We give a simple proof that assuming the Exponential Time Hypothesis (ETH), determining the winner of a Rabin game cannot be done in time $2^{o(k \log k)} \cdot n^{O(1)}$, where $k$ is the number of pairs of vertex subsets involved in the winning condition and $n$ is the vertex count of the game graph. While this result follows from the lower bounds provided by Calude et al [SIAM J. Comp. 2022], our reduction is simpler and arguably provides more insight into the complexity of the problem. In fact, the analogous lower bounds discussed by Calude et al, for solving Muller games and multidimensional parity games, follow as simple corollaries of our approach. Our reduction also highlights the usefulness of a certain pivot problem -- Permutation SAT -- which may be of independent interest.

翻译：我们给出一个简单证明：假设指数时间假说（ETH）成立，则判断Rabin博弈胜者不能在时间 $2^{o(k \log k)} \cdot n^{O(1)}$ 内完成，其中 $k$ 为获胜条件中涉及的顶点子集对的数量，$n$ 为博弈图的顶点数。虽然该结果可由Calude等人[SJAM J. Comp. 2022]提供的下界推导得出，但我们的归约更为简洁，且能更深入地揭示该问题的复杂度本质。事实上，Calude等人讨论的关于求解Muller博弈与多维奇偶博弈的类似下界，均可作为本方法的简单推论。我们的归约还凸显了某个关键问题——置换SAT——的实用性，该问题或具有独立研究价值。