Solution of Exercise Sheet 12

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1 Complexity Theory (fall 2016) Solution of Exercise Sheet 12 Dominique Unruh Problem 1: Deterministic PCPs PCP stands for probabilistically checkable proofs. We will now show that the probabilistic part is very important for the definition. Namely, we show that PCP(0, log) = P and PCP(0, poly) = NP. (a) Show that P PCP(0, log). (In fact, it is even in PCP(0, 0).) Fix some L P. We need to show that L PCP(0, log). That is, we need to show that there is a sound and complete (0, O(log))-PCP-verifier V. This verifier is constructed as follows: Upon input x, the verifier uses r = 0 bits of randomness, and queries q = 0 of the proof π. Then the verifier outputs 1 iff x L. (This can be decided in deterministic polynomial-time since L P.) The verifier is complete, since for any x L, V (x) returns 1 (independent of π). And for any, V (x) returns 1 with probability (independent of π). Thus V is sound. (b) Show that NP PCP(0, poly). Fix some L NP. We need to show that L PCP(0, poly). That is, we need to show that there is a sound and complete (0, poly)-pcp-verifier V. Since L is in NP, there is a polynomial-time Turing machine M and a polynomial p, such that for any x L, there exists a w {0, 1} p( x ) such that M(x, w) = 1. And for any x / L and all w {0, 1} p( x ) we have M(x, w) = 0. From M, we construct a (0, p)-pcp-verifier V. Upon input x, V uses 0 bits of randomness and queries the first p( x ) bits of the proof π. Let w denote those bits. Then V outputs 1 iff M(x, w) = 1. V is a (0, p)-pcp-verifier. We show that V is complete: Fix x L. Then there is a w {0, 1} p( x ) such that M(x, w) = 1. Let π := w. Then V (x) returns 1 given access to the proof π. Thus we have that for x L, there exists a proof π such that V (x) returns 1 with probability 1. This shows that V is complete.

2 We show that V is sound: Fix x / L. Then for any w {0, 1} p( x ), M(x, w) = 1. Thus for any π, V returns 1 with probability (since M(x, w) invoked by V never returns 1). This shows that V is sound. Thus there is a sound and complete (0, p)-pcp-verifier V for L. Since p is a polynomial, this implies that L PCP(0, poly). (c) Consider a sound and complete (0, q)-pcp-verifier V for some language L (i.e., a deterministic verifier). Construct a polynomial-time deterministic algorithm V (x, w) such that: If x L then w {0, 1} q( x ). V (x, w) = 1 If x / L then w {0, 1} q( x ). V (x, w) = 0 Hint: V already kind-of satisfies this, except that V gets a longer input than q bits. But V only looks at q bits of that input. V (x) outputs q = q( x ) indices i 1,..., i q (deterministically) and then expects π := π i1... π iq and from this computes its output bit b. (Without loss of generality, we assume that the indices contain no duplicates.) The verifier V (x, w) simulates V (x), but when V (x) expects π := π i1... π iq, V supplies π := w instead. Since V is polynomial-time, so is V. If x L, then by completeness of V, there is a π such that V (x) = 1 with probability 1. Let i 1... i q be the indices of the bits that V queries on input x. (Those are well-defined, because V uses 0 random bits, so the indices are determined by x.) Let w := π i1... π iq. Then V (x, w) returns the same as V (x) would return given π. Thus V (x, w) = 1. If x / L, soundness of V implies that for any π, V (x) returns 0. (Soundness says that this happens with probability 1 2, but since V is deterministic, this means it happens always.) We show that for any w {0, 1} q( x ), V (x, w) returns 0. Fix some such w. Let i 1... i q be the indices of the bits that V queries on input x. (Those are well-defined, because V uses 0 random bits, so the indices are determined by x.) Let π be a bitstring such that π ij = w j for j = 1,..., q and π i = 0 everywhere else. Then V (x, w) returns the same as V (x) with proof π. And V (x) with proof π returns 0. Thus V (x, w) = 0 for all w {0, 1} q( x ). (d) Show that PCP(0, poly) NP. Hint: This follows quite easily from (c). 2

3 Let L PCP(0, poly). By definition, this implies that there is a (0, q)- PCP-verifier for some polynomial q. By (c), this implies that there is a polynomial-time algorithm V such that: If x L then w {0, 1} q( x ). V (x, w) = 1 If x / L then w {0, 1} q( x ). V (x, w) = 0 By definition of NP this means that L NP. (e) Show that PCP(0, log) P. Hint: Given the verifier V from (c), how can you can whether V (x, w) = 1 for all w? How long does that take? Let L PCP(0, log). By definition, this implies that there is a (0, c)- PCP-verifier for some polynomial q = c log n. (Where n denotes the input length.) By (c), there is a polynomial-time algorithm V such that: x L iff w {0, 1} q( x ). V (x, w) = 1 Let M be the algorithm that on input x runs V (x, w) for all w {0, 1} q( x ) and returns 1 iff at least one invocation of V returned 1. Then M(x) = 1 iff x L. There are 2 c log n = n c bitstrings w {0, 1} q( x ), i.e., polynomially-many of them. And V also runs in polynomial time. Hence M runs in polynomial time. Thus there is a deterministic polynomial-time algorithm M deciding L, hence L P. Problem 2: Approximating MAX-SAT is NP-hard In the following, let q 2 be an integer. (a) Let ϕ be a qcsp. That is, ϕ = ϕ 1 ϕ t, where each ϕ t is a formula containing at most q different variables. No question no solution. (This was only accidentally marked as a separate item.) (b) Show that ϕ can be transformed into a qcnf formula with at most 2 q t clauses (in polynomial time in the size of ϕ) such that ϕ is satisfiable iff ϕ is satisfiable. Note: If you are troubled whether a runtime factor of 2 q is OK for a polynomial-time algorithm, recall that q is a constant in this context, so 2 q is, as well. 3

