Ù     #  Uniquify it\n          #   → [[1],[1,2],[1,2],[1,2],[2,1],[2,1],[2,1],[2]]\n     g    #  Pop and push the length to get the amount of unique values\n          #   → [1,2,2,2,2,2,2,1]\n      ¹Q  #  Check if its equal to the first input `b`\n          #   → [0,1,1,1,1,1,1,0]\n   }O     # After the map: check how many were truthy by taking the sum\n          #  → 6\n          # (which is output implicitly as result)\n```\n[Answer]\n# [Vyxal](https://github.com/Vyxal/Vyxal) `s`, 7 bytes\n```\nɾ↔vUvL=\n```\n[Try it Online!](https://vyxal.pythonanywhere.com/#WyJzIiwiIiwiyb7ihpR2VXZMPSIsIiIsIjNcbjQiXQ==) or [Run all the test cases below 6](https://vyxal.pythonanywhere.com/#WyIiLCJAZjoyfCIsIjrJviTiiIfihpR2VXZMPeKIkSIsIjtcblxuNTrhuopzOsabw7dAZjs7WsabYCA9PiBgajvigYsiLCIiXQ==)\n## How?\n```\nɾ↔vUvL=\nɾ       # List in the range [1, (implicit) first input `b`]\n ↔      # Get all possible combinations of this list of size (implicit) second input `a`\n  vU    # For each item, uniquify\n    vL  # For each item, get the length\n      = # Is each item equal to the (implicit) first input `b`?\n        # `s` flag sums the top of the stack\n```\nOther 7-byters:\n```\nɾ↔ƛUL;=\nɾ↔ƛUL¹=\n```\n### Flagless:\n# [Vyxal](https://github.com/Vyxal/Vyxal), 8 bytes\n```\nɾ↔vUvL=∑\n```\n[Try it Online!](https://vyxal.pythonanywhere.com/#WyIiLCIiLCLJvuKGlHZVdkw94oiRIiwiIiwiM1xuNCJd)\n[Answer]\n# [Python](https://www.python.org), ~~97~~ ~~95~~ ~~88~~ 84 bytes\n*-2 bytes thanks to [@Number Basher](https://codegolf.stackexchange.com/users/111945/number-basher)*\n*-4 bytes thanks to [@Kevin Cruijssen](https://codegolf.stackexchange.com/users/52210/kevin-cruijssen)*\n```\nfrom itertools import*\nf=lambda a,b:sum(b==len({*x})for x in product(*[range(b)]*a))\n```\n[Attempt This Online!](https://ato.pxeger.com/run?1=NY6xDoIwFEV3vqLja1MHYTEk_IUbYWiFahPa1zweCcb4JS4s-k_-jSWRO9zpnJv7-qQ73zCu63tmdzh9z44wCM8DMeI4CR8SEqvCNaMJtjfCaFtPcwDbNOMQ4aGWp3RIYhE-ikTYzxcG1ZKJ1wGs7JSR8r993Ljsb2QLpa6khkqXucvcXV2InEQ-MjjI3C7u534)\n[Answer]\n# [Charcoal](https://github.com/somebody1234/Charcoal), 23 bytes\n```\nＦ⊕η⊞υ¬ιＦＮＵＭυ×λ⁺κ§υ⊖λＩ⊟υ\n```\n[Try it online!](https://tio.run/##Rcy9DgIhEATg/p6Cckmw0fIqczZXeKHwBRDWQOTnArvGt0dJTJxy5stYb6otJvb@KFXAmm3FhJnQgZdSaG4eWImtEAQp5@mndqaN0x0rfNHV7EtJyWQ36C0kbBCV0JEbPJU405odvsd2wf99lCPzpGvIBItpBLrswKPs/SSO/fCKHw \"Charcoal – Try It Online\") Link is to verbose version of code. Explanation:\n```\nＦ⊕η⊞υ¬ι\n```\nCreate a list of one `1` and `k` `0`s corresponding to a 0-indexed first row of the table in OEIS linked by @alephalpha.\n```\nＦＮ\n```\nRepeat `n` times:\n```\nＵＭυ×λ⁺κ§υ⊖λ\n```\nUpdate the row using the recurrence relation given in OEIS but also @m90's answer.\n```\nＩ⊟υ\n```\nOutput the last value calculated.\n]"}}},{"rowIdx":13661,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n Questions without an **objective primary winning criterion** are off-topic, as they make it impossible to indisputably decide which entry should win.\nClosed 7 years ago.\n[Improve this question](/posts/10777/edit) \n[I asked this question](https://softwareengineering.stackexchange.com/questions/187205/how-to-write-useful-java-programs-without-using-mutable-variables)  on [programmers](https://softwareengineering.stackexchange.com/) 10 days back but wasn't able to get the answer that I wanted. \nWrite program in java to print the squares of first 25 integers(1-25) without using any variables.\nI am not looking for any theory explaining concepts about java or functional programming. I just want to know if its possible in java, the way the author has described in the [original context.](http://pragprog.com/magazines/2013-01/functional-programming-basics)\n--------Edit--------------------\nActually making it mutable variable makes it quite easy. The original context was \n> \n> (take 25 (squares-of (integers)))\n>  Notice that it has no variables. Indeed, it has nothing more than three functions and one constant. Try writing the squares of integers in Java without using a variable. Oh, there’s probably a way to do it, but it certainly isn’t natural, and it wouldn’t read as nicely as my program above.\n> \n> \n> \nSo try writing same program without using any variables.\n      \n[Answer]\nObligatory cheat:\n```\npublic class PrintTwentyFiveSquares {\n    public static void main() {\n        System.out.println(\"1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625\");\n    }\n}\n```\n[Answer]\nUgly hack in 216 characters, this does not even use method args:\n```\nclass M{static void m(){System.out.println((Thread.currentThread().getStackTrace().length-2)*(Thread.currentThread().getStackTrace().length-2));if(Thread.currentThread().getStackTrace().length-2<25)m();}static {m();}}\n```\nFormatted, this looks like:\n```\nclass M {\n    static void m()\n    {\n        System.out.println((Thread.currentThread().getStackTrace().length-2)*(Thread.currentThread().getStackTrace().length-2));\n        if (Thread.currentThread().getStackTrace().length-2 < 25) m();\n    }\n    public static void main(String[] a)\n    {\n        m();\n    }\n    // We can comply fully with the question (no variables),\n    // but then cannot avoid the \"java.lang.NoSuchMethodError: main\":\n    static \n    {\n        m();\n    }\n}\n```\nFor those who don't understand: It takes the stacks length squares it. This works because I'm recursively calling `m()`, always increasing the stack by 1. Maybe this could be written even shorter.\n[Answer]\nHow about this?\n```\npublic class Squares {\n    public static void printSquares(final int n) {\n        if (n > 1) {\n            printSquares(n - 1);\n        }\n        System.out.println(n * n);\n    }\n    public static void main(final String... args) {\n        printSquares(25);\n    }\n}\n```\nNow that you completely changed the question, my solution depends on whether you consider function arguments to be variables or not. I'd say it still stands.\nI should add that the Closure solution depends on the pre-availability of several functions in the standard library. The same functions could be written in Java, leading to a nearly identical main program:\n```\nimport java.util.AbstractList;\nimport java.util.List;\npublic class Squares2 {\n    public static List integers() {\n        return new AbstractList() {\n            @Override\n            public Integer get(final int index) {\n                return index + 1;\n            }\n            @Override\n            public int size() {\n                return Integer.MAX_VALUE;\n            }\n        };\n    }\n    public static List squaresOf(final List l) {\n        return new AbstractList() {\n            @Override\n            public Integer get(final int index) {\n                return l.get(index) * l.get(index);\n            }\n            @Override\n            public int size() {\n                return l.size();\n            }\n        };\n    }\n    public static List take(final int n, final List l) {\n        return l.subList(0, n);\n    }\n    public static void main(final String... args) {\n        System.out.println(take(25, squaresOf(integers())));\n    }\n}\n```\n]"}}},{"rowIdx":13662,"cells":{"text":{"kind":"string","value":"[Question]\n      [\nJack is a little businessman. He found out a way to earn money by buying electricity on days when it's cheap and selling it when it's much more expensive. He stores the electricity in a battery he made by himself.\n### Challenge\nYou are given `N` (if required), the number of days Jack knows the cost of electricity for, and `X`, the amount of money Jack has available to invest in electricity, and the value(buy/sell value) of the electricity for `N` days. Your job is to determine when Jack should buy and when he should sell electricity in order to earn as much money as possible and simply print the largest possible sum of money he could have afterwards.\nThe value of the electricity is always a positive integer but depending on the amount of money Jack has, the amount of electricity and money he has may be floating point numbers.\n### Examples\n(I/O need not be exactly as shown):\n1.\n```\n4 10\n4 10 5 20\n```\nOutput: `100`  \n- because he buys electricity on the 1st day and the sells it on the 2nd and buys it on the 3rd and sells it on the 4th day.  \n2.\n```\n3 21  \n10 8 3\n```\nOutput: `21`  \n- because it's better if he doesn't buy/sell any electricity.\n3.\n```\n3 10\n8 10 14\n```\nOutput: `17.5`  \n- because he buys electricity on the 1st day, but he sells it on the 3rd day.\n4.\n```\n5 10\n4 2 10 5 20\n```\nOutput: `200`  \n- much like example 1, but this time Jack waits for the price to drop to `2` before making a purchase.\n---\nThe program with the shortest amount of code wins! This is a code golf problem.\nNote:You can shorten the input, as long as it still makes sense.\n      \n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), ~~8~~ 7 [bytes](https://github.com/DennisMitchell/jelly/wiki/Code-page)\n-1 byte thanks to Mr. Xcoder (use the `Ɲ` quick which performs `2\\`)\n```\n÷@Ɲ»1P×\n```\nA dyadic link taking a list of the prices on the left and the initial capital on the right which returns the maximal capital.\n**[Try it online!](https://tio.run/##y0rNyan8///wdodjcw/tNgw4PP3///8mOgpGOgqGBjoKpkCWwX9DAwA \"Jelly – Try It Online\")** or see the [test-suite](https://tio.run/##y0rNyan8///wdodjcw/tNgw4PP3/4eX6j5rW/P8fHR1toqNgaKCjYKqjYGQQC2LH6nApREeDxCx0FIyBQkaGECELiEpDEyRlQN1GGAbEAgA \"Jelly – Try It Online\").\n### How?\nJack can sell every day on which the preceding day had a lower price and buy on those days (even if it is one on which he sold). When he does so his capital increases by `priceToday ÷ priceYesterday`, such gains multiply his initial capital, while any time the price falls he may simply forego the loss and multiply his capital by `1` instead by not transacting.\n```\n÷@Ɲ»1P× - Link: list, prices; number, initial capital  e.g. [4,2,10,5,20], 10\n  Ɲ     - pairwise overlapping reduce left (prices) by:\n÷@      -   division with swapped arguments                 [0.5,5,0.5,4]\n    1   - literal one                                       1\n   »    - maximum (vectorises)                              [1,5,1,4]\n     P  - product                                           20\n      × - multiply by right (initial capital)               200\n```\n[Answer]\n# [Haskell](https://www.haskell.org/), 41 bytes\n```\nl#n=foldr((*).max 1)n$zipWith(/)(tail l)l\n```\n[Try it online!](https://tio.run/##DcexCoAgFAXQX7mQg4aURo1@R0M0CBU9epmYQ/Tz1nbO7u9jZS6Fq@C2i5ckZa2a0z@wKoiX4kh5l62S2RODFZfTU4BDTBQyBKZeo9OwRmP4ZWZUf8oH \"Haskell – Try It Online\")\n[Answer]\n# [Coconut](http://coconut-lang.org/), ~~51~~ 45 bytes\nthanks to @totallyhuman for -5 bytes\n```\nd->m->reduce((*),map(max$(1)..(/),d[1:],d))*m\n```\n[Try it online!](https://tio.run/##RY5BCoMwFESvkoWL/@VrTZqCFJqLpFlIEqGLqKiBLHr3NLqwm4HHzDBjZztPcc/j651do0KjVu@i9QA1UhgWCEOqgGPbwg3Jaf405BDrkDXwjrSkog8SnUFiIDjpwj3dTzwC/RHg8mJJ4l8x7KtYWakgsUaxEZLmBosWDy9vWT/TfmL5usXg8w8 \"Coconut – Try It Online\")\nUses [Jonathan Allan's algorithm](https://codegolf.stackexchange.com/a/155470/64121).\nCoconut extends Python with several functional programming constructs. \n```\nd->m->                      lambda function taking two arguments, money and prices, in currying syntax\n    reduce((*),              product of...\n           map(               map\n               max$(1)..(/),   (x,y)->max(1, x/y)\n                              over\n               d[1:],d         consecutive pairs of prices\n               )\n          )*m                multiply with the initial money\n```\n[Answer]\n# [Husk](https://github.com/barbuz/Husk), 7 bytes\n```\nΠ:mY1Ẋ/\n```\n[Try it online!](https://tio.run/##yygtzv7//9wCq9xIw4e7uvT///8fbaJjpGNooGOqY2QQ@9/QAAA \"Husk – Try It Online\")\n### How?\n```\nΠ:mY1Ẋ/ – Full program.\n     Ẋ  – For each pair of adjacent elements from the input list...\n      / – ... Divide the second by the first.\n  m     – For each quotient...\n   Y1   – ... Get the maximum between it and 1.\n :      – Append the second input.\nΠ       – Get the product of the resulting list.\n```\n[Answer]\n# Perl, 37 bytes\nIncludes `+4` for `-ap`\nOn STDIN first give a line with the list of prices, then a line with the starting capital\n`electricity.pl`:\n```\n#!/usr/bin/perl -ap\n$\\=<>;s%\\d+%$\\*=$&>$'||$'/$&%eg}{\n```\nYou can then for example run it as:\n```\n(echo 4 2 10 5 20; echo 10) | electricity.pl; echo\n```\n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E), 9 bytes\n```\nRü/εXM}P*\n```\n[Try it online!](https://tio.run/##MzBNTDJM/f8/6PAe/XNbI3xrA7T@/482NNCx0DE00THSMY/lMjQFAA \"05AB1E – Try It Online\")\nUses the same method as in [Jonathan Allan's Jelly answer](https://codegolf.stackexchange.com/a/155470/47066)\n**Explanation**\n```\nR           # reverse input list of buy/sell values\n ü/         # pairwise division\n   εXM}     # max of each element and 1\n       P    # product\n        *   # multiply by initial capital\n```\n[Answer]\n# [Stax](https://github.com/tomtheisen/stax), ~~21~~ 20 bytes\n```\nàq'4¢<╪╦ÉdεÖ!