modular inverse fermat little theorem Algorithm
If a is not divisible by p, Fermat's small theorem is equivalent to the statement that ap − 1 − 1 is an integer multiple of p. Fermat’s small theorem is the basis for the Fermat primality test and is one of the fundamental outcomes of elementary number theory. Pierre de Fermat first stated the theorem in a letter dated October 18, 1640, to his friend and confidant Frénicle de Bessy.et en tous nombres premiers; de quoiHis formulation is equivalent to the following: If p is a prime and a is any integer not divisible by p, then a p − 1 − 1 is divisible by p.
/*
* C++ Program to find the modular inverse using Fermat's Little Theorem.
* Fermat's Little Theorem state that => ϕ(m) = m-1, where m is a prime number.
*
* (a * x) ≡ 1 mod m.
* x ≡ (a^(-1)) mod m.
*
* Using Euler's theorem we can modify the equation.
*
* (a^ϕ(m)) ≡ 1 mod m (Where '^' denotes the exponent operator)
* Here 'ϕ' is Euler's Totient Function. For modular inverse existence 'a' and 'm' must be relatively primes numbers.
* To apply Fermat's Little Theorem is necessary that 'm' must be a prime number.
* Generally in many competitive programming competitions 'm' is either 1000000007 (1e9+7) or 998244353.
*
* We considered m as large prime (1e9+7).
* (a^ϕ(m)) ≡ 1 mod m (Using Euler's Theorem)
* ϕ(m) = m-1 using Fermat's Little Theorem.
* (a^(m-1)) ≡ 1 mod m
* Now multiplying both side by (a^(-1)).
* (a^(m-1)) * (a^(-1)) ≡ (a^(-1)) mod m
* (a^(m-2)) ≡ (a^(-1)) mod m
*
* We will find the exponent using binary exponentiation. Such that the algorithm works in O(log(m)) time.
*
* Example: -
* a = 3 and m = 7
* (a^(-1) mod m) is equivalent to (a^(m-2) mod m)
* (3^(5) mod 7) = (243 mod 7) = 5
* Hence, ( 3^(-1) mod 7 ) = 5
* or ( 3 * 5 ) mod 7 = 1 mod 7 (as a*(a^(-1)) = 1)
*/
#include<iostream>
#include<vector>
// Recursive function to calculate exponent in O(log(n)) using binary exponent.
int64_t binExpo(int64_t a, int64_t b, int64_t m) {
a %= m;
int64_t res = 1;
while (b > 0) {
if (b%2) {
res = res * a % m;
}
a = a * a % m;
// Dividing b by 2 is similar to right shift.
b >>= 1;
}
return res;
}
// Prime check in O(sqrt(m)) time.
bool isPrime(int64_t m) {
if (m <= 1) {
return false;
} else {
for (int i=2; i*i <= m; i++) {
if (m%i == 0) {
return false;
}
}
}
return true;
}
int main() {
int64_t a, m;
// Take input of a and m.
std::cout << "Computing ((a^(-1))%(m)) using Fermat's Little Theorem";
std::cout << std::endl << std::endl;
std::cout << "Give input 'a' and 'm' space separated : ";
std::cin >> a >> m;
if (isPrime(m)) {
std::cout << "The modular inverse of a with mod m is (a^(m-2)) : ";
std::cout << binExpo(a, m-2, m) << std::endl;
} else {
std::cout << "m must be a prime number.";
std::cout << std::endl;
}
}
