Lecture 31

Measuring the sizes of fractal sets in Rk (cont’d)

In the previous lecture, we examined the following generator in R2:

1

3ll

1

3l

1

3l

1

3l

This generator removes the middle third of a line segment of length l and replaces it with two segments

of length l3. Starting with the set J0 = [0, 1], the unit interval on the x-axis, repeated application of

this generator produces curves Jn of increasing length which, in the limit n → ∞ converge to the “von

Koch curve” C shown below.

The von Koch curve

We showed that the von Koch curve C had infinite length and zero area. We also showed that it had

a fractal dimension,

D =ln 4

ln 3≈ 1.26 . (1)

Since 1 < D < 2, the curve C can be viewed as “thicker” than a simple curve of dimension 1 and

“thinner” than a planar area of dimension 2.

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We now consider a generalization of the von Koch curve generator as shown below.

rll

rl

rl rl

This generator, to be denoted as Gr in order to explicitly state its dependence upon the parameter r,

replaces a line segment of length l with four shorter line segments of length rl as shown above. The

feasible range of the parameter is 1

4≤ r ≤ 1

2. To see this, note the following:

• If r < 1

4, then the four shrunken segments, each of length rl would have total length 4rl < l.

If the first segment started at the left endpoint of the original line segment, the fourth segment

could not reach the right endpoint.

• When r = 1

4, the line segment is replaced with four line segments of one-quarter its length

touching each other. The result is the original line segment. As such, the limiting curve Cr is

the original seed line segment.

• If r > 1

2, then the first and last segments would have to overlap since their combined length is

2rl > l. Otherwise, the distance between the left endpoint of the first segment and the right

endpoint of the fourth segment would be greater than l. We don’t want to allow such overlapping

since the resulting curve would have a “loop” in it.

Finally, we mention that the parameter value r = 1

3corresponds to the standard von Koch curve

discussed earlier.

We now consider the limiting set of points obtained from the iteration of Gr. Without loss of

generality, we assume that our initial “seed” set is I0 = [0, 1] from which the sequence In+1 = Gr(In)

is then generated. Each set In is a piecewise linear curve. From the action of the generator, it should

not be difficult to see that the length Ln of In is related to the length Ln+1 of In+1 as follows,

Ln+1 = 4r Ln . (2)

Since L0 = 1, it follows that

Ln = (4r)nL0 = (4r)n . (3)

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When r = 1

4, Ln = 1 for all n ≥ 0, as discussed earlier. When r > 1

4, 4r > 1 which implies that

Ln → ∞ as n → ∞. In other words, apart from the trivial case r = 1

4, all generalized von Koch curves

have infinite length.

In the limit n → ∞, the sequence of piecewise linear curves In converges to a curve Cr. Some

limiting sets for various r values are shown on the next page. As r increases, the curves Cr demonstrate

a greater degree of jaggedness or “fractality.” In each case, Cr is self-similar and may be expressed

as a union of four contracted copies of itself:

Cr =⋃

k=1

Cr,i. (4)

Each copy Cr,k is obtained by shrinking Cr toward the origin (0,0) by a factor of r and then

rotating and translating it appropriately. Perhaps the most striking limiting “curve” Cr occurs in the

case r = 1

2. The limiting set is, in fact, the triangular region shown on the next page. Our curve Cr

is, in fact, a “space-filling” curve. The reader is encouraged to work out the first few iterates In to see

how these sets approach the limiting area. (See the book, The Fractal Geometry of Nature by Benoit

Mandelbrot, for more examples of generators that yield space-filling curves.)

The next step is to determine the fractal dimension D of the limiting curve Cr. The self-similar

nature of Cr described above will be very useful here. For an ǫ > 0, suppose that we can tile the

curve Cr with N(ǫ) circular discs of diameter ǫ. Also assume that ǫ is sufficiently small so that we

may make the approximation

LD(Cr) ≈ N(ǫ)ǫD . (5)

(This approximation may be made as closely as desired by choosing ǫ sufficiently close to 0.) We

sketch this covering below.

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Generalized von Koch curve: r = 0.3

Generalized von Koch curve: r = 0.4

Generalized von Koch curve: r = 0.5

402

Now construct four shrunken copies of the above figure, i.e. the curve Cr as well as the N(ǫ) discs,

with contraction factor r, and arrange these copies so that the the curve Cr is reconstructed. The

resulting figure, shown below, shows curve Cr being covered with 4N(ǫ) circular discs of diameter rǫ.

The D-dimensional volume of Cr may therefore be approximated as

LD(Cr) ≈ 4N(ǫ)(rǫ)D (6)

We now equate the two length estimates,

N(ǫ)ǫD = 4N(ǫ)(rǫ)D . (7)

Dividing both sides by N(ǫ)ǫD yields the relation

4rD = 1 , (8)

which may be rewritten as

r−D = 4 =⇒

(

1

r

)D

= 4 . (9)

Solving for D yields

Dr =ln 4

ln 1

r

,1

4≤ r ≤

1

2. (10)

This is the fractal dimension of the generalized von Koch curve Cr.

We could also have derived the above result in a quick way by using a scaling result from the previous

lecture which we repeat here:

N(rǫ) = N(ǫ)r−D . (11)

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The number of N(rǫ) balls needed to cover the set is 4 times the number of N(ǫ) balls, i.e.,

N(rǫ) = 4N(ǫ) . (12)

Substituting this into (11) yields,

4N(ǫ) = N(ǫ)r−D , (13)

which yields (8).

Special cases:

1. r = 1

4: D = 1; the generator replaces a line with itself, as discussed earlier.

2. r = 1

3: D = ln 4

ln 3; the limiting curve is the von Koch curve studied in the previous lecture.

3. r = 1

2: D = 2; the limiting set is the triangular region shown in the figure.

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Further generalizations of the von Koch curve generator

The factor “4” appearing in the expression for the fractal dimension Dr is due to the fact that Cr is

composed of four contracted copies of itself. Now consider an even more generalized generator, shown

below, in which a line segment is replaced by n copies of itself that are contracted by a factor r.

n

l rlrl

rl rl

rl

1

Iteration of this generator will yield, in the limit, a fractal object – call it C – which can be expressed

as a union of n contracted copies of itself. From an argument that is almost identical to the one used

above, the “4” in Eq. (6) is replaced by “n” so that the fractal dimension of the attractor curve C is

given by

D(n, r) =lnn

ln 1

r

. (14)

Note the relationship between n, r and D may also be left in the following form:

nrD = 1 . (15)

Let’s now step back back and derive the results in Eq. (14) and (15) in the same way that we did

for the four-copy case. Suppose that we can cover the fractal curve with N(ǫ) circular boxes or disks of

radius ǫ. We can imagine ǫ to be sufficiently, in fact, infinitesimally, small so that the D-dimensional

measure of the fractal curve C is well approximated by (up to a multiplicative constant),

LD(C) ∼= N(ǫ)ǫD . (16)

(Here, we are ignoring the π/4 factor that gives the area of a circular disk of radius ǫ. Even if we used

it, it will eventually disappear. We could also have used rectangles with sides of length ǫ.) In fact, it

may make things a little less confusing if we simply write that

LD(C) = limǫ→0

N(ǫ)ǫD . (17)

(Although that being said, the limit itself may be infinite!)

