Shamir s Theorem. Johannes Mittmann. Technische Universität München (TUM)
|
|
|
- Kristopher Moore
- 5 years ago
- Views:
Transcription
1 IP = PSPACE Shamir s Theorem Johannes Mittmann Technische Universität München (TUM) 4 th Joint Advanced Student School (JASS) St. Petersburg, April 2 12, 2006 Course 1: Proofs and Computers Johannes Mittmann IP = PSPACE JASS
2 Outline 1 Introduction 2 Polynomial Space Quantified Satifiability PSPACE-Completeness 3 Shamir s Theorem Arithmetization Reduction to a Finite Field Polynomials and Simple Expressions The Interactive Protocol Johannes Mittmann IP = PSPACE JASS
3 Introduction History Papadimitriou (1983): IP PSPACE. Toda (1989): PH P #P. Nisan (Nov. 27, 1989): PH MIP. Lund, Fortnow, Karloff, Nisan (Dec. 13, 1989): PH IP. Shamir (Dec. 26, 1989): PSPACE IP. Figure: Adi Shamir Johannes Mittmann IP = PSPACE JASS
4 Polynomial Space Polynomial Space PSPACE = k>0 SPACE(n k ). L NL P NP PSPACE. Johannes Mittmann IP = PSPACE JASS
5 Polynomial Space Quantified Satifiability Quantified Boolean Formulas Definition Let X = {x 1, x 2,... } be an alphabet of Boolean variables. They can take the two truth values true and false. A quantified Boolean expression (QBF) φ is defined inductively by 1 a Boolean variable x i, or an expression of the form 2 φ 1 3 φ 1 φ 2 4 φ 1 φ 2 (negation), (disjunction), (conjunction), 5 x i φ 1 (existential quantification), 6 x i φ 1 (universal quantification), where φ 1 and φ 2 are quantified Boolean expressions. Johannes Mittmann IP = PSPACE JASS
6 Polynomial Space Quantified Satifiability Proposition Let φ and ψ be QBFs. Then 1 (φ ψ) φ ψ. (De Morgan s Laws) 2 (φ ψ) φ ψ. 3 ( x i φ) x i φ. 4 ( x i φ) x i φ. 5 ( φ) φ. Johannes Mittmann IP = PSPACE JASS
7 Polynomial Space Quantified Satifiability Proposition Let φ and ψ be QBFs. Then 1 (φ ψ) φ ψ. (De Morgan s Laws) 2 (φ ψ) φ ψ. 3 ( x i φ) x i φ. 4 ( x i φ) x i φ. 5 ( φ) φ. 6 x i (φ ψ) ( x i φ) ( x i ψ). 7 x i (φ ψ) ( x i φ) ( x i ψ). Johannes Mittmann IP = PSPACE JASS
8 Polynomial Space Quantified Satifiability Proposition Let φ and ψ be QBFs. Then 1 (φ ψ) φ ψ. (De Morgan s Laws) 2 (φ ψ) φ ψ. 3 ( x i φ) x i φ. 4 ( x i φ) x i φ. 5 ( φ) φ. 6 x i (φ ψ) ( x i φ) ( x i ψ). 7 x i (φ ψ) ( x i φ) ( x i ψ). 8 If x i does not appear free in ψ, x i (φ ψ) ( x i φ) ψ. 9 If x i does not appear free in ψ, x i (φ ψ) ( x i φ) ψ. Johannes Mittmann IP = PSPACE JASS
9 Polynomial Space Quantified Satifiability Proposition Let φ and ψ be QBFs. Then 1 (φ ψ) φ ψ. (De Morgan s Laws) 2 (φ ψ) φ ψ. 3 ( x i φ) x i φ. 4 ( x i φ) x i φ. 5 ( φ) φ. 6 x i (φ ψ) ( x i φ) ( x i ψ). 7 x i (φ ψ) ( x i φ) ( x i ψ). 8 If x i does not appear free in ψ, x i (φ ψ) ( x i φ) ψ. 9 If x i does not appear free in ψ, x i (φ ψ) ( x i φ) ψ. 10 If xj does not appear in φ, x i φ x j φ[x i x j ]. Johannes Mittmann IP = PSPACE JASS
10 Polynomial Space Quantified Satifiability Prenex Normal Form φ = x 1 x 2 x 3... Q n x n ψ, Q n {, }. Johannes Mittmann IP = PSPACE JASS
11 Polynomial Space Quantified Satifiability Prenex Normal Form φ = x 1 x 2 x 3... Q n x n ψ, Q n {, }. Proposition Any QBF φ can be transformed to an equivalent one in prenex normal form. Johannes Mittmann IP = PSPACE JASS
12 Polynomial Space Quantified Satifiability Quantified Satisfiability QSAT = { φ : φ is a true QBF in prenex CNF }. Johannes Mittmann IP = PSPACE JASS
13 Polynomial Space PSPACE-Completeness PSPACE-Completeness Theorem (Stockmeyer/Meyer 1973) QSAT is PSPACE-complete. Johannes Mittmann IP = PSPACE JASS
14 Polynomial Space PSPACE-Completeness PSPACE-Completeness Theorem (Stockmeyer/Meyer 1973) QSAT is PSPACE-complete. Proof. It suffices to show: 1 QSAT PSPACE. 2 For all L PSPACE : L log QSAT. Johannes Mittmann IP = PSPACE JASS
15 Polynomial Space PSPACE-Completeness Proof sketch of (1). φ = x 1 x 2 x 3... Q n x n ψ(x 1,..., x n ) x 1 = false x 1 = true x 2 x 3... ψ(0, x 2,... ) x 2 x 3... ψ(1, x 2,... ). x 2 = false x 2 = true.. x 3... ψ(0, 0, x 3,... ) x 3... ψ(0, 1, x 3,... ). Q n Q n... ψ(0,..., 0) ψ(0,..., 1) ψ(0,..., 1, 0) ψ(0,..., 1, 1) Johannes Mittmann IP = PSPACE JASS
16 Polynomial Space PSPACE-Completeness Proof sketch of (1). φ = x 1 x 2 x 3... Q n x n ψ(x 1,..., x n ) x 1 = false x 1 = true x 2 x 3... ψ(0, x 2,... ) x 2 x 3... ψ(1, x 2,... ). x 2 = false x 2 = true.. x 3... ψ(0, 0, x 3,... ) x 3... ψ(0, 1, x 3,... ).... ψ(0,..., 0) ψ(0,..., 1) ψ(0,..., 1, 0) ψ(0,..., 1, 1) Johannes Mittmann IP = PSPACE JASS
17 Polynomial Space PSPACE-Completeness Algorithm: Truth(φ) 1: if φ is quantifier-free then 2: return truth value of φ 3: end if 4: denote φ = Q 1 x 1... Q n x n ψ(x 1,..., x n ) 5: b 0 Truth(Q 2 x 2... Q n x n ψ(false, x 2,..., x n )) 6: b 1 Truth(Q 2 x 2... Q n x n ψ(true, x 2,..., x n )) 7: if Q 1 = then 8: return b 0 b 1 9: else 10: return b 0 b 1 11: end if Johannes Mittmann IP = PSPACE JASS
18 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. Johannes Mittmann IP = PSPACE JASS
19 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. Johannes Mittmann IP = PSPACE JASS
20 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. For input x consider the configuration graph of M. Johannes Mittmann IP = PSPACE JASS
21 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. For input x consider the configuration graph of M. 2 m configurations, where m = O(n k ). Johannes Mittmann IP = PSPACE JASS
22 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. For input x consider the configuration graph of M. 2 m configurations, where m = O(n k ). Reachability method: ψ i (X, Y) is true configuration Y can be reached from configuration X in 2 i steps, for i = 0,..., m. Johannes Mittmann IP = PSPACE JASS
23 Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. For input x consider the configuration graph of M. 2 m configurations, where m = O(n k ). Reachability method: ψ i (X, Y) is true configuration Y can be reached from configuration X in 2 i steps, for i = 0,..., m. Required QBF: ψ m (A, B). Johannes Mittmann IP = PSPACE JASS
24 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Johannes Mittmann IP = PSPACE JASS
