# Topological Complexity and Degrees of Discontinuity by findpdf

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Topological Complexity and Degrees of
Discontinuity

Peter Hertling

Institut für Theoretische Informatik und Mathematik
Universität der Bundeswehr München, Germany

International Conference on
Inﬁnity in Logic and Computation
Cape Town, 3–5 November 2007
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

Introduction
Level of Discontinuity       Tests in Computation Trees    Degrees of Discontinuity

Introduction
Goal
Analyse the discontinuities appearing in computation problems
Level of Discontinuity       Tests in Computation Trees    Degrees of Discontinuity

Introduction
Goal
Analyse the discontinuities appearing in computation problems

Motivation
Discontinuities cause problems when computing real number
functions.
Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

Introduction
Goal
Analyse the discontinuities appearing in computation problems

Motivation
Discontinuities cause problems when computing real number
functions.
• Discontinuities in
• numerical computation: instabilities.
• computational geometry: degenerate conﬁgurations.
Level of Discontinuity             Tests in Computation Trees      Degrees of Discontinuity

Introduction
Goal
Analyse the discontinuities appearing in computation problems

Motivation
Discontinuities cause problems when computing real number
functions.
• Discontinuities in
• numerical computation: instabilities.
• computational geometry: degenerate conﬁgurations.
• In Computable Analysis (Turing machine model, computing
Computable functions are continuous.
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Introduction
Goal
Analyse the discontinuities appearing in computation problems

Motivation
Discontinuities cause problems when computing real number
functions.
• Discontinuities in
• numerical computation: instabilities.
• computational geometry: degenerate conﬁgurations.
• In Computable Analysis (Turing machine model, computing
Computable functions are continuous.
Levels of discontinuity are topological levels of
noncomputability.
Level of Discontinuity     Tests in Computation Trees   Degrees of Discontinuity

Overview

I Level of Discontinuity
= Number of Tests in Continuous Computation Trees
Level of Discontinuity           Tests in Computation Trees         Degrees of Discontinuity

Overview

I Level of Discontinuity
= Number of Tests in Continuous Computation Trees
• Level of Discontinuity
• Hausdorff’s Reducible Sets/Difference Hierarchy
• Schwarz Genus
Level of Discontinuity           Tests in Computation Trees       Degrees of Discontinuity

Overview

I Level of Discontinuity
= Number of Tests in Continuous Computation Trees
• Level of Discontinuity
• Hausdorff’s Reducible Sets/Difference Hierarchy
• Schwarz Genus
• The Number of Tests in Computation Trees
• Degenerate Conﬁgurations in Computational Geometry
• Problems from Algebraic Topology and from Algebraic
Complexity Theory
• The Topological Complexity of Zero Finding for Continuous
Functions in Various Settings
Level of Discontinuity           Tests in Computation Trees       Degrees of Discontinuity

Overview

I Level of Discontinuity
= Number of Tests in Continuous Computation Trees
• Level of Discontinuity
• Hausdorff’s Reducible Sets/Difference Hierarchy
• Schwarz Genus
• The Number of Tests in Computation Trees
• Degenerate Conﬁgurations in Computational Geometry
• Problems from Algebraic Topology and from Algebraic
Complexity Theory
• The Topological Complexity of Zero Finding for Continuous
Functions in Various Settings
II Continuous Reducibility of Functions
→ Reﬁnement of the Level of Discontinuity
• Degrees of Discontinuity
Level of Discontinuity              Tests in Computation Trees   Degrees of Discontinuity

Level of Discontinuity of a Function
Level of Discontinuity              Tests in Computation Trees   Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =   {(0, 0)}
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =   {(0, 0)}
C3 (f ) =
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =   {(0, 0)}
C3 (f ) =   ∅
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =   {(0, 0)}
C3 (f ) =   ∅

Lev(f ) := min{α | Cα (f ) = ∅},
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) =   R2
C1 (f ) =   {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) =   {(0, 0)}
C3 (f ) =   ∅
Lev(f ) =

Lev(f ) := min{α | Cα (f ) = ∅},
Level of Discontinuity              Tests in Computation Trees            Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) = R2
C1 (f ) = {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) = {(0, 0)}
C3 (f ) = ∅
Lev(f ) = 3

Lev(f ) := min{α | Cα (f ) = ∅},
Level of Discontinuity              Tests in Computation Trees               Degrees of Discontinuity

Level of Discontinuity of a Function
Let X , Y be topological spaces, f :⊆ X → Y a (possibly partial)
function.
Cα (f ) := dom(f ) ∩             {x ∈ Cβ (f ) | f |Cβ (f ) is discontinuous in x}
β<α
Example

C0 (f ) = R2
C1 (f ) = {(x, y ) ∈ R2 |
y = 0 ∨ y > 0 ∧ x = 0}
C2 (f ) = {(0, 0)}
C3 (f ) = ∅
Lev(f ) = 3

Lev(f ) := min{α | Cα (f ) = ∅},                lev(f , x) := min{α | x ∈ Cα (f )}.
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

Properties
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Properties

• If Lev(f ) ≥ α then Lev(f ) = α + Lev(f |Cα (f ) ).
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Properties

• If Lev(f ) ≥ α then Lev(f ) = α + Lev(f |Cα (f ) ).
• Lev(f ◦ g) ≤ Lev(f ) · Lev(g).
Level of Discontinuity          Tests in Computation Trees   Degrees of Discontinuity

Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations
Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations

A function δ :⊆ Σω → X is an admissible representation
iff δ is surjective, continuous, and for every cont. f :⊆ Σω → X
there is some cont. h :⊆ Σω → Σω with f = δ ◦ h
(use adm. repr. of X for computations over X via oracle Turing
machines).
Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations

A function δ :⊆ Σω → X is an admissible representation
iff δ is surjective, continuous, and for every cont. f :⊆ Σω → X
there is some cont. h :⊆ Σω → Σω with f = δ ◦ h
(use adm. repr. of X for computations over X via oracle Turing
machines).
Let X , Y be T0 -spaces with countable bases, δX :⊆ Σω → X
and δY :⊆ Σω → Y admissible representations of X and Y .
Level of Discontinuity                 Tests in Computation Trees                Degrees of Discontinuity

• The level is invariant under admissible representations

A function δ :⊆ Σω → X is an admissible representation
iff δ is surjective, continuous, and for every cont. f :⊆ Σω → X
there is some cont. h :⊆ Σω → Σω with f = δ ◦ h
(use adm. repr. of X for computations over X via oracle Turing
machines).
Let X , Y be T0 -spaces with countable bases, δX :⊆ Σω → X
and δY :⊆ Σω → Y admissible representations of X and Y .

