Motivated by Barrington's innovating result, Cai and Furst CF87, CF91] introduced the notion of bottleneck Turing machines. A width-k bottleneck Turing machine ...
On Serializable Languages Mitsunori Ogihara1 Department of Computer Science University of Rochester Rochester, NY 14627
Technical Report 519 June, 1994
1 Supported in part by the NSF and the JSPS under grant NSF-INT-9116781/JSPS-ENG-207.
Abstract Cai and Furst introduced the notion of bottleneck Turing machines and showed that the languages recognized by width-5 bottleneck Turing machines are exactly those in PSPACE. Computational power of bottleneck Turing machines with width fewer than 5 is investigated. It is shown that width-2 bottleneck Turing machines capture polynomial-time many-one closure of nearly near-testable sets. For languages recognized by bottleneck Turing machines with intermediate width 3 and 4, some lower- and upper-bounds are shown.
1 Introduction Branching program is one of the most interesting topics in complexity theory. For k 2, a width-k branching program for n-bit inputs is a sequence of instructions f(pi; fi ; gi)gmi=1 such that for each i; 1 i m, 1 pi n and fi ; gi 2 Fk , where Fk is the monoid consisting of all mappings of [k] = f1; ; kg to itself. Given an input x 2 = f0; 1g of length n, for each i, let hi = fi if the pi -th bit of x is a 1 and gi otherwise. Let H = hm h1 , where the multiplication is from right to left. We say that the program accepts the input if the resulting mapping H xes 1 (maps 1 to 1). Recently, Barrington[Bar89] has shown that a language belongs to NC1 if and only if it is recognized by a width-5 polynomial-length branching programs. Regarding complexity of languages recognized by polynomial-size branching programs with fewer width, Barrington and Therien [BT88] have shown that they belong to ACC(6), the class of languages recognized by constant depth, polynomial-size, unbounded fan-in circuits with MOD(6). Motivated by Barrington's innovating result, Cai and Furst [CF87, CF91] introduced the notion of bottleneck Turing machines. A width-k bottleneck Turing machine is a deterministic Turing machine with a read/write \safe-storage" that is capable to hold an integer in [k] = f1; ; kg and with a read-only polynomial-length \binary counter," of which content is referred as \phase." With the input on the input tape, the safe-storage as well as the binary counter is set to 1, and the machine starts phase 0. In each phase, after a polynomial number of steps, the machine is obliged to store a number in safe-storage, then, the counter is incremented by 1, the machine is reset except for the safe-storage and the counter, and the machine proceeds to the next phase. The entire computation by the machine is terminated when it nishes its nal phase, in which the counter holds its maximum value. The machine is said to accept the input if the safe-storage holds 1 after the nal phase. It is not hard to see that the computation of a width-k bottleneck Turing machine in one phase can be viewed as a mapping in Fk . So, for a width-k bottleneck TM M , de ne f to be a function of to Fk such that f (x; y) = if the computation of M on x in phase y behaves as . Let p be the polynomial such that the length of the counter on inputs of length n is p(n). Then, for every x, x is accepted by M if and only if
f (x; 1p(jxj)) f (x; 0p(jxj))(1) = 1:
(1)
We call f and p, respectively, the associate polynomial-time function and the associate polynomial for M . 1
Let SFk denote the class of languages recognized by width-k bottleneck Turing machines. Obviously, SFk PSPACE for any k. By following Barrington's construction [Bar89], Cai and Furst [CF91] showed that PSPACE SF5.1 An immediate question follows from this result is whether the inclusion holds for SFk with k 4. Barrington and Therien's [BT88] observation that branching programs over nite solvable monoids capture only languages in ACC suggests that SF4 is contained in the counting polynomial-time hierarchy [Wag86, PP S . In fact, we locate the class in a nite level of the hierarchy. Hence, Tor91] CH = k PP | {z }
k
we cannot hope PSPACE = SF4 unless PSPACE collapses to a nite level of the counting polynomial-time hierarchy. We also locate SF2 and SF3 class in the counting polynomialtime hierarchy and gives some upper- and lower-bounds on the complexity of these SFk classes. Cai and Furst showed PSPACE = SF5 by demonstrating that QBF, complete language for PSPACE, is accepted by a width-5 bottleneck Turing machine whose task in each phase is executing a permutation over [5]. This motivates us to consider a bottleneck Turing machine whose associate function has range Sk , where Sk is the permutation group over [k]. We call such a machine a permutation bottleneck Turing machine and let perm -SFk denote the class of languages recognized by width-k permutation bottleneck Turing machines. Cai and Furst's result PSPACE = perm -SF5 = SF5 leads us to a question of whether SFk = perm -SFk for k < 5. We give a complete characterization of languages in perm -SFk , k 4, in terms of MODk P [CH90] classes; that is, perm -SF2 = P, perm -SF3 = 3 PP , and perm -SF4 = P3 P P (for the de nition of these classes, see Section 2). From these characterizations and the location of SFk classes in the counting polynomial-time hierarchy, it follows that SF2 6= perm -SF2 unless p2 P, that SF2 6= perm -SF2 unless p2 3 PP , and that SF4 6= perm -SF4 unless PH P3 P P . In an attempt to clarify the di erence between SFk and perm -SFk , we de ne another subclass of SFk . Call a mapping in Fk a shrink if it maps some a to some b 6= a but xes all indices other than a. We use [a ! b] to denote the mapping and Rk to denote the set of all shrinks in Fk . It is easy to see that every mapping in Fk is decomposed as a product of at most k ? 1 shrinks and one permutation. Consider a bottleneck Turing machine whose associate function has range Rk [ fIk g, where Ik is the identity. We call
1 We note here that the range of safe-storages for width-k bottleneck TMs has been changed to [0; k] in [CF91]. Consequently, the class SF +1 in [CF87] is represented as SF in [CF91]. In this paper, we employ
the notation in [CF87].
