CS151 Complexity Theory Lecture 16 May 28 2019 May 28 2019 PCP PCPrnqn set of languages L with ppt verifier V that has r qrestricted access to a string proof V tosses ID: 763785
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CS151Complexity Theory Lecture 16 May 28, 2019
May 28, 2019 PCP PCP[r(n),q(n)] : set of languages L with p.p.t. verifier V that has (r, q)-restricted access to a string “proof”V tosses O(r(n)) coins V accesses proof in O(q(n)) locations (completeness) x L proof such that Pr[V(x, proof) accepts] = 1(soundness) x L proof* Pr[V(x, proof*) accepts] ½ 2
May 28, 2019 PCP Two observations: PCP[1, poly n] = NPproof?PCP[log n, 1] NPproof?The PCP Theorem (AS, ALMSS): PCP[log n, 1] = NP. 3
May 28, 2019 The PCP Theorem Two major components: NP PCP[log n, polylog n] (“outer verifier”)we will prove this from scratch, assuming low-degree test, and self-correction of low-degree polynomialsNP PCP[n2, 1] (“inner verifier”)we will prove (low-degree test on Problem Set) 4
May 28, 2019 The inner verifier Theorem : NP PCP[n2, 1]Proof (first steps): 1. Quadratic Equations is NP-hard 2. PCP for QE: proof = all quadratic functions of a soln. x verification = check that a random linear combination of equations is satisfied by x (if prover keeps promise to supply all quadratic fns of x)5
Quadratic Equations quadratic equation over F 2: i<j ai,j Xi Xj + i bi Xi + c = 0language QUADRATIC EQUATIONS (QE)= { systems of quadratic equations over F2 that have a solution (assignment to the X variables) } May 28, 20196
Quadratic EquationsLemma : QE is NP-complete. Proof: clearly in NP; reduce from CIRCUIT SATcircuit C an instance of CIRCUIT SATQE variables = variables + gate variables May 28, 2019 gi z g i z 1 z 2 g i – z = 1 g i – z 1 z 2 = 0 g i z 1 z 2 g i – (1-z 1 )(1-z 2 ) = 1 … and g out = 1 7
Quadratic Functions Code intended proof :F the field with 2 elementsgiven x Fn, a solution to instance of QEfx: Fn F2 all linear functions of xfx(a) = i ai xigx: Fn x n F2 includes all quadratic fns of xgx(A) = i, j A[i,j]xixjKEY: can evaluate any quadratic function of x with a single evaluation of fx and gx May 28, 20198
PCP for QE If prover keeps promise to supply all quadratic fns of x, a solution of QE instance… Verifier’s action:query a random linear combination R of the equations of the QE instanceCompleteness: obviousSoundness: x fails to satisfy some equation; imagine picking coeff. for this one lastPr[x satisfies R] = 1/2 May 28, 2019 9
PCP for QE To “enforce promise”, verifier needs to perform:linearity test: verify f, g are (close to) linearself-correction: access the linear f’, g’ that are close to f, g [so f’ = Had(u) and g’ = Had(V)] consistency check: verify V = u u May 28, 2019x Fn soln fx (a) = i a i x i Had(x) g x (A) = i , j A[ i,j ] x i x j Had(x x ) 10
PCP for QE Linearity test: given access to h:Fm Fpick random a,b; check if h(a) + h(b) = h(a+b) ; repeat O(1) timesdo this for functions f and g supplied by proverTheorem [BLR]: h linear prob. success = 1; prob. success 1 – linear h’ s.t.Pra [h’(a) = h(a)] 1 – O() May 28, 2019 x F n soln f x (a) = i a i x i Had(x) g x (A) = i , j A[ i,j ] x i x j Had(xx) 11
PCP for QE Self-correction: given access to h:FmF close to linear h’; i.e.,Pra [h’ (a) = h(a)] 1 – O()to access h’(a), pick random b; computeh(b) + h(a+b)with prob. at least 1 – 2O(), h(b) = h’(b) and h(a+b) = h’(a+b); hence we compute h’(a) May 28, 2019 x F n soln f x (a) = i a i x i Had(x) g x (A) = i , j A[ i,j ] x i xj Had(xx) 12
