Redundancy in Complete Sets

Transcription

1 Redundancy n Complete Sets Chrstan Glaßer, A. Pavan, Alan L. Selman, Lyu Zhang July 6, 2005 Abstract We show that a set s m-autoreducble f and only f t s m-mtotc. Ths solves a long standng open queston n a surprsng way. As a consequence of ths uncondtonal result and recent work by Glaßer et al. [11], complete sets for all of the followng complexty classes are m-mtotc: NP, conp, P, PSPACE, and NEXP, as well as all levels of PH, MODPH, and the Boolean herarchy over NP. In the cases of NP, PSPACE, NEXP, and PH, ths at once answers several well-studed open questons. These results tell us that complete sets share a redundancy that was not known before. We dsprove the equvalence between autoreducblty and mtotcty for all polynomaltme-bounded reducbltes between 3-tt-reducblty and Turng-reducblty: There exsts a sparse set n EXP that s polynomal-tme 3-tt-autoreducble, but not weakly polynomal-tme T-mtotc. In partcular, polynomal-tme T-autoreducblty does not mply polynomal-tme weak T-mtotcty, whch solves an open queston by Buhrman and Torenvlet. We generalze autoreducblty to defne poly-autoreducblty and gve evdence that NPcomplete sets are poly-autoreducble. 1 Introducton It s a well known observaton that for many nterestng complexty classes, all known complete sets contan redundant nformaton. For example, consder SAT. Gven a boolean formula φ one can produce two dfferent formulas φ 1 and φ 2 such that the queston of whether φ s satsfable or not s equvalent to the queston of whether φ 1 or φ 2 are satsfable. Thus φ 1 and φ 2 contan nformaton about φ. Another example s the Permanent. Gven a matrx M, we can reduce the computaton of the permanent of M to computng the permanent of M +R, M +2R,...,M +nr, where R s a randomly chosen matrx. Thus nformaton about the permanent of M s contaned Lehrstuhl für Informatk IV,Unverstät Würzburg. Emal: glasser@nformatk.un-wuerzburg.de. Department of Computer Scence, Iowa State Unversty. Research supported n part by NSF grants CCR and CCF Emal: pavan@cs.astate.edu Department of Computer Scence and Engneerng, Unversty at Buffalo. Research supported n part by NSF grant CCR Emal: selman@cse.buffalo.edu Department of Computer Scence, Unversty at Buffalo. Emal:lzhang7@cse.buffalo.edu 1

2 n a few random lookng matrces. We nterpret ths as SAT and Permanent contan redundant nformaton. In ths paper we study the queston of how much redundancy s contaned n complete sets of complexty classes. There are several ways to measure redundancy. We focus on the two notons autoreducblty and mtotcty. Trakhtenbrot [16] defned a set A to be autoreducble f there s an oracle Turng machne M such that A = L(M A ) and M on nput x never queres x. For complexty classes lke NP and PSPACE refned measures are needed. In ths sprt, Ambos-Spes [2] defned the noton of polynomaltme autoreducblty and the more restrcted form m-autoreducblty. A set A s polynomal-tme autoreducble f t s autoreducble va a oracle Turng machne that runs n polynomal-tme. A s m-autoreducble f A s polynomal-tme many-one reducble to A va a functon f such that f(x) x for every x. Both notons demand nformaton contaned n A(x) to be present among strngs dfferent from x. In the case of m-autoreducblty, the redundancy n A s even more apparent f a set A s m-autoreducble, then x and f(x) have the same nformaton about A. A stronger form of redundancy s descrbed by the noton of mtotcty whch was ntroduced by Ladner [13] for the recursve settng and by Ambos-Spes [2] for the polynomal-tme settng. A set A s m-mtotc f there s a set S P such that A, A S, and A S are polynomal-tme many-one equvalent. Thus f a set s m-mtotc, then A can be splt nto two parts such that both parts have exactly the same nformaton as the orgnal set has. Ambos-Spes [2] showed that f a set s m-mtotc, then t s m-autoreducble and he rased the queston of whether the converse holds. In ths paper we resolve ths queston and show that every m-autoreducble set s m-mtotc. Ths s our man result. Snce ts proof s very nvolved, we present our man combnatoral dea wth help of a smplfed graph problem whch wll be descrbed n Secton 3. Ths smplfcaton drops many of the mportant detals from our formal proof, but stll contans the combnatoral core of the problem. Our man result s all the more surprsng, because t s known [2] that polynomal-tme T-autoreducblty does not mply polynomal-tme T- mtotcty. We mprove ths and dsprove the equvalence between autoreducblty and mtotcty for all polynomal-tme-bounded reducbltes between 3-tt-reducblty and Turng-reducblty: There exsts a sparse set n EXP that s polynomal-tme 3-tt-autoreducble, but not weakly polynomaltme T-mtotc. In partcular, polynomal-tme T-autoreducble does not mply polynomal-tme weakly T-mtotc. Ths result settles another open queston rased by Buhrman and Torenvlet [8]. Our man result relates local redundancy to global redundancy n the followng sense. If a set A s m-autoreducble, then x and f(x) contan the same nformaton about A. Ths can be vewed as local redundancy. Whereas f A s m-mtotc, then A can be splt nto two sets B and C such that A, B, and C are polynomal-tme many-one equvalent. Thus the sets B and C have exactly the same nformaton as the orgnal set A. Ths can be vewed as global redundancy n A. Our man result states that local redundancy s the same as global redundancy. As a consequence of ths result and recent work of Glaßer et al. [11], we can show that all complete sets for many nterestng classes such as NP, PSPACE, NEXP, and levels of PH are m-mtotc. Thus they all contan redundant nformaton n a strong sense. Ths resolves several long standng open questons rased by Ambos-Spes [2], Buhrman, Hoene, and Torenvlet [7], and Buhrman and Torenvlet [8]. 2

