Domain%20Theory,%20Computational%20Geometry%20and%20Differential%20Calculus

1
Domain Theory, Computational Geometry and
Differential Calculus

• Abbas Edalat
• Imperial College London
• www.doc.ic.ac.uk/ae
• With contributions from Andre Lieutier, Ali
  Khanban,
• Marko Krznaric and Dirk Pattinson
• First SQU Workshop on Topology and its
  Applications
• December 2004

2
First rudimentary notion of a real number

• In his Commentaries on the Difficulties in the
  Postulates of
  Euclid's Elements'', Omar Khayyam, the 11th
  century Persian mathematician and
  poet, first showed the equivalence of Euclid's
  notion of ratios with that of continued
  fractions.
• Then, in a stroke of genius, he defined two
  ratios as equal
  when they can be expressed by the ratio of
  integer numbers with as great a degree of
  accuracy as we like.''
• Three centuries later, Ghiasseddin Jamshid
  Kashani, another Persian mathematician, devised
  the first fixed point technique for computation
  in analysis in the beginning of the 15th century
  
• He used a cubic polynomial in a recursive
  scheme to approximate thesine of 1 correctly up
  to 17 decimal places

3
Computational Model for Classical Spaces

• Research project since 1993 Reconstruct
  mathematical analysis in an order-theoretic
  setting
• Embed classical spaces into the set of maximal
  elements of suitable partially ordered sets,
  called domains

4
Computational Model for Classical Spaces

• Applications
• Fractal Geometry
• Measure Integration Theory Generalized Riemann
  Integration
• Topological Representation of Spaces
• Exact Real Arithmetic
• Computational Geometry and Solid Modelling
• Differential Calculus
• Quantum Computation

5
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 supi?I ai
• The way-below relation in a dcpo is defined bya
  b iff for all directed subsets ai i?I ,
  the relation b ? supi?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
• A dcpo is bounded complete if every bounded
  subset has a lub
• A continuous Scott Domain is an ?-continuous
  bounded complete dcpo

6
Continuous functions

• The Scott topology of a dcpo has as closed
  subsets downward closed subsets that are closed
  under the lub of directed subsets, usually only
  T0.

• Proposition. If D is a dcpo then fD? D is Scott
  continuous iff
• (i) f is monotone, i.e. f(x) ? f(y) if x ? y,
  and
• (ii) f preserves sups of directed subsets, i.e.
  for any directed set A?D, we have supa?A f(a)
  f(supA)

• Fact. The Scott topology on a continuous dcpo A
  with basis B has basic open sets a ? A b a
  for each b ? B.

7
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 supi?I ai ?i?I ai
• a b ? ao ? b
• (IR, ?) is ?-continuouscountable basis p,q
  p lt q p, q ? Q
• (IR, ?) is, thus, a continuous Scott domain.
• Scott topology has basis?a b ao ? b

8
Continuous Functions

• Scott continuous f0,1n ? IR is given by lower
  and upper semi-continuous functions f -, f
  0,1n ? R with f(x)f -(x),f (x)
• f 0,1n ? R, f ? C00,1n, has continuous
  extension i(f ) 0,1n ? IR
  x ? f (x)

• Scott continuous maps 0,1n ? IR with
  f ? g ? ?x ? R . f(x) ? g(x)is another
  continuous Scott domain.
• The function space 0,1n ? IR is a continuous
  Scott domain. ? C00,1n ?
  ( 0,1n ? IR), with f ? i(f)
  is a topological
  embedding into a proper subset of maximal
  elements of 0,1n? IR .
• We identify x and x, also f and i(f)

9
Step Functions

• Lubs of finite and bounded collections of single-
  step functions f
  sup1?i?n(ai ? bi) are called step functions with
  N(f) n (number of pieces)
• Step functions with ai, bi rational intervals
  give a basis for 0,1n ? IR. They
  are used to approximate C0 functions.

