On the Strongly Bounded Turing Degrees of the Computably

On the Strongly Bounded Turing Degrees of theComputably Enumerable Sets

Klaus Ambos-Spies

University of HeidelbergInstitut fur Informatik

Mal’tsev Meeting 2011Novosibirsk, 11 October 2011

Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 1 / 29

Strongly Bounded Turing Reducibilities

Notions of bounded Turing reducibilities have been introduced by imposingbounds b(x) on the admissible sizes of the oracle queries in a Turingreduction A(x) = ΦB(x):

b(x) computable: weak truth-table (wtt) or bounded Turing (bT)

b(x) = id(x) = x : identity bounded Turing (ibT)

b(x) = id(x) + c = x + c : strong weak truth-table (cl) or computableLipschitz (cl)

We refer to ibT and cl as the strongly bounded Turing (sbT) reducibilities.


Origins and Applications

ibT-Reducibility was introduced by Soare (2004) in the context ofsome applications of computability theory to differential geometry(Nabutovski and Weinberger).

cl-Reducibility was introduced by Downey, Hirschfeldt and LaForte(2001) in the context of computable randomness.

Note that, for a set A which is cl-reducible to a set B, the Kolmogorovcomplexity of A � n is bounded by the Kolmogorov complexity of B � nup to an additive constant.

Moreover, Downey, Hirschfeldt and LaForte have shown that, on the

computably enumerable (c.e.) sets, cl-reducibility coincides with Solovay

reducibility which may be viewed as a relative measure of the speed by

which a real number can be effectively approximated by rational numbers.


Strongly bounded Turing reductions and c.e. sets

Here we will discuss the strongly bounded Turing reducibilities on thecomputably enumerable (c.e.) sets and the corresponding degreestructures.

For any reducibility r , the r -degree of a set A is defined by

degr (A) = {B : B =r A}

An r -degree is called computably enumerable (c.e.) if it contains ac.e. set.

The partial ordering of the c.e. r -degrees is denoted by (Rr ,≤)(where degr (A) ≤ degr (B) if A ≤r B).


Examples of ibT-Reductions among c.e. sets

Some typical, frequently used examples of ibT-reductions on the c.e. setsare the following.

Permittingx ∈ Aat s ⇒ ∃ y ≤ x (y ∈ Bat s)

SplittingA = B ∪ C ⇒ B ≤ibT A and C ≤ibT A

In fact, for r = ibT, cl,

degr (A) = degr (B) ∨ degr (C )


Relations to the Classical Strong Reducibilities

The strongly bounded Turing reducibilities are incomparable (evenincompatible) with the truth-table type reducibilites.

THEOREM (AS ta). There are noncomputable c.e. sets A and Bsuch that

I for r ∈ {1,m,btt, tt}, A <r BI for r ′ ∈ {ibT, cl}, B <r ′ A

Just as the classical strong reducibilities, the strongly bounded Turingreducibilities are more sensitive to structural properties of c.e. setsthan the weak reducibilities (bT, T). An example:

THEOREM (AS). There is a noncomputable c.e. set which is notcl-equivalent to any simple set (whereas any noncomputable c.e. set isbT-equivalent to a simple set).


Strongly Bounded Turing Reducibilites and ComputableInvariance

The strongly bounded Turing reducibilities are not computablyinvariant.

THEOREM (AS). For any noncomputable c.e. set A there are c.e.sets A+ and A− such that the following hold.

I A, A+ and A− are computably isomorphic.

I For r ∈ {ibT, cl}, A− <r A <r A+.


Separation of ibT, cl and bTTHEOREM (Downey, Hirschfeldt and Laforte 2001, Barmpalias and Lewis2006). Let A be a noncomputable c.e. set.

For A + 1 = {x + 1 : x ∈ A},

A + 1 <ibT A whereas A + 1 =cl A.

For 2A = {2x : x ∈ A},

2A <cl A whereas 2A =bT A.

PROOF.

A ≤ibT A + 1 ⇒ A selfreducible(i.e., A � x computes A(x))

⇒ A computable

(And, similarly, for A ≤cl 2A.)


The Partial Orderings of the C.E. ibT- and cl-Degrees((RibT,≤) and (Rcl,≤)) and Their Theories

THEOREM (AS). For r = ibT, cl,

Th(Rr ,≤) is undecidable.

Th(Rr ,≤) realizes infinitely many 1-types (hence is countablycategorical).

So - just as the other degree structures of the c.e. sets, for which thecorresponding results have been previously proven - the partial orderings ofthe c.e. sbT-degrees are rich and complicated structures.