3 Our result can also be vewed as a step towards understandng the somorphsm conecture [5]. Ths conecture states that all NP-complete sets are somorphc to each other. In spte of several years of research, we do not have any concrete evdence ether n support or aganst the somorphsm conecture 1. It s easy to see that f the somorphsm conecture holds for classes such as NP, PSPACE, and EXP, then complete sets for these classes are m-autoreducble as well as m-mtotc. Gven our current nablty to make progress about the somorphsm conecture, the next best thng we can hope for s to make progress on statements that the somorphsm conecture mples. We note that ths s not an entrely new approach. For example, f the somorphsm conecture s true, then NP-complete sets cannot be sparse. Ths motvated researchers to consder the queston of whether complete sets for NP can be sparse. Ths lne of research led to the beautful results of Mahaney [14] and Ogwara and Watanabe [15] who showed that complete sets for NP cannot be sparse unless P = NP. Our results show that another consequence of somorphsm, namely NP-complete sets are m-mtotc holds. Note that ths s an uncondtonal result. Buhrman et al. [6] and Buhrman and Torenvlet [9, 10] argue that t s crtcal to study the notons of autoreducblty and mtotcty. They showed that resolvng questons regardng autoreducblty of complete sets leads to uncondtonal separaton results. For example, consder the queston of whether truth-table complete sets for PSPACE are non-adaptve autoreducble. An affrmatve answer separates NP from NL, whle a negatve answer separates the polynomal-tme herarchy from PSPACE. They argue that ths approach does not have the curse of relatvzaton and s worth pursung. We refer the reader to the recent survey by Buhrman and Torenvlet [10] for more detals. In Secton 5, we extend the noton of autoreducblty and defne poly-autoreducblty. A motvaton for ths s to understand the somorphsm conecture and the noton of paddablty. Recall that the somorphsm conecture s true f and only f all NP-complete sets are paddable. Paddablty mples the followng: If L s paddable, then gven x and a polynomal p, we can produce p( x ) dstnct strngs such that f x s n L, then all these strngs are n L and f x s not n L, then none of these strngs are n L. Autoreducblty mples that gven x we can produce a sngle strng y dfferent from x such that L(x) = L(y). A natural queston that arses s whether we can produce more strngs whose membershp n L s the same as the membershp of x n L. Ths leads us to the noton of f(n)-autoreducblty: A set L s f(n)-autoreducble, f there s a polynomal-tme algorthm that on nput x outputs f( x ) dstnct strngs (dfferent from x) whose membershp n L s the same as the membershp of x n L. It s obvous that paddablty mples poly-autoreducblty. The queston of whether NP complete sets are poly-autoreducble s weaker than the queston of whether NP-complete sets are paddable. We provde evdence for poly-autoreducblty of NP-complete sets. We show that f one-way permutatons exst, then NP-complete sets are log-autoreducble. Moreover, f one-way permutatons and quck pseudo-random generators exst, then NP-complete sets are poly-autoreducble. We also show that f NP-complete sets are poly-autoreducble, then they have nfnte subsets that can be decded n lnear-exponental tme. 1 It s currently beleved that f one-way functons exst, then the somorphsm conecture s false. However, we do not have a proof of ths. 3

4 1.1 Prevous Work The queston of whether complete sets for varous classes are autoreducble has been studed extensvely [17, 4, 6]. Begel and Fegenbaum [4] showed that Turng complete sets for the classes that form the polynomal herarchy, Σ P,ΠP, and P, are Turng autoreducble. Thus, all Turng complete sets for NP are Turng autoreducble. Buhrman et al. [6] showed that Turng complete sets for EXP and EXP are autoreducble, whereas there exsts a Turng complete set for EESPACE that s not Turng auto-reducble. Regardng NP, Buhrman et al. [6] showed that truth-table complete sets for NP are probablstc truth-table autoreducble. Recently, Glaßer et al. [11] showed that complete sets for classes such as NP, PSPACE, Σ P are m-autoreducble. Buhrman, Hoene, and Torenvlet [7] showed that EXP complete sets are weakly many-one mtotc. Ths result was recently mproved ndependently by Kurtz [10] and Glaßer et al. [11]. Glaßer et al. also showed that NEXP complete sets are weakly m-mtotc and PSPACE-complete sets are weak Turng-mtotc. 2 Prelmnares We use standard notaton and assume famlarty wth standard resource-bounded reductons. We consder words n lexcographc order. All used reductons are polynomal-tme computable. Defnton 2.1 ([2]) A set A s polynomally T-autoreducble (T-autoreducble, for short) f there exsts a polynomal-tme-bounded oracle Turng machne M such that A = L(M A ) and for all x, M on nput x never queres x. A set A s polynomally m-autoreducble (m-autoreducble, for short) f A p ma va a reducton functon f such that for all x, f(x) x. Defnton 2.2 ([2]) A recursve set A s polynomal-tme T-mtotc (T-mtotc, for short) f there exsts a set B P such that A p T A B p T A B. A s polynomal-tme m-mtotc (m-mtotc, for short) f there exsts a set B P such that A p m A B p m A B. Defnton 2.3 ([2]) A recursve set A s polynomal-tme weakly T-mtotc (weakly T-mtotc, for short) f there exst dsont sets A 0 and A 1 such that A 0 A 1 = A, and A p T A 0 p T A 1. A s polynomal-tme weakly m-mtotc (weakly m-mtotc, for short) f there exst dsont sets A 0 and A 1 such that A 0 A 1 = A, and A p m A 0 p m A 1. Defnton 2.4 Let f be a functon from N to N. A set L s f(n)-autoreducble, f there s a polynomal-tme algorthm A that on nput x outputs y 1,y 2,,y m such that f( x ) = m, f x L, then {y 1,y 2,,y m } L, and f x / L, then {y 1,y 2,,y m } L =. A set s poly-autoreducble, f t s n k -autoreducble for every k 1. A language s DTIME(T(n))-complex f L does not belong to DTIME(T(n)) almost everywhere; that s, every Turng machne M that accepts L runs n tme greater than T( x ), for all but 4

5 fntely many words x. A language L s mmune to a complexty class C, or C-mmune, f L s nfnte and no nfnte subset of L belongs to C. A language L s b-mmune to a complexty class C, or C-b-mmune, f both L and L are C-mmune. Balcázar and Schönng [3] proved that for every tme-constructble functon T, L s DTIME(T(n))-complex f and only f L s b-mmune to DTIME(T(n)). 3 m-autoreducblty equals m-mtotcty It s easy to see that f a nontrval language L s m-mtotc, then t s m-autoreducble. If L s m-mtotc, then there s a set S P such that L S p m L S va some f and L S p m L S va some g. On nput x, the m-autoreducton for L works as follows: If x S and f(x) / S, then output f(x). If x / S and g(x) S, then output g(x). Otherwse, output a fxed element from L {x}. So m-mtotcty mples m-autoreducblty. The man result of ths paper shows that surprsngly the converse holds true as well,.e., m-mtotcty and m-autoreducblty are equvalent notons. Theorem 3.1 Let L be any set such that L 2. L s m-autoreducble f and only f L s m- mtotc. Before proceedng to the proof we frst menton the man deas and the ntuton behnd the proof and descrbe the combnatoral core of the problem. Assume that L s m-autoreducble va reducton functon f. Gven x, the repeated applcaton of f yelds a sequence of words x,f(x),f(f(x)),..., whch we call the traectory of x. These traectores ether are nfnte or end n a cycle of length at least 2. Note that as f s an autoreducton, x f(x). At frst glance t seems that m-mtotcty can be easly establshed by the followng dea: In every traectory, label the words at even postons wth + and all other words wth. Defne S to be the set of strngs whose label s +. Wth ths defnton of S t seems that f reduces L S to L S and L S to L S. However, ths labelng strategy has at least two problems. Frst, t s not clear that S P; because gven a strng y, we have to compute the party of the poston of y n a traectory. As traectores can be of exponental length, ths mght take exponental tme. The second and more fundamental problem s the followng: The labelng generated above s nconsstent and not well defned. For example, let f(x) = y. To label y whch traectory should we use? The traectory of x or the traectory of y? If we use traectory of x, y gets a label of +, whereas f we use the traectory of y, then t gets a label of. Thus S s not well defned and so ths dea does not work. It fals because the labelng strategy s a global strategy. To label a strng we have to consder all the traectores n whch x occurs. Every sngle x gves rse to a labelng of possbly nfntely many words, and these labelngs may overlap n an nconsstent way. We resolve ths by usng a local labelng strategy. More precsely, we compute a label for a gven x ust by lookng at the neghborng values x, f(x), and f(f(x)). It s mmedately clear that such 5

