NP with Small Advice

Lance Fortnow Adam R. Klivans∗ Department of Computer Science Department of Computer Science University of Chicago The University of Texas at Austin Chicago, IL 60637 Austin, TX 78712 [email protected] [email protected]

Abstract • It is known that if EXP cannot be computed by nondeterministic polynomial-size circuits then it We prove a new equivalence between the non-uniform is possible to obtain similar derandomizations of and uniform complexity of exponential time. We show AM [18, 21, 25]. Shaltiel and Umans [24] were NP that EXP ⊆ NP/log if and only if EXP = P|| . Our equiv- the first to prove that if EXP 6⊂ NP/poly then alence makes use of a recent result due to Shaltiel and AM ⊆ NSUBEXP for infinitely many input lengths. Umans showing EXP in PNP implies EXP in NP/poly. || Collapses

Perhaps less well known than the above derandom- 1. Introduction izations are equally important results showing that uni- form complexity classes such as EXP or NEXP collapse Let A and B be uniform complexity classes such that if they are contained in smaller, non-uniform classes: B ⊆ A. If A seems much “larger” than B then it is of- • Babal et al. [2] showed that EXP ⊆ P/poly implies ten the case that we can prove that B is strictly con- that EXP = MA, improving on work due to Meyer tained in A, e.g. let B = P and A = EXP. Is the same [17] who first proved that EXP ⊆ P/poly implies true if we consider a non-uniform analogue of B? That EXP = ΣP. is to say, augment B by giving it access to some ad- 2 vice string b such that b depends only on the length of • Impagliazzo et al. [13] improved the above collapse x; can we still separate A from B/b? If not, can we de- and showed that NEXP ⊆ P/poly if and only if rive interesting consequences on A if it is contained in NEXP = EXP = MA. This result is crucial to Ka- B/b, i.e. can we show that A collapses to some smaller banets and Impagliazzo’s breakthrough paper [15] complexity class? showing that derandomizing BPP implies proving These questions are of central importance in com- circuit lower bounds. putational complexity theory, particularly in the area If we pay particular attention to MA, then the above of derandomization, where both separations of uni- separations and collapses match up nicely– if EXP ⊆ form from non-uniform classes or collapses of uni- P/poly then EXP collapses to MA, and if EXP 6⊂ P/poly form classes have important consequences: then MA can be derandomized (and will be contained in NSUBEXP). Separations The same is not true, however, for AM. Separat- • If EXP 6⊂ P/poly, then Babai et al. [2], building on ing EXP from NP/poly implies that AM is contained in the “Hardness versus Randomness” paradigm [22], non-deterministic, sub-exponential time [24]. Placing P have shown that BPP is contained in subexpo- EXP ⊆ NP/poly, however, implies only that EXP = Σ3 , 1 nential time and that MA is contained in non- the third level of the polynomial-time hierarchy . deterministic subexponential time (both contain- Is it true that EXP ⊆ NP/poly implies that EXP = ments are for infinitely many input lengths). AM? If so, combining this fact with the above deran-

∗ Work done while visiting the Toyota Technological Institute at 1 Actually one can prove that under the assumption that EXP ⊆ P Chicago. NP/poly, EXP ⊆ ZPPΣ2 [7] NP domization of AM [24] would yield a rare uncondi- (it is also known that NEXP 6⊂ P|| [11]). We can con- tional derandomization of AM, namely that AM is con- sider, however, the consequences of NEXP being con- tained in Σ2 − SUBEXP, the subexponential time ana- tained in randomized complexity classes that take ad- P P logue of Σ2 (AM is currently only known to be in Π2 )– vice (such classes have been a focus of research inter- see Gutfreund et al. [12] for a discussion. Shaltiel and est as of late [4, 10]). We observe that the techniques Umans [25] have asked if EXP ⊆ NP/ log implies that of Impagliazzo et al. [13] can be used to prove that EXP = AM, as even this is not known. NEXP ⊆ BPP/ log implies NEXP = BPP, strengthen- ing a result of Trevisan and Vadhan [27]. 