Your task is to compute the greatest common divisor (GCD) of two given integers in as few bytes of code as possible.

You may write a program or function, taking input and returning output via any of our accepted standard methods (including STDIN/STDOUT, function parameters/return values, command-line arguments, etc.).

Input will be two non-negative integers. You should be able to handle either the full range supported by your language's default integer type, or the range [0,255], whichever is greater. You are guaranteed that at least one of the inputs will be non-zero.

You are not allowed to use built-ins that compute either the GCD or the LCM (least common multiple).

Standard rules apply.

Test Cases

0 2     => 2
6 0     => 6
30 42   => 6
15 14   => 1
7 7     => 7
69 25   => 1
21 12   => 3
169 123 => 1
20 142  => 2
101 202 => 101
  • 1
    \$\begingroup\$ If we're allowing asm to have inputs in whatever registers are convenient, and the result in whatever reg is convenient, we should definitely be allowing functions, or even code fragments (i.e. just a function body). Making my answer a complete function would add about 4B with a register calling convention like MS's 32bit vectorcall (one xchg eax, one mov, and a ret), or more with a stack calling convention. \$\endgroup\$ Commented Apr 8, 2016 at 23:05
  • \$\begingroup\$ @PeterCordes Sorry, I should have been more specific. You can totally just write the bear necessary code but if you would be so kind as to include a way to run said code it would be nice. \$\endgroup\$ Commented Apr 9, 2016 at 18:59
  • \$\begingroup\$ So count just the gcd code, but provide the surrounding code so people can verify / experiment / improve? BTW, your test-cases with zero as one of the two inputs break our x86 machine code answers. div by zero raises a hardware exception. On Linux, your process gets a SIGFPE. \$\endgroup\$ Commented Apr 9, 2016 at 19:30
  • 3
    \$\begingroup\$ @CodesInChaos Memory and time limitations are usually ignored as long as the algorithm itself can in principle handle all inputs. The rule is just meant to avoid people hardcoding arbitrary limits for loops that artificially limits the algorithm to a smaller range of inputs. I don't quite see how immutability comes into this? \$\endgroup\$ Commented Apr 10, 2016 at 14:10
  • 4
    \$\begingroup\$ gcd(0,n) is error not n \$\endgroup\$
    – user58988
    Commented Mar 18, 2019 at 12:33

85 Answers 85


Retina, 16

^(.+)\1* \1+$

This doesn't use Euclid's algorithm at all - instead it finds the GCD using regex matching groups.

Try it online. - This example calculates GCD(8,12).

Input as 2 space-separated integers. Note that the I/O is in unary. If that is not acceptable, then we can do this:

Retina, 30

^(.+)\1* \1+$

Try it online.

As @MartinBüttner points out, this falls apart for large numbers (as is generally the case for anything unary). At very a minimum, an input of INT_MAX will require allocation of a 2GB string.

  • 2
    \$\begingroup\$ I would like to vote for this more \$\endgroup\$
    – MickyT
    Commented Apr 8, 2016 at 1:51
  • \$\begingroup\$ Should be fine with the number range now. I've changed the spec (with the OPs permission) to require only the language's natural number range (or [0,255] if that's more). You'll have to support zeroes though, although I think changing your +s to *s should do. And you can significantly shorten the last stage of the long code by reducing it to 1. \$\endgroup\$ Commented Apr 10, 2016 at 9:14
  • 2
    \$\begingroup\$ For future reference, I just found an alternative 16-byte solution that works for an arbitrary number of inputs (including one) so it might be more useful in other contexts: retina.tryitonline.net/… \$\endgroup\$ Commented Sep 23, 2016 at 11:15
  • 2
    \$\begingroup\$ Just noticed that neither your solutions nor the one in my comment above needs the ^, because it's impossible for the match to fail from the starting position. \$\endgroup\$ Commented Dec 18, 2017 at 11:01

i386 (x86-32) machine code, 8 bytes (9B for unsigned)

+1B if we need to handle b = 0 on input.

amd64 (x86-64) machine code, 9 bytes (10B for unsigned, or 14B 13B for 64b integers signed or unsigned)

10 9B for unsigned on amd64 that breaks with either input = 0

Inputs are 32bit non-zero signed integers in eax and ecx. Output in eax.

