516
\$\begingroup\$

So... uh... this is a bit embarrassing. But we don't have a plain "Hello, World!" challenge yet (despite having 35 variants tagged with , and counting). While this is not the most interesting code golf in the common languages, finding the shortest solution in certain esolangs can be a serious challenge. For instance, to my knowledge it is not known whether the shortest possible Brainfuck solution has been found yet.

Furthermore, while all of Wikipedia (the Wikipedia entry has been deleted but there is a copy at archive.org ), esolangs and Rosetta Code have lists of "Hello, World!" programs, none of these are interested in having the shortest for each language (there is also this GitHub repository). If we want to be a significant site in the code golf community, I think we should try and create the ultimate catalogue of shortest "Hello, World!" programs (similar to how our basic quine challenge contains some of the shortest known quines in various languages). So let's do this!

The Rules

  • Each submission must be a full program.
  • The program must take no input, and print Hello, World! to STDOUT (this exact byte stream, including capitalization and punctuation) plus an optional trailing newline, and nothing else.
  • The program must not write anything to STDERR.
  • If anyone wants to abuse this by creating a language where the empty program prints Hello, World!, then congrats, they just paved the way for a very boring answer.

    Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.

  • Submissions are scored in bytes, in an appropriate (pre-existing) encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score - if in doubt, please ask on Meta.
  • This is not about finding the language with the shortest "Hello, World!" program. This is about finding the shortest "Hello, World!" program in every language. Therefore, I will not mark any answer as "accepted".
  • If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck-derivatives like Alphuck), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf - these are still useful to this question as it tries to compile a catalogue as complete as possible. However, do primarily upvote answers in languages where the authors actually had to put effort into golfing the code.

For inspiration, check the Hello World Collection.

The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where 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:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If 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:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](https://esolangs.org/wiki/Fish), 121 bytes

/* Configuration */

var QUESTION_ID = 55422; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8478; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, 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: true,
    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 = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    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 = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

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,
      });
    else console.log(body);
  });
  
  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('<a>'+lang+'</a>').text();
    
    languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) 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;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 500px;
  float: left;
}

table thead {
  font-weight: bold;
}

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/all.css?v=ffb5d0584c5f">
<div id="language-list">
  <h2>Shortest Solution 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>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>

\$\endgroup\$
21
  • 4
    \$\begingroup\$ @isaacg No it doesn't. I think there would be some interesting languages where it's not obvious whether primality testing is possible. \$\endgroup\$ Aug 28, 2015 at 13:56
  • 7
    \$\begingroup\$ If the same program, such as "Hello, World!", is the shortest in many different and unrelated languages, should it be posted separately? \$\endgroup\$ Aug 28, 2015 at 15:33
  • 2
    \$\begingroup\$ @mbomb007 Well it's hidden by default because the three code blocks take up a lot of space. I could minify them so that they are a single line each, but I'd rather keep the code maintainable in case bugs come up. \$\endgroup\$ Aug 28, 2015 at 19:34
  • 8
    \$\begingroup\$ @ETHproductions "Unlike our usual rules, feel free to use a language (or language version) even if it's newer than this challenge." Publishing the language and an implementation before posting it would definitely be helpful though. \$\endgroup\$ Aug 29, 2015 at 23:01
  • 2
    \$\begingroup\$ @MartinEnder ... Almost. If two BF solutions have the same size, the one with smaller lexicographical order will take smaller number of bytes in Unary. Of course the smallest Unary solution translated to BF is guaranteed to be smallest. \$\endgroup\$
    – DELETE_ME
    May 20, 2018 at 10:20

978 Answers 978

1
6 7
8
9 10
33
4
\$\begingroup\$

unc, 38 bytes

ZNVa[]<<chgf[L'uRYYb~ JbeYQ#']:if 5:>>
\$\endgroup\$
4
\$\begingroup\$

Rust, 34 bytes

fn main(){print!("Hello, World!")}
\$\endgroup\$
4
\$\begingroup\$

Nim, 20 19 bytes

echo"Hello, World!"

Saved one byte thanks to sp3000!

\$\endgroup\$
2
  • \$\begingroup\$ You can drop the space in between for the first one, I think :) \$\endgroup\$
    – Sp3000
    Aug 28, 2015 at 15:09
  • \$\begingroup\$ Indeed, saved me one byte! \$\endgroup\$
    – kvill
    Aug 28, 2015 at 15:18
4
\$\begingroup\$

Emily, 22 bytes

println"Hello, World!"

This is a nice little language I stumbled upon recently.

\$\endgroup\$
4
\$\begingroup\$

Applescript, 15 bytes

"Hello, World!"