F
A function F :⊆ Σω → Σω real-                            Σω      −→   Σω
izes f if for all p ∈ dom(f δX )
δX ↓            ↓ δY
f δX (p) = δY F (p).
X   −→   Y
f
Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations
Level of Discontinuity           Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations

Theorem
Level of Discontinuity            Tests in Computation Trees     Degrees of Discontinuity

• The level is invariant under admissible representations

Theorem
• (global)
Lev(f ) ≤ α ⇐⇒ there exists F realizing f with Lev(F ) ≤ α.
Level of Discontinuity             Tests in Computation Trees    Degrees of Discontinuity

• The level is invariant under admissible representations

Theorem
• (global)
Lev(f ) ≤ α ⇐⇒ there exists F realizing f with Lev(F ) ≤ α.
• (local)
For every function F realizing f and all α,
Cα (f ) ⊆ δX (Cα (F )).
Level of Discontinuity              Tests in Computation Trees       Degrees of Discontinuity

• The level is invariant under admissible representations

Theorem
• (global)
Lev(f ) ≤ α ⇐⇒ there exists F realizing f with Lev(F ) ≤ α.
• (local)
For every function F realizing f and all α,
Cα (f ) ⊆ δX (Cα (F )).
If Lev(f ) is deﬁned then there exists a function F realizing f
with Cα (f ) = δX (Cα (F )) for all α.
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Level of Discontinuity              Tests in Computation Trees              Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Let X be a topological space, M ⊆ X . Hausdorff’s residues:

M
                                       if α = 0
Rα (M) :=       R (M) ∩ Rβ (M) \ Rβ (M) if α = β + 1
 β
β<α Rβ (M)             if α is a limit number.

Level of Discontinuity              Tests in Computation Trees              Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Let X be a topological space, M ⊆ X . Hausdorff’s residues:

M
                                       if α = 0
Rα (M) :=       R (M) ∩ Rβ (M) \ Rβ (M) if α = β + 1
 β
β<α Rβ (M)             if α is a limit number.


Lemma
Rα (M) = M ∩ C2·α (cfM ).
Level of Discontinuity              Tests in Computation Trees              Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Let X be a topological space, M ⊆ X . Hausdorff’s residues:

M
                                       if α = 0
Rα (M) :=       R (M) ∩ Rβ (M) \ Rβ (M) if α = β + 1
 β
β<α Rβ (M)             if α is a limit number.


Lemma
Rα (M) = M ∩ C2·α (cfM ).

Hausdorff called a set M reducible if there exists an α with
Rα (M) = ∅.
Level of Discontinuity              Tests in Computation Trees              Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Let X be a topological space, M ⊆ X . Hausdorff’s residues:

M
                                       if α = 0
Rα (M) :=       R (M) ∩ Rβ (M) \ Rβ (M) if α = β + 1
 β
β<α Rβ (M)             if α is a limit number.


Lemma
Rα (M) = M ∩ C2·α (cfM ).

Hausdorff called a set M reducible if there exists an α with
Rα (M) = ∅.
Equivalent to: Lev(cfM ) is deﬁned.
Level of Discontinuity              Tests in Computation Trees              Degrees of Discontinuity

Hausdorff’s Reducible Sets/Difference Hierarchy
Let X be a topological space, M ⊆ X . Hausdorff’s residues:

M
                                       if α = 0
Rα (M) :=       R (M) ∩ Rβ (M) \ Rβ (M) if α = β + 1
 β
β<α Rβ (M)             if α is a limit number.


Lemma
Rα (M) = M ∩ C2·α (cfM ).

Hausdorff called a set M reducible if there exists an α with
Rα (M) = ∅.
Equivalent to: Lev(cfM ) is deﬁned.
For subsets M of a Polish space:
M is in Fσ ∩ Gδ ⇐⇒ M is reducible ⇐⇒ Lev(cfM ) is deﬁned.
Level of Discontinuity         Tests in Computation Trees      Degrees of Discontinuity

The Schwarz Genus
Let p : X → Y be continuous and surjective. A function
s :⊆ Y → X is a section if p(s(y )) = y for all y ∈ dom(s).
Level of Discontinuity             Tests in Computation Trees    Degrees of Discontinuity

The Schwarz Genus
Let p : X → Y be continuous and surjective. A function
s :⊆ Y → X is a section if p(s(y )) = y for all y ∈ dom(s).

minLev(p) := min{Lev(s) | s : Y → X is a total section},
g(p) := min{k | (∃ open V1 , . . . , Vk ⊆ Y ) ( k Vi = Y ∧
i=1
(∀i ≤ k ) (∃ cont. section si : Vi → X ))}.
g(p) is called Schwarz genus.
Level of Discontinuity             Tests in Computation Trees     Degrees of Discontinuity

The Schwarz Genus
Let p : X → Y be continuous and surjective. A function
s :⊆ Y → X is a section if p(s(y )) = y for all y ∈ dom(s).

minLev(p) := min{Lev(s) | s : Y → X is a total section},
g(p) := min{k | (∃ open V1 , . . . , Vk ⊆ Y ) ( k Vi = Y ∧
i=1
(∀i ≤ k ) (∃ cont. section si : Vi → X ))}.
g(p) is called Schwarz genus.

Proposition
• Always minLev(p) ≤ g(p).
• Let Y be a locally connected metric space and p : X → Y
a covering map. If g(p) or minLev(p) is ﬁnite, then

g(p) = minLev(p).
Level of Discontinuity      Tests in Computation Trees   Degrees of Discontinuity

The Number of Tests in Computation Trees
Level of Discontinuity       Tests in Computation Trees     Degrees of Discontinuity

The Number of Tests in Computation Trees
A Computation Tree over the real numbers is a tree containing
unary and binary nodes where
Level of Discontinuity                     Tests in Computation Trees        Degrees of Discontinuity

The Number of Tests in Computation Trees
A Computation Tree over the real numbers is a tree containing
unary and binary nodes where
• unary nodes contain operations of the form
“xi := f (xi1 , . . . , xik )” where
• xi1 , . . . , xik : values computed earlier,
• xi : value computed in the current node,
• f : some arithmetic operation, often only constants,
id, +, −, ∗, /, also exp, log, | · |, and others,
• binary nodes contain tests of the form “xi ◦ 0” with
◦ ∈ {>, <, ≥, ≤},
• the last node (leaf) on each path is unary.
Level of Discontinuity                     Tests in Computation Trees        Degrees of Discontinuity

The Number of Tests in Computation Trees
A Computation Tree over the real numbers is a tree containing
unary and binary nodes where
• unary nodes contain operations of the form
“xi := f (xi1 , . . . , xik )” where
• xi1 , . . . , xik : values computed earlier,
• xi : value computed in the current node,
• f : some arithmetic operation, often only constants,
id, +, −, ∗, /, also exp, log, | · |, and others,
• binary nodes contain tests of the form “xi ◦ 0” with
◦ ∈ {>, <, ≥, ≤},
• the last node (leaf) on each path is unary.
The computation
Level of Discontinuity                     Tests in Computation Trees        Degrees of Discontinuity

The Number of Tests in Computation Trees
A Computation Tree over the real numbers is a tree containing
unary and binary nodes where
• unary nodes contain operations of the form
“xi := f (xi1 , . . . , xik )” where
• xi1 , . . . , xik : values computed earlier,
• xi : value computed in the current node,
• f : some arithmetic operation, often only constants,
id, +, −, ∗, /, also exp, log, | · |, and others,
• binary nodes contain tests of the form “xi ◦ 0” with
◦ ∈ {>, <, ≥, ≤},
• the last node (leaf) on each path is unary.
The computation
• starts in the root with the input in some registers x1 , . . . , xn .
• ends if a leaf is reached. Then the value in that leaf is the
result of the computation.
Level of Discontinuity      Tests in Computation Trees   Degrees of Discontinuity

Computation trees over the real numbers are the common
computation model in Computational Geometry.
Level of Discontinuity        Tests in Computation Trees    Degrees of Discontinuity

Computation trees over the real numbers are the common
computation model in Computational Geometry.

