On the Complexity of Parity Games

A. Beckmann, F. Moller
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引用次数: 5

Abstract

Parity games underlie the model checking problem for the modal µ-calculus, the complexity of which remains unresolved after more than two decades of intensive research. The community is split into those who believe this problem - which is known to be both in NP and coNP - has a polynomial-time solution (without the assumption that P = NP) and those who believe that it does not. (A third, pessimistic, faction believes that the answer to this question will remain unknown in their lifetime.) In this paper we explore the possibility of employing Bounded Arithmetic to resolve this question, motivated by the fact that problems which are both NP and coNP, and where the equivalence between their NP and coNP description can be formulated and proved within a certain fragment of Bounded Arithmetic, necessarily admit a polynomial-time solution. While the problem remains unresolved by this paper, we do proposed another approach, and at the very least provide a modest refinement to the complexity of parity games (and in turn the µ-calculus model checking problem): that they lie in the class PLS of Polynomial Local Search problems. This result is based on a new proof of memoryless determinacy which can be formalised in Bounded Arithmetic. The approach we propose may offer a route to a polynomial-time solution. Alternatively, there may be scope in devising a reduction between the problem and some other problem which is hard with respect to PLS, thus making the discovery of a polynomial-time solution unlikely according to current wisdom.
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论奇偶性游戏的复杂性
奇偶博弈是模态微演算的模型检验问题的基础,经过二十多年的深入研究,其复杂性仍未得到解决。社区分为两派,一派认为这个问题——已知在NP和coNP中都有——具有多项式时间解(不假设P = NP),另一派则认为不是。(第三个悲观派认为,在他们有生之年,这个问题的答案将是未知的。)在本文中,我们探讨了用有界算术来解决这个问题的可能性,其动机是这样一个事实,即NP和coNP问题,并且它们的NP和coNP描述之间的等价可以在有界算术的某个片段内表述和证明,必然承认多项式时间解。虽然本文仍未解决这个问题,但我们确实提出了另一种方法,并且至少提供了对奇偶对策复杂性的适度改进(反过来是微微积分模型检查问题):它们位于多项式局部搜索问题的PLS类中。这一结果是基于一种新的无记忆确定性证明,该证明可以在有界算术中形式化。我们提出的方法可能为多项式时间解提供一条途径。或者,在这个问题和其他一些关于PLS的困难问题之间设计一个简化可能是有余地的,因此根据当前的智慧,发现一个多项式时间的解决方案是不太可能的。
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