arXiv cs/01020192001-02-22cs.CLcs.CC

OT中的易约束和硬约束排序：算法和复杂性

Easy and Hard Constraint Ranking in OT: Algorithms and Complexity

Jason Eisner

备注：: 12 pages, online proceedings version (small corrections and clarifications to printed version)
发表信息：: Jason Eisner, Lauri Karttunen and Alain Theriault (eds.), Finite-State Phonology: Proceedings of the 5th Workshop of the ACL Special Interest Group in Computational Phonology (SIGPHON), pp. 22-33. Luxembourg, August 2000

摘要：

我们考虑的问题，排名一组OT约束的方式与数据一致。我们加快Tesar和Smolensky的RCD算法是线性的约束的数量。这找到了一个排名，所以每个证明形式x_i击败或平局一个特定的竞争对手y_i。我们还推广了RCD，使每个x_i击败或追平所有可能的竞争对手。唉，这个更现实的学习版本没有多项式算法，除非P=NP！事实上，甚至一代人也没有。因此，我们无法在蛮力的基础上进行定性的改进：如果曲面形式被完全观测到，那么仅仅检查一个（给定的）排序与给定的形式是否一致就是coNP完全的，否则就是Delta_2^p完全的。事实上，OT生成是OptP完全的。至于排序，如果形式被完全观察到，则确定是否存在任何一致的排序是coNP困难的（但在Delta_2 ^p中），如果不存在，则Sigma_2^p是完全的。最后，我们表明，生成和排名更容易在派生理论：在P，NP-完全。

英文摘要：

We consider the problem of ranking a set of OT constraints in a manner consistent with data. We speed up Tesar and Smolensky's RCD algorithm to be linear on the number of constraints. This finds a ranking so each attested form x_i beats or ties a particular competitor y_i. We also generalize RCD so each x_i beats or ties all possible competitors. Alas, this more realistic version of learning has no polynomial algorithm unless P=NP! Indeed, not even generation does. So one cannot improve qualitatively upon brute force: Merely checking that a single (given) ranking is consistent with given forms is coNP-complete if the surface forms are fully observed and Delta_2^p-complete if not. Indeed, OT generation is OptP-complete. As for ranking, determining whether any consistent ranking exists is coNP-hard (but in Delta_2^p) if the forms are fully observed, and Sigma_2^p-complete if not. Finally, we show that generation and ranking are easier in derivational theories: in P, and NP-complete.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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英文摘要：

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