A complexity theory has been built on this model by Blum,
Shub and Smale (1989).
Level of Discontinuity        Tests in Computation Trees        Degrees of Discontinuity

Computation trees over the real numbers are the common
computation model in Computational Geometry.

A complexity theory has been built on this model by Blum,
Shub and Smale (1989).

In practice problematic: the comparisons “xi ≥ 0”, etc
(unstable).
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

Continuous Computation Trees
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
Level of Discontinuity            Tests in Computation Trees   Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.
Level of Discontinuity            Tests in Computation Trees   Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.

Let T be a continuous computation tree.
Level of Discontinuity             Tests in Computation Trees   Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.

Let T be a continuous computation tree.

Size(T ) := the number of its leaves
=   1 + the number of comparison nodes
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.

Let T be a continuous computation tree.

Size(T ) := the number of its leaves
=   1 + the number of comparison nodes
size(T , x) := the number of leaves that can be reached
if x may be disturbed slightly at the beginning
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.

Let T be a continuous computation tree.

Size(T ) := the number of its leaves
=   1 + the number of comparison nodes
size(T , x) := the number of leaves that can be reached
if x may be disturbed slightly at the beginning
size (T , x) := the number of leaves that can be reached
if x may be disturbed slightly at any time
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

Continuous Computation Trees
Similar to computation trees as above but
• over arbitrary topological spaces X , Y ,
• with arbitrary continuous operations,
• tests of the form “x ∈ O” for open sets O.

Let T be a continuous computation tree.

Size(T ) := the number of its leaves
=   1 + the number of comparison nodes
size(T , x) := the number of leaves that can be reached
if x may be disturbed slightly at the beginning
size (T , x) := the number of leaves that can be reached
if x may be disturbed slightly at any time

Clear: size(T , x) ≤ size (T , x) ≤ Size(T )
Level of Discontinuity        Tests in Computation Trees     Degrees of Discontinuity

For a continuous computation tree T let fT be the function
computed by T .
Level of Discontinuity             Tests in Computation Trees       Degrees of Discontinuity

For a continuous computation tree T let fT be the function
computed by T .

Theorem
For any continuous computation tree T

lev(fT , x) ≤ size(T , x) ≤ size (T , x)

for all x ∈ X and
Lev(fT ) ≤ Size(T ).
Level of Discontinuity             Tests in Computation Trees       Degrees of Discontinuity

For a continuous computation tree T let fT be the function
computed by T .

Theorem
For any continuous computation tree T

lev(fT , x) ≤ size(T , x) ≤ size (T , x)

for all x ∈ X and
Lev(fT ) ≤ Size(T ).

Theorem
Let f be a function with Lev(f ) < ω. Then there exists a
continuous computation tree T with

lev(fT , x) = size(T , x) = size (T , x)

for all x ∈ X and
Lev(fT ) = Size(T ).
Level of Discontinuity    Tests in Computation Trees   Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry
Level of Discontinuity        Tests in Computation Trees   Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Level of Discontinuity        Tests in Computation Trees   Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
inherent                     algorithm-induced
and
degeneracies x               degeneracies x
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
inherent                     algorithm-induced
and
degeneracies x               degeneracies x

Can be modeled by:
Level of Discontinuity         Tests in Computation Trees          Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
inherent                     algorithm-induced
and
degeneracies x               degeneracies x

Can be modeled by:

lev(f , x) > 1   and         size(T , x) > 1 or size (T , x) > 1
Level of Discontinuity         Tests in Computation Trees          Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
inherent                     algorithm-induced
and
degeneracies x               degeneracies x

Can be modeled by:

lev(f , x) > 1   and         size(T , x) > 1 or size (T , x) > 1

Yap: “It seems that induced degeneracies subsume inherent
degeneracies”.
Level of Discontinuity          Tests in Computation Trees           Degrees of Discontinuity

Degenerate Conﬁgurations in Computational
Geometry

Let f be a geometric function to be computed.
Yap (1990) distinguishes between
inherent                      algorithm-induced
and
degeneracies x                degeneracies x

Can be modeled by:

lev(f , x) > 1    and         size(T , x) > 1 or size (T , x) > 1

Yap: “It seems that induced degeneracies subsume inherent
degeneracies”.

Our Theorem:      lev(fT , x) ≤ size(T , x) ≤ size (T , x).
Level of Discontinuity       Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity
of a problem P over the real numbers:
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity
of a problem P over the real numbers:
comptotal (P)
ARI
:= minARI−trees T Size(T ) − 1
Level of Discontinuity       Tests in Computation Trees    Degrees of Discontinuity

Topological Complexity
of a problem P over the real numbers:
comptotal (P)
ARI
:= minARI−trees T Size(T ) − 1
= the minimum of the total number of comparisons in the
tree,
where the minimum is taken over all computation trees that
solve the problem using a certain set ARI of operations.
Level of Discontinuity        Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity
of a problem P over the real numbers:
comptotal (P)
ARI
:= minARI−trees T Size(T ) − 1
= the minimum of the total number of comparisons in the
tree,
where the minimum is taken over all computation trees that
solve the problem using a certain set ARI of operations.

Another variant:
Level of Discontinuity        Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity
of a problem P over the real numbers:
comptotal (P)
ARI
:= minARI−trees T Size(T ) − 1
= the minimum of the total number of comparisons in the
tree,
where the minimum is taken over all computation trees that
solve the problem using a certain set ARI of operations.