≤♠┬☻∟In\n```\n[Run it](http://stax.tomtheisen.com/#c=c%7C%29s%5CDc%7B%7CMms%7Bhm%5C%7BE%3A_m%3A**&i=10%2C+%5B8%2C10%2C14%5D)\nThis is the ascii representation of the same program:\n```\nc|)s\\Dc{|Mms{hm\\{E:_m:**\n```\nI'm sure there is room for golfing here, but I wanted to give Stax a spin. Here's the breakdown.\n```\n                            Input stack: initial funds, array of prices\nc                           Copy the array of prices\n |)                         Rotate the top copy one place right\n   s                        Swap the two arrays on the main stack\n    \\                       Zip the arrays to get consecutive price pairs\n     D                      Drop the first pair (don't need [price N, price 1])\n      c                     Copy the result\n       {|Mm                 Map max: Keep only the max of each pair (copy 1)\n           s                Swap copy 1 and copy 2 on the main stack\n            {hm             Map first: Keep only the first of each pair (copy 2)\n               \\            Zip copy 1 and copy 2\n                {E:_m       Map explode, divide: Transform pairs with float division\n                     :*     Array product: Multiply all resulting quotients\n                       *    Multiply by other input (initial funds)\n```\n[Answer]\n# [Ruby](https://www.ruby-lang.org/), 57 bytes\n```\n->a,x{x*a.each_cons(2).map{|t|t.max*1.0/t[0]}.reduce(:*)}\n```\n[Try it online!](https://tio.run/##ZY7RCsIgGIXvfYr/sg0ztQUjWC8iEuYcCeXG5mAx9@zm2k3U3eE75//P6cfbKzZV3F8UnuYpV8Qofb/q1g07npGn6ubgg09iyhmhBy@oXEhv6lGb3TnPlmhdN3qoQCAQosDAKIYTBk7lqiVe8cpKDMeEONtQuSVZ8RVL1/zvAZKfSVC3ELreajNgsM56qx4BAaTyARrx40hkXB3jGw \"Ruby – Try It Online\")\nA lambda taking an array of prices `a` and the starting capital `x`. The most interesting thing I found here is that it's faster to treat the pairs of prices as a tuple `t`, whereas my first instinct was `.map{|n,m|[m,n].max*1.0/m}`.\n```\n->a,x{\n  x *                       # Multiply the starting capital by...\n  a.each_cons(2).map{ |t|   # For each consecutive pair of prices\n    t.max*1.0 / t[0]        # The larger of that pair, divided by the first\n  }.reduce(:*)              # All multiplied together\n}\n```\n]"}}},{"rowIdx":13663,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n### Challenge:\nTo take two points on the plane and figure out the position of the point that's a) on the line that connects them, and b) equally distant from each of the original points.\n### Example:\n```\n(0,0), (3,3) -> (1.5,1.5)\n(-1,-1), (1,1) -> (0,0)\n(-7,3), (2,-2) -> (-2.5,0.5)\n(3, 1), (-5,9) -> (-1,5)\n```\n### Rules:\n* The input and output can be a tuple, array, list, or any other type of sequence.\n* There must be two inputs, each representing a point.\n* The output has to support floating point numbers.\nShortest solution in bytes wins.\n      \n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), ~~7~~ 2 [bytes](https://github.com/DennisMitchell/jelly/wiki/Code-page)\n-5 bytes (ha ha, yes I am an idiot) thanks to FryAmTheEggman!\n```\n+H\n```\nTakes two lists each of length d and returns one list of length d where d is the dimensions of the Cartesian space (question is posed for 2 dimensions, but should work for any d).\n**[Try it online!](https://tio.run/nexus/jelly#@6/t8f///2hdcx3j2P/RRjq6RrEA)**\n### How?\n```\n+H - Main link: a, b (lists of coordinates)\n+  - add (vectorises across the dimensions)\n H - halve (vectorises across the result)\n```\n[Answer]\n# Mathematica, 4 bytes\n```\nMean\n```\nWorks on lists of vectors: for example, `Mean[{{-7, 3}, {2, -2}}]` returns `{-5/2, 1/2}`. (If the input consists of floating-point numbers, so will the output.)\n[Answer]\n# [Python 3](https://docs.python.org/3/), ~~39~~ 35 bytes\n```\nlambda*x:[sum(a)/2for a in zip(*x)]\n```\n[Try it online!](https://tio.run/nexus/python3#S1OwVYj5n5OYm5SSqFVhFV1cmquRqKlvlJZfpJCokJmnUJVZoKFVoRn7v6AoM69EI02Di0tBIVrXXEfBOFZHIdpIR0HXKJaLS1PzPwA \"Python 3 – TIO Nexus\")\n[Answer]\n## [Haskell](https://www.haskell.org/), ~~22~~ 19 bytes\n```\nzipWith$((/2).).(+)\n```\nCan be called with `(zipWith$((/2).).(+)) (p1::[a]) (p2::[a])` where `a` is any type deriving `Fractional`. Works in arbitrary dimensions.\n[Answer]\n# TI-BASIC, 8 bytes\n```\n.5(L₁+L₂\n```\nTakes input on List 1 and List 2. Works for all dimensions.\n[Answer]\n## [Julia](https://julialang.org/) 0.6, 13 bytes\n```\n(x->x/2)∘(+)\n```\n[Answer]\n# JavaScript , 30 bytes\nvalid for any number of dimensions\n```\na=>b=>a.map((c,i)=>(c+b[i])/2)\n```\n[Try it online!](https://tio.run/nexus/javascript-node#S7P9n2hrl2Rrl6iXm1igoZGsk6lpa6eRrJ0UnRmrqW@k@T85P684PydVLyc/XSNNI9pQxyRWUyPaSMcgVlOT6z8A)\n[Answer]\n# R, 22 bytes\nAs an unnamed function that takes 2 vectors as parameters. Simply returns the mean of input vectors the same as a lot of answers.\n```\nfunction(a,b)((a+b)/2)\n```\n[Try it online!](https://tio.run/nexus/r#@6/BlVaal1ySmZ@nkaiTpKmhkaidpKlvpMmlqZGsYahjpKmTrGGqY6qp@f8/AA \"R – TIO Nexus\")\nAnd 16 bytes using the pryr library\n```\npryr::f((a+b)/2)\n```\n[Answer]\n# Common Lisp, ~~70~~ ~~54~~ 49 bytes\n*Thanks to @Julian Wolf for helping to save a bunch of bytes!*\n```\n(defun f(a b)(mapcar(lambda(c d)(/(+ c d)2))a b))\n```\n[Try it online!](https://tio.run/nexus/clisp#dYzBCoMwEETvfsUeZxCtSQ9@z2oMCFHE2u9Ps5JeCj3Mso99Oxlhie9dIlQmYtNj1hNJtykoZgnEA63Y4klTmHGc634JoiCtr0uGfrCw4rNgCcnm1@xcudn4usbujzvWnsreXv3t5g8)\n[Answer]\n# Math.JS, 14 bytes\n```\nf(a,b)=(a+b)/2\n```\nCreates function `f` which takes 2 One Dimensional Matricies as input.\nAs expected, this just adds the two, then halfs the result.\n[Try it!](http://a-ta.co/math/f(a%2Cb)%3D(a%2Bb)%2F2%0A%0Aprint(%0A%20%20%20%20f(%5B0%2C0%5D%2C%5B3%2C3%5D)%0A))\n[Answer]\n## [Python](https://www.python.org/) + [NumPy](http://www.numpy.org/), 18 bytes\n```\nlambda a,b:(a+b)/2\n```\nWorks in two dimensions for points inputted as imaginary numbers, or (since OP specifies that inputs must be taken as sequences) in arbitrary dimensions for points inputted as NumPy arrays.\n[Try it online!](https://tio.run/nexus/python3#VcxBCsMgEIXhfU4xyxk6tmoWhUBPYlyMFEGoRiRZ5PRWSjZ9y@/BH@EFa/9IDm8B4bCg3AI97JRy3doO5cj17HFrkCAVcA41a2KceSbP4FAZVmaAYXPBc1yMlpUl75cJxmpLZceIv9pdWpMTk9Oe@F@MJ6L@BQ \"Python 3 – TIO Nexus\")\n[Answer]\n# Python + NumPy, 17 bytes\n```\nlambda a:sum(a)/2\n```\nA slightly less boring version of the other python answer. Takes args as 1 numpy array, containing both pairs.\n[Try it online!](https://tio.run/nexus/python3#Zc2xCsMwDATQPV@hUQK5jZ2hEOiXpB5UikFQO0ZNhny9GzK0lN508DguwRVu7Sn5/hCQ8bVmFDqHTnOdbYGy5rq1NBsoaIGpg08m7LknxoEHigw/4jw7v5tn/2@XfcAY2AWKX4nj0atpWTDhcXwSM9lQiai9AQ \"Python 3 – TIO Nexus\")\n[Answer]\n# Julia, 13 bytes\n```\na->b->(a+b)/2\n```\nCall with currying syntax, i.e. `(a->b->(a+b)/2)([1,2])([3,4])`.\n]"}}},{"rowIdx":13664,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, [visit the help center](/help/reopen-questions).\n \n \nClosed 11 years ago.\nThere are two objects:\n```\nObject1, Object2\n```\nYou can easily swap them by:\n```\nObject3 = Object1\nObject1 = Object2\nObject2 = Object3\n```\nHowever, you must solve this problem by using no temporary object and Object1 should be changed to Object 2. (and vice versa)\n      \n[Answer]\n## Python\nI'll say it first.\n```\nobject1, object2 = object2, object1\n```\n[Answer]\n## Mathematica\nA standard feature:\n```\na = 1; b = 2;\nPrint[a, \" \", b];\n{a, b} = {b, a};\nPrint[a, \" \", b];\n```\nOutput\n```\n(*\n1 2\n2 1\n*)\n```\n[Answer]\n**PostScript**\nIn PostScript, it's common to use the stack instead of named variables, so a simple\n```\nexch\n```\nwould do the job. But if you have variables, you can do it like\n```\n/a 1 def\n/b 2 def\n/a b /b a def def\n```\n[Answer]\nIf you have access to the memory address of these objects, you can swap them by xor-ing their addresses.\n```\n// return 0 for success, 1 for error\nint swap(void *object1, void *object2, int size)\n{\n    for (int i = 0; i < size; i++)\n    {\n        char *x = (char *)(object1 + i);\n        char *y = (char *)(object2 + i);\n        if (x == y)\n            return 1;\n        *x ^= *y;\n        *y ^= *x;\n        *x ^= *y;\n    }\n    return 0;\n}\n```\nper dmckee suggestion for clarity\n[Answer]\n## Perl\n```\n($object1, $object2) = ($object2, $object1);\n```\nUsing Perl's list assertions feels like cheating, but it's a pretty common feature.\n[Answer]\n# PHP\nObjects / arrays / resources:\n```\nlist($a, $b) = [$b, $a];\n```\nIntegers / booleans / strings of equal length:\n```\n$a ^= $b ^= $a ^= $b;\n```\n[Answer]\n## Forth\n```\nswap\n```\nAs i have to include more than 30 characters, here's a link to a [Forth primer](http://galileo.phys.virginia.edu/classes/551.jvn.fall01/primer.htm)\n]"}}},{"rowIdx":13665,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n## Challenge\nGiven an input `x`, create a directory (AKA folder) in the file system with the name `x`.\nLanguages that are not able to create file system directories are excluded from this challenge.\nThe directory may be created anywhere, as long as the location is writable to the user running the program. Assume that the program is run as a non-privileged user on a modern Linux (or Windows if needed) desktop system.\n### Clarifications\n* Assume that the current directory is `$HOME`/`%USERPROFILE%`.\n* Assume that no file or directory exists with the name given.\n* Assume that the filename will only contain the characters `abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789`.\n* Assume that the filename will always be 8 characters long or shorter.\n## Rules\n* The directory must actually be created. You can't just say \"I made it in /dev/null\".\n* Standard loopholes apply.\n* This is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\").\n      \n[Answer]\n# MATLAB/Octave, 6 bytes\n```\n@mkdir\n```\nCreates a function handle named `ans` to the built-in function `mkdir` which accepts the folder to be created as a string\n```\nans('folder_to_create')\nans('/absolute/path/to/folder/to/create')\n```\n[Answer]\n# [Julia 0.5](http://julialang.org/), 5 bytes\n```\nmkdir\n```\n[Try it online!](https://tio.run/##yyrNyUw0/Z@mYPs/Nzsls@i/UoWSQo2dQhpXUWpiClBAQxPETcksLshJrPwPAA \"Julia 0.5 – Try It Online\")\n[Answer]\n## Java, 33 31 bytes\n-2 bytes thanks to @Phoenix - removing curly braces around lambda.\n```\ns->new java.io.File(s).mkdir();\n```\nIn the form of a `Consumer` where `s` is the name of the folder to create.\n[Answer]\n# [Ruby](https://www.ruby-lang.org/), 11 + 1 = 12 bytes\n+1 for `-n`\n```\nDir.mkdir$_\n```\n[Try it online!](https://tio.run/##KypNqvz/3yWzSC83OyWzSCX@f0FpSbECSCA1r6QoM7VYQ11PXfN/Wn7@v/yCksz8vOL/unkA \"Ruby – Try It Online\")\n[Answer]\n# [Bash](https://www.gnu.org/software/bash/), 8 bytes\n```\nmkdir $1\n```\n[Try it online!](https://tio.run/##S0oszvj/Pzc7JbNIQcXwf07x//9p@fkA \"Bash – Try It Online\")\n[Answer]\n# [Python](https://docs.python.org/3/), 18 bytes\n```\nimport os\nos.mkdir\n```\n[Answer]\n## Batch, 6 bytes\n```\n@md %*\n```\n~~Try it online!~~\n[Answer]\n# [Go](https://golang.org/), 46 bytes\n```\nimport.\"os\"\nfunc f(n string){Mkdir(n,ModeDir)}\n```\n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), 14 bytes\n```\nRun[\"md \"<>#]&\n```\n[Try it online!](https://tio.run/##y00syUjNTSzJTE78r/E/qDQvWik3RUHJxk45Vu2/ZrRSYlKyUux/AA \"Wolfram Language (Mathematica) – Try It Online\")\n---\nThe [trivial approach](https://codegolf.stackexchange.com/a/123911/69850) is longer. Surprise! (demonstrates that Mathematica is very verbose)\nWorks on Windows only, where `md` is the command to create a directory. See [this answer](https://codegolf.stackexchange.com/a/154632/69850) or [this answer](https://codegolf.stackexchange.com/a/123931/69850).\n---\nExplanation:\n```\nRun[\"md \"<>#]&   Anonymous function.\n    \"md \"        String literal \"md \". (with a trailing space to separate\n                   the command and the parameter (directory name)\n           #     The input.\n         <>      `StringJoin` together.\nRun[        ]    Runs as a command.\n```\n[Answer]\n# Mathematica, 15 bytes\n```\nCreateDirectory\n```\n[Answer]\n# C#, 35 bytes\n```\nSystem.IO.Directory.CreateDirectory\n```\nDoes what it says on the tin.\n[Answer]\n# Batch, 2 bytes\n```\nmd\n```\n`md` means **m**ake **d**irectory. Here it is in action:\n```\nC:\\tmp>set f=md\nC:\\tmp>%f% foo\nC:\\tmp>dir f*.*\n Volume in drive C is OS\n Volume Serial Number is 6CC3-3025\n Directory of C:\\tmp\n02-Feb-18  09:56              foo\n               0 File(s)              0 bytes\n               1 Dir(s)  378,122,883,072 bytes free\n```\n[Answer]\n# [APL (Dyalog Unicode)](https://www.dyalog.com/), 6 bytes\nPrefix function.\n```\n⎕MKDIR\n```\n[Try it online!](https://tio.run/##SyzI0U2pTMzJT///P@1R24RHfVN9vV08g4A8BfW0/Hx1rkdtE4GCwR4K6jnFCro5BeoA \"APL (Dyalog Unicode) – Try It Online\")\n[Answer]\n# tcl, don't know how to count\nMy count could be:\n* 13, if everything on `file mkdir $x` is accounted.