Fast Prime Factorization Algorithm

问题

I'm writing a code in C that returns the number of times a positive integer can be expressed as sums of perfect squares of two positive integers.

R(n) is the number of couples (x,y) such that x² + y² = n where x, y, n are all 
non negative integers.

To compute R(n), I need to first find the prime factorization of n.

The problem is that I've tried a lot of algorithm for prime factorization that I can use on C but I need my code to be as fast as possible, so I would appreciate it if anyone can give me what he/she considers as the fastest algorithm to compute the prime factorization of a number as large as 2147483742.

回答1:

What an odd limit; 2147483742 = 2^31 + 94.

As others have pointed out, for a number this small trial division by primes is most likely fast enough. Only if it isn't, you could try Pollard's rho method:

/* WARNING! UNTESTED CODE! */
long rho(n, c) {
    long t = 2;
    long h = 2;
    long d = 1;

    while (d == 1) {
        t = (t*t + c) % n;
        h = (h*h + c) % n;
        h = (h*h + c) % n;
        d = gcd(t-h, n); }

    if (d == n)
        return rho(n, c+1);
    return d;
}

Called as rho(n,1), this function returns a (possibly-composite) factor of n; put it in a loop and call it repeatedly if you want to find all the factors of n. You'll also need a primality checker; for your limit, a Rabin-Miller test with bases 2, 7 and 61 is proven accurate and reasonably fast. You can read more about programming with prime numbers at my blog.

But in any case, given such a small limit I think you are better off using trial division by primes. Anything else might be asymptotically faster but practically slower.

EDIT: This answer has received several recent upvotes, so I'm adding a simple program that does wheel factorization with a 2,3,5-wheel. Called as wheel(n), this program prints the factors of n in increasing order.

long wheel(long n) {
    long ws[] = {1,2,2,4,2,4,2,4,6,2,6};
    long f = 2; int w = 0;

    while (f * f <= n) {
        if (n % f == 0) {
            printf("%ld\n", f);
            n /= f;
        } else {
            f += ws[w];
            w = (w == 10) ? 3 : (w+1);
        }
    }
    printf("%ld\n", n);

    return 0;
}

I discuss wheel factorization at my blog; the explanation is lengthy, so I won't repeat it here. For integers that fit in a long, it is unlikely that you will be able to significantly better the wheel function given above.

回答2:

There's a fast way to cut down the number of candidates. This routine tries 2, then 3, then all the odd numbers not divisible by 3.

long mediumFactor(n)
{
    if ((n % 2) == 0) return 2;
    if ((n % 3) == 0) return 3;
    try = 5;
    inc = 2;
    lim = sqrt(n);
    while (try <= lim)
    {
       if ((n % try) == 0) return try;
       try += inc;
       inc = 6 - inc;  // flip from 2 -> 4 -> 2
    }
    return 1;  // n is prime
}

The alternation of inc between 2 and 4 is carefully aligned so that it skips all even numbers and numbers divisible by 3. For this case: 5 (+2) 7 (+4) 11 (+2) 13 (+4) 17

Trials stop at sqrt(n) because at least one factor must be at or below the square root. (If both factors were > sqrt(n) then the product of the factors would be greater than n.)

Number of tries is sqrt(m)/3, where m is the highest possible number in your series. For a limit of 2147483647, that yields a maximum of 15,448 divisions worst case (for a prime near 2147483647) including the 2 and 3 tests.

If the number is composite, total number of divisions is usually much less and will very rarely be more; even taking into account calling the routine repeatedly to get all the factors.

来源：https://stackoverflow.com/questions/12756335/fast-prime-factorization-algorithm

标签

algorithm

prime-factoring