CS 579: Computational Complexity. Lecture 10 IP=PSPACE, Part 2 - PowerPoint PPT Presentation

CS 579: Computational Complexity. Lecture 10 IP=PSPACE, Part 2 Alexandra Kolla

Today  Proof: PSPACE ⊆ IP.  Discuss MIP, PCP.

PSPACE - complete language: TQBF  For 3CNF boolean formula we may think of the satisfiability problem as determining the truth value of the statement: ∃𝑦 1 ∃𝑦 2 … ∃𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 )  Can generalize this idea to allow universal quantifiers, e,g. ∀𝑦 1 ∃𝑦 2 (𝑦 1 ⋁𝑦 2 )⋀(𝑦 1 ⋁𝑦 2 )

PSPACE - complete language: TQBF  Consider the language of all true quantified boolean formulas: TQBF={ Φ : Φ is a true quantified boolean formula }  TQBF is PSPACE-complete  Thus, if we have an interactive proof recognizing TQBF, we have it for all PSPACE.

Arithmetization of TQBF  We consider that all quantified boolean formulas are given to us as: Φ = ∀𝑦 1 ∃𝑦 2 ∀𝑦 3 … ∀𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 ) Where 𝜚 is 3 CNF formula.  Similar ideas as #P ⊆IP  First, arithmetize the formula and then the prover convinces verifier that the arithmetized formula evaluates to 1.  In what follows, random elements are drawn from field F p , for large enough p

Arithmetization of TQBF  Formula 𝜚 with m clauses on variables 𝑦 1 , … , 𝑦 𝑜 .  p large prime  Translate 𝜚 to a polynomial F over the field (mod p) as follows:  𝑨 1 ∨ 𝑨 2 ∨ 𝑨 3 → 1 − (1 − 𝑨 1 )(1 − 𝑨 2 )(1 − 𝑨 3 )  F is the product of all the m expressions corresponding to the m clauses.

Arithmetization of TQBF  Each literal has degree 3, so F has degree at most 3m.  For a zero-one assignment, F evaluates to zero if this assignment does not satisfy 𝜚 , and to 1 otherwise.  Read quantifiers from left to right and consider the expression ∀𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 )  This expression has n-1 free variables and for each substitution of values to the variables its is either true or false.  We are looking for a polynomial with the same behavior.

Arithmetization of TQBF  Write new polynomial G x 1 , … , x n−1 = P ∀𝑦 𝑜 𝐺 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 = F x 1 , … , x n−1 , 1 ⋅ 𝐺(𝑦 1 , … , 𝑦 𝑜−1 , 0)  In a similar manner, we want to find a polynomial representation of ∃𝑦 𝑜−1 ∀𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 )  Write new polynomial P ∃𝑦 𝑜−1 𝐻 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜−1 = 1 − (1 − G x 1 , … , x n−2 , 0) ⋅ (1 − 𝐻(𝑦 1 , … , 𝑦 𝑜−2 , 1))

Arithmetization of TQBF  Denote the polynomial for ∃𝑦 𝑜−1 ∀𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 ) P ∃𝑦 𝑜−1 𝑄 ∀𝑦 𝑜 𝐺 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜  Turn 3CNF formula ϕ into F as in last lecture.  Replace ∃𝑦 𝑗 with P ∃𝑦 𝑗  Replace ∀𝑦 𝑗 with P ∀𝑦 𝑗  Final expression always evaluates to 0 or 1. It evaluates to 1 iff the quantified boolean formula Φ is true.

Arithmetization of TQBF  Final arithmetic expression: P ∀𝑦 1 P ∃𝑦 2 … 𝑄 ∀𝑦 𝑜 𝐺 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜  Let’s try to (naively) mimic the protocol from last lecture.

Φ Φ P V 𝐻 1 𝑦 1 = P ∃𝑦 2 … 𝑄 ∀𝑦 𝑜 𝐺 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 Check 𝐻 1 0 ⋅ 𝐻 1 1 = 1? Pick random 𝑠 1 Compute 𝛾 1 = 𝐻 1 (𝑠 1 ) 𝑠 1 Check 1 − (1 − 𝐻 2 0 ) ⋅ 1 − 𝐻 2 1 = 𝐻 2 𝑦 2 = P ∀𝑦 3 … 𝑄 ∀𝑦 𝑜 𝐺 𝑠, 𝑦 2 , … , 𝑦 𝑜 𝐻 1 𝑠 1 ? Pick random 𝑠 2 Compute 𝛾 2 = 𝐻 2 (𝑠 2 ) 𝑠 2 Check 𝛾 𝑜 = 𝐺(𝑠 1 , 𝑠 2 , … , 𝑠 𝑜 ) . . .

Naïve protocol  Problem is that the degree of the polynomial in the end can be exponential.  Prover would have to send exponentially many coefficients but verifier will not be able to read them all.  Need to ensure that the degree of any variable in any intermediate stage of the transformation never goes above two.

Revised solution  At every stage of the transformation, we have some polynomial 𝐾 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 where some variables might have degree bigger than two.  We can’t expect to transform it into J’ where the degree of all variables is at most two and they evaluate the same at every point.  We only need J, J’ to agree on 0 -1 assignments.

Revised solution  Key observation 𝑦 𝑙 = 𝑦, 𝑦 = 0 𝑝𝑠 𝑦 = 1 for all positive integers k.  J’ can be obtained by J by erasing all exponents. 4 + 5𝑦 1 𝑦 2 3 + 𝑦 2 3 𝑦 2 6 𝑦 3 2  E.g. 𝐾(𝑦 1 , 𝑦 2 , 𝑦 3 ) = 𝑦 1 replace by 𝐾′(𝑦 1 , 𝑦 2 , 𝑦 3 ) = 6𝑦 1 𝑦 2 + 𝑦 2 𝑦 3  Define new operator 𝑆𝑦 𝑗 which reduces the exponent of 𝑦 𝑗 to 1 at all occurances.

Revised solution  Formally, we have 𝑆𝑦 𝑗 𝐾 𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 = 𝑦 𝑗 ⋅ 𝐾 𝑦 1 , … , 𝑦 𝑗−1 , 1, 𝑦 𝑗+1 , … , 𝑦 𝑜 + 1 − 𝑦 𝑗 ⋅ 𝐾 𝑦 1 , … , 𝑦 𝑗−1 , 0, 𝑦 𝑗+1 , … , 𝑦 𝑜  𝐾′(𝑦 1 , … , 𝑦 𝑜 ) = 𝑆𝑦 1 𝑆𝑦 2 … 𝑆𝑦 𝑜 𝐾(𝑦 1 , … , 𝑦 𝑜 )

Revised solution  We now arithmetize the quantified boolean formula of the form Φ = ∀𝑦 1 ∃𝑦 2 ∀𝑦 3 … ∀𝑦 𝑜 𝜚(𝑦 1 , 𝑦 2 , … , 𝑦 𝑜 ) into  E = P ∀𝑦 1 𝑆𝑦 1 𝑄 ∃𝑦 2 𝑆𝑦 1 𝑆𝑦 2 … 𝑄 ∀𝑦 𝑜 𝑆𝑦 1 … 𝑆𝑦 𝑜 𝐺(𝑦 1 , … , 𝑦 𝑜 )