基于2个输入的伪随机数生成器 [英] A pseudo-random number generator based on 2 inputs

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 xx哈希：

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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