## 32bit code, signed integers:  eax, ecx
08048420 <gcd0>:
 8048420:       99                      cdq               ; shorter than xor edx,edx
 8048421:       f7 f9                   idiv   ecx
 8048423:       92                      xchg   edx,eax    ; there's a one-byte encoding for xchg eax,r32.  So this is shorter but slower than a mov
 8048424:       91                      xchg   ecx,eax    ; eax = divisor(from ecx), ecx = remainder(from edx), edx = quotient(from eax) which we discard
    ; loop entry point if we need to handle ecx = 0
 8048425:       41                      inc    ecx        ; saves 1B vs. test/jnz in 32bit mode
 8048426:       e2 f8                   loop   8048420 <gcd0>
08048428 <gcd0_end>:
 ; 8B total
 ; result in eax: gcd(a,0) = a

This loop structure fails the test-case where ecx = 0. (div causes a #DE hardware execption on divide by zero. (On Linux, the kernel delivers a SIGFPE (floating point exception)). If the loop entry point was right before the inc, we'd avoid the problem. The x86-64 version can handle it for free, see below.

Mike Shlanta's answer was the starting point for this. My loop does the same thing as his, but for signed integers because cdq is one byter shorter than xor edx,edx. And yes, it does work correctly with one or both inputs negative. Mike's version will run faster and take less space in the uop cache (xchg is 3 uops on Intel CPUs, and loop is really slow on most CPUs), but this version wins at machine-code size.

I didn't notice at first that the question required unsigned 32bit. Going back to xor edx,edx instead of cdq would cost one byte. div is the same size as idiv, and everything else can stay the same (xchg for data movement and inc/loop still work.)

Interestingly, for 64bit operand-size (rax and rcx), signed and unsigned versions are the same size. The signed version needs a REX prefix for cqo (2B), but the unsigned version can still use 2B xor edx,edx.

In 64bit code, inc ecx is 2B: the single-byte inc r32 and dec r32 opcodes were repurposed as REX prefixes. inc/loop doesn't save any code-size in 64bit mode, so you might as well test/jnz. Operating on 64bit integers adds another one byte per instruction in REX prefixes, except for loop or jnz. It's possible for the remainder to have all zeros in the low 32b (e.g. gcd((2^32), (2^32 + 1))), so we need to test the whole rcx and can't save a byte with test ecx,ecx. However, the slower jrcxz insn is only 2B, and we can put it at the top of the loop to handle ecx=0 on entry:

## 64bit code, unsigned 64 integers:  rax, rcx
0000000000400630 <gcd_u64>:
  400630:       e3 0b                   jrcxz  40063d <gcd_u64_end>   ; handles rcx=0 on input, and smaller than test rcx,rcx/jnz
  400632:       31 d2                   xor    edx,edx                ; same length as cqo
  400634:       48 f7 f1                div    rcx                      ; REX prefixes needed on three insns
  400637:       48 92                   xchg   rdx,rax
  400639:       48 91                   xchg   rcx,rax
  40063b:       eb f3                   jmp    400630 <gcd_u64>
000000000040063d <gcd_u64_end>:
## 0xD = 13 bytes of code
## result in rax: gcd(a,0) = a

Full runnable test program including a main that runs printf("...", gcd(atoi(argv[1]), atoi(argv[2])) ); source and asm output on the Godbolt Compiler Explorer, for the 32 and 64b versions. Tested and working for 32bit (-m32), 64bit (-m64), and the x32 ABI (-mx32).

Also included: a version using repeated subtraction only, which is 9B for unsigned, even for x86-64 mode, and can take one of its inputs in an arbitrary register. However, it can't handle either input being 0 on entry (it detect when sub produces a zero, which x - 0 never does).

