Euclid’s Algorithm when % and / operations are costly
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); } |
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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); } |
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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); } |
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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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