\n* 12, if `file mkdir $`as how I have seen in some answers `x` is not taken into account\n* 11, because in tcl whenever you use the value of `x` , you type `$x`, so only `file mkdir`  is taken into account\n* 10, if is `file mkdir` by removing the trailing space from previous.\nSource:\n```\nfile mkdir $x\n```\n[demo](http://www.tutorialspoint.com/execute_tcl_online.php?PID=0Bw_CjBb95KQMRXZqNEpXWnpJUW8) — Please reserve it for question asker. Look at files tree at the left where there is only `root` and `main.tcl` . Click the `Execute` button, it will run a script command on the green area; In the top of the files tree there is a `Refresh` button symbolized by an icon, click it and see `folder1` and `folder2` appearing.\n]"}}},{"rowIdx":13666,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**This question already has answers here**:\n \n \n[Find the Factorial!](/questions/607/find-the-factorial)\n (216 answers)\n \nClosed 9 years ago.\n# Challenge\nThe objective is simple, in the shortest amount of bytes, write a function or program that takes a number and prints its [double factorial](http://en.wikipedia.org/wiki/Double_factorial).\nThe input must be taken from stdin or a function argument, and it must print the resulting double factorial to stdout.\nInbuilts are disallowed\n      \n[Answer]\n# Haskell 21\n```\np n=product[n,n-2..1]\n```\n[Answer]\n# CJam, ~~13~~ 12 bytes\n```\nri,:)1~%1+:*\n```\nExpanded code:\n```\nri            \"Read the input as a string and convert it to integer\";\n  ,           \"Create an array of 0 to input integers - 1\";\n   :)         \"Increment each element in the array\";\n     1~       \"Put 1 and do bitwise not to get -2\";\n       %      \"Take every other integer starting from end\";\n        1+    \"Add 1 to the array to account for result of 0 being 1\";\n          :*  \"Reduce product to get double factorial\";\n```\nAt the end of the code, anything on stack is automatically printed to STDOUT in CJam.\n[Try it online here](http://cjam.aditsu.net/)\n[Answer]\n# JavaScript, ES6, ~~38~~ 36 bytes\nWow this is huge!\n```\nf=a=>alert((g=n=>n>1?n*g(n-2):1)(a))\n```\n```\n\n```\nOf course I am just counting JS code. HTML is just for stack snippet demo.\nRun it in latest Firefox browser. \n[Answer]\n# Matlab 16\n```\n@(n)prod(n:-2:1)\n```\nIt is pretty sraight forward.\n[Answer]\n## r, 30\n```\nf=function(n)prod(seq(n,1,-2))\n```\n[Answer]\n# [Pyth](https://github.com/isaacg1/pyth): 11 character\n```\nu*GhH%2_UQ1\n```\n`Q` is the input value. `UQ` creates the list `[0, 1, 2, ..., Q-1]`, `_` inverts it, and `%2` removes every other element `[Q-1, Q-3, Q-5, ..., (1 or 0)]`. `u*GhH...1` reduces the list by multiplying the list elements (increased by 1) to `1`. \nTry it here: [Pyth Compiler/Executor](https://pyth.herokuapp.com/)\n]"}}},{"rowIdx":13667,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**This question already has answers here**:\n \n \n[We're no strangers to code golf, you know the rules, and so do I](/questions/6043/were-no-strangers-to-code-golf-you-know-the-rules-and-so-do-i)\n (73 answers)\n \nClosed 5 years ago.\nWrite a function or program that outputs the following lines `` time:\n```\nHello World. Dlrow Olleh. Hwlrod Eoll.\nBig Brown Fox. Xof Nworb Gib. Bbfgox Iro.\n```\n**EDIT**: It's \"*B* rown *F* ox\" not \"brown fox\".\nThis is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\") so the program with the shortest bytecount wins!\n      \n[Answer]\n## vim - 82 strokes\n```\noHello World. Dlrow Olleh. Hwlrod Eoll.\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.\n```\nType the number of repeats you want before entering this.\n[Answer]\n**Python ~~110~~ 105 Characters**\n```\nexec\"print'Hello World. Dlrow Olleh. Hwlrod Eoll.\\\\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.';\"*input()\n```\n[Answer]\n# Ruby - 84 chars\n```\n\"Hello World. Dlrow Olleh. Hwlrod Eoll.\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.\"*n\n```\n[Answer]\n## JavaScript - 100 characters\n```\nwhile(n--)print('Hello World. Dlrow Olleh. Hwlrod Eoll.\\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.')\n```\n[Answer]\n# D: 127 Characters\n```\nauto f(int n){return join(repeat(\"Hello World. Dlrow Olleh. Hwlrod Eoll.\\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.\",n),\"\\n\");}\n```\nMore Legibly:\n```\nauto f(int n)\n{\n    return join(repeat(\"Hello World. Dlrow Olleh. Hwlrod Eoll.\\nBig brown fox. Xof Nworb Gib. Bbfgox Iro.\", n), \"\\n\");\n}\n```\n[Answer]\n# PowerShell: 95\n```\n\"Hello World. Dlrow Olleh. Hwlrod Eoll.`nBig Brown Fox. Xof Nworb Gib. Bbfgox Iro.`n\"*(read-host)\n```\nRead-Host prompts the user for input, then the string is multiplied by that value and printed. This will have an extra newline at the end, but I don't see that as a constraint in the challenge.\n[Answer]\n# [Whitespace](https://web.archive.org/web/20150618184706/http://compsoc.dur.ac.uk/whitespace/tutorial.php), ~~761~~ 724 bytes\n```\n[S S S N\n_Push_0][S N\nS _Duplicate_0][T   N\nT   T   _Read_input_as_integer][T   T   T   _Retrieve_heap_at_0][S N\nS _Duplicate][N\nS S N\n_Create_Label_LOOP][S N\nN\n_Discard_top][S S S T   N\n_Push_1][T  S S T   _Subtract][S N\nS _Duplicate_input][N\nT   T   T   N\n_Jump_to_Label_EXIT_if_negative][S S S N\n_Push_0][S S T  T   S T T   S S T   N\n_Push_-89_\\n][S S T T   T   S T S T N\n_Push_-53_.][S S S T    T   S S N\n_Push_12_o][S S S T T   T   T   N\n_Push_15_r][S S T   T   T   S T S N\n_Push_-26_I][S S T  T   S S S S T   T   N\n_Push_-67_space][S S S T    S T S T N\n_Push_21_x][S S S T T   S S N\n_Push_12_o][S S S T S S N\n_Push_4_g][S S S T  T   N\n_Push_3_f][S S T    T   N\n_Push_1_b][S S T    T   S S S S T   N\n_Push_-33_B][S T    S S T   T   S N\n_Copy_6th_space][S S T  T   T   S T S T N\n_Push_-53_.][S S T  T   N\n_Push_1_b][S S S T  T   S N\n_Push_6_i][S S T    T   T   T   S S N\n_Push_-28_G][S T    S S T   S S N\n_Copy_4th_space][S S T  T   N\n_Push_-1_b][S S S T T   T   T   N\n_Push_15_r][S S S T T   S S N\n_Push_12_o][S S S T S T S S N\n_Push_20_w][S S T   T   S T S T N\n_Push_-21_N][S T    S S T   S T N\n_Copy_5th_space][S S S T    T   N\n_Push_3_f][S S S T  T   S S N\n_Push_12_o][S S T   T   S T T   N\n_Push_-11_X][S T    S S T   T   N\n_Copy_3rd_space][S S T  T   T   S T S T N\n_Push_-53_.][S S S T    S T S T N\n_Push_21_x][S S S T T   S S N\n_Push_12_o][S S T   T   T   T   S T N\n_Push_-29_F][S T    S S T   S S N\n_Copy_4th_space][S S S T    S T T   N\n_Push_11_n][S S S T S T S S N\n_Push_20_w][S S S T T   S S N\n_Push_12_o][S S S T T   T   T   N\n_Push_15_r][S S T   T   S S S S T   N\n_Push_-33_B][S T    S S T   S T N\n_Copy_5th_space][S S S T    S S N\n_Push_4_g][S S S T  T   S N\n_Push_6_i][S T  S S T   T   N\n_Copy_3rd_B][S S T  T   S T T   S S T   N\n_Push_-89_\\n][S S T T   T   S T S T N\n_Push_-53_.][S S S T    S S T   N\n_Push_9_l][S N\nS _Duplicate_9_l][S S S T   T   S S N\n_Push_12_o][S S T   T   T   T   T   S N\n_Push_-30_E][S T    S S T   S S T   N\n_Copy_9th_space][S S S T    N\n_Push_1_d][S T  S S T   T   N\n_Copy_3rd_o][S S S T    T   T   T   N\n_Push_15_r][S S S T S S T   N\n_Push_9_l][S S S T  S T S S N\n_Push_20_w][S S T   T   T   S T T   N\n_Push_-27_H][S T    S S T   T   S N\n_Copy_6th_space][S S T  T   T   S T S T N\n_Push_-53_.][S S S T    S T N\n_Push_5_h][S S S T  S N\n_Push_2_e][S S S T  S S T   N\n_Push_9_l][S N\nS _Duplicate_9_l][S S T T   S T S S N\n_Push_-20_O][S T    S S T   T   S N\n_Copy_6th_space][S S S T    S T S S N\n_Push_20_w][S S S T T   S S N\n_Push_12_o][S S S T T   T   T   N\n_Push_15_r][S S S T S S T   N\n_Push_9_l][S S T    T   T   T   T   T   N\n_Push_-31_D][S T    S S T   S T N\n_Copy_5th_space][S S T  T   T   S T S T N\n_Push_-53_.][S S S T    N\n_Push_1_d][S S S T  S S T   N\n_Push_9_l][S S S T  T   T   T   N\n_Push_15_r][S S S T T   S S N\n_Push_12_o][S S T   T   T   S S N\n_Push_-12_W][S T    S S T   T   S N\n_Copy_6th_space][S S S T    T   S S N\n_Push_12_o][S S S T S S T   N\n_Push_9_l][S N\nS _Duplicate_9_l][S S S T   S N\n_Push_2_e][S S T    T   T   S T T   N\n_Push_-27_H][N\nS S S N_Create_Label_PRINTER\n][S N\nS _Duplicate][N\nT   S N_Jump_to_Label_LOOP_if_0\n][S S S T   T   S S S T T   N_Push_99\n][T S S S _Add][T   N\nS S _Print_as_character][N\nS N\nS N\n_Jump_to_Label_PRINTER]\n```\nLetters `S` (space), `T` (tab), and `N` (new-line) added as highlighting only.  \n`[..._some_action]` added as explanation only.\n[Try it online](https://tio.run/##dVJBDgIxCDwPr@APvsgYE72ZaOLzK0xLod11m1AWBoYh/T6en/v7db3dW1NVsQOBf/TdmPWYGf@zjBCoBoFH6ZpPj6F@IzMdzShzB3DUyDREi2cw6qMKUdqjrEX0iCS2/pgjR9eYoOCoaHBuqraJyYvCqyja9HQRRVE01akkOU@XyojoOgCm8A7KLrkAVawLKBL1oLFfKyNG7Sz5K1FzcKAqrSTLVNjWvxPl85jyiwoZD1YS7WEMGqZau/wA) (containing the raw spaces, tabs and new-lines).\n**Explanation in pseudo-code:**\n```\ninteger i = input_as_integer\nStart LOOP\n  i = i-1\n  if i is negative:\n    EXIT\n  Call function PRINT\nfunction PRINT\n  Print \"Hello World. Dlrow Olleh. Hwlrod Eoll.\\n\n         Big Brown Fox. Xof Nworb Gib. Bbfgox Iro.\\n\n         \"\n  Go to the next iteration of LOOP\n```\n**Some things I did to golf it:**\n* `SNS` (Duplicate) is used for repeated characters (`ll`)\n* 99 is subtracted from every ASCII-value, which we'll add in the PRINT function to reduce loads of bytes. In Whitespace an integer/character is formed by using `SS` (Stack Manipulation: Push Number) + `S` or `T` (Positive or Negative) + some `T` / `S` (Binary representation of the number where `T=1` and `S=0`) + `N`. So having lower binary values and adding the 99 in the `PRINT` function is shorter than using the large binary values directly.\n* Since creating the number for a space is 11 bytes (`SSTTSSSSTTN`), I copied the spaces from the last occurrences to the top of the stack instead. For example, after the first space is created in `\\n.orI xogfbB`, I then use `STSSTTSN` (copy 6th to top) instead to copy the previous space. The largest distant between two spaces is 9, and since `STSSTSSTN` (copy 9th to top) is still 2 bytes shorter than `SSTTSSSSTTN` (create space) I've done this for all spaces to save bytes.\n* I did something similar as with the copying of spaces above for every character that has a distance of 3 or less. This applies to the `B` in `Big B` and the `o` in `od Eo` (Note: The distance between both `b` in `b. Bb` is also three, but creating the number for `b` (`SSTTN`) is actually shorter than copying the 3rd value to the top of the stack (`STSSTTN`).)\n[Answer]\n## Game Maker Language, 107\n```\nrepeat argument0{return \"Hello World. Dlrow Olleh. Hwlrod Eoll.#Big Brown Fox. Xof Nworb Gib. Bbfgox Iro.\"}\n```\n[Answer]\n# Java 8, 121 bytes\n```\nn->{for(;n-->0;System.out.println(\"Hello World. Dlrow Olleh. Hwlrod Eoll.\\nBig Brown Fox. Xof Nworb Gib. Bbfgox Iro.\"));}\n```\n**Explanation:**\n[Try it online.](https://tio.run/##LY@9TsNAEIR7nmKU6lz4REN1IoXFT1JgihQgAcXZPptLNrvW@RIHRX52c5BIu8XMaLXfbO3R5tI73ja7uSY7DHixns83gOfoQmtrh/JPAkfxDWqVfHBmkjWlTTNEG32NEox7zJwvz60EZTjPl7dm8zNEt9dyiLoP6ZRYLVaOSPAmgRqNBwoy4pXIfWusxqQaPAqR/uTCdyhSyniSk8a7tChHCRWefaVRVG0nJ6yD6EWWmWk2F5z@UFHCuVL9Q@9TJbWJ6X/38WWzSx3Wtbq79pjmXw)\n```\nn->{           // Method with integer parameter and no return-type\n  for(;n-->0;  //  Loop `n` times\n    System.out.println(\"Hello World. Dlrow Olleh. Hwlrod Eoll.\\n\n                        Big Brown Fox. Xof Nworb Gib. Bbfgox Iro.\"));}\n               //   And print this text + a trailing new-line that many times\n```\n]"}}},{"rowIdx":13668,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/201120/edit).\nClosed 3 years ago.\n[Improve this question](/posts/201120/edit) \n# Challenge\nYour task is to generate a string using this sequence:\n```\n1,1,2,1,1,2,3,2,1,1,3,2,4,2,3,1,....\n```\nWhich is more understandable in this format:\n```\n1\n1 2 1\n1 2 3 2 1\n1 3 2 4 2 3 1\n1 4 3 2 5 2 3 4 1\n```\nThe pattern increases the number of Digits in the string by two for each increment in the Value of N\n**For each integer value up to the total N, *aside from 1*** the string is centred on the current value of N and bookended by the previous values of N in decreasing order on the left, and increasing order on the right - with the exception of the value 1, which bookends the entire string.\n# Input\nNumber of steps\n# Output\nThe series separated by comma\n      \n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E), 7 bytes\nPort of the Jelly answer.\n```\nL¦ÁR1šû\n```\n[Try it online!](https://tio.run/##yy9OTMpM/f/f59Cyw41BhkcXHt79/7@hAQA \"05AB1E – Try It Online\")\n# Explanation\n```\nL       1 .. N\n ¦      Tail\n  Á     Shift right\n   R    Reverse\n    1š  Prepend 1\n      û Palindromize\n```\n# [05AB1E](https://github.com/Adriandmen/05AB1E), 9 bytes\nIf the whole format is necessary, here's one with 9 bytes. (Question: Does 05AB1E have any sort of implicit range?)