Decomposability and co-modular indices of tournaments

Abstract

Given a tournament $T$, a module of $T$ is a subset $X$ of $V(T)$ such that for $x, y\in X$ and $v\in V(T)\setminus X$, $(x,v)\in A(T)$ if and only if $(y,v)\in A(T)$. 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. The decomposability index of $T$, denoted by $\delta(T)$, is the smallest number of arcs of $T$ that must be reversed to make $T$ indecomposable. The first author conjectured that for $n \geq 5$, we have $\delta(n) = \left\lceil \frac{n+1}{4} \right\rceil$, where $\delta(n)$ is the maximum of $\delta(T)$ over the tournaments $T$ with $n$ vertices. In this paper we prove this conjecture by introducing the co-modular index of a tournament $T$, denoted by $\Delta(T)$, as the largest number of disjoint co-modules of $T$, where a co-module of $T$ is a subset $M$ of $V(T)$ such that $M$ or $V(T) \setminus M$ is a nontrivial module of $T$. We prove that for $n \geq 3$, we have $\Delta(n) = \left\lceil \frac{n+1}{2} \right\rceil$, where $\Delta(n)$ is the maximum of $\Delta(T)$ over the tournaments $T$ with $n$ vertices. Our main result is the following close relationship between the above two indices: for every tournament $T$ with at least $5$ vertices, we have $\delta(T) = \left\lceil \frac{\Delta(T)}{2} \right\rceil$. As a consequence, we obtain $\delta(n) = \left\lceil \frac{\Delta(n)}{2} \right\rceil = \left\lceil \frac{n+1}{4} \right\rceil$ for $n \geq 5$, and we answer some further related questions.