Honors Thesis Bounded Query Functions With Limited Output Bits II

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1 Honors Thesis Bounded Query Functions With Limited Output Bits II Daibor Zeený University of Maryand, Batimore County May 29, 2007 Abstract We sove some open questions in the area of bounded query function casses with imited output bits. In particuar, we demonstrate a coapse of the Poynomia Hierarchy to Σ P 3 as a consequence of both PF NP [4] 3 PF NP[3] 3 and PF NP [5] 2 PF NP[3] 2. We aso generaize the resut for the former condition, and discuss some imitations of our generaization. 1 Introduction The study of bounded query casses gives some insight into the structure of the Poynomia Hierarchy. One way of studying bounded query casses is via the comparison of different orace access mechanisms. An orace Turing machine can ask its queries one at a time, and use answers to previous queries to determine the next query. We refer to such queries as seria or adaptive. If, instead, the Turing machine asks a its queries at once, and receives a the answers at once, we ca the queries parae. The cass of anguages decided by deterministic poynomia time Turing machines that make k seria queries to an NP orace is caed P NP[k], and if the deterministic poynomia time Turing machines make parae queries instead, we get the cass P NP []. We can repace P with PF to get casses of functions that are computed by the Turing machines mentioned earier. If we add a subscript to PF, for exampe PF NP [4] 3, we get casses of functions whose outputs have a imited ength. We give precise definitions of these compexity casses in the next section. Our work is motivated by some earier resuts. Beige [2] shows that P NP[k] = P NP [2k 1]. This is equivaent to saying that PF NP[k] 1 = PF NP [2k 1] 1. Beige, Kummer, and Stephan [3] show that this equaity doesn t hod for any number of output bits. If we disregard the number of output bits, Beige, Kummer, and Stephan [3] show that PF NP[k] = PF NP [2k 1] impies P = NP. Chang and Squire [7] demonstrate a coapse of the Poynomia Hierarchy to Σ P 3 for the specia case when PFNP[3] 2 = PF NP [3] 2, and generaize this resut. Their generaization eaves some open questions, some of which we resove in this paper. We show that PF NP [5] 2 PF NP[3] 2 and PF NP [4] 3 PF NP[3] 3 both cause a coapse of the Poynomia Hierarchy to Σ P 3. We aso generaize the proof of the coapse of PH to ΣP 3 for the case PFNP [4] 3 PF NP[3] 3, and get a generaization that encompasses more cases than the one done by Chang and Squire [7]. We gain this improvement by modifying the proof technique used by Chang and Squire [7]. Just ike they did, we find a function Q that can be computed using a PFm NP [] machine, and assume it can be computed by some PFm NP[k] machine. We use this assumption to get a P/poy m -reduction from BL to cobl, which causes a coapse of the Booean Hierarchy, which, in turn, causes a coapse of the Poynomia Hierarchy [10, 12]. We aso use the advisees technique which was first used by Amir, Beige, and Gasarch [1] to find one incorrect vaue of χ SAT k, which yieds a poynomia time agorithm for satisfiabiity [3] that wi be necessary to compete the reduction from BL to cobl. 1

2 Before we present our proofs, we cover some necessary definitions, notations and resuts in Section 2. After that, in Section 3, we sove the two open questions posed by Chang and Squire, and make a generaization in Section 4. Finay, we discuss some imitations of our work in Section 5. 2 Preiminaries We present some definitions necessary for the understanding of bounded query functions. The reader shoud be famiiar with basic compexity casses such as P, PF, NP, and P/poy. Knowedge of the NPcompete anguage SAT and poynomia-time many-one reductions ( P m-reductions) is aso assumed. We aso assume that the reader is famiiar with orace Turing machines and some compexity casses utiizing them, such as P NP and the Poynomia Hierarchy (PH). The reader can find these definitions in [9, 11]. Some other definitions are given ater in this section. First, we introduce some notation. If x is a string, we wi use x to denote the ength of the string. For a set S, we wi use S to denote the cardinaity of S. We use the notation x 1,...,x k to denote a k-tupe of strings. We use χ SAT (and χ SAT k in the mutivaued case) for the characteristic function of SAT. Aso, uness stated otherwise, a ogarithms have base 2. There are two main ways how an orace Turing machine can ask its queries. The queries can be asked in a series. We ca such queries seria or adaptive (these two terms wi be used interchangaby). Another way is to ask a the queries a at once (in parae), in which case we ca the queries parae. When a Turing machine uses adaptive queries, it uses answers to a previous queries, as we as the input string, to determine the next query. The computation of an orace Turing machine that uses adaptive queries can be viewed as a tree. We ca this tree the orace query tree. The orace query tree induced by the computation of a PF NP[2] 2 machine (one that computes a function with a two-bit output and makes two adaptive queries to its SAT orace) is shown in Figure 1. Chang and Squire [7] order outputs in the orace query tree induced by the computation of an orace Turing machine M on input x. The ordering of the outputs for the orace query tree in Figure 1 woud be 00,11,10,11. This ordering is caed the output sequence of M on input x. In genera, an output in eaf 1 comes before the output in eaf 2 in the output sequence if the query where paths to the two eaves spit is answered no on the path to eaf 1 and yes on the path to eaf 2. We borrow the notation from [7] and use OUT(M,x) = 00,11,10,11 to denote that the output sequence of M on input x is 00,11,10,00. Consider the output sequence of an orace Turing machine that makes adaptive queries. A subsequence of its output sequence consisting of a string of a zeros foowed by a string of a ones, or vice versa, is caed a mind change. In Figure 1, the first and second eaf form a mind change, and aso the second and the fourth eaf form a mind change. If we chain these two together, we can say that the machine makes two mind changes or that there is a chain of two mind changes in the output sequence of M on input x. An orace Turing machine that uses parae access to its orace wi compute a its orace queries and a truth tabe. It wi ask the queries at once and get the answers at once. It wi then consut the truth tabe to determine its next course of action. Now that we have compared the two basic orace access mechanisms, we can define compexity casses that wi be the objects of our study. Definition 1 (Bounded Query Function Casses) Let k,, m be positive integers. We define the foowing compexity casses: PF NP[k] is the cass of functions that can be computed by a poynomia-time Turing machine with an NP orace using at most k seria orace queries. PF NP [] is simiar to PF NP[k], but a orace queries have to be made at once (in parae) and a the answers wi be received at once. 2

