Arthur and Merlin as Oracles

Transcription

1 Update Meeting on Algorithms and Complexity, IMSc Arthur and Merlin as Oracles Lecturer: Venkat Chakravarthy Scribe: Bireswar Das In this talk we will discuss about two results 1) BPP NP P pram 2) S P 2 PprAM and some of the corollaries and extensions of the results Definition 1 (Promise Problems) A promise problem is a pair of sets (L +, L ) such that L +, L {0, 1} and L + L = φ When an oracle for a promise problem (L +, L ) is queried a query q it does the following: if q L + then it answers yes, if q L then it answers no and if q / L + L then it may answer arbitrarily An algorithm that queries a promise problem (L +, L ) must be robust enough to tolerate any answer for a query q that is not in L + L Definition 2 (Promise MA (prma)) A promise problem (L +, L ) is in the class prma if there exists a polynomial time computable predicate A and polynomials p and q with the property that x L + = ( y {0, 1} p( x ) )( z {0, 1} q( x ) )[A(x, y, z) = 1], and x L = ( y {0, 1} p( x ) )Pr z {0,1} q( x )[A(x, y, z) = 1] 1 2 Definition 3 (Promise AM (pram)) A promise problem (L +, L ) is in the class pram if there exists a polynomial time computable predicate A and polynomials p and q with the property that x L + = ( y {0, 1} p( x ) )( z {0, 1} q( x ) )[A(x, y, z) = 1], and x L = Pr y {0,1} p( x )[ z {0, 1} q( x ) A(x, y, z) = 1] 1 2 We will use the following result about approximate counting due to Stockmeyer [St83] Let C be a Boolean circuit with n + m input bits Let X = {x {0, 1} n y {0, 1} m C(x, y) = 1} Let 0 δ < 1 be a parameter The problem of approximate counting is to compute a number e such that (1 δ) X e X Theorem 4 (Stockmeyer) Approximate counting can be performed in P pram (in time poly( 1 δ )?) 1 BPP NP P pram Next we are going to prove that BPP NP P pram Let L be a language in BPP NP A BPP NP algorithm for L, on input x will have the following structure: Choose a random string y {0, 1} p( x ), where p is a polynomial Construct the queries Φ(y) = {φ 1,, φ K } ( Since the queries are made to NP oracle, we can assume without loss of generality that each φ i is a Boolean formula) Send Φ(y) to the oracle and get the answers Compute and output the answer 1

2 We might think of this algorithm as a circuit M that has access to an NP oracle and if x L then Pr y [M(x, y) = 1] 3/4 and Pr y [M(x, y) = 0] 1/4 and if x / L then Pr y [M(x, y) = 0] 3/4 and Pr y [M(x, y) = 1] 1/4 So, to decide if x L it is enough to compute or even approximate the size of the set {y M(x, y) = 1} But unfortunately testing whether a y (or Φ(y) = {φ 1,,φ K }) leads to 1 or 0 is not an NP computation Before we go into the details of the proof of BPP NP pram, we will amplify the success probability of the BPP NP algorithm for L We will assume that the new algorithm A with amplified success probability makes K queries and the error probability is bounded by 1 8K Let us denote the set of random string leading to wrong answer by BAD and the set of random strings leading to correct answer by GOOD We will have the following property BAD number of random strings 8K GOOD (1 1 8K ) number of random strings 3 number of random strings 4 Let y {0, 1} m be a random string used by the algorithm A on input x Let M = 2 m Let Φ(y) = {φ 1,, φ K } be the queries made by the algorithm Let N(y) be the number of satisfiable formulas in Φ(y) Clearly 0 N(y) K For j = 0,,K we define the jthe slice S j = {y {0, 1} m y GOOD and N(y) = j} Notice that the slices form a partition of the set GOOD We also define the sets Cr 1 and C0 r for r = 0,,K These sets are defined using simulation (possibly incorrect) of A on the input x The set Cr 1 consists of all random strings y such that the following NP computation accepts 1 Φ(y) = {φ 1,,φ K } 2 Guess l r formulas in Φ(y) 3 Verify that the guessed formulas are satisfiable 4 Assume that the rest K l formulas are unsatisfiable 5 Simulate A on x with these assumptions on the queries If A outputs 1 then accept The set Cr 0 is define similarly It consists of all random strings y such that the following NP computation accepts 1 Φ(y) = {φ 1,,φ K } 2 Guess l r formulas in Φ(y) 3 Verify that the guessed formulas are satisfiable 4 Assume that the rest K l formulas are unsatisfiable 5 Simulate A on x with these assumptions on the queries If A outputs 0 then accept It is easy to see that Lemma 5 The sets C 1 r and C0 r are in NP Note that for x L if y S j for some r j K then y Cr 1 Thus we have C1 r S r + + S K Also notice that if y S j for j = 0,,r then y / Cr 0 Hence, for x L C0 r S r S K + BAD If x / L then the role of Cr 1 and Cr 0 switches and we have Cr 0 S r + + S K and Cr 1 S r S K + BAD If x L then Cr 1 C0 r S r BAD and Cr 0 C1 r BAD If x / L then C0 r C1 r S r BAD and Cr 1 Cr 0 BAD Since the total number of slices is K + 1 and the total number of good strings is at least 3 4 M, by pigeon hole principle there is an index r 3 such that S r 4(K+1) M 3 8K M For x L this implies that Cr 1 C0 r S r 3 BAD 8K M 1 8K M = M 4K Thus we get 2

