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关于k-连通图的问题

[英]Question about k-Connected Graphs

原文 2011-05-07 02:47:57 4 1 java

Given an undirected Graph, G, is there any standard algorithm to find the value of k, where (k-1) represents the number of vertices whose removal results in a graph that is still connected and the removal of the kth vertex makes the graph disconnected? 给定无向图G，是否有任何标准算法可以找到k的值，其中（k-1）表示顶点的数量，这些顶点的去除结果仍然是连接的图，而第k个顶点的去除使得该图断开连接？

Thank you! 谢谢！ Hop 跳

1 个解决方案

I don't know of any standard algorithm, but for a graph to have this property, every pair of vertices must have >= k independent paths between them (its a simple proof by contradiction to see that this is the case). 我不知道任何标准算法，但是要使图具有此属性，每对顶点之间必须具有> = k条独立的路径（这是一个矛盾的简单证明，事实就是如此）。

So a potential algorithm would be to check that for all pairs of vertices in your graph there are at least K independent paths. 因此，一种潜在的算法将是检查图中的所有顶点对是否至少有K条独立的路径。 To find this you can use a Maximum Flow algorithm . 为了找到这个，您可以使用最大流量算法。 Unfortunately doing this trivially will probably take a long time. 不幸的是，琐碎地这样做可能会花费很长时间。 Ford-Fulkerson network flow takes O(EV) time (on the graph you would use for this), and there are O(V^2) pairs of nodes to check. Ford-Fulkerson网络流需要O（EV）时间（在用于此的图形上），并且有O（V ^ 2）对节点需要检查。 So worst case time is approx. 所以最坏的情况是大约 O(V^5). O（V ^ 5）。