Normally a fairly verbose language, for this one this is all that is required.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ It's nice to see these verbose languages -- Applescript, PHP, PowerShell, etc. -- getting the better of lots of other languages for once. :) \$\endgroup\$ Aug 28, 2015 at 15:53
  • \$\begingroup\$ @TimmyD That doesn't mean that they are good languages. It's just that they are better at some things. \$\endgroup\$
    – galexite
    Aug 28, 2015 at 16:53
  • 1
    \$\begingroup\$ @georgeunix Oh, without a doubt. Every language has its pluses and minuses. There are plenty of things I'd change about PowerShell, but there's nothing else I'd rather use to script and config Exchange. Even if it had commands and functionality to do so, I don't think I could use Pyth or CJam or whatnot on a day-to-day basis instead. I was just meaning "It's nice to see non-golfing languages toward the top of the lowest-byte-count list for a change." \$\endgroup\$ Aug 28, 2015 at 17:06
  • \$\begingroup\$ I understand @TimmyD \$\endgroup\$
    – galexite
    Aug 28, 2015 at 17:07
4
\$\begingroup\$

VBScript, 28 Bytes

WScript.Echo "Hello, World!"

This (should be) the shortest that prints to STDOUT (i.e., the command prompt window), when executed via command prompt wscript .\hello-world.vbs or cscript //nologo .\hello-world.vbs (the //nologo is necessary to prevent copyright info from being displayed). If you just double-click it, you'll get a pop-up message box instead, similar to the shorter example, below, at 22 bytes:

MsgBox "Hello, World!"

When executed, this second option will output a pop-up message box displaying the text inside the quotes. Since it's not technically STDOUT, and we do have a legitimate way to display STDOUT, we'll count the longer version instead.

\$\endgroup\$
4
\$\begingroup\$

Ook!, 949 Bytes

Just translated one of the Brainfuck answers here.

Ook! Ook! Ook. Ook? Ook. Ook? Ook. Ook. Ook. Ook.
Ook. Ook? Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook?
Ook. Ook. Ook. Ook? Ook! Ook! Ook! Ook! Ook! Ook!
Ook! Ook! Ook. Ook? Ook! Ook! Ook! Ook! Ook! Ook?
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook. Ook! Ook? Ook. Ook? Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook? Ook.
Ook! Ook! Ook? Ook! Ook! Ook! Ook? Ook. Ook. Ook.
Ook? Ook! Ook. Ook? Ook. Ook? Ook! Ook! Ook! Ook!
Ook! Ook! Ook! Ook! Ook! Ook. Ook. Ook? Ook. Ook.
Ook. Ook. Ook. Ook. Ook! Ook. Ook. Ook? Ook! Ook!
Ook! Ook. Ook! Ook. Ook. Ook. Ook. Ook. Ook. Ook.
Ook! Ook. Ook. Ook? Ook. Ook? Ook. Ook? Ook! Ook.
Ook? Ook. Ook! Ook. Ook? Ook. Ook! Ook. Ook? Ook.
Ook! Ook. Ook. Ook. Ook. Ook. Ook. Ook. Ook! Ook.
Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook! Ook!
Ook! Ook! Ook! Ook. Ook? Ook. Ook! Ook! Ook! Ook.
Ook. Ook? Ook. Ook? Ook. Ook? Ook. Ook. Ook! Ook.
\$\endgroup\$
4
  • 2
    \$\begingroup\$ I always smile when I see this language. \$\endgroup\$ Aug 28, 2015 at 16:36
  • 2
    \$\begingroup\$ This might be a case of "If your language of choice is a trivial variant of another (potentially more popular) language...". The shortest Ook! program will always be the translation of the shortest BF program (because each BF character is converted to the same length in Ook). So if Ook has a separate answer it needs to be updated every time someone finds a new shortest Brainfuck solution. (Ultimately, it's your call though if Ook should remain separate.) \$\endgroup\$ Aug 28, 2015 at 17:06
  • 3
    \$\begingroup\$ I love reading the "Code" in a manner as if two people were talking to each other :) \$\endgroup\$
    – MrPaulch
    Aug 29, 2015 at 11:12
  • \$\begingroup\$ @MrPaulch I agree. The fact that first O is capitalized makes it sound like a fast Oh, okay with vocalized differences for punctuation in my mind. \$\endgroup\$
    – mbomb007
    Sep 1, 2015 at 15:04
4
\$\begingroup\$

///, 13

Hello, World!

can't get much simpler than this

\$\endgroup\$
4
\$\begingroup\$

Octave, 19 bytes

disp"Hello, World!"
\$\endgroup\$
4
\$\begingroup\$

Vim, 17 bytes

iHello, World!{ESC}ZZ

Where {ESC} is a raw escape byte \x1b.

This will switch to insert mode (i), write Hello, World!, leave it (ESC), and save+quit (ZZ). An environment like vimgolf or anarchy golf has to do the output part for you, as Vim is, of course, just a text editor.

\$\endgroup\$
3
  • \$\begingroup\$ @LegionMammal978 the other seems to be Vimscript mislabelled as Vim. \$\endgroup\$
    – primo
    Feb 26, 2016 at 18:33
  • 3
    \$\begingroup\$ I think generally the standard is that vim solutions don't have to save and quit, they can just display the text onscreen at the end. (That's what mine have all done). This would allow you to take 3 bytes off. \$\endgroup\$
    – DJMcMayhem
    Jun 29, 2016 at 17:33
  • \$\begingroup\$ Way longer and wrong output, but more interesting: :h_4<CR>/"H<CR>ly2wZZp \$\endgroup\$
    – BlackCap
    Oct 12, 2017 at 19:22
4
\$\begingroup\$

PARI/GP, 22 bytes

print("Hello, World!")
\$\endgroup\$
4
\$\begingroup\$

3var, 65 bytes

iiisa-<*>P/>is+iP>PPm-iiiPi<O/<m/>+<O+d<+<O+><kkkOP->siskkkOP</>P

Here's a 3var program found by brute force. Note that this might not be optimal since I assume that we'll only ever need numbers in the range 0-150, for efficiency reasons. I'll probably address this in a later edit.

