On the Strongly Bounded Turing Degrees of the Computably
Transcript of 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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 2 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 3 / 29
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).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 4 / 29
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 )
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 5 / 29
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).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 6 / 29
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+.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 7 / 29
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.)
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 8 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 9 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 10 / 29
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)).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 11 / 29
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).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 12 / 29
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?
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 13 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 15 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 16 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 17 / 29
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.
•
• ••
•
•
••
•
•
66666666666
�����������
�����������
66666666666 66666666666
|||||||||
�����������
BBBBBBBBB
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 ...
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 18 / 29
A Nonembeddable Lattice for (RT,≤)
THEOREM (Lachlan and Soare 1980). The lattice S8 cannot beembedded into (RT,≤).
•
•• •
•
• •
•
::::::::::
����������
����������
::::::::::
8888888
�������
�������
8888888
(I.e., for any embedding of the M5 into the c.e. Turing degrees, the top isnonbranching.)
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 19 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 20 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 21 / 29
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,≤).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 22 / 29
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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 23 / 29
(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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 24 / 29
(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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 25 / 29
(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)
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 26 / 29
(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,≤).
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 27 / 29
(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.
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 28 / 29
THANK YOU!
Klaus Ambos-Spies (Heidelberg) Strongly Bounded Turing Degrees Novosibirsk 2011 29 / 29