For any graph $G=(V,E)$ let $\tau(G)$ be the minimum cardinality of a [vertex cover](http://en.wikipedia.org/wiki/Vertex_cover) of $G$. 


As noted [here](https://math.stackexchange.com/questions/1123167/chromatic-number-and-vertex-covering-number), we have $\tau(G) \geq \chi(G) - 1$ for all finite graphs $G$. I'm interested in graphs $G$ with $\tau(G) = \chi(G) - 1$. I found only examples of such graphs where $\chi(G) = \omega(G)$ (where $\omega(G)$ is the clique number of $G$).


**Question**: Is there a graph $G$ with $\omega(G) < \chi(G)$ and $\tau(G) = \chi(G) - 1$?



When the vertex covering number is smaller than the chromatic number
Expanding a bit the construction given by Leen Droogendijk on MSE, let $C$ be a vertex cover of minimal size. Then we can colour $G$ by colouring $C$ (considered as an induced subgraph), and giving all vertices of $V-C$ an extra colour. This shows
$$\chi(G)\le1+\chi(C),$$
hence $\tau(G)=\chi(G)-1$ can only happen if $\chi(C)=|C|$, i.e., $C$ is a clique.


Furthermore, if $v\in V-C$ is not connected by an edge to some $u\in C$, we can colour $v$ the same as $u$ (here we use that vertices from $C$ have pairwise distinct colours). Thus, we can save the extra colour altogether unless some $v\in V-C$ is connected to all vertices in $C$, in which case $C\cup\{v\}$ is a clique of size $|C|+1=\chi(G)$. All in all, $\tau(G)=\chi(G)-1$ implies $\chi(G)=\omega(G)$.