GNU C inline asm source for the 32bit version (compile with gcc -m32 -masm=intel)

int gcd(int a, int b) {
    asm (// ".intel_syntax noprefix\n"
        // "jmp  .Lentry%=\n" // Uncomment to handle div-by-zero, by entering the loop in the middle.  Better: `jecxz / jmp` loop structure like the 64b version
        ".p2align 4\n"                  // align to make size-counting easier
         "gcd0:   cdq\n\t"              // sign extend eax into edx:eax.  One byte shorter than xor edx,edx
         "        idiv    ecx\n"
         "        xchg    eax, edx\n"   // there's a one-byte encoding for xchg eax,r32.  So this is shorter but slower than a mov
         "        xchg    eax, ecx\n"   // eax = divisor(ecx), ecx = remainder(edx), edx = garbage that we will clear later
         "        inc     ecx\n"        // saves 1B vs. test/jnz in 32bit mode, none in 64b mode
         "        loop    gcd0\n"
         : /* outputs */  "+a" (a), "+c"(b)
         : /* inputs */   // given as read-write outputs
         : /* clobbers */ "edx"
    return a;

Normally I'd write a whole function in asm, but GNU C inline asm seems to be the best way to include a snippet which can have in/outputs in whatever regs we choose. As you can see, GNU C inline asm syntax makes asm ugly and noisy. It's also a really difficult way to learn asm.

It would actually compile and work in .att_syntax noprefix mode, because all the insns used are either single/no operand or xchg. Not really a useful observation.

  • 2
    \$\begingroup\$ @MikeShlanta: Thanks. If you like optimizing asm, have a look at some of my answers over on stackoverflow. :) \$\endgroup\$ Commented Apr 8, 2016 at 3:19
  • 2
    \$\begingroup\$ @MikeShlanta: I found a use for jrcxz after all in the uint64_t version :). Also, didn't notice that you'd specified unsigned, so I included byte counts for that, too. \$\endgroup\$ Commented Apr 8, 2016 at 22:54
  • \$\begingroup\$ Why couldn't you use jecxz in the 32-bit version to the same effect? \$\endgroup\$ Commented Aug 13, 2016 at 13:32
  • 1
    \$\begingroup\$ @CodyGray: inc/loop is 3 bytes in the 32-bit version, but 4B in the 64-bit version. That means that in the 64-bit version only, it doesn't cost extra bytes to use jrcxz and jmp instead of inc / loop. \$\endgroup\$ Commented Aug 13, 2016 at 14:43
  • \$\begingroup\$ Can't you point to the middle as entry? \$\endgroup\$
    – l4m2
    Commented Dec 18, 2017 at 20:31

Hexagony, 17 bytes



  ? ' ?
 > } ! @
< \ = % )
 > { \ .
  ( . .

Try it online!

Fitting it into side-length 3 was a breeze. Shaving off those two bytes at the end wasn't... I'm also not convinced it's optimal, but I'm sure I think it's close.


Another Euclidean algorithm implementation.

The program uses three memory edges, which I'll call A, B and C, with the memory pointer (MP) starting out as shown:

enter image description here

Here is the control flow diagram:

enter image description here

Control flow starts on the grey path with a short linear bit for input:

?    Read first integer into memory edge A.
'    Move MP backwards onto edge B.
?    Read second integer into B.

Note that the code now wraps around the edges to the < in the left corner. This < acts as a branch. If the current edge is zero (i.e. the Euclidean algorithm terminates), the IP is deflected to the left and takes the red path. Otherwise, an iteration of the Euclidean algorithm is computed on the green path.

We'll first consider the green path. Note that > and \ all acts as mirrors which simply deflect the instruction pointer. Also note that control flow wraps around the edges three times, once from the bottom to the top, once from the right corner to the bottom row and finally from the bottom right corner to the left corner to re-check the condition. Also note that . are no-ops.

That leaves the following linear code for a single iteration:

{    Move MP forward onto edge C.
'}   Move to A and back to C. Taken together this is a no-op.
=    Reverse the direction of the MP so that it now points at A and B. 
%    Compute A % B and store it in C.
)(   Increment, decrement. Taken together this is a no-op, but it's
     necessary to ensure that IP wraps to the bottom row instead of
     the top row.

Now we're back where we started, except that the three edges have changed their roles cyclically (the original C now takes the role of B and the original B the role of A...). In effect, we've relpaced inputs A and B with B and A % B, respectively.

Once A % B (on edge C) is zero, the GCD can be found on edge B. Again the > just deflects the IP, so on the red path we execute:

}    Move MP to edge B.
!    Print its value as an integer.
@    Terminate the program.

