Question

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.

Answer 1

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

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.

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