Loading...

PPT – Domain-theoretic Models of Differential Calculus and Geometry PowerPoint presentation | free to download - id: 19eb2a-ZDc1Z

The Adobe Flash plugin is needed to view this content

Domain-theoretic Models of Differential

Calculus and Geometry

(I) Domain Theory and Differential

Calculus (II) Domain-theoretic Solution of

Differential Equations (III) Domain-theoretic

Model of Geometry and Solid Modelling

- Abbas Edalat
- Imperial College London
- www.doc.ic.ac.uk/ae
- BRICS, February 2003
- Contributions from Andre Lieutier, Ali Khanban,

Marko Krznaric

Computational Model for Classical Spaces

- A research project since 1993
- Reconstruct some basic mathematics
- Embed classical spaces into the set of maximal

elements of suitable domains

Computational Model for Classical Spaces

- Previous Applications
- Fractal Geometry
- Measure Integration Theory
- Topological Representation of Spaces
- Exact Real Arithmetic

Part (I) Domain Theory and Differential Calculus

- Synthesize Differential Calculus developed by

Newton and Leibnitz in the 17th century with

Computer Science developed in the 20th century

Non-smooth Mathematics

Smooth Mathematics

- Geometry
- Differential Topology
- Manifolds
- Dynamical Systems
- Mathematical Physics
- .
- .
- All based on differential

calculus

- Set Theory
- Logic
- Algebra
- Point-set Topology
- Graph Theory
- Model Theory
- .
- .

A Domain-Theoretic Model for Differential

Calculus

- Indefinite integral of a Scott continuous

function - Derivative of a Scott continuous function
- Fundamental Theorem of Calculus for

interval-valued functions - Domain of C1 functions
- Domain of Ck functions

Continuous Scott Domains

- A directed complete partial order (dcpo) is a

poset (A, ?) , in which every directed set ai

i?I ? A has a sup or lub ?i?I ai - The way-below relation in a dcpo is defined by
- a b iff for all directed subsets ai i?I

, the relation b ? ?i?I ai

implies that there exists i ?I such that a ? ai - If a b then a gives a finitary approximation to

b - B ? A is a basis if for each a ? A , b ? B b

a is directed with lub a - A dcpo is (?-)continuous if it has a (countable)

basis - The Scott topology on a continuous dcpo A with

basis B has basic open sets a ? A b a for

each b ? B - A dcpo is bounded complete if every bounded

subset has a lub - A continuous Scott Domain is an ?-continuous

bounded complete dcpo

The Domain of nonempty compact Intervals of R

- Let IR a,b a, b ? R ? R
- (IR, ?) is a bounded complete dcpo with R as

bottom ?i?I ai ?i?I ai - a b ? ao ? b
- (IR, ?) is ?-continuous
- countable basis p,q p lt q p, q ? Q
- (IR, ?) is, thus, a continuous Scott domain.
- Scott topology has basis ?a b ao ? b

Continuous Functions

- Scott continuous maps 0,1 ? IR with

f ? g ? ?x ? R . f(x) ? g(x) is another

continuous Scott domain. - ? C00,1 ? ( 0,1 ? IR), with f ?

If is a topological embedding into a proper

subset of maximal elements of 0,1 ? IR .

Step Functions

- Lubs of finite and bounded collections of single-

step functions

- ?1?i?n(ai ? bi)
- are called step function.
- Step functions with ai, bi rational intervals,

give a basis for 0,1 ? IR

Step Functions-An Example

R

b3

a3

b1

b2

a1

a2

0

1

Refining the Step Functions

R

b3

a3

b1

a1

b2

a2

0

1

Operations in Interval Arithmetic

- For a a, a ? IR, b b, b ? IR, and ?

, , ? we have a b xy x ?

a, y ? b - For example
- a b a b, a b

The Basic Construction

- What is the indefinite integral of a single step

function a?b ?

- We expect ? a?b ? (0,1 ? IR)

- For what f ? C10,1, should we have If ? ? a?b

?

- Intuitively, we expect f to satisfy

Interval Derivative

Definition of Interval Derivative

- f ? (0,1 ? IR) has an interval derivative b ?