Another variant:
path
compARI (P)
:= the minimum of the maximum number of comparisons on
a computation path in the tree,
where the minimum is taken over all computation trees that
solve the problem using a certain set ARI of operations.
Level of Discontinuity               Tests in Computation Trees      Degrees of Discontinuity

For all computation trees:
path
compARI (P) ≥ log2 (comptotal (P)) + 1) .
ARI
Level of Discontinuity               Tests in Computation Trees      Degrees of Discontinuity

For all computation trees:
path
compARI (P) ≥ log2 (comptotal (P)) + 1) .
ARI

In fact, trees with ARI = {cont. op.} or ARI = {+, −, ∗, /, | · |}
path
compARI (P) = log2 (comptotal (P)) + 1)
ARI
Level of Discontinuity               Tests in Computation Trees      Degrees of Discontinuity

For all computation trees:
path
compARI (P) ≥ log2 (comptotal (P)) + 1) .
ARI

In fact, trees with ARI = {cont. op.} or ARI = {+, −, ∗, /, | · |}
path
compARI (P) = log2 (comptotal (P)) + 1)
ARI

Hence, in these cases, comptotal (P) is ﬁner and therefore more
ARI
interesting!
X ! Computation Trees
and their disjunction t1 _ t2 : Tests in fTRUE FALSEg are also total topological tests.
Level of Discontinuity                                                          Degrees of Discontinuity
Lemma 2.10 Let t1 t2 : X ! fTRUE FALSEg be total topological tests and T1, T2, T3
CCTs. Then the two CCTs in Figure 1 are equivalent to each other and the two CCTs in
Figure 2 are via:
equivalent to
Balancing at each point. each other. Furthermore, in both cases the branching number
is the same

?
H                                               ?         H
; H t1 HH +                      ()                     ; H t1 ^ tHH +
?       HH         ?
H                               ? HH               2
?
T1             ; H t2 HH +                     ; H tHHH +                       T3
?      H H             ?           ? HH1       ?
T2                     T3          T1                     T2

Figure 1: Equivalent CCTs, Part 1
Figure: Equivalent computation trees, Part 1

?
H          ()
and similarly in the other direction.                           ? H
; H t1 HH +                        ; H t1 _ tHH +
?
H      HH               ?           ?      HH     2
?
; H t2 HH +                 T3          T1          ; H tHHH +
?       HH             ?                              ? HH1        ?
T1                   T2                                 T2                      T3
X ! Computation Trees
and their disjunction t1 _ t2 : Tests in fTRUE FALSEg are also total topological tests.
Level of Discontinuity                                                          Degrees of Discontinuity
Lemma 2.10 Let t1 t2 : X ! fTRUE FALSEg be total topological tests and T1, T2, T3
CCTs. Then the two CCTs in Figure 1 are equivalent to each other and the two CCTs in
Figure 2 are via:
equivalent to
Balancing at each point. each other. Furthermore, in both cases the branching number
is the same

?
H                                               ?          H
; H t1 HH +                      ()                      ; H t1 ^ tHH +
?       HH         ?
H                               ? HH                2
?
T1             ; H t2 HH +                     ; H tHHH +                       T3
?      H H             ?           ? HH1       ?
T2                     T3          T1                     T2

Figure 1: Equivalent CCTs, Part 1
Figure: Equivalent computation trees, Part 1

?
H          ()
and similarly in the other direction.                H             ?
; H t1 HH +               ; H t1 _ tHH +
? HH
Possible for computation trees ?
HHH +                     ? HH 2
using the operations     ?
H
{+, −, ∗, /, | · |} because with |T3| one can compute;min and H +
; H t2
?       HH             ?
·        T1             H t1 H max.
?      HH               ?
T1                   T2                                 T2                      T3
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Level of Discontinuity       Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
Level of Discontinuity               Tests in Computation Trees   Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Level of Discontinuity             Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Level of Discontinuity             Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
Level of Discontinuity               Tests in Computation Trees        Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Level of Discontinuity             Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Level of Discontinuity             Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: n − 1, even if arbitrary cont. op.’s are allowed.
Level of Discontinuity              Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: n − 1, even if arbitrary cont. op.’s are allowed.
3. Input: a vector (x1 , . . . , xn ) ∈ Rn with xi = xj for i = j.
Level of Discontinuity              Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: n − 1, even if arbitrary cont. op.’s are allowed.
3. Input: a vector (x1 , . . . , xn ) ∈ Rn with xi = xj for i = j.
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Level of Discontinuity              Tests in Computation Trees          Degrees of Discontinuity

Topological Complexity of Problems over the Reals:
Examples
Sorting real numbers. Three versions:
1. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a vector (xπ(1) , . . . , xπ(n) ) ∈ Rn where π is a
permutation of {1, . . . , n} such that xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
2. Input: a vector (x1 , . . . , xn ) ∈ Rn .
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: n − 1, even if arbitrary cont. op.’s are allowed.
3. Input: a vector (x1 , . . . , xn ) ∈ Rn with xi = xj for i = j.
Output: a permutation π of {1, . . . , n} such that
xπ(1) ≤ . . . ≤ xπ(n) .
Top. Compl.: 0, if arbitrary cont. op.’s are allowed.
Level of Discontinuity      Tests in Computation Trees   Degrees of Discontinuity

Algebraic Topology
Level of Discontinuity        Tests in Computation Trees      Degrees of Discontinuity

Algebraic Topology

Smale (1987) invented the notion topological complexity and
asked for the topological complexity of the following problem:
Level of Discontinuity          Tests in Computation Trees        Degrees of Discontinuity

Algebraic Topology

Smale (1987) invented the notion topological complexity and
asked for the topological complexity of the following problem:
Input: a number δ > 0 and a vector (an−1 , . . . , a0 ) ∈ Cn such
that the polynomial p(z) := z n + an−1 · z n−1 + . . . + a1 · z + a0
has n pairwise different complex zeros.
Level of Discontinuity          Tests in Computation Trees        Degrees of Discontinuity

Algebraic Topology

Smale (1987) invented the notion topological complexity and
asked for the topological complexity of the following problem:
Input: a number δ > 0 and a vector (an−1 , . . . , a0 ) ∈ Cn such
that the polynomial p(z) := z n + an−1 · z n−1 + . . . + a1 · z + a0
has n pairwise different complex zeros.
Output: a vector (ζ1 , . . . , ζn ) ∈ Cn approximating the zeros
z1 , . . . , zn of p in the sense |ζi − zi | ≤ δ for all i.
Level of Discontinuity          Tests in Computation Trees        Degrees of Discontinuity

Algebraic Topology

Smale (1987) invented the notion topological complexity and
asked for the topological complexity of the following problem:
Input: a number δ > 0 and a vector (an−1 , . . . , a0 ) ∈ Cn such
that the polynomial p(z) := z n + an−1 · z n−1 + . . . + a1 · z + a0
has n pairwise different complex zeros.
Output: a vector (ζ1 , . . . , ζn ) ∈ Cn approximating the zeros
z1 , . . . , zn of p in the sense |ζi − zi | ≤ δ for all i.

Comment:       The top.
compl. of Smale’s prob-
lem for sufﬁciently small
δ is equal to the top.
compl. of the problem to
sort n pairwise different
complex numbers.
Level of Discontinuity          Tests in Computation Trees                 Degrees of Discontinuity

Algebraic Topology

Smale (1987) invented the notion topological complexity and
asked for the topological complexity of the following problem:
Input: a number δ > 0 and a vector (an−1 , . . . , a0 ) ∈ Cn such
that the polynomial p(z) := z n + an−1 · z n−1 + . . . + a1 · z + a0
has n pairwise different complex zeros.
Output: a vector (ζ1 , . . . , ζn ) ∈ Cn approximating the zeros
z1 , . . . , zn of p in the sense |ζi − zi | ≤ δ for all i.

