Local search to find minimum degree spanning tree 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 $\Delta(T) &gt; \Delta^*$, 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 $T$?
No, we cannot always find an edge $e$ to add when $\Delta(T) &gt; \Delta^*$.

 

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.