40
\$\begingroup\$

This challenge, while probably trivial in most "standard" languages, is addressed to those languages which are so esoteric, low-level, and/or difficult to use that are very rarely seen on this site. It should provide an interesting problem to solve, so this is your occasion to try that weird language you've read about!

The task

Take two natural numbers a and b as input, and output two other numbers: the result of the integer division a/b, and the remainder of such division (a%b).

This is : shortest answer (in bytes), for each language, wins!

Input/Output

  • 0<=a<=255, 1<=b<=255. Each of your inputs (and outputs too) will fit in a single byte.
  • You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable (e.g. no printing the two results together without a delimiter)

Examples

a,b->division,remainder
5,7->0,5
5,1->5,0
18,4->4,2
255,25->10,5

Note: Builtins that return both the result of the division and the remainder are forbidden. At least show us how your language deals with applying two functions to the same arguments.

Note 2: As always, an explanation of how your code works is very welcome, even if it looks readable to you it may not be so for someone else!


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 = 114003; // 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
  • \$\begingroup\$ Can I reverse the arguments, i.e. instead of providing a b providing b a instead? \$\endgroup\$ – Erik the Outgolfer Mar 27 '17 at 13:29
  • \$\begingroup\$ @EriktheOutgolfer: You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable \$\endgroup\$ – Emigna Mar 27 '17 at 13:30
  • \$\begingroup\$ @Emigna Yeah, I was not sure if reversing didn't make them indistinguishable though. \$\endgroup\$ – Erik the Outgolfer Mar 27 '17 at 13:32
  • \$\begingroup\$ @EriktheOutgolfer if you know that they need to be reversed you have no problem in distinguishing them :) \$\endgroup\$ – Leo Mar 27 '17 at 13:45
  • \$\begingroup\$ Unfortunately, the BF algorithm doesn't work if the divisor is 1. \$\endgroup\$ – mbomb007 Mar 27 '17 at 13:45

79 Answers 79

1 2
3
1
\$\begingroup\$

NO!, 118 bytes

It is too insane to do this challenge with NO! which is why it's non-competing (not really, the language is too young to compete in the bloodthirsty arena known as PPCG)

NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOO?yes!yess
NOOOOOOOOOOOOOOOOO?yes!yess
NOOOOOOOO?nooo!noooo

Please say you don't want an explanation (one added anyway)

NOOOOOOOOOOOO?                   Create an integer from 
              NOOOOOOOOOOO                              STDIN

NOOOOOOOOOOOO?                   Create an integer from 
              NOOOOOOOOOOO                              STDIN

NOOOOO?                          Divide               by 
       yes!                             line 1 result
           yess                                          line 2 result

NOOOOOOOOOOOOOOOOO?              Modulo               by
                   yes!                 line 1 result
                       yess                              line 2 result

NOOOOOOOO?                       Output               and
          nooo!                         line 3 result
               noooo                                      line 4 result

A yes is used to denote a line number (number determined by number of ss) but the output command takes either numbers or line numbers as arguments and noo!nooo is shorter than yess!yesss

\$\endgroup\$
1
\$\begingroup\$

Triangular, 21 19 bÿtes

$\:A@$U.vS/p%_<%mp<

Prints the result of division, a newline, then the remainder. Formats into this triangle:

     $
    \ :
   A @ $
  U . v S
 / p % _ <
% m p < ÿ ÿ

ÿ is the no-op that is automaticallÿ inserted when there is no code to fill the smallest possible triangle.

Trÿ it online!

Explanation:

I don't know whÿ \ is the first character... Removed it and managed to save two bÿtes. Still don't know whÿ it was there in the first place. ಠ_ಠ

Directionals:

     .
    \ .
   . . .
  . . v .
 / . . . <
. . . <

This is how the interpreter sees the code, without directionals:

$:$S_%pUA@pm%
  • $: reads integer input to the stack and duplicates it. Stack contains input1 input1.
  • $S reads another input to the stack, then stashes it in memorÿ. Stack contains input1 input1 input2, and memorÿ contains input2.
  • _ divides the top two stack values. Now the stack contains input1 input1/input2.
  • % prints the top of stack as an integer.
  • p pops the top of stack. Now the stack contains input1.
  • U pulls memorÿ onto the stack. Now the stack contains input1 input2.
  • A pushes 10 to the stack. Stack contains input1 input2 10.
  • @ prints it as ASCII (newline).
  • p pops it - stack contains input1 input2.
  • m performs modulus on the top two stack values. Stack now contains input1%input2.
  • % prints the top of stack as an integer.
  • the IP then runs off the plaÿing field, and the program terminates.
\$\endgroup\$
1
\$\begingroup\$