IR at a ? I0,1 if ?x1, x2 ? ao, - b(x1 x2) ? f(x1) f(x2).

- The tie of a with b, is
- ?(a,b) f ?x1,x2 ? ao. b(x1 x2) ?

f(x1) f(x2)

For Classical Functions

Thus, ?(a,b) is our candidate for ? a?b .

Properties of Ties

- ?(a1,b1) ? ?(a2,b2) iff a2 ? a1 b1 ? b2
- ?ni1 ?(ai,bi) ? ? iff ai?bi 1? i ? n

bounded. - ?i?I ?(ai,bi) ? ? iff ai?bi i?I

bounded iff ?J ?finite I ?i?J

?(ai,bi) ? ? - In fact, ?(a,b) behaves like a?b

we call ?(a,b) a single-step tie.

The Indefinite Integral

- ? (0,1 ? IR) ? (P(0,1 ? IR), ? )
- ( P

the power set constructor)

- ? a?b ?(a,b)
- ? ?i ?I ai ? bi ?i?I ?(ai,bi)
- ? is well-defined and Scott continuous.
- But unlike the classical case, the indefinite

integral ? is not 1-1.

Example

- (0,1/2 ? 0) ? (1/2,1 ? 0) ? (0,1 ?

0,1) - ?(0,1/2 , 0) ? ? (1/2,1 , 0) ? ? (0,1 ,

0,1) - ?(0,1 , 0)
- ? 0,1 ? 0

The Derivative

Examples

The Derivative Operator

- (0,1 ? IR) ? (0,1 ? IR) is monotone

but not continuous. Note that the classical

operator is not continuous either. - (a?b) ?x . ?
- is not linear! For f x ? x

-1,1 ? IR

g x ? x -1,1 ? IR - (fg) (0) ? (0) (0)

Domain of Ties, or Indefinite Integrals

- Recall ? (0,1 ? IR) ? (P(0,1 ? IR), ? )

- Domain of ties ( T0,1 , ? )
- Theorem. ( T0,1 , ? ) is a continuous Scott

domain.

The Fundamental Theorem of Calculus

- Define (T0,1 , ?) ? (0,1 ? IR)
- ? ? ? f ?

?

Fundamental Theorem of Calculus

- For f, g ? C10,1, let f g if f g r, for

some r ? R. - We have

F.T. of Calculus Isomorphic version

- For f , g ? 0,1 ? IR, let f g if f

g a.e. - We then have

A Domain for C1 Functions

- What pairs ( f, g) ? (0,1 ? IR)2 approximate a

differentiable function?

Function and Derivative Consistency

- Define the consistency relation Cons ? (0,1 ?

IR) ? (0,1 ? IR) with (f,g) ? Cons if

(?f) ? (? g) ? ?

- In fact, if (f,g) ? Cons, there are least and

greatest functions h with the above properties in

each connected component of dom(g) which

intersects dom(f) .

Consistency for basis elements

- (?i ai?bi, ?j cj?dj) ? Cons is a finitary

property

- We will define L(f,g), G(f,g) in general and

show that - (f,g) ? Cons iff L(f,g) ?G(f,g).
- Cons is decidable on the basis.

- Up(f,g) (fg , g) where fg t ?

L(f,g)(t) , G(f,g)(t)

Function and Derivative Information

Updating

Consistency Test and Updating for (f,g)

- Let O be a connected component of dom(g) with

O ? dom(f) ? ?. For x , y ? O

define

- Define L(f,g)(x) supy?O?dom(f)(f (y)

d(x,y)) and G(f,g)(x)

infy?O?dom(f)(f (y) d(x,y)) - Theorem. (f, g) ? Con iff ?x ? O. L (f, g) (x) ?

G (f, g) (x).

Updating Linear step Functions

- For (f, g) (?1?i?n ai?bi , ?1?j?m cj?dj)

with f linear g standard, - the rational endpoints of ai and cj induce a

partition y0 lt y1 lt y2 lt lt yk of

the connected component O of dom(g).

- Hence L(f,g) is the max of k2 linear maps.
- Similarly for G(f,g)(x).

