Degreewidth: a New Parameter for Solving Problems on Tournaments

In the paper, we define a new parameter for tournaments called degreewidth which can be seen as a measure of how far is the tournament from being acyclic. The degreewidth of a tournament $T$ denoted by $\Delta(T)$ is the minimum value $k$ for which we can find an ordering $\langle v_1, \dots, v_n \rangle$ of the vertices of $T$ such that every vertex is incident to at most $k$ backward arcs (\textit{i.e.} an arc $(v_i,v_j)$ such that $j<i$). Thus, a tournament is acyclic if and only if its degreewidth is zero. Additionally, the class of sparse tournaments defined by Bessy et al. [ESA 2017] is exactly the class of tournaments with degreewidth one. We first study computational complexity of finding degreewidth. Namely, we show it is NP-hard and complement this result with a $3$-approximation algorithm. We also provide a cubic algorithm to decide if a tournament is sparse. Finally, we study classical graph problems \textsc{Dominating Set} and \textsc{Feedback Vertex Set} parameterized by degreewidth. We show the former is fixed parameter tractable whereas the latter is NP-hard on sparse tournaments. Additionally, we study \textsc{Feedback Arc Set} on sparse tournaments

Proof of a tournament partition conjecture and an application to 1-factors with prescribed cycle lengths

In 1982 Thomassen asked whether there exists an integer f(k,t) such that every strongly f(k,t)-connected tournament T admits a partition of its vertex set into t vertex classes V_1,...,V_t such that for all i the subtournament T[V_i] induced on T by V_i is strongly k-connected. Our main result implies an affirmative answer to this question. In particular we show that f(k,t) = O(k^7 t^4) suffices. As another application of our main result we give an affirmative answer to a question of Song as to whether, for any integer t, there exists an integer h(t) such that every strongly h(t)-connected tournament has a 1-factor consisting of t vertex-disjoint cycles of prescribed lengths. We show that h(t) = O(t^5) suffices.Comment: final version, to appear in Combinatoric