k
k
2
such a machine a shrink bottleneck Turing machine and by shr -SFk denote the class of all languages recognized by width-k shrink bottleneck Turing machines. Interestingly, we show that, witdh-(k + 1) shrink bottleneck Turing machines can simulate width-k `general' bottleneck Turing machines; that is, SFk shr -SFk+1 . But, we show this is perhaps the limit of the power of one additional width given to shrink bottleneck Turing machines. For, we show shr -SFk+1 and SFk have almost the same computational power, and if we allow bottleneck Turing machines to access polynomial-size `advice' then SF3 and shr -SF4, even perm -SF4 have exactly the same computational power, and SF2 and shr -SF3 have exactly the same computational power. We also show complete characterizations of width-2 classes; that is, shr -SF2 = p2 = PNP and SF2 = P==OptP, where P==OptP is the class of languages recognized by a Planguage with some input-dependent advice from an OptP-function [KT94]. The latter result is interesting in the light of self-reducible structure of sets|the nearly near-testable sets [HH91]. A language A is nearly near-testable [HH91] if there exists a polynomial-time algorithm that, given x, either outputs the membership of x or outputs the parity of A (x) and A (x? ), where x? is the predecessor of x. Hemachandra and Hoene [HH91] showed that the pm -closure of the nearly near-testable sets coincides with P==OptP. The former characterization|p2 = perm -SF2|strengthens the result by Cai and Furst [CF91] that the Boolean hierarchy over NP is in SF2 . The p2 = shr -SF2 result leads us to a question about the smallest width for bottleneck Turing machines to capture levels of the polynomial-time hierarchy. We show that shr -SF3 captures p2 and p2 and that shr -SF4 captures PH.
2 De nitions and Basic Properties of SF Classes All our laguages are over alohabet = f0; 1g. We assume the reader's familiarith with basic notions from complexity theory.
2.1 Complexity Classes We de ne the complexity classes we will be concerned with. For k 2, MODk P [CH90] is the class of languages L for which there exists a polynomial time-bounded nondeterministic Turing machine M such that for every x, it holds that x 2 L if and only if the number of accepting computation paths of M on x is not a multiple of k. Especially, MOD2P is denoted by P [GP86, PZ83]. We will also be specially interested in MOD3 P. Purely as a convention, we will be using 3 P to denote MOD3 P throughout this paper. For a class of 3
languages C , de ne C (respectively, 3 C ) be the class of languages L for which there exist a polynomial p and a language A 2 C such that for every x, x 2 L if and only if the number of y 2 p(jxj) such that xy 2 A if odd (respectively, not divisible by 3). The following properties are well-known.