25 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Bad idea: ψ i+1 = Z [ ψ i (A, Z) ψ i (Z, B) ]. Johannes Mittmann IP = PSPACE JASS
26 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Bad idea: ψ i+1 = Z [ ψ i (A, Z) ψ i (Z, B) ]. Savitch s trick: ψ i+1 = Z X Y [( (X = A Y = Z) (X = Z Y = B) ) ψ i (X, Y) ]. Johannes Mittmann IP = PSPACE JASS
27 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Bad idea: ψ i+1 = Z [ ψ i (A, Z) ψ i (Z, B) ]. Savitch s trick: ψ i+1 = Z X Y [( (X = A Y = Z) (X = Z Y = B) ) ψ i (X, Y) ]. Convert to prenex DNF. Johannes Mittmann IP = PSPACE JASS
28 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Bad idea: ψ i+1 = Z [ ψ i (A, Z) ψ i (Z, B) ]. Savitch s trick: ψ i+1 = Z X Y [( (X = A Y = Z) (X = Z Y = B) ) ψ i (X, Y) ]. Convert to prenex DNF. L log QSAT. Johannes Mittmann IP = PSPACE JASS
29 Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Bad idea: ψ i+1 = Z [ ψ i (A, Z) ψ i (Z, B) ]. Savitch s trick: ψ i+1 = Z X Y [( (X = A Y = Z) (X = Z Y = B) ) ψ i (X, Y) ]. Convert to prenex DNF. L log QSAT. PSPACE = copspace. Johannes Mittmann IP = PSPACE JASS
30 Shamir s Theorem Theorem (Shamir 1992) IP = PSPACE. Johannes Mittmann IP = PSPACE JASS
31 Shamir s Theorem Theorem (Shamir 1992) IP = PSPACE. Proof. : Optimal prover strategy: Traverse the tree of all possible interactions between Alice and Bob. Compute the probabilities of acceptance. Johannes Mittmann IP = PSPACE JASS
32 Proof Outline Proof. : It suffices to show QSAT IP, since IP is closed under reductions, and QSAT PSPACE-complete. Johannes Mittmann IP = PSPACE JASS
33 Proof Outline Proof. : It suffices to show QSAT IP, since IP is closed under reductions, and QSAT PSPACE-complete. Outline: Arithmetization: φ A φ. Reduction to F p. Polynomials and simple expressions. Interactive protocol that decides QSAT. Johannes Mittmann IP = PSPACE JASS
34 Arithmetization Arithmetization Quantified Boolean expression φ Arithmetization A φ Conversion rules: QBF φ Arithmetization A φ true 1 false 0 x i X z i Z x i 1 z i ψ 1 ψ 2 A ψ1 + A ψ2 ψ 1 ψ 2 A ψ1 A ψ2 x i ψ 1 z i =0 A ψ x i ψ 1 z i =0 A ψ Johannes Mittmann IP = PSPACE JASS
35 Arithmetization Example φ = x 1 [ x1 x 2 x 3 (x 1 x 2 ) x 3 ]. A φ = 1 z 1 =0 A φ is called Σ-Π expression. [ (1 z 1 ) z 2 =0 z 3 =0 ] (z 1 z 2 + z 3 ). Johannes Mittmann IP = PSPACE JASS
36 Arithmetization Lemma Let φ be a closed QBF with negation only over variables. Then φ is true A φ > 0. Johannes Mittmann IP = PSPACE JASS
37 Arithmetization Lemma Let φ be a closed QBF with negation only over variables. Then φ is true A φ > 0. Proof. Induction on the structure of a (not necessarily closed) QBF φ. φ = x i : φ is true x i = true A φ = z i = 1 > 0. Johannes Mittmann IP = PSPACE JASS
38 Arithmetization Proof (continued). φ = x i : φ is true x i = false A φ = 1 z i = 1 0 > 0. Johannes Mittmann IP = PSPACE JASS
39 Arithmetization Proof (continued). φ = x i : φ is true x i = false A φ = 1 z i = 1 0 > 0. φ = ψ 1 ψ 2 : φ is true ψ 1 is true or ψ 2 is true A ψ1 > 0 or A ψ2 > 0 A φ = A ψ1 + A ψ2 > Johannes Mittmann IP = PSPACE JASS
40 Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ) [ A φ = zk + (1 z k ) ] = 2 2k 1. z 1 =0 z 2 =0 z k 1 =0 z k =0 Johannes Mittmann IP = PSPACE JASS
41 Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ) [ A φ = zk + (1 z k ) ] = 2 2k 1. z 1 =0 z 2 =0 z k 1 =0 z k =0 exponentially many bits needed! Johannes Mittmann IP = PSPACE JASS
42 Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ) [ A φ = zk + (1 z k ) ] = 2 2k 1. z 1 =0 z 2 =0 z k 1 =0 z k =0 exponentially many bits needed! Idea: reduction to a finite field. Johannes Mittmann IP = PSPACE JASS
43 Reduction to a Finite Field Lemma Let A φ be a Σ-Π expression of length n. Then A φ 2 2n. Johannes Mittmann IP = PSPACE JASS
44 Reduction to a Finite Field Lemma Let A φ be a Σ-Π expression of length n. Then A φ 2 2n. Proof. Induction on the structure of φ. φ = ( )x i : A φ Johannes Mittmann IP = PSPACE JASS
45 Reduction to a Finite Field Lemma Let A φ be a Σ-Π expression of length n. Then A φ 2 2n. Proof. Induction on the structure of φ. φ = ( )x i : A φ φ = ψ 1 ψ 2, {, }: A φ 2 2l 2 2m = 2 2l +2 m 2 2n (l + m n). Johannes Mittmann IP = PSPACE JASS
46 Reduction to a Finite Field Lemma Let A φ be a Σ-Π expression of length n. Then A φ 2 2n. Proof. Induction on the structure of φ. φ = ( )x i : A φ φ = ψ 1 ψ 2, {, }: A φ 2 2l 2 2m = 2 2l +2 m 2 2n (l + m n). φ = Qx i ψ, Q {, }: A φ 2 2m 2 2m = 2 2 2m = 2 2m+1 2 2n (m < n). Johannes Mittmann IP = PSPACE JASS
47 Reduction to a Finite Field Number of Primes Lemma For n 3, n π(n) n. Johannes Mittmann IP = PSPACE JASS
48 Reduction to a Finite Field Number of Primes Lemma For n 3, n π(n) n. Proof. π(n) n p n p 1 p n n i=2 i 1 i = n/ n n. Johannes Mittmann IP = PSPACE JASS
49 Reduction to a Finite Field The Chinese Remainder Theorem Theorem (Sun Tzu) Let a 1,..., a k Z, and let p 1,..., p k N >0 be pairwise coprime. Then the system of simultaneous congruences x = a 1 (mod p 1 ). x = a k (mod p k ) has a unique solution x modulo k i=1 p i. Johannes Mittmann IP = PSPACE JASS
50 Reduction to a Finite Field Reduction to F p Proposition For every Σ-Π expression A 0 of length n, there exists a prime p [ 2 n, 2 3n] such that A 0 (mod p). Johannes Mittmann IP = PSPACE JASS
51 Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. Johannes Mittmann IP = PSPACE JASS
52 Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. k = π ( 2 3n) π ( 2 n) 2 3n 2 n > 2 n. Johannes Mittmann IP = PSPACE JASS
53 Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. k = π ( 2 3n) π ( 2 n) 2 3n 2 n > 2 n. Suppose A = 0 (mod p i ) i. Johannes Mittmann IP = PSPACE JASS
54 Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. k = π ( 2 3n) π ( 2 n) 2 3n 2 n > 2 n. Suppose A = 0 (mod p i ) i. Chinese Remainder Theorem: A = 0 mod k p i > 2 2n. i=1 Johannes Mittmann IP = PSPACE JASS
55 Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. k = π ( 2 3n) π ( 2 n) 2 3n 2 n > 2 n. Suppose A = 0 (mod p i ) i. Chinese Remainder Theorem: A = 0 mod k p i > 2 2n. i=1 A 2 2n = A = 0, contradiction. Johannes Mittmann IP = PSPACE JASS