T-SQL, 153 169 bytes

Someone mentioned worst language for golfing?


Creates a table valued function that uses a recursive query to work out the common divisors. Then it returns the maximum. Now uses the euclidean algorithm to determine the GCD derived from my answer here.

Example usage

    ) TestSet(A, B)

A           B           D
----------- ----------- -----------
15          45          15
45          15          15
99          7           1
4           38          2

(4 row(s) affected)
  • 1
    \$\begingroup\$ Jesus that is verbose. \$\endgroup\$
    – Cyoce
    Commented Apr 9, 2016 at 7:46

32-bit little-endian x86 machine code, 14 bytes

Generated using nasm -f bin

d231 f3f7 d889 d389 db85 f475

    gcd0:   xor     edx,edx
            div     ebx
            mov     eax,ebx
            mov     ebx,edx
            test    ebx,ebx
            jnz     gcd0
  • 4
    \$\begingroup\$ I got this down to 8 bytes by using cdq and signed idiv, and one-byte xchg eax, r32 instead of mov. For 32bit code: inc/loop instead of test/jnz (I couldn't see a way to use jecxz, and there's no jecxnz). I posted my final version as a new answer since I think the changes are big enough to justify it. \$\endgroup\$ Commented Apr 8, 2016 at 2:05

Jelly, 7 bytes


Recursive implementation of the Euclidean algorithm. Try it online!

If built-ins weren't forbidden, g (1 byte, built-in GCD) would achieve a better score.

How it works

ṛß%ðḷṛ?  Main link. Arguments: a, b

   ð     Convert the chain to the left into a link; start a new, dyadic chain.
 ß       Recursively call the main link...
ṛ %        with b and a % b as arguments.
     ṛ?  If the right argument (b) is non-zero, execute the link.
    ḷ    Else, yield the left argument (a).
  • \$\begingroup\$ That almost feels like cheating haha, I may have to specify that answers can't use butlins... \$\endgroup\$ Commented Apr 7, 2016 at 18:14
  • 17
    \$\begingroup\$ If you decide to do so, you should do it quickly. It would currently invalidate three of the answers. \$\endgroup\$
    – Dennis
    Commented Apr 7, 2016 at 18:16
  • \$\begingroup\$ Notice he specified length is in bytes -- those characters are mostly > 1 byte in UTF8. \$\endgroup\$
    – cortices
    Commented Apr 8, 2016 at 3:45
  • 11
    \$\begingroup\$ @cortices Yes, all code golf contests are scored in bytes by default. However, Jelly doesn't use UTF-8, but a custom code page that encodes each of the 256 characters it understands as a single byte. \$\endgroup\$
    – Dennis
    Commented Apr 8, 2016 at 4:31

Python 3, 31

Saved 3 bytes thanks to Sp3000.

g=lambda a,b:b and g(b,a%b)or a
  • 3
    \$\begingroup\$ In Python 3.5+: from math import*;gcd \$\endgroup\$
    – Sp3000
    Commented Apr 7, 2016 at 18:11
  • \$\begingroup\$ @Sp3000 Nice, I didn't know they had moved it to math. \$\endgroup\$ Commented Apr 7, 2016 at 18:12
  • 1
    \$\begingroup\$ While you're at it: g=lambda a,b:b and g(b,a%b)or a \$\endgroup\$
    – Sp3000
    Commented Apr 7, 2016 at 18:15
  • \$\begingroup\$ @Sp3000 Thanks! I just finished a recursive solution, but that's even better than what I had. \$\endgroup\$ Commented Apr 7, 2016 at 18:16
  • \$\begingroup\$ Built-ins for GCD and LCM are disallowed, so the 2nd solution wouldn't be valid. \$\endgroup\$
    – mbomb007
    Commented Oct 4, 2016 at 14:38

Haskell, 19 bytes

a#b=b#rem a b

Usage example: 45 # 35-> 5.

Euclid, again.

PS: of course there's a built-in gcd, too.