3var is a Deadfish variant which has, well, three variables A, B and R. The relevant commands are:

Command              A           B           other
-----------------------------------------------------------
Increment            i           a
Decrement            d           k
Square               s           m
Output as char       P           O
Copy from R          >           <
Set R = A+B                                  +
Set R = A-B                                  -
Set R = A*B                                  *
Set R = A div B                              /

And here's a trace:

Line           A     B     R     Output
----------------------------------------------------------------
iiisa-         9     1     8
<*             9     8     72
>P             72    8     72    H
/>             9     8     72    H
is+            100   8     108   H
iP             101   8     108   He
>PP            108   8     108   Hell
m-             108   64    44    Hell
iiiP           111   64    44    Hello
i<O            112   44    44    Hello,
/<m            112   4     2     Hello,
/>             28    4     28    Hello,
+<O            28    32    32    Hello, 
+d<            27    60    60    Hello, 
+<O            27    87    87    Hello, W
+><            114   114   114   Hello, W
kkkOP          114   111   114   Hello, Wor
->             3     111   3     Hello, Wor
siskkkOP       100   108   3     Hello, World
</             100   3     33    Hello, World
>P             33    3     33    Hello, World!
\$\endgroup\$
1
  • \$\begingroup\$ @LegionMammal978 Do you mean esolang wiki? The site seems fine to me... \$\endgroup\$
    – Sp3000
    Sep 28, 2015 at 2:39
4
\$\begingroup\$

4, 117 bytes

3.6000160103602136033260433605446067260787008070200908000120902111120111011015065095105105115055035075115125105085044

How it works

Generating characters with a code point below 100 is straightforward.

I've managed to create the others (derol) with three assignments and five additions/subtractions, which I believe is optimal.

3.            Begin the program.
  6 00 01     Set cell[ 0] to 1.
  6 01 03     Set cell[ 1] to 3.
  6 02 13     Set cell[ 2] to 13.
  6 03 32     Set cell[ 3] to 32 = ' '.
  6 04 33     Set cell[ 4] to 33 = '!'.
  6 05 44     Set cell[ 5] to 44 = ','.
  6 06 72     Set cell[ 6] to 72 = 'H'.
  6 07 87     Set cell[ 7] to 87 = 'W'.
  0 08 07 02  Set cell[ 8] to cell[ 7] + cell[2] =  87 + 13 = 100 = 'd'.
  0 09 08 00  Set cell[ 9] to cell[ 8] + cell[0] = 100 +  1 = 101 = 'e'.
  0 12 09 02  Set cell[12] to cell[ 9] + cell[2] = 101 + 13 = 114 = 'r'.
  1 11 12 01  Set cell[11] to cell[12] - cell[1] = 114 -  3 = 111 = 'o'.
  1 10 11 01  Set cell[10] to cell[11] + cell[1] = 111 -  3 = 108 = 'l'.
  5 06        Print cell[ 6] = 'H'.
  5 09        Print cell[ 9] = 'e'.
  5 10        Print cell[10] = 'l'.
  5 10        Print cell[10] = 'l'.
  5 11        Print cell[11] = 'o'.
  5 05        Print cell[ 5] = ','.
  5 03        Print cell[ 3] = ' '.
  5 07        Print cell[ 7] = 'W'.
  5 11        Print cell[11] = 'o'.
  5 12        Print cell[12] = 'r'.
  5 10        Print cell[10] = 'l'.
  5 08        Print cell[ 8] = 'd'.
  5 04        Print cell[ 4] = '!'.
4             End the program.
\$\endgroup\$
4
\$\begingroup\$

Q, 16 bytes

1"Hello, World!"

Just Y to go and we have the alphabet :)

Bit of a late update, but thanks to Mauris, we now have at least one language for every letter of the alphabet :D

Thanks @AaronDavies

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Not quite the requested output. I think you want 1"Hello, World!"; (doesn't include the trailing newline; to add one, change the 1 to a -1). Note, also valid for k. \$\endgroup\$ Aug 29, 2015 at 23:19
4
\$\begingroup\$

123, 282 267 bytes

22221121121112112222222211211112111211211222222221121121133121121312121122222222111211213
31211213122222222111211332113312112222221112112331123322222221111211111211222221111211211
22222222112111112112112112222222111112112112112222213312112131222222221121113321133121121

The newlines are only for cosmetic purposes. I'm fairly sure that this is not optimal.

