Strongly Bounded Turing Reducibilities and Computably Enumerable Sets. 4. Strongly Bounded T- Reducibilities

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1 Strongly Bounded Turing Reducibilities and Computably Enumerable Sets Prof. Dr. Klaus Ambos-Spies Sommersemester Strongly Bounded T- Reducibilities Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 1 / 16

2 Outline In this section we begin with our closer analysis of the strongly bounded Turing reducibilities on the c.e. sets. We start with some basic observation. Then we discuss the relations to the permitting technique. Finally we make some remarks on the role played by cl-reducibility in the context of algorithmic randomness. 1 sbt-reducibilities: some basic facts 2 sbt-reducibilities and permitting 3 cl-reducibility and algorithmic randomness (postponed) Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 2 / 16

3 4.1 SbT-reducibilities: Some basic facts Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 3 / 16

4 The sbt-reducibilities (definition reviewed) 1 A set A is bounded-turing (bt-) reducible or weak-truth-table (wtt-) reducible to a set B (A bt B) if there is a Turing functional Φ such that A = Φ B and the use function ϕ B of Φ B is bounded by some computable function f. 2 A set A is identity-bounded-turing (ibt-) reducible to a set B (A ibt B) if there is a Turing functional Φ such that A = Φ B and the use function ϕ B of Φ B is bounded by the identity function f (x) = x. 3 A set A is (i + k)-bt-reducible to a set B (A (i+k)bt B) if if there is a Turing functional Φ such that A = Φ B and the use function ϕ B of Φ B is bounded by f (x) = x + k. 4 A set A is computable Lipschitz (cl-) reducible to a set B (A cl B) if there is a Turing functional Φ such that A = Φ B and the use function ϕ B of Φ B is bounded by f (x) = x + k for some k 0. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 4 / 16

5 Summary of basic facts For r = ibt, cl the following hold. 1 r is reflexive and transitive. So r-equivalence and r-degrees can be defined as usual and we can let (R r, ) be the partial ordering of the c.e. r-degrees. Obviously, (R r, ) has a least element, the degree of the computable sets. 2 In contrast, for k 2, (i + k)-bt-reducibility is not transitive. In general, A (i+k)bt B & B (i+k )bt C A (i+k+k )bt C holds and, in general, this is optimal. This can be shown by observing that, for noncomputable A, A (i+k)bt A + k but A (i+k)bt A + k 1 (k 1). 3 r is closed under finite variations. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 5 / 16

6 4.2 SbT-reducibilities and permitting Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 6 / 16

7 Permitting and Generalized Permitting Definition Let A and B be c.e. sets and k 0. Then A T B by k-permitting if there are computable enumerations {A s } s 0 and {B s } s 0 of A and B, respectively, such that x s (x A at s y x + k (y B at s )) (1) holds. In particular, A T B by permitting if A T B by 0-permitting. In the following we will show that k-permitting and the (i + k)btreducibility on the c.e. sets are closely related. One direction is straightforward. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 7 / 16

8 k-permitting (i + k)bt- reducible Lemma Let A and B be c.e. sets such that A T B by k-permitting. Then A (i+k)bt B. In particular, if A and B are c.e. sets such that A T B by permitting then A ibt B. Conversely, any (i + k)bt-reduction from a c.e. set A to a c.e. set B can be represented by a k-permitting if we replace A and B by some ibt-equivalent subsets. In the following we state this observation in a somewhat more general form. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 8 / 16

9 (i + k)bt- reducible k-permitting Lemma (Representation Lemma) Let A and B 0,..., B m (m 0) be noncomputable c.e. sets and let k 0 such that B j (i+k)bt A for j m. There are c.e. subsets  of A and ˆB j of B j, and computable 1-1 enumerations {a(n)} n 0 and {b j (n)} n 0 of  and ˆB j, respectively, such that, for j m, (i)  = ibt A and ˆB j = ibt B j and (ii) n (a(n) min{b 0 (n),..., b m (n)} + k) Note that the above implies that ˆB j T  by k-permitting via the enumerations {ˆB j,s } s 0 and {Â} s 0 given by ˆB j,s = {b j (0),..., b j (s 1)} and  s = {a(0),..., a(s 1)}. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 9 / 16

