Complexity Theory. Jörg Kreiker. Summer term Chair for Theoretical Computer Science Prof. Esparza TU München

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1 Complexity Theory Jörg Kreiker Chair for Theoretical Computer Science Prof. Esparza TU München Summer term 2010

2 Lecture 16 IP = PSPACE

3 3 Goal and Plan Goal IP = PSPACE Plan 1. PSPACE IP by showing QBF IP 2. IP PSPACE by computing optimal prover strategies in polynomial space

4 4 Agenda arithmetization of Boolean formulas arithmetization of quantified formulas by linearization interactive protocol for QBF Tomorrow optimal prover strategy to show IP PSPACE a note on grpah isomorphism summary: interactive proofs incl further reading and context outlook: approximation and PCP theorem evaluation

5 5 Proof Idea Show that QBF IP. This implies PSPACE IP because QBF is PSPACE-complete IP closed under polynomial reductions Technique Turn formulas into polynomials, similar to reduction from 3SAT to ILP: arithmetization.

6 6 Setting let Φ = Q 1 x 1... Q n x n ϕ(x 1,..., x n ) be a quantified boolean formula, where ϕ is in 3CNF with m clauses Φ is either true or false running example: Φ = = x y (x y) (x y), where the body is written ϕ = deciding truth value of Φ is PSPACE-complete

7 Observation x y is satisfiable iff x y = 1 for x, y {0, 1} x is satisfiable iff 1 x = 1 x y is satisfiable iff x + y 1 note that x y x y x y x y x y is satisfiable iff x + y xy = 1

8 8 Arithmetization of Boolean formulas For Boolean formula ϕ(x 1,..., x n ) we define ari ϕ (x 1,..., x n ) such that ϕ(x 1,..., x n ) is satisfiable iff ari ϕ (x 1,..., x n ) is 1 for satisfying assignment of x i to true/false and the corresponding x i.

9 9 Arithmetization of Boolean formulas ari xi (x 1,..., x n ) = x i ari ϕ (x 1,..., x n ) = 1 ari ϕ (x 1,..., x n ) ari ϕ1 ϕ 2 (x 1,..., x n ) = ari ϕ1 (x 1,..., x n ) ari ϕ2 (x 1,..., x n ) ari ϕ1 ϕ 2 (x 1,..., x n ) = ari ϕ1 (x 1,..., x n ) + ari ϕ2 (x 1,..., x n ) ari ϕ1 (x 1,..., x n ) ari ϕ2 (x 1,..., x n ) Example ari ϕ= (x, y) = (x + (1 y) x(1 y)) ((1 x) + y (1 x)y) = (1 y + xy) (1 x + xy) = 1 x y + 3xy xy 2 x 2 y + x 2 y 2 =: f = (x, y)

10 10 Observation degree of arithmetization is 3m crucial for polynomial representation of formulas

11 What about quantification? Intuition universal quantification corresponds to conjunction corresponds to multiplication existential quantification corresponds to disjunction corresponds to addition ari xi.ϕ(x 1,..., x i,..., x n ) equals ari ϕ (x 1,..., 0,..., x n ) ari ϕ (x 1,..., 1,..., x n ) ari xi.ϕ(x 1,..., x i,..., x n ) equals ari ϕ (x 1,..., 0,..., x n ) + ari ϕ (x 1,..., 1,..., x n ) ari ϕ (x 1,..., 0,..., x n ) ari ϕ (x 1,..., 1,..., x n )

12 12 Running Example Example ari Φ= (x, y) = ari y.ϕ= (0, y) ari y.ϕ= (1, y) = (f = (0, 0) + f = (0, 1) f = (0, 0)f = (0, 1))... =... = 1

13 13 Lessons learnt Φ = is true degree of polynomial might get exponential in m coefficients too Rescue over {0, 1} we have x c = x gives rise to linearization to get rid of large coefficients: compute over some sufficiently small finite field

14 Agenda arithmetization of Boolean formulas arithmetization of quantified formulas by linearization interactive protocol for QBF

15 15 Linearization Linearization means reducing all exponents in polynomial to 1. L y (f(x, y)) = f(x, 1) y + f(x, 0) (1 y) L y (f(x, y) is linear in y L y (f(x, y)) is equivalent to f(x, y) over {0, 1} 2 Example L y (f = (x, y)) = L y (1 x y + 3xy xy 2 x 2 y + x 2 y 2 ) = (1 y)(1 x) + y ( x + 3x x x 2 + x 2 ) = 1 x y + 2xy

16 16 General form L j (f(x 1,..., x j,..., x n )) = f(x 1,..., 1,..., x k )x j + f(x 1,..., 0,..., x k )(1 x j ) Arithmetization 1. arithmetize Boolean body of formula 2. linearize all variables 3. for innermost quantifier apply ari x (resp. ari x) 4. linearize all but x 5. repeat from 3.

