Maker-Breaker resolving game played on corona products of graphs

TL;DR

This paper studies the Maker-Breaker resolving game on corona product graphs, determining the game's outcome based on properties of the component graphs and providing specific results for paths and cycles.

Contribution

It characterizes the game outcome on corona products of graphs, especially for paths and cycles, and extends understanding of the game on graphs with diameter at most two.

Findings

If o(H) in {N, S}, then o(G⊙H) = S.

o(G⊙P_k) = S if k=5; otherwise R.

o(G⊙C_k) = S if k=3; otherwise R.

Abstract

The Maker-Breaker resolving game is a game played on a graph $G$ by Resolver and Spoiler. The players taking turns alternately in which each player selects a not yet played vertex of $G$. The goal of Resolver is to select all the vertices in a resolving set of $G$, while that of Spoiler is to prevent this from happening. The outcome $o(G)$ of the game played is one of $\mathcal{R}$, $\mathcal{S}$, and $\mathcal{N}$, where $o(G)=\mathcal{R}$ (resp.\ $o(G)=\mathcal{S}$), if Resolver (resp.\ Spoiler) has a winning strategy no matter who starts the game, and $o(G)=\mathcal{N}$, if the first player has a winning strategy. In this paper, the game is investigated on corona products $G\odot H$ of graphs $G$ and $H$. It is proved that if $o(H)\in\{\mathcal{N}, \mathcal{S}\}$, then $o(G\odot H) = \mathcal{S}$. No such result is possible under the assumption $o(H) = \mathcal{R}$. It is proved that $o(G\odot P_k) = \mathcal{S}$ if $k=5$, otherwise $o(G\odot P_k) = \mathcal{R}$, and that $o(G\odot C_k) = \mathcal{S}$ if $k=3$, otherwise $o(G\odot C_k) = \mathcal{R}$. Several results are also given on corona products in which the second factor is of diameter at most $2$.