Modular arithmetic is the branch of arithmetic mathematics related with the “mod” functionality. Basically, modular arithmetic is related with computation of “mod” of expressions. Expressions may have digits and computational symbols of addition, subtraction, multiplication, division or any other. Here we will discuss briefly about all modular arithmetic operations.
It states that, for any pair of integers a and b (b is positive), there exist two unique integers q and r such that:
a = b x q + r where 0 <= r < b
Example: If a = 20, b = 6 then q = 3, r = 2 20 = 6 x 3 + 2
Rule for modular addition is:
(a + b) mod m = ((a mod m) + (b mod m)) mod m
Example:
(15 + 17) % 7
= ((15 % 7) + (17 % 7)) % 7
= (1 + 3) % 7
= 4 % 7
= 4
The same rule is to modular subtraction. We don’t require much modular subtraction but it can also be done in the same way.
The Rule for modular multiplication is:
(a x b) mod m = ((a mod m) x (b mod m)) mod m
Example:
(12 x 13) % 5
= ((12 % 5) x (13 % 5)) % 5
= (2 x 3) % 5
= 6 % 5
= 1
The modular division is totally different from modular addition, subtraction and multiplication. It also does not exist always.
(a / b) mod m is not equal to ((a mod m) / (b mod m)) mod m.
This is calculated using the following formula:
(a / b) mod m = (a x (inverse of b if exists)) mod m
The modular inverse of a mod m exists only if a and m are relatively prime i.e. gcd(a, m) = 1. Hence, for finding the inverse of an under modulo m, if (a x b) mod m = 1 then b is the modular inverse of a.
Example: a = 5, m = 7 (5 x 3) % 7 = 1 hence, 3 is modulo inverse of 5 under 7.
Finding a^b mod m is the modular exponentiation. There are two approaches for this – recursive and iterative. Example:
a = 5, b = 2, m = 7
(5 ^ 2) % 7 = 25 % 7 = 4
There is often a need to efficiently calculate the value of xn mod m. This can be done in O(logn) time using the following recursion:
It is important that in the case of an even n, the value of xn/2 is calculated only once.
This guarantees that the time complexity of the algorithm is O(logn) because n is always halved when it is even.
The following function calculates the value of xn mod m:
int modpower(int x, int n, int m) { if (n == 0) return 1%m; long long u = modpower(x,n/2,m); u = (u*u)%m; if (n%2 == 1) u = (u*x)%m; return u; } |
C++
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int modpower( int x, int n, int m)
{
if (n == 0)
return 1%m;
long long u = modpower(x,n/2,m);
u = (u*u)%m;
if (n%2 == 1)
u = (u*x)%m;
return u;
}
int main()
{
cout<<modpower(5,2,7)<<endl;
return 0;
}
|
C
#include <stdio.h>
int modpower( int x, int n, int m)
{
if (n == 0)
return 1 % m;
long long u = modpower(x, n / 2, m);
u = (u * u) % m;
if (n % 2 == 1)
u = (u * x) % m;
return u;
}
int main()
{
printf ( "%d\n" , modpower(5, 2, 7));
return 0;
}
|
Java
import java.util.*;
class GFG {
public static int modpower( int x, int n, int m) {
if (n == 0 )
return 1 % m;
long u = modpower(x, n / 2 , m);
u = (u * u) % m;
if (n % 2 == 1 )
u = (u * x) % m;
return ( int )u;
}
public static void main(String[] args) {
System.out.println(modpower( 5 , 2 , 7 ));
}
}
|
Python
def modpower(x, n, m):
if n = = 0 :
return 1 % m
u = modpower(x, n / / 2 , m)
u = (u * u) % m
if n % 2 = = 1 :
u = (u * x) % m
return u
print (modpower( 5 , 2 , 7 ))
|
Javascript
function modpower(x, n, m) {
if (n == 0) {
return 1 % m;
}
let u = modpower(x, Math.floor(n / 2), m);
u = (u * u) % m;
if (n % 2 == 1) {
u = (u * x) % m;
}
return u;
}
console.log(modpower(5, 2, 7));
|
output:
4
Time complexity: O(logn), because n is always halved when it is even.
Fermat’s theorem states that
xm−1 mod m = 1
when m is prime and x and m are coprime. This also yields
xk mod m = xk mod (m−1) mod m.
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Last Updated :
24 Apr, 2023
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