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 = Z(2) and v(0) is any vertex, Maker has a winning strategy whenever p >= 2q, while Breaker has a winning strategy whenever 2p <= 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. (C) 2020 Elsevier Inc. All rights reserved.