Non-Uniform Reductions

Transcription

1 Non-Uniform Reductions Harry Buhrman Benjamin Hescott Steven Homer Leen Torenvliet January 19, 2007 Abstract Reductions and completeness notions form the heart of computational complexity theory. Recently non-uniform reductions have been naturally introduced in a variety of settings concerning completeness notions for NP and other classes. We follow up on these results by strengthening some of them. In particular, we show that under certain well studied hypotheses: many-one complete sets for NP are length increasing complete with a single bit of advice, strengthening a result of Hitchcock and Pavan [HP06]. 1-truth-table complete sets for NP are many-one complete with a single bit of advice. We tighten another result of [HP06] and give a trade-off between the amount of advice that is needed for the reduction and its honesty, i.e., how length decreasing a reduction is for the many-one complete degree of NEXP. We also construct an oracle relative to which this trade-off is optimal. For uniform reductions, the trade-off only yields exponential honesty and the oracle witnesses an exponentially length decreasing many-one complete set for NEXP. Next, we start a more systematic study of non-uniform reductions and show, among other things, that in certain cases the non-uniformity can be removed at the cost of more queries. For example, for EXP we show that complete sets under many-one reductions that use O(logn) bits of advice are still Turing complete without advice. In line with Post s program in complexity theory [BT05] we connect such uniformization properties to the separation of complexity classes. 1 Introduction Reductions and completeness are two of the original concepts in complexity and form part of the core of the field to this day. Determining whether a problem is complete for a particular class is central in determining the computational complexity of that problem. In a broader perspective, questions of whether different types of reductions are the same or different, and whether completeness notions induced by these reductions are the same or different have been assiduously explored, e.g.,[lls75, BHT91, Wat87, BST93a]. Answers to these questions are related to the central open problems of the field. In some cases, inequality of reductions on certain complexity classes implies inequality of complexity classes [LM94]. In others the collapse of degrees to an isomorphism type under some notion of reduction yields the inequality of classes as well [BH77, FFK92, HH91, HS92, KMR89, You90, AB93, All88]. This makes the properties of reductions, for example whether they CWI;Kruislaan 409;1098 SJ Amsterdam;the Netherlands; buhrman@cwi.nl Boston University;Computer Science Department;111 Cummington St; Boston, MA Boston University;Computer Science Department;111 Cummington St; Boston, MA ILLC;Plantage Muidergracht 24;Amsterdam;the Netherlands; leen@science.uva.nl 1

2 are length-increasing, 1-1, etc. important objects of study. Finally, certain properties of complete sets in different complexity classes might have implications for the question of equality of these classes [AS84, BvMFT00, BT04]. The lion s share of investigations of reductions has been into uniform reductions. Now, nonuniformity has entered the realm of the reduction. As with the definitions of non-uniform complexity classes, by means of advice classes, (P/poly and families of circuits of a certain size), one can define reductions that are computable by means of additional advice or by polynomial size circuits. Allender et. al. [ABK + 02] have shown that the set R of Kolmogorov incompressible strings, with respect to exponential time Kolmogorov complexity, is complete for EXP with respect to polynomial time (truth-table) reductions that have a polynomial amount of advice. Moreover the advice is indispensable: R is not complete with respect to uniform Turing reductions [BM95]. For a strengthening of this see the thesis of Ronneburger [Ron04]. Agrawal [Agr02], while studying the isomorphism conjecture [BH77] for NP complete sets, used non-uniform reductions with advice. He shows under the assumption that a certain type of one-way function exists, that all many-one complete sets for NP are 1-1 and length increasing complete for reductions that use some amount of non-uniform advice. Recently Hitchcock and Pavan [HP06] showed under a different assumption, namely that NP is not small (does not have resource bounded measure zero), that every many-one complete set for NP is length increasing complete with reductions that use a polynomial amount of advice. Moreover they also show that for NEXP the many-one complete sets are length increasing complete with reductions that use a polynomial amount of advice. In this paper we improve the results of Hitchcock and Pavan. In particular we show that under the weaker assumption that there exists a DTIME(2 nc ) bi-immune set in NP, that one bit of advice suffices to show that many-one complete sets are length increasing complete 1. We also improve their results for NEXP. We again reduce the number of advice bits needed to make the reductions length increasing annd show that our result is optimal relative to some oracle. In particular this yields an oracle where the many-one complete sets for NEXP are not (uniformly) length increasing complete, in fact they are only exponentially honest, matching the best known result due to Ganesan and Homer [GH88]. Surprisingly, this shows that non-relativizing techniques are needed to settle this question. Another structural property known for EXP [HKR93] and for NEXP [BST93b] is the following. Every 1-truth-table complete set is many-one complete. We show that under the hypothesis that NP contains a set that is NTIME(2 nc ) co-ntime(2 nc ) bi-immune every 1-truth-table complete set for NP is many-one complete with 1 bit of advice 2. We also construct, extending an earlier result of [BT99], an oracle world where such a bi-immune set in NP exists. The results above warrant a more systematic investigation into non-uniform reductions. In particular is the amount of advice needed for the non-uniform reductions above optimal? In general when does advice yield additional power? In the second part of this paper we make a beginning of such a study. We first show that for EXP constant query reductions that have advice are strictly more powerful than their uniform counterparts. For example we show that the completeness notion induced by many-one reductions 1 Technically we show something slightly weaker. We show that the many-one reduction is either length increasing or accepts/rejects without querying a string. The result does hold for 1-tt reductions. 2 Again we technically prove this for the weaker version of many-one reduction that can also accept/reject without producing a query. 2