56 Polynomials and Simple Expressions Functional and Randomized Form Definition Let A be a Σ-Π expression. The functional form A is defined by eliminating the leftmost 1 1 z i =0 symbol in A, and can be considered as a polynomial q(z i ) Z[z i ]. z i =0 or The randomized form of A is A (z i = r), where r R F p is a random number supplied by the verifier. Johannes Mittmann IP = PSPACE JASS
57 Polynomials and Simple Expressions Problem: Exponentially High Degree Example φ = x 1 x 2... x k (x 1 x 2 x k ) A φ = (z 1 + z z k ). z 1 =0 z 2 =0 z k =0 = deg q(z 1 ) = 2 k 1. Johannes Mittmann IP = PSPACE JASS
58 Polynomials and Simple Expressions Problem: Exponentially High Degree Example φ = x 1 x 2... x k (x 1 x 2 x k ) A φ = (z 1 + z z k ). z 1 =0 z 2 =0 z k =0 = deg q(z 1 ) = 2 k 1. dense polynomial of exponentially high degree! Johannes Mittmann IP = PSPACE JASS
59 Polynomials and Simple Expressions Problem: Exponentially High Degree Example φ = x 1 x 2... x k (x 1 x 2 x k ) A φ = (z 1 + z z k ). z 1 =0 z 2 =0 z k =0 = deg q(z 1 ) = 2 k 1. dense polynomial of exponentially high degree! Idea: elimination of universal quantifiers. Johannes Mittmann IP = PSPACE JASS
60 Polynomials and Simple Expressions Simple Expressions Definition A QBF φ is called simple, if any occurrence of a variable is separated by at most one universal quantifier from its point of quantification. Johannes Mittmann IP = PSPACE JASS
61 Polynomials and Simple Expressions Simple Expressions Definition A QBF φ is called simple, if any occurrence of a variable is separated by at most one universal quantifier from its point of quantification. Example φ = x 1 x 2 x 3 [ (x1 x 2 ) x 4 (x 2 x 3 x 4 ) ] is simple. ψ = x 1 x 2 [ (x1 x 2 ) x 3 ( x 1 x 3 ) ] is not simple. Johannes Mittmann IP = PSPACE JASS
62 Polynomials and Simple Expressions Lemma Let φ be a simple QBF of length n, and let q(z i ) be the polynomial of the functional form of A φ. Then deg q(z i ) 2n. Johannes Mittmann IP = PSPACE JASS
63 Polynomials and Simple Expressions Lemma Let φ be a simple QBF of length n, and let q(z i ) be the polynomial of the functional form of A φ. Then Proof. deg q(z i ) 2n. The degree of quantifier-free subexpressions in z i is bounded its size. Johannes Mittmann IP = PSPACE JASS
64 Polynomials and Simple Expressions Lemma Let φ be a simple QBF of length n, and let q(z i ) be the polynomial of the functional form of A φ. Then Proof. deg q(z i ) 2n. The degree of quantifier-free subexpressions in z i is bounded its size. Summations cannot change the degree. Johannes Mittmann IP = PSPACE JASS
65 Polynomials and Simple Expressions Lemma Let φ be a simple QBF of length n, and let q(z i ) be the polynomial of the functional form of A φ. Then Proof. deg q(z i ) 2n. The degree of quantifier-free subexpressions in z i is bounded its size. Summations cannot change the degree. Products can at most double the degree. Johannes Mittmann IP = PSPACE JASS
66 Polynomials and Simple Expressions Lemma Let φ be a simple QBF of length n, and let q(z i ) be the polynomial of the functional form of A φ. Then Proof. deg q(z i ) 2n. The degree of quantifier-free subexpressions in z i is bounded its size. Summations cannot change the degree. Products can at most double the degree. In simple expressions this can happen at most once. Johannes Mittmann IP = PSPACE JASS
67 Polynomials and Simple Expressions Lemma Any QBF φ can be transformed in logarithmic space to an equivalent simple expression. Johannes Mittmann IP = PSPACE JASS
68 Polynomials and Simple Expressions Lemma Any QBF φ can be transformed in logarithmic space to an equivalent simple expression. Proof. Consider φ =... Qx i... x j ψ(x i ), Q {, }. Johannes Mittmann IP = PSPACE JASS
69 Polynomials and Simple Expressions Lemma Any QBF φ can be transformed in logarithmic space to an equivalent simple expression. Proof. Consider φ =... Qx i... x j ψ(x i ), Transformation: Q {, }. φ =... Qx i... x j x i (x i x i ) ψ(x i ) [ =... Qx i... x j x i (xi x i ) ( x i x i ) ] ψ(x i ). Johannes Mittmann IP = PSPACE JASS
70 Polynomials and Simple Expressions Lemma Any QBF φ can be transformed in logarithmic space to an equivalent simple expression. Proof. Consider φ =... Qx i... x j ψ(x i ), Transformation: Q {, }. φ =... Qx i... x j x i (x i x i ) ψ(x i ) [ =... Qx i... x j x i (xi x i ) ( x i x i ) ] ψ(x i ). O(n 2 ) steps. Johannes Mittmann IP = PSPACE JASS
71 The Interactive Protocol Protocol Setup Prover Verifier Choose p [ 2 n, 2 3n]. Compute a A φ (mod p). p, a Verify a 0, p [ 2 n, 2 3n], and p Primes. Johannes Mittmann IP = PSPACE JASS
72 The Interactive Protocol Split Step A = A 1 + A 2 or A = A 1 A 2, where A 2 starts with the leftmost 1 z i =0 or 1 z i =0 symbol: A 2 = 1... or A 2 = z i = z i =0 Johannes Mittmann IP = PSPACE JASS
73 The Interactive Protocol Simplification Step Prover Verifier Compute q(z i ) of A. q(z i ) A A 2. a a a 1 (mod p), or a a/a 1 (mod p). Verify a = q(0) + q(1) (mod p), or a = q(0) q(1) (mod p). r Choose r R F p. A A (z i = r) (mod p). a q(r) (mod p). Johannes Mittmann IP = PSPACE JASS
74 The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 [ (1 z 1 ) z 2 =0 z 3 =0 ] (z 1 z 2 + z 3 ), Johannes Mittmann IP = PSPACE JASS
75 The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z [ (1 z 1 ) + 1 z 2 =0 z 3 =0 1 1 z 2 =0 z 3 =0 1 (z 1 z 2 + z 3 ), ] (z 1 z 2 + z 3 ), Johannes Mittmann IP = PSPACE JASS
76 The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z [ (1 z 1 ) + Verify a = 2 = 2 1 = q(0) q(1). 1 z 2 =0 z 3 =0 1 1 z 2 =0 z 3 =0 1 (z 1 z 2 + z 3 ), ] (z 1 z 2 + z 3 ), Johannes Mittmann IP = PSPACE JASS
77 The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z [ (1 z 1 ) + Verify a = 2 = 2 1 = q(0) q(1). 1 z 2 =0 z 3 =0 A A (z 1 = 3) = (1 3) + a q(3) = z 2 =0 z 3 =0 1 (z 1 z 2 + z 3 ), 1 z 2 =0 z 3 =0 ] (z 1 z 2 + z 3 ), 1 (3z 2 + z 3 ), Johannes Mittmann IP = PSPACE JASS
78 The Interactive Protocol Example (Round 2) 1 1 A A 2 = (3z 2 + z 3 ), z 2 =0 z 3 =0 a a a 1 = 10 ( 2) = 12. Johannes Mittmann IP = PSPACE JASS
