Miller-Rabin 确定性素数检验 (C)

Question

我发现了一个 C 语言程序，它实现了 Miller-Rabin 素数测试的确定性变体 here。但是，修改后的代码（如下所示）在处理大于 2^32 的数字时不起作用，即使我使用 unsigned long long 数据类型来存储我的数字。它应该能够容纳最多 2^64 的整数。哪里出错了？

简而言之：我的问题是我的代码正确地确定了一个数字是否为质数，但前提是它小于 2^32，但事实并非如此，因为我最多可以存储 2^ 64


unsigned long long power(unsigned long long a, unsigned long long n, unsigned long long mod)
{
    unsigned long long power = a;
    unsigned long long result = 1;

    while (n)
    {
        if (n & 1)
            result = (result * power) % mod;
        power = (power * power) % mod;
        n >>= 1;
    }
    return result;
}

// n−1 = 2^s * d with d odd by factoring powers of 2 from n−1
unsigned long long witness(unsigned long long n, unsigned long long s, unsigned long long d, unsigned long long a)
{
    unsigned long long x = power(a, d, n);
    unsigned long long y;

    while (s) {
        y = (x * x) % n;
        if (y == 1 && x != 1 && x != n-1)
            return 0;
        x = y;
        --s;
    }
    if (y != 1)
        return 0;
    return 1;
}

/*
 * if n < 1,373,653, it is enough to test a = 2 and 3;
 * if n < 9,080,191, it is enough to test a = 31 and 73;
 * if n < 4,759,123,141, it is enough to test a = 2, 7, and 61;
 * if n < 1,122,004,669,633, it is enough to test a = 2, 13, 23, and 1662803;
 * if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11;
 * if n < 3,474,749,660,383, it is enough to test a = 2, 3, 5, 7, 11, and 13;
 * if n < 341,550,071,728,321, it is enough to test a = 2, 3, 5, 7, 11, 13, and 17.
 */

int is_prime_mr(unsigned long long n)
{
    if (((!(n & 1)) && n != 2 ) || (n < 2) || (n % 3 == 0 && n != 3))
        return 0;
    if (n <= 3)
        return 1;

    unsigned long long d = n / 2;
    unsigned long long s = 1;
    while (!(d & 1)) {
        d /= 2;
        ++s;
    }
    unsigned long long a1 = 4759123141;
    unsigned long long a2 = 1122004669633;
    unsigned long long a3 = 2152302898747;
    unsigned long long a4 = 3474749660383;
    if (n < 1373653)
        return witness(n, s, d, 2) && witness(n, s, d, 3);
    if (n < 9080191)
        return witness(n, s, d, 31) && witness(n, s, d, 73);
    if (n < a1)
        return witness(n, s, d, 2) && witness(n, s, d, 7) && witness(n, s, d, 61);
    if (n < a2)
        return witness(n, s, d, 2) && witness(n, s, d, 13) && witness(n, s, d, 23) && witness(n, s, d, 1662803);
    if (n < a3)
        return witness(n, s, d, 2) && witness(n, s, d, 3) && witness(n, s, d, 5) && witness(n, s, d, 7) && witness(n, s, d, 11);
    if (n < a4)
        return witness(n, s, d, 2) && witness(n, s, d, 3) && witness(n, s, d, 5) && witness(n, s, d, 7) && witness(n, s, d, 11) && witness(n, s, d, 13);
    return witness(n, s, d, 2) && witness(n, s, d, 3) && witness(n, s, d, 5) && witness(n, s, d, 7) && witness(n, s, d, 11) && witness(n, s, d, 13) && witness(n, s, d, 17);
}

int main()
{
  unsigned long long in1 = 4294967291;
  unsigned long long in2 = 4294967311;
  int a = is_prime_mr(in1); //4294967291 is the last prime below 2^32, should return 1 and does so correctly
  printf("%d\n",a);
  int b = is_prime_mr(in2); //4294967311 is the first prime after 2^32, should also return 1 but returns 0
  printf("%d",b);
  return 0;
}

Answer 1

正如@Michael Dorgan 所建议的，问题出在两个 64 位整数的乘法上，使用 powermod.

的不同实现解决了这个问题

//computes (a*b) (mod n)
unsigned long long mul_mod(unsigned long long a, unsigned long long b, unsigned long long m)
{
   unsigned long long d = 0, mp2 = m >> 1;
   if (a >= m) a %= m;
   if (b >= m) b %= m;
   for (int i = 0; i < 64; ++i)
   {
       d = (d > mp2) ? (d << 1) - m : d << 1;
       if (a & 0x8000000000000000ULL)
           d += b;
       if (d >= m) d -= m;
       a <<= 1;
   }
   return d;
}
//computes (a^b) (mod m)
unsigned long long powermod(unsigned long long a, unsigned long long b, unsigned long long m)
{
    unsigned long long r = m==1?0:1;
    while (b > 0) {
        if (b & 1)
            r = mul_mod(r, a, m);
        b = b >> 1;
        a = mul_mod(a, a, m);
    }
    return r;
}

我的代码现在适用于最大 2^64 的整数。

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