Ohrimenko+ / Barnsley. Superfractals

Figure 1.33 Interacting level sets of shortest-distance functions. The level sets are for a fern, a Sierpinski triangle and a square. See the main text.

E x e r c i s e 1.12.22 Show that

dH (C D, E F) ≤ max{dH (C, E), dH (D, F)} for all C, D, E, F H(X).

E x e r c i s e 1.12.23 Show that

A BB (DB (A)) for all A, B H(X).

The furthest-distance function

In this subsection we discuss optimization problems associated with the Hausdorff distance.

In applications of the Hausdorff metric on H(R2) to pattern matching and fractal approximation, we are led to consider the minimization of dH (A, B), where the ‘target’ set A is held fixed and the set B depends upon certain parameters. The goal is to adjust the parameters so that B is as close as possible to A. For example, as illustrated in Figure 1.34, A might be a set that looks like a leaf, B0 might represent the silhouette of another leaf and B might represent B0 translated by the vector (x, y), namely

B = B(x, y) = {(x + x0, y + y0) : (x0, y0) B0} for all (x, y) R2.

We wish to search for the minimum value of f and the locations (x, y) at which this minimum is achieved.

There are many approaches to optimization problems that might be applied, but the problem is that finding the Hausdorff distance is computationally expensive and we do not have neat formulas or approximations for f (x, y) with which to work. Also, we might wish to compare A with many different leaves B0. How do we start to think about developing efficient algorithms to approach this type of problem?

Further insight into the behaviour of the Hausdorff distance on H(R2), apropos this question, is provided by the furthest-distance function. Let x X and A H(X). Then

dH ({x}, A) = max DA({x}), D{x}(A) = D{x}(A).

We refer to FA(x) := D{x}(A) as the furthest-distance function for A. We have

FA(x) = max{d(x, a) : a A}.

Imagine that A R2 represents a leaf and that an (infinitesimally small) ant is located at a point x R2. Then the path of steepest descent for FA(x), which cuts the level sets of FA(x) at right angles, where they are differentiable curves, represents the route to be followed by the ant to decrease the Hausdorff distance most rapidly. Then, usually, by following a path of steepest descent of FA(x) the ant will arrive at some ‘central’ point on the leaf such that the Hausdorff distance between the leaf and the ant is a minimum. This is in contrast to what happens when the ant follows a path of steepest descent of DA(x), which may lead the ant to a point on the boundary of the leaf nearest to its starting point. When driving to a city, the distance to the city specified on road signs is often the distance to

Figure 1.34 How do we adjust parameters such as x and y , which represent the positions of members of a set B = B (x , y ), so as to minimize the Hausdorff distance dH ( A , B (x , y ))?

the city centre rather than to the boundary of the city. But when driving to France from Italy, the distances quoted on road signs are to the border.

E x e r c i s e 1.12.24 Figure 1.35 shows some level sets for both the shortestdistance function (in red) and the furthest-distance function (in black) associated with a line segment L. The number on a contour gives the value of the corresponding distance function. Neatly draw a few paths of steepest descent, for both DL (x) and FL (x). Where do the paths of steepest descent for FL (x) terminate?

Given the two functions DA(x) and FA(x) for some fixed A H(X), the following theorem provides a simple upper bound to dH (A, B). This upper bound can be evaluated using only extrema of the two functions for x B H(X). This makes it easier to compare approximate distances for different values of B.

T h e o r e m 1.12.25 Let (X, d) be a metric space and let (H(X), dH ) denote the space of compact nonempty subsets of X together with the Hausdorff metric. Then

Figure 1.35 Level sets for the shortest-distance function DL (x ) (in red) and the furthest-distance function FL (x ) (in black) for the set L R2. The set L is the line segment, shown in green, at the centre. The underlying metric is the euclidean distance. Contours are labelled with corresponding distances. This array of level sets can be used to estimate the Hausdorff distance from a set B that is overlayed on the contours. Imagine that the ant is very small, relative to the spacing of the contours, and located as indicated by the arrow. Then you can estimate dH (ant, L ) very accurately! Can you find an upper bound for dH (leaf, L )?

P r o o f This follows from

dH (A, B) = max{DA(B), DB (A)},

Thus, the Hausdorff distance from A to B is bounded by the larger of the maximum value (on B) of the shortest-distance function DA(x) and the minimum value (on B) of the furthest-distance function FA(x).

