Python Integer Partitioning with given k partitions

Question

I'm trying to find or develop Integer Partitioning code for Python.

FYI, Integer Partitioning is representing a given integer n as a sum of integers smaller than n. For example, an integer 5 can be expressed as 4 + 1 = 3 + 2 = 3 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1

I've found a number of solutions for this. http://homepages.ed.ac.uk/jkellehe/partitions.php and http://code.activestate.com/recipes/218332-generator-for-integer-partitions/

However, what I really want is to restrict the number of partitions.

Say, # of partition k = 2, a program only need to show 5 = 4 + 1 = 3 + 2,

if k = 3, 5 = 3 + 1 + 1 = 2 + 2 + 1

Answer 1

First I want to thanks everyone for their contribution. I arrived here needing an algorithm for generating integer partitions with the following details :

Generate partitions of a number into EXACTLY k parts but also having MINIMUM and MAXIMUM constraints.

Therefore, I modified the code of "Snakes and Coffee" to accommodate these new requirements:

def partition_min_max(n, k, l, m):
    ''' n is the integer to partition, k is the length of partitions, 
    l is the min partition element size, m is the max partition element size '''
    if k < 1:
        raise StopIteration
    if k == 1:
        if n <= m and n>=l :
            yield (n,)
        raise StopIteration
    for i in range(l,m+1):
        for result in partition_min_max(n-i, k-1, i, m):
            yield result+(i,)



>>> x = list(partition_min_max(20 ,3, 3, 10 ))
>>> print(x)
>>> [(10, 7, 3), (9, 8, 3), (10, 6, 4), (9, 7, 4), (8, 8, 4), (10, 5, 5), (9, 6, 5), (8, 7, 5), (8, 6, 6), (7, 7, 6)]

Answer 2

Building upon previous answer with maximum and minimum constraints, we can optimize it be a little better . For eg with k = 16 , n = 2048 and m = 128 , there is only one such partition which satisfy the constraints(128+128+...+128). But the code searches unnecessary branches for an answer which can be pruned.

def partition_min_max(n,k,l,m):
#n is the integer to partition, k is the length of partitions, 
#l is the min partition element size, m is the max partition element size
    if k < 1:
        return
    if k == 1:
        if n <= m and n>=l :
            yield (n,)
        return
    if (k*128) < n: #If the current sum is too small to reach n
        return
    if k*1 > n:#If current sum is too big to reach n
        return
    for i in range(l,m+1):
        for result in partition_min_max(n-i,k-1,i,m):                
            yield result+(i,)

Answer 3

def part(n, k):
    def _part(n, k, pre):
        if n <= 0:
            return []
        if k == 1:
            if n <= pre:
                return [[n]]
            return []
        ret = []
        for i in range(min(pre, n), 0, -1):
            ret += [[i] + sub for sub in _part(n-i, k-1, i)]
        return ret
    return _part(n, k, n)

Example:

>>> part(5, 1)
[[5]]
>>> part(5, 2)
[[4, 1], [3, 2]]
>>> part(5, 3)
[[3, 1, 1], [2, 2, 1]]
>>> part(5, 4)
[[2, 1, 1, 1]]
>>> part(5, 5)
[[1, 1, 1, 1, 1]]
>>> part(6, 3)
[[4, 1, 1], [3, 2, 1], [2, 2, 2]]

UPDATE

Using memoization:

def part(n, k):
    def memoize(f):
        cache = [[[None] * n for j in xrange(k)] for i in xrange(n)]
        def wrapper(n, k, pre):
            if cache[n-1][k-1][pre-1] is None:
                cache[n-1][k-1][pre-1] = f(n, k, pre)
            return cache[n-1][k-1][pre-1]
        return wrapper

    @memoize
    def _part(n, k, pre):
        if n <= 0:
            return []
        if k == 1:
            if n <= pre:
                return [(n,)]
            return []
        ret = []
        for i in xrange(min(pre, n), 0, -1):
            ret += [(i,) + sub for sub in _part(n-i, k-1, i)]
        return ret
    return _part(n, k, n)

Answer 4

I've written a generator solution

def partitionfunc(n,k,l=1):
    '''n is the integer to partition, k is the length of partitions, l is the min partition element size'''
    if k < 1:
        raise StopIteration
    if k == 1:
        if n >= l:
            yield (n,)
        raise StopIteration
    for i in range(l,n+1):
        for result in partitionfunc(n-i,k-1,i):
            yield (i,)+result

This generates all the partitions of n with length k with each one being in order of least to greatest.

Just a quick note: Via cProfile, it appears that using the generator method is much faster than using falsetru's direct method, using the test function lambda x,y: list(partitionfunc(x,y)). On a test run of n=50,k-5, my code ran in .019 seconds vs the 2.612 seconds of the direct method.