Updating Algorithm

Updating Algorithm (left to right)

f

1

1

Updating Algorithm (left to right)

Updating Algorithm (right to left)

Updating Algorithm (right to left)

Updating Algorithm (similarly for upper one)

Output of the Updating Algorithm

The Domain of C1 Functions

- Lemma. Cons ? (0,1 ? IR)2 is Scott closed.
- Theorem. D1 0,1 (f,g) ? (0,1?IR)2 (f,g)

? Cons is a continuous Scott domain, which can

be given an effective structure.

- Theorem. ??? C10,1 ? C00,1 ? (0,1 ? IR)2

restricts

to give a topological embedding - D1c ? D1

(with C1 norm) (with Scott topology)

Higher Interval Derivative

Higher Derivative and Indefinite Integral

- For f 0,1 ? IR we define 0,1 ?

IR by - Then ?f ? ?2(a,b) a?b
- ? (0,1 ? IR) ? (P(0,1 ? IR), ? )
- ? a?b ? (a,b)
- ? ?i ?I ai ? bi ?i?I ? (ai,bi)
- ? is well-defined and Scott continuous.

(2)

(2)

2

(2)

2

(2)

Domains of C 2 functions

- Theorem. Cons (f0,f1,f2) is decidable on basis

elements. - (The present algorithm to check seems to be

NP-hard.) - D2 (f0,f1,f2) ? (I0,1?IR)3 Cons

(f0,f1,f2)

- Theorem. ????? restricts to give a topological

embedding D2c ? D2

Domains of C k functions

- The decidability of Cons on basis elements for k

? 3 is an open question.

- Dk (fi)0?i?k ? (0,1?IR)k1 Cons

(fi)0?i?k

Part (II) Domain-theoretic Solution of

Differential Equations

- Develop proper data types for ordinary

differential equations. - Solve initial value problem up to any given

precision.

Picards Theorem

Picards Solution Reformulated

- Apv (f,g) ? (f , ?t. v(t,f(t)))

A domain-theoretic Picards theorem

- To obtain Picards theorem with domain theory, we

have to make sure that derivative updating

preserves consistency. - (f , g) is strongly consistent, (f , g) ?S-Cons,

if - ? h ? g we have (f ,

h) ? Cons - Q(f,g)(x) supy?O?Dom(f) (f (y) d(x,y))
- R(f,g)x) infy?O?Dom(f) (f (y)

d(x,y)) - Theorem. If f , f , g, g 0,1 ? R are

bounded and g, g are continuous a.e. (e.g. for

polynomial step functions f and g), then (f,g) is

strongly consistent iff for any connected

component O of dom(g) with O ? dom(f) ? ? ,

we have ?x ? O. Q(f,g)(x),

R(f,g)(x) ? f (x) , f (x) - Thus, on basis elements strong consistency is

decidable.

A domain-theoretic Picards theorem

- Consider any initial value f ? 0,1 ? IR with
- (f, ?t. v (t , f(t) )

) ? S-Cons - Then the continuous map Up ? Apv has a least

fixed point above (f, ?t.v (t , f(t))) given by - (fs, gs) ?n ?0 (Up ? Apv )n

(f, ?t.v (t , f(t) ) )

The Classical Initial Value Problem

- Suppose v Ih for a continuous h -1,1 ? R

? R which satisfies the Lipschitz property around

(t0,x0) (0,0). - Then h is bounded by M say in a compact rectangle

K around the origin. We can choose positive a ? 1

such that -a,a ?-Ma,Ma ? K. - Put f ?n ?0 fn where fn -a/2n,a /2n ?

-Ma/2n , Ma/ 2n

- Then (f , -a,a ? -M ,M ) ? S-Cons, hence

(f, ?t. v(t , f(t) ) ) ? S-Cons since

(-a,a ? -M ,M ) ?

?t. v (t , f(t) ) - Theorem. The domain-theoretic solution

(fs, gs) ?n ?0 (Up ? Apv )n (f, ?t. v (t ,

f(t) ))

gives the unique classical solution through (0,0).

Computation of the solution for a given precision

? gt0

- We express f and v as lubs of step functions
- f ?n ?0 fn

v ?n ?0 vn - Putting Pv Up ? Apv the solution is obtained

as

- For all n ?0 we have un- ? un1- ? un1 ?

un with un - un- ? 0 - Compute the piecewise linear maps un- , un until

- the first n ?0 with un - un- ? ?

