35
\$\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!


Leaderboard

Here is a Stack Snippet to generate an overview of winners by language.

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 leaderboard snippet:

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

var QUESTION_ID=114003,OVERRIDE_USER=62393;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={};e.forEach(function(e){var o=e.language;/<a/.test(o)&&(o=jQuery(o).text().toLowerCase()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link,uniq:o}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.uniq>s.uniq?1:e.uniq<s.uniq?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#language-list{padding:10px;width:290px;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="//cdn.sstatic.net/codegolf/all.css?v=617d0685f6f3"> <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><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>

\$\endgroup\$
  • \$\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

71 Answers 71

23
\$\begingroup\$

BitCycle, 146 79 64 bytes

Just realized a whole section of my original code was unneccessary. Huge reduction!

v  <>!
A\B^^=
? D^>^<
 >\v^~ D@
 >/ C/
  > C ^
A/B v
^   <
?  D^

The program takes input in unary from the command line, with the divisor first. It outputs the quotient and remainder in unary, separated by a 0. For example, here's a=11, b=4, a/b=2, a%b=3:

C:\>python bitcycle.py divmod.btc 1111 11111111111
110111

Ungolfed, in action

Here's my ungolfed version computing a=3, b=5 with animation turned on (sorry about the glitchiness):

Divmod running in BitCycle

Attempt at an explanation

The explanation applies to the ungolfed version. Before you tackle it, I highly recommend you read the Esolangs page to get a feel for how the language works.

The algorithm goes like this:

  • Run an outer loop until the program is terminated.
    • Run an inner loop over the bits of the divisor, pairing them off with bits from the dividend.
      • If all bits of the divisor have matching dividend bits, output a single bit.
      • If not all bits of the divisor have matching dividend bits, output the separator 0 followed by what dividend bits there were, then terminate.

The heart of the code is the relationships among the collectors (the uppercase letters). Since there are multiple separate collectors with each letter, let's refer to them as A1, A2, B1, B2, etc., numbering from top to bottom.

  • A1 and A2 hold the divisor and dividend, respectively, at the beginning of the main loop.
  • The inner loop peels off one bit at a time from the divisor and the dividend.
    • The rest of the divisor, if any, always goes into B1.
    • If both the divisor and dividend were nonempty, one bit goes into C1 and one into C3. The rest of the dividend goes into B2.
    • If only the divisor was nonempty, we've reached the end of the dividend, and it's time to print the remainder. The bit from the divisor goes into C2.
    • If only the dividend was nonempty, we've reached the end of the divisor; it's time to process the bits in C3 or C2 for output. The rest of the dividend goes into C4.
  • If there are any bits in the B collectors, they cycle their contents back around to the A collectors and continue in the inner loop.
  • Once the A and B collectors are all empty, the C collectors open and we proceed to the processing stage:
    • C1 and C4 dump their contents (the divisor and the remaining dividend, respectively) into D1 and D3.
    • If C2 is empty, we're still printing the quotient.
      • The contents of C3 go up to the top right = switch. The first 1 bit passes straight through to ! and is output.
      • When the 1 bit passes through, it activates the switch to point rightward, which sends all the subsequent bits off the board.
    • If C2 is not empty, we're printing the remainder.
      • The first bit of C2 is negated to a 0 and passed through the switch. The 0 goes on to ! and is output.
      • When the 0 bit passes through, it activates the switch to point leftward. Now all the bits from C3 go leftward from the switch and are redirected around into the !, outputting the entire remainder.
      • A copy of the first bit from C2 is also sent into D2.
  • Now the D collectors open.
    • If there is anything in D2, that means we just printed the remainder. The bit from D2 hits the @, which terminates the program.
    • Otherwise, the contents of D1 and D3 loop back into A1 and A2 respectively, and the main loop starts over.
\$\endgroup\$
  • \$\begingroup\$ that is awesome \$\endgroup\$ – Evan Carslake Mar 30 '17 at 17:33
  • \$\begingroup\$ "The program takes input in unary from the command line": That looks like binary to me? \$\endgroup\$ – therealfarfetchd Mar 30 '17 at 18:36
  • \$\begingroup\$ Whoops. Because the output looked like binary, I though the input should be too. Then I read the text. Nevermind. :P \$\endgroup\$ – therealfarfetchd Mar 31 '17 at 15:37
15
\$\begingroup\$

brainfuck, 43 41 bytes

,<,[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]>>.>>>.

