This paper studies a notion called polynomial-time membership comparable sets. For a function g, a set A is polynomial-time g-membership comparable if there ...
Polynomial-Time Membership Comparable Sets Mitsunori Ogihara1 Department of Computer Science University of Rochester Rochester, NY 14627
Technical Report 552 December, 1994
1 Previously known as Mitsunori Ogiwara. Supported in part by the NSF and the JSPS under grant NSF-INT-9116781/JSPS-ENG-207.
Abstract This paper studies a notion called polynomial-time membership comparable sets. For a function g , a set A is polynomial-time g -membership comparable if there is a polynomialtime computable function f such that for any x1; ; xm with m g (maxfjx1j; ; jxmjg), outputs b 2 f0; 1gm such that (A(x1); ; A(xm)) 6= b. The following is a list of major results proven in the paper. 1. Polynomial-time membership comparable sets construct a proper hierarchy according to the bound on the number of arguments. 2. Polynomial-time membership comparable sets have polynomial-size circuits. 3. For any function f and for any constant c > 0, if a set is pf (n)-tt-reducible to a Pselective set, then the set is polynomial-time (1 + c) log f (n)-membership comparable. 4. For any C chosen from fPSPACE; UP; FewP; NP; C= P; PP; MOD2 P; MOD3 P; g, if C P-mc(c log n) for some c < 1, then C = P. As a corollary of the last two results, it is shown that if there is some constant c < 1 such that all of C are polynomial-time nc -truth-table reducible to some P-selective sets, then C = P, which resolves a question that has been left open for a long time.
1 Introduction Given two strings x and y , can we tell which is more likely to be in a set A? Jockusch [Joc68] de ned a set A to be semirecursive if there is a recursive function f such that for all x and y , (i) f (x; y ) 2 fx; y g and (ii) if fx; y g\ A 6= ;, then f (x; y ) 2 A. We call the function f a selector for A. Selman [Sel79] considered a polynomial-time analogue of semirecursive sets and de ned a set A to be P-selective if A has a polynomial-time computable selector. P-selective sets have been widely studied [Sel82b, Sel79, Sel82a, AH92, Ko83, Tod91, LS93, BvHT93, NOS93, HNOS93, HHO+ 93, HNOS94]. Recently, there have been some remarkable results about P-selective sets. Buhrman, van Helden, and Torenvliet [BvHT93] have shown that a set is in P if and only if it is pT -self-reducible and P-selective, while previously known characterization is A 2 P if and only if A is pptt-self-reducible and P-selective [Sel82b]. Naik et al [NOS93, HNOS93] have proven, by constructing P-selective sets with certain properties, that NP search problems are not reducible to corresponding decision problems unless some implausible collapses of exponential-time complexity classes occur. Hemachandra et al [HHO+ 93] studied internal structure of the class of sets pT -reducible to P-selective sets and introduced the notion of FC -selectivity for various function classes FC . Hemaspaandra, Naik, Ogiwara, and Selman [HNOS94] have studied sets with nondeterministically polynomial-time computable selectors, and proven that if there is an NP-function that computes satisfying assignments uniquely, then the polynomial-time hierarchy [MS72, Sto77], collapses to its second level p2. For a set A, let us identify A and its characteristic function. For any x and y , there are four possible values of (A(x); A(y )). By mapping a pair (x; y ) to y , a selector for A declares that \x 2 A ! y 2 A," equivalently, \(A(x); A(y )) 6= (1; 0)." Thus one can view a selector for A as a function f that maps (x; y ) to b 2 f0; 1g2 such that (A(x); A(y )) 6= b, where b is always either 01 or 10. An interesting and fundamental question arising from this observation is \how strong is the restriction b 2 f01; 10g?"; that is, if we allow f to map to 00 or 11, then \how largely does the complexity of A increase?" Let us consider this in a more generalized setting and de ne the notion of polynomial-time membership comparable sets. Let N ( N+ ) denote the set of all nonnegative (positive) integers. Call a function g : N ! N+ polynomially bounded if there is a polynomial p such that for every n, g(n) p(n), and polynomial-time computable if there is a polynomial-time bounded machine that, on input x, outputs 1g(jxj).
De nition 1.1 Let g : N ! N+ be monotone nondecreasing, polynomial-time com1
putable, and polynomially bounded. 1. A function f is called a g -membership comparing function (a g -mc-function, in short) for A if for every x1; ; xm with m g (maxfjx1j; ; jxmjg),
f (x1; ; xm) 2 f0; 1gm and (A(x1); ; A(xm)) 6= f (x1; ; xm): 2. A set A is polynomial-time g -membership comparable if there exists a polynomial-time computable g -mc-function for A. 3. P-mc(g ) denotes the class of all polynomial-time g -membership comparable sets. Crucial property of mc-functions is that they exclude one value out of 2m possible values. The notion|excluding possible values of (A(x1); ; A(xm))|has already appeared in the literature. For a xed k 1, the function that, on input x1 ; ; xk , outputs A(x1) A(xk ) has been called FkA ; some notions related to FkA have been introduced and studied (see [AG88, ABG90, Bei91, BGGO88]). If one can always reduce 2k possible values of FkA to m < 2k , then FkA is said to be computable by a set of m polynomial-time functions [ABG90] and said to be m-enumerable [CH89]. A set A is non-p-superterse [Bei91] if for some k 1, there is a polynomial-time algorithm that computes FkA using k ? 1 adaptive queries to some set X . So, for a non-p-superterse set A, FkA is 2k?1 -enumerable for some k 1. Polynomial-time membership comparable sets are more general than these notions in the sense that (i) k can be increased according to the length of the input and that (ii) only one value is required to be excluded. Purely as a convention, when a function g with the set of real numbers as its range is used, we will be identifying g and n:[maxf1; bg (n)cg]. We use P-mc(const), P-mc(log), S S and P-mc(poly) respectively to denote fP-mc(k) : k 1g, fP-mc(f ) : f = O(log n)g, S and fP-mc(p) : p is a polynomialg, where log is base 2. It is possible for many di erent (indeed, even an uncountable number of) sets to be in P-mc(poly) via the same function g . See Proposition 4.1 for an example of when this actually happens. In Section 2, we study basic properties of polynomial-time membership comparable sets. It is easily observed that the smallest P-mc-class, namely P-mc(1), is equal to P. Noting that P-Sel, the class of P-selective sets, is a subclass of P-mc(2) and that P P-Sel [Sel79], we have P-mc(1) P-mc(2). We show that the inequality holds for arbitrary k; that is, P-mc(k) P-mc(k + 1) for any k 1. More generally, we prove that for any f and g such that f (n) < g(n) for nitely many n, P-mc(g) contains a set not in P-mc(f ). 2
