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<metadata xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:dc="http://purl.org/dc/elements/1.1/"><dc:title>Maker-Breaker resolving game played on lexicographic products of graphs</dc:title><dc:creator>Savitha,	K. S.	(Avtor)
	</dc:creator><dc:creator>Klavžar,	Sandi	(Avtor)
	</dc:creator><dc:creator>James,	Tijo	(Avtor)
	</dc:creator><dc:subject>Maker-Breaker game</dc:subject><dc:subject>metric dimension</dc:subject><dc:subject>resolving set</dc:subject><dc:subject>Maker-Breaker resolving game</dc:subject><dc:subject>lexicographic product of graphs</dc:subject><dc:description>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 a complete graph, 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.</dc:description><dc:date>2026</dc:date><dc:date>2026-03-23 14:25:41</dc:date><dc:type>Neznano</dc:type><dc:identifier>28549</dc:identifier><dc:identifier>UDK: 519.17</dc:identifier><dc:identifier>ISSN pri članku: 1018-6301</dc:identifier><dc:identifier>DOI: 10.1007/s41980-026-01046-x</dc:identifier><dc:identifier>COBISS_ID: 272611587</dc:identifier><dc:language>sl</dc:language></metadata>