Comment:       The top.
compl. of Smale’s prob-
Cn
lem for sufﬁciently small
πn              d   pn
δ is equal to the top.                                      d
compl. of the problem to                 ©
∼
=

d
sort n pairwise different              Pn '                    Cn /Sn
ρn
complex numbers.
Level of Discontinuity     Tests in Computation Trees   Degrees of Discontinuity

Theorem (Vassiliev 1988)
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Theorem (Vassiliev 1988)
Let

Dp (n) := the sum of the digits of n, written
in base p, for a prime number p,
U(n) := n + 1 − min{Dp (n) | primes p}.

Then for all sufﬁciently small δ > 0:

U(n) − 1 ≤ comptotal op. (Smale’s problem) ≤ n − 1
cont.
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Theorem (Vassiliev 1988)
Let

Dp (n) := the sum of the digits of n, written
in base p, for a prime number p,
U(n) := n + 1 − min{Dp (n) | primes p}.

Then for all sufﬁciently small δ > 0:

U(n) − 1 ≤ comptotal op. (Smale’s problem) ≤ n − 1
cont.

Remark: If n is a prime power then U(n) = n, hence, the top.
compl. of Smale’s problem is n − 1.
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Theorem (Vassiliev 1988)
Let

Dp (n) := the sum of the digits of n, written
in base p, for a prime number p,
U(n) := n + 1 − min{Dp (n) | primes p}.

Then for all sufﬁciently small δ > 0:

U(n) − 1 ≤ comptotal op. (Smale’s problem) ≤ n − 1
cont.

Remark: If n is a prime power then U(n) = n, hence, the top.
compl. of Smale’s problem is n − 1.
Question: What is top. complexity of Smale’s problem if n is not
a prime power?
Level of Discontinuity        Tests in Computation Trees     Degrees of Discontinuity

Question: What is the top. complexity of Smale’s problem if n is
not a prime power? Is it then n − 1 as well?
Level of Discontinuity            Tests in Computation Trees        Degrees of Discontinuity

Question: What is the top. complexity of Smale’s problem if n is
not a prime power? Is it then n − 1 as well?

Theorem (De Concini et al 2003)
• For n = 6 it is not 5 but 4,
• If n = 3 · 2m , for some positive integer m, then it is also
< n − 1.
Level of Discontinuity      Tests in Computation Trees   Degrees of Discontinuity

Another Example from Algebraic Topology
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.

Borsuk’s Antipodal Theorem
Level of Discontinuity         Tests in Computation Trees     Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.

Borsuk’s Antipodal Theorem
For n ≥ 1, for every continuous function f : S n → Rn there
exists at least one point x ∈ S n with f (−x) = f (x).
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.

Borsuk’s Antipodal Theorem
For n ≥ 1, for every continuous function f : S n → Rn there
exists at least one point x ∈ S n with f (−x) = f (x).

A function f : S n → Rm is called odd if f (−x) = −f (x) for all
x ∈ Sn.
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.

Borsuk’s Antipodal Theorem
For n ≥ 1, for every continuous function f : S n → Rn there
exists at least one point x ∈ S n with f (−x) = f (x).

A function f : S n → Rm is called odd if f (−x) = −f (x) for all
x ∈ Sn.

Easily shown to be
Equivalent to Borsuk’s Antipodal Theorem:
Level of Discontinuity                Tests in Computation Trees         Degrees of Discontinuity

Another Example from Algebraic Topology
S n := {x ∈ Rn+1 | ||x|| = 1}.

Borsuk’s Antipodal Theorem
For n ≥ 1, for every continuous function f : S n → Rn there
exists at least one point x ∈ S n with f (−x) = f (x).

A function f : S n → Rm is called odd if f (−x) = −f (x) for all
x ∈ Sn.

Easily shown to be
Equivalent to Borsuk’s Antipodal Theorem:
For n ≥ 1,

min{Lev(f ) | f : S n → {−1, 1} odd} = n + 1.
Level of Discontinuity    Tests in Computation Trees   Degrees of Discontinuity

Another Example:
Level of Discontinuity      Tests in Computation Trees   Degrees of Discontinuity

Another Example: Problem: compute f := cfRn .
+
Level of Discontinuity            Tests in Computation Trees     Degrees of Discontinuity

Another Example: Problem: compute f := cfRn .
+

Proposition
path
comptotal (f ) = compARI (f ) = 1
ARI
for {+, −, ∗, /, | · |} ⊆ ARI ⊆ {continuous functions}.
Level of Discontinuity               Tests in Computation Trees   Degrees of Discontinuity

Another Example: Problem: compute f := cfRn .
+

Proposition
path
comptotal (f ) = compARI (f ) = 1
ARI
for {+, −, ∗, /, | · |} ⊆ ARI ⊆ {continuous functions}.
• Every computation tree using arbitrary continuous
functions needs at least 1 test.
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Another Example: Problem: compute f := cfRn .
+

Proposition
path
comptotal (f ) = compARI (f ) = 1
ARI
for {+, −, ∗, /, | · |} ⊆ ARI ⊆ {continuous functions}.
• Every computation tree using arbitrary continuous
functions needs at least 1 test.
• There exists a computation tree using only 1 test and the
arithmetic operations +, −, ∗, /, | · |:
test whether min{x1 , . . . , xn } > 0.
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Another Example: Problem: compute f := cfRn .
+

Proposition
path
comptotal (f ) = compARI (f ) = 1
ARI
for {+, −, ∗, /, | · |} ⊆ ARI ⊆ {continuous functions}.
• Every computation tree using arbitrary continuous
functions needs at least 1 test.
• There exists a computation tree using only 1 test and the
arithmetic operations +, −, ∗, /, | · |:
test whether min{x1 , . . . , xn } > 0.