3 with one bit of advice is incomparable with the 2-truth-table complete degree for EXP. Analogous to uniform reductions, we show that complete sets for EXP with respect to many-one reductions that use c-bits of advice are 1-1 and length increasing complete with c-bits of advice. These results require new techniques as the original ideas for uniform reductions cannot be used directly. Second we show that non-uniformity can be removed at the price of more queries. In particular we show that sets that are many-one complete with respect to reductions that use O(log n) bits of advice are Turing-complete with respect to uniform reductions for EXP. For the delta levels of the polynomial hierarchy we show that truth-table reductions suffice in order to uniformize the reduction. Such results can be seen as a Karp-Lipton [KL80] type of result for reductions. We use ideas and techniques from the paper that studies auto-reducibility of complete sets [BvMFT00]. We have the added bonus that these theorems do not relativize and hope that these results could be used in a nonrelativizing proof that separates complexity classes. In particular it follows from our results that solving the question of whether many-one complete sets with 1 bit of advice are uniformly truth-table complete will separate complexity classes (EXP from EXPSPACE or PH from EXP). Hence understanding questions like these has deep implications for complexity theory. 2 Notation We adopt the standard definitions for well known complexity classes, P, NP, PSPACE, EXP, NEXP. For more information on these classes see Balcázar and J. Díaz and J. Gabarró [BDG88] and Homer and Selman[HS01]. We use the standard definitions for polynomial time reductions, i.e. we say A, many-one reduces to set B, denoted A m B if there is a polynomial time computable function f where x A f(x) B. The reduction is considered a 1-1 reduction, p 1 1, if in addition the function f is one to one. A many-one reduction is length increasing, p m,li, if in addition, x, f(x) > x. Two sets A and B are polynomial time isomorphic, A p B, if there is a bijection, f : Σ Σ, where f and f 1 are polynomial-time computable and x A f(x) B. If we relax the definition of many-one reduction to allow the machine computing the reduction to either accept or reject without producing a query we say this as an extended many-one reduction, and denote it pˆm. More formally A pˆm B if there exists a polynomial time Turing machine M with an output tape, where on input x, M does one of the following: M outputs f(x) Σ and x A f(x) B, M outputs ACCEPT and x A, or M outputs REJECT and x / A. A polynomial reduction is nonuniform if it is in P/poly, i.e. A P/poly m B if f PF and h poly where A = {x f(x,h( x )) B}, here poly = {g n, g(n) p(n) for some polynomial p}. Many times the amount of advice is more or less restrictive, this restriction is on the function f in the definition above. Namely, for one bit of advice the range of f is {0,1}. We denote a one-bit nonuniform many one reduction, p/1 m. A reduction from set A to B is considered to be adaptive if the membership of a string x A is decided by a polynomial-time Turing machine which has oracle access to B. Here the computation is allowed to adapt to membership queries to B, by basing future queries on previous answers. This reduction is denoted A T B and is commonly referred to as a Turing reduction. It is natural to consider reductions which are computed with oracle access to the set B, but are not adaptive. We say A reduces to B nonadaptively, denoted, A p tt B if there exist a polynomial-time function, r and Turing machine M where r(x) = (q 1,q 2,...,q m ) and M(x,[q 1 B],[q 2 B],...[q m B]) accepts if and only if x A, these reductions are commonly referred to as truth table reductions. In this work we consider the measure hypothesis on NP formulated by Lutz which states that 3

4 NP does not have p-measure 0, µ(np) 0. This is equivalent to assuming that there is no polynomial time martingale which succeeds on every language in NP. A martingale is a function d : Σ [0, ) where w Σ,2d(w) = d(w0) + d(w1). A martingale succeeds on language L if lim sup n d(a n ) =. This can be thought of as a betting strategy on the languages in NP, i.e. is there a procedure where one can bet on the membership of strings in an NP language and win? Informally this relates to whether or not NP is a large subset of E. A consequence of µ(np) 0 is the existence of a p-random set within NP. A set is considered t(n)-random if no O(t(n)) martingale succeeds on it [ASTZ94]. A set is C-immune if it does not have any infinite subsets in C. A set is C-bi-immune if it and its complement are both C-immune. The measure hypothesis on NP also implies the existence of a DTIME(2 nǫ )-bi-immune set within NP [HP04]. 3 Advice to Strengthen Reductions When does advice help to strengthen reductions? In this section we will discuss and show that in certain cases reductions can be massaged to have a certain property. We study here the two cases where we require that the reductions are length increasing, and do not ask negative queries, the latter in the case of truth-table reductions. 3.1 Length increasing reductions Hitchcock and Pavan [HP06] recently showed that reductions can be made length increasing under the assumption that NP is not small. We improve upon the advice needed in their paper in the following theorem. Theorem 1 µ(np) 0 = Every p m-complete set is p/o(loglog n) m,li -complete. Proof. Choose ǫ > 0 and let R be a n 1+ǫ -random set in NP. For every n let x n 0,...xn 2log n+1 be the lexicographically first 2log n + 1 strings of length n. We claim that ( n)( i log n + 1)[R(0 n ) R(x i ) = 1]. Suppose not, then the following betting strategy wins infinitely on R. Start with reserving 1 2n of the starting capital for use at length n. Now bet evenly on {0 n } and use the 2 outcome of R(0 n ) to bet on x n 0,...xn 2log n+1. The capital at length n is then 2log n+1 /2n 2 > 1. To identify an index i for which R(0 n ) R(x i ) = 1 we need about log log n bits. Let D = { x,y,φ x = y = φ R(x) + R(y) + SAT(φ) 2}. p/log log n By the observation above, SAT m,li D. Now let A be a p m-complete set in NP. We show that the reduction M from D to A must be honest. Combining a padded version of the reduction from SAT to D and the reduction from D to A then gives a length increasing reduction from SAT to A using O(log log n) bits of advice the advice is needed to give the length of the padded string which might be polynomially longer than the input. Suppose not, then for every k there are infinitely many x,y,φ such that M( x,y,φ ) < x,y,φ 1/k. The set A is in NP so it is DTIME(2 nc )-computable for some c. Hence for infinitely many x,y,φ it is DTIME(2 n ) computable whether x,y,φ is in D. This gives a betting strategy since M( x,y,φ ) / A implies x / R or y / R and M( x,y,φ ) A implies x R or y R. Assume x is lexicographically less than or equal to y. Given the outcome of M( x,y,φ ) A use 4

