Expander Graphs and Their Applications (X) Yijia Chen Shanghai Jiaotong University

Transcription

1 Expander Graphs and Their Applications (X) Yijia Chen Shanghai Jiaotong University

2 Review of the Previous Lecture

3 Probabilistic Verifier Definition A verifier V is a probabilistic polynomial time algorithm with access to an input x Σ and a string τ Σ of (internal) random binary bits. Furthermore, V has access to a proof π Σ. V will either accept or reject the input x, depending on (x, τ, π). We require that V is non-adaptive, i.e., it first reads the input x and the random bits τ, and then decides which positions in the proof τ it wants to query. That is, the positions V queries do not depend on the answers that V got from previous queries. The result of V s computation on x, τ and π is denoted by V (x, τ, π). As usual 1 means accepting, 0 for rejecting.

4 Probabilistic Verifier (cont d) 1. V is polynomial time in x. 2. The query positions depend on and only on x and τ (non-adaptive), so we might view the query positions as a function p V : Σ Σ N, such that p V (x, τ) is computable in time polynomial in x. We may assume p V is given explicitly with the verifier V. 3. The run of V (x, τ, π) that does not involve π might also depend on the random string τ. 4. We can divide the computation of V (x, τ, π) into 3 parts: (i) Compute from x and τ a sequence of positions p = p 1 p 2... p l = p V (x, τ). (ii) Read the letters on the positions p of π to form a string π p := π(p 1 )π(p 2 )... π(p l ), where π(p) denotes the letter in the p-th position of π. (iii) Compute an answer from x, τ, and π p.

5 (r(n), q(n))-restricted Verifier Definition Let r, q : N N be two monotone functions. An (r(n), q(n))-restricted verifier is a verifier that for inputs of length n uses at most O(r(n)) random bits and queries at most O(q(n)) bits from the proof. 1. If V is (r(n), q(n))-restricted, then for any input x Σ, Then τ = O(r(n)) and p V (x, τ) = O(q(n)). 2. There is no restriction on the length of the proof π.

6 PCP Classes Definition Let r, q : N N be two monotone functions. The class PCP(r(n), q(n)) consists of all languages Q where there exists an (r(n), q(n))-restricted verifier V such that for all x x Q π Pr[V (x, τ, π) = 1] = 1, τ x Q π Pr τ [V (x, τ, π) = 1] < 1 2. Here in Pr τ [...] the probability is taken over all random strings τ of length O(r(n)). Remark. 1/2 in the above definition can be replaced by any 0 < ε < 1. Definition Let R, Q N N be two classes of monotone functions. PCP(R, Q) := PCP(r(n), q(n)). r R,q Q

7 Let poly denote the class of all polynomials over N, i.e. N[x]. Theorem PCP(poly, 0) = corp, PCP(0, poly) = NP.

8 Max-dSat

9 Max-dSat Let d N. Max-dSat Input: An α dcnf. Solution: An assignment V for α. Cost: the number of clauses in α that V satisfies. Goal: max.

10 Max-dSat Let d N. Max-dSat Input: An α dcnf. Solution: An assignment V for α. Cost: the number of clauses in α that V satisfies. Goal: max. Definition For every dcnf-formula α and an assignment V for α, sat(α, V) := the number of clauses in α that V satisfies

11 Max-dSat Let d N. Max-dSat Input: An α dcnf. Solution: An assignment V for α. Cost: the number of clauses in α that V satisfies. Goal: max. Definition For every dcnf-formula α and an assignment V for α, sat(α, V) := the number of clauses in α that V satisfies and optsat(α) := max sat(α, V). V

12 Hardness of Approximating Max-dSat Definition Let A be an algorithm that on every dcnf-formula α always output an assignment for α which we denote by V α. The performance ratio of A is optsat(α) max α dcnf sat(α, V. α)

13 Hardness of Approximating Max-dSat Definition Let A be an algorithm that on every dcnf-formula α always output an assignment for α which we denote by V α. The performance ratio of A is optsat(α) max α dcnf sat(α, V. α) Theorem For some constants d N and r > 1, there is no polynomial time approximation algorithm for Max-dSat with performance ratio r, unless P = NP.

