# Maker-Breaker Percolation Games II: Escaping to Infinity

## TL;DR

This paper analyzes a positional game on infinite graphs where Maker tries to keep a vertex in an infinite component while Breaker aims to disconnect it, establishing winning strategies for specific graph types and parameters.

## Contribution

It determines precise winning conditions for Maker and Breaker on the 2D lattice and infinite regular trees, extending understanding of percolation games.

## Key findings

- Maker wins if p ≥ 2q on Z² lattice.
- Breaker wins if 2p ≤ q on Z² lattice.
- Complete characterization of winning strategies on infinite regular trees.

## Abstract

Let $\Lambda$ be an infinite connected graph, and let $v_0$ be a vertex of $\Lambda$. We consider the following positional game. Two players, Maker and Breaker, play in alternating turns. Initially all edges of $\Lambda$ are marked as unsafe. On each of her turns, Maker marks $p$ unsafe edges as safe, while on each of his turns Breaker takes $q$ unsafe edges and deletes them from the graph. Breaker wins if at any time in the game the component containing $v_0$ becomes finite. Otherwise if Maker is able to ensure that $v_0$ remains in an infinite component indefinitely, then we say she has a winning strategy. This game can be thought of as a variant of the celebrated Shannon switching game. Given $(p,q)$ and $(\Lambda, v_0)$, we would like to know: which of the two players has a winning strategy? Our main result in this paper establishes that when $\Lambda = \mathbb{Z}^2$ and $v_0$ is any vertex, Maker has a winning strategy whenever $p\geq 2q$, while Breaker has a winning strategy whenever $2p\leq q$. In addition, we completely determine which of the two players has a winning strategy for every pair $(p,q)$ when $\Lambda$ is an infinite $d$-regular tree. Finally, we give some results for general graphs and lattices and pose some open problems.