Year: 2024
Type: article
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 $$\mathcal {R}$$ R , $$\mathcal {S}$$ S , and $$\mathcal {N}$$ N , where $$o(G)=\mathcal {R}$$ o ( G ) = R (resp. $$o(G)=\mathcal {S}$$ o ( G ) = S ), if Resolver (resp. Spoiler) has a winning strategy no matter who starts the game, and $$o(G)=\mathcal {N}$$ o ( G ) = N , if the first player has a winning strategy. In this paper, the game is investigated on corona products $$G\odot H$$ G ⊙ H of graphs G and H. It is proved that if $$o(H)\in \{\mathcal {N}, \mathcal {S}\}$$ o ( H ) ∈ { N , S } , then $$o(G\odot H) = \mathcal {S}$$ o ( G ⊙ H ) = S . No such result is possible under the assumption $$o(H) = \mathcal {R}$$ o ( H ) = R . It is proved that $$o(G\odot P_k) = \mathcal {S}$$ o ( G ⊙ P k ) = S if $$k=5$$ k = 5 , otherwise $$o(G\odot P_k) = \mathcal {R}$$ o ( G ⊙ P k ) = R , and that $$o(G\odot C_k) = \mathcal {S}$$ o ( G ⊙ C k ) = S if $$k=3$$ k = 3 , otherwise $$o(G\odot C_k) = \mathcal {R}$$ o ( G ⊙ C k
Source: Aequationes Mathematicae
Language: English
Subfield: Computational Theory and Mathematics
Field: Computer Science
Domain: Physical Sciences
Open Access status: hybrid
Estimated APC paid: $2,990
OpenAlex ID: W4403739647