5 1/3d of the capital on x R and, if necessary, the remaining 2/3d of the capital on y R. In either case the capital grows to 4/3. With a weaker assumption, namely that NP has a DTIME(2 nc ) bi-immune set, we arrive at an even stronger conclusion. We can make reductions length increasing with the help of only one bit of advice. Win some, loose some, the reductions in the following theorem drop from many-one reductions to extended many-one reductions, ones that also are allowed to accept/reject without querying a string (as previously defined.) Currently, we are unable to show this for many-one reductions. Theorem 2 If for some c > 0 ( R NP) R is DTIME(2 nc )-bi-immune, then A is pˆm -complete A is pˆm,l.i. -complete. Proof. Suppose that A is in NTIME(n d ). Let { φ SAT 0 D = { φ,b φ R if b = 0 φ SAT 0 φ R if b = 1 } It is easy to see that SAT p m,li D. In fact, by paddability of SAT, we can force this reduction to be length increasing by a factor of any fixed polynomial. To prove the theorem, it therefore suffices to prove that D pˆm,honest A. Let M be the reduction from D to A. The following holds for all φ by definition of D. φ / SAT = [0 φ R M( φ,0 ) A] φ SAT = [0 φ R M( φ,1 ) A] Suppose that M is not honest, i.e, ( k)( φ)[ M( φ,b ) < φ,b 1 k. Then at least one of the following two cases must be true. case 1: ( k)( φ)[ M( φ,0 ) < φ,0 1 k] or case 2: ( k)( φ)[ M( φ,1 ) < φ,1 1 k]. Then by the DTIME(2 nc )-bi-immunity of R and the fact that M( φ,b ) in A is NTIME(n d ) and hence DTIME(2 nd ) computable if M( φ,b ) < φ,b c d we can conclude for almost all of these φ in the first case that φ SAT or in the second case that φ / SAT. So for a suitable k whenever M( φ,b ) < φ,b 1/k, we can decide membership of φ in SAT in P. In the remaining cases M is honest. Hitchcock and Pavan [HP06] showed that for nondeterministic exponential time, no assumption is needed to make the many-one complete sets length increasing complete via reductions that use a polynomial amount of advice. Next we will improve this result to n log n bits of advice. Theorem 3 For every constant c, NEXP many-one complete sets are complete with length increasing nonuniform many one reductions which use n clog n bits of advice. Proof. Let A be any p m NEXP complete set and let K be the standard 1-1, length increasing, paddable, many-one complete set computable in NTIME(2 n ). Let p : Σ Σ Σ be a polynomial 5

6 time padding function for K where x r x K p(x, r) K, and let f be a polynomial time 1-1 reduction from K to A. Since f and p are both 1-1, then for every n there is some r n Σ n+1 where f(p(x,r n )) > x for all x Σ n. Let r be r n with the last clog n bits removed. Now consider the function p which takes x, r as input and tries all possible n bit strings, r i, that are extensions consistent with r and outputs f(p(x,r i )) for the first r i where f(p(x,r i )) > x. We know that such an extension exists and that it will be polynomial in n to find such an r i. Let g(x) = p (x,r ) be the reduction from K to A. We remark here that the same proof yields a trade-off between the amount of advice and the honesty of the reduction. For any function g, let M be a many-one reduction. We call M g- honest if for almost all x : M(x) > g( x ). Note that log n-honest sets are sometimes also called exponentially honest. Theorem 4 For any c let g be the function that determines the honesty of the reduction. The many-one complete sets for NEXP are g( x )-honest for reductions that use g( x ) c log n bits of advice. Corollary 5 ([GH88]) For any c, the many-one complete sets for NEXP are clog n-honest complete. The amount of advice used in the previous theorem is optimal relative to some oracle. We will first construct an oracle relative to which NEXP has a set that is many-one complete, but only under length decreasing reductions. We will then point out how to adapt this construction to obtain the optimality of Theorem 4. Theorem 6 There exists an oracle witnessing that nondeterministic exponential time has a many one complete set that is not complete with length increasing reductions. Proof. Let K A = {< i,x,n > Ni A (x) accepts within n steps } be the standard complete set for NEXP A under P A m reductions. We will initially define a set C, first by constructing two sets A and D in stages and letting C = K A \ D where D and A will be defined in stages below. Part I - Diagonalization Let {fi A } be an enumeration of the polynomial time many-one reductions that have access to oracle A. Let Mi A denote the deterministic polynomial time Turing machine with oracle A that computes function fi A. Without loss of generality we will assume our enumeration is such that the running time of Mi A is bounded by n i. In the following stage construction Q i is used to keep track of the queries asked by the reduction f i at stage i, D i to keep potential values of the reduction queried to A during the computation as well as the value of the reduction when it is finished. Stage 0 D =,Q =,A =,b(0) = 2. Stage i Set counter = 1 i log b(i),d i =,Q i =. Simulate M A i (0b(i) ) where b(i) = 2 2b(i 1). Whenever M A i queries a string, q j, do the following: If q j < b(i) then answer yes if q j A, no otherwise. Add q j to Q i and continue. 6

7 Otherwise q j b(i), add q j to Q i, and If q j A Then answer yes, decrement counter, and continue. Else If q j =< 0,s,x > for some s, s = i log b(i), and s counter. Then for t = 0 i log b(i) to counter Add < 0,t,x > to A t = t + 1 Add x to D i, decrement counter, answer yes, and continue. Else Answer no, decrement counter, and continue. When M A i finishes its computation, let w = f i (0 b(i) ). If w > b(i) add it to D i, and for t = 0 i log b(i) to counter Add < 0,t,w > to A t = t + 1 Set D = D D i, and Q = Q Q j. At the end of this part of the construction A contains blocks of elements of the form < 0,0 k,q >... < 0,counter,q > where q is a length increasing query or the value of a reduction computed by machine M k on input 0 k whose length is greater than or equal to k. Now we initially define C as K A \ D. We are not able to fix C as A is not fixed, and therefore neither is K A. However, given this C it follows that for any query, q, q > b(i), by reduction fi A with input 0 b(i), q / C; moreover if f i (0 b(i) ) > b(i) then f i (0 b(i) ) / C. We will ensure that this property continues to hold as we change C and A in Part II. Part II - Encoding It may be that C is not hard for NEXP A under P A reductions, even though K A remains complete. For any element x K A D we will need to add a witness to x K A back into C whose length is shorter than x. The construction ensures that every element x removed from C is coded back into A in the form < 0,0 i log b(i),x >. We will replace x in C with < λ,1,counter > where counter is the largest ilog(n) bit string where < 0,counter,x > A. As C has not yet been fixed we will need to keep track of these strings in another set, T, which we now define. Let T = For every x K A D we will do a search in A to find the value of the counter when x was added to D i. Find the largest i, with b(i) n, given n = x. Notice this can be done in polynomial time as i = (log (n) 1) 2 Let t = 0 i log b(i). 7