This uses a modified version of my destructive modulus algorithm on Esolangs.

The program reads two bytes – d and n, in that order – from STDIN and prints two bytes – n%d and n/d, in that order – to STDOUT. It requires a brainfuck interpreter with a doubly infinite or circular tape, such as the one on TIO.

Try it online!

How it works

Before the program starts, all cells hold the value 0. After reading d from STDIN (,), moving one step left (<) and reading n from STDIN (,), the tape looks as follows.

       v
A      B      C      D      E      F      G      H      J
0      n      d      0      0      0      0      0      0

Next, assuming that n > 0, we enter the while loop

[>->+<[>]>>>>+<<<[<+>-]<<[<]>-]

which transforms the tape as follows.

First, >->+< advances to cell C and decrements it, then advances to cell D and increments it, and finally goes back to cell C. What happens next depends on whether the value of cell C is zero or not.

  • If cell C hold a positive value, [>] (go right while the cell is non-zero) will advance to cell E.

    >>>>+<<< advances to cell J to increment it, then goes back to cell F.

    Since cell F will always hold 0, the while loop [<+>-] is skipped entirely, and << goes back to cell D.

    Finally, since neither D nor C hold 0, [<] (go left while the cell is non-zero) will retrocede to cell A.

  • If cell C holds 0, the loop [>] is skipped entirely; >>>>+<<< advances to cell G to increment it, then goes back to cell D.

    At this point, D will hold d (in fact, the sum of the values in C and D will always be d), so [<+>-] (while D is positive, increment C and decrement D) will set C to d and D to 0.

    Finally, << retrocedes to cell B, [<] (go left while the cell is non-zero) further left to cell A.

In both cases, >- advances to cell B and decrements it, and the loop starts over unless this zeroes it out.

After k iterations, the tape looks as follows.

       v
A      B      C      D      E      F      G      H      J
0      n-k    d-k%d  k%d    0      0      k/d    0      k-k/d

After n iterations B is zeroed out and we break out of the loop. The desired values (n%d and n/d) will be stored in cells D and G, so >>.>>>. prints them.

\$\endgroup\$
12
\$\begingroup\$

Funciton, 224 108 bytes

Byte count assumes UTF-16 encoding with BOM.

 ┌──┬───┐
┌┴╖╓┴╖ ┌┴╖
│%╟║f╟┐│÷╟┘
╘╤╝╙─╜│╘╤╝
 └────┴─┘

Try it online!

The above defines a function f, which takes two integers and returns both their division and their product (functions in Funciton can have multiple outputs as long as the sum of inputs and outputs doesn't exceed 4).

Using two input values for multiple purposes is actually quite trivial: you simply split off the connector with a T-junction at the value will be duplicated along both branches, which we can then feed separately to the built-ins for division and modulo.

It actually took me twice as long to figure out how to display the result to the user than just to implement the solution.

Also, Funciton has a built-in divmod, ÷%, and amusingly the built-ins ÷ and % that my solution uses are implemented in terms of ÷%. However, my function f above isn't quite identical to ÷%: I had to swap the order of the inputs and although it seems like it should be easy to change that, so far I haven't been able to do so without increasing the byte count.

\$\endgroup\$
10
\$\begingroup\$

JavaScript (ES6), 17 bytes

Thanks to @Arnauld for golfing off one byte

x=>y=>[x/y|0,x%y]

Receives input in format (x)(y)

Gets floor of x/y by performing bitwise or
Gets remainder by x%y
Puts both values in an array so that they can both be returned

Try it online!

