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# Lecture1 - PCPs and Inapproximability The PCP Theorem and...

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PCPs and Inapproximability The PCP Theorem and GAP-MAX-E3SAT My T. Thai 1

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1 Recap 1.1 NP-Completeness Language L is in NP iff there is a polynomial time deterministic verifier V (say a Turning machine) and an arbitrarily powerful prover P , which the following properties: Completeness: For every x L , P can write a proof π of length poly( | x | ) that V accepts Soundness: For every x / L , no matter what π proof P writes, V rejects. For example, consider the Vertex Cover (VC) problem. Input x = ( G, k ) where G is a graph and k is a positive integer number. (Note that here is the decision problem of VC). The question here is that if there exists a VC of G with at most k vertices. The prover P will write a proof π which is a subset S of vertices. The verifier C can verify if π is a VC or not. Thus VC is in NP. 1.2 PCP The first question arising from the above definition is: What if we change the requirement of verifier V ? What if V produces one-sided-error? That is, if x is a NO instance, V still output YES. What if V can read a certain amount of π , not the whole π as above? Randomness complexity : The total bits of randomness (usually is O (log n )) that V can uses, called r . In the case of NP as above, the randomness complexity is 0. Query complexity : The total bits of π that V can read, called q . that is, V uses r bits of randomness to choose q random locations in π and reads the bits in these q randomly chosen locations. Such a verifier V is called ( r, q )-restricted verifier. More formally: Definition 1 V is said an ( r, q ) -restricted verifier V if V is a randomized poly-time algorithm with randomness complexity r and query complexity q . 2
(Note that r and q can be a function of the input size, not necessary a con- stant.) Given 0 s c 1, Π PCP c,s [ r, q ] if there exists an ( r, q )-restricted verifier satisfying the following conditions: Completeness: If x is a YES-instance, then there exists a proof π such that Prob[ V ( x, π ) = Y ES ] c Soundness: If x is a NO-instance, then for any π , Prob[ V ( x, π ) = Y ES ] s When c = 1, and s = 1 / 2, for simplicity, we write PCP[ r, q ] instead of PCP 1 , 1 / 2 [ r, q ] Now, can you show that NP = PCP 1 ,s [0 , poly ( n )] for all s < 1? Can you show that PCP [ O (log n ) , O (log n )] NP ? Note that when r = O (log n ), the the proof-checking can be derandomized, that is, V can be simulated by a polynomial time deterministic verifier that simulates the computation of V on each of the 2 r = n O (1) possible random inputs and then computes the probability that V ( x, π ) accepts, then accepts if and only if this probability is one. 2 The PCP Theorem Theorem 1 The PCP Theorem: There exists two constant numbers c 1 and c 2 such that NP = PCP [ c 1 log n, c 2 ] . For simplicity, NP = PCP [ O (log n ) , O (1)] Theorem 2 Hastad (1997) Theorem [2] For every constant , δ > 0 , NP = PCP 1 - , 1 / 2+ δ [ O (log n ) , 3] That is, for every constant , δ > 0, there is a poly-size PCP for NP that reads just three random bits and test their XOR. Its completeness is 1 - and its soundness is 1 / 2 + . This result has imperfect completeness. However, if

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