6 a strategy s well-defned and therefore defnes a consstent labelng. We also should guarantee that ths local strategy strctly alternates labels,.e., x gets + f and only f f(x) gets. Such an alternaton of labels would help us to establsh the m-mtotcty of L. Thus our goal wll be to fnd a local labelng strategy that has a nce alternaton behavor. However, we settle for somethng less. Instead of requrng that the labels strctly alternate, we only requre that gven x, at least one of f(x),f(f(x)),,f m (x) gets a label that s dfferent from the label of x, where m s polynomally bounded n the length of x. Ths suffces to show m-mtotcty. The most dffcult part n our proof s to show that there exsts a local labelng strategy that has ths weaker alternaton property. We now formulate the core underlyng problem. To keep ths proof sketch smpler, we make several assumptons and gnore several techncal but mportant detals. If we assume (for smplcty) that on strngs x / 1 the autoreducton s length preservng such that f(x) > x, then we arrve at the followng graph labelng problem. Core Problem: Let G n be a drected graph wth 2 n vertces such that every strng of length n s a vertex of G n. Assume that 1 n s a snk, that nodes u 1 n have outdegree 1, and that u < v for edges (u,v). For u 1 n let s(u) denote u s unque successor,.e., s(u) = v f (u,v) s an edge. Fnd a strategy that labels each node wth ether + or such that: () Gven a node u, ts label can be computed n polynomal tme n n. () There exsts a polynomal p such that for every node u, at least one of the nodes s(u),s(s(u)),...,s p(n) (u) gets a label that s dfferent from the label of u. We exhbt a labelng strategy wth these propertes. To defne ths labelng, we use the followng dstance functon: d(x,y) df = log y x (our formal proof uses a varant of ths functon). The core problem s solved by the followng local strategy. 0 // Strategy for labelng node x 1 let y = s(x) and z = s(y). 2 f d(x, y) > d(y, z) then output 3 f d(x, y) < d(y, z) then output + 4 r := d(x,y) 5 output + ff x/2 r+1 s even Clearly, ths labelng strategy satsfes condton (). We gve a sketch of the proof that t also satsfes condton (). Defne m = 5n and let u 1,u 2,...,u m be a path n the graph. It suffces to show that not all the nodes u 1,u 2,...,u m obtan the same label. Assume that ths does not hold, say all these nodes get label +. So no output s made n lne 2 and therefore, the dstances d(u,u +1 ) do not decrease. Note that the dstance functon maps to natural numbers. If we have more than n ncreases, then the dstance between u m 1 and u m s bgger than n. Therefore, u m u m 1 > 2 n+1, whch s mpossble for words of length n. So along the path u 1,u 2,...,u m there exst at least m n = 4n postons where the dstance stays the same. By a pgeon hole argument there exst 6

7 four consecutve such postons,.e., nodes v = u, w = u +1, x = u +2, y = u +3, z = u +4 such that d(v, w) = d(w, x) = d(x, y) = d(y, z). So for the nputs v, w, and x, we reach lne 4 where the algorthm wll assgn r = d(v,w). Observe that for all words w 1 and w 2, the value d(w 1,w 2 ) allows an approxmaton of w 2 w 1 up to a factor of 2. More precsely, w v, x w, and y x belong to the nterval [2 r,2 r+1 ). It s an easy observaton that ths mples that not all of the followng values can have the same party: v/2 r+1, w/2 r+1, and x/2 r+1. Accordng to lne 5, not all words v, w, and x obtan the same label. Ths s a contradcton whch shows that not all the nodes u 1,u 2,...,u m obtan the same label. Ths proves () and solves the core of the labelng problem. The labelng strategy allows the defnton of a set S P such that whenever we follow the traectory of x for more than 5 x steps, then we fnd at least one alternaton between S and S. Ths establshes m-mtotcty for L. Now we gve a formal proof of Theorem 3.1. The dyadc representaton of natural numbers provdes a one-one correspondence between words over Σ = {0, 1} and natural numbers. Ths correspondence translates operatons and relatons over natural numbers to operaton and relatons over words. We denote the absolute value of an nteger by abs(x). Ths avods a conflct between the notaton of the length of a word w and the notaton of the absolute value of the nteger represented by w. Moreover, log(x) denotes x s logarthm to base 2. We use the followng proposton. Proposton 3.2 Let L be any set such that L 2. L s m-mtotc f and only f there exst a total g PF and a set S P such that for all x, 1. x L g(x) L, and 2. x S g(x) / S. Proof. Choose dstnct words w 1,w 2 L. If L s m-mtotc, then there exsts S P such that L S p ml S va some g 1 PF and L S p ml S va some g 2 PF. We may assume that w 1 S and w 2 S; otherwse the set S {w 1 } {w 2 } can be used nstead of S. Observe that the followng functon g satsfes the statements 1 and 2 from the proposton. g 1 (x) : f x S and g 1 (x) S w g(x) = df 2 : f x S and g 1 (x) S g 2 (x) : f x S and g 2 (x) S : f x S and g 2 (x) S w 1 Now assume there exst a total g PF and an S P that satsfy the statements 1 and 2. It follows that L S p ml S and L S p ml S, both va g. The followng functon reduces L to L S. g x : f x S (x) = df g(x) : f x S 7

8 The followng functon reduces L S to L. g (x) = df x w 1 : f x S : f x S Ths shows L p ml S p ml S and hence L s m-mtotc. Proof. (Theorem 3.1) If L s m-mtotc, then there exst S P and f 1,f 2 PF such that L S p ml S va f 1 and L S p ml S va f 2. By assumpton, there exst dfferent words v,w L. The followng functon s an m-autoreducton for L. f 1 (x) : f x S and f 1 (x) / S f (x) = df f 2 (x) : f x / S and f 2 (x) S mn({v,w} {x}) : otherwse For the other drecton, let us assume that L s m-autoreducble and let f PF be an m- autoreducton for L. Choose k 1 such that f s computable n tme n k + k. Usng Proposton 3.2, we show L s m-mtotcty as follows: We construct a total g PF and an S P such that (x L g(x) L) and (x S g(x) / S). Let t be a tower functon defned by: t(0) = 0 and t( + 1) = t() k + k for 0. Defne the nverse tower functon as t 1 (n) = mn{ t() n}. Note that t 1 PF. We partton the set of all words accordng to the party of the nverse tower functon of ther lengths. Note that S 0,S 1 P. S 0 df = {x t 1 ( x ) 0(mod 2)} S 1 df = {x t 1 ( x ) 1(mod 2)} The followng dstance functon for natural numbers x and y plays a crucal role n our proof. d(x,y) df = sgn(y x) log(abs(y x)). Ths functon s computable n polynomal tme. We defne a set S (whch wll be used as separator for L) by the followng algorthm whch works on nput x. 0 // Algorthm for set S 1 y := f(x), z := f(f(x)) 2 f y > x then (accept ff x S 0 ) 3 f z > y then (accept ff y S 1 ) 4 f x = z then (accept ff x > f(x)) 5 // here x, y, and z are parwse dfferent 6 f d(x, y) > d(y, z) then reect 7 f d(x, y) < d(y, z) then accept 8 r := d(x,y) 9 accept ff y/2 abs(r)+1 s even 8