Still the partial orderings of the c.e. ibT/cl-degrees look quite differentthan the partial orderings of the c.e. bT/T-degrees.


ibT/cl vs. bT/T: First Examples of Similarities andElementary Differences

For r ∈ {ibT, cl,bT,T}, the partial ordering (Rr ,≤) possesses a leastelement, namely the degree 0 of the computable sets.

For r ∈ {bT,T}, the partial ordering (Rr ,≤) possesses a greatestelement, namely the degree 0′r of the r -complete sets (= the r -degreeof the halting problem).

For r ∈ {ibT, cl}, the partial ordering (Rr ,≤) does not possess anymaximal elements (Barmpalias 05 (for cl)).

PROOF for ibT. For noncomputable A, A <ibT A− 1 whereA− 1 = {x − 1 : x ≥ 1 & x ∈ A}.

PROOF for cl (AS, Ding, Fan, Merkle; Belanger). For noncomputablec.e. A, fix an infinite computable subset R of A, let A = A \ R and letA be the compressed version of A obtained by mapping ω \ R onto ω.Then A =ibT A <cl A.


ibT/cl vs. bT/T: Further Elementary Differences andSimilarities

For r ∈ {bT,T}, the partial ordering (Rr ,≤) is an upper semi-lattice:

degr (A) ∨ degr (B) = degr (A⊕ B)

where A⊕ B = {2n : n ∈ A} ∪ {2n + 1 : n ∈ B}.

For r ∈ {ibT, cl}, the partial ordering (Rr ,≤) is not a u.s.l.(Barmpalias 05, Fan and Lu 05).

In fact, Barmpalias 05 and Fan and Lu 05 proved that there aremaximal pairs in the partial ordering (Rr ,≤), i.e., c.e. r -degrees a andb such that there is no c.e. r -degree c such that a,b ≤ c.

For r ∈ {ibT, cl,bT,T}, the partial ordering (Rr ,≤) is not a lowersemi-lattice (Lachlan, Jockusch, Downey and Hirschfeldt (andothers)).


ibT/cl vs. bT/T: Further Elementary Differences andSimilarities

For r ∈ {ibT, cl,bT,T}, every nonzero degree splits, i.e., isjoin-reducible (Sacks’s Splitting Theorem)

For r ∈ {bT, ibT, cl}, every (incomplete) degree is branching, i.e., ismeet-reducible (Ladner and Sasso; AS, Bodewig, Kraling, Yu).

Whereas there are nonbranching c.e. T-degrees (Lachlan).

For r ∈ {bT,T}, (Rr ,≤) is dense (Sacks’s Density Theorem).

Whereas, for r ∈ {ibT, cl}, (Rr ,≤) is not dense (Barmpalias andLewis 10, Day ta).


Elementary Differences: Recent Research

The preceding examples show that there are numerous striking differencesbetween the partial orderings of the c.e. ibT- and cl-degrees and thepartial orderings of the c.e. bT- and T-degrees. On the other hand wehaven’t seen any elementary differences between the partial orderings ofthe c.e. ibT-degrees and the c.e. cl-degrees.

In the remainder of my talk I want to present some recent results andopen problems related to the following questions.

On what (logical) complexity level do differences between ibT/cl andbT/T occur?

I Embedding Problem (for finite lattices)

Are the partial orderings (RibT,≤) and (Rcl,≤) elementarilyequivalent?


The ∃∗- and ∀∗∃∗-Theories

For r ∈ {ibT, cl,bT,T}, the ∃∗- (or Σ1-) Theories coincide.

This follows from Sacks’s Theorem on independent sequences in thec.e. Turing degrees which implies that all finite partial orderings canbe embedded (as partial orderings) into (Rr ,≤).

This determines ∃∗ − Th(Rr ,≤) and shows that this theory isdecidable.

On the other hand:

For r ∈ {ibT, cl} and r ′ ∈ {bT,T}, the ∀∃-Theories of (Rr ,≤) and(Rr ′ ,≤) differ by the nonexistence/existence of a greatest degree.

The decidability of ∀∗∃∗ − Th(Rr ,≤) seems to be open in all cases.