3 Start here no q in SAT? 2 yes q in SAT? 1 q in SAT? 3 no yes no yes mind changes Figure 1: The orace query tree induced by the computation of a PF SAT[2] 2 machine. On the path that outputs 11, the machine does some computations and then asks whether q 2 is a satisfiabe booean formua. The SAT orace says that it is not. The machine uses this answer in its computation and eventuay computes another orace query, q 1. The orace says that the query is satisifiabe. The machine uses this answer and continues computing unti it outputs 11. The two arrows aso indicate the two mind changes made by this machine. PF NP[k] m is a cass of functions that output m bits and make at most k seria queries to an NP orace. A simiar definition can be made for function casses that make parae queries and have a imited amount of output bits. The anguage BL k is commony used as the P m-compete anguage for the k-th eve of the Booean Hierarchy. Individua eves of the Booean Hierarchy, as we as some of their compete anguages, can be found for exampe in [6]. Even more information about the Booean Hierarchy can be found in [4, 5]. The compete anguages contain k-tupes of booean formuas that satisfy the predicates shown beow. BL 1 = SAT BL 2k = { x 1,...,x 2k x 1,...,x 2k 1 BL 2k 1 x 2k SAT} BL 2k+1 = { x 1,...,x 2k+1 x 1,...,x 2k BL 2k x 2k+1 SAT} cobl k = { x 1,...,x k x 1,...,x k / BL k } We can expand the definitions above to get a concrete exampe of a P m-compete anguage for the fourth eve of the Booean Hierarchy. The anguage BL 4 is a set of 4-tupes of booean formuas x 1,x 2,x 3,x 4 that satisfy the predicate [(x 1 SAT x 2 / SAT) x 3 SAT] x 4 / SAT. We can see that a 4-tupe x 1,x 2,x 3,x 4 is in BL 4 if the rightmost satisfiabe formua in x 1,x 2,x 3,x 4 has an odd index. This hods for any BL k. The compement of BL 4 is the anguage cobl 4 that contains 4-tupes of booean formuas that satisfy the predicate [(x 1 / SAT x 2 SAT) x 3 / SAT] x 4 SAT. We ca a k-tupe F 1,...,F k of booean formuas nested if there exists an index i such that F 1,...,F i are a satisfiabe and F i+1,...,f k are a unsatisfiabe. We can convert any k-tupe F 1,...,F k of booean formuas into a nested one by etting F j = r j F r. Notice that the rightmost satisfiabe formua in F 1,...,F k has an odd index if and ony if the rightmost satisfiabe formua in F 1,...,F k has an odd index. Hence, F 1,...,F k BL k F 1,...,F k BL k. 3

4 We define ODD SAT k to be the anguage of k-tupes of booean formuas such that an odd number of formuas in the k-tupe is satisfiabe. Notice that for nested k-tupes F 1,...,F k, F 1,...,F k BL k F 1,...,F k cobl k. We define a function Q 43 PF NP [4] 3 that takes 4-tupes of booean formuas to bit strings of ength 3. The first bit of the function Q 43 wi be 1 if F 1,F 2,F 3,F 4 BL 4, and 0 otherwise. For nested inputs, we wi aso require the vaue of Q 43 for nested sequences F 1,F 2,F 3,F 4 to be 111 or 000. We are not interested in the exact vaues of Q 43 in a cases. Tabe 1 summarizes the cases we are interested in. We wi use this function ater in our proof. We wi assume that it can be computed by a PF NP[3] 3 machine and show that this wi et us reduce BL 4 to cobl 4, a key step in our proofs. F i SAT? F 1 F 2 F 3 F 4 Q Tabe 1: The vaues of Q 43 ( F 1,F 2,F 3,F 4 ) for some combinations of satisfiabiities of the four formuas. The ony restriction on vaues of Q 43 for other combintations of satisfiabiities of the four formuas is that the first bit of Q 43 indicate whether F 1,F 2,F 3,F 4 BL 4. We present two more resuts we wi use in our proofs. The first one is a modification of a emma used by Chang and Squire [7]. The second one is a specia case of a resut about enumerabiity proved by Beige, Kummer, and Stephan [3, Lemma 4.2]. We don t prove either of those. We prove a more genera version of Lemma 2 ater as Lemma 9, and we refer the reader to [3] for a proof of Theorem 3. Lemma 2 Let F 1,F 2,F 3,F 4 be a nested 4-tupe of booean formuas. Suppose that a PFNP[3] 3 machine M computes Q 43 and that OUT(M, F 1,F 2,F 3,F 4 ) doesn t contain 000,111,000,111, 000 as a subsequence. Then we can compute in poynomia time a 4-tupe G 1,G 2,G 3,G 4 of booean formuas such that F 1,F 2,F 3,F 4 BL 4 G 1,G 2,G 3,G 4 cobl 4. Theorem 3 If there exists a function f PF that outputs k bits, and χ SAT k ( x 1,...,x k ) f( x 1,...,x k ) for any k-tupe x 1,...,x k of booean formuas, then P = NP. 3 Proofs of Open Probems We prove that PF NP [4] 3 PF NP[3] 3 impies the coapse of the Poynomia Hierarchy to its third eve. The proof is simiar to the one in [7], but uses a different definition of advisees and uses it to eiminate one possibiity for χ SAT 2 ( H 1,H 2 ). Theorem 4 PF NP [4] 3 PF NP[3] 3 = PH = Σ P 3 4

5 Proof: It is easy to see that Q 43 can be computed by a poynomia time Turing machine N that makes at most 4 parae queries to an NP orace. We assume that it is aso computabe by a poynomia time Turing machine M that makes at most 3 seria queries to an NP orace, and show that in that case it is possibe to reduce BL 4 to cobl 4 using a poynomia-time many-one reduction with poynomia advice. By Kadin [10], this impies that SAT NP/poy, which causes a coapse of the Poynomia Hierarchy by Yap [12]. To compete the proof, it is necessary to construct a poynomia-time many-one reduction with poynomia advice that reduces BL 4 to cobl 4. In other words, the proof is competed by showing that if Q 43 PF NP[3] 3, BL 4 P/poy m cobl 4. Suppose that a PF NP[3] 3 machine M can compute Q 43. We use this fact to give a P/poy m reduction h from BL 4 to cobl 4. The reduction h gets a 4-tupe of booean formuas F 1,F 2,F 3,F 4 as input. First, it wi construct the nested version of the input, F 1,F 2,F 3,F 4. Let H 1,H 2 be booean formuas. We define the set ADVISEES( H 1,H 2 ) as the set of 4-tupes of booean formuas x 1,x 2,x 3,x 4 such that at east one of the foowing hods: 1. χ SAT 2 ( H 1,H 2 ) = 11 and OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) doesn t contain 000,111,000,111,000 as a subsequence. 2. χ SAT 2 ( H 1,H 2 ) = 10 and at east one of the eements of OUT 2 = {000,111,101,010} is not present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). 3. χ SAT 2 ( H 1,H 2 ) = 01 and at east one of the eements of OUT 1 = {000,010,100,011} is not present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). 4. χ SAT 2 ( H 1,H 2 ) = 00 and at east one of the eements of OUT 0 = {000,110,001} is not present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). Notice that h can compute OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) in poynomia time because there are ony a constant number of computation paths M can take. Suppose a 4-tupe F 1,F 2,F 3,F 4 is an advisee of H 1,H 2. For each of the four cases, we describe how to construct in poynomia time a 4-tupe G 1,G 2,G 3,G 4 of booean formuas such that F 1,F 2,F 3,F 4 BL 4 G 1,G 2,G 3,G 4 cobl If χ SAT 2 ( H 1,H 2 ) = 11, we can use Lemma 2 to get the desired 4-tupe of booean formuas. 2. There are four possibiities when χ SAT 2 ( H 1,H 2 ) = 10. (a) If 000 is missing in the output sequence, at east one of F 1,F 2,F 3,F 4 is satisfiabe. This is because if a were unsatisfiabe, M woud have to output 000 because it computes Q 43. However, 000 is not in the output sequence of M, so M cannot output it. This means that F 1,F 2,F 3,F 4 BL 4 (F 2 F 4 ) / SAT (F 3 SAT F 4 / SAT). We rewrite this as F 4 / SAT (F 3 SAT F 2 / SAT). We can fit this in the norma form for cobl 4 and et G 1 = F 2, G 2 = F 3, G 3 = F 4, G 4 = FALSE. We can use K-map simpification to make this concusion. Figure 2 iustrates the thought process. (b) If 111 is missing in the output sequence, χ SAT 4 ( F 1,F 2,F 3,F 4 ) xx00 where xx is not 00. It cannot be the case, then, that (F 1 F 2 ) SAT and (F 3 F 4 ) / SAT. If it were, we woud have χ SAT 4 ( H 1 F 1,H 2 F 2,F 3,F 4 ) = 1000, which woud require M to output 111 that is missing in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). It foows that F 1,F 2,F 3,F 4 BL 4 F 3 SAT F 4 / SAT, and we et G 1 = TRUE, G 2 = F 3, G 3 = F 4, G 4 = FALSE. Once again, we can use K-maps to see that. (c) If 101 is missing in the output sequence, χ SAT 4 ( F 1,F 2,F 3,F 4 ) xx10, so it is not the case that F 3 SAT and F 4 / SAT. Hence, F 1,F 2,F 3,F 4 BL 4 F 1 SAT (F 2 F 4 ) / SAT. We et G 1 = TRUE, G 2 = F 1, G 3 = F 2 F 4, G 4 = FALSE. 5