Proposition 2.1 1. [PZ83] P(P) = P(P) = P = P. 2. [BGH90] 3 P(3P) = P(3P) = 3 P = 3 P. 3. [BGH90] For every L 2 3 P, L 2 3 P is witnessed by a polynomial p and A 2 P such that for every x, k fy 2 p(jxj) j xy 2 Ag k 0; 1 (mod 3). 4. [TO92] P(PH) BPP(P) and 3 P(PH) BPP(3 P). T 5. [Lau83, Sip83] P(BPP) = BPP and BPP p2 p2 . For a class C , de ne 9 C (8 C ) to be the class of languages L for which there exist a polynomial p and a language A 2 C such that for every x, x 2 L if and only if for some y 2 p(jxj), xy 2 A (x 2 L if and only if for every y 2 p(jxj), xy 2 A). It is well-known for every k 1, that pk = 9 pk?1 and pk = 8 pk?1 . OptP [Kre88] is the class of functions f for which there exist a polynomial p and a set A 2 P such that for all x, it holds that
f (x) = maxfy 2 p(jxj) j xy 2 Ag if such a y exists and 0p(jxj) otherwise. For a class of languages C , C ==OptP [KT94] is the class of sets L for which there a set A 2 C and a function f 2 OptP such that for all x, it holds that x 2 L $ xh(x) 2 A:
2.2 Basic Properties of SF-classes For k 2, let [k] denote f1; ; kg. Fk denotes the monoid consisting of all mappings of [k] to itself with the identity mapping Ik . Sk denotes the permutation group over [k]. For a; b 2 [k] with a 6= b, (a b) denotes the transpose between a and b, and [a ! b], called a shrink, denotes the mapping that xes c 6= a and maps a to b, where a and b are refereed to as root([a ! b]) and val([a ! b]), respectively. Rk denotes f[a ! b] j a; b 2 f1; ; kgg. It is easy to see that each ' 2 Fk can be decomposed as k?1 1 with 1; ; k?1 2 Rk [fIk g and 2 Sk . For = [a ! b] 2 Rk and 2 Sk , de ne = = [ (a) ! (b)]. It is worth noting that = (= ) . 4
Let f be a function of to Fk and let x 2 and S be a nite set. Then we write f [x; S ] to denote the product f (x; ym ) f (x; y1), where y1 ; ; ym is the enumeration of all elements in S in increasing order. By convention, de ne f [x; ;] = Ik . Let L 2 SFk via a polynomial p and f 2 PF so that for every x, x 2 L if and only if f [x; p(jxj)] = f (x; 1p(jxj)) f (x; 0p(jxj)) xes 1. For any a 6= 1 2 [k], de ne g(x; y) = (1 a)f (x; y) if y = 1p(jxj) and f (x; y) otherwise. Then x 2 L if and only if g [x; p(jxj)](1) = a. Moreover, for any a; b 2 [k]; a 6= b, de ne h(x; y) = f (x; y) if y = 1p(jxj) and f (x; y) otherwise, where is a mapping that maps 1 to a and everything else to b. It is easy to see that is decomposed as a product of mappings in Rk and that x 2 L if and only if h [x; p(jxj)](1) = a and x 62 L if and only if h [x; p(jxj)](1) = b. Thus, we have the following characterization of SF-classes.
Proposition 2.2 Let k 2. 1. A set L is in SFk (shr -SFk ) if and only if there exist a; b 2 [k]; a 6= b, a polynomial p, and a polynomial-time computable function f : ! Fk ( f : ! Rk ) such that for every x 2 , it holds that
x 2 L $ f [x; p(jxj)](1) = a and x 62 L $ f [x; p(jxj)](1) = b: 2. A set L is in perm -SFk if and only if there exist a 2 [k], a polynomial p, and a polynomial-time computable function f : ! Sk such that for every x 2 , it holds that x 2 L $ f [x; p(jxj)](1) = a:
Proposition 2.3 For any k 2, both SFk and shr -SFk are closed under complement. Note that, it does not follow from the above discussion that perm -SFk is closed under complement. But, we shall see later, perm -SF-classes are all closed under pT -reductions.
Proposition 2.4 For any k, SFk , perm -SFk , and shr -SFk are all closed-under pm reductions.
Proof Let L pm A via function g and let A 2 perm -SFk via a polynomial p and a function f 2 PF. There is a polynomial q such that for every x, jg(x)j q(jxj). De ne r(n) = p(q(n)) and de ne de ne h(x; y ) as follows:
5
If jyj = r(jxj) and of the form 0lz for some z; jzj = p(g(x)), then h(x; y) = f (g(x); z). Otherwise, h(x; y) = Ik . It is not hard to see that h [x; r(jxj)] = f [g (x); jg(x)j]. So, x 2 L if and only if h [x; r(jxj)](1) = 1. This implies L 2 perm -SFk . Similar constructions show that both SFk are shr -SFk are closed under pm -reductions. 2 Next we prove two lemmas, which will be used later.