PCP for QE Consistency check: given access to linear functions f’ = Had(u) and g’ = Had(V)pick random a, b Fn; check that f’(a)f’(b) = g’(abT)completeness: if V = u uf’(a)f’(b) = (iaiui )(ibiui) = i,j aibjV[i,j] = g’(abT) May 28, 2019 x F n soln f x (a) = i a i x i Had(x) g x (A) = i , j A[ i,j ] x i x j Had(xx) 13
PCP for QE Consistency check: given access to linear functions f’ = Had(u) and g’ = Had(V)soundness: claim that if V u u Pr[(iaiui )(ibiui) = i,j aibjV[i,j] ] 3/4Pr[(uuT)b Vb] 1/2Pr[aT(uu T)b a T Vb ] ½ ½ = ¼ May 28, 2019 i,j s.t. uu T and V differ in entry ( i,j ); pick b j last i s.t. ( uu T )b and Vb differ in entry i ; pick ai last x Fn soln fx(a) = i ai xi Had(x) gx(A) = i, j A[i,j]xix j Had(xx) 14
May 28, 2019 The outer verifier Theorem : NP PCP[log n, polylog n]Proof (first steps):define: Polynomial Constraint Satisfaction (PCS) problemprove: PCS gap problem is NP-hard 15
May 28, 2019 NP PCP[log n, polylog n] MAX-k-SATgiven: k-CNF output: max. # of simultaneously satisfiable clausesgeneralization: MAX-k-CSPgiven:variables x1, x2, …, xn taking values from set Sk-ary constraints C1, C2, …, Ctoutput: max. # of simultaneously satisfiable constraints 16
May 28, 2019 NP PCP[log n, polylog n] algebraic version: MAX-k-PCS given:variables x1, x2, …, xn taking values from field Fqn = qm for some integer m k-ary constraints C1, C2, …, Ctassignment viewed as f:(Fq)m Fq output: max. # of constraints simultaneously satisfiable by an assignment that has deg. ≤ d 17
May 28, 2019 NP PCP[log n, polylog n] MAX-k-PCS gap problem: given:variables x1, x2, …, xn taking values from field Fqn = qm for some integer m k-ary constraints C1, C2, …, Ctassignment viewed as f:(Fq)m F q YES: some degree d assignment satisfies all constraints NO: no degree d assignment satisfies more than ( 1- ) fraction of constraints 18
May 28, 2019 NP PCP[log n, polylog n] Lemma: for every constant 1 > ε > 0, the MAX-k-PCS gap problem with t = poly(n) k-ary constraints with k = polylog(n) field size q = polylog(n) n = qm variables with m = O(log n / loglog n) degree of assignments d = polylog (n) gap is NP -hard. 19
May 28, 2019 NP PCP[log n, polylog n] t = poly(n) k-ary constraints with k = polylog(n) field size q = polylog(n) n = qm variables with m = O(log n / loglog n) degree of assignments d = polylog(n)check: headed in right direction O(log n) random bits to pick a constraintquery assignment in O( polylog (n)) locations to determine if it is satisfied completeness 1; soundness 1- (if prover keeps promise to supply degree d polynomial) 20
May 28, 2019 NP PCP[log n, polylog n] Proof of Lemmareduce from 3-SAT3-CNF φ(x1, x2,…, xn) can encode as :[n] x [n] x [n] x {0,1}3{0,1}(i1, i2, i3, b1, b2, b3) = 1 iff φ contains clause (xi 1 b 1 x i 2 b 2 x i 3 b 3 ) e.g. ( x 3 x 5 x 2 ) ( 3,5,2,1,0,1) = 1 21
May 28, 2019 NP PCP[log n, polylog n] pick H Fq with {0,1} H, |H| = polylog n pick m = O(log n/loglog n) so |H|m = nidentify [n] with Hm:Hm x Hm x Hm x H3 {0,1} encodes φ assignment a: H m {0,1} Key : a satisfies φ iff i 1 ,i 2 ,i 3 ,b 1 ,b 2 ,b 3 (i1 ,i2,i3,b1 ,b2,b 3) = 0 or a(i1)=b1 or a(i2)=b2 or a(i3)=b3 22
May 28, 2019 NP PCP[log n, polylog n] :Hm x Hm x Hm x H3 {0,1} encodes φa satisfies φ iff i1,i2,i3,b1,b2,b3 ( i 1 ,i 2 ,i 3 ,b 1 ,b 2 ,b 3 ) = 0 or a(i 1 )=b 1 or a(i 2 )=b 2 or a(i 3 )=b 3 extend to a function ’:(F q)3m+3 Fq with degree at most |H| in each variablecan extend any assignment a:Hm {0,1} to a’:(Fq )m Fq with degree |H| in each variable 23