Therefore, P-mc-classes construct a proper hierarchy according to the bound on the number of arguments. As P-mc(2)-sets can be viewed as less restrictive P-selective sets, one might expect that P-mc-sets do not go far beyond P-Sel. In Section 3, we consider the question of how they are related to each other. We seek to prove inclusions between reducibility classes of P-Sel and P-mc-classes. For a reducibility pr and a class C , let Rr (C ) denote the class of all sets that are pr -reducible to some set in C . Basically, P-mc(2) properly includes P-Sel: there is a tally set in P-mc(2) ? P-Sel. Furthermore, we prove for any function f , that Rf (n)-tt (P-Sel) P-mc((1 + c) log f (n)) for any constant c > 0, which yields Rbtt(P-Sel) P-mc(const) and Rtt(P-Sel) P-mc(log). We also study the question of the other direction, namely, whether P-mc-sets are polynomial-time reducible to P-selective sets. We show that P-mc(poly) P=poly, where P=poly is the class of all sets having polynomial-size circuits. Then, since RT (P-Sel) = P=poly [Ko83], we have P-mc(poly) RT (P-Sel). The pT -reducibility in this inclusion is optimal, for, we show Rtt(P-Sel) P-mc(poly). On the other hand, converse statement \P=poly P-mc(poly)" does not appear to hold. But we can show at least that the question is very subtle. For, P=poly 6 P-mc(poly) implies P 6= NP. Thus, proving that P-mc(poly) P=poly is at least as hard as proving P 6= NP. We conjecture that the converse inclusion does not hold. It is well-known that P=poly = Rtt (SPARSE) = Rtt(TALLY), where SPARSE (TALLY) denotes the class of all sparse (tally) sets. As it holds that P-mc(poly) P=poly, every P-mc-set is ptt -reducible to some tally set. In Section 4, we attempt to nd close relationships between P-mc-sets and sparse sets as well as tally sets. We prove that R1-tt (SPARSE) P-mc(poly), and thus, Rtt(P-mc(poly)) = P=poly. Interestingly, this contrasts that Rtt(P-Sel) P=poly [HHO+ 93]. In order to study relationships between P-mc-sets and tally sets, binary real numbers will be useful since the characteristic sequence of a tally set can be viewed as a binary real number in [0; 1) (see [Sel79]). We show for any binary real number r 2 [0; 1), that Pre x [r], the set of all pre xes of r, is in P-mc(2), while there is a binary real number r such that Pre x [r] 62 P-Sel. We add a few words about relationships between P-mc(poly) and TALLY. Noting that P-mc(poly) P=poly and TALLY P-mc(poly), we show for every A 2 P-mc(poly), that there is a tally set T in P-mc(poly) \ p3 (A) such that A ptt T . On the other hand, any tally set is pT -equivalent to some P-mc(2)-set. Thus, every set A in P-mc(poly) is pT -reducible to some set in P-mc(2) \ p3(A). As a consequence, we have P=poly = RT (SPARSE \ 3
P-mc(2)); that is, pT -reducibility to sparse sets in P-mc(2) completely characterizes P=poly. Selman [Sel79] showed that SAT is P-selective if and only if SAT is in P. Can we prove a similar result for P-mc-sets? We consider this question in Section 5. Noting that P-mc(poly) P=poly and that NP P=poly implies PH = p2 ([KL80]), one can easily observe that SAT 2 P-mc(poly) only if PH = p2 . But we have a stronger collapse for some P-mc-class. We prove NP P-mc(c log n) for some c < 1 if and only if P = NP. The proof technique we develop enables us to resolve some open questions. The rst bonus we get is the following result: If for some constant c < 1, NP Rnc -tt (P-Sel), then P = NP. It has been one of the most important questions whether NP Rr (P-Sel) =) P = NP holds for a reducibility pr [Sel82b, Tod91, HHO+ 93, TTW94]. Selman extended his rst result to pptt-reducibility by observing that Rptt(P-Sel) = Rm (P-Sel) [Sel82b]. As for \nonpositive" truth-table reducibilities, there have appeared some observations. Toda [Tod91] proved that NP Rtt(P-Sel) implies P = FewP and RP = NP. Hemachandra et al [HHO+ 93] noticed that NP R1-tt (P-Sel) implies P = NP. Thierauf, Toda, and Watanabe [TTW94] proved that if NP Rbtt(P-Sel), then NP is in deterministic subexponentialtime. Nonetheless, it has been open for a long time whether NP Rtt(P-Sel) implies P = NP, and even whether NP Rbtt(P-Sel) implies P = NP. We not only give an armative answer to the latter, but improves the upper bound on the number of queries to nc for any constant c < 1. We note here that the same result has been independently proven by Beigel, Kummer, and Stephan [BKS94] and by Agrawal and Arvind [AA94]. Another bonus we get is about complexity of functions that are polynomial-time computable with access to sets in NP. Krentel [Kre88] showed for any function f (n) 21 log n, SAT that if FPSAT f (n)-T = FP(f (n)?1)-T , then P = NP. He asked whether the same statement holds for a larger function f . Krentel's proof directly applies to the case f (n) c log n for some constant c < 1. Related to this, Beigel [Bei88] asked a question of whether SAT FPSAT f (n)-tt FP(f (n)?1)-T . For the case f (n) c log n with c < 1, Beigel [Bei88] showed for any p1-tt -hard set A for NP, that if FPAf(n)-tt FPX(f (n)?1)-T for some X , then RP = NP and P = UP. Regarding general O(log n) case, Amir, Beigel, and Gasarch [ABG90] showed for p p X any function f (n) = O(log n), that if FPSAT f (n)-tt FP(f (n)?1)-T for some X , then 3 = 3 . In this paper, we prove that the conclusion of Beigel's result can be strengthen to P = NP. The proof techniques we develop can be applied to complexity classes other than NP. We prove for any class C chosen from fUP; FewP; C=P; PP; PSPACE; MOD2 P; MOD3P; g, that if C P-mc(c log n) for some c < 1, then C = P; and thus, if C Rnc -tt (P-Sel) for some c < 1, then C = P. 4
2 Basic Properties of Polynomial-time Membership Comparable Sets In this section, we prove some fundamental properties of polynomial-time membership comparable sets. The reducibility notions we will use are due to [LLS75]. First we state some rather trivial properties.
