Let $G$ be an undirected, connected graph, and $s,t$ non-adjacent vertices in $G$. Denote by $k_{st}(G)$ the $st$-connectivity of $G$. That is, $k_{st}(G)$ is the size of any minimum $st$-separator of $G$.

(*) It can be shown that if a vertex $v$ belongs to any minimum $st$-separator of $G$, then $k_{st}(G-v)=k_{st}(G)-1$.

How do you prove the following: Let $S$ be a minimal $st$-separator of $G$ that is not a minimum $st$-separator of $G$. Prove that $S$ contains at least one vertex $v\in S$ such that $k_{st}(G-v)=k_{st}(G)$.

It seems very intuitive that this would hold, especially since (*) can be shown. However, I have yet to find a formal proof...


1 Answer 1


It does not hold, as can be seen from the red separator in this example.


Furthermore, a vertex in a minimum separator can be separated from $s$ and $t$ by a minimal separator:

second couterexample

  • $\begingroup$ Thank you for the reply! Here is a follow-up question: Let $u\in V(G)$ such that $u$ belongs to some minimum $st$-separator of $G$. Let $S$ be any minimal $st$-separator that does not contain $u$ (i.e., $u\notin S$). Also, let $C_s(S)$ and $C_t(S)$ denote the connected components containing $s$ and $t$ respectively in $G-S$. Is it necessarily the case that $u\in C_s(S)\cup C_t(S)$ ? $\endgroup$
    – BBK
    Apr 6 at 8:41
  • $\begingroup$ Do you know Menger's theorem? If yes, what can be said about the relationship of any minimum separator with any maximum set of disjoint $st$-path? $\endgroup$
    – Guyslain
    Apr 6 at 11:08
  • $\begingroup$ Yes. The $st$-connectivity of $G$ is the maximum number of internally-disjoint $st$-paths in $G$. I don't see how this implies that for any minimal $st$-separator $S$, every vertex $u\notin S$ that belongs to some minimum $st$-separator necessarily belongs to either $C_s(S)$ or $C_t(S)$. $\endgroup$
    – BBK
    Apr 6 at 13:39
  • $\begingroup$ Use the fact (an easy consequence of Menger's theorem) that for any k disjoint paths $P_1,\ldots,P_k$ and any $st$-separator $S$ of size $k$, $P_i \cap S$ = 1 for each $i$ (each path $P_i$ contains exactly one vertex of $S$). $\endgroup$
    – Guyslain
    Apr 7 at 5:19
  • $\begingroup$ If $S$ is a minimum $st$-sep then this is indeed the case, because every path meets exactly one vertex from $S$. But a minimal separator can be larger than $k$, and hence cut such a path in more than one position, sending some vertices to connected components other than $C_s(S)$ or $C_t(S)$. Can it be shown that for vertices that belong to a minimum $st$-sep this cannot happen ? $\endgroup$
    – BBK
    Apr 7 at 7:11

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