244
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Using your language of choice, golf a quine.

A quine is a non-empty computer program which takes no input and produces a copy of its own source code as its only output.

No cheating -- that means that you can't just read the source file and print it. Also, in many languages, an empty file is also a quine: that isn't considered a legit quine either.

No error quines -- there is already a separate challenge for error quines.

Points for:

  • Smallest code (in bytes)
  • Most obfuscated/obscure solution
  • Using esoteric/obscure languages
  • Successfully using languages that are difficult to golf in

The following Stack Snippet can be used to get a quick view of the current score in each language, and thus to know which languages have existing answers and what sort of target you have to beat:

var QUESTION_ID=69;
var OVERRIDE_USER=98;

var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(index){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+index+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}
function commentUrl(index,answers){return"https://api.stackexchange.com/2.2/answers/"+answers.join(';')+"/comments?page="+index+"&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(data){answers.push.apply(answers,data.items);answers_hash=[];answer_ids=[];data.items.forEach(function(a){a.comments=[];var id=+a.share_link.match(/\d+/);answer_ids.push(id);answers_hash[id]=a});if(!data.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(data){data.items.forEach(function(c){if(c.owner.user_id===OVERRIDE_USER)
answers_hash[c.post_id].comments.push(c)});if(data.has_more)getComments();else if(more_answers)getAnswers();else process()}})}
getAnswers();var SCORE_REG=(function(){var headerTag=String.raw `h\d`
var score=String.raw `\-?\d+\.?\d*`
var normalText=String.raw `[^\n<>]*`
var strikethrough=String.raw `<s>${normalText}</s>|<strike>${normalText}</strike>|<del>${normalText}</del>`
var noDigitText=String.raw `[^\n\d<>]*`
var htmlTag=String.raw `<[^\n<>]+>`
return new RegExp(String.raw `<${headerTag}>`+String.raw `\s*([^\n,]*[^\s,]),.*?`+String.raw `(${score})`+String.raw `(?=`+String.raw `${noDigitText}`+String.raw `(?:(?:${strikethrough}|${htmlTag})${noDigitText})*`+String.raw `</${headerTag}>`+String.raw `)`)})();var OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(a){return a.owner.display_name}
function process(){var valid=[];answers.forEach(function(a){var body=a.body;a.comments.forEach(function(c){if(OVERRIDE_REG.test(c.body))
body='<h1>'+c.body.replace(OVERRIDE_REG,'')+'</h1>'});var match=body.match(SCORE_REG);if(match)
valid.push({user:getAuthorName(a),size:+match[2],language:match[1],link:a.share_link,})});valid.sort(function(a,b){var aB=a.size,bB=b.size;return aB-bB});var languages={};var place=1;var lastSize=null;var lastPlace=1;valid.forEach(function(a){if(a.size!=lastSize)
lastPlace=place;lastSize=a.size;++place;var answer=jQuery("#answer-template").html();answer=answer.replace("{{PLACE}}",lastPlace+".").replace("{{NAME}}",a.user).replace("{{LANGUAGE}}",a.language).replace("{{SIZE}}",a.size).replace("{{LINK}}",a.link);answer=jQuery(answer);jQuery("#answers").append(answer);var lang=a.language;lang=jQuery('<i>'+a.language+'</i>').text().toLowerCase();languages[lang]=languages[lang]||{lang:a.language,user:a.user,size:a.size,link:a.link,uniq:lang}});var langs=[];for(var lang in languages)
if(languages.hasOwnProperty(lang))
langs.push(languages[lang]);langs.sort(function(a,b){if(a.uniq>b.uniq)return 1;if(a.uniq<b.uniq)return-1;return 0});for(var i=0;i<langs.length;++i)
{var language=jQuery("#language-template").html();var lang=langs[i];language=language.replace("{{LANGUAGE}}",lang.lang).replace("{{NAME}}",lang.user).replace("{{SIZE}}",lang.size).replace("{{LINK}}",lang.link);language=jQuery(language);jQuery("#languages").append(language)}}
body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

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3
  • 6
    \$\begingroup\$ Do you not mean, "Golf you a quine for greater good!"? \$\endgroup\$ Commented May 3, 2011 at 2:49
  • 64
    \$\begingroup\$ @muntoo it's a play on "Learn you a Haskell for Great Good". \$\endgroup\$ Commented May 3, 2011 at 2:52
  • 26
    \$\begingroup\$ Did anybody notice that this is question 69? \$\endgroup\$
    – aidan0626
    Commented Oct 24, 2020 at 22:47

455 Answers 455

1
9 10
11
12 13
16
2
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ACL2, 41 bytes

(let((q"(let((q~x0))(cw q q))"))(cw q q))
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2
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Vim, 27 bytes

ii^V^V^V^[BDuplxbbpp^[BDuplxbbpp

^V being CTRL+V and ^[ being ESC.

