A DNC function that computes no effectively bi-immune set

Transcription

1 A DNC function that computes no effectively bi-immune set Achilles A. Beros Laboratoire d Informatique de Nantes Atlantique, Université de Nantes July 5, 204

2 Standard Definitions Definition f is diagonally non-computable if f (e) φ e (e). A is immune if A contains no infinite c.e. set. A is effectively immune if there is a computable (total) function, g such that W e A W e < g(e). A is bi-immune if A and A are both immune. A is effectively bi-immune (EBI) if A and A are both effectively immune. Note: If g and h witness effective immunity for A and A, respectively, then the function defined by k(x) = max{g(x), h(x)} witness effective immunity for both of them.

3 The Question and Answer In 989, Carl Jockusch proved that every DNC function computes an effectively immune set. In 202, Carl Jockusch and Andrew Lewis-Pye proved that every DNC function computes a bi-immune set. Can these results be combined? Jockusch and Lewis-Pye pose this question in their paper: Does every DNC function compute an effectively bi-immune set?

4 The Question and Answer In 989, Carl Jockusch proved that every DNC function computes an effectively immune set. In 202, Carl Jockusch and Andrew Lewis-Pye proved that every DNC function computes a bi-immune set. Can these results be combined? Jockusch and Lewis-Pye pose this question in their paper: Does every DNC function compute an effectively bi-immune set? The answer: No.

5 Results Theorem (Beros) There is a DNC function that computes no effectively bi-immune set. Corollary There is a 0 2 tree every branch of which is a DNC function that computes no effectively bi-immune set. Theorem (Kučera) Every 0 2 DNC degree computes a promptly simple set. Corollary There is a DNC function that computes no effectively bi-immune set and bounds a promptly simple set.

6 Overview Producing a DNC function is very easy: build a tree such that every level is a branching level.

7 Overview Producing a DNC function is very easy: build a tree such that every level is a branching level. For h, a branch of such a tree, to compute no EBI we must guarantee that, for every e N and for every computable function h, there is a c.e. set W a such that W a > h(a) and either W a W h e or W a W h e. This is the requirement for e and h.

8 Overview Producing a DNC function is very easy: build a tree such that every level is a branching level. For h, a branch of such a tree, to compute no EBI we must guarantee that, for every e N and for every computable function h, there is a c.e. set W a such that W a > h(a) and either W a W h e or W a W h e. This is the requirement for e and h. If the current outcome of a requirement is that there is a W a W f e, then the outcome can only be injured by a higher priority requirement as the computation is already complete.

9 Overview Producing a DNC function is very easy: build a tree such that every level is a branching level. For h, a branch of such a tree, to compute no EBI we must guarantee that, for every e N and for every computable function h, there is a c.e. set W a such that W a > h(a) and either W a W h e or W a W h e. This is the requirement for e and h. If the current outcome of a requirement is that there is a W a W f e, then the outcome can only be injured by a higher priority requirement as the computation is already complete. If the current outcome is that there is a W a W f e, then the outcome requires preservation from damage by lower priority requirements.

10 Overview Each requirement is assigned a set of roots. The roots can change a finite number of times, but they will always form a maximal antichain. The requirement can have different outcomes for each root.

11 Overview Each requirement is assigned a set of roots. The roots can change a finite number of times, but they will always form a maximal antichain. The requirement can have different outcomes for each root. The roots along a branch of the tree will adhere to the priority order and the roots of different requirements are disjoint. Root nodes for 3 requirements: Root of the 0 th requirement Root of the st requirement Root of the 2 nd requirement

12 Fix a root, σ, for the e, h requirement. For a finite set S we say σ accepts S for e if S W σ e. Conversely, σ preserves S for e if S W σ e.

13 Fix a root, σ, for the e, h requirement. For a finite set S we say σ accepts S for e if S W σ e. Conversely, σ preserves S for e if S W σ e. If W is a finite family, then a selection of W is a set that consists of one member from each member of W.