Lemma 2.5 Let k 2. S 1. (9 SFk ) (8 SFk ) SFk+1 . S 2. (9 shr -SFk ) (8 shr -SFk ) shr -SFk+1 . Proof Let L 2 9 SFk via a polynomial p and a set A 2 SFk . Let A 2 SFk via a polynomial-time computable function f and a polynomial q such that for every x 2 , x 2 A $ f [x; q(jxj)](1) = 1: For each x 2 , y 2 p(jxj), and z 2 q(jxyj), de ne g (x; yz ) = [1 ! 4]f (xy; z ) if z = 1q(jxyj) and f (xy; z ) otherwise. De ne r(n) = p(n) + q (n + p(n)). Then, g is polynomial-time computable, and for every x 2 ,
x 2 L $ g [x; r(jxj)](1) = 4: So, L 2 SFk+1 . Thus, 9 SFk SFk+1 . Since SFk and SFk+1 are closed under complement by Proposition 2.3, we have 8 SFk SFk+1 . The proof for shr -SF is quite similar, and thus omitted. 2
Lemma 2.6 For k 2, SFk ==OptP SFk and shr -SFk ==OptP shr -SFk . Proof Let L 2 SFk ==OptP via A 2 SFk and g 2 OptP so that for every x, x 2 L if and only if xg (x) 2 L. Let g 2 OptP via a polynomial p and B 2 P so that g (x) = maxfy 2 p(jxj) j xy 2 Bg. Without loss of generality, we may assume for every x 2 , that x0p(jxj) 2 B. Now suppose that A 2 SFk via a polynomial r and f 2 PF such that for every x 2 , x 2 L $ f [x; r(jxj)](1) = 1: For each x 2 , y 2 p(jxj), b 2 , and z 2 q(jxyj) , de ne h(x; ybz ) to be 6
Ik if xy 62 B, [k ! 1][(k ? 1) ! 1] [2 ! 1] if xy 2 B and b = 0, and f (xy; z) if xy 2 B and b = 1. De ne s(n) = p(n) + 1 + q (n + p(n)). Then for every x 2 , h [x; s(jxj)] = h(x; g(x)11s(jxj)) h(x; g(x)10s(jxj))[k ! 1] [2 ! 1] = f [xg (x); q(jxg(x)j)][k ! 1] [2 ! 1] Therefore, h [x; s(jxj)](1) = 1 if and only if x 2 L, which implies L 2 SFk . Hence, L 2 SFk . This proves the lemma. The same proof works for shr -SFk . 2
Lemma 2.7 For any k 2, SFk shr -SFk+1 . Proof For a 6= b 2 [k], let = [k + 1 ! a][a ! b][b ! k + 1]. It is easy to see that (a) = b, (b) = a, and (c) = c for every c 2 [k] ? fa; bg. So, behaves as the transpose (a b) over [k]. Thus, every mapping in Fk is decomposed as a product of mappings in Rk+1 . Therefore, SFk shr -SFk+1 . 2 Lemma 2.8 For any k 2, shr -SFk+1 P(SFk (NP(shr -SFk S NP))). Proof Let L 2 shr -SFk+1 be witnessed by a polynomial p and and f 2 PF. For each i= 6 j 2 [k + 1], de ne r(x; i; j ) = ([i ! j ] h [x; p(jxj)])(j ): Note for every i = 6 j , that h [x; p(jxj)](j ) 2 fi; j g if and only if r(x; i; j ) = i, and therefore, for every i; j; l that are all distinct, that h [x; p(jxj)](j ) = i if and only if r(x; i; j ) = j and r(x; l; j ) = i. Note for every 2 Fk and i; j 2 [k], that (i) = j if and only if (i) = 6 l for p ( j x j ) every l = 6 j . Thus, h[x; ] can be easily determined from r(x; i; j )'s. So, it suces to S
show that r(x; i; j ) is polynomial-time computable with oracle in SFk (NP(shr -SFk NP)). By symmetry, we only prove this for r(x; k + 1; 1). De ne q (n) = p(n) + 1 and de ne g (x; 0y ) = f (x; y ) and g (x; 1y ) = [k + 1 ! 1] if y = p(jxj) and Ik+1 otherwise. Clearly, g 2 PF and g [x; q(jxj)] = [k +1 ! 1] f [x; p(jxj)]. Let Y be the set of all y 2 q(jxj) such that root(g (x; y )) = k + 1. For each y 2 Y , let y 0 denote the largest z 2 Y smaller than y if such z exists, and let start (y ) be successor of y 0 if y 0 is de ned and 0q(jxj) otherwise. Then, since 1q(jxj) 2 Y ,
g [x; q(jxj)] = (g(x; ym) g(x; start(ym ))) (g(x; y1) g(x; pred(y1))); 7
where y1 ; ; ym is the enumeration of all y 2 Y in increasing order. De ne h(x; y ) = g(x; y) g(x; start (y)) if y 2 Y and Ik+1 otherwise. Clearly, g [x; q(jxj)] = h [x; q(jxj)]. Now h(x; y ) is computed in the following way: Since the membership in Y can be tested in polynomial-time, determining h(x; y ) for y 62 Y is easy. So, suppose y 2 Y . For each z; start (y) z y, let (z) = g 0(x; pred (z)) g 0(x; start (y)), where pred (z) is predecessor of z and g 0(x; w) = g (x; w) if g (x; w) 2 Rk and Ik+1 otherwise. Note that g (x; y ) is the unique mapping in the product with val = k + 1. So, for any a 2 [k],
if g(x; pred (y)) g(x; start (y)) maps a to k + 1, then h(x; y)(a) = val(g(x; y)); otherwise, h(x; y)(a) = (z)(a). The rst condition is met if and only if