9 Observe that S P. We wll show L p ml S p ml S whch mples that L s m-mtotc. Clam 3.3 Let y be any word and let m = y. If [0,6m + 3], f (y) f +1 (y), then there exsts [0,6m + 3] such that f (y) S f +1 (y) / S. Proof. Assume the clam does not hold. Moreover, assume that for all [0,6m + 4], f (y) S. For the other case (.e., for all [0,6m + 4], f (y) / S) one can argue analogously. Consder the algorthm for S. Fact 1: For [0,6m + 2], the algorthm on nput f (y) stops ether n lne 7 or n lne 9. Assume there exsts [0,6m + 2] such that the algorthm on nput f (y) stops n lnes 2 or 3. In ths case, f (y) < f +1 (y) or f +1 (y) < f +2 (y) whch contradcts our assumpton. Assume there exsts [0,6m + 2] such that the algorthm on nput f (y) stops n lnes 4. By assumpton of the clam, f (y) f +1 (y) f +2 (y). Moreover, f (y) = f +2 (y), snce we stop n lne 4. So f (y) = f +1 (y). Therefore, on both nputs, f (y) and f +1 (y), the algorthm stops n lne 4. Note that f (y) f +1 (y), snce f s an m-autoreducton. Hence by lne 4, f (y) S f +1 (y) / S, whch contradcts our assumpton. Assume there exsts [0,6m + 2] such that the algorthm on nput f (y) stops n lnes 6. So f (y) / S whch contradcts the assumpton. Ths proves Fact 1. J K = df { [0,6m + 2] on nput f (y) the algorthms for S stops n lne 7} = df { [0,6m + 2] on nput f (y) the algorthms for S stops n lne 9} By Fact 1, J K = {0,...,6m + 2}. From the algorthm we see the followng. J, d(f (y),f +1 (y)) < d(f +1 (y),f +2 (y)) (1) K, d(f (y),f +1 (y)) = d(f +1 (y),f +2 (y)) (2) Case 1: J > 2m. Together wth (1) and (2) ths shows It follows that or d(f 6m+3 (y),f 6m+4 (y)) d(f 0 (y),f 1 (y)) > 2m. (3) d(f 6m+3 (y),f 6m+4 (y)) > m (4) d(f 0 (y),f 1 (y)) < m. (5) Assume that (4) holds. By the assumpton of the clam, f 6m+3 (y) and f 6m+4 (y) are words of length m. So the length of abs(f 6m+4 (y) f 6m+3 (y)) s m. From the dyadc representaton of numbers t follows that log(abs(f 6m+4 (y) f 6m+3 (y))) < m + 1 and therefore, d(f 6m+3 (y),f 6m+4 (y)) m. Ths s a contradcton, snce we assumed that (4) holds. 9

10 Assume now that (5) holds. Agan, f 0 (y) and f 1 (y) are words of length m. So the length of abs(f 0 (y) f 1 (y)) s m. It follows that log(abs(f 1 (y) f 0 (y))) < m + 1 and therefore, d(f 0 (y),f 1 (y)) m. Ths s a contradcton, snce we assumed that (5) holds. Case 2: J 2m. Note that [0,6m + 2] contans 6m + 3 elements whle J contans at most 2m elements. So there exsts [0, 6m] such that, + 1, + 2 K. A look at the algorthm tells us the followng. d(f (y),f +1 (y)) = d(f +1 (y),f +2 (y)) = d(f +2 (y),f +3 (y)) = d(f +3 (y),f +4 (y)) (6) Defne r as the number shown n (6), and let z 1 df = f (y), z 2 df = f +1 (y), z 3 df = f +2 (y), and z 4 df = f +3 (y). Recall that z 1,z 2,z 3 S and that on nput of these words, the algorthm stops n lne 9. Therefore, the followng must hold. a 1 df = z 2 /2 abs(r)+1 s even (7) a 2 df = z 3 /2 abs(r)+1 s even (8) a 3 df = z 4 /2 abs(r)+1 s even (9) Case 2a: r = 0. Here z 2 z 4, snce otherwse on nput z 2 the algorthm stops n lne 4 whch contradcts Fact 1. Also, z 2 z 3 and z 3 z 4, snce f s an m-autoreducton. From (6) and from the defnton of the dstance functon d we obtan, ether z 2 = z 3 1 = z 4 2, or z 4 = z 3 1 = z 2 2. So z 4 z 2 equals 2 or 2, and hence a 3 a 1 equals 1 or 1. The latter contradcts the observatons (7) and (9). Case 2b: r > 0. Here we have z 1 < z 2 < z 3 < z 4 and therefore, a 1 a 2 a 3. Assume a 1 = a 3. Snce d(z 2,z 3 ) = r, t holds that log(abs(z 3 z 2 )) r and hence, z 3 z 2 2 r. The same argument shows z 4 z 3 2 r. So z 4 z r+1 = z abs(r)+1 and hence, a 3 a The latter contradcts the assumpton a 1 = a 3. So assume a 1 < a 3 whch mples a 3 a 1 2, snce both values are even. Snce a 2 s even as well, we obtan a 2 a 1 2 or a 3 a 2 2. If a 2 a 1 2, then z 3 z 2 > 2 r+1 and so d(z 2,z 3 ) > r. If a 3 a 2 2, then z 4 z 3 > 2 r+1 and so d(z 3,z 4 ) > r. Both conclusons contradct (6). Case 2c: r < 0. Here we have z 1 > z 2 > z 3 > z 4 and therefore, a 1 a 2 a 3. Assume a 1 = a 3. Snce d(z 2,z 3 ) = r, t holds that log(abs(z 3 z 2 )) abs(r) and hence, z 2 z 3 2 abs(r). The same argument shows z 3 z 4 2 abs(r). So z 2 z abs(r)+1 and hence, a 1 a The latter contradcts the assumpton a 1 = a 3. So assume a 1 > a 3 whch mples a 1 a 3 2, snce both values are even. Snce a 2 s even as well, we obtan a 1 a 2 2 or a 2 a 3 2. If a 1 a 2 2, then z 2 z 3 > 2 abs(r)+1 and so d(z 2,z 3 ) < abs(r) = r. If a 2 a 3 2, then z 3 z 4 > 2 abs(r)+1 and so d(z 3,z 4 ) < abs(r) = r. Both conclusons contradct (6). Ths proves Clam