The other one is beating mine, but it took a while and I didn't think it was possible.

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2
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QBIC, 8 bytes

?A+@?A+@

I was trying to do this, but I accidentally golfed an actual quine. Whadda you know?

Explanation

?           Print
 A            A$
  +         concatenated with
   @       a string literal containing
    ?A+@   "?A+@"

When A$ is used the first time, it might not seem to have a value yet, but it already contains the string literal ?A+@ because the QBIC interpreter first scans the code, sees the @, looks up what the first available string variable is (it's A$, because it hasn't been used by other QBIC language features yet), and it then extracts the definition A$ = "?A+@" to the top of the file, and inserts A$ at the place where it found the literal.

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2
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JScript, 175 bytes

WScript.Echo((k="WScript.Echo((k=\"%Z\").replace(/%[Z]/,k.replace(/[\"\\\\]/g,function(e){return\"\\\\\"+e})))").replace(/%[Z]/,k.replace(/["\\]/g,function(e){return"\\"+e})))

JScript is Microsoft's implementation of the JavaScript language. On a microsoft console, you can invoke the program as <name>.js, and this will output to a popup. To output to the console, one must use:

cscript //E:JScript //nologo <name>.js

and add a trailing CRLF to the source code.

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2
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Foam, 15 bytes

[. <' |: ~|]: ~

This prints itself with a trailing newline. Without a trailing newline:

[. .' |: ~|]: ~
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2
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Gaia, 10 bytes

“:ṙpp”:ṙpp

Try it online!

Explanation

“:ṙpp”      Push this string.
      :     Copy it.
       ṙ    Get the string representation of it.
        p   Print the string representation.
         p  Print the string.
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2
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PHP, 50 chars

<?printf($s='<?printf($s=%c%s%1$c,39,$s);',39,$s);

I realized I could save four chars by enabling short tags (<?=sprintf becomes <?printf)

I saved four more bytes by reusing one of the values in my printf.

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2
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Proton, 32 30 bytes

-2 bytes thanks to ppperry.

s='s=%rprint(s%%s)'print(s%s)

Try it online!

I swear, if I get a bounty for this...

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2
  • \$\begingroup\$ Hey, I promised a bounty, so if nobody beats it, good for you I guess :P \$\endgroup\$
    – hyper-neutrino
    Commented Aug 17, 2017 at 19:39
  • 1
    \$\begingroup\$ You can save two bytes by deleting the semicolons \$\endgroup\$ Commented Aug 18, 2017 at 22:18
2
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MaybeLater, 74 bytes

x="write(('x='+chr(34))+x+(chr(34))+x)"write(('x='+chr(34))+x+(chr(34))+x)

Try it online!

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2
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Wumpus, 9 bytes

"#34#9&o@

Try it online!

Explanation

This is a fairly standard Fungeoid quine. However, as opposed to most other Fungeoids, Wumpus's grid doesn't wrap around, so the IP actually bounces back and forth through the code:

"#34#9&o@o&9#32#"
     This pushes the individual code points of this string to the stack.
#34  Push 34.
#9   Push 9.
&o   Print 9 characters from the top of the stack.
@    Terminate the program.

There are several other ways to do this in 9 bytes, e.g. this one which generates the " from the # instead:

"#9[(~&o@

I haven't yet found a way to get it down to 8 bytes though (it might be possible: if there's a way to generate the 34 in three bytes that doesn't end in a digit, we could get rid of the # in front of the 9).

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2
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Rust, 108 characters

macro_rules!f(()=>("macro_rules!f(()=>({:?}));fn main(){{print!(f!(),f!())}}"));fn main(){print!(f!(),f!())}

This is a suboptimal solution, but it's so close to the current shortest solution that I wanted to post it anyway as it uses a completely different strategy. I think it can be optimized by using macro keyword instead of verbose macro_rules! when it becomes stable (which would reduce this to 96 characters).

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2
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SmileBASIC, 66 54 bytes

?MID$(("+CHR$(34))*3,19,54)?MID$(("+CHR$(34))*3,19,54)

This will work in most BASIC dialects.