\n```\nLεL¦ÁR1šû\n```\n[Try it online!](https://tio.run/##yy9OTMpM/f/f59xWn0PLDjcGGR5deHj3//@GBgA \"05AB1E – Try It Online\")\n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), ~~10~~ 10 ?8 bytes\n```\nḊṙ-U1;ŒB\n```\n[Try it online!](https://tio.run/##ATEAzv9qZWxsef//4biK4bmZLVUxO8WSQv874oCcIC0@IOKAnTvDh@G5viQKMTDDh@KCrFn/ \"Jelly – Try It Online\")\nI’m assuming you want the \\$n\\$th row of the pyramid. If you actually want the full sequence for the \\$n\\$th step, then it would be [10 bytes](https://tio.run/##ATMAzP9qZWxsef//4biK4bmZLVUxO8WSQilG/zvigJwgLT4g4oCdO8OH4bm@JAoxMMOH4oKsWf8).\nA monadic link taking an integer and returning a list of integers.\n## Explanation\n```\nḊ        | List 1..(n-1)\n ṙ-      | Rotate right 1\n   U     | Reverse list\n    1;   | Prepend 1\n      ŒB | Bounce (mirror\n```\n[Answer]\n# [Python 3.8 (pre-release)](https://docs.python.org/3.8/), ~~86~~ ~~81~~ ~~58~~ 53 bytes\n```\nlambda n:[[1],[1,*(l:=range(2,n))[::-1],n,*l,1]][n>1]\n```\n[Try it online!](https://tio.run/##JcoxDoMwDADAva/wGEfu4LIgS/QjJkOqNhApmChi4fWhEvNdPY91t2Gsradp7iVun28EE1UOpEzeFZlatOXnXmSIKvL8i5EvxCGovTn0tDfIkA3uyMSM8gCoLdvhfHIZsV8 \"Python 3.8 (pre-release) – Try It Online\") Hats off to @Noodle9 for saving me a *whooping* 23 bytes and thanks to @squid for saving me 5 bytes!\n[Answer]\n# [C# (Visual C# Interactive Compiler)](http://www.mono-project.com/docs/about-mono/releases/5.0.0/#csc), 145 bytes\n```\nn=>{var s= \"\";for(int i=0;i++2?i-j+2:1)+\",\";s+=i;for(int j=i;j-->1;)s+=\",\"+(j>1?i-j+1:j);s+=ilists:join(\" \",[integer_to_list(X)||X<-lists:seq(2,I-1)]).\nx(0)->\"\";x(I)->x(I-1)++\"\\n\"++if I>1->\"1 \"++lists:reverse(q(I))++\" \"++integer_to_list(I)++\" \"++q(I)++\" 1\";1<2->\"1\"end.\n```\n[Try it online!](https://tio.run/##XY5BDoMgEEX3noKwglCMuFTjnhuYWGNMOxoaixVI68K7U6jtprOYzOT9@X/AzIOeONiLUQ/nE78SSXk9K@tscVuUJhjhU6u0gwlM75Y@ItLQfW8qfsgsrCQ/SS5oR9NkI1kwwLjcPk6hB8AYPmvMmBqRrEXAAoXtODfwBGOBxOQojOQ/T/7A@h0FLkWVRyMM@pr6@xBebTuKeJ2gUGopxpdRDshGREZp6t8 \"Erlang (escript) – Try It Online\")\n[Answer]\n# JavaScript (ES6), ~~68~~ 65 bytes\nReturns the whole sequence after \\$n\\$ iterations.\n```\nf=n=>--n?[...f(n),1,...(g=k=>k>1?[k,...g(k-1),k]:[n+1])(n),1]:[1]\n```\n[Try it online!](https://tio.run/##HcpNCoAgEEDh68yQCrNoE6gHERdRKTUyRkXXt5/d@@Bt4z2e07Hul5Y6L60lK9ZpLT4YYxIIKlJvQbZsHTvygT9nYE2oOA5BOor4jy8otqnKWctiSs2QoEdsDw \"JavaScript (Node.js) – Try It Online\")\n---\n# JavaScript (ES6), ~~53~~ 46 bytes\nReturns the \\$n\\$th row of the pyramid.\n```\nn=>(g=k=>k>1?[k%n||1,...g(k-1),k%n||1]:[n])(n)\n```\n[Try it online!](https://tio.run/##JcqxCoMwEIDh3ae4pXCHMZDBxfbig4iDqAk2cle0uNQ@eyp0@uHjfw7HsI/b8npXotOcA2dhj5ET@@Rd26WbnKcz1tqIqXJk/tA3nfSEQjnohgIM7g4CD4b6alkSfAqAUWXXdbarRhQD4fqp@OYf \"JavaScript (Node.js) – Try It Online\")\n]"}}},{"rowIdx":13669,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n*Create a program that calculates my reading speed!*\n# The challenge\nWrite a program which takes some text as input and waits for a *start* signal from the user to display the input. When to user gives a second *end* signal, the program should output the *reading speed* of the user, which is defined as the number of words in the input divided by the time between both signals in minutes.\nThe word count of the input text is determined by splitting the text at white space.\n## Input\n* A block of text via any standard input means.\n* A *start* and *end* signal by the user, either a click or pressing a certain key of your choice.\n## Output\n* The block of input text.\n* The *reading speed* of the user as defined above.\n## Rules\n* The start and end signal may **not** be running and ending the program.\n* The text may wrap or the user may scroll.\n* The reading speed value may be returned in any of the standard methods in the meta post found [here](https://codegolf.meta.stackexchange.com/questions/2447/default-for-code-golf-input-output-methods), though the text must be displayed beforehand to the user, e.g. by printing to `stdout`.\n* This is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"), so the shortest code in bytes wins.\n      \n[Answer]\n# MATL, 24 bytes\n```\nO$Y.Y`OYDtDYbn60*O$Y.Z`/\n```\n**Explanation**\n```\nO$Y.    % Wait for the user's keypress\nY`      % Start the timer\n        % Implicitly grab the input\nOYD     % Converts control characters and creates a string\ntD      % Duplicate the string and display\nYbn     % Split the string into words and count them\n60*     % Multiply by 60 to convert words per second to words per minute\nO$Y.    % Wait for user to press a key\nZ`      % Get the elapsed time in seconds\n/       % Divide the number of words (times 60) by the time in seconds\n        % Implicitly display the result\n```\n[Answer]\n# Octave, 59 bytes\n```\npause;tic;disp(x=input(''));pause;numel(strsplit(x))*60/toc\n```\n**Explanation**\n```\npause               % Wait for a user's keypress\ntic                 % Start the timer\ndisp(..)            % Display paragraph\npause;              % Wait for user's keypress\nnumel(strsplit(x))  % Determine number of words\n*60                 % Multiply by 60 to convertwords per second to words per minute \n/toc                % Divide by the total time that it took (in seconds)\n```\n[Answer]\n# Processing Js 77 bytes\n```\nvar i=\"\",b=second();println(i);i.split(\" \");keyTyped=function(){println(i.length/(second()-b));};\n```\n[Try it out](https://www.khanacademy.org/computer-programming/new-program/5790491865317376)\n]"}}},{"rowIdx":13670,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/31520/edit).\nClosed 9 years ago.\n[Improve this question](/posts/31520/edit) \nHere is one more \"Challenge\"! Today i had a meeting with a friend, and mostly we are a little bit crazy.. He bets, that it is not possible to create a Fibonacci Sequence without using loops.\nNo static output like:\n```\nreturn \"1, 2, 3, 5, 8, 13, ..\"; \n```\n**GOAL / Requirements**\n* Use only 1 int var\n* Return the first 5 (or more) numbers\n* Name your function \"fibonacci\"\n* Call no other functions except \"fibonacci\"\n* No loops are allowed\n* Have fun ;)\n**Scoring**\nPopularity contest (Most upvotes minus downvotes)\n**End Date**\nJune 17. 2014\n**No special language required** \n      \n[Answer]\n# Python\n```\ndef fibonacci(n):\n    if n > 1:\n        return fibonacci(n-1) + [(1.618033988749895**n/2.23606797749979 + 0.2) // 1]\n    else:\n        return [1]\n```\nCall `fibonacci(n)` to generate a list up to the nth element of the Fibonacci series. Sample output of `fibonacci(20)`:\n```\n>>> fibonacci(20)\n[1, 1.0, 2.0, 3.0, 5.0, 8.0, 13.0, 21.0, 34.0, 55.0, 89.0, 144.0, 233.0, 377.0, 610.0, 987.0, 1597.0, 2584.0, 4181.0, 6765.0]\n```\nThe correctness is limited by the precision of the double precision floating point.\n[Answer]\n**EXCEL / NUMBERS / $spreadsheetapp**\nI think I saw this somewhere here, but I can't remember. Nevertheless, credits go to unknown author. Just Fill in the first 2 fields, do a simple addition and then it's time to drag, baby!\n![enter image description here](https://i.stack.imgur.com/orVGL.jpg)\n[Answer]\n# [Kona](https://github.com/kevinlawler/kona)\n```\nfibonacci:{((x-1)(|+\\)\\1 1)[;1]}\n```\nExecute in shell as `fibonacci 12` to get the first 12 Fibonacci numbers.\n**Explanation**:\n* `1 1` is a vector\n* `(|+\\)\\` generates a vector of vectors by summing the elements of the vector\n* `(x-1)` reduces 12 to 11 (due to 0 indexing) which is applied to the summation to repeat `x-1` times\n* `[;1]` returns/prints the second element of each vector in the larger vectors.\nIf I neglected the `[;1]` portion, you can see what happens to the vector:\n```\n> fib 1\n,1 1\n> fib 2\n(1 1\n 2 1)\n> fib 3\n(1 1\n 2 1\n 3 2)\n```\nAnd so on.\n[Answer]\n# JavaScript (ECMAScript 6)\n```\nfibonacci=n=>n<2?[0,1]:[...fibonacci(n-1),fibonacci(n-1)[n-1]+fibonacci(n-2)[n-2]]\n```\nor more efficiently\n```\nfibonacci=n=>n<2?[0,1]:(i=fibonacci(n-1),[...i,i[n-1]+i[n-2]])\n```\nor even more efficently (but with a loop instead of recursion)\n```\nfibonacci=n=>[i[m]=m<2?+m:i[m-1]+i[m-2]for(m in i=[...Array(n)])]\n```\n]"}}},{"rowIdx":13671,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/175908/edit).\nClosed 5 years ago.\n[Improve this question](/posts/175908/edit) \n## FreeChat Online\nIt is 1985, and you are a humble Russian potato farmer. Sadly, your glorious homeland is going through a time of crisis and it needs **your** help! The KGB, with the \"assistance\" of your wife and children, has convinced you to build their up-and-coming government *FreeChat Online* service for high ranking soviet officials. Luckily, the Russian government has provided you with several items seized through routine surveillance of the US's technological innovationsâ€”a cutting edge Macintosh XL, and the prototype source code for LucasArts's *Habitat*, an advanced communication service with cutting edge graphics. Your task is simple: create a few modifications for the Russian governmentâ€¦\nDue to Communism's tendency to coalesce toward collectivistic cultures, the Russian government wants this chat to embody their perfect worldâ€”one without silly notions of self or individualism. Alas, Macintosh compatibility leads you astray. Due to problems in the prototype, every other sentence in the chat buffer is corrupted, except for the ending period. Lucky for you, the Russian people make classification easy for you.\n**CHALLENGE**:\nFor a String `s` in the input, there are 1 or more sentences, separated by `\". \"`(period, TWO spaces) The 2nd, 4th, 6th, and so on sentences are corrupted. The last sentence may have nothing, `\".\"`, `\". \"`, or `\". \"` as its end. Here's some examples of an input:\n`\"L'etat c'est moi (I am the state) - King Louis XIV\"`\n|----------------one sentence, ended by nothing--------------|\n`\"Hello World. GojdGojog. How are you. \"`\n|--three sentences, one corrupted, ended by `\". \"`--|\n`\"Hi. gfhyG. My favorite food is Baseball. jijhojh. Foobar. \"`\n|---------------Five sentences, two corrupted, ended by `\". \"`-----------|\nYou must process this string, replacing any capitalized word besides the first word of a sentence, words in parentheses *(no nested, but there can be multiple sets in a sentence)*, or corrupted sentences with the string `\"The State\"`.\n**Examples**:\n`\"L'etat c'est moi (I am the state) - King Louis XIV\"`\n--->`\"L'etat c'est moi (I am the state) - The State The State The State\"`\n`\"Hello World. GojdGojog. How are you. \"`\n--->`\"Hello The State. GojdGojog. How are you. \"`\n`\"Hi. gfhyG. My favorite food is Baseball. jijhojh. Foobar. \"`\n--->`\"Hi. gfhyG. My favorite food is The State. jijhojh. Foobar. \"`\n**Rules**:\n1. [Standard Loopholes](https://codegolf.meta.stackexchange.com/questions/1061/loopholes-that-are-forbidden-by-default) are prohibited\n2. ASCII Text Input\n3. If the input doesn't match the specifications set above, do whatever you please with the output!\n4. This is code-golf, so the shortest(valid)answer (in each language) wins!\n5. Bonus points for a one-liner ðŸ˜Ž (Not really tho, just cool points)\n      \n[Answer]\n# [Retina 0.8.2](https://github.com/m-ender/retina/wiki/The-Language/a950ad7d925ec9316e3e2fb2cf5d49fd15d23e3d), 77 bytes\n```\n(?<=^(([^.]*)\\. +([^.]*)\\. +)*[^ .][^.]*)(?{var S=s.split(\".  \",-1);for(int l=S.length,i=0;i{                       // Method with String as both parameter and return-type\n  var S=s.split(\".  \",-1); //  Split the input by \".  \" to get the list of sentences\n                           //  (the -1 is in case the input ends in a trailing \".  \")\n  for(int l=S.length,      //  Amount of sentences\n      i=0;i<-[.\\s]>+/ The State/}\n```\n[Try it online!](https://tio.run/##TU1NT4NAFLz7K8YGBWK7m3jwUBETTWwb6alGTcSYpSywZPE1u6AhyG/H1Xjw8DJvJvNxkEZfTE2P0wJXmIbdsuSpjZZJF0eLF5ba1/gsuh6OSyMPGDzODkY2ot1XbE9NFnhv4cn5OPcZ4M95GnylIXfoch4fY46HSmLXilbycbKiRxHMEl86Abe@tC0aUgg2EA1a57Q/zhAL3Kv3Egl1yuJ58zgLL4/@wmupNeGJjM7d5Irq3B2V7l/TJ4SR6Klz7H9EOV4WVb9yuHWS@CCjWomCKIdbuBFWZkJrhlrVFdUVwx1RJgz7rZm@AQ \"Perl 6 â€“ Try It Online\")\nRegex based solution.\n### Explanation:\n```\n S:g/                  # Do a global substitution\n     \\s<:Lu><-[.\\s]>+  # On all words starting with capital letters\n                 # Where:\n                  !grep {                       } # None of:\n                                                 ,'.  '   # Sentence terminators\n                                                 ,/\\(|\\)/ # Parenthesises\n                                                 ,/\\.\\s$/ # Is first word of sentence?\n                                    .comb($_)%2   # Have an odd amount of matches\n                         $/.prematch              # From the string before\n                                                           / The State/  # And replace\n```\n]"}}},{"rowIdx":13672,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/52428/edit).\nClosed 8 years ago.\n[Improve this question](/posts/52428/edit) \nThe challenge is to write a program which outputs all characters of the Unicode table from n to z range.\nFor code \"0000\" is the first in order and you must skip all blank Unicode combinations.\nFor example, let n = 134 and z = 300, you must save or print all unicode characters between 0134 and 0300.