Find power of power under mod of a prime

Given four numbers A, B, C and M, where M is prime number. Our task is to find ABC (mod M).

Example:

Input  : A = 2, B = 4, C = 3, M = 23
Output : 6
243(mod 23) = 6

A Naive Approach is to calculate res = BC and then calculate Ares % M by modular exponential. The problem of this approach is that we can’t apply directly mod M on BC, so we have to calculate this value without mod M. But if we solve it directly then we will come up with the large value of exponent of A which will definitely overflow in final answer.

An Efficient approach is to reduce the BC to a smaller value by using the Fermat’s Little Theorem, and then apply Modular exponential.

According the Fermat's little
a(M - 1) = 1 (mod M) if M is a prime.

So if we rewrite BC as x*(M-1) + y, then the
task of computing ABC becomes Ax*(M-1) + y
which can be written as Ax*(M-1)*Ay.
From Fermat's little theorem, we know Ax*(M-1) = 1.
So task of computing ABC reduces to computing Ay

What is the value of y?
From BC = x * (M - 1) + y,
y can be written as BC % (M-1)

We can easily use the above theorem such that we can get
A ^ (B ^ C) % M = (A ^ y ) %  M

Now we only need to find two things as:-
1. y = (B ^ C) % (M - 1)
2. Ans = (A ^ y) % M

C++

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// C++ program to find (a^b) mod m for a large 'a'
#include<bits/stdc++.h>
using namespace std;
  
// Iterative Function to calculate (x^y)%p in O(log y)
unsigned int power(unsigned int x, unsigned int y, 
                                   unsigned int p)
{
    unsigned int res = 1;      // Initialize result
  
    x = x % p;  // Update x if it is more than or
                // equal to 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;
}
  
unsigned int Calculate(unsigned int A, unsigned int B,
                       unsigned int C, unsigned int M)
{
    unsigned int res, ans;
  
    // Calculate B ^ C (mod M - 1)
    res = power(B, C, M-1);
  
    // Calculate A ^ res ( mod M )
    ans = power(A, res, M);
  
    return ans;
}
  
// Driver program to run the case
int main()
{   // M must be be a Prime Number
    unsigned int A = 3, B = 9, C = 4, M = 19;
  
    cout << Calculate(A, B, C, M);
  
    return 0;
}

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Java














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// Java program to find (a^b)  


// mod m for a large 'a' 


import java.util.*; 


  


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;  


  


    while (y > 0) { 


          


    // If y is odd, multiply x with result 


    if (y % 2 != 0) 


        res = (res * x) % p; 


  


    // y must be even now 


    y = y >> 1; // y = y/2 


    x = (x * x) % p; 


    } 


    return res; 


} 


  


static int Calculate(int A, int B, int C, int M) { 


    int res, ans; 


  


    // Calculate B ^ C (mod M - 1) 


    res = power(B, C, M - 1); 


  


    // Calculate A ^ res ( mod M ) 


    ans = power(A, res, M); 


  


    return ans; 


} 


  


// Driver code 


public static void main(String[] args) { 


      


    // M must be be a Prime Number 


    int A = 3, B = 9, C = 4, M = 19; 


  


    System.out.print(Calculate(A, B, C, M)); 


} 


} 


  


// This code is contributed by Anant Agarwal. 



















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Python














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# Python program to calculate the ans 


def calculate(A, B, C, M): 


  


    # Calculate B ^ C (mod M - 1) 


    res = pow(B, C, M-1) 


  


    # Calculate A ^ res ( mod M ) 


    ans = pow(A, res, M) 


  


    return ans 


  


# Driver program to run the case 


A = 3


B = 9


C = 4


  


# M must be Prime Number 


M = 19


print( calculate(A, B, C, M) ) 



















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C#














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// C# program to find (a^b) mod m for 


// a large 'a' 


using System; 


  


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;  


      


        while (y > 0) { 


              


            // If y is odd, multiply x  


            // with result 


            if (y % 2 != 0) 


                res = (res * x) % p; 


          


            // y must be even now 


            y = y >> 1; // y = y/2 


            x = (x * x) % p; 


        } 


        return res; 


    } 


      


    static int Calculate(int A, int B,  


                          int C, int M) 


    { 


        int res, ans; 


      


        // Calculate B ^ C (mod M - 1) 


        res = power(B, C, M - 1); 


      


        // Calculate A ^ res ( mod M ) 


        ans = power(A, res, M); 


      


        return ans; 


    } 


      


    // Driver code 


    public static void Main() 


    { 


          


        // M must be be a Prime Number 


        int A = 3, B = 9, C = 4, M = 19; 


      


        Console.Write(Calculate(A, B, C, M)); 


    } 


} 


  


// This code is contributed by nitin mittal. 



















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PHP














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<?php 


// PHP program to find  


// (a^b) mod m for a  


// large 'a' 


  


// Iterative Function 


// to calculate (x^y)%p 


// in O(log y) 


function power($x, $y, $p) 


{ 


    $res = 1; // Initialize result 


  


    $x = $x % $p; // Update x if it  


                  // is more than or  


                  // equal to 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; 


} 


  


function Calculate($A, $B, $C, $M) 


{ 


    $res; $ans; 


  


    // Calculate B ^ C 


    // (mod M - 1) 


    $res = power($B, $C, $M - 1); 


  


    // Calculate A ^  


    // res ( mod M ) 


    $ans = power($A, $res, $M); 


  


    return $ans; 


} 


  


// Driver Code 


  


// M must be be  


// a Prime Number 


$A = 3; $B = 9;  


$C = 4; $M = 19; 


  


echo Calculate($A, $B,   


               $C, $M); 


  


// This code is contributed 


// by ajit 


?> 



















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Output:




18



Time Complexity: O(log(B) + log(C))

Auxiliary space: O(1)


This article is contributed by  Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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