Now recall that each component of the fractal curve C is a contracted (and rotated, translated)

copy of C. We now shrink C, along with the N(ǫ) ǫ-disks that cover it, by a factor of r. The result

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is a shrunken copy of C that is covered with N(ǫ) disks of radii r′ = rǫ. We replace each segment

of curve C (as well as the ǫ-disks that cover it) with the shrunken copy of C with rǫ-disks. The net

result is that we have reconstructed curve C, but it is now covered with nN(ǫ) disks of radius r′ = rǫ.

The approximation to the D-dimensional measure of the curve is

LD(C) ∼= N(rǫ)(rǫ)D = nN(ǫ)(rǫ)D . (18)

But we can once again take limits,

LD(C) = limǫ→0

N(rǫ)(rǫ)D = limǫ→0

nN(ǫ)(rǫ)D . (19)

We now combine the two limiting results for LD(C),

limǫ→0

N(ǫ)ǫD = limǫ→0

nN(ǫ)(rǫ)D . (20)

Notice that we can divide out N(ǫ)ǫD from both sides to obtain

nrD = 1 , (21)

from which it follows that

D(n, r) =lnn

ln 1

r

. (22)

And if that was too confusing, just keep in mind the basic scaling result

N(rǫ) = N(ǫ)r−D , (23)

that we discussed in the previous lecture. (It’s actually an approximation - the equality holds in the

limit ǫ → 0+.)

One may generalize the above generator Gr to employ scaling factors ri that are not necessarily

equal to each other, as sketched below.

rn−1ll r1l r2l

r3l rnl

This generalization makes it possible to generate a much wider class of fractal curves. Some examples

are shown in the figure on the next page. On the left of each of the first rows is shown the generator

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G. On the right is the resulting fractal curve. The second and third sets show that when two lines of

the generator are superimposed, tree-like patterns result. As we travel from the leftmost point of the

generator (0,0) to the rightmost point (1,0), the fractal curve always lies to our left. As such, the two

superimposed lines have different orientations, and a copy of the curve lies on each side.

In the final entry, the generator and fractal curve are shown on the same plot. The generator is

slightly more complicated, in the sense that the path from (0,0) to (1,0) is not straight. The fractal

curve is much more “cloudlike.” In fact, a slight perturbation of this result, shown in the next figure,

produces a curve which intersects back on itself infinitely often, giving rise to a more “curly” fractal

set.

A fractal set C obtained from a generalized generator will once again be self-similar, i.e., it will be

a union of n contracted copies Ci of itself. The major difference is that each copy Ci will be obtained

by contracting C by the factor ri. The fractal dimension D of these generalized curves is a more

complicated function of the scaling functions ri as compared to the homogeneous case ri = r studied

earlier. As such, we cannot use the quick formula that we derived earlier, i.e.,

D =ln (number of copies)

ln (1/contraction factor)=

lnn

ln 1

r

. (24)

Here we peform a quick derivation of the fractal dimension for the generalized attractor with contrac-

tion factors ri, 1 ≤ i ≤ n.

Once again, let us assume that ǫ is sufficiently small so that we can cover the entire curve C with

N(ǫ) circular disks of radius ǫ. The D-dimensional measure of the curve is well approximated by

LD(C) = N(ǫ)ǫD . (25)

Now make n copies of this curve, along with the ǫ-radius disks on it. Each copy, along with the ǫ-disks

will be contracted by the factor ri. Now replace each piece of the fractal curve C with the appropriate

ri-contracted copy of C with the disks. The curve C will remain the same. The first piece of C will be

covered with with N(ǫ) disks of radius r1ǫ, the second piece with N(ǫ) disks of radius r2ǫ, etc., until

we arrive at the nth piece covered with N(ǫ) disks of radius rnǫ. As a result, we now have another

estimate of the D-dimensional measure,

LD(C) = N(ǫ)(r1ǫ)D +N(ǫ)(r2ǫ)

D + · · · +N(ǫ)(rnǫ)D . (26)

If we now combine these two estimates for LD(C), and understand that they are equal in the limit

ǫ → 0+, we obtain, after cancelling the N(ǫ) and ǫD factors from both sides, the following equation

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Some examples of generalized von Koch curves

Generator G with four copies – the result is a modified von Koch curve

Generator G with four copies – the result resembles a collection of trees

Generator G with seven copies – another tree-like scene

Rather than being dendritic, the fractal curve is cloudlike. Here we show the generator and the

curve on the same plot.

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Generator G producing a more “curly” fractal set.

for D,

rD1 + rD2 + · · ·+ rDn = 1. (27)

Note that in the special case of a homogeneous fractal, i.e., ri = r, the above result reduces to our

earlier result, Eq. (21).

We claim that the fractal dimension D is the unique solution to this equation and leave this as

an exercise for the reader. The basic steps are as follows:

• For a fixed set of scaling factors, 0 < ri < 1 for 1 ≤ i ≤ n, n ≥ 2, define the following function

of D:

f(D) = rD1 + rD2 + · · · rDn . (28)

• Note that f(0) = n.

• Also note that f(D) → 0+ as D → ∞ since each of the ri lie between 0 and 1.

• Show that f ′(D) < 0 for D > 0, which implies that f(D) is strictly decreasing for D > 1.

• From the above information, it should follow that the graph of f(D) vs. D should intersect the

line y = 1 only once. At this value of D, Eq. (27) is satisfied.

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The “Sierpinski triangle” or “Sierpinski gasket”

We now conclude this section with a classic fractal set in the plane, the so-called “Sierpinski triangle”.

We begin with an equilateral triangle and define a generator G that removes the middle quarter of

triangular regions (open sets) as shown below:

The action of the generator on the triangular region may also be interpreted geometrically as follows:

G produces three contracted copies of the region by shrinking in the x and y directions by a factor

of r = 1

2and then placing the three copies so that they touch at the midpoints of the triangular

boundary. The repeated action of G produces the following sets:

In the limit, the sequence of sets converges to the following fractal set S, the Sierpinski triangle, shown

in the figure below.

Sierpinski triangle

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The set S is easily seen to be a union of three contracted copies of itself, with contraction factor

r = 1

2. A measurement procedure involving circular discs of diameter ǫn = 1

2nshows that S has zero

two-dimensional area and infinite one-dimensional length. Let us now recall the basic scaling result

for self-similar fractals, relating the number of disks needed to cover a set with fractal dimension N ,

N(rǫ) = N(ǫ)r−D . (29)

In the case of the Sierpinski triangle,

r =1

2(30)

and

N(1

2ǫ) = 3N(ǫ) , (31)

This comes from the fact that if you use N(ǫ) disks of radius ǫ to cover the entire triangle, and

then make three shrunken copies (by factor r = 1

2, you can cover each of the three subsets of the tri-

angle with N(ǫ) disks of radius 1

2ǫ. As a result, you need 3N(ǫ) disks of radius 1

2ǫ to cover the triangle.