79 The Interactive Protocol Example (Round 2) 1 1 A A 2 = (3z 2 + z 3 ), z 2 =0 z 3 =0 a a a 1 = 10 ( 2) = A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z z 2. Johannes Mittmann IP = PSPACE JASS
80 The Interactive Protocol Example (Round 2) 1 1 A A 2 = (3z 2 + z 3 ), z 2 =0 z 3 =0 a a a 1 = 10 ( 2) = A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z z 2. Verify a = 12 = = q(0) + q(1). Johannes Mittmann IP = PSPACE JASS
81 The Interactive Protocol Example (Round 2) 1 1 A A 2 = (3z 2 + z 3 ), z 2 =0 z 3 =0 a a a 1 = 10 ( 2) = A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z z 2. Verify a = 12 = = q(0) + q(1). 1 A A (z 2 = 2) = (z 3 + 6), z 3 =0 a q(2) = = 42. Johannes Mittmann IP = PSPACE JASS
82 The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. Johannes Mittmann IP = PSPACE JASS
83 The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z Johannes Mittmann IP = PSPACE JASS
84 The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z Verify a = 42 = 6 7 = q(0) q(1). Johannes Mittmann IP = PSPACE JASS
85 The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z Verify a = 42 = 6 7 = q(0) q(1). A A (z 3 = 5) = = 11, a q(5) = = 11. Johannes Mittmann IP = PSPACE JASS
86 The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z Verify a = 42 = 6 7 = q(0) q(1). A A (z 3 = 5) = = 11, a q(5) = = 11. Verify A = 11 = a, and accept. Johannes Mittmann IP = PSPACE JASS
87 The Interactive Protocol Correctness of the Interactive Proof Theorem 1 When φ is true and Alice is honest, Bob will always accept the proof. 2 When φ is false, Bob accepts the proof with negligible probability. Johannes Mittmann IP = PSPACE JASS
88 The Interactive Protocol Correctness of the Interactive Proof Theorem 1 When φ is true and Alice is honest, Bob will always accept the proof. 2 When φ is false, Bob accepts the proof with negligible probability. Proof of completeness. Alice is able to provide all the polynomials. Bob will always accept. Johannes Mittmann IP = PSPACE JASS
89 The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Johannes Mittmann IP = PSPACE JASS
90 The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Alice must supply a wrong polynomial q (z i ) in the i-th round. Johannes Mittmann IP = PSPACE JASS
91 The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Alice must supply a wrong polynomial q (z i ) in the i-th round. 0 q(z i ) q (z i ) has at most 2n roots. Johannes Mittmann IP = PSPACE JASS
92 The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Alice must supply a wrong polynomial q (z i ) in the i-th round. 0 q(z i ) q (z i ) has at most 2n roots. Probability of a false positive: Pr[error in the i-th round] 2n p 2n 2 n. Johannes Mittmann IP = PSPACE JASS
93 The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Alice must supply a wrong polynomial q (z i ) in the i-th round. 0 q(z i ) q (z i ) has at most 2n roots. Probability of a false positive: After m n rounds: Pr[error in the i-th round] 2n p Pr[error] = 1 1 2n 2 n. m Pr[no error in the i-th round] i=1 ( 1 2n 2 n ) n. Johannes Mittmann IP = PSPACE JASS
94 References For further reading: L. J. Stockmeyer and A. R. Meyer: Word problems requiring exponential time. Proc. 5th ACM Symp. on the Theory of Computing, pp. 1-9, L. Babai: and the unexpected power of interaction. Structure in Complexity Theory Conf., pp , Adi Shamir: IP=PSPACE. Journal of the ACM, 39 (4), pp , Christos H. Papadimitriou: Computational Complexity. Addison Wesley, Reading, Johannes Mittmann IP = PSPACE JASS
95 Exercise Exercise Exercise Show that #P IP. Hint. Arithmetization of the #P-complete counting problem #SAT. Johannes Mittmann IP = PSPACE JASS
96 Exercise Solution Arithmetization: 3-CNF φ Arithmetization A φ x i X z i Z ψ 1 A ψ ψ 1 ψ 2 1 (1 A ψ1 )(1 A ψ2 ) ψ 1 ψ 2 A ψ1 A ψ2 Then φ(x 1,..., x n ) has exactly K satisfying assignments, iff 1 1 K = A φ (z 1,..., z n ). z 1 =0 z n=0 Johannes Mittmann IP = PSPACE JASS
IP=PSPACE. Adi Shamir. Applied Mathematics Department The Weizmann Institute of Science Rehovot, Israel
P=PSPACE Abstract n this paper we prove that when both randomization and interaction are allowed, the proofs which can be verified in polynomial time are exactly those proofs which can be generated with
Lecture 19: Interactive Proofs and the PCP Theorem
Lecture 19: Interactive Proofs and the PCP Theorem Valentine Kabanets November 29, 2016 1 Interactive Proofs In this model, we have an all-powerful Prover (with unlimited computational prover) and a polytime
Lecture 14: IP = PSPACE
IAS/PCMI Summer Session 2000 Clay Mathematics Undergraduate Program Basic Course on Computational Complexity Lecture 14: IP = PSPACE David Mix Barrington and Alexis Maciel August 3, 2000 1. Overview We
1 PSPACE-Completeness
CS 6743 Lecture 14 1 Fall 2007 1 PSPACE-Completeness Recall the NP-complete problem SAT: Is a given Boolean formula φ(x 1,..., x n ) satisfiable? The same question can be stated equivalently as: Is the
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 16 IP = PSPACE 3 Goal and Plan Goal IP = PSPACE Plan 1. PSPACE IP by showing QBF
Interactive Proofs 1
CS294: Probabilistically Checkable and Interactive Proofs January 24, 2017 Interactive Proofs 1 Instructor: Alessandro Chiesa & Igor Shinkar Scribe: Mariel Supina 1 Pspace IP The first proof that Pspace
The Proof of IP = P SP ACE
The Proof of IP = P SP ACE Larisse D. Voufo March 29th, 2007 For a long time, the question of how a verier can be convinced with high probability that a given theorem is provable without showing the whole
Introduction to Interactive Proofs & The Sumcheck Protocol
CS294: Probabilistically Checkable and Interactive Proofs January 19, 2017 Introduction to Interactive Proofs & The Sumcheck Protocol Instructor: Alessandro Chiesa & Igor Shinkar Scribe: Pratyush Mishra
Complexity Theory VU , SS The Polynomial Hierarchy. Reinhard Pichler
Complexity Theory Complexity Theory VU 181.142, SS 2018 6. The Polynomial Hierarchy Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität Wien 15 May, 2018 Reinhard
Outline. Complexity Theory EXACT TSP. The Class DP. Definition. Problem EXACT TSP. Complexity of EXACT TSP. Proposition VU 181.
