Modular Exponentiation (Power in Modular Arithmetic)
Given three numbers x, y and p, compute (xy) % p.
Examples :
Input: x = 2, y = 3, p = 5 Output: 3 Explanation: 2^3 % 5 = 8 % 5 = 3. Input: x = 2, y = 5, p = 13 Output: 6 Explanation: 2^5 % 13 = 32 % 13 = 6.
We have discussed recursive and iterative solutions for power.
Below is discussed iterative solution.
C
/* Iterative Function to calculate (x^y) in O(log y) */int power(int x, unsigned int y){ int res = 1; // Initialize result while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = res*x; // y must be even now y = y>>1; // y = y/2 x = x*x; // Change x to x^2 } return res;} |
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Efficient Approach:
The problem with above solutions is, overflow may occur for large value of n or x. Therefore, power is generally evaluated under modulo of a large number.
Below is the fundamental modular property that is used for efficiently computing power under modular arithmetic.
(ab) mod p = ( (a mod p) (b mod p) ) mod p For example a = 50, b = 100, p = 13 50 mod 13 = 11 100 mod 13 = 9 (50 * 100) mod 13 = ( (50 mod 13) * (100 mod 13) ) mod 13 or (5000) mod 13 = ( 11 * 9 ) mod 13 or 8 = 8
Below is the implementation based on above property.
C++
// Iterative C++ program to compute modular power #include <iostream>using namespace std;/* Iterative Function to calculate (x^y)%p in O(log y) */int power(long long x, unsigned int y, int p) { int res = 1; // Initialize result x = x % p; // Update x if it is more than or // equal to p if (x == 0) return 0; // In case x is divisible by p; while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res*x) % p; // y must be even now y = y>>1; // y = y/2 x = (x*x) % p; } return res; } // Driver code int main() { int x = 2; int y = 5; int p = 13; cout << "Power is " << power(x, y, p); return 0; } // This code is contributed by shubhamsingh10 |
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C
// Iterative C program to compute modular power
#include <stdio.h>
/* Iterative Function to calculate (x^y)%p in O(log y) */
int power(int x, unsigned int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0) return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Driver program to test above functions
int main()
{
int x = 2;
int y = 5;
int p = 13;
printf("Power is %u", power(x, y, p));
return 0;
}
Java
// Iterative Java program to
// compute modular power
import java.io.*;
class GFG {
/* Iterative Function to calculate
(x^y)%p in O(log y) */
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
if (x == 0) return 0; // In case x is divisible by p;
while (y > 0)
{
// If y is odd, multiply x
// with result
if((y & 1)==1)
res = (res * x) % p;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Driver Program to test above functions
public static void main(String args[])
{
int x = 2;
int y = 5;
int p = 13;
System.out.println("Power is " + power(x, y, p));
}
}
// This code is contributed by Nikita Tiwari.
Python3
# Iterative Python3 program# to compute modular power# Iterative Function to calculate# (x^y)%p in O(log y) def power(x, y, p) : res = 1 # Initialize result # Update x if it is more # than or equal to p x = x % p if (x == 0) : return 0 while (y > 0) : # If y is odd, multiply # x with result if ((y & 1) == 1) : res = (res * x) % p # y must be even now y = y >> 1 # y = y/2 x = (x * x) % p return res # Driver Codex = 2; y = 5; p = 13print("Power is ", power(x, y, p))# This code is contributed by Nikita Tiwari. |
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C#
// Iterative C# program to
// compute modular power
class GFG
{
/* Iterative Function to calculate
(x^y)%p in O(log y) */
static int power(int x, int y, int p)
{
// Initialize result
int res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
if (x == 0)
return 0;
while (y > 0)
{
// If y is odd, multiply
// x with result
if ((y & 1) == 1)
res = (res * x) % p;
// y must be even now
// y = y / 2
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Driver Code
public static void Main()
{
int x = 2;
int y = 5;
int p = 13;
System.Console.WriteLine("Power is " +
power(x, y, p));
}
}
// This code is contributed by mits
PHP
<?php// Iterative PHP program to // compute modular power// Iterative Function to // calculate (x^y)%p in O(log y) function power($x, $y, $p){ // Initialize result $res = 1; // Update x if it is more // than or equal to p $x = $x % $p; if ($x == 0) return 0; while ($y > 0) { // If y is odd, multiply // x with result if ($y & 1) $res = ($res * $x) % $p; // y must be even now // y = $y/2 $y = $y >> 1; $x = ($x * $x) % $p; } return $res;}// Driver Code$x = 2;$y = 5;$p = 13;echo "Power is ", power($x, $y, $p);// This code is contributed by aj_36?> |
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Output
Power is 6
Time Complexity of above solution is O(Log y).
Modular exponentiation (Recursive)
This article is contributed 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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