8 While < 0,t,x > A t = t + 1. Add < λ,1,t > to T We will eventually define C = K A \ D T, but to show that C is in NEXP A we need a way to decide each < λ,1,t > in T using an NEXP A machine. Informally we will do this by encoding a string z in A that represents the original x in K A D as well as the counter value used to replace it in C. We will encode this strings into A using another stage construction. Stage i 1. If D i = end stage i, otherwise let X = D i K A. For every d j D i find the maximum t j with < 0,t j,d j > A by doing a binary search in A starting at < 0,λ,d j > 2. While X : pick the least x j X. Fix its leftmost accepting path, P j, of N A (x j ) (Recall that N is the machine computing K A.) Let Q xj be the set of queries asked on P j. Let Q = Q Q xj Now find the least string z where: b(i) < z < b(i) 5i t, t = log b(i), < 1,t,z >/ {A i D i Q} Add < 1,t j,z > to A, where t j is the corresponding counter value for x j found in step 1. X = {D i K A } \ x j In order to prove that we have enough space to encode, we need to prove that within the while loop in Part II we can always find a z of proper length while avoiding queries, and all previous encodings. Finally, we must show that the while loop will end. The intuition of why the while loop will end is that as we add elements to A it may be the case that some element d j in D i is now in K A, but this can happen at most n i times, i.e. D i K A. To show that we have enough space to find a z we need only look at the size of A D i Q. Lemma 7 A i D i Q cn i 2 ni at any stage i above, for some constant c. Proof. First consider the sizes of A i and D i. Each of these sets will be at most 2n 2i in size. For the size of A i, we know at most n i queries will be made by the f i reduction during the diagonalization in Part I, at most n i strings will be added for each of these queries, and we add at most one string to A i during each iteration of the loop above. Given that once a string is encoded its membership in K A will not change, the loop is bounded by the size of D i. Since D i is at most n i, the maximum number of queries that Mi A can make on input 0 b(i), the two sets are less than 2n 2i. Now we need only show that Q will be bounded as well. We know that the queries added from the diagonalization are bounded by n i and since D i is bounded by n i we have at most n i accepting paths to fix, and since the largest string in D i is at most n i, the size of Q is at most n i 2 ni + n i. Therefore there is some c where A D i Q cn i 2 ni. Lemma 8 During the encoding at stage i, a z will always be found. 8

9 Proof. We know that there are 2 n5i +1 2 n+1 strings whose length is between b(i) and b(i) 5i. Finding an appropriate z is equivalent to finding 2n i consecutive strings none of which are in A D i Q. Given the size of A D i Q we know that there are more than 2n5i +1 2 n+1 possible z s, since this number is exponential in n i we have more than enough z s to encode at each iteration. Moreover the loop will end, for as the number of possible candidates is no more than n i, we can at each iteration find another z, if needed. Finally we have finished the construction of A. Therefore K A is defined and we now can formally define C = K A \ D T. Given this set C we can proceed with the final proof. Lemma 9 C is in NEXP A cn i 2 ni Proof. On input x, we decide if x C by, If x is not of the form < a,b,c > for some a,b,c Σ reject If a λ AND b 1 simulate N A (x). Nondeterministically choose a string z with length between 2 c and 2 c 5 with z =< 1,c,z > for some z. Accept if z A. Lemma 10 C is complete for NEXP A under P A many one reductions. Proof. Given some L in NEXP A, with f L the P A reduction from L to K A consider the following reduction from L to C. On input x, Calculate f L (x). Given f L (x), calculate i, the largest integer with b(i) f L (x). If < 0,0 i b(i),f L (x) > A Do a binary search to find the largest i b(i) length string, t, between 0 i b(i) and 1 i b(i) with < 0,t,f L (x) > A Output < λ,1,t > Else output f L (x). C is not complete for NEXP A under l.i. reductions. Assume not, since 0 NEXP A there is a reduction f from 0 to C. The reduction f is polynomial so it appeared as some f j in the enumeration from Part I, therefore element 0 b(j) will map to an element not in C. 9

10 The above construction can be adapted to reductions that have k bits of advice by the following observation. In stage i of the construction we diagonalize against many-one reduction M i, resulting in the coding of n i strings from K A in strings of length log(n i ) = ilog n. Instead of diagonalizing against just M i the same construction can be used to diagonalize against 2 k different advices for M i resulting in a coding of 2 k n i strings from K A in strings of length log(2 k n i ) = k + ilog n. This exactly matches the bound from Theorem 4. The oracle above works for nondeterministic polynomial time as well as nondeterministic exponential time. The only significant change in the constructions above is that we pad the counter so that the replacement in C, < λ,1,t > is only polynomially smaller in length than the original element taken from C. This is done by adding leading zeroes to the counter so that its length grows from i b(i) bits to b(i) 1 2i bits. We omit the full proof here, but state the result as a theorem. Theorem 11 There exists an oracle witnessing that NP has a many one complete set that is not complete with length increasing reductions truth-table versus many-one reductions Sometimes 1-tt reductions can be converted into many-one reductions. Under a rather strong assumption we can prove this theorem for 1-tt reductions that use one bit of advice. Theorem 12 Let A be p 1 tt complete for NP. If there exists a set R in NP that is NTIME(2cn ) co-ntime(2 cn )-bi-immune, then A is also p/1 m -complete for NP. Proof. Let A be p 1 tt complete and let D = { φ,b { 0 φ R φ SAT if b = 0 0 φ R φ SAT if b = 1 } Since R NP, D NP and so D p 1 tt A. Obviously SAT p/1 m D. The single bit of advice needed is whether 0 φ R. Let M be the p 1 tt reduction from D to A. On input φ,0 M produces a string z. Then, depending on the program of M, φ,0 D z A or φ,0 D z / A. In the first case we say M is of type m (many-one) on input φ, 0 and in the second case that M is of type m on input φ,0. We now claim that the bi-immunity of R implies M can be of type m for only finitely many unsatisfiable φ. Or, in other words, ( φ / SAT)[type(M(φ,0)) = m] B, B =,B NTIME(2 2n ) co-ntime(2 2n ),B R B R. We first prove this claim. Consider C = {0 n φ / SAT, φ = n,type(m(φ,0)) = m} C = [ C R = C R = ] Assume C R =. Set B = C R. Now B NTIME(2 2n ) by the following algorithm. On input 0 n guess φ with φ = n and φ / SAT; Check that type(m(φ,0)) = m and verify 0 n R using the reduction to A. It is also the case that B is in NTIME(2 2n ) by the following algorithm. 10

