Question

I’m searching for an algorithm that generates all permutations of fixed-length partitions of an integer. Order does not matter.

For example, for n=4 and length L=3:

[(0, 2, 2), (2, 0, 2), (2, 2, 0),
 (2, 1, 1), (1, 2, 1), (1, 1, 2),
 (0, 1, 3), (0, 3, 1), (3, 0, 1), (3, 1, 0), (1, 3, 0), (1, 0, 3),
 (0, 0, 4), (4, 0, 0), (0, 4, 0)]

I bumbled about with integer partitions + permutations for partitions whose length is lesser than L; but that was too slow because I got the same partition multiple times (because [0, 0, 1] may be a permutation of [0, 0, 1] 😉

Any help appreciated, and no, this isn’t homework — personal interest 🙂

Answer 1

Okay. First, forget about the permutations and just generate the partitions of length L (as suggested by @Svein Bringsli). Note that for each partition, you may impose an ordering on the elements, such as >. Now just “count,” maintaining your ordering. For n = 4, k = 3:

(4, 0, 0)
(3, 1, 0)
(2, 2, 0)
(2, 1, 1)

So, how to implement this? It looks like: while subtracting 1 from position i and adding it to the next position maintains our order, subtract 1 from position i, add 1 to position i + 1, and move to the next position. If we’re in the last position, step back.

Here’s a little python which does just that:

def partition_helper(l, i, result):
    if i == len(l) - 1:
        return 
    while l[i] - 1 >= l[i + 1] + 1:
        l[i]        -= 1
        l[i + 1]    += 1
        result.append(list(l))
        partition_helper(l, i + 1, result)

def partition(n, k):
    l = [n] + [0] * (k - 1)
    result = [list(l)]
    partition_helper(l, 0, result)
    return result

Now you have a list of lists (really a list of multisets), and generating all permutations of each multiset of the list gives you your solution. I won’t go into that, there’s a recursive algorithm which basically says, for each position, choose each unique element in the multiset and append the permutations of the multiset resulting from removing that element from the multiset.