On a Quantitative Notion of Uniformity? Appeared In: MFCS'95, LNCS 969, 169{178, Springer{Verlag, 1995.

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1 169 On a Quantitative Notion of Uniformity? (Extended Abstract) Appeared In: MFCS'95, LNCS 969, 169{178, Springer{Verlag, Susanne Kaufmann??, Martin Kummer Institut fur Logik, Komplexitat und Deduktionssysteme, Universitat Karlsruhe, D-7618 Karlsruhe, Germany. fkaufmann; kummerg@ira.uka.de 1 Introduction Recent work on \Bounded Query Classes" in complexity theory and computability theory (see [5] for a survey) has sparked renewed interest in quantitative aspects of computability theory. A central result in this eld is the Nonspeedup Theorem [1, Theorem 9], which states that if any n parallel queries to a set A can be computed with n sequential queries to some oracle, then A must be recursive. More formally, for A! (! is the set of all natural numbers) and xed n 1, let Cn A(x 1; : : :; x n ) = ( A (x 1 ); : : :; A (x n )) denote the n-fold characteristic function of A. If there is an oracle B! such that C A n can be computed with n sequential queries to B, then A is recursive. (The bound on the number of queries is tight, i.e., n cannot be replaced by n + 1.) In [11] the analogous result for the n-place cardinality function # A n (x 1 ; : : :; x n ) = jfi : x i Agj is obtained (Cardinality Theorem). There is also a classical result of Trakhtenbrot [18] with a similar avor, concerning approximative computation of n distinct parallel queries to A: A set A is called (m; n)-recursive for 1 m n, if there is a recursive function f :! n! f0; 1g n (an \(m; n)-operator") such that for all pairwise distinct x 1 ; : : :; x n, f(x 1 ; : : :; x n ) and Cn A (x 1 ; : : :; x n ) agree in at least m components. Trakhtenbrot's Theorem states that every (m; n)-recursive set is recursive if m > n. (Again, this is tight, i.e., the theorem does not hold for m = b nc.) A typical point of these results is the nonuniform nature of their proofs: A set A is shown to be decidable, but no eective method for deciding A is provided. Beigel et al. [1] and Kinber [9, 10] proved that this nonuniformity is inevitable. More precisely, Kinber [10] proved that, given an index for f, there is no algorithm that computes the characteristic function (up to nitely many errors) of some set A which is (m; n)-recursive via f (if such A exists), even if m = n? 1. Thus, the following questions naturally arise: Do there exist two algorithms such that for every index at least one of them is successful? In general, what is the least number k depending on m; n such that there are k algorithms and for every index at least one of them is successful? This number can also be considered in two other ways. In a probabilistic interpretation, it tells us that A can be computed from an index of an (m; n)-operator with probability 1=k.? The complete version of this paper will appear in Fundamenta Informaticae.?? Supported by the Deutsche Forschungsgemeinschaft (DFG) grant Me 67/4-.

