CODEWITHSUNDEEP
arrayspython-3.xsetpartitioning
elnaz irannezhad
elnaz irannezhad

Reputation: 339

partition of a set or all possible subgroups of a list

Let's say I have a list of [1,2,3,4] I want to produce all subsets of this set which covers all members once, the result should has 15 lists which the order isn't important, instead t provides all possible subgroups:

>>>>[[1,2,3,4]]
[[1][2][3][4]]
[[1,2],[3][4]]
[[1,2],[3,4]]
[[1][2],[3,4]]
[[1,3],[2][4]]
[[1,3],[2,4]]
[[1][3],[2,4]]
[[1],[2,3][4]]
[[1,4],[2,3]]
[[1][2,3,4]]
[[2][1,3,4]]
[[3][1,2,4]]
[[4][1,2,3]]

This is a set partitioning problem or partitions of a set which is discussed here, but the response made me confused as it just suggests recalling permutations, but I don't know how! and another response also doesn't include [1,3] Meanwhile I should solve this problem for high numbers and the result should provide Bell Number Sorry I'm quite new to python and became confused. Could anyone make t clear for me?

Upvotes: 2

Views: 1538

Answers (1)

tommy.carstensen
tommy.carstensen

Reputation: 9622

Instead of doing all permutations and remove the duplicates, which was my initial thought, then you can use this recursive function, which I found here and here:

def partitions(set_):
    if not set_:
        yield []
        return
    for i in range(int(2**len(set_)/2)):
        parts = [set(), set()]
        for item in set_:
            parts[i&1].add(item)
            i >>= 1
        for b in partitions(parts[1]):
            yield [parts[0]]+b

l = [1, 2, 3, 4]
for p in reversed(sorted(partitions(l))):
    print(p)
print('The Bell number is', len(list(partitions(l))))

It prints:

[{1, 2, 3, 4}]
[{1, 2, 4}, {3}]
[{1, 4}, {2, 3}]
[{1, 4}, {3}, {2}]
[{2, 4}, {1, 3}]
[{2, 4}, {3}, {1}]
[{1, 3, 4}, {2}]
[{2, 3, 4}, {1}]
[{3, 4}, {1, 2}]
[{3, 4}, {2}, {1}]
[{4}, {1, 2, 3}]
[{4}, {1, 3}, {2}]
[{4}, {2, 3}, {1}]
[{4}, {3}, {1, 2}]
[{4}, {3}, {2}, {1}]
The Bell number is 15

Upvotes: 1

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