基于2个输入的伪随机数生成器

问题描述

我需要基于2个输入值X和Y生成一个伪随机数。给定相同的X和Y值，我需要得到相同的结果。结果应该在0到1之间（含0和1）。

I need to generate a pseudo-random number based on 2 input values X and Y. Given the same X and Y values I need to get the same result. The result should be between 0 and 1 inclusive.

到目前为止，我已经做到了：

So far I have this:

const int a = 0x7fffffff / 48271;
const int b = 0x7fffffff % 48397;

float rand(int x, int y) {
    float seed, result;

    seed = x ^ ((y << 1) & 0x2AAAAAAA) ^ ((y >> 1) & 0x33333333);
    result = 48353 * (seed % a) - b * (seed / a);

    return (result);
}

这给了我一个结果，但不是我想要的。我从网上看到的随机事物中整理了一下，所以不知道它是否真的有用。

It's giving me a result but not what I'm looking for. I've cobbled it together from random things I've seen on the net, so no idea if it's really any good.

推荐答案

从：

float rand(uint32_t x, uint32_t y) {
  /* mix around the bits in x: */
  x = x * 3266489917 + 374761393;
  x = (x << 17) | (x >> 15);

  /* mix around the bits in y and mix those into x: */
  x += y * 3266489917;

  /* Give x a good stir: */
  x *= 668265263;
  x ^= x >> 15;
  x *= 2246822519;
  x ^= x >> 13;
  x *= 3266489917;
  x ^= x >> 16;

  /* trim the result and scale it to a float in [0,1): */
  return (x & 0x00ffffff) * (1.0f / 0x1000000);
}

通常的想法是服从 x 和 y 进行各种1：1转换，并将它们混合在一起以在结果中均匀分布所有输入位。然后将结果浮点数设为[0,1）。我已经从可能的输出中排除了1.0，因为包含它可能有点儿麻烦。

The general idea is to subject x and y to a variety of 1:1 transforms and to mix those together to distribute all of the input bits evenly(ish) throughout the result. Then the result in floating-point to [0,1). I've excluded 1.0 from the possible outputs because including it turns out to be kind of fiddly.

与任何奇数的乘积（无符号溢出）为1：1之所以进行转换，是因为奇数均是具有2的幂的互质数（ uint32_t 的范围限制）。不幸的是，乘法只允许低阶位影响高阶位。它不允许高位影响低位。为了弥补这一点，我们有一些 x ^ = x>> k 项，以将高位混合到低位。

Multiplication by any odd number, with unsigned overflow, is a 1:1 transform because odd numbers are all co-prime with powers of two (the range limit of a uint32_t). Unfortunately multiplication only allows low order bits to affect high order bits; it doesn't allow high bits to affect low. To make up for that, we have a few x ^= x >> k terms, to mix high bits into low positions.

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