2 170 From the point of view of descriptional complexity, it tells us that dlog ke bits of additional information (to specify one of the k possibilities) are needed to compute A from an index of an (m; n)-operator. For a unied treatment of these questions we move to the domain of innite binary trees: It is known that each of the above theorems is equivalent to the statement that a certain natural class of recursive or r.e. trees has only recursive branches. These trees are also interesting in their own right. Owings [15] pointed out that the Nonspeedup Theorem is equivalent to the statement that every branch of an r.e. tree of bounded width is recursive. In this version the question of uniformity amounts to asking whether there is an algorithm that computes a branch from an index of the tree. While there is no uniformity in this strong sense, we get nontrivial positive results by weakening the concept of uniformity as indicated above: For some xed number k we want an algorithm which, given an index of a tree, outputs a list of k programs such that at least one of them computes a branch of the tree up to nitely many errors. Now the goal is to determine the least k for which this works. We show that, for r.e. trees of width at most n, the least possible k is n? 1. For trees arising from Trakhtenbrot's Theorem with parameters m; n we determine k = n? m + 1 as the optimal value. In addition, several other, related classes of trees will be investigated. The proofs of the upper bounds require re- nements of the proofs of the original theorems. The lower bounds are shown by suitable diagonalizations using the recursion theorem. Due to space limitations all proofs have been omitted, except, for illustration, a proof sketch of the upper bound of Theorem 13. The results of Sections 4, 5 originally appeared in [8] where additional material can be found. Very recently our results have been applied in inductive inference [4]. Notation and Denitions The notation is standard (see e.g., the textbooks [14, 16, 17]).! = f0; 1; ; : : :g. ' i is the i-th partial recursive function in the standard enumeration, W i! is the i-th r.e. set in the standard enumeration (W i = dom(' i )). For a given r.e. set A, let fa s g s! denote a recursive enumeration of A such that fa s g s! is a strong array, A s A s+1, and S s! A s = A, cf. [17, II..7]. P 1 is the set of all partial recursive one-place functions. R 1 is the set of all total recursive one-place functions. For functions f and g let f = g denote that f and g agree almost everywhere, i.e., (9x 0 )(8x x 0 )[f(x) = g(x)]. f y denotes the restriction of f to arguments x y. A is the characteristic function of A. We identify A with A, e.g., we write A(x) instead of A (x). f0; 1g is the set of nite strings of 0s and 1s. is the empty string. jtj denotes the length of string t. For instance, jj = 0. For strings t 1 and t we write t 1 t if t 1 is an initial segment of t. Let t(x) = b if x < jtj and b is the (x + 1)-th symbol of t. For t 1 ; t f0; 1g n, t 1 = e t means that t 1 and t disagree in at most e components. The concatenation of t 1 and t is denoted by t 1 t. A tree T is a subset of f0; 1g which is closed under initial segments. Note

3 171 that all of our trees are binary. t T is called a node of T. Let cod : f0; 1g!! be a recursive bijection. We often identify a tree T with the set fcod(t) : t T g. Then we say that T is r.e. if W i = fcod(t) : t T g for some i. Such an i is called a 1 -index of T. We say that a tree T is recursive if T is a recursive function, in which case i is called a 0 -index of T if ' i = T. f f0; 1g! is a branch of T if every nite initial segment of f is a node of T. We also say that A! is a branch of T if A is a branch of T. [T ] is the set of all branches of T. The degrees of branches of recursive trees were investigated intensively; see [7] as the basic reference. The width w(t ) of a tree T is the maximum number of nodes on any level, i.e., w(t ) = maxfjt \ f0; 1g n j : n 0g. B n = f0; 1g n is the full binary tree of depth n. A mapping g : B n! T is an embedding of B n into T if (8t)[jtj < n! [g(t0) g(t)0 ^ g(t1) g(t)1]]. rk(t ), the rank of T, is the maximal n such that B n is embeddable into T. For A; B! let A4B denote the symmetric dierence of A and B, i.e., A4B = (A n B) [ (B n A). For M! let (4M) = maxfja4bj : A; B Mg. For a tree T we write (4T ) instead of (4[T ]). A Weak Notion of Uniformity In this section we develop our notion