Let Λ be an infinite connected graph, and let $v_0$ be a vertex of Λ. We consider the following positional game. Two players, Maker and Breaker, play in alternating turns. Initially all edges of Λ 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 $(\mathrm{\Lambda },v_0)$, we would like to know: which of the two players has a winning strategy?

Our main result in this paper establishes that when $\mathrm{\Lambda }={\mathbb{Z}}^2$ and $v_0$ is any vertex, Maker has a winning strategy whenever $p\ge 2q$, while Breaker has a winning strategy whenever $2p\le q$. In addition, we completely determine which of the two players has a winning strategy for every pair $(p,q)$ when Λ is an infinite d-regular tree. Finally, we give some results for general graphs and lattices and pose some open problems.

令Λ为无限连通图，令 $v_0$是Λ的顶点。我们考虑以下位置博弈。Maker 和 Breaker 两个玩家轮流玩。最初 Λ 的所有边都被标记为不安全的。在她的每个回合中，Maker 将p 个不安全的边标记为安全，而在他的每个回合中，Breaker 将q 个不安全的边从图中删除。如果在游戏中的任何时间包含包含的组件，Breaker 获胜$v_0$变得有限。否则，如果 Maker 能够确保$v_0$无限期地保持在无限分量中，那么我们说她有一个获胜策略。这个游戏可以被认为是著名的香农切换游戏的变体。给定的$(磷,q)$ 和 $(\mathrm{\Lambda },v_0)$，我们想知道：两位选手中哪一位有制胜策略？

我们在本文中的主要结果表明，当 $\mathrm{\Lambda }={\mathbb{Z}}^2$ 和 $v_0$ 是任何顶点，Maker 有一个获胜策略 $磷\ge 2q$，而 Breaker 有一个获胜策略，无论何时 $2磷\le q$. 此外，我们完全确定两个玩家中的哪一个对每对都有获胜策略$(磷,q)$当 Λ 是无限d -正则树时。最后，我们给出了一般图和格的一些结果，并提出了一些未解决的问题。