\nThe output of the program does not HAVE to output them, for example using std::cout can be used but if you have an array with all the saved characters then that still qualifies.(You don't have to print them out)\nThe shortest program in bytes wins.\n      \n[Answer]\n# Pyth - 16 bytes\nMaps `C`har-of over the range, and detects unprintables by if the string `\"\\x\"` is inside its repr because that is how python represents unprintables like `\\x00`.\n```\nf!}\"\\\\x\"`TCMrQvw\n```\n[Try it here online](http://pyth.herokuapp.com/?code=f!%7D%22%5C%5Cx%22%60TCMrQvw&input=134%0A300&debug=1).\n]"}}},{"rowIdx":13673,"cells":{"text":{"kind":"string","value":"[Question]\n      [\nYou need to find the sum of minimum values of two list (same size) of small numbers. In a C like language, that could be expressed as:\n`v = min(a[0],b[0]) + min(a[1],b[1])+ min(a[2],b[2]) ...`\nThe range of values is 0..7 and the 2 list can have up to 7 values. If you like, all the values in each list can be stored as a single 32 bit number.\nMy goal is avoiding branches in code, so the rules are\n* Single line expression,\n* No function calls or method calls (except for basic operations like sum, if your language mandate this),\n* No if statements, including things that compile to if statements such as ternary operators or \"greater than\" operators\n**Edit 1** To clarify, what I want is 'no branches in code'. So, no if, no calls (function of methods). Loops with fixed ranges (0..7) could be OK because can be unrolled. Sorry if was not crystal clear about this.   \nAll what your language can do inline is OK. So if a functional language has a strange syntax that resembles a function call but is inlined, that's OK. \n**Shortest code wins.**\n  \n  \nAs a side note: this challenge is related to a quasi-real problem: a fast scoring function in mastermind game.\n      \n[Answer]\n## C/C++ with SSE intrinsics (11 instructions)\nBecause everything's better with SIMD...\n```\ninline int foo(const int16_t *a, const int16_t *b) \n{\n    __m128i va = _mm_loadu_si128((__m128i *)a);\n    __m128i vb = _mm_loadu_si128((__m128i *)b);\n    __m128i vmin = _mm_min_epi16(va, vb);\n    vmin = _mm_add_epi32(_mm_unpacklo_epi16(vmin, vmin), _mm_unpackhi_epi16(vmin, vmin));\n    vmin = _mm_add_epi32(vmin, _mm_srli_si128(vmin, 4));\n    vmin = _mm_add_epi32(vmin, _mm_srli_si128(vmin, 8));\n    return _mm_cvtsi128_si32(vmin);\n}\n```\nNo branches, loops or function calls. This might not be the shortest *source* code, but it compiles to just 11 SSE instructions, so I claim shortest *generated* code.\n[Answer]\n**vb.net**\n```\nv = a.Zip(b,Function(p,q) Math.Min(p,q)).Sum\nv = a.Zip(b,AddressOf Math.Min).Sum \n```\nEdit:\n A no function call version. Just a single expression.\n```\nDim v = {a(0), a(0), b(0)}(1 + ((a(0) - b(0)) / ((a(0) - b(0)) * (a(0) - b(0)) ^ (1 / 2)))) +\n        {a(1), a(1), b(1)}(1 + ((a(1) - b(1)) / ((a(1) - b(1)) * (a(1) - b(1)) ^ (1 / 2)))) +\n        {a(2), a(2), b(2)}(1 + ((a(2) - b(2)) / ((a(2) - b(2)) * (a(2) - b(2)) ^ (1 / 2)))) +\n        {a(3), a(3), b(3)}(1 + ((a(3) - b(3)) / ((a(3) - b(3)) * (a(3) - b(3)) ^ (1 / 2)))) +\n        {a(4), a(4), b(4)}(1 + ((a(4) - b(4)) / ((a(4) - b(4)) * (a(4) - b(4)) ^ (1 / 2)))) +\n        {a(5), a(5), b(5)}(1 + ((a(5) - b(5)) / ((a(5) - b(5)) * (a(5) - b(5)) ^ (1 / 2)))) +\n        {a(6), a(6), b(6)}(1 + ((a(6) - b(6)) / ((a(6) - b(6)) * (a(6) - b(6)) ^ (1 / 2)))) +\n        {a(7), a(7), b(7)}(1 + ((a(7) - b(7)) / ((a(7) - b(7)) * (a(7) - b(7)) ^ (1 / 2))))\n```\nNote: Above fails in a(x)=b(x) as a(x)-b(x) = 0 which results in divide by zero .\n```\n((a(0) - b(0)) / ((a(0) - b(0)) * (a(0) - b(0)) ^ (1 / 2)))\n delta = a(x)-b(x)\n   abs = (delta * delta)^(1/2)  ; Sqr(delta^2) \n  sign = delta / abs            ; This is where the 0/0 happens. \n index = sign + 1               ; Since sign in normalise to -1 to +1\n                                ; when need offset it by +1 for 0-Indexed arrays.\n```\nVersion using < in mathematical sense. (akin to @Heiko Oberdiek entry)\n```\nDim v = {a(0),b(0)}((a(0)$=)*$=for 0..7\n```\nIt is a single line expression without functions and if statements.\nIn an updated question, loops are excluded because of the end condition. In this case the `for` operator unfolds an array. It is not clear, whether this is acceptable or not, thus it is something for the OP to decide.\n(Of course, `for` could be avoided by making the summation explicit, but such a solution would be too boring for me).\nThe input numbers are expected in arrays `@a` and `@b`. The sum is stored in `$v`. (If it is used a second time, `$v` needs to be reset to zero.)\nThe minimum summand for each value in the index array `0..7` is calculated and added to the result variable `$v`. As far as I have understood the question, the `for` operator is not excluded.\nThe result of the comparison operators `<=` and `>` are not used inside an `if` condition, but in a numerical context.\n**Ungolfed with test:**\n```\n@a = (1,7,3,4,5,0,3);\n@b = (3,1,4,1,5,2,6);\n$v = 0;\nfor $_ (0..7) {\n    $A = $a[$_];\n    $B = $b[$_];\n    $v += ($A <= $B) * $A\n        + ($A > $B) * $B\n}\n# The use of $= instead of $B in the golfed version saves one byte.\nprint  \"Result: $v\\n\";\n```\n**Result:** 14\n[Answer]\n# C, ~~81~~ 68\nNot the shortest, but I found this interesting, so here's mine. Thanks to @edc65 for the improvement, which is not only shorter, but also faster.\n```\nint s,i;for(s=i=0;i<7;s+=(((x[i]-y[i])>>31)&(x[i]-y[i]))+y[i],i++);\n```\nJust &, shift, multiplication and addition, no division. Replace 0x80000000 and 31 with the appropriate constants for non-4-bytes ints. Loop can be unrolled (and gcc with -O3 unrolls it for me.) Usage:\n```\nint main(void) {\n  signed int x[7] = {1,2,3,4,5,6,7};\n  signed int y[7] = {9,0,5,7,1,2,7};\n  int s,i;\n  for (s=0,i=0;i!=7;s+=(((x[i]-y[i])&0x80000000)>>31)*(x[i]-y[i])+y[i],i++);\n  return s; \n}\n```\nproduces\n```\ngcc min_bare.c && ./a.out ; echo $?\n18\n```\nLet's compare with the forked if version (b)\n```\nif x[i] > y[i]\n  s+=y[i]\nelse\n  s+=x[i]\nend\n```\nand with (c)\n```\ns+= ((x[i] > y[i]) ? y[i] : x[i]);\n```\nLet's benchmark with (taking the arrays from stdin)\n```\nint main(void) {\n  int l,i,j,s;\n  scanf(\"%d\",&l);\n  signed int x[l], y[l];\n  for (i=0; i!=l; scanf(\"%d\",&(x[i++])));\n  for (i=0; i!=l; scanf(\"%d\",&(y[i++])));\n  for (j=0; j!=99999999;j++) {\n    for (s=0,i=0;i!=l;i++) {\n      // put code to test here\n    }\n  }\n  printf(\"%d\\n\",s);\n  return 0;\n}\n```\nThe if version (b) takes ~4.3s, version (c) ~3.4s, the original ~~~5.3s~~ ~4.3s; all with -O3. So don't optimize prematurely ; )\n[Answer]\n## JavaScript - 84 ~~89~~ ~~97~~ ~~98~~\n**The code:**\n```\nfor(s=i=7;i--;s+=x*r/r|0+y*l/l|0){x=~-a[i];y=~-b[i];l=0xff&1<<(x-y);r=0xff&1<<(y-x)}\n```\n**49** if the comparison operators are allowed in numerical context:\n```\nfor(s=i=7;i--;s+=(x=~-a[i])-(x-y)*(y a=[1,4,3];b=[2,3,5];\n[ 2, 3, 5 ]\n> for(s=i=7;i--;s+=x*r/r|0+y*l/l|0,--s){x=a[i];y=b[i];l=0xff&1<<(x-y);r=0xff&1<<(y-x)}\n4\n> s\n7\n```\n[Answer]\n# Perl, 42\n```\n$s+=($a[$_],$b[$_])[$b[$_]<$a[$_]]for 0..7\n```\nI expect scripting/C-like languages with sigil-free syntax will fare slightly better in terms of their code-golf scores. Loops over the indices and creates a list with the element from `@a` and `@b`. The index is determined by the result of the `<` comparison.\nJust for kicks, one could write the following to make the logic scale for any number of elements by selecting the upper limit for the index variable to be the larger of `$#a` or `$#b`:\n```\n$s+=($a[$_],$b[$_])[$b[$_]<$a[$_]]for 0..($#a,$#b)[@a<@b]\n```\n[Answer]\n## J (9)\n```\nv=:+/a<.b\n```\nSomeone is going to need to help me out on this one. Does this match the rules?\n```\nSingle line expression\n```\nCheck.\n```\nNo function calls or method calls (except for basic operations like sum, if your language mandate this)\n```\nThe things J is made of are called `verbs` (it is basically impossible to write any J without verbs), are those considered functions?\n```\nNo if statements, including things that compile to if statements such as ternary operators or \"greater than\" operators\n```\n`<.` is basically equal to `min`, and `+/` is basically equal to `sum`. Are those considered to `compile to if statements`?\n[Answer]\n# Golfscript, 13\n```\n0\\~zip{$0=+}/\n```\nThe arguments have to be passed to the program as a list of the two lists (e.g. `[ [0 1 2] [4 2 0] ]`). \nI'm really not sure if it matches the rules, though (you said you won't forbid basic list operations? - zip in this case).\n### Explanation:\n* Golfscript pushes the arguments it get as a string into the stack before the program runs.\n* `0\\` pushes 0 and switching top elements of the stack (will be used for the sum).\n* `~` executes the code in the string, which is just pushing the array down the stack.\n* `zip` transposes array's rows with columns (e.g. `[[0 1 2][4 2 0]] -> [[0 4][1 2][2 0]]`).\n* `{...}/` executes a block (`{...}`) over each element of the array.\n* `$0=+` sorts each array in the big array gets the first element and adds to the \"sum\".\nAt the end of the program only the sum is left on stack and the stack values are printed.\n]"}}},{"rowIdx":13674,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**This question already has answers here**:\n \n \n[Converting a string to lower-case (without built-in to-lower functions!)](/questions/12760/converting-a-string-to-lower-case-without-built-in-to-lower-functions)\n (63 answers)\n \nClosed 6 years ago.\nGiven a string that consists entirely of printable ASCII, convert all lowercase letters to their uppercase equivalent. Non-lowercase characters are left untouched.\nI am not going to outright ban builtin solutions, but if you are going to post a solution that uses the builtin, please also post one that doesn't in your answer.\nThis is a [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\") so shortest code wins, standard rules apply.\n## Test cases\n```\ninput => output\nHello, World! => HELLO, WORLD!\nTEST => TEST\n1gN0r3 50M3 cH4r4C73r5 => 1GN0R3 50M3 CH4R4C73R5\n```\nGOOD LUCK GOLFERS!\n      \n[Answer]\n# [Japt](https://github.com/ETHproductions/japt/), [05AB1E](http://github.com/Adriandmen/05AB1E) & [2sable](https://github.com/Adriandmen/2sable) (and probably others), 1 byte\n```\nu\n```\n**Try it:** [Japt](https://ethproductions.github.io/japt/?v=1.4.5&code=dQ==&input=IkhlbGxvLCBXb3JsZCEi)  |  [05AB1E](https://tio.run/##MzBNTDJM/f@/9P9/w3Q/gyJjBVMDX2OFZA@TIhNnc@MiUwA)  |  [2sable](https://tio.run/##MypOTMpJ/f@/9P9/w3Q/gyJjBVMDX2OFZA@TIhNnc@MiUwA)\nThanks to [Mr. Xcoder](https://codegolf.stackexchange.com/users/59487/mr-xcoder) for pointing out the polyglots.\n---\n## Japt Without Built-in, 16 bytes\n```\nc_¨#a©Z§#z?Z-H:Z\n```\n[Try it](https://ethproductions.github.io/japt/?v=1.4.5&code=Y1+oI2GpWqcjej9aLUg6Wg==&input=IkhlbGxvLCBXb3JsZCEi)\n[Answer]\n# Pyth, 3 bytes\n```\nrQ1\n```\nEquivalent to `str.upper` in python.\n[Answer]\n# [Perl 5](https://www.perl.org/), 6 bytes\n**5 bytes code + 1 for `-p`.**\n```\n$_=uc\n```\n[Try it online!](https://tio.run/##K0gtyjH9/18l3rY0@f9/j9ScnHwdhfD8opwURa4Q1@AQLsN0P4MiYwVTA19jhWQPkyITZ3PjItN/@QUlmfl5xf91CwA \"Perl 5 – Try It Online\")\n[Answer]\n# JavaScript (ES6), 18 bytes\n```\ns=>s.toUpperCase()\n```\n[Answer]\n# Mathematica, 11 bytes\n```\nToUpperCase\n```\n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), 2 bytes\n```\nŒu\n```\n[Try it online!](https://tio.run/##y0rNyan8///opNL///8bpvsZFBkrmBr4Giske5gUmTibGxeZAgA \"Jelly – Try It Online\")\n[Answer]\n# [Python 3](https://docs.python.org/3/), 9 bytes\n```\nstr.upper\n```\n**[Try it online!](https://tio.run/##K6gsycjPM/6fZhvzv7ikSK@0oCC16H9BUWZeiUaahpJHak5Ovo5CeH5RToqikqbmfwA \"Python 3 – Try It Online\") or [Verify all test cases.](https://tio.run/##K6gsycjPM/6fZhvzv7ikSK@0oCC16H9BUWZeiUaahpJHak5Ovo5CeH5RToqikqYmF1wmxDU4BEXAMN3PoMhYwdTA11gh2cOkyMTZ3LjIFKjkPwA)**\n---\n# [Python 3](https://docs.python.org/3/), 50 bytes\nSaved many bytes thanks to @FelipeNardiBatista.\n```\nlambda k:''.join(chr(ord(c)-32*(c>'_'))for c in k)\n```\n[Try it online!](https://tio.run/##K6gsycjPM/6fZhvzPycxNyklUSHbSl1dLys/M08jOaNII78oRSNZU9fYSEsj2U49Xl1TMy2/SCFZITNPIVvzf0FRZl6JRpqGUqKjR2pOTr6OQnh@UU5KVZSikqYmF1w2xDU4BEXAMN3PoMhYwdTA11gh2cOkyMTZ3LjIFKjkPwA \"Python 3 – Try It Online\")\n]"}}},{"rowIdx":13675,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n Questions without an **objective primary winning criterion** are off-topic, as they make it impossible to indisputably decide which entry should win.\nClosed 8 years ago.\n[Improve this question](/posts/48718/edit) \nWrite a program that takes as close to 1 second Wall-time sharp to execute in your language.\nAnything goes except to have the processor sleep / pause / wait, by either internal or relying on external processes.