似乎不太可能有一个小于O(n*|V|^2)
的解决方案,所以这里是一个Python中的方法,似乎并不太可怕。
# some basic topologies
def lineE(N):
return(set((i,i+1) for i in range(N-1)))
def ringE(N):
return(lineE(N).union([(0,N-1)]))
def fullE(N):
return(set([(i,j) for i in range(N) for j in range(i)]))
# propagate visitors from x to y
def propagate(V, curr, x, y, d):
nexty = set()
for cx in curr[x]:
if not cx in V[y]["seen"]:
V[y]["seen"].add(cx)
V[y]["foaf"][d] = V[y]["foaf"].get(d,0) + 1
nexty.add(cx)
return(nexty)
# run for D iterations
def mingle(N, E, D):
V = dict((i, {"seen":set([i]), "foaf":{0:1}}) for i in range(N))
curr = dict((i, set([i])) for i in range(N))
for d in range(1, min(D+1, N)):
next = dict((i, set()) for i in range(N))
for (h, t) in E:
next[t] = next[t].union(propagate(V, curr, h, t, d))
next[h] = next[h].union(propagate(V, curr, t, h, d))
curr = next
return(V)
尝试它具有10个节点和距离3,
N=10
D=3
for (topology, E) in [("line", lineE(N)), ("ring", ringE(N)), ("full", fullE(N))]:
V = mingle(N, E, D)
print "\n", topology
for v in V:
print v, V[v]["foaf"]
我们得到
line
0 {0: 1, 1: 1, 2: 1, 3: 1}
1 {0: 1, 1: 2, 2: 1, 3: 1}
2 {0: 1, 1: 2, 2: 2, 3: 1}
3 {0: 1, 1: 2, 2: 2, 3: 2}
4 {0: 1, 1: 2, 2: 2, 3: 2}
5 {0: 1, 1: 2, 2: 2, 3: 2}
6 {0: 1, 1: 2, 2: 2, 3: 2}
7 {0: 1, 1: 2, 2: 2, 3: 1}
8 {0: 1, 1: 2, 2: 1, 3: 1}
9 {0: 1, 1: 1, 2: 1, 3: 1}
ring
0 {0: 1, 1: 2, 2: 2, 3: 2}
1 {0: 1, 1: 2, 2: 2, 3: 2}
2 {0: 1, 1: 2, 2: 2, 3: 2}
3 {0: 1, 1: 2, 2: 2, 3: 2}
4 {0: 1, 1: 2, 2: 2, 3: 2}
5 {0: 1, 1: 2, 2: 2, 3: 2}
6 {0: 1, 1: 2, 2: 2, 3: 2}
7 {0: 1, 1: 2, 2: 2, 3: 2}
8 {0: 1, 1: 2, 2: 2, 3: 2}
9 {0: 1, 1: 2, 2: 2, 3: 2}
full
0 {0: 1, 1: 9}
1 {0: 1, 1: 9}
2 {0: 1, 1: 9}
3 {0: 1, 1: 9}
4 {0: 1, 1: 9}
5 {0: 1, 1: 9}
6 {0: 1, 1: 9}
7 {0: 1, 1: 9}
8 {0: 1, 1: 9}
9 {0: 1, 1: 9}
这似乎是正确的。另外,在我的笔记本电脑上运行带有100000个节点的距离为100的简单拓扑大约需要一分钟。当然,如果你有一个密集的图表(如fullE
),这将炸毁。
N=100000
D=100
for (topology, E) in [("line", lineE(N)), ("ring", ringE(N))]:
V = mingle(N, E, D)
[所有最短路径问题(http://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm)基本上是你的距离和计数分组后寻求的东西,你也许可以真的比O(| V |^3)好得多。 – Nuclearman 2013-04-09 19:41:40
我的广度优先搜索是O(| E |),在我的情况下它等于O(| V |)。我必须为每个节点都做,所以我目前的复杂度是O(| V |²)。我现在正在使用并行计算来加速这个过程,但其他建议是最受欢迎的! – 2013-04-11 12:03:19
它应该仍然是O(| V | * | E |),在最坏的情况下它是O(| V |^3)。但是,如果你在说| V |接近| E |,那么考虑到有可能需要列出最短路径的顶点的可能组合O(| V |^2),那么可能没有太多的工作要做。尽管如果大多数顶点的度数为2或更小,那么简单地列出最长路径(或足够长的路径)并从中提取最短路径可能是实用的。 – Nuclearman 2013-04-11 23:49:05