Talk:Kruskal's algorithm

Many problems with this proof. Confusion between vertices and edges, etc., etc.. --zero 05:56, 13 Oct 2003 (UTC)

There was actually also a problem with the running time. Kruskal's can run in O(|E| log |V|) where E is the set of edges and V the set of vertices. It should be obvious that |E| >= |V| in any graph that connects all vertices (though not necessarily fully connected).