Question

I would like to find all the possible combinations of weighted elements in a set where the sum of their weights is exactly equal to a given weight W

Say I want to select k elements from the set { 'A', 'B', 'C', 'D', 'E' } where weights = {'A':2, 'B':1, 'C':3, 'D':2, 'E':1} and W = 4.

Then this would yield :
('A','B','E')
('A','D')
('B','C')
('B','D','E')
('C','E')

I realize the brute force way would be to find all permutations of the given set (with itertools.permutations) and splice out the first k elements with a weighted sum of W. But I’m dealing with at least 20 elements per set, which would be computationally expensive.

I think using a variant of knapsack would help, where only weight (not value) is considered and where the sum of weights must be equal to W (not inferior).

I want to implement this in python but any cs-theory hints would help. Bonus points for elegance!

Answer 1

Looping through all n! permutations is much too expensive. Instead, generate all 2^n subsets.

from itertools import chain, combinations

def weight(A):
    return sum(weights[x] for x in A)

# Copied from example at http://docs.python.org/library/itertools.html
def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in xrange(len(s) + 1))

[x for x in powerset({'A', 'B', 'C', 'D', 'E'}) if weight(x) == W]

yields

[('A', 'D'), ('C', 'B'), ('C', 'E'), ('A', 'B', 'E'), ('B', 'E', 'D')]

This can be converted to sorted tuples by changing the return part of the list comprehension to tuple(sorted(x)), or by replacing the list call in powerset with one to sorted.