Let $G=(V,E)$ be a simple, undirected graph.


We call a partition ${\cal P}$ of a non-empty subset of $V$ a *Hadwiger partition* if


1. every block (member of ${\cal P}$) is non-empty and connected, and
2. if $x, y \in {\cal P}$ are distinct blocks then there are $v\in x$ and $w \in y$ such that $\{v,w\} \in E$.


The *Hadwiger number* $\eta(G)$ is the maximum number of blocks a Hadwiger partition can have.


Let $n=\eta(G)$ and suppose that every Hadwiger partition of $G$ into $n$ blocks contains at least one block consisting of 1 vertex only ("singleton block"). Does this imply that there is $v\in V$ such that $\eta(G\setminus\{v\}) = n-1$?



Hadwiger partitions where one block is always a singleton
The graph $G=2K\_n$ is a counterexample. Or, if $G$ is supposed to be connected, then $G=2K\_n+e\ $ (that's $2K\_n$ with an additional edge) is a counterexample for $n\ge3.$