Complexity Theory Complexity Theory Outline Complexity Theory VU 181.142, SS 2018 6. The Polynomial Hierarchy Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität
2 Evidence that Graph Isomorphism is not NP-complete
Topics in Theoretical Computer Science April 11, 2016 Lecturer: Ola Svensson Lecture 7 (Notes) Scribes: Ola Svensson Disclaimer: These notes were written for the lecturer only and may contain inconsistent
Reasoning with Quantified Boolean Formulas
Reasoning with Quantified Boolean Formulas Martina Seidl Institute for Formal Models and Verification Johannes Kepler University Linz 1 What are QBF? Quantified Boolean formulas (QBF) are formulas of propositional
Theorem 11.1 (Lund-Fortnow-Karloff-Nisan). There is a polynomial length interactive proof for the
Lecture 11 IP, PH, and PSPACE May 4, 2004 Lecturer: Paul Beame Notes: Daniel Lowd 11.1 IP and PH Theorem 11.1 (Lund-Fortnow-Karloff-Nisan). There is a polynomial length interactive proof for the predicate
Lecture 8. MINDNF = {(φ, k) φ is a CNF expression and DNF expression ψ s.t. ψ k and ψ is equivalent to φ}
6.841 Advanced Complexity Theory February 28, 2005 Lecture 8 Lecturer: Madhu Sudan Scribe: Arnab Bhattacharyya 1 A New Theme In the past few lectures, we have concentrated on non-uniform types of computation
POLYNOMIAL SPACE QSAT. Games. Polynomial space cont d
T-79.5103 / Autumn 2008 Polynomial Space 1 T-79.5103 / Autumn 2008 Polynomial Space 3 POLYNOMIAL SPACE Polynomial space cont d Polynomial space-bounded computation has a variety of alternative characterizations
Database Theory VU , SS Complexity of Query Evaluation. Reinhard Pichler
Database Theory Database Theory VU 181.140, SS 2018 5. Complexity of Query Evaluation Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität Wien 17 April, 2018 Pichler
Lecture : PSPACE IP
IITM-CS6845: Theory Toolkit February 16, 2012 Lecture 22-23 : PSPACE IP Lecturer: Jayalal Sarma.M.N. Scribe: Sivaramakrishnan.N.R. Theme: Between P and PSPACE 1 Interactive Protocol for #SAT In the previous
Algorithms & Complexity II Avarikioti Zeta
Algorithms & Complexity II Avarikioti Zeta March 17, 2014 Alternating Computation Alternation: generalizes non-determinism, where each state is either existential or universal : Old: existential states
Quantified Boolean Formulas Part 1
Quantified Boolean Formulas Part 1 Uwe Egly Knowledge-Based Systems Group Institute of Information Systems Vienna University of Technology Results of the SAT 2009 application benchmarks for leading solvers
Author s address: Applied Mathematics Department, The Weizmann Institute of Science, Rehovot,
1P = IW?ACE ADI SHAMIR The Welzrnann Institute of Science, Reholot, Israel Abstract. In this paper, it is proven that when both randomization and interaction are allowed, the proofs that can be verified
CS151 Complexity Theory. Lecture 14 May 17, 2017
CS151 Complexity Theory Lecture 14 May 17, 2017 IP = PSPACE Theorem: (Shamir) IP = PSPACE Note: IP PSPACE enumerate all possible interactions, explicitly calculate acceptance probability interaction extremely
LOGIC PROPOSITIONAL REASONING
LOGIC PROPOSITIONAL REASONING WS 2017/2018 (342.208) Armin Biere Martina Seidl biere@jku.at martina.seidl@jku.at Institute for Formal Models and Verification Johannes Kepler Universität Linz Version 2018.1
198:538 Complexity of Computation Lecture 16 Rutgers University, Spring March 2007
198:538 Complexity of Computation Lecture 16 Rutgers University, Spring 2007 8 March 2007 In this lecture we discuss Shamir s theorem that PSPACE is the set of languages that have interactive proofs with
Comp487/587 - Boolean Formulas
Comp487/587 - Boolean Formulas 1 Logic and SAT 1.1 What is a Boolean Formula Logic is a way through which we can analyze and reason about simple or complicated events. In particular, we are interested
CSE200: Computability and complexity Space Complexity
CSE200: Computability and complexity Space Complexity Shachar Lovett January 29, 2018 1 Space complexity We would like to discuss languages that may be determined in sub-linear space. Lets first recall
Lecture 8: Complete Problems for Other Complexity Classes
IAS/PCMI Summer Session 2000 Clay Mathematics Undergraduate Program Basic Course on Computational Complexity Lecture 8: Complete Problems for Other Complexity Classes David Mix Barrington and Alexis Maciel
Propositional Logic: Evaluating the Formulas
Institute for Formal Models and Verification Johannes Kepler University Linz VL Logik (LVA-Nr. 342208) Winter Semester 2015/2016 Propositional Logic: Evaluating the Formulas Version 2015.2 Armin Biere
CS151 Complexity Theory. Lecture 13 May 15, 2017
CS151 Complexity Theory Lecture 13 May 15, 2017 Relationship to other classes To compare to classes of decision problems, usually consider P #P which is a decision class easy: NP, conp P #P easy: P #P
CSE 555 HW 5 SAMPLE SOLUTION. Question 1.
