Let $G=(V,E)$ be a finited connected graph, $V\neq \emptyset$. Let $[V]^2 := \big\{ \{v,w\}: v, w \in V\text{ and } v\neq w\big\}$. Given $F\subseteq [V]^2$ we say that $F$ is a vertex-disjoint extension of $E$ if

  • $F\supseteq E$, and
  • $f_1\neq f_2\in (F\setminus E)$ implies $f_1\cap f_2 = \emptyset$.

Given $n\in\mathbb{N}$, is there a connected graph $G=(V,E)$ such that for every vertex-disjoint extension $F$ of $E$ we have $\text{diam}(V,F) \geq n$?


Yes. Let $G=(V,E)=P_k$, the $k$-point path, where $k=3\cdot2^{n-1}-1$, and let $F$ be any vertex-disjoint extension of $E$; I claim that the graph $H=(V,F)$ has radius $\operatorname{rad}(H)\ge n$.

Assume for a contradiction that $\operatorname{rad}(H)\le n-1$. Since $\Delta(H)\le3$, by the answer to this question it follows that $$k=|V|\le1+3(1+2+4+\cdots+2^{n-2})=3\cdot2^{n-1}-2=k-1,$$ which is absurd.

  • $\begingroup$ Thanks for this nice argument! You wrote "==" --> C programmer? :) $\endgroup$ – Dominic van der Zypen Apr 10 '15 at 7:59

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.