Suppose for a graph $G=(V,E)$ and a spanning tree T of G, $\Delta(T)$ is the largest degree of a vertex in T, and let $\Delta^*$ be the smallest such quantity over all spanning trees of $G$.

We have the following local search procedure which can changes spanning tree $T$ into a different spanning tree $T'$: We find an edge $e$ not in $T$ and add it to $T$. This results in a cycle $C$ - call its vertices vertices $V(C)$. We then delete an edge in $C$ incident to a vertex in $V(C)$ with highest degree.

My question is this: if $\Delta(T) > \Delta^*$, can we always find an edge $e$ to add, such that the maximum degree of vertices in $V(C)$ is strictly less in $T'$ than it is in $T$?

  • $\begingroup$ How about just using a greedy approach, modifying Prim's algorithm to add an edge according to minimal degree of the origin vertex? I suspect that that will give minimal maximal degree... $\endgroup$
    – vonbrand
    Nov 22, 2015 at 1:46
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    Nov 22, 2015 at 2:20
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    – Raphael
    Nov 22, 2015 at 8:48

1 Answer 1


No, we cannot always find an edge $e$ to add when $\Delta(T) > \Delta^*$.

An counterexample to show local search to find minimum degree spanning tree might not https://graphonline.ru/en/?graph=qUQjSqORjZZwsRMY

The graph above was drawn at https://graphonline.ru/en

Here is an counterexample. Let $G$ be the 3-regular graph on the right side that has 14 vertices and 21 edges. Let $T$ be the spanning tree that consists of all the edges in bold. That is, $T$ has all the edges in $G$ except the innermost 7 edges and the edge at the top right. The graph on the left side shows a hamiltonian path of $G$. That is, $\Delta(T)=3$ and $\Delta^*=2$.

Let $e$ be an edge not in $T$. If we add $e$ to $T$, the resulting cycle $C$ contains at least two non-adjacent vertices of degree 3 in $T$. If we further delete any edge in $C$ incident to a vertex in $C$ with the highest degree 3, we obtain graph $T'$. There is at least one vertex of degree 3 in $T'$. That is, the maximum degree of vertices in $V(C)$ in $T'$ is 3.


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