如何从一个范围内生成一个随机整数 [英] How to generate a random integer number from within a range
问题描述
这是之前发布的问题的后续:
This is a follow on from a previously posted question:
我希望能够在特定范围内生成一个随机数，例如 1 到 6 来模拟骰子的侧面.
I wish to be able to generate a random number from within a particular range, such as 1 to 6 to mimic the sides of a die.
我该怎么做?
推荐答案
到目前为止所有的答案在数学上都是错误的.返回 rand() % N 不会统一给出 [0, N) 范围内的数字，除非 N 将区间的长度划分为rand() 返回(即 2 的幂).再者，不知道rand()的模数是否独立:有可能是0, 1, 2, ...，这是一致的，但不是很随意.唯一似乎合理的假设是 rand() 给出泊松分布:任何两个相同大小的非重叠子区间都是等可能且独立的.对于有限的一组值，这意味着均匀分布，并确保 rand() 的值很好地分散.
All the answers so far are mathematically wrong. Returning rand() % N does not uniformly give a number in the range [0, N) unless N divides the length of the interval into which rand() returns (i.e. is a power of 2). Furthermore, one has no idea whether the moduli of rand() are independent: it's possible that they go 0, 1, 2, ..., which is uniform but not very random. The only assumption it seems reasonable to make is that rand() puts out a Poisson distribution: any two nonoverlapping subintervals of the same size are equally likely and independent. For a finite set of values, this implies a uniform distribution and also ensures that the values of rand() are nicely scattered.
这意味着改变rand()范围的唯一正确方法是将其划分为多个框；例如，如果 RAND_MAX == 11 并且您想要 1..6 的范围，您应该将 {0,1} 分配给 1, {2,3} 到 2，依此类推.它们是不相交、大小相等的区间，因此均匀且独立地分布.
This means that the only correct way of changing the range of rand() is to divide it into boxes; for example, if RAND_MAX == 11 and you want a range of 1..6, you should assign {0,1} to 1, {2,3} to 2, and so on. These are disjoint, equally-sized intervals and thus are uniformly and independently distributed.
使用浮点除法的建议在数学上是合理的，但原则上存在舍入问题.或许 double 的精度足以让它工作；也许不是.我不知道，也不想弄清楚；无论如何，答案取决于系统.
The suggestion to use floating-point division is mathematically plausible but suffers from rounding issues in principle. Perhaps double is high-enough precision to make it work; perhaps not. I don't know and I don't want to have to figure it out; in any case, the answer is system-dependent.
正确的方法是使用整数算法.也就是说，您需要类似以下内容:
The correct way is to use integer arithmetic. That is, you want something like the following:
#include <stdlib.h> // For random(), RAND_MAX
// Assumes 0 <= max <= RAND_MAX
// Returns in the closed interval [0, max]
long random_at_most(long max) {
unsigned long
// max <= RAND_MAX < ULONG_MAX, so this is okay.
num_bins = (unsigned long) max + 1,
num_rand = (unsigned long) RAND_MAX + 1,
bin_size = num_rand / num_bins,
defect = num_rand % num_bins;
long x;
do {
x = random();
}
// This is carefully written not to overflow
while (num_rand - defect <= (unsigned long)x);
// Truncated division is intentional
return x/bin_size;
}
循环是获得完美均匀分布所必需的.例如，如果给你从 0 到 2 的随机数，而你只想要从 0 到 1 的随机数，你就一直拉直到你没有得到 2；不难检查这是否以相同的概率给出 0 或 1.这种方法也在 nos 在他们的答案中给出的链接中进行了描述，尽管编码不同.我正在使用 random() 而不是 rand() 因为它具有更好的分布(如 rand() 的手册页所述)).
The loop is necessary to get a perfectly uniform distribution. For example, if you are given random numbers from 0 to 2 and you want only ones from 0 to 1, you just keep pulling until you don't get a 2; it's not hard to check that this gives 0 or 1 with equal probability. This method is also described in the link that nos gave in their answer, though coded differently. I'm using random() rather than rand() as it has a better distribution (as noted by the man page for rand()).
如果你想获得默认范围[0, RAND_MAX]之外的随机值，那么你必须做一些棘手的事情.也许最权宜之计是定义一个函数 random_extended() 拉 n 位(使用 random_at_most())并返回 [0, 2**n)，然后应用 random_at_most() 和 random_extended() 代替 random() (和 2**n - 1 代替 RAND_MAX) 以提取小于 2**n 的随机值，假设您有一个数字可以保存这样一个值的类型.最后，当然，您可以使用 min + random_at_most(max - min) 获取 [min, max] 中的值，包括负值.
If you want to get random values outside the default range [0, RAND_MAX], then you have to do something tricky. Perhaps the most expedient is to define a function random_extended() that pulls n bits (using random_at_most()) and returns in [0, 2**n), and then apply random_at_most() with random_extended() in place of random() (and 2**n - 1 in place of RAND_MAX) to pull a random value less than 2**n, assuming you have a numerical type that can hold such a value. Finally, of course, you can get values in [min, max] using min + random_at_most(max - min), including negative values.
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