Euclid's Algorithm when % and / operations are costly
Last Updated :
27 Feb, 2023
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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);
}
// 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);
}
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == b)
return a;
return (a > b) ? gcd(a - b, b) : gcd(a, b - a);
}
// This code is contributed by subham348.
# Recursive function to return gcd of a and b
def gcd(a, b):
if (a == b):
return a
return gcd(a-b, b) if (a > b) else gcd(a, b-a)
# This code is contributed by subham348.
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == b)
return a;
return (a > b) ? gcd(a - b, b) : gcd(a, b - a);
}
// This code is contributed by subham348.
// Recursive function to return gcd of a and b
function gcd(a, b)
{
if (a === b)
return a;
return (a > b)? gcd(a-b, b): gcd(a, b-a);
}
// This code is contributed by subham348.
Time Complexity: O(max(a, b))
Auxiliary Space: O(1)
Version 2 (Using modulo operator)
// Function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to return gcd of a and b
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
// Function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
// This code is contributed by subham348.
// Function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
// This code is contributed by subham348.
// Function to return gcd of a and b
function gcd(a, b)
{
if (a === 0)
return b;
return gcd(b%a, a);
}
// This code is contributed by subham348.
# Python3 Function to return gcd of a and b
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
# This code is contributed by phasing17
Time Complexity: O(log(max(a, b)))
Auxiliary Space: O(1)
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);
}
// 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);
}
// Java program to implement
// the above approach
import java.util.*;
class GFG {
static 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);
}
// Driver Code
public static void main(String[] args)
{
System.out.println(gcd(54, 36));
}
}
// This code is contributed by phasing17
def gcd(a, b):
# Base cases
if (b == 0 or 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 and (b & 1) == 0):
return gcd(a >> 1, b >> 1) * 2
# If a is even and b is odd, divide a by 2
if ((a & 1) == 0 and (b & 1) != 0):
return gcd(a >> 1, b)
# If a is odd and b is even, divide b by 2
if ((a & 1) != 0 and (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 gcd(a-b, b) if a > b else gcd(a, b-a)
# This code is contributed by Vikram_Shirsat
// C# program to implement
// the above approach
using System;
class GFG
{
// 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 code is contributed by code_hunt.
// Efficient JavaScript program when % and / are not allowed
function gcd(a, 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 code is contributed by phasing17
Time Complexity: O(log(max(a, b)))
Auxiliary Space: O(1)
This article is compiled by Shivam Agrawal.
