Maker-Breaker渗透游戏1:穿越网格

A. Day, Victor Falgas‐Ravry
{"title":"Maker-Breaker渗透游戏1:穿越网格","authors":"A. Day, Victor Falgas‐Ravry","doi":"10.1017/S0963548320000097","DOIUrl":null,"url":null,"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.","PeriodicalId":10503,"journal":{"name":"Combinatorics, Probability and Computing","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2018-10-11","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"8","resultStr":"{\"title\":\"Maker–Breaker percolation games I: crossing grids\",\"authors\":\"A. Day, Victor Falgas‐Ravry\",\"doi\":\"10.1017/S0963548320000097\",\"DOIUrl\":null,\"url\":null,\"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.\",\"PeriodicalId\":10503,\"journal\":{\"name\":\"Combinatorics, Probability and Computing\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2018-10-11\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"8\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Combinatorics, Probability and Computing\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.1017/S0963548320000097\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Combinatorics, Probability and Computing","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.1017/S0963548320000097","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 8

摘要

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。
Maker–Breaker percolation games I: crossing grids
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.
求助全文
通过发布文献求助,成功后即可免费获取论文全文。 去求助
来源期刊
自引率
0.00%
发文量
0
×
引用
GB/T 7714-2015
复制
MLA
复制
APA
复制
导出至
BibTeX EndNote RefMan NoteFirst NoteExpress
×
提示
您的信息不完整,为了账户安全,请先补充。
现在去补充
×
提示
您因"违规操作"
具体请查看互助需知
我知道了
×
提示
确定
请完成安全验证×
copy
已复制链接
快去分享给好友吧!
我知道了
右上角分享
点击右上角分享
0
联系我们:info@booksci.cn Book学术提供免费学术资源搜索服务,方便国内外学者检索中英文文献。致力于提供最便捷和优质的服务体验。 Copyright © 2023 布克学术 All rights reserved.
京ICP备2023020795号-1
ghs 京公网安备 11010802042870号
Book学术文献互助
Book学术文献互助群
群 号:481959085
Book学术官方微信