Complicated Complementations
Complicated Complementations Harry Buhrman CWI PO Box 94079 1090 GB Amsterdam The Netherlands
Leen Torenvliety University of Amsterdam Department of Computer Science Plantage Muidergracht 24 1018 TV Amsterdam The Netherlands November 12, 1998 Abstract
Kolmogorov complexity has proven to be a very useful tool in simplifying and improving proofs that use complicated combinatorial arguments. In this paper we use Kolmogorov complexity for oracle construction. We obtain separation results that are much stronger than separations obtained previously even with the use of very complicated combinatorial arguments. Moreover the use of Kolmogorov arguments almost trivializes the construction itself. In particular we construct relativized worlds where: 1. NP \ CoNP 2= P=poly. 2. NP has a set that is both simple and NP \ CoNP-immune. 3. CoNP has a set that is both simple and NP \ CoNP-immune. 4. p2 has a set that is both simple and p2 \ p2 -immune.
1 Introduction Some complexity classes are closed under complementation and some are not. Obviously all deterministic complexity classes are closed under complementation, but for most nondeterministic complexity classes the closure remains a big open question. When results in the real world seem hard to obtain, oracle construction is a very useful tool to guide research (see [For94]). For nondeterministic complexity classes, oracles that force closure under complementation are relatively easy to construct. Oracles that force non-closure seem much harder. The oracles given by Baker, Gill and Solovay [BGS75] relative to which P = NP, not only give closure under complementation of NP, but also of all subsequent classes in the polynomial time hierarchy. The oracle given later by Baker and Selman [BS79], that separates the second level of the polynomial time hierarchy is far more complicated. Yao [Yao85], building on an earlier established relationship between the polynomial time hierarchy and circuits by Furst, Saxe and Sipser [FSS84], was the rst to build an oracle relative to which the polynomial time hierarchy is in nite. Ko [Ko90] proved the existence of oracles relative to which the levels of the hierarchy are separated by immunity and Bruschi [Bru92] showed an oracle relative to which complementary levels of the hierarchy are separated by simple sets. Of a di erent order is the combination of simplicity and immunity. A simple set has to be very dense, because its complement must be immune. An immune set has to be very thin because no set belonging to some complexity class may grow in nite within it. The combination of these two seems to require a di erent (more complicated) technique. Before this paper only separation on the rst level achieving both simplicity and immunity was known [TV89]. In that paper, an oracle is constructed relative to which NP has a set that is both simple and P-immune. The question arises which separations are possible and which are not. Email: [email protected]. Partially supported by the Dutch foundation for scienti c research (NWO) by SION project 612-34-002, and by the European Union through NeuroCOLT ESPRIT Working Group Nr. 8556, and HC&M grant nr. ERB4050PL93-0516. y Research done while on leave at CWI. Email: [email protected].
1
In this paper we further explore the question of how strong separations can be achieved in relativized worlds concentrating on the polynomial time hierarchy. How far can a class in the hierarchy and its complement exactly be separated? We prove here that an oracle A can be constructed such that which NPA has a set that is both simple and NPA \ CoNPA -immune. Note that this result is notationally almost the same, but much stronger than (NP \ CoNP)A -immune, since the latter is the class of languages recognized by pairs of NP machines that recognize each others complement relative to every oracle. Through complementation, the same result can be achieved for CoNP. I.e., we construct an oracle relative to which CoNP has a set that is simple and NP \ CoNP-immune. We prove an analogous result for the second level that solves a stronger version of an open problem posed in [Bru92]. The strongest separation possible is that of a class, like NP, to have a set that is both simple and CoNP-immune. Our result is just an inch away from that. In the past, strengthening separation results like that of Baker and Selman, seemed hard to obtain because the combinatorics needed to prove the simple separation results were already prohibitively complicated. We overcome this problem by using Kolmogorov complexity and the incompressibility method. Fortnow and Laplante [FL95] used this method to redo some lower bound results in circuit complexity. A question directly related to complementation in levels of the polynomial time hierarchy is that of NP having small circuits. The famous Karp-Lipton theorem [KL80] shows that in any world where NP has small circuits the second level of the hierarchy is closed under complementation. The question whether the converse is true remains open. Can there be a world in which the hierarchy collapses, yet NP is not in P=poly. This question was rst put by Fortnow [For98]. In Section 3 we address the question how close to P can classes not having polynomial size circuits be. We construct an oracle relative to which NP \ CoNP does not have polynomial size circuits. In Section 4 we construct the oracles for the rst and second levels of the hierarchy. In Section 5, we discuss possible extensions of the theorems presented here and open problems.