\$\endgroup\$
10
\$\begingroup\$

APL (Dyalog), 5 bytes

-2 bytes thanks to @ngn

⌊÷,|⍨

This is an atop (2-train) of a fork (3-train), where the atop's right tine is a derived function (the result of an operator applied to a function):

       result 
         ↑┌──────────┐
         ││    ┌────┐│┌──────┐ (derived function)
         │↓    │    ↓│↓      │╱
       ┌───┐ ┌───┐ ┌───┐ ╔═══╤═══╗
       │ ⌊ │ │ ÷ │ │ , │ ║ | │ ⍨ ║ 
       └───┘ └───┘ └───┘ ╚═══╧═══╝
              ↑ ↑         ↑     ↑ ╲
left argument ┴─────────────────┘  (operator)
                └─────────┴ right argument

 floor of

÷ division

, catenated to

| division remainder
 with swapped arguments (APL modulus is "backwards")

Try it online!

\$\endgroup\$
  • \$\begingroup\$ How did you make that cool diagram? \$\endgroup\$ – emiflake Mar 27 '17 at 20:19
  • 2
    \$\begingroup\$ @WolfgangTS Painstakingly. Dyalog APL comes with ability to make basic tree diagrams of tacit functions. Try it online! I started with that... \$\endgroup\$ – Adám Mar 27 '17 at 20:23
  • \$\begingroup\$ Ouch, looks very difficult. I don't have the patience for that I'm afraid, haha \$\endgroup\$ – emiflake Mar 27 '17 at 21:48
  • \$\begingroup\$ shorter: ⌊÷,|⍨ \$\endgroup\$ – ngn Mar 28 '17 at 11:05
  • \$\begingroup\$ @ngn Ouch, you got me. Happy to see that you're still here. \$\endgroup\$ – Adám Mar 28 '17 at 11:05
7
\$\begingroup\$

Brachylog, 6 bytes

{÷|%}ᶠ

Try it online!

Explanation

We abuse the metapredicate ᶠ findall to apply two different predicates to the Input list of two arguments:

{   }ᶠ         Findall for the Input [A,B] :
 ÷               Integer division
  |              Or…
   %             Modulo
\$\endgroup\$
7
\$\begingroup\$

MATL, 12 10 bytes

Qt:ie=&fhq

Input is a, then b. Output is remainder, then quotient.

Try it online!

Explanation

This avoids both modulo and division. Instead it uses array reshaping:

  1. Build an array of a+1 nonzero elements.
  2. Reshape as a 2D array of b rows. This automatically pads with zeros if needed.
  3. The row and column indices of the last nonzero entry, minus 1, are respectively the remainder and quotient.

Consider for example a=7, b=3.

Q    % Input a implicitly. Push a+1
     % STACK: 8
t:   % Duplicate. Range from 1 to that
     % STACK: 8, [1 2 3 4 5 6 7 8]
ie   % Input b. Reshape as a matrix with b rows (in column major order)
     % STACK: 8, [1 4 7;
                  2 5 8]
                  3 6 0]
=    % Compare for equality
     % STACK: [0 0 0;
               0 0 1;
               0 0 0]
&f   % Row and column indices (1-based) of nonzero element
     % STACK: 2, 3
hq   % Concatenate. Subtract 1. Implicitly display
     % STACK: [1 2]
\$\endgroup\$
  • 2
    \$\begingroup\$ This is a fine example of lateral thinking, nice work! \$\endgroup\$ – Leo Mar 27 '17 at 14:34
6
\$\begingroup\$

Mathematica, 20 18 bytes

⌊#/#2⌋@Mod@##&

Minor abuse of the flexible output rules: the result is given as div[mod], which will remain unevaluated. The individual numbers can be extracted with result[[0]] and result[[1]].

And hey, it's only one byte longer than the ridiculously named built-in QuotientRemainder.

