English

Identifying an Honest ${\rm EXP}^{\rm NP}$ Oracle Among Many

Computational Complexity 2015-04-08 v2

Abstract

We provide a general framework to remove short advice by formulating the following computational task for a function ff: given two oracles at least one of which is honest (i.e. correctly computes ff on all inputs) as well as an input, the task is to compute ff on the input with the help of the oracles by a probabilistic polynomial-time machine, which we shall call a selector. We characterize the languages for which short advice can be removed by the notion of selector: a paddable language has a selector if and only if short advice of a probabilistic machine that accepts the language can be removed under any relativized world. Previously, instance checkers have served as a useful tool to remove short advice of probabilistic computation. We indicate that existence of instance checkers is a property stronger than that of removing short advice: although no instance checker for EXPNP{\rm EXP}^{\rm NP}-complete languages exists unless EXPNP=NEXP{\rm EXP}^{\rm NP} = {\rm NEXP}, we prove that there exists a selector for any EXPNP{\rm EXP}^{\rm NP}-complete language, by building on the proof of MIP=NEXP{\rm MIP} = {\rm NEXP} by Babai, Fortnow, and Lund (1991).

Cite

@article{arxiv.1502.07258,
  title  = {Identifying an Honest ${\rm EXP}^{\rm NP}$ Oracle Among Many},
  author = {Shuichi Hirahara},
  journal= {arXiv preprint arXiv:1502.07258},
  year   = {2015}
}

Comments

20 pages; a simplified proof for the main theorem

R2 v1 2026-06-22T08:37:56.159Z