10 Representation lemma: simple form Here we will only prove the special case of the Representation Lemma for one B-set and for k = 0. The proof of the general lemma which is similar is given in the lecture notes. Lemma (Representation Lemma (Special Case)) Let A and B be noncomputable c.e. sets such that B ibt A. There are c.e. subsets  of A and ˆB of B, and computable 1-1 enumerations {a(n)} n 0 and {b(n)} n 0 of  and ˆB, respectively, such that, (i)  = ibt A and ˆB = ibt B and (ii) n (a(n) b(n)) Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 10 / 16

11 Proof (1) Fix computable enumerations {A s } s 0 and {B s } s 0 of A and B, respectively, such that A s+1 A s and B s+1 B s (s 0). Let Φ be an ibt-functional such that B = Φ A. Define the length (of agreement) function l: l(s) = max{x s : y < x (B s (y) = Φ As s (y))}. Note that lim s ω l(s) = ω. So there are infinitely many expansionary stages s, i.e., stages s such that t < s (l(t) < l(s)). Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 11 / 16

12 Proof (2) Call an expansionary stage s critical if B s l(s) B l(s). Note that by noncomputability of B there are infinitely many critical expansionary stages. Say that t > s witnesses criticalness of s if B s l(s) B t l(s). Note that if criticalness of s is witnessed by t then criticalness of s is witnessed by all t t. Moreover, the set of all pairs (s, t) such that s is critical and criticalness of s is witnessed by t is computably enumerable. It follows that we can define a computable ascending sequence of expansionary stages s 0 < s 1 < s 2 <... such that s n is critical and s n+1 witnesses criticalness of s n. Now let a(n) = µ x (x A sn+1 \ A sn ) b(n) = µ x (x B sn+1 \ B sn ) Â = {a(n) : n 0} ˆB = {b(n) : n 0} Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 12 / 16

13 Proof (3): Correctness We first observe that a(n) and b(n) are well defined and that a(n) b(n) (i.e., (ii)) holds. Then, obviously, a(n) and b(n) are total computable and 1-1 functions, and  and ˆB are subsets of A and B, respectively. So it will only remain to verify (i). By choice of s n and s n+1, So b(n) exists and b(n) < l(s n ). B sn l(s n ) B sn+1 l(s n ) Since s n and s n+1 are expansionary, it follows that b(n) < l(s n ) < l(s n+1 ) and 0 = B sn (b(n)) = Φ Asn s n (b(n)) & 1 = B sn+1 (b(n)) = Φ As n+1 (b(n)). So Φ Asn s n (b(n)) Φ As n+1 s n+1 (b(n)). Since Φ is an ibt-functional, it follows that A sn b(n) + 1 A sn+1 b(n) + 1. So a(n) = µ x (x A sn+1 \ A sn ) is defined and a(n) b(n). s n+1 Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 13 / 16

14 Proof (4): Correctness (continued) Proof of (i)  = ibt A and ˆB = ibt B: Let Then and  n = {a(0),..., a(n 1)} & ˆB n = {b(0),..., b(n 1)}. µ x (x  n+1 \  n ) = µ x (x A sn+1 \ A sn ) µ x (x ˆB n+1 \ ˆB n ) = µ x (x B sn+1 \ B sn ). So  = ibt A and ˆB = ibt B by permitting. q.e.d. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 14 / 16

15 Splitting (1) We have seen already: If we split a c.e. set A into two disjoint c.e. sets A 0 and A 1 then the Turing degree of A is the least upper bound (join) of the Turing degrees of the parts A 0 and A 1 (see the T-Splitting Lemma). Conversely, given any join deg T (A) = deg T (B) deg T (C), this join can be represented by the splitting of the disjoint union B C into the sets B and C. While these observations easily carry over to the bt-degrees, for the strongly bounded Turing degrees we obtain only one direction. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 15 / 16

16 Splitting (2) Lemma (Splitting Lemma) Let A 0,..., A m (m 1) be pairwise disjoint c.e. sets and let A = A 0 A m. Then, for r {ibt, cl}, deg r (A) = deg r (A 0 ) deg r (A m ). Proof. For j m, A j ibt A 0 A m = A by permitting. So, given B such that A j (i+kj )bt B, it suffices to show that A 0 A m (i+k)bt B where k = max k j. But this is obviously true. Strongly Bounded T-Reducibilities (SS 2011) Strongly Bounded T- Reducibilities Chapter 4 16 / 16