From Eqs. (29)-(31), we have

3N(ǫ) = N(ǫ)

(

1

2

)

−D

, (32)

or

3 = 2D , (33)

which implies that

D =ln 3

ln 2∼= 1.58496 (34)

Note that 1 < D < 2 – the Sierpinski triangle, as expected, is “thicker” than a line but “thinner” than

a two-dimensional region in the plane.

Another look at fractal dimensions in terms of scaling: Let’s examine this result one more

time to see why the fractal dimension D lies between the values of 1 and 2. The Sierpinski triangle

is a union of three contracted copies of itself. Each shrunken copy is obtained by contracting the

triangle by a factor of r = 1

2in the x- and y-directions. Since the Sierpinski triangle is a union of

three contracted copies of itself, this means that the “size” of each copy is one-third the “size” of

the entire Sierpinski triangle. (By “size”, we can mean “mass” or “measure”, or whatever.)

411

Now constrast this fact with the way that “normal” Euclidean objects scale, using the scaling factor

r = 1

2that was used for the Sierpinski triangle:

1. Consider a line of length 1 in the plane: say, the interval [0, 1] on the x-axis. Let us now make

two contracted copies of this interval, namesly [0, 12] and [1

2, 1]. Both of then are contractions of

the interval [0, 1] by a factor of r = 1

2. And the length of each of these copies is one-half the

length of the interval [0, 1]. This is simple!

2. Now consider a square with sides of length 1, say the interval [0, 1]2. If we shrink the sides of

the square by factor 1

2, we obtain a square that is one-quarter the area of the square. As such,

we’ll need four of these copies to reconstruct the original square. This is simple as well.

3. But when we shrink the Sierpinski triangle in the x- and y-directions by the factor r = 1

2, we

produce a smaller triangle which is one-third of the size of the original triangle: Not one-half,

as was the case for lines, nor one-quarter, which was the case of regions, but one-third, which

lies between one-half and one-quarter. Because the Sierpinski triangle has “holes” in it, it scales

neither as a one-dimensional nor as a two-dimensional object. It scales as a D-dimensional

object, where D = ln 3/ ln 2.

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Lecture 32

Measuring the sizes of fractal sets in Rk (cont’d)

Other fractals obtained by dissection

There are many generalizations of the Sierpinski triangle in R2 – a few of these are presented in

Problem Set No. 6. One can also extend the method to higher dimensions, i.e., a “Sierpinski gasket”

in R3. A sketch of this fractal, taken from B. Mandelbrot’s book, The Fractal Geometry of Nature, is

presented in the figure below.

Three-dimensional Sierpinski gasket, from The Fractal Geometry of Nature, by B. Mandelbrot.

The 3D Sierpinski gasket is self-similar: It may be viewed as a union of 4 contracted copies of itself.

In this case, the contraction factor is r = 1

2: Each copy is obtained from the original set by shrinking

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it in three directions by a factor of 1

2. As such, the 3D gasket is a homogeneous fractal and we may

use the same formula used earlier for our generalized von Koch curves, i.e.,

D =ln (number of copies)

ln (1/contraction factor)=

lnn

ln 1

r

=ln 4

ln 2= 2 . (35)

In the event that you are not sufficiently comfortable with this quick result, let’s derive it by going

back to our fundamental scaling relation,

N(rǫ) = N(ǫ)r−D . (36)

Here, N(ǫ) will be the number of (3D) cubes with sides of length ǫ needed to cover the 3D Sierpinski

gasket. As mentioned above, the scaling factor is r = 1

2. If we shrink such a covering of the 3D gasket

by a factor of 1

2and place four copies of this shrunken covering at the appropriate places, we obtain a

covering of the 3D gasket with cubes with sides of length 1

2ǫ. As such, the number of cubes with sides

of length 1

2ǫ needed to cover the entire 3D gasket is

N(1

2ǫ) = 4N(ǫ) , (37)

From Eqs. (36) and (37), we then have that

4N(ǫ) = N(ǫ)

(

1

2

)

−D

. (38)

Division by N(ǫ) yields

4 = 2D , (39)

which then gives us the result for D obtained earlier, i.e.,

D =ln 4

ln 2= 2 . (40)

Isn’t this a rather pecular result. The 3D Sierpinski gasket, a set in R3 scales as a two-dimensional

object. Normally, when we think of a two-dimensional object, we think of a planar area or a surface.

414

Sierpinski carpet and Menger sponge

Consider the following generator in R2, which operates on regions in the plane. It removes the

“middle-ninth” of a square region:

G

ℓl3

Repeated application of this procedure yields, in the limit, the so-called two-dimensional “Sier-

pinski carpet,” an approximation of which is shown in the figure below.

The “Sierpinski carpet”

The Sierpinski carpet is self-similar: It may be expressed as a union of 8 copies of contracted copies of

itself. In this case, the contraction factor is r = 1

3. As such, the fractal dimension D of the Sierpinski

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carpet is

D =lnn

ln 1

r

=ln 8

ln 3∼= 1.89279 . (41)

An illustration of a three-dimensional version of the Sierpinski carpet, known as the “Menger sponge,”

taken from Mandelbrot’s book, Fractal Geometry of Nature is shown below.

Three-dimensional “Menger sponge,” from The Fractal Geometry of Nature by B. Mandelbrot.

The Menger sponge is a union of 27−7 = 20 contracted copies of itself, with contraction factor r = 1

3.

The fractal dimension of the sponge is therefore

D =lnn

ln 1

r

=ln 20

ln 3∼= 2.727 . (42)

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A final and interesting example: A simulation of a “strange attractor”

Consider an infinite set S of lines that is distributed in the plane as shown in the figure below. This

distribution of lines may be viewed as a simulation of a “strange attractor” – recall our brief study of

the strange attractor of the H’enon map in the plane. A line perpendicular to these lines intersects the

ensemble along a ternary Cantor set. For simplicity, we consider lines of length 1 and let the width of

the ensemble to be 1. In what follows, we shall compute the fractal dimension D of this ensemble.

ǫn

1

3n

blocks

12n

blocks

ǫn =1

3n

ǫn

Covering a Cantor set of lines with ǫ2n-boxes

First of all, the total length L of this infinite set S of lines of length 1 is ∞. We now compute the

area A of this ensemble. As was the case for the ternary Cantor set, it is convenient to consider square

blocks with sides of length ǫn =1

3n.

