


























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.
此内容由惯性聚合(RSS阅读器)自动聚合整理,仅供阅读参考。 原文来自 — 版权归原作者所有。