Here is a slightly more readable (and also runnable) version:

H 22221121121112112
e 2222222112111121112112112
l 22222221121121133121121312
l 12112
o 22222221112112133121121312
, 2222222111211332113312112
  2222211121123311233
W 222222211112111112112
o 222211112112112
r 2222222112111112112112112
l 222222111112112112112
d 2222133121121312
! 2222222112111332113312112
1

I started out by constructing an optimal linear code (i.e. one which doesn't use 3s which allow for loop). That is quite simple: for each character, determine which bytes to flip from the last one. Move to the right-most character that has to be flipped (with a series of 2s), then move back to the left with 1 for each byte that has to be flipped and 121 for each byte that shouldn't be flipped. Finally move to the writing index -2 and print the character with 21. Repeat. At the very end, move to index -1 with a trailing 1 in order for the program to terminate.

This jumble of 1s and 2s was generated with this CJam script, which you can run online here:

0c"Hello, World!"+2ew::^{
_{2b8Ue[1a/W<1a*_,'2*'1@W%{'1"121"?}/"12"}{;"12112"}?
}/
'1

Then I removed some repetition of ones and twos by inserting loops by hand. 3 works as follows: if the instruction pointer is to the left of index 0, skip the 3. Otherwise, jump to the previous 3 if the current bit is 1 or jump ahead to the next 3 if the bit is 0. So simple loops, repeating a code segment x can be constructed as 33x33 or 33x3 (depending on whether the termination condition is "current bit is zero" or "moved to a negative index"). Then I started enumerating some relevant simple loops and when they are applicable. I've been using these loops only when moving back through the bits to change one character code to the next. If we can use a loop here depends both on the current state of a bit a and the target state b. I'll be denoting this combined state of each position as [a b]. Now here are the relevant loops and the required position patterns in a regex-like syntax:

121:    (^|[0 0]|[0 1]) ([1 1])+ [0 0]
112:    (^|[1 1]) ([0 0])+ ([0 1]|[1 1])
211:    ([0 0]|[0 1]) ([1 1])+ [0 0] ([0 0]|[1 1])
121121: ([0 0]|[0 1]) ([1 1] ([1 1]|[0 0]))+ [0 0]

Listing out the combined states for each character, we can annotate the potential loops and how many bytes they'll save (each ___ annotates the character above; sometimes multiple loops are possible):

H [[0 0] [0 1] [0 0] [0 0] [0 1] [0 0] [0 0] [0 0]]
e [[0 0] [1 1] [0 1] [0 0] [1 0] [0 1] [0 0] [0 1]]
l [[0 0] [1 1] [1 1] [0 0] [0 1] [1 1] [0 0] [1 0]]
        __________________121 -2
  ________________________121121 -3
l [[0 0] [1 1] [1 1] [0 0] [1 1] [1 1] [0 0] [0 0]]
o [[0 0] [1 1] [1 1] [0 0] [1 1] [1 1] [0 1] [0 1]]
        __________________121 -2
              __________________211 -2
  ________________________121121 -3
, [[0 0] [1 0] [1 1] [0 0] [1 1] [1 1] [1 0] [1 0]]
        ________________________211 -2
  [[0 0] [0 0] [1 1] [0 0] [1 0] [1 0] [0 0] [0 0]]
  ____________112 -2
W [[0 0] [0 1] [1 0] [0 1] [0 0] [0 1] [0 1] [0 1]]
o [[0 0] [1 1] [0 1] [1 0] [0 1] [1 1] [1 1] [1 1]]
r [[0 0] [1 1] [1 1] [0 1] [1 0] [1 0] [1 1] [1 0]]
l [[0 0] [1 1] [1 1] [1 0] [0 1] [0 1] [1 0] [0 0]]
d [[0 0] [1 1] [1 1] [0 0] [1 0] [1 1] [0 0] [0 0]]
        __________________121 -2
  ________________________121121 -3
! [[0 0] [1 0] [1 1] [0 0] [0 0] [1 0] [0 0] [0 1]]
        ________________________211 -2

Now I just picked the most profitable loop in each case and inserted it into the code.

I'm fairly certain that one could find a couple more loops that I've overlooked. But I also think that it's possible to find a significantly shorter solution that isn't based on anything a human would come up with. So far I have no idea how to efficiently search for such a solution automatically though, so I'll leave it at that for now.

\$\endgroup\$
4
\$\begingroup\$

KEMURI, 65 bytes

`^^^^"^^'"'^'"'"^^`^^'^''^"^^^^^^''^'"''"^^`^^^^^'"^^'^'^''^'^'^|

There's a KEMURI to C compiler available here if you'd like to test.

KEMURI is stack-based, and has the following 6 instructions:

~    Pop byte and push its NOT
^    Pop two bytes and push their XOR
"    Duplicate top of stack
'    Rotate top three of stack (top becomes third)
`    Push the ASCII values of "Hello, world!"
|    Output stack as ASCII

Note that ` pushes "Hello, world!" with a lowercase w. This means that the shortest "Hello, world!" program is

`|

but that doesn't mean that the best "Hello, World!" program, with an uppercase w, will be particularly short.

