Digraph Coloring and Distance to Acyclicity

In \(k\)-Digraph Coloring we are given a digraph and are asked to partition its vertices into at most \(k\) sets, so that each set induces a DAG. This well-known problem is \(\mathsf{NP}\)-hard, as it generalizes (undirected) \(k\)-Coloring, but becomes trivial if the input digraph is acyclic. This poses the natural parameterized complexity question what happens when the input is “almost” acyclic. In this paper we study this question using parameters that measure the input’s distance to acyclicity in either the directed or the undirected sense. It is already known that, for all \(k \geq 2\), \(k\)-Digraph Coloring is \(\mathsf{NP}\)-hard on digraphs of DFVS at most \(k+4\). We strengthen this result to show that, for all \(k \geq 2\), \(k\)-Digraph Coloring is \(\mathsf{NP}\)-hard for DFVS \(k\).

Refining our reduction we obtain two further consequences:

(i) for all \(k \geq 2\), \(k\)-Digraph Coloring is \(\mathsf{NP}\)-hard for graphs of feedback arc set (FAS) at most \(k2\); interestingly, this leads to a dichotomy, as we show that the problem is \(\mathsf{FPT}\) by \(k\) if FAS is at most \(k^2−1\);

(ii) \(k\)-Digraph Coloring is \(\mathsf{NP}\)-hard for graphs of DFVS \(k\), even if the maximum degree \(\Delta\) is at most \(4k−1\); we show that this is also almost tight, as the problem becomes \(\mathsf{FPT}\) for DFVS \(k\) and \(\Delta \leq 4k−3\).

We then consider parameters that measure the distance from acyclicity of the underlying graph. We show that \(k\)-Digraph Coloring admits an \(\mathsf{FPT}\) algorithm parameterized by treewidth, whose parameter dependence is \((tw!)k^{tw}\). Then, we pose the question of whether the \(tw!\) factor can be eliminated. Our main contribution in this part is to settle this question in the negative and show that our algorithm is essentially optimal, even for the much more restricted parameter treedepth and for \(k=2\). Specifically, we show that an \(\mathsf{FPT}\) algorithm solving \(2\)-Digraph Coloring with dependence \(td^{o(td)}\) would contradict the ETH.