Proposition 2.1
1. P = P-mc(1).
2. For any f and g such that f (n) g (n) for all but nitely many n, P-mc(f ) P-mc(g ). 3. If A p1-tt B and B is in P-mc(f (n)), then A 2 P-mc(f (p(n))) for some polynomial p. Especially, for any k 1, P-mc(k) is closed under p1-tt -reductions.
Proof [(1)] The statement holds because for any set A, A is in P if and only if Ac is
in P if and only if there is a polynomial-time computable function f such that for all x, f (x) = Ac (x). [(2)] Let f and g be as in the hypothesis and let n0 be such that for all n > n0 , f (n) g (n). Let A 2 P-mc(f ) and let T be the set of all strings in A of length at most n0 . Let x1 ; ; xm and n be such that maxfjx1j; ; jxm jg = n and m g (n). If there is some i such that jxij n0, then, (A(x1); ; A(xm)) 6= 0i?1b0m?i?1, where b = 0 if xi 2 T and 1 otherwise. If for every i, jxi j > n0 , then (A(x1 ); ; A(xm)) 6= f (x1 ; ; xf (n))0m?f (n) . Therefore, A is in P-mc(g ). [(3)] Let A p1-tt B via a machine M and B 2 P-mc(f ) via h. Let p be a polynomial bounding the run-time of M . For each x, let Q(x) denote the unique query of M on x. Let W0 (W1) be the set of all x such that M on x rejects (accepts) no matter what the answer from the oracle is. Let R0 (R1) be the set of all x such that M on x accepts if and only if Q(x) is not in the oracle (Q(x) is in the oracle). Note that W0 ; W1; R0; R1 2 P. Let x1 ; ; xm and n be such that maxfjx1j; ; jxmjg = n and m f (p(n)). De ne h0 (x1; ; xm) as follows:
(Case 1) fx1; ; xmg \ (W0 [ W1) 6= ;: Let j be the smallest i such that xi 2 W0 [ W1. De ne h0 (x1 ; ; xm ) = 0j ?1 b0m?j , where b = 1 if xj 2 W0 and 0 otherwise. (Case 2) fx1; ; xmg \ (W0 [ W1) = ;: 5
(Subcase 2a) There is some (i; j ); i < j , such that Q(xi) = Q(xj ): Let (k; l) be the smallest such pair. De ne h0(x1 ; ; xm) = 0l?1 1n?l+1 if either xk ; xl 2 R0 or xk ; xl 2 R1, and 0k?1 1n?k+1 otherwise. (Subcase 2b) For every i; j; i < j , Q(xi) =6 Q(xj ): Let b1 bm = h(x1; ; xm). De ne h0 (x1; ; xm ) = c1 cm , where for every i; 1 i m, ci = bi if xi 2 R1 and 1 ? bi otherwise. It is easy to observe that h0 behaves correctly. 2 Proposition 2.1 (2) states that P-mc(f ) P-mc(g ) if for all but nitely many n, f (n) g(n). Interestingly, as we shall show below, P-mc(f ) di ers from P-mc(g) if for in nitely many n, f (n) 6= g (n). We note here that the proof we develop has a avor similar to that of [AG88, Theorem 2].
Theorem 2.2 Let f and g be monotone nondecreasing, polynomial-time computable, polynomially bounded functions that map N to N+ . Suppose for in nitely many n, it holds that g (n) > f (n). Then there is a set A 2 P-mc(g ) n P-mc(f ). Proof Let f and g as in the hypothesis of the theorem. Since f is polynomially bounded, there is some n0 such that for every n n0 , f (n) < 2n . De ne a sequence fli gi1 as follows: l1 = minfn n0 j g(n) > f (n)g; and for i > 1, li = minfn 22li?1 j g(n) > f (n)g. By our hypothesis, fligi1 is a sequence of integers for which g is larger than f . Let h1 ; h2; be an enumeration of all polynomial-time computable functions such that for each i, hi is computable in time ni + i. We construct A in stages. At stage s, we diagonalize against hs by putting at most f (ls) string of length ls into A. Note that there are more than f (ls ) strings of length ls because ls > n0 . The construction at stage s proceeds as follows: For each i; 1 i f (ls), let wi be the i-th smallest string of length ls . Let b = hs (w1; ; wf (ls) ). If b 62 f0; 1gf (ls) , then hs is already not an f -mc function. So, we proceed to the next stage adding no new elements to A. If b 2 f0; 1gf (ls) , then for each i, we put wi into A if and only if the i-th bit of b is a 1. This yields hs (w1; ; wf (ls) ) = A(w1) A(wf (ls) ), so hs cannot be an f -mc function for A. Clearly, this construction establishes A 62 P-mc(f ). Next we de ne a g -mc function for A. Let y1 ; ; ym be such that m = g(maxfjy1j; ; jymjg). Without loss of generality, we may assume that jy1j jymj. Let be the largest i such that li jym j. There are the following four possible cases: 6
(a) For some i, jyij 62 fl1; ; lg: Clearly, for any such i, yi 62 A. De ne (y1; ; ym) = 1m .
(b) For some i; j; i < j , it holds that yi = yj : Clearly, for such i and j , yi 2 A if and only if yj 2 A. So, let s be the smallest i such that yi appears in yi+1 ; ; ym . De ne
(y1; ; ym) = 0s?1 1m?s+1 . (c) jy1j = = jymj = l and y1; ; ym are all distinct: By de nition, it holds that m > f (l). Since A \ l has at most f (l) elements, some yi is not in A. De ne
(y1; ; ym) = 1m. (d) jy1j; ; jymj 2 fl1; ; lg, jymj = l, jy1j = ls for some s < , and y1; ; ym are all distinct: We simulate the construction of A at stage s to compute Ac (y1) and de ne
(y1; ; ym) = Ac (y1)1m?1 .