11 Clam 3.4 There exsts a total r PF such that L p ml va r and for every x, 1. f(r(x)) r(x) or 2. x S r(x) / S. Proof. For every x, let r(x) df = f (x) where s the smallest number such that f +1 (x) f (x) or (x S f (x) / S). We wll prove that such exsts. Consder the followng algorthm whch works on nput x. 0 // Algorthm for functon r 1 z := x 2 whle ( f(z) > z and (x S z S)) 3 // here z < x k +k 4 z := f(z) 5 end 6 output z Observe that ths algorthm computes the functon r. We prove the nvarant n lne 3, whch wll guarantee that the loop n the algorthm halts wthn polynomal steps n x. Assume that at some pont ths nvarant does not hold. We consder the frst tme when ths happens. In ths case, we must have reached lne 3 before, snce otherwse x x k + k whch s not possble. Let z denote the value of varable z when lne 3 was reached last tme. So z = f(z ). Note that the followng nequaltes hold, snce otherwse the algorthm stops earler. Moreover, x < f(x) (10) z < f(z ) (11) z < f(z) (12) x < z (13) z < x k + k, (14) snce otherwse already z volates the nvarant, whch contradcts the fact that wth z we chose the earlest volaton of the nvarant. From (13) and (14) we obtan t 1 ( x ) t 1 ( z ) t 1 ( x k + k) = t 1 ( x ) + 1. (15) From (10) t follows that on nput x, the algorthm for S stops n lne 2. We see the same for z and z usng (11) and (12). Ths mples the followng. x S x S 0 (16) z S z S 0 (17) z S z S 0 (18) 11

12 Note that x S z S z S, (19) snce otherwse the algorthm for r stops earler. Together wth (16), (17), and (18) ths shows and therefore, Now (15) mples t 1 ( x ) = t 1 ( z ) and we obtan x S 0 z S 0 z S 0 (20) t 1 ( x ) t 1 ( z ) t 1 ( z ) (mod 2). (21) t 1 ( z ) = t 1 ( x ) < t 1 ( x k + k) t 1 ( z ). (22) From (21) and (22) t follows that t 1 ( z ) t 1 ( z ) 2. Therefore, f(z ) > z k + k. Ths contradcts f s computaton tme and proves the nvarant n lne 3. From the nvarant we mmedately obtan that every sngle step of the algorthm can be carred out n tme polynomal n x. Each executon of lne 4 ncreases the length of z. By our nvarant, the algorthm must termnate wthn x k + k teratons of the loop. Ths shows that r s total and polynomal-tme computable. Snce r s defned by repeated applcatons of f, and snce f s an autoreducton of L, we obtan L p ml va r. The statements 1 and 2 of the clam follow mmedately from lne 2 of the algorthm. Ths proves Clam 3.4. Choose a functon r accordng to Clam 3.4. Defne a functon g by the followng algorthm whch works on nput x. Below we wll show that g satsfes the condtons n Proposton // Algorthm for functon g 1 y := r(x), m := y 2 f y < f(y) then return y 3 // here y f(y) 4 z := y 5 for := 0 to 6m +3 6 // here z = f (y), z m, and for all 0, f (y) f +1 (y) 7 f f(z) < f(f(z)) then 8 f (f(z) S x S) then return z else return f(z) 9 endf 10 z := f(z) 11 next 12 // here for all 0 6m +3, f (y) f +1 (y) 13 z := y 14 for := 0 to 6m // here z = f (y) and z m 16 f z S f(z) / S then 17 f (z S x S) then return f(z) else return z 18 endf 19 z := f(z) 20 next 21 // ths lne s never reached 12

13 Clam 3.5 The statements clamed n the comments of the algorthm for g hold true. Proof. Clearly, the condton n lne 3 holds. Observe that whenever we reach lne 6, then z = f (y) and z f(z). Therefore, the condton n lne 6 holds. It follows that f we reach lne 12, then we must have passed lne 6 for = 6m+3. Ths shows the condton n lne 12. Whenever we reach lne 15 t holds that z = f (y). From the condton n lne 12 t follows that z m n lne 15. Fnally we argue that we do not reach lne 21. Assume that we reach lne 12. By the condton n lne 12, we satsfy the assumpton of Clam 3.3. Therefore, there exsts [0,6m + 3] such that f (y) S f +1 (y) / S. So for =, the condton n lne 16 s true and therefore, the algorthm stops before reachng lne 21. Clam 3.6 g s a total functon n PF and L p ml va g. Proof. We mmedately see that g s total, snce lne 21 s never reached. We argue that g PF. Recall that f and r are total functons n PF, and recall that S P. So steps 1 4 are computable n polynomal tme n x. Note that m s polynomally bounded n x. By the remark n lne 6, the loop 5 11 needs only polynomal tme n x. The remark n lne 15 mples the same for the loop Ths shows g PF. We show L p ml va g. Observe that n any case the algorthm returns f (y) for a sutable 0. By Clam 3.4, x L y = r(x) L. Snce f s an autoreducton of L, we obtan x L g(x) = f (y) L. Clam 3.7 For every x, x S g(x) / S. Proof. Consder the computaton of the algorthm for g on nput x. Case 1: The output s made n lne 2. So we have f(r(x)) > r(x). From Clam 3.4 t follows x S g(x) = r(x) / S. Case 2: The output s made n lne 8. By lnes 6 and 7, f (y) f +1 (y) and f +1 (y) < f +2 (y). Therefore, f we look at the algorthm for S, then we see that on nput f (y) the algorthm stops n step 3, whle on nput f +1 (y) the algorthm stops n step 2. It follows that f (y) S f +1 (y) S 1 and f +1 (y) S f +1 (y) S 0. So z = f (y) S f(z) / S and therefore, by lne 8 of the algorthm for g, x S g(x) / S. 13

14 Case 3: The output s made n lne 17. From lne 16 t follows that x S g(x) / S. The Clams 3.6 and 3.7 allow the applcaton of Proposton 3.2. Hence L s m-mtotc. Call a set L nontrval f L 2 and L 2. Corollary 3.8 Every nontrval set that s many-one complete for one of the followng complexty classes s m-mtotc. NP, conp, P, PSPACE, EXP, NEXP any level of PH, MODPH, or the Boolean herarchy over NP Proof. Glaßer et al. [11] showed that all many-one complete sets of the above classes are m- autoreducble. By Theorem 3.1, these sets are m-mtotc. Corollary 3.9 A nontrval set L s NP-complete f and only f L s the unon of two dsont P-separable NP-complete sets. So unons of dsont P-separable NP-complete sets form exactly the class of NP-complete sets. What class s obtaned when we drop P-separablty? Does ths class contan a set that s not NP-complete? In other words, s the unon of dsont NP-complete sets always NP-complete? We leave ths as an open queston. Ambos-Spes [2] defned a set A to be ω-m-mtotc f for every n there exsts a partton (Q 1,...,Q n ) of Σ such that the followng sets are polynomal-tme many-one equvalent: A,A Q 1,...,A Q n. Corollary 3.10 Every nontrval nfnte set that s many-one complete for a class mentoned n Corollary 3.8 s ω-m-mtotc. 4 3-tt-Autoreducblty does not mply Weak T-Mtotcty In ths secton we prove a theorem that shows n a strong way that T-autoreducble does not mply weakly T-mtotc. Hence, our man theorem cannot be generalzed. Lemma 4.1 Let l,m 0 and let k (l + 2) 2m. If Q 1,...,Q k are sets of cardnalty l and f n 1,...,n k are parwse dfferent numbers, then there exst parwse dfferent ndces 1,..., m such that for all s,t [1,m], s t n s / Q t. 14

