Making a tournament indecomposable by one subtournament-reversal operation

TL;DR

This paper investigates how reversing a small number of arcs or a small subtournament in a tournament can make it indecomposable, establishing bounds and relationships between these operations.

Contribution

It introduces the concept of making a tournament indecomposable via a single subtournament reversal and relates this to the minimal arc reversals needed.

Findings

Reversing a small subtournament can make a tournament indecomposable.

The minimal number of arc reversals is half the size of the smallest subtournament reversal.

Established bounds for the number of arc reversals needed to achieve indecomposability.

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