For a finer analysis we consider existential sentences in richer languagesobtained by adding (relational) symbols for joins and meets (and 0).(NB. The existential theories in these augmented languages are subtheoriesof the forall-exists-theories in the language of partial orderings).Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 14 / 29

The ∃∗-Theories in the languages L(≤,∨,∧) andL(≤,∨,∧, 0)

A decision procedure for ∃∗ − Th(Rr ,≤,∨,∧) and ∃∗ − Th(Rr ,≤,∨,∧, 0)requires an answer to the following

EMBEDDING PROBLEM FOR (Rr ,≤). Which finite lattices can beembedded into (Rr ,≤) as lattices (by a map which preserves the leastelement)?

The Embedding Problem for (RbT,≤) is solved, while the EmbeddingProblem for (RT,≤) is one of the longlasting open problems about thisdegree structure. The Embedding Problem for the strongly boundedTuring reducibilities has been attacked only very recently.


The Embedding Problem for Distributive Lattices

The Embedding Problem for Distributive Lattices has been completelysolved by Lachlan-Thomason and Lerman by a rather straightforwardvariant of the minimal pair technique.

THEOREM (Lachlan, Thomason, Lerman). The countable atomlessBoolean algebra can be embedded into (RT,≤) (as a lattice) by a mapwhich preserves the least element.

Inspection of the proof (together with some general observations) showsthat the result holds for the other reducibilities (considered here) too.

COROLLARY. For r ∈ {ibT, cl, bT,T}, every countable distributivelattice can be lattice-embedded into (Rr ,≤) by a map which preserves theleast element.


The Embedding Problem for (RbT,≤)

Since Lachlan has shown that the u.s.l. of the c.e. bT-degrees isdistributive, the universal embedding result for distributive latticescompletely solves the Embedding Problem for (RbT,≤):

THEOREM (Lachlan, Thomason, Lerman). For a finite (or countable)lattice L the following are equivalent.

1 L is distributive.

2 L is embeddable into (RbT,≤).

3 L is embeddable into (RbT,≤) by a map which preserves the leastelement.

In contrast to (RbT,≤), however, the partial orderings (RT,≤), (RibT,≤)and (Rcl,≤) are nondistributive. So there the solution of the EmbeddingProblem requires the by far more sophisticated analysis of thenondistributive lattices.


Embedding Nondistributive Lattices into (RT,≤)

THEOREM (Lachlan 1972). The nondistributive 5-element lattices M5

(modular) and N5 (nonmodular) are embeddable into (RT,≤) by mapspreserving 0.

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Since any nondistributive lattice contains a copy of the M5 or N5,Lachlan’s result was considered for a while as an indication that all finitelattices can be embedded. But ...


A Nonembeddable Lattice for (RT,≤)

THEOREM (Lachlan and Soare 1980). The lattice S8 cannot beembedded into (RT,≤).

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(I.e., for any embedding of the M5 into the c.e. Turing degrees, the top isnonbranching.)


Current Status of the Embedding Problem for (RT,≤)

AS and Lerman (1986, 1989) have given very general embeddabilityand nonembeddability criteria (EC, NEC).

Lempp and Lerman (1997) give a nonembeddable 20-element latticenot covered by the previous NEC.

Lerman (2006) gives a Π02-characterization of the finite embeddable

lattices without “critical triples”.

Despite these successes, the Embedding Problem for (RT,≤) is stillopen.

All lattices which have been proven to be embeddable are embeddablepreserving the least element. So it has been conjectured thatembeddability and embeddability preserving 0 coincide.


The Embedding Problem for the C.E. sbT-Degrees: AFirst StepFirst results on the embeddability of nondistributive lattices into thepartial orderings (Rr ,≤) for the strongly bounded Turing reducibilitiesr = ibT, cl have been obtained by AS, Bodewig, Kraling, and Yu:

1 The nondistributive nonmodular lattice N5 is embeddable into(Rr ,≤). (So, in particular, the partial ordering (Rr ,≤) is neitherdistributive nor modular.)

2 Every c.e. r -degrees is the least element of an embedding of the2-atom Boolean Algebra.

3 The finite lattices embeddable into (Rr ,≤) differ from the finitelattices embeddable into (RT,≤). Namely, by the precedingobservation M5 is embeddable into (Rr ,≤) if and only if S8 isembeddable into (Rr ,≤).

4 The nondistributive modular lattice M5 cannot be embedded into(Rr ,≤) by a map which preserves the least element.


The Embedding Problem for the C.E. sbT-Degrees: MoreRecent Work

AS and Wang (in 2010) have given a (sufficient) criterion for thenonembedabbility by maps preserving 0 (which is based on the dual of“critical triples” which were previously introduced in the context ofNEC).

