On P-selective Sets and EXP Hard Sets. Bin Fu. Yale University, New Haven, CT and. UMIACS, University of Maryland at College Park, MD 20742

Transcription

1 On P-selective Sets and EXP Hard Sets Bin Fu Department of Computer Science, Yale University, New Haven, CT and UMIACS, University of Maryland at College Park, MD May

2 Abbreviated form of the title: P-selective Sets and EXP Hard Sets Mailing Address: Bin Fu 5119 Hollywood Road College Park, MD

3 Abstract Let PSel be the class of all P-selective sets. We show that DTIME(2 nk+a ) 6 P n k?t(psel) for all k; a > 0. It implies EXP 6 P n c?t(psel) for all c > 0. This greatly improves Toda's result that EXP 6 P tt (PSel) since P tt (PSel) is equal to P O(log n)?t (PSel). We construct an oracle A such that DTIME(2 nk ) A P A (PSel). We also show that the symmetric dierence of a n k+1+a -T E-P m-hard set and a P-selective set is a E- P 2?T-hard. This generalizes a result by Rao [18] who showed the symmetric dierence of a E- P m-hard set and a P-selective set is exponentially dense. 3

4 Symbols used in the paper: a; b; c; d; e; f; g; h; i; j; k; l; m; n; o; p; q; r; s; t; u; v; w; x; y; z A; B; C; D; E; F; G; H; I; J; K; L; M; N; O; P; Q; R; S; T; U; V; W; X; Y; Z [; \; ;?; ; ; ; ; ; 9; 8; ; ; 4; k; [; ](; ); f:g; (); =);?!; 6=; =; 2; ; ;; ; 6; 8; ^ 4

5 1. Introduction The study of reducibilities of complexity classes to the low sets is an important topics in complexity theory. It has long and rich history. One of the most notable low notions is P-selectivity. It was introduced by Selman [20, 22, 21] as a complexity-theoretic analog to the semi-recursive sets [12] from recursive theory. He used it to distinguish polynomial time m-reducibility from Turing-reducibility in NP under the assumption E6=NE, and proved that if every set in NP is positively truth-table reducible to P- selective sets then P=NP. Ko [14] extended the notion to weakly P-selectivity and showed that no NP-hard set is polynomial time disjunctively reducible to weakly P- selective sets unless P=NP. The class of languages that are polynomial time Turing reducible to P-selective sets are the same as P/poly [20]. Separating complexity class from P/poly is one of the most fundamental problems in complexity theory. It is still open if NEXP 6P/poly. The line of research about the consequence of NP sets reducing to a P-selective set is very active in the recent years. It has attracted a great deal of interest and have been many advances in understanding P-selectivity. Buhrman, Torenvliet and Emde Bosa [6] generalized Selman's result that if a NP-hard set under Turing positive reductions is reducible to a P-selective set then P=NP. Beigel [2] and Toda [26] showed that if a NP- P tt-hard set is P-selective then P=FewP and NP=RP. Thierauf, Toda and Watanabe [24] showed that if a NP- P btt-hard set is P-selective then NP DTIME(2 no(1=p log n)). Agrawal and Arvind [1], Beigel, Kummer and Stephan [3], and Ogiwara [17] independently proved that if a NP- P btt-hard set is P-selective, then P=NP. Cai, Naik and Selman showed that if a NP- P tt-hard set is P-selective then then NP DTIME(2 no(1=(log n)k )) for all k > 0. Toda [26] showed that EXP is not polynomial time truth-table reducible to P- selective sets. Recently Buhrman and Longre [5], and Wang [27] independently proved that the class of languages that are polynomial time truth-table reducible to P-selective sets has measure in E. P tt (PSel) is the same as P O(log n)-t(psel). We show that EXP 6 P n c?t(psel) for all c > 0. This improves the Toda's result that EXP 6 P tt (PSel). We construct an oracle A such that DTIME(2 nk ) A P A (PSel). Since P n k +1+a-T T(PSel) = P/poly, it is impossible to prove the no P-Selective set is E- P T-hard unless the most fundamental problem in complexity theory can be solved. This indicates our result is almost optimal by the relativizable methods. We also study the symmetric dierence between hard set with P-selective sets. Yesha [29] rst studied the symmetric dierence between NP- P tt-hard sets and the sets in P. Schoning showed that no the symmetric dierence between a E- P T-hard and a set in P is of exponential density. The symmetric dierence of NP-hard sets and P-selective was investigated by Fu and Li [9]. Rao showed that the symmetric dierence of a EXP- P tt-hard set and a P-selective set is of exponential density. We study the symmetric dierence between a EXP- P m-hard set and a P-selective set from the stability point of view in [23] We generalize Rao's result by proving that the symmetric dierence of a E- P m-hard set and a P-selective set is a E- P 2?T-hard. 1