  • \$\begingroup\$ you should explain the trick that reverses the input order to avoid the conditional check \$\endgroup\$ Commented Apr 9, 2016 at 1:13
  • \$\begingroup\$ @proudhaskeller: what trick? Everybody uses this algorithm, i.e. stopping on 0 or going on with the modulus. \$\endgroup\$
    – nimi
    Commented Apr 9, 2016 at 23:59
  • \$\begingroup\$ Nevrmind, everybody is using the trick \$\endgroup\$ Commented Apr 10, 2016 at 7:10
  • \$\begingroup\$ This, less golfed, is almost exactly what's in Prelude \$\endgroup\$ Commented Jun 1, 2016 at 5:21

MATL, 11 9 bytes

No one seems to have used brute force up to now, so here it is.


Input is a column array with the two numbers (using ; as separator).

Try it online! or verify all test cases.


t     % Take input [a;b] implicitly. Duplicate
s     % Sum. Gives a+b
:     % Array [1,2,...,a+b]
\     % Modulo operation with broadcast. Gives a 2×(a+b) array
a~    % 1×(a+b) array that contains true if the two modulo operations gave 0
f0)   % Index of last true value. Implicitly display

Julia, 21 15 bytes


Recursive implementation of the Euclidean algorithm. Try it online!

If built-ins weren't forbidden, gcd (3 bytes, built-in GCD) would achieve a better score.

How it works

a\b=             Redefine the binary operator \ as follows:
    a>0?     :       If a > 0:
        b%a\a        Resursively apply \ to b%a and a. Return the result.
              b      Else, return b.

C, 38 bytes

g(x,y){while(x^=y^=x^=y%=x);return y;}
  • 1
    \$\begingroup\$ You need to include the definition of the function in your bytecount. \$\endgroup\$
    – Riker
    Commented Dec 29, 2017 at 0:24
  • 1
    \$\begingroup\$ @Riker sorry for that, I add the definition and update the count \$\endgroup\$
    – How Chen
    Commented Dec 29, 2017 at 0:32
  • \$\begingroup\$ You can save two bytes by naming the function just g instead of gcd. \$\endgroup\$
    – Steadybox
    Commented Dec 29, 2017 at 0:37
  • \$\begingroup\$ @Steadybox ok, yes, first time join this community:) \$\endgroup\$
    – How Chen
    Commented Dec 29, 2017 at 0:38
  • 1
    \$\begingroup\$ Welcome to PPCG! \$\endgroup\$
    – Riker
    Commented Dec 29, 2017 at 0:38

TI-Basic, 27 bytes

Prompt A,B:While B:B→T:BfPart(A/B→B:T→A:End:A

35 byte recursive solution without gcd( or lcm( built-ins (requires 2.53 MP operating system or higher, must be named prgmG):

If Ans(2:Then:{Ans(2),remainder(Ans(1),Ans(2:prgmG:Else:Disp Ans(1:End

You would pass arguments to the recursive variant as {A,B} so for example {1071, 462}:prgmG would yield 21.

TI-Basic, 10 bytes

Prompt A,B:gcd(A,B

Not allowed due to new rule forbidding gcd built-ins

17 byte solution without gcd( built-in

Prompt A,B:abs(AB)/lcm(A,B

Not allowed due to new rule forbidding lcm built-ins

  • \$\begingroup\$ Color me impressed. \$\endgroup\$ Commented Apr 7, 2016 at 18:57
  • \$\begingroup\$ You should probably mention that the last one needs to be saved as prgmG. \$\endgroup\$ Commented Apr 7, 2016 at 19:48

C, 28 bytes

A rather straightforward function implementing Euclid's algorithm. Perhaps one can get shorter using an alternate algorithm.

g(a,b){return b?g(b,a%b):a;}

If one writes a little main wrapper

int main(int argc, char **argv)
  printf("gcd(%d, %d) = %d\n", atoi(argv[1]), atoi(argv[2]), g(atoi(argv[1]), atoi(argv[2])));

then one can test a few values:

$ ./gcd 6 21
gcd(6, 21) = 3
$ ./gcd 21 6
gcd(21, 6) = 3
$ ./gcd 6 8
gcd(6, 8) = 2
$ ./gcd 1 1
gcd(1, 1) = 1
$ ./gcd 6 16
gcd(6, 16) = 2
$ ./gcd 27 244
gcd(27, 244) = 1

Labyrinth, 18 bytes

( =

Terminates with an error, but the error message goes to STDERR.

