C# ModInverse Function

Question

Is there a built in function that would allow me to calculate the modular inverse of a(mod n)? e.g. 19^-1 = 11 (mod 30), in this case the 19^-1 == -11==19;

Answer 1

The BouncyCastle Crypto library has a BigInteger implementation that has most of the modular arithmetic functions. It's in the Org.BouncyCastle.Math namespace.

Answer 2

int modInverse(int a, int n) 
{
    int i = n, v = 0, d = 1;
    while (a>0) {
        int t = i/a, x = a;
        a = i % x;
        i = x;
        x = d;
        d = v - t*x;
        v = x;
    }
    v %= n;
    if (v<0) v = (v+n)%n;
    return v;
}

Answer 3

Here is a slightly more polished version of Samuel Allan's algorithm. The TryModInverse method returns a bool value, that indicates whether a modular multiplicative inverse exists for this number and modulo.

public static bool TryModInverse(int number, int modulo, out int result)
{
    if (number < 1) throw new ArgumentOutOfRangeException(nameof(number));
    if (modulo < 2) throw new ArgumentOutOfRangeException(nameof(modulo));
    int n = number;
    int m = modulo, v = 0, d = 1;
    while (n > 0)
    {
        int t = m / n, x = n;
        n = m % x;
        m = x;
        x = d;
        d = checked(v - t * x); // Just in case
        v = x;
    }
    result = v % modulo;
    if (result < 0) result += modulo;
    if ((long)number * result % modulo == 1L) return true;
    result = default;
    return false;
}

Answer 4

Since .Net 4.0+ implements BigInteger with a special modular arithmetics function ModPow (which produces “X power Y modulo Z”), you don't need a third-party library to emulate ModInverse. If n is a prime, all you need to do is to compute:

a_inverse = BigInteger.ModPow(a, n - 2, n)

For more details, look in Wikipedia: Modular multiplicative inverse, section Using Euler's theorem, the special case “when m is a prime”. By the way, there is a more recent SO topic on this: 1/BigInteger in c#, with the same approach suggested by CodesInChaos.

Answer 5

There is no library for getting inverse mod, but the following code can be used to get it.

// Given a and b->ax+by=d
long[] u = { a, 1, 0 };
long[] v = { b, 0, 1 };
long[] w = { 0, 0, 0 };
long temp = 0;
while (v[0] > 0)
{
    double t = (u[0] / v[0]);
    for (int i = 0; i < 3; i++)
    {
        w[i] = u[i] - ((int)(Math.Floor(t)) * v[i]);
        u[i] = v[i];
        v[i] = w[i];
    }
}
// u[0] is gcd while u[1] gives x and u[2] gives y. 
// if u[1] gives the inverse mod value and if it is negative then the following gives the first positive value
if (u[1] < 0)
{
        while (u[1] < 0)
        {
            temp = u[1] + b;
            u[1] = temp;
        }
}