11 On input 0 n check that for every φ of length n either φ SAT or that type(m(φ,0)) = m, or that there exists a φ / SAT of length n for which type(m(φ,0)) = m but m(φ,0) A (which means 0 n / R). We conclude that under the assumption of the theorem there can only be finitely many unsatisfiable φ such that M(φ,0) is of type m. A similar proof shows that there can be only finitely many satisfiable φ such that M(φ,1) is of type m. From this we can build our many-one reduction. The premise of the previous theorem seems rather strong. Yet it is not impossible that NP does have such sets. To provide evidence for this statement we construct precisely such a set in the following theorem. Theorem 13 For every constant c, there exists an oracle A such that NP has a set that is NTIME(2 nc ) co-ntime(2 nc )-immune. Proof. We use a construction appearing in [BT99] in which the oracle is created from an infinite Kolmogorov random string. The language that will have the property stated in the theorem will be the following. For all oracles X define D X = {x ( y)[ y = n 2c xy X}. Let Y be an infinite string that has the property that ( n)[k(y [1..n] ) > n c ]. We will construct A by stages. A s+1 is the oracle defined by the end of stage s and A s+1 A s forall s. The oracle A may have 2 n strings of length n + n 2c. We now describe the initial oracle A 0. To encode a string x into D, we need a sequence of x 2c bits. For this encoding we use substrings of Y. Up to x we have used bits of Y for i< x 2i = 2 x strings, so if we let the substrings of Y that encode strings of length x start at the 2 x x 2c th bit of Y, there will never be a conflict (two strings in A taking their encoding from the same substring of Y ). Moreover, we have sequence from the 2 x x 2c th bit up until the 2 x +1 x 2c th bit of substrings of length x 2c that can be used to encode strings of length x. The encoding of strings of length x + 1 then starts at the 2 x +1 x + 1 2c th bit of Y, which is quite a bit further along. For A 0 up to the ith string of of length 2 m m 2c + im 2c bits. The incompressibility of Y implies that for some constant d and all m and i it holds that K(Y m,i A 0 ) 2 m m 2c + im 2c d. This is the property we will use to prove correctness of our construction. At each stage s we will decide the membership of the string x s in D A, where x s is the sth string in the lexicographical ordering of Σ. Deciding membership of x s is deciding whether to remove x s y from A where x s y is the only string that is an extension of x s currently in A. The Kolmogorov property that makes the entire construction work is that an NP machine that rejects some string x s with oracle A s must also reject x s with oracle A or else it will allow us to describe some initial segment of Y using significantly less bits than the length of this segment. Let {M i } i be an enumeration of all nondeterministic Turing machines, where 2 nc is the time bound on inputs of length n for all machines in the enumeration. The construction maintains two sets of requirements. First, a set U s of yet unsatisfied requirements to which occasionally a new element is added and from which satisfied requirements are removed. Second, a set V s in which satisfied requirements are kept. Sometimes a satisfied requirement will be moved from V s to U s at which time it will become unsatisfied again. Every requirement corresponds to a language in NTIME(2 nc ) co-ntime(2 nc ), represented by nondeterministic 2 nc -time bounded machines machines M i and M j. Even requirements R 2 i,j will represent the need to establish a nonempty intersection of L(M i ) L(M j ) with D, whereas odd requirements R 2 i,j +1 will represent the need to establish a nonempty intersection of L(M i ) L(M j ) with D. length m we use a substring of Y, say Y m,i A 0 11

12 We will prove that the construction of A meets the following requirements for all i,j. 1. R 2 i,j : L(M A i ) = = [[L(MA i ) L(MA j )] [L(MA i ) D A ]]. 2. R 2 i,j +1 : L(M A i ) = = [[L(MA i ) L(MA j )] [L(MA i ) D A ]]. There are two ways to fulfill these requirements. Either maintain a difference between L(M i ) and L(M j ), i.e., the pair M i,m j does not represent a language in NTIME(2 cn ) co-ntime(2 n ), or maintain a string in the intersection of L(M i ) and D, respectively D. A requirement R e with e = 2 i,j, or e = 2 i,j + 1 is active at stage s if x s L(M As i ). Now the construction can be described as follows. stage s: If no requirement is active then A s+1 = A s ; V s+1 = V s ; U s+1 = U s ; Else let e be the least active requirement in U s ; If e is even then U s+1 = U s {e}; V s+1 = V s e,x s ;A s+1 = A s ; Else If { i,x V s (i < e) (2 x c > x s )} = then V s+1 = V s { i,x V s i > e} U s+1 = U s {i (i > e) ( x)[ i,x V s ]} {e}; A s+1 = A s { x s,y s }; Else A s+1 = A s ; V s+1 = V s ; U s+1 = U s ; If s = 2 k then U s+1 = U s+1 {k}. End of Construction We will now prove correctness of our construction in a series of lemmas. The key lemma in the proof is the following. Lemma 14 ( s)( e < log s)[x s / L(M As e ) = x s / L(M A e )] Proof. Suppose not. Let e and s be such that e < log s and x s L(Me A ) L(Me As ). Let s be minimal such that s s and x s L(M A s +1 e ) L(M A s e ). By construction A s A s +1 = {x s y s }. The string x s y s must be queried in any accepting computation of M A s +1 e on input x s. Otherwise M A s e would also have an accepting computation on input x s contradicting the assumption. On input x s, machine M e can only query strings of length less than or equal to 2 xs c. Moreover {y y A 0 A s +1} < log s since for every such y there is an odd index in U t (U t+1 V t+1 ) for some t < s and there are no more than log s indices in {U t t s }. Let n be such that n+n 2c > 2 xs c. We will show how to construct the first 2 n n 2c bits of Y using significantly less bits and hence arrive at a contradiction. Suppose that we have 2 n n 2c x s 2c bits of Y that describe the initial segment of Y used to encode A 0 for all substrings of Y up to strings of length n + n 2c except y s. Furthermore suppose that we have a list of at most log s strings x i1,...,x ik that says which x ij y ij are in A 0 A s. Note that no strings greater than x s y s are in A 0 A s. Finally let q < 2 xs c be the index of the query x s y s in the leftmost accepting computation of M A s +1 e (x s ). Then we can construct the first 2 n n 2c bits of Y from 2 n n 2c x s 2c + log s x s + x s c + O(1) bits of information. We arrive at a contradiction for all but finitely many s, since the complexity assumption on Y is violated. Lemma 15 ( e)[ {s ( x)[ e,x V s+1 V s ]} < ] 12