10
Step Functions-An Example in R
R
b3
a3
b1
b2
a1
a2
0
1
11
Refining the Step Functions
R
b3
a3
b1
a1
b2
a2
0
1
12
Kleenes Fixed Point Theorem

• Theorem. If D is a dcpo with bottom (least
  element) d0 and if f D ? D is Scott continuous
  then f has a least fixed point given by supi?N f
  i(d0)

13
Initial Value Problems in Domain

• Consider initial value problems given by

• where y(y1,,yn) -a,a?Rn and v
  -K,Kn?-M,Mn are continuous..

• ODE solving
• Mathematical Analysis no direct computational
  content
• Numerical Analysis no correctness guarantee
• Interval Analysis no convergence guarantee
• Computable Analysis no data types
• Domain Theory provides a method which
• is based on proper data types
• has guaranteed speed of convergence
• is directly implementable on a digital computer

14
Picard's Classical Theorem

• Standard assumption v -K, Kn ? -M, Mn and
  aM K.
• Theorem Suppose v is Lipschitz v(x) - v(y)
  L x y
• for some L gt 0 and all x, y ?-K, Kn. then,
  there exists a unique
• y -a, a ? -K, Kn with

• Proof For y -a, a ? -K, Kn let Pv(y) -a,
  a ? -K, Kn with

• By Banach's fixed point theorem, the sequence yk
  defined by
• y0 any initial continuous function, and yk1
  Pv(yk) converges to
• the unique fixed point.

15
Translation to Domain Theory Road map

• Formulate the problem as fixed point equation
  over
  -a,a?I-K,Kn
• Apply Kleene's Theorem to obtain the least
  fixed point
• Relate domain-theoretic fixed points to the
  classical problem
• Show that the computations restrict to bases of
  the involved domains.
• Furthermore, we give
  (i)
  Estimates on the speed of convergence
  (ii) Estimates on the
  (algebraic) complexity of the problem.

16
Translation to Domain Theory Solutions

• Interval valued approximations to the solution
• S y -a, a ? I-K, Kn y Scott
  continuous where
• I-K, Kn A A ?-K, Kn is a compact
  rectangle is the sub-domain of rectangles
  contained in -K, Kn with reverse inclusion.
• The order on S is pointwise
• If f, g ?S then f ? g iff f(x) ? g(x) for all x
  ? -a, a.

17
Translation into Domain Theory Vector Field

• Computing v(y) requires vector fields with
  interval input
• V u I-K, Kn ? I-M, Mn u Scott
  continuous with pointwise ordering
• Relationship to Classical Vector Fields
  Extensions
• u ?V extends a classical v -K, Kn ? -M, Mn
  if
• u(x) v(x) for all x ? -K, Kn.
• The greatest or canonical extension of v is given
  by
• Iv I-K, Kn ? I-M, Mn
  with ((Iv)(A))i viA

18
The Domain Theoretic Picard Operator

• Let u ?V.
• The domain-theoretic Picard operator PuS?S is
  given by

• where for f f- -, f -a, a ? I-K, K

• The monotone convergence theorem shows that Pu is
  Scott continuous.

19
Relationship to the Classical Problem

• Assume u is an extension of v -K, Kn ? -M,
  Mn.
• Suppose y is the least fixed point of Pu
• Every solution f satisfies y ? f
• If y y-, y and y- y, then y- is the
  unique solution.
• That is, we look for fixed points of width 0
• For (A1, ..,An) ? I-K, Kn, we write
  
  
  Ai Ai-, Ai and
  define the width of A by
• w(A)maxAi-Ai- 1in
• w(f)supw(f(x)) x?-a,a

20
The Lipschitz Case

• Recall The classical proof assumes that v is
  Lipschitz.
• Suppose u is interval Lipschitz, that is
• w(u(x)) L w(x) for all x ?I-K, Kn.
• Then w(Pu(y)) a L w(y) for all y ?S.
• Assuming aL lt 1 (which can be actually removed),
  we obtain
• Theorem
• y0 -K, Kn and yk1 Pu(yk). Theny
  sup k?N yk satisfies Pu(y) y and w(y) 0.
• In particular, y - y is the unique solution.