2 Preliminaries 2.1 De nitions
We assume the reader familiar with standard notions in structural complexity theory. However we will use some maybe less common notions which we de ne in this section as follows:
Simplicity and Immunity De nition 2.1 Let C be a class of languages. A co-in nite language L 2 C is called simple, or C -simple if (8L0 2 C )[j L0 j = 1 =) L0 \ L = 6 ;]. De nition 2.2 Let C be a class of languages. An in nite language L is called C -immune if (8L0 2 C )[j L0j = 6 ;]. 1 =) L0 \ L = Complexity Classes We assume the reader familiar with all standard complexity classes such as P, NP, CoNP etc. We recall the de nitions of the following classes. De nition 2.3 A language L is in P=poly if there exists a function f : N ! , a polynomial p and a set B in P such that (8n)[jf (n)j p(n)] ^ (8x)[x 2 L () <x; f (jxj)> 2 B ]. De nition 2.4 The Polynomialp Hierarchy consists of a sequence of complexity classes pi , pi and pi , p p p p where 0 = 0 = P, i+1 = P i , i+1 = NP i and pi = Copi . Kolmogorov Complexity We use two versions of Kolmogorov complexity in this paper. The standard version C and the pre x free version K . We refer to [LV93] for an extensive description of their properties. Here we need the following results. Theorem 2.5 ([LV93] Theorem 2.13) Let f (n) be a recursive function such that P1n=1 2?f (n) < 1 is recursively convergent. If an in nite binary sequence ! is random with respect to the uniform measure then C (!1:n jn) > n ? f (n) from some n onwards. Which means that there is an in nite string such that from some n onwards, the rst n bits cannot be described using less than e.g., n ? log n ? 2 log log n bits. For pre x free descriptions the situation is even 2
more tight.
Theorem 2.6 ([LV93] Theorem 3.3) An in nite binary sequence ! is random with respect to the uniform measure if and only if there is a constant c such that K (!1:n) n ? c, for all n. Which means that we can pick an in nite binary string such that for some constant c and all n there is no pre x free description of the rst n bits of this string using less than n ? c bits.
2.2 Notations
Substrings For a string x, the i-th bit of x will be denoted by x[i]. The i-th bit of the j -th string in
some sequence looks like xj [i]. Substrings of string x can now be described using the same notation. The ith through j th bit substring of x will be denoted x[i::j]. The cited theorems from [LV93] use a di erent notation.
Quanti ers Apart from the usual universal and existential quanti ers we simplify notations by using the following extended versions. 8k Here k can be any function. The meaning of a quanti cation 8k x is \for all x in of length less than or equal to k." The existential version of this quanti er is 9k . 1 8 This quanti cation means \for almost all" or \for all but nitely many." Its complementation 91 then means \there exist in nitely many."
p;A Oracles The classes in the Polynomial Hierarchy relativized to oracle A will be denoted as p;A i , i and
p;A i .
3 NP, Co-NP and NP/poly The famous Karp-Lipton theorem [KL80] states that the polynomial hierarchy collapses in any world where NP has small circuits. Therefore any world in which the hierarchy extends at least up to the second level is a world in which NP is not in P=poly. The converse is however still an open question. Is it perhaps always the case that NP is in P=poly whenever p2 = p2 , or are there perhaps worlds in which the hierarchy collapses and yet NP does not have small circuits? A related question is the following. How close to P can sets not having small circuits actually be? The current state of a airs in the real world can be found in [BFT98]. They give a nonrelativizing proof that MAEXP, the exponential time version of MA does not have small circuits. Theorem 3.1 There exists a set A such that NPA \ CoNPA 2= PA=poly. Proof: For all x in and j 2jxj in this proof let xj denote the j th string in the lexicographical ordering of all strings of length jxj. De ne DA = fx j (9y)[jyj = jxj2 ^ y[jyj] = 1 ^ x = xj ^ <j; y> 2 A]g. Clearly, DA 2 NPA . Furthermore DA will be nonempty only at certain easily recognizable lengths n. At these lengths for all j 2n there will be a pair <j; y> in A. Hence it will also be the case that DA 2 CoNPA . The set A is constructed by stages. At stage s we choose a new length ns = 2ns?1 . This gives us for almost all s that ns+1 > (ns )s so that that changes made to A at stage s cannot disrupt earlier stages and also (ns )s+2 > (s + 3)(ns )s+1 and 2ns > (s + k)(ns )s+k for any xed k, which we need later on in the proof. Then at stage i we ful ll the requirement: (8z )[jz j = ni =) [(9x 2 n )[x 2 DA () <x; z> 2= L(MiA)]]]. If this condition is ful lled for all i then there can be no polynomial p, function f and set B in P. Such that (8x)[x 2 DA () <x; f (jxj)> 2 B ]. This means that DA 2= PA=poly. We now come to a description of the construction. stage 0: n0 = 2; A0 = ;; stage s: ns = 2ns?1 ; ms = 2ns ; Choose a string Ys such that Ys = y1 : : : yms where jyi j = n2s and C (Ys ) jYs j. De ne As = f<j; yj > j j ms g.