13 Proof. Whenever e,x is in V s+1 V s there is a smaller index that is moved from U s+1 either to oblivion or to V s+1. This means that 0 can enter V s for only one 1 and there is no t such that 0,x is in V t V t+1. By induction for i < e, let #i be the number of times that index i is in V t+1 V t for some t and x. Then it is clear that #e e 1 i=0 (1 + #i) which is finite for all e. Corollary 16 ( e)[( s)[e U s ] = ( s)[e U s ]] Lemma 17 If L(M A e ) = then L(MA e ) D A Proof. Let e be such that L(Me A ) = and L(Me A ) D A =. If 2e in V s+1 then L(M A s+1 e ) D A s+1. Since L(Me A ) D A = there are infinitely many s such that 2e / V s. It follows from Corollary 16 that 2e U s for almost all s. Consequently there are infinitely many s such that x s L(Me A ) where 2e U s and e < log s. At stage s there are two possibilities. 1. x s L(M As e ) 2. x s / L(M As e ). In case 1, since x s / D A there is some e < e such that 2e +1 U s+1 U s. Hence case 1 can appear only finitely often. Case 2 must appear infinitely often. Consider a stage where case 2 appears. Then we have e < log s and x s L(Me A) L(MAs e ), so it follows from Lemma 14 that this can also appear only finitely often, a contradiction. Lemma 18 If L(M A i ) = L(MA i ) and L(MA i ) = then L(MA i ) D A. Proof. We will first argue that for almost all i and j, if 2 i,j + 1 is in U s (U s+1 V s+1 ) then L(M A i ) D A. Note that a particular language in NTIME(2 nc ) co-ntime(2 nc ) is represented by infinitely many pairs i,j. Suppose that for some (large) s the index 2 i,j +1 is in U s (U s+1 V s+1 ). This means that x s D A and x s / L(M As j ). It follows from Lemma 14 that x s / L(M A j ) and hence from the assumption L(M A i ) = L(MA j ) that x s L(M A i ). If ( s)( x)[ e,x V s ] then let t be maximal so that e / V t. Since A t = A t+1, it is then the case that x t L(M A t+1 i ) L(M A t+1 j ) and moreover since e,x t V u for all u t + 1, subsequent changes to the oracle are made only at lengths greater than 2 xt c whence x t L(Mi A) L(MA j ). In this case there is nothing to prove. Finally, let e = 2 i,j + 1 be such that L(Mi A) = L(MA j ) and L(MA i ) = and L(MA i ) D A and moreover s [e U s ]. By Corollary 16 also ( s)( e e)[e U s = e U s+1 ], i.e., no requirement of higher priority will be satisfied at stage s. Consider only two stages where x s L(Mi A) and ( e e)[e U s = e U s+1 ]. At any of these stages there are two possibilities. 1. x s L(M As i ) 2. x s / L(M As i ). In the first case 2 i,j + 1 / U s+1 contradicting the assumption and in the second case x s L(Mi A) L(MAs i ) so this case can occur only finitely often according to Lemma 14 The following lemma concludes the proof. 13

14 Lemma 19 D A = and D A =. Proof. This is immediate from the fact that both sets have nonempty intersections with languages in NTIME(2 nc ) co-ntime(2 nc ) at infinitely many different points x s. 4 When Does Advice Help? As we stated in the introduction, reductions with advice seem to inhabit a higher level of complexity than do uniform reductions. In the next theorem we show a simple diagonalization that demonstrates incomparability of uniform 2-tt reductions and many-one reductions with advice. Theorem 20 There exists a set A EXP that is p 2 tt-hard for EXP but not p/1 m -hard for EXP, and there exists a set A EXP that is p/1 m -hard for EXP but not p 2 tt-hard for EXP. Proof.(sketch) This is a fairly simple diagonalization. We start with an enumeration {M i } i of all polynomial time transducers that take one bit of advice and look at the outcome of the computations on input 0 n for n spaced exponentially far apart. Our exponential-time computable set D that we use for diagonalization is a subset of {0} {1}, where strings of this type appear with exponential gaps between them. Our complete set A is a subset of { b,x b {0,1}}, and the 2 tt reduction from K to A on input x computes 0,x A 1,x A and accepts if the outcome is 1. K is the standard EXP complete set. On suitable inputs 0 n we diagonalize against M i. If M i accepts or rejects without querying, or queries a part of A that has already been set, i.e. exponentially smaller than the input, we diagonalize by letting M i (0 b i ) compute the wrong answer. If the queries are not of the form b,x then we do the same, so for the rest of the proof we assume that M i on input 0 b (i) computes a query that has not yet been set of the form b,x, given advice 0 or 1. There are several possibilities. 1. M i computes b 1,x with advice 0 and b 2,y with advice 1 where x y. Then we set 0 b(i) D, b 1,x / A, and b 2,y / A, furthermore, set 1 b 1,x in A if x K, and set 1 b 2,y in A if y K. 2. M i computes b,x both with advice 0 and 1. Then we set 0 b i D and set 1 b,x to reflect x K correctly. 3. M i computes b,x with advice 0 and 1 b,x with advice 1. Now we set 0 b(i) D and compute M i (1 b(i) ) with advice 0 and 1. The case of small query strings is treated in the same way, and the cases above are treated in the same way. If we cannot already diagonalize on 1 b(i) alone then it must be the case that M i (1 b(i) ) computes b y,y and 1 b y,y. There are three cases. (a) x = y and b = b y. In this case we set 0 b(i) D and 1 b(i) / D. Set b,x and 1 b,x consistent with x K. (b) b = 1 b y. In this case we set 0 b(i) D, 1 b i D and M 0 (0 b(i) ) = M 1 (1 b(i) ) / A. (c) x y. In this case we set 0 b(i) D and 1 b(i) D and let M 0 (0 b(i) ) / A and M 1 (1 b(i) ) / A. 14

