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Modular exponentiation (Recursive)

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Given three numbers a, b and c, we need to find (ab) % c
Now why do ā€œ% cā€ after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store such a large number.
Examples: 
 

Input : a = 2312 b = 3434 c = 6789
Output : 6343

Input : a = -3 b = 5 c = 89 
Output : 24

Auxiliary Space: O(1)
 

The idea is based on below properties.
Property 1: 
(m * n) % p has a very interesting property: 
(m * n) % p =((m % p) * (n % p)) % p
Property 2: 
if b is even: 
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ? this suggests divide and conquer 
if b is odd: 
(a ^ b) % c = (a * (a ^( b-1))%c
Property 3: 
If we have to return the mod of a negative number x whose absolute value is less than y: 
then (x + y) % y will do the trick
Note: 
Also as the product of (a ^ b/2) * (a ^ b/2) and a * (a ^( b-1) may cause overflow, hence we must be careful about those scenarios 
 

C++




// Recursive C++ program to compute modular power
#include <bits/stdc++.h>
using namespace std;
 
int exponentMod(int A, int B, int C)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
 
    // If B is odd
    else {
        y = A % C;
        y = (y * exponentMod(A, B - 1, C) % C) % C;
    }
 
    return (int)((y + C) % C);
}
 
// Driver code
int main()
{
    int A = 2, B = 5, C = 13;
    cout << "Power is " << exponentMod(A, B, C);
    return 0;
}
 
// This code is contributed by SHUBHAMSINGH10


C




// Recursive C program to compute modular power
#include <stdio.h>
 
int exponentMod(int A, int B, int C)
{
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
 
    // If B is even
    long y;
    if (B % 2 == 0) {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
 
    // If B is odd
    else {
        y = A % C;
        y = (y * exponentMod(A, B - 1, C) % C) % C;
    }
 
    return (int)((y + C) % C);
}
 
// Driver program to test above functions
int main()
{
   int A = 2, B = 5, C = 13;
   printf("Power is %d", exponentMod(A, B, C));
   return 0;
}


Java




// Recursive Java program
// to compute modular power
import java.io.*;
 
class GFG
{
static int exponentMod(int A,
                       int B, int C)
{
         
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    long y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return (int)((y + C) % C);
}
 
// Driver Code
public static void main(String args[])
{
    int A = 2, B = 5, C = 13;
    System.out.println("Power is " +
                        exponentMod(A, B, C));
}
}
 
// This code is contributed
// by Swetank Modi.


Python3




# Recursive Python program
# to compute modular power
def exponentMod(A, B, C):
     
    # Base Cases
    if (A == 0):
        return 0
    if (B == 0):
        return 1
     
    # If B is Even
    y = 0
    if (B % 2 == 0):
        y = exponentMod(A, B / 2, C)
        y = (y * y) % C
     
    # If B is Odd
    else:
        y = A % C
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C
    return ((y + C) % C)
 
# Driver Code
A = 2
B = 5
C = 13
print("Power is", exponentMod(A, B, C))
     
# This code is contributed
# by Swetank Modi.


C#




// Recursive C# program
// to compute modular power
class GFG
{
static int exponentMod(int A, int B, int C)
{
         
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    long y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return (int)((y + C) % C);
}
 
// Driver Code
public static void Main()
{
    int A = 2, B = 5, C = 13;
    System.Console.WriteLine("Power is " +
                    exponentMod(A, B, C));
}
}
 
// This code is contributed
// by mits


PHP




<?php
// Recursive PHP program to
// compute modular power
function exponentMod($A, $B, $C)
{
    // Base cases
    if ($A == 0)
        return 0;
    if ($B == 0)
        return 1;
     
    // If B is even
    if ($B % 2 == 0)
    {
        $y = exponentMod($A, $B / 2, $C);
        $y = ($y * $y) % $C;
    }
     
    // If B is odd
    else
    {
        $y = $A % $C;
        $y = ($y * exponentMod($A, $B - 1,
                               $C) % $C) % $C;
    }
     
    return (($y + $C) % $C);
}
 
 
// Driver Code
$A = 2;
$B = 5;
$C = 13;
echo "Power is " . exponentMod($A, $B, $C);
 
// This code is contributed
// by Swetank Modi.
?>


Javascript




<script>
 
// Recursive Javascript program
// to compute modular power
 
// Function to check if a given
// quadrilateral is valid or not
function exponentMod(A, B, C)
{
     
    // Base cases
    if (A == 0)
        return 0;
    if (B == 0)
        return 1;
     
    // If B is even
    var y;
    if (B % 2 == 0)
    {
        y = exponentMod(A, B / 2, C);
        y = (y * y) % C;
    }
     
    // If B is odd
    else
    {
        y = A % C;
        y = (y * exponentMod(A, B - 1,
                             C) % C) % C;
    }
     
    return parseInt(((y + C) % C));
}
 
// Driver code
var A = 2, B = 5, C = 13;
document.write("Power is " +
               exponentMod(A, B, C));
     
// This code is contributed by Khushboogoyal499
 
</script>


Output: 

Power is 6

 

Time Complexity : O(logn)

Auxiliary Space: O(logn)

Iterative modular exponentiation.
 


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Last Updated : 06 Mar, 2023
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