Now with analytic functions .....
Level of Discontinuity       Tests in Computation Trees   Degrees of Discontinuity

Problem: compute f := cfRn .
+
Level of Discontinuity            Tests in Computation Trees     Degrees of Discontinuity

Problem: compute f := cfRn .
+

Theorem
path
comptotal (f ) = compARI (f ) = n for ARI ⊆ {analytic functions}.
ARI
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Problem: compute f := cfRn .
+

Theorem
path
comptotal (f ) = compARI (f ) = n for ARI ⊆ {analytic functions}.
ARI
• There exists a computation tree using only n tests and the
comparisons xi > 0 for i = 1, . . . , n.
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Problem: compute f := cfRn .
+

Theorem
path
comptotal (f ) = compARI (f ) = n for ARI ⊆ {analytic functions}.
ARI
• There exists a computation tree using only n tests and the
comparisons xi > 0 for i = 1, . . . , n.
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Problem: compute f := cfRn .
+

Theorem
path
comptotal (f ) = compARI (f ) = n for ARI ⊆ {analytic functions}.
ARI
• There exists a computation tree using only n tests and the
comparisons xi > 0 for i = 1, . . . , n.
• Rabin’s Theorem (1972) (proof corrected by Montaña et
al(1994)): Every computation tree using only analytic
functions as arithmetic operations needs at least n
comparisons on the longest path in the tree!
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Problem: compute f := cfRn .
+

Theorem
path
comptotal (f ) = compARI (f ) = n for ARI ⊆ {analytic functions}.
ARI
• There exists a computation tree using only n tests and the
comparisons xi > 0 for i = 1, . . . , n.
• Rabin’s Theorem (1972) (proof corrected by Montaña et
al(1994)): Every computation tree using only analytic
functions as arithmetic operations needs at least n
comparisons on the longest path in the tree!

Note: A short proof for Rabin’s theorem was given by Vassiliev
in 1997 using the Newton polyhedron of a power series.
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings
Level of Discontinuity             Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .

A real number x ∈ [0, 1] is a zero of a function f : [0, 1] → R if
f (x) = 0.
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .

A real number x ∈ [0, 1] is a zero of a function f : [0, 1] → R if
f (x) = 0. For ε > 0, an ε–approximation of a number x ∗ is a
number x with |x − x ∗ | ≤ ε.
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .

A real number x ∈ [0, 1] is a zero of a function f : [0, 1] → R if
f (x) = 0. For ε > 0, an ε–approximation of a number x ∗ is a
number x with |x − x ∗ | ≤ ε.

Problem
Fix some class G ∈ {F , Fnd , Finc } and some ε > 0.
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .

A real number x ∈ [0, 1] is a zero of a function f : [0, 1] → R if
f (x) = 0. For ε > 0, an ε–approximation of a number x ∗ is a
number x with |x − x ∗ | ≤ ε.

Problem
Fix some class G ∈ {F , Fnd , Finc } and some ε > 0.
Input: A function f ∈ G, given: an oracle for function values.
Level of Discontinuity              Tests in Computation Trees         Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

F   := {f ∈ C[0, 1] | f (0) · f (1) < 0} ,
Fnd := {f ∈ F | f is nondecreasing} ,
Finc := {f ∈ F | f is increasing} .

A real number x ∈ [0, 1] is a zero of a function f : [0, 1] → R if
f (x) = 0. For ε > 0, an ε–approximation of a number x ∗ is a
number x with |x − x ∗ | ≤ ε.

Problem
Fix some class G ∈ {F , Fnd , Finc } and some ε > 0.
Input: A function f ∈ G, given: an oracle for function values.
Output: An ε-approximation of a zero of f .
Level of Discontinuity            Tests in Computation Trees      Degrees of Discontinuity

The Topological Complexity of Zero Finding for
Continuous Functions in Various Settings

Results for the three function classes F , Fnd , and Finc and for
different classes ARI of allowed arithmetic operations:
• ARI = {+, −, ∗, /, exp, log} and
ARI = {all operations satisfying a Hölder condition on each
bounded subset of their domain}:
z
[Novak, Wo´ niakowski 1996].
• {+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}:
[H 1996].
• ARI = {+, −, ∗, /}: [H 2002]
Level of Discontinuity   Tests in Computation Trees   Degrees of Discontinuity

Bisection
Level of Discontinuity              Tests in Computation Trees   Degrees of Discontinuity

Bisection

• Maximum number of tests on a path:
(log2 (1/(2ε)) ∼ log2 (1/(2ε)).
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Bisection

• Maximum number of tests on a path:
(log2 (1/(2ε)) ∼ log2 (1/(2ε)).
• Total number of tests: 1/(2ε) − 1
Level of Discontinuity        Tests in Computation Trees    Degrees of Discontinuity

Trisection
An algorithm for computing an ε-approximation of the zero of
f ∈ Finc , using {+, −, ∗, /, | · |} and no tests:
Level of Discontinuity                 Tests in Computation Trees           Degrees of Discontinuity

Trisection
An algorithm for computing an ε-approximation of the zero of
f ∈ Finc , using {+, −, ∗, /, | · |} and no tests:

Start with [a, b] := [0, 1] and repeat the following loop
max{0, − log3/2 (2ε) } times.
Output: midpoint of the last interval [a, b].

begin {loop}
c := (b − a)/3;
for j = 0, . . . , 3 do begin xj := a + j · c;
zj := f (xj ) end;
for j = 1, 2 do rj := max{0, −zj+1 · zj−1 };
x := (x1 r1 + x2 r2 )/(r1 + r2 );
a := x − c; b := x + c
end {loop}
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

Zero Finding for Increasing Functions

Theorem
For G = Finc and 0 < ε < 1/2.
Level of Discontinuity             Tests in Computation Trees       Degrees of Discontinuity

Zero Finding for Increasing Functions

Theorem
For G = Finc and 0 < ε < 1/2.
• For
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}
comptotal (Finc , ε) = 0 .
ARI
Level of Discontinuity             Tests in Computation Trees       Degrees of Discontinuity

Zero Finding for Increasing Functions

Theorem
For G = Finc and 0 < ε < 1/2.
• For
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}
comptotal (Finc , ε) = 0 .
ARI
• [Novak, Wo´ niakowski 1996] For
z
ARI = {+, −, ∗, /, exp, log}
comptotal (Finc , ε) = 0 .
ARI
Level of Discontinuity             Tests in Computation Trees       Degrees of Discontinuity

Zero Finding for Increasing Functions

Theorem
For G = Finc and 0 < ε < 1/2.
• For
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}
comptotal (Finc , ε) = 0 .
ARI
• [Novak, Wo´ niakowski 1996] For
z
ARI = {+, −, ∗, /, exp, log}
comptotal (Finc , ε) = 0 .
ARI
• For ARI = {+, −, ∗, /}
comptotal (Finc , ε) = 1
ARI
Level of Discontinuity         Tests in Computation Trees   Degrees of Discontinuity

Results for comptotal (G, ε)
ARI
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}
G = Finc
G = Fnd
G=F
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}
G = Finc                              0
G = Fnd
G=F
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}
G = Finc                              0
G = Fnd                      log2 (ε−1 + 2) − 2
G=F
Level of Discontinuity                 Tests in Computation Trees       Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}
G = Finc                              0
G = Fnd                      log2 (ε−1 + 2) − 2
G=F                          log2 (ε−1 + 2) − 2
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0
G = Fnd                      log2 (ε−1 + 2) − 2
G=F                          log2 (ε−1 + 2) − 2
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2
G=F                          log2 (ε−1 + 2) − 2
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1

Results for comppath (G, ε):
ARI
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1

Results for comppath (G, ε): ∼ log2 (comptotal (G, ε)).
ARI                      ARI
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1

Results for comppath (G, ε): ∼ log2 (comptotal (G, ε)).
ARI                      ARI

Bisection algorithm:
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1

Results for comppath (G, ε): ∼ log2 (comptotal (G, ε)).
ARI                      ARI

Bisection algorithm: number of tests: 1/(2ε) − 1 ≈ 1/(2ε).
Level of Discontinuity                 Tests in Computation Trees         Degrees of Discontinuity

Results for comptotal (G, ε)
ARI

{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arb. cont. op.}   ARI = {+, −, ∗, /}
G = Finc                              0                              1
G = Fnd                      log2 (ε−1 + 2) − 2                 ∼ log2 (1/ε)
G=F                          log2 (ε−1 + 2) − 2                  1/(2ε) − 1

Results for comppath (G, ε): ∼ log2 (comptotal (G, ε)).
ARI                      ARI

Bisection algorithm: number of tests: 1/(2ε) − 1 ≈ 1/(2ε).