Mathematica, actually has a neat way to apply multiple functions to the same input, but it's three bytes longer:

Through@*{Quotient,Mod}
\$\endgroup\$
  • 1
    \$\begingroup\$ You know it's bad when you language creates built-ins that simply combines built-ins… \$\endgroup\$ – Fatalize Mar 27 '17 at 13:08
  • 1
    \$\begingroup\$ @Fatalize Is it? I find divmod built-ins quite useful, and Mathematica is by far not the only language to have one. \$\endgroup\$ – Martin Ender Mar 27 '17 at 13:11
  • 8
    \$\begingroup\$ @Fatalize, a lot of the same work is required to compute quotients as is required to compute remainders. If both results are to be used, a properly-engineered quotRem builtin can save significant time over calling quot and rem separately. \$\endgroup\$ – Julian Wolf Mar 27 '17 at 14:26
6
\$\begingroup\$

05AB1E, 5 bytes

÷²¹%‚

Try it online!

05AB1E has a bug, so implicit input doesn't work :( Emigna noted that inputs are often pushed in reverse.

\$\endgroup\$
  • \$\begingroup\$ You could do ÷²¹%) for 5 bytes. \$\endgroup\$ – Emigna Mar 27 '17 at 13:23
  • \$\begingroup\$ @Emigna I dunno if it's valid though. Wait, how did that work? \$\endgroup\$ – Erik the Outgolfer Mar 27 '17 at 13:26
  • 1
    \$\begingroup\$ I don't see why it wouldn't be valid. It works because implicit inputs are pushed to the stack in the reverse order of what you'd assume in cases like this. \$\endgroup\$ – Emigna Mar 27 '17 at 13:27
  • \$\begingroup\$ @Emigna I have asked OP if I can reverse the arguments. \$\endgroup\$ – Erik the Outgolfer Mar 27 '17 at 13:31
  • 2
    \$\begingroup\$ I take You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable to mean that you can decide that the inputs are taken as divisor, dividend. You could just specify "Input are taken as divisor, dividend" in the answer and they'll be clearly distinguishable :) \$\endgroup\$ – Emigna Mar 27 '17 at 13:34
6
\$\begingroup\$

Jellyfish, 14 bytes

p
m
,|S
% i
Ei

Try it online!

Explanation

Jellyfish is a beautiful language when it comes to applying multiple functions to the same input. The language is 2D and all binary functions look south for one input and east for another. So by approaching one value from the west and from the north, we can feed it to two functions without having to duplicate it in the code.

The two is in the program are replaced with the two input values when the program starts. Now % is division. It takes one input directly from the east, and when going south it hits the E which redirects that search east as well. So both inputs get fed to % as arguments.

| is the built-in for modulo, which basically does the same thing, but ends up looking south for both in puts.

We concatenate both results into a pair with ,. Then m is the floor function (which we need because % is floating-point division) and finally we print the result with p.

\$\endgroup\$
6
\$\begingroup\$

Cubix, 12 13 bytes

;W@o,I|\S%;O

Which maps onto the following cube

    ; W
    @ o
, I | \ S % ; O
. . . . . . . .
    . .
    . .

Try it here

Explanation with steps as executed
,I|I, - starts with an superflous integer divide, gets the first integer from input, reflects back and gets the next integer from input, then divides again
O; - Output the result of the integer division and pop it
% - do the mod. This could be done later, but ended up here
S\o - Add space character to stack, redirect up and output space
W; - Shift left and pop the space from the stack
O|@ - Output the mod previously calculated, pass through the horizontal reflector and halt.

\$\endgroup\$
  • \$\begingroup\$ Beat me by two minutes. Nice answer! \$\endgroup\$ – Luke Mar 27 '17 at 19:17
  • \$\begingroup\$ @Luke Thanks, thought I could get another one off, but proving elusive \$\endgroup\$ – MickyT Mar 27 '17 at 19:22
6
\$\begingroup\$

Brain-Flak, 56 54 bytes

({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

Try it online!