(*) (9z 2 q(jxj))(9b 2 [k])[start (y) z y; (z)(a) = b, and g(x; z) = [a ! k + 1]]: Since start (y ) is polynomial-time computable with an NP-oracle and, once start (y ) is computed, (z ) is polynomial-time computable with oracle in shr -SFk , (*) is a condition in NPNP[shr -SF , and thus, h 2 PF(NPNP[shr -SF ). Note that if f (x; y ) 6= Ik , then f (x; y)([k +1]) [k]. So, we may view h(x; y) as a mapping over [k]. Therefore, r(x; k +1; 1) can be computed in polynomial-time with oracle in (SFk )h SFk (NPNP[shr -SF ). This proves the lemma. 2 Lemma 2.9 For any k 2, SFk P(shr -SFperm -SF ). k
k
k
k
k
Proof Suppose that L 2 SFk via a polynomial p and f . Recall that any ' 2 F3 is decomposed as m 1 with 2 S3 , m k ? 1, and m ; ; 1 2 Rk . De ne the following functions:
f 0 (x; y) = f (x; y) if f (x; y) 62 Rk and Ik otherwise; g(x; y) = f [x; fz 2 p(jxj) j z yg]; and h(x; y) = f 0(x; y)=g(x; y) if h(x; y) 2 Rk and Ik otherwise: 0
Then for every x 2 , it holds that:
f [x; p(jxj)] = h[x; p(jxj)] g(x; 0p(jxj)): Note that h has range Rk [fIk g and is polynomial-time computable with oracle g and that g is a product of polynomial-time computable function with range Sk . So, f [x; p(jxj)] is -SF . This proves the lemma. 2 polynomial-time computable with oracle in shr -SFperm k k
8
3 Computational Power of Width-2 Bottleneck Turing Machines Theorem 3.1 perm -SF2 = P. Proof Let L 2 P = P via a polynomial p and a set A 2 P. For x and y 2 p(jxj), de ne f (x; y ) = (1 2) if xy 2 A and I2 otherwise. Then f [x; p(jxj)] = (1 2) if and only if x 2 L. Thus, P SF2 . Let L 2 SF2 via a polynomial p and f 2 PF. Clearly, f [x; p(jxj)] = (1 2) if and only if k fy 2 p(jxj) j f (x; y ) = (1 2)g k is odd. The latter condition can be tested by a P-predicate. So, L 2 P. 2 Theorem 3.2 shr -SF2 = p2. Proof Let L 2 shr -SF2 via a polynomial p and f 2 PF so that for every x, f [x; p(jxj)](1) = 1. Since the membership in L depends only on the image of 1 by the product, we may assume that f (x; 0p(jxj)) = [2 ! 1]. De ne T (x) to be the largest y 2 p(jxj) such that f (x; y ) 2 R2. Then T is a total function and for every x, f [x; p(jxj)] = f (x; T (x)). Clearly, T 2 PFNP . Thus, L 2 p2 . For the converse direction, since shr -SF2 is closed under pm -reductions by Proposition 2.4, it suces to show that the following pm -complete language MaxSat for p2 [Kre88] is in shr -SF2:
MaxSat = fx j x is a Boolean formula and the largest satisfying assignment for x is odd g.
Note that every assignment for a formula x can be encoded into a string of length jxj. So, de ne p(n) = n and de ne f (x; y ) to be (i) [1 ! 2] if jxj = jy j, y is a satisfying assignment for x, and y is odd, (ii) [2 ! 1] if jxj = jy j, y is a satisfying assignment for x, and y is even, and (iii) I2 otherwise. Then, clearly, f [x; p(jxj)](1) = 2 if and only if x 2 MaxSat . Since, shr -SF2 is closed under complement, we have MaxSat 2 shr -SF2 . 2
Theorem 3.3 SF2 = P==OptP. Proof P==OptP SF2 follows from Lemma 2.6 and Theorem 3.1. For the other direction, let L 2 SF2 via a polynomial p and f 2 PF such that for every x 2 , it holds that x 2 L $ f [x; p(jxj)](1) = 1. As in the proof of Theorem 3.2, we may assume that f (x; 0p(jxj)) = [2 ! 1]. De ne T (x) = maxfy 2 p(jxj) j f (x; y) 2 R2g. Then T is a total function in OptP. De ne A to be the set of all xy , jy j = p(jxj), such that 9
f (x; y) 2 R2 and f (x; y) = [1 ! 2] if and only if the number of z 2 p(jxj) such that z > y and f (x; z) = (1 2) is odd. Noting that PP = P, it is not hard to see that A 2 P. Moreover, since f (x; T (x)) f (x; 0p(jxj)) = f (x; T (x)), f [x; p(jxj)](1) = 1 if and only if xT (x) is in A. Thus, L 2 P==OptP. This proves the theorem. 2
4 Computational Power of Width-3 Bottleneck Turing Machines Barrington [Bar89, Theorem 6] showed that languages recognized by polynomial-length branching programs over S3 are AC0 -reducible to the MOD(6) functions. Based on the analysis, we obtain the following result.
