Modular exponentiation (Recursive)
Given three numbers a, b and c, we need to find (ab) % c
Now why do “% c” 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
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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 Codepublic 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 CodeA = 2B = 5C = 13print("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 Codepublic 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 // quadilateral 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 codevar A = 2, B = 5, C = 13; document.write("Power is " + exponentMod(A, B, C)); // This code is contributed by Khushboogoyal499</script> |
Power is 6
Iterative modular exponentiation.
