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We consider playing the game of Tic‐Tac‐Toe on block designs BIBD and transversal designs TD. Players take turns choosing points and the first player to complete a block wins the game. We show that triple systems, BIBD, are a first‐player win if and only if . Further, we show that for , TD is a first‐player win if and only if . We also consider a weak version of the game, called Maker–Breaker, in which the second player wins if they can stop the first player from winning. In this case, we adapt known bounds for when either the first or second player can win on BIBD and TD, and show that for Maker–Breaker, BIBD is a first‐player win if and only if . We show that TD is a second‐player win, and so the second player can force a draw in the regular game by playing the same strategy.
We consider playing the game of Tic‐Tac‐Toe on block designs BIBD and transversal designs TD. Players take turns choosing points and the first player to complete a block wins the game. We show that triple systems, BIBD, are a first‐player win if and only if . Further, we show that for , TD is a first‐player win if and only if . We also consider a weak version of the game, called Maker–Breaker, in which the second player wins if they can stop the first player from winning. In this case, we adapt known bounds for when either the first or second player can win on BIBD and TD, and show that for Maker–Breaker, BIBD is a first‐player win if and only if . We show that TD is a second‐player win, and so the second player can force a draw in the regular game by playing the same strategy.
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 ) = R . Several results are also given on corona products in which the second factor is of diameter at most 2.
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