CSE 555 HW 5 SAMPLE SOLUTION Question 1. Show that if L is PSPACE-complete, then L is NP-hard. Show that the converse is not true. If L is PSPACE-complete, then for all A PSPACE, A P L. We know SAT PSPACE
Umans Complexity Theory Lectures
Umans Complexity Theory Lectures Lecture 12: The Polynomial-Time Hierarchy Oracle Turing Machines Oracle Turing Machine (OTM): Deterministic multitape TM M with special query tape special states q?, q
Space is a computation resource. Unlike time it can be reused. Computational Complexity, by Fu Yuxi Space Complexity 1 / 44
Space Complexity Space is a computation resource. Unlike time it can be reused. Computational Complexity, by Fu Yuxi Space Complexity 1 / 44 Synopsis 1. Space Bounded Computation 2. Logspace Reduction
Ταουσάκος Θανάσης Αλγόριθμοι και Πολυπλοκότητα II 7 Φεβρουαρίου 2013
Ταουσάκος Θανάσης Αλγόριθμοι και Πολυπλοκότητα II 7 Φεβρουαρίου 2013 Alternation: important generalization of non-determinism Redefining Non-Determinism in terms of configurations: a configuration lead
Lecture 26. Daniel Apon
Lecture 26 Daniel Apon 1 From IPPSPACE to NPPCP(log, 1): NEXP has multi-prover interactive protocols If you ve read the notes on the history of the PCP theorem referenced in Lecture 19 [3], you will already
Chapter 9. PSPACE: A Class of Problems Beyond NP. Slides by Kevin Wayne Pearson-Addison Wesley. All rights reserved.
Chapter 9 PSPACE: A Class of Problems Beyond NP Slides by Kevin Wayne. Copyright @ 2005 Pearson-Addison Wesley. All rights reserved. 1 Geography Game Geography. Alice names capital city c of country she
15-855: Intensive Intro to Complexity Theory Spring Lecture 7: The Permanent, Toda s Theorem, XXX
15-855: Intensive Intro to Complexity Theory Spring 2009 Lecture 7: The Permanent, Toda s Theorem, XXX 1 #P and Permanent Recall the class of counting problems, #P, introduced last lecture. It was introduced
Computational Complexity IV: PSPACE
Seminar on Theoretical Computer Science and Discrete Mathematics Aristotle University of Thessaloniki Context 1 Section 1: PSPACE 2 3 4 Time Complexity Time complexity of DTM M: - Increasing function t:
Quantified Boolean Formulas: Complexity and Expressiveness
Dagstuhl SAT Interactions 2012 Quantified Boolean Formulas: Complexity and Expressiveness Uwe Bubeck University of Paderborn 20.11.2012 Outline Introduction Free Variables and Equivalence Complexity Expressiveness
Part 1: Propositional Logic
Part 1: Propositional Logic Literature (also for first-order logic) Schöning: Logik für Informatiker, Spektrum Fitting: First-Order Logic and Automated Theorem Proving, Springer 1 Last time 1.1 Syntax
9. PSPACE. PSPACE complexity class quantified satisfiability planning problem PSPACE-complete
9. PSPACE PSPACE complexity class quantified satisfiability planning problem PSPACE-complete Lecture slides by Kevin Wayne Copyright 2005 Pearson-Addison Wesley Copyright 2013 Kevin Wayne http://www.cs.princeton.edu/~wayne/kleinberg-tardos
CS156: The Calculus of Computation
CS156: The Calculus of Computation Zohar Manna Winter 2010 It is reasonable to hope that the relationship between computation and mathematical logic will be as fruitful in the next century as that between
CSC 2429 Approaches to the P versus NP Question. Lecture #12: April 2, 2014
CSC 2429 Approaches to the P versus NP Question Lecture #12: April 2, 2014 Lecturer: David Liu (Guest) Scribe Notes by: David Liu 1 Introduction The primary goal of complexity theory is to study the power
First-Order Logic. 1 Syntax. Domain of Discourse. FO Vocabulary. Terms
First-Order Logic 1 Syntax Domain of Discourse The domain of discourse for first order logic is FO structures or models. A FO structure contains Relations Functions Constants (functions of arity 0) FO
Lecture 12: Interactive Proofs
princeton university cos 522: computational complexity Lecture 12: Interactive Proofs Lecturer: Sanjeev Arora Scribe:Carl Kingsford Recall the certificate definition of NP. We can think of this characterization
Real Interactive Proofs for VPSPACE
Brandenburgische Technische Universität, Cottbus-Senftenberg, Germany Colloquium Logicum Hamburg, September 2016 joint work with M. Baartse 1. Introduction Blum-Shub-Smale model of computability and complexity
Introduction to Complexity Theory. Bernhard Häupler. May 2, 2006
Introduction to Complexity Theory Bernhard Häupler May 2, 2006 Abstract This paper is a short repetition of the basic topics in complexity theory. It is not intended to be a complete step by step introduction
Rational Proofs with Multiple Provers. Jing Chen, Samuel McCauley, Shikha Singh Department of Computer Science
Rational Proofs with Multiple Provers Jing Chen, Samuel McCauley, Shikha Singh Department of Computer Science Outline of the Talk RATIONAL INTERACTIVE PROOFS with MULTI-PROVERs Interactive Proofs [GMR,
Interactive protocols & zero-knowledge
Interactive protocols & zero-knowledge - interactive protocols formalize what can be recognized by polynomial time restricted verifiers in arbitrary protocols - generalizes NP - zero-knowledge formalizes
Space and Nondeterminism
CS 221 Computational Complexity, Lecture 5 Feb 6, 2018 Space and Nondeterminism Instructor: Madhu Sudan 1 Scribe: Yong Wook Kwon Topic Overview Today we ll talk about space and non-determinism. For some
SOLUTION: SOLUTION: SOLUTION:
Convert R and S into nondeterministic finite automata N1 and N2. Given a string s, if we know the states N1 and N2 may reach when s[1...i] has been read, we are able to derive the states N1 and N2 may
Informal Statement Calculus
FOUNDATIONS OF MATHEMATICS Branches of Logic 1. Theory of Computations (i.e. Recursion Theory). 2. Proof Theory. 3. Model Theory. 4. Set Theory. Informal Statement Calculus STATEMENTS AND CONNECTIVES Example
9. PSPACE 9. PSPACE. PSPACE complexity class quantified satisfiability planning problem PSPACE-complete
Geography game Geography. Alice names capital city c of country she is in. Bob names a capital city c' that starts with the letter on which c ends. Alice and Bob repeat this game until one player is unable
Conjunction: p q is true if both p, q are true, and false if at least one of p, q is false. The truth table for conjunction is as follows.
