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The Riis Complexity Gap for QBF Resolution

机译:QBF分辨率的Riis复杂度差距

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We give an analogue of the Riis Complexity Gap Theorem for Quanti fied Boolean Formulas (QBFs). Every fi rst-order sentence without finite models gives rise to a sequence of QBFs whose minimal refutations in tree-like Q-Resolution are either of polynomial size (if has no models) or at least exponential in size (if has some in finite model). However, di fferently from the translations to propositional logic, the translation to QBF must be given additional structure in order for the polynomial upper bound to hold in tree-like Q-Resolution. This extra structure is not needed in the system tree-like Exp+Res, where we see the complexity gap on a natural translation to QBF.
机译:我们给出了量化布尔公式(QBFs)的Riis复杂性缺口定理的类似物。每个没有有限模型的一阶句子都会产生一系列QBF,这些QBF在树状Q-Resolution中的最小反驳要么是多项式大小(如果没有模型),要么至少是指数大小(如果在有限模型中有一些) )。但是,与对命题逻辑的翻译不同,对QBF的翻译必须赋予其他结构,以使多项式上限保持在树状Q分辨率中。系统树状的Exp + Res不需要这种额外的结构,在该结构中,我们看到了自然转换为QBF的复杂性差距。

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