Theorem 4.1 perm -SF3 = 3 P. Proof Let L 2 3 P via a polynomial p and A 2 P. Since P(P) = P, by Proposition 2.1, we may assume for every x, that k fy 2 p(jxj) j xy 2 Ag k is congruent
to either 0 or 1 modulo 3. By the proof of Theorem 3.1, there exist a polynomial q and g1; g2 2 PF such that for every x, x 2 A if and only if g1 [x; q(jxj)] = (1 2) if and only if g2 [x; q(jxj)] = (2 3) and x 62 A if and only if g1 [x; q(jxj)] = g2 [x; q(jxj)] = I3. De ne r(n) = p(n) + 1 + q(n + p(n)) and h(x; y0z) = g1(xy; z) and h(x; y1z) = g2(x; y1z) for y 2 p(jxj) and z 2 q(jxyj) . Then,
h [x; r(jxj)] = ?
( ) ! 1 2 3 kfy2 jxy2Agk; 231 p jxj
which is 12 23 31 if x 2 A and I3 otherwise. So, h [x; r(jxj)](1) = 2 if x 2 A and 1 otherwise. Thus, L 2 perm -SF3 . The proof for the other direction is similar to that of [Bar89, Theorem 6]. Let L 2 perm -SF3 via a polynomial p and f 2 PF such that for every x 2 ,
x 2 L $ f [x; p(jxj)](1) = 1: Let a = (1 2) and b = (2 3). It is easy to see that S3 = fI3; a; b; ab; ba; abag. Noting that I3 = a2 , without loss of generality, we may assume that for every x 2 , it holds that 10
(i) for every y 2 p(jxj), f (x; y) 2 fa; b; I3g and (ii) there is a xed constant k such that for every k consecutive y1; ; yk 2 p(jxj), at least one of f (x; y1); ; f (x; yk) is in fa; bg. De ne Q(x; y ) = 1 if the number of z 2 p(jxj); z y , such that f (x; y ) = 6 I3 is odd and 0 otherwise. Clearly, Q is a predicate in P. De ne top(x) to be the largest y 2 p(jxj) such that f (x; y ) = 6 I4 and (x) = f (x; top(x)) if Q(x; top(x)) = 1 and I4 otherwise. By (ii) above, top 2 PF, and thus, 2 PF(P). De ne pre(x; y ) to be the largest z < y such that f (x; y ) = 6 I4 if such z exists and unde ned otherwise. By (ii) above, pre 2 PF. Note that if Q(x; y ) = 1 and f (x; y ) = 6 I4, then pre(x; y) is de ned. So, de ne (x; y) to be d 2 f0; 1; 2g such that f (x; y)f (x; pre(x; y)) = (ab)d if f (x; y) = 6 I3 and Q(x; y) = 0; and P 0 otherwise. Then, 2 PF(P). Now de ne (x) = ( y;jyj=p(jxj) (x; y )) mod 3. Then, f [x; p(jxj)] = (x) (ab)(x), so f [x; p(jxj)] is polynomial-time computable with oracle in 3 P. This implies L 2 P(3 P) = 3 P. 2 Theorem 4.2
1. shr -SF3 P(p2) BPP(P).
S S 2. p2 p2 (P==OptP) shr -SF3 .