May 28, 2019 NP PCP[log n, polylog n] ’:(Fq)3m+3 Fq encodes φa’:(Fq)m Fq s.a. iff (i1,i2 ,i3,b 1 ,b 2 ,b 3 ) H 3m+3 ’ (i 1 ,i 2 ,i 3 ,b 1 ,b 2 ,b 3 ) = 0 or a ’ (i 1 )=b 1 or a’(i2 )=b2 or a’(i 3)=b3 define: pa’:(Fq)3m+3 Fq from a’ as follows pa’(i1,i2,i3,b1,b2,b3) = ’(i1,i 2,i3,b1,b2,b3)(a’(i1) - b1 )(a’(i2) - b 2 )(a’(i3) - b3) a’ s.a. iff (i1,i2,i3,b1,b2,b3) H3m+3pa’(i1,i2,i3,b1,b2,b3) = 0 24
May 28, 2019 NP PCP[log n, polylog n] ’:(Fq)3m+3 Fq encodes φa’:(Fq)m Fq s.a. iff (i1,i2,i 3,b1 ,b 2 ,b 3 ) H 3m+3 p a ’ (i 1 ,i 2 ,i 3 ,b 1 ,b 2 ,b 3 ) = 0 note: deg (p a ’ ) ≤ 2(3m+3)|H| start using Z as shorthand for (i1,i2,i3,b1,b2 ,b3) another way to write “a’ s.a.” is: exists p0:(Fq)3m+3 Fq of degree ≤ 2(3m+3)|H| p0 (Z) = pa’(Z) Z (Fq)3m+3p0(Z) = 0 Z H3m+3 25
May 28, 2019 NP PCP[log n, polylog n] Focus on “p0(Z) = 0 Z H3m+3”given: p0:(Fq)3m+3 Fqdefine: p1(x1, x2, x3, …, x 3m+3) =Σ h j H p 0 ( h j , x 2 , x 3 , …, x 3m+3 ) x 1 j Claim : p 0 (Z)=0 Z H 3m+3 p 1 (Z)=0 Z FqxH3m+3-1Proof () for each x2, x3, …, x3m+3 H3m+3-1, resulting univariate poly in x1 has all 0 coeffs. 26
May 28, 2019 NP PCP[log n, polylog n] Focus on “p0(Z) = 0 Z H3m+3”given: p0:(Fq)3m+3 Fqdefine: p1(x1, x2, x3, …, x 3m+3) =Σ h j H p 0 ( h j , x 2 , x 3 , …, x 3m+3 ) x 1 j Claim : p 0 (Z)=0 Z H 3m+3 p 1 (Z)=0 Z Fq xH3m+3-1Proof () for each x2, x3, …, x3m+3 H 3m+3-1, univariate poly in x1 is 0 has all 0 coeffs. deg(p 1) ≤ deg(p0) + |H|27
May 28, 2019 NP PCP[log n, polylog n] given: p1:(Fq)3m+3 Fqdefine: p2(x1, x2, x3, …, x3m+3) =Σhj Hp1(x1 , hj , x 3 , x 4 , …, x 3m+3 ) x 2 j Claim: p 1 (Z)=0 Z F q x H 3m+3-1 p 2 (Z)=0 Z ( F q ) 2 x H 3m+3-2Proof: same. deg(p2) ≤ deg(p1) + |H|28
May 28, 2019 NP PCP[log n, polylog n] given: pi-1:(Fq)3m+3 Fqdefine: pi(x1, x2, x3, …, x3m+3) =Σhj Hpi-1(x1 , x2, …, x i-1 , h j , x i+1 , x i+2 , …, x 3m+3 ) x i j Claim: p i-1 (Z)=0 Z ( F q ) i-1 x H 3m+3-(i-1) p i(Z)=0 Z (Fq)i x H3m+3-iProof: same. deg(pi) ≤ deg(pi-1) + |H|29
May 28, 2019 NP PCP[log n, polylog n] define degree 3m+3+2 poly. δi:Fq Fq so that δi(v) = 1 if v = iδi(v) = 0 if 0 ≤ v ≤ 3m+3+1 and v ≠ idefine Q:Fq x (Fq)3m+3 Fq by:Q(v, Z) = Σ i=0…3m+3δ i (v)p i (Z) + δ 3m+3+1 (v)a ’ (Z) note: degree of Q is at most 3(3m+3)|H| + 3m + 3 + 2 < 10m|H| 30
May 28, 2019 NP PCP[log n, polylog n] Recall: MAX-k-PCS gap problem: given:variables x1, x2, …, xn taking values from field Fqn = qm for some integer m k-ary constraints C1, C2, …, Ctassignment viewed as f:(Fq)m F q YES: some degree d assignment satisfies all constraints NO: no degree d assignment satisfies more than ( 1- ) fraction of constraints 31
May 28, 2019 NP PCP[log n, polylog n] Instance of MAX-k-PCS gap problem:set d = 10m|H| given assignment Q:Fq x (Fq)3m+3 Fqexpect it to be formed in the way we have described from an assignment a:Hm {0,1} to φnote to access a’(Z) , evaluate Q(3m+3+1, Z) p a ’ (Z) formed from a ’ and ’ (formed from φ ) to access p i (Z) , evaluate Q( i , Z) 32
May 28, 2019 NP PCP[log n, polylog n] Instance of MAX-k-PCS gap problem:set d = 10m|H| given assignment Q:Fq x (Fq)3m+3 Fqexpect it to be formed in the way we have described from an assignment a:Hm {0,1} to φconstraints: Z ( F q ) 3m+3 (C 0 , Z ): p 0 (Z) = p a ’ (Z) 0<i≤3m+2 ( C i , Z ): p i (z 1, z2, …, z i, zi+1, …, z3m+3) = Σh jH pi-1(z1, z2, …, zi-1, hj, z i+1, …, zk)zij (C3m+3,Z): p3m+3 (Z) = 0 33
