Maker-Breaker resolving game played on corona products of graphs

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 $$\mathcal {R}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>R</mml:mi> </mml:math> , $$\mathcal {S}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>S</mml:mi> </mml:math> , and $$\mathcal {N}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mi>N</mml:mi> </mml:math> , where $$o(G)=\mathcal {R}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>R</mml:mi> </mml:mrow> </mml:math> (resp. $$o(G)=\mathcal {S}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>S</mml:mi> </mml:mrow> </mml:math> ), if Resolver (resp. Spoiler) has a winning strategy no matter who starts the game, and $$o(G)=\mathcal {N}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>N</mml:mi> </mml:mrow> </mml:math> , if the first player has a winning strategy. In this paper, the game is investigated on corona products $$G\odot H$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:mi>H</mml:mi> </mml:mrow> </mml:math> of graphs G and H . It is proved that if $$o(H)\in \{\mathcal {N}, \mathcal {S}\}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>H</mml:mi> <mml:mo>)</mml:mo> <mml:mo>∈</mml:mo> <mml:mo>{</mml:mo> <mml:mi>N</mml:mi> <mml:mo>,</mml:mo> <mml:mi>S</mml:mi> <mml:mo>}</mml:mo> </mml:mrow> </mml:math> , then $$o(G\odot H) = \mathcal {S}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:mi>H</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>S</mml:mi> </mml:mrow> </mml:math> . No such result is possible under the assumption $$o(H) = \mathcal {R}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>H</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>R</mml:mi> </mml:mrow> </mml:math> . It is proved that $$o(G\odot P_k) = \mathcal {S}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:msub> <mml:mi>P</mml:mi> <mml:mi>k</mml:mi> </mml:msub> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>S</mml:mi> </mml:mrow> </mml:math> if $$k=5$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>k</mml:mi> <mml:mo>=</mml:mo> <mml:mn>5</mml:mn> </mml:mrow> </mml:math> , otherwise $$o(G\odot P_k) = \mathcal {R}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:msub> <mml:mi>P</mml:mi> <mml:mi>k</mml:mi> </mml:msub> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>R</mml:mi> </mml:mrow> </mml:math> , and that $$o(G\odot C_k) = \mathcal {S}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:msub> <mml:mi>C</mml:mi> <mml:mi>k</mml:mi> </mml:msub> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mi>S</mml:mi> </mml:mrow> </mml:math> if $$k=3$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>k</mml:mi> <mml:mo>=</mml:mo> <mml:mn>3</mml:mn> </mml:mrow> </mml:math> , otherwise $$o(G\odot C_k) = \mathcal {R}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>o</mml:mi> <mml:mo>(</mml:mo> <mml:mi>G</mml:mi> <mml:mo>⊙</mml:mo> <mml:msub> <mml:mi>C</mml:mi> <mml:mi>k</mm