6 2. Preliminaries We use = f0; 1g as our alphabet. By \string" we mean an element of, jxj denotes the length of x. We use lexicographic order on. For any strings x and y, x is smaller than y (write x < y) if either jxj < jyj, or jxj = jyj and there exists some k, 1 k jxj, such that (8i : 1 i < k[x i = y i ] and [x k = 0 and y k = 1]), where x i is the ith symbol of the string x. For two strings x; y, x y if x is an initial segement of y. For S, the cardinality of S is denoted by ksk. Set S =n (S n ) consists of all words of length = n( n) in S. In particular, let n = fx : x 2 and jxj = ng and n = fx : x 2 and jxj ng. For a language A, A(x) is the characteristic function of A. N = f0; 1; 2; g. For real number x, bxc is the largest integer x. Denition 1. [20] A set A is P-selective if there is a total polynomial time function g so that, for every x; y 2, g(x; y) 2 fx; yg, and if x 2 A or y 2 A, then g(x; y) 2 A. We call g as the selector function for A. The above g induces a linear order g that x g y if there exists z 1 ; ; z t such that g(x; z 1 ) = x; g(z i ; z i+1 ) = z i ; and g(z t ; y) = z t. The P-selective set A is an initial segment of the linear order g. Denition 2. E = S 1 k=1 DTIME(2kn+k ). EXP = S 1 k=1 DTIME(2nk +k ). Denition 3. A P T -reduction of A to B is a polynomial-time oracle Tur ing machine M such that for each x 2 P, x 2 A () M B accepts x. For a function g : N?! N, a P -reduction of A to B is a polynomial-time oracle Turing machine M such g(n)?t that A P T B is witnessed by M and M will not query the oracle more that g(n) times for each input with length n. Denition 4. Let H, C be a class of languages and P r be a type of reductions. P r (H) is the class of all languages which are P r -reducible to H. P r (C) is the class of languages which are P r -reducible to some languages in C. If C P r (H), then we say H is C- P-hard. r Denition 5. For a function d from N to N, DENSITY(d(n)) is the class of all languages that has density bounded by d(n). If there exists a polynomial p(n) such that k A n k< p(n) for all n, then we say A is sparse. \SPARSE" represents the class of all sparse languages. 2

7 Hartmanis [11] introduced a \generalized Kolmogorov complexity measure". We employ this tool in the proof of our theorems. We consider standard deterministic time- bounded Turing machines that act as transducers. We assume a standard enumeration of such transducers, say N 1 ; N 2 ;. For each i, let f i be the function computed by N i, and let T i be the running time of the transducer N i. We assume that the enumeration has the property that there exists a universal Turing transducer N u and a description function d with the following property: for every i > 0 there is a constant c i such that for all x 2, (a)d(i) is not a prex of d(j) if i 6= j, (b) f u (d(i)x) = f i (x), and (c)t u (d(i)x) c i T i (x) log T i (x) + c i. We then dene the following classes of strings. Denition 6. K i [g; t] = fy : 9x[jxj g(jyj); f i (x) = y and T i (x) t(jyj)]g: K[g; t] = K u [g; t], where u denotes the index of the universal Turing machine. Let f u be the transducer for the universal Turing machine. For two integers m; n, dene K[m; n] = fy : 9x[jxj m; f u (x) = y, and T i (x) n]g. Lemma 7 ([11]). Let i be any index and let g(n) and t(n) be time-constructible functions. Then there exists c > 0 such that K i [g(n); t(n)] K u [g(n)+c; ct(n) log t(n)+c]. 3. E- P T-hard sets The function cod :?! is dened by cod(x) = 1a 1 1a 2 1a r 0 2, where x = a 1 a 2 a r 2. The function bin : N?! is dened so that for each i 2 N, bin(i) is the binary expression of i (for example, bin(5) = 101 and bin(8) = 1000). For tuple hx 1 ; ; x n i, we use cod(bin(jx 1 j))x 1 cod(bin(jx n j))x n to encode it into a string. It is very easy to see that hx 1 ; ; x t i can be encoded and decoded in polynomial time. The length of cod(bin(jx 1 j))x 1 cod(bin(jx n j))x n is bounded by nx i=1 jx i j + 4n(1 + log( nx i=1 jx i j)): The following proposition is easy to verify (because uvwvx can be compressed into cod(bin(juj))cod(bin(jvj))cod(bin(jwj))uvwz). Proposition 8 ([8]). There exists a polynomial p(n) such that for all large n, if there exist strings 2 n and u; v; w; x to satisfy = uvwvx and jvj > 7 log n, then 2 K[n? 1; p(n)]. 3