Try it online!

This doesn't feel quite optimal yet, but I'm not seeing a way to compress the loop below 3x3 at this point.


This uses the Euclidean algorithm.

First, there's a linear bit to read input and get into the main loop. The instruction pointer (IP) starts in the top left corner, going east.

?    Read first integer from STDIN and push onto main stack.
}    Move the integer over to the auxiliary stack.
     The IP now hits a dead end so it turns around.
?    Read the second integer.
     The IP hits a corner and follows the bend, so it goes south.
:    Duplicate the second integer.
)    Increment.
     The IP is now at a junction. The top of the stack is guaranteed to be
     positive, so the IP turns left, to go east.
"    No-op.
%    Modulo. Since `n % (n+1) == n`, we end up with the second input on the stack.

We now enter a sort of while-do loop which computes the Euclidean algorithm. The tops of the stacks contain a and b (on top of an implicit infinite amount of zeros, but we won't need those). We'll represent the stacks side to side, growing towards each other:

    Main     Auxiliary
[ ... 0 a  |  b 0 ... ]

The loop terminates once a is zero. A loop iteration works as follows:

=    Swap a and b.           [ ... 0 b  |  a 0 ... ]
{    Pull a from aux.        [ ... 0 b a  |  0 ... ]
:    Duplicate.              [ ... 0 b a a  |  0 ... ]
}    Move a to aux.          [ ... 0 b a  |  a 0 ... ]
()   Increment, decrement, together a no-op.
%    Modulo.                 [ ... 0 (b%a)  |  a 0 ... ]

You can see, we've replaced a and b with b%a and a respectively.

Finally, once b%a is zero, the IP keeps moving east and executes:

{    Pull the non-zero value, i.e. the GCD, over from aux.
!    Print it.
     The IP hits a dead end and turns around.
{    Pull a zero from aux.
%    Attempt modulo. This fails due to division by 0 and the program terminates.

Cubix, 10 12 bytes


Try it here

This wraps onto the cube as follows:

    ? v
    % u
I I / ; O @ . .
. . . . . . . .
    . .
    . .

Uses the Euclidean Method.

II Two numbers are grabbed from STDIN and put on the stack
/ Flow reflected up
% Mod the Top of Stack. Remainder left on top of stack
? If TOS 0 then carry on, otherwise turn right
v If not 0 then redirect down and u turn right twice back onto the mod
/ If 0 go around the cube to the reflector
; drop TOS, O output TOS and @ end

  • \$\begingroup\$ I just wrote up a 12-byte Cubix answer, then started scrolling through the answers to see if I needed to handle both 0,x and x,0... then I came across this. Nice one! \$\endgroup\$ Commented Oct 4, 2016 at 15:05

C#, 24 bytes


Brachylog, 5 bytes


Try it online!

Takes input as a list through the input variable and outputs an integer through the output variable. If you use it as a generator, it actually generates all common divisors, it just does so largest first.

         The output is
    ×    the product of the elements of
  ⊇      the maximal sublist
   ᵛ     which is shared by
ḋ        the prime factorization
 ˢ       s of the elements of the input which have them.

Windows Batch, 76 bytes

Recursive function. Call it like GCD a b with file name gcd.

if %2 equ 0 (set f=%1
goto d)
set/a r=%1 %% %2
call :g %2 %r%
echo %f%

MATL, 7 bytes


Try it Online!


Since we can't explicitly use the builtin GCD function (Zd in MATL), I have exploited the fact that the least common multiple of a and b times the greatest common denominator of a and b is equal to the product of a and b.

p       % Grab the input implicitly and multiply the two elements
G       % Grab the input again, explicitly this time
1$Zm    % Compute the least-common multiple
/       % Divide the two to get the greatest common denominator
  • \$\begingroup\$ You can save one byte with two separate inputs: *1MZm/ \$\endgroup\$
    – Luis Mendo
    Commented Apr 8, 2016 at 18:59

ARM machine code, 12 bytes:


gcd: cmp r0, r1
     sublt r0, r0, r1
     bne gcd

Currently can't compile this, but each instruction in ARM takes 4 bytes. Probably it could be golfed down using THUMB-2 mode.