May 28, 2019 NP PCP[log n, polylog n] given Q:Fq x (Fq)3m+3 Fq of degree d = 10m|H| constraints: Z (Fq)3m+3 (C0,Z): p 0(Z) = p a ’ (Z) ( C i , Z ): p i (z 1 , z 2 , …, z i , z i+1 , …, z 3m+3 ) = Σ h j H p i-1(z1, z 2, …, zi-1, h j, zi+1, …, zk)zij (C3m+3,Z): p3m+3(Z) = 0Schwartz-Zippel : if any one of these sets of constraints is violated at all then at least a (1 – 12m|H|/q) fraction in the set are violated Key: all low-degree polys 34
May 28, 2019 NP PCP[log n, polylog n] Proof of Lemma (summary):reducing 3-SAT to MAX-k-PCS gap problemφ(x1, x2,…, xn) instance of 3-SATset m = O(log n/loglog n)H Fq such that |H|m = n (|H| = polylog n, q | H| 3 ) generate |F q | 3m+3 = poly(n) constraints: C Z = i =0…3m+3+1 C i, Z each refers to assignment poly Q and φ (via p a ’ ) all polys degree d = O( m|H |) = polylog neither all are satisfied or at most d/q = o(1) << ε 35
May 28, 2019 NP PCP[log n, polylog n] O(log n) random bits to pick a constraintquery assignment in O(polylog(n)) locations to determine if constraint is satisfiedcompleteness 1soundness (1-) if prover keeps promise to supply degree d polynomialprover can cheat by not supplying proof in expected form 36
May 28, 2019 NP PCP[log n, polylog n] Low-degree testing:want: randomized procedure that is given d, oracle access to f:(Fq)m Fqruns in poly(m, d) timealways accepts if deg(f) ≤ drejects with high probability if deg(f) > d too much to ask. Why? 37
May 28, 2019 NP PCP[log n, polylog n] Definition: functions f, g are δ-close ifPrx[f(x) ≠ g(x)] δLemma: δ > 0 and a randomized procedure that is given d, oracle access to f:(Fq)m Fqruns in poly(m, d) timeuses O(m log | Fq |) random bits always accepts if deg (f) ≤ d rejects with high probability if f is not δ -close to any g with deg (g) ≤ d 38
May 28, 2019 NP PCP[log n, polylog n] idea of proof:restrict to random line Lcheck if it is low degreealways accepts if deg(f) ≤ dother direction more complex (Fq) m 39
May 28, 2019 NP PCP[log n, polylog n] can only force prover to supply function f that is close to a low-degree polynomialhow to bridge the gap?recall low-degree polynomials form an error correcting code (Reed-Muller)view “close” function as corrupted codeword40
May 28, 2019 NP PCP[log n, polylog n] Self-correction:want: randomized procedure that is given x, oracle access to f:(Fq)m (Fq) that is δ-close to a (unique) degree d polynomial gruns in poly(m, d) timeuses O(m log |Fq|) random bitswith high probability outputs g(x) 41
May 28, 2019 NP PCP[log n, polylog n] Lemma: a randomized procedure that is given x, oracle access to f:(Fq)m (Fq) that is δ-close to a (unique) degree d polynomial gruns in poly(m, d) timeuses O(m log |Fq|) random bitsoutputs g(x) with high probability 42
May 28, 2019 NP PCP[log n, polylog n] idea of proof:restrict to random line L passing through xquery points along lineapply error correction(Fq)m 43
May 28, 2019 NP PCP[log n, polylog n] Putting it all together:given L NP and an instance x, verifier computes reduction to MAX-k-PCS gap problemprover supplies proof in form f:(Fq)m (Fq) (plus some other info used for low-degree testing)verifier runs low-degree testrejects if f not close to some low degree function gverifier picks random constraint Ci; checks if sat. by g uses self-correction to get values of g from f accept if C i satisfied; otherwise reject 44