End of Construction
3
A will be the union of all As . We will now prove that A ful lls all requirements. First note one caveat about notations. Although xj denotes the j th string in the lexicographical ordering of the strings of length jxj j, this is not the case for yj . At each stage s the string yj is the j th substring of Ys . De ne a string ds of length ms as ds [j] = yj [n2s ]. I.e., ds is the string formed from the last bits of the substrings of Ys . Then clearly C (ds ) jds j and DA \ns = ds |i.e., the bits of the characteristic string of DA at length n coincide with the bits of ds . Furthermore for any string x = xj that is not in DA \ ns by de nition yj [n2s ] = 0 and <j; yj > 2 A. Hence it is also clear that DA 2 NPA \ CoNPA . For the remainder of the proof let n = ns , d = ds and m = ms . We claim the following. Claim 3.2 For almost all s and all z, if jzj ns then there are less than ns di erent j such that Ms queries <j; yj > on inputs <x1 ; z>; : : : ; <xm ; z>. Proof: Suppose that there is a z0 of length ns such that Ms queries ns of these strings. We will show how to describe Ys with less than m n2 bits and hence arrive at a contradiction. Let V be a set of pairs <j; yj >, where yj is the j th substring of Ys , de ned as follows. A pair is in V if and only if <j; yj > is queried in the kth step of the computation of MsA on input <xi ; z0 >. Let V be ordered such that < i i (ns + 1) + k < i0 (ns + 1) + k0 . Let Vns consist of the rst ns strings in V . We can recover ns pairs <j; yj > from Vns by running Ms on the appropriate inputs without the help of A. Notably all queries not appearing in Vns are answered NO in this recovery procedure. Furthermore Vns can be described using less than (s + 1)ns+1 bits. Using this recovery procedure for ns substrings of Ys , we need the following for a total description of Ys . 1. ns bits to encode z0 2. (s + 1)ns+1 to help recover ns substrings of Ys as described above. 3. (2n ? ns )n2 bits for the description of the other strings of Ys . Which gives a total cost of (2n ? ns )n2 + (s + 2)ns+1 + O(1) bits which is less than 2n n2 for almost all s, a contradiction. 2 Now suppose there is an s and a z of length ns such that Ms recognizes precisely those strings <x; z> such that x 2 DA \ n . Then, since Ms cannot query more than ns strings in As as per Claim 3.2, it requires ns (n + s log n) to describe the appearance of those queries in a simulation of the computation of Ms on inputs x1 ; : : : ; xm (as all other queries are answered NO). Hence as the outcomes of these computations coincide with d, it requires (s + 1)ns+1 + O(1) bits to describe d. Since C (d) 2n this gives a contradiction for almost all s. 2 Theorem 3.1 Some remarks can be made on the issue raised in the beginning of this section. The diagonalization language used in Theorem 3 is in NP \ CoNP. Therefore this theorem provides a world in which NP does not have small circuits. Whether this is also a new world remains the question, since it is unclear whether in this world p2 = p2 . To diagonalize against all possible polynomial size advices, we had to build an oracle of super-polynomial density. This would not be necessary if we would diagonalize against advices of xed length (NP 2= P=nk ). We can in that case let DA at length n consist of a subset of the lexicographically rst nk+1 strings. Having an oracle of xed polynomial density would allow us to create a world in which NP = CoNP by starting with an oracle A where PA; = PSPACEA; and then create an oracle B for which NPAB 2= PAB =nk . Since an NP machine can now query the entire oracle B at length n, we have that in this world both NP = CoNP and NP 2= P=nk . However, Homer and Mocas [HM95] showed that EXP 2= P=nk and of course EXP is closed under complementation, so that the statement NP = CoNP ^ NP 2= P=nk is true in any world in which NP = EXP.
4 The Polynomial Hierarchy
4.1 Very Strong Separations on the First Level
In this section we will construct an oracle B relative to which NPA has a set that is both simple and NPA \ CoNPA -immune. By complementation of the Kolmogorov argument used in Theorem 4.1, we can also construct an oracle B relative to which CoNPB has a set that is both simple and NPB \ CoNPB -immune. Previously, the strongest known separation on this level was the existence of an oracle C such that NPC 4
has a set that is both simple and PC -immune [TV89]. This construction involved complicated searches and priority arguments. Contrary to this, our construction here is an \in place" diagonalization. I.e., at each stage s the construction decides whether to add the string xs to the witness language and no strings smaller than xs are considered after stage s. Theorem 4.1 There is an oracle A such that NPA has a set that is simple and NPA \ CoNPA-immune. Proof: The language that will have this property is DA = fx j (9y)[jyj = jxj2 ^xy 2 A]g. Theorem 2.6 gives us an in nite random string Y that has the property for some constant c and all n that K (Y [1::n]) > n ? c. We will construct A by stages. As+1 is the oracle de ned by the end of stage s and As+1 As for all s. The oracle A may have 2n strings of length n + n2 . We de ne A0 from substrings of Y as follows. Let xmi be the i-th string of length m in and let yim be the substring of length m2 of Y that starts at the (m2 ? 4m + 6)2m + (i ? 1) m2 ? 5th bit of Y . Note that the string that describes A0 m , say YAm0 , has length (2m2 ? 4m + 6)2m ? 6 and therefore for some c and all m it holds that K (YAm0 ) (2m2 ? 