15 Proof. The proof s by nducton on n = l + m such that the nducton base covers all cases where l = 0 or m = 0. For these cases the lemma holds trvally. In partcular, ths covers the case n = 1. Assume there exsts n 1 such that the lemma holds for all l and m such that l = 0 or m = 0 or l + m n. Now we prove t for l and m such that l 1, m 1, and l + m = n + 1. Case 1: There exst at least k k l 1 ndces > 1 such that n 1 Q. Let k = k k l 1 and choose parwse dfferent ndces 1,..., k such that for all, 1 and n 1 Q. Let l = l 1 and let R = Q {n 1 } and r = n for 1 k. Observe l 0 and m 1. We estmate k as follows. k k k l 1 (l + 2) 2m (l + 2) 2m l 1 (snce (a b a a b b) for a,b 1) (l + 2) 2m (follows from (24) n the estmaton below) (23) For (23) the followng estmaton s needed. l + 1 l + 1 (l + 2) 2m 1 1 (l + 2 1) (l + 1) 2m 1 1 (l + 1) (l + 2) 2m 1 1 (l + 1) 2m 1 (snce (l + 2) 2m 1 1 1) (l + 2) 2m 1 [(l + 2) 2m 1 1 ] (l + 2) 2m 1 [(l + 1) 2m 1 ] (l + 2) 2m (l + 2) 2m 1 (l ) (l + 1) 2m 1 1 [(l + 1) 2m 1 ] (l + 2) 2m (l + 2) 2m (l + 1) 2m + (l + 1) 2m 1 (l + 2) 2m (l + 2) 2m l 1 (l + 2) 2m (snce (l + 1) 2m 1 l + 1) (24) Note that l + m = n. Also, R 1,...,R k are sets of cardnalty l and r 1,...,r k are parwse dfferent numbers. By nducton hypothess there exst parwse dfferent ndces 1,..., m such that for all s,t [1,m], (s t r s / R t ). For all s [1,m], r s n 1. Therefore, for all s,t [1,m], (s t r s / R t {n 1 }) and hence (s t n s / Q t ). So the lemma s satsfed by the ndces 1, 2,..., m. Case 2: There exst less than k k l 1 ndces > 1 such that n 1 Q. So there exst more than k + l ndces > 1 such that n 1 / Q. Snce Q 1 l, there exst more than k ndces > 1 such that n 1 / Q and n / Q 1. Hence there exst at least k df = k such ndces. So we can choose parwse dfferent ndces 1,..., k such that for all, 1 n 1 / Q n / Q 1. (25) Let m = m 1 and let R = Q and r = n for 1 k. Note that l 1 and m 0. Observe that k k (l + 2) 2m = (l + 2) 2m 15

16 and l + m = n. Also, R 1,...,R k are sets of cardnalty l and r 1,...,r k are parwse dfferent numbers. So by nducton hypothess there exst parwse dfferent ndces 1,..., m such that for all s,t [1,m ], s t r s / R t and hence s t n s / Q t. From (25) t follows that the lemma s satsfed by the ndces 1, 1, 2,..., m. Theorem 4.2 There exsts L SPARSE EXP such that L s 3-tt-autoreducble, but L s not weakly T-mtotc. Proof. Defne a tower functon by t(0) = 4 and t(n + 1) = t(n). For any word s, let W(s) = {s00,s01,s10,s11}. We wll defne L such that t satsfes the followng: () If w L, then there exsts n such that w = t(n). () For all n and all s Σ t(n) 2, the set W(s) L ether s empty or contans exactly two elements. It s easy to see that such an L s 3-tt-autoreducble: On nput w, determne n such that w = t(n). If such n does not exst, then reect. Otherwse, let s be w s prefx of length w 2. Accept f and only f the set L (W(s) {w}) contans an odd number of elements. Ths s a 3-tt-autoreducton. We turn to the constructon of L. Let M 1,M 2,... be an enumeraton of determnstc, polynomaltme-bounded Turng machnes such that the runnng tme of M s n +. Let, be a parng functon such that x,y > x + y. We construct L stagewse such that n stage n we determne whch of the words of length t(n) belong to L. In other words, at stage n we defne a set W n Σ t(n), and fnally we defne L to be the unon of all W n. We start by defnng W 0 =. Suppose we are at stage n > 0. Let m = t(n) and determne and such that n =,. If such and do not exst, then let W n = and go to stage n+1. Otherwse, and exst. In partcular, + < log log m. Let O df =W 0 W n 1 be the part of L that has been constructed so far. Let O 1,O 2,...,O l be the lst of all subsets of O (lexcographcally ordered accordng to ther characterstc sequences). Snce O Σ t(n 1) we obtan O 2 t(n 1)+1. Therefore, l 2 2t(n 1) t(n 1) = log log t(n) = log log m. (26) 16

17 We gve some ntuton for the clam below. If L s weakly T-mtotc, then n partcular, there exsts a partton L = L 1 L 2 such that L 2 p T L 1 va some machne M. Hence O L 1 must appear (say as O k ) n our lst of subsets of O. The followng clam makes sure that we can fnd a lst of words s 1,...,s l of length m 2 such that for all k [1,l] t holds that f the partton of L s such that O L 1 = O k, then M on nput of a strng from {s k 00,s k 01,s k 10,s k 11} does not query the oracle for words from W(s r ) f r k. Hence, f M queres a word of length m that does not belong to {s k 00,s k 01,s k 10,s k 11}, then t always gets a no answer. So the followng s the only nformaton about the partton of L that can be exploted by M : the partton of O = Σ <t(n) L the partton of W(s k ) L In partcular, M cannot explot nformaton about the partton of W(s r ) L for r k. Ths ndependence of M makes our dagonalzaton possble. Clam 4.3 There exst parwse dfferent words s 1,...,s l Σ m 2 such that for all k,r [1,l], k r, and all y W(s k ), nether M O O k (y) nor M O k (y) query the oracle for words n W(s r ). Proof. For s Σ m 2, let Q df s ={s Σ m 2 k [1,l], y W(s), q W(s ) such that q s quered by M O O k (y) or M O k (y)}. Observe that for every s Σ m 2, Q s 4l[(m + ) + (m + )] 4(log log m)[m log log m + log log m] log log m 8(log log m)m 2log log m m 2 log2 m 2. (27) We dentfy numbers n [1,2 m 2 ] wth strngs n Σ m 2. Consdered n ths way, each Q s s a subset of [1,2 m 2 ]. By (27), Q 1,Q 2,...,Q 2 m 2 are sets of cardnalty 2 log2 m 2. Clearly, 1,2,...,2 m 2 are parwse dfferent numbers. By (26), 2 m 2 (2 log2 m ) log m (2 log2 m ) 2l. Therefore, we can apply Lemma 4.1. We obtan ndces s 1,...,s l such that for all k,r [1,l], r k s r / Q sk. (28) Assume there exst k,r [1,l], k r, and y W(s k ) such that some q W(s r ) s quered by M O O k (y) or M O k (y). Hence s r Q sk. Ths contradcts (28) and fnshes the proof of Clam