3 Proposition 6 1 x L = r ( C 1 r C0 r M 4K ) 2 x / L = r( C 1 r C0 r M 8K ) With the help of Proposition 6 we can now give a P pram algorithm for L for r [0, K] do approximately count Cr 0 and C1 1 r with the parameter δ = 32K with the help of a PprAM subroutine for approximate counting Let e 0 and e 1 be the approximate value of Cr 0 and C1 r respectively if e 1 e 0 3M 16K reject x accept x We notice that by Theorem 4 (1 1 32K ) C1 r e 1 Cr 1 1 and (1 32K ) C0 r e 0 Cr 0 Thus we have (1 1 32K )( C1 r C0 r ) 1 32K C0 r e 1 e 0 ( C 1 r C0 r ) K C0 r Using Proposition 6 and the fact that Cr 0 M we get e 1 e 0 3M 16K when x L and e 1 e 0 < 3M 16K when x / L Thus, we have proved: Theorem 7 BPP NP P pram 2 S P 2 P pram Definition 8 A language L is in the complexity class S P 2 if there is a polynomial-time predicate A and two polynomials p and q such that If x L then y {0, 1} p( x ) such that z {0, 1} q( x ), A(x, y, z) = 1 If x / L then z {0, 1} q( x ) such that y {0, 1} p( x ), A(x, y, z) = 0 Cai [Cai07] proved that the class S P 2 is contained in ZPPNP It contains the classes P NP and BPP (see Figure 1) The complexity class S P 2 has an easy formulation in terms of succinctly encoded bipartite tournaments described below BPP P NP P S 2 ZPP NP P pram Figure 1: The complexity class S P 2 Σ 2 p Π 2 p BPP NP Definition 9 (Bipartite Tournament) A bipartite tournament is a directed bipartite graph G = (R C, E) such that for all r R and c C either (r, c) E or (c, r) E 3

