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I am working on the following exercise:

Consider an undirected graph $G = (V,E)$. Let $T^* = (V,E_{T^*})$ be a $MST$ and let $e$ be an edge in $E_{T^*}$. We define the set of all values that can be assigned to $w_e$ such that $T^*$ remains a MST as $I_e$.

  1. Show that $I_e$ is an interval.
  2. Devise an efficient algorithm to calculate $I_e$ for a given edge $e$.
  3. Devise an efficient algorithm that determines all $I_e$ in one step. It should be more efficient than repeatedly using the algorithm from 2.

I did the following:

  1. Consider an edge $e$ in $E_{T^*}$. Delete it from $G$ and find the new lowest weighted edge connecting the resulting components, say $e'$. The upper bound for $I_e$ is $w(e')$. The lower bound would be $-\infty$.
  2. Use the algorithm sketched in 1.
  3. I do not know what to do here. Could you help me?
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The solution in short. For each edge $e$ not in the tree, check the path between its endpoints in the tree. The weight of each edge on this path is upper-bounded buy the weight of $e$. Keep track of the smallest upper-bound for each edge on the tree while iterating over all edges not in the tree. Following is a sketch of the correctness.

Let us call an edge in a simple cycle Heavy for this cycle, if its weight is maximum among all edges in the cycle. We claim that each edge $e$ in $G\setminus T^*$ is Cycle heavy for some cycle in $G$.

Proof. By adding the edge $e$ to the tree, it constructs a cycle with a path in the tree between its endpoints. If there is an edge in the cycle with strictly heavier weight, we can remove this edge and keep $e$ constricting a lighter tree which contradicts the assumption that the given tree is minimum.

On the other hand, it is not hard to prove that an edge is in an MST of a given graph if it is not heavy for any simple cycle containing this edge. The correctness of the sketched algorithm follows directly from this statement.

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