21
Approximations of the Vector Field

• Proposition The map P V?(S?S) with u? Pu is
  Scott continuous.
• This allows us to use approximations of the
  vector field to obtain the solution

• . Then y sup k ?N yk satisfies Pu(y) y
  and w(y) 0.
• Speed of Convergence
• If aL lt c lt 1 and d(u, uk) lt 2Mck (c - aL), then
  w(yk) ck w(y0), where

22
Restrinting to a Base

• Let a,K,M?Q.
• Denote the class of rational step functions
  I-K,Kn?I-M,Mn by VQ.
• We also have the class SL Q of rational piecewise
  linear step functions
  f-a,a?I-K,Kn

We again put N(f) number of pieces in f
In the above example N(f)3
23
Computing with SLQ and VQ

• Pu restricts to the base
• Suppose u ?VQ and y ? SLQ then
• The map u(y) is piecewise constant, hence Pu(y)?
  SLQ Pu(y)? SLQ .
• Pu(y) can be computed in time O(N(u)N(y)), i.e.
• N(Pu(y)) ? O(N(u)N(y)).

24
Summary

• Suppose u supk uk with uk ?VQ , y0 -K, Kn
  and then

• yk ?SLQ for all k ? N
• y supk yk has width 0 and is the unique
  solution
• w(yk) ?O(ck) if d(u, uk) ? O(ck).
• Given an elementary function u, one can use
  continued fractions for to obtain uk with the
  above properties..
• These properties are preserved in an
  implementation using rational arithmetic.

25
PART II A Domain-Theoretic Model for
Differential Calculus

• Overall Aim Synthesize Computer Science with
  Differential Calculus
• Plan of the talk
• Primitives of continuous interval-valued function
  in Rn
• Derivative of a continuous function in Rn
• Fundamental Theorem of Calculus for
  interval-valued functions in Rn
• Domain of C1 functions in Rn
• Inverse and implicit functions in domain theory

26
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
• Recall that for real x, we identify x with x.

27
Basic Construction n1

• What is a primitive map 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
  ?
• f should satisfy

28
Interval Lipschitz contant

• Assume f ? C10,1, a ? I0,1, b ? IR.
• Suppose ?x ? ao . b- ? f ' (x) ? b.
• We think of b-, b as an interval Lipschitz
  constant for f at a.

• Note that ?x ? ao . b- ? f '(x) ? b
• iff ?x1, x2 ? ao x1 gt x2 ,
• b- (x1 x2) ? f(x1) f(x2) ? b(x1
  x2), i.e.
• b(x1 x2) ? f(x1) f(x2) f(x1)
  f(x2)

29
Definition of Interval Lipschitz constant

• f ? (0,1 ? IR) has an interval Lipschitz
  constantb ? 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)

30
For Classical Functions

• Let f ? C10,1 the following are equivalent
  
• f ? ?(a,b)
• ?x ? ao . b- ? f '(x) ? b
• ?x1,x2 ?0,1, x1,x2 ? ao.
• b(x1 x2) ? f (x1) f (x2)
• a?b ? f '

Thus, ?(a,b) is our candidate for ? a?b .
31
Set of primitive maps

• ? (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.
• ? f is always a non-empty set.

32
The Derivative
33
Examples
34
Fundamental Theorem of Calculus

• f ??g iff g
  (interval version)
• If g?C0 then f ??g iff g
  (classical version)

?
35
Idea of Domain for C1 Functions

• If h ?C10,1 , then( h , h' ) ? (0,1 ? IR) ?
  (0,1 ? IR)

• We can approximate ( h, h' ) in (0,1 ? IR)2
• i.e. ( f, g) ? ( h ,h' ) with f ? h and g
  ?h'

• What pairs ( f, g) ? (0,1 ? IR)2 approximate a
  differentiable function?