4 A bipartite tournament G = (R C, E) can be represented by a {0, 1}-matrix M of dimension R C, where the rows of M are indexed by the elements of R and the columns of M are indexed by the elements of C If (r, c) E then M[r, c] = 1 (in this case we say that row r beats column c), if (c, r) E then M[r, c] = 0 (and we say that column c beats row r) A row r that beats all the columns is said be a row side champion Similarly, a column that beats all the rows is called a column side champion Definition 10 (S 2 -type tournament) A bipartite tournament that either has a row side or column side tournament is called an S 2 -type tournament The matrix corresponding to a S 2 -type tournament is called an S 2 -type matrix Definition 11 (S 2 -Basic Problem) Input: An S 2 -type matrix Problem: Decide which side has a champion If the S 2 -type matrix is given explicitly then the above problem clearly admits a polynomial time algorithm But we can encode the S 2 -type matrix M succinctly by a circuit C as follows: The circuit takes a row r and a column c as input and outputs 1 if and only if M[r, c] = 1 We can see that a small circuit can encode a big S 2 -type matrix Definition 12 (S 2 -Basic Problem) Input: An S 2 -type matrix encoded as a circuit C Problem: Decide which side has a champion It is easy to see that solving S 2 -basic problem is enough to decide the membership problem of a language in S P 2 Given an S 2-type matrix the problem of finding a champion is a very hard problem as it is suggested by the following theorem Theorem 13 (Chakravarthy, Roy) If there exists a P pram algorithm for finding a champion in a succinctly encoded tournament, then PH collapses to BPP NP Definition 14 (Collective Champion (CC)) A set of rows Y is a row side CC, if every column z is beaten by at least one row from Y A set of columns Z is a column side CC, if every row y is beaten by at least one column from Z If a S 2 -type matrix has a row side CC then it must have a row side champion, because it cannot have a column side champion Similarly, if a S 2 -type matrix has a column side CC then it must have a column side champion Theorem 15 Given a succinctly encoded S 2 -type matrix M there is a P pram algorithm to find a row side or a column side CC The algorithm to find a CC uses two P pram algorithms as subroutine One algorithm is for finding a row side CC of size n from a S 2 -type matrix M of size 2 m 2 n which has a row side champion The other one is for finding a column side CC of size m from a S 2 -type matrix of size 2 m 2 n which has a column side champion When we run both the algorithms on the input S 2 -type matrix M, one of them must output a correct CC The other one may output arbitrarily But once we have a small set of rows or columns we can check if it is indeed a CC by querying an NP oracle Next we describe the algorithm to find a row side CC of size n from a S 2 -type 2 m 2 n matrix M which has a row side champion The algorithm proceeds as follows: Find a row r 1 that beats half the columns Find a row r 2 that beats half the remaining unbeaten columns, next find a row r 3 that beats half the remaining unbeaten columns and so on Since the input matrix has a row side champion the row r 1 exists If r 1 has not beaten all the rows then r 2 also exists and so on When the process ends we will have a row side CC of size at most n More precisely the algorithm looks as follows: Let Y := φ, UNBEATEN = [2 n ] 4

5 While UNBEATEN φ do Find a row r that beats at least half the columns in UNBEATEN Let BEATEN(r) be the set of columns in UNBEATEN that gets beaten by r Let Y := Y {r} and UNBEATEN := UNBEATEN BEATEN(r) Output Y Clearly Y must be a row side CC of size at most n One can check that the set UNBEATEN can be succinctly encoded as a circuit if we have the succinct encoding of M In each step the algorithm tries to find a good row Next we define the goodness µ of a row r µ(r) := BEATEN(r) UNBEATEN In each step the goal of the algorithm is to find a row r with µ(r) 1/2 A row r is said to extend a string α if α is a prefix of r We define the goodness µ of a string α as follows: µ(α) := max{µ(r) r extends α} We are going to find the row r that beats atleast half of the columns in UNBEATEN by extending the empty string in an iterative process Notice that µ(empty string) = 1 The first step is to find a bit b 1 such that µ(b 1 ) 1 ǫ Suppose we have found a prefix b 1 b 2 b k 1 of length k 1 with µ(b 1 b 2 b k 1 ) 1 (k 1)ǫ Next we are going to find a bit b k such that µ(b 1 b 2 b k 1 b k ) 1 (k 1)ǫ ǫ = 1 kǫ Notice that either µ(b 1 b 2 b k 1 0) 1 (1 k)ǫ or µ(b 1 b 2 b k 1 1) 1 (k 1)ǫ If we choose ǫ = 1 2n then after n steps we will have µ(b 1 b n ) 1 nǫ = 1/2 Suppose we have found a prefix α with µ(α) ρ The problem of finding the row now reduces to the problem of finding a bit b such that µ(αb) ρ ǫ We will have either µ(α0) ρ or µ(α1) ρ If b = 0 satisfies the condition that µ(α0) ρ then there exist a row r extending α0 such that µ(r) ρ If b = 0 does not satisfy the condition that µ(αb) ρ ǫ then for all row r extending α0, µ(r) < ρ ǫ So, we have to distinguish between the following two cases: 1 r such that {z UNBEATEN r beats z and extends α0} ρ UNBEATEN 2 r, {z UNBEATEN r beats z and extends α0} < (ρ ǫ) UNBEATEN The size of U can be approximated in P pram We can aslo distinguish between the two cases above with just one query to a P pram oracle Hence we obtain Theorem 16 S P 2 P pram 3 Corollaries Here we list some of the corollaries of the results BPP NP P pram Proposition 17 P pram BPP NP Corollary 18 BPP NP = P pram Proposition 19 P pram P ΣP 2 Theorem 20 BPP NP P ΣP 2 5

