The PCP theorem by gap amplification

  • Authors:
  • Irit Dinur

  • Affiliations:
  • The Hebrew University, Jerusalem, Israel

  • Venue:
  • Journal of the ACM (JACM)
  • Year:
  • 2007

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Abstract

The PCP theorem [Arora and Safra 1998; Arora et. al. 1998] says that every language in NP has a witness format that can be checked probabilistically by reading only a constant number of bits from the proof. The celebrated equivalence of this theorem and inapproximability of certain optimization problems, due to Feige et al. [1996], has placed the PCP theorem at the heart of the area of inapproximability. In this work, we present a new proof of the PCP theorem that draws on this equivalence. We give a combinatorial proof for the NP-hardness of approximating a certain constraint satisfaction problem, which can then be reinterpreted to yield the PCP theorem. Our approach is to consider the unsat value of a constraint system, which is the smallest fraction of unsatisfied constraints, ranging over all possible assignments for the underlying variables. We describe a new combinatorial amplification transformation that doubles the unsat-value of a constraint-system, with only a linear blowup in the size of the system. The amplification step causes an increase in alphabet-size that is corrected by a (standard) PCP composition step. Iterative application of these two steps yields a proof for the PCP theorem. The amplification lemma relies on a new notion of “graph powering” that can be applied to systems of binary constraints. This powering amplifies the unsat-value of a constraint system provided that the underlying graph structure is an expander. We also extend our amplification lemma towards construction of assignment testers (alternatively, PCPs of Proximity) which are slightly stronger objects than PCPs. We then construct PCPs and locally-testable codes whose length is linear up to a polylog factor, and whose correctness can be probabilistically verified by making a constant number of queries. Namely, we prove SAT ∈ PCP 1/2,1[log2(n⋅poly log n), O(1)].