A zero one law for RP Russell Impagliazzo
Philippe Mosery
every x in a small range whether N accepts x when fed with random string t. We know this happens with nonnegligeable probability and that every such t has high circuit complexity. Thus we obtain a zero error probabilistic algorithm to find a high circuit complexity Boolean function. Using Impagliazzo and Widgerson’s [4] pseudorandom generator we thus obtain ZPP = BPP. In [7] Lutz asked what derandomization results could be derived from the plausible hypothesis ”NP has p-measure non zero”. In [1] V. Arvind and J. Ko¨ bler proved that under this hypothesis, partial derandomization of AM was possible. More precisely they proved that NP not having pmeasure zero implies NP= log n = AM, where NP= log n denotes non-uniform NP with logarithmic size advice. Using different techniques, we obtain total derandomization of AM under the same assumption, i.e. if NP has p-measure non zero then NP = AM.
Abstract In this paper we show that if RP has p-measure non zero then ZPP = EXP. As corollaries we obtain a zero-one law for RP, and that both probabilistic classes ZPP and RP have the same p-measure. Finally we prove that if NP has p-measure non zero then NP = AM.
1. Introduction It is an open question whether a zero-one law holds with respect to the resource bounded measure developed by Lutz [6], i.e. we do not know whether every complexity classes contained in EXP has either measure zero or one. It was shown in [9] that a zero-one law holds for BPP with respect to Lutz’s p-measure [6], i.e. BPP either has p-measure zero or else has p-measure one. It was also shown in [9] that every subclass of BPP closed under complement such as ZPP satisfies a zero-one law. It was thus left open whether RP also satisfies a zero-one law, since RP is not known to be closed under complement. We show that this is the case. In fact we prove something stronger, namely that if RP has p-measure non zero, then ZPP = EXP. As Corollaries we obtain a zero-one law for RP, and that both probabilistic classes ZPP and RP have the same p-measure. To this end we use Impagliazzo and Widgerson’s [4] pseudorandom generator, which transforms a Boolean string with high circuit complexity into a pseudorandom generator. The idea of the proof is the following: first we show that if RP has p-measure non zero, then there is a language L in RP and a small range, such that for almost every word lengths there is an element of L which is in the range, and for almost every element x of L, every random string t making the probabilistic Turing machine N accept x (where N is a probabilistic Turing machine deciding L), when viewed as a Boolean function, has large circuit complexity. Thus we can use L to find a Boolean function with high circuit complexity, by simply choosing a random string t, and checking for
2. Preliminaries We use standard notation for traditional complexity classes; see for instance [3] and [2], or [11]. Let us fix some notations in strings and languages. Let s0 ; s1 ; : : : be the standard enumeration of the strings in f0; 1g in lexicographical order, where s0 = denotes the empty string. A sequence is an element of f0; 1g1. If w is a string or a sequence and 0 i < jwj then w[i] and w[si ] denotes the ith bit of w. Similarly w[i : : : j ] and w[si : : : sj ] denote the ith through j th bits. We identify language L with its characteristic function L , where L is the sequence such that L[i] = 1 iff si 2 L. If w1 is a string and w2 is a string or a sequence extending w1 , we write w1 v w2 . We will need the following definition of the relativized hardness of a pseudorandom generator. Definition 1 Let A be any language. The hardness H A (Gm;n ) of a random generator Gm;n : f0; 1gm ?! f0; 1gn, is defined as the minimal s such that there exists an n-input circuit C with oracle gates to A, of size at most s, for which:
CSE Department, UC, San Diego, La Jolla, CA. email:
[email protected]. Research supported by NSF Award CCR-0098197 y Address: Computer Science Department, University of Geneva, email: moser cui.unige.ch
1. j x2fPr [ C ( G ( x )) = 1] ? Pr [ C ( y ) = 1] j m m n s 0;1g y2f0;1g