4m + 6)2m ? c. Note also that DA0 = . At each stage s we will decide the membership of the string xs in DA , where xs is the sth string in the lexicographical ordering of . Deciding membership of xs is deciding whether to remove xs y from A where xs y is the only string that is an extension of xs currently in A. The Kolmogorov property that makes the entire construction work (as expressed in Lemma 4.2) is that an NP machine that rejects some string xs with oracle As must also reject xs with oracle A or else it will allow us to describe some initial segment of Y using signi cantly less bits than the length of this segment. Let (Me )e be an enumeration of all nondeterministic polynomial time bounded Turing machines, where ne is the time bound on inputs of length n. The construction maintains two sets of requirements. First, a set Us of yet unsatis ed requirements to which occasionally a new element is added and from which satis ed requirements are removed. Second, a set Vs in which satis ed requirements are kept. Sometimes a satis ed requirement will be moved from Vs to Us at which time it will become unsatis ed. Every requirement either corresponds to a language in NPA or to a language in NPA \ CoNPA . Requirement R2e will represent the language L(MeA) and requirement R2e+1 will represent the language L(MiA) \ L(MjA ), where = e. We will prove that the construction of A meets the following requirements for all e. 1. R2e : j L(MeA)j = 1 =) L(MeA) \ DA 6= ;. 2. R2+1 : j L(MiA )j = 1 =) [[L(MiA ) 6= L(MjA)] _ [L(MiA) \ DA 6= ;]]. Requirements of type 1 ensure that DA has a nonempty intersection with every in nite language in NPA , and requirements of type 2 ensure that DA has a nonempty intersection with every in nite language in NPA \ CoNPA . A requirement R2e is active at stage s if 2e 2 Us and xs 2 L(MeAs ). A requirement R2e+1 is active at stage s if 2e + 1 2 Us , e = and xs 2 L(MiAs ). Now the construction can be described as follows: stage s: If no requirement is active As+1 = As ; Vs+1 = Vs ; Us+1 = Us; Else let e be the least active requirement in Us . If e is even then Us+1 = Us ? feg; Vs+1 = Vs [ f<e; xs >g; As+1 = As ; Else If MjAs (xs ) = 1 then Vs+1 = Vs [ f<e; xs >g; As+1 = As ; Us+1 = Us Else If f 2 Vs j i < e ^ jxji > jxs jg = ; then Vs+1 = Vs ? f 2 Vs j i > eg; Us+1 = Us [ fi j i > e ^ (9x)[ 2 Vs ]g ? feg; As+1 = As ? f<xs ; ys >g; Else As+1 = As ; Vs+1 = Vs ; Us+1 = Us ; If s = 2k then Us+1 = Us+1 [ fkg
End of Construction
Now we prove that all requirements are satis ed in the limit. First we show the following key lemma about the construction. Lemma 4.2 (81s)(8e < log s)[xs 2= L(MeAs ) =) xs 2= L(MeA)]. Proof: Suppose not. Let e and s be such that e < log s and xs 2 L(MeA) ? L(MeAs ). Let s0 be minimal such that s0 s and xs 2 L(MeAs0 +1 ) ? L(MeAs0 ). By construction As0 ? As0 +1 = fxs0 ys0 g. The string 5
xs0 ys0 must be queried in any accepting computation of MeAs0 +1 on input xs . Otherwise MeAs0 would also have an accepting computation on input xs contradicting the assumption. On input xs machine Me can only query strings of length less than or equal to jxs je , which is less than or equal to jxs jlog s . Moreover j fy j y 2 A0 ? As0 +1 gjj log s0 , since for every such y an odd index is in Ut ? (Ut+1 [ Vt+1 ) for some t s0 and there are no more than log s0 indices in [fUt j t s0 g. Let n be such that n + n2 > jxs je . We will show how to reconstruct the rst (2n2 ? 4n +6)2n ? 6 bits of Y using signi cantly less bits and hence arrive at a contradiction. Suppose that we have (2n2 ? 4n + 6)2n ? 6 ?jxs0 j2 bits of Y that describe all substrings of Y up to length n2 , except ys0 . Furthermore suppose that we have a list of at most log s0 strings xi1 ; : : : ; xik that says which xij yij are in A0 ? As0 . Note that no strings greater than xs0 ys0 are in A0 ? As0 . Finally let q < jxs je be the index of the query xs0 ys0 in the leftmost accepting computation of MeAs+1 (xs ). Then we can reconstruct the rst (2n2 ? 4n + 6)2n ? 6 bits of Y from (2n2 ? 4n + 6)2n ? 6 ? jxs0 j2 + log s0 jxs j + e(log jxs j + 1) + O(1) bits of information. We arrive at a contradiction for all but nitely many s since the complexity assumption on Y is violated. 2
The following lemma has a trivial claim. We isolate it nonetheless to avoid duplication in the proofs to come. Lemma 4.3 (8e)[j fs j (9x)[<e; x> 2 Vs+1 ? Vs ]gjj < 1]. Proof: Whenever <e; x> is in Vs+1 ? Vs there is a smaller index that is moved from Us+1 either to oblivion or to Vs+1 . This means that 0 can enter Vs for only one s and there is no t such that is in Vt ? Vt+1 . By induction forP i
Leen Torenvliety University of Amsterdam Department of Computer Science Plantage Muidergracht 24 1018 TV Amsterdam The Netherlands November 12, 1998 Abstract
Kolmogorov complexity has proven to be a very useful tool in simplifying and improving proofs that use complicated combinatorial arguments. In this paper we use Kolmogorov complexity for oracle construction. We obtain separation results that are much stronger than separations obtained previously even with the use of very complicated combinatorial arguments. Moreover the use of Kolmogorov arguments almost trivializes the construction itself. In particular we construct relativized worlds where: 1. NP \ CoNP 2= P=poly. 2. NP has a set that is both simple and NP \ CoNP-immune. 3. CoNP has a set that is both simple and NP \ CoNP-immune. 4. p2 has a set that is both simple and p2 \ p2 -immune.