36
Function and Derivative Consistency

• Define the consistency relationCons ? (0,1 ?
  IR) ? (0,1 ? IR) with(f,g) ? Cons if
  (?f) ? (? g) ? ?

• Proposition
• (f,g) ? Cons iff there is a continuous h
  dom(g) ? R
• with f ? h and 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) .

37
Consistency Test for (f,g)

• For x ? dom(g), let g(x) g- (x), g(x)
  where
• g - , g dom(g) ?R are lower and upper
  semi-continuous. Similarly we define f -, f
  dom(f) ?R. Write f f , f .

• Let O be a connected component of dom(g) with
  O ? dom(f) ? ?. For
  x , y ? O, let

38
Similarly, let

• The lower envelop t(f,g)(x) inf y?O?dom(f) (f
  (y) T (f,g) (x,y))
• Proposition. t(f,g) is the greatest function
  with
• t(f,g) ? f and g ?
• Theorem. (f, g) ? Con iff for all component O of
  dom(g) with O? dom(f) ? ? and all x ? O. s (f,
  g) (x) ? t (f, g) (x).

39
Consistency for basis elements

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

40
Function and Derivative Information
41
Least and greatest functions
42
Decidability of consistency

• For (f, g) (?1?i?n ai?bi , ?1?j?m cj?dj),
• 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 s(f,g) is the max of k2 rational linear
  maps.
• Similarly for t(f,g)(x).
• Thus, s(f,g)? t(f,g) is decidable.
• Theorem. Consistency is decidable.

43
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.? C00,1 ? D1 0,1
  
• f ? (f , ) is
  topological embedding, giving a
• computational model for continuous functions
  and their differential properties. It restricts
  to give an embedding of C10,1.

44
Functions of several varibales

• (IR)1 n row n-vectors with entries in IR
• For dcpo A, let(An)s smash product of n
  copies of Ax?(An)s if x(x1,..xn) with xi
  non-bottomor xbottom
• We work with (IR1 n)s

45
Definition of Interval Lipschitz constant

• f ? (0,1n ? IR) has an interval Lipschitz
  constantb ? (IR1xn)s in a ? I0,1n if ?x, y ?
  ao,
• b(x y) ? f(x) f(y).

• The tie of a with b, is
• ?(a,b) f ?x,y ? ao. b(x y) ? f(x)
  f(y)

• Proposition. If f??(a,b), then f(x) ? Maximal
  (IR) for x ? ao and
• for all x,y ? ao. f(x)-f(y)? k x-y
  with kmax i (bi, bi-)

46
For Classical Functions

• Let f ? C10,1n the following are
  equivalent
• f ? ?(a,b)
• ?x ? ao . b- ? f (x) ? b
• ? x,y ? ao.
  b(x y)
  ? f (x) f (y)
• a?b ? f '

Thus, ?(a,b) is our candidate for ? a?b .
47
Set of primitive maps

• ? (0,1n ? IR) ? (P(0,1n ? (IR1xn)s), ? )
• ( P
  the power set constructor)

• ? a?b ?(a,b)
• ? supi ?I ai ? bi ?i?I ?(ai,bi)
• ? is well-defined and Scott continuous.
• ? g can be the empty set for 2 ? n
• Eg. g(g1,g2), with g1(x , y) y , g2(x,
  ,y)0

48
The Derivative
49
Fundamental Theorem of Calculus

• f ??g iff g ?
  (interval version)
• If g?C0 then f ??g iff g
  (classical version)

50
Relation with Clarkes gradient

• For a locally Lipschitz f 0,1n ? IR
• ? f (x)convex lim mf (xm) x m?x
• It is a non-empty compact convex subset of Rn
• Theorem
• For locally Lipschitz f 0,1n ? IR, the
  following holds
• The domain-theoretic derivative at x , i.e.
  is the smallest
• n-dimensional rectangle with sides parallel
  to the coordinate planes that contains ? f (x)
• In dimension one the two coincide.

