Euclid's Algorithm when % and / operations are costly Last Updated : 27 Feb, 2023 Comments Improve Suggest changes Like Article Like Report Euclid's algorithm is used to find GCD of two numbers. There are mainly two versions of algorithm. Version 1 (Using subtraction) C++ // 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); } C // 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); } Java // 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. Python3 # 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. C# // 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. JavaScript // 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) C++ // 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); } C // Function to return gcd of a and b int gcd(int a, int b) { if (a == 0) return b; return gcd(b%a, a); } Java // 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. C# // 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. JavaScript // 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 # 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. C++ // 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); } C // 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 // 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 Python3 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# // 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. JavaScript // 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. Comment More infoAdvertise with us Next Article Euclid's Algorithm when % and / operations are costly K kartik Follow Improve Article Tags : Bit Magic Mathematical DSA Practice Tags : Bit MagicMathematical Similar Reads Bitwise Algorithms Bitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit 4 min read Introduction to Bitwise Algorithms - Data Structures and Algorithms Tutorial Bit stands for binary digit. A bit is the basic unit of information and can only have one of two possible values that is 0 or 1. In our world, we usually with numbers using the decimal base. 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