std :: mersenne_twister_engine和随机数生成 [英] std::mersenne_twister_engine and random number generation

问题描述

如果执行以下操作，将产生什么分布(均匀分布，泊松分布，正态分布等)?输出似乎表明分布均匀.但是，为什么我们需要std::uniform_int_distribution?

What is the distribution (uniform, poisson, normal, etc.) that is generated if I did the below? The output appears to indicate a uniform distribution. But then, why do we need std::uniform_int_distribution?

int main()
{
  std::mt19937_64 generator(134);
  std::map<int, int> freq;
  const int size = 100000;
  for (int i = 0; i < size; ++i) {
    int r = generator() % size;
    freq[r]++;
  }
  for (auto f : freq) {
    std::cout << std::string(f.second, '*') << std::endl;
  }
  return 0;
}

谢谢！

推荐答案

因为generator()是[generator.min(), generator.max()]上的均匀分布，所以generator() % n不是[0, n)上的均匀分布(除非generator.max()是假设generator.min()== 0)n的精确倍数.

Because while generator() is an uniform distribution over [generator.min(), generator.max()], generator() % n is not a uniform distribution over [0, n) (unless generator.max() is an exact multiple of n, assuming generator.min() == 0).

让我们举个例子:min() == 0，max() == 65'535和n == 7.

Let's take an example: min() == 0, max() == 65'535 and n == 7.

gen()将给出范围为[0, 65'535]的数字，并且在以下范围内:

gen() will give numbers in the range [0, 65'535] and in this range there are:

• 9'363编号，例如gen() % 7 == 0
• 9'363编号，例如gen() % 7 == 1
• 9'362编号，例如gen() % 7 == 2
• 9'362编号，例如gen() % 7 == 3
• 9'362编号，例如gen() % 7 == 4
• 9'362编号，例如gen() % 7 == 5
• 9'362编号，例如gen() % 7 == 6

• 9'363 numbers such that gen() % 7 == 0
• 9'363 numbers such that gen() % 7 == 1
• 9'362 numbers such that gen() % 7 == 2
• 9'362 numbers such that gen() % 7 == 3
• 9'362 numbers such that gen() % 7 == 4
• 9'362 numbers such that gen() % 7 == 5
• 9'362 numbers such that gen() % 7 == 6

如果您想知道这些数字是从哪里得到的，请像这样:65'534是7(65'534 = 7 * 9'362)的精确倍数.这意味着在[0, 65'533]中，确切地是9'362个数字，它们通过执行gen() % 7映射到每个{0, 1, 2, 3, 4, 5, 6}.这样，将65'534映射到0，将65'535映射到1

If you are wondering where did I get these numbers think of it like this: 65'534 is an exact multiple of 7 (65'534 = 7 * 9'362). This means that in [0, 65'533] there are exactly 9'362 numbers who map to each of the {0, 1, 2, 3, 4, 5, 6} by doing gen() % 7. This leaves 65'534 who maps to 0 and 65'535 who maps to 1

因此，您看到对[0, 1]的偏爱要比对[2, 6]的偏爱，即

So you see there is a bias towards [0, 1] than to [2, 6], i.e.

• 0和1出现的几率(9'363 / 65'536 ≈ 14.28680419921875 %)比
• 2，3，4，5和6(9'362 / 65'536 ≈ 14.2852783203125‬ %).

• 0 and 1 have a slightly higher chance (9'363 / 65'536 ≈ 14.28680419921875 %)‬ of appearing than
• 2, 3, 4, 5 and 6 (9'362 / 65'536 ≈ 14.2852783203125‬ %).

std::uniformn_distribution不存在此问题，并使用了一些数学上的可可性，可能会从生成器中获取更多随机数以实现真正的均匀分布.

std::uniformn_distribution doesn't have this problem and uses some mathematical woodo with possibly getting more random numbers from the generator to achieve a truly uniform distribution.

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