Making a Tournament Indecomposable by One Subtournament-Reversal Operation

Abstract

Given a tournament T, a module of T is a subset M of V(T) such that for $x, y\in M$ and $v\in V(T)\setminus M$, $(v,x)\in A(T)$ if and only if $(v,y)\in A(T)$. The trivial modules of T are $\varnothing$, $\{u\}$ $(u\in V(T))$ and V(T). The tournament T is indecomposable if all its modules are trivial; otherwise it is decomposable. Let T be a tournament with at least five vertices. In a previous paper, the authors proved that the smallest number $\delta (T)$ of arcs that must be reversed to make T indecomposable satisfies $\delta (T) \le \left\lceil \frac{v(T)+1}{4} \right\rceil$, and this bound is sharp, where $v(T) = |V(T)|$ is the order of T. In this paper, we prove that if the tournament T is not transitive of even order, then T can be made indecomposable by reversing the arcs of a subtournament of T. We denote by $\delta '(T)$ the smallest size of such a subtournament. We also prove that $\delta (T) = \left\lceil \frac{\delta '(T)}{2} \right\rceil$.