-2 bytes thanks to Wheat Wizard

Explanation

The current best known integer division and modulo in Brain-Flak are very similar (in fact the currently used integer division is just a modification I made on feersum's modulo).

Comparison of modulo and integer division:
Modulo:   ({}(<>))<>     {   (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><  {}   {} >)

Conveniently, the integer division program uses only the third stack for storing data while the modulo program uses only the normal two stacks for storing data. Thus by simply running them both at the same time they do not collide at each other.

Combination of modulo and integer division:
Modulo:   ({}(<>))<>     {   (({})){({}[()])<>}{} }{}<> ([{}()]{})
Division: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><  {}   {} >)

Combined: ({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><([{}()]{})>)

Finally, both the integer division and modulo programs used in this combination were designed to be stack clean (not leave garbage on the stacks/not depend on the (non)existence of values on the stacks other than their input) but that is not necessary for this problem. Thus we can save two bytes by not bothering to pop the zero at the end of the main loop and another two bytes by not pushing zero at the start, instead relying on the zero padding on the bottom of the stacks.

This gives us the final program:
({}<>)<>([()]{()<(({})){({}[()])<>}{}>}<><([{}()]{})>)

For the explanation for the integer division program see feersum's answer

Integer Division Explanation Coming Soon...

\$\endgroup\$
5
\$\begingroup\$

Java 8, 18 Bytes

(a,b)->a/b+","+a%b

This is a lambda expression of the type BiFunction<Integer, Integer, String>.

I'm surprised... this is actually a fairly concise solution for Java. Go lambda expressions!

\$\endgroup\$
5
\$\begingroup\$

Brain-Flak, 168 148 110 bytes

I guess I should have checked the Wiki first

(({})(<({}(<(({})<>)>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>)

Format:

Input:    Output:
A (18)    remainder (2)
B (4)     division  (4)

Try it online!

(({})(<           # Copy A
({}(<             # Pick up A
(({})<>)          # Copy B to the other stack
>))               # Put A on top of a 0 on the second stack
                  # At this point the stacks look like this:   A
                                                               0
                                                             B B
                                                               ^

<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>) # Positive division from the wiki
>))                                          # Put down A on top of a 0
                                             # The stack now: A
                                                              0
                                                            Div B
                                                              ^

<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<>)     # Modulo from the wiki
\$\endgroup\$
5
\$\begingroup\$

sed, 36 bytes

35 bytes of code, +1 for the -r flag.

:a;s/^(1+)( 1*)\1/\1\2x/;ta;s/.* //

Takes input in unary, space-separated, with the smaller number first. Outputs as unary, with the quotient first in 1s and the remainder second in xs. (If this isn't acceptable, let me know and I'll change it to space-separated 1s like the input.)

Explanation

:a;                                  Define label a
   s/            /     /;            Perform this substitution:
     ^(1+)                           Match the first unary number...
          ( 1*)                      ... followed by a space and 0 or more 1s...
               \1                    ... followed by the the first group again
                  \1\2x              Keep the first two parts unchanged; replace the third
                                     with an x
                         ta;         If the substitution succeeded, goto a
                            s/.* //  After the loop is over, remove the first number
\$\endgroup\$
5
\$\begingroup\$

Excel 2013, 31 30 26 bytes

=INT(A1/B1)&","&MOD(A1;B1)

Explanation

Input is in cell A1 and B1. This simply returns the return values of the FLOOR and MOD function, which are for flooring the division and for the remainder. These values are separated by a comma.

\$\endgroup\$
  • \$\begingroup\$ I think you mean cell A1 and B1 not A1 and A2 \$\endgroup\$ – fəˈnɛtɪk Mar 27 '17 at 13:57
  • \$\begingroup\$ Yes, thanks. Fixed now \$\endgroup\$ – Luke Mar 27 '17 at 14:09
  • \$\begingroup\$ Save 1 byte with FLOOR(A1/B1;1) instead of QUOTIENT(A1;B1) \$\endgroup\$ – Engineer Toast Mar 30 '17 at 12:36
  • \$\begingroup\$ Because the input is always a Natural number, I think you can replace FLOOR(A1/B1;1) with `INT(A1/B1)' to save 4 more bytes \$\endgroup\$ – Wernisch Sep 20 '17 at 13:39
4
\$\begingroup\$

Python 2, 20 bytes

lambda a,b:(a/b,a%b)

Try it online!