1 Introduction Some complexity classes are closed under complementation and some are not. Obviously all deterministic complexity classes are closed under complementation, but for most nondeterministic complexity classes the closure remains a big open question. When results in the real world seem hard to obtain, oracle construction is a very useful tool to guide research (see [For94]). For nondeterministic complexity classes, oracles that force closure under complementation are relatively easy to construct. Oracles that force non-closure seem much harder. The oracles given by Baker, Gill and Solovay [BGS75] relative to which P = NP, not only give closure under complementation of NP, but also of all subsequent classes in the polynomial time hierarchy. The oracle given later by Baker and Selman [BS79], that separates the second level of the polynomial time hierarchy is far more complicated. Yao [Yao85], building on an earlier established relationship between the polynomial time hierarchy and circuits by Furst, Saxe and Sipser [FSS84], was the rst to build an oracle relative to which the polynomial time hierarchy is in nite. Ko [Ko90] proved the existence of oracles relative to which the levels of the hierarchy are separated by immunity and Bruschi [Bru92] showed an oracle relative to which complementary levels of the hierarchy are separated by simple sets. Of a di erent order is the combination of simplicity and immunity. A simple set has to be very dense, because its complement must be immune. An immune set has to be very thin because no set belonging to some complexity class may grow in nite within it. The combination of these two seems to require a di erent (more complicated) technique. Before this paper only separation on the rst level achieving both simplicity and immunity was known [TV89]. In that paper, an oracle is constructed relative to which NP has a set that is both simple and P-immune. The question arises which separations are possible and which are not. Email: [email protected]. Partially supported by the Dutch foundation for scienti c research (NWO) by SION project 612-34-002, and by the European Union through NeuroCOLT ESPRIT Working Group Nr. 8556, and HC&M grant nr. ERB4050PL93-0516. y Research done while on leave at CWI. Email: [email protected].
1
In this paper we further explore the question of how strong separations can be achieved in relativized worlds concentrating on the polynomial time hierarchy. How far can a class in the hierarchy and its complement exactly be separated? We prove here that an oracle A can be constructed such that which NPA has a set that is both simple and NPA \ CoNPA -immune. Note that this result is notationally almost the same, but much stronger than (NP \ CoNP)A -immune, since the latter is the class of languages recognized by pairs of NP machines that recognize each others complement relative to every oracle. Through complementation, the same result can be achieved for CoNP. I.e., we construct an oracle relative to which CoNP has a set that is simple and NP \ CoNP-immune. We prove an analogous result for the second level that solves a stronger version of an open problem posed in [Bru92]. The strongest separation possible is that of a class, like NP, to have a set that is both simple and CoNP-immune. Our result is just an inch away from that. In the past, strengthening separation results like that of Baker and Selman, seemed hard to obtain because the combinatorics needed to prove the simple separation results were already prohibitively complicated. We overcome this problem by using Kolmogorov complexity and the incompressibility method. Fortnow and Laplante [FL95] used this method to redo some lower bound results in circuit complexity. A question directly related to complementation in levels of the polynomial time hierarchy is that of NP having small circuits. The famous Karp-Lipton theorem [KL80] shows that in any world where NP has small circuits the second level of the hierarchy is closed under complementation. The question whether the converse is true remains open. Can there be a world in which the hierarchy collapses, yet NP is not in P=poly. This question was rst put by Fortnow [For98]. In Section 3 we address the question how close to P can classes not having polynomial size circuits be. We construct an oracle relative to which NP \ CoNP does not have polynomial size circuits. In Section 4 we construct the oracles for the rst and second levels of the hierarchy. In Section 5, we discuss possible extensions of the theorems presented here and open problems.
