Combinatorics in Python

Question

I have a sort of a one level tree structure as:

alt text

Where p are parent nodes, c are child nodes and b are hypothetical branches.

I want to find all combinations of branches under the constraint that only one parent can branch to only one child node, and two branches can not share parent and/or child.

E.g. if combo is the set of combinations:

combo[0] = [b[0], b[3]]
combo[1] = [b[0], b[4]]
combo[2] = [b[1], b[4]]
combo[3] = [b[2], b[3]]

I think that's all of them. =)

How can this be achived automaticly in Python for arbitrary trees of this structures i.e. the number of p:s, c:s and b:s are arbitrary.

EDIT:

It is not a tree but rather a bipartite directed acyclic graph

Your image suggests that there are branches available from every parent to every child. Do you assume this? — dhill, Nov 04 '10 at 10:37
@dhill - Does it? Parent node p1 does not branch to child c0. — Theodor, Nov 04 '10 at 10:40
@Space_C0wb0y - A what now? Please paste a Wikipedia reference or similar. ;) As for data structure, consider b as an object with variables p and c, like `b[0].p = 0`. — Theodor, Nov 04 '10 at 10:45
@Theodor: I meant a [bipartite](http://en.wikipedia.org/wiki/Bipartite_graph) [directed acyclic graph](http://en.wikipedia.org/wiki/Directed_acyclic_graph). — Björn Pollex, Nov 04 '10 at 10:46
I am verily confused: In your example graphic, branches b3 and b2 share the same parent and children, which is converse to the rule "two branches can not share parent and/or child". Any chance of updating the graphic ? — Richard Green, Nov 04 '10 at 11:06
@Richard - The branches in the graph are _hypothetical_ branches. I'll add this in the question. — Theodor, Nov 04 '10 at 11:56
@Richard b3 and b2 have mirror parent, children. What are you talking about? — aaronasterling, Nov 04 '10 at 11:58
@Muhammad - MS Visio 2010. If your a student you can sometimes get a free student licence from MSDN. — Theodor, Nov 11 '10 at 11:30

aaronasterling · Accepted Answer · 2010-11-04T12:13:15.313

5

Here's one way to do it. There are lot's of micro-optimizations that could be made but their efficacy would depend on the sizes involved.

import collections as co
import itertools as it

def unique(list_):
    return len(set(list_)) == len(list_)

def get_combos(branches):
    by_parent = co.defaultdict(list)

    for branch in branches:
        by_parent[branch.p].append(branch)

    combos = it.product(*by_parent.values())

    return it.ifilter(lambda x: unique([b.c for b in x]), combos)

I'm pretty sure that this is at least hitting optimal complexity as I don't see a way to avoid looking at every combination that is unique by parent.

edited Nov 04 '10 at 12:13

answered Nov 04 '10 at 11:11

aaronasterling

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I think you meant to have the arg to get_combos be branches, otherwise for branch in branches will throw an exception. – Philip Starhill Nov 04 '10 at 12:11
@Philip Good looking out. Fixed. – aaronasterling Nov 04 '10 at 12:13
1

what is `branches` meant to be? what python type? what contents? – abcd Sep 02 '15 at 02:58
^ i now see that this is addressed in the comments. still, it'd be nice to have it stated in the answer. – abcd Sep 02 '15 at 03:11

score 4 · Answer 2 · edited Nov 04 '10 at 10:45

4

Look at itertools combinatoric generators:

product()
permutations()
combinations()
combinations_with_replacement()

Looks like you can write an iterator to achieve what you want.

edited Nov 04 '10 at 10:45

Björn Pollex

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answered Nov 04 '10 at 10:43

Paulo Scardine

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Combinatorics in Python

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