CODEWITHSUNDEEP
c++algorithmprimes
Becks
Becks

Reputation: 468

Deterministic Miller-Rabin implementation

I'm trying to implement a primality check function with a deterministic Miller-Rabin algorithm but the results are not always correct: when checking first 1,000,000 numbers it only founds 78,495 instead of 78,498.

This is obtained using [2, 7, 61] as a base which, according to wikipedia, should always be correct for values up to 4,759,123,141.
The interesting thing is that the 3 missing primes are exactly the ones componing the base (2, 7 and 61).

Why is this happening? The code I'm using is the following: 

T modular_power(T base, T exponent, T modulo) {
    base %= modulo;
    T result = 1;

    while (exponent > 0) {
        if (exponent % 2 == 1)
            result = (result * base) % modulo;
        base = (base * base) % modulo;
        exponent /= 2;
    }

    return result;
}

bool miller_rabin(const T& n, const vector<T>& witnesses) {
    unsigned int s = 0;
    T d = n - 1;
    while (d % 2 == 0) {
        s++;
        d /= 2;
    }

    for (const auto& a : witnesses) {
        if (modular_power<T>(a, d, n) == 1)
            continue;

        bool composite = true;
        for (unsigned int r = 0; r < s; r++) {
            if (modular_power<T>(a, (T) pow(2, r) * d, n) == n - 1) {
                composite = false;
                break;
            }
        }

        if (composite)
            return false;
    }

    return true;
}

bool is_prime(const T& n) {
    if (n < 4759123141)
        return miller_rabin(n, {2, 7, 61});
    return false; // will use different base
}

Upvotes: 2

Views: 988

Answers (1)

user555045
user555045

Reputation: 64913

Miller-Rabin indeed does not work when the base and the input are the same. What happens in that case is that ad mod n is zero (because a mod n is zero, so this is really raising zero to some irrelevant power), and the rest of the algorithm is unable to "escape" from zero and concludes that you're dealing with a composite.

As a special case of that, Miller-Rabin never works with an input of 2, because there is no base that can be selected. 2 itself is useless, so is 1, that leaves nothing.

Upvotes: 3

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