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2
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Nim, 95 bytes

import strutils;let s="import strutils;let s=$#;echo s%s.repr[^49..^1]";echo s%s.repr[^49..^1]

Try it online!

After scrolling through the leaderboard snippet, I was surprised to see that Nim hasn't been represented here yet. So, let's fix it!

This follows the standard pattern of constructing and printing a string with a quoted representation of itself inserted in the middle. Unfortunately, Nim has a few features that make this golf-unfriendly:

  • String interpolation with % is not available unless you import strutils.
  • repr not only quotes strings, but also prepends them with a hex number (seemingly, memory address), like this: 0x40a6d0"my string". Therefore, we have to extract the right part.

Still, not so bad overall, as it is significantly shorter than the version currently posted at Rosetta Code, which actually doesn't even work without tweaks in recent versions of the language.

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1
  • 1
    \$\begingroup\$ var x="var x=;echo x[0..5],x.repr[14..^1],x[6..^1]";echo x[0..5],x.repr[14..^1],x[6..^1] \$\endgroup\$
    – ASCII-only
    Commented Aug 27, 2019 at 3:25
2
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Excel, 17 Bytes

In cell A1...

=FunctionText(A1)

I discovered this yesterday by accident. Feels a bit cheaty somehow though. I appreciate this answer much more, but this does seem to be a quine :D

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2
  • \$\begingroup\$ Wouldn't this count as reading its own source? \$\endgroup\$ Commented May 3, 2022 at 19:01
  • \$\begingroup\$ I guess that depends on what the 'source' of excel formulas is, if it's the formula text, then yes, otherwise it could be the source code of Excel? I'm not sure. \$\endgroup\$ Commented May 4, 2022 at 20:16
2
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Cardinal, 10 bytes

",-#) %8-$

Try it online!

This should be a quine, except for a small bug in the interpreter which causes the ( command to throw an error. For completeness' sake, here's a version where a # is placed in the position the ( is accidentally calling. Note that the space is actually a NUL character. Bug is fixed, yay!

"% (#-,0-$ also works with just one pointer.

How It Works:

The % creates two pointers, going left and right (the ones going vertical don't matter). The right pointer is delayed by the 8 for three steps, which lets the left pointer execute ,-#) which changes the active value to #, decrements it and prints the ". Then the right pointer starts again, decrementing the active value to -1. $ sets the pointer location to 0,-1, which then runs " over the rest of the code, printing it.

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3
  • 1
    \$\begingroup\$ You can download fixed and recompiled version (including source code) of the interpreter from my Github repo: github.com/m-lohmann/Cardinal. The original interpreter had several errors that needed fixing. \$\endgroup\$
    – M L
    Commented Jun 2, 2018 at 0:08
  • \$\begingroup\$ I don't think your last edit worked properly, the URLs are identical. \$\endgroup\$ Commented Sep 5, 2018 at 18:11
  • \$\begingroup\$ @Orjan lol, it seems I had already made the exact change in a previous edit. \$\endgroup\$
    – Jo King
    Commented Sep 7, 2018 at 13:30
2
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Backhand, 21 bytes

"#v{<@^:[ba+0v|{$:o[}

Try it online!

This is my new 1D language Backhand. It's a little bit more complicated than the typical wrapping string literal quine.

Explanation:

The program initially starts with a step count of 3.

"      Start string literal, stepping 3 places at a time
       This pushes the source code, but all jumbled up :(

See my Hello, World! answer to see what you have to do to push a string normally.

"      End the garbage string literal
  v{   Step left and decrease the step count to 2
    <  Change direction to left
  v    Decrease the step count to 1
 #     No-op
"      Start string literal
       Now the step count is 1, so it actually pushes the source code
 #v{<@^:[ba+0v|{$:o[}    Push to stack going right and bounce off the end
 #v{<@^:[ba+0v|{$:o[     Push to stack going left
"      End string literal
  v    Decrease the step count to 0
  v    Decrease the step count to -1
       Now the pointer is technically going right, but with a step count of -1
"      Push the source code again...
  v    Decrease the step count to -2
    <  Change direction to left (step count is still negative, so it goes right)

      ^           Increase step count to -1
       :[         Dupe the top of stack (# 35) and decrement to 34 (")
         ba+      Add 10 and 11 to push 21 as the counter
            0     Push 0
             v    Decrease step count to -2
               {  Step left
              |   Pop 0 and continue moving left
                $ o    Swap the top two items and print the character
                   [}  Step right (against the wall so it bounces) and decrement the counter
              |{ :     Duplicate the counter and reflect if it is non-zero
                       Repeat this 21 times to print the source code
        [ a 0  Garbage
      ^        Increase the step counter to -1
     @         Terminate program
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2
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Pascal (FPC), 103 bytes

const s=#39'const s=#39;begin write(s[2..12],s,s[10..50])end.'#39;begin write(s[2..12],s,s[10..50])end.