Chapter 1 Logic 1.1 Introduction and Definitions Definitions. A sentence (statement, proposition) is an utterance (that is, a string of characters) which is either true (T) or false (F). A predicate is
6.841/18.405J: Advanced Complexity Wednesday, February 12, Lecture Lecture 3
6.841/18.405J: Advanced Complexity Wednesday, February 12, 2003 Lecture Lecture 3 Instructor: Madhu Sudan Scribe: Bobby Kleinberg 1 The language MinDNF At the end of the last lecture, we introduced the
COMPLEXITY THEORY. Lecture 17: The Polynomial Hierarchy. TU Dresden, 19th Dec Markus Krötzsch Knowledge-Based Systems
COMPLEXITY THEORY Lecture 17: The Polynomial Hierarchy Markus Krötzsch Knowledge-Based Systems TU Dresden, 19th Dec 2017 Review: ATM vs. DTM Markus Krötzsch, 19th Dec 2017 Complexity Theory slide 2 of
Lecture 7: The Polynomial-Time Hierarchy. 1 Nondeterministic Space is Closed under Complement
CS 710: Complexity Theory 9/29/2011 Lecture 7: The Polynomial-Time Hierarchy Instructor: Dieter van Melkebeek Scribe: Xi Wu In this lecture we first finish the discussion of space-bounded nondeterminism
De Morgan s a Laws. De Morgan s laws say that. (φ 1 φ 2 ) = φ 1 φ 2, (φ 1 φ 2 ) = φ 1 φ 2.
De Morgan s a Laws De Morgan s laws say that (φ 1 φ 2 ) = φ 1 φ 2, (φ 1 φ 2 ) = φ 1 φ 2. Here is a proof for the first law: φ 1 φ 2 (φ 1 φ 2 ) φ 1 φ 2 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0 a Augustus DeMorgan
Lecture 11: Proofs, Games, and Alternation
IAS/PCMI Summer Session 2000 Clay Mathematics Undergraduate Program Basic Course on Computational Complexity Lecture 11: Proofs, Games, and Alternation David Mix Barrington and Alexis Maciel July 31, 2000
Complete problems for classes in PH, The Polynomial-Time Hierarchy (PH) oracle is like a subroutine, or function in
Oracle Turing Machines Nondeterministic OTM defined in the same way (transition relation, rather than function) oracle is like a subroutine, or function in your favorite PL but each call counts as single
6.841/18.405J: Advanced Complexity Wednesday, April 2, Lecture Lecture 14
6.841/18.405J: Advanced Complexity Wednesday, April 2, 2003 Lecture Lecture 14 Instructor: Madhu Sudan In this lecture we cover IP = PSPACE Interactive proof for straightline programs. Straightline program
Outline. Complexity Theory. Introduction. What is abduction? Motivation. Reference VU , SS Logic-Based Abduction
Complexity Theory Complexity Theory Outline Complexity Theory VU 181.142, SS 2018 7. Logic-Based Abduction Reinhard Pichler Institut für Informationssysteme Arbeitsbereich DBAI Technische Universität Wien
Lecture 20: PSPACE. November 15, 2016 CS 1010 Theory of Computation
Lecture 20: PSPACE November 15, 2016 CS 1010 Theory of Computation Recall that PSPACE = k=1 SPACE(nk ). We will see that a relationship between time and space complexity is given by: P NP PSPACE = NPSPACE
Worst-Case Upper Bound for (1, 2)-QSAT
Worst-Case Upper Bound for (1, 2)-QSAT Minghao Yin Department of Computer, Northeast Normal University, Changchun, China, 130117 ymh@nenu.edu.cn Abstract. The rigorous theoretical analysis of the algorithm
Interactive protocols & zero-knowledge
Interactive protocols & zero-knowledge - interactive protocols formalize what can be recognized by polynomial time restricted verifiers in arbitrary protocols - generalizes NP - zero-knowledge formalizes
A brief introduction to Logic. (slides from
A brief introduction to Logic (slides from http://www.decision-procedures.org/) 1 A Brief Introduction to Logic - Outline Propositional Logic :Syntax Propositional Logic :Semantics Satisfiability and validity
Bounded-width QBF is PSPACE-complete
Bounded-width QBF is PSPACE-complete Albert Atserias Universitat Politècnica de Catalunya Barcelona, Spain atserias@lsi.upc.edu Sergi Oliva Universitat Politècnica de Catalunya Barcelona, Spain oliva@lsi.upc.edu
CS Lecture 29 P, NP, and NP-Completeness. k ) for all k. Fall The class P. The class NP
CS 301 - Lecture 29 P, NP, and NP-Completeness Fall 2008 Review Languages and Grammars Alphabets, strings, languages Regular Languages Deterministic Finite and Nondeterministic Automata Equivalence of
A New 3-CNF Transformation by Parallel-Serial Graphs 1
A New 3-CNF Transformation by Parallel-Serial Graphs 1 Uwe Bubeck, Hans Kleine Büning University of Paderborn, Computer Science Institute, 33098 Paderborn, Germany Abstract For propositional formulas we
Complexity Theory. Knowledge Representation and Reasoning. November 2, 2005
Complexity Theory Knowledge Representation and Reasoning November 2, 2005 (Knowledge Representation and Reasoning) Complexity Theory November 2, 2005 1 / 22 Outline Motivation Reminder: Basic Notions Algorithms
Technische Universität München Summer term 2010 Theoretische Informatik August 2, 2010 Dr. J. Kreiker / Dr. M. Luttenberger, J. Kretinsky SOLUTION
Technische Universität München Summer term 2010 Theoretische Informatik August 2, 2010 Dr. J. Kreiker / Dr. M. Luttenberger, J. Kretinsky SOLUTION Complexity Theory Final Exam Please note : If not stated
Principles of Knowledge Representation and Reasoning
Principles of Knowledge Representation and Reasoning Complexity Theory Bernhard Nebel, Malte Helmert and Stefan Wölfl Albert-Ludwigs-Universität Freiburg April 29, 2008 Nebel, Helmert, Wölfl (Uni Freiburg)
CSE 1400 Applied Discrete Mathematics Definitions
CSE 1400 Applied Discrete Mathematics Definitions Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Arithmetic 1 Alphabets, Strings, Languages, & Words 2 Number Systems 3 Machine
Definition 1. NP is a class of language L such that there exists a poly time verifier V with
Lecture 9: MIP=NEXP Topics in Pseudorandomness and Complexity Theory (Spring 2017) Rutgers University Swastik Kopparty Scribe: Jinyoung Park 1 Introduction Definition 1. NP is a class of language L such
Validating QBF Invalidity in HOL4
Interactive Theorem Proving (ITP) 14 July, 2010 Quantified Boolean Formulae Quantified Boolean Formulae Motivation System Overview Related Work QBF = propositional logic + quantifiers over Boolean variables
x 2 +2 x 2 x 3 2x 2 +2x +1 (mod 5).
A DETAILED PROOF THAT IP=PSPACE B.J. MARES Abstract. I will define IP, the class of interactive proofs, and present a thorough proof that IP=PSPACE, assuming minimal prior knowledge. Such a presentation
Lecture 9: Polynomial-Time Hierarchy, Time-Space Tradeoffs
CSE 531: Computational Complexity I Winter 2016 Lecture 9: Polynomial-Time Hierarchy, Time-Space Tradeoffs Feb 3, 2016 Lecturer: Paul Beame Scribe: Paul Beame 1 The Polynomial-Time Hierarchy Last time
Winter 2011 Josh Benaloh Brian LaMacchia
Winter 2011 Josh Benaloh Brian LaMacchia Fun with Public-Key Tonight we ll Introduce some basic tools of public-key crypto Combine the tools to create more powerful tools Lay the ground work for substantial
CS 580: Algorithm Design and Analysis. Jeremiah Blocki Purdue University Spring 2018
CS 580: Algorithm Design and Analysis Jeremiah Blocki Purdue University Spring 2018 Chapter 9 PSPACE: A Class of Problems Beyond NP Slides by Kevin Wayne. Copyright @ 2005 Pearson-Addison Wesley. All rights
Lecture 17: Interactive Proof Systems
Computational Complexity Theory, Fall 2010 November 5 Lecture 17: Interactive Proof Systems Lecturer: Kristoffer Arnsfelt Hansen Scribe: Søren Valentin Haagerup 1 Interactive Proof Systems Definition 1.
