Discard some sets in list with all possible combinations

Question

Let’s say I have 5 points in graph, v = [0,1,2,3,4].

The list with all possible combinations would look like this:

[(), (0,), (1,), (2,), (3,), (4,), (0, 1), (0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4),
(2, 3), (2, 4), (3, 4), (0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 3), (0, 2, 4), (0, 3, 4),
 (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4), (0, 1, 2, 3), (0, 1, 2, 4), (0, 1, 3, 4),
 (0, 2, 3, 4), (1, 2, 3, 4), (0, 1, 2, 3, 4)]

Now lets say I have the following edges:

[(0, 3), (1, 2), (2, 3), (2, 4), (3, 4)]

Lets take the (0,3) for example. As you can see that is a edge in the graph so all the combinations with 0 and 3 should be discarded, like (0,3), (0,1,3), etc.

How can I make a code that uses a condition to verify if the possible combination as 2 adjacent vertices?

The code to generate the possible combinations is the following:

list_combinations = list()
for n in range(len(N_vert) + 1):
    list_combinations += list(combinations(vert, n))    
    n_comb = len(list_combinations)

Asked By: Goncalo Freitas

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Source

Answer 1

You could check, for each combination, is one edge is a subset of it

vert = [0, 1, 2, 3, 4]
edges = [(0, 3), (1, 2), (2, 3), (2, 4), (3, 4)]
edges = [set(edge) for edge in edges]

list_combinations = []
for n in range(len(vert) + 1): # start loop at 1 to remove the empty tuple
    for comb in combinations(vert, n):
        if not any(edge.issubset(comb) for edge in edges):
            list_combinations.append(comb)

print(list_combinations)
# [(), (0,), (1,), (2,), (3,), (4,), (0, 1), (0, 2), (0, 4), (1, 3), (1, 4), (0, 1, 4)]

Answered By: azro

Discard some sets in list with all possible combinations

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