对于大学项目我正在尝试实施Bron–Kerbosch algorithm,即列出给定图表中的所有最大派系。
我正在尝试实现第一个算法(没有透视),但我的代码在Wikipedia's example上测试后没有产生所有答案,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], [])
为什么我只能得到答案的一部分?
答案 0 :(得分:8)
Python正在变得困惑,因为你正在修改它正在迭代的列表。
更改
for vertex in p:
到
for vertex in p[:]:
这将导致它迭代p的副本。
答案 1 :(得分:8)
正如@VaughnCato正确地指出错误正在迭代P[:]
。我认为值得注意的是,你可以“产生”这个结果,而不是打印,如下所示(在这个重构的代码中):
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]]
如果有人在将来寻找Bron-Kerbosch算法实现......
答案 2 :(得分:0)
Wikipedia中两种形式的Bron-Kerbosch算法的实现:
无需旋转
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}]
可旋转
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}]