6 (a) K-map for BL D D (b) We know that χ SAT 4 ( F 1, F 2, F 3, F 4 ) is not 0000, so we repace that fied in the K-map with a don t care. The two highighted (in bodface and with arger font) groups cover a the ones. Figure 2: Using k-maps to find formuas G 1,G 2,G 3,G 4 such that F 1,F 2,F 3,F 4 BL 4 G 1,G 2,G 3,G 4 cobl 4. Coumns correspond to vaues of χ SAT 2 ( F 1,F 2 ), rows correspond to vaues of χ SAT 2 ( F 3,F 4 ). We see that F 1,F 2,F 3,F 4 BL 4 if and ony if (F 3 SAT F 4 / SAT) (F 1 SAT F 4 / SAT). (d) If 010 is missing in the output seqeunce, it cannot be the case that F 4 SAT, which means that F 1,F 2,F 3,F 4 BL 4 F 3 SAT (F 1 SAT F 2 / SAT). This means that G 1 = TRUE, G 2 = F 1, G 3 = F 2, G 4 = F There are four possibiities when χ SAT 2 ( H 1,H 2 ) = 01. (a) If 000 is missing in the output sequence, χ SAT 4 ( F 1,F 2,F 3,F 4 ) x000. That means that at east one of F 2,F 3,F 4 is satisfiabe. Hence, F 1,F 2,F 3,F 4 BL 4 F 3 SAT F 4 / SAT. Then et G 1 = TRUE, G 2 = F 3, G 3 = F 4, G 4 = FALSE. (b) If 010 is missing in the output sequence, that means that χ SAT 4 ( F 1,F 2,F 3,F 4 ) x100, so it cannot be the case that F 2 SAT and F 3 F 4 / SAT. Then F 1,F 2,F 3,F 4 BL 4 (F 3 SAT F 4 / SAT) (F 1 SAT F 4 / SAT). This is equivaent to saying that F 1,F 2,F 3,F 4 BL 4 F 4 / SAT ((F 1 F 3 ) SAT), and we et G 1 = TRUE, G 2 = F 3, G 3 = F 4, G 4 = FALSE. (c) If 100 is missing in the output sequence, it foows that χ SAT 4 ( F 1,F 2,F 3,F 4 ) xx10, so it cannot be the case that F 3 SAT and F 4 / SAT. In that case, F 1,F 2,F 3,F 4 BL 4 F 1 SAT (F 2 F 4 ) / SAT. Hence, we et G 1 = TRUE, G 2 = F 1, G 3 = F 2 F 4, G 4 = FALSE. (d) If 011 is missing in the output sequence, it cannot be the case that F 4 SAT. Then F 1,F 2,F 3,F 4 BL 4 F 3 SAT (F 1 SAT F 2 / SAT), and we et G 1 = TRUE, G 2 = F 1, G 3 = F 2, G 4 = F There are three possibiities when χ SAT 2 ( H 1,H 2 ) = 00. (a) If 000 is missing in the output sequence, χ SAT 4 ( F 1,F 2,F 3,F 4 ) xx00, so it is not the case that F 3 F 4 / SAT. Then F 1,F 2,F 3,F 4 BL 4 F 4 / SAT, so we et G 1 = F 4, and G 2 = G 3 = G 4 = FALSE. (b) If 110 is missing in the output sequence, it foows that χ SAT 4 ( F 1,F 2,F 3,F 4 ) xx10, so it cannot be the case that F 3 SAT and F 4 / SAT. In that case, F 1,F 2,F 3,F 4 BL 4 F 1 SAT (F 2 F 4 ) / SAT. Hence, we et G 1 = TRUE, G 2 = F 1, G 3 = F 2 F 4, G 4 = FALSE. (c) If 001 is missing in the output seqeunce, it cannot be the case that F 4 SAT. Then F 1,F 2,F 3,F 4 BL 4 F 3 SAT (F 1 SAT F 2 / SAT), and we et G 1 = TRUE, G 2 = F 1, G 3 = F 2, G 4 = F 3. Let H 1,H 2 and F 1,F 2,F 3,F 4 be given. Furthermore, suppose that F 1,F 2,F 3,F 4 is not an advisee of H 1,H 2. Then it must be the case that 6

