Let $G = (V, E)$ be a weighted, undirected graph, with $f: E \to \mathbb{R}$ its weight function. Given a path $P = (e_1, \dots, e_k)$, we call $\operatorname{bwd}(P) = \min_{1 \le i \le k} f(e_i)$ the "path-bandwidth" of $P$. Let $\mathcal{P}_G(v, w)$ be the set of paths between two vertices $v, w \in V(G)$. Given two vertices $v, w \in V$, we call $\operatorname{bwd}_G(v, w) = \operatorname{max}_{P \in \mathcal{P}_G(v, w)}{\operatorname{bwd}(P)}$ the bandwidth in $G$ between $v$ and $w$. That is, the maximum path-bandwidth of any path in $G$ between $v$ and $w$.

If $T$ is a spanning tree of $G$, $\operatorname{bwd}_T(v, w) = \operatorname{bwd}(P)$ where $P$ is the path from $v$ to $w$ in $T$. We call $T$ a "maximum pairwise bandwidth spanning tree" (MPBST) when $\operatorname{bwd}_T(v, w) = \operatorname{bwd}_G(v, w)$, for all $v, w \in V$.

It is easy to show that every maximum spanning tree is a MPBST.

Experimentally the converse seems to hold: Every MPBST is a maximum spanning tree. Is this true, and if so, why?

I tried proving this using induction in the minimum symmetric difference between an MPBST and the set of all maximum spanning trees of $G$, to show via edge swaps I can make an MPBST of the same weight as the original one, but closer and closer in symmetric difference to this set. I could not finish the argument unless I assumed all weights were distinct.

  • $\begingroup$ If you can prove the case when all weights are distinct, applying the usual perturbation technique can show the general case. $\endgroup$
    – John L.
    Feb 28, 2022 at 7:15

1 Answer 1


Yes, it is true that every MPBST is a maximum spanning tree.

Here is a characterization of maximum spanning Tree(MST). A proof for that characterization can be found here.

Given a spanning Tree $T$ of $G$, $T$ is an MST $\iff$ for any edge $uv\notin T$, $uv$ has the minimal weight among edges in the cycle created by adding $uv$ to $T$.

Suppose $S$ is an MPBST. Let us prove $S$ is an MST.

Let edge $uv\notin S$. Let $P$ be the path from $v$ to $u$ in $S$, i.e., the cycle created by adding $uv$ to $S$ is $P$ followed by edge $uv$.

Since $((u,v))$ is a path from $u$ to $v$, $\operatorname{bwd}_G(u, v)\ge\operatorname{bwd}_{((u,v))}(u,v)=w((u,v))$.

Since $S$ is an MPBST, $$\operatorname{bwd}(\text{the path from }v\text{ to }u\text{ in }S)=\operatorname{bwd}_S(u,v) = \operatorname{bwd}_G(u, v)\ge w((u,v)),$$ i.e., edge $(u,v)$ has the minimal weight among edges in the cycle created by adding $uv$ to $S$. By the characterization of MST, $S$ is an MST.

  • 1
    $\begingroup$ Thanks for the characterization! Indeed it's very useful in this case. $\endgroup$ Feb 28, 2022 at 16:12

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