Find power of power under mod of a prime

3.6

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

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)*Ayx*(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++

// 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;
}

Python

# 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) )
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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