{ "id": "1907.00210", "version": "v1", "published": "2019-06-29T14:37:56.000Z", "updated": "2019-06-29T14:37:56.000Z", "title": "Maker-Breaker Percolation Games II: Escaping to Infinity", "authors": [ "A. Nicholas Day", "Victor Falgas-Ravry" ], "comment": "19 pages, 1 figure", "categories": [ "math.CO" ], "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.", "revisions": [ { "version": "v1", "updated": "2019-06-29T14:37:56.000Z" } ], "analyses": { "subjects": [ "05C57", "91A43", "05D99" ], "keywords": [ "maker-breaker percolation games", "winning strategy", "unsafe edges", "turns breaker", "breaker wins" ], "note": { "typesetting": "TeX", "pages": 19, "language": "en", "license": "arXiv", "status": "editable" } } }