Proof By Lemma 2.8, together with Theorems 3.2 and 3.3, we get shr -SF3 NP SF2 (NPNP[shr -SF2 ) (P==OptP)NP 2 = (P==OptP)2 P3 BPP(P). [
p
p
p
For the second part, the rst inclusion follows from Lemma 2.5 and Theorem 3.2, the second inclusion is dual of the rst, and the last inclusion follows from Lemma 2.7 and Theorem 3.3. 2
Theorem 4.3 1. ( 3 P)==OptP SF3. S S S 2. NP(P) coNP(P) p2 p2 SF3 . 3. SF3 BPP( 3 P). Proof By Theorem 4.1, for any A 2 3 P, there exist a function f and a polynomial q such that for every x, f [x; q(jxj)] = (2 3)(1 2) if x 2 L and I3 otherwise. De ne f 0 (x; y) = [3 ! 2]f (x; y) if y = 1q(jxj), f (x; y)[3 ! 1] if y = 0q(jxj), and f (x; y) otherwise. ? ? ? Then f [x; q(jxj)] = 12 21 32 if x 2 A and 11 22 31 otherwise. Note that 11 22 31 behaves as ? the identity over f1; 2g and 12 21 32 as (1 2). So, by following a discussion similar to the 0
proof of Theorem 3.3, we can de ne a polynomial p and a function g such that for every x, 11
g [x; q(jxj)] = ?12 21 32 if x 2 L and ?11 22 31 otherwise. Thus, 3 P SF3. Then, from
Lemma 2.6, the rst inclusion follows. The second part follows from Theorem 3.3 via Lemma 2.5. The third part is proven as follows: SF3 shr -SF3(perm -SF3) (by Lemma 2.9) P(p3(3 P)) (by Theorems 4.2 and 4.1) BPP( 3 P) (by Proposition 2.1). 2
5 Computational Power of Width-4 and Width-5 Bottleneck Turing Machines
Theorem 5.1
perm -SF4 = 3 P:
Proof Let L 2 (3 P) via polynomial p and A 2 3 P. By the proof of Thoerem 4.1, there exist a polynomial q and a function f00 such that for every x, f00 [x; q(jxj)] = (2 3)(1 2) if x 2 A and I4 otherwise. Similarly, we can construct f01 ; f10; f11 such that for every x,
f01 [x; q(jxj)] = (1 3)(3 4), f10 [x; q(jxj)] = (1 2)(1 3), and f11 [x; q(jxj)] = (3 4)(2 3); and
f01 [x; q(jxj)] = f10 [x; q(jxj)] = f11 [x; q(jxj)] = I4 otherwise. De ne r(n) = q (n) + 2 and de ne g (x; bcy ) = fbc (x; y ). Then g [x; r(jxj)] = (1 3)(2 4) if x 2 A and I4 otherwise. Clearly, ((1 3)(2 4))2 = I4. So, by following a proof similar to that of Theorem 3.1, we have L 2 perm -SF4. For the other inclusion, suppose that L 2 perm -SF4 via a polynomial p and f such that for every x 2 , x 2 L $ f [x; p(jxj)](1) = 1: Let a = (1 2), b = (2 3), and c = (3 4). Then every mapping in S4 can be expressed as a product of length of nite length consisting only of a; b; and c. Furthermore, let = ba, = b, and = a c. It holds that 3 = I4, a = , b = , and c = . Thus, without loss of generality, we may assume that for every x; y , f (x; y ) 2 f ; ; ; I4g. For each x and y 2 p(jxj), let P (x; y ) = f (x; y ) f (x; 0p(jxj)). We show that P (x; y ) is expressed as a product of the form e(x;y) H (x; y ), where H (x; y ) is a product consisting only of and . Note the following properties: 12
= 2 , 2 = , = ( )2, and 2 = 2 . Let pred (y ) be predecessor of y , and de ne e(x; y ) and H (x; y ) as follows:
(i) e(x; y) = e(x; pred (y)) + 1 mod 3 if f (x; y) = , e(x; pred (y))2 mod 3 if f (x; y) = , and e(x; pred (y )) otherwise; and
(ii) H (x; y) = H (x; pred (y)), where = if f (x; y) = and e(x; pred (y)) = 1, if f (x; y ) = and e(x; pred (y )) = 2, and f (x; y ) otherwise.
