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 $\emptyset$, $\{u\}$ $(u\in V(T))$ 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 $\delta(T) \leq \left\lceil \frac{v(T)+1}{4} \right\rceil$, 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 $\delta(T) = \left\lceil \frac{\delta'(T)}{2} \right\rceil$.