有一个谷歌搜索发现大量有关生成的整数n成m部分所有可能的分区，但我还没有发现有关抽样n的均匀分布的随机划分成m部分东西。

A Google search reveals plenty about generating all possible partitions of an integer n into m parts, but I haven't found anything about sampling a uniformly distributed random partition of n into m parts.

推荐答案

下面是一些code，做它。这是O（ N 的 2 ）的第一次调用它，但它构建了一个高速缓存，以使后续调用是O（ N 的）。

Here is some code that does it. This is O(n2) the first time you call it, but it builds a cache so that subsequent calls are O(n).

import random

cache = {}

def count_partitions(n, limit):
    if n == 0:
        return 1
    if (n, limit) in cache:
        return cache[n, limit]
    x = cache[n, limit] = sum(count_partitions(n-k, k) for k in range(1, min(limit, n) + 1))
    return x

def random_partition(n):
    a = []
    limit = n
    total = count_partitions(n, limit)
    which = random.randrange(total)
    while n:
        for k in range(1, min(limit, n) + 1):
            count = count_partitions(n-k, k)
            if which < count:
                break
            which -= count
        a.append(k)
        limit = k
        n -= k
    return a

如何工作的：我们可以计算出一个整数的多个分区的 N 的还有在O（ N 的 2 ）的时间。作为一个副作用，这产生的O号的表（ N 的 2 ），我们就可以用它来生成 K 个分区 N 的，对于任何整数的 K 的，在O（ N 的）时间。

How this works: We can calculate how many partitions of an integer n there are in O(n2) time. As a side effect, this produces a table of size O(n2) which we can then use to generate the kth partition of n, for any integer k, in O(n) time.

因此​​，让我们的总的=分区的数量。从0选择一个随机数的 K 的到的总的 - 1.生成的 K 个分区

So let total = the number of partitions. Pick a random number k from 0 to total - 1. Generate the kth partition.