18 Let s 1,...,s l Σ m 2 be the words assured by Clam 4.3. We defne W n such that for every k [1,l] we defne a set V k W(s k ), and fnally we defne W n to be the unon of all V k. The cardnalty of each V k s ether 0 or 2. Fx some k [1,l] and let Q k df = O O k. Case 1: M Q k (s k 00) accepts or M O k (s k 00) accepts. Defne V df k =. Case 2: M Q k (s k 00) and M O k (s k 00) reect. Case 2a: For all y {s k 01,s k 10,s k 11}, M Q k {s k 00} (y) reects. Defne V k as a subset of W(s k ) such that V k = 2, s k 00 V k, and s k 01 V k M O k {s k 00} (s k 01) reects. Case 2b: For all y {s k 01,s k 10,s k 11}, M O k {s k 00} (y) reects. Defne V k as a subset of W(s k ) such that V k = 2, s k 00 V k, and s k 01 V k M Q k {s k 00} (s k 01) reects. Case 2c: y {s k 01,s k 10,s k 11} and z {s k 01,s k 10,s k 11} such that M Q k {s k 00} (y) accepts and (z) accepts. Choose v W(s k ) {s k 00,y,z} and defne V df k ={s k 00,v}. M O k {s k 00} Ths fnshes the constructon of V k. We defne W n df = k [1,l] V k. Fnally, L s defned as the unon of all W n. Note that by the constructon, W n Σ t(n) whch shows (). Observe that the constructon also ensures (). We argue for L EXP: Snce l log log m, there are not more than 2m log log m possbltes to choose the strngs s 1,...,s l. For each such possblty we have to smulate O(l 2 ) computatons M (y) and M (y). Ths can be done n exponental tme n m. For the defnton of each V k we have to smulate a constant number of computatons M (y) and M (y). Ths shows that L s prntable n exponental tme. Hence L EXP. From the constructon t follows that L Σ m 2l 2log log m. In partcular, L SPARSE. It remans to show that L s not weakly T-mtotc. Assume L s weakly T-mtotc. So L can be parttoned nto L = L 1 L 2 (a dsont unon) such that () L 1 p T L 2 va machne M and (v) L 2 p T L 1 va machne M. 18

19 Let n =,, m = t(n), and O = W 0 W n 1,.e., O = L Σ <t(n). Let O 1,O 2,...,O l be the lst of all subsets of O (agan lexcographcally ordered accordng to ther characterstc sequences). Let s 1,...,s l and V 1,...,V l be as n the defnton of W n. Choose k [1,l] such that L 1 Σ <t(n) = O k. Let Q k = O O k. So L 2 Σ <t(n) = Q k. Clearly, V k must be defned accordng to one of the cases above. Assume V k was defned accordng to Case 1: So V k = and n partcular, s k 00 / L 1. Wthout loss of generalty assume that M Q k (s k 00) accepts. M L 2 (s k 00) has runnng tme m + < m m +m < t(n+1). Hence M L 2 (s k 00) behaves lke M L 2 Σ t(n) (s k 00). Snce s k was chosen accordng to Clam 4.3, for all r [1,l] {k}, M Q k (s k 00) does not query the oracle for words n W(s r ). Note that W(s k ) L = V k =. Therefore, M L 2 (s k 00) behaves lke M L 2 Σ <t(n) (s k 00). The latter accepts, but s k 00 / L 1. Ths contradcts (). M Q k (s k 00) whch s the same as Assume V k was defned accordng to Case 2: So V k = {s k 00,u} where u {s k 01,s k 10,s k 11}. Assume V k L 1. Then as above, M (s k 00) wth oracle L 2 behaves the same way as M (s k 00) wth oracle Q k. The latter reects, because we are n Case 2. So s k 00 / L 1 whch contradcts our assumpton. Analogously the assumpton V k L 2 mples a contradcton. Therefore, ether (s k 00 L 1 u L 2 ) or (u L 1 s k 00 L 2 ). (29) Assume V k was defned accordng to Case 2a: So for all y {s k 01,s k 10,s k 11}, M Q k {s k 00} (y) reects. In partcular, M Q k {s k 00} (u) reects. Assume u L 1 and s k 00 L 2. So M L 2 (u) reects, snce t behaves the same way as M Q k {s k 00} (u). By () ths contradcts u L 1. Therefore, by (29) we must have s k 00 L 1 and u L 2. In Case 2a, V k s defned such that s k 01 V k M O k {s k 00} (s k 01) reects. Note that M O k {s k 00} (s k 01) and M L 1 (s k 01) behave the same way. Hence, s k 01 V k M L 1 (s k 01) reects. If s k 01 V k, then u = s k 01 and hence M L 1 (u) reects. Ths contradcts (v). Otherwse, f s k 01 / V k, then M L 1 (s k 01) accepts and hence u = s k 01 / V k. Ths contradcts the assumpton u V k. Assume V k was defned accordng to Case 2b: Here we obtan contradctons analogously to Case 2a. Assume V k was defned accordng to Case 2c: Choose y and z such that M Q k {s k 00} (y) accepts and M O k {s k 00} (z) accepts. So u {s k 01,s k 10,s k 11} {y,z}. Assume s k 00 L 2. Hence M L 2 (y) and M Q k {s k 00} (y) behave the same way showng that M L 2 (y) accepts. So y L 1 whch contradcts the defnton of V k. Assume s k 00 L 1. Hence M L 1 (z) and M O k {s k 00} (z) behave the same way showng that M L 1 (z) accepts. So z L 2 whch contradcts the defnton of V k. Ths fnshes Case 2. From the fact that all possble cases led to contradctons, we obtan that the ntal assumpton was false. Hence, L s not weakly T-mtotc. 19

20 5 Poly-Autoreducblty In ths secton we consder the queston of whether NP-complete sets are f(n)-autoreducble, for some growng functon f. We frst start wth the followng lemma. Lemma 5.1 Let L be an NP-complete language. For every polynomal q(.) there s a polynomaltme algorthm A such that A on nput x, x = n, ether decdes the membershp of x n L or outputs strngs y 1,,y m such that x L {y 1,y 2,,y m } L, x / L {y 1,y 2,,y m } L =, m = q(n), and x y 1, y 2 y m. Proof. Let R(.,.) be a polynomal-tme search predcates assocated wth L. Gven x, let w x be the lexcographcally maxmum wtness of x. Wthout loss of generalty, assume that for every x L, every wtness of x s of length p( x ), for some polynomal p. Consder the followng set n NP. L = { x,y x L, y = p( x ),y w x }. Snce L s NP complete, there s a polynomal-tme reducton f from L to L. We now descrbe the algorthm A. Input x, x = n. Let m = p(n). f 1 m s a wtness of x, Accept. l = 0 m If f( x,l ) = f( x,1 m ), Reect. Q = {f( x,l )}. Whle Q q(n) + 1 do By dong a bnary search fnd a strng a such that l a 1 m and, f( x,a ) Q and f( x,a + 1 ) / Q. Set l = a + 1. f a s a wtness of x, then Accept. f f( x,l ) = f( x,1 m ), Reect. Q = Q {f( x,l )}. Output the frst q(n) elements of Q {x}. Clam 5.2 When the algorthm halts, for every strng y Q, x L y L. 20