In each case, it holds that (y1; ; ym ) 6= A(y1 ) A(ym ). So, is a g -mc function for A. It remains to show that is polynomial-time computable. Since f and g are both polynomial-time computable and fligi1 is a strictly increasing sequence, and l1; ; l are computable in time polynomial in jym j. So, one can easily compute the value of for the cases (a), (b), and (c). Now suppose that the case (d) holds. Since s < , it holds that s jy1 j log log jym j. Since hs is computable in time ns + s and f is polynomially bounded, for some xed constant k, the construction at stage s can be simulated in time O(s(jy1jk )s) O(jy1j2jy1jk ) O(22jy1j ) O(jymj). Thus, whether y1 2 A can be tested in time O(jym j). Therefore, is polynomial-time computable. This proves the theorem. 2
Corollary 2.3 If f (n) > g(n) for in nitely many n and g(n) > f (n) for in nitely many n, then by the above theorem, P-mc(f ) and P-mc(g ) are incomparable.
Corollary 2.4 P-mc-classes construct a proper hierarchy according to the bound on the number of arguments; namely, P = P-mc(1) P-mc(2) P-mc(3) P-mc(k) P-mc(k + 1) P-mc(const) P-mc(log) P-mc(poly).
3 Relationships with P-selective Sets In this section, we study relationships between polynomial-time membership comparable sets and P-selective sets. First of all, by de nition, P-Sel is a subclass of P-mc(2).
Proposition 3.1 P-Sel P-mc(2). 7
The above inclusion is proper.
Theorem 3.2 There is a tally set T 2 P-mc(2) ? P-Sel. Proof Let f1; f2; be an enumeration of all polynomial-time computable arity-2 functions.
Let fi be computable in time pi (n) = ni + i. De ne (0) = 1 and (n) = 2(n?1) for n > 0, and de ne (n) = (4n). For each i 1, let ui = 0 (i). Our set T is constructed in stages. At stage i, we do the following:
(*) If fi(u2i; u2i+1) = u2i, then put u2i+1 into T . Otherwise, put u2i into T . Clearly, for any i, either fi is not a selector function or there exist some x 2 T and y 2 T c such that either fi (x; y ) = y or fi (y; x) = y . So, T 62 P-Sel. We need to show that T 2 P-mc(2). Note for any i, that the construction up to stage i can be simulated in time i pi (2 (2i + 1)) < (2(i + 1)). De ne a function g as follows: Let x = 0s and y = 0t be tally strings.
(Case 1) s 6= (k) for any k: De ne g(x; y) = 10. (Case 2) s = (k) for some k and t 6= (l) for any l: De ne g(x; y) = 01. (Case 3) s = (k) for some k and t = (l) for some l: (Subcase A) bk=2c < bl=2c: Simulate the construction of T up to stage bk=2c to test whether x 2 T . De ne g (x; y ) = 00 if x 2 T and 10 otherwise. (Subcase B) bk=2c > bl=2c: Simulate the construction of T up to stage bl=2c to test whether y 2 T . De ne g (x; y ) = 00 if y 2 T and 01 otherwise. (Subcase C) bk=2c = bl=2c: De ne g(x; y) = 11 if x =6 y and 01 otherwise. It is easy to see that g witnesses that T 2 P-mc(2). This proves the theorem. 2
Are reducibility classes of P-Sel included in P-mc(poly)? The following theorem answers the question.
Theorem 3.3 Let f : N ! N+ be a monotone nondecreasing function. Let L be pf (n)-ttreducible to a P-selective set. Then, L 2 P-mc((1 + c) log f (n)) for any constant c > 0. The proof of the theorem is based on Lemma 3.4 below, due to Toda, stating that, given a P-selective set A and a nite set Q, one can compute a linear order over Q such 8
that A \ Q is the initial segment of the order. Originally, Jockusch ([Joc68]. Attributed to Appel and McLaughlin.) proved that being semi-recursive is equivalent to being the initial segment of a recursive linear ordering. Regarding P-selective sets, which are de ned as the polynomial-time analogue of semi-recursive sets, Selman [Sel82b] showed that the initial segment of a polynomial-time linear order is a P-selective set. Ko [Ko83] showed that being P-selective is equivalent to being the union of initial segments of polynomial-time preorder.
Lemma 3.4 [Tod91] Let A be P-selective. There is a polynomial-time algorithm that, given a nite set Q , outputs an enumeration y1 ; ; ykQk of elements in Q such that there exists some m; 0 m k Q k, such that A \ Q = fyi j 1 i mg. Now we prove the above theorem. Proof of Theorem 3.3 Let f and L be as in the hypothesis. Let L pf (n)-tt A via a machine M and let A be P-selective. Let c > 0 and de ne h(n) = b(1 + c) log f (n)c. Let n and x1; ; xh(n) be such that n = maxfjx1j; ; jxmjg). For each i; 1 i h(n), let Qi denote the set of all queries of M on xi , and let R = Q1 [ [ Qh(n) . Since f is monotone nondecreasing, k Qi k f (n), so, for suciently large n, it holds that
k R k h(n)f (n) 2log h(n)+log f (n) 2h(n) ? 2:
P
By Lemma 3.4, we can compute, in time polynomial in y2R jy j, and thus, in time polynomial in jxj, an enumeration y1 ; ; ykRk of elements in R such that for some m; 0 m k R k, R \ A = fyi j 1 i mg. Now for each m; 0 m k R k, let Bm = fyi j 1 i mg and for each j; 1 j h(n), let bm;j = 1 if M Bm on xj accepts and 0 otherwise. Clearly, there is some m such that for every j; 1 j h(n), L(xj ) = bm;j . Since k R k 2h(n) ? 2, there is some v 2 f0; 1gh(n) such that v 6= bm;1 bm;h(n) for any m. Let v0 be the smallest such v and de ne r(x1; ; xh(n)) = v0 . It is easy to see that r witnesses that A 2 P-mc(h). This proves the theorem. 2
Corollary 3.5 Pbtt(P-Sel) P-mc(const) and Ptt(P-Sel) P-mc(log). A function h is said to be polynomially length-bounded if there is a polynomial p such that for every x, jh(x)j p(jxj).
De nition 3.6 [KL80] A set L is in P=poly if there exist a polynomially length-bounded function h and a set A 2 P such that for every x, it holds that x 2 L $ (x; h(0jxj)) 2 A: 9
Ko [Ko83] showed that P-selective sets have polynomial-size circuits. Noting for a Pselective set A, a nite set W , and a string x, that W is partitioned into two sets W1 ; W2, such that x 2 A =) W1 A and x 2 Ac =) W2 Ac , Ko developed a divide-and-conquer method to nd polynomially length-bounded advice. Such a method is, however, hard to nd for P-mc(2)-sets, for, the set W is now partitioned into four sets W1; ; W4 such that x 2 A =) W1 A, x 2 A =) W2 Ac , x 2 Ac =) W3 Ac , and x 2 Ac =) W4 Ac . Nonetheless, very surprisingly, P-mc(poly)-sets have polynomial-size circuits, which is stated below.