15 In the reverse case we have to deal with an enumeration of polynomial time truth-tables. We use the same complete set, but use the advice to either produce 0,x or 1,x on input x. There are again some cases to consider, but in this case we only need a subset of {0} to diagonalize. Yet reductions with advice and the induced completeness notions show familiar properties. Sets complete for EXP are complete under length-increasing, one-one reductions and are dense. Theorem 21 If A is complete for EXP under p/1 m -reductions, then A is complete for EXP under length-increasing p/1 m -reductions as well. Proof. Let {M i } i be an enumeration of polynomial time many-one reductions that use one bit of advice. Let D be defined as follows. For b {0,1} if M b ( b,i,x ) x then let b,i,x D iff x / A. Else let b,i,x D iff x K. Since A is complete for EXP, one of the reductions, say j must compute the reduction from D to A. For this reduction, it is never the case that the M j ( b,i,x ) x when b is the correct advice for b,i,x. Hence the reduction f(x) = M j ( b,i,x ) where b is the correct advice for b,i,x is a p/1 m length-increasing reduction from K to A. Theorem 22 Let A be complete for EXP under p/1 m reductions, then A is dense. Proof. We construct a set W in EXP as follows. For b {0,1} and x {0,1}. If there is an x < x such that Mi b( b,i,x ) = Mb i ( b,i,x ), then b,i,x W b,i,x / W. Otherwise b,i,x W. The set W is exponential time computable and dense (it has at least one of 0,i,x, 1,i,x for every i and x). The reduction from W to A cannot have collisions for the correct advice, hence A must also be dense. Theorem 23 Let A be complete in EXP under p/1 m -reductions, then A is also complete under p/1 m,1 1,li -reductions. Proof. Let K be the canonical EXP-complete set. Let K 0 = { 0 n,x n N,x K}. Clearly, K 0 is p m,1 1,l.i. -complete in EXP. We show a p/1 m,1 1,li -reduction from K 0 to A. Let {M i } i be an enumeration of polynomial time many-one reductions that take one bit advice. Let Mi b (x) denote the output of machine i on input x using advice b. First we use the constructions of Theorem 21 and Theorem 22 to find a reduction M k from K 0 to A that is both length-increasing, and 1-1 on strings of the same length. That is, if b is the correct advice for x then ( x)[ Mk b(x) > x ] and ( x,y)[ x = y Mk b(x) Mb k (y)]. Without loss of generality, M k runs in time n k for almost all inputs, and hence for almost all x, y and b if y > x k then Mk b(x) Mb k (y). Now let y 0(x) = min{y ( l)[y = 0 2(k+1)l x x]} and define f b (x) = Mk b (y 0(x)). First note that f is polynomial time computable. For every x the length of y 0 (x) is at most 2 k x, so that the computation of M k on y 0 (x) takes at most 2 k2 x k time, which is polynomial in x. Next, if x x then either y 0 (x) = y 0 (x ) in which case f b (x) f b (x ), for b the correct advice for length y 0 (x) by properties of M k, or y 0 (x) y 0 (x ) but then, without loss of generality, assume 15

16 y 0 (x ) to be the longer string y 0 (x ) y 0 (x) k, then f b (x) f b (x ) for b the correct advice for length y 0 (x) and b the correct advice for length y 0 (x ) by the fact that M k is length increasing and n k -time computable. In both cases f is 1 1. The advice given to f on input x is the advice belonging to M k on input y 0 (x), which depends only on x. However, unlike the uniform case, EXP has a p/1 m -complete set that is P-bi-immune. Theorem 24 There exists a p/1 m -complete set in EXP that is P-bi-immune. Proof. (sketch). For immunity we construct a set A complete under p/1 m -reductions. Let {M i } i be an enumeration of polynomial time computable sets. If at some length n we find that one of the machines M i with i < n accepts a string {b,x} for b {0,1} then we leave all strings {b,x} out of A for this length and put {1 b,x} in A iff x K. This construction is easily adapted to produce a bi-immune set by installing odd and even requirements and forcing a nonempty intersection with A as well as with A by priority. At the cost of more queries, in particular at the cost of adaptive queries, there does exist a relation between completeness under reductions that use advice and uniform reductions as the following theorem shows. Theorem 25 Every set A that is p/1 m complete in EXP is also p T-complete in EXP. Proof.(sketch) For this proof we use the representation of exponential time machines with tableaux. Let A be some m p/1 complete set. Suppose that we want to compute x B for B some complete set. Let B = L(M A/1 ) Consider an exponential size rectangle that is an encoding of of the computation x B. We call this rectangle the tableau of the computation x B. Say, a bit in this computation has position i,j, where 1 i,j 2 p(n), and assume that the final bit in this computation is 0 or 1 representing reject and accept respectively. Every bit in the tableau can be computed using M A/1. By padding i,j, where 1 i,j 2 p(n), we can ensure that input x,i,j is of the same length for every x,i,j. Then M A/1, given the right advice bit, computes every bit of the tableau of x B correctly. We will call the tableau representing the result of the computation using advice 0 the 0-tableau and the tableau representing the result of the computation using advice 1 the 1-tableau. It follows that if the final bits in the 0-tableau and the 1 tableau are the same, then we know the answer to x B. If the final bits are not the same, then there must be an inconsistency for exactly one of the two tableaux (the one computed with the incorrect advice). Given the correct advice bit we can in exponential time recover the first inconsistency in the tableau computed using the reduction to A with the incorrect advice. This is done as follows. We suppose that 0 is the correct advice, and using this advice, we retrieve the first sequence of bits that is inconsistent when the 1-tableau for x B is computed. Then, using advice 1, we recompute these bits. If 0 is indeed the correct advice, then, since the computation using 1 as advice is exponential time, we must indeed find an inconsistent sequence of bits. On the other hand, if 1 is the correct advice, such a sequence of bits does not exist and using 1 as the advice, we find a consistent sequence of bits on the indicated position (as we would find on any position). With a slightly more complicated proof the above theorem can be extended to reductions using up to clog n advice. The different advice strings then give rise to a polynomial number of (possibly 16

