Talk:Strongly connected component

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There is also an algorithm called SCC that computes strongly connected components in graphs, by taking the inverse of a graph and working on the transpose G^t (V, E^-1) of G. Would it help if I put the pseudocode algorithm in here? Jontce 11:08, 17 May 2005 (UTC)[reply]
Isn't this a dictionary definition? At least, it should be linked to something else. m.e. 10:40, 28 May 2004 (UTC)[reply]

This page does not render correctly in IE7

Question about the algorithm

In step 1, DFS is called on a specific vertex v.

It needs to visit every vertex in the (weakly) connected component that v belongs to, to compute its finishing time.

In a directed graph, how can we choose the start node v for DFS in such a way as to visit every node in the weakly connected component of v and not add to the total complexity of the algorithm? This does not seem obvious to me. Thanks if you can explain this.

- Panayk (talk) 16:40, 18 January 2008 (UTC)[reply]

Nevermind. DSF is called on more than one starting vertices, taking care not to call it on (or revisit during a call) an already visited vertex. Perhaps we should mention that.

- Panayk (talk) 18:16, 18 January 2008 (UTC)[reply]

Algorithm

The algorithm section seems to be lifted word-for-word from CLR... —Preceding unsigned comment added by 67.114.158.146 (talk) 08:05, 9 April 2008 (UTC)[reply]

Is the image correct?

I see that C and H are not connected, is that correct? —Preceding unsigned comment added by 128.61.23.186 (talk) 19:36, 5 June 2008 (UTC)[reply]

yes, it is ok. They don't need to be directly connected by edges. --Fioravante Patrone en (talk) 20:08, 21 August 2008 (UTC)[reply]

I think he's right, since the definition is a PATH between two points. C can get to H and vice versa over a path. It's the fact that you can go both ways that makes that group strongly connected Delvarworld (talk) 08:22, 17 April 2011 (UTC)[reply]