E x e r c i s e 1.12.26 Use Figure 1.35 to obtain an upper bound for dH (L , leaf ); see the figure caption.

E x e r c i s e 1.12.27 In Figure 1.36 trace the paths of shortest descent for both ants, with respect to DL (x) and FL (x).

Figure 1.36 This figure is similar to Figure 1.35, but here the underlying metric is dmax. Sketch some paths of steepest descent. What route will each ant follow to decrease dH (ant, L ) as rapidly as possible? Where will each ant end up?

E x e r c i s e 1.12.28 Find the minimum value of f (x, y) in Equation (1.12.6) and a value of (x, y) at which it occurs, when A = {(x, y) R2 : x2 + y2 = 1, x ≥ 0} and B = {(x, y) R2 : x2 + y2 = 12 } {(0, y) R2 : − 12 ≤ y ≤ − 14 }.

Hausdorff distances on code spaces

Here we consider distance functions on H( ), the set of nonempty compact subsets of a code space. The code space may be the metric space ( , d ), where d is defined in Equation (1.6.1), or ( , d|A|), where d|A| is defined in Equation (1.6.6), or more generally ( , dξ ), where ξ : → R2 is an embedding and

dξ (σ1, σ2) = |ξ (σ1) − ξ (σ2)| for all σ1, σ2 ,

as in Theorem 1.5.5.

When the underlying metric is obtained by embedding, as in the case of d|A| and more generally dξ , it is possible to make ‘pictures’ of the associated shortestdistance functions and to think quite geometrically and ‘optically’ about the metric, as illustrated in Figures 1.37 and 1.38.

In Figure 1.37 we consider the code space ( , dξ ), where = {0,1,2,3} and

for all (x, y) R2, ω {0, 1, 2, 3}, where [x] denotes the greatest integer less than or equal to the real number x. We defer until Chapter 4 a proof that ξ is indeed an embedding function and a more precise discussion of such embeddings. What matters here is that the embedded set ξ ( ) looks like the set of green points in the bottom left panel of Figure 1.37. What you cannot see is that each small green rectangle represents many more green rectangles organized in the same sort of way as those green rectangles that you can see, and so on. ξ ( ) is in fact of the form C × C R2, where C R is a classical Cantor set.

The top right panel of Figure 1.37 represents the embedded set ξ (S), where S . The top left panel shows level sets of Dξ (S)(x) for x R2. Of course, this top left panel is not a picture of the level sets of DS (σ ) for σ , because most points x on level sets of Dξ (S)(x) do not correspond to points in . But the level sets of Dξ (S)(x) accurately describe the distances from points in to points in S for all x Rξ , the range of ξ , because

Dξ (S)(ξ (σ )) = DS (σ ) for all σ .

The function Dξ (S) : R2 → [0, ∞) is a continuous extension of Dξ (S) : Rξ R2 → [0, ∞) to all R2.

The bottom right panel shows the level sets of Dξ (S)(x), for x , superimposed on ξ ( ). Since this panel contains in principle the points of both ξ (S) and ξ ( ), we can use it to estimate Dξ (S)( ) and hence, since D (ξ (S)) = 0, dH ( , S). But our purpose, of course, is not so much to do this as it is to enable us to think geometrically and visually about code-space metrics.

In Figure 1.38 the level sets of Dξ (S)(x), in the right-hand panel, may be compared with the level sets of Dξ ( )(x), in the left-hand panel, for x R2. In this

example = {0,1} {0,1}, where {0,1} and {0,1} are the code spaces defined in Section 1.4, and a different embedding function ξ : → R2 is used, similar

to the one used in Figure 1.15. The points of ξ ( {0,1}) are situated at the branch points, also called nodes, of a tree-like structure in R2 much the same as that seen in Figure 1.15 and the points of ξ ( {0,1}) are located on the canopy of the tree-like structure. Although it does not appear to be so, the canopy ξ ( {0,1})

Figure 1.37 Illustrations relating to the shortest-distance function for a subset of the codespace ={0,1,2,3}. The bottom left panel shows an embedded set ξ ( ) of the code space in R2, where ξ : → R2 is the embedding function. The top right panel shows the embedded set ξ (S) of a compact subset S . The top left panel shows level sets of Dξ (S)(x ). The bottom right panel shows the level sets of Dξ (S)(x ) superimposed on ξ ( ). Assuming that the width of each band of level sets is one unit, can you estimate Dξ (S)( )?

is totally disconnected. You can deduce the locations of some of the points of ξ ( {0,1}) because they are at the centres of concentric circles formed by level sets. Comparison between the two images in Figure 1.38 enables us, as in the previous example, to estimate dH (S, ) by making use of the fact that Dξ (S)(ξ (σ )) = DS (σ ) for all σ .

