Past exam of the graph theory course of the University of Oxford / 2011 / 2 / b

Take a proper colouring $c:V(G)\to \{1,\dots ,k\}$ and orient every edge from its lower-coloured endpoint towards its higher-coloured endpoint. Along a directed path the colour strictly increases, so such a graph path has at most $k$ vertices. A directed cycle is impossible for the same reason. Thus

For each vertex $v$, let $c(v)$ be the maximum number of vertices in a directed path ending at $v$. These numbers lie between one and $\ell$.

If $u\to v$ is an edge, a longest graph path ending at $u$ cannot already contain $v$, since that would give a directed cycle on appending $u\to v$. Appending it therefore creates a graph path ending at $v$, so $c(v)\ge c(u)+1$. Adjacent vertices consequently receive different colours. HenceAcyclicity is exactly what justifies extending the graph path without repeating a vertex.

Part (ii) shows that $\chi (G)\le \lambda (D)$ for every acyclic orientation $D$, so $\chi (G)$ is at most their minimum. Part (i), applied to an optimal colouring, supplies an acyclic orientation with $\lambda (D)\le \chi (G)$. Combining the inequalities givesThe parameter counts vertices on a graph path; if graph path length were instead counted in edges, the corresponding minimum would be $\chi (G)-1$.

These constructions prove chromatic number from acyclic longest paths.