\nIt should be interesting to see that adding one character more somewhere will more easily bring you to the correct time (though obviously at the cost of a character). Think for example to have a for loop run 10000 instead of 1000 times which \"costs\" 1 character.\n**Execution time** will be measured online by ideone.\nI guess for an \"official time\", I'll run a benchmark of say 100 times. Next to a low amount of characters and close to 1.000 seconds, low variance should also earn you street creds.\nIt might actually be fun to come up with an idea for how much \"worth\" a character gets in terms of getting closer to 5.00 seconds. \nMeasures:\n1. `abs(5 - execution_time)` (lower better)\n2. Number of characters (lower better)\n3. Variability (lower better)\n4. Cleverness (higher better)\n5. Fun (higher better)\n      \n[Answer]\n# Linux shell - 0.999 seconds (19 characters)\n```\n$ ping -c 2 127.0.0.1\nPING 127.0.0.1 (127.0.0.1) 56(84) bytes of data.\n64 bytes from 127.0.0.1: icmp_seq=1 ttl=64 time=0.057 ms\n64 bytes from 127.0.0.1: icmp_seq=2 ttl=64 time=0.049 ms\n--- 127.0.0.1 ping statistics ---\n2 packets transmitted, 2 received, 0% packet loss, time 999ms\nrtt min/avg/max/mdev = 0.049/0.053/0.057/0.004 \n```\nI got 1000ms on the dot a few times, but 999ms was more common. Perhaps relying on ping's inherent waits between packets to decrease network congestion may cross a line, but in my defense, I didn't write a sleep or busy-wait in my code... I just really wanted a N=2 sample size for my pings. You never know when your sneaky computer might respond to one message and not another!\n[Answer]\n# Perl: 1.0031 seconds, 16 characters\n(An unhandled signal terminates the process, so this code was measured from “outside” with `bash`'s `time`. Result includes the `perl` interpreter's startup time too. :()\n```\nalarm 1;1while 1\n```\n# Perl: 1.000059 seconds, 42 characters\n(This code was surrounded with `Time::HiRes::gettimeofday()` calls for more accurate measurement.)\n```\neval{$SIG{ALRM}=sub{die};alarm 1;1while 1}\n```\n(No, [`alarm`](http://perldoc.perl.org/functions/alarm.html) is not a `sleep`. Not works on ideone.)\n[Answer]\n# Perl - ~~1.0065~~ ~~1.0032~~ 1.0025 seconds\n```\nfor ($i = 0;$i < 11489000;$i++) {\n \"Hey, I can see my post from here!\";\n}\n```\nTested using the command `time perl onesecond.pl`, and averaged.  \nThere is also another Perl answer, so it looks like I may have to step up my game...\n[Answer]\n# **The Hexadecimal Stacking Pseudo-Assembly Language**\nProbably the longest program you'll see here at 120. Ranges from (usually; there are some rare cases which are well outside this range for some reason) 0.93 to 1.03 seconds run time (I rigged up a simple timer to my copy of the java interpreter). Pushes 40,960 33s to a stack, prints them all as chars, then halts with status 418.\n```\n20A000\n400000\n000000\n200021\n400100\n410000\n400200\n200001\n400000\n410200\n400000\n220000\n420000\n400000\n030000\n010001\n010000\n000001\n140100\n0401A2\n```\n]"}}},{"rowIdx":13676,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**This question already has answers here**:\n \n \n[Filter an array so that duplicate items are removed](/questions/201449/filter-an-array-so-that-duplicate-items-are-removed)\n (26 answers)\n \nClosed 2 years ago.\n## Task:\nRemove all duplicates from one list of integers. A list is simply a sequence of connected values that allows the same values to be stored at different positions in this sequence.\nIf an item was found to be the same value as another item in this list, keep the first occurence of the item and remove the second (third and so on...) occurence of the item.\n## Input/Output\nYou may write a function or full program that takes a list of integers as an input and returns a list without duplicates as an output.\n## Rules:\n* This is a [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\") challenge, so shortest answer wins!\n      \n[Answer]\n## APL (1)\n```\n∪\n```\n`∪` takes a list and returns the unique elements in a list, i.e.:\n```\n    ∪ 1 2 3 2 3 4 5\n1 2 3 4 5\n```\nif you need I/O it's 2:\n```\n∪⎕\n```\n[Answer]\n## GolfScript (2 bytes)\n```\n.&\n```\nTakes input on stack, leaves output on stack.\n[Answer]\n# Red, 8 bytes\n```\nunique s\n```\nassuming list is in word 's\n[Answer]\n# Python 2: 12 bytes\n```\nprint set(l)\n```\nNo input spec, so I'm assuming the list is held in a variable called `l`.\n[Answer]\n# Python - 12 bytes\n```\nlist(set(l))\n```\nAssuming the list is given in the variable `l`.\n[Answer]\n# Ruby, 3 characters\n```\na|a\n```\nAssumes the list is stored in variable `a`. Also works with `&` instead of `|`.\nIf the original list has to be altered (instead of keeping the original the same and creating a new list with duplicates removed), then 4 characters:\n```\na|=a\n```\n[Answer]\n## Mathematica, 7 bytes or 3 characters\nAssumes input in variable `x`\n```\nUnion@x\n```\nIf we're counting in Mathematica characters, I can do 3 (the second input line):\n![enter image description here](https://i.stack.imgur.com/9bGp9.png)\n]"}}},{"rowIdx":13677,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/247341/edit).\nClosed 1 year ago.\n[Improve this question](/posts/247341/edit) \nIn this challenge, you will create the shortest graphical program that displays exactly \"**Hello, world!**\"\nYou will need **no input**.\nYou might be asking yourself, \"what does he mean by graphical?\" Well, by graphical, I mean **not writing to stdout.** For example, look at my ES6 answer, which makes an alert box pop up rather than the traditional hello world where you do write to stdout:\n```\nalert('Hello, world!')\n```\nThat doesn't write to stdout. Which is what I am challenging you to do today.\n### Formatting\nThis is an example post in markdown for the ES6 snippet I posted above:\n```\n# JavaScript (ES6), 22 bytes\n    alert('Hello, world!')\n```\nView an [example](https://codegolf.stackexchange.com/a/247342/111787) post.\n      \n[Answer]\n# Minecraft Command Blocks, 47 bytes\n```\nsummon cow ~ ~ ~ {CustomName:'\"Hello, world!\"'}\n```\nSpawns a cow with the name `Hello, world!`[![cow with name](https://i.stack.imgur.com/hUNL9.png)](https://i.stack.imgur.com/hUNL9.png)\n[Answer]\n# JavaScript (ES6), 22 bytes\n```\nalert('Hello, world!')\n```\nThis is an example for my question.\n[Answer]\n# PHP (8.1), 20 bytes\n```\necho'Hello, world!';\n```\n]"}}},{"rowIdx":13678,"cells":{"text":{"kind":"string","value":"[Question]\n      [\nNotice: This question originally had people write comments with a shorter program, rather than new answers. This has been changed so that people with shorter programs can get reputation too. Hopefully this doesn't become unmanageable...\nThere have been a couple of [OEIS-related](https://codegolf.stackexchange.com/questions/188142/cops-the-hidden-oeis-substring) [questions](https://codegolf.stackexchange.com/questions/188147/reverse-your-code-reverse-the-oeis) recently, so here's another:\nPick a sequence from the [Online Encyclopedia of Integer Sequences](https://oeis.org), and write a full program or function which computes the sequence in one of the following ways:\n* It prints every term in the sequence in order\n* It takes an index in the sequence as input, and returns/outputs the term at that index.\nYou will be trying to pick the sequence with the longest possible shortest possible program/function to compute that sequence (see the example below if that's confusing). You must include which sequence you are using in your answer.\nTry to make your program/function as short as possible. **If anyone can find a shorter program which computes the same sequence, they should add a another answer with their shorter program**. If you do find a shorter program than someone else, you should refer to their answer in yours. For all answers, a [Try it Online](https://tio.run) link would be nice. If someone else has written a shorter program than yours, it would be nice if you put a link to it in your answer.\nOf course, none of the programs should access the internet or the OEIS in any way, and they should all theoretically work on arbitrarily large inputs (you can ignore things like the maximum size of an integer type).\nThe sequence with the longest shortest program in bytes is the winner.\n## For example,\nLet's say I pick the sequence [A000027](https://oeis.org/A000027), the positive integers, and I submit this answer:\n# Python, 49 bytes, [A000027](https://oeis.org/A000027)\n```\nx = \"o\"\nwhile True:\n    print len(x)\n    x += \"o\"\n```\nObviously this is not a very short program for computing that sequence, so someone else (let's call them foo) might come along and add this answer:\n# Haskell, 2 bytes, [A000027](https://oeis.org/A000027):\n```\nid\n```\n(`id` is the identity function, so it will just return whatever you pass into it, and because the `n`th positive integer is just `n`, it will compute that sequence).\nThen, the person who posted the Python solution should edit their answer:\n# Python, 49 bytes, [A000027](https://oeis.org/A000027)\nSuperseded by foo's answer (link).\n```\nx = \"o\"\nwhile True:\n    print len(x)\n    x += \"o\"\n```\nAs long as no one else finds a shorter program, this sequence would get a score of `2` (because `id` is two bytes), and the sequence with the highest score wins.\n---\n[Current best sequence](https://codegolf.stackexchange.com/a/188238/62859): [A014715](https://oeis.org/A014715), 252 bytes\n      \n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), ~~270~~ 252 bytes [A014715](https://oeis.org/A014715)\n**252 bytes, by Expired Data**\n```\nRealDigits[NSolve[3x-6==Plus@@({6,-12,4,-7,7,-1,0,-5,2,4,12,-2,-7,-12,7,10,4,-3,-9,7,0,8,-14,3,-9,-2,3,10,2,6,-1,-10,3,-1,-7,7,-7,12,5,-8,-6,-10,8,8,7,3,-9,-1,-6,-6,2,3,10,2,-3,-5,-2,1,1,1,1,1,-1,-2,-2,1,2,1,0,-1}x^2~Range~71),x,#][[-1,-1,-1]]][[1,#]]&\n```\n[Try it online!](https://tio.run/##PY7dCsIwDIUfRhCFFJrup/Ni4IXXInpZKhTZ5mCboFUG4l69JtsYaSHny8lpW@fvRet8fXOhzMO5cM2hrmr/MsfLo/kUJupFmuen5v3a7zffFAQqiEFo0NSCBJEAA6JCMea5BpRsikDsSEjICMcwSnJFPFbAWXQkc5wTNQclIGghHWcZlZ43cYQpLAn8QMKJuBSb1IT40v/w11/VcHZdVQwat9DDyhojJitaSwIJ2XU4PevOm9KglNaG8Ac \"Wolfram Language (Mathematica) – Try It Online\")\nI've never golfed in mathematica before, but the sequence seemed like an obvious contender. The code is more or less a copy of the example on the OEIS page.\nThe sequence is the decimal expansion of Conway's constant, which itself is the root of this polynomial:\n\\$x^{71}-x^{69}-2x^{68}-x^{67}+2x^{66}+2x^{65}+x^{64}-x^{63}-x^{62}-x^{61}-x^{60}-x^{59}+2x^{58}+5x^{57}+3x^{56}-2x^{55}-10x^{54}-3x^{53}-2x^{52}+6x^{51}+6x^{50}+x^{49}+9x^{48}-3x^{47}-7x^{46}-8x^{45}-8x^{44}+10x^{43}+6x^{42}+8x^{41}-5x^{40}-12x^{39}+7x^{38}-7x^{37}+7x^{36}+x^{35}-3x^{34}+10x^{33}+x^{32}-6x^{31}-2x^{30}-10x^{29}-3x^{28}+2x^{27}+9x^{26}-3x^{25}+14x^{24}-8x^{23}-7x^{21}+9x^{20}+3x^{19}-4x^{18}-10x^{17}-7x^{16}+12x^{15}+7x^{14}+2x^{13}-12x^{12}-4x^{11}-2x^{10}+5x^{9}+x^{7}-7x^{6}+7x^{5}-4x^{4}+12x^{3}-6x^{2}+3x-6\\$\n[Answer]\n# Python 3, 3 bytes, [A055642](https://oeis.org/A055642)\nFinally, a ridiculously short answer that isn't just A000027 or some other stupidly simple sequence.\n```\nlen\n```\nThis is a function. Number should be inputted as a string of its digits. Undefined behavior for non-integers and other strings.\n[Answer]\n# [cQuents](https://github.com/stestoltz/cQuents), 1 byte, [A000027](https://oeis.org/A000027)\n```\n$\n```\n[Try it online!](https://tio.run/##Sy4sTc0rKf7/X@X/fwA \"cQuents – Try It Online\")\nJust to have a baseline answer in case every other answer is cracked.\n[Answer]\n# [Wolfram Language (Mathematica)](https://www.wolfram.com/wolframscript/), 490,453 bytes [A002205](https://oeis.org/A002205)\n```\nFromDigits[ToCharacterCode[StringReplace[\"$H\\.06,\\.1dVngsIno\\.12RT\",\":3:\"->FromCharacterCode[13]]]-1,127]\n```\nWhere \\.FF is a single byte of that value.\n[Try it online!](https://tio.run/##VVTpVzkNFD5IxBwlsi8ZIYXGvg@N7CVMw8SMabI12ZmE0J/e6/fx/Xyf59x7nuWOaPajO6JZpk3/lefMmG3@ZeaT0QPTZ9hF82WS@qDndJvtzlOTTreJsgdIv9qdDul2twnyogbTKAD1YrVyUupPbMPjZbvxgCIRu0H/i6hljxvkWfZDENqWd@lpNR2jT56UExoUr/RYeM6PeCOp3kL1WzWF7d7lPa4krc3zxqqnYCFTSvOLp1briK0rScSfIKpRHNLf6e9B/McSjFoIcG1QQB1dyD64KKBM/eFV7iojGdPiuYnpN5LrcvLICfzYQvQJW82LcOAzRe9V/fPkjqM@LPU8u1K41KlvSwwu6nAdAZJtxroTOe/N3C0BEqBx/LW0vyQ4klXUCbdmGo6/IAnzYiiyLF4mbIkn5RrDs55FR/17mvIycFbu774V6W3O796LMNXUQQWPN5XPzEkyXLDg2E1NOW7xbIH9xcO0bYGgBicsGq2S@7E5a3C857k8hCBgcuAs@cyWHc45/ox7n3xsNDx6lvrHS8wSX9bedi4gPp5rq@vCh9BXqWXOhdaiTbLNTapeN70ccM1GgogFBjppHL3qYhJcjDas/SsCtF6KLcHrs8Ko4vPE8/ooz64PcdbIlwv4mFWDT5RgK0B8dd5jTOPu1PDCYvOzl4USty9zWF8W@UZayVcdGugSyZhABRRDBAglcpV76@KIjWA2VpUb@LUu9iaFirw2@E75W3mzA2@P2DjcjqfHVqpy0YzzjejjEz9WANCP2f17wHeFtcUdDUMQp5OOSHcwY8pQw8XKK9FE3DnbbtYumChNg3LUpxGu@HrUmzsDw7PqRB459igRJsO/i3v7RiAKEGALObDPvRrk9k5Vz06zFmgEx9vtgHXqOAyagr5HmMQaFUpwV4D5moYj9u6667wv7ffBzEyxMzoFrjNciO3/pfD28TKDMCKVTCUmiIChaulDOh0U9HA5J8lb8uzitKeAYgHuZvBOgDEfWq@VOnzfmp2uc8XWa7/klvHzfZWPSyW7hW34kherA6gJIkuB@bBk4eQhcRSmozNOfO95WzqmMqtJKv8UpWX@mvmalxaqenRst4KzALL5bu7gWHaOZ7v5G/KMuYVSFkE5tDALxQatIZLWc2tpIOG2PK/lcptUHcgrqxttk/KcYI6Of6s4HhIEOVstHASxlOAoASp2g/HL8bcna1NTGnIxM8LfJRUHQy8ZQyQcVavtERucDOmvTfrXnIElj7ZnRUTwBTrAgySgE/73Bv7ffreXJEmn2@H2BMk/8u8/ \"Wolfram Language (Mathematica) – Try It Online\") - Limited to 2000 digits, I don't think the dev would appriciate ~1.5M of data coming through.