Tuesday, March 25, 2008

* Collapse of PH
* Probabilistic analog of NP: Interactive Proofs
* IP is subset of PSPACE

COLLAPSE OF PH

A set A is in Sigma_k^P iff there is a deterministic poly-time machine
M and constant c such that

A = {x : Exists y_1 All y_2 Exists y_3 ... M(x, y_1, ..., y_k)
accepts}

Theorem: If NP = co-NP, then PH = NP

INTERACTIVE PROOFS: PROBABILISTIC ANALOG OF NP

We have a prover and a verifier.  Both of them have an input tape, a
worktape, and a shared communication tape.  Prover is a deterministic
machine with no bounds on its running time.  Verifier is a
probabilistic poly-time Turing machine.

Alternately, the prover and verifier exchange information through the
shared tape.  The prover can only write a polynomial amount of
information on the tape.  The goal of the prover is to convince the
verifier that the given string is in the language with high
probability.  The number of rounds allowed is polynomial.

We say that a language L is in IP if there exists an IP protocol such
that:

Completeness: There exists a prover such that w is in L => Pr[Verifier
accepts w] >= 2/3

Soundness: For all provers, if w is not in L => Pr[Verifier rejects w]
>= 2/3

Again, the 2/3 is not significant.  We can bring this up to 1 - 2^{-n}.

Theorem: NP is subset of IP.

Proof: For a language in NP, the prover supplies a certificate (by
exhaustingly going over all possibilities, if needed) to the
verifier.  The verifier can check in poly-time (no random bits
needed).  For string not in language, there is no poly-length
certificate, so no prover will be able to provide this.

End Proof

Lemma: Graph nonisomorphism is in IP.

Proof: Given G_1 and G_2, the verifier selects a random permutation of
the nodes, applies it to one of the two graphs G_i in random and
passes this graph, call it H, to the prover.  The prover is asked to
prove or disprove that H is isomorphic to G_1.

If G_1 is not isomorphic to G_2, then with probability 1/2, H is not
isomorphic to G_1, and the prover can easily check this and say no.
If G_1 is isomorphic to G_2, then under no circumstance can any prover
distinguish between the two.  

Suppose we run this for k rounds.  The verifier's protocol is that if
the graph sent was the same as G_1 and the prover says "not
isomorphic", then immediately reject.  If whenever the graph sent was
different, the prover says "not isomorphic", then accept.

An honest prover would ensure that nonisomorphic graphs will always be
accepted.  What is the probability that isomporphic graphs will be
accepted?  It says "not isomorphic" for precisely the cases where the
two graphs sent were different -- this is at most 1/2^k.

End Proof