of uniformity, provide a formal denition, and summarize our results. As a starting point consider the following search problem for a class A of recursive (or r.e.) trees. Search Problem for A Input: A 0 -index (or 1 -index) of a tree T A with [T ] 6= ;. Output: An index i of some branch A of T (i.e., ' i = A). This makes sense only if T has at least one recursive branch. For the classes A which we investigate in this paper, we have [T ] R 1, i.e., every branch is recursive. However, it turns out that the search problem for A is still algorithmically unsolvable, even if we allow the function ' i to make nitely many errors (i.e., ' i = A instead of ' i = A). Therefore we are led to the following weak version of the search problem. Weak k-search Problem for A Input: An index of a tree T A with [T ] 6= ;. Output: A list of k indices i 1 ; : : :; i k such that there is some 1 j k and some branch A of T with ' ij = A. In this version one has to produce k possibilities such that at least one of them is correct up to nitely many errors. A partial recursive solution g of the Weak k-search Problem for A is called a k-selector for A. Our goal is to determine the least k such that there is a k-selector for A. The following denition states all this formally. Denition1. Let A be a class of recursive [r.e.] trees. A function g P 1 is a k-selector for A if, whenever i is a 0 -index [ 1 -index] of T A with [T ] 6= ;,

4 17 there exist i 1 ; : : :; i k such that g(i) = hi 1 ; : : :; i k i and there is some 1 j k, and some A [T ], with ' ij = A. The least k such that there is a k-selector for A is denoted by sel(a). If no such k exists we write sel(a) = 1. Remarks: (1) We may assume that g is total recursive: By the s-m-n theorem there is a total recursive function h such that ' h(i;j) = ' ij if g(i) = hi 1 ; : : :; i k i^ 1 j k, otherwise ' h(i;j) is undened everywhere. Therefore we may replace g by the total recursive function g 0 (i) = hh(i; 1); : : :; h(i; k)i. () Also note that we get the same notion even if we only require g(i) to be an r.e. index of a nite set of size at most k which contains a correct index of a branch. We just dene h such that ' h(i;j) = ' z if z is the j-th element enumerated into W g(i) ; then we dene g 0 as above. We now introduce the classes of trees which are central to our investigation: Denition. D n := ft : T is a recursive tree and (4T ) ng. B n := ft : T is an r.e. tree with w(t ) ng. B r n := ft : T is a recursive tree with w(t ) ng. E n := ft : T is an r.e. tree with rk(t ) ng. E r n := ft : T is a recursive tree with rk(t ) ng. Trakhtenbrot [18], Owings[15], and Kummer [11, Lemma 1] proved that [T ] R 1 for T D n ; B n ; E n, respectively. It is easy to see that B n is a subclass of E n. D n is incomparable to the other classes. The weak search problems for D n ; B n, and E n turn out to be markedly dierent. For instance, while the weak search problems for D n and B n are eectively solvable, this is not the case for E n. Our quantitative results are summarized in the following table: A D n B n B r n E n E r n sel(a) d n+1 e n? 1 n 1 n + 1 It follows at once that sel(a) = 1 if A is the class of all recursive trees with (4T ) nite ( S n D n), the class of all recursive trees of nite width ( S n Br n ), or the class of all recursive trees of nite rank ( S n E n). r The results concerning D n, B n, E n are connected with the uniformity of Trakhtenbrot's Theorem, the Nonspeedup Theorem, and the Cardinality Theorem, respectively. We treat these classes and explain the connections in sections 3, 4, 5 below. Remarks: (1) The fact that sel(d 0 ) = 1 is well known: If a recursive tree has exactly one branch then this branch is (uniformly) recursive [17, VI.5.1]. () We do not consider the r.e. analogue of D n, because, even if an r.e. tree has a unique branch, this branch need not be recursive. Owings [15, p. 766] gave the following example: Let T := ft f0; 1g : (9s)[t (K s (0); : : :; K s (s))]g where fk s g s! is a recursive enumeration of the halting problem K. Then [T ] = fkg. (3) Note that j[t ]j n for every T Bn. r On the other hand, it is easy to see that,

