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I need your help proving the following: G=(V,E) is an undirected connected graph with non-negative weights on the edges. Let T be a MST of G and T' be a spanning tree of G (not a minimum one) so it holds that Weight(T') > Weight(T). Prove that the weight of the haviest edge in T' isn't smaller than the weight of heaviest edge in T.

I'm not sure how to approch this, may if we let e(u,v) - heaviest edge on T and e'(u',v') - heaviest edge on T' and then if we look at the cut defined by (u,u') we can use Kruskal algorithem and show that e' isn't chosen to be in T or something in this direction...

Thank you,

Noam
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2 Answers2

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Assume the contrary -- there exists a weighted undirected graph with a minimum spanning tree T and a spanning tree T' such that the heaviest edge of T is heavier than the heaviest edge of T', i.e., the heaviest edge of T is heavier than every edge in T'. Consider the cut induced by deleting the heaviest edge h of T. Since T' is connected, some edge in T' crosses this cut. If we add this edge to T - h, we get a spanning tree that is lighter than T, which is a minimum spanning tree. Contradiction.

David Eisenstat
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I'd take another direction. For simplicity, let all edge weights be distinct, so that MST is unique. Consider the heaviest edge e in MST, and the way Kruskal algorithm constructs this MST. Turns out the last added edge is the heaviest edge in the resulting MST.

Now look at the situation just before adding the last edge. We have a forest consisting of two trees. As we are going with Kruskal algorithm, there are currently no cheaper ways than e to connect these two trees. Therefore, any edge between them in any other spanning tree has weight at least as large as e has.

Edges of equal weight require some care, or perhaps a clever idea, to be handled properly.

Gassa
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