To aid our search for the best "Hello, World!", here are a couple of observations:

  • | empties the stack, so we will only need it exactly once, as the very last character in the program.
  • ~ is useless, since NOT will flip the most significant bit to 1, which no printable ASCII character needs.
  • We will never need to duplicate with " if the top two stack elements are the same, since:
    • Rotating three identical elements is a no-op.
    • The only way to reduce the stack size is with ^ XOR. XOR of two identical elements just introduces a 0 and XOR 0 is a no-op.
    • "Hello, World!" contains neither a triple letter nor ASCII 0.

This means that we only need to look at the four instructions ^"'`. To piece together the "Hello, World!", I looked at programs which contain a single `, at the very start. This gives a bunch of "jigsaw pieces" which we can fit together to form the whole message. There's no guarantee that this approach is optimal, but the search space is pretty big, so any better solution will probably need to be a bit more intelligent.

The pieces I managed to obtain were (<sp> is trailing space):

World!   `^^^^^"^^^|
orld!    `^^^^^^^"^^|
rld!     `^^^^^^^^"^^|
ld!      `^^^^^^^^^"^^|
d!       `"^^^^^^^^^^^^|
World!   `^^^^^"^'"'^'^|
<sp>     `^"^^^^^^^^^^^^|
!        `^^^^^^^^^^^"^^|
H        `^^^^^^^^^^'^"^^|
,        `^^^^^^^^^^''"^^^|
d        `^^^^^^^^^^''^"^^|
o        `^^^^''^"^^^^^^^^|
e        `"^^^^^^'"^^^^^^^^|
W        `^"^^^^^^'"^^^^^^^|
l        `^^"^^^^'"^^^^^^^^|
, World! `^^^^"^^'"'^'"'"^^|
ld       `^^^^^^^^^"^^''"^^|
ll       `^^"^^^^'"^^^^^^^^"|
r        `^^^^^^'"^^^^'^"^^^|
rl       `^^^^^^'"^^^^''"^^''^|
He       `^^^^^'"^^'^'^''^'^'^|
el       `^^"^^^^^^^^''"^^''"^^|
o,       `^^'^"^^^^"'^'^'^'^'^'^|
Wo       `^^^^"^^^'"^^'^''^''^''^|
,<sp>    `^^^^^'^'^'^''^''^'"'^'^|
lo       `^^^^^^^"^^'"^^''"^^''"^^|
or       `^^^^^^'^"^^''"^^''"^^''^|
llo      `^^'^''^"^^^^^^''^'"''"^^|
ell      `^^"^^^^^^^^''"^^'"''"^^''|
 W       `^"^^^^^^''^''^''^''^''^"'^|

The program at the top of the post was formed by combining the He, llo and , World! pieces.

\$\endgroup\$
4
\$\begingroup\$

Piet, 132 codels

On a 4x33 grid. On the last few commands I had to stretch to reach the end, meaning it could be golfed a little more (it probably fits on a 4x31 grid). Here it is, with codel size 10:

Piet code

I made it in a rectangular space to minimize the number of time I needed to flip the pointer. The stack is based on numbers 36 and 108 that are constantly being duplicated or rolled to produce the new letters.

Made and tested on PietDev.

\$\endgroup\$
4
\$\begingroup\$

Commodore Basic, 16 bytes

In order to input this program, you'll need to switch your Commodore 64 to character set 2 by pressing <SHIFT> + <C=>.

1?"hELLO wORLD!

The Commodore home computers come with two character sets: "unshifted" mode, which is derived from ASCII-1963 and so lacks lower-case letters, and "shifted" mode, which has both lower- and upper-case letters, but in the opposite order from modern ASCII-1967-derived encodings. Any "Hello, World!' program that produces the requested byte stream on a Commodore will look funny on the Commodore's screen. In the interests of not having to look up a half-dozen obscure Unicode characters, I've chosen to write my program in "shifted" mode, which merely has reversed case.

As a side note, the Commmodore Basic interpreter (and presumably many other Microsoft Basic variants) will let you omit the trailing quotation mark if a string extends to the end of the current source line.

\$\endgroup\$
4
\$\begingroup\$

Actionscript 3.0, 23 22 bytes

trace("Hello, World!")
\$\endgroup\$
3
  • \$\begingroup\$ Does actionscript 3.0 REQUIRE semi-colons? I stopped after 2.0 but I don't think it did back then. \$\endgroup\$ Sep 29, 2015 at 21:27
  • 1
    \$\begingroup\$ @AlbertRenshaw is correct, it doesn't require a semi-colon in 3.0. \$\endgroup\$ Oct 2, 2015 at 14:17
  • \$\begingroup\$ This works in the Actions pane of the Flash IDE, but Flash Builder or the Actionscript compiler, you need a full class definition. \$\endgroup\$
    – Brian
    Jan 15, 2016 at 18:54
4
\$\begingroup\$

Gol><>, 16 bytes

"!dlroW ,olleH"H

Try it online.

I've really enjoyed golfing in ><>, but unfortunately I've found that ><> lacks several features, e.g. STDIN integer input, which prevent it from being competitive in challenges it otherwise would be. Gol><> is designed to (hopefully) be an easier-to-use variation of ><>. I worked on it earlier in the year, around when the language showcase was happening, but took a break and only picked it up again recently. It's starting to stabilise, so I thought it'd be a good time to post a first answer.