Polynomial Hierarchy
Polynomial Hierarchy A polynomial-bounded version of Kleene s Arithmetic Hierarchy becomes trivial if P = NP. Karp, 1972 Computational Complexity, by Fu Yuxi Polynomial Hierarchy 1 / 44 Larry Stockmeyer
Approximate Verification
Approximate Verification Michel de Rougemont Université de Paris-II Ekaterinenburg-April 2014 i ii Contents 1 Probabilistic Classes 3 1.1 Probabilistic decision classes...................................
Real Interactive Proofs for VPSPACE
eal Interactive Proofs for VPSPACE Martijn Baartse 1 and Klaus Meer 2 1 Computer Science Institute, BTU Cottbus-Senftenberg Platz der Deutschen Einheit 1 D-03046 Cottbus, Germany 2 Computer Science Institute,
INAPPROX APPROX PTAS. FPTAS Knapsack P
CMPSCI 61: Recall From Last Time Lecture 22 Clique TSP INAPPROX exists P approx alg for no ε < 1 VertexCover MAX SAT APPROX TSP some but not all ε< 1 PTAS all ε < 1 ETSP FPTAS Knapsack P poly in n, 1/ε
Lecture 22: PSPACE
6.045 Lecture 22: PSPACE 1 VOTE VOTE VOTE For your favorite course on automata and complexity Please complete the online subject evaluation for 6.045 2 Final Exam Information Who: You On What: Everything
PSPACE COMPLETENESS TBQF. THURSDAY April 17
PSPACE COMPLETENESS TBQF THURSDAY April 17 Definition: Language B is PSPACE-complete if: 1. B PSPACE 2. Every A in PSPACE is poly-time reducible to B (i.e. B is PSPACE-hard) QUANTIFIED BOOLEAN FORMULAS
Tecniche di Verifica. Introduction to Propositional Logic
Tecniche di Verifica Introduction to Propositional Logic 1 Logic A formal logic is defined by its syntax and semantics. Syntax An alphabet is a set of symbols. A finite sequence of these symbols is called
First-Order Logic First-Order Theories. Roopsha Samanta. Partly based on slides by Aaron Bradley and Isil Dillig
First-Order Logic First-Order Theories Roopsha Samanta Partly based on slides by Aaron Bradley and Isil Dillig Roadmap Review: propositional logic Syntax and semantics of first-order logic (FOL) Semantic
1 First-order logic. 1 Syntax of first-order logic. 2 Semantics of first-order logic. 3 First-order logic queries. 2 First-order query evaluation
Knowledge Bases and Databases Part 1: First-Order Queries Diego Calvanese Faculty of Computer Science Master of Science in Computer Science A.Y. 2007/2008 Overview of Part 1: First-order queries 1 First-order
6.045 Final Exam Solutions
6.045J/18.400J: Automata, Computability and Complexity Prof. Nancy Lynch, Nati Srebro 6.045 Final Exam Solutions May 18, 2004 Susan Hohenberger Name: Please write your name on each page. This exam is open
THE BLACK-BOX QUERY COMPLEXITY OF POLYNOMIAL SUMMATION
THE BLACK-BOX QUERY COMPLEXITY OF POLYNOMIAL SUMMATION Ali Juma, Valentine Kabanets, Charles Rackoff, and Amir Shpilka Abstract. For any given Boolean formula φ(x 1,..., x n ), one can efficiently construct
Complexity. Complexity Theory Lecture 3. Decidability and Complexity. Complexity Classes
Complexity Theory 1 Complexity Theory 2 Complexity Theory Lecture 3 Complexity For any function f : IN IN, we say that a language L is in TIME(f(n)) if there is a machine M = (Q, Σ, s, δ), such that: L
Solution of Exercise Sheet 12
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
Chapter 2. Reductions and NP. 2.1 Reductions Continued The Satisfiability Problem (SAT) SAT 3SAT. CS 573: Algorithms, Fall 2013 August 29, 2013
Chapter 2 Reductions and NP CS 573: Algorithms, Fall 2013 August 29, 2013 2.1 Reductions Continued 2.1.1 The Satisfiability Problem SAT 2.1.1.1 Propositional Formulas Definition 2.1.1. Consider a set of
Theory of Computation Space Complexity. (NTU EE) Space Complexity Fall / 1
Theory of Computation Space Complexity (NTU EE) Space Complexity Fall 2016 1 / 1 Space Complexity Definition 1 Let M be a TM that halts on all inputs. The space complexity of M is f : N N where f (n) is
CMSC Discrete Mathematics SOLUTIONS TO SECOND MIDTERM EXAM November, 2005
CMSC-37110 Discrete Mathematics SOLUTIONS TO SECOND MIDTERM EXAM November, 2005 Instructor: László Babai Ryerson 164 e-mail: laci@cs This exam contributes 20% to your course grade. 1. (6 points) Let a
Propositional Logic: Models and Proofs
Propositional Logic: Models and Proofs C. R. Ramakrishnan CSE 505 1 Syntax 2 Model Theory 3 Proof Theory and Resolution Compiled at 11:51 on 2016/11/02 Computing with Logic Propositional Logic CSE 505
Chapter 1 : The language of mathematics.
MAT 200, Logic, Language and Proof, Fall 2015 Summary Chapter 1 : The language of mathematics. Definition. A proposition is a sentence which is either true or false. Truth table for the connective or :
Topic 1: Propositional logic
Topic 1: Propositional logic Guy McCusker 1 1 University of Bath Logic! This lecture is about the simplest kind of mathematical logic: propositional calculus. We discuss propositions, which are statements
Notes on Complexity Theory Last updated: October, Lecture 6
Notes on Complexity Theory Last updated: October, 2015 Lecture 6 Notes by Jonathan Katz, lightly edited by Dov Gordon 1 PSPACE and PSPACE-Completeness As in our previous study of N P, it is useful to identify
1 Randomized Computation
CS 6743 Lecture 17 1 Fall 2007 1 Randomized Computation Why is randomness useful? Imagine you have a stack of bank notes, with very few counterfeit ones. You want to choose a genuine bank note to pay at
-bit integers are all in ThC. Th The following problems are complete for PSPACE NPSPACE ATIME QSAT, GEOGRAPHY, SUCCINCT REACH.
CMPSCI 601: Recall From Last Time Lecture 26 Theorem: All CFL s are in sac. Facts: ITADD, MULT, ITMULT and DIVISION on -bit integers are all in ThC. Th The following problems are complete for PSPACE NPSPACE