Equivalence between pathbreadth and strong pathbreadth

TL;DR

This paper proves that the strong pathbreadth of any graph is at most four times its pathbreadth, establishing an equivalence between these two graph parameters.

Contribution

It demonstrates that strong pathbreadth is bounded above by four times the pathbreadth, confirming a conjecture related to graph decompositions.

Findings

Proved that (G) \u2264; 4 (G) for all graphs G.

Established an explicit bound linking pathbreadth and strong pathbreadth.

Confirmed a conjecture inspired by prior work in graph theory.

Abstract

We say that a given graph $G = (V, E)$ has \emph{pathbreadth} at most $\rho$, denoted $\pb(G) \leq \rho$, if there exists a Roberston and Seymour's path decomposition where every bag is contained in the $\rho$-neighbourhood of some vertex. Similarly, we say that $G$ has \emph{strong pathbreadth} at most $\rho$, denoted $\spb(G) \leq \rho$, if there exists a Roberston and Seymour's path decomposition where every bag is the complete $\rho$-neighbourhood of some vertex. It is straightforward that $\pb(G) \leq \spb(G)$ for any graph $G$. Inspired from a close conjecture in [Leitert and Dragan, COCOA'16], we prove in this note that $\spb(G) \leq 4 \cdot \pb(G)$.