\nI took a while to search though the OEIS 'webcam', but I paused and looked at XKCD, where it mentioned the book \"A Million Random Digits with 100,000 Normal Deviates\". And what do you know, it's A002205. A pretty old entry too.\nAccording to dieharder, a program for finding if a sequence of numbers is random, the sequence is pretty random, so I doubt there's going to be an algorithm to generate a given number. This makes it more of a Kolmogorov-complexity problem. Since every number is between 0 and 9 inclusive, then there are Log2[10] bits of information in a digit, or `1,000,000 * Log2[10]/Log2[256] bytes` of information total, or just a tad under 415,242 bytes.\nMy final solution was to take the entire 1,000,000 digit number in as a number (big-int in other languages), convert it to base 127, and then put it in the range 1-127. `1,000,000 * Log2[10]/Log2[127] = 475,330 bytes`. Character 13 is the Carriage Return, which is converted to a Line Feed on most systems, so it's replaced with the trigraph :3:, since all the digraphs are taken. Mathematica was surprisingly happy to convert a 1 million digit number to base 127. The answer can be optimized, since TIO allows the entire 127+ unicode space, but using this method the results would only reduce the score by 1%.\nHere is the encoder I used, where `str` is a string containing all the digits:\n```\nStringReplace[FromCharacterCode[IntegerDigits[FromDigits[str],127]+1],{\"\\\\\"->\"\\\\\\\\\",\"\\\"\"->\"\\\\\\\"\",FromCharacterCode[13]->\":3:\"}]\n```\nBZIP2 can compress the data down to 431,122 bytes, which is pretty close, but I felt just BZipping the data and calling it an answer was a bit cheap since it's not a 'language', as much as those answers have been fun on other questions.\n]"}}},{"rowIdx":13679,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/183600/edit).\nClosed 4 years ago.\n[Improve this question](/posts/183600/edit) \n## Codegolf challenge - mapping values in a multidimensional array\n```\n[\n  [0, 0, 0],\n  [0, 0, 0],\n  [0, 0, 0]\n]\n```\nThis **2D** array is sick of being so **false**, and wants to find the **truth** about life, once and **for** all. Your task is to write a map function to act as a highway to all the values in this array.\nAn example implementation of this super function would be if the above array is defined as `arr` below:\n```\narr.multimap(x => {\n  return 1;\n}\n```\nWhich would result in the array being:\n```\n[\n  [1, 1, 1],\n  [1, 1, 1],\n  [1, 1, 1]\n]\n```\nThis is a 2D number example, yours has to work in any dimension, and not be dimensionally-bias.\n---\n## The rules\n* Must work in any type of dimensional array (1D, 2D, 3D, etc..)\n* Least bytes, wins!\n* Must only have one callback function which accounts for any level of dimensional interaction\n* Meta-languages get fewer points\n* Can simply be a function which applies the logic to an array given\n* Can be any arbitrary initial value, as long as you can replace it on-command\n      \n[Answer]\n# [JavaScript (Node.js)](https://nodejs.org), ~~35~~ ~~27~~ 17 bytes\n*-18 thanks to @Arnauld and @Shaggy*\n```\nf=a=>a?a.map(f):1\nconsole.log(f([\n  [0, 0, 0],\n  [0, [0, 0, 0], 0],\n  [[0, 0, [0]], 0, 0]\n]))\n```\n[Answer]\n# [Jelly](https://github.com/DennisMitchell/jelly), 1 byte\n```\n‘\n```\n[Try it online!](https://tio.run/##y0rNyan8//9Rw4z///9HcykoRBvoKIBQrA6UgxCACUJFog1iY6FSXLEA \"Jelly – Try It Online\")\nReplaces `n` with `n + 1`.\nBecause Jelly things.\nNote that `[x]` displays as `x` in Jelly (single-element lists don't show as lists); the internal data is still a single-element list. You can make it show using Python's `str`: [Try it online!](https://tio.run/##y0rNyan8//9Rw4yjkx7unPH///9oLgWFaAMdBRCK1YFyEAIwQahItEFsLFSKKxYA \"Jelly – Try It Online\").\n[Answer]\n# [Perl 6](https://github.com/nxadm/rakudo-pkg), 4 bytes\n```\n!<<*\n```\n[Try it online!](https://tio.run/##K0gtyjH7n1upoJamYPtf0cZG6781F1dxYqVCmkZ0tIEOEMYCCSAzFkiASCCMjdW0hqkByoGlwDIgFToQJZrW/wE \"Perl 6 – Try It Online\")\nReplaces `0`s with `True`. *Extremely* short for a non-esolang.\n```\n   *  # From the input\n!     # Boolean not \n <<   # Each element recursively\n```\n[Answer]\n# [Python 2](https://docs.python.org/2/), ~~29~~ ~~26~~ 25 bytes\n*-3 bytes thanks to @JonathanAllan*\n*-1 byte thanks to @JoKing*\n```\nf=lambda a:a<1or map(f,a)\n```\nReplaces `0` with `True`\n[Try it online!](https://tio.run/##K6gsycjPM/r/P802JzE3KSVRIdEq0cYwv0ghN7FAI00nUfN/QVFmXolCUWpBkUaaRjSXgkK0gY4CCMXqQDkIAZggVCTaIDYWIgVkRwOZ0bGxsVyxmpr/AQ)\n[Answer]\n# [Japt](https://github.com/ETHproductions/japt), 6 bytes\n```\n?¡ßX:1\n```\n[Try it](https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&code=P6HfWDox&input=W1swLDAsMF0sWzAsWzAsMCwwXSwwXSxbWzAsMCxbMF1dLDAsMF1dCi1R)\n```\n¬Íª¡ßX\n```\n[Try it](https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&code=rM2qod9Y&input=W1swLDAsMF0sWzAsWzAsMCwwXSwwXSxbWzAsMCxbMF1dLDAsMF1dCi1R)\n[Answer]\n# JavaScript, 36 bytes\n```\nt=>JSON.stringify(t).replace(/0/g,1)\n```\n[Try it online!](https://tio.run/##y0osSyxOLsosKNHNy09J/Z9m@7/E1s4r2N9Pr7ikKDMvPTOtUqNEU68otSAnMTlVQ99AP13HUPN/cn5ecX5Oql5OfrpGmkY0l4JCtIGOAgjF6kA5CAGYIFQk2iA2FirFFaup@R8A \"JavaScript (Node.js) – Try It Online\")\n]"}}},{"rowIdx":13680,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/174488/edit).\nClosed 5 years ago.\n[Improve this question](/posts/174488/edit) \nInput: A valid 6 HEX digit color code wihout a # prefix  \nValid:\n> \n> * 000000\n> * FF00FF\n> * ABC123\n> \n> \n> \nYour job is to find to which one of these:\n```\nRed:   FF0000\nGreen: 00FF00\nBlue:  0000FF\nWhite: FFFFFF\nBlack: 000000\n```\nthe input color is the closest too, in this case you try to only match it to black or white when the RGB is the same value for all 3 (the color is grey, or black or white ofc), if there is a \"tie\" you can pick whichever\nThe output is one of the aforementioned color codes.\nExamples:\n```\nInput: FF0000, Output: FF0000\nInput: F10000, Output: FF0000\nInput: 22FF00, Output: 00FF00\nInput: FFFF00, Output: FF0000 or 00FF00\nInput: 111112, Output: 0000FF\nInput: 010101, Output: 000000\nInput: A0A0A0, Output: FFFFFF\n```\nThis is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"), so fewest bytes will win!\n      \n[Answer]\n# [Python 3](https://docs.python.org/3/), ~~163 162~~ 150 bytes\n*-1 byte thanks to Kevin Cruijssen*\n*-12 bytes by some small tricks (thanks to HyperNeutrino for the* `\"0F\"[bool]` *idea)*\n```\ndef f(i):m=i[:2],i[2:4],i[4:];a,b,c=(e==max(m)for e in m);return\"\".join(\"0F\"[e]*2for e in([i[:2]>\"7E\"]*3if a&b&c else([a,b,c]if a^b^c else[a,1^a,0])))\n```\n[Try it online!](https://tio.run/##RZDRasMgFIbv@xQHLxotdhhbGFgs9GK@hJiRWMMciymJg@3pM09K2H8u/PiU49HHb/4Y02lZ7qGHnkamBh2tko5HK9UZl7Nyl5Z33GsatB7aHzqwfpwgQEwwsMsU8veUCHn5HGOiRBhigzvI7Qi1a78reX0j7nCKPbT7bu8hfM2B2rWxQ9l0zVMWVzctF44xtuQw53ffzmEGDbYyRpRUHCpTbyQl2tWZjYRAfjpkpBoj190aC@kmsCq3w2mzxxf936h2UPKYYso0ew4EjlcgvHxT9mW0Pw \"Python 3 – Try It Online\")\nI was able to find a solution without turning the three hex values to integers (mainly because 'A' comes later in the ASCII table than '9', and so '9'<'A' is true). I am not really happy with some parts, but this was the shortest I could find.\nBecause this solution uses a different approach than the other python3 solution, I'm posting this as a seperate answer.\n[Answer]\n# JavaScript (ES6), 94 bytes\n```\ns=>'00FF000000FFFFFF'.substr(([r,g,b]=s.match(/../g),r+g==g+b?r<'8'?4:10:r>b?r>g&&2:b>g&&6),6)\n```\n[Try it online!](https://tio.run/##hY5RC4IwFIXf@xXig9twbVNCQprSi38ielDTVZiLzfr7q7mQUKJzHi7n8nHuvZbPUtfqch/WvTw1puVG8wwwVhRsVDEKEP2o9KAgPCgscHXkmtzKoT5DSggVCKtQcC7CKlc7sAX5Jo1YqrJ3zEQQxGllR4Jwgkwtey27hnRSwBb67o6PkEep58JqjkR/kTi2@w/inl@0FF@Ia/Gk@gFHVvHUZ6E5wiLrL2TZsmfW00kr8wI \"JavaScript (Node.js) – Try It Online\")\n### How?\nWe could convert a bitmask to upper case hexadecimal with leading zeros using:\n```\n.slice(1).toString(16).toUpperCase()\n```\nBut that would cost 36 bytes, and building such a bitmask is not especially cheap either.\nIt's significantly shorter to extract the relevant substring from `\"00FF000000FFFFFF\"`, using the following offsets:\n```\n pattern | offset | 00FF000000FFFFFF\n---------+--------+------------------\n 000000  |    4   |     ^^^^^^\n FFFFFF  |   10   |           ^^^^^^\n FF0000  |    2   |   ^^^^^^\n 00FF00  |    0   | ^^^^^^\n 0000FF  |    6   |       ^^^^^^\n```\n[Answer]\n# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), ~~32~~ ~~31~~ 30 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)\n```\n3äHDËiнžy@Di'F}6∍ë¾5∍sZk·„FFsǝ\n```\nNot really happy with it, so will see if I can golf it further..\nWhen two `RGB` parts are the largest and equal, it will use the first one (i.e. `11ABAB` will result in `00FF00`). When all three parts are equal it will output `000000` when it's the range `[00,7F]` and `FFFFFF` for the range `[80,FF]`.\n[Try it online](https://tio.run/##yy9OTMpM/f/f@PASD5fD3ZkX9h7dV@ngkqnuVmv2qKP38OpD@0yBdHFU9qHtjxrmubkVH5/7/7@FAQgCAA) or [verify all test cases](https://tio.run/##yy9OTMpM/V9TVmmvpPCobZKCkn3lf@PDSzxcDndnXth7dF@lg0umulut2aOO3sOrD@0zBdLFUdmHtj9qmOfmVnx87v9anf9ubgZAwGVgAGJwgdhubmAKyHMDAy4LMOByMwQLGhmBVYJkgJQhCBhxGRqBaCAPRHMZGIIgl6MBCAIFHZ0cnbjcDIDQkMvNCEwZWIIgF1iFI5e5GwhyWRiAIAA).\n**Explanation:**\n```\n3ä                  # Split the (implicit) input into three parts\n  H                 # Convert each part from Hexadecimal to a base-10 decimal\n   D                # Duplicate this list\n    Ëi              # If all three values are equal:\n      н             #  Leave only the first item of the list\n       žy@          #  Check if it's larger than or equal to 128 (\"80\" in hex)\n          D         #  Duplicate this truthy/falsey result\n           i  }     #  If the result is truthy:\n            'F     '#   Push \"F\"\n                    #  (Implicit else:)\n                    #   (Take the duplicated falsey 0)\n               6∍   #  Lenghten it to size 6\n     ë              # Else:\n      ¾5∍           #  Push 0, lenghtened to size 5: \"00000\"\n         s          #  Swap so the duplicated list is at the top of the stack\n          Z         #  Take the max (without popping)\n           k        #  Take the index of this max in the list\n            ·       #  Double this index\n             „FFsǝ  #  Replace the \"0\" in \"00000\" at this doubled index with \"FF\"\n                    # (And output the top of the stack implicitly as result)\n```\n[Answer]\n# [Python 3](https://docs.python.org/3/), ~~219~~ ~~112~~ 110 bytes\n```\ndef f(s):h=max([0,1,2],key=lambda i:s[i*2:][:2]);return[\"00\"*h+\"FF\"+\"00\"*(2-h),\"F0\"[s[:2]<\"80\"]*6][s==s[:2]*3]\n```\n[Try it online!](https://tio.run/##HYq7DoMgGIV3nsIwAaXJDyatpXXowhN0Iww2YiBVaxTT@vS2cL7hXHKmLfr3WO5767qiIwtVvh6aLzHABZeWv9xW983wbJsiqMUEJpU1Slp6nV1c59FgAMz8AWuNDzkTefSUYw3YLOl5wxVgy07WLHWdF1ba/eND74rHvDpVTHMYI@lIGKc1EkrprjX8hbTIJmXqSOtsIkkiEAl0hwSqIIHOl8QP \"Python 3 – Try It Online\")\n-107 bytes by not being bad; drew a bit of inspiration from @Heitera by avoiding input integer parsing entirely.  \n-2 bytes thanks to @Heitera\n]"}}},{"rowIdx":13681,"cells":{"text":{"kind":"string","value":"[Question]\n      [\nYour challenge: Connect to the Online Encyclopedia of Integer Sequences [OEIS.org](http://oeis.org), access any random valid sequence, output it's ID, and ouput all of the members of the sequence provided by the OEIS entry. The *only* website you're allowed to connect to is OEIS.org.\nHints: Sequences are found in URLs such as , however a more easily parseable version is found at URLs such as . Every present *and future* sequence matches the regex `\"[A-Z]\\d{6}\"`. (The regex is there to describe what an ID looks like. There are a lot of invalid sequences that match it, too). The webcam () feature of OEIS generates random sequences, but you'll have to find a way to change the dropdown menu from 'best sequences' to 'all sequences' to use it.\nExample output:\n```\nA280629\n1, 1, 7, 139, 6913, 508921, 57888967, 9313574419, 1984690709953, 547467006437041, 188946742298214727, 79783392959511537499, 40498043815904027702593, 24314800861291379306213161, 17047720745682515427867108487, 13802952030641885344209574247779, 12780883488499783875309105315925633, 13420910251496135926622603184056054881, 15863354775169518855398667975850797997447, 20966527201075972453953302254528386060431659\n```\nEvery existing sequence has to have a chance of being outputed, it does not have to be an equal chance.\nYour program should continue to function properly regardless of future sequences being added. Even though all IDs presently start with A, there will eventually be a sequence B000001 after all the A sequences run out. You can assume that there will be no more sequences after Z999999 is defined, but you do have to assume that it *will* eventually be defined.