C中线性同余发生器的快速模乘模素数

Question

我正在尝试实现一个以梅森素数 (2³¹-1) 作为模数的随机数生成器。以下工作代码基于几个相关帖子：

How do I extract specific 'n' bits of a 32-bit unsigned integer in C?
Fast multiplication and subtraction modulo a prime

然而，

它不适用于 uint32_t hi, lo;，这意味着我不了解问题的签名与未签名方面。

根据上面的#2，我期待答案是 (hi+lo)。也就是说，我不明白为什么需要下面的语句。

   if (x1 > r)
        x1 += r + 2;

有人可以澄清我困惑的根源吗？
代码本身可以改进吗？
生成器应该避免使用 0 或 2³¹-1 作为种子吗？
素数 (2^p-k) 的代码会发生怎样的变化？

原码

#include <inttypes.h>
// x1 = a*x0 (mod 2^31-1)
int32_t lgc_m(int32_t a, int32_t x)
{
    printf("x %"PRId32"\n", x);
    if (x == 2147483647){
    printf("x1 %"PRId64"\n", 0); 
        return (0);
    }
    uint64_t  c, r = 1;
    c = (uint64_t)a * (uint64_t)x;
    if (c < 2147483647){
        printf("x1 %"PRId64"\n", c); 
        return (c);
    }
    int32_t hi=0, lo=0;
    int i, p = 31;//2^31-1
    for (i = 1; i < p; ++i){
       r |= 1 << i;
    }
   lo = (c & r) ;
   hi = (c & ~r) >> p;
   uint64_t x1 = (uint64_t ) (hi + lo);
   // NOT SURE ABOUT THE NEXT STATEMENT
   if (x1 > r)
        x1 += r + 2; 
   printf("c %"PRId64"\n", c);
   printf("r %"PRId64"\n", r);
   printf("\tlo %"PRId32"\n", lo);
   printf("\thi %"PRId32"\n", hi);
   printf("x1 %"PRId64"\n", x1); 
   printf("\n" );
   return((int32_t) x1);
}

int main(void)
{
    int32_t r;
    r = lgc_m(1583458089, 1);
    r = lgc_m(1583458089, 2000000000);
    r = lgc_m(1583458089, 2147483646);
    r = lgc_m(1583458089, 2147483647);
    return(0);
}

Answer 1

Sue 提供了这个解决方案：

With some experimentation (new code at the bottom), I was able to use uint32_t, which further suggests that I do not understand how the signed integers work with bit operations.

The following code uses uint32_t for input as well as hi and lo.
 #include <inttypes.h>
  // x1 = a*x0 (mod 2^31-1)
 uint32_t lgc_m(uint32_t a, uint32_t x)
  {
    printf("x %"PRId32"\n", x);
    if (x == 2147483647){
    printf("x1 %"PRId64"\n", 0); 
        return (0);
    }
    uint64_t  c, r = 1;
    c = (uint64_t)a * (uint64_t)x;
    if (c < 2147483647){
        printf("x1 %"PRId64"\n", c); 
        return (c);
    }
    uint32_t hi=0, lo=0;
    int i, p = 31;//2^31-1
    for (i = 1; i < p; ++i){
       r |= 1 << i;
    }
   hi = c >> p;
   lo = (c & r) ;
   uint64_t x1 = (uint64_t ) ((hi + lo) );
   // NOT SURE ABOUT THE NEXT STATEMENT
   if (x1 > r){
       printf("x1 - r = %"PRId64"\n", x1- r);
           x1 -= r; 
   }
   printf("c %"PRId64"\n", c);
   printf("r %"PRId64"\n", r);
   printf("\tlo %"PRId32"\n", lo);
   printf("\thi %"PRId32"\n", hi);
   printf("x1 %"PRId64"\n", x1); 
   printf("\n" );
   return((uint32_t) x1);
  }

  int main(void)
 {
    uint32_t r;
    r = lgc_m(1583458089, 1583458089);
    r = lgc_m(1583458089, 2147483645);
    return(0);
  }
The issue was that my assumption that the reduction will be complete after one pass. If (x > 2³¹-1), then by definition the reduction has not occurred and a second pass is necessary. Subtracting 2³¹-1, in that case does the trick. In the second attempt above, r = 2^31-1 and is therefore the modulus. x -= r achieves the final reduction.

Perhaps someone with expertise in random numbers or modular reduction could explain it better.

Cleaned function without printf()s.
uint32_t lgc_m(uint32_t a, uint32_t x){
    uint64_t c, x1, m = 2147483647; //modulus: m = 2^31-1
    if (x == m)
        return (0);
    c = (uint64_t)a * (uint64_t)x;
    if (c < m)//no reduction necessary
        return (c);
    uint32_t hi, lo, p = 31;//2^p-1, p = 31 
    hi = c >> p;
    lo = c & m;
    x1 = (uint64_t)(hi + lo);
    if (x1 > m){//one more pass needed 
       //this block can be replaced by x1 -= m;
        hi = x1 >> p;
        lo = (x1 & m);
        x1 = (uint64_t)(hi + lo);
    }
   return((uint32_t) x1);
}

Answer 2

下面的if语句

if (x1 > r)
    x1 += r + 2;

应该写成

if (x1 > r)
    x1 -= r;

两个结果都是相同的模 2^31:

x1 + r + 2 = x1 + 2^31 - 1 + 2 = x1 + 2^31 + 1
x1 - r = x1 - (2^31 - 1) = x1 - 2^31 + 1

第一个解决方案溢出 int32_t 并假设从 uint64_t 到 int32_t 的转换是模 2^31。虽然许多 C 编译器以这种方式处理转换，但这并不是 C 标准强制要求的。实际结果是实现定义的。

第二种解决方案避免了溢出并适用于 int32_t 和 uint32_t。

您还可以为 r 使用整数常量：

uint64_t r = 0x7FFFFFFF; // 2^31 - 1

或者干脆

uint64_t r = INT32_MAX;

编辑： 对于 2^p-k 形式的质数，您必须使用带 p 位的掩码并使用

计算结果

uint32_t x1 = (k * hi + lo) % ((1 << p) - k)

如果k * hi + lo能溢出一个uint32_t（也就是(k + 1) * (2^p - 1) >= 2^32），就得用64位算术：

uint32_t x1 = ((uint64_t)a * x) % ((1 << p) - k)

根据平台的不同，后者可能更快。

C中线性同余发生器的快速模乘模素数

Fast modular multiplication modulo prime for linear congruential generator in C

c

random

primes

modular-arithmetic

原码