Similarly to ><>, " is a string parsing operator which pushes chars one at a time until it reaches a closing ". H then halts the program, outputting the stack until it is empty.

Even without H, Gol><> can still output the stack in a relatively short way. l pushes the length of the stack, o outputs a char from the stack and R pops a number n, repeating the next instruction n times. Thus, an equivalent program would be

"!dlroW ,olleH"lRo;

where ; terminates the program with no output.

\$\endgroup\$
2
  • \$\begingroup\$ Another solution would be S"Hello, World!". \$\endgroup\$ Nov 9, 2015 at 12:39
  • \$\begingroup\$ @LegionMammal978 Indeed :) (although you'll need a ; at the end or it'll print forever) \$\endgroup\$
    – Sp3000
    Nov 9, 2015 at 12:43
4
\$\begingroup\$

Vitsy, 18 16 bytes

"!dlroW ,olleH"Z

"!dlroW ,olleH"     Push Hello, World! to the stack.
               Z    Push the entire stack to STDOUT - equivalent to l\O

Output:

Hello, World!

Z is new syntax - it was not made for this question.

\$\endgroup\$
4
\$\begingroup\$

Dogescript, 42 37 bytes

plz console.loge with "Hello, World!"

Translates to console.log("Hello, World!").

\$\endgroup\$
4
  • \$\begingroup\$ plz console.loge with "Hello, World!"is shorter, but admittedly less wow \$\endgroup\$
    – kvill
    Aug 31, 2015 at 15:21
  • 1
    \$\begingroup\$ Is the plz necessary? I thought you could just console.loge. \$\endgroup\$
    – Alex A.
    Sep 2, 2015 at 5:59
  • \$\begingroup\$ @AlexA. doesn't work without plz in the online interpreter, in accordance with the specs \$\endgroup\$
    – kvill
    Sep 2, 2015 at 19:56
  • \$\begingroup\$ That's what I was thinking... \$\endgroup\$ Sep 3, 2015 at 10:57
4
\$\begingroup\$

Purple, 62 bytes

AA1AA1AA1bA1b1Bo1bb1bbibb1Bi1b     
 ! d l r o W   , o l l e H

Purple in a Nutshell:

Purple is a self-modifying language in the same sense that self-modifying brainfuck is: The code is executed from the same array that contains data, which is infinite and otherwise initialized with zeroes. It has one instruction with three arguments: subtract the third argument from the second and store it in the first. It has two registers, a and b, which can be dereferenced as A and B to get the contents of that memory address. It also has i, the instruction pointer, o which represents the outside world (i.e., stdout in the first argument, stdin in either of the other two), and the literal 1, which cannot be the first argument.

It is as hard to read and write as it looks.

This Program:

It may seem strange that I'm entering a program that is almost more not-code than code in a contest for "shortest program", but it would be REALLY DIFFICULT to do it in less. The reason is that, when doing loops in Purple, it requires the least effort to jump to memory location 3 (because you just set i to 0), but this means you have exactly one instruction to initialize the loop. This means we need to set A to the location of the first character to be printed in a single instruction. Otherwise, we'd have to do a lot of extra work to jump somewhere else at the end of each loop. But since a starts out at zero, the only positive value we can set it to in a single instruction is 64. (i.e. the contents of the zeroth cell--the "A" itself, which is ASCII 65, minus one.)

Obviously, we're going to want to iterate backwards over the string since

  • Iterating forwards means we have to put the string AFTER position 64, thereby making the program longer.
  • It takes one fewer instruction to decrement the pointer than to increment it.

And we can shave bytes off the end of the program by decrementing the pointer before we print. In fact, we have enough space between the cell 64 and the end of the program to decrement twice between each address to be printed. Thus, the first character we need to print can be at character 62, hence, exactly 62 bytes long.

Here's The Nitty:

AA1               Set the first cell to 64
AA1AA1            This is the entry point for the loop. M[0]=M[0]-2
bA1               Point b the cell to the left of what cell 0 points to.
b1B               Set b to one more than the opposite of the character there.
o1b               Output the character M[0] pointed to (one more than the opp. of b)
b1b               Set b to the just output character.
bib               Subtract the just output character from the IP (24)
                  Until we hit the newline (ascii 10), this yields a negative.
b1B               Set b to 1 minus what b was pointing to.
                  Negative addresses are initialized to zero, so until we hit the newline
                  this will set b to 1. When we hit the newline, b will be pointing to
                  the 11th character ("1"), and this will set it to -48.
i1b               Set the instruction pointer to 1-b. 
                  Until the newline, this sets i=0, jumping back the beginning of the loop.
                  After the newline, this sets i=49, where it finds the 
                  non-instruction "W  ", and Purple halts without error in such a case.

The rest of the program is the string itself and arbitrary padding to position the characters in the right place.

EDIT: Figured out how to save 30 bytes on this program, and updated all explanations to match the new version.

\$\endgroup\$
4
\$\begingroup\$

TeaScript, 12 bytes

(using ISO/IEC 8859 character encoding)

D`HÁM, Wld!

