Maker-Breaker domination game on Cartesian products of graphs

TL;DR

This paper analyzes the Maker-Breaker domination game on Cartesian product graphs, determining the winner and game length for various graph classes, including paths, stars, and complete bipartite graphs, and provides exact formulas for certain cases.

Contribution

It characterizes the winner and game parameters of the Maker-Breaker domination game on Cartesian products of specific graph classes, extending previous understanding of the game's dynamics.

Findings

Dominator wins on P_m × P_n for all m,n.

Exact game length for P_3 × P_n with 3 ≤ n ≤ 5.

Dominator wins on G × H if G and H have nontrivial path covers.

Abstract

The Maker-Breaker domination game is played on a graph $G$ by two players, called Dominator and Staller. They alternately select an unplayed vertex in $G$. Dominator wins the game if he forms a dominating set while Staller wins the game if she claims all vertices from a closed neighborhood of a vertex. The game is called \emph{D-game} if Dominator starts the game and it is an \emph{S-game} when Staller starts the game. If Dominator is the winner in the D-game (or the S-game), then $\gmb(G)$ (or $\gmb'(G)$) is defined by the minimum number of moves of Dominator to win the game under any strategy of Staller. Analogously, when Staller is the winner, $\gsmb(G)$ and $\gsmb'(G)$ can be defined in the same way. We determine the winner of the game on the Cartesian product of paths, stars, and complete bipartite graphs, and how fast the winner wins. We prove that Dominator is the winner on $P_m \square P_n$ in both the D-game and the S-game, and $\gmb(P_m \square P_n)$ and $\gmb'(P_m \square P_n)$ are determined when $m=3$ and $3 \le n \le 5$. Dominator also wins on $G \square H$ in both games if $G$ and $H$ admit nontrivial path covers. Furthermore, we establish the winner in the D-game and the S-game on $K_{m,n} \square K_{m',n'}$ for every positive integers $m, m',n,n'$. We prove the exact formulas for $\gmb (G)$, $\gmb'(G)$, $\gsmb(G)$, and $\gsmb'(G)$ where $G$ is a product of stars.