From the previous recipe http://code.activestate.com/recipes/577764-combinations-of-a-sequence-without-replacement-usi/ we can easily build a generator that return all the K-combinations without replacement.
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Copy to clipboard1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 | #!/usr/bin/env python
__version__ = "0.8"
"""xcombinations.py
This recipe shows a way to solve the K-combination problems without replacement
of a N items sequence using the dynamic programming technique.
Keywords: combination, binomial coefficient, generator
See also: http://code.activestate.com/recipes/577764-combinations-of-a-sequence-without-replacement-usi/
See also: http://code.activestate.com/recipes/190465-generator-for-permutations-combinations-selections/
See also: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/105962
See also: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/66463
See also: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/66465
"""
def xcombinations(s, K):
"""
On entry: s sequence of items; K size of the combinations
Returns: generator of the K-combinations without replacement of
s.
"""
N = len(s)
assert K<=N, 'Error K must be less or igual than N'
S = [[] for i in range(K) ]
for n in range(1,N+1):
newS = [[] for i in range(K) ]
for k in range(max(1, n-(N-K)), min(K+1, n+1)):
if k == K:
for el in S[k-1]:
yield el+[s[n-1]]
else:
newS[k].extend(S[k])
if len(S[k-1])==0:
newS[k].append([s[n-1]])
else:
newS[k].extend( [el+[s[n-1]] for el in S[k-1]] )
S = newS
if __name__ == '__main__':
s = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
for c in xcombinations(s, 5):
print c
|
For each S(n, K) where n=0...N appears the partial solutions of S(N, K). So, it returns a combination every time it's available. Now the vector S and newS, representing the matrix solutions, are no longer with size K because they doesn't need to store all the combinations up to the final execution of the function.