Surprise: There is an exponentially better algorithm with
number of tests only log2 (ε−1 + 2) − 2 ≈ log2 (1/(2ε)).
Level of Discontinuity            Tests in Computation Trees                Degrees of Discontinuity

Test-optimal algorithm
for continuous f : [0, 1] → R with f (0) < 0 < f (1), and
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}:

1         x       0     if x < 0
sig : R \ {0} → {0, 1},      sig(x) :=          · (1 +     )=
2        |x|      1     if x > 0
Level of Discontinuity         Tests in Computation Trees       Degrees of Discontinuity

Test-optimal algorithm
for continuous f : [0, 1] → R with f (0) < 0 < f (1), and
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}:
Level of Discontinuity         Tests in Computation Trees       Degrees of Discontinuity

Test-optimal algorithm
for continuous f : [0, 1] → R with f (0) < 0 < f (1), and
{+, −, ∗, /, | · |} ⊆ ARI ⊆ {arbitrary continuous operations}:
Level of Discontinuity        Tests in Computation Trees   Degrees of Discontinuity

Degrees of Discontinuity
Let V , W , X , Y be topological spaces,
f :⊆ V → W and g :⊆ X → Y be functions.
Level of Discontinuity                 Tests in Computation Trees   Degrees of Discontinuity

Degrees of Discontinuity
Let V , W , X , Y be topological spaces,
f :⊆ V → W and g :⊆ X → Y be functions.
Deﬁnition
f ≤0 g : ⇐⇒    (∃ cont. B) (∀x ∈ dom f )
f (x) = gB(x)
Level of Discontinuity                 Tests in Computation Trees      Degrees of Discontinuity

Degrees of Discontinuity
Let V , W , X , Y be topological spaces,
f :⊆ V → W and g :⊆ X → Y be functions.
Deﬁnition
f ≤0 g : ⇐⇒    (∃ cont. B) (∀x ∈ dom f )
f (x) = gB(x)
f ≤1 g : ⇐⇒    (∃ cont. A, B) (∀x ∈ dom f )
f (x) = AgB(x)
(Weihrauch 1992)
Level of Discontinuity                 Tests in Computation Trees      Degrees of Discontinuity

Degrees of Discontinuity
Let V , W , X , Y be topological spaces,
f :⊆ V → W and g :⊆ X → Y be functions.
Deﬁnition
f ≤0 g : ⇐⇒    (∃ cont. B) (∀x ∈ dom f )
f (x) = gB(x)
f ≤1 g : ⇐⇒    (∃ cont. A, B) (∀x ∈ dom f )
f (x) = AgB(x)
(Weihrauch 1992)
f ≤2 g : ⇐⇒    (∃ cont. A, B) (∀x ∈ dom f )
f (x) = A(x, gB(x))
(Hirsch 1993, Weihrauch 1992)
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

Lemma
• The relations ≤i are reﬂexive and transitive.
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

Lemma
• The relations ≤i are reﬂexive and transitive.
• f ≤0 g ⇒ f ≤1 g and f ≤1 g ⇒ f ≤2 g.
Level of Discontinuity            Tests in Computation Trees   Degrees of Discontinuity

Lemma
• The relations ≤i are reﬂexive and transitive.
• f ≤0 g ⇒ f ≤1 g and f ≤1 g ⇒ f ≤2 g.
• If f ≤i g then Lev(f ) ≤ Lev (g).
Level of Discontinuity         Tests in Computation Trees     Degrees of Discontinuity

Continuous Reductions Directly over R

Theorem
For i = 0, 1 the ≤i -relation is not well-founded on
{F : R → {0, 1} | Lev(f ) = 2}.
Level of Discontinuity               Tests in Computation Trees            Degrees of Discontinuity

Continuous Reductions over Σω

Theorem (H 1993, Selivanov 2007)
• With any function f :⊆ Σω → {0, . . . , k − 1} one can
associate a forest (= set of trees) F(f ) describing the
discontinuities of f and deﬁne a relation ≤0 on trees and
forests so that

f ≤0 g ⇐⇒ F(f ) ≤0 G(g)

for any f , g :⊆ Σω → {0, . . . , k − 1} of level < ω1 .
Level of Discontinuity               Tests in Computation Trees            Degrees of Discontinuity

Continuous Reductions over Σω

Theorem (H 1993, Selivanov 2007)
• With any function f :⊆ Σω → {0, . . . , k − 1} one can
associate a forest (= set of trees) F(f ) describing the
discontinuities of f and deﬁne a relation ≤0 on trees and
forests so that

f ≤0 g ⇐⇒ F(f ) ≤0 G(g)

for any f , g :⊆ Σω → {0, . . . , k − 1} of level < ω1 .
• Similarly for ≤1 and ≤2 for functions of ﬁnite level
(works probably also for functions of level < ω1 : work in
progress).
Level of Discontinuity              Tests in Computation Trees                  Degrees of Discontinuity

Trees Measuring the Discontinuities of Functions
Example
For f : {0, 1}ω → {2, 3, 7} deﬁned by

7 if p = 0ω

f (p) := 3 if (∃i ∈ N) p = 0i 1ω

2 otherwise.


F(f ) =   {   2, 3, 7, 3,                  7,            7           }.
 t                  d
   t                  d
2          2         3   2    3    3

2
Level of Discontinuity           Tests in Computation Trees   Degrees of Discontinuity

The ≤0 Relation on Trees and Forests

For trees T1 and T2 :

T1 ≤0 T2 : ⇐⇒ (∃f : T1 → T2 ) f preserves ancestors
and node values.
Level of Discontinuity              Tests in Computation Trees    Degrees of Discontinuity

The ≤0 Relation on Trees and Forests

For trees T1 and T2 :

T1 ≤0 T2 : ⇐⇒ (∃f : T1 → T2 ) f preserves ancestors
and node values.