Built-in, 6 bytes

divmod

Try it online!

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4
\$\begingroup\$

Jelly, 3 bytes

:,%

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Darn, I was just going to post this, I tested it a few extra times and got ninja'd :( good job! \$\endgroup\$ – HyperNeutrino Mar 27 '17 at 13:06
4
\$\begingroup\$

OIL, 134 106 103 102 bytes

Takes the input from stdin, the two numbers seperated by a newline. Outputs the result of the integer division, then a newline, and then the remainder.

This is one of the most complicated OIL programs I've ever written, as OIL lacks builtins for division, remainder, addition, substraction, and so on. It works with the primitive way of doing division: repeated nested decrementation.

I present the code in an annotated format, with comments in the style of scripting languages. Before executing, the comments have to be removed.

5  # read input into lines 0 and 2

5
2
0  # the zero to compare to (nops)
1  # make a backup of the second input at line 3
2
3
10 # check if the second input is 0. %
4
2
24 # if so, jump to 24 (marked with §)
13 # else, go on
10 # check if the first input is zero &
4

31 # if so, jump to 31 (marked with $)
18 # else, go on
9  # decrement both numbers

9
2
6  # jump to line 8 (marked with %)
8
8  # increment the value in line 1 (initially a zero) §
1
1  # "restore the backup"
3
2
6  # jump to line 13 (marked with &)
13
10 # is the second number zero? $
4
2
42 # if so, jump to 42 (marked with +)
36 # else go on
9  # decrement both the second number and the backup
2
9
3
6  # jump to 31 (marked with $)
31
4  # print the division +
1
11 # a newline
4
3  # and the remainder (from the backup)

edit: Shaved off 3 more bytes by moving a "constant" to a one-digit location (less bytes to reference), and then implicit-ing 2 zero-locations (By using an empty line instead. One of them I could have done before).

edit: And another byte by making the initial zero implicit. We really only need a single literal zero.

\$\endgroup\$
  • \$\begingroup\$ Great work! That's exactly the kind of answer I hoped this challenge would receive :) Just a note: you are guaranteed that the divisor will always be strictly positive, so you don't need to check for a division by 0 ;) \$\endgroup\$ – Leo Mar 27 '17 at 14:12
  • \$\begingroup\$ @Leo I'm guaranteed that the divisor will always be strictly positive at the beginning. It won't work if I take the division by zero part out, this case can happen even when the "actual" division is a normal one. If I remember correctly this occurs when the remainder is zero. \$\endgroup\$ – L3viathan Mar 27 '17 at 14:38
  • \$\begingroup\$ I'm talking about the check on line 4, not the one on line 12... Doesn't it execute just once at the start of the program? \$\endgroup\$ – Leo Mar 27 '17 at 14:41
  • \$\begingroup\$ @Leo Done, almost 30 characters less, thanks! \$\endgroup\$ – L3viathan Mar 27 '17 at 15:02
4
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Retina, 14 bytes

Let's abuse the input/output formats!

(.*)¶(\1)*
$#2

Takes input as b\na, in unary, using for unary digit any single non-digit, non-newline character. Outputs the quotient in decimal, immediately followed by the remainder in unary, using the same character as the input.

Try it online!

(.*) ¶(\1)* matches the first number , then a newline (¶ is Retina's shorthand for \n), then the first number again as many times as possible. The number of matches of the second group will be the result of the division, and the part not matched will be the remainder.

With $#2, we replace everything that was matched in the previous line with the number of captures of the second group, and get then our result.