14 Proof.

15 Proof. Let Q Σ be an NP-complete language.

16 Proof. Let Q Σ be an NP-complete language. By the PCP Theorem, there is a (log n, 1)-restricted verifier V accepting Q:

17 Proof. Let Q Σ be an NP-complete language. By the PCP Theorem, there is a (log n, 1)-restricted verifier V accepting Q: for x Σ, V uses c log x random bits and makes d oracle queries for two constants c, d N.

18 Proof. Let Q Σ be an NP-complete language. By the PCP Theorem, there is a (log n, 1)-restricted verifier V accepting Q: for x Σ, V uses c log x random bits and makes d oracle queries for two constants c, d N. (R1) x Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] = 1, (R2) x / Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] < 1 2.

19

20 We define the algorithm A(x, τ, a 1, a 2,..., a d ):

21 We define the algorithm A(x, τ, a 1, a 2,..., a d ): 1. p = p 1... p d p V (x, τ).

22 We define the algorithm A(x, τ, a 1, a 2,..., a d ): 1. p = p 1... p d p V (x, τ). 2. Simulate V (x, π) for an imaginary π by replacing each π pi by a i for 1 i d.

23 We define the algorithm A(x, τ, a 1, a 2,..., a d ): 1. p = p 1... p d p V (x, τ). 2. Simulate V (x, π) for an imaginary π by replacing each π pi by a i for 1 i d. where τ {0, 1} c log x,

24 We define the algorithm A(x, τ, a 1, a 2,..., a d ): 1. p = p 1... p d p V (x, τ). 2. Simulate V (x, π) for an imaginary π by replacing each π pi by a i for 1 i d. where τ {0, 1} c log x, and a 1, a 2,..., a d {0, 1}.

25 We define the algorithm A(x, τ, a 1, a 2,..., a d ): 1. p = p 1... p d p V (x, τ). 2. Simulate V (x, π) for an imaginary π by replacing each π pi by a i for 1 i d. where τ {0, 1} c log x, and a 1, a 2,..., a d {0, 1}. Clearly A is polynomial time in x.

26

27 We reduce Q to Max-dSat.

28 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying

29 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied.

30 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. First let P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) },

31 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. First let P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) }, i.e. P x is the set of positions that V makes queries.

32 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying First let x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) }, i.e. P x is the set of positions that V makes queries. Obviously P x is computable in time polynomial in x.

33 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying First let x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) }, i.e. P x is the set of positions that V makes queries. Obviously P x is computable in time polynomial in x. In addition P x d x c.

34 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying First let x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) }, i.e. P x is the set of positions that V makes queries. Obviously P x is computable in time polynomial in x. In addition P x d x c. For each p P x, we introduce a propositional variable X p.

35 We reduce Q to Max-dSat. Given an x Σ, we will construct a dcnf formula α x satisfying First let x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied. P x := { p for some τ {0, 1} c log x p appears in the sequence p V (x, τ) }, i.e. P x is the set of positions that V makes queries. Obviously P x is computable in time polynomial in x. In addition P x d x c. For each p P x, we introduce a propositional variable X p. Then, every proof π can be viewed as an assignment V π with V π(x p) := π(p).

36

37 For every τ {0, 1} c log x, we define a dcnf formula α x,τ :

38 For every τ {0, 1} c log x, we define a dcnf formula α x,τ : For every a 1,..., a d {0, 1}, and p = p 1... p d = p V (x, τ),

39 For every τ {0, 1} c log x, we define a dcnf formula α x,τ : For every a 1,..., a d {0, 1}, and p = p 1... p d = p V (x, τ), if A(x, τ, a 1,..., a d ) rejects, then α x,τ contains the clause T 1 T 2 T d where T i := { X pi if a i = 0, X pi if a i = 1.

40 For every τ {0, 1} c log x, we define a dcnf formula α x,τ : For every a 1,..., a d {0, 1}, and p = p 1... p d = p V (x, τ), if A(x, τ, a 1,..., a d ) rejects, then α x,τ contains the clause T 1 T 2 T d where T i := { X pi if a i = 0, X pi if a i = 1. α x,τ has at most 2 d clauses.

41 For every τ {0, 1} c log x, we define a dcnf formula α x,τ : For every a 1,..., a d {0, 1}, and p = p 1... p d = p V (x, τ), if A(x, τ, a 1,..., a d ) rejects, then α x,τ contains the clause T 1 T 2 T d where T i := { X pi if a i = 0, X pi if a i = 1. α x,τ has at most 2 d clauses. More importantly: Let π be a proof with corresponding assignment V π (F1) if V (x, τ, π) accepts, then V π satisfies α x,τ, (F2) if V (x, τ, π) rejects, then V π makes at least one of the clauses in α x,τ false.

