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

• f the satisfiability problem as

determining the truth value of the statement: ∃𝑦1∃𝑦2 … ∃𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜)

 Can generalize this idea to allow universal

quantifiers, e,g. ∀𝑦1∃𝑦2(𝑦1⋁𝑦2)⋀(𝑦1⋁𝑦2)

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 = ∀𝑦1∃𝑦2∀𝑦3 … ∀𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜) 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 Fp, for large enough p

Arithmetization of TQBF

 Formula 𝜚 with m clauses on variables

𝑦1, … , 𝑦𝑜.

 p large prime  Translate 𝜚 to a polynomial F over the

field (mod p) as follows:

 𝑨1 ∨ 𝑨2 ∨ 𝑨3 → 1 − (1 − 𝑨1)(1 − 𝑨2)(1 −

𝑨3)

 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

• therwise.

 Read quantifiers from left to right and consider

the expression ∀𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜)

 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 x1, … , xn−1 = P

∀𝑦𝑜𝐺 𝑦1, 𝑦2, … , 𝑦𝑜

= F x1, … , xn−1, 1 ⋅ 𝐺(𝑦1, … , 𝑦𝑜−1, 0)

 In a similar manner, we want to find a

polynomial representation of ∃𝑦𝑜−1∀𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜)

 Write new polynomial

P

∃𝑦𝑜−1𝐻 𝑦1, 𝑦2, … , 𝑦𝑜−1

= 1 − (1 − G x1, … , xn−2, 0) ⋅ (1 − 𝐻(𝑦1, … , 𝑦𝑜−2, 1))

Arithmetization of TQBF

 Denote the polynomial for

∃𝑦𝑜−1∀𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜) P

∃𝑦𝑜−1𝑄∀𝑦𝑜𝐺 𝑦1, 𝑦2, … , 𝑦𝑜

 Turn 3CNF formula ϕ into F as in last

lecture.

 Replace ∃𝑦𝑗 with P

∃𝑦𝑗

 Replace ∀𝑦𝑗 with P

∀𝑦𝑗

 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

∀𝑦1P ∃𝑦2 … 𝑄∀𝑦𝑜𝐺 𝑦1, 𝑦2, … , 𝑦𝑜

 Let’s try to (naively) mimic the protocol from

last lecture.

P V Φ Φ 𝐻1 𝑦1 = P

∃𝑦2 … 𝑄∀𝑦𝑜𝐺 𝑦1, 𝑦2, … , 𝑦𝑜

Check 𝐻1 0 ⋅ 𝐻1 1 = 1? Pick random 𝑠

1

Compute 𝛾1 = 𝐻1(𝑠

1)

𝑠

1

𝐻2 𝑦2 = P

∀𝑦3 … 𝑄∀𝑦𝑜𝐺 𝑠, 𝑦2, … , 𝑦𝑜

𝑠2 Check 1 − (1 − 𝐻2 0 ) ⋅ 1 − 𝐻2 1 = 𝐻1 𝑠

1 ?

Pick random 𝑠2 Compute 𝛾2 = 𝐻2(𝑠2) . . . Check 𝛾𝑜 = 𝐺(𝑠

1, 𝑠2, … , 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 𝑦1, 𝑦2, … , 𝑦𝑜 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 𝑦𝑙 = 𝑦, 𝑦 = 0 𝑝𝑠 𝑦 = 1 for

all positive integers k.

 J’ can be obtained by J by erasing all

exponents.

 E.g. 𝐾(𝑦1, 𝑦2, 𝑦3) = 𝑦1

3𝑦2 4 + 5𝑦1𝑦2 3 + 𝑦2 6𝑦3 2

replace by 𝐾′(𝑦1, 𝑦2, 𝑦3) = 6𝑦1𝑦2 + 𝑦2𝑦3

 Define new operator 𝑆𝑦𝑗 which reduces the

exponent of 𝑦𝑗 to 1 at all occurances.

Revised solution

 Formally, we have

𝑆𝑦𝑗𝐾 𝑦1, 𝑦2, … , 𝑦𝑜 = 𝑦𝑗 ⋅ 𝐾 𝑦1, … , 𝑦𝑗−1, 1, 𝑦𝑗+1, … , 𝑦𝑜 + 1 − 𝑦𝑗 ⋅ 𝐾 𝑦1, … , 𝑦𝑗−1, 0, 𝑦𝑗+1, … , 𝑦𝑜

 𝐾′(𝑦1, … , 𝑦𝑜) = 𝑆𝑦1𝑆𝑦2 … 𝑆𝑦𝑜𝐾(𝑦1, … , 𝑦𝑜)

Revised solution

 We now arithmetize the quantified boolean formula

• f the form Φ = ∀𝑦1∃𝑦2∀𝑦3 … ∀𝑦𝑜𝜚(𝑦1, 𝑦2, … , 𝑦𝑜)

into

 E =

P

∀𝑦1𝑆𝑦1𝑄∃𝑦2𝑆𝑦1𝑆𝑦2 … 𝑄∀𝑦𝑜𝑆𝑦1 … 𝑆𝑦𝑜𝐺(𝑦1, … , 𝑦𝑜)