4 For each i, we compute the set X i of assignments a 1,..., a q to the variables occurring in ϕ i such that ϕ i (a 1,..., a q ) = 1. Let ˆϕ i := (x a1 1 xaq q ) (a 1,...,a q) X i where x j denotes the j-th of the q variables in X i, and x a j j stands for x j if a j = 1 and for x j if a j = 0. We have that (x a 1 1 xaq q ) = 1 iff (x 1,..., x q ) = (a 1,..., a q ). Thus ˆϕ i = 1 iff (x 1,..., x q ) X i iff ϕ(x 1,..., x q ) = 1. Thus ˆϕ i ϕ i (in the sense that they have the same result given the same variable assignments). Let ϕ i := (x 1 a1 1 x 1 aq q ). (a 1,...,a q) X i Then ϕ i ˆϕ i (by applying De Morgan s laws). And thus ϕ i ϕ i ϕ i. Thus ϕ i is in qcnf and has at X i 2 q clauses. Let ϕ := ϕ 1... ϕ t. Then ϕ is in qcnf and has at most t2 q clauses. And ϕ ϕ 1 ϕ t = ϕ. Thus ϕ is satisfiable iff ϕ is satisfiable. The steps of the construction above also give rise to a polynomial-time algorithm for converting φ into ϕ. (c) Show: If val(ϕ) 1 ε (for any assignment at least an ε-fraction of the clauses is violated), then val(ϕ ) 1 ε/2 q (for any assignment at least an ε/2 q -fraction of the clauses is violated). To show this, it is sufficient to show: If a = (a 1,..., a n ) is an assignment that violates at least an ε-faction of the t clauses of ϕ, then a violates at least an ε/2 q fraction of the clauses of ϕ. (The claim from (c) then follows directly.) Thus, fix such an assignment a. Then there are at least εt clauses in ϕ that are violated. Let I denote the indices of these clauses. (I.e., I εt and for all i I, ϕ i is violated. Since we showed in (b) that ϕ i ϕ i, we have that ϕ i is also violated by a for all i I. ϕ i is the conjunction of clauses, so if ϕ i is violated, then one of its clauses is violated. Thus in each ϕ i (i I), at least one clause is violated. Thus in ϕ, at least I clauses are violated. Thus the fraction of clauses that is violated in ϕ is at least (where N denotes the total number of clauses in ϕ): I N εt N t2 q N εt t2 q = ε/2q. 4

5 (d) Assume there is a polynomial-time algorithm A that approximates MAX-SAT up to a factor of ρ := 1 2 q 2. (That is, given a formula ϕ in CNF it returns a value δ such that ρ val(ϕ ) δ val(ϕ ).) Show that there is a polynomial-time algorithm B such that for any qcsp ϕ, we have: If val(ϕ) = 1, then B(ϕ) = 1. If val(ϕ) 1 2, then B(ϕ) = 0. Note: If you manage to prove this with another approximation factor ρ, that is also fine, as long as 0 < ρ < 1. Let g denote the function that maps ϕ to ϕ as in (b). Let B(ϕ) returns 1 if A(f(ϕ)) ρ. Clearly B is polynomial-time since A and g are. If val(ϕ) = 1, then by definition ϕ is satisfiable. By (b), this implies that ϕ := f(ϕ) is satisfiable. Hence val(ϕ ) = 1. By assumption on A, this means that A(ϕ ) ρval(ϕ ) = 1. Thus B returns 1. If val(ϕ) 1 2 = 1 1 2, then by (c), val(ϕ ) /2q = 1 2 q 1 < 1 2 q 2 = ρ. Thus B returns 0. (e) Assume that P NP. Show that an constant 0 < ρ < 1 such that no polynomial-time algorithm A approximates MAX-SAT up to a factor of ρ := 1 2 q 2. Hint: Use a theorem from the lecture here, together with (d). In the lecture, we showed that there exists a q N and a polynomial-time computable function f that outputs qcsp instances such that: If x SAT, then val(f(x)) = 1, and if x / SAT, then val(f(x)) 1 2. For this value q, we show that there is no polynomial-time algorithm A approximates MAX-SAT up to a factor of ρ: Assume such an algorithm A exists. By (d), there is a polynomial-time algorithm B taking qcsps as input, that has the properties described in (d). Let C(x) := B(f(x)). Then C is polynomial-time since B and f are. If x SAT, then val(f(x)) = 1, then C(x) = B(f(x)) = 1. If x / SAT, then val(f(x)) 1 2, then C(x) = B(f(x)) = 0. Thus C is a polynomial-time algorithm that decides SAT. Since SAT is NP-hard, this implies that P = NP, in contradiction to the assumption from the problem statement. Thus the assumption that A exists was false. 5