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

Choose a spanning tree of the connected graph and root it at $x$. List vertices in nonincreasing order of their distance from $x$ in this tree, breaking ties arbitrarily. The root has distance zero and is last, so $v_n=x$.

Every other vertex has a parent one step closer to the root. That parent is adjacent to it and appears later in the ordering. HenceFor a graph of one vertex the condition is vacuous and the same ordering works.

Write $\Delta =\Delta (G)$, which is at least two because $x$ has the two specified neighbours. Give $y$ and $z$ the same colour, say colour one; this is proper because they are nonadjacent.

Apply part (i) to the connected graph $G\setminus \{y,z\}$, with $x$ last. Colour its vertices greedily in the resulting order using a palette of $\Delta$ colours. Every vertex before $x$ has a neighbour later in this order, so at most $\mathrm{deg}(v)-1\le \Delta -1$ of its neighbours have already received colours. At least one palette colour is available.

When $x$ is reached, all its neighbours are coloured, but the two neighbours $y,z$ have the same colour. Thus there are at most $\mathrm{deg}(x)-1\le \Delta -1$ distinct forbidden colours. Colour $x$ with a remaining one. This constructs a proper colouring withThe common colour on the two initial vertices replaces the usual need for an uncoloured neighbour at the last greedy step.

This is greedy colouring with two precoloured neighbours.

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.

Fix a proper colouring with exactly $k$ nonempty colour classes. For any class $C_i$, suppose every vertex in it misses at least one of the other $k-1$ colours in its neighbourhood. Recolour each of these vertices by one of its missing colours. No edge joins two vertices of $C_i$, so these simultaneous recolourings cannot conflict with one another, and the chosen colours do not conflict with vertices outside $C_i$.

This would eliminate colour $i$ and produce a $(k-1)$-colouring, contradicting optimality. Therefore each colour class contains a vertex adjacent to at least one vertex of every other colour. That vertex has degree at least $k-1$. Choosing one from each class gives $k$ distinct vertices, provingFor $k=1$, a nonempty graph already has a vertex of degree at least zero, so the conclusion holds as well.