2
$\begingroup$

Be $G$ a connected graph with weights associated to its edges. Be $T(G)$ the graph that has the spanning trees of $G$ as vertex, and two spanning trees are adjacent to each other if and only if each one of them has only one edge that the other doesn't have. To each vertex of $T(G)$ we associate the weight of the spanning tree that correspond to $G$.

Be $T_1$ any vertex of $T(G)$, that is, any spanning tree of $G$. Proof that $T_1$ is a local minimum if and only if $T_1$ is global minimum.

To put it in another words, proof that $T_1$ has weight minor or equal than all its neighbours in $T(G)$ if and only if $T_1$ is an minimum spanning tree.

$\endgroup$
3
$\begingroup$

Let $T$ be a local minimum spanning tree that is not a global minimum. Let $e_1,\ldots,e_m$ the edges of $T$ ordered by non-decreasing weight. Let $T'$ be a minimum weight spanning tree that coincides with $T$ on the largest possible begin segment, so we may assume $T'$ has edges $e_1,\ldots,e_{n-1},f_n,\ldots,f_m$ (again ordered by non-decreasing weight) and $f_n\ne e_n$. Let $w$ be the weight function in our graph.

We claim that $w(f_n)<w(e_n)$: indeed, if $w(e_n)\leq w(f_n)$ then $e_1,\ldots,e_{n-1},e_n$ would be a valid startup for Kruskal's algorithm and would lead to a minimum weight spanning tree that coincides with $T$ on a larger initial segment.

$T+f_n$ contains exactly one cycle $C$. The edges of $C$ different from $f_n$ cannot all be in $\{e_1,\ldots,e_{n-1}\}$ or we would have a cycle in $T'$. So we find at least one edge $f$ on $C$ that is different from $e_1,\ldots,e_{n-1}$ and $f_n$. But then $w(f)\geq w(e_n)>w(f_n)$, so $T+f_n-f$ is a spanning tree with a smaller total weight that is a neighbour of $T$ in the spanning tree graph. Contradiction.

(I left some small details for you to prove. Let me know if they cause trouble).

Note that this proves the slightly stronger statement, that each non-minimal tree has a neighbour of strictly smaller total weight.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.