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The Maker-Breaker percolation game on a random board

Vojtěch Dvořák, Adva Mond, Victor Souza

Abstract

The $(m,b)$ Maker-Breaker percolation game on $(\mathbb{Z}^2)_p$, introduced by Day and Falgas-Ravry, is played in the following way. Before the game starts, each edge of $\mathbb{Z}^2$ is removed independently with probability $1-p$. After that, Maker chooses a vertex $v_0$ to protect. Then, in each round Maker and Breaker claim respectively $m$ and $b$ unclaimed edges of $G$. Breaker wins if after the removal of the edges claimed by him the component of $v_0$ becomes finite, and Maker wins if she can indefinitely prevent Breaker from winning. We show that for any $p < 1$, Breaker almost surely has a wining strategy for the $(1,1)$ game on $(\mathbb{Z}^2)_p$. This fully answers a question of Day and Falgas-Ravry, who showed that for $p = 1$ Maker has a winning strategy for the $(1,1)$ game. Further, we show that in the $(2,1)$ game on $(\mathbb{Z}^2)_p$ Maker almost surely has a winning strategy whenever $p > 0.9402$, while Breaker almost surely has a winning strategy whenever $p < 0.5278$. This shows that the threshold value of $p$ above which Maker has a winning strategy for the $(2,1)$ game on $\mathbb{Z}^2$ is non-trivial. In fact, we prove similar results in various settings, including other lattices and biases $(m,b)$. These results extend also to the most general case, which we introduce, where each edge is given to Maker with probability $α$ and to Breaker with probability $β$ before the game starts.

The Maker-Breaker percolation game on a random board

Abstract

The Maker-Breaker percolation game on , introduced by Day and Falgas-Ravry, is played in the following way. Before the game starts, each edge of is removed independently with probability . After that, Maker chooses a vertex to protect. Then, in each round Maker and Breaker claim respectively and unclaimed edges of . Breaker wins if after the removal of the edges claimed by him the component of becomes finite, and Maker wins if she can indefinitely prevent Breaker from winning. We show that for any , Breaker almost surely has a wining strategy for the game on . This fully answers a question of Day and Falgas-Ravry, who showed that for Maker has a winning strategy for the game. Further, we show that in the game on Maker almost surely has a winning strategy whenever , while Breaker almost surely has a winning strategy whenever . This shows that the threshold value of above which Maker has a winning strategy for the game on is non-trivial. In fact, we prove similar results in various settings, including other lattices and biases . These results extend also to the most general case, which we introduce, where each edge is given to Maker with probability and to Breaker with probability before the game starts.
Paper Structure (20 sections, 23 theorems, 75 equations, 6 figures, 1 algorithm)

This paper contains 20 sections, 23 theorems, 75 equations, 6 figures, 1 algorithm.

Key Result

Theorem 1.1

Almost surely Breaker has a winning strategy for the $(1,1)$ Maker-Breaker percolation game on $(\mathbb{Z}^2)_p$, for any $0 \leqslant p < 1$.

Figures (6)

  • Figure 1: The known regions of the phase diagram for the $(1,1)$ game on $(\mathbb{Z}^2)_{\alpha,\beta}$. Almost surely, Breaker wins for points coloured red and Maker wins for points coloured blue.
  • Figure 2: The known regions of the phase diagram for the $(2,1)$ game on the left, and for the $(1,2)$ game on the right, both played on $(\mathbb{Z}^2)_{\alpha,\beta}$.
  • Figure 3: Clumps of size $1$ and $2$ produced by \ref{['alg:core-construction']} on a configuration with $p = 0.8$. Vertices are numbered as they appear.
  • Figure 4: Disjoint triples of edges in $L_{v_0}^N$ used in Maker's strategy.
  • Figure 5: Example of fully claimed $L_{v_0}^N$ in the protective game. Dual paths claimed by Breaker shown in red.
  • ...and 1 more figures

Theorems & Definitions (41)

  • Theorem 1.1
  • Theorem 1.2
  • Theorem 1.3
  • Theorem 1.4
  • Corollary 1.5
  • Theorem 1.6
  • Theorem 1.7
  • Theorem 1.8
  • Lemma 2.1: Harris' inequality
  • Proposition 3.1
  • ...and 31 more