17 disagreeing) tableax. With help of the one correct advice, an exponential time computation, and thus a polynomial time reduction, can point out the first errors in the other computations, thus providing proof for the correct result. The tableau technique is a nonrelativizing technique. It is no surprise that this theorem does not relativize. We make this explicit in the following theorem. Theorem 26 There exists an oracle A and an set B that is p/1,a m -complete for EXP A but not p,a T -complete for EXPA. Proof.(sketch) For this proof we use the construction of an oracle appearing in [BvMFT00]. Here, a set A is constructed such that EXP A has a Turing complete sets that is not autoreducible. In particular there is a subset of S of {0 n } such that A S is no longer Turing complete. The Turing reduction that makes A complete is a very simple one. If 0 x A then x K 0,x A else x K 1,x A. Of course this is a p/1 m reduction and A S remains p/1 m complete under this reduction The fact that many-one completeness with one bit of advice can be simulated using more queries, extends to the delta levels of the polynomial hierarchy. Here the reduction does not even have to be adaptive as the following theorem shows. Theorem 27 For every k and every set A in p k, if A is p/1 m -complete, then A is p tt -complete. Proof. We give the proof here for P NP. The proof for higher levels is analogous, but more complicated and will appear in the final paper. Let L 1 = { φ,i the i-th bit in the lexicographically least assignment that makes φ true is 1}. Then clearly, since ODDMAXBIT is complete for P NP, so is L 1. Let A be a p/1 m complete set for P NP. We show how to compute φ ODDMAXBIT using a truth table reduction to A. Suppose then that M computes the reduction from L 1 to A given advice b, and let M b ( φ,i ) be the output of the reduction, i.e. for the correct advice b it holds that M b ( φ,i ) A if and only if φ,i L 1. Without loss of generality, we assume that φ has n variables and use a padded version of L 1, such that φ,0,..., φ,n are all of the same length, i.e. use the same advice. Now compute ω 0 = M 0 φ,0,..., M 0 φ,n = M 1 φ,0,..., M 1 φ,n ω 1 and check that either ω 0 or ω 1 makes φ true. At least one of these must be true as 0 or 1 must be the correct advice and return the bits of the lexicographically least assignment. This implies that the other advice either returns an assignment that does not make φ true, or an assignment that is not lexicographically smaller. In any case, we now have the lexicographically least assignment in hand. This has some interesting consequences. We don t know whether theorem 25 can be strengthened to truth-table reductions. However there are two cases. 1. If p/1 m -completeness implies p tt-completeness on EXP then EXP EXPSPACE. 2. If p/1 m -completeness does not imply p tt-completeness on EXP, then PH EXP. 17

18 5 Further Research We have initiated here the systematic study of non-uniform reductions. Many of the results here concern reductions which use only 1 bit of advice. In Theorem 12, we do not know that even this one bit of advice in necessary, Can it be eliminated? In places where more advice bits are needed can they be reduced? In general, how much stronger are non-uniform reductions than their uniform counterparts. Another, more technical question about reductions left open here, is whether the extended many-one reductions used in Theorem 2 are really needed? This may be more difficult than it first seems as Theorem 11 presents an oracle where length increasing many-one reductions do not exist for NP-complete sets. Can we add an immunity assumption to this oracle showing that nonrelativizing techniques would be required to prove Theorem 2 for standard many-one reductions. Another line of research is to strengthen some of the theorems concerning NP by weakening the strong hypotheses used in their proof. In particular, are the immunity and measure assumptions necessary for Theorems 2 and 12? Part of this paper was motivated by the recent work of Hitchcock and Pavan [HP04]. That paper features and compares a number of these strong hypotheses, and it would be interesting to determine the minimal hypotheses needed in our results here. Acknowledgments We thank John Hitchcock and Eric Allender for useful discussions and feedback. We thank John Hitchcock for his observation that the hypothesis in our length increasing result for NP can be weakened to bi-immunity instead of measure zero. References [AB93] [ABK + 02] [Agr02] [All88] [AS84] [ASTZ94] M. Agrawal and S. Biswas. Polynomial isomorphism of 1-L complete sets. In Proc. Structure in Complexity Theory 7th Annual Conference, pages 75 80, San Diego, California, IEEE Computer Society Press. Eric Allender, Harry Buhrman, Michal Koucký, Dieter van Melkebeek, and Detlef Ronneburger. Power from random strings. In FOCS, pages IEEE Computer Society, Manindra Agrawal. Pseudo-random generators and structure of complete degrees. In IEEE Conference on Computational Complexity, pages , E. Allender. Isomorphisms and 1-L reductions. J. Computer and System Sciences, 36(6): , K. Ambos-Spies. p-mitotic sets. In E. Börger, G. Hasenjäger, and D. Roding, editors, Logic and Machines, Lecture Notes in Computer Science 177, pages Springer- Verlag, K. Ambos-Spies, S.A. Terwijn, and X. Zheng. Resource bounded randomness and weakly complete problems. Theoretical Computer Science, 172((1-2)): ,