5 173 for every recursive tree T with at most n branches, one can uniformly compute a recursive tree T 0 B r n with [T ] = [T 0 ]. Thus our results on the uniformity of B r n transfer to the class of all recursive trees with at most n branches. In particular, sel(a) = 1 for the class A of all recursive trees with nitely many branches. (4) The result that every branch of a tree of bounded width is recursive appears implicitly already in [13, p. 55 f.] where it is credited to A. R. Meyer. It is the key to the proof that an innite sequence f0; 1g! is recursive if the lengthconditional Kolmogorov complexity of all its initial segments is bounded by a constant. 3 Uniformity of Trakhtenbrot's Theorem In this section we investigate the uniformity of Trakhtenbrot's Theorem in relation to the class D n. Let us rst introduce some further notation. For any recursive function f :! n! f0; 1g n we call A! (m; n)-recursive via f i for all pairwise distinct x 1 ; : : :; x n we have C A n (x 1 ; : : :; x n ) = n?m f(x 1 ; : : :; x n ). Let T T m;n (f) denote the class of all sets that are (m; n)-recursive via f. 3.1 Previous Results We can state Trakhtenbrot's Theorem as follows: Theorem 3. (Trakhtenbrot [18]) Let n < m n and let f :!n! f0; 1g n be a recursive function. Then every set in T T m;n (f) is recursive. A more easily accessible reference for the proof of Theorem 3 is [6, Section 1]. A proof also falls out of Theorem 7 and Lemma 6 below. Theorem 3 also follows directly from Lemma 6 and the fact that every recursive tree with only countably many branches (but at least one) has a recursive branch [14, V.5.7] (see also the Tree Lemma of Owings in [15]). The question of uniformity for Theorem 3 leads to the following weak search problem (for xed k; m; n with n < m n). Weak k-uniformity Problem for Trakhtenbrot's Theorem with Parameters m; n Input: An index of f :! n! f0; 1g n with T T m;n (f) 6= ;. Output: A list of k indices i 1 ; : : :; i k such that ' ij = A for some 1 j k and some A T T m;n (f). This problem is uniformly equivalent to a weak search problem for a suitable class of recursive trees which we now dene (cf. [6, p. 683], [18]). It consists of all nite initial segments that are consistent with the (m; n)-operator f. Denition4. For any recursive function f :! n! f0; 1g n and m n let T m;n (f) := ft f0; 1g : (8x 1 < < x n < jtj)[(t(x 1 ); : : :; t(x n )) = n?m f(x 1 ; : : :; x n )]g: T m;n (f) is called an (m; n)-recursive tree (via f). Let A m;n denote the class of all (m; n)-recursive trees.

6 174 Note that the branches of T m;n (f) are exactly the sets which are (m; n)-recursive via f, i.e., [T m;n (f)] = T T m;n (f). From m; n and an index of f :! n! f0; 1g n we can uniformly compute a 0 -index of T m;n (f). Conversely, from m; n and a 0 -index of T m;n (f) we can compute an index of g :! n! f0; 1g n such that T m;n (g) = T m;n (f). (Dene g by induction on k for all n-tuples x 1 < < x n with x n = k.) Therefore, the Weak k-uniformity Problem for Trakhtenbrot's Theorem with Parameters m; n is uniformly equivalent to the Weak k-search Problem for A m;n. Kinber proved the rst lower bound on sel(a m;n ). His result can be restated as follows: Theorem 5. (Kinber [9, 10]) sel(a m;n ) > 1 for n < m < n. The connection between A m;n and D n is given by the following observation of Trakhtenbrot, which is essential for his proof of Theorem 3. Lemma 6. (Trakhtenbrot [18]) A m;n D (n?m) for n < m n. Remarks: (1) By an argument in [6, p. 683], we also have A m;n E r, for n?1 n < m n. () Fix m; n; k with n < m < n and k 1. It follows from Kinber's result that there is no k-selector g for A m;n which, given an innite tree T A m;n, produces at least one index i j such that ' ij [T ]: Otherwise, there would be a 1-selector for A m;n which rst computes g(i) = hi 1 ; : : :; i k i and then outputs an index of the following function f i : f i (x): Search for 1 j k with ' ij (z) # for all z x+1 and ' ij (x+1) T. Then let f i (x) := ' ij (x). Note that for suciently large x we have ' ij (x + 1) T only if ' ij is a branch of T. By Lemma 6, all branches of T which are equal to ' ij for some 1 j k agree from some point on. Hence f i makes only nitely many errors and gives us a 1-selector. By a similar argument, there can be no k-selector g and a xed error bound e! such that one of the ' ij computes a branch up to e errors. This shows that it would have been too restrictive not to allow a nite (and unbounded) number of errors in the denition of a k-selector. 3. Main Results The next result gives the exact value of sel(a m;n ): Theorem 7. sel(a m;n ) = n? m + 1 for n < m n. We shall organize the proof of this theorem in such a way that we get in addition the exact value of sel(d n ): Theorem 8. sel(d n ) = d n+1 e for n 0.