When the underlying metric is d it is hard to make illustrations similar to Figures 1.37 and 1.38, because generally there exist large sets of equidistant points in the metric space ( , d ). For example, when = {0,1} there exists a set containing 2m points, each of which is at a distance 1/2m from all the other points

Figure 1.38 Can you spot the differences and thereby estimate the Hausdorff distance between the two embedded sets? This figure shows the level sets of the shortest-distance function for embeddings of the code space {0,1} {0,1} and a subset of the code space. The embedding function here is similar to the one used in Figure 1.15.

in the set, for all m = 1, 2, . . . This implies that there does not exist an embedding ξ : {0,1} → Rn such that d (σ, ω) = deuclidean(ξ (σ ), ξ (ω)) for all σ, ω {0,1} and all n = 1, 2, . . . ; if the latter were the case then there would exist in Rn a set containing more than n + 1 points, each of which is at unit euclidean distance from all the other points in the set. The latter statement is not true, as demonstrated in Exercise 1.5.17. See also Figure 1.12.

In this sense we can think of the space ( A, d ) as being very high dimensional, whereas we can think of ( A, d|A|) as being contained in a one-dimensional space. Recall that d|A| is defined in Equation (1.6.6) by means of an embedding of in R. Despite this difference, remember that, as asserted in Theorem 1.9.6, the natural topology on ( , d ) is the same as the natural topology on ( , d|A|).

E x e r c i s e 1.12.29 Show that in the code space ( {0,1}, d ) the furthestdistance function F {0,1} (σ ) of {0,1} is constant for all σ {0,1}. Show too that

in ( {0,1}, d|A|) we have F {0,1} (σ ) = max{ξ (σ ), ξ (111 · · · ) − ξ (σ )}, where ξ is the embedding function defined in Equation (1.6.6).

E x e r c i s e 1.12.30 Prove that in the metric space ( {0,1}, d ) there exists a set containing 2m points, each of which is at a distance 1/2m from all the other points in the set, for all m = 1, 2, . . .

1.13The metric spaces (H(X), dH ), (H(H(X)), dH(H)), . . .

In this section we investigate some properties of the space (H(X), dH ) that it inherits from the underlying metric space (X, d). The space (H(X), dH ) is a very natural setting in which to study fractal sets. As we will see in Chapter 4, sequences of approximations to fractal sets may be described as Cauchy sequences of points in (H(X), dH ). Thus the existence of limits of such sequences, the fractals themselves, depends upon the the completeness of the space (H(X), dH ). Similarly, the existence of superfractal sets depends upon the completeness of the space (H(H(X)), dH(H)), as we will see in Chapter 5. So we begin by showing that (H(X), dH ) inherits the property of completeness from the space (X, d).

The completeness of H(X)

The statements and proofs of Theorems 1.13.1 and 1.13.2 follow closely [9], p. 34, Lemma 7.2 and p. 35, Theorem 7.1.

T h e o r e m 1.13.1 (Extension lemma) Let (X, dX ) be a complete metric space and let { An H(X)}∞n=1be a Cauchy sequence in (H(X), dH ). Suppose that {xn j An j }∞j=1 is a Cauchy sequence in (X, dX ), where {n j }∞j=1 is an increasing sequence of positive integers. Then there exists a Cauchy sequence {xn An }∞n=1 in (X, dX ) for which {xn j An j }∞j=1 is a subsequence.

P r o o f Let n0 = 0. For each j {1, 2, 3, . . . } and n {n j−1 + 1, . . . , n j }

choose xn An such that DAn (xn j ) = dX (xn , xn j ). Then {xn j An j }∞j=1 is a subsequence of {xn An }∞n=1. To show that the latter is a Cauchy sequence let > 0

be given. There is an integer N1 > 0 such that whenever nk , nl ≥ N1 we have dX (xnk , xnl ) ≤ /3. Also, there is an integer N2 > 0 such that whenever m, n ≥ N2 we have dH (An , Am ) ≤ /3.