1. For each horizontal row of blocks covering the solid lines in the x-direction, we need 3n blocks.

2. We need 2n blocks in each vertical row to cover the Cantor set (as was the case for the one-

dimensional ternary Cantor set – there, we needed only segments of length ǫn).

We therefore need N(ǫn) = 3n × 2n = 6n blocks of area ǫ2n. This yields the following estimate, An, of

the area A:

An = N(ǫn) ǫ2n = 6n ǫ2n = 6n

(

1

3n

)2

=

(

2

3

)n

. (43)

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Clearly,

A = limn→∞

An = 0 . (44)

The Cantor set ensemble of lines has zero area.

We now compute the fractal dimension D of this ensemble in two ways:

1. Computing the D-dimensional measure of the set and adjusting the exponent ap-

propriately to obtain a non-zero and non-infinite “size”: Once again, we employ N(ǫn)

measuring blocks with side-length ǫn. The D-dimensional measure will be given by

LD(S) = limn→∞

N(ǫn) ǫDn

= limn→∞

6n(

1

3n

)D

= limn→∞

(

6

3D

)n

. (45)

This limit is non-zero and non-infinite when

3D = 6 =⇒ D =ln 6

ln 3. (46)

2. Scaling: Suppose that we require N(ǫ) square blocks of side-length ǫ to cover the set S – for

example N(ǫn) = 3n × 2n = 6n, as found earlier. We now consider square blocks of side-length

ǫ3. In order to cover the lines horizontally, we’ll need three times the number of blocks in the

horizontal direction. In order to cover the Cantor set, we’ll ned two times the number of blocks

in the vertical direction. Therefore,

N

(

1

3ǫ

)

= 6N(ǫ) . (47)

We now employ the scaling equation,

N(rǫ) = N(ǫ)r−D , (48)

to deduce that(

1

3

)

−D

= 6 , (49)

which implies that

3D = 6 =⇒ D =ln 6

ln 3. (50)

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Both methods yield the same result.

Let’s now examine the result for the fractal dimension D of the set S:

D =ln 6

ln 3

=ln 3 + ln 2

ln 3

= 1 +ln 2

ln 3. (51)

We see that D is the sum of the dimension D1 = 1 of the (horizontal) line and the dimension D2 =ln 2

ln 3

of the Cantor set.

We may view the set S as a kind of “product” of a one-dimensional horizontal line with a vertical

Cantor set. The dimension of the resulting set is the sum of the dimensions of the two components.

If we think of a two-dimensional region, say, [0, 1]2 as the “product” of a one-dimensional horizontal

line and a one-dimensional vertical line, then the sum of these two dimensions yields the dimension of

the product set.

419

A short note on fractal-like sets occurring in nature

Benoit Mandelbrot’s classic work, The Fractal Geometry of Nature centered, as its name suggests, on

the idea that fractal patterns occur in nature – not just here and there, but everywhere, e.g., clouds,

rock, soil, river/tributary patterns, etc..

Diffusion-limited aggregates: Fractal-like sets occurring in nature

An excerpt from the article entitled, “Diffusion-Limited Aggregation: A Model for Pattern Formation,”

by Thomas C. Halsey from the Physics Today online archive at

http://web.archive.org/web/20070405094836/http://aip.org/pt/vol-53/iss-11/p36.html

The basic concept

To understand the basics, consider colloidal particles undergoing Brownian motion in some fluid, and

let them adhere irreversibly on contact with one another. Suppose further that the density of the

colloidal particles is quite low, so one might imagine that the aggregation process occurs one particle

at a time. We are then led to the following model.

Fix a seed particle at the origin of some coordinate system. Now introduce another particle at a

large distance from the seed, and let it perform a random walk. Ultimately, that second particle will

either escape to infinity or contact the seed, to which it will stick irreversibly. Now introduce a third

particle into the system and allow it to walk randomly until it either sticks to the two-particle cluster

or escapes to infinity. Clearly, this process can be repeated to an extent limited only by the modeler’s

patience and ingenuity (the required computational resources grow rapidly with n, the number of

particles).

The clusters generated by this process are both highly branched and fractal. The cluster’s fractal

structure arises because the faster growing parts of the cluster shield the other parts, which therefore

become less accessible to incoming particles. An arriving random walker is far more likely to attach

to one of the tips of the cluster shown in the left figure (a) below than to penetrate deeply into one of

the cluster’s ”fjords” without first contacting any surface site. Thus the tips tend to screen the fjords,

a process that evidently operates on all length scales. Figure (b) at the right shows the ”equipotential

lines” of walker probability density near the cluster, confirming the unlikelihood of random walkers

penetrating the fjords.

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A theoretical cluster obtained by a computer simulation of this aggregation process is shown below.

The fractal dimension D of both the theoretical as well as the experimental cluster in three-

dimensions is roughly D = 2.5.

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Estimating the (global) fractal dimension D of sets/objects

Given a set S ∈ Rk, let N(ǫ) be the minimum number of k-dimensional balls of diameter ǫ in R

k

needed to “cover” S, i.e., S is contained in a union of these balls. If S has fractal dimension D, then

N(ǫ) ∼= Aǫ−D = A

(

1

ǫ

)D

as ǫ → 0+ . (52)

For mathematical objects, the approximation improves as ǫ → 0+. Now take logarithms of oth sides

of the above equation (it doesn’t matter what base you choose for the logarithm),

logN(ǫ) ∼= D log

(

1

ǫ

)

+ logA . (53)

Rearranging,

D ∼=logN(ǫ)

log 1

ǫ

−logA

log 1

ǫ

. (54)

As ǫ → 0+, 1

ǫ→ ∞ which implies that log 1

ǫ→ ∞ which, finally, implies that the second term in the

above equation goes to zero. The mathematical definition of D is therefore

D = limǫ→0+

logN(ǫ)

log 1

ǫ

, (55)

provided that the limit exists. For natural objects, one will not be able to let ǫ → 0+. As such, we

determine N(ǫ) over a range of “reasonable” ǫ-values.

Basic idea:

1. Work with a finite set of epsilon values, i.e., ǫ1 > ǫ2 > · · · ǫn, e.g.,

ǫk = rk , (56)

where 0 < r < 1 is a convenient scaling factor, possibly relevant to the problem.

On a computer, this is often done with a digital image by using dyadic squares of pixels. For

example, if our set S of concern is shown in a 512 × 512-pixel digital image, we let ǫ1 = 512,

ǫ2 = 512/2 = 256, ǫ3 = 256/2 = 128, · · · , ǫ9 = 1. In other words, a single pixel represents the

smallest ǫ-tile used to cover the set S. N(ǫ9) will be the number of pixels needed to cover S.

N(ǫ8) will be the number of (nonoverlapping) 2× 2-pixel blocks needed to cover S, and so on.

2. Let N(ǫk) denote the number of ǫk-balls (tiles) needed to cover S.

422

3. Plot logN(ǫk) vs. log1

ǫkfor 1 ≤ k ≤ n.