Theorem 3.7 P-mc(poly) P=poly. The proof of the above theorem is essentially the same as that of [ABG90, Theorem 10], so we omit the proof here. As a matter of fact, in [ABG90], Amir, Beigel, and Gasarch showed for any k 1, that P-mc(k) P=poly developing an algorithm to construct an advice string of length O(kn2 ) for n . Thus, even if k is a function of n that is polynomially bounded, their construction still works. It is well-known that P=poly = Rtt(TALLY) = Rtt(SPARSE) and that TALLY RT (P-Sel) [Sel79]. So, every set in P-mc(poly) is pT -reducible to some P-selective set.
Corollary 3.8 P-mc(poly) RT (P-Sel). Since Rtt(P-Sel) P-mc(log) and P-mc(log) 6= P-mc(poly), the above inclusion is optimal.
Corollary 3.9 P-mc(poly) 6 Rtt(P-Sel). The converse of Theorem 3.7 does not appear to hold. The question of whether the converse holds is very subtle. For, proving P-mc(poly) 6= P=poly is at least as hard as proving P 6= NP.
Theorem 3.10 If P = NP, then P=poly P-mc(poly). Proof Suppose P = NP. Let L 2 P=poly. Since P=poly = Rtt(TALLY), there is a tally
set T and a polynomial time-bounded deterministic oracle Turing machine M such that for every x, x 2 L if and only if M T on x accepts. Let p be a polynomial bounding the run-time of M . Without loss of generality, we may assume for every x and oracle X , that any query of M X on x is in f0; ; 0p(jxj)g. Let x1 ; ; xm be strings of length at most 10
n with m = p(n) + 1. Let T1; ; Tk be an enumeration of all subsets of f01; ; 0p(n)g, where k = 2p(n). Note that evert Ti can be represented by a string of length p(n). There is some i; 1 i k, such that for every j; 1 j m, M T on xj accepts if and only if M Ti on xj accepts. For each i; 1 i k, and j; 1 j m, let b(i; j ) = 1 if M Ti on xj accepts and 0 otherwise. For each i; 1 i k, let ci = (b(i; 1); ; b(i; m)). Clearly, there is some i; 1 i k, such that ci = (L(x1); ; L(xm )). Since m = p(n) + 1, there is some c 2 f0; 1gm such that c = 6 ci for any i. Let c^ be the smallest such c and de ne g(x1; ; xm) = c^. Then, c^ = 6 (L(x1); ; L(xm)). By our supposition that P = NP, as representation of Ti ranges over strings of length p(n), the above c^ can be computed in time polynomial in n, so g is polynomial-time computable. Therefore, L 2 P-mc(p + 1). Hence, P=poly P-mc(poly). 2 We conjecture that P-mc(poly) is a proper subclass of P=poly.
4 Relationships with Sparse and Tally Sets In this section we study relationships between polynomial-time membership comparable sets and sparse sets as well as tally sets. Since P=poly = Rtt(SPARSE) = Rtt(TALLY), by Theorem 3.7, it holds that P-mc(poly) Rtt(SPARSE) and P-mc(poly) Rtt(TALLY). Moreover, as P=poly = RT (P-Sel) and P-Sel P-mc(2), we have SPARSE RT (P-mc(poly)). But, in fact, it holds that SPARSE P-mc(poly).
Proposition 4.1 SPARSE P-mc(poly). Proof Let S be a sparse set. There is a polynomial p such that for every n, k S \ n k< p(n). De ne g to be a function that, given x1 ; ; xm with m p(n) and n = maxfjx1j; ; jxmjg, outputs 1m . The function g is polynomial-time computable. For every x1 ; ; xm with m p(n) and n = maxfjx1j; ; jxmjg, it cannot happen that x1; ; xm 2 S because k S \ n k< p(n). Thus, g witnesses that S 2 P-mc(poly). 2 Thus, it is possible for an uncountable number of sets to be in P-mc(poly) via the same function g . By Proposition 2.1 (3), we have the following theorem.
Theorem 4.2 R1-tt(SPARSE) P-mc(poly). For any tally set T , let r(T ) denote T (0)T (00)T (000) . The string r(T ) can be viewed as a binary real number. For a binary real number r(T ) 2 [0; 1), de ne Left-Cut [r] to be 11
the set of binary strings w smaller than or equal to r(T ) in dictionary order and Pre x [r] to be the set of all initial bits of r(T ). Selman [Sel79] showed the following.
Theorem 4.3 For any binary real number r(T ) 2 [0; 1), the following properties hold. 1. Left-Cut [r] is P-selective. 2. Left-Cut [r] pptt T and T pT Left-Cut [r]. 3. If Pre x [r] is P-selective, then T 2 P. 4. Pre x [r] ptt T and T pT Pre x [r]. How complex is Pre x [r]? We show below that Pre x [r] is polynomial-time 2membership comparable.
Theorem 4.4 For any r 2 [0; 1), Prefix[r] is in P-mc(2). Proof Let r 2 [0; 1). Let x; y be distinct two strings. Then the following properties hold: If x is a pre x of y, then y 2 Prefix[r] ! x 2 Prefix[r]. If y is a pre x of x, then x 2 Prefix[r] ! y 2 Prefix[r]. If x is not a pre x of y and y is not a pre x of x, then at most one of x and y is in Prefix[r]. De ne g (x; y ) = 01 if the rst condition is satis ed, 10 if the second condition is satis ed, and 11 otherwise. Clearly, g witnesses that Prefix[r] 2 P-mc(2). 2
Corollary 4.5 For any tally set T , there is a sparse set S 2 P-mc(2) such that T pT S and S ptt T . Since there is a tally set not in P, we have the following corollary.
Corollary 4.6 There is a tally set T such that Pre x [r(T )] is in P-mc(2) ? P-Sel. It is well-known that for every set A 2 P=poly, there is a tally set T 2 p3 (A) such that A ptt T (see for example, [Sch86]). By Proposition 4.1, TALLY P-mc(poly). So, we have the following corollary. 12
Corollary 4.7 For every A 2 P-mc(poly), there is a tally set T in P-mc(poly) \ p3(A) such that A ptt T . Moreover, by Corollary 4.5, for every tally set T , there is a sparse set S 2 P-mc(2) such that T and S are pT -equivalent. Therefore, we have the following corollary.