Try it online!

s is the string that the output is generated from. In Pascal, subtrings can be easily extracted with [from..to] syntax. #39 is replacement for ' using its ASCII codepoint. As seen in this program, sequences of character codepoints can be glued together with the rest of the string delimited with 's at any time. s consists of characters before and after 's concatenated together. #39 occurs immediately before first ' and after second ' so it can be put in s only once and used in both substrings in the output.

 
The version that may be more suitable in modified, quine-like programs is at 106 bytes:

const s='const s=;begin write(s[1..8],#39,s,#39,s[9..52])end.';begin write(s[1..8],#39,s,#39,s[9..52])end.

Try it online!

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2
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Clean, 123 102 bytes

module q;import StdEnv;Start=(s,q,s,q);q=inc'!';s="module q;import StdEnv;Start=(s,q,s,q);q=inc'!';s="

Save as q.icl and compile with -b -nt.

Saved 21 bytes thanks to Οurous.

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2
  • \$\begingroup\$ This can be improved to 102: module q;import StdEnv;Start=(s,q,s,q);q=inc'!';s="module q;import StdEnv;Start=(s,q,s,q);q=inc'!';s=" \$\endgroup\$
    – Οurous
    Commented Feb 13, 2019 at 7:09
  • \$\begingroup\$ @Οurous nice idea, thanks. \$\endgroup\$
    – user42682
    Commented Feb 13, 2019 at 8:36
2
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Brachylog v2, 12 bytes

"~k;?w₁";?w₁

Try it online!

Full program. Essentially a translation of Fatalize's (non-builtin) Brachylog v1 answer, although it also uses different SWI-Prolog formatting sequences, or rather, a single different one, which saves about 20 bytes (both [34:s, both :34]s, and both ~cs). It seems to have existed back in 2016, but it was probably bugged or something. The last two bytes saved come from using the implicit input, which Brachylog being Brachylog is useful even when the program receives no input, because it's a variable (so instead of explicitly unifying the string with S, we just let it be implicitly unified with ?).

          w     Print
"~k;?w₁"        "~k;?w₁"
                which is the input
           ₁    formatted with
        ;?      the input.
 ~k             (so that the ~k is replaced with the input's canonical representation, i.e. in quotes)

We don't actually need to use implicit input--"~kgjw₁"gjw₁ works just as well (and might even translate back to v1)--but doing so regardless manages to both more closely mirror the structure of the original and feel cleverer.

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2
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Tcl, 47 bytes

puts [join {p \{ \}]} {uts [join {p \{ \}]} }]

Based on Joe Miller's quine on this page.

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2
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Java, 515 bytes

In order to make this quine, I made strings representing every part of the code that needed to be printed. Then, I created and printed a string representing the code by adding the strings together.

This is my first time playing code golf, let me know what you think of my quine!

interface a{static void main(String[]a){String ce="interface a{static void main(String[]a){String ",e="=",c="\",",q="\"",cq="\\",ec=";System.out.print(ce+cc+ee+e+q+ce+c+ee+e+q+e+c+cc+e+q+cq+c+c+qq+e+q+cq+q+c+cc+qq+e+q+cq+cq+c+ee+cc+e+q+ec+c+ee+qq+e+q+eq+c+ee+ee+e+q+ee+c+cc+cc+e+q+cc+c+qq+qq+e+q+qq+q+ec+eq",eq=");}}",ee="e",cc="c",qq="q";System.out.print(ce+cc+ee+e+q+ce+c+ee+e+q+e+c+cc+e+q+cq+c+c+qq+e+q+cq+q+c+cc+qq+e+q+cq+cq+c+ee+cc+e+q+ec+c+ee+qq+e+q+eq+c+ee+ee+e+q+ee+c+cc+cc+e+q+cc+c+qq+qq+e+q+qq+q+ec+eq);}}

in readable form:

interface a
{
    static void main(String[] a)
    {
        String ce = "interface a{static void main(String[]a){String ",
            e = "=",
            c = "\",",
            q = "\"",
            cq = "\\",
            ec = ";System.out.print(ce+cc+ee+e+q+ce+c+ee+e+q+e+c+cc+e+q+cq+c+c+qq+e+q+cq+q+c+cc+qq+e+q+cq+cq+c+ee+cc+e+q+ec+c+ee+qq+e+q+eq+c+ee+ee+e+q+ee+c+cc+cc+e+q+cc+c+qq+qq+e+q+qq+q+ec+eq",
            eq = ");}}",
            ee = "e",
            cc = "c",
            qq = "q";
        System.out.print(ce+cc+ee+e+q+ce+c+ee+e+q+e+c+cc+e+q+cq+c+c+qq+e+q+cq+q+c+cc+qq+e+q+cq+cq+c+ee+cc+e+q+ec+c+ee+qq+e+q+eq+c+ee+ee+e+q+ee+c+cc+cc+e+q+cc+c+qq+qq+e+q+qq+q+ec+eq);
    }
}
\$\endgroup\$
2
  • \$\begingroup\$ If you want to try shortening this further, have a look into using the printf function \$\endgroup\$
    – Jo King
    Commented Jun 28, 2019 at 6:41
  • \$\begingroup\$ I tried that, but I found print easier. I guess I'll try again. \$\endgroup\$ Commented Jul 1, 2019 at 2:22
2
\$\begingroup\$

Pip, 26 23 bytes

Y\"O"Y"ORPyy\"O"Y"ORPyy

Try it online!

\$\endgroup\$
2
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C (gcc), 85 bytes

#define q(k)main(){puts(#k"\nq("#k")");}
q(#define q(k)main(){puts(#k"\nq("#k")");})

Try it online!

The q() macro expands into a program that prints out its argument on the first line, and prints out the argument called by q() itself in the second line. So:

#define q(k)main(){puts(#k"\nq("#k")");}
q(foo)

would expand into:

main(){puts("foo""\nq(""foo"")");}

and after string literal concatenation, becomes:

main(){puts("foo\nq(foo)");}

And executing and running the program would produce:

foo
q(foo)

Replacing foo with the macro definition itself results in the quine.

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2
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33, 26 bytes

"34cktptptptp"34cktptptptp

Try it online!

Explanation:

"34cktptptptp"             (The instructions)
              34ck         (Load 34 (") into destination string)
                  tp       (Print it)
                    tp     (Print the instructions)
                      tptp (Repeat)
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2
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Corea, 11 bytes

<0C>;"<0C>;

Try it online!

Alternatively, <0C>>"<0C>>

<0C>;"<0C>;
<0C>;            set the contents to that literal string
     "           start command sequence
      <          push a copy of the contents to the stack
       0C        push a quote "
         >       append that to the contents
          ;      and append the original copy and stop command sequence
\$\endgroup\$
2
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Alchemist, 299 bytes

_->14733272090064622117723033634640281434301133345502153207896692199003336573010981052872005814325038964478266287468505274190371239580629370336756929651609657090021232407437472153714372752689076920028135a+Out_"_->"+Out_a+d
d+0e->118b
a+b->o
118o->c+d
0a+0e+c->Out'o+e
e+b->e+o
e+d+c->e+d+a
0c+e+d->d

Try it online!

I'm posting this as a separate answer to my existing one since it uses newer features (character output). This functions much the same but with less logic regarding the modulo values, countered partially by the number being in base 118 instead of base 9. Here is an encoder that can be used to encode the large number up front.

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2
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Poetic, 2100 1896 bytes

-204 bytes by using ASCII 255 and 1 instead of ASCII 122 and 123.

ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿÿ ÿÿÿÿÿ ÿ ÿÿÿÿÿÿ ÿ ÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿ ÿÿ ÿÿÿ ÿ ÿÿÿÿÿÿ ÿÿÿ ÿ ÿÿÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿÿ ÿÿ ÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿ ÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿ ÿÿÿÿÿÿ ÿ ÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿ ÿ ÿ ÿÿÿÿÿ ÿ ÿÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿÿÿÿ ÿÿÿÿÿ ÿÿ ÿÿÿ ÿÿÿ ÿ ÿÿÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿÿ ÿÿ ÿÿ ÿÿÿÿÿ ÿÿÿ ÿ ÿÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿÿ ÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿÿ ÿ ÿÿÿÿÿÿ ÿ ÿ ÿÿÿÿÿ ÿ ÿÿÿÿÿÿÿ ÿÿÿÿÿÿ ÿ ÿÿÿÿ ÿ ÿÿ ÿÿÿÿÿ ÿ ÿÿÿ ÿÿ ÿÿÿÿÿÿÿ ÿÿÿÿÿÿ ÿÿ ÿÿ 

(All of the space characters are actually ASCII code 1, or SOH. StackExchange doesn't seem to like unprintables all that much.)