\nAn invalid sequence, which your program should never output, is one that doesn't exist. You may treat sequences with no given members (such as [A280611](http://oeis.org/A280611) either as invalid or give just the sequence ID and nothing else or a falsy value, but you must be consistent.\nThis is [code-golf](/questions/tagged/code-golf \"show questions tagged 'code-golf'\"), fewest bytes wins!\n      \n[Answer]\n## Perl, 93 bytes\n```\n$u=chr(65+rand 26).sprintf(\"%06d\",rand 1e6),$_=`curl oeis.org/$u`while!/tt>(.*?)(.*?) /dev/null\n```\nExplanations:  \n-`$u=chr(65+rand 26).sprintf(\"%06d\",rand 1e6)` generates a random name,  \n-`$_=curl oeis.org/$u` tries to retrieve the sequence,  \n-`/tt>(.*?)` is present on the page (it's only present on valid pages). If so, it stores the sequence's elements in `$1`. And if not, the `while` keeps looping.  \n-`say\"$u\\n$1\"` outputs the sequence's name and its elements.\n[Answer]\n## Python 3, 221,208 bytes\n```\nfrom requests import*\nfrom random import*\nimport re\nr=randint\ng=lambda x:re.findall(r'(?<=tt>)[^<]*',get('https://oeis.org/'+x).text)\na=''\nwhile not g(a):a=chr(r(65,90))+str(r(0,999999))\nprint(a+'\\n'+g(a)[0])\n```\nI'm sure there's a few tricks I missed (and requests is an external lib), but this was a fun first go at python code golf\n[Answer]\n# JavaScript, 104 bytes\nBecause [someone commented](https://codegolf.stackexchange.com/questions/106068/output-a-random-oeis-sequence#comment257935_106068) that this isn't possible in JS!\nMust be run from the root directory of OEIS ([consensus](https://codegolf.meta.stackexchange.com/questions/13938/if-we-can-require-a-particular-domain-can-we-require-a-particular-path)). Returns a 2-element array, with the first element being the sequence number and the second the sequence itself.\n```\nf=async n=>(s=/tt>[^<]+/.exec(await(await fetch(n+=Math.random()*1e6|0)).text()))?[n,s[0].slice(3)]:f`A`\n```\nTo test it, go to the [OEIS homepage](http://oeis.org/) (or any sequence's page), open your browser's console and copy & paste the code below:\n```\nf=async n=>(s=/tt>[^<]+/.exec(await(await fetch(n+=Math.random()*1e6|0)).text()))?[n,s[0].slice(3)]:f`A`\nawait f()\n```\n## 97 bytes\nThis *should* work and could maybe be golfed a little further but I'm on my phone so can't test it properly.\n```\nf=async n=>(s=/(?<=tt>)[^<]+/.exec(await(await fetch(n+=Math.random()*1e6|0)).text()))?[n,s]:f`A`\n```\n]"}}},{"rowIdx":13682,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/17270/edit).\nClosed 9 years ago.\n[Improve this question](/posts/17270/edit) \nRay tracers are easy to write. At each pixel on the screen, you shoot a ray into the scene and see what it hits. For the most basic ray tracer, the ray acquires the color of the object it hit.\nIf you're doing shading, then you don't color the pixel immediately, instead you try to trace a ray back to the *light source*. The shaded color is then the (color of the surface)\\*(cosine of the angle (dot product) of ray-to-the-lightsource has with the surface normal). If the light source cannot be reached, then the object is in shadow.\nThis challenge is to write the raytracer in any language you can. The winner is the most **succinct** code, while still remaining readable (see here for more [elegant definitions](https://softwareengineering.stackexchange.com/a/97914/13572)).\n**Requirements**:\n* Ray trace spheres\n* Shading\n* Shadows\n* At least 1 bounce interreflection.\n* You must display your result using pixels *or* write an image file. Don't draw into the console.\n**Bonus**:\n* Include ability to ray trace general polygons\n* Translucency\n**Superbonus**:\n* General BRDFs\n* Subsurface scattering (BSSRDF)\n**Bad examples**:\n* **Don't do this:** \n* **And don't do this:** \nThe first is not readable. The 2nd is just a really *compact* C++ program.\n*This is a hard project*, so you might want to bookmark it and come back later. I'll leave this open for about a month before awarding the checkmark.\n# Resources:\n* [http://www.cs.utah.edu/~shirley/irt/](http://www.cs.utah.edu/%7Eshirley/irt/)\n(more will be added later)\n      \n[Answer]\nThis needs a little more work to fulfill the requirements (shading, interreflection), but mostly works. It generates the pixels line-by-line in the callback procedure to the `image` operator. The readability comes at the expense of being very wasteful with memory, creating and discarding temporary arrays with abandon.\n# Postscript 90 lines\nUses an external [matrix library](http://web.archive.org/web/20160404161539/https://code.google.com/p/xpost/downloads/detail?name=mat.ps&can=2&q=).\n```\n(mat.ps) run  % load matrix library\n/div { dup 0 eq { pop 10000 }{ div } ifelse } bind def  % make divide-by-zero safe\n/tan { dup sin exch cos div } def \n%[vector] scalar {op}  .  [vector']\n%create a new vector and fill with the scalar,\n%call matrix library's vector-op\n/sop {\n    exch % [] {} s\n    [ exch % [] {} [ s \n    3 index length 1 sub{dup}repeat % [] {} [ s*n-1\n    ] exch % [] [] {}  \n    vop \n} def \n/normalize { dup mag {div} sop } def \n<< % parameters\n    /Cam    [ 0 0 0 ]   % camera location\n    /Lookat [ 0 0 -1 ]  % camera lookat point\n    /Up     [ 0 1 0 ]   % up vector\n    /fovx   45          % horizontal field of view\n    /W      320         % image width\n    /H      200         % image height\n    /Sc     [ 1 1 -5 ]  % sphere center\n    /Sr     2           % sphere radius\n>> begin\n<< % dependent constants\n    /Left Lookat Cam {sub} vop Up cross\n    /buf W string\n    /fovy H W div fovx mul \n    /ang 0\n    /pos 3\n>> {def} forall  % add these values to existing dictionary\n% place the light in the scene according to the /ang parameter\n/setLight { /Light [ ang cos pos mul 3 ang sin pos mul ] def } def \n0 10 360 { /ang exch def   % for-loop sets /ang, calls image and showpage\nsetLight                   % place the light\n150 200 translate          % position lower left corner of image on page\n/y 0 def                   % initial y-value\nW H 8                      % put width height depth on stack  for `image` call\n[ 1 0 0 1 0 0 ]            % 1-to-1 matrix, so width*height points is the image size\n{                          % image procedure yields a string with one image row \n    0 1 W 1 sub { /x exch def                   % for-loop iterates over x-values\n        Left   x 2 mul W sub W div fovx tan mul  {mul} sop  % construct vector from eye through center of pixel\n        Up     y 2 mul H sub H div fovy tan mul  {mul} sop \n        Lookat {add} vop {add} vop \n        Cam {sub} vop \n        normalize /R exch def \n        Cam Sc {sub} vop   % use the quadratic formula to check for real intersection\n        dup R dot 2 mul /B exch def \n        dup dot Sr dup mul sub /C exch def \n        B dup mul C 4 mul sub /disc exch def  % discriminant, the part under the radical-sign\n        % ray hits sphere? no:0 yes:calculate\n        disc 0 lt { 0 }{\n            disc sqrt /sdisc exch def  % complete the quadratic formula to yield t value\n            B neg sdisc sub .5 mul /t exch def\n            t 0 le { B neg sdisc add .5 mul /t exch def } if\n            R t {mul} sop\n            Cam {add} vop /ri exch def  %intersection point\n            ri Sc {sub} vop\n            1 Sr div {mul} sop %normalize\n            /rn exch def  %surface normal\n            Light\n            ri\n            {sub} vop %normalize\n            rn dot  % normal .dot. light-ray\n            1 add .3 mul\n            %.3 mul\n            dup 0 lt {pop 0} if  % clip value into 8-bit range\n            dup 1 gt {pop 1} if\n            255 mul truncate cvi\n        } ifelse   % pixel_color = 0..255\n        buf x 3 2 roll put  % put pixel value in string\n    } for\n    /y y 1 add def  % increase y for next row\n    buf  % yield string buffer to image\n}\n%exec  % image doesn't report internal errors, debug the procedure with `exec`\nimage\nshowpage\n} bind for\n```\nCan generate a numbered series of images using ghostscript. And ImageMagick's `convert` can compile them into an animated gif.\n![enter image description here](https://i.stack.imgur.com/Pe9g3.gif)\n]"}}},{"rowIdx":13683,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/15317/edit).\nClosed 8 years ago.\n[Improve this question](/posts/15317/edit) \nIn case the title was confusing, I mean something like the following C code:\n```\nmain(){}znva(){}\n```\nWhen rot13 encoded, the code becomes\n```\nznva(){}main(){}\n```\nand still runs.\nBut that isn't a valid answer, because of these rules:\n* Your code must produce output. Code that doesn't produce output is disqualified.\n* The program can't be empty. That's disqualified too.\n* The program can't be the same or similar to the output. This means that you can't use Golfscript or PHP.\nIn a valid answer, the program produces some output without being rot13'd, such as:\n```\nHello, World!\n```\nAnd when the program is rot13'd, the output changes to the following:\n```\nUryyb, Jbeyq!\n```\nwhich is the rot13 of `Hello, World!`.\nHave fun!\n      \n[Answer]\n## Ruby\nCheaty answer:\n```\n$><<'Hello, world!'\n```\nSlightly less cheaty answer:\n```\ns='Hello, world!'\n$-I?puts(s):chgf(s)\n```\nGolf (15 chars)\n```\n$-I?p(:a):c(:a)\n```\n[Answer]\n## Perl, ~~25~~ ~~14~~ 12 characters\nI believe this adheres to all of the requirements:\n```\n$_=a;fnl;say\n```\nOutputs `a`. After `rot13` the program looks like this:\n```\n$_=n;say;fnl\n```\nand outputs `n`.\n[Answer]\n## Brainf\\*\\*k\nSince you never said the output had to actually be affected by rot13, this is totally a valid answer:\n```\n-.\n```\nNot sure why it was downvoted, although it is kind of a cheap answer.\n[Answer]\n### PHP 6\n```\nH U ]]&&rpub $@||echo $@; };_ Hello, Jbeyq.\n```\n]"}}},{"rowIdx":13684,"cells":{"text":{"kind":"string","value":"[Question]\n      [\nThe task is simple: Create an if-statement from scratch in x86 assembly that runs in as few clock-cycles as possible.\n* Must be x86 assembly.\n* Can't use any of the *conditional* jump/branch instructions (e.g. `je`, `jne`, `jz`, `jg`, `jge`, `jl`, `jle`).\n* **Can** use `jmp` to jump to existing blocks of code.\n* **Can** use `cmp` if necessary.\n* Can't call any external libraries or functions.\n* Can't use `call`.\n* Free to use any of the other available instructions in the [Intel docs](http://www.intel.com/content/dam/www/public/us/en/documents/manuals/64-ia-32-architectures-software-developer-instruction-set-reference-manual-325383.pdf).\n* No C.\n* Any of the ~7 Intel Core architectures are acceptable (listed on [page 10](http://www.agner.org/optimize/instruction_tables.pdf)). Ideally Nehalem (Intel Core i7).\n* The `if` statement needs to check arbitrary conditions. Basically replicating the functionality of any one of the `je`, `jge`, etc. *conditional* jumps.\nThe winning answer is one that uses the fewest clock cycles possible. That is, it can have the most operations per second. Approximate clock cycle summaries are here: . Calculating clock cycles can happen after the answer is posted.\n      \n[Answer]\nAnd then I realised that we can be faster, and use no additional registers:\n```\ncmp eax, ebx\nmov eax, offset false_case\nmov ebx, offset true_case\ncmove eax, ebx ;or cmovb, etc.\njmp eax\n```\n[Answer]\n```\nxor ecx, ecx\nmov edx, 1\ncmp eax, ebx\ncmove ecx, edx ;or cmovb, etc.\njmp [false + ecx*4]\nfalse dd offset false_case\ntrue  dd offset true_case\n```\nAccepts eax and ebx for the comparison.  \nReplace the cmov with the corresponding check to perform.\n]"}}},{"rowIdx":13685,"cells":{"text":{"kind":"string","value":"[Question]\n      [\n**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.\n \n \n---\n**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/161669/edit).\nClosed 5 years ago.\n[Improve this question](/posts/161669/edit) \nRecreational languages win at code golf too much. This challenge is simple, when using higher level functions. \n**The task is as follows:**\n**Given any valid url, which is guaranteed to result in a valid HTML document** *(for example , which has 9 images)*, count the number of images displayed from the resultant webpage.\nOnly images visible in the HTML code are counted, marked by the HTML `` tag.\nWebsites that can continually load images (Google Images) are not valid tests.\n      \n[Answer]\n# JavaScript, 61 bytes\nAs per [this consensus](https://codegolf.meta.stackexchange.com/a/13626/58974), the code needs to run under the same domain as the page being requested, to avoid CORS issues. Returns a Promise conatining the count.\n```\nu=>fetch(u).then(r=>r.text()).then(t=>t.split`fetch(`https://crossorigin.me/`+u).then(r=>r.text()).then(t=>t.split`o.innerText=x);b.addEventListener(`click`,_=>f(i.value).then(x=>o.innerText=x),0)\n```\n```\n
\n```\n[Answer]\n# CJam, 10 bytes\n```\nlg\"` tag is contained by `

\"+e.body.replace(OVERRIDE_REG,\"\")+\"

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