Maker-Breaker resolving game played on lexicographic products of graphs
Savitha K S, Sandi Klavžar, Tijo James
TL;DR
Problem: determine who wins the Maker-Breaker resolving game on lexicographic products $G\circ H$ and compute MB-resolving numbers in key cases. Approach: establish general bounds and layer-based arguments, then treat complete-factor and path/cycle second factors with twin considerations. Key findings: if $o(H)\in\{\mathcal{N},\mathcal{S}\}$, then $o(G\circ H)=\mathcal{S}$; for $G\circ K_n$ with $n\ge3$ Spoiler wins in two moves and $S_{MB}=2$; for $K_2$-factors outcomes depend on true/false twins; for $G\circ P_n$ and $G\circ C_n$ many even-length cases yield Resolver wins while odd-length cases depend on twin structure, with precise results for small lengths and a general $o=\mathcal{R}$ when twins are absent. Significance: delivers a systematic taxonomy of MBRG outcomes on lexicographic products, guiding both theoretical understanding and potential applications in network monitoring and metric-dimension-inspired strategies.
Abstract
In the Maker-Breaker resolving game, two players named Resolver and Spoiler alternately select unplayed vertices of a given graph $G$. The aim of Resolver is to select all the vertices of some resolving set of $G$, while Spoiler aims to select at least one vertex from every resolving set of $G$. In this paper, this game is investigated on the lexicographic product of graphs. It is proved that if Spoiler has a winning strategy on a graph $H$ no matter who starts the game, or if the first player has a winning strategy on $H$, then Spoiler always has a winning strategy on $G\circ H$. Special attention is paid to lexicographic products in which the second factor is either complete, or a path, or a cycle. For instance, in $G\circ P_{2\ell}$ and in $G\circ C_{2\ell}$, Resolver always wins, while in $G\circ P_{2\ell+1}$ and in $G\circ C_{2\ell+1}$ the same conclusion holds provided $G$ is free from false twins. On the other hand, Spoiler always wins on $G\circ P_5$. In most of the cases, the corresponding Maker-Breaker resolving number is also determined.