\$\endgroup\$
  • \$\begingroup\$ Haha, quite right, I clearly shouldn't be writing programs that late in the evening. \$\endgroup\$ – FryAmTheEggman Mar 30 '17 at 18:37
4
\$\begingroup\$

ArnoldC, 286 283 bytes

HEY CHRISTMAS TREE c
YOU SET US UP 0
HEY CHRISTMAS TREE d
YOU SET US UP 0 
GET TO THE CHOPPER c
HERE IS MY INVITATION a
HE HAD TO SPLIT b
ENOUGH TALK
GET TO THE CHOPPER d
HERE IS MY INVITATION a
I LET HIM GO b
ENOUGH TALK
TALK TO THE HAND c
TALK TO THE HAND d
YOU HAVE BEEN TERMINATED

Try it online!

How It Works

HEY CHRISTMAS TREE c      //DECLARE VARIABLE c = 0
YOU SET US UP 0
HEY CHRISTMAS TREE d      //DECLARE VARIABLE d = 0
YOU SET US UP 0

GET TO THE CHOPPER c      /*
HERE IS MY INVITATION a      SET c = a/b
HE HAD TO SPLIT b         
ENOUGH TALK                */

GET TO THE CHOPPER d      /*
HERE IS MY INVITATION a      SET d = a mod b
I LET HIM GO b
ENOUGH TALK                */

TALK TO THE HAND c        // PRINT c
TALK TO THE HAND d        // PRINT d
YOU HAVE BEEN TERMINATED  //END

Output Format

a/b
a mod b
\$\endgroup\$
3
\$\begingroup\$

Labyrinth, 11 bytes

?:?:}/!\{%!

Try it online!

Explanation

?:   Read a and duplicate.
?:   Read b and duplicate.
}    Move a copy of b over to the auxiliary stage.
/    Compute a/b.
!    Print it.
\    Print a linefeed.
{    Get back the other copy of b.
%    Compute a%b.
!    Print it.

The IP then hits a dead end, turns around and the program terminates due to the attempted division by zero when % is executed again.

\$\endgroup\$
3
\$\begingroup\$

C, 32 bytes

f(a,b){printf("%d %d",a/b,a%b);}

Try it online!

\$\endgroup\$
3
\$\begingroup\$

><>, 27 26 16 + 1 = 17 bytes

:r:{%:n','o-$,n;

Note

  • Input using the -v flag, see TIO for an example.
  • This outputs the remainder first, then a comma and lastly the integer division.

Try it online!

Explanation

Note that the stack starts as A, B, where A and B represent the first and second input, because of the -v flag used.

:r:{%:n','o-$,n; # Explanation
:r:{             # Do some stack modifications to prepare it for
                 #    next part
                 #    (Stack: B, A, A, B)
    %            # Take the modulo of the top two items
                 #    (Stack: B, A, A%B)
     :           # Duplicate it
                 #    (Stack: B, A, A%B, A%B)
      n          # Pop once and output as number
                 #    (Stack: B, A, A%B)
       ','o      # Print separator
                 #    (Stack: B, A, A%B)
           -     # Subtract the modulo from the first input
                 #    (Stack: B, A-A%B)
            $,   # Swap the two inputs so they are back in order
                 #     and divide, so we get an integer
                 #    (Stack: floor(A/B))
              n; # Output that as a number and finish.
\$\endgroup\$
  • \$\begingroup\$ How can you provide input values up to 255? \$\endgroup\$ – Leo Mar 27 '17 at 14:43
  • \$\begingroup\$ Just use higher ASCII/Unicode values. That way, į becomes 255. \$\endgroup\$ – Luke Mar 27 '17 at 14:46
  • \$\begingroup\$ Ok, nice :) By the way, wouldn't it be shorter to take input numbers from command line directly with the -v flag? \$\endgroup\$ – Leo Mar 27 '17 at 15:18
  • \$\begingroup\$ It would, but I couldn't get that to work on TIO, so I settled with this solution. It would save 8 bytes - 1 (for the -v flag). \$\endgroup\$ – Luke Mar 27 '17 at 15:23
  • \$\begingroup\$ Here you are :) \$\endgroup\$ – Leo Mar 27 '17 at 15:42
3
\$\begingroup\$

C, 21 bytes

#define f(a,b)a/b,a%b

A macro that replaces f(a,b) with the 2 terms comma separated. Though you'd better be passing it to a function or else there's no way to pick the 2 apart.

