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Jan 11, 2010 at 0:55 comment added Hans-Peter Stricker Even if those graphs were biconnected, their biconnectivity would have to be proved, but would not be the defining property, for which I am looking for a name.
Jan 11, 2010 at 0:36 comment added David Eppstein A figure-eight graph has every vertex in a cycle but is not biconnected.
Jan 11, 2010 at 0:32 comment added Klingonesque of course, this works only for connected graphs.
Jan 11, 2010 at 0:31 history answered Klingonesque CC BY-SA 2.5