8 Theorem 9. Let f(n) be a n O(1) time computable nondecreasing function from N to N. if f(n) 6= O(log n) and f(n) n c 0 for some constant c 0 > 0, then for all a > 0. DTIME(2 (3+a)f (n) ) \ DENSITY(f(n)) 6 P f (n)?t (PSel) Proof: Suppose DTIME(2 (3+a)f (n) ) \ DENSITY(f(n)) 6 P f (n)?t (PSel) for some a > 0. We will derive a contradiction. In this chapter we give informal description about the idea of proof. The idea is to use the resource bounded Kolmogorov complexity point of view. A selective set is a initial segment of a linear order g. For a P-selective set S, if A P T S via oracle Turing machine M that makes small number f(n) of queries. Let's consider all of strings queried by M with inputs of length n. Among them there is a string y that is at the boundary of S. In other words, for every string u queried by M, u g y =) u 2 S and y < g u =) u 62 S. Let y be i-th query by M S with input x 0. Let the initial i?1 answers be a 1 a i?1 we can get y in polynomial time from x 0 and a 1 a i?1. So, we can know all of the queries answers for the other inputs of length n. We can generate all of the strings in A =n by x 0 and a 1 a i?1. If f(n) is small, the information content of x 0 and a 1 a i?1 is small. We will construct a set A that has high information content for its strings of length n. Without lose generality, we suppose a < 1. Let d = 400c 0 =a r = [d log n] + 1 k = (1 + a=6)f(n) [ ]: r Let n be the least string in kr such that is not in K(kr? 1; 2 (1+a=4)f (n) ). n can be found in time 2 kr?1 2 (1+a=4)f (n) 2 kr 2 (3+a=2)f (n) : Let n = u 1 u k, where each ju i j = r. By Proposition 8, u i 6= u j for i 6= j. Dene u 0 i = u i 10 n?ju ij?1. Let A n = fu 0 1; ; u 0 kg and L = [ 1 n=1a n. Since n can be found in 2 (3+a=2)f (n) time, L 2 DTIME(2 (2+a)f (n) ). By the assumption, Let L 2 P n k?t(s) via polynomial time Turing machine M, where S is a P-selective set with selector function g. Suppose M runs in n c time and g is n c time computable. Lemma 10. Assume Q = fs 1 ; ; s t g n. With time n c+1 k Q k we can built a chain s i1 g s i2 g g s it ; where s i1 ; s i2 ; ; s it is a permutation of s 1 ; ; s t. 4

9 Proof: For t = 1, it is trivial. Suppose we have built s i1 g s i2 g g s im for fs 1 ; s 2 ; ; s m g, where m < t. If there is a j between 1 and m such that s m+1 = g(s m+1 ; s ij ). Let j 0 be the least j that s m+1 = g(s m+1 ; s ij ). Put s m+1 between s ij?1 and s ij. Otherwise, put s m+1 right after s im. Since g is computable in time n c, it is easy to verify the time bound for generating the chain. Denition 11. query(m; x; a 1 a t ) is the (t + 1)-th query by M with input x assuming the answers to the initial t questions are a 1 ; ; a t. Let Q 0 = fquery(m; x; a 1 a t ) : (t f(n)) ^ 9y(x = y10 n?r?1 ^ jyj = r)g: Since n c is the time bound for M, Q 0 nc. By the Lemma 10 we can get a chain for Q 0 Let x 0 = v10 n?r?1 2 n such that i-th query of M S (x 0 ) is a string y that has the following property: for every query u of M S (z) with z 2 n, if u is in the left side of y in the chain then u 2 S,and if that u is in the right side of y in the chain then u 62 S. Since S is an initial segment of g, x 0 and y do exist. Let a 1 a i be the rst i answers from the set S. Clearly y = query(m; x 0 ; a 1 a i?1 ). Let = hv; a 1 a i ; l 1 ; ; l k i: Lemma 12. can generate n in O(2 (1+a=4)f (n) ) time. Proof: We rst generate set Q 0. By the denition Q 0 can be generated in 2 2 r 2 f (n)+1 n c time. We generate the chain for the set Q 0. By Lemma 10 we can get the chain for Q 0 in time n c2 +c k Q 0 k. Since k Q 0 k 2 r 2 f (n)+1, the time for generating the chain is bounded by n c2 +c 2 r 2 f (n)+1 : Suppose a i = 1 (the case 0 is similar). For each z = w10 n?r?1 2 n, run M(z). When M(z) queries if string u is in S, we can determine the answer by checking the relative position with y in the chain (If it is at the left side of u, the answer is yes. Otherwise the answer is no. The time to check the position at the chain is k Q 0 k). After generating the chain for Q 0, we can generate all of the strings in A n 5