4. A least-squares fit to this data will produce a straight line with slope m that will be an approx-

imation to D.

We illustrate this idea with a couple of examples below.

Experiment No. 1

Shown in the figure below is a 512 × 512-pixel digital image of a modified Sierpinski triangle. (It

is obtained in the same way as before, i.e., by taking the middle-quarters. But here, we started with

a right triangle as initial “seed.”) Our fractal set S consists of all pixels which are black.

A simple computer program was written (in MATLAB) to perform “box counting” on this set using

boxsizes of ǫ = 1, 2, 4, 8, 16 and 32. (It is not very useful to use higher values of ǫ, since very few boxes

are used The numbers of tiles N(ǫ) needed to cover the set S for these ǫ-values are shown in the table

below.

423

ǫ N(ǫ)

32 81

16 243

8 729

4 2187

2 6561

1 19683

A plot of the values of N(ǫ) vs. log(1/ǫ) from the above table is shown below.

-5 -4 -3 -2 -1 0

log (1/ )

0

1

2

3

4

5

6

7

8

9

10

log

N(

)

The plot seems to indicate that the points lie very close to a straight line. Indeed, a least-squares

fitting of the data yields a slope of

D ≈ 1.5860 , (57)

which agrees extremely well with the theoretical result discussed earlier,

D =ln 3

ln 2≈ 1.5860 . (58)

424

Experiment No. 2

The result of the above experiment may seem rather contrived since we used a “perfect” fractal,

i.e., the Sierpinski gasket, even though it was a finite-dimensional representation of the gasket. Let

us now consider a more realistic example, as presented in the figure below. It is a 512 × 512 digital

image of the vasculature (blood vessels) in a tiny section of a a mouse ear obtained by a rather and

novel method of imaging being developed at UW:

http://www.photomedicinelabs.com

The computer program used in Experiment No. 1 was used once again to perform a box-counting

analysis of the mouse ear image. As before, box sizes of ǫ = 1, 2, 4, 8, 16 and 32 were used. The results

are shown below.

ǫ N(ǫ)

32 218

16 631

8 1435

4 2970

2 5657

1 9568

425

A plot of the values of N(ǫ) vs. log(1/ǫ) from the above table is shown below.

-5 -4 -3 -2 -1 0

log (1/ )

0

1

2

3

4

5

6

7

8

9

10

log

N(

)

The points could be imagined to lie roughly on a straight line. If we fit a straight line using

least-squares to all six points, the value of the slope is

D ∼= 1.16 , (59)

which suggests that the network of blood vessels is slightly thicker than a simple curve. This estimate,

however, might seem a little too low, given the complexity of the network.

If we look at the plot of N(ǫ) vs. log(1/ǫ) values, we see that it flattens out as log(1/ǫ) approaches

0, i.e., ǫ approaches 1. It may well be that using too fine a grid, i.e., boxes of lengths 2 and 1, is

detecting only the curves. One can see this effect in the table of N(ǫ) values. As ǫ decreases from 4

to 2 to 1, the values of N(ǫ) are not increasing as rapidly as they were for higher values of ǫ.

For this reason, it may be desirable to use only the larger boxes, say, ǫ = 32, 16 and 8. If we look

at only the first three points (from the left) of the plot, they seem to lie close to a line with higher

slope. Indeed, if we perform a least-squares fit on only these three points, we obtain a slope of

D ∼= 1.35 , (60)

which is a significant increase from the previous value.

426

This motivates the use of even larger box sizes. If we use ǫ = 64 and 128 sized boxes, we obtain the

following additional counts:

ǫ N(ǫ)

128 16

64 60

A plot of the logarithms of these values, along with those of the three points used earlier, is shown

below. A least-squares fit through these five points yields a line with slope,

D ∼= 1.64 . (61)

-5 -4 -3 -2 -1 0

log (1/ )

0

1

2

3

4

5

6

7

8

9

10

log

N(

)

Our box-counting method seems to be detecting a set which is thicker than a curve but thinner

than a planar area. That being said, the above analysis may seem somewhat unsettling to the reader

since a definitive result is not obtained. Is the fractal dimension D of the vasculature closer to 1.2?

Or is it closer to 1.6. It might seem that D is closer to 1.6 but we can’t be too precise. One reason is

that we don’t have enough data to answer this question. But the other point that should be made is

that the “fractal scaling” exhibited by the vasculature is dependent upon the scales that we use, i.e.,

the sizes of boxes used, in our covering. Unlike mathematical objects, we cannot go to infinitely small

scales. We have to acknowledge that the scaling properties discussed earlier for mathematical fractals

hold only over a finite range of scales for objects in the real world.

427

Iterated function systems (IFS) and the construction of fractal sets

We now discuss a very convenient and powerful method of constructing and analyzing self-similar

fractal sets – the so-called method of iterated function systems, abbreviated as IFS. The IFS

method will involve the use of a set contractive mappings on the line R, or in the plane R2 or in

3D space R3. as opposed to the generators discussed in the previous lectures. In this way, one doesn’t

have to keep track of line segments and then operating on them. A remarkable aspect of the IFS

method is that one can generate very good pictures of fractals by simply applying the maps in an IFS

in a random manner, as was discussed in Problem Set No. 5.

We’ll illustrate the idea with a simple example – the one, in fact, that was used in Problem Set

No. 5. From that example, we’ll move on to a more general discussion, and then look a little more

closely at the mathematics behind IFS.

Consider the following two maps, f1 and f2, which map the interval I = [0, 1] into itself: For any

x ∈ [0, 1], f1(x) ∈ [0, 1] and f2(x) ∈ [0, 1]:

1. f1(x) = 1

3x with fixed point x̄1 = 0. The iteration dynamics associated with f1(x) is quite

straightforward: For any x0 ∈ [0, 1], xn = fn1 (x0) =

1

3nx0 → 0 as n → ∞.

2. f2(x) =1

3x+ 2

3with fixed point x̄2 = 1. The iteration dynamics associated with f2(x) is also quite

straightforward: The distance between xn+1 = f2(xn) and the fixed point x̄2 = 1 is one-third

the distance between the point xn and x̄2. As such, xn = fn2 (x0) → 1 as n → ∞.

We now examine the (repeated) action of the maps f1 and f2 on the interval I = [0, 1]. Instead of

looking at where each map fi sends a point x ∈ [0, 1], we consider where each map fi will send sets

of points in [0,1]. It is instructive to examine the action of each of the above maps on the interval

[0,1]. To do this, let us define the following associated interval- or set-valued mappings:

Definition: For i ∈ {1, 2} and any subset S ⊆ [0, 1], we denote

f̂i(S) = {fi(x)|x ∈ S} , (62)

In other words, f̂i(S) is the set of all points obtained by applying the mapping fi(x) to all

points x ∈ S.

428

To illustrate:

1. f̂1([0, 1]) = [0, 13]. In other words, f̂1 “shrinks” the interval [0,1] to [0, 1

3].