1.1. Our Results 1.2. Related Work We give a new collapse for exponential time if it is computed by a nondeterministic, slightly non-uniform The first important collapse of a uniform class con- complexity class. More precisely we show that if EXP ⊆ tained in a non-uniform class is due to Karp and Lip- NP/ log then EXP = PNP, i.e. EXP is computed by a ton [17] who showed that NP ⊆ P/poly implies that || P polynomial-time turing machine with non-adaptive ac- PH = Σ2 and that NP ⊂ P/ log implies P = NP. cess to an NP-oracle. Further, we can also prove the For exponential time, aside from the collapse results converse: mentioned above due to Babai et al. [2] and Impagli- azzo et al. [13], Buhrman and Homer [8] showed that if Theorem 1 The following are equivalent. EXPNP ⊆ EXP/poly then EXPNP = EXP and Buhrman, Fortnow, and Pavan [7] showed a weak relativization of 1. EXP ⊆ PNP || Impagliazzo et al. [13], namely that for any A ∈ EXP, A A A 2. EXP ⊆ NP/ log NEXP ⊆ PA/poly implies NEXP = EXP and if A P A A is complete for Σk then NEXP ⊆ P /poly implies The forward direction of our equivalence makes use NEXPA = EXP = MAA. of a new hardness amplification result due to Shaltiel Buhrman, Chang and Fortnow [6] give an equiva- and Umans. They prove that if EXP 6⊂ NP/poly then lence of a non-uniform collapse to NP and a uniform NP EXP 6⊂ P|| /poly. The contrapositive gives a partial inclusion. collapse of exponential time which we show how to strengthen via a non-standard method of computing Theorem 4 (Buhrman-Chang-Fortnow) The fol- NP lowing are equivalent. advice. As a result we obtain EXP ⊆ P|| implies EXP ⊆ NP/ log, improving on the conclusion EXP ⊆ 1. coNP ⊆ NP/1 AM/ log obtained by Shaltiel and Umans [25]. 2. The polynomial-time hierarchy collapses to Dp The backwards direction requires two collapses. First we prove that if EXP ⊆ NP/ log then EXP = PNP, where Dp is the set of languages that are the difference of and then we use the fact that the ODDMAXBIT func- two NP languages. tion is complete for PNP to show how the above advice Buhrman, Chang and Fortnow also generalize Theo- strings can be computed and verified non-adaptively. rem 4 to show that coNP in NP/k if and only if the We also prove variations of Theorem 1 for other polynomial-time hierarchy collapses to the 2kth level classes. of the Boolean hierarchy where the first level of the Theorem 2 The following are equivalent. Boolean hierarchy is NP and the i + 1st level is the set of differences of sets in NP and the sets in the ith level. NP 1. PSPACE ⊆ P|| This extension only works for finite k but Buhrman, 2. PSPACE ⊆ NP/ log Fortnow and Chang conjecture that it extends to k = O(log n). Theorem 3 The following are equivalent. Conjecture 5 (Buhrman-Chang-Fortnow) The #P NP following are equivalent. 1. P ⊆ P|| 1. coNP ⊆ NP/ log 2. P#P ⊆ NP/ log 2. The polynomial-time hierarchy collapses to PNP Is it possible to prove something similar to Theo- || P rem 1 for NEXP? We show that, in fact, such a state- Since EXP in NP/poly implies EXP ⊆ Σ3 [1, 29], ment is vacuously true for NEXP since one can sepa- Theorem 4 implies EXP ⊆ NP/1 if and only if EXP = rate NEXP from NP/ log outright via diagonalization Dp. Likewise Conjecture 5 implies Theorem 1 so we can view our Theorem 1 as a partial resolution of Conjec- game where on input x, Arthur sends a random chal- ture 5. lenge r to Merlin who responds with y; Arthur then probabilistically verifies y to determine acceptance of 2. Preliminaries x (see the survey by Kabanets [14]).

