Best way to find if two numbers are coprime?

Question

Two numbers a and b are said to be co-prime if gcd(a,b) is 1.

Now my question is about the algorithm to find out if given two numbers are co-prime.

Comments

You need to find the best algo to get the GCD of two numbers.

Answer 1

Consider Binary GCD Algorithm and if GCD is 1 then numbers are co-prime:
(Source http://en.wikipedia.org/wiki/Binary_GCD_algorithm)

The algorithm reduces the problem of finding the GCD by repeatedly applying these identities:

1. gcd(0, v) = v, because everything divides zero, and v is the largest number that divides v. Similarly, gcd(u, 0) = u. gcd(0, 0) is not typically defined, but it is convenient to set gcd(0, 0) = 0.
2. If u and v are both even, then gcd(u, v) = 2·gcd(u/2, v/2), because 2 is a common divisor.
3. If u is even and v is odd, then gcd(u, v) = gcd(u/2, v), because 2 is not a common divisor. Similarly, if u is odd and v is even, then gcd(u, v) = gcd(u, v/2).
4. If u and v are both odd, and u ≥ v, then gcd(u, v) = gcd((u − v)/2, v). If both are odd and u < v, then gcd(u, v) = gcd((v − u)/2, u). These are combinations of one step of the simple Euclidean algorithm, which uses subtraction at each step, and an application of step 3 above. The division by 2 results in an integer because the difference of two odd numbers is even.[3]
5. Repeat steps 2–4 until u = v, or (one more step) until u = 0. In either case, the GCD is 2kv, where k is the number of common factors of 2 found in step 2.

The algorithm requires O(n2)[4] worst-case time, where n is the number of bits in the larger of the two numbers. Although each step reduces at least one of the operands by at least a factor of 2, the subtract and shift operations take linear time for very large integers (although they're still quite fast in practice, requiring about one operation per word of the representation).

unsigned int gcd(unsigned int u, unsigned int v)
{
    // simple cases (termination)
    if (u == v)
        return u;

    if (u == 0)
        return v;

    if (v == 0)
        return u;

    // look for factors of 2
    if (~u & 1) // u is even
    {
        if (v & 1) // v is odd
            return gcd(u >> 1, v);
        else // both u and v are even
            return gcd(u >> 1, v >> 1) << 1;
    }

    if (~v & 1) // u is odd, v is even
        return gcd(u, v >> 1);

    // reduce larger argument
    if (u > v)
        return gcd((u - v) >> 1, v);

    return gcd((v - u) >> 1, u);
}