2. f̂2([0, 1]) = [23, 1]. In other words, f̂1 “shrinks” the interval [0, 1] to [2

3, 1].

It might help to use some pictures. Let’s consider map f1 for the moment. If we let I0 = [0, 1], then

the interval I1 = f̂1(I0) = [0, 13] is sketched below. We then apply f̂1 to the set I1 to produce the set

I2 = f̂1(I1) = [0, 19].

0 1 0

f̂1

I0 I1

00

0

f̂1

f̂1

I1

I2

0

1

3

1

3

1

9

1

27

1

9

I3

I2

In general, n applications of the map f̂1 to the interval I0 produces the set

In = f̂1(I0) =

[

0,1

3n

]

. (63)

Clearly, the intervals In are shrinking in length. Moreover, they appear to be approaching a limit, i.e.,

limn→∞

In = limn→∞

[

0,1

3n

]

= {0} . (64)

They are approaching the single point {0}. Note that this point is the fixed point of f1. We may view

this single point as a set or an interval consisting of one point.

Let’s now consider the map f2. If we let I0 = [0, 1], then the interval I1 = f̂2(I0) = [23, 1] is sketched

below. We then apply f̂2 to the set I1 to produce the set I2 = f̂1(I1) = [89, 1]. This procedure is shown

graphically below.

I2

0 1

I0 I1

1

311

1

I3

8

9

126

27

I1

1

31

I2

18

9

f̂2

f̂2

f̂2

429

In general, n applications of the map f̂2 to the interval I0 produces the set

In = f̂1(I0) =

[

1−1

3n, 1

]

=

[

3n − 1

3n, 1

]

. (65)

Once again, the intervals In are shrinking in length. Moreover, they appear to be approaching a limit,

i.e.,

limn→∞

In = limn→∞

[

1−1

3n, 1

]

= {1} . (66)

They are approaching the single point {1}, which is the fixed point of f2. We may once again view

this single point as a set or an interval consisting of one point.

In summary:

1. Repeated application of the set-valued map f̂1 to the interval I = [0, 1] produces a set of intervals

In which are “shrinking” toward the fixed point x̄1 of f1.

2. Repeated application of the set-valued map f̂2 to the interval I = [0, 1] produces a set of intervals

In which are “shrinking” toward the fixed point x̄2 of f2.

Now, instead of considering the action of each of the set-valued maps f̂1 and f̂2 separately, let’s com-

bine their actions as if they were components of a “Parallel Machine”. We’ll do this as follows:

For any subset S ∈ [0, 1], define that following set-valued mapping,

f̂(S) = f̂1(S) ∪ f̂2(S) . (67)

In other words, the set-valued mapping is the union of the actions of the individual maps f̂1 and f̂2.

Let’s examine the action of f̂ on the set I0 = [0, 1]:

I1 = f̂([0, 1]) = f̂1([0, 1]) ∪ f̂2([0, 1])

=

[

0,1

3

]

∪

[

2

3, 1

]

. (68)

This is shown graphically below:

430

0 1

I0

111

30 2

3

f̂

f̂2(I0)f̂1(I0)

I1

Now apply f̂ to the set I1:

I2 = f̂(I1) = f̂1(I1) ∪ f̂2([I1)

=

([

0,1

9

]

∪

[

2

3,1

3

])

∪

([

2

3,7

9

]

∪

[

8

9, 1

])

=

[

0,1

9

]

∪

[

2

3,1

3

]

∪

[

2

3,7

9

]

∪

[

8

9, 1

]

. (69)

This is shown graphically below:

0 1

I0

1f̂

f̂2(I0)f̂1(I0)

f̂

0 1

3

2

31

I1

I1

I2

0 11

9

2

9

1

3

2

3

7

9

8

9

We see that the repeated application of the “parallel machine” f̂ performs the “middle-

thirds dissection procedure” that was employed in the construction of the ternary Cantor

set C in [0, 1]:

^ ^

I = [0,1]0

I = f ( I )1 0

2 1

C

I = f ( I )

1 0 2 0f ( I ) f ( I )

1 1 2 1f ( I ) f ( I )

1 2f ( C ) f ( C )

^

^

^ ^

^ ^

If we consider the following iteration process involving the parallel map f̂ :

In+1 = f̂(In) = f̂1(In) ∪ f̂2(In) , (70)

then it appears that

limn→∞

In = C , ternary Cantor set in [0,1] . (71)

431

But the story is not over! As shown in the figure above, the ternary Cantor set C is a “fixed

point” of the parallel operator f̂ , i.e.,

C = f̂(C) = f̂1(C) ∪ f̂2(C) . (72)

In other words, f̂ maps the set C to itself.

Recalling that the set-valued maps f̂1 and f̂2 “shrink” sets, we see that:

The ternary Cantor set C is a union of shrunken copies of itself.

We actually noticed this earlier, but we didn’t really specify the maps that produced these copies.

Now, we can identify the maps that produce the copies – the two maps f̂1 and f̂2 that comprise the

“parallel operator” f̂ .

What has been discussed above is a particular example of an iterated function system (IFS). For

the moment, we shall provide a “working definition” of an IFS in Rn. But we must first provide a

couple of other useful definitions.

Distance function/metric in Rn: Let x, y ∈ R

n. We shall let d(x, y) denote the (Euclidean)

distance between x and y. In the special case n = 1,

d(x, y) = |x− y| , x, y ∈ R . (73)

For n ≥ 1, where x = (x1, x2, · · · , xn) and y = (y1, y2, · · · , yn),

d(x, y) = ‖x− y‖2 =

[

∑

k=1

(xk − yk)2

]1/2

, x, y ∈ Rn . (74)

Here, ‖ · ‖2 denotes the Euclidean norm in Rn. (Note: Other distance functions/metrics can be used

but we’ll use the Euclidean distance for simplicity.)

Definition (Contraction Mapping): Let D ⊆ Rn and f : D → D. We say that f is a contraction

mapping on D if there exists a constant 0 ≤ C < 1 such that

d(f(x), f(y)) ≤ Cd(x, y) for all x, y ∈ D . (75)

The smallest C for which the above inequality holds is called the contraction factor of f .

432

In other words, a contraction mapping f maps any two distinct points x and y closer to each other.

Examples: Most of our discussion of IFS will be limited to R and R2, so the following examples

should be sufficient.

1. The maps f1 and f2 examined earlier are contraction mappings on R. Consider f1(x) =1

3x: For

any x, y ∈ R,

d(f1(x), f1(y)) = |f1(x)− f1(y)|

=

∣

∣

∣

∣

1

3x−

1

3y

∣

∣

∣

∣

=1

3|x− y|

=1

3d(x, y) . (76)

In other words, the distance between f1(x) and f2(y) is always one-third the distance between

x and y. The contraction factor for f1(x) =1

3x is C = 1

3.