7 1. χ SAT 2 ( H 1,H 2 ) = 11 impies that 000,111,000,111,000 appears as a subsequence in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). We say that the presence of 000,111,000,111,000 in the output sequence is an indicator of 11 as the vaue of χ SAT 2 ( H 1,H 2 ), or simpy the indicator of 11. Note that the presence of 000,111,000,111, 000 in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) does not impy that χ SAT 2 ( H 1,H 2 ) = 1,1. It merey indicates a possibiity 2. χ SAT 2 ( H 1,H 2 ) = 10 impies that a eements of OUT 2 = {000,111,101,010} are present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). The presence of a eements of OUT 2 in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) is an indicator of χ SAT 2 ( H 1,H 2 ) = 01 impies that a eements of OUT 1 = {000,010,100,011} are present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). The presence of a eements of OUT 1 in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) is an indicator of χ SAT 2 ( H 1,H 2 ) = 00 impies that a eements of OUT 0 = {000,110,001} are present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). The presence of a eements of OUT 0 in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) is an indicator of 00. Notice that for any pair of H 1,H 2 and F 1,F 2,F 3,F 4, it cannot be the case that OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ) contains 000,111,000,111, 000 as a subsequence and aso contains a of the eements of a of the three sets OUT i. This is because there are ony eight computation paths in the orace query tree of M, and M woud have to output 000 or 111 on five of them, which woud eave us with ony three paths for the remaining six bit strings of ength 3. We define INCORRECT(H 1,H 2,F 1,F 2,F 3,F 4 ) to be a bit string of ength 2 whose indicator is not present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ). If more than one indicator is not present, we pick an arbitrary bit string of ength 2 whose indicator is not present in the output sequence. Notice that if F 1,F 2,F 3,F 4 is not an advisee of H 1,H 2, the indicator of the correct vaue of χ SAT 2 ( H 1,H 2 ) is present in OUT(M, H 1 F 1,H 2 F 2,F 3,F 4 ), so if F 1,F 2,F 3,F 4 is not an advisee of H 1,H 2, INCORRECT(H 1,H 2,F 1,F 2,F 3,F 4 ) χ SAT 2 ( H 1,H 2 ). We now construct poynomia advice that wi et us reduce BL 4 to cobl 4 using a P/poy m reduction. We assume that an OR of 4 formuas of ength n has ength at most 9n, and that an OR of 2 formuas of ength n has ength at most 5n. We make this assumption because the formuas we wi dea with wi be ORs of either 2 or 4 (or ess) formuas. The construction starts with sets S 0 consisting of a 4-tupes of booean formuas of ength at most 9n and T 0 consisting of a 2-tupes of booean formuas of ength at most 5n, and proceeds in steps, starting with step 0. We describe the i-th step of the construction. Step i: For each H 1,H 2 T i, find A i ( H 1,H 2 ) = ADVISEES( H 1,H 2 ) S i. There are two cases to consider. 1. There is a 2-tupe H 1,H 2 T i such that A i S i /32. We pick one such tupe H 1,H 2, and put H 1,H 2 and χ SAT 2 ( H 1,H 2 ) in the advice. We aso remove H 1,H 2 from T i to form T i+1, and remove A i ( H 1,H 2 ) from S i to get S i+1. If S i+1 16, we put a 4-tupes x 1,x 2,x 3,x 4 S i+1 and χ SAT 4 ( x 1,x 2,x 3,x 4 ) in the advice. After that, we terminate the advice construction, and indicate in the advice that the construction terminated in case 1. We et S = S i+1 and T = T i For a 2-tupes H 1,H 2 T i, A i < S i /32. This means that for a H 1,H 2 T i, Prob x Si [x ADVISEES( H 1,H 2 )] < 1/32. Then there exists a sequence s with a poynomia number of 4-tupes x 1,x 2,x 3,x 4 S i such that for a 2-tupes H 1,H 2 T i, we have INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for ess than 1/4 of the 4-tupes in s. We put x 1,x 2,x 3,x 4 and χ SAT 2 ( x 1,x 2,x 3,x 4 ) in the advice for each eement of the poynomia-size sequence s and terminate the advice construction. We indicate in the advice that its construction terminated in case 2. We et S = S i and T = T i. 7

8 We show that the poynomia sequence s of eements of S i mentioned in the previous paragraph exists. First note that T 2 10n. For a eements H 1,H 2 T we have Prob x S [x ADVISEES( H 1,H 2 )] < 1/32 by construction. We show that there exists a sequence of 44n tupes in S such that for a H 1,H 2 T, we have INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for at most 11n of those 4-tupes. Let H 1,H 2 be given. Suppose we pick uniformy at random 44n + 4 eements x 1,x 2,x 3,x 4 of S and compute INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) for each of them. We are interested in an upper bound on the probabiity that INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for 11n + 1 or more of the eements we picked. The foowing theorem can be found in [8] as Theorem C.2. Theorem 5 The probabiity that at east k out of n Bernoui trias are successfu is at most ( n k) p k, where p is the probabiity of success for an individua tria. Using Theorem 5, we see that the probabiity that INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for 11n + 1 or more of the eements we picked is at most ( )( 44n n ) 11n+1 ( ) 1 55n n+4 = 2 ( ) 1 11n+1. 2 Then the probabiity that for a particuar sequence of 44n + 4 eements of S there exists a 2-tupe H 1,H 2 such that INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for at east 11n+1 4-tupes x 1,x 2,x 3,x 4 in that sequence is at most 2 10n ( 1 2) 11n+1 = 2 10n 11n 1 = ( ) 1 n+1 < for a n 1. This impies that there exists a sequence of 4-tupes from S of ength 44n + 4 such that for every H 1,H 2 / T, INCORRECT(H 1,H 2,x 1,x 2,x 3,x 4 ) = χ SAT 2 ( H 1,H 2 ) for at most 11n of the 4-tupes in that sequence. We now show how to use the advice to reduce BL 4 to cobl 4 in poynomia time. Our goa is to construct a 4-tupe G 1,G 2,G 3,G 4 of booean formuas such that x 1,x 2,x 3,x 4 BL 4 if and ony if G 1,G 2,G 3,G 4 cobl 4. Assume that a PF NP[3] 3 machine M computes Q 43. We can assume without oss of generaity that a formuas in x 1,x 2,x 3,x 4 have the same ength, n. On input x 1,x 2,x 3,x 4 : 1. Construct the nested version of x 1,x 2,x 3,x 4, x 1,x 2,x 3,x For each 2-tupe H 1,H 2 in the advice, compute OUT(M, H 1 x 1,H 2 x 2,x 3,x 4 ). If x 1,x 2,x 3,x 4 turns out to be an advisee of some H 1,H 2 in the advice, we can construct G 1,G 2,G 3,G 4 using the method we discussed earier. 3. If advice construction terminated in case 1 of the construction and we didn t constrcut G 1,G 2,G 3,G 4 in the previous step, this means that x 1,x 2,x 3,x 4 is in the advice together with χ SAT 4 ( x 1,x 2,x 3,x 4 ). In that case we output a trivia G 1,G 2,G 3,G 4 (G 1 = G 2 = G 3 = G 4 = TRUE if x 1,x 2,x 3,x 4 BL 4 and G 1 = G 2 = G 3 = TRUE, G 4 = FALSE if x 1,x 2,x 3,x 4 / BL 4 ). 4. If advice construction terminated in case 2 of the construction, we first check if at east one of the eements in the 4-tupe x 1,x 2,x 3,x 4 is in the advice string as one of the formuas. If it is, we aso know whether it s satisfiabe or not, and we have enough information to construct G 1,G 2,G 3,G 4. 8

