Weakly connected graph: If there is a single vertex without incoming edges, is it a mother vertex?

Question

This is just a request for confirmation. I have a simple directed graph which is weakly connected. When I require there to be exactly one vertex with indegree == 0, does it follow that all nodes in the graph are reachable from that vertex?

I think yes: When I contract the graph (replace all strongly connected components with a single vertex), the result will be a DAG. All vertices with indegree == 0 will be "roots" of that DAG. Ex hypothesi, I have only one such vertex, hence the DAG is a tree. (And it will be a single tree, not a forest, because I start with a single weakly connected component.) q. e. d. Am I correct, or have I missed something?

score 1 · Accepted Answer · answered Aug 17 '20 at 07:51

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It seems that the graph A -> B <- C <-> D is a graph where A has an indegree of 0 but one cannot reach C from A.

answered Aug 17 '20 at 07:51

Kevin

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Thanks. So I think I must check whether the condensed graph is a tree, – Sebastian Aug 17 '20 at 07:58
Or rather, it suffices to check that the condensed graph has only vertex with an indegree of 0. – Sebastian Aug 17 '20 at 09:32

Weakly connected graph: If there is a single vertex without incoming edges, is it a mother vertex?

1 Answers1