Now consider f2(x) =1

3x+ 2

3: For any x, y ∈ R,

d(f2(x), f2(y)) = |f2(x)− f2(y)|

=

∣

∣

∣

∣

(

1

3x+

2

3

)

−

(

1

3y +

2

3

)∣

∣

∣

∣

=1

3|x− y|

=1

3d(x, y) . (77)

The contraction factor for f2(x) =1

3x+ 2

3is also C = 1

3.

It will be useful to work with smaller subsets D ⊂ R over which these maps are contractions.

With reference to the Cantor set example studied earlier, we can establish that each of the fi

maps the interval [0, 1] to itself:

(a) For x ∈ [0, 1], f1(x) =1

3x ∈ [0, 1]. Therefore, f1 maps [0,1] to itself.

(b) For x ∈ [0, 1], f2(x) =1

3x+ 2

3∈ [0, 1]. Therefore, f2 maps [0,1] to itself.

We could also use the set-valued versions of these maps to arrive at these results:

433

(a) f̂1 : [0, 1] → [0, 13] ⊂ [0, 1]. Therefore f1 maps [0,1] to itself.

(b) f̂1 : [0, 1] → [23, 1] ⊂ [0, 1]. Therefore f2 maps [0,1] to itself.

Therefore f1 and f2 are contraction maps on the set D = [0, 1].

In general, the following affine map on R,

f(x) = ax+ b , (78)

is a contraction mapping on R if |a| < 1 in which case its contraction factor is C = |a|. For all

x, y ∈ R,

|f(x)− f(y)| = |a| |x− y| , (79)

i.e., the distance between f(x) and f(y) is exactly |a| times the distance between x and y. For

nonlinear maps, such an equality will not exist and the best we can do is to find an inequality

the distances. More on this later.

That being said, in most of the examples and applications examined for the remainder of this

secction, affine maps will be employed.

2. Consider the following class of affine transformations in the plane R2,

f(x) = Ax+ b , (80)

or, in coordinate form,

f(x1, x2) =

a11 a12

a21 a22

x1

x2

+

b1

b2

. (81)

Note that for x,y ∈ R2,

f(x)− f(y) = A(x− y) . (82)

This implies that

d(f(x), f(y)) = ‖f(x)− f(y)‖2

= ‖A(x− y)‖2

≤ ‖A‖2 d(x,y) . (83)

Here, ‖A‖2 denotes the (Euclidean) matrix norm of A, which you may have encountered

in a course on linear algebra. It suffices to state here that a sufficient condition for the affine

mapping f to be contractive is if |λ1| < 1 and |λ2| < 1, where the λi are the eigenvalues of A.

434

Lecture 33

Iterated Function Systems (IFS) and the construction of fractal sets

(cont’d)

We are now in a position to formally define “iterated function systems” (IFS).

Definition (Iterated function system): Let f = {f1, f2, · · · , fN} denote a set of N contraction

mappings on a closed and bounded subset D ⊂ Rn, i.e., for each k ∈ {1, 2, · · ·N}, fk : D → D and

there exists a constant 0 ≤ Ck < such that

d(fk(x), fk(y)) ≤ Ck d(x, y) for all x, y ∈ D . (84)

Associated with this set of contraction mappings is the “parallel set-valued mapping” f̂ , defined as

follows: For any subset S ⊂ D,

f̂(S) =

n⋃

k=1

f̂k(S) , (85)

where the f̂k denote the set-valued mappings associated with the mappings fk. The set of maps f

with parallel operator f̂ define an N -map iterated function system on the set D ⊂ Rn.

We now state the main result associated regarding N -map Iterated Function Systems as defined above.

Theorem: There exists a unique set A ⊂ D which is the “fixed point” of the parallel IFS operator

f̂ , i.e.,

A = f̂(A) =

N⋃

k=1

f̂k(A) . (86)

Moreover, if you start with any set S0 ∈ D (even a single point x0 ∈ D), and form the iteration

sequence,

Sn+1 = f̂(Sn) , (87)

then the sequence of sets {Sn} converges to the fixed-point set A ⊂ D. For this reason, A is known

as the attractor of the IFS.

435

From Eq. (86) is that the set A is self-similar, i.e., A is the union of N geometrically-contracted

copies of itself. We shall be examining this property in a number of examples below.

There are actually two practical consequences of the above result, depending upon one’s perspective:

1. If you have a set of contraction maps f = {fk}Nk=1

, you can quickly construct “pictures” of the

attractor set A. We’ll discuss this in more detail a little later.

2. Given a self-similar, possibly fractal, set S, one may be able to determine the maps {fk} com-

prising the IFS f for which S is the attractor. This will then allow us to construct the “pictures”

of S mentioned in 1. above without having to use “generators.”

Comment: There may be some concern – and rightfully so – about the exact meaning of the state-

ment, “then the sequence of sets {Sn} converges to the fixed point set A ⊂ D”. Recall that whenever

we state that a sequence “converges”, we are implying that elements of the sequence are “getting

closer” to a limit. As such, a distance function, or metric, is needed. In this case, we need a distance

function, or metric, between sets. In this case, an appropriate distance function between (compact)

sets is the so-called Hausdorff metric. It will be defined in a later lecture (purely for supplementary

reasons).

Examples of IFS and their (fractal) attractors

Cantor-like sets on [0,1]

At this point, it is instructive to look at some examples. We have already shown how the Cantor set

C can be viewed as the attractor of a two-map IFS on the space D = [0, 1], namely, the two maps,

f1(x) =1

3x , f2(x) =

1

3x+

2

3. (88)

Recall that both of these maps are contraction maps on [0, 1] with contractivity factors equal to1

3.

This is due to the factor1

3which multiplies x in each map. What would happen if we changed this

factor? For example, consider the two maps,

f1(x) =1

4x , f2(x) =

1

4x+

3

4. (89)

436

Once again, we have f1(0) = 0 and f2(1) = 1. The action of the IFS parallel map f̂ composed of these

two maps on the interval I0 = [0, 1] is as follows:

f̂([0, 1]) = f̂1([0, 1])⋃

f̂2([0, 1])

=

[

0,1

4

]

⋃

[

3

4, 1

]

= I1 . (90)

The action of f̂ on an interval is to remove the middle one-half of the interval:

1

I0

1

4

1

4

I1

f̂1(I0) f̂2(I0)

An application of the IFS parallel map on I1 yields the following result,

f̂(I1) = f̂1(I1)⋃

f̂2(I1)

=

[

0,1

16

]

⋃

[

3

16,1

4

]

⋃

[

3

4,13

16

]

⋃

[

15

16, 1

]

= I2 . (91)

Repeated action of this IFS produces a nested set of sets In which, in the limit n → ∞, converges to

a Cantor-like set – we’ll call it C1/4 – that lies in [0, 1].