Try It Online

\$\endgroup\$
3
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Haskell, 21 bytes

a#b=(div a b,mod a b)

Try it online! Example usage: 13#2 returns (6,1). Yes, this is pretty boring, however slightly more interesting than the divMod build-in which works the same.

While we are at it, there is also quot, rem and quotRem which behave the same on natural numbers as div, mod and divMod. However, for negative inputs the result of mod has the same sign as the divisor, while the result of rem has the same sign as the dividend. Or, as it is put in the Prelude documentation, quot is integer division truncated toward zero and div is integer division truncated toward negative infinity.


How about no div or mod build-ins?

No build-ins, 36 32 31 bytes

a#b|a<b=(a,0)|m<-a-b=(+1)<$>m#b

Try it online! Example usage: 13#2 returns (1,6), that is the mod result is first and the div result second. If a is smaller b, then a mod b is a and a div b is 0, so (a,0) is returned. Otherwise recursively compute mod and div of a-b and b, add 1 to the division result and keep the remainder.

Adding 1 to the division result is achieved by using <$>, which is commonly used as map to map functions over lists, but works on tuples too, however the function is applied to the second tuple element only.

Edit: Saved one byte thanks to xnor!

\$\endgroup\$
  • 2
    \$\begingroup\$ Your second solution can shave a byte using <$> on a tuple to act on its second element :a#b|a<b=(a,0)|m<-a-b=(+1)<$>m#b. \$\endgroup\$ – xnor Mar 28 '17 at 0:20
3
\$\begingroup\$

SWI Prolog, 109 bytes

p(A):-print(A).
d(F,S,0,F):-S>F.
d(F,S,D,R):-G is F-S,d(G,S,E,R),D is E+1.
d(F,S):-d(F,S,D,R),p(D),p(-),p(R).

Output:

?- d(255,25).
10-5
true .
?- d(5,7).
0-5
true .

Description:

Simple recursive algorithm without builtin division or modulo. It simply counts "how many times fits the Second number into the First one?" and reports the result (unified to D) with the remainder (R).

//edit: removed unnecessary spaces

\$\endgroup\$
  • \$\begingroup\$ Welcome to PPCG! I have never used Prolog before, but noticed there are spaces around :- in the last line but not on the others. Are they required there for some reason? The same goes for E + 1 while F-S suggest there are no spaces needed. \$\endgroup\$ – Laikoni Mar 31 '17 at 11:20
  • \$\begingroup\$ Laikoni: You are definitely right! I've just removed the spaces and updated final byte count. \$\endgroup\$ – Jan Drozen Mar 31 '17 at 12:19
2
\$\begingroup\$

Gol><>, 11 bytes

I:I:@$S,N%h

Online interpreter

\$\endgroup\$
2
\$\begingroup\$

MATL, 5 bytes

/k&G\

Try it out at MATL Online!

Explanation

        % Implicitly grab the two inputs as numbers
/       % Divide them
k       % Round down the result
&G      % Grab the two inputs again
\       % Compute the remainder
\$\endgroup\$
2
\$\begingroup\$

Ouroboros, 15 bytes

r.r.@/Inao\%n1(

Takes the numbers in reverse order (e.g. 10 42). Try it here.

Explanation

r.r.             Read a number, duplicate, read a number, duplicate
    @            Rotate a copy of the first number to the top of the stack
     /I          Divide and truncate to integer
       n         Output as number
        ao       Push 10 and output as character (newline)
          \%     Swap the remaining two values and take the mod
            n    Output as number
             1(  Push 1 and swallow that many characters from the end of the program,
                 halting execution
\$\endgroup\$

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