Shamir s Theorem. Johannes Mittmann. Technische Universität München (TUM)

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 2006 1

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 2006 2

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 2006 3

Polynomial Space Polynomial Space PSPACE = k>0 SPACE(n k ). L NL P NP PSPACE. Johannes Mittmann IP = PSPACE JASS 2006 4

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 2006 5

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 2006 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. 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 2006 6

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 2006 7

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 2006 7

Polynomial Space Quantified Satifiability Quantified Satisfiability QSAT = { φ : φ is a true QBF in prenex CNF }. Johannes Mittmann IP = PSPACE JASS 2006 8

Polynomial Space PSPACE-Completeness PSPACE-Completeness Theorem (Stockmeyer/Meyer 1973) QSAT is PSPACE-complete. Johannes Mittmann IP = PSPACE JASS 2006 9

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 2006 9

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 2006 10

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 2006 10

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 2006 11

Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. Johannes Mittmann IP = PSPACE JASS 2006 12

Polynomial Space PSPACE-Completeness Proof sketch of (2). Let L PSPACE. L is decidable by a Turing machine M. Johannes Mittmann IP = PSPACE JASS 2006 12

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 2006 12

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 2006 12

Polynomial Space PSPACE-Completeness Proof sketch of (2). ψ 0 (A, B) can be written in DNF. Johannes Mittmann IP = PSPACE JASS 2006 13

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 2006 13

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 2006 13

Shamir s Theorem Theorem (Shamir 1992) IP = PSPACE. Johannes Mittmann IP = PSPACE JASS 2006 14

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 2006 14

Proof Outline Proof. : It suffices to show QSAT IP, since IP is closed under reductions, and QSAT PSPACE-complete. Johannes Mittmann IP = PSPACE JASS 2006 15

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 2006 15

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 2006 16

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 ) + 1 1 z 2 =0 z 3 =0 ] (z 1 z 2 + z 3 ). Johannes Mittmann IP = PSPACE JASS 2006 17

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 2006 18

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 2006 18

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 2006 19

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 > 0.... Johannes Mittmann IP = PSPACE JASS 2006 19

Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ). 1 1 1 1 [ 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 2006 20

Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ). 1 1 1 1 [ 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 2006 20

Arithmetization Problem: Exponential Arithmetizations Example φ = x 1 x 2 x k 1 x k (x k x k ). 1 1 1 1 [ 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 2006 20

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 2006 21

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 1 2 21. Johannes Mittmann IP = PSPACE JASS 2006 21

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 1 2 21. 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 2006 21

Reduction to a Finite Field Number of Primes Lemma For n 3, n π(n) n. Johannes Mittmann IP = PSPACE JASS 2006 22

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 2006 22

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 2006 23

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 2006 24

Reduction to a Finite Field Proof. Denote by p 1,..., p k all primes in [ 2 n, 2 3n]. Johannes Mittmann IP = PSPACE JASS 2006 25

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 2006 25

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 2006 25

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 2006 25

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 2006 25

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 2006 26

Polynomials and Simple Expressions Problem: Exponentially High Degree Example φ = x 1 x 2... x k (x 1 x 2 x k ). 1 1 1 A φ = (z 1 + z 2 + + z k ). z 1 =0 z 2 =0 z k =0 = deg q(z 1 ) = 2 k 1. Johannes Mittmann IP = PSPACE JASS 2006 27

Polynomials and Simple Expressions Problem: Exponentially High Degree Example φ = x 1 x 2... x k (x 1 x 2 x k ). 1 1 1 A φ = (z 1 + z 2 + + 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 2006 27

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 2006 28

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 2006 29

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 2006 29

Polynomials and Simple Expressions Lemma Any QBF φ can be transformed in logarithmic space to an equivalent simple expression. Johannes Mittmann IP = PSPACE JASS 2006 30

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 2006 30

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 2006 30

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 2006 31

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 =0 1... z i =0 Johannes Mittmann IP = PSPACE JASS 2006 32

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 2006 33

The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 [ (1 z 1 ) + 1 1 z 2 =0 z 3 =0 ] (z 1 z 2 + z 3 ), Johannes Mittmann IP = PSPACE JASS 2006 34

The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z 2 1 + 1. [ (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 2006 34

The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z 2 1 + 1. [ (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 2006 34

The Interactive Protocol Example (Round 1) A A φ = a 2. 1 z 1 =0 A = (1 z 1 ) + q(z 1 ) = z 2 1 + 1. [ (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) = 10. 1 1 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 2006 34

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 2006 35

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. 1 A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z 2 2 + 3z 2. Johannes Mittmann IP = PSPACE JASS 2006 35

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. 1 A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z 2 2 + 3z 2. Verify a = 12 = 0 + 12 = q(0) + q(1). Johannes Mittmann IP = PSPACE JASS 2006 35

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. 1 A = (3 z 2 + z 3 ), z 3 =0 q(z 2 ) = 9z 2 2 + 3z 2. Verify a = 12 = 0 + 12 = q(0) + q(1). 1 A A (z 2 = 2) = (z 3 + 6), z 3 =0 a q(2) = 9 4 + 3 2 = 42. Johannes Mittmann IP = PSPACE JASS 2006 35

The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. Johannes Mittmann IP = PSPACE JASS 2006 36

The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z 3 + 6. Johannes Mittmann IP = PSPACE JASS 2006 36

The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z 3 + 6. Verify a = 42 = 6 7 = q(0) q(1). Johannes Mittmann IP = PSPACE JASS 2006 36

The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z 3 + 6. Verify a = 42 = 6 7 = q(0) q(1). A A (z 3 = 5) = 5 + 6 = 11, a q(5) = 5 + 6 = 11. Johannes Mittmann IP = PSPACE JASS 2006 36

The Interactive Protocol Example (Round 3) 1 A (z 3 + 6), z 3 =0 a 42. A = z 3 + 6, q(z 3 ) = z 3 + 6. Verify a = 42 = 6 7 = q(0) q(1). A A (z 3 = 5) = 5 + 6 = 11, a q(5) = 5 + 6 = 11. Verify A = 11 = a, and accept. Johannes Mittmann IP = PSPACE JASS 2006 36

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 2006 37

The Interactive Protocol Proof of soundness. Suppose A φ = 0 and still Alice claims an a 0. Johannes Mittmann IP = PSPACE JASS 2006 38

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 2006 38

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 2006 38

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 2006 38

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, 1973. L. Babai: E-mail and the unexpected power of interaction. Structure in Complexity Theory Conf., pp. 30-44, 1990. Adi Shamir: IP=PSPACE. Journal of the ACM, 39 (4), pp. 869-877, 1992. Christos H. Papadimitriou: Computational Complexity. Addison Wesley, Reading, 1994. Johannes Mittmann IP = PSPACE JASS 2006 39

Exercise Exercise Exercise Show that #P IP. Hint. Arithmetization of the #P-complete counting problem #SAT. Johannes Mittmann IP = PSPACE JASS 2006 40

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 2006 41