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(1) Let's say $T_1$ has $r$ number of edges and $T_2$ has $s$ number of edges; where $r \ne s$ but two trees has same weight since they both are MST.

I kind feel this kind not be true because to span $n$ vertices with minimum number of edges, we need $n-1$ edges. so MST must have at least $n-1$ edges; if some MST has more than $n-1$ edges, then there must exists a cycle which leads to non-MST.

However, I don't know how to show that if the number of edges is greater or equal to the number of vertices of an an undirected, connected graph $G$, then $G$ has a cycle.

What if the graph is directed? Is this hold as well?

Any hint?

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The answer is no. When running an algorithm for finding MST, It is first of all a tree, it has |v| - 1 edges $exactly$.

If there are more than |v| - 1 edges, than there're two vertices that their path isn't simple, which is simply shown with induction. Meaning if there's a path from s to t with more than |v| - 1 edges, one of the vertices will be repeated.

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