Implementing Bron–Kerbosch algorithm in python

Question

for a college project I\'m trying to implement the Bron–Kerbosch algorithm, that is, listing all maximal cliques in a given graph.

I\'m trying to implement the first alg

Answer 1

Implementation of two forms of the Bron-Kerbosch Algorithm from Wikipedia:

Without pivoting

algorithm BronKerbosch1(R, P, X) is
    if P and X are both empty then:
        report R as a maximal clique
    for each vertex v in P do
        BronKerbosch1(R ⋃ {v}, P ⋂ N(v), X ⋂ N(v))
        P := P \ {v}
        X := X ⋃ {v}

adj_matrix = [
    [0, 1, 0, 0, 1, 0],
    [1, 0, 1, 0, 1, 0],
    [0, 1, 0, 1, 0, 0],
    [0, 0, 1, 0, 1, 1],
    [1, 1, 0, 1, 0, 0],
    [0, 0, 0, 1, 0, 0]]

N = {
    i: set(num for num, j in enumerate(row) if j)
    for i, row in enumerate(adj_matrix)
}

print(N)
# {0: {1, 4}, 1: {0, 2, 4}, 2: {1, 3}, 3: {2, 4, 5}, 4: {0, 1, 3}, 5: {3}}

def BronKerbosch1(P, R=None, X=None):
    P = set(P)
    R = set() if R is None else R
    X = set() if X is None else X
    if not P and not X:
        yield R
    while P:
        v = P.pop()
        yield from BronKerbosch1(
            P=P.intersection(N[v]), R=R.union([v]), X=X.intersection(N[v]))
        X.add(v)

P = N.keys()
print(list(BronKerbosch1(P)))
# [{0, 1, 4}, {1, 2}, {2, 3}, {3, 4}, {3, 5}]

With pivoting

algorithm BronKerbosch2(R, P, X) is
    if P and X are both empty then
        report R as a maximal clique
    choose a pivot vertex u in P ⋃ X
    for each vertex v in P \ N(u):
        BronKerbosch2(R ⋃ {v}, P ⋂ N(v), X ⋂ N(v))
        P := P \ {v}
        X := X ⋃ {v}

---

import random  

def BronKerbosch2(P, R=None, X=None):
    P = set(P)
    R = set() if R is None else R
    X = set() if X is None else X
    if not P and not X:
        yield R
    try:
        u = random.choice(list(P.union(X)))
        S = P.difference(N[u])
    # if union of P and X is empty
    except IndexError:
        S = P
    for v in S:
        yield from BronKerbosch2(
            P=P.intersection(N[v]), R=R.union([v]), X=X.intersection(N[v]))
        P.remove(v)
        X.add(v)

print(list(BronKerbosch2(P)))
# [{0, 1, 4}, {1, 2}, {2, 3}, {3, 4}, {3, 5}]