On the inversion number of oriented graphs

TL;DR

This paper investigates the inversion number of oriented graphs, establishing bounds, conjecturing additive properties for graph joins, proving special cases, and analyzing the computational complexity of inversion number decision problems.

Contribution

It introduces bounds on inversion numbers, proposes a conjecture on their additive property under graph joins, proves cases of this conjecture, and analyzes the NP-completeness of the inversion number decision problem.

Findings

Inversion number bounds relate to cycle transversals and packings.

Conjecture on inversion number additivity for graph joins is supported in specific cases.

Deciding if inversion number ≤ 1 is NP-complete, implying NP-completeness for all k if the conjecture holds.

Abstract

Let $D$ be an oriented graph. The inversion of a set $X$ of vertices in $D$ consists in reversing the direction of all arcs with both ends in $X$. The inversion number of $D$, denoted by ${\rm inv}(D)$, is the minimum number of inversions needed to make $D$ acyclic. Denoting by $\tau(D)$, $\tau' (D)$, and $\nu(D)$ the cycle transversal number, the cycle arc-transversal number and the cycle packing number of $D$ respectively, one shows that ${\rm inv}(D) \leq \tau' (D)$, ${\rm inv}(D) \leq 2\tau(D)$ and there exists a function $g$ such that ${\rm inv}(D)\leq g(\nu(D))$. We conjecture that for any two oriented graphs $L$ and $R$, ${\rm inv}(L\rightarrow R) ={\rm inv}(L) +{\rm inv}(R)$ where $L\rightarrow R$ is the dijoin of $L$ and $R$. This would imply that the first two inequalities are tight. We prove this conjecture when ${\rm inv}(L)\leq 1$ and ${\rm inv}(R)\leq 2$ and when ${\rm inv}(L) ={\rm inv}(R)=2$ and $L$ and $R$ are strongly connected. We also show that the function $g$ of the third inequality satisfies $g(1)\leq 4$. We then consider the complexity of deciding whether ${\rm inv}(D)\leq k$ for a given oriented graph $D$. We show that it is NP-complete for $k=1$, which together with the above conjecture would imply that it is NP-complete for every $k$. This contrasts with a result of Belkhechine et al. which states that deciding whether ${\rm inv}(T)\leq k$ for a given tournament $T$ is polynomial-time solvable.