@
1
The following result states that a pseudorandom generator can be constructed from any Boolean function with high circuit complexity. Theorem 1 (Impagliazzo-Widgerson [4]) There is a polynomial-time computable function F : f0; 1g f0; 1g ! f0; 1g, with the following properties. For every > 0, there exists a; b 2 N such that F : f0; 1gna f0; 1gb log n ! f0; 1gn, and if r is the truth table of a (a log n)-variables Boolean function of Boolean circuit complexity at least na , then the function Gr (s) = F (r; s) is a generator, mapping f0; 1gb log n into f0; 1gn, which has hardness H (Gr ) > n. Klivans and Melkebeek [5] noticed that Theorem 1 relativizes; i.e. for any language A, there is a deterministic polynomial time procedure that converts the truth table of a Boolean function that is hard to compute for circuits having oracle gates for A, into a pseudorandom generator that is pseudorandom for circuits with A oracle gates. More precisely, Theorem 2 (Klivans-Melkebeek) Let A be any language. There is a polynomial-time computable function F : f0; 1g f0; 1g ! f0; 1g, with the following properties. For every > 0, there exists a; b 2 N such that F : f0; 1gna f0; 1gb log n ! f0; 1gn, and if r is the truth table of a (a log n)-variables Boolean function of A-oracle circuit complexity at least na , then the function Gr (s) = F (r; s) is a generator, mapping f0; 1gb log n into f0; 1gn, which has hardness H A(Gr ) > n.
2.1. Lutz’s p-measure In this section we describe the fragment of Lutz’s measure theory for the class E that we will need. For a more detailed presentation of this theory we refer the reader to the survey by Lutz [8]. The measure on E is obtained by imposing appropriate resource-bound on a game theoretical characterization of the classical Lebesgue measure. A martingale is a function d : f0; 1g ! [0; 1[ such that,
d(w) = d(w0) +2 d(w1) for every w 2 f0; 1g. d is a p-martingale if d is computable in time polynomial in jwj. This definition can be motivated by the following betting game in which a gambler puts bets on the successive membership bits of a hidden language A. The game proceeds in infinitely many rounds where at the end of round
n, it is revealed to the gambler whether sn 2 A or not. The game starts with capital 1. Then, in round n, depending on the first n ? 1 outcomes w = A [0 : : : n ? 1], the gambler bets a certain fraction w d(w) of his current capital d(w), that the nth word sn 2 A, and bets the remaining capital (1 ? w )d(w) on the complementary event sn 62 A. The
game is fair, i.e. the amount put on the correct event is doubled, the one put on the wrong guess is lost. The value of d(w), where w = A [0 : : : n] equals the capital of the gambler after round n on language A. The player wins on a language A if he manages to make his capital arbitrarily large during the game. We say that a martingale d succeeds on a language A, if d(A) := lim supw ne , where e = ac. Check whether there exists x in the range s2m to s2m +m?1 such that M (x; t) = 1, where t is chosen randomly. We know that there are many such witnesses t for at least one such x, and that each has circuit complexity at least m1=2 > ne . Thus, we can use Theorem 1 to construct from t a pseudorandom generator from O(log n) bits to nc bits secure against circuits of size nc . Use the outputs of this pseudorandom generator to simulate the Turing machine N for L0 . This has expected poly-time, and never errs for sufficiently large inputs, so L0 2 ZPP. ut Corollary 1 If EXP.
RP has p-measure non-zero, then ZPP =
Proof. If RP has p-measure non-zero then BPP = EXP by ut Theorem 4. By Theorem 5 we have ZPP = EXP. As a consequence we obtain that both probabilistic classes ZPP and RP have the same p-measure. Corollary 2
ZPP
and
p (ZPP) = p (RP).
RP
have the same p-measure i.e.
As a Corollary we obtain the following zero-one law for
RP.
Corollary 3 one.
RP has either p-measure zero
or p-measure
4. NP not being small implies total derandomization of AM The next result derives total derandomization for from the hypothesis NP has p-measure non-zero.
AM
Theorem 6 If AM.