Compresses Hello, World!, decompresses with D (æ) function

\$\endgroup\$
3
  • \$\begingroup\$ This is 12 bytes, I think; there should be an unprintable in Wl. \$\endgroup\$ Jan 8, 2016 at 2:58
  • \$\begingroup\$ @ETHproductions you're right, whoops. I guess SE kills unprintables \$\endgroup\$
    – Downgoat
    Jan 8, 2016 at 2:58
  • 2
    \$\begingroup\$ Perhaps a hexdump would be useful (via xxd or similar). \$\endgroup\$
    – primo
    Jan 8, 2016 at 4:25
4
\$\begingroup\$

BTClang, 53 bytes

My newest invention! BTClang is short for Bitcoin language. Although it has nothing to do with bitcoins, it shares some similiarities with this language. Code:

4|$&2h
2|A%
3|Im!
3|%([F
2|!4P
2|"Cv
3|zJO
1|!M
2|!&r

Explanation:

First of all, each line of the code consists of a number, a pipe and a key. The process goes as following for the example 2|5C. We take the key (5C), and generate the SHA256-hash of it. We get this:

ad5d3cc03d8b60e308b22e27fe4bbccae6a83d5496bc5e2a36aeb76eae51aeb0

The number before the pipe says how many hexadecimal number we want to extract from the end of the hash. This number is 2, so we take two 2-digit hexadecimal numbers from the end of the hash.

We are left with ae and b0. Converting these to integers will result into 174 and 176. These will be processed with the formula n % 94 + 32, so when this is converted to a character, the character will always be a printable ASCII character with 31 < ord < 128. The hashtags are replaced with newlines.

174 % 94 + 32 = 112 (p)
176 % 94 + 32 = 114 (r)

And so on...

The final translation of the code is print("Hello, World!"), which is then evaluated as normal Python. Although this is a solution, I am pretty sure this can be golfed further. It just takes a lot of computational power...

(By the way, you can try to find sets of characters yourself with the BTClang_miner)

\$\endgroup\$
3
  • \$\begingroup\$ How do you choose which hexadecimal numbers? \$\endgroup\$ Apr 8, 2016 at 0:32
  • 1
    \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ You mean from the hash? The numbers chosen are all taken from the end of the hash. So from ad5d3cc03d8b60e308b22e27fe4bbccae6a83d5496bc5e2a36aeb76eae51aeb0, the last two hexadecimal numbers are ae and b0. \$\endgroup\$
    – Adnan
    Apr 8, 2016 at 14:15
  • \$\begingroup\$ Oh, I'm an idiot. :| \$\endgroup\$ Apr 8, 2016 at 14:41
4
\$\begingroup\$

Scratch, 15 bytes

Script
(scoring used)
Makes the sprite say "Hello, World!" Can't get much simpler than that.

\$\endgroup\$
4
  • \$\begingroup\$ You already welcomed me in Output the Current Time \$\endgroup\$ May 20, 2016 at 15:25
  • \$\begingroup\$ Sorry, the review link said you were new. \$\endgroup\$ May 20, 2016 at 15:35
  • \$\begingroup\$ Is this different from codegolf.stackexchange.com/a/76182/6691 ? \$\endgroup\$
    – b_jonas
    Apr 17, 2017 at 13:31
  • \$\begingroup\$ Well, this is awkward. \$\endgroup\$ Apr 17, 2017 at 18:06
4
\$\begingroup\$

Go, 64 61 bytes

3 bytes thanks to George Gibson

package main
import."fmt"
func main(){Print("Hello, World!")}

Go requires an import to print to standard output, unfortunately. No trailing newline.

\$\endgroup\$
5
  • \$\begingroup\$ package main;fund main(){print("Hello, world!")} \$\endgroup\$ Sep 6, 2015 at 16:53
  • 1
    \$\begingroup\$ @eric_lagergren That prints to STDERR, not STDOUT. See here \$\endgroup\$
    – isaacg
    Sep 6, 2015 at 22:04
  • \$\begingroup\$ Oh gotcha. Never looked it up because I never use it. Thanks. \$\endgroup\$ Sep 6, 2015 at 22:06
  • \$\begingroup\$ Save 3 bytes by importing fmt into the global namespace with import."fmt" then just call Print("Hello, World!"). \$\endgroup\$
    – user53406
    Jun 5, 2016 at 9:45
  • \$\begingroup\$ @GeorgeGibson Thanks, that's a nice trick. \$\endgroup\$
    – isaacg
    Jun 5, 2016 at 17:24
4
\$\begingroup\$

Emojicode, 37 bytes

🏁🍇😀🔤Hello, World!🔤🍉
\$\endgroup\$
4
\$\begingroup\$

Golfuck, 39 bytes

jrseeqzjzzzsvDsj*aaa*r"s*hB(FsxahB(z*sh

Credit to primo, this is his answer, but in Golfuck.