2 Preliminaries 2.1 De nitions
We assume the reader familiar with standard notions in structural complexity theory. However we will use some maybe less common notions which we de ne in this section as follows:
Simplicity and Immunity De nition 2.1 Let C be a class of languages. A co-in nite language L 2 C is called simple, or C -simple if (8L0 2 C )[j L0 j = 1 =) L0 \ L = 6 ;]. De nition 2.2 Let C be a class of languages. An in nite language L is called C -immune if (8L0 2 C )[j L0j = 6 ;]. 1 =) L0 \ L = Complexity Classes We assume the reader familiar with all standard complexity classes such as P, NP, CoNP etc. We recall the de nitions of the following classes. De nition 2.3 A language L is in P=poly if there exists a function f : N ! , a polynomial p and a set B in P such that (8n)[jf (n)j p(n)] ^ (8x)[x 2 L () <x; f (jxj)> 2 B ]. De nition 2.4 The Polynomialp Hierarchy consists of a sequence of complexity classes pi , pi and pi , p p p p where 0 = 0 = P, i+1 = P i , i+1 = NP i and pi = Copi . Kolmogorov Complexity We use two versions of Kolmogorov complexity in this paper. The standard version C and the pre x free version K . We refer to [LV93] for an extensive description of their properties. Here we need the following results. Theorem 2.5 ([LV93] Theorem 2.13) Let f (n) be a recursive function such that P1n=1 2?f (n) < 1 is recursively convergent. If an in nite binary sequence ! is random with respect to the uniform measure then C (!1:n jn) > n ? f (n) from some n onwards. Which means that there is an in nite string such that from some n onwards, the rst n bits cannot be described using less than e.g., n ? log n ? 2 log log n bits. For pre x free descriptions the situation is even 2
more tight.
Theorem 2.6 ([LV93] Theorem 3.3) An in nite binary sequence ! is random with respect to the uniform measure if and only if there is a constant c such that K (!1:n) n ? c, for all n. Which means that we can pick an in nite binary string such that for some constant c and all n there is no pre x free description of the rst n bits of this string using less than n ? c bits.
2.2 Notations
Substrings For a string x, the i-th bit of x will be denoted by x[i]. The i-th bit of the j -th string in
some sequence looks like xj [i]. Substrings of string x can now be described using the same notation. The ith through j th bit substring of x will be denoted x[i::j]. The cited theorems from [LV93] use a di erent notation.
Quanti ers Apart from the usual universal and existential quanti ers we simplify notations by using the following extended versions. 8k Here k can be any function. The meaning of a quanti cation 8k x is \for all x in of length less than or equal to k." The existential version of this quanti er is 9k . 1 8 This quanti cation means \for almost all" or \for all but nitely many." Its complementation 91 then means \there exist in nitely many."
p;A Oracles The classes in the Polynomial Hierarchy relativized to oracle A will be denoted as p;A i , i and
p;A i .
3 NP, Co-NP and NP/poly The famous Karp-Lipton theorem [KL80] states that the polynomial hierarchy collapses in any world where NP has small circuits. Therefore any world in which the hierarchy extends at least up to the second level is a world in which NP is not in P=poly. The converse is however still an open question. Is it perhaps always the case that NP is in P=poly whenever p2 = p2 , or are there perhaps worlds in which the hierarchy collapses and yet NP does not have small circuits? A related question is the following. How close to P can sets not having small circuits actually be? The current state of a airs in the real world can be found in [BFT98]. They give a nonrelativizing proof that MAEXP, the exponential time version of MA does not have small circuits. Theorem 3.1 There exists a set A such that NPA \ CoNPA 2= PA=poly. Proof: For all x in and j 2jxj in this proof let xj denote the j th string in the lexicographical ordering of all strings of length jxj. De ne DA = fx j (9y)[jyj = jxj2 ^ y[jyj] = 1 ^ x = xj ^ <j; y> 2 A]g. Clearly, DA 2 NPA . Furthermore DA will be nonempty only at certain easily recognizable lengths n. At these lengths for all j 2n there will be a pair <j; y> in A. Hence it will also be the case that DA 2 CoNPA . The set A is constructed by stages. At stage s we choose a new length ns = 2ns?1 . This gives us for almost all s that ns+1 > (ns )s so that that changes made to A at stage s cannot disrupt earlier stages and also (ns )s+2 > (s + 3)(ns )s+1 and 2ns > (s + k)(ns )s+k for any xed k, which we need later on in the proof. Then at stage i we ful ll the requirement: (8z )[jz j = ni =) [(9x 2 n )[x 2 DA () <x; z> 2= L(MiA)]]]. If this condition is ful lled for all i then there can be no polynomial p, function f and set B in P. Such that (8x)[x 2 DA () <x; f (jxj)> 2 B ]. This means that DA 2= PA=poly. We now come to a description of the construction. stage 0: n0 = 2; A0 = ;; stage s: ns = 2ns?1 ; ms = 2ns ; Choose a string Ys such that Ys = y1 : : : yms where jyi j = n2s and C (Ys ) jYs j. De ne As = f<j; yj > j j ms g.
