The Maker-Breaker Largest Connected Subgraph Game

Given a graph $G$ and $k \in \mathbb{N}$, we introduce the following game played in $G$. Each round, Alice colours an uncoloured vertex of $G$ red, and then Bob colours one blue (if any remain). Once every vertex is coloured, Alice wins if there is a connected red component of order at least $k$, and otherwise, Bob wins. This is a Maker-Breaker version of the Largest Connected Subgraph game introduced in [Bensmail et al. The Largest Connected Subgraph Game. {\it Algorithmica}, 84(9):2533--2555, 2022]. We want to compute $c_g(G)$, which is the maximum $k$ such that Alice wins in $G$, regardless of Bob's strategy. Given a graph $G$ and $k\in \mathbb{N}$, we prove that deciding whether $c_g(G)\geq k$ is PSPACE-complete, even if $G$ is a bipartite, split, or planar graph. To better understand the Largest Connected Subgraph game, we then focus on {\it A-perfect} graphs, which are the graphs $G$ for which $c_g(G)=\lceil|V(G)|/2\rceil$, {\it i.e.}, those in which Alice can ensure that the red subgraph is connected. We give sufficient conditions, in terms of the minimum and maximum degrees or the number of edges, for a graph to be A-perfect. Also, we show that, for any $d \geq 4$, there are arbitrarily large A-perfect $d$-regular graphs, but no cubic graph with order at least $18$ is A-perfect. Lastly, we show that $c_g(G)$ is computable in linear time when $G$ is a $P_4$-sparse graph (a superclass of cographs).