2.1. Complexity Classes 2.3. Alternation and Games

We assume the reader is familiar with complex- We will make use of the characterization of PSPACE k k ity classes P = ∪kDTIME(n ), NP = ∪kNTIME(n ), due to Chandra, Kozen, and Stockmeyer as a game nk nk [9]. Chandra et al. showed that PSPACE is equivalent EXP = ∪kDTIME(2 ), NEXP = ∪kNTIME(2 ), k to the following two person game: on input x, play- PSPACE = ∪kDSPACE(n ) as well as notions of ora- cle turing machines and the polynomial-time hierarchy ers alternate announcing bits for a polynomial num- (see e.g. [3] for further explanations). ber of rounds and a polynomial-time computable judge The non-uniform class NP/ log is the set of languages chooses a winner based on x and the announced bits: L such that there exists a language A in NP and a func- Theorem 7 (Chandra-Kozen-Stockmeyer) A ∗ tion a : N → Σ with |a(n)| = O(log n) such that for language L is in PSPACE if there exists a polynomial- ∗ all x in Σ , x is in L if and only if (x, a(|x|)) is in A. time relation R on 2k + 1 strings where k = nO(1) and NP/poly has the same definition except that we allow players P1 and P2 such that |a(n)| = O(nk) for some k. Similarly NP can be re- placed with any machine based complexity class, e.g. • On round i for i odd, P1 takes as input x and all BPP/ log is the set of languages accepted by a BPP strings from previous rounds and ouputs string xi. machine augmented with an advice string of length • On round j for j even, P2 takes as input x and all O(log n) which depends only on the input length. strings from previous rounds and outputs string yj. The class PNP consists of the languages accepted in • After k rounds, the input x is in the language L if polynomial-time with oracle access to some NP lan- and only if R(x, x , y , x , y , . . . , x , y ) is true. guage. Since SAT, the set of satisfiable Boolean for- 1 1 2 2 k k mula, is NP-complete, we can use SAT as the oracle Furthermore, each player Pi requires only PSPACE language. We will make use of the following theorem to output his/her string for each round. Hence we say giving a natural complete language for PNP: each player has a strategy computable in PSPACE.

Theorem 6 (Krentel [19]) Let φ(x1, . . . , xn) be a Boolean formula. Let a be the lexicographically small- 3. The Proof est satisfying assignment for φ, if there is one. The prob- In this section we give the proof of Theorem 1 show- lem of determining whether the nth bit of a is equal to ing the following are equivalent: one is many-one complete for PNP. 1. EXP ⊂ NP/log The above language is often referred to as ODD- NP MAXBIT. 2. EXP ⊂ P|| We use the following nice result of Shaltiel and NP NP The class P|| (sometimes written Ptt ) is the set Umans [25]: of languages accepted in polynomial-time with non- Theorem 8 (Shaltiel-Umans) If EXP ⊆ PNP/poly adaptive oracle access to SAT; in other words all || queries must be made before any the oracle returns then EXP ⊆ NP/poly. any answers. The proof of the above theorem makes use of the fact that EXP has a low-degree extension f, and if this ex- NP 2.2. Randomized Classes tension is computable in P|| then for each query q made by the oracle-machine, one can give an ad- We also assume the reader is familiar with random- vice p equal to the fraction of x’s resulting in a q(x) ized complexity classes such as BPP and MA, the set which should be answered as true by the NP or- of languages accepted by a Merlin-Arthur game where acle. For any x, it then suffices to choose a ran- on input x, Merlin, the prover, sends a single message dom low-degree curve through x and guess wit- y and Arthur (the verifier) probabilistically verifies the nesses for a p fraction of points on this curve. purported proof y to determine membership of x. AM is the set of languages accepted by an Arthur-Merlin Proof of Theorem 1: (2 ⇒ 1) ai is chosen from IN and ao is chosen from OUT: simu- late players P1 and P2 where P1’s strategy is computed Fix an EXP-complete language L. By Theorem 8, L using advice ai and P2’s strategy is simulated using ad- is in NP/poly. Fix the appropriate NP-machine M and vice ao. Since each strategy is in PSPACE ⊆ NP/ log, NP let an be the lexicographically smallest