21 Proof. If x / L, then none of x,c,0 m c 1 m, belong to L. Observe that the algorthm places a strng y n Q only f y = f( x,a ) where 0 m a 1 m. Snce f s a many-one reducton from L to L no strng from Q belongs to L. So from now we assume x L. We prove the clam by nducton. Intally, Q = {f( x,0 m }. Clearly, x L x,0 m L. Snce f s a many-one reducton L to L, the clam holds ntally. Assume that the clam holds before an teraton of whle loop. The whle loop fnds a node a such that f( x,a ) Q, but f( x,a + 1 ) / Q. Snce f( x,a ) Q, x L, by the nducton hypothess f( x,a ) L. Thus x,a L whch mples a w x. At ths pont the algorthm checks f a s a wtness of x. If a s a wtness of x, then t accepts and halts. If a s not a wtness, then we have a+1 w x. Thus x,a+1 L. Thus f( x,a+1 ) L. Snce the algorthm places f( x,a + 1 ) n Q at ths step, after the teraton of the whle loop the clam holds. Clam 5.3 If the algorthm Accepts or Reects x, then the algorthm s correct. Proof. The algorthm accepts x only when t fnds a wtness of x. Thus f the algorthm accepts x, then x L. The algorthm reects when f( x,l ) = f( x,1 m ). Note that f( x,l ) Q, thus by prevous clam x L f and only f f( x,l ) L. Observe that snce 1 m s not a wtness of x, and 1 m s the largest strng at length m, x,1 m / L. Thus f( x,1 m ) / L. Snce f( x,l ) = f( x,1 m ), the clam follows. Observe that algorthm places a strng f( x,l ) n Q only f f( x,l ) f( x,1 m ). Thus f( x,1 m ) s not placed n Q durng any teraton. So the bnary search step always fnds a wth desred propertes. Every teraton of the whle loop adds a new strng to Q or decdes the membershp of x n L. Thus the algorthm halts n polynomal tme and when t halts all elements n Q are dstnct. Ths fnshes the proof of the lemma. The above lemma comes close to showng that NP-complete sets are poly-autoreducble, except for a small caveat. Let L be any NP-complete language. If the algorthm from Lemma 5.1 nether accepts x or reects x, then t produces polynomally many equvalent strngs. However, to show L s poly-autoreducble, we must produce polynomally-many equvalent strngs even when the algorthm accepts or reects. Ths bols down to the followng problem: Let L be an NP-complete language. Gven 0 n as nput, n polynomal tme output polynomally many dstnct strngs such that all of them are n L. Smlarly, output polynomally many dstnct strngs such that none of them are n L. Below, we show that f one-way permutatons exst, then we can acheve ths task. We start wth a result by Agrawal [1]. Defnton 5.4 Let f be a many-one reducton from A to B. We say f s g(n)-sparse, f for every n, no more than g(n) strngs of length n are mapped to a sngle strng va f. Lemma 5.5 ([1]) If one-way permutatons exst, then NP-complete sets are complete wth respect reductons that are 2 n /2 nγ sparse. Here γ s a fxed constant less than 1. 21

22 Lemma 5.6 Let L be NP-complete. If one-way permutatons exst, then there exsts a polynomaltme algorthm that on nput 0 n outputs log n dstnct strngs n L and log n strngs out of L. Proof. By Lemma 5.5, there s a 2 n /2 nγ sparse reducton f from Σ to L. Thus f(σ log n ) 2 (log n)γ 2 log log n = log n. Thus by applyng f to every strng n Σ log n, we obtan at least log n dstnct strngs. Snce f s a reducton from Σ to L, these strngs are n L. It s obvous that ths can be done n polynomaltme. To generate strngs out of L, consder a 2 n /2 nγ -sparse reducton from to L. If we consder probablstc algorthms, then we obtan a stronger consequence. Lemma 5.7 Let L be NP-complete. Assume one-way permutatons exst. For every polynomal q, there exsts a polynomal-tme probablstc algorthm B that on nput 0 n outputs q(n) dstnct strngs from L and q(n) dstnct strngs from L. Proof. Let L be any NP-complete language. Consder a 2 n /2 nγ -sparse reducton from Σ to L. Consder the followng probablstc procedure. 1. Input 0 n, S = 2. Repeat Steps 3 and 4 q(n) tmes 3. Randomly pck a strng y Σ n. 4. If f(y) / f(s), S = S {y}. 5. Output S. We clam that the above procedure outputs a set S of sze q(n), wth hgh probablty. Consder Steps 3 and 4 durng an teraton, let S = k at ths tme. Snce at most 2 n /2 nγ strngs from Σ n are mapped to same strng, there exst at most k2 n /2 nγ strngs y from Σ n such that f(y) f(s). Thus the probablty that we do not add a new strng S durng Steps 3 and 4 s at most k/2 nγ. Thus the probablty that we fal to add a strng durng one of the q(n) teratons s at most q(n)k/2 nγ < 1/4. Thus the above algorthm outputs a set of sze q(n) wth hgh probablty. By the constructon of S no two strngs from S are mapped to same strng va f, thus f(s) = q(n). Snce f s a reducton from Σ to L, f(s) L. To generate strngs from L, we consder a reducton from to L. If we assume quck pseudo-random generators exst, then we can derandomze the above procedure. Lemma 5.8 Let L be any NP-complete language. If one-way permutatons and quck pseudorandom generators exst, then for every polynomal q(n), there s a polynomal-tme algorthm that on nput 0 n outputs q(n) many dstnct strngs from L and q(n) many dstnct strngs out of L. 22

23 Combnng Lemmas 5.1 and 5.6, we obtan the followng result. Theorem 5.9 If one-way permutatons exst, then every NP-complete language s log n- autoreducble. Combnng Lemmas 5.1 and 5.8, we obtan the followng result. Theorem 5.10 If one-way permutatons and quck pseudo-random generators exst, NP-complete sets are poly-autoreducble. Fnally, we consder another hypothess from whch poly-autoreducblty of NP-complete sets follows. Theorem 5.11 If there exsts a UP machne M that accepts 0 such that no P-machne can compute nfntely many acceptng computatons of M(0 n ), then NP-complete sets are polyautoreducble. Proof. Let L be any NP-complete language. We show that f the hypothess s true, then there s a polynomal-tme algorthm that on nput 0 n outputs polynomally many strngs from L and polynomally strng from L. Ths combned wth Lemma 5.1 shows that NP-complete sets are polyautoreducble. We frst show how to produce polynomally many strngs from L. The hypothess mples that there s a polynomal-tme decdable predcate R(.,.) and a polynomal p such that for every n, there exsts a unque strng w n and R(0 n,w n ) holds. Moreover, no polynomal-tme algorthm, on nput 0 n, can output w n for nfntely many n. Consder the followng language L = { 0 n,y y = p(n),y w n }. Snce L s n UP, there s a many-one reducton f from L to L. Now consder the behavor algorthm A, descrbed n Lemma 5.1, on nput 0 n. Snce 0 n belongs to 0, the algorthm never reects. The algorthm accepts only when t fnds a wtness w n for 0 n. However, our hypothess says that no polynomal-tme algorthm can output nfntely wtnesses w n. Thus ths algorthm must output a set Q of cardnalty q(n). Usng smlar arguments as n Lemma 5.1, we can show that for every y Q, 0 n 0 y L. Thus the algorthm outputs q(n) strngs from L. To output strngs from L, consder the many-one reducton g from L to L and proceed as before. Snce L s n UP CoUP such a reducton exsts. Thus there exsts a polynomal-tme algorthm that outputs polynomally many strngs from L and polynomally many strngs from L. 23