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The Maker-Breaker Largest Connected Subgraph Game

Julien Bensmail, Foivos Fioravantes, Fionn Mc Inerney, Nicolas Nisse, Nacim Oijid

Abstract

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).

The Maker-Breaker Largest Connected Subgraph Game

Abstract

Given a graph and , we introduce the following game played in . Each round, Alice colours an uncoloured vertex of 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 , 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 , which is the maximum such that Alice wins in , regardless of Bob's strategy. Given a graph and , we prove that deciding whether is PSPACE-complete, even if 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 for which , {\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 , there are arbitrarily large A-perfect -regular graphs, but no cubic graph with order at least is A-perfect. Lastly, we show that is computable in linear time when is a -sparse graph (a superclass of cographs).
Paper Structure (13 sections, 17 theorems, 4 equations, 4 figures)

This paper contains 13 sections, 17 theorems, 4 equations, 4 figures.

Key Result

Lemma 2.1

If $H$ is a (not necessarily proper) subgraph of a graph $G$ with connected components $H_1,\dots,H_k$, then $c_g(H)=\max\left\{c_g(H_1),\dots,c_g(H_k)\right\} \le c_g(G)$.

Figures (4)

  • Figure 1: Illustration of the construction in the proof of Theorem \ref{['theorem:pspace-planar']}.
  • Figure 2: Cases 1.(a) and 1.(b)i. in the proof of Theorem \ref{['thm:dense']}.
  • Figure 3: Examples of $d$-regular A-perfect graphs constructed in the proof of Lemma \ref{['theorem regular n']}.
  • Figure 4: Illustration of the infinite Hexagonal grid $H_{\infty}$ in the proof of Proposition \ref{['prop:hexGrid']}. The connected red subgraphs are vertex-disjoint $6$-cycles covering the vertices of $H_{\infty}$. The black edges induce a matching of $H_{\infty}$.

Theorems & Definitions (33)

  • Lemma 2.1
  • proof
  • Corollary 2.2
  • proof
  • Lemma 2.3
  • proof
  • Theorem 3.1
  • proof
  • Corollary 3.2
  • proof
  • ...and 23 more