# 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 $$\delta (T) = \left\lceil \frac{\delta '(T)}{2} \right\rceil$$.

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Correspondence to Houmem Belkhechine.

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