7 175 With Lemma 6 the plan for the proofs of Theorems 7, 8 is clear: We rst show the upper bound sel(d n ) d n+1 e. Then, using Lemma 6, we get the upper bound of Theorem 7: sel(a m;n ) sel(d (n?m) ) n? m + 1. Second, we show the lower bound sel(a m;n ) n? m + 1. Then, by Lemma 6, we get the lower bound of Theorem 8: k + 1 sel(a k+1;k+1 ) sel(d k ) sel(d k+1 ). 3.3 Total Selectors We would now like to point out a direct proof for sel(a m;n ) n? m + 1 which yields a stronger result and provides new insights into the structure of trees that arise from Trakhtenbrot's Theorem. To this end let sel t (A) be the least number k for which there is a k-selector for A such that at least one of the indices which it produces on input T A computes a branch of T almost everywhere and computes a total function. Clearly, sel t (A) sel(a). The next result shows that the bound in Theorem 7 also holds for total selectors. An analogous result holds for the tree classes B n ; B r n; E n ; E r n which we consider in the next sections. Theorem 9. sel t (A m;n ) = n? m + 1 for n < m n. The lower bound follows from Theorem 7. For the upper bound we need the following combinatorial notion. Denition10. A tree T has dimension d 0 (dim(t ) = d 0 ) if d 0 is the maximal number d such that there exist x 1 < < x d with (8v f0; 1g d )(9A [T ])[v = C A d (x 1; : : :; x d )]. If d 0 does not exist then dim(t ) := 1. Remark: In more technical terms dim(t ) is the Vapnik-Chervonenkis dimension of the concept class [T ]!. See [3] for more information on this notion; in [, Section ], [1, Theorem 6, Proposition 7] further recursion theoretic applications are given. Note that for every T A m;n we have dim(t ) n? m: If dim(t ) > n? m then choose x 1 < < x n?m+1 witnessing this fact. Suppose that T is an (m; n)-recursive tree via f. Let (b 1 ; : : :; b n ) = f(x 1 ; : : :; x n?m+1 ; x n?m+1 + 1; : : :; x n?m+1 +m?1). Now we can choose a branch A of T such that 1?A(x i ) = b i for 1 i n? m + 1. But this means that A is not (m; n)-recursive via f, a contradiction. The next lemma shows how the knowledge of the dimension of T can help to compute a branch of T. Lemma 11. Let T be an innite recursive tree with dim(t ) = d and (4T ) nite. Then, uniformly in a 0 -index of T, there is f K such that f= A for some A [T ], and in addition f is total.