NP
has p-measure non-zero, then
NP =
Proof. The proof follows from the following Lemma. Lemma 2 If NP has p-measure non-zero then there is a language L in NP and a nondeterministic polynomial time Turing machine M deciding L such that: For all but finitely many n 2 N , there is an element in L \ f0; 1gn, and for all but finitely many x 2 L, if t is a witness such that M (x; t) = 1, then the Boolean function on log jtj variables whose truth table is t has circuit complexity greater than jxj1=2 , for circuits with oracle gates for SAT. Proof (of Lemma). The first property is easily verified by constructing the appropriate martingale d1 as in Lemma 1. For the second property consider the following martingale d. Fix an enumeration of nondeterministic Turing machines so that Mi always halts within nj steps where j log i. On input w where w is the characteristic sequence of some language L up to string x, where x has size l (i.e. w = L[s0 : : : x]), d simulates the first log l nondeterministic machines on all nondeterministic witnesses for inputs of length at most log l. Let Mi be the first such machine that agrees with L NP-wise on all such inputs, i.e. decides L correctly (NP-wise) on inputs smaller than log l. d then constructs all circuits with oracle gates for SAT of size smaller than jxj1=2 on log i log jxj inputs, and computes the truth table t for each. If Mi (x; t) = 1 for any such t, d bets half its current capital that x 2 L. Otherwise, d does not bet. Claim 2 Let L 2 NP. Let Mi0 be the first nondeterministic poly-time machine deciding L. If there are infinitely many x 2 L and infinitely many t such that Mi0 (x; t) = 1 and t viewed as a function of log jtj inputs has circuit complexity less than jxj1=2 , then L 2 S 1 [d]. Proof (of Claim). After a finite amount of time d will always pick i = i0 . Thereafter we have that for each x such that there exists a nondeterministic witness with low circuit complexity t (such that Mi0 (x; t) = 1), d will always bet half its current capital that x 2 L. Since Mi0 decides NPwise, this bet will be correct, and d’s total stake will be multiplied by 3=2. Otherwise, d does not bet, so will never lose (once i = i0 ). So if there are infinitely many such x, d’s capital grows unbounded. Moreover d can be computed in polynomial time, since on input w as above, there are log l machines to simulate on all inputs of size log l, where each machine runs in time smaller than nlog log l , with n log l, log log l l to be simulated by trying all noni.e. takes time 2log deterministic witnesses. Next d needs to construct 2l circuits of size at most l1=2 and construct their truth table.
Since evaluating a SAT gate of size at most l1=2 takes time 2O(l) , d(w) can be computed in time polynomial in jwj. This proves Lemma 2, since if NP does not have pmeasure zero, there must be a language L 2 NP such that d1 + d’s capital does not grow unbounded on L. The proof of Theorem 6 then follows, let L and M be as in Lemma 2. Let L0 2 AM be any language and N a probabilistic nondeterministic Turing machine deciding it, and assume that on input of size n, N runs in time nc . For = 1 let a and b be as in Theorem 2 and pick m so that m1=2 > ne , where e = ac. Nondeterministically guess an input x of size m and nondeterministically guess a witness t for machine M on input x. Check whether M (x; t) = 1 ? we know that there is at least one such witness for at least one such x, and every such witness has circuit complexity at least m1=2 = nac ? if so use Theorem 2 to construct from t a pseudorandom generator from O(log n) bits to nc bits secure against circuits of size nc with oracle gates to SAT. Use the outputs of this pseudorandom generator to simulate the nondeterministic Turing machine N for L0 . This has expected nondeterministic polynomial time, and never errs for sufficiently large inputs, so L0 2 NP. ut Corollary 4 If NP has p-measure non-zero, then BPP has p-measure zero, unless NP = EXP. Proof. Suppose NP has p-measure non-zero, then NP = AM by Theorem 6, which implies BPP NP. Thus if NP 6= EXP, Theorem 4 implies that BPP has p-measure
ut
zero.
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P = BPP E XOR
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BPP
