Let $(X, \mathcal{F})$ be a hypergraph. The Maker-Breaker game on $(X, \mathcal{F})$ is a combinatorial game between two players, Maker and Breaker. Beginning with Maker, the players take turns claiming vertices from $X$ that have not yet been claimed. Maker wins if she manages to claim all vertices of some hyperedge $F \in \mathcal{F}$. Breaker wins if he claims at least one vertex in every hyperedge. M. L. Rahman and Thomas Watson proved in 2021 that, even when only Maker-Breaker games on 6-uniform hypergraphs are considered, the decision problem of determining which player has a winning strategy is PSPACE-complete. They also showed that the problem is NL-hard when considering hypergraphs of rank 5. In this paper, we improve the latter result by showing that deciding who wins Maker-Breaker games on 5-uniform hypergraphs is still a PSPACE-complete problem. We achieve this by polynomial transformation from the problem of solving the generalized geography game on bipartite digraphs with vertex degrees 3 or less, which is known to be PSPACE-complete.

翻译：设 $(X, \mathcal{F})$ 为一超图。$(X, \mathcal{F})$ 上的 Maker-Breaker 游戏是 Maker 与 Breaker 两名玩家间的组合博弈。游戏从 Maker 开始，双方轮流选取 $X$ 中尚未被占有的顶点。若 Maker 成功占有某条超边 $F \in \mathcal{F}$ 的所有顶点，则其获胜。若 Breaker 在每条超边中均占有至少一个顶点，则其获胜。M. L. Rahman 与 Thomas Watson 于 2021 年证明，即使仅考虑 6-一致超图上的 Maker-Breaker 游戏，判定哪一方玩家拥有必胜策略的决策问题仍是 PSPACE 完全的。他们还指出，当考虑秩为 5 的超图时，该问题是 NL 难的。本文中，我们改进了后一结果，证明判定 5-一致超图上 Maker-Breaker 游戏的胜者仍是 PSPACE 完全问题。我们通过从求解顶点度数不超过 3 的二部有向图上的广义地理游戏问题（已知为 PSPACE 完全问题）进行多项式变换来实现这一证明。