9 5. If none of the previous four cases appy, we use the advice string to decide the membership of each of x 1,x 2,x 3,x 4 in SAT in poynomia time using Theorem 3. A we need to show is that our advice enabes us to compute a function f( y 1,y 2 ) that outputs 2 bits and f( y 1,y 2 ) χ SAT 2 ( y 1,y 2 ) for any booean formuas y 1, y 2 of ength at most 5n. Either y 1,y 2 is in the advice string together with its characteristic function, or it had ess than S advisees in S at the end of advice construction. In the former case, we know χ SAT 2 ( y 1,y 2 ) and can pick any arbitrary vaue that s not χ SAT 2 ( y 1,y 2 ) as f( y 1,y 2 ). In the atter case, we compute INCORRECT(y 1,y 2,F 1,F 2,F 3,F 4 ) for each 4-tupe in the advice. By construction, any vaue that appears as the output of INCORRECT for at east 1/4 of the 4-tupes in the advice cannot be the correct vaue of χ SAT 2 ( y 1,y 2, so we pick any such vaue as f( y 1,y 2 ). Since the advice has poynomia ength and we can compute indicators in poynomia time, we can compute f in poynomia time. This competes the proof. Now we turn our attention to the other open probem proposed by Chang and Squire [7]. We wi show that the containment PF NP [5] 2 PF NP[3] 2 aso impies a coapse of the Poynomia Hierarchy. This proof wi be even more simiar to the one done by Chang and Squire [7] than the one we just did. Theorem 6 PF NP [5] 2 PF NP[3] 2 = PH = Σ P 3 Proof Sketch: We wi modify the proof of Theorem 4. We wi define the first bit of the function Q 52 ( x 1,x 2,x 3,x 4,x 5 ) to be 1 if x 1,x 2,x 3,x 4,x 5 BL 5 and 0 otherwise. The second bit wi indicate whether x 1,x 2,x 3,x 4,x 5 ODD SAT 5. This function is computabe by a PF NP [5] 2 machine N. We wi assume that it is aso computabe by a PF NP[3] 2 machine M. Since this function is different from the one used in the proof of Theorem 4, we have to rephrase Lemma 2. Lemma 7 Suppose that a PF NP[3] 2 machine M can compute Q 52 for nested inputs F 1,F 2,F 3,F 4,F 5. If its output sequence on input F 1,F 2,F 3,F 4,F 5 doesn t contain 00,11,00,11,00,11 as a subsequence, there exists a 5-tupe G 1,G 2,G 3,G 4,G 5 of booean formuas computabe in poynomia time such that F 1,F 2,F 3,F 4,F 5 BL 5 G 1,G 2,G 3,G 4,G 5 cobl 5. Instead of using 2-tupes H 1,H 2 of booean formuas, we wi have singe booean formuas H as advisors of 5-tupes of booean formuas. We wi say that x 1,x 2,x 3,x 4,x 5 of booean formuas is an advisee of H if at east one of the foowing happens. 1. H SAT and OUT(M, H F 1,F 2,F 3,F 4,F 5 ) doesn t contain 00,11,00,11,00, 11 as a subsequence. 2. H / SAT and OUT(M, H F 1,F 2,F 3,F 4,F 5 ) contains 00,11,00,11,00, 11 as a subsequence. We show how construct a 5-tupe G 1,G 2,G 3,G 4,G 5 that beongs to cobl 5 if and ony if F 1,F 2,F 3,F 4,F 5 BL 5, where F 1,F 2,F 3,F 4,F 5 is an advisee of H. 1. If H SAT is an advisor of F 1,F 2,F 3,F 4,F 5, we use Lemma If H / SAT is an advisor of F 1,F 2,F 3,F 4,F 5, we notice that the output sequence of M on input H F 1,F 2,F 3,F 4,F 5 contains at most two terms that are neither 00 nor 11. Without oss of generaity assume there are two of them, and ca them a and b. Notice that Q 52 ( H F 1,F 2,F 3,F 4,F 5 ) cannot be 11 because if the rightmost satisfiabe formua in Q 52 ( H F 1,F 2,F 3,F 4,F 5 ) has an even index, then, since F 1,F 2,F 3,F 4,F 5 is nested, H F 1,F 2,F 3,F 4,F 5 / ODDSAT 5. Now there are three cases to consider. 9

10 (a) If a = b = 10, we know that H F 1,F 2,F 3,F 4,F 5 / ODDSAT 5 because a eaves of the orace query tree induced by the computation of M on input H F 1,F 2,F 3,F 4,F 5 that hod the right vaue of Q 52 have 0 as its second bit, which means that the rightmost satisfiabe formua in H 1,H 2,H 3,H 4,H 5 doesn t have an even index. This means that F 1,F 2,F 3,F 4,F 5 BL 5 if and ony if at east one of F 1,F 3,F 5 is satisfiabe. We et G 1 = G 2 = G 3 = TRUE, G 4 = F 1 F 3 F 5, and G 5 = FALSE. (b) If a = b = 01, it must be the case that F 4 / SAT F 5 SAT because 10 is not present in the output sequence. Then F 1,F 2,F 3,F 4,F 5 BL 5 if and ony if [F 1 SAT (F 2 F 4 ) / SAT] F 3 SAT, and we et G 1 = TRUE, G 2 = F 1, G 3 = F 2 F 4, G 4 = F 3, and G 5 = FALSE. (c) If a = 10 and b = 01 or vice versa, 01 appears ony once in the output sequence of M on input H F 1,F 2,F 3,F 4,F 5. The machine M asks queries q 1, q 2, and q 3 on the computation path of on which 01 is output. For simpicity, ca this path p. Some of the queries on path p are answered yes, some no. We take a the yes queries and AND them together to get a booean formua τ, and OR a the no queries together to get a booean formua ϕ. The machine M wi foow path p if τ SAT and ϕ / SAT. Notice that if p is the path that M takes, H F 1,F 2,F 3,F 4,F 5 ODDSAT 5, which means F 1,F 2,F 3,F 4,F 5 / BL 5, so F 1,F 2,F 3,F 4,F 5 BL 5 ony if p is not the path that M takes on input F 1,F 2,F 3,F 4,F 5. If p is not the path that M takes, we re in case (a), and F 1,F 2,F 3,F 4,F 5 BL 5 if and ony if F 1 F 3 F 5 SAT. Then F 1,F 2,F 3,F 4,F 5 BL 5 if and ony if (τ / SAT ϕ SAT) (F 1 F 3 F 5 ) SAT, which happens if and ony if (τ / SAT (F 1 F 3 F 5 ) SAT) (ϕ (F 1 F 3 F 5 )) SAT. We et G 1 = TRUE, G 2 = F 1 F 3 F 5, G 3 = τ, G 4 = ϕ (F 1 F 3 F 5 ), and G 5 = FALSE. The advice is constructed the same way as in the proof of Theorem 4. It is used differenty, however. Instead of eminating one vaue of χ SAT 2 ( H 1,H 2 ) as we did in the proof of Theorem 4, we wi use the idea of Chang and Squire [7], and notice that if H is not an advisor of F 1,F 2,F 3,F 4,F 5, H SAT if and ony if OUT(M, H F 1,F 2,F 3,F 4,F 5 ) contains 00,11,00,11,00,11 as a subsequence. The poynomia-ength sequence of 5-tupes in the advice wi, instead, have the property that if H is not in the advice, H wi be satisfiabe if and ony if for a three-fourths majority of 5-tupes in the advice, OUT(M, H F 1,F 2,F 3,F 4,F 5 ) contains 00,11,00,11,00, 11 as a subsequence. An argument simiar to the one made in the previous proof can be made to show that such poynomia-size set of 5-tupes exists. 4 Generaization The technique used in the proof of Theorem 4 in the previous section can be generaized to prove the foowing theorem. Theorem 8 For a k, and m such that 1 < m k < 2 k 1, if > 2 k 2 m + 1 then PFm NP [] PFm NP[k] PH = Σ P 3. Proof: We need to make a few modifications to the proof of Theorem 4 from the previous section. The necessary modifications are isted beow. 1. Define a function Q simiar to Q Generaize Lemma Generaize the definition of ADVISEES() using Q and the PF NP[k] m machine M that can compute it. 4. Show how to reduce BL to cobl when some -tupe of booean formuas is an advisee. 10