End of Construction
3
A will be the union of all As . We will now prove that A ful lls all requirements. First note one caveat about notations. Although xj denotes the j th string in the lexicographical ordering of the strings of length jxj j, this is not the case for yj . At each stage s the string yj is the j th substring of Ys . De ne a string ds of length ms as ds [j] = yj [n2s ]. I.e., ds is the string formed from the last bits of the substrings of Ys . Then clearly C (ds ) jds j and DA \ns = ds |i.e., the bits of the characteristic string of DA at length n coincide with the bits of ds . Furthermore for any string x = xj that is not in DA \ ns by de nition yj [n2s ] = 0 and <j; yj > 2 A. Hence it is also clear that DA 2 NPA \ CoNPA . For the remainder of the proof let n = ns , d = ds and m = ms . We claim the following. Claim 3.2 For almost all s and all z, if jzj ns then there are less than ns di erent j such that Ms queries <j; yj > on inputs <x1 ; z>; : : : ; <xm ; z>. Proof: Suppose that there is a z0 of length ns such that Ms queries ns of these strings. We will show how to describe Ys with less than m n2 bits and hence arrive at a contradiction. Let V be a set of pairs <j; yj >, where yj is the j th substring of Ys , de ned as follows. A pair is in V if and only if <j; yj > is queried in the kth step of the computation of MsA on input <xi ; z0 >. Let V be ordered such that < i i (ns + 1) + k < i0 (ns + 1) + k0 . Let Vns consist of the rst ns strings in V . We can recover ns pairs <j; yj > from Vns by running Ms on the appropriate inputs without the help of A. Notably all queries not appearing in Vns are answered NO in this recovery procedure. Furthermore Vns can be described using less than (s + 1)ns+1 bits. Using this recovery procedure for ns substrings of Ys , we need the following for a total description of Ys . 1. ns bits to encode z0 2. (s + 1)ns+1 to help recover ns substrings of Ys as described above. 3. (2n ? ns )n2 bits for the description of the other strings of Ys . Which gives a total cost of (2n ? ns )n2 + (s + 2)ns+1 + O(1) bits which is less than 2n n2 for almost all s, a contradiction. 2 Now suppose there is an s and a z of length ns such that Ms recognizes precisely those strings <x; z> such that x 2 DA \ n . Then, since Ms cannot query more than ns strings in As as per Claim 3.2, it requires ns (n + s log n) to describe the appearance of those queries in a simulation of the computation of Ms on inputs x1 ; : : : ; xm (as all other queries are answered NO). Hence as the outcomes of these computations coincide with d, it requires (s + 1)ns+1 + O(1) bits to describe d. Since C (d) 2n this gives a contradiction for almost all s. 2 Theorem 3.1 Some remarks can be made on the issue raised in the beginning of this section. The diagonalization language used in Theorem 3 is in NP \ CoNP. Therefore this theorem provides a world in which NP does not have small circuits. Whether this is also a new world remains the question, since it is unclear whether in this world p2 = p2 . To diagonalize against all possible polynomial size advices, we had to build an oracle of super-polynomial density. This would not be necessary if we would diagonalize against advices of xed length (NP 2= P=nk ). We can in that case let DA at length n consist of a subset of the lexicographically rst nk+1 strings. Having an oracle of xed polynomial density would allow us to create a world in which NP = CoNP by starting with an oracle A where PA; = PSPACEA; and then create an oracle B for which NPAB 2= PAB =nk . Since an NP machine can now query the entire oracle B at length n, we have that in this world both NP = CoNP and NP 2= P=nk . However, Homer and Mocas [HM95] showed that EXP 2= P=nk and of course EXP is closed under complementation, so that the statement NP = CoNP ^ NP 2= P=nk is true in any world in which NP = EXP.
4 The Polynomial Hierarchy
4.1 Very Strong Separations on the First Level
In this section we will construct an oracle B relative to which NPA has a set that is both simple and NPA \ CoNPA -immune. By complementation of the Kolmogorov argument used in Theorem 4.1, we can also construct an oracle B relative to which CoNPB has a set that is both simple and NPB \ CoNPB -immune. Previously, the strongest known separation on this level was the existence of an oracle C such that NPC 4