Corollary 4.8 For every A 2 P-mc(poly), there is a sparse set S in P-mc(2) \ p3(A) such that A pT S . As SPARSE P-mc(poly), we have the following.
Corollary 4.9 For every sparse set S , there is a sparse set S 0 2 P-mc(2) \ p3(S ) such that S pT S 0. Therefore, P=poly = Rtt(SPARSE) = RT (SPARSE \ P-mc(2)):
5 Polynomial-Time Membership Comparable Hard Sets In this section, we show for some complexity classes C , that if for some c < 1, C P-mc(c log n), then C = P. We note here that the author has been recently informed that some of the results in this section had been independently proven by Agrawal and Arvind [AA94] and by Beigel, Kummer, and Stephan [BKS94]. We start by considering NP.
Theorem 5.1 If NP P-mc(c log n) for some c < 1, then P = NP. Proof Suppose that NP P-mc(c log n) for some c < 1. Take a as a natural number such that c < 1 ? 1=a. Consider an NP-complete set SAT. Without loss of generality, we may assume that
for every formula ', each truth assignment for ' is encoded into a string of length j'j1=2a. Let ' be a formula and Y be a set of pre xes of truth assignments for '. Call Y good for ' if Y contains a pre x of a satisfying assignment for '. By our assumption, there is an encoding '#Y of ' and Y such that if k Y k j'j ? 1, then j'#Y j = j'j1+1=a. De ne A = f'#Y j Y is good for 'g. Obviously, A 2 NP, so, A 2 P-mc(c log n). Let f be an membership comparing function witnessing this property. 13
Let ' be a formula with j'j = r = 2d and Z = fy1 ; ; yr?1g be a set of r ? 1 many pre xes of truth assignments for '. Suppose that the sets of truth assignments represented by these pre xes are disjoint; that is, for any i; j; 1 i < j r ? 1, yi is not a pre x of yj and yj is not a pre x of yi . For each i; 1 i d, let Yi denote the set of all yj ; 1 j r ? 1, such that the i-th bit of j 's binary representation in f0; 1gd is a 1; that is, j 's representation is of the form b1 bd with bi = 1. Note for any i; 1 i d, that k Yi k= 2d?1 = r=2, so j'#Yi j = j'j1+1=a = r1+1=a = 2d(1+1=a), and thus, c log j'#Yij < cd(1 + 1=a) < (1 ? 1=a2)d < d. So, given '#Y1 ; ; '#Yd as arguments, f must exclude one possibility of (A('#Y1); ; A('#Yd)); that is, f ('#Y1; ; '#Yd) maps to some b = b1 bd 2 f0; 1gd so that (A('#Y1); ; A('#Yd)) 6= b. Suppose that b = 0d . Then, it holds (A('#Y1); ; A('#Yd)) 6= 0d . So, at least one of '#Y1 ; ; '#Yd is in A, and thus, at least one of Y1 ; ; Yd is good for '. Therefore, ' is satis able. On the other hand, suppose that b 2 f0; 1gd ? f0d g. Let t be the number whose binary representation is b = b1 bd . We show that if Z is good then Z ?fyt g is good. Assume, by way of contradiction, that Z is good but Z ? fytg is not good. Then for every i, '#Yi 2 A if and only if yt 2 Yi . On the other hand, for every i, yt 2 Yi if and only if the i-th bit of t, which is bi, is a 1. Therefore, for every i, '#Yi 2 A if and only if bi = 1. This implies (A('#Y1); ; A('#Yd )) = b1 bd = b, which contradicts f 's declaration that (A('#Y1); ; A('#Yd)) 6= b. So, if Z is good, then Z ?fyt g is good. Moreover, if Z ?fyt g is good, then since it is a subset of Z , Z is good, too. Therefore, in this case, it holds that Z is good if and only if Z ? fytg is good. De ne PRUNE to be a procedure that, given ' and Z as above, (i) computes b = f ('#Y1 ; ; '#Yd) and (ii) outputs `YES' if b is all 0 and outputs Z ? fyt g otherwise. Clearly, PRUNE is a polynomial-time procedure and for every ' and Z , it holds that:
if PRUNE outputs `YES', then ' 2 SAT, and if PRUNE outputs a set Z 0, then Z 0 is good if and only if Z is good. Moreover, when Z consists of truth assignments for ', whether Z is good can be tested in time polynomial in j'j. Now consider a decision procedure that, given a formula ', behaves as follows: (0) Initially, set Z to fg.
(1) According to k Z k, do the following. 14
(1a) k Z k< j'j ? 1: Let z be the smallest y 2 Z in canonical lexicographic order
and replace z with z 0 and z 1. (1b) k Z k= j'j ? 1: Call PRUNE ('; Z ). If PRUNE outputs `YES', then accept '. Otherwise, replace Z with the output of PRUNE .
(2) If not all y 2 Z are truth assignments for ', then goto (1). Otherwise, accept ' if and only if Z is good.
It is not hard to see that the procedure is polynomial-time bounded and accepts ' if and only if ' is satis able. Therefore, SAT 2 P. This proves the theorem. 2
Remark 1 We note here that the above proof does not seem to work for the case c 1,
even if we wish to prove a consequence weaker than P = NP. Let ' be of length 2d . Suppose we wish to preserve k Z k 2H (d) ? 1 for some function H . (Note that H (d) = O(d) if we wish to develop a polynomial-time algorithm.) In order to eliminate one pre x from Z , we construct subsets Y1; ; YH (d) of Z , each consisting of 2H (d)?1 pre xes. Let t be the average length of pre xes in Z . Then, '#Yi 's must encode at least 2d + 2H (d)?1t bits in average. Thus, for some i, it holds j'#Yi j (2d + 2H (d)?1t), where is a constant depending only on the size of the encoding alphabet. Now, the number of arguments we must give to f is at least
c log maxfj'#Y1j; ; j'#YH (d)jg c log( (2d + 2H (d)?1t)) > c log + c log(2d + 2H (d)?1t) > cH (d) + c log(2d?H (d) + t=2) As t cannot be bounded by any constant, we may assume log(2d?H (d) + t=2) 1. So, we need more than cH (d)+ c arguments, but if c 1, this is impossible, because we have only H (d) arguments. The same arguments apply for the proof of Theorem 5.5. Next we consider subclasses of NP; namely, UP and FewP, which are de ned by Valiant [Val76] and Allender [All85], respectively. For a polynomial time-bounded nondeterministic Turing machine M , let #accM denote the function that maps x to the number of accepting computation paths of M on input x. A set L is in UP (respectively, FewP) if there is a polynomial time-bounded nondeterministic Turing machine M witnessing L 2 NP such that for every x, #accM (x) 1 (respectively, for every x, #accM (x) p(jxj) for some polynomial depending only on M ). By Cook's reduction [Coo71] and padding arguments 15
(see for example, [BH77]), for any NP-acceptor M and for any a 2 N+ , one can construct a polynomial-time computable function f such that the following conditions are satis ed:
f () is a set of formulas, and is in P; for every x, #accM (x) equals the number of satisfying assignments for f (x); and for every x, truth assignments for f (x) are of length jf (x)j1=2a. De ne S = f () \ SAT and de ne A as in the proof of Theorem 5.1 with S in place of SAT. Then L(M ) pm S; S pm A, and S; A 2 UP if L(M ) 2 UP (S; A 2 FewP if L(M ) 2 FewP). Thus, we can use our technique to prove results similar to that of Theorem 5.1 for UP and FewP.