Example

v is approximated by a sequence of step

functions, v0, v1, v ?i vi

.

t

The initial condition is approximated by

rectangles ai?bi (1/2,9/8) ?i ai?bi,

v

t

Solution

At stage n we find un - and un

.

Solution

At stage n we find un - and un

.

Solution

At stage n we find un - and un

.

un - and un tend to the exact solution f t ?

t2/2 1

Computing with polynomial step functions

Part III A Domain-Theoretic Model of Geometry

- To develop a Computable model for Geometry and

Solid Modelling, so that

- the model is mathematically sound, realistic

- the basic building blocks are computable

- it bridges theory and practice.

Why do we need a data type for solids?

- Answer To develop robust algorithms!
- Lack of a proper data type and use of real RAM

in which comparison of real numbers is decidable

give unreliable programs in practice!

The Intersection of two lines

- With floating point arithmetic, find the point P

of the intersection L1 ? L2. Then

min_dist(P, L1) gt 0, min_dist(P,

L2) gt 0.

The Convex Hull Algorithm

With floating point we can get

The Convex Hull Algorithm

A, B C nearly collinear

With floating point we can get (i) AC,

or

The Convex Hull Algorithm

A, B C nearly collinear

With floating point we can get (i) AC,

or (ii) just AB, or

The Convex Hull Algorithm

A, B C nearly collinear

With floating point we can get (i) AC,

or (ii) just AB, or (iii) just BC, or

The Convex Hull Algorithm

A, B C nearly collinear

With floating point we can get (i) AC,

or (ii) just AB, or (iii) just BC, or (iv) none

of them.

The quest for robust algorithms is the most

fundamental unresolved problem in solid modelling

and computational geometry.

A Fundamental Problem in Topology and Geometry

- Subset A ? X topological space. Membership

predicate ?A X ? tt, ff - is continuous iff A is both open and closed.

- In particular, for A ? Rn, A ? ?, A ? Rn ?A

Rn ? tt, ff is not continuous. - Most engineering is done, however, in Rn.

Non-computability of the Membership Predicate

- There is discontinuity at the boundary of the

set.

False

True

Non-computable Operations in Classical CG SM

- ?A Rn ? tt, ff not continuous means it is not

computable, even for simple objects like

A0,1n. - x ? A is not decidable even for simple objects

for A 0,?) ? R, we just have the

undecidability of x ? 0. - The Boolean operation ? is not continuous, hence

noncomputable, wrt the natural notion of topology

on subsets ? C(Rn) ? C(Rn) ? C(Rn), where

C(Rn) is compact subsets with the Hausdorff

metric.

Intersection of two 3D cubes

Intersection of two 3D cubes

Intersection of two 3D cubes

This is Really Ironical!

- Topology and geometry have been developed to

study continuous functions and transformations on

spaces. - The membership predicate and the binary operation

for ? are the fundamental building blocks of

topology and geometry. - Yet, these fundamental functions are not

continuous in classical topology and geometry.

Elements of a Computable Topology/Geometry

- The membership predicate ?A X ? tt, ff fails

to be continuous on ?A, the boundary of A. - For any open or closed set A, the predicate x ?

?A is non-observable, like x 0.

- ?A is now a continuous function.

Elements of a Computable Topology/Geometry

- Note that ?A?B iff int Aint B int

Acint Bc, i.e. sets with the same

interior and exterior have the same membership

predicate. - We now change our view In analogy with classical

set theory where every set is completely

determined by its membership predicate, we define

a (partial) solid object to be given by any

continuous map - f X ? tt, ff ?
- Then f 1tt is open its called the interior

of the object. f 1ff is open its called

the exterior of the object.

Partial Solid Objects

- We have now introduced partial solid objects,

since X \ (f 1tt ? f

1ff)

may have non-empty interior. - We partially order the continuous functions f, g

X ? tt, ff ? f ? g ? ?x ? X . f(x) ?

g(x) - f ? g ? f 1tt ? g 1tt f 1ff

? g 1ff Therefore, f ? g means g has more

information about an idealized real solid object.