They also have given an example of a finite (nonmodular) latticewhich can be embedded into (Rr ,≤) (for r = ibT, cl) but whichcannot be embedded by a map preserving 0.

More recently (in 2011), the question of the embeddability of themodular lattice M5 has been resolved:

THEOREM (AS, Bodewig, Kraling, and Wang). M5 is embeddableinto (RibT,≤) and (Rcl,≤).


The Embedding Problem for the C.E. sbT-Degrees: OpenProblems

Since we couldn’t find any principal obstacles to embedding latticesinto the partial orderings (RibT,≤) and (Rcl,≤), we conjecture:

CONJECTURE. Any finite lattice can be embedded into (RibT,≤)and (Rcl,≤).

A characterization of the lattices which can be embedded preserving 0seems to be more difficult.

OPEN PROBLEM. What finite lattices can be embedded into(RibT,≤) and (Rcl,≤) by maps preserving 0.


(RibT,≤) vs. (Rcl,≤): an Elementary Difference

It seems that (RibT,≤) and (Rcl,≤) share the most basic algebraicproperties.

All of the discussed proofs of structural properties of (RibT,≤) could beextended to (Rcl,≤) though the arguments became somewhat (or, in veryfew cases, more than somewhat) more technical and involved.

In fact, in many cases, the proof of the results for (Rcl,≤) could bereduced to the case of (RibT,≤) (e.g. maximal and minimal pairs in bothstructures coincide and (not directly related to the degree structure) a c.e.set is cl-equivalent to a simple set if and only if it is ibT-equivalent to asimple set).

There is a concept, however, which is important for the analysis ofibT-reducibility and which has no direct counterpart in cl-reducibility,namely finite (or bounded) shifts.


(RibT,≤) vs. (Rcl,≤): Finite Shifts vs. Computable shifts

As we have observed already, for any noncomputable set A,A + 1 <ibT A. By iteration,

· · · <ibT A + 2 <ibT A + 1 <ibT A <ibT A− 1 <ibT A− 2 <ibT . . .

The role played by the finite shifts for ibT is played by the unboundedcomputable shifts for cl:

Af = {f (x) : x ∈ A}

where f is computable, strictly increasing, and f (n)− n→∞:Af <cl A.

In contrast to the finite shifts A + k which are discrete and invertible,the unbounded computable shifts are dense and, in general, notinvertible.


(RibT,≤) vs. (Rcl,≤): Finite Shifts and Cupping

The observation that, for any noncomputable set A, A + 1 <ibT Acan be strengthened as follows.

FINITE-SHIFT LEMMA (AS; special case). Let A and B be c.e. setssuch that A + 1 ∩ B = ∅ and A ≤ibT A + 1 ∪ B. Then A ≤ibT B.

DEFINITION. Let (P,≤) be a partial ordering and let a ∈ P. Anelement b ≤ a of P is a-cuppable if a = b ∨ c for some c < a; and bis a-noncuppable.

The Finite-Shift Lemma implies

THEOREM (AS). Let a > 0 be a c.e. ibT-degree. Then any degreeb ≤ a such that b 6≤ a + 1 is a-cuppable.

In other words, there is a degree c < a (namely c = a + 1) such that

∀ b ∈ NCUP(a) (b ≤ c)


(RibT,≤) vs. (Rcl,≤): Cupping in the cl-Degrees

For the c.e. ibT-degrees we have observed: For any a > 0 there is adegree c < a (namely c = a + 1) such that ∀ b ∈ NCUP(a) (b ≤ c).

For the cl-degrees the corresponding fact fails:

THEOREM (AS, Bodewig, Fan, Kraling). There is a c.e. cl-degreea > 0 such that, for any c < a there is a a-noncuppable degree b < asuch that b 6≤ c.

COROLLARY. Th(RibT,≤) 6= Th(Rcl,≤).In fact, Σ5 − Th(RibT,≤) 6= Σ5 − Th(Rcl,≤).


(RibT,≤) vs. (Rcl,≤): Open Problems

We have shown that

Σ5 − Th(RibT,≤) 6= Σ5 − Th(Rcl,≤).

On the other hand, as mentioned before,

Σ1 − Th(RibT,≤) = Σ1 − Th(Rcl,≤).

OPEN PROBLEM. Up to what level Σk do the theories of (RibT,≤)and (Rcl,≤) agree?

OPEN PROBLEM. Is the partial ordering of all (not necessarily c.e.)ibT-degrees elementarily equivalent to the partial ordering of allcl-degrees.


THANK YOU!


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On the Strongly Bounded Turing Degrees of the Computably