Implicit, 9 bytes

$$|/%;_@9

Try it online!

$$|/%;_@9
$$         read two integer inputs
  |        duplicate stack
   /       division
    %      print
     ;     pop
      _    modulus
       @9  print HTAB (delimiter)
\$\endgroup\$
1
\$\begingroup\$

Whispers v2, 77 bytes

> Input
> Input
>> 1÷2
>> ⌊3⌋
>> 1%2
>> Output 4
>> Output 5
>> Then 6 7

Try it online!

\$\endgroup\$
0
1
\$\begingroup\$

Keg, 12 10 bytes

Other than the creator, I think only I am actively using it. Therefore it is rarely seen on this site; in addition it is esoteric.

Takes 2 inputs separated with newlines.

¿¿:^:"$/"%

Explanation

¿¿#          Takes 2 integer inputs: e.g.     stack [1, 2]
  :#         Duplicate the top item.          stack [1, 2, 2]
   ^#        Reverse the stack.               stack [2, 2, 1]
    :#       Duplicate the stack.             stack [2, 2, 1, 1]
     "#      Right shifts (put top to bottom) stack [1, 2, 2, 1]
      $#     Swap top two items               stack [1, 2, 1, 2]
       /#    Divide top two                   stack [1, 2, 0.5]
         "#  Right shifts                     stack [0.5, 1, 2]
          %# Modulos top two items            stack [0.5, 1]

# Stack is implicitly outputted

TIO

\$\endgroup\$
1
\$\begingroup\$

33, 11 bytes

OcOcsdoilro

Try it online!

Explanation

OcOc        | Harvest input from user
    s       | Store "a" for later use
     doi    | Print the integer division followed by a newline
        l   | Get "a" back
         ro | Print the remainder of the division
\$\endgroup\$
1
\$\begingroup\$

Vyxal, 3 bytes

₍ḭ%

Try it Online!

Try a test suite!

The header and footer are for formatting purposes. Otherwise takes arguments in reversed order.

Explained

₍ḭ%
₍   # parallel apply the next two commands and collect into a list:
 ḭ% # integer divide and modulo
\$\endgroup\$
1
  • \$\begingroup\$ What is the input format, if I just want to use it without the test suite? \$\endgroup\$ – Deadcode Apr 6 at 3:01
0
\$\begingroup\$

Wonder, 13 bytes

@@[/#1#0%#1#0

Usage:

((@@[/#1#0%#1#0)10)5

Simply a curried lambda that returns a list with both the division and modulus. I'm looking for a golfier way to do this, something that doesn't involve lambdas.

Be sure to do tK1000 beforehand to view list items in the output.

\$\endgroup\$
0
\$\begingroup\$

PHP, 35 Bytes

[,$a,$b]=$argv;echo$a/$b^0,_,$a%$b;
\$\endgroup\$
0
\$\begingroup\$

Forth, 25 bytes

: f 2dup / -rot mod ;

Try it online

Input like a b, output like remainder quotient.

\$\endgroup\$
0
\$\begingroup\$

Elixir, 24 bytes

&{div(&1,&2),rem(&1,&2)}

Anonymous function which uses the capture operator and returns a tuple containing the results.

Full program with test cases (and yes, the . in the function call is required!):

f =
&{div(&1,&2),rem(&1,&2)}

# test cases:
IO.inspect f.(5,7)      # 0,5
IO.inspect f.(5,1)      # 5,0
IO.inspect f.(18,4)     # 4,2
IO.inspect f.(255,25)   # 10,5

Try it online on Elixir Playground.

\$\endgroup\$
0
\$\begingroup\$

CJam, 7 bytes

q_~/n~%

Try it online!

n = print with trailing newline

-2 thanks to Martin Ender.
-3 thanks to Peter Taylor and Basic Sunset.I was thinking eval first then the rest.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ You can save a few bytes by duplicating the input string before evaling it: q_~/n~% \$\endgroup\$ – Business Cat Mar 27 '17 at 14:12
  • \$\begingroup\$ It's worth noting that without the rule against divmod, this would be q~mdp (5 bytes): Try it online! \$\endgroup\$ – Deadcode Apr 5 at 23:30
0
\$\begingroup\$

