在python中实现Bron–Kerbosch算法

Question

对于一个大学项目，我正在尝试实现Bron–Kerbosch algorithm，即在给定图中列出所有最大集团。

[我正在尝试实现第一个算法（不进行透视），但是在Wikipedia's example上对其进行测试后，我的代码没有得到所有答案，到目前为止，我的代码是：

# dealing with a graph as list of lists 
graph = [[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]]


#function determines the neighbors of a given vertex
def N(vertex):
    c = 0
    l = []
    for i in graph[vertex]:
        if i is 1 :
         l.append(c)
        c+=1   
    return l 

#the Bron-Kerbosch recursive algorithm
def bronk(r,p,x):
    if len(p) == 0 and len(x) == 0:
        print r
        return
    for vertex in p:
        r_new = r[::]
        r_new.append(vertex)
        p_new = [val for val in p if val in N(vertex)] # p intersects N(vertex)
        x_new = [val for val in x if val in N(vertex)] # x intersects N(vertex)
        bronk(r_new,p_new,x_new)
        p.remove(vertex)
        x.append(vertex)


    bronk([], [0,1,2,3,4,5], [])

任何帮助我为什么只得到答案的一部分？

Answer 1

Python感到困惑，因为您正在修改要迭代的列表。

更改

for vertex in p:

to

for vertex in p[:]:

这将导致它遍历p的副本。

您可以在http://effbot.org/zone/python-list.htm上了解有关此内容的更多信息。

Answer 2

正如@VaughnCato正确指出，该错误正在P[:]上进行迭代。我认为值得一提的是，您可以按照以下方式（在此重构代码中）“屈服”此结果，而不是打印：

def bronk2(R, P, X, g):
    if not any((P, X)):
        yield R
    for v in P[:]:
        R_v = R + [v]
        P_v = [v1 for v1 in P if v1 in N(v, g)]
        X_v = [v1 for v1 in X if v1 in N(v, g)]
        for r in bronk2(R_v, P_v, X_v, g):
            yield r
        P.remove(v)
        X.append(v)
def N(v, g):
    return [i for i, n_v in enumerate(g[v]) if n_v]

In [99]: list(bronk2([], range(6), [], graph))
Out[99]: [[0, 1, 4], [1, 2], [2, 3], [3, 4], [3, 5]]

如果将来有人在寻找Bron–Kerbosch算法的实现...

Answer 3

Wikipedia中两种形式的Bron-Kerbosch算法的实现：

无枢轴旋转

算法 BronKerbosch1（R，P，X）是如果 P 和 X都是空的[[then：报告R作为最大集团for each顶点v in P doBronKerbosch1（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}]

有旋转
算法 BronKerbosch2（R，P，X）是if P

和X均为空then报告R作为最大集团在P⋃X中选择一个枢轴点ufor each顶点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}]