A note on the partition dimension of Cartesian product graphs

TL;DR

This paper investigates the partition dimension of Cartesian product graphs, establishing upper bounds based on the partition and metric dimensions of the component graphs, thus advancing understanding of graph resolving partitions.

Contribution

It provides new bounds on the partition dimension of Cartesian product graphs in terms of the component graphs' partition and metric dimensions.

Findings

pd(G×H) ≤ pd(G) + pd(H)

pd(G×H) ≤ pd(G) + dim(H)

pd(G×H) ≤ dim(G) + dim(H) + 1

Abstract

Let $G=(V,E)$ be a connected graph. The distance between two vertices $u,v\in V$, denoted by $d(u, v)$, is the length of a shortest $u-v$ path in $G$. The distance between a vertex $v\in V$ and a subset $P\subset V$ is defined as $min\{d(v, x): x \in P\}$, and it is denoted by $d(v, P)$. An ordered partition $\{P_1,P_2, ...,P_t\}$ of vertices of a graph $G$, is a \emph{resolving partition}of $G$, if all the distance vectors $(d(v,P_1),d(v,P_2),...,d(v,P_t))$ are different. The \emph{partition dimension} of $G$, denoted by $pd(G)$, is the minimum number of sets in any resolving partition of $G$. In this article we study the partition dimension of Cartesian product graphs. More precisely, we show that for all pairs of connected graphs $G, H$, $pd(G\times H)\le pd(G)+pd(H)$ and $pd(G\times H)\le pd(G)+dim(H).$ Consequently, we show that $pd(G\times H)\le dim(G)+dim(H)+1.$