advice string the entire simulation is computable in P . such that for all x of length n, x is in L iff M(x, an) Since some advice string a is the correct advice accepts. string, either a will be in IN and P1 using this ad- Fix n. Let bi be the ith bit of an. We can compute bi vice will defeat P2 using any advice from OUT or vice in time exponential in n so by assumption computing versa. If the good advice string is in IN (and hence NP bi is in P|| . Let Qi be the set of queries to SAT made causes P1 to always beat P2), then we know x is in L NP S and we will accept correctly. If we discover a to be in by the P|| algorithm to compute bi. Let Q = i Qi. Let r be the number of formulas in Q that are satisfi- OUT we reject. able. r is our O(log n) bits of advice. Our NP/ log algorithm works as follows on input x Proof of Lemma 10: of length n: guess a subset S of r formulas in Q and From Theorem 6, we know that the ODDMAXBIT guess and verify their satisfying assignments. For each language consisting of the set of formulas whose lexi- NP i, simulate the P|| algorithm to compute bi answering cographically minimum satisfying assignment sets the each query yes if it is in S and no otherwise. From the last variable to true is complete for PNP. Hence, it b ’s we now have a . Now output M(x, a ). NP i n n suffices to give a P|| algorithm for deciding ODD- MAXBIT. (1 ⇒ 2) Given a formula φ of n variables, let ai be the set- ting of the ith variable in the minimum satisfying as- This direction follows by combining the following signment (ai = 0 if there is no satisfying assignment). NP two lemmas: We can compute ai in P and thus in NP/ log. Hence, given the correct advice we can compute a with one Lemma 9 If EXP ⊆ NP/ log then EXP ⊆ PNP. i query to NP. NP NP NP Lemma 10 If P ⊆ NP/ log then P = P|| . For each possible advice string b, we compute a1, . . . , an via n parallel queries to NP (we can do this Proof of Lemma 9: since each bit is computable by assumption by one in- dependent query to NP). Given all of these purported It is known that if EXP is in NP/log then EXP = minimum assignments, we find the lexicographi- PSPACE. This follows, for example, from the fact that cally minimum assignment a0 that satisfies φ. Since if EXP ⊂ PA/poly then EXP ⊆ MAA, i.e. a relativized at least one advice is correct a0 is the minimum sat- version of a collapse due to Babai et al. observed by isfying assignment and the last bit of a0 gives us the Buhrman et al. [2, 7]. Choosing A = NP places EXP ⊆ answer to the ODDMAXBIT question. MANP ⊆ PSPACE. By Theorem 7, we can view PSPACE as a interac- tive game between two players and a polynomial-time 3.1. Extending the Proof to PSPACE and computable judge (recall each player’s strategy is com- P#P putable in PSPACE and thus NP/ log by assumption). Let L be a PSPACE-complete language and fix an input The proof of Theorem 8 in Shaltiel and Umans [25] NP x. We will give an P algorithm to determine whether extends to PSPACE and P#P. x is in L. O(1) NP Let T be the set of all n advice strings and let Theorem 11 (Shaltiel-Umans) If PSPACE is in P|| M be the NP advice taking machine deciding L. For #P NP #P then PSPACE is in NP/poly. If P is in P|| then P is each advice string a ∈ T , simulate M(x, a) and di- in NP/poly. vide T into two groups labeled IN and OUT depend- ing on whether M(x, a) accept or rejects. Since one ad- To prove Theorem 2 note that the proof of Theo- vice string gives the correct answer, if either IN or OUT rem 1 goes through directly using PSPACE instead of is empty then we know whether x is in L. This simu- EXP. NP #P NP lation can be carried out in P . To prove Theorem 3 that P is in P|| if and only if Otherwise, IN and OUT are both non-empty. Do the P#P is in NP/ log we need a little more work. To show following for each pair of advice strings ai and ao where the “if” direction we first need the following lemma. Lemma 12 If P#P is in NP/poly then for every L in contained in either class. Fu, Li and Zhong [11] showed #P NP P there exists an NP machine M and a sequence of ad- that NEXP 6⊆ P|| . This result and Theorem 1 does vice strings a1,... where not immediately imply that NEXP is not contained in 1. For all x, x is in