11 5. Revise the construction of the advice. 6. Argue that if advice construction terminates in case 2, that is, with some -tupes that neither have an advisor in the advice nor are in the advice themseves, we can sti, given an -tupe of formuas x 1,...,x, eiminate one possibiity for χ SAT ( x 1,...,x ). We carry out these steps in the order isted. We start with a generaization of the function Q 43. We define the function Q to be a function mapping -tupes F 1,...,F of booean formuas to binary strings of ength m. If an -tupe F 1,...,F of booean formuas is nested, a bits of the output are the same and indicate whether F 1,...,F BL. For non-nested inputs, Q outputs the ast m bits of χ SAT ( F 1,...,F ). The second part can be done because > m, and we do it because we want Q to be onto, as we see ater. It is aso easy to see that Q PFm NP []. Finay, et s be a bit string of ength m. Notice that a non-nested k-tupes F 1,...,F for which χ SAT m ( F m+1,...,f ) = s wi map to the same string of ength m under Q, which is another desired property of Q. We show that Q is an onto mapping. Pick a bit string s of ength m. We wi find a non-nested -tupe of booean formuas F such that Q(F) = s. There exists an m-tupe F 1,...,F m of booean formuas such that ( F 1,...,F m ) = s. We append this m-tupe at the end of an ( m)-tupe FALSE,...,FALSE. Now F = Q( FALSE,...,FALSE,F 1,...,F m ) is an -tupe of booean formuas that maps to s if F is non-nested, which happens if at east one of the formuas in F 1,...,F m is satisfiabe. If a formuas in F 1,...,F m are unsatisfiabe, F is a nested -tupe of unsatisfiabe booean formuas. Then Q(F) = s = because F / BL. χ SAT m Lemma 9 (Generaization of Lemma 2) Let F 1,...,F be a nested -tupe of booean formuas. Suppose that a PFm NP[k] machine M computes Q and that OUT(M, F 1,...,F ) doesn t contain as a subsequence a sequence of ength + 1 with a chain of mind changes starting with a string of a zeros. Then we can compute in poynomia time an -tupe G 1,...,G of booean formuas such that F 1,...,F BL G 1,...,G cobl. Proof: A specia case of the proof was sketched in [7]. We give a proof for the genera case. Nodes in the orace query tree correspond to queries made by M. If a query is answered no, M wi take the path that goes to the eft chid of the query that was answered no. Otherwise it wi take the path that goes right. We wi abe the queries using an inorder traversa of the orace query tree. The eftmost query wi be caed q 1 ; the rightmost one, q 2 k 1. We wi abe each eaf in the orace query tree with the index of the query with the highest index that was answered yes on the path to that eaf. Figure 3 demonstrates this notation using a PF NP[3] 2 machine as an exampe. q 2 no q 4 yes q 6 Output: Changed: Labe: no q 1 q 3 q 5 yes q Figure 3: We abe eaves of the orace query tree induced by the computation of a PF NP[3] 2 machine based on the index of the highest query that was answered yes on the path to that eaf. If a query is answered yes, the machine foows the path that goes to the right from the node representing that query. Otherwise it foows the path that goes eft. Since ony outputs 00 and 11 make sense on nested inputs, we change outputs other than those to the nearest output that is 00 or 11, breaking ties arbitrariy. 11

12 Figure 3 aso shows outputs made by the PF NP[3] 2 machine. However, M can output ony a-zero or a-one strings on nested inputs. Therefore, without oss of generaity, we can make any output that is neither a zeros nor a ones the same as the a-zero or a-one output that s in the nearest eaf, breaking ties arbitrariy. This is aso demonstrated in Figure 3. We can do this because we know that M wi not foow any path eading to an output that is neither a zeros nor a ones. This aso doesn t affect the ength of the ongest chain of mind changes. As a consequence, we now have bocks of a-zero or a-one strings in the modified eaves of the orace query tree. Define ϕ i to be the AND of a queries q j that are answered yes on the path to eaf i. We define ϕ 0 = TRUE because a queries on the path to eaf 0 are answered no. Finay, we define φ i = j iϕ j. To give some exampes, consider the orace query tree in Figure 3. We wi have ϕ 7 = q 4 q 6 q 7, ϕ 6 = q 4 q 6, ϕ 5 = q 4 q 5, φ 6 = (q 4 q 6 q 7 ) (q 4 q 6 ), and φ 5 = (q 4 q 6 q 7 ) (q 4 q 6 ) (q 4 q 5 ). Notice that F 1,...,F BL ony if M outputs a string of a ones. We can aso say that F 1,...,F BL ony if M foows a path where we set the output to be a ones (even though it may have been something ese originay). Reca that there is no chain of mind changes starting with an a-zeros string in the output sequence of M. Suppose that a chains of mind changes in the output sequence of M start with an a-ones string. Aso reca that a strings in the output sequence that were not a-zero or a-one were changed to the nearest a-zero or a-one string. This creates + 1 custers of identica strings in the output sequence of M. Then F 1,...,F BL if M foows a path that eads to some custer of a-one strings. Suppose that the first eaf in a custer of a-one strings has index i and that the ast eaf in that custer has index I. Then M takes a path eading to that custer if φ i SAT and φ I+1 / SAT. In Figure 3, the machine M wi foow a path that eads to the ony custer of a-one outputs if φ 2 SAT and φ 4 / SAT. If i = 0, we drop the φ i SAT requirement, and if I = 2 k 1, we drop the φ I+1 / SAT requirement. There are /2 1 custers of a-one strings for which neither i = 0 nor I = 2 k 1. The first custer of a-one strings has i = 0. If is even, there is aso an additiona custer of a-one strings with I = 2 k 1. We give each custer (from eft to right) a number j {0,1,..., /2 1} if is odd and j {0,1,..., /2 } if is even, and ca i j the eftmost eaf in custer j and I j the rightmost eaf in custer j. If is odd, F 1,...,F BL if φ I0 +1 / SAT (φ i1 SAT φ I1 +1 / SAT) (φ i /2 1 SAT φ I /2 1 +1), which can be rewritten as ( [ ) (φ I0+1 / SAT φ SAT) φ i1 I1+1 / SAT] φ i /2 1 SAT φ I / (1) because φ 0,...,φ 2 k 1 is nested. If is even, we get that F 1,...,F BL if [( [ ) ] (φ I0+1 / SAT φ SAT) φ i1 I1+1 / SAT] φ i /2 1 SAT φ I / φ i /2. (2) But then if is odd, F 1,...,F BL if and ony if φ I0 +1,φ i1,φ I1 +1,...,φ i /2 1,φ I / cobl. Simiary, if is even, F 1,...,F BL if and ony if φ I0 +1,φ i1,φ I1 +1,...,φ i /2 1,φ I /2 1 +1,φ i /2 cobl. Hence, we et G 1 = φ I0 +1, G 2 = φ i1, and so on unti G = φ I / if is odd and G = φ i /2 if is even. If the ongest chain of mind changes is shorter than, we remove parts of the expressions in (1) or (2) and the formuas G 1,...,G can sti be constructed. Fix an -tupe H 1,...,H of booean formuas. Let χ SAT ( H 1,...,H ) = t and et s 1,...,s r be a possibe strings that can be output by Q on inputs F 1 H 1,...,F H where F 1,...,F is the 12