\$\endgroup\$
4
\$\begingroup\$

Addict, 248 bytes

Addict is my new Turing-tarpit esolang, based on PRINDEAL.

a A
 i 1
 i 1
 d
a B
 A 1
 A 1
 d
a C
 B 1
 B 1
 d
a D
 C 1
 C 1
 d
a E
 D 1
 D 1
 d
E H
E H
C H
c H
E e
E e
E e
B e
i e
c e
E l
E l
E l
C l
B l
c l
c l
E o
E o
E o
D o
d o
c o
E c
C c
B c
c c
E s
c s
D H
d H
c H
c o
A o
i o
c o
c l
d e
c e
i s
c s

Test it online here!

Primer on addict

  • All memory is stored in variables. Variables can hold only non-negative integers; all variables start out at 0.
  • Addict has 4 built-in commands: decrement, increment, print a charcode, and take a charcode from input.
  • You can define your own commands with alias. This has very strict syntax:

a commandname
 command1
 command2
 command3

This creates a new command called commandname. Whenever commandname is called, the following process happens:

  • command1 is called.
  • If command1 succeeded, command2 is run.
  • If command1 failed, command3 is run.

See the GitHub repo for more information about Addict.


Act I

The first part of the program defines five commands: A, B, C, D, and E. Each one has this format:

a A
 i 1
 i 1
 d

This defines a command A which adds two to the input through the following process:

  • increment the 1st input.
  • If this succeeded, increment again. (i always succeeds unless it has no argument.)
  • Otherwise, decrement nothing. (This never gets run for the above reason.)

The next command defined is B, which adds 4 to the input:

a B
 A 1
 A 1
 d
  • Run A on the 1st input. (Always succeeds.)
  • If this succeeded, run A again. (Always gets run.)
  • Otherwise, decrement nothing. (Never gets run.)

Through the same process, C adds 8, D adds 16, and E adds 32.

Act II

The rest of the program is devoted to outputting Hello, World! in as few bytes as possible. The charcodes we need to output are 72 101 108 108 111 44 32 87 111 114 108 100 33, in that order. The shortest method I have found to output them all is to use six variables:

  • H to output 72 and 87
  • e to output 101 and 100
  • l to output 108
  • o to output 111 and 114
  • c to output 44
  • s to output 32 and 33

Here's a table of commands, and the values of the variables after each command:

Command  Output   H   e   l   o   c   s
E H              32   0   0   0   0   0
E H              64   0   0   0   0   0
C H              72   0   0   0   0   0
c H      H       72   0   0   0   0   0
E e              72  32   0   0   0   0
E e              72  64   0   0   0   0
E e              72  96   0   0   0   0
B e              72 100   0   0   0   0
i e              72 101   0   0   0   0
c e      e       72 101   0   0   0   0
E l              72 101  32   0   0   0
E l              72 101  64   0   0   0
E l              72 101  96   0   0   0
C l              72 101 104   0   0   0
B l              72 101 108   0   0   0
c l      l       72 101 108   0   0   0
c l      l       72 101 108   0   0   0
E o              72 101 108  32   0   0
E o              72 101 108  64   0   0
E o              72 101 108  96   0   0
D o              72 101 108 112   0   0
d o              72 101 108 111   0   0
c o      o       72 101 108 111   0   0
E c              72 101 108 111  32   0
C c              72 101 108 111  40   0
B c              72 101 108 111  44   0
c c      ,       72 101 108 111  44   0
E s              72 101 108 111  44  32
c s      (space) 72 101 108 111  44  32
D H              88 101 108 111  44  32
d H              87 101 108 111  44  32
c H      W       87 101 108 111  44  32
c o      o       87 101 108 111  44  32
A o              87 101 108 113  44  32
i o              87 101 108 114  44  32
c o      r       87 101 108 114  44  32
c l      l       87 101 108 114  44  32
d e              87 100 108 114  44  32
c e      d       87 100 108 114  44  32
i s              87 100 108 114  44  33
c s      !       87 100 108 114  44  33

If you can find any way to golf this program, please let me know!

\$\endgroup\$
6
  • \$\begingroup\$ Wow, this is a great language. Nice work! \$\endgroup\$ Sep 27, 2016 at 18:58
  • \$\begingroup\$ Seeing that 108, 111 and 114 are part of the output, maybe an alias to add 3 might help? \$\endgroup\$ Sep 27, 2016 at 18:58
  • \$\begingroup\$ @MartinEnder Thanks for the suggestion. An alias to add N will cost at least 17 bytes, so it'd need to save at least 5 lines (4 bytes each) to be worth it. (I originally had an alias F to add 64, but I only used it 4 times, so getting rid of it saved 1 byte.) \$\endgroup\$ Sep 27, 2016 at 19:01
  • \$\begingroup\$ @ConorO'Brien Thanks, I'm glad you like it! After spending a few months designing it, writing sample programs, and wishing I had time to code it, it took me about 4 hours to code: by far my shortest start-to-finish esolang implementation. ;) \$\endgroup\$ Sep 27, 2016 at 19:08
  • \$\begingroup\$ @Martin Using aliases for adding 2 3 6 12 24 48 seems to be about 7 bytes longer, but perhaps there's a different optimal set of aliases. I might write a brute-forcer when I have time. \$\endgroup\$ Sep 29, 2016 at 1:59
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