51
Idea of Domain for C1 Functions

• If h ?C10,1n , then( h , h' ) ? (0,1n ? IR)
  ? (0,1n ? IR)ns

• We can approximate ( h, h ) in
• (0,1n ? IR) ? (0,1n ? IR)ns
• i.e. ( f, g) ? ( h ,h ) with f ? h and g
  ?h

• What pairs ( f, g) ? (0,1n ? IR) ? (0,1n ?
  IR)ns
• approximate a differentiable function?

52
Function and Derivative Consistency

• Define the consistency relationCons ? (0,1n ?
  IR) ? (0,1n ? IR)ns 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) .

53
Basis elements

• Definition. g0,1n ? (IRn)s the domain of g is
  
  dom(g) x g(x) non-bottom
• Basis element (f, g1,g2,.,gn) ? (0,1n? IR) ?
  (0,1n? IR)ns
• Each f, gi 0,1n ? IR is a rational step
  function.
• Recall the intersection of a closed and an open
  set is a crescent.
• dom(g) is partitioned by disjoint crescents in
  each of which g is a constant rational interval. .

Eg. For n2 A step function gi with four single
step functions with two horizontal and two
vertical rectangles as their domains and a hole
inside, and with eight vertices.
54
Corners and their coaxial points

• corners

• verticescorners ? coaxial points

• their coaxial points

55
Decidability of Consistency

• (f,g) ? Cons if (?f) ? (? g) ? ?
• First we check if g is integrable, i.e. if ? g ?
  ?
• In classical calculus, g0,1n ? Rn will be
  integrable by Greens theorem iff for any
  piecewise smooth closed non-intersecting path
• p0,1? 0,1n with p(0)p(1)

• We generalize this to the type g0,1n ? (IRn)s

56
Interval-valued path integral

• For v?IRn , u?Rn define the interval-valued
  scalar product

57
Generalized Greens Theorem

• Theorem. ? g ? ? iff for all piecewise smooth
  non-intersecting path p0,1? dom(g) with
  p(0)p(1), we have zero-containment

• For step functions, we can and will replace
  smooth with linear.
• Then, the lower and upper path integrals will
  depend piecewise linearly on the coordinates of
  nodes the path, so their extreme values will be
  reached when these nodes are at the corners of
  dom(g) or their coaxial points.
• Since there are finitely many of these extreme
  paths, zero containment can be decided in finite
  time for all paths.

58
Minimal surface for g

• Step function g0,1n ? (IRn)s with ? g ? ? .
  Let O be a component of dom(g) . Let x,y?cl(O).
• Consider the following supremum over all
  piecewise linear paths p in cl(O) with p(0 ) y
  and p(1) x.

59
Maximal surface for g

• Step function g0,1n ? (IRn)s . Let O be a
  component of dom(g) . Let x,y?cl(O).
• The following infimum is attained over all
  piecewise linear paths p in cl(O) with p(0 ) y
  and p(1) x.

60
Minimal surface for (f,g)

• (f,g)?(0,1n ? IR) ? (0,1n ? IR)ns rational
  step function
• Assume we have determined that ? g ? ?
• Put

• Proposition.

Theorem. s(f,g) dom(g) ?R is the least
continuous function with f - ? s(f,g) and g ?

61
Maximal surface for (f,g)
Theorem. t(f,g) dom(g) ?R is the least
continuous function with t(f,g) ? f and g ?

Theorem. Consistency is decidable. Proof In
s(f,g)?t(f,g) we compare two rational
piecewise-linear surfaces, which is decidable.
62
The Domain of C1 Functions

• Lemma. Cons ? (0,1n ? IR)? (0,1n ? IR)ns is
  Scott closed.
• Theorem.D1 0,1n (f,g) (f,g) ? Cons is a
  continuous Scott domain that can be given an
  effective structure.

63
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
64
Solution
At stage n we find un - and un
.
65
Solution
At stage n we find un - and un
.
66
Solution
At stage n we find un - and un
.
un - and un tend to the exact solutionf t ?
t2/2 1
67
Computing with polynomial step functions
68
Some Open Problems

• Topologically, what sort of a set is ?(a,b) ?
• Topologically, what sort of a set is C0 in the
  set of maximal elements of D00,1?IR?
  Similarly, for C1 as a subset of maximal elements
  of D1.
• For a domain of twice differentiable functions,
  is consistency decidable? (It is true for n1.)
• Decidability of consistency for 3rd and higher
  derivatives (not known even for n1).
• Construct a domain for analytic functions.
• For solving PDEs, obtain a domain-theoretic
  version of the finite difference and finite
  element methods.

69
Part III A Domain-Theoretic Model of Geometry

• To develop a model for Computational Geometry
  and Solid Modelling, so that

• the model is mathematically sound, realistic

• the basic building blocks are computable

• it bridges theory and practice.

70
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!

71
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.

72
The Convex Hull Algorithm
With floating point we can get
73
The Convex Hull Algorithm
A, B C nearly collinear
With floating point we can get (i) AC,
or
74
The Convex Hull Algorithm
A, B C nearly collinear
With floating point we can get (i) AC,
or (ii) just AB, or
75
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
76
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.
77
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.

78
Non-computability of the Membership Predicate

• There is discontinuity at the boundary of the
  set.

False
True
79
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.

80
Intersection of two 3D cubes
81
Intersection of two 3D cubes
82
Intersection of two 3D cubes
83
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.

84
Computable Topology and 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.

85
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 ?
• Thenf 1tt is open its called the interior
  of the object. f 1ff is open its called
  the exterior of the object.

86
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 functionsf, g
  X ? tt, ff ? f ? g ? ?x ? X . f(x) ?
  g(x)
• f ? g ? f 1tt ? g 1tt f 1ff
  ? g 1ffTherefore, f ? g means g has more
  information about an idealized real solid object.

87
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

88
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 (i.e. a continuous
  Scott domain) with (U1, U2) ltlt (V1, V2) iff
  the closures of U1 and U2 are compact subsets of
  V1 and V2 respectively.

89
Examples

• A x?R2 ? x 1 ? 1, 2represented in the
  model byArep (int A, int Ac)
• ( x ? x lt 1, R2 \ A )is a classical (but
  non-regular) solid object.

90
Boolean operations and predicates

• Theorem All these operations are Scott
  continuous and preserve classical solid objects.

91
Subset Inclusion

• Subset inclusion is Scott continuous.

92
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.

93
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) supnK ß(n) ( ?n p1 ( K ß(n) )
, ?n p2 ( K ß(n) ) )
94
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.

95
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.

96
Quantative Measure of Convergence

• In our present model for computable solids, we
  require a quantitative measure for the
  convergence of the basis elements to a computable
  solid.
• We will enrich the notion of domain-theoretic
  computability in two different ways to include a
  quantitative measure of convergence.

97
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

98
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.

99
Boolean Intersection is not Hausdorff computable
is Hausdorff computable.
However Q?(0,1 ? 0) r,1 ? 0 ? R2is
not Hausdorffcomputable.
100
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.

101

• Hausdorff computable ? Lebesgue
  computableComplement 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.

102
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.

103
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.

104
Summary

• 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

105
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

106
The Outer Convex Hull Algorithm
107
The Inner Convex Hull Algorithm
108
The Convex Hull Algorithm
109
The Convex Hull Algorithm
110
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

111
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.

112
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.
113
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.

114
Partial Perpendicular Bisector of Two Partial
Points
115
PPBs for Three Partial Points
116
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.
117
Partial Circles
With more exact partial points, the boundaries of
the interior and exterior of the partial circle
get closer to each other.
118
Partial Circles

• In the limit, the area between the interior and
  exterior of the partial circle, and the Hausdorff
  distance between their boundaries, tends to zero.