13 nested version of some -tupe F 1,...,F. The presence of a the strings s 1,...,s r in OUT(M, F 1 H 1,...,F H ) is an indicator that χ SAT ( H 1,...,H ) = t. As in the previous section, the indicator that χ SAT ( H 1,...,H ) = is the occurrence of a chain of mind changes, starting with , in OUT(M, F 1 H 1,...,F H ). We say that F 1,...,F is an advisee of H 1,...,H if the indicator of the true vaue of χ SAT ( H 1,...,H ) is not present in OUT(M, F 1 H 1,...,F H ) in its entirety. Now we need to show how to find an -tupe G 1,...,G of booean formuas such that F 1,...,F BL G 1,...,G cobl, given that F 1,...,F is an advisee of H 1,...,H. Fix a nested -tupe x 1,...,x and suppose that Q( x 1 H 1,...,x H ) = s is not in OUT(M, F 1 H 1,...,F H ). Let j be the argest integer such that H j x j is satisfiabe, and et J be the smaest integer greater than j such that H j SAT. Tabe 2 iustrates the definitions of j and J. Index (i) x i SAT? H i SAT? Tabe 2: Exampes of the definitions of j and J from the paragraph above. The highest integer such that H j x j is 4. The eftmost integer greater than j such that H J SAT is 7. Hence, we have j = 4, J = 7. Suppose that j doesn t exist for any F 1,...,F. Then χ SAT ( H 1,...,H ) = , and H 1,...,H cannot be an advisor for any -tupe of formuas. If it were, M woudn t compute Q because the indicator that χ SAT ( H 1,...,H ) = is the singeton set { } and M has to be abe to output the correct vaue of Q. Hence, we can assume that j aways exists. First assume that J exists. Notice that J 1 because that woud make i 0, which cannot happen. Reca that we defined s = Q( x 1 H 1,...,x H ) for some k-tupe x 1,...,x and used the tupe x 1,...,x to define j and J. Since s is not in OUT(M, F 1 H 1,...,F H, it must be the case that either F j F J 1 / SAT or that F J F SAT. Assume that s does not appear in OUT(M, F 1 H 1,...,F H ). Suppose that F j F J 1 SAT and F J F / SAT. Then because F 1,...,F is nested, we must have F j SAT. By definition of j and because we assume that a of F J,...,F are unsatisfiabe, H j+1 F j+1,...,h F are a unsatisfiabe. Aso, we have H i F i SAT H i x i SAT for a i { m+1,...,} by the choice of j and J, and because x 1,...,x and F 1,...,F are nested. Then Q( H 1 F 1,...,H F ) = Q( H 1 x 1,...,H x ) = s. This is a contradiction because M computes Q and s is not in the output sequence of M on input H 1 F 1,...,H F, so M can t output s. Hence, either F j F J 1 / SAT or F J F SAT. Therefore, the index of the rightmost satisfiabe formua in F 1,...,F is either ess than j or at east J, which means that the membership of F 1,...,F in BL does not depend on the satisfiabiity of a formuas in F 1,...,F. This is sufficient for a reduction from BL to cobl, which is what we show next. Now we find a reduction for the genera case when j < and when J exists and is greater than 1. Since either F j F J 1 / SAT or F J F SAT hods, it must be the case that F 1,...,F BL if and ony if or i odd i<j ( ) (F r / SAT) J r [ ( (F i SAT) ( i odd J+1<i i<r )] (F r / SAT) [ ( (F i SAT) i<r (3) (F r / SAT))] ). (4) Assume that j, J and are a even. We wi dea with a other possibiities ater. We notice that both (3) and (4) hod ony if F / SAT. Then at east one of (3) and (4) hods ony if F / SAT and if at east 13

14 one of the two statements beow is true. [ ( (F i SAT) i odd i<j ( ) (F r / SAT) J r 2 ( i odd J+1<i i<r 2 )] (F r / SAT) [ ( (F i SAT) i<r 2 (5) (F r / SAT))] ) (6) Notice that (6) is true if F 1 SAT. We can then say that at east one of (3) and (4) hods if F / SAT and if either (5) hods or the condition beow hods. ( ) ( [ ( (F 1 SAT) (F r / SAT) (F i SAT) (F r / SAT))] ) (7) J r 2 i<r 2 i odd J+1<i 2 Then F 1,...,F BL if and ony if [F / SAT (F 1 SAT ((5) hods (7) hods))]. We can further expand the condition (5) hods (7) hods to concude that F 1,...,F BL if and ony if F / SAT (F 1 SAT (F J+3 SAT X) ) where X is true if ( i odd i<j [ ( (F i SAT) (F r / SAT))] ) i<r J+2 ( (F r / SAT) J r J+2 ). (8) We can further rewrite (8) as ( ( ) ) ( [ ] ) F r / SAT F j 1 SAT F j 2 / SAT ((F 1 SAT) (TRUE / SAT)) (9) j r J+2 Putting this a together, we get F 1,...,F BL if and ony if F / SAT (F 1 SAT (F J+3 SAT (9) hods) ). This is a statement about membership in cobl. Hence, we have a reduction from BL to cobl. We can make a simiar argument for other combinations of parities of j, J and, and aso when J doesn t exist at a. We have > 2 k 2 m + 1. A PFm NP[k] machine M has 2 k computation paths, which means that its output sequence has 2 k terms. If the output sequence of M contained indicators of a possibe vaues of χ SAT ( H 1,...,H ) on input H 1 x 1,...,H x, it woud have to output a strings of ength m because Q is onto and every bit string of ength m beongs to at east one indicator (take x 1,...,x = TRUE,...,TRUE ; then χ SAT ( H 1 x 1,...,H x ) = χsat ( H 1,...,H )). Its output sequence woud have to have at east m 2 > 2 k 2 m m 2 = 2 k terms, which can t happen. The + 1 comes from the chain of mind changes indicating χ SAT ( H 1,...,H ) = , the remaining 2 m 2 are the remaining bit strings of ength m that are not present in the indicator of This impies that the indicator of at east one vaue of χ SAT won t be present in its entirety in the output sequence of M on any input. We have to modify the construction of the advice. We now put an -tupe of booean formuas from T i (the set of potentia advisors that have not been put in the advice yet) into the advice if T i is an advisor for at east S i eements of S i (reca that S i is the set of -tupes of booean formuas for which we have not found an advisor yet during the construction of the advice). When advice construction terminates in case 2, a eements of T i wi advise ess than S i -tupes in S i. 14

