# 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 $\Delta(T) > \Delta^*$, 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 $T$?

• 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... – vonbrand Nov 22 '15 at 1:46
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No, we cannot always find an edge $$e$$ to add when $$\Delta(T) > \Delta^*$$.
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.