De ne (x; y ) = ab if f (x; y ) = , b if f (x; y ) = , and I3 otherwise. Then, for every x; y ,
e(x; y) (x; y) (x; 0p(jxj))(3) (mod 3): So, by Theorem 4.1, e 2 PF(3 P). On the other hand, note that the mu in (ii) above can be expressed as a product of four mappings in f ; ; I4g so that not all of the four mappings are I4. So, de ne q(n) = p(n) + 2 and de ne so that (x; y00); (x; y01); (x; y10); (x; y11) are such four mappings representing . Then 2 PF(3 P) and [x; q(jxj)] = H (x; 1p(jxj)). De ne Q(x; y) = 1 if the number of z 2 q(jxj); z y, such that (x; y) 6= I4 is odd and 0 otherwise. Clearly, Q is a predicate in 3 P. De ne top(x) to be the largest y 2 q(jxj) such that (x; y) 6= I4 and (x) = (x; top(x)) if Q(x; top(x)) = 1 and I4 otherwise. By our de nition of , top 2 PF(3 P), and thus, 2 PF( 3 P). De ne start (x; y ) to be the largest z < y such that (x; y) 6= I4 if such z exists and unde ned otherwise. By our de nition of , start 2 PF(3 ). Now de ne (x; y) 2 f0; 1; 3g as follows:
If (x; y) 6= I4, (x; y) = and (x; pre(x; y)) = , then (x; y) = 1. If (x; y) 6= I4, (x; y) = and (x; pre(x; y)) = , then (x; y) = 1. Otherwise, (x; y) = 0. Then, 2 PF( 3 P) and
P
[x; q(jxj)] = (x)( )( 13
= ( ) (x;y))mod4 ;
y;jyj
q jxj
which is polynomial-time computable with oracle in 3 P. So, f [x; p(jxj)] is polynomial-time computable with oracle in 3 P, and therefore, L 2 3 P. 2
Theorem 5.2
1. shr -SF4 BPP( 3 P).
S S S S 2. p3 p3 NP(P) coNP(P) 3 P shr -SF4 .
Proof By Lemma 2.8 and Theorems 4.2 and 4.3, we get shr -SF4 P(BPP( 3 P(NP(NP [ P(p3)))));
which yields, by Proposition 2.1, shr -SF4 BPP( 3 P). For the second part, the rst four containment results follow from Lemma 2.5 and Theorem 4.2 (2), and the last follows from Lemma 2.7 and Theorem 4.3. 2
Theorem 5.3 1. ( 3 3 P)==OptP SF4. 2. SF4 BPP 3 3 P. S S S S S 3. p3 p3 p2(P) p2 (P) NP(3 P) coNP(3 P) SF4 . Especially, PH SF4 . Proof Let C denote 3 P. By Theorem 4.3, for every? A 2 C , there exist a polynomial ?1 2 3 4 1 2 4 p ( j x j ) p and f 2 PF such that for every x, f [x; ] = 2 1 2 = 2 1 3 2 if x 2 A and ?1 2 4 ? = 11 22 33 41 otherwise. Similarly, we get f 0 2 PF such that for every x, f [x; p(jxj)] = 1 2 1 ?2 3 4 ?1 2 3 4 ? 2 3 4 ?1 2 3 4 p(jxj) 3 2 3 = 1 3 2 3 if x 2 A and 2 3 2 = 1 2 3 2 otherwise. Apply f?[x; ], and then, f [x; p(jxj)], to [4 ! 1]. This de nes a new product, which is = 12 23 31 42 if x 2 A and ? = 11 22 33 41 otherwise. Since 3 = , for any set B 2 3 C via some polynomial and A, there exist a polynomial q and a function g 2 PF such that for every x, g [x; q(jxj)] = if x 2 B and otherwise. ? ? Let = [3 ! 1] and = [3 ! 1] . Then = 12 21 31 42 and = 11 22 31 41 . Since 2 = , for a set L 2 3 C via some polynomial and B , there exist a polynomial r and a function h 2 PF such that for every x, g [x; q(jxj)] = if x 2 B and otherwise. This implies 3 3 P SF4. So, the rst part is established. 0
0
The second part follows from Lemma 2.8 and Theorems 4.1 and 4.3, which locates SF4 in P(BPP( 3 P(3 P))) BPP( 3 3 P). For the third part, the rst two containment results follow from Theorem 5.2 and the others from Theorem 4.3 and Lemma 2.5. The PH-containment is by Proposition 2.1. 2 14
Theorem 5.4 1. shr -SF5 BPP( 3 3 3 P). 2. 3 3 P shr -SF5. Proof The rst part follows from Lemma 2.8 and Theorems 5.2 and 5.3 via Proposition 2.1.
The second part follows from Lemma 2.7 and Theorem 5.3. 2 For a class of sets C , C =poly [KL80] is the class of sets L for which there exist a set A 2 C and a function h : ! such that for all x, it holds that x 2 L $ xh(0jxj) 2 A. It is wellknown [Sch87] for any class C closed under ppos -reductions [Sel82], that BPP(C ) P=poly . The complexity bounds we have obtained for SF-classes allow us to obtain the following classi cation in the light of polynomially length bounded advice.
Corollary 5.5
p2=poly = shr -SF2 =poly P=poly = perm -SF2=poly = SF2=poly = shr -SF3=poly (3 P)=poly = perm -SF3 =poly ( 3 P)=poly = SF3 =poly = shr -SF4 =poly = perm -SF4 =poly ( 3 3 P)=poly = SF4 =poly shr -SF5 =poly ( 3 3 3 P)=poly PSPACE=poly = perm -SF5 =poly = SF5 =poly:
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