  • \$\begingroup\$ Nice job man anyone who does this in machine code gets serious props from me. \$\endgroup\$ Commented Apr 8, 2016 at 20:15
  • \$\begingroup\$ This appears to be an attempt at Euclid's algo using only subtraction, but I don't think it works. If r0 > r1 then sublt will do nothing (lt predicate is false) and bne will be an infinite loop. I think you need a swap if not lt, so the same loop can do b-=a or a-=b as needed. Or a negate if the sub produced carry (aka borrow). \$\endgroup\$ Commented Apr 9, 2016 at 18:55
  • \$\begingroup\$ This ARM instruction set guide actually uses a subtraction GCD algorithm as an example for predication. (pg 25). They use cmp r0, r1 / subgt r0, r0, r1 / sublt r1, r1, r0 / bne gcd. That's 16B in ARM instructions, maybe 12 in thumb2 instructions? \$\endgroup\$ Commented Apr 9, 2016 at 19:21
  • 1
    \$\begingroup\$ On x86, I managed 9 bytes with: sub ecx, eax / jae .no_swap / add ecx,eax / xchg ecx,eax / jne. So instead of a cmp, I just sub, then undo and swap if the sub should have gone the other way. I tested this, and it works. (add won't make jne exit at the wrong time, because it can't produce a zero unless one of the inputs was zero to start with, and we don't support that. Update: we do need to support either input being zero :/) \$\endgroup\$ Commented Apr 9, 2016 at 19:25
  • \$\begingroup\$ For Thumb2, there's an ite instruction: if-then-else. Should be perfect for cmp / sub one way / sub the other way. \$\endgroup\$ Commented Apr 9, 2016 at 23:23

Racket (Scheme), 44 bytes

Euclid implementation in Racket (Scheme)

(define(g a b)(if(= 0 b)a(g b(modulo a b))))

Edit: Didn't see @Numeri 's solution lol. Somehow we got the exact same code independently

  • \$\begingroup\$ Does this work in both? \$\endgroup\$ Commented Jun 2, 2016 at 15:42
  • \$\begingroup\$ @NoOneIsHere yeah it works in both \$\endgroup\$
    – Simon Zeng
    Commented Jun 3, 2016 at 12:53

><>, 32 bytes


Accepts two values from the stack and apply the euclidian algorithm to produce their GCD.

You can try it here!

For a much better answer in ><>, check out Sok's !

  • 1
    \$\begingroup\$ I found a new language today :) \$\endgroup\$
    – nsane
    Commented Oct 1, 2016 at 20:19

R, 39 33 bytes

Surprised not to see an R answer on here yet. A recursive implementation of the Euclidean algorithm. Saved 2 bytes due to Giuseppe.


And here is a vectorized version (35 bytes) which works well for problems like Natural pi calculation.

  • 1
    \$\begingroup\$ "if" rather than ifelse for -2 bytes, but you lose the vectorization... \$\endgroup\$
    – Giuseppe
    Commented Sep 5, 2017 at 2:29
  • \$\begingroup\$ @Giuseppe Thanks, good catch! \$\endgroup\$
    – rturnbull
    Commented Sep 6, 2017 at 22:15

ReRegex, 23 bytes

Works identically to the Retina answer.

^(_*)\1* \1*$/$1/#input

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Proton, 21 bytes


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Shush you, this is totally not a port of the Python answer.

  • 1
    \$\begingroup\$ Conditional operator is shorter: f=(x,y)=>y? f(y,x%y):x (yes the space is necessary due to an interpreter bug) \$\endgroup\$ Commented Aug 18, 2017 at 20:14
  • \$\begingroup\$ ...Right, I need to start thinking Proton, not Python. Thanks! \$\endgroup\$ Commented Aug 18, 2017 at 20:16

Add++, 26 bytes


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Generates the range \$1, 2, 3, ..., \max(a, b)\$, then filters out elements that don't divide either \$a\$ or \$b\$, Finally, we return the maximum value of the filtered elements.