42

43 Let α x := τ {0,1} c log x α x,τ

44 Let α x := τ {0,1} c log x α x,τ α x has at most x c 2 d clauses.

45 Let α x := τ {0,1} c log x α x,τ α x has at most x c 2 d clauses. We need to show:

46 Let α x := τ {0,1} c log x α x,τ α x has at most x c 2 d clauses. We need to show: x Q α x is satisfiable, x / Q for any assignment V, at least 2 d 1 -fraction of clauses are not satisfied.

47 Let x Q.

48 Let x Q. By (R1) x Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] = 1, we have a proof π such that V (x, τ, π) always accepts for any τ {0, 1} c log x.

49 Let x Q. By (R1) x Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] = 1, we have a proof π such that V (x, τ, π) always accepts for any τ {0, 1} c log x. Thus V π satisfies all α x,τ by (F1) if V (x, τ, π) accepts, then V π satisfies α x,τ,

50 Let x Q. By (R1) x Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] = 1, we have a proof π such that V (x, τ, π) always accepts for any τ {0, 1} c log x. Thus V π satisfies all α x,τ by and hence α x, i.e., (F1) if V (x, τ, π) accepts, then V π satisfies α x,τ, x Q α x is satisfiable.

51

52 Let x / Q.

53 Let x / Q. For any assignment V of α x, it is easy to see we can view it as some V π for an appropriate proof π.

54 Let x / Q. For any assignment V of α x, it is easy to see we can view it as some V π for an appropriate proof π. The by (R2) x / Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] < 1 2. we have for more than half of all possible τ {0, 1} c log x, V (x, τ, π) rejects.

55 Let x / Q. For any assignment V of α x, it is easy to see we can view it as some V π for an appropriate proof π. The by (R2) x / Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] < 1 2. we have for more than half of all possible τ {0, 1} c log x, V (x, τ, π) rejects. For any τ with V (x, τ, π) rejecting, then (F2) if V (x, τ, π) rejects, then V π makes at least one of the clauses in α x,τ false. implies that at least one clause in α x,τ is false under V π = V.

56 Let x / Q. For any assignment V of α x, it is easy to see we can view it as some V π for an appropriate proof π. The by (R2) x / Q π Pr τ {0,1} c log x [V (x, τ, π) = 1] < 1 2. we have for more than half of all possible τ {0, 1} c log x, V (x, τ, π) rejects. For any τ with V (x, τ, π) rejecting, then (F2) if V (x, τ, π) rejects, then V π makes at least one of the clauses in α x,τ false. implies that at least one clause in α x,τ is false under V π = V. Recall α x,τ contains at most 2 d many clauses. Thus in total we have more than d = 2 d 1 fraction of clauses in α x are not satisfied under V.

57

58 Let r := d 1.

59 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r.

60 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q:

61 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q: Let x Σ.

62 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q: Let x Σ. We compute the dcnf-formula α x as described above.

63 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q: Let x Σ. We compute the dcnf-formula α x as described above. Assume α x has m clauses.

64 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q: Let x Σ. We compute the dcnf-formula α x as described above. Assume α x has m clauses. If x Q, then α x is satisfiable.

65 Let 1 r := 1 2. d 1 We claim Max-dSat cannot have a polynomial time approximation algorithm with performance ratio r. Assume B is such an algorithm. We would use it to decide Q: Let x Σ. We compute the dcnf-formula α x as described above. Assume α x has m clauses. If x Q, then α x is satisfiable. Thus B would find an assignment that satisfies at least m = m (1 2 d 1 ) r clauses of α x.

66

67 If x / Q, then less than m (1 2 d 1 ) clauses of α x can be satisfied by any assignment.

68 If x / Q, then less than m (1 2 d 1 ) clauses of α x can be satisfied by any assignment. In particular the assignment that B outputs on α x cannot satisfy m (1 2 d 1 ) clauses.

69 If x / Q, then less than m (1 2 d 1 ) clauses of α x can be satisfied by any assignment. In particular the assignment that B outputs on α x cannot satisfy m (1 2 d 1 ) clauses. In summary: x Q B(α x) satisfies m (1 2 d 1 ) clauses; x / Q B(α x) satisfies < m (1 2 d 1 ) clauses.