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We have a connected, undirected graph $G$, and suppose we have a set of tree edges $T_1,\cdots,T_n$ of $G$ which encompass all possible spanning trees of $G$. Now suppose that we construct a set of edges $A$ such that $A \cap T_i \neq \emptyset$ for all $1 \leq i \leq n$.

$A$ is not necessarily a unique set, for an edge $e$ to qualify being a member of $A$ it must merely be part of at least one of $T_i$, so at its largest $A$ will just be a set of all edges of $G$.

How to show that this set $A$ contains a subset which is the cutset of $G$?

John Doe
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  • Are you using a strict or a loose definition of "cutset"? – Misha Lavrov Oct 26 '23 at 15:49
  • @MishaLavrov The definition taught to me is that an edge cutset is any set of edges, the removal of which increases the number of components of the graph. In this case as I am dealing with a connected graph, a cutset just splits the graph in 2 or more, but the minimum is 2. Not sure if this helps you understand better – John Doe Oct 26 '23 at 16:06

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If we delete all edges in $A$ from $G$, then the resulting graph $G-A$ cannot be connected. (If $G-A$ were connected, it would have a spanning tree $T$, but then $T$ would also be a spanning tree of $G$ that has no edges in common with $A$.)

By the loose definition, that makes $A$ a cutset, so we're done.

Some people take the definition of a cutset more strictly, either as

  • the set of all edges between two disjoint sets $S, T \subseteq V(G)$ with $S \cup T = V(G)$, or even
  • the above, but with the additional requirement that $G[S]$ and $G[T]$ are connected.

If $G-A$ is disconnected, it is possible to prove that $A$ contains a cutset by either of the stricter definitions, too. (More precisely, if $B \subseteq A$ is a minimal subset of $A$ such that $G-B$ is not connected, then $B$ satisfies both of the stricter definitions.) So we win in any case.

Misha Lavrov
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