Abbas Edalat

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Abbas Edalat

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Title: Abbas Edalat

1
Interval Derivative of Functions

Abbas Edalat
Imperial College London
www.doc.ic.ac.uk/ae

2
The Classical Derivative

Let f a,b ? R be a real-valued function.
The derivative of f at x is defined as

when the limit exists (Cauchy 1821).
If the derivative exists at x then f is
continuous at x.
However, a continuous function may not be
differentiable at a point x and there are indeed
continuous functions which are nowhere
differentiable, the first constructed by
Weierstrass

with 0 ltblt 1 and a an odd positive integer.
3
Non Continuity of the Derivative

The derivative of f may exist in a neighbourhood
O of x but the function

may be discontinuous at x, e.g.
we have
4
A Continuous Derivative for Functions?

A computable function needs to be continuous with
respect to the topology used for approximation.
Can we define a notion of a derivative for real
valued functions which is continuous with respect
to a reasonable topology for these functions?

5
Dinis Four Derivates of a Function (1890)

Upper right derivate at x
Upper left derivate at x
Lower right derivate at x
Lower left derivate at x

Clearly, f is differentiable at x iff its four
derivates are equal, the common value will
then be the derivative of f at x.

6
Example
7
Interval Derivative

Let IR a,b a, b ? R ? R and consider
(IR, ?) with R as bottom.

The interval derivative of f c,d ? R is
defined as

if both limits are finite otherwise
8
Example

We have already seen that

We have

Thus

9
Interval-Limit of Functions

Then

The interval limit of f is defined as

10
Examples
11
Interval-limit of Interval-valued Functions

Let and
be extended real-valued functions
with .
Consider the interval-valued function

The interval-limit of f is now defined as

12
Continuity of the Interval Derivative

Theorem. Given any

the interval limit
is Scott continuous.

Corollary. The interval derivative of f c,d
? R

is Scott continuous.
13
Computational Content of the Interval Derivative

Definition. (AE/AL in LICS02) We say f c,d ?
R has interval Lipschitz constant
in an open interval if
The set of all functions with interval
Lipschitz constant b at a is called the tie of a
with b and is denoted by .

Recall the definition of single-step function

Theorem. For f c,d ? R we have

14
Fundamental Theorems of Calculus

Continuous function versus continuously
differentiable function

for continuous f for continuously differentiable
F

Lebesgue integrable function versus absolutely
continuous function

for any Lebesgue integrable f iff F is
absolutely continuous
15
Locally Lipschitz functions

The interval derivative induces a duality between
locally Lipschitz maps versus bounded integral
functions and their interval limits.

A map f (c,d) ? R is locally Lipschitz if it is
Lipschitz in a neighbourhood of each
.

A locally Lipschitz map f is differentiable a.e.
and

The interval derivative of a locally Lipschitz
map is never bottom

16
Primitive of a Scott Continuous Map

Given Scott continuous
is there

with

In other words, does every Scott continuous
function has a primitive with respect to the
interval derivative?
For example, is there a function f with
?

17
Total Splittings of Intervals

A total splitting of 0,1 is given by disjoint
measurable subsets A, B with
such that for any interval
we have

where is the Lebesgue measure.

It follows that A and B are both dense with empty
interior.

Non-example

18
Construction of a Total Splitting

Construct a fat Cantor set in 0,1 with

In the open intervals in the complement of
construct
countably many Cantor sets
with

In the open intervals in the complement of
construct
Cantor sets with
.

Continue to construct with

19
Primitive of a Scott Continuous Function

To construct with
for a given

Take any total splitting (A,B) of 0,1 and put

Theorem.

20
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
.
.

21
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

22
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

23
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

24
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 ?
Ifis a topological embedding into a proper
subset of maximal elements of 0,1 ? IR .

25
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

26
Step Functions-An Example
R
b3
a3
b1
b2
a1
a2
0
1
27
Refining the Step Functions
R
b3
a3
b1
a1
b2
a2
0
1
28
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

29
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

30
Interval Derivative
31
Definition of Interval Derivative

f ? (0,1 ? IR) has an interval derivativeb ?
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)

32
For Classical Functions
Thus, ?(a,b) is our candidate for ? a?b .
33
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.

34
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.

35
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

36
The Derivative
37
Examples
38
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
0,1 ? IR
g x ? x 0,1 ? IR
(fg) (0) ? (0) (0)

39
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.

40
The Fundamental Theorem of Calculus

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

41
Fundamental Theorem of Calculus

For f, g ? C10,1, let f g if f g r, for
some r ? R.
We have

42
F.T. of Calculus Isomorphic version

For f , g ? 0,1 ? IR, let f g if f
g a.e.
We then have

43
A Domain for C1 Functions

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

44
Function and Derivative Consistency

Define the consistency relationCons ? (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) .

45
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)

46
Function and Derivative Information
47
Updating
48
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).

49
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).

50
Updating Algorithm
51
Updating Algorithm (left to right)
f
1
1
52
Updating Algorithm (left to right)
53
Updating Algorithm (right to left)
54
Updating Algorithm (right to left)
55
Updating Algorithm (similarly for upper one)
56
Output of the Updating Algorithm
57
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)

58
Higher Interval Derivative
59
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)
60
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

61
Domains of C k functions

The decidability of Cons on basis elements for k
? 3 is an open question.

Dk (fi)0?i?k ? (I0,1?IR)k1 Cons
(fi)0?i?k

62
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.

63
Picards Theorem
64
Picards Solution Reformulated

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

65
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 (y)
Thus, on basis elements strong consistency is
decidable.

66
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) ) )

67
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).

68
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- ? ?

69
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
70
Solution
At stage n we find un - and un
.
71
Solution
At stage n we find un - and un
.
72
Solution
At stage n we find un - and un
.
un - and un tend to the exact solutionf t ?
t2/2 1
73
Computing with polynomial step functions
74
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.

75
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!

76
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.

77
The Convex Hull Algorithm
With floating point we can get
78
The Convex Hull Algorithm
A, B C nearly collinear
With floating point we can get (i) AC,
or
79
The Convex Hull Algorithm
A, B C nearly collinear
With floating point we can get (i) AC,
or (ii) just AB, or
80
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
81
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.
82
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.

83
Non-computability of the Membership Predicate

There is discontinuity at the boundary of the
set.

False
True
84
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.

85
Intersection of two 3D cubes
86
Intersection of two 3D cubes
87
Intersection of two 3D cubes
88
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.

89
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.

90
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.

91
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.

92
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

93
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.

94
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.

95
Boolean operations and predicates

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

96
Subset Inclusion

Subset inclusion is Scott continuous.

97
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.

98
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 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) ) )
99
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.

100
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.

101
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.

102
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 r C ? Dr D ? Cr
where Dr x ? y ? D.
x-y ? r

103
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.

104
Boolean Intersection is not Hausdorff computable
is Hausdorff computable.
However Q?(0,1 ? 0) r,1 ? 0 ? R2is
not Hausdorffcomputable.
105
Lebesgue Computability

(A , B) ? S k, kd is Lebesgue computable iff
there exists an effective chain K ß(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.

106
Hausdorff and Lebesgue computability

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.

107
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.

108
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.

109
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

110
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

111
The Outer Convex Hull Algorithm
112
The Inner Convex Hull Algorithm
113
The Convex Hull Algorithm
114
The Convex Hull Algorithm
115
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) y?(R2)m, yi?Ti, 1 ? i ? m
Im(x) ?Hm (y) y?(R2)m, yi?Ti, 1 ? i ? m

116
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.

117
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.
118
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.

119
Partial Perpendicular Bisector of Two Partial
Points
120
PPBs for Three Partial Points
121
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.
122
Partial Circles
With more exact partial points, the boundaries of
the interior and exterior of the partial circle
get closer to each other.
123
Partial Circles