[英]Maximum Simple Cycle in Directed Graph
你好,
我构造了一个有向图G =(v,e),我想在此图中找到最大的简单循环,在该循环中,边权重被相乘而不是相加。 我这样做的最初方法是构造一个新图G =(v,e'),其中e'(i,j)= 1 / e(i,j)/ min(e'),然后应用Floyd-Warshall在此图上查找所有最短路径。 我的想法是,在反转图形之后,最大路径将变为最小,并且如果我们将其除以最小值,则所有边缘权重都将为“正”(> = 1,因为我们是相乘而不是相加)。 但是,当我运行算法时(下面是我的Python代码),它似乎不起作用,我想知道这是因为我的算法根本不起作用,还是因为我的代码错误。
#construct G' = (V,dist) for our modified Floyd-Warshall algorithm
# edge weights are initially the inverse of w(u,v). They are then all divided
# by the minimum value to ensure all are >= 1
dist = {}
nxt = {}
minimum = float("inf")
for i in e:
dist[i] = {}
nxt[i] = {}
for j in e[i]:
dist[i][j] = 1/e[i][j]
nxt[i][j] = j
if dist[i][j] < minimum:
minimum = dist[i][j]
for i in dist:
for j in dist[i]:
dist[i][j] /= minimum
# Perform Floyd-Warshall
for k in v:
for i in v:
for j in v:
try:
one = dist[i][j]
two = dist[i][k]
three = dist[k][j]
except KeyError:
continue
if one > two * three:
dist[i][j] = two * three
nxt[i][j] = nxt[i][k]
# Find the shortest cycle using shortest paths
minimum = float("inf")
for i in v:
for j in v:
if i == j:
continue
try:
one = dist[i][j]
two = dist[j][i]
except KeyError:
continue
if one * two < minimum:
minimum = one * two
pair = [i,j]
def Path(u,v):
if nxt[u][v] == None:
return []
path = [u]
while u != v:
u = nxt[u][v]
path.append(u)
return path
# Format the cycle for output
p1 = Path(pair[0],pair[1])
p2 = Path(pair[1],pair[0])
p = p1 + p2[1:]
print(p)
# Find the total value of the cycle
value = 1
for i in range(len(p)-1):
value *= e[p[i]][p[i+1]]
print('The above cycle has a %f%% weight.' % ((value-1)*100))
我用图形G =(V,E)测试了上面的示例,其中
V = {a,b,c,d}, and
E = {
(a,b): 1/0.00005718 * 0.9975,
(a,c): 1/0.03708270 * 0.9975,
(a,d): 18590.00000016 * 0.9975,
(b,a): 0.00010711 * 0.9975,
(b,c): 0.00386491 * 0.9975,
(c,a): 0.03700994 * 0.9975,
(c,b): 1/18590.00000017 * 0.9975,
(c,d): 688.30000000 * 0.9975,
(d,a): 1/18590.00000017 * 0.9975,
(d,c): 1/688.30000000 * 0.9975
}
上图的输出是周期[a,d,a]
最好,权重为86.385309%。 但是,正如我们所看到的,周期[a,b,c,a]
权重为148.286055%,这要好得多,这使我相信算法错误或在某处出错。
任何建议深表感谢!
我认为问题不是实现,而是算法。 确实采用以下具有四个顶点a,b,c和d以及以下边的示例:
W(A,B)= 10/3
瓦特(B,C)= 10
瓦特(C,d)= 5
瓦特(d,A)= 10/3
瓦特(d,B)= 5
然后,您的算法将返回有向循环(b,c,d,b),而最佳解决方案是(a,b,c,d,a)。
此外,您还应该知道问题可能是NP完全的,因为最长路径问题是NP完全的(即使Shortest Path problem
可以在多项式上解决),所以只有少数希望有一个如此简单的算法为您的问题。
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