L if and only if M(x, a ) accepts, NP/ log since we do not know how to directly show |x| NEXP in NP/ log implies NEXP = EXP. Instead we 2. For all n, |an| is bounded by a fixed polynomial in n, prove the separation directly. and n i Theorem 14 NEXP 6⊂ NP/ log. 3. The language D = {1 0 | the ith bit of an is one} is in P#P. Proof: Proof: Assume by way of contradiction that NEXP ⊆ NP/ log. Then by a padding argument, NEEXP ⊆ Valiant [28] showed that Permanent (computing NEXP/poly. I.e. non-deterministic doubly exponential the ith bit of the permanent of a given 0-1 matrix) #P time is contained in a non-uniform analogue of NEXP. is Turing-complete for P . Similar to EXP, if the But now we apply the assumption that NEXP ⊆ Permanent is in PA/poly then the Permanent is in A NP/ log again and obtain NEEXP ⊆ NP/poly. Via a MA [2, 7]. Setting A = SAT we have Permanent in standard diagonalization argument one can show that the polynomial-time hierarchy. #P even EEXP, deterministic doubly exponential time, Let L be in P and let M be an NP machine such does not have non-deterministic polynomial-size cir- that there exists a sequence of polynomially-long ad- cuits. This is because in doubly exponential time we vice strings b1,... where x in L if and only if M(x, b|x|) can enumerate over all say quasipolynomial-size non- accepts. Consider the language D consisting of the deterministic circuits. strings 1n0i where the ith bit of the lexicographically least advice that computes L correctly on all inputs on Roy [23] improves Theorem 14. First need we need length n is one. We can define D with a few quanti- the following result by Buhrman [5]. fiers over L and L is reducible to the permanent which Theorem 15 (Buhrman) is in the polynomial-time hierarchy. This puts D in #P NP the polynomial-time hierarchy and thus in P be- EXP|| ⊆ NEXP/poly. cause of Toda’s theorem [26] that every language in Proof: the polynomial-time hierarchy is in P#P. Fix an EXP machine M and let Qn be the union of #P NP #P the set of all queries to SAT made by M(x) on all x of We can now prove that P in P|| implies P in NP/ log using the same techniques as the proof of The- length n. Our advice an is |Q ∩ SAT|. orem 1 using Theorem 11 and Lemma 12. Our NEXP machine on input x and advice a|x| works Since PNP ⊆ P#P, the other direction of Theorem 3 as follows: Compute Q|x| and guess which of the a|x| follows from the appropriate analog of Lemma 9. formula are satisfiable and guess their satisfying as- signments. If the advice is correct and we have guessed #P #P NP Lemma 13 If P ⊆ NP/ log then P ⊆ P . the assignments then we know which queries made by Proof: M(x) are satisfiable and we just simulate M(x) with those answers. Fix a P#P complete language L. If L is in NP/ log p then L is in Σ3, the third level of the polynomial- p Theorem 16 (Roy) time hierarchy [20, 29]. We can view Σ3 as a three NP round game between two players and a polynomial- NEXP 6⊆ P|| / log . time judge. Each player’s strategy is computable in Proof: the polynomial-time hierarchy and thus in P#P by NP NP Toda [26]. We can now show L is in P by the same ar- Assume NEXP ⊆ P|| / log. By padding we have gument as the proof of Lemma 9. NP 2O(n) NEE ⊆ EXP|| /poly where NEE = NTIME(2 ). By Theorem 15 and the assumption we have NP NP 4. On the Non-Uniform Complexity of NEE ⊆ EXP|| /poly ⊆ NEXP/poly = P|| /poly NEXP and thus NEE has polynomial-size circuits with NP E gates. However Σ4 ⊆ NEE cannot have polynomial-size We would like to to extend the equivalence in Theo- circuits with NP gates using diagonalization techniques rem 1 to hold for NEXP. We can do so, but the equiv- due to Kannan [16]. alence holds vacuously in the sense that NEXP is not 4.1. NEXP versus randomized, non-uniform [2] L. Babai, L. Fortnow, N. Nisan, and A. Wigderson. BPP classes has subexponential simulations unless EXPTIME has publishable proofs. Computational Complexity, 3:307– In light of the fact that NEXP is known to not be in 318, 1993. NP/ log, it seems natural to consider the consequences [3] J. Balc´azar,J. D´ıaz,and J. Gabarr´o. Structural Com- of NEXP being contained in BPP/ log or MA/ log. Sep- plexity I. 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