Theorem 5.2 If UP P-mc(c log n) for some c < 1, then UP = P. Therefore, if UP Rnc-tt (P-Sel) for some c < 1, then UP = P.
Theorem 5.3 If FewP P-mc(c log n) for some c < 1, then FewP = P. Therefore, if UP Rnc -tt(P-Sel) for some c < 1, then FewP = P. Next we consider counting complexity classes C= P and PP, and PSPACE. A set A is in PP [Gil77, Sim75] (C= P [Wag86, Sim75]) if there exist some machines M and N such that for every x, x 2 A if and only if #accM (x) #accN (x) (#accM (x) = #accN (x)).
Theorem 5.4 Let C be in fPP; C=P; PSPACEg. If C P-mc(c log n) for some c < 1, then C = P. Proof Note that NP is a subset of either C or co-C . Since C and co-C are both contained
in P-mc(c log n), by Theorem 5.1, P = NP. Hence P = p2 . Since each of C= P, PP, and PSPACE has `one word-decreasing self-reducible' pm -complete sets [OL93] and if a `worddecreasing self-reducible' set A is in P=poly, then p2(A) p2 [KL80], we have C p2. This establishes that C = P. 2 Let k 2. A set A is in MODk P [CH90] if there is some machine M such that for every x, x 2 A if and only if #accM (x) 6 0 modulo k. The argument for C= P; PP, and PSPACE cannot be applied to MODk P, for, it is not known whether NP or coNP is included in MODk P. So, we need to develop a direct proof.
Theorem 5.5 Let k 2. If MODk P P-mc(c log n) for some c < 1, then MODk P = P. 16
Proof Let k 2 and suppose MODk P P-mc(c log n) for some c < 1. Let a be a natural number such that c < 1 ? 1=a. Without loss of generality, we may assume for any formula ' of propositional logic and any truth assignment y for ', that jyj = j'j1=2a. For a formula ' and y; jyj j'j1=2a, let ('; y) =k fyz : yz is a satisfying assignment for 'g k modulo k. Note for every ' and y , ('; y ) 2 f0; ; k ? 1g. De ne L0 = f' : ('; ) 6= 0g and L1 = f('; l) : ('; ) = lg. It is well-known that L0 is pm -complete for MODk P and it is clear that L0 is p(k?1)-dtt -reducible to L1. Below, we will show that L1 is in P. Let ' be a formula and let Y be a set of pairs (y1 ; m1); ; (yd ; md) such that for all i; 1 i d, jyij j'j1=2a and mi 2 f0; ; k ? 1g. Call Y good for ' if for every i; 1 i d, (x; yi ) = mi . By our assumption on the length of formulas and their truth assignments, there is an encoding '#Y such that if k Y k j'j=2, then j'#Y j = j'j1+1=a. De ne A to be the set of all '#Y such that Y is good for '. Since MODk P is closed under pdtt reductions [Her90], and thus, co-MODk P is closed under pctt -reductions by symmetry, we have A 2 co-MODk P. So, by our supposition, Ac is in P-mc(c log n). Let f be a function witnessing this property. We will show that L1 2 UP. Let ' be a formula such that j'j = r = 2d and let Z = f(yi; zi ) : 1 i r ? 1g be such that for every i; 1 i r ? 1, jyij r1=2a and mi 2 f0; ; k ? 1g. Moreover, suppose for every i; j; 1 i < j r ? 1, that yi is not a pre x of yj and yj is not a pre x of yi . For each l; 1 l d, let Yl be the set of all (yi ; zi) such that the l-th bit of i's binary representation in f0; 1gd is a 1; that is, i's representation is of the form b1 bd with bl = 1. It is easy to see that each Yl contains exactly r=2 pairs, and thus, that j'#Yl j = r1+1=a = 2d(1+1=a). Let b = f ('#Y1; ; '#Yd ). Since c log j'#Ylj cd(1 + 1=a) < d(1 ? 1=a2 ) < d, b must be of length d and di er from (Ac ('#Y1 ); ; Ac ('#Yd )). Suppose that b 2 0d . Then there is some l such that '#Yl 2 Ac , so Yl is not good for ', and thus Z is not good for ' because each Yl is a subset of Z . On the other hand, suppose that b 2 f0; 1gd ? f0d g. For each i, let bi denote the i-th bit of b. Let t be the number whose binary representation is b. We show that Z is good if Z ? f(yt ; mt)g is good. Assume, by way of contradiction, that Z ? f(yt ; mt)g is good and Z is not good. By de nition, for every l, Yl is good if and only if (yt ; mt) 62 Yl . On the other hand, for every l, (yt ; mt) 2 Yl if and only if the l-th bit of t's binary representation, which is bl, is a 1. So, for every l, Yl is good if and only if bl = 0. 17
Thus, (Ac ('#Y1); ; Ac('#Yd )) = b1 bd = b, which contradicts f 's declaration that (Ac ('#Y1 ); ; Ac ('#Yd )) 6= b. Therefore, Z is good if Z ? f(yt; mt)g is good. Moreover, if Z is good, then obviously, for any nonempty subset Y of Z , Y is good. Hence, Z is good if and only if Z ? f(yt ; mt)g is good. Now de ne PRUNE to be a procedure that, given x and Z as above, computes b = f ('#Y1; ; '#Yd) and outputs `NO' if the value is all 0 and outputs Z 0 = Z ? f(yt ; mt)g otherwise. Then the following properties hold:
PRUNE is a polynomial-time procedure; if PRUNE outputs `NO', then Z is not good; and if PRUNE outputs Z 0, then Z is good if and only if Z 0 is good. Moreover, if Z consists only of pairs of the form (y; m) with y being a truth assignment for ', then whether Z is good can be easily tested. Because Z is good if and only if for every (y; m) 2 Z , it holds that m = 1 if y is a satisfying assignment for x and m = 0 otherwise. Now de ne M to be a nondeterministic Turing machine that, on input ('; l), behaves as follows:
(0) Initially, set Z to f(; l)g. (1) According to k Z k, do one of the following. (1a) k Z k= j'j ? 1: Call PRUNE ('; Z ). If PRUNE outputs `NO,' then reject and
halt. Otherwise, set Z to the output of PRUNE . (1b) k Z k< j'j ? 1: Let Z = f(yi; mi) : 1 i dg and let yl be the smallest in fy1; ; ydg in canonical lexicographic order. Nondeterministically guess n0; n1 2 f0; ; k ? 1g such that n0 + n1 mr modulo k and replace (yr ; mr) with two elements (yr 0; n0) and (yr 1; n1).
