Write the shortest iterative function to find gcd of two numbers (assume 32 bit numbers).


  • Arithmetic, Comparison and Logical Operators are not allowed (except equality comparison (a==b) operator).
  • Function should work for both positive and negative numbers
  • No inbuilt functions allowed.
  • All bitwise operators allowed.
  • \$\begingroup\$ What is allowed in the loop head? Only zero-test? \$\endgroup\$ Oct 19, 2013 at 6:21
  • \$\begingroup\$ @JanDvorak Can be anything, but within the limit of the rules \$\endgroup\$
    – Coding man
    Oct 19, 2013 at 6:27
  • \$\begingroup\$ ok, does equality test count as comparison, or do we have to do a poor-man's XOR followed by a zero-check? \$\endgroup\$ Oct 19, 2013 at 6:39
  • \$\begingroup\$ @JanDvorak No comparison operator allowed. You can use your second idea. \$\endgroup\$
    – Coding man
    Oct 19, 2013 at 6:44
  • 1
    \$\begingroup\$ @JanDvorak Unfortunately XOR is also not allowed and has to be replaced by (a|b)&!(a&b). Together with no comparisons allowed this seems rather arbitrary to me and disqualifies lots of languages. Not interesting. \$\endgroup\$
    – Howard
    Oct 19, 2013 at 9:01

1 Answer 1


Python - 196 bytes

def A(a,b):
 while b:a,b=a&~b|~a&b,(a&b)<<1
 return a
def G(a,b,i=0):
 if a>>31:a=A(~a,1)
 if b>>31:b=A(~b,1)
 return a<<i

Sample usage:

from random import randint
for i in range(10):
  a = randint(-2147483647,2147483647)
  b = randint(-2147483647,2147483647)
  print 'gcd(%d, %d) = %d'%(a, b, G(a,b))

Sample output:

gcd(-36916085, -1872111029) = 1
gcd(1355889652, 1816917540) = 188
gcd(-366482295, 1612196424) = 9
gcd(836632083, -1156302534) = 3
gcd(1223074731, -1299765354) = 123
gcd(-1154829176, 522085100) = 4
gcd(-1673024403, 1589241938) = 1
gcd(-1871498822, -1089342630) = 2
gcd(1653429392, 2095617430) = 2
gcd(1525670601, -1985869899) = 39

Implementation Notes

  • A -> a function which adds two integers.
  • a&~b|~a&b -> a^b
  • if a>>31 -> if a<0
  • a=A(~a,1) -> a=-a (taken together, if a>>31:a=A(~a,1) -> a=abs(a))
  • ~a&1 -> a%2==0 a.k.a. a is even.

The function G begins by removing powers of 2 from both a and b (and shifting the result if both are even). When both a and b are odd, it continues on with a=b, b=(a+b)/2. This works because gcd(a, b) = gcd(a, a+b), and a+b is necessarily even. This would terminate noticeably sooner using a comparison and a subtraction (subtracting the smaller value from the larger), but neither are available.

  • \$\begingroup\$ great!!, first to answer \$\endgroup\$
    – Coding man
    Oct 19, 2013 at 9:15
  • 1
    \$\begingroup\$ you are using recursion, please do it iteratively.. \$\endgroup\$
    – Coding man
    Oct 19, 2013 at 9:32
  • 1
    \$\begingroup\$ please, verify if the description was ambiguous "Write the shortest iterative function to find gcd of two numbers". If ambiguous, recursion will be fine and I will update the question else you have to update the code :P \$\endgroup\$
    – Coding man
    Oct 19, 2013 at 9:49
  • 1
    \$\begingroup\$ I apologize. I must have skipped over the word 'iterative' while reading the description. \$\endgroup\$
    – primo
    Oct 19, 2013 at 10:02

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