I'm trying to prove a statement "given a graph G=(V,T) and that no cycle C exists that contains only edge "e" and other edges that their weight is smaller than that of "e", prove that "e" must be in some mst". I tried to proof by contradiction, but I'm not sure what I can say about "e" when it never belongs to any mst.

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    $\begingroup$ Do you know kruskal's algorithm? What would it mean for $e$ to never be part of an MST? Well, when you consider it, its two endpoints are always already in the same component. How can that be? $\endgroup$
    – Pål GD
    Mar 9 at 18:35
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    $\begingroup$ Thank you, I didn’t think about that. $\endgroup$
    – CSstudent
    Mar 9 at 19:21

1 Answer 1


Suppose that you run Kruskal's algorithm, which computes an MST by repeatedly adding the cheapest edge that does not create a cycle.

In any run of Kruskal, when we consider your edge $e = uv$, it turns out that $u$ and $v$ are already in the same connected component.

That means that there is a path from $u$ to $v$ where every edge is strictly cheaper than $e$.

Hence $e$ must be the heaviest edge in a cycle.


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