ARMv7 (OakSim), 28 bytes


0x00010000: 01 00 80 E0 00 10 81 E0 01 00 50 E1 01 00 40 C0 ..........P...@.
0x00010010: 00 10 41 B0 FB FF FF 1A 1E FF 2F E1 00 00 00 00 ..A......./.....


Callable function, expects the arguments in r0 and r1, output is in r0.

Expects address of caller stored in lr. This is the standard method of procedure calling as per the ATPCS (ARM Thumb Procedure Call Standard).

    add r0, r0, r1        /* Make sure r0 is nonzero */
    add r1, r1, r0        /* Make sure r1 is nonzero */
                          /* We're basically doing the inverse afterwards */

    loop:                 /* main loop: */
        cmp r0, r1        /*     Compare r0 and r1 */
        subgt r0, r0, r1  /*     If r0 >  r1: r0 = r0 - r1 */
        sublt r1, r1, r0  /*     If r0 <  r1: r1 = r1 - r0 */
        bne loop          /*     If r0 != r1: jump back to main loop */

    bx lr                 /* return to caller */

Example call

b caller       /* jump to the caller */
               /* omitted to save space */
    mov r0, 30 /* first operand  = 30 */
    mov r1, 42 /* second operand = 42 */
    mov lr, pc /* link register  = program counter */

    b gcd      /* branch to function */

Forth (gforth), 37 34 bytes

: f dup if tuck mod recurse then ;

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Same as the code below, but using recursion instead. Turns out that the combination if then recurse is shorter than begin while repeat by 3 chars, even though the word recurse looks quite bulky.

Previous solution, 37 bytes

: f begin dup while tuck mod repeat ;

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Input is two single-cell integers, output is one double-cell integer.

How it works

A "cell" means a space for one item on the stack. A double-cell integer in Forth takes two cells, where the most significant cell is on the top. For this challenge, the GCD of two single-cell integers always fits in a cell, so the upper cell is always 0.

: f ( a b -- d ) \ Define a function f which takes two singles and gives a double
  begin          \ Start a loop
    dup while    \   While b is not zero...
    tuck         \   Copy b under a ( stack: b a b ) and
    mod          \   Calculate a%b  ( stack: b a%b )
                 \   That is, change ( a b ) to ( b a%b )
  repeat ;       \ End loop
                 \ The result is ( gcd 0 ) which is gcd in double-cell

Logy, 64 23 bytes



gcd[X, Y] -> Y < 1 & X | gcd[Y, X%Y];

EDIT: Removed way too many bytes because there is no need for a full program


bash, 37


g(){ (($2))&&g $2 $[$1%$2]||echo $1;}

But for making useful function, I prefer assigning variable in order to avoid forks. So I will prefer this for loop:

for((i=$1,l=$2;l;k=l,l=i%l,i=k)){ :;}

Note: both routines use 37 characters!

Test cases

for pair in 0\ 2 6\ 0 30\ 42 15\ 14 7\ 7 69\ 25 21\ 12 169\ 123 20\ 142 101\ 202;do
    printf '%4d %4d => %3d\n' $pair $(g $pair)
   0    2 =>   2
   6    0 =>   6
  30   42 =>   6
  15   14 =>   1
   7    7 =>   7
  69   25 =>   1
  21   12 =>   3
 169  123 =>   1
  20  142 =>   2
 101  202 => 101

As a dedicated function:

gcd() {
    if [[ $1 == -v ]];then local -n i=$2;shift 2;else local i;fi
    local l=$2 k
    for((i=$1;l;k=l,l=i%l,i=k)){ :;}
    [[ ${i@A} != i=* ]] || echo "$i"

Test cases

for pair in 0:2 6:0 30:42 15:14 7:7 69:25 21:12 169:123 20:142 101:202;do
    IFS=: read -ra vals <<<$pair
    gcd -v res ${vals[@]}
    printf '    %4d %4d => %3d\n' ${vals[@]} $res
   0    2 =>   2
   6    0 =>   6
  30   42 =>   6
  15   14 =>   1
   7    7 =>   7
  69   25 =>   1
  21   12 =>   3
 169  123 =>   1
  20  142 =>   2
 101  202 => 101

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