Math++, 18 bytes

?>a
?>b
_(a/b)
a%b
\$\endgroup\$
0
\$\begingroup\$

ForceLang, 106 bytes

def N io.readnum()
def S set
S W gui.show
S a N
S b N
W S c math.floor a.mult b.pow -1
S c c.mult b
W a+-c
\$\endgroup\$
0
\$\begingroup\$

Lua, 40 Bytes

a=arg;io.write(a[1]//a[2],",",a[1]%a[2])
\$\endgroup\$
1
  • \$\begingroup\$ You can any format you like for output, as long as the two numbers are distinguishable. a=arg;print(a[1]//a[2],a[1]%a[2]) is 33 bytes: Try it online! \$\endgroup\$ – Deadcode Apr 5 at 23:37
0
\$\begingroup\$

Pyth, 7 bytes

/QKE%QK

Online interpreter

\$\endgroup\$
1
  • \$\begingroup\$ It's worth noting that without the rule against divmod, this would be .DE (3 bytes): Try it online! \$\endgroup\$ – Deadcode Apr 5 at 23:36
0
\$\begingroup\$

Sinclair ZX80 16-bit Integer BASIC (4K ROM), ~152 BASIC bytes used (listing)

 1 LET R=0
 2 PRINT "ENTER A NUMBER THEN DIVISOR"
 3 INPUT Z
 4 INPUT D
 5 PRINT Z;"/";D;"=";
 6 GO SUB 9
 7 PRINT R;" REMAINDER ";Z
 8 STOP
 9 FOR I=1 TO 0
10 LET Z=Z-D
11 LET R=R+1
12 LET I=(Z>D)+(Z=D)
13 NEXT I
14 RETURN

Some notes:

Because the ZX80 (with the old ROM) can only handle 16-bit signed integer numbers, your range is -32768 to +32767 for your inputs. There is no native modulo function, nor does there seem to be a >= (if there is then I can't find it) - hence why this had to be split in line 12.

I haven't handled the division by zero in the program, but I guess that it entering 0 as your divisor will keep the sub-routine from line 9 in an infinite loop (as it should), except for the 16-bit integer range, so Z/0 should be Z r ∞.

The byte count is only an approximation for now.

\$\endgroup\$
0
\$\begingroup\$

Taxi, 1189 bytes

"," is waiting at Writer's Depot.Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.Pickup a passenger going to What's The Difference.Pickup a passenger going to Divide and Conquer.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:n 2 r 2 r 1 r.Pickup a passenger going to Trunkers.Go to Trunkers:e 1 r 3 r 1 l.Pickup a passenger going to Cyclone.Go to Cyclone:w 2 r.Pickup a passenger going to Multiplication Station.Pickup a passenger going to Multiplication Station.Go to Multiplication Station:s 1 l 2 r 4 l.Pickup a passenger going to What's The Difference.Go to Cyclone:s 1 r 2 l 2 r.Pickup a passenger going to The Babelfishery.Go to What's The Difference:n 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:e 3 r.Pickup a passenger going to Post Office.Go to Writer's Depot:n 1 l 1 l 2 l.Pickup a passenger going to Post Office.Go to The Babelfishery:n 1 r 2 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

Try it online!


With line breaks:

"," is waiting at Writer's Depot.
Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Cyclone.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l 1 l 2 r.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to Divide and Conquer.
Pickup a passenger going to Divide and Conquer.
Go to Divide and Conquer:n 2 r 2 r 1 r.
Pickup a passenger going to Trunkers.
Go to Trunkers:e 1 r 3 r 1 l.
Pickup a passenger going to Cyclone.
Go to Cyclone:w 2 r.
Pickup a passenger going to Multiplication Station.
Pickup a passenger going to Multiplication Station.
Go to Multiplication Station:s 1 l 2 r 4 l.
Pickup a passenger going to What's The Difference.
Go to Cyclone:s 1 r 2 l 2 r.
Pickup a passenger going to The Babelfishery.
Go to What's The Difference:n 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:e 3 r.
Pickup a passenger going to Post Office.
Go to Writer's Depot:n 1 l 1 l 2 l.
Pickup a passenger going to Post Office.
Go to The Babelfishery:n 1 r 2 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.
\$\endgroup\$
0
\$\begingroup\$

W r, 3 bytes

/@m

Explanation

abab    % Implicit ops
    /   % Divide a by b.                Stack: a b (a/b)
     @  % Roll down to show 2 operands. Stack: (a/b) a b
      m % Modulo the operands.          Stack: (a/b) (a%b)

% Output the whole stack
```
\$\endgroup\$
1 2
3

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