The Geometric (Solid) Domain of X

- The geometric (solid) domain S (X) of X is the

poset (X ? tt, ff ?, ? ) - S(X) is isomorphic to the poset SO(X) of pairs of

disjoint open sets (O1,O2) ordered componentwise

by inclusion

Properties of the Geometric (Solid) Domain

- Theorem For a second countable locally compact

Hausdorff space X (e.g. Rn), S(X) is bounded

complete and ?continuous with (U1, U2) ltlt (V1,

V2) iff the closures of U1 and U2 are compact

subsets of V1 and V2 respectively.

Examples

- A x?R2 ? x 1 ? 1, 2 represented in the

model by Arep (int A, int Ac) - ( x ? x lt 1, R2 \ A ) is a classical (but

non-regular) solid object.

Boolean operations and predicates

- Theorem All these operations are Scott

continuous and preserve classical solid objects.

Subset Inclusion

- Subset inclusion is Scott continuous.

General Minkowski operator

- For smoothing out sharp corners of objects.
- SbRn (A, B) ? SRn Bc is bounded ?(Ø,Ø).
- All real solids are represented in SbRn.
- Define _?_ SRn ? SbRn ? SRn

((A,B) , (C,D)) ? (A ? C , (Bc ? Dc)c)

where A ? C ac a? A, c? C - Theorem _?_ is Scott continuous.

An effectively given solid domain

- The geometric domain SX can be given effective

structure for any locally compact second

countable Hausdorff space, e.g. Rn, Sn, Tn,

0,1n. - Consider XRn. The set of pairs of disjoint open

rational bounded polyhedra of the form K (L1 ,

L2) , with L1 ? L2 ?, gives a basis for

SX. - Let Kn (p1 ( K n ) , p2 ( K n) ) be an

enumeration of this basis.

- (A, B) is a computable partial solid object if

there exists a total recursive function ßN?N

such that ( K ß(n) ) n ?0 is an increasing

chain with

(A , B) ( ?n p1 ( K ß(n) ) , ?n p2 (

K ß(n) ) )

Computing a Solid Object

- In this model, a solid object is represented by

its interior and exterior.

- The interior and the exterior
- are approximated by two
- nested sequence of rational polyhedra.

Computable Operations on the Solid Domain

- F (SX)n ? SX or F (SX)n ? tt,

ff ? - is computable if it takes computable sequences

of partial solid objects to computable sequences. - Theorem All the basic Boolean operations and

predicates are computable wrt any effective

enumeration of either the partial rational

polyhedra or the partial dyadic voxel sets.

Quantative Measure of Convergence

- In our present model for computable solids,

there is no quantitative measure for the

convergence of the basis elements to a computable

solid. - We will enrich the notion of domain-theoretic

computability to include a quantitative measure

of convergence.

Hausdorff Computability

- We strengthen the notion of a computable solid by

using the Hausdorff distance d between compact

sets in Rn. - d(C,D) min rgt0 C ? Dr D ? Cr

where Dr x ? y ? D.

x-y ? r

Hausdorff computability

- Two solid objects which have a small Hausdorff

distance from each other are visually close. - The Hausdorff distance gives a natural

quantitative measure for approximation of solid

objects. - However, the intersection or union of two

Hausdorff computable solid objects may fail to be

Hausdorff computable. - Examples of such failure are nontrivial to

construct.

Boolean Intersection is not Hausdorff computable

is Hausdorff computable.

However Q?(0,1 ? 0) r,1 ? 0 ? R2 is

not Hausdorff computable.

Lebesgue Computability

- (A , B) ? S k, kd is Lebesgue computable iff

there exists an effective chain Kß(n) of basis

elements with ß N?N a total recursive

function such that - (A , B) ( ?n p1 ( K ß(n) ) , ?n

p2 ( K ß(n) ) ) - µ(A) - µ(p1 ( K ß(n) ) ) lt 1/2 n µ(B)

- µ(p2 ( K ß(n) ) ) lt 1/2 n - A computable function is Lebesgue computable if

it preserves Lebesgue computable sequences. - Theorem Boolean operations are Lebesgue

computable.