15 When advice construciton terminates in case 1, we get a reduction using the argument presented earier. Suppose advice construction terminates in case 2 and that we fix an -tupe H 1,...,H of booean 1 formuas in T i. We can find the probabiity that at east q(n) out of q(n) randomy picked -tupes 2(2 1) from S i are advisees of H 1,...,H. This aso gives us an upper bound on the probabiity that at east q(n)/2 out of q(n) randomy picked -tupes from S i are advisees of H 1,...,H. Aso suppose that p (n) is the maximum ength of an OR of booean formuas of ength n. We can pick q(n) so that the probabiistic argument in the proof of Theorem 4 sti goes through, i.e., we can find a sequence of q(n) -tupes from S i such that if H 1,...,H / T i, OUT(M, H 1 x 1,...,H x ) doesn t contain the indicator of the correct vaue of χ SAT ( H 1,...,H ) for ess than q(n)/2 of those -tupes x 1,...,x. This wi ensure that for at east one of the incorrect vaues of χ SAT ( H 1,...,H ), the indicator of this vaue wi not be present in OUT(M, H 1 x 1,...,H x ) in at east q(n)/2 cases. We wi pick this indicated vaue as an incorrect vaue of χ SAT ( H 1,...,H ), and get a poynomia time agorithm for satisfiabiity. Now we have a the machinery that s necessary to compete the proof of Theorem 8. The agorithm that gives a reduction resembes the one given for the specia case PF NP [4] 3 PF NP[3] 3 presented in the previous section. It uses a generaized version of Q 43, Q, a generaized Lemma 9 in pace of Lemma 2, and a different definition of advisees, where both advisors and advisees are -tupes of booean formuas. Advice construction is the same, ony the fraction of S i that shoud be advised by a formua that s being added to the advice is different. If an -tupe F 1,...,F is an advisee of some -tupe H 1,...,H in the advice string, we have a genera method to reduce BL to cobl. If advice construction terminates in case 2, we wi have a way of eiminating one possibiity for χ SAT ( H 1,...,H ) in poynomia time. This wi give us a way to decide satisfiabiity in poynomia time, which is sufficient for a reduction from BL to cobl. 5 Summary We have resoved two open questions posed by Chang and Squire [7]. We showed as Theorems 4 and 6 that both PF NP [4] 3 PF NP[3] 3 and PF NP [5] 2 PF NP[3] 2 impy a coapse of the Poynomia Hierarchy to Σ P 3. We aso proved a more genera theorem of which PFNP [4] 3 PF NP[3] 3 = PH = Σ P 3 is a specia case. There are sti some unknown cases, however. To see that, we make the foowing observation. Observation 10 PF NP [] m PF NP[k] m whenever k og( + 1) + m. Proof Sketch: Suppose a PFm NP [] machine M computes a function f. Let x be an input to M. Using binary search, it takes og(+1) queries to an NP orace to find out how many of the queries are answered yes by M s orace. Now we can use the census trick to find any one bit of the output made by M on input x using one orace query. Therefore, if we make m more queries, we can find out what each of the m bits output by M on input x is. Observation 10 impies that if < 2 k m, PFm NP [] PFm NP[k]. We aso know from our generaization that if > 2 k 2 m + 1 (and a the other conditions), PH = Σ P 3. Notice that if k,m 2, there is a gap between 2 k m and 2 k 2 m + 1. In other words, if k,m 2, there exists an integer between 2 k m and 2 k 2 m + 1. Tabe 3 demonstrates this caim for some pairs of k,m 2. It is not hard to see that the caim hods for a k,m 2. We do not know what the consequences are when PFm NP [] PFm NP[k] and 2 k m 2 k 2 m + 1. We have shown a coapse of the Poynomia Hierarchy for one case in this gap by showing that PF NP [5] 2 PF NP[3] 2 = PH = Σ P 3 as Theorem 6. The remaining cases in this gap remain open, however. A new technique or perhaps a modification of the current technique is necessary to generaize the resut proved in Theorem 6. 15

16 m\k (a) Highest vaues of that satisfy < 2 k m m\k (b) Smaest vaues of that satisfy > 2 k 2 m + 1 Tabe 3: We see that for a pairs of k,m 2 such that k > m shown in this tabe, the difference between the argest vaue of such that < 2 k m and the smaest vaue of such that > 2 k 2 m + 1 is at east 2. Acknowedgements I woud ike to thank Dr. Richard Chang for supervising my work on this paper, for discussing it with me, for expaining some of the background to me, for reviewing this paper, and for his support in genera. Our discussions gave me many vauabe insights that heped me when I was writing this paper. References [1] Amihood Amir, Richard Beige, and Wiiam I. Gasarch. Some connections between bounded query casses and non-uniform compexity. In Proceedings of the 5th Structure in Compexity Theory Conference, pages , Washington, DC, USA, IEEE Computer Society. [2] Richard Beige. Bounded queries to SAT and the booean hierarchy. Theoretica Computer Science, 84(2): , [3] Richard Beige, Martin Kummer, and Frank Stephan. Approximabe sets. Information and Computation, 120(2): , [4] Jin-Yi Cai, Thomas Gundermann, Juris Hartmanis, Lane A. Hemachandra, Vivian Seweson, Kaus Wagner, and Gerd Wechsung. The booean hierarchy I: Structura properties. Society for Industria and Appied Mathematics Journa on Computing, 17(6): , [5] Jin-Yi Cai, Thomas Gundermann, Juris Hartmanis, Lane A. Hemachandra, Vivian Seweson, Kaus Wagner, and Gerd Wechsung. The booean hierarchy II: Appications. Society for Industria and Appied Mathematics Journa on Computing, 18(1):95 111, [6] Richard Chang and Jim Kadin. The booean hierarchy and the poynomia hierarchy: A coser connection. SIAM Journa on Computing, 25(2): , [7] Richard Chang and Jon Squire. Bounded query functions with imited output bits. In CCC 01: Proceedings of the 16th Annua Conference on Computationa Compexity, page 90, Washington, DC, USA, IEEE Computer Society. [8] Thomas H. Cormen, Chares E. Leiserson, Ronad L. Rivest, and Cifford Stein. Introduction to Agorithms. MIT Press, [9] Lane Hemaspaandra and Mitsunori Ogihara. Compexity Theory Companion. Springer, [10] Jim Kadin. The poynomia time hierarchy coapses if the booean hierarchy coapses. Society for Industria and Appied Mathematics Journa on Computing, 17(6): , [11] Christos H. Papadimitriou. Computationa Compexity. Addison-Wesey,