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Euclid’s Algorithm when % and / operations are costly
  • Difficulty Level : Medium
  • Last Updated : 27 Mar, 2017

Euclid’s algorithm is used to find GCD of two numbers.

There are mainly two versions of algorithm.
Version 1 (Using subtraction)

// Recursive function to return gcd of a and b
int gcd(int a, int b)
    if (a == b)  
       return a;
    return (a > b)? gcd(a-b, b): gcd(a, b-a);

Version 2 (Using modulo operator)

// Function to return gcd of a and b
int gcd(int a, int b)
    if (a == 0) 
       return b;
    return gcd(b%a, a);

Which of the above two is more efficient?
Version 1 can take linear time to find the GCD, consider the situation when one of the given numbers is much bigger than the other. Version 2 is obviously more efficient as there are less recursive calls and takes logarithmic time.

Consider a situation where modulo operator is not allowed, can we optimize version 1 to work faster?

Below are some important observations. The idea is to use bitwise operators. We can find x/2 using x>>1. We can check whether x is odd or even using x&1.

gcd(a, b) = 2*gcd(a/2, b/2) if both a and b are even.
gcd(a, b) = gcd(a/2, b) if a is even and b is odd.
gcd(a, b) = gcd(a, b/2) if a is odd and b is even.

Below is C++ implementation.

// Efficient C++ program when % and / are not allowed
int gcd(int a, int b)
    // Base cases
    if (b == 0 || a == b) return a;
    if (a == 0) return b;
    // If both a and b are even, divide both a
    // and b by 2.  And multiply the result with 2
    if ( (a & 1) == 0 && (b & 1) == 0 )
       return gcd(a>>1, b>>1) << 1;
    // If a is even and b is odd, divide a by 2
    if ( (a & 1) == 0 && (b & 1) != 0 )
       return gcd(a>>1, b);
    // If a is odd and b is even, divide b by 2
    if ( (a & 1) != 0 && (b & 1) == 0 )
       return gcd(a, b>>1);
    // If both are odd, then apply normal subtraction 
    // algorithm.  Note that odd-odd case always 
    // converts odd-even case after one recursion
    return (a > b)? gcd(a-b, b): gcd(a, b-a);

This article is compiled by Shivam Agrawal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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