Maker–Breaker percolation games I: crossing grids

A. Day, Victor Falgas‐Ravry
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引用次数: 8

Abstract

Abstract Motivated by problems in percolation theory, we study the following two-player positional game. Let Λm×n be a rectangular grid-graph with m vertices in each row and n vertices in each column. Two players, Maker and Breaker, play in alternating turns. On each of her turns, Maker claims p (as yet unclaimed) edges of the board Λm×n, while on each of his turns Breaker claims q (as yet unclaimed) edges of the board and destroys them. Maker wins the game if she manages to claim all the edges of a crossing path joining the left-hand side of the board to its right-hand side, otherwise Breaker wins. We call this game the (p, q)-crossing game on Λm×n. Given m, n ∈ ℕ, for which pairs (p, q) does Maker have a winning strategy for the (p, q)-crossing game on Λm×n? The (1, 1)-case corresponds exactly to the popular game of Bridg-it, which is well understood due to it being a special case of the older Shannon switching game. In this paper we study the general (p, q)-case. Our main result is to establish the following transition. If p ≥ 2q, then Maker wins the game on arbitrarily long versions of the narrowest board possible, that is, Maker has a winning strategy for the (2q, q)-crossing game on Λm×(q+1) for any m ∈ ℕ. If p ≤ 2q − 1, then for every width n of the board, Breaker has a winning strategy for the (p, q)-crossing game on Λm×n for all sufficiently large board-lengths m. Our winning strategies in both cases adapt more generally to other grids and crossing games. In addition we pose many new questions and problems.
Maker-Breaker渗透游戏1:穿越网格
摘要受渗流理论问题的启发,我们研究了以下两个人的位置博弈。设Λm×n为一个矩形网格图,每行有m个顶点,每列有n个顶点。两名玩家,制造者和破坏者,轮流玩。在她的每一轮中,Maker声称p(尚未被认领)的棋盘边Λm×n,而在他的每一轮中,Breaker声称q(尚未被认领)的棋盘边并摧毁它们。如果Maker成功地占领了连接左侧和右侧的交叉路径的所有边缘,那么她就赢了游戏,否则Breaker就赢了。我们称这个博弈为Λm×n上的(p, q)交叉博弈。给定m, n∈n,对于哪些对(p, q) Maker在Λm×n上的(p, q)交叉博弈中有制胜策略?(1,1)-的情况正好对应于流行的桥牌游戏,它很容易理解,因为它是较老的香农交换游戏的特殊情况。本文研究了一般的(p, q)-情况。我们的主要结果是建立以下过渡。如果p≥2q,则Maker在任意长版本的最窄棋盘上获胜,即对于任意m∈n,在Λm×(q+1)上的(2q, q)交叉博弈中,Maker有一个获胜策略。如果p≤2q−1,则对于棋盘的每一个宽度n,对于所有足够大的棋盘长度m, Breaker在Λm×n上的(p, q)交叉博弈中都有一个获胜策略。在这两种情况下,我们的获胜策略更普遍地适用于其他网格和交叉博弈。此外,我们提出了许多新的问题和问题。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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