Let G be undirected graph, G=(V,E), and all edge weights are distinct. Consider an edge e=(u,v)∈E that wasn't included in the solution obtained from applying Kruskal Algorithm to G. Prove that this edge isn't in any Minimimum Spanning Tree of G.

  • $\begingroup$ Do you perhaps also have that all edge weights are distinct? $\endgroup$ – JimN Sep 6 '20 at 23:05
  • $\begingroup$ Yes, all edge weights are distinct. $\endgroup$ – Kate Austen Sep 6 '20 at 23:32
  • $\begingroup$ You should probably include that In your first sentence in your question statement $\endgroup$ – JimN Sep 7 '20 at 0:01
  • $\begingroup$ If this is quoted content, please attribute it properly - hyperlink welcome. Consider using a block quote (see, e.g, post editor tool bar "quotation mark button (")). $\endgroup$ – greybeard Sep 7 '20 at 14:17

Without additional assumptions about the edge weights that's false. A simple counterexample is a triangle graph where all edge weights are equal.


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