8 176 The totality of the selector function is not guaranteed by the proof of the upper bound of Theorem 8, in which the correct f 0 a may be undened at two places. We close this section by showing that non-totality is essential for the upper bound sel(d n ) d n+1e: By the next result, sel t(d n ) is twice as large as sel(d n ). Theorem 1. sel t (D n ) = n + 1 for n 0. 4 Uniformity of the Nonspeedup Theorem - Trees of Bounded Width In this section we present optimal bounds for trees of bounded width. The main results are: Theorem 13. sel(b n ) = n? 1 for n 1. Theorem 14. sel(bn r ) = n for n 1. Proof sketch of the upper bound sel(b n ) n? 1: We proceed in two steps. First, we will show that sel(b n ) n(n+1) ; this will just be a uniform version of a proof of Owings in [15, p. 764]. Second, we will take a closer look at the selector that we have found and dene a rened version which, somewhat surprisingly, reduces the quadratic bound to a linear bound and gives us sel(b n ) n? 1. (1) sel(b n ) n(n+1) : For a tree T B n let L(T; k) := T \f0; 1g k (the k-th level of T ), a 0 := j[t ]j, and dene the supremal width as b 0 := maxfl : (9 1 k)[jl(t; k)j = l]g. We assume that a 0 > 0. Clearly, a 0 b 0 n. We prove that there is a uniform procedure f a0;b0 which, given a 0 ; b 0, and a 1 -index of T, computes a branch of T up to nitely many errors. Since there are n(n+1) possible pairs (a 0 ; b 0 ) this gives us an n(n+1) -selector for B n. () sel(b n ) n? 1: The idea is to \amalgamate" several f a;b -functions into one function. If (a; b) 6= (a 0 ; b 0 ) but dom(f a;b ) is nite then f a;b does not hurt in the amalgamation. Therefore we have to consider when dom(f a;b ) is nite. Claim: If (a 0 ; b 0 ) f(a; b); (a + c; b + c)g for some a; b; c > 0 then exactly one of dom(f a;b ) and dom(f a+c;b+c ) is innite. By the Claim, we are led to dene n?1 functions g a;b, for a = 1; and a b n, as follows: g a;b (x): Dovetail the computations of f a;b (x); f a+;b+1 (x); : : :; f a+c;b+c (x), where c is maximal with a + c b + c. As soon as a converging computation with output y appears, let g a;b (x) := y.

9 177 5 Uniformity of the Cardinality Theorem - Trees of Bounded Rank In this section we prove optimal bounds for trees of bounded rank and discuss the connection with the uniformity of the Cardinality Theorem. Theorem 15. sel(e 1 ) = 1. Theorem 16. sel(en r ) = n + 1 for n 0. The trees that arise in the Cardinality Theorem (CT) are a proper subclass of the r.e. trees of bounded rank. CT can be stated as follows [11]: Let A! and n 1. If there is a recursive function g :! n!! such that for every x 1 < < x n, (1) W g(x1;:::;xn) f0; 1; : : :; ng (note that it is a proper inclusion), () # A n (x 1; : : :; x n ) W g(x1;:::;xn), then A is recursive. For a given g as above we can dene a uniformly r.e. tree T g such that the branches of T g are just the sets A which satisfy conditions (1), (): T g = ft f0; 1g : (8x 1 < < x n )[x n < jtj! P n t(x i=1 i) W g(x1;:::;xn)]g: We say that T g arises from CT with parameter n. Let C n denote the class of all trees that arise from CT with parameter n. CT is proved by showing that the trees in C n have bounded rank. However, there are trees of bounded rank which do not belong to any C n. The class C is studied in greater detail in [8, 15]. It is shown in [8] that E 1 6 C E, and that, in contrast to Theorem 15, sel(c ) is nite (in fact, sel(c ) 6). It is not known whether sel(c n ) is nite for all n. From Theorem 16 we can derive a positive result for the uniformity of a weak version of CT (proved in [15]), in which W g(x1;:::;xn) is replaced by D g(x1;:::;xn) (D i is the i-th nite set in a canonical enumeration of all nite sets): Let Cn r denote the class of all recursive trees arising from this version of CT with parameter n. It is shown in [11] that the rank of every tree in C n is bounded by 4 n?. In particular, Cn r E r 4 n?. Thus, by Theorem 16, we get sel(cn r ) 4n? 1. 6 Conclusion We have introduced a new notion of uniformity to investigate the problem of how to compute a branch of a tree. We have obtained precise quantitative descriptions of the borderline between uniformity and nonuniformity for several interesting classes of trees. Furthermore, we were able to apply these ndings to central results of quantitative computability theory such as Trakhtenbrot's Theorem and the Nonspeedup Theorem. In this paper we have studied our quantitative notion of uniformity in a domain which in itself had a quantitative structure, namely classes of trees parameterized by natural numbers. It would be interesting to study this notion in other areas as well.

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