For forests F1 and F2 :

F1 ≤0 F2 : ⇐⇒ (∀T1 ∈ F1 ) (∃T2 ∈ F2 ) T1 ≤0 T2 .
Level of Discontinuity           Tests in Computation Trees      Degrees of Discontinuity

The ≤2 Relation on Trees and Forests

For trees T1 and T2 :

T1 ≤2 T2 : ⇐⇒       (∃f : T1 → T2 ) f preserves ancestors and
if two nodes on a path in T1 have different values
then there images have different values as well.
Level of Discontinuity              Tests in Computation Trees      Degrees of Discontinuity

The ≤2 Relation on Trees and Forests

For trees T1 and T2 :

T1 ≤2 T2 : ⇐⇒          (∃f : T1 → T2 ) f preserves ancestors and
if two nodes on a path in T1 have different values
then there images have different values as well.

For forests F1 and F2 :

F1 ≤2 F2 : ⇐⇒ (∀T1 ∈ F1 ) (∃T2 ∈ F2 ) T1 ≤2 T2 .
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Corollary
For any n ∈ N≥2 and k ∈ N, the number of ≡i -classes of
functions f :⊆ Σω → {0, . . . , k − 1} of level ≤ n is ﬁnite.
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Corollary
For any n ∈ N≥2 and k ∈ N, the number of ≡i -classes of
functions f :⊆ Σω → {0, . . . , k − 1} of level ≤ n is ﬁnite.

But for ≤0 and ≤1 it grows quickly for growing k .
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Corollary
For any n ∈ N≥2 and k ∈ N, the number of ≡i -classes of
functions f :⊆ Σω → {0, . . . , k − 1} of level ≤ n is ﬁnite.

But for ≤0 and ≤1 it grows quickly for growing k .

Not so for ≤2 once k ≥ n!
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Corollary
For any n ∈ N≥2 and k ∈ N, the number of ≡i -classes of
functions f :⊆ Σω → {0, . . . , k − 1} of level ≤ n is ﬁnite.

But for ≤0 and ≤1 it grows quickly for growing k .

Not so for ≤2 once k ≥ n!

Theorem
For any n ∈ N, the number of ≡2 -classes of functions f deﬁned
on a subset of Σω and with discrete range of level ≤ n is ﬁnite.
Level of Discontinuity          Tests in Computation Trees       Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Corollary
For any n ∈ N≥2 and k ∈ N, the number of ≡i -classes of
functions f :⊆ Σω → {0, . . . , k − 1} of level ≤ n is ﬁnite.

But for ≤0 and ≤1 it grows quickly for growing k .

Not so for ≤2 once k ≥ n!

Theorem
For any n ∈ N, the number of ≡2 -classes of functions f deﬁned
on a subset of Σω and with discrete range of level ≤ n is ﬁnite.

Work in progress: also for functions f :⊆ Σω → Σω ?
Level of Discontinuity        Tests in Computation Trees      Degrees of Discontinuity

Continuous Degrees: Finite Reﬁnement of the Level

Theorem
With any function f :⊆ Σω → Y , Y an arbitrary discrete space,
one can associate a forest (= set of trees) F (f ) describing the
discontinuities of f and deﬁne a relation ≤2 on trees and forests
so that
f ≤2 g ⇐⇒ F (f ) ≤2 G (g)
for any f :⊆ Σω → Y , f :⊆ Σω → Z (Y , Z discrete spaces) of
level < ω.

Work in progress: also for functions f :⊆ Σω → Σω ?
Level of Discontinuity             Tests in Computation Trees         Degrees of Discontinuity

The Degrees are invariant under admissible
representations

Theorem
Let δ :⊆ Σω → X be an admissible representation,
Y a discrete space,
f :⊆ X → Y a function.
•
F (f ) ≡2 F (f δ).

• Let Y be countable, ν :⊆ {0, 1}∗ → Y be a notation of Y .
• For every (δ, ν)-realization F of f one has F (f ) ≤2 F (F ).
• There is a (δ, ν)-realization F of f with F (f ) ≡2 F (F ).
Level of Discontinuity         Tests in Computation Trees    Degrees of Discontinuity

A Stronger Version of Smale’s Problem

Smale’s Problem
How many tests does an algorithm need which, given a
complex polynomial of degree n with pairwise different zeros,
computes a vector of zeros of the polynomial?
Level of Discontinuity         Tests in Computation Trees     Degrees of Discontinuity

A Stronger Version of Smale’s Problem

Smale’s Problem
How many tests does an algorithm need which, given a
complex polynomial of degree n with pairwise different zeros,
computes a vector of zeros of the polynomial?

Equivalent to:
What is the minimum level of discontinuity of any function
f : {z ∈ Cn | (∀i, j)(i = j ⇒ zi = zj } → Sn
with (∀z ∈ dom(f )) (∀γ ∈ Sn ) f (γz)γz = f (z)z?
Level of Discontinuity         Tests in Computation Trees     Degrees of Discontinuity

A Stronger Version of Smale’s Problem

Smale’s Problem
How many tests does an algorithm need which, given a
complex polynomial of degree n with pairwise different zeros,
computes a vector of zeros of the polynomial?

Equivalent to:
What is the minimum level of discontinuity of any function
f : {z ∈ Cn | (∀i, j)(i = j ⇒ zi = zj } → Sn
with (∀z ∈ dom(f )) (∀γ ∈ Sn ) f (γz)γz = f (z)z?

More difﬁcult problem
Determine the possible ≤2 -minimal forests of such functions!
Level of Discontinuity          Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems
Level of Discontinuity            Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
Level of Discontinuity                Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
• These notions give rise to questions in set theory and
algebraic topology,
Level of Discontinuity                Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
• These notions give rise to questions in set theory and
algebraic topology,
• Number of tests in computation trees with restricted set of
arithmetic operations: algebraic complexity theory.
Level of Discontinuity                Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
• These notions give rise to questions in set theory and
algebraic topology,
• Number of tests in computation trees with restricted set of
arithmetic operations: algebraic complexity theory.
• Can also be analyzed for algorithms in numerical
mathematics (information-based complexity).
Level of Discontinuity                Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
• These notions give rise to questions in set theory and
algebraic topology,
• Number of tests in computation trees with restricted set of
arithmetic operations: algebraic complexity theory.
• Can also be analyzed for algorithms in numerical
mathematics (information-based complexity).
• Continuous reducibilities lead to natural reﬁnements of the
level.
Level of Discontinuity                Tests in Computation Trees   Degrees of Discontinuity

Summary and open problems

• Closely related:
• Level of discontinuity
• Schwarz genus
• number of tests in computation trees
• These notions give rise to questions in set theory and
algebraic topology,
• Number of tests in computation trees with restricted set of
arithmetic operations: algebraic complexity theory.
• Can also be analyzed for algorithms in numerical
mathematics (information-based complexity).
• Continuous reducibilities lead to natural reﬁnements of the
level.
• Develop practical theory for handling discontinuities!
E.g., in computational geometry.

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