Try it online!

Poetic is an esolang I made in 2018 for a class project. It's basically brainfuck with word-lengths instead of symbols.

This isn't a terribly Poetic program (it's only two distinct characters), but it's the most compact solution for writing commands that I was able to come up with.

Basically, the core of the program is code that takes tape values corresponding to Poetic commands in a line, adds values to the beginning that would put these values on the tape when executed, and then outputs the values as Poetic commands. The initial values on the tape are, of course, an encoded version of that code. (Standard stuff, surely, but it took me a while to wrap my head around it.)

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  • \$\begingroup\$ Unless Poetic uses a custom code page/encoding, this is actually 3312 bytes as ÿ is 2 bytes in TUF-8 \$\endgroup\$ Commented Aug 9, 2020 at 23:16
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Symbolic Raku, 36 bytes

$_={$_~"<$_>)"}(<$_={$_~"<$_>)"}(>)

Try it online!

Explanation:

$_=                                   # Set the output to
   {          }(                  )   # The result of the code block
                <                >    # With this string:
                 $_={$_~"<$_>)"}(        # The first half of the program
    $_~                               # Concatenate this string with
       "<$_>)"                        # The quoted string, and the extra bracket
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Ral, 1389 1043 bytes

1111111101011101011101010101111101011101010101111101010101111101011101010101111101010101111101011101010101111000111000111000111000111000110101010000110101110111011101011101010101111101011101010101111101011101010101111101011101010101111101011101010101111101011011010101111111111101011101010101111101011101010101111101010101111101011101010101111101011101010101111000111000111000111000110101010000111011010101010000110101111111010111011101011101010101111101010101111101010101111101010101111101011000111101011000111000110101110101010101111101011000111111111101011101010101111101010101111101011101010101111101011101010101111000111000111000110101010000110101111011010101011011010000111000111000111011111011111011010000111000110101111011110000110101111101010101111011011101010101111101010101111101010101111101010101111101010101111101010101111101011101010101111101011101010101111000111000111000111000111111:++:++:+:+:+:+:+:+-:+:0=:10-==110-*-:0*111:++:++:+:++?1+:*:11+1+:+:+:+:++./:0*-0*1111:++:++:+:++:++?:-+:++:++:++:++:++.:0*11111:++:+:++:+:++:+++?

Try it online!

Explanation

All characters in Ral (except the no-op) are in the range 32-63, which is 001xxxxx in binary. By omitting the leading zeroes, every character in the code can be stored as a group of 6 bits. (A hexad? hextet? Hextet will do.) The one is still included in the payload to make decoding easier.

Here is a short summary of what each part of the code does. I have omitted the value juggling on the stack and inner workins of the loops from the explanation, as they would make the explanation too complicated:

111111...100011                Push the payload data to the stack.

1111:++:++:+:+:+:+:+:+-:+:0=   Load the value 894, which is the length of the payload
                               and is used as a base pointer for the jump destinations.

:10-==110-*-:0*111:++:++:+:++? Move the payload data to memory.

1+:*:11+1+:+:+:+:++.           Loop through the payload, printing 48+n for each value
/:0*-0*1111:++:++:+:++:++?     and push each value to the stack again.

:-+:++:++:++:++:++.            Decode and print each hextet.
:0*11111:++:+:++:+:++:+++?
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Python 3, 190 216 bytes

s='s=\\\'\'+s.replace(\'\\\\\',\'\\\\\\\\\').replace(\'\\\'\',\'\\\\\\\'\')+\'\\\'\\nprint(\\\'\'+s+\'\\\')'
print('s=\''+s.replace('\\','\\\\').replace('\'','\\\'')+'\'\nprint(\''+s+'\')')

Try it online!

An old solution of mine following the standard pattern. It only uses the most common commands though.

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  • \$\begingroup\$ Nice answer! I have edited in a link to the online interpreter, in case people want to test out your code. \$\endgroup\$ Commented Jun 1, 2020 at 10:48
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    \$\begingroup\$ You can save a few bytes by removing the redundant spaces. Try it online! \$\endgroup\$ Commented Jun 1, 2020 at 10:54
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