7

Suppose we have a graph G. How can we find a spanning tree that minimizes the maximum weight of all the edges in the tree? I am convinced that by simply finding an MST of G would suffice, but I am having a lot of trouble proving that my idea is actually correct. Can anyone show me a proof sketch or give me some hints as to how to construct the proof? Thanks!

David Richerby
  • 82,470
  • 26
  • 145
  • 239
Aden Dong
  • 1,151
  • 2
  • 11
  • 24

2 Answers2

5

What follows is taken from Tsuyoshi Ito's comment.

If you know Kruskal’s algorithm for the minimum spanning tree, it is an easy exercise to show that the output of Kruskal’s algorithm is a minimum bottleneck spanning tree. (I think that it is easier than showing that the output of Kruskal’s algorithm is a minimum spanning tree.)

Community
  • 1
Yuval Filmus
  • 280,205
  • 27
  • 317
  • 514
2

Given a graph $G(V, A)$, we know that any spanning tree contains an edge in every cutset. Let $S_{min}^{max}$ and $S$ be the minimax weight spanning tree of $G$ and minimum weight spanning tree of $G$ resp. Any edge $e \in S$ is associated with a cutset $C$. Corresponding to cutset $C$,$S_{min}^{max}$ must also contain an edge, say $e'$. Then if $w(e') < w(e)$, we know that replacing $e$ with $e'$ in $S$ will produce a new spanning tree with lower overall weight, thus contradicting our assumption of optimality of $S$. Hence, we can see that the weight of every edge in $S$ is no greater than the weight of a corresponding edge (obtained from the cutset) in $S_{min}^{max}$. Since, we have compared every edge in $S$ but perhaps possibly may not have compared every edge of $S_{min}^{max}$ in the above process for comparison, the max weight edge in $S_{min}^{max}$ has to be atleast as much as much as the max weight in $S$. Hence we can conclude that $S$ itself produces a minimax weight spanning tree.

csTheoryBeginner
  • 393
  • 1
  • 12