17 17 Recursive definition of general arithmetization f n,n (x 1,..., x n ) := ari ϕ (x 1,..., x n ) f i,i (x 1,..., x i ) := f i+1,0 (x 1,..., x i, 0)f i+1,0 (x 1,..., x i, 1) if x i+1 universal f i,i (x 1,..., x i ) := f i+1,0 (x 1,..., x i, 0) + f i+1,0 (x 1,..., x i, 1) f i+1,0 (x 1,..., x i, 0)f i+1,0 (x 1,..., x i, 1) if x i+1 existential f i,j (x 1,..., x i ) = L j+1 (f i,j+1 (x 1,..., x i ))

18 18 Observations there are O(n 2 ) functions f, functions f n, have degree at most 3m all other functions have degree of each variable at most 2 f 0,0 = 1 iff Φ QBF

19 19 Agenda arithmetization of Boolean formulas arithmetization of quantified formulas by linearization interactive protocol for QBF

20 20 Protocol intuition V accepts if f 0,0 = 1 P needs to convince V of that fact by iterating over all f i,j V challenges P by choosing random values from a finite field P inserts these values into polynomials and return linear function V checks that functions adhere to recursive scheme

21 Initialization verifier and prover agree on prime p such that 12 Φ 2 < p 24 Φ 2 all polynomials will be computed in Z/pZ this is a range, where linear functions can be polynomially represented and evaluated start: P sends f 0,0, the prime and the primality proof if f 0,0 = 1 then iterate from i = 1 and j = 0 until both reach n; otherwise reject O(n 2 ) rounds

22 Quantor case j = 0 V asks for f i,0 (r 1,..., r i 1, x i ) P sends f i,0 (r 1,..., r i 1, x i ) if x i is universally quantified, V checks whether f i,0 (r 1,..., r i 1, 0)f i,0 (r 1,..., r i 1, 1) p f i 1,i 1 (r 1,..., r i 1 ) if x i is existentially quantified, V checks f i,0 (r 1,..., r i 1, 0) + f i,0 (r 1,..., r i 1, 1) f i,0 (r 1,..., r i 1, 0)f i,0 (r 1,..., r i 1, 1) p f i 1,i 1 (r 1,..., r i 1 ) V picks random number r i Z/pZ and set j to 1

23 23 Linearization case j > 0 V asks fo f i,j (r 1,..., x j,..., r i ) P sends f i,j (r 1,..., x j,..., r i ) V checks (1 r j )f i,j (r 1,..., 0,..., r i )+ r j f i,j (r 1,..., 1,..., r i ) p f i,j 1 (r 1,..., r i ) V picks r j at random and increases j (or sets j to 0 and increases i)

24 Finally... P tests whether ari ϕ (r 1,..., r n ) p f n,n (r 1,..., r n )

25 25 Observations P only sends linear functions total message length still polynomial V can compute linear functions in Z/pZ if Φ QBF P can always convince V by sending correct polynomials perfect completeness we have public coins

26 26 What if Φ QBF? An honest prover admits this fact. A cheating prover can try to send forged polynomials g i,j (x) instead of f i,j (x 1,..., x,..., x i ). For soundness P must fail to convince V with high probability.

27 Soundness P can cheat in round (i,j) iff f i,j (x 1,..., x,..., x i ) g i,j (x) p 0 that is: iff V by chance picks a root r k of a polynomial probability to do so in round (i, j) is q i,j deg(f i,j )/p since polynomials of degree n have at most n roots f n, have degree at most 3m f i<n, have degree at most 2 there are (n + 1)(n + 2)/2 polynomials, n + 1 large ones Pr[P cheats] Σ n i=1 Σi j=0 q i,j 3m(n+1) p 4 Φ 2 p + n(n+1) p 1/3

28 28 Agenda arithmetization of Boolean formulas arithmetization of quantified formulas by linearization interactive protocol for QBF Tomorrow optimal prover strategy to show IP PSPACE a note on grpah isomorphism summary: interactive proofs incl further reading and context outlook: approximation and PCP theorem evaluation

Complexity Theory. Jörg Kreiker. Summer term Chair for Theoretical Computer Science Prof. Esparza TU München

Complexity Theory. Jörg Kreiker. Summer term Chair for Theoretical Computer Science Prof. Esparza TU München Complexity Theory Jörg Kreiker Chair for Theoretical Computer Science Prof. Esparza TU München Summer term 2010 Lecture 4 NP-completeness Recap: relations between classes EXP PSPACE = NPSPACE conp NP conp