(2) If there is some (y; m) in Z such that y is not a truth assignment for ', then goto (1). Otherwise, accept ' if and only if Z is good.
Suppose that M on input ('; l) is at the start of step (1) with Z . Suppose that W is good. If M is to enter (1a), then Z is replaced with a good one, and if M is to enter (1b), then clearly, there uniquely exists a guess of (n0 ; n1), for which Z is substituted with a good one. On the other hand, suppose that Z is not good. If M is to enter (1a), then M either rejects 18
or substitutes Z with a one that is not good, and if M is to enter (1b), then for every guess of (n0 ; n1), M substitutes Z with a one that is not good. So, if ('; l) is in L1 , then there exists a unique path leading to step (2) with a good Z ; and if ('; l) 62 L1 , there is no such path. Therefore, if ('; l) 2 L1, there uniquely exists a path leading to acceptance, and if ('; l) 62 L1 , then there exist no such paths. This implies L1 2 UP. So, MODk P UP. Since UP MODk P, by Theorem 5.2, we have UP = P. Hence, MODk P = P. This proves the theorem. 2 The proof techniques we have developed enable us to resolve some open questions. Selman [Sel79] showed if NP Rm (P-Sel), then P = NP. It has been studied whether a similar statement holds for more exible reducibilities. But, it has been open for a long time whether NP Rbtt(P-Sel) implies P = NP. By Theorems 3.3 and 5.1, we give an armative answer to this question.
Corollary 5.6 NP Rbtt(P-Sel) implies P = NP. In fact, NP Rnc-tt(P-Sel) implies P = NP for any c < 1.
Theorem 5.1 yields another consequence. For a set A and a function f : N ! N, let FPAf(n)-T (FPAf(n)-tt) denote the class of functions that are polynomial-time computable with at most f (jxj) adaptive (nonadaptive) queries to A. Krentel [Kre88] showed for any SAT f (n) 12 log n, that FPSAT f (n)-T FP(f (n)?1)-T if and only if P = NP. Krentel asked whether a similar result holds for a larger function f . Beigel [Bei88] strengthened the bound to c log n for any constant c < 1. He further asked a similar question with FPSAT f (n)-tt in place of p SAT FPf (n)-T and showed for any 1-tt -hard set A for NP, for any constant c < 1, and for any f such that f (n) c log n, that if FPAf(n)-tt FPX(f (n)?1)-T for some X , then RP = NP and P = UP. We prove that the conclusion of Beigel's result can be strengthened to P = NP.
Theorem 5.7 Let f (n) c log n for some constant c < 1. Let B be p1-tt-hard for NP. If for some set X it holds that FPBf(n)-tt FPX(f (n)?1)-T , then P = NP. Proof The proof is quite similar to that of Theorem 5.1. Let f , c, B, and X be as in the hypothesis. Without loss of generality, we may assume that f (n) = b(1 ? 1=a) log nc for
some natural number a. De ne the notion of `good' sets, the encoding x#Z , and the set A as in the proof of Theorem 5.1. For every ' and a set Z of at most j'j? 1 pre xes of truth assignments for ', j'#Z j = j'j1+1=a. De ne g (n) = f (n1+1=a). Then, g (n) < log n for all n. Let h be a function that, given ' and a set Z of at most j'j? 1 pre xes of truth assignments for ', outputs (A('#Y1 ); ; A('#Yg(j'j))), where Yi 's are subsets of Z de ned in the proof 19
of Theorem 5.1. Since j'#Z j = j'j1+1=a, h 2 FPAf(n)-tt, and thus, h 2 FPBf(n)-tt . So, by our supposition, h 2 FPX(f (n)?1)-T . Let M be a machine witnessing that h 2 FPX(f (n)?1)-T . For every ' and Z , there are 2g(j'j)?1 possible outputs of M . Since g (n) < log n, all such values can be computed in time polynomial in j'j. Moreover, since there are 2g(j'j) possible values of (A('#Y1); ; A('#Yg(j'j))), we can compute, in time polynomial in j'j, a value v 2 f0; 1gg(j'j), which is not equal to (A('#Y1); ; A('#Yg(j'j))). Now let h0 be a function that, given '#Z , maps to v 0log j'j?g(j'j). Clearly, h0 is polynomial-time computable, and (A('#Y1); ; A('#Ylog j'j )) 6= h0 ('#Z ). Therefore, as in the proof of Theorem 5.1, we can de ne a polynomial-time decision procedure for SAT. This proves the theorem. 2 The above two results can be applied to other complexity classes. Corollary 5.8 Let C be a class chosen from f PSPACE, UP, FewP, C= P, PP, MOD2P, MOD3P, g. 1. If C Rnc -tt (P-Sel) for some c < 1, then C = P. 2. Let H be p1-tt -hard for A and f (n) c log n for some c < 1. If FPHf(n)-tt FPX(f (n)?1)-T for some X , then C = P.
Acknowledgment The author would like to thank Lane Hemaspaandra for useful com-
ments.
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