has a set that is both simple and PC -immune [TV89]. This construction involved complicated searches and priority arguments. Contrary to this, our construction here is an \in place" diagonalization. I.e., at each stage s the construction decides whether to add the string xs to the witness language and no strings smaller than xs are considered after stage s. Theorem 4.1 There is an oracle A such that NPA has a set that is simple and NPA \ CoNPA-immune. Proof: The language that will have this property is DA = fx j (9y)[jyj = jxj2 ^xy 2 A]g. Theorem 2.6 gives us an in nite random string Y that has the property for some constant c and all n that K (Y [1::n]) > n ? c. We will construct A by stages. As+1 is the oracle de ned by the end of stage s and As+1 As for all s. The oracle A may have 2n strings of length n + n2 . We de ne A0 from substrings of Y as follows. Let xmi be the i-th string of length m in and let yim be the substring of length m2 of Y that starts at the (m2 ? 4m + 6)2m + (i ? 1) m2 ? 5th bit of Y . Note that the string that describes A0 m , say YAm0 , has length (2m2 ? 4m + 6)2m ? 6 and therefore for some c and all m it holds that K (YAm0 ) (2m2 ? 4m + 6)2m ? c. Note also that DA0 = . At each stage s we will decide the membership of the string xs in DA , where xs is the sth string in the lexicographical ordering of . Deciding membership of xs is deciding whether to remove xs y from A where xs y is the only string that is an extension of xs currently in A. The Kolmogorov property that makes the entire construction work (as expressed in Lemma 4.2) is that an NP machine that rejects some string xs with oracle As must also reject xs with oracle A or else it will allow us to describe some initial segment of Y using signi cantly less bits than the length of this segment. Let (Me )e be an enumeration of all nondeterministic polynomial time bounded Turing machines, where ne is the time bound on inputs of length n. The construction maintains two sets of requirements. First, a set Us of yet unsatis ed requirements to which occasionally a new element is added and from which satis ed requirements are removed. Second, a set Vs in which satis ed requirements are kept. Sometimes a satis ed requirement will be moved from Vs to Us at which time it will become unsatis ed. Every requirement either corresponds to a language in NPA or to a language in NPA \ CoNPA . Requirement R2e will represent the language L(MeA) and requirement R2e+1 will represent the language L(MiA) \ L(MjA ), where = e. We will prove that the construction of A meets the following requirements for all e. 1. R2e : j L(MeA)j = 1 =) L(MeA) \ DA 6= ;. 2. R2+1 : j L(MiA )j = 1 =) [[L(MiA ) 6= L(MjA)] _ [L(MiA) \ DA 6= ;]]. Requirements of type 1 ensure that DA has a nonempty intersection with every in nite language in NPA , and requirements of type 2 ensure that DA has a nonempty intersection with every in nite language in NPA \ CoNPA . A requirement R2e is active at stage s if 2e 2 Us and xs 2 L(MeAs ). A requirement R2e+1 is active at stage s if 2e + 1 2 Us , e = and xs 2 L(MiAs ). Now the construction can be described as follows: stage s: If no requirement is active As+1 = As ; Vs+1 = Vs ; Us+1 = Us; Else let e be the least active requirement in Us . If e is even then Us+1 = Us ? feg; Vs+1 = Vs [ f<e; xs >g; As+1 = As ; Else If MjAs (xs ) = 1 then Vs+1 = Vs [ f<e; xs >g; As+1 = As ; Us+1 = Us Else If f 2 Vs j i < e ^ jxji > jxs jg = ; then Vs+1 = Vs ? f 2 Vs j i > eg; Us+1 = Us [ fi j i > e ^ (9x)[ 2 Vs ]g ? feg; As+1 = As ? f<xs ; ys >g; Else As+1 = As ; Vs+1 = Vs ; Us+1 = Us ; If s = 2k then Us+1 = Us+1 [ fkg
End of Construction
Now we prove that all requirements are satis ed in the limit. First we show the following key lemma about the construction. Lemma 4.2 (81s)(8e < log s)[xs 2= L(MeAs ) =) xs 2= L(MeA)]. Proof: Suppose not. Let e and s be such that e < log s and xs 2 L(MeA) ? L(MeAs ). Let s0 be minimal such that s0 s and xs 2 L(MeAs0 +1 ) ? L(MeAs0 ). By construction As0 ? As0 +1 = fxs0 ys0 g. The string 5
xs0 ys0 must be queried in any accepting computation of MeAs0 +1 on input xs . Otherwise MeAs0 would also have an accepting computation on input xs contradicting the assumption. On input xs machine Me can only query strings of length less than or equal to jxs je , which is less than or equal to jxs jlog s . Moreover j fy j y 2 A0 ? As0 +1 gjj log s0 , since for every such y an odd index is in Ut ? (Ut+1 [ Vt+1 ) for some t s0 and there are no more than log s0 indices in [fUt j t s0 g. Let n be such that n + n2 > jxs je . We will show how to reconstruct the rst (2n2 ? 4n +6)2n ? 6 bits of Y using signi cantly less bits and hence arrive at a contradiction. Suppose that we have (2n2 ? 4n + 6)2n ? 6 ?jxs0 j2 bits of Y that describe all substrings of Y up to length n2 , except ys0 . Furthermore suppose that we have a list of at most log s0 strings xi1 ; : : : ; xik that says which xij yij are in A0 ? As0 . Note that no strings greater than xs0 ys0 are in A0 ? As0 . Finally let q < jxs je be the index of the query xs0 ys0 in the leftmost accepting computation of MeAs+1 (xs ). Then we can reconstruct the rst (2n2 ? 4n + 6)2n ? 6 bits of Y from (2n2 ? 4n + 6)2n ? 6 ? jxs0 j2 + log s0 jxs j + e(log jxs j + 1) + O(1) bits of information. We arrive at a contradiction for all but nitely many s since the complexity assumption on Y is violated. 2
The following lemma has a trivial claim. We isolate it nonetheless to avoid duplication in the proofs to come. Lemma 4.3 (8e)[j fs j (9x)[<e; x> 2 Vs+1 ? Vs ]gjj < 1]. Proof: Whenever <e; x> is in Vs+1 ? Vs there is a smaller index that is moved from Us+1 either to oblivion or to Vs+1 . This means that 0 can enter Vs for only one s and there is no t such that is in Vt ? Vt+1 . By induction forP i