So we assume that nk , nl ≥ N1 and that m, n ≥ N2. Then we note that, by the triangle inequality,

# $ # $ # $ dX (xm , xn ) ≤ dX xm , xnk + dX xnk , xnl + dX xnl , xn .

Let k, l be such that m {nk−1 + 1, . . . , nk }, n {nl−1 + 1, . . . , nl } and let m, n ≥ max{N1, N2}. Then dX (xm , xnk ) = DAm (xnk ) ≤ DAm (Ank ) ≤ dH (An , Ank ) ≤ /3; similarly, dX (xnl , xn ) ≤ /3. Since we also have dX (xnk , xnl ) ≤ /3 it follows

that dX (xm , xn ) ≤ for all m, n ≥ max{N1, N2}.

The following result provides not only a general condition under which (H(X), dH ) is complete but also a characterization of the limits of Cauchy sequences in H(X).

T h e o r e m 1.13.2 Let (X, dX ) be a complete metric space. Then (H(X), dH ) is a complete metric space. Moreover, if { An H(X)}∞n=1 is a Cauchy sequence

then A := limn→∞ An can be characterized as

A = x X : there is a Cauchy sequence {xn An }∞n=1 that converges to x .

(1.13.1)

P r o o f Let { An H(X)}∞n=1 and let A be defined by Equation (1.13.1). We prove that (i) A = ; (ii) A is closed and hence complete; (iii) for given > 0 there is an N such that for n ≥ N we have A BAn ( ); (iv) A is totally bounded and hence by (ii) is compact; (v) A = limn→∞ An .

Proof of (i): We establish the existence of a Cauchy sequence {xn An }∞n=1 in X. We can select an increasing sequence of positive integers {Nn }∞n=1 such that dH (An , Am ) ≤ 1/2i for m, n > Ni . Choose xN1 AN1 . Then since dH (AN1 , AN2 ) ≤ 1/2 we can find xN2 AN2 such that dX (xN1 , xN2 ) < 1/2. Now

existence of a convergent sequence {xn An }∞n=1. Since X is complete the limit exists and, by the definition of A, Equation (1.13.1), it belongs to A.

Proof of (ii): To show that A is closed, suppose that {ai A}i∞=1 converges to a point a X. We need to show that a A. Hence we can find an increasing sequence of integers {Nn }∞n=1 such that dX (aNn , a) < 1/n. Also, since ai A it follows from the definition of A that there is a sequence {ai,n An }∞n=1 that converges to ai for each i. And so we can find an increasing sequence of integers

such that dX (aNn ,Mn , aNn ) < 1/n. It follows that dX (aNn ,Mn , a) < 2/n. Hence the sequence {xNn = aNn ,Mn ANn }∞n=1 is a Cauchy sequence convergent to

a. By Theorem 1.13.1 it can be extended to a sequence {xn An }∞n=1 convergent to a and, by the definition of A, Equation (1.13.1), it follows that a belongs to A.

Proof of (iii): Let > 0. Then there exists N such that n, m ≥ N implies that dH (An , Am ) ≤ and, as in Equation (1.12.5), Am BAn ( ). Let a A and let {am Am }∞m=1 be a sequence that converges to a. Then we must have am BAn ( ) whenever n, m ≥ N . But BAn ( ) is closed because An is compact. So a BAn ( ) whenever n ≥ N and therefore A BAn ( ) for all n ≥ N .

Proof of (iv): Suppose that A is not totally bounded. Then for some > 0 we can find a sequence of points {xi A}i∞=1 such that dX (xi , x j ) ≥ whenever i =j. From (iii) we have that A BAn ( /3) for some large enough n. It follows that for each xi we can find a corresponding yi An such that dX (xi , yi ) ≤ /3. Since An is compact some subsequence {yi j }∞j=1 of {yi }i∞=1 converges. So we can find points y j1 and y j2 such that dX (y j1 , y j2 ) < /3. But then it follows that

# $ # $ # $ # $

dX x j1 , x j2 ≤ dX x j1 , y j1 + dX y j1 , y j2 + dX y j2 , x j2 < .

This is a contradiction. So A is totally bounded. Since A is also complete, by (ii), it must be compact.