10 in n c+1 2 r k Q 0 j k n c+1 2 r 2 r 2 f (n)+1 time. Let v 1 10 n?r?1 ; ; v 1 10 n?r?1 be all of the elements generated in A n. Suppose v 1 < v 2 < < v k. Let l i be the location of v i in n (l i = j if v i = u j ). Clearly v 1 ; ; v k is a permutation of u 1 ; ; u k. With v 1 ; ; v k and l 1 ; ; l k we can get n in j n j 2 time. The total time is bounded by 2 2 r 2 f (n)+1 n c + n c2 +c 2 r 2 f (n)+1 + n c+1 2 r 2 r 2 f (n)+1 + j n j 2 (1+a=4)f (n) 2 for all large n. Lemma 13. jj (1 + a=25)f(n). Proof: jvj = r ja 1 a i j = i d log n + 1 2d log n: f(n): jl i j log j n j log(1 + a=6)f(n) log 2f(n) 2c 0 log n: k (1 + a=6)f(n) + 1 r 4f(n) d log n : jvj + ja 1 a i j + jl 1 j + + jl k j 2d log n + f(n) + 4f(n) d log n 2c 0 log n f(n) + a 50 f(n): The length of tuple hv; a 1 a i ; l 1 ; ; l k i is bounded by f(n) + a 4f(n) f(n) + 4( d log n ) (1 + log(f(n) + a 50 f(n))) f(n) + a f(n) for all large n: 25 By Lemma 12, Lemma 13 and Lemma 7, n is compressible in time 2 (1+a=2)f (n). This is a contradiction. Corollary 14. E 6 P cn?t (PSel) for every c > 0. 6

11 Corollary 15. EXP 6 P n c?t(psel) for every c > 0. Corollary 16. DTIME(2 nk+a ) 6 P n k?t(psel) for every a > 0. Corollary 17. SPARSE 6 P n c?t(psel) for every c > 0. Theorem 18. For any c > 0, P n c?t(psel) \ EXP has measure 0 in EXP. Theorem 19. For any c > 0, P cn?t (PSel) \ E has measure 0 in E. 4. Oracle In this section we constrcut an oracle to indicate the limit of relativizable method. Theorem 20. For every k > 1 and a > 0, there exists an oracle A such that DTIME(2 nk ) A P A n k+1+a?t (PSel). Proof: For any oracle B, let K B = fx10 i : M B i accepts x in 2 jxjk steps g. K B is computable in 2 nk time relative oracle B. There is a oracle Turing machine M such that M B accepts K B and M runs in 2 nk time. Dene Query(M; B; x) be the set of all strings that are queried by M B with input x. Construction of set A: Stage 0: Set s 0 = ;, A 0 = ; and P 0 = ;. Stage n: Let Q n = [ Query(M; A n?1 ; x) x2 n P n = P n?1 [ Q n : Since M runs in 2 nk time, k Query(M; A n?1 ; x) k 2 nk for any x 2 n. Therefore, j n k+ a 2 k k Q n k 2 n+1 2 nk = 2 nk +n+1 k P n k nx i=1 k Q i k = n 2 nk +n+1 : There exists y n 2 such that hy n ; xi 62 P n for any x 2 n. Let s n = s n?1 y n. Let G n be the set of all hy n ; xi such that x 2 K A n?1. Set A n = A n?1 [ G n. End of Stage n. Dene A = S 1 G n=1 n. End of the consruction of set A. 7

