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This is an exercise problem (Ex.3) from the excellent lecture note by Jeff Erickson Lecture 20: Minimum Spanning Trees [Fa’13] .

Prove that an edge-weighted graph $G$ has a unique minimum spanning tree if and only if the following conditions hold

  • For any partition of the vertices of $G$ into two subsets, the minimum-weight edge with one endpoint in each subset is unique.

  • The maximum-weight edge in any cycle of $G$ is unique.

Consider the "$\Rightarrow$" direction and the following graph $G$.

mst

$G$ has a unique MST. However, for the partition $\{A\}$ and $\{B,C\}$, the minimum-weight crossing edge is not unique.

Did I misunderstand some points? Or if there are flaws in the theorem, how can we fix it?

hengxin
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    Yes, this seems to be a mistake. Try to figure out what version of the exercise is correct. For example, it seems that the second condition is indeed necessary. – Yuval Filmus Dec 08 '14 at 19:54
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    Unless I misunderstand, the second condition is not necessary either. Consider the graph {(A,B,1), (A,C,1), (A,D,1), (B,D,10), (D,C,10)}. It also has a minimum spanning tree composed of edges connected to A. But there is a cycle with 2 maximum weight edges (and the first condition is not met either). CC @YuvalFilmus – babou Jun 01 '15 at 11:04
  • @jeffe, what do you think? ;) – Luke Mathieson Jun 01 '15 at 13:04
  • I think the second should be in "in any chordless cycle" (so a minimal cycle in the sense that it doesn't include smaller ones as induced subgraphs). The first condition seems significantly wrong. For example take $G$ to be any tree where all edge weights are $1$, then $G$ has a unique MST (itself), but any partition with more than one edge crossing it has several minimum weight edges. – Luke Mathieson Jun 01 '15 at 13:15
  • Actually, scrap my idea about the cycle condition, if we a cycle with edge weights all "big" (say 2 for argument's sake), then add a vertex connected to every vertex in the cycle, where all these edges have weight "small" (~1), then there is a unique MST, but the original cycle certainly doesn't have a unique maximum weight edge, and also has no chords. – Luke Mathieson Jun 01 '15 at 13:26
  • @LukeMathieson MST is such a deceptively simple concept that I often get simple, plausible, but wrong proofs about it. – hengxin Jun 01 '15 at 13:34
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    Oops! Yes, this is a bug. (Note to self: Change every instance of "Prove" to "Prove or disprove".) – JeffE Oct 29 '15 at 23:32
  • @JeffE Thanks for your reply. Both my teacher and I (as a TA) enjoy your lecture notes. Thanks for your efforts. – hengxin Oct 30 '15 at 03:07

1 Answers1

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Answer my own question by simply copying the comment made by @JeffE, the author of the lecture note：

Oops! Yes, this is a bug. (Note to self: Change every instance of "Prove" to "Prove or disprove".) – JeffE

hengxin
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