Hausdorff and Lebesgue computability

- Hausdorff computable ? Lebesgue

computable Complement of a Cantor set with

Lebesgue measure 1 r with r lim rn left

computable but non-computable real.

- start with

- stage 1

- stage 2

- At stage n remove 2n open mid-intervals of length

sn/2n.

Hausdorff and Lebesgue computability

- Lebesgue computable ? Hausdorff computable
- Let 0 lt rn ? Q with rn ? r, left

computable, non-computable 0 lt r lt 1.

Hausdorff and Lebesgue Computable Objects

- Hausdorff computable ? Lebesgue computable
- Lebesgue computable ? Hausdorff computable
- Theorem A regular solid object is computable

iff it is Hausdorff computable. - However A computable regular solid object may

not be Lebesgue computable.

Conclusion

- Our model satisfies
- A well-defined notion of computability
- Reflects the observable properties of geometric

objects - Is closed under basic operations
- Captures regular and non-regular sets
- Supports a methodology for designing robust

algorithms

Data-types for Computational Geometry and Systems

of Linear Equations

- The Convex Hull

- Voronoi Diagram or the Post Office problem

- Delaunay Triangulation
- The Partial Circle through three partial points

The Outer Convex Hull Algorithm

The Inner Convex Hull Algorithm

The Convex Hull Algorithm

The Convex Hull Algorithm

The Convex Hull map

- Let Hm (R2)m ? C(R2) be the classical convex

Hull map, with C(R2) the set of compact subsets

of R2, with the Hausdorff metric. - Let (IR2, ? ) be the domain of rectangles in R2.

- For x(T1,T2,,Tm)?(IR2)m, define
- Cm (IR2)m ? SR2, Cm(x)

(Im(x),Em(x)) with - Em(x)?(Hm (y))c y?(R2)m, yi?Ti, 1 ? i ?

m - Im(x) ?(Hm (y))0 y?(R2)m, yi?Ti, 1 ? i ?

m

The Convex Hull is Computable!

- Proposition Em(x)(H4m((Ti1,Ti2,Ti3,Ti4))1?i?m)c

Im(x)Int(?Hm((Tin))1?i?m)

n1,2,3,4). - Theorem The map Cm (IR2)m ? SR2 is Scott

continuous, Hausdorff and Lebesgue computable. - Complexity
- Em(x) is O(m log m).
- Im(x) is also O(m log m).
- We have precisely the complexity of the

classical convex hull algorithm in R2 and R3.

Voronoi Diagrams

- We are given a finite number of points in the

plane.

- Divide the plane into components closest to

these points.

- The problem is equivalent to the Delaunay

triangulation of the points - (1) Triangulate the set of given points so

that the interior of the circumference circles do

not contain any of the given points.

(2) Draw the perpendicular bisectors of

the edges of the triangles.

Voronoi Diagram Partial Circles

- The centre of the circle through the three

vertices of a triangle is the intersection of the

perpendicular bisectors of the three edges of the

triangle.

- The partial circle of three partial points in the

plane is obtained by considering the Partial

Perpendicular Bisector of two partial points in

the plane.

Partial Perpendicular Bisector of Two Partial

Points

PPBs for Three Partial Points

Partial Circles

Each partial circle is defined by its interior

and exterior. The exterior (interior) consists of

all those points of the plane which are outside

(inside) all circles passing through any three

points in the three rectangles.

The exterior is the union of the exteriors of the

three red circles.

The Interior is the intersection of the interiors

of the three blue circles.

Partial Circles

With more exact partial points, the boundaries of

the interior and exterior of the partial circle

get closer to each other.

Partial Circles

- The limit of the area between the interior and

exterior of the partial circle, and the Hausdorff

distance between their boundaries, is zero.

- We get a Scott continuous map C (IR2)3?SR2
- We obtain a robust Voronoi algorithm.

Current and Further Work

- Solving Differential Equations with Domains
- Differential Calculus with Several Variables
- Implicit and Inverse Function Theorems
- Reconstruct Geometry and Smooth Mathematics with

Domain Theory - Continuous processes, robotics,

THE END http//www.doc.ic.ac.uk/ae