6 Theorem 21 BPP NP[log] P ΣP 2 [log] It is not known if BPP NP P ΣP 2 or BPP P NP Karp and Lipton [KL80] showed that if NP has polynomial size circuit then PH collapses to Σ p 2 Πp 2 Köbler and Watanabe [KW98] improved the result by showing that under the assumption, PH = ZPP NP Sengupta (see [Cai07]) further improved the result by showing that under the same assumption PH collapses to S P 2 Chakravarthy and Roy [CR08] showed that if NP P/ploy then PH collapses to P prma This result is an improvement over the result by Köbler and Watanabe because P prma ZPP NP The result is incomparable with Sengupta s result because we do not know if P prma S P 2 (Though it is known that PMA S P 2 ) The result gives the corollary that P prma SIZE(n k ) But Santhanam [Sa07] recently proved that prma SIZE(n k ) Under the hypothesis NP P/poly, Bshouty et al [BCG+96] proved that SAT circuits can be learnt in ZPP NP Chakravarty and Roy showed that under the same hypothesis SAT circuits can be learnt in P prma This an improvement over the previous result because P prma ZPP NP 4 Extensions 41 λ-strong CC Definition 22 For a S 2 -type matrix a set of rows X is called λ-strong CC if for every column c number of rows in X beating c X λ Theorem 23 (Chakravarthy, Roy) Given a S 2 -type matrix a λ-strong CC can be computed in P pram 42 Succinct Zero Sum Games The payoff matrix of a two player zero sum game with {0, 1}-payoff can be encoded succinctly with a Boolean circuit Let the two players be R and C The rows of the payoff matrix are indexed by the strategies of R, and the columns are indexed by the strategies of C The Boolean circuit encoding the payoff matrix can compute the payoff when R chooses strategy r and C chooses strategy c where the inputs to the circuit are r and c Problem Find near optimal strategies Fortnow et al [FIKU05] gave a ZPP NP algorithm for this problem Chakravarthy and Roy gave a P pram algorithm for this problem 5 Open Problems 1 P prma S P 2? 2 BPP NP P pram? (P pram P Σp 2 ) 3 Conditional discrepancy set in P pram? References [BCG+96] Nader H Bshouty, Richard Cleve, Ricard Gavald, Sampath Kannan, Christino Tamon: Oracles and Queries That Are Sufficient for Exact Learning J Comput Syst Sci 52(3): (1996) [Cai07] Jin-yi Cai: S P 2 is subset of ZPP NP J Comput Syst Sci 73(1): (2007) 6

7 [CR08] V Chakaravarthy, S Roy Finding Irrefutable Certificates for S p 2 Symposium on Theoretical Aspects of Computer Science, 2008 via Arthur and Merlin 25th [FIKU05] Lance Fortnow, Russell Impagliazzo, Valentine Kabanets, Christopher Umans: On the Complexity of Succinct Zero-Sum Games IEEE Conference on Computational Complexity 2005: [KL80] [KW98] Richard M Karp, Richard J Lipton: Some Connections between Nonuniform and Uniform Complexity Classes STOC 1980: Johannes Köbler, Osamu Watanabe: New Collapse Consequences of NP Having Small Circuits SIAM J Comput 28(1): (1998) [Sa07] Rahul Santhanam: Circuit lower bounds for Merlin-Arthur classes STOC 2007: [St83] Larry J Stockmeyer: The Complexity of Approximate Counting (Preliminary Version) STOC 1983: