Set Inversion Fractals

Transcription

1 Set Inversion Fractals by Bryson Boreland A Thesis Presented to The University of Guelph In partial fulfilment of requirements for the degree of Master of Science in Applied Mathematics Guelph, Ontario, Canada c Bryson Boreland, August, 2016

2 ABSTRACT SET INVERSION FRACTALS Bryson Boreland University of Guelph, 2016 Advisor: Dr. Herb Kunze In 2000, Frame and Cogevina introduced a method for constructing fractals using circle inversion maps. The literature focuses on the graphical aspect of such fractals, without presenting a careful development of the underlying mathematical framework. In this thesis, we present such a framework, making a strong connection to iterated function systems (IFS) theory. Our nal result establishes that the set valued system of circle inversion maps induced by a collection of possibly overlapping circles in the plane has a unique set attractor. We then establish a similar mathematical framework in the setting of non-touching star-shaped sets. We present graphical examples for both settings using the chaos game. Finally, fractals literature develops the well-known concept of local iterated function systems with grey-level maps, with applications to image processing. We follow this path to establish a framework that uses local circle inversion maps as the functions. We demonstrate the results with examples.

3 iii Acknowledgements To my advisor and friend, Herb Kunze, for always challenging me and motivating me to work harder. You have helped shape who I am and although we might have cheered for different sides of the octagon at times, we still always had fun together. To the Province of Ontario (Ontario Graduate Scholarship) and the University of Guelph for the financial support that made this opportunity possible. To my Parents, Pam and Chris, for getting me to where I am today. Your love and support never goes unnoticed. Without your hard work and example of what it means to serve Him, I wouldn t be the man that I am today. Thank you for editing my proofs. To my siblings, Dustin, Brittany and Taylor, for celebrating my successes with me and pretending to understand me when I answered the question, What is your math thesis even about?. I m looking forward to the next chapter of my life and having the three of you around to enjoy it with me. To Dave, Taylor, Pat, Chris and Nate, for the important roles they have played in my life. To Gord, for teaching me that the Great Plains begin at the hundredth meridian and for uniting an entire country with your music. To my beautiful Fiancée and best friend, Jennifer, for being with me every step of the way and your constant support. You always helped me to relax when it was crunch time and make me laugh when I needed it. I am excited for the next chapter

4 iv of our lives together and really hope your mom let s us take Willow into that next chapter with us.

5 v Contents Abstract Acknowledgements List of Figures ii iii vi 1 Introduction 1 2 Background Metric Spaces Fixed Point Theory Iterated Function Systems The Chaos Game Collections of Touching and Non-Touching Circles Inversion in a Circle Mathematical Framework for Two Non-Touching Circles Collections of Circles Application of the Chaos Game Collections of Star-Shaped Sets Star-Shaped Set A New Metric for Star-shaped Sets Application of the Chaos Game D Signal and Function Approximation 69 6 Summary and Future Work 90

6 vi List of Figures 2.1 The set attractor A of the 3 maps define in Example 1 called the Sierpinski Gasket ỹ satisfying (3.3) is called the inverse of x, denoted T ( x) = ỹ x 3 is the inverse of x 3 with respect to C 2, denoted T 2 ( x 3 ) = x (a) Semi infinite ray (b) Infinite ray (c) Interior along ray and (d) Exterior along ray (a) Ray from x to ỹ (b) Ray from õ 1 to b 1 and (c) Ray from b 1 to b A system of three circles with the intersection of all of them equal to the empty set Set attractors after 5,000 iterations of the chaos game An illustration of 2 star-shaped sets with their kernels highlighted plus one not a star-shaped set (a) Two points within red star-shaped set (b) Two star-shaped sets and (c) Two inverted points with respect to the blue star-shaped set Set attractors after 5,000 iterations of the chaos game The three Sierpinski Gasket IFS maps with three circle inversion maps overlapping the Sierpinski triangles The two bottom Sierpinski Gasket IFS maps with one inversion map for a triangle taking the place of the top Sierpinski triangle The three Sierpinski Gasket IFS maps with three circle inversion maps completely contained outside of the three Sierpinski triangles The left picture is the target signal and the right picture is the two shrunken copies of the target The left picture includes the grey level maps and the right picture is overlaps added together Parallelogram formed by 0 β i 1 and 0 α i + β i

7 vii 5.4 X = [0, 1] divided up into 4 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent X = [0, 1] divided up into 8 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent X = [0, 1] divided up into 16 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent A picture of u(x) in blue and T (u(x)) in red with 4 children A picture of u(x) in blue and T (u(x)) in red with 8 children A picture of u(x) in blue and T (u(x)) in red with 16 children iterations of T with 4 children iterations of T with 8 children iterations of T with 16 children Set attractors after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game

8 1 Chapter 1 Introduction Circle inversion was introduced in Apollonius of Pregas book, Plane Loci, and has drawn interest in geometry. Loosely, the action of a circle map, which we mathematize later, is to move points along radial rays in such a way that the distance to the circle s centre is inverted. Mandelbrot introduced the early concepts of fractals in the 1970s and, in The Fractal Geometry of Nature, Mandelbrot discussed successive inversion with respect to a family of N circles. The literature applies the chaos game to circle inversion maps [1, 2] to explore the graphical aspect, while casually stating that there are contraction maps involved. In [3], the author extends the concept to star-shaped sets and uses the chaos game to iterate star-shaped set inversion maps to generate images. This paper is focussed on graphical images, once again with no mathematics being presented; the author merely writes that the maps are contractive. Our survey of the literature found no mathematical framework to support the work in [3], not even in the case of circles in

9 2 the plane with inversions performed with respect to centres. As we will see in Chapter 3, a circle inversion map is only contractive on part of its domain, making clear the need for care. In this thesis, we build the needed rigorous mathematical framework for the circle inversion and star-shaped set inversion settings. The structure of this thesis is as follows. In Chapter 2, we present a brief background of concepts on metric spaces, fixed point theory and iterated function systems (IFS). An example of an affine IFS is given at the end of the chapter for readers new to this material. In Chapter 3, we establish contractivity results for a system of two circle inversion maps. We then extend these results to a general system of non-touching circles and then finally to a general system of possibly overlapping circles. Once these results are established, we also show the existence of a unique set attractor to the system of set-valued circle inversion maps. Approximations of this fractal set can be drawn using the chaos game; we provide some examples at the end of Chapter 3. In Chapter 4, we introduce a counterexample to show that star-shaped set inversion maps are not contractive with respect to the Euclidean metric. Once a new metric is defined we can establish contractivity results as well as show the existence of a unique set attractor with respect to this new metric. Finally, in Chapter 5, we apply the circle inversion map to 1-D signal and function approximation. The framework involves constructing a fixed collection of circle inversion maps with domains and ranges both contained in the domain of the given

10 3 function of interest. The endeavor is more complicated than the traditional IFS approach, but it is an interesting mathematical exercise and it is fun.

11 4 Chapter 2 Background The concepts in this background section are discussed in further detail in [4]. 2.1 Metric Spaces A metric space is a set of like objects with a metric that defines the distance between two elements of the set. Definition 1. A metric d on a set X is a real valued function d(x, y) defined for all x, y X such that i. d(x, y) 0 (non-negativity) ii. d(x, y) = 0 if and only if x = y (identity) iii. d(x, y) = d(y, x) (symmetry) iv. d(x, y) d(x, z) + d(z, y) (triangle inequality)

12 5 The pair (X, d) is called a metric space. Definition 2. A sequence {x n } in a metric space (X, d) is a Cauchy sequence if for every ε > 0, N > 0 such that d(x n, x m ) < ε, where n, m > N. Definition 3. A sequence {x n } in a metric space (X, d) converges to x means that lim d(x n, x) = 0. n Theorem 1. Every convergent sequence in (X, d) is a Cauchy sequence. Proof. Suppose x n is a sequence that converges to x. Then by the definition of limit, ε > 0, N > 0 such that d(x n, x) < ε 2 whenever n > N. This means when n, m > N we have, d(x n, x m ) d(x n, x) + d(x, x m ) = d(x n, x) + d(x m, x) < ε 2 + ε 2 = ε {x n } is Cauchy. Definition 4. A metric space (X, d) is called complete if every Cauchy sequence in X converges to an element of X.

13 6 2.2 Fixed Point Theory Definition 5. A function f : X X is a contraction map on (X, d) if c [0, 1) such that d(f(x), f(y)) c d(x, y), x, y X, where c is called the contraction factor. (f shrinks the distances between points in X.) It is worth noting that it is easy to show that every contraction mapping is continuous. For convenience we denote by Con(X) the set of all contraction maps on X: Con(X) = {f : X X c [0, 1) s.t. d(f(x), f(y)) c d(x, y)}. A fundamental principle that is used in fractal-based methods is Banach s Fixed Point Theorem. Theorem 2. (Banach s Fixed Point Theorem, [4], Page 366) Let (X, d) be a complete metric space and f Con(X) with contraction factor c [0, 1). Then a unique x X such that f( x) = x where x is the unique fixed point of f and for any x X, lim d(f n (x), x) 0. n If we iterate a contraction map f on a complete metric space X, the result is straightforward. If we start at any point x 0 X, the iteration sequence {x n } generated by iterating f upon x 0 ; i.e. x n+1 = f(x n ), converges in the metric d to the fixed point x = f( x).

14 7 2.3 Iterated Function Systems We wish to extend the concepts of the previous section to encompass particular kinds of sets, thinking of these sets as points in an appropriate metric space. Definition 6. Let the Hausdorff set be defined by H(X) = {A X : A and A is compact }, the set of all nonempty compact subsets of X. Now that we have defined the space we are working on, we can define a metric on this space. Definition 7. For A, B H(X) we define the Hausdorff metric by { d H (A, B) = max sup x A inf y B d(x, y), sup y B inf x A } d(x, y). Theorem 3. Let (X, d) be a complete metric space. Let H(X) be the Hausdorff set of X. Then (H(X), d H ) is a complete metric space. Proof. The completeness proof is given in ([5], Theorem 1, Page 37). Remark 1. It is typical to use the same letter d in three different ways: point-to-point distance, point-to-set distance and set-to-set distance. The following discussion uses this notation: d(x, B) := inf d(x, y) y B

15 8 is the distance from the point x to the compact set B and d(a, B) := sup x A inf y B d(x, y) is the distance from the compact set A to B. The following four lemmas and additional theorem are needed in order to prove an upcoming theorem. Lemma 1. Let A, B, C H(X). Then d(a B, C) = max{d(a, C), d(b, C)} Proof. d(a B, C) = sup d(x, C) x A B { = max sup x A d(x, C), sup d(x, C) x B = max{d(a, C), d(b, C)}. } Lemma 2. Let A, B, C H(X). Then d(c, A B) min{d(c, A), d(c, B)}

16 9 Proof. d(c, A B) = sup d(x, A B) x C = sup x C inf y A B d(x, y). But, the infimum on A may be bigger than that on A B so we have sup x C inf y A B d(x, y) sup inf d(x, y) x C y A = d(c, A) and the same argument on B gives sup x C inf y A B d(x, y) sup inf d(x, y) x C y B = d(c, B). Thus, d(c, A B) min{d(c, A), d(c, B)} Lemma 3. Let A, B, C, D H(X). Then d H (A B, C D) max{d H (A, C), d H (B, D)} Proof. We start by applying Lemma 1, d H (A B, C D) = max{d(a B, C D), d(c D, A B)} = max{d(a, C D), d(b, C D), d(c, A B), d(d, A B)}

17 10 Now, use Lemma 2 to obtain the following four results d(a, C D) min{d(a, C), d(a, D)} d(a, C), d(b, C D) min{d(b, C), d(b, D)} d(b, D), d(c, A B) min{d(c, A), d(c, B)} d(c, A), d(d, A B) min{d(d, A), d(d, B)} d(d, B), d H (A B, C D) = max{d(a, C D), d(b, C D), d(c, A B), d(d, A B)} max{d(a, C), d(b, D), d(c, A), d(d, B)} max{max{d(a, C), d(c, A)}, max{d(b, D), d(d, B)}} = max{d H (A, C), d H (B, D)}. Lemma 4. Let A i, B i H(X), i = 1,..., N. Then ( N N d H A i, i=1 i=1 B i ) max 1 i N {d H(A i, B i )} Proof. This Lemma is the generalized form of Lemma 3 and can be easily proven by induction. Having established the completeness of (H(X), d H ) and some properties of d H, we now look at contraction mappings on this space.

18 11 Theorem 4. Let (X, d) be a complete metric space. Define the complete metric space (H(X), d H ). Let f Con(X) and define ˆf(S) = {f(x), x S}. Then i. ˆf : H(X) H(X), and ii. ˆf is a contraction mapping on (H(X), dh ). Proof. Proof of (i): Let A H(X), A is nonempty and compact which means that f(a) is nonempty. Let {x n } n=1 be an infinite sequence of points in A and let y n = f(x n ). Since A is compact, there exists an infinite subsequence of {x n } n=1, denoted {x im } m=1, which has a limit point x A. Since f is a contraction, we know it is continuous. Thus, it follows that: lim i m m = lim i m ) m = ( ) f lim i m, by continuity m ( = f x ). Since x A, we know that f ( x ) f(a). The infinite sequence {yn } n=1 in ˆf(A) contains a subsequence which converges to a point in ˆf(A) which implies that ˆf(A) is compact. Hence, ˆf : H(X) H(X).

19 12 Proof of (ii): d H ( ˆf(A), ˆf(B) ) { = max { = max d ( ) ( )} ˆf(A), ˆf(B), d ˆf(B), ˆf(A) sup inf x ˆf(A) y ˆf(B) d(x, y), sup inf y ˆf(B) x ˆf(A) d(y, x) { } = max sup inf d(f(a), f(b)), sup inf d(f(b), f(a)) a A b B b B a A { } max sup inf c fd(a, b), sup inf c fd(b, a) a A b B b B a A { } = c f max sup inf d(a, b), sup inf d(b, a) a A b B b B a A } = c f max {d(a, B), d(b, A)} = c f d H (A, B) where c f [0, 1) is the contraction factor of f. Definition 8. Let (H(X), d H ) be a complete metric space with S H(X) and define f i Con(X) with contraction factor c i [0, 1), for 1 i N. Then the set-valued maps defined by ˆf i (S) = {f i (x) x S}, and ˆ f(s) = N i=1 ˆf i (S) are called a contractive iterated function system, or IFS, on the metric space H(X). Our final theorem is due to Hutchinson, [6]. Theorem 5. Let (H(X), d H ) be a complete metric space. Let f i Con(X) with contraction factor c i [0, 1), for 1 i N. Then the set-valued mapping ˆf :

20 13 H(X) H(X) with action defined by ˆ f(s) = N i=1 ˆf i (S), S H(X) is contractive on (H(X), d H ), with contraction factor c = max 1 i N c i Proof. d H (ˆf(A), ˆf(B) ) = d H ( N i=1 max 1 i N d H ˆf i (A), N i=1 ˆf i (B) ) ( ˆf(A), ˆf(B) ), by Lemma 4 max c i d H (A, B), by Theorem 4 1 i N = c d H (A, B) Banach s Fixed Point Theorem can be applied to give the following corollary. Corollary 1. There exists a unique non-empty compact subset A H(X) such that i. ˆf(A) = N i=1 ˆf i (A) = A ii. d H (ˆf n (S), A ) 0 as n, S H(X). A is called the set attractor. As an example, consider the interval I 0 = [0, 1] and remove the open middle third interval to produce I 1 = [ 0, 1 3] [ 2 3, 1]. Once again remove the open middle third interval from the two closed intervals in I 1 to produce I 2 = [ 0, 1 9] [ 2 9, 3 9] [ 6 9, 7 9] [ 8 9, 1].

21 14 We can continue removing the middle third from each interval n times and see that the interval I n approaches a set attractor A. This attractor is known as the Cantor set. Using our previously established results above we can write this process using set-valued mappings. ˆf 1 (S) = {f 1 (x), x S} and ˆf 2 (S) = {f 2 (x), x S} where f 1 (x) = 1x and f 3 2(x) = 1x + 2. We can see that 3 3 [ ˆf 1 (I 0 ) = ˆf 1 ([0, 1]) = 0, 1 ] and 3 ˆf 2 (I 0 ) = ˆf 2 ([0, 1]) = and so I 1 can be written as I 1 = ˆf 1 (I 0 ) ˆf 2 (I 0 ) = We can continue in this fashion to produce [ 0, 1 ] [ ] 2 3 3, 1. [ ] 2 3, 1 I 0 = [0, 1] I 1 = ˆf 1 (I 0 ) ˆf 2 (I 0 ) = ˆf(I 0 ) I 2 = ˆf 1 (I 1 ) ˆf 2 (I 1 ) = ˆf(I 1 ). I n = ˆf 1 (I n 1 ) ˆf 2 (I n 1 ) = ˆf(I n 1 ), where our IFS is ˆf = { ˆf 1, ˆf 2 }. Applying Corollary 1 from earlier, we know our attractor A exists and satisfies A = ˆf(A) = ˆf 1 (A) ˆf 2 (A).

22 15 The set attractor A or Cantor set is composed of self-similar copies of itself. Fractals that are composed of self-similar copies of itself are called self affine. 2.4 The Chaos Game Following Section 2.3, having defined an N-map IFS with contraction mappings f i, i = 1,..., N, and attractor A, we could use the Deterministic Algorithm to construct an approximation of A. Such an algorithm would proceed as follows: given an N-map IFS with contraction maps ˆf = { ˆf 1,..., ˆf N }, the algorithm will give us ˆf(A) = A. We begin by letting S = { x 0 } be the set containing our starting point and applying ˆ f to give the first N points of the attractor as follows ˆ f(s) = N f i ( x 0 ) = { ˆf 1 ( x 0 ),..., ˆf N ( x 0 ))}. i=1 Now, in order to get a better approximation of A, we can compose ˆf n times so that every contraction map is applied to every possible point. It looks like the following ˆ f n (S) = { ˆf i1 ( x 0 ) ˆf in ( x 0 ))} for all possible (i 1,..., i n ), i j {1,..., N}. Notice that if S = { x 0 } is chosen to be in A, then all points in ˆf n (S) are in A. The major disadvantage to approximating the set attractor with this method is that it can be very computationally expensive. Instead, we can use the Chaos Game ([4], Ch. 6) to plot approximations of the set attractor A of ˆf. Choose an initial point x 0 X. With probabilities p i such N that p i = 1, select one of the contraction maps ˆf 1,..., ˆf N. Apply the map to i=1

23 16 x 0 to produce x 1 X. Continue randomly selecting contraction maps to produce the sequence { x n } n=0. If x 0 is not in A, the points in the sequence of points that is generated nevertheless grows arbitrarily close to A. In fact, the sequence { x n } n=0 is dense on the attractor, (Elton s Ergodic Theorem, [5]). An example of an affine IFS generated by the Chaos Game is the Sierpinski Gasket illustrated in the following example. Example 1. Let X = [0, 1] [0, 1] and S H(X). We will define the maps for the Sierpinski Gasket and build the fixed point set A of the 3-map IFS ˆf(A) = 3 for all (x, y) S. ˆf 1 (S) = ˆf 2 (S) = ˆf 3 (S) = x + 0, y 0 x 1 + 2, y 0 x + y , i=1 ˆf i (A). Figure 2.1 illustrates an approximation of the Sierpinski Gasket, the set attractor of this IFS. It was drawn using the chaos game. A small solid circle is drawn at each iteration point.

24 17 Figure 2.1: The set attractor A of the 3 maps define in Example 1 called the Sierpinski Gasket. With reference to Figure 2.1, we see that ˆf 1 (A) is the blue triangle on the lower left, ˆf2 (A) is the green triangle on the lower right, and ˆf 3 (A) is the upper orange 3 triangle. As expected, the figure illustrates that A = ˆf i (A). i=1

25 18 Chapter 3 Collections of Touching and Non-Touching Circles 3.1 Inversion in a Circle We open this chapter by introducing our fundamental notation and definitions for circle inversion maps. We then consider the setting of two non-touching circles before moving on to general systems, and we also finally allow for touching and overlapping circles. Let C R 2 be a solid circle with centre õ and boundary C; then the following property is true, as a result of the Jordan curve theorem. Property 1. C divides R 2 into 3 pieces the interior of the circle, C int the boundary of the circle, C where C = C int C the exterior of the circle, C ext In the plane, the continuous boundary of C can be parametrized in terms of polar

26 19 coordinates. Given C and õ, C can be represented as: α(t) = r w(t) + õ = r(cos(t), sin(t)) + õ, t [0, 2π] (3.1) where r > 0 is the radius and w(t) represents the unit radial vector with angle t. Definition 9. Given C in R 2 with fixed centre õ and radius r we define the map T on R 2 \ {õ} as follows. For x R 2 \ {õ}, T ( x) is the unique point along the radial ray from õ through x satisfying T ( x) õ x õ = Π( x) õ 2 = r 2, x R 2 \ {õ}, (3.2) where is the Euclidean norm and Π( x) is the projection of x onto C along the radial ray from õ to x. Alternatively, we can write equation (3.2) as d 2 (T ( x), õ)d 2 ( x, õ) = d 2 2(Π( x), õ) = r 2, x R 2 \ {õ}, (3.3) where d 2 ( x, ỹ) = x ỹ. Now that the inversion map has been defined in equation (3.3) the following lemma divides the action of T into three cases. Lemma 5. Given a circle C with inversion map T we have the following: i. T : C int \ {õ} C ext, ii. T : C ext C int, and iii. T : C C.

27 20 Proof. Follows from the definition of T and Property 1. When a point ỹ = T ( x) satisfies (3.3), ỹ is called the inverse of x; in this case, T (ỹ) = x, as well. It is true that T ( b) = b, for b C. A visualization of circle inversion is given in Figure 3.1. Figure 3.1: ỹ satisfying (3.3) is called the inverse of x, denoted T ( x) = ỹ. The upcoming definition generalizes equation (3.3) in order for the inversion map to be applied to a collection of circles. Definition 10. Given a collection of possibly overlapping circles C i, i = 1,..., N, with centres õ i and radii r i we define the map T i on R 2 \ {õ i } as follows. For x R 2 \ {õ i }, T i ( x) is the unique point along the radial ray from õ i through x satisfying T i ( x) õ i x õ i = Π i ( x) õ i 2 = r 2 i, x R 2 \ {õ i }, x o i, (3.4) where Π i ( x) is the projection of x onto C i along the radial ray from õ i to x. A visualization of circle inversion on a collection of circles is given in Figure 3.2.

28 21 Figure 3.2: x 3 is the inverse of x 3 with respect to C 2, denoted T 2 ( x 3 ) = x 3. We can build (3.4) when x and T i ( x) satisfy the expression T i ( x) = õ i + Π i( x) õ i 2 x õ i 2 ( x õ i ) T i ( x) õ i = Π i( x) õ i 2 x õ i 2 x õ i T i ( x) õ i x õ i 2 = Π i ( x) õ i 2 x õ i T i ( x) õ i x õ i = Π i ( x) õ i 2 Any point x R 2 can be represented in terms of the parametrization for the circle C i : x = ar i w(t) + õ i for some a 0, t [0, 2π], (3.5) where, as a reminder, w(t) = (cos(t), sin(t)). If a < 1 then x C i,int, if a = 1 then x C i and if a > 1 then x C i,ext. Notice that Equation (3.5) is just rescaled polar coordinates with õ i as the origin.

29 22 For x R 2 \ {õ i }, we have a > 0, and we find that T i ( x) = T i (ar i w(t) + õ i ) = Π i(ar i w(t) + õ i )) õ i 2 (ar ar i w(t) + õ i õ i 2 i w(t) + õ i õ i ) + õ i = r i w(t) 2 ar i w(t) (ar 2 i w(t)) + õ i = r2 i w(t) 2 r 2 i a2 w(t) 2 (ar i w(t)) + õ i = 1 a r i w(t) + õ i. (3.6) Thus, we see clearly why T i is called a circle inversion map: under the action of T i, the radial scaling factor a i > 0 becomes a radial scaling factor of 1 a i. 3.2 Mathematical Framework for Two Non-Touching Circles We now develop a mathematical framework for a system of two non-touching circles. We begin by observing properties along the radial rays and defining some notation for use in upcoming results. Given a circle C 1 in R 2 with centre õ 1, let Rõ1 (õ 1, b 1 ) = semi-infinite ray from õ 1 through b 1 C 1, with endpoint õ 1 Rõ1, (õ 1, b 1 ) = infinite line through õ 1 and b 1 Rõ1,int(õ 1, b 1 ) = Rõ1 (õ 1, b 1 ) C 1 Rõ1,ext(õ 1, b 1 ) = Rõ1 (õ 1, b 1 ) \ {Rõ1,int(õ 1, b 1 ), b 1 } The figure below illustrates each section of the radial ray appearing in the order they were defined.

30 23 (a) (b) (c) Figure 3.3: (a) Semi infinite ray (b) Infinite ray (c) Interior along ray and (d) Exterior along ray. (d) We now present a lemma that states the properties of an inversion map T along

31 24 the radial ray of a circle. Lemma 6. For C 1 R 2 with centre õ 1, b 1 C 1, ã R 2 \ {õ 1 } and T 1 the inversion map for C 1, we have the following: i. T 1 : Rõ1,int(õ 1, b 1 ) \ {õ 1, b 1 } Rõ1,ext(õ 1, b 1 ), ii. T 1 : Rõ1,ext(õ 1, b 1 ) Rõ1,int(õ 1, b 1 ), iii. T 1 ( b 1 ) = b 1, iv. T 2 1 : Rõ1,int(õ 1, b 1 ) \ {õ 1 } Rõ1,int(õ 1, b 1 ), v. T 2 1 : Rõ1,ext(õ 1, b 1 ) Rõ1,ext(õ 1, b 1 ), and vi. T 2 1 (ã) = ã. Proof. (i), (ii) and (iii) follow from Lemma 5. (iv) and (v) follow from the fact that T i maps the radial ray Rõ1 (õ 1, b 1 ) to itself and by iterating Lemma 5 a second time. (vi) follows from iterating Lemma 5 twice as well as looking at our inversion equation (3.4). It is easy to see that a point ã R 2 \{õ 1 } only has one possible choice to satisfy equation (3.4) and vice versa. Thus, Lemma 5 guarantees that if we start outside C 1, we will land outside after two iterations of T 1 and equation (3.4) guarantees iterating T 1 twice will map back to the point at which we started. Now, given two non-touching circles C 1 and C 2 in R 2 with centres õ 1 and õ 2,

32 25 respectively, define (C i ) int, (C i ) ext, C i, T i, b i and projection maps Π i. Let Rõ1 õ 2 ( x, ỹ) = portion of the open line segment between õ 1 and õ 2 from x to ỹ Rõ1 õ 2 (õ i, b i ) = Rõ1 õ 2 (õ 1, õ 2 ) C i Rõ1 õ 2 ( b 1, b ( 2 ) = Rõ1 õ 2 (õ 1, õ 2 )\ Rõ1 õ 2 (õ 1, b 1 ) Rõ1 õ 2 (õ 2, b ) 2 ) Similar to the previously defined notation, the figure below illustrates these three line segments in the order they were defined. The subsequent lemma presents the properties of T on these line segments.

33 26 (a) (b) (c) Figure 3.4: (a) Ray from x to ỹ (b) Ray from õ 1 to b 1 and (c) Ray from b 1 to b 2. Lemma 7. For i, j {1, 2} with i j, and the preceding setup, we have i. T i : Rõi õ j (õ j, b i ) Rõi õ j (õ i, b i ) ii. T i : Rõi õ j (õ i, b i ) Rõi õ j (õ j, b i )

34 27 Proof. (i) If we are given a point x on the line Rõi õ k (õ j, b i ), j i, then x (C i ) ext. If we apply T i to x (C i ) ext, then by Lemma 6(ii) we know T i must take x to (C i ) int along the line Rõi õ j (õ i, b i ). (ii) If we are given a point x on the line Rõi õ j (õ i, b i ), j i, then x (C i ) int. If we apply T i to x (C i ) int, then by Lemma 6(i) we know T i must take x to x (C i ) ext along the line Rõi õ j (õ j, b i ). Theorem 6. Let C 1 and C 2 be two non-touching circles in R 2 with centres õ 1, õ 2, respectively. Then there exists a c [0, 1) such that, d 2 (T 1 ( x 1 ), T 1 ( x 2 )) c d 2 ( x 1, x 2 ), x 1, x 2 Rõ1 õ 2 (õ 2, b 2 ) Proof. Introduce the x-axis so that Rõ1 õ 2 (õ 2, b 2 ) lies along it. For x 1 Rõ1 õ 2 (õ 2, b 2 ), T 1 ( x 1 ) is defined by T 1 ( x 1 ) õ 1 x 1 õ 1 = b 1 õ 1 2. Since we oriented the x-axis in this way, x 1, b 1 and õ 1 correspond to numbers along the x-axis; we think of T 1 as mapping numbers to numbers on the axis and therefore we can drop the tildes in the following argument. Without loss of generality let o 1 < o 2 along the x-axis which implies that b 1 < b 2. So, T 1 (x 1 ) o 1 = (b 1 o 1 ) 2 x 1 o 1 for x 1 R o1 o 2 (o 2, b 2 ).

35 28 Observe that, x 1 > o 1 T 1 (x 1 ) > o 1 then, T 1 (x 1 ) = o 1 + (b 1 o 1 ) 2 (x 1 o 1 ). Now, T 1(x 1 ) = (b 1 o 1 ) 2 (x 1 o 1 ) 2 = T 1(x 1 ) (b 1 o 1 ) 2 (b 2 o 1 ), x 2 1 R o1 o 2 (o 2, b 2 ). By the Mean Value Theorem we get, T 1 (x 1 ) T 1 (x 2 ) = T 1(ξ) x 1 x 2, ξ R o1 o 2 (o 2, b 2 ) (b 1 o 1 ) 2 (b 2 o 1 ) 2 x 1 x 2 = c x 1 x 2, 0 < c < 1. Returning to the vector notation, we have proved there exists a c [0, 1) such that, d 2 (T 1 ( x 1 ), T 1 ( x 2 )) c d 2 ( x 1, x 2 ), x 1, x 2 Rõ1 õ 2 (õ 2, b 2 ) Theorem 6 proves contractivity of the circle inversion map along the radial ray connecting the centres of two circles. Using our notation defined in the previous chapter this means T Con(X), where X = Rõ1 õ 2 (õ 2, b 2 ). Theorem 6 also leads to the following significant corollary.

36 29 Corollary 2. There exists a c [0, 1) such that d 2 ((T 1 T 2 )( x 1 ), (T 1 T 2 )( x 2 )) c d 2 ( x 1, x 2 ), x 1, x 2 Rõ1 õ 2 (õ 1, b 1 ), where we note that T 1 T 2 : Rõ1 õ 2 (õ 1, b 1 ) Rõ1 õ 2 (õ 1, b 1 ). Proof. The result follows from two applications of Theorem 6. In Corollary 2, (Rõ1 õ 2 (õ 1, b 1 ), ) is a complete metric space and T 1 T 2 is a contraction map from (Rõ1 õ 2 (õ 1, b 1 ), ) to itself. Banach s Fixed Point Theorem applies: there exists a unique x on the radial ray in C 1 such that (T 1 T 2 )( x) = x. A similar statement can be made for T 2 T 1 and the radial ray in C 2. Beginning at any point on the ray from õ 1 to õ 2, repeated alternating application of T 1 and T 2 approaches the two cycle consisting of these two fixed points. Up to this point we have only looked at cases dealing with circles in R 2 but all previous definitions and results can be applied to spheres in R 3 and hyperspheres in their associated dimension. In each n-dimensional setting we can work in a plane containing x 1, x 2 and õ Collections of Circles We present the following result in R 2 with the understanding that this may be a two dimensional subspace of a higher dimensional setting. In this result we consider a collection of possibly touching or overlapping circles in the plane.

37 30 Lemma 8. Given a circle C 1 with centre õ 1, radius r 1 and inversion map T 1, d 2 (T 1 ( x 1 ), T 1 ( x 2 )) = 1 a 1 a 2 d 2 ( x 1, x 2 ), x 1, x 2 R 2 \ {õ 1 }, where, for j = 1, 2, x j = a j r 1 w 1 (t j ) + õ 1 for some t j [0, 2π], a j > 0. Proof. We consider two cases. Case 1: x 1, x 2 and õ 1 are co-linear lying on the ray R. Then in a plane containing R, we can parametrize the boundary of C 1 as in (3.1). For b C 1, for some fixed t 1 [0, 2π] we have, b = Π1 ( x 1 ) = Π 1 ( x 2 ) = α 1 (t 1 ) and x j = a j r 1 w(t 1 ) + õ 1, a j > 0, j = 1, 2 where w(t 1 ) = (cos(t 1 ), sin(t 1 )). Using this parametrization of the boundary we have, d 2 (T 1 ( x 1 ), T 1 ( x 2 )) = T 1 ( x 1 ) T 1 ( x 2 ) ( ) = Π1 ( x 1 ) õ 1 2 ( x x 1 õ õ 1 ) + õ 1 ( Π1 ( x 2 ) õ 1 2 ( x x 2 õ õ 1 ) + õ 1) ( ) ( ) b õ 1 2 = x 1 õ 1 ( x b õ õ 1 ) x 2 õ 1 ( x 2 2 õ 1 ) ( ) ( = r1 w(t 1 ) 2 a 1 r 1 w(t 1 ) (a r1 w(t 1 ) 2 1r 2 1 w(t 1 )) a 2 r 1 w(t 1 ) (a 2r 2 1 w(t 1 ))) ( ) ( = 1 1 r 1 w(t 1 ) r 1 w(t 1 )) a 1 a ( 2 = 1 1 ) r 1 w(t 1 ) a 1 a 2.

38 31 We arrive at d 2 (T 1 ( x 1 ), T 1 ( x 2 )) = = = = 1 (a 1 a 2 )r 1 w(t 1 ) a 1 a 2 1 a 1 r 1 w(t 1 ) a 2 r 1 w(t 1 ) a 1 a 2 1 a 1 r 1 w(t 1 ) + o 1 (a 2 r 1 w(t 1 ) + o 1 ) a 1 a 2 1 d 2 ( x 1, x 2 ). a 1 a 2 Case 2: x 1, x 2 and õ 1 are not co-linear. Once again the boundary C 1 can be parametrized as in (3.1). There exists t 1, t 2 [0, 2π], t 2 > t 1, such that Π 1 ( x j ) = α 1 (t j ) and x j = a j r 1 (cos(t j ), sin(t j )) + õ 1, j = 1, 2. Now, d 2 2( x 1, x 2 ) = (a 1 r 1 cos(t 1 ) a 2 r 1 cos(t 2 ), a 1 r 1 sin(t 1 ) a 2 r 1 sin(t 2 )) 2 = a 2 1r1 2 cos 2 (t 1 ) + a 2 2r1 2 cos 2 (t 2 ) 2a 1 a 2 r1 2 cos(t 1 ) cos(t 2 ) + a 2 1r1 2 sin 2 (t 1 ) + a 2 2r1 2 sin 2 (t 2 ) 2a 1 a 2 r1 2 sin(t 1 ) sin(t 2 ) = a 2 1r1 2 + a 2 2r1 2 2a 1 a 2 r1 2 cos(t 2 t 1 ). (3.7) By (3.6), T 1 ( x j ) = 1 a j r 1 w(t j ) + õ 1.

39 32 Thus we have, d 2 2(T 1 ( x 1 ), T 1 ( x 2 )) = ( 1 r 1 cos(t 1 ) 1 r 1 cos(t 2 ), 1 r 1 sin(t 1 ) 1 ) 2 r 1 sin(t 2 ) a 1 a 2 a 1 a 2 = 1 r a r 1 a r1 2 cos(t 2 t 1 ) 2 a 1 a 2 = 1 (a 2 a 1r 2 1a a 2 2r1 2 2a 1 a 2 r1 2 cos(t 2 t 1 )) 2 = 1 d 2 a 2( x 2 1a 2 1, x 2 ), 2 which upon taking the square root gives the desired result. Remark 2. Notice that in Lemma 8 we are no longer restricting ourselves to points along the radial ray connecting centres of circles but rather any point in the plane, excluding the centre of the circle with respect to which we invert. We will use Lemma 8 and the following definition when establishing the subsequent contractivity result. Definition 11. Given a collection of non-touching circles C i, i = 1,..., N, with centres õ i and radii r i, for x = a j r i w(t) + õ i R 2, a j > 0, t [0, 2π], we define a i,min = min {a j} > 1 x C j,j i corresponding to the radial scaling of the closest point to õ i in all other circles C j, j i. Theorem 7. T i : C j C i, i j, is contractive. That is, there exists a c [0, 1) such that, d 2 (T i ( x 1 ), T i ( x 2 )) c d 2 ( x 1, x 2 ), x 1, x 2 C j, i j.

40 33 Proof. By Lemma 8 we have that Using the previous definition we get d 2 (T i ( x 1 ), T i ( x 2 )) = 1 a 1 a 2 d 2 ( x 1, x 2 ). d 2 (T i ( x 1 ), T i ( x 2 )) 1 a 2 i,min d 2 ( x 1, x 2 ) = c d 2 ( x 1, x 2 ). This theorem is important in establishing that the inversion map T i corresponding to circle C i is contractive on the exterior, C i,ext, and, as we see after additional thought, expansive on the interior, C i,int. Since T i : C i,int C i,ext is expansive, we redefine the map as follows in order to enforce that T i is non-expansive on the interior of C i. Furthermore, this redefined map will work in the setting of possibly overlapping circles. To facilitate this result, we must also redefine the space X that we are working in. We say circles are overlapping even when they merely touch, i.e. when they share a common boundary point. Definition 12. Given a collection of possibly overlapping circles C i, i = 1,..., N, with centres õ i, define X = N i=1 C i

41 34 and the intersection of circle C i with all other circles as We require that I i = N (C i C j ). j=1 j i N C i =. That is, there is no point in all circles. Now, relative to each C i, define i=1 T i : X X by T i ( x) = T i ( x) if x C j, j i x if x C i We must also redefine a i,min in order to deal with possible intersections of circles. Definition 13. Given a collection of possibly overlapping circles C i, i = 1,..., N, with centres õ i and radii r i, for x = a j r i w i (t) + õ i R 2, a j > 0, t [0, 2π], we define the following two cases: (a) if C j C i = then a i,j,inf = min {a j } > 1 x C j, x C i j i corresponding to the radial scaling of the closest point to õ i in all other circles C j with C j C i =, j i. (b) if C j C i then a i,j,inf = inf {a j } = 1 x C j, x C i j i

42 35 corresponding to the radial scaling of the point on the boundary of C i in C j C i. We can combine the cases to conclude = 1 if C j C i a i,j,inf = > 1 if C j C i =. Theorem 8. T i : X X satisfies (i) d 2 (T i ( x 1 ), T i ( x 2 )) c d 2 ( x 1, x 2 ) for some c [0, 1), x 1 C j, x 2 C k, C j C i = and C k C i =, j, k i (ii) d 2 (T i ( x 1 ), T i ( x 2 )) = d 2 ( x 1, x 2 ) x 1 C j, x 2 C k, C j C i or C k C i, x 1, x 2 I i, j, k i (iii) d 2 (T i ( x 1 ), T i ( x 2 )) = d 2 ( x 1, x 2 ) x 1, x 2 C i (iv) d 2 (T i ( x 1 ), T i ( x 2 )) c d 2 ( x 1, x 2 ) for some c [0, 1), x 1 C i, x 2 C j, C j C i =, j i (v) d 2 (T i ( x 1 ), T i ( x 2 )) = d 2 ( x 1, x 2 ) x 1 C i, x 2 C j, C j C i, x 2 I i, j i Proof. Proof of (i): C j C i = and C k C i = so the result follows from Theorem 7 and our a i,j,inf > 1 which implies c < 1. Proof of (ii): C j C i or C k C i so we have a i,j,inf = 1 which implies c = 1 and the result follows.

43 36 Proof of (iii): The result follows immediately since T i is the identity map in this case. To prove (iv) and (v) we first consider a 2 > 1 > a 1 0, and, for values t 1, t 2 [0, 2π], d 2 2( x 1, x 2 ) = r 2 i ( a a 2 2 2a 1 a 2 cos(t 2 t 1 ) ), (3.8) using (3.7), and d 2 2(T i ( x 1 ), T i ( x 2 )) = d 2 2( x 1, T i ( x 2 )) ( = r i a 1 cos(t 1 ) 1 cos(t 2 ), a 1 sin(t 1 ) 1 ) 2 sin(t 2 ) a 2 a ( 2 = ri 2 a a ) 1 cos(t a 2 2 t 1 ) 2 a 2 = r2 ( i a 2 a 2 2 a a 1 a 2 cos(t 2 t 1 ) ). (3.9) 2 The bracketed expression in (3.9) is less than or equal to the bracketed expression in (3.8) provided that a 2 2a a a 2 2, where a 2 > 1 > a 1 0. (3.10) Hence, letting y = a 2 1 and z = a 2 2, we consider the function f(y, z) = y yz + z. If we show that f(y, z) 1 for y [0, 1) and z > 1, then (3.10) holds. We write f(y, z) = y yz + z = y(z 1) + (z 1) + 1 = (z 1)(1 y) + 1 and we see that f(y, z) > 1 for y < 1 and z > 1, proving the result. Using the above, in (iv) we have C j C i = so c = 1 a i,j,inf < 1, proving our statement in case (iv) of the theorem. In (v), C j C i so we have c = 1 a i,j,inf = 1 and the result follows.

44 37 We have been able to prove that our newly defined T i in the case of possibly overlapping circles is non-expansive and can now move to compositions of the T i map. First, we will define the following for use in our upcoming result. Definition 14. Given a collection of possibly overlapping circles C i, i = {1,..., N}, let σ i {1,..., N} for each i. Let σ be a sequence of distinct digits (σ 1, σ 2,..., σ M ), where M N. Now, define the set of the intersection of M overlapping circles as P M = σ s.t. length(σ)=m ( M ) C σi. In other words, P M is the set of all points in M circles. Now, i=1 M max = max{m P M } is the maximum number of overlaps with the requirement that M max < N. That is, there is a point x X such that x is in M max distinct circles, and there is no x X such that x is in more than M max distinct circles. Now, using Theorem 8 and Definition 14 we can prove the following contractivity result involving compositions of circle inversion maps. Theorem 9. Let i 1,..., i Mmax+1 {1,..., N} be distinct. Assume C i1 is a circle that does not touch at least one other circle. Then (T i1 T i2 T i1 T immax+1 ): X X, is contractive: d 2 ((T i1 T i2 T i1 T immax+1 )( x 1 ), (T i1 T i2 T i1 T immax+1 )( x 2 )) c d 2 ( x 1, x 2 ) for some c [0, 1), x 1, x 2 X

45 38 Proof. Let s start by considering the simple case M max = 1: d 2 ((T i1 T i2 )( x 1 ), (T i1 T i2 )( x 2 )) = d 2 (T i1 (T i2 ( x 1 )), T i1 (T i2 ( x 2 ))). Since M max = 1 we know that the requirement in Definition 12 is satisfied because there are only two circles to choose from. Now, if (a) T i2 ( x 1 ), T i2 ( x 2 ) C i1 or (b) T i2 ( x 1 ) C i1, T i2 ( x 2 ) C i1 then by Theorem 8 (i) and (iv) respectively, we have d 2 (T i1 (T i2 ( x 1 )), T i1 (T i2 ( x 2 ))) c d 2 ( x 1, x 2 ) but, if (c) T i2 ( x 1 ), T i2 ( x 2 ) C i1 then by Theorem 8 (iii) we get d 2 (T i1 (T i2 ( x 1 )), T i1 (T i2 ( x 2 ))) = d 2 (T i2 ( x 1 ), T i2 ( x 2 )) and since T i2 ( x 1 ), T i2 ( x 2 ) C i1 and M max = 1 then we know that T i2 ( x 1 ), T i2 ( x 2 ) C i2 and we get d 2 (T i2 ( x 1 ), T i2 ( x 2 )) 1 a i2,j,inf d 2 ( x 1, x 2 ) = c d 2 ( x 1, x 2 ). Now, we consider the case M max = k: d 2 ((T i1 T i2 T i1 T ik+1 )( x 1 ), (T i1 T i2 T i1 T ik+1 )( x 2 )) = d 2 (T i1 (T i2 T i1 (T ik+1 ( x 1 ))), T i1 (T i2 T i1 (T ik+1 ( x 2 )))) and similarly to the case of M max = 1 we have one of two possible cases assuming that C i1 and C i2 do not touch. If (a) (T i2 T i1 T ik+1 )( x 2 ), (T i2 T i1 T ik+1 )( x 2 ) C i1 or

46 39 (b) (T i2 T i1 T ik+1 )( x 2 ) C i1, (T i2 T i1 T ik+1 )( x 2 ) C i1 then d 2 (T i1 (T i2 T i1 T ik+1 ( x 1 )), T i1 (T i2 T i1 T ik+1 ( x 2 ))) c d 2 ((T i2 T i1 T ik+1 )( x 1 ), (T i2 T i1 T ik+1 )( x 2 )) = c d 2 ( x 1, x 2 ). However, if (T i2 T ik+1 )( x 2 ), (T i2 T ik+1 )( x 2 ) C i1 or C i1 C im then we strip away the two outermost maps by Theorem 8 (ii), (iii) or (v), depending on the case we are in, to get d 2 (T i1 (T i2 T i1 T ik+1 ( x 1 )), T i1 (T i2 T i1 T ik+1 ( x 2 ))) = d 2 ((T i1 T i3 T i1 T ik+1 )( x 1 ), (T i1 T i3 T i1 T ik+1 )( x 2 )) = d 2 (T i1 (T i3 T i1 T ik+1 ( x 1 )), T i1 (T i3 T i1 T ik+1 ( x 2 ))) and once again have the same two cases to deal with. We continue in this fashion and strip off the outermost maps until we are in a position to take out a contraction factor and give the result. In the longest case possible without obtaining a contraction factor we would strip off all of the outermost maps until we are only left with T i1 T ik+1. Since k = M max and there are k + 1 maps, we know that C i1 C ik+1 = and T ik+1 ( x 1 ), T ik+1 ( x 2 ) are contained in C i1 if all other maps have been stripped away. Thus we have d 2 (T i1 T ik+1 ( x 1 ), T i1 T ik+1 ( x 2 )) 1 a ik+1,j,inf d 2 ( x 1, x 2 ) = c d 2 ( x 1, x 2 ).

47 40 This shows that the case of M max = k holds. Therefore, we have proved the result holds for all k. Remark 3. If circle C i1 does not overlap C ij, then any composition of maps that contains the sub-composition T i1 T ij is a contraction. The idea is that such a composition will access Theorem 8 (i) or (iv) to give contractivity. By Theorem 9 and Banach s Fixed Point Theorem, for each j we can say that there exists a unique x k in C i1 such that (T i1 T ij )( x k ) = x k for each choice of k. In fact, by Corollary 2 we know that each such fixed point lies on the appropriate radial ray, Rõi1 õ ij (õ i1, b i1 ). We now formulate this multi-circle framework in the context of sets. For a set S R 2, define (as usual) ˆ T i (S) = {T i ( x), x S} Recall from Section 2.3 our definitions for a Hausdorff set, Hausdorff metric and that (H(X), d H ) is complete for our upcoming results related to the multi-circle framework stated above. Theorem 10. Following Remark 3, let C i1 and C ij, j 1, be two non-overlapping circles. Then 1. ˆT i1 ˆT ij : H(X) H(X), and 2. ˆT i1 ˆT ij is contractive on (H(X), d H ).

48 41 Proof. The first claim follows because T i1 T ij : H(X) H(X). The proof relies on the continuity of the map ([5], Lemma 2, page 80). By Theorem 9, we can denote by c j the contractivity factor with respect to d 2 of ˆT i1 ˆT ij. Let A, B H(X) then, d H ((ˆT i1 ˆT ) ij (A), (ˆT i1 ˆT ) ) ij (B) = max sup inf d 2 ( x 1, x 2 ), x 1 (T i1 T ij )(A) x 2 (T i1 T ij )(B) sup inf d 2 ( x 1, x 2 ) = max c j max x 1 (T i1 T ij )(B) x 2 (T i1 T ij )(A) { sup inf d 2 x 1 A x 2 B sup x 1 B { sup (( T i1 T ij ) ( x1 ), ( T i1 T ij ) ( x2 ) ), inf d (( ) 2 T i1 T ij ( x1 ), ( ) T i1 T ij ( x2 ) )} x 2 A x 1 A = c j d H (A, B). inf d 2( x 1, x 2 ), sup x 2 B x 1 B } inf d 2( x 1, x 2 ), by Theorem 9 x 2 A We reach our final theorem of the chapter, which establishes the contractivity of the union of the maps in Theorem 10. Theorem 11. For A H(X), define ˆT (A) = 1. ˆT : H(X) H(X), and N j=2 C ij C i1 = (ˆT i1 ˆT ij) (A). Then 2. ˆT is contractive on (H(X), d H ).

49 42 Proof. The first claim follows from Theorem 10 because the union is finite. For the second claim we have d H (ˆT (A), ˆT (B) ) = d H N j=2 C ij C i1 = (ˆT i1 ˆT ij) (A), N j=2 C ij C i1 = (ˆT i1 ˆT ij) (B) max d H ((ˆT i1 ˆT ) ij (A), 2 j N C ij C i1 = (ˆT i1 ˆT ) ij) (B), property of d H max c j d H (A, B), by Theorem 10 2 j N C ij C i1 = = c d H (A, B). Finally, by Theorem 11 and Banach s Fixed Point Theorem we can conclude that there exists a unique fixed point A H(X), in fact with A C i1, the set attractor of ˆT, satisfying ˆ T (A) = A. At this point, we could play the chaos game with the maps T i1 T ij, C i1 C ij, j 1 to generate an approximation of the attractor A (in C i1 ) of the operator in Theorem 11. In fact, if there are several circles that do not overlap at least one other circle, i 1 in Theorem 11 can take on several values. The union of the corresponding maps in

50 43 Theorem 11 has a set attractor living in the union of all of these circles. We can use the chaos game to draw an approximation. However, it is desirable to avoid selecting pairwise compositions when playing the chaos game. For example, with the 3-circle setup in Figure 3.5, there are no pairwise 3 compositions that fit the above setup, even though C i =. With this in mind, we present one final construction. i=1 Figure 3.5: A system of three circles with the intersection of all of them equal to the empty set. Let X i = X \ C i = C c i, the complement of C i in the universe X = T i : X i C i is contractive by Theorem 8 (since T i = T i on X). N C i. Then i=1 In addition, ˆT i : H(X i ) H(C i ) is contractive by Theorem 10 (replacing ˆT i1 ˆT ij with ˆT i ). Finally, define ˆT (A) = N i=1 A H(X i ) ˆT i (A), A H(X).

51 44 The domain of ˆT is N X i = i=1 = N i=1 C c i ( N ) c C i, by de Morgan s Law i=1 = ( ) c = X. Thus ˆT : H(X) H(X) and ˆT is contractive, using Theorem 11 with ˆT i1 ˆT ij replaced by T i. As a result, T has a unique attractor A H(X). We can draw an approximation of A by playing the chaos game with the maps T i, provided we don t apply T i to points inside C i, or equivalently by playing the chaos game with T i, defined to be the identity on C i in Definition Application of the Chaos Game In [3], the Chaos Game from 2.4 is modified in a few ways. They omit the first 20 points from the plot: since the starting point, x 0, is outside of all circles, the first few points will lie outside of the approximation of the attractor. They also only select a single circle inversion map T i at each step, and they discard the selection if the current iterate is in C i, driven by the casual observation that T i is not a contraction in this case. We note that the framework described in the preceding paragraph doesn t discard selections; if it were implemented in terms of selecting single maps T i, then

52 45 an iterate in C i would not move if T i is selected. To avoid the repeated point in our sequence, we can discard it, arriving at the same Chaos Game implemented in [3]. Examples of fractals generated by a multi-circle system are given in Figure 3.6. The fractals in each of the six setups are produced by including different amounts of circles, circles of different radii and changing the placements of the circles.

53 Figure 3.6: Set attractors after 5,000 iterations of the chaos game. 46

54 47 Chapter 4 Collections of Star-Shaped Sets 4.1 Star-Shaped Set We open this chapter by introducing the concepts of a star-shaped set, related definitions associated with parametrizing boundaries, and star-shaped set inversion maps. Once notation and definitions are established we look to extend contractivity results to the setting of systems of star-shaped sets. Definition 15. A closed, bounded n-dimensional subset S of R n is star-shaped if S contains at least one interior point õ with the property that each ray emanating from õ pierces the continuous boundary of S, S, at exactly one point. The set of all such points õ is called the kernel of S, denoted kernel(s). In Chapter 3, we discussed circles and circle inversion maps. In R 2, circles are special star-shaped sets for which the kernel consists of all interior points; the same statement is true for other familiar convex shapes, such as ellipses, triangles and

55 48 rectangles. In Figure 4.1, we illustrate the definition, including non-convex shapes that are star-shaped. Figure 4.1: An illustration of 2 star-shaped sets with their kernels highlighted plus one not a star-shaped set. For a star-shaped set S, the continuous boundary of S can be parametrized in terms of polar coordinates. Given S R 2 and any õ kernel(s), S can be represented as: α(t) = r(t) w(t) + õ, t [0, 2π] (4.1) where w(t) = (cos(t), sin(t)) is the angular component of α(t) and r(t) is the radius at angle t. Definition 16. Given S in R 2 with õ a chosen fixed point in kernel(s), we define the map T on R 2 \ {õ} as follows. For x R 2 \ {õ}, T ( x) is the unique point along the radial ray from õ through x satisfying T ( x) õ x õ = Π( x) õ 2, x R 2 \ {õ} (4.2)

56 49 where is the Euclidean norm and Π( x) is the projection of x onto S along the radial ray from õ to x. We call õ the point of inversion. For a given x represented as x = ar(t 1 ) w(t 1 ) + õ, t 1 [0, 2π] and a > 0, we can write the projection of x as Π( x) = r(t 1 ) w(t 1 ) + õ. In the case of the circle of Chapter 3, õ is the centre and r(t) = r, the radius of the circle. As a result, in Definition 9 of the inversion map, we find r 2 on the right hand side. Compare to (4.2), where, upon using the preceding representation for x, we get (r(t 1 )) 2 on the right. The variability of the radius complicates both the definition of the inversion map and the construction needed to establish its contractivity. These complications even surface in a system of circles when we merely make the point of inversion of one circle something other than the circle s centre. We illustrate the situation in the following example. Example 2. Define two ellipses S 1 and S 2 with inversion points õ 1 kernel(s 1 ) and õ 2 kernel(s 2 ), where both õ 1 and õ 2 are not the centres. Let x 1, x 2 S 1. Then if we apply T 2, the inversion map with respect to S 2, the points T 2 ( x 1 ) and T 2 ( x 2 ) will be inside S 2. In Figure 4.2, we present a particular example. Here õ 1 = ( 0.5, 10.5), ( ) 2 2x ( y ) 2, õ 2 = (4.1, 2), S 1 is a circle, S 2 is the ellipse 1 = x1 = (0.5, 11) 10 and x 2 = (0.45, 11). It is easy to calculate d 2 ( x 1, x 2 ) = 0.05 (see Figure 4.2(a)). We have used Maple

57 50 to perform the inversion of x 1 and x 2 via map T 2 into ellipse S 2. The image points are drawn in Figure 4.2(b), where the dots overlap each other. Figure 4.2(c) presents a zoomed-in version of the image points, with distance between them calculated to be d 2 (T 2 ( x 1 ), T 2 ( x 2 )) = (a) (b) (c) Figure 4.2: (a) Two points within red star-shaped set (b) Two star-shaped sets and (c) Two inverted points with respect to the blue star-shaped set. We remind the reader that [3] merely writes that the inversion maps are contractive. Example 2 illustrates that the maps are not necessarily contractive with respect to the Euclidean metric, even when points are moved from the exterior of the starshaped set into the interior. In the next section, we introduce a modified metric to obtain contractivity, and we subsequently construct a metric that works for a system of star-shaped sets.

58 A New Metric for Star-shaped Sets In Definition 9, the projection is defined as Π i : R 2 \ {õ i } S i but now we redefine Π i : R 2 S i by r i (t) w(t) + õ i Π i ( x) = r i (0) w(0) + õ i if x õ i if x = õ i where x R 2, x = ar i (t) w(t) + õ i for some t [0, 2π], and in the case x = õ i the angle 0 is chosen arbitrarily. Definition 17. Given x 1 = a 1 r i (t 1 ) w(t 1 ) + õ i and x 2 = a 2 r i (t 2 ) w(t 2 ) + õ i we define ( ) x 1 õ i d S,i ( x 1, x 2 ) = d 2 Π i ( x 1 ) õ i, x 2 õ i. Π i ( x 2 ) õ i Remark 4. Geometrically, the transformation x x õ i Π i ( x) õ i moves the centre of inversion õ i to the origin and scales the boundary of star-shaped set S i by the distance from the inversion point to the boundary along the radial ray so that we are working with the unit circle at the origin. Note that we could have defined the transformation to be x x õ i Π i ( x) õ i + õ i so that instead of working on the unit circle centred at the origin we are centred at the point of inversion. Theorem 12. d S,i as defined in Definition 17 satisfies all properties of a metric on R 2.

59 52 Proof. Let x, ỹ, z R 2. (i) non-negativity: ( ) x õ i d S,i ( x, ỹ) = d 2 Π i ( x) õ i, ỹ õ i Π i (ỹ) õ i 0, since d 2 is a metric. (ii) identity: ( ) x õ i d S,i ( x, ỹ) = 0 d 2 Π i ( x) õ i, ỹ õ i = 0 Π i (ỹ) õ i d 2 (a 1 w(t 1 ), a 2 w(t 2 )) = 0 d 2 ((a 1 cos(t 1 ), a 1 sin(t 1 )), (a 1 cos(t 1 ), a 1 sin(t 1 ))) = 0 (a 1 cos(t 1 ) a 2 cos(t 1 )) 2 + (a 1 sin(t 1 ) a 2 sin(t 1 )) 2 = 0 a 1 cos(t 1 ) a 2 cos(t 2 ) = 0 and a 1 sin(t 1 ) a 2 sin(t 2 ) = 0 a 1 cos(t 1 ) = a 2 cos(t 2 ) and a 1 sin(t 1 ) = a 2 sin(t 2 ). This implies that a 1 w(t 1 ) = a 2 w(t 2 ) which means a 1 = a 2. Since a 1 = a 2 we have cos(t 1 ) = cos(t 2 ) and sin(t 1 ) = sin(t 2 ) and now need to show that t 1 = t 2. Without loss of generality we will pick t 2 t 1, t 1, t 2 [0, 2π]. Now, we have (i) t 1 = t 2 + 2nπ or (ii) t 1 + t 2 = 2π (iii) t 1 = t 2 + 2nπ or (iv) t 1 + t 2 = 3π

60 53 where the first case is derived from the graph of cosine and the second case from the graph of sine. Case 1 (i) and (iii): Pick n = 0, then we have t 1 = t 2. Case 2 (i) and (iv): Rearrange (iv) for t 1 to get t 1 = 3π t 2 and substitute in to (i). We have 3π t 2 = t 2 + 2nπ 2t 2 = 3π 2nπ. If we pick n = 0 then t 2 = 3π 2 and if we plug back in to (iv) we have 3π 2 + t 1 = 3π t 1 = 3π 2 t 1 = t 2. Thus t 1 = t 2 whenever t 1 = t 2 = 3π 2. Case 3 (ii) and (iii): Rearrange (ii) for t 2 to get t 2 = 2π t 1 and substitute in to (iii). We have t 1 = 2π t 1 + 2nπ t 1 = (n + 1)π. If we pick n = 0 then t 1 = π and if we plug back in to (ii) we have π + t 2 = 2π t 2 = π t 1 = t 2. Thus t 1 = t 2 whenever t 1 = t 2 = π. Case 4 (ii) and (iv): In this case, there are no possibilities to have t 1 = t 2.

61 54 Putting all four cases together we conclude that t 1 = t 2. Now, from the other direction we have x = ỹ Π i ( x) = Π i (ỹ) x õ i Π i ( x) õ i = ỹ õ i Π i (ỹ) õ i ( ) x õ i d 2 Π i ( x) õ i, ỹ õ i Π i (ỹ) õ i d S,i ( x, ỹ) = 0. = 0, since d 2 is a metric (iii): symmetry ( ) x õ i d S,i ( x, ỹ) = d 2 Π i ( x) õ i, ỹ õ i Π i (ỹ) õ i ( ) ỹ õ i = d 2 Π i (ỹ) õ i, x õ i, since d 2 is a metric Π i ( x) õ i = d S,i (ỹ, x). (iv): triangle inequality d S,i ( x, z) = d 2 ( x õi Π( x) õ i, d 2 ( x õi Π( x) õ i, p ) z õ i Π( z) õ i ) ( + d 2 p, z õ i Π( z) õ i ), p R 2 since d 2 is a metric. We can represent any p R 2 as p = relationship is 1-to-1 ỹ R 2. So, ỹ õ i Π(ỹ) õ i for some ỹ R 2 and, in fact, this d S,i d 2 ( x õi Π( x) õ i, = d S,i ( x, ỹ) + d S,i (ỹ, z). ) ( ỹ õ i ỹ õi + d 2 Π(ỹ) õ i Π(ỹ) õ i, ) z õ i Π( z) õ i

62 55 Now that we have established our new metric d S,i, we can use it to extend Lemma 8 to the setting of star-shaped sets. Lemma 9. Given a star-shaped set S 1 with õ 1 kernel(s 1 ) and inversion map T 1 we have, d S,1 (T 1 ( x 2 ), T 1 ( x 2 )) = 1 a 1 a 2 d S,1 ( x 1, x 2 ), x 1, x 2 R 2. Proof. There exists t 1, t 2 [0, 2π], t 2 > t 1, such that x j = a j r 1 (t j ) w(t j ) + õ 1 and Π 1 ( x j ) = r 1 (t j ) w(t j ) + õ i, j = 1, 2. First we need to establish the following, ( ) d 2 S,1( x 1, x 2 ) = d 2 x 1 õ 1 2 Π 1 ( x 1 ) õ 1, x 2 õ 1 Π 1 ( x 2 ) õ 1 ( = d 2 a1 r 1 (t 1 ) w(t 1 ) + õ 1 õ 1 2 r 1 (t 1 ) w(t 1 ) + õ 1 õ 1, a ) 2r 1 (t 2 ) w(t 2 ) + õ 1 õ 1 r 1 (t 2 ) w(t 2 ) + õ 1 õ 1 ( = d 2 a1 r 1 (t 1 ) w(t 1 ) 2, a ) 2r 1 (t 2 ) w(t 2 ) r 1 (t 1 ) r 1 (t 2 ) = d 2 2(a 1 w(t 1 ), a 2 w(t 2 )) = a a 2 2 2a 1 a 2 cos(t 2 t 1 ) = d 2 2( x 1, x 2 ), by Equation (3.7) from Section 3.3. (4.3)

63 56 Taking the square root of both sides we obtain d S,1 ( x 1, x 2 ) = d 2 ( x 1, x 2 ). Now, ( ) d 2 S,1(T 1 ( x 1 ), T 1 ( x 2 )) = d 2 T 1 ( x 1 ) õ 1 2 Π 1 (T 1 ( x 1 )) õ 1, T 1 ( x 2 ) õ 1 Π 1 (T 1 ( x 2 )) õ 1 ( 1 ) 1 = d 2 a 1 r 1 (t 1 ) w(t 1 ) + õ 1 õ 1 2 r 1 (t 1 ) w(t 1 ) + õ 1 õ 1, a 2 r 1 (t 2 ) w(t 2 ) + õ 1 õ 1 r 1 (t 2 ) w(t 2 ) + õ 1 õ 1 ( 1 ) 1 = d 2 a 1 r 1 (t 1 ) w(t 1 ) a 2, 2 r 1 (t 2 ) w(t 2 ) r 1 (t 1 ) r 1 (t 2 ) ( 1 = d 2 2 w(t 1 ), 1 ) w(t 2 ) a 1 a 2 ( 1 = cos(t 1 ) 1 ) 2 ( 1 cos(t 2 ) + sin(t 1 ) 1 ) 2 sin(t 2 ) a 1 a 2 a 1 a 2 = cos(t a 2 1 a 2 2 t 1 ) 2 a 1 a 2 1 = (a 2 a 2 1a a 2 2 2a 1 a 2 cos(t 2 t 1 )) 2 1 = d 2 a 2( x 2 1a 2 1, x 2 ) 2 1 = d 2 a S,1( x 2 1a 2 1, x 2 ), by (4.3) 2 which upon taking the square root gives the desired result. Recalling Definition 18 for a i,min, it is easy to see that this definition can be applied to the case of non-touching star-shaped sets. We can use this definition in the case of star-shaped sets to prove the following theorem. We recast Definition 18 for a i,min in terms of star-shaped sets. Definition 18. Given a collection of non-touching star-shaped sets S i, i = 1,..., N, with points of inversion õ i and radii r i (t), for x = a j r i (t) w(t) + õ i R 2, a j > 0, t [0, 2π], we define a i,min = min {a j} > 1 x S j,j i

64 57 corresponding to the radial scaling of the closest point to õ i in all other star-shaped sets S j, j i. We can use this definition in the proof of the following theorem. Theorem 13. T i : S j S i with S j S j =, i j, is contractive with respect to d S,i. That is, c [0, 1) such that d S,i (T i ( x 1 ), T i ( x 2 )) c d S,i ( x 1, x 2 ), x 1, x 2 S j, i j Proof. d S,i (T i ( x 1 ), T i ( x 2 )) = 1 d S,i ( x 1, x 2 ), by Lemma 9 a 1 a 2 1 d S,i ( x 1, x 2 ), by Definition 18 a 2 i,min = c d S,i ( x 1, x 2 ). Example 3. We can now use the new metric to calculate the distance between the points used in Example 2. Recall x 1 = (0.5, 11), x 2 = (0.45, 11) and õ 2 = (4.1, 2). After some work in Maple we arrive at d S,i ( x 1, x 2 ) = and d S,i (T 2 ( x 1 ), T 2 ( x 2 )) =

65 58 Applying the inversion map results in the distance shrinking with respect to d S,i, as expected by Theorem 13. We define the following for a collection of non-touching star-shaped sets in order to prove an upcoming theorem. Definition 19. Given a collection of non-touching star-shaped sets S i, i = 1,..., N, with points of inversion õ i kernel(s i ), define N X = i=1 and relative to each S i, we can define S i T i : X X by T i ( x) = T i ( x) if x S j, j i x if x S i Theorem 14. T i : X X satisfies (i) d S,i (T i ( x 1 ), T i ( x 2 )) c d S,i ( x 1, x 2 ) for some c [0, 1), x 1 S j, x 2 S k, j, k i (ii) d S,i (T i ( x 1 ), T i ( x 2 )) = d S,i ( x 1, x 2 ) x 1, x 2 S i (iii) d S,i (T i ( x 1 ), T i ( x 2 )) c d S,i ( x 1, x 2 ) for some c [0, 1), x 1 S i, x 2 S j, j i

66 59 Proof. Proof of (i): The result follows from Theorem 13. Proof of (ii): The result follows immediately since T i is the identity map in this case. Proof of (iii): With x 1 S i and x 2 S j, we have a 2 > 1 > a 1 0, and, for values t 1, t 2 [0, 2π], d 2 S,i( x 1, x 2 ) = ( a a 2 2 2a 1 a 2 cos(t 2 t 1 ) ), (4.4) using (4.3), and ( d 2 S,i T i ( x 1 ), T i ( x 2 ) ) = d 2 S,i ( x1, T i ( x 2 ) ) ( ) = d 2 x1 õ i 2 Π( x 1 ) õ i, T i ( x 2 ) õ i Π(T i ( x 2 )) õ i ( = d 2 2 a 1 w(t 1 ), 1 ) w(t 2 ) a 2 = a 1 cos(t 1 ) 1 cos(t 2 ), a 1 sin(t 1 ) 1 sin(t 2 ) a 2 a 2 ( = a a ) 1 cos(t a 2 2 t 1 ) 2 a 2 2 = 1 a 2 2 ( a 2 2 a a 1 a 2 cos(t 2 t 1 ) ). (4.5) The bracketed expression in (4.5) is less than or equal to the bracketed expression in (4.4) provided that a 2 2a a a 2 2, where a 2 > 1 > a 1 0. (4.6) Hence, letting y = a 2 1 and z = a 2 2, we consider the function f(y, z) = y yz + z. If we show that f(y, z) 1 for y [0, 1) and z > 1, then (4.6) holds. We write f(y, z) = y yz + z = y(z 1) + (z 1) + 1 = (z 1)(1 y) + 1

67 60 and we see that f(y, z) > 1 for y < 1 and z > 1, proving the result. This means that the statement in (iii) of the theorem holds for c = 1 a 2. Now, suppose we have a collection of star-shaped sets {S i } N i=1 with points of inversion õ i kernel(s i ) and maps T i. Each S i has an associated metric d S,i with respect to which T i is contractive, but d S,i depends on õ i and Π i. We combine all of these different metrics together to produce a single metric with respect to which all of the inversion maps are contractive. Definition 20. Given a collection of non-touching star-shaped sets S 1,..., S N with points of inversion õ 1,..., õ N we define for x, ỹ R 2 d S,i ( x, ỹ) d S ( x, ỹ) = max d S,j ( x, ỹ) j if x, ỹ S i otherwise. Theorem 15. d S as defined in Definition 20 satisfies all properties of a metric in R 2. Proof. Let x, ỹ, z R 2. (i) non-negativity: d S ( x, ỹ) = d S,k ( x, ỹ), for some k 0, since d S,k is a metric.

68 61 (ii) identity: d S ( x, ỹ) = 0 d S,k ( x, ỹ) = 0, for some k x = ỹ, since d S,k is a metric. and x = ỹ d S,k ( x, ỹ) = 0, for all appropriate k since each d S,k is a metric d S ( x, ỹ) = 0. (iii): symmetry d S ( x, ỹ) = d S,k ( x, ỹ), for some k = d S,k (ỹ, x), since d S,k is a metric = d S (ỹ, x). (iv): triangle inequality d S = d S,k ( x, z), for some k d S,k ( x, ỹ) + d S,k (ỹ, z), since d S,k is a metric = d S,k ( x, ỹ) + d S,k (ỹ, z) if x, ỹ, z S k = max d S,j ( x, ỹ) + d S,k (ỹ, z) j d S,k ( x, ỹ) + max d S,j (ỹ, z) j if x S k, ỹ, z S k if x, ỹ S k, z S k max j = d S ( x, ỹ) + d S (ỹ, z) d S,j ( x, ỹ) + max d S,j (ỹ, z) j if x, ỹ, z S k

69 62 Now that we have established that d S is a metric on R 2 we can use it to prove contractivity as in Theorem 14 and extend the contractivity result to the composition of star-shaped set inversion maps. Theorem 16. T i : X X satisfies (i) d S (T i ( x 1 ), T i ( x 2 )) c d S ( x 1, x 2 ) for some c [0, 1), x 1 S j, x 2 S k, j, k i (ii) d S (T i ( x 1 ), T i ( x 2 )) = d S ( x 1, x 2 ) x 1, x 2 S i (iii) d S (T i ( x 1 ), T i ( x 2 )) c d S ( x 1, x 2 ) for some c [0, 1), x 1 S i, x 2 S j, j i Proof. Proof of (i): d S (T i ( x 1 ), T i ( x 2 )) = d S,i (T i ( x 1 ), T i ( x 2 )) c d S,i ( x 1, x 2 ), by Theorem 14 (i) d S,i ( x, ỹ) c max d S,j ( x 1, x 2 ) j = c d S ( x 1, x 2 ). if x 1, x 2 S i otherwise The proof of (ii) follows immediately since T i is the identity map in this case.

70 63 Proof of (iii): d S (T i ( x 1 ), T i ( x 2 )) = d S,i (T i ( x 1 ), T i ( x 2 )) c d S,i ( x 1, x 2 ), by Theorem 14 (iii) d S,i ( x, ỹ) c max d S,j ( x 1, x 2 ) j = c d S ( x 1, x 2 ). if x 1, x 2 S i otherwise Theorem 17. For i j, i, j {1,..., N}, (T i T j ): X X, is contractive: d S ((T i T j )( x 1 ), (T i T j )( x 2 )) c d S ( x 1, x 2 ) for some c [0, 1), x 1, x 2 X Proof. If x 1, x 2 C k, k j, then the result follows by applying Theorem 16 (i) twice. If x 1, x 2 C j, then the result follows by applying Theorem 16 (ii) and Theorem 16 (i). If x 1 C j, x 2 C k, k j, then the result follows by apply Theorem 16 (iii) and Theorem 16 (i). As in the case of circles we will now formulate this multiple star-shaped set framework in the context of sets in the Hausdorff metric space, (H(X), d H ). Recall for a set S R 2 we defined ˆ T i (S) = { T i ( x), x S }.

71 64 Now, we want to show that ˆT for a collection of star-shaped sets has a unique fixed point A H(X) just as in the setting of circles. The result follows from the same path as in Section 3.3 where we prove ˆT is contractive on (H(X), d H ) but instead we now use the metric d S to induce d H. Theorem 18. For i j, 1. ˆT i ˆT j : H(X) H(X), and 2. ˆT i ˆT j is contractive on (H(X), d H ). Proof. The first claim follows because T i T j : H(X) H(X). The proof relies on the continuity of the map ([5], Lemma 2, page 80). By Theorem 17, we can denote by c ij the contractivity factor with respect to d S of T i T j. Let A, B H(X) then, d H ((ˆT i ˆT ) j (A), (ˆT i ˆT ) ) j (B) = max sup inf d S ( x 1, x 2 ), { sup x 1 (ˆT i ˆT j )(A) x 2 (ˆT i ˆT j )(B) sup inf x 1 (ˆT i ˆT j )(B) x 2 (ˆT i ˆT j )(A) = max inf d S( ( ) T i T j ( x1 ), ( ) T i T j ( x2 )), x 1 A x 2 B sup inf d S( ( ) T i T j ( x1 ), ( } ) T i T j ( x2 )) x 1 B x 2 A { } c ij max sup inf d S( x 1, x 2 ), sup inf d S( x 1, x 2 ), by Theroem 17 x 2 B x 2 A x 1 A = c ij d H (A, B). x 1 B d S ( x 1, x 2 )

72 65 We reach our final theorem of the chapter, which establishes the contractivity of the union of the maps in Theorem 18. Theorem 19. For A H(X), define ˆT (A) = 1. ˆT : H(X) H(X), and N i,j=1 i j (ˆT i ˆT j ) (A). Then 2. ˆT is contractive on (H(X), d H ). Proof. The first claim follows from Theorem 18 because the union is finite. For the second claim we have d H (ˆT (A), ˆT (B) ) = d H N i,j=1 i j max 1 i,j N d H (ˆT i ˆT j ) (A), N i,j=1 i j ((ˆT i ˆT j ) (A), (ˆT i ˆT ) j (B) (ˆT i ˆT j ) (B) ), property of d H max c ij d H (A, B), by Theorem 18 1 i,j N = c d H (A, B). Finally, by Theorem 19 and Banach s Fixed Point Theorem we can conclude that there exists a unique fixed point A H(X), the set attractor of ˆT, satisfying ˆ T (A) = A. With this final result established we can approximate the set attractor of a collection of star-shaped sets using the Chaos Game exactly as we did in Section 3.3 with a collection of circles.

73 Application of the Chaos Game As you will see below the four fractals are produced with collections of nontouching star-shaped sets. The mathematical framework has not yet been developed for overlapping star-shaped sets but are shown later in Chapter 6 to illustrate much more interesting fractals generated by the Chaos Game. Figure 4.3: Set attractors after 5,000 iterations of the chaos game. An interesting application is to combine both affine IFS maps and star-shaped set inversion maps. Since we use affine IFS maps that are contractive with respect to the

74 67 Euclidean metric, we further redefine the d S metric by including d 2 : d S( x, ỹ) = max{d S ( x, ỹ), d 2 ( x, ỹ)}. In the following examples, we play the chaos game with equal probability of either applying the affine IFS map or applying the inversion map to a point. Recall the Sierpinski Gasket from Chapter 2 and consider the hybridization of these two types of contraction maps below in Figures 4.4, 4.5 and 4.6. Figure 4.4: The three Sierpinski Gasket IFS maps with three circle inversion maps overlapping the Sierpinski triangles.

75 68 Figure 4.5: The two bottom Sierpinski Gasket IFS maps with one inversion map for a triangle taking the place of the top Sierpinski triangle. Figure 4.6: The three Sierpinski Gasket IFS maps with three circle inversion maps completely contained outside of the three Sierpinski triangles.

76 69 Chapter 5 1-D Signal and Function Approximation In this chapter we apply the 1-D circle inversion map framework to function approximation. In Chapters 3-4, we considered the direct problem of rendering an approximation of a set attractor via the chaos game. In contrast, the function approximation problem is an inverse problem: loosely we wish to find a system of circle inversion maps with a fixed point, that closely approximates some target 1-D function. First, we introduce the Collage Theorem and some other well-established fundamental concepts, before applying these ideas to circle inversion maps. Theorem 20. (Collage Theorem, [4], Page 23) Let (X, d) be a complete metric space and T Con(X) with contraction factor c [0, 1) and fixed point x X. Then for any x X, we have d(x, x) 1 d(x, T (x)). 1 c

77 70 Remark 5. For a fixed point approximation inverse problem, we want the approximation error d(x, x) to be as small as possible. The Collage Theorem allows us to control the approximation error by minimizing the collage distance d(x, T (x)) subject to c << 1. The domain of a nice 1-D function is a closed interval (or union of closed intervals), which we naturally think of as a set. We plan to represent this set by the set attractor of a system of circle inversion maps on the real line. In order to represent the function, to each point of the set attractor we associate a function value. Since these ideas apply to digital images and signals, the function values are usually thought of as grey level values of a single-colour image. As a result, the new maps we introduce to assign function values to points in the set attractor are called grey level maps. We illustrate the classical formulation of iteration function systems with grey level maps (IFSM) in the following example. Example 4. Consider the two-map IFS {w 1, w 2 } = { 6 10 x, 6 10 x } on X = [0, 1]. We see that ŵ 1 : [0, 1] [0, 0.6] and ŵ 2 : [0, 1] [0.4, 1]. We now take our target signal u(x) and shrink it to produce two copies. One copy uses ŵ 1 to produce a signal with domain X 1 = [0, 0.6] and the other uses ŵ 2 to produce a signal with domain X 2 = [0.4, 1], as below

78 71 Figure 5.1: The left picture is the target signal and the right picture is the two shrunken copies of the target. Notice that in the picture we have X = X 1 X 2 = [0, 1] and the following equations a 1 (x) = u(w 1 1 (x)), x ŵ 1 (X), a 2 (x) = u(w 1 2 (x)), x ŵ 2 (X). Now, we modify the function values by using grey level maps. For this example, we multiply each value of a 1 (x) by 1 and add 1, and multiple each value of a 2 2 2(x) by 3. 4 Then we must have some procedure to deal with overlaps. For example, we can add values in the overlap regions; this region is [0.4, 0.6] in this example. Next, we combine these newly transformed values and obtain a function defined on all of X = [0, 1].

79 72 Figure 5.2: The left picture includes the grey level maps and the right picture is overlaps added together. In this picture, we have the following new equations b 1 (x) = φ 1 (a 1 (x)) = φ 1 (u(w 1 1 (x))), x ŵ 1 (X) b 2 (x) = φ 2 (a 2 (x)) = φ 2 (u(w 1 2 (x))), x ŵ 2 (X) where φ 1 = 1 2 t and φ 2 = 3 4 t. The function v(x) = (T u)(x) is produced by the two-map IFS with grey level maps, or two-map IFSM, as defined below b 1 (x) for x ŵ 1 (X) \ ŵ 2 (X) = [0, 0.4), v(x) = (T u)(x) = b 1 (x) + b 2 (x) for x ŵ 1 (X) ŵ 2 (X) = [0.4, 0.6], b 2 (x) for x ŵ 2 (X) \ ŵ 1 (X) = (0.6, 1]. Referring to the notation in the example, the goal of the function approximation inverse problem is to find choices of the maps in the formulation so that v = T (u) lies

80 73 close to u. That is, we want the collage distance d(u, T (u)) to be very small so that, by the Collage Theorem, the fixed point of T will lie close to u. Notice that in the statement of the Collage Theorem, we used x to denote an element of the space X. Of course, when working with functions, x is the independent domain variable. We use the letter u to represent the function, and the map T acts on both the domain and the range values. To make this situation as easily manageable as possible and to help guarantee contractivity of T, one typically chooses a fixed collection of maps w i, i = 1,..., N, with known contraction factors. The grey level maps are chosen to be affine; the two parameters in such a map will be chosen to minimize the collage distance. We can generalize our IFSM method by considering the application of contractive maps w i on subsets of X, called parent blocks. We call this generalization a local IFSM (LIFSM). The parent blocks are then mapped to smaller subsets using the w i maps; these subsets are called child blocks. The difference between IFSM and LIFSM is that the method of IFSM uses a union of contracted copies of the entire signal to approximate the signal whereas LIFSM uses a union of copies of subsets of the signal to approximate the signal. We will formulate a local IFSM by first considering some function u(x) for x in X = [0, 1]. In order to use a N-map IFS, X is first divided into parent intervals P j, j = 1,..., M, and non-overlapping child intervals Q i, i = 1,..., N, where all child intervals are smaller than the parent intervals. Each parent P j is mapped to each

81 74 child Q i by our circle inversion map w ij : P j Q i, i = 1,..., N, j = 1,..., M, so that ŵ(x) = i,j ŵ ij (X) = X. Now, we will add in our affine grey level maps φ i (t) = α i t + β i, i = 1,..., N in order to get a local IFSM. On each child interval Q i the fractal transform of u(x) is (T u)(x) = φ i (u(w 1 ij (x))) = α iu(w 1 ij (x)) + β i, x Q i. Comparing to Example 4, note that the preceding statement takes advantage of the lack of overlap regions. The squared L 2 collage distance for child i with parent j is ij = ((T u)(x) u(x)) 2 dx = Q i (α i u(w 1 ij (x)) + β i u(x)) 2 dx. Q i We can now use calculus to minimize ij and find the associated α i and β i. This is easy to do since ij is a quadratic function of two unknown parameters. Thus we solve the following linear system of two equations ij α i = 0 and ij β i = 0. (5.1)

82 75 We can now use Equation (5.1) to find our α i and β i as follows, 0 = ij α i [ 0 = 2 (αi u(w 1 ij (x)) + β i u(x)) u(w 1 ij (x))] dx, using chain rule Q i [ 0 = αi (u(w 1 ij (x)))2 + β i u(w 1 ij (x)) u(x)u(w 1 ij (x))] dx Q i 0 = α i (u(w 1 ij (x)))2 dx + β i u(w 1 ij (x)) dx u(x)u(w 1 ij (x)) dx Q i Q i Q i 0 = α i u(w 1 ij (x)) 2 + β 2 i u(w 1 ij (x)) u(x), u(w 1 ij (x)) u(x), u(w 1 ij (x)) = α i u(w 1 ij (x)) 2 + β 2 i Next we look at the partial derivative with respect to β i, u(w 1 ij (x)) (5.2) 0 = ij β i [ 0 = 2 (αi u(w 1 ij (x)) + β i u(x)) 1 ] dx, using chain rule Q i 0 = α i u(w 1 ij (x)) dx + β i 1 dx u(x) dx Q i Q i Q i 0 = α i u(w 1 ij (x)) + βi Q i u(x) u(x) = α i u(w 1 ij (x)) + βi Q i. (5.3) Now, we can apply Cramer s Rule to (5.2) and (5.3) to solve for α i and β i : u(x), u(w 1 ij (x)) u(w 1 ij (x)) u(x) Q i α i = u(w 1 ij (x)) 2 2 u(w 1 ij (x)) u(w 1 ij (x)) Q i u(x), u(w 1 ij α i = (x)) Q i u(w 1 ij (x)) u(x) u(w 1 ij (x)) 2 Q 2 i u(w 1 ij (x)) 2

83 76 and β i = β i = u(w 1 ij (x)) 2 2 u(x), u(w 1 ij (x)) u(w 1 ij (x)) u(x) u(w 1 ij (x)) 2 2 u(w 1 ij (x)) u(w 1 ij (x)) Q i 1 u(w 1 ij (x)) 2 2 u(w 1 ij (x)) u(x), u(w 1 ij (x)) u(w 1 ij (x)) u(w 1 ij (x)) 2 Q 2 i u(w 1 ij (x)) 2. In the case that the function u(x) is a signal with prescribed range [0, 1], for example, we must also ensure that T u has the same range. We see that we obtain the constraints 0 β i 1 and 0 α i + β i 1. Thus, we seek to minimize ij with respect to α i and β i on the parallelogram below. Figure 5.3: Parallelogram formed by 0 β i 1 and 0 α i + β i 1. We can use calculus to find the minimum of α i and β i in this parallelogram. If

84 77 the resultant minimizing values of α i or β i do not satisfy the constraints, we look for the minimum value of ij along the boundary of the parallelogram. In order to find the best suited parent, α i and β i for each child Q i, we must find the minimum collage distance for each j. The collection of all of these parameters yields the fractal transform containing the best child/parent pairings and best approximates our target signal. In our case, the w ij maps will be circle inversions and so this method of approximating signals is called (by us) local circle inversion systems with grey level maps, denoted LCISM. In a more general setting, we will now prove that T is contractive in the L 2 metric and use this result later to calculate our contraction factor. Consider an N- map LCISM defined on X = [0, 1] with circle inversion maps w ij (x), i = 1,..., N, j = 1,..., M, and grey level maps φ i (t) = α i t + β i, i = 1,..., N. As before we have the fractal transform (T u)(x) = N i=1 α i u(w 1 ij (x)) + β i. Let u, v L 2 ([0, 1]), then we have for each child i [ (T i (u) T i (v)) 2 dx = αi u(w 1 ij (x)) + β i α i v(w 1 ij (x)) β 2 i] dx Q i Q i [ = αi 2 u(w 1 ij (x)) v(w 1 ij (x))] 2 dx Q i [ = αi 2 u(w 1 ij (x)) v(w 1 ij (x))] 2 dx. Q i

85 78 Now, if we let x = w ij (y), each point x in Q i is sent to a point y in P j, say. Referring to page 28, w ij(y) (b i o i ) 2 (b j o i ) 2 =: c i, where o i is the centre of the circle inducing inversion map T i = w ij, and either 1. b i is the right-most boundary point of Q i and b j is the left-most boundary point of P j, or 2. b i is the left-most boundary point of Q i and b j is the right-most boundary point of P j. In fact, in either case, we can say that T i (b j ) = b i. So, Q i (T i (u) T i (v)) 2 dx = α 2 i [ u(w 1 ij Q i and by the change of variable (COV) theorem we have ( (T (u) T (v)) 2 dx Q i and we get our final result ) 1 2 (x)) v(w 1 ij (x))] 2 dx ] = αi [c 2 i [u(y) v(y)] 2 dy, COV theorem P j [ ci d 2 2(u, v) ], since P j X α 2 i α i c i d 2 (u, v) d 2 (T u, T v) = N ( (T (u) T (v)) 2 dx i=1 Q i [ N ] ci α i d 2 (u, v). i=1 ) 1 2

86 79 N Provided that ci α i < 1, T is a contraction. In fact, the same condition holds i=1 for affine IFS with c i the contraction factor of w i. The following example uses this method to approximate the target function u = sin(πx) on [0, 1]. Example 5. First, we divide X = [0, 1] into some number of non-overlapping child intervals Q i. In Figures we illustrate the cases of 4,8 and 16 children. In the Figures, the child intervals are drawn in green. Notice that the union of these intervals is [0, 1]. We next decide on the size relationship between parents and children; we decide that parents are double the size of children, so that the circle inversion maps from parent to children have contraction factor 1. 2 Next, for each child, we generate a collection of parents. In Figures , we draw the parent intervals in red (to the right of the child) and blue (to the left of the child). We also draw the inversion point for each parent (as a diamond). To help with the visualization, the parents are each drawn at a different vertical-axis value. Notice that the parents cannot overlap children, since the circle inversion map is only contractive when sending the exterior of the circle to the interior (and the interior is the child). In the affine IFS framework, parents and children can overlap. We decided to slide the parent intervals by one quarter the length of the child intervals, as can be seen in Figures

87 Figure 5.4: X = [0, 1] divided up into 4 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent. 80

88 Figure 5.5: X = [0, 1] divided up into 8 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent. 81

89 82 Figure 5.6: X = [0, 1] divided up into 16 child intervals where child intervals are green, left parent intervals are blue and right parent intervals are red. The blue and red diamonds are the centres of inversion for each parent. Each child and parent represents a circle in 1-D with the circle inversion contraction map taking points from a given parent to a point of its associated child. That is, each map w i takes either a red parent interval or a blue parent interval to the green child interval. Think of the picture as looking at the circles from a side view rather than looking from a bird s eye view.

90 83 Now that we have established children intervals, parent intervals and the circle inversion maps we can do some work in Maple to obtain the following table for the case of 16 children, Child Minimal Collage Distance Best α i Best β i Parent Q P 53, Right Q P 8, Right Q P 45, Right Q P 41, Right Q P 34, Right Q P 24, Right Q P 13, Right Q P 3, Right Q P 3, Left Q P 13, Left Q P 24, Left Q P 33, Left Q P 41, Left Q P 45, Left Q P 8, Left Q P 53, Left In the parent column, the left or right indicates which side of the child the par-

91 84 ent lies on. To get the squared L 2 collage distance we need to sum all the minimum collage distances for each child interval together. (d(u, T u)) 2 = 16 i=1 Q i ((T u)(x) u(x)) 2 dx = In the following three pictures, T (u(x)) is on top of u(x) and we can see that the approximations of u(x) improve when more children are covering X. Figure 5.7: A picture of u(x) in blue and T (u(x)) in red with 4 children.

92 85 Figure 5.8: A picture of u(x) in blue and T (u(x)) in red with 8 children. Figure 5.9: A picture of u(x) in blue and T (u(x)) in red with 16 children.

93 86 We can start with a function u 0 (x) = 0, say, and iterate T : u n (x) = (T n u 0 ) (x) to create a sequence of signals {u n (x)} that converge to the fixed point of T, ū. After iterating, the ū is our approximation of our target signal u(x). Figures 5.10, 5.11 and 5.12 below demonstrate that as we used more children to cover the space [0, 1], the approximation of the target improves. Each case is iterated 7 times but we see that in the first set of graphs with 4 children the final iteration is much farther away from approximating the target compared to using 8 or 16 children as in the bottom two sets of graphs.

94 Figure 5.10: 7 iterations of T with 4 children. 87

95 Figure 5.11: 7 iterations of T with 8 children. 88

96 Figure 5.12: 7 iterations of T with 16 children. 89

97 90 Chapter 6 Summary and Future Work In this thesis we have developed a rigourous mathematical framework for circle inversion maps and star-shaped set inversion maps. The framework for circle inversion maps allows for collections of overlapping circles of different sizes so long as the intersection of all the circles is the empty set. For star-shaped sets we were able to establish results in the non-overlapping case but we demonstrated that set attractors exist even in the overlapping case. Future work includes: 1. Establishing the star-shaped set framework for overlapping sets. This work would seem to entail developing a metric that can accommodate overlaps. The metric we defined in Definition 17 relied on the fact that in the non-overlapping case if two points are in the same star-shaped set then neither of them is in another star-shaped set. This property appeared to be essential in the contractivity proof. Thus, the goal is to develop a metric without this flaw. As we can see in the Figures below, the fractal attractors generated in the

98 91 setting of overlapping sets are far more interesting. 2. Extending LCISM to 2-D images. 3. Exploring and comparing how LCISM and LIFSM perform for image compression, denoising, interpolation and other imaging operations. 4. Extending the examples we consider to three dimensional graphics.

99 Figure 6.1: Set attractors after 5,000 iterations of the chaos game. 92

100 Figure 6.2: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 93

101 Figure 6.3: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 94

102 Figure 6.4: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 95

103 Figure 6.5: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 96

104 Figure 6.6: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 97

105 Figure 6.7: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 98

106 Figure 6.8: Set attractors with shifted centres of inversion after 5,000 iterations of the chaos game. 99