14 Fix a root, σ, for the e, h requirement. For a finite set S we say σ accepts S for e if S W σ e. Conversely, σ preserves S for e if S W σ e. If W is a finite family, then a selection of W is a set that consists of one member from each member of W. σ is -bad relative to W, m, e if σ accepts a selection of W for e. σ is (n + )-bad relative to W, m, e if σ has m immediate extensions that are n-bad relative to W, m, e.

15 n-bad for m = 3: A tree.

16 n-bad for m = 3: Suppose these nodes are -bad.

17 n-bad for m = 3: Three nodes are 2-bad

18 n-bad for m = 3: One node is 3-bad

19 n-bad for m = 3: The root is 4-bad

20 A diagonalization point, d, is a code such that we can control the enumeration of W d (found using the recursion theorem).

21 A diagonalization point, d, is a code such that we can control the enumeration of W d (found using the recursion theorem). Let a N be a diagonalization point. Wait for h(a).

22 A diagonalization point, d, is a code such that we can control the enumeration of W d (found using the recursion theorem). Let a N be a diagonalization point. Wait for h(a). Pick further diagonalization points b 0,..., b h(a). Wait for h(b i ).

23 A diagonalization point, d, is a code such that we can control the enumeration of W d (found using the recursion theorem). Let a N be a diagonalization point. Wait for h(a). Pick further diagonalization points b 0,..., b h(a). Wait for h(b i ). Choose disjoint subsets of W σ e of size max{h(b i ) : i h(a)}. Each W bi is set equal to one of the sets for i h(a).

24 A diagonalization point, d, is a code such that we can control the enumeration of W d (found using the recursion theorem). Let a N be a diagonalization point. Wait for h(a). Pick further diagonalization points b 0,..., b h(a). Wait for h(b i ). Choose disjoint subsets of W σ e of size max{h(b i ) : i h(a)}. Each W bi is set equal to one of the sets for i h(a). Search for m sibling nodes, τ, immediately below σ that are k-bad relative to {W b0,..., W bh(a) }, m, e for some k and some well-chosen large value of m.

25 If we cannot find enough siblings, then we can avoid one of the W bi, preserving it in W σ e.

26 If we cannot find enough siblings, then we can avoid one of the W bi, preserving it in W σ e. If we can, then a lemma guarantees we can choose a highly branching tree below σ with enough redundancy to be safe from updated computations by roots that are prefixes of σ (unless outcomes change) and every branch accepts a particular selection of {W b0,..., W bh(a) }.

27 If we cannot find enough siblings, then we can avoid one of the W bi, preserving it in W σ e. If we can, then a lemma guarantees we can choose a highly branching tree below σ with enough redundancy to be safe from updated computations by roots that are prefixes of σ (unless outcomes change) and every branch accepts a particular selection of {W b0,..., W bh(a) }. Set W a to be the selection of points accepted by every branch. W a = h(a) + and W a is contained in the set computed by e with an oracle extending any branch in the trees.

28 Large Lemma For a finite set of finite sets, W, a string σ and e, k, y N, either:. there is a y-branching tree of depth k below σ, or 2. there are m = (y ) ( Π A W A ) + that are i-bad relative to W, m, e for i k.

29 Large Lemma For a finite set of finite sets, W, a string σ and e, k, y N, either:. there is a y-branching tree of depth k below σ, or 2. there are m = (y ) ( Π A W A ) + that are i-bad relative to W, m, e for i k. This allows us to predict the maximum amount of injury that higher priority requirements that are preserving a set can do to lower priority requirements that have accepted a set.

30 Large Lemma For a finite set of finite sets, W, a string σ and e, k, y N, either:. there is a y-branching tree of depth k below σ, or 2. there are m = (y ) ( Π A W A ) + that are i-bad relative to W, m, e for i k. This allows us to predict the maximum amount of injury that higher priority requirements that are preserving a set can do to lower priority requirements that have accepted a set. Using this, the strategy does not accept unless it has enough branching to be unharmed by the maximum amount of injury all higher priority requirements can inflict (before changing outcome).

31 Thank You