12 The following claim is easy to verify by the construction of A. Claim M A n?1 (x) = M A (x) for all x 2 n. From the above the construction we get innite strings s 1 s 2 : : : s n : : : : Let s 1 be the string of length 1 such that every s i is an initial segement of it. Let S = fx : x s 1 g. Thus, s 1 is the boundary of the P-selective set S under the lexicographic order. Since s 1 is the boundary of S, 1 s 1 i 1 2 S. Assume z n 2 n is an initial segement of s 1. One of z n 0 and z n 1 is an initial segement of s 1. Furthermore, z n 1 2 S ) z n 1 s 1 and z n 1 62 S ) z n 0 s 1. Therefore, the rst n bits of s 1 can be found in polynomial time by making n queries to set S. Since jy n j = j n k+ a 2 k, js n j = nx i=1 jy i j nx j k i k+ a 2 i=1 n j n k+ a 2 n k+1+ a 2 : So, we can nd y n by making no more than n k+1+ a 2 queries to S. For each x 2 n, hy n ; xi 2 A () x 2 K A. So, K A 2 P A n k+1+ a (S). For every 2?T language L accepted by M A i in time 2 nk, we have x 2 L () x10 i 2 K A. So, L 2 P A (S). n k+1+a?t Form the Theorem 20, we know the Corollary 16 is almost the optimal result that we can get by the relativizable method. 5. Symmetric Dierences Let g(n) be a time constructible super-polynomial function from N to N. A g;p m - reduction from A to B is a function f such that for some polynomial p(n), f is computable in g(n)-time, jf(x)j p(jxj) and x 2 A () f(x) 2 B for each input x. Theorem 21. Let g(n) be a time constructible super polynomial function from N to N. Let C be a class of languages which has P m-complete set and is closed under g;p m - reductions, H be P m-hard for C, A be a P-selective set, then A 4 H is P 2?T-hard for C. k 8

13 Proof: Let f 1 ; f 2 ;, be an eective enumeration of all P m-reductions and K is a -complete set for C. Let f be the selector functor function for A. P m Construction of L Input hi; xi Within g(jhi; xij)=2 steps, compute f i (hi; x; 0i) = y 0 ; f i (hi; xi) = y 1 and f i (hi; x; 2i) = y 2 and let y i0 f y i1 f y i2, where i 0 ; i 1 ; i 2 is a permutation of 0; 1; 2. If the above computation can be nished in the given time, then Accept hi; x; i 1 i and ( for w in fhi; x; i 0 i; hi; x; i 2 ig accepts w i x 2 K). End of the construction. Since C is closed under g;p m -reductions, it is easy to see that L 2 C. Since H is C- P m -hard, so L P m H via f j for some j. Lemma 22. For almost all x, let f j (hj; x; 0i) = y 0,f j (hj; x; 1i) = y 1, f j (hj; x; 2i) = y 2 and y i0 f y i1 f y i2, if y i1 62 A 4 H, then x 2 K () y i2 2 A 4 H, and; if y i1 2 A 4 H, then x 2 K () y i0 62 A 4 H. Proof: Since hj; x; i 1 i 2 L, we have y i1 2 H. If y i1 2 A 4 H,then y i1 62 A. It implies y i2 62 A since y i1 f y i2. Hence, x 2 K () y i2 2 A 4 H. If y i1 62 A 4 H, then y i1 2 A. It implies y i0 2 A since y i0 f y i1. So, x 2 K () y i0 62 A 4 H. From the above discussion, we have K P 2?T A. So, A is C- P 2?T -hard. Corollary 23. The symmetric dierence of a DTIME(n O(log n) )- P m-hard set and a P-selective set is DTIME(n O(log n) )- P 2?T-hard. Corollary 24. The symmetric dierence of a E- P m-hard set and a P-selective set is E- P 2?T-hard. Corollary 25. The symmetric dierence of a NE- P m-hard set and a P-selective set is NE- P 2?T-hard. We know that every E- P 2?T-hard is of exponential density [25]. The Corollary 24 generalizes a result by Rao [18] who showed the symmetric dierence of a E- P m-hard set a P-selective set is of exponential density. We note that the symmetric dierence of a NP- P m-hard set and a P-selective set is NP- P T-hard. 9

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