Question

After reading this question. I was wondering is it possible using O(1) space can we generate a random permutation of the sequence [1…n] with a uniform distribution using something like double hashing?

I tried this with a small example for the sequence [1,2,3,4,5] and it works. But it fails for scale for larger sets.

int h1(int k) {
    return 5 - (k % 7);
}

int h2(int k) {
    return (k % 3) + 1;
}

int hash(int k, int i) {
    return (h1(k) + i*h2(k)) % size;
}

int main() {
    for(int k = 0; k < 10; k++) {
        std::cout << "k=" << k <<  std::endl;
        for(int i = 0; i < 5; i++) {
            int q = hash(k, i);
            if(q < 0) q += 5;
            std::cout << q;
        }
        std::cout << std::endl;
    }
}

Answer 1

You can try another approach.

1. Take arbitrary integer number P that GCD(P, N) == 1 where GCD(P,
N) is greatest common divisor of P and N (e.g. GCD(70, 42) == 14,
   GCD(24, 35) == 1).
2. Get sequence K[i] ::= (P * i) mod N + 1, i from 1 to N
3. It’s proven that sequence K[i] enumerates all numbers between
   1 and N with no repeats (actually K[N + 1] == K[1] but that is not a problem because we need only first N numbers).

If you can efficiently generate such numbers P with uniform distribution (e.g. with a good random function) with using Euclidean algorithm to calculate GCD in O(log(N)) complexity you’ll get what you want.