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

119
Part IVInverse and Implicit Function
TheoremsDomain of Manifolds

• The mains tool in multi-variable calculus.
• In CAD curves and surfaces are built from the
  implicit function theorem, f(x,y,z) 0 for a
  C1 map f R3? R

• A domain-theoretic framework can be
• the basis of a robust CAD.

• Implicit function theorem follows from
• the inverse function theorem, both
  classically and
  domain-theoretically.

120
Calculus of operations in D1

• For a vector function f (fi)i
  (fi)1in we write f?D1
  (0,1n ? Rn) if fi?D1 (0,1n ? R) for 1 i
  n
• The following operations are well-defined and
  Scott continuous
• - ? - D1 (0,1n ? IRn) D1 (0,1n ? IRn)
  ? D1 (0,1n ? IRn)
• 
  ( f , g ) ?
  f ? g
• with (f ? g)i ( fi0 ? gi0 ,
  fi1 ? gi1 ,.., fin ? gin )
• - ? - D1 (Rn ? IRn) D1 (0,1n ? IRn) ? D1
  (0,1n ? IRn)
• 
  ( f , g ) ?
  f?g

• This extends the higher dimensional chain rule.

121
Inverse Constructor (function part)

• Idea First find the inverse of a map which
  differs from the identity map by a contraction
• Let A-a,an and B-ca,can for agt 0 and 0lt c
  1/2
• V (A?IB)(B?IB)?(B?IB)
• (f,g) ? - If ? (idg) ,
  where id is the identity map.
• Theorem. Suppose f-a,an?Rn has Lipschitz
  constant nclt1 wrt the max norm and f(0)0. Then
• V(f , .) (B?IB)?(B?IB) has a unique fixed point
  g satisfying g - g and g(0)0
• idf -a,an?Rn has inverse idg
• In fact g - f ? (idg) implies (id
  f)?(idg) id g g id
  
  

122
Inverse Constructor (derivative part)

• Recall A-a,an and B-ca,can for agt 0 and 0lt
  c 1/2
• The classical functional (f, g)? - f ? (idg)
  C1C1?C1 is extended to
• T D1(A ? IB) D1( B ? IB ) ? D1( B?IB )
  (f , g ) ? ( V((fi0)i , (gi0)i )
  , S ((fij)ij ,(gi0)i, (gij)ij) )
• with
• S (A ? IB) (B ? IB) (B? IRn) ? (B ? IRn)
• (h , u ,w) ? - ( Ih ? (id u) ) ? (?x.id
  w)

123
Invesre Constructor (derivative part)

• Suppose f-a,an?Rn has Lipschitz constant nc lt
  1 wrt the max norm and f(0)0. Then the functional

has a unique fixed point ((gi0)i,(gij)ij)
where (gi0)i is the unique fixed point of V(f,-)
and (gij)ij is the unique fixed point of

• If f '(x) exists for some x?-a,an, so does
  (gi0)i '(x) and we have (gi0)i ' (x) (gij)ij
  (x)

124
Mean Differential

• Lemma u -a,an ? Rn has Lipschitz constant
  nclt1 wrt the max norm iff ui?d(-a,an ,-c,c)
  for 1 i n

125
Inverse Function theorem

• The map u has a Lipschitz inverse in a
  neighbourhood of 0 .
• Given an increasing sequence of linear step
  functions converging to u, we can effectively
  obtain an increasing sequence of linear step
  functions converging to u-1
• If furthermore u is C1 and given also an
  increasing sequence of linear step functions
  converging to u' we can also effectively obtain
  an increasing sequence of polynomial step
  functions converging to (u-1)'

126
Current and Further Work

• A domain (w-continuous dcpo)for Lipschitz
  manifolds with basisgiven by rational piecewise
  linearmanifold (i.e., polyhedra).

• Construct the kernel of a robust CAD, based on
  domain-theoretic approximations to curves and
  surfaces obtained by the implicit function
  theorem.

127
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• http//www.doc.ic.ac.uk/ae