This limiting set is clearly not the ternary Cantor set – which would be called C1/3 to be

consistent – but it is a Cantor-like set. In fact, using the methods from the previous on fractal

dimension, it should not be too difficult to see that the fractal dimension of this set is

D =log (no. of copies)

log (1/scaling factor)=

log 2

log 4=

1

2. (92)

We may easily generalize this dissection procedure by considering the following two maps on [0, 1],

f1(x) = rx , f2(x) = r(x− 1) + 1 = rx+ (1− r) , (93)

437

where

0 < r <1

2. (94)

Note that

f1(0) = 0 , f2(1) = 1 . (95)

The action of the associated IFS parallel map f̂ on I0 = [0, 1] is as follows,

f̂([0, 1]) = f̂1([0, 1])⋃

f̂2([0, 1])

= [0, r]⋃

[1− r, 1]

= I1 . (96)

This IFS produces a dissection of [0,1] composed of two intervals of length r, as shown below.

1

I0

r r

I1

f̂1(I0) f̂2(I0)

We saw this iteration procedure earlier in the course in connection with shifted Tent Maps on [0,1].

It produces a set of nested sets In which, in the limit n → ∞, converge to a Cantor-like set Cr in

[0,1]. Once again using the methods of the previous section on fractal dimension, the dimension of

this Cantor set is found to be

D =log (no. of copies)

log 1

r

=log 2

log 1

r

. (97)

Special cases: Note that if we allow the parameter r to be 1

2, then no dissection is performed, i.e.,

f̂([0, 1]) = f̂1([0, 1])⋃

f̂2([0, 1])

=

[

0,1

2

]

⋃

[

1

2, 1

]

= [0, 1] . (98)

438

In other words, we have simply regenerated the interval [0, 1]. As such, the interval [0, 1] is the fixed

point attractor of the IFS operator f̂ . From Eq. (97), the (fractal) dimension of this set is

D =log 2

log 1

1/2

= 1 , (99)

which is consistent with the fact that the attractor is the interval [0, 1].

Let us now examine the case, 1

2< r < 1. (Note that if r = 1, the maps are no longer contractiive.)

Here, the two sets

f̂1([0, 1]) = [0, r] , f̂2([0, 1]) = [1− r, 1] , (100)

intersect. There is no dissection of the interval [0, 1]. The union of these sets produces [0, 1]. Unfor-

tunately, the formula for the fractal dimension is not applicable here since the two sets overlap. The

dimension for the attractor set [0, 1] is 1 for all 1

2≤ r < 1.

The dissection procedure discussed above can be further generalized by allowing the contraction factors

of the two maps to be independent, i.e.,

f1(x) = rx , f2(x) = s(x− 1) + 1 = sx+ (1− s) , (101)

where

0 < r < 1 , 0 < s < 1 and r + s < 1 . (102)

Note that

f1(0) = 0 , f2(1) = 1 . (103)

The action of the associated IFS parallel map f̂ on I0 = [0, 1] is as follows,

f̂([0, 1]) = f̂1([0, 1])⋃

f̂2([0, 1])

= [0, r]⋃

[1− s, 1]

= I1 . (104)

This IFS produces a dissection of [0,1] composed of two intervals, one of length r and the other of

length s as sketched below. (We also saw this iteration procedure in Lecture 22.)

439

1

I0

r s

I1

f̂1(I0) f̂2(I0)

Repeated application of this IFS map will produce a nested set of intervals In which, in the limit

n → ∞, converges to a Cantor-like set Crs ⊂ [0, 1]. From our previous discussion of the fractal

dimension, in particular, the generalized von Koch curves, the fractal dimension Drs of the Cantor-

like set Crs will be the unique solution of the equation,

rD + sD = 1 . (105)

In general, there is no closed form expression for the solution of this equation. In the special case that

0 < r = s < 1

2, however, the above equation becomes

2rD = 1 =⇒ D =log 2

log 1

r

, (106)

which is the result obtained earlier.

We don’t have to limit ourselves to two-map IFS on [0,1] to produce Cantor-like sets – we can

easily have three or more. For example, let us return to the two-map IFS examined earlier,

f1(x) =1

4x , f2(x) =

1

4x+

3

4, (107)

and add a third contractive map, f3 : [0, 1] → [0, 1], to produce the following three-map IFS,

f1(x) =1

4x , f2(x) =

1

4x+

3

4, f3(x) =

1

3x+

1

3, (108)

The map f3(x) is a contractive map with contraction factor C3 = 1

3. You can also check that it has

a fixed point in [0,1], namely, x̄3 = 1

2. The action of the IFS parallel map f̂ composed of these three

440

maps on the interval I0 = [0, 1] is as follows:

f̂([0, 1]) = f̂1([0, 1])⋃

f̂2([0, 1])⋃

f̂3([0, 1])

=

[

0,1

4

]

⋃

[

3

4, 1

]

⋃

[

1

3,2

3

]

= I1 . (109)

Graphically, the result is as follows:

1

I0

1

4

1

3

1

4

I1

f̂1(I0) f̂3(I0) f̂2(I0)

The reader may recognize this as a particular example of the “r-s-t”-type dissection procedure exam-

ined in Problem Set No. 5. Here, r = 1

4, s = 1

3and t = 1

4so that r + s + t = 5

6< 1. Here, we shall

simply state that repeated application of the IFS “parallel operator” f̂ produces a sequence of sets,

In, n ≥ 1, each of which is a union of 3n closed subintervals. (What is the total length of the set In?)

In the limit n → ∞, the sets In converge to a Cantor-like set on [0,1].

441

von Koch curve

In a previous lecture, we showed that the von Koch curve, shown below,

von Koch curve

may be produced by the repeated action of the following generator G,

1

3ll

1

3l

1

3l

1

3l

starting with the set I0 = [0, 1]. We now show how this fractal curve can be generated by an IFS. We

assume, once again, that the leftmost point of the curve is situated at (0,0) and rightmost point at (1,0).

Recall once again that the von Koch curve C is self-similar, i.e., it may be expressed as a union of

four contracted copies of itself. Each contracted copy is one-third the size of C. This self-similarity

could be seen to be built into the set from the generator G that was used to construct it.

The first copy, which starts at the point (0,0), is obtained by shrinking the von Koch curve by a factor

of 1

3toward (0,0) using the following affine map in R

2:

f1(x, y) =

1

30

0 1

3

x

y

. (110)

The second copy, which starts at the point (13, 0) and ends at (1

2,√

3

6), may be obtained by first

shrinking the von Koch curve toward (0,0) with contraction factor 1

3, then rotating it by an angle

442

π3and finally translating it in the x-direction by 1

3. All of these actions are accomplished with the

following affine map,

f2(x, y) =1

3

1

2−

√

3

2√

3

2

1

2

x

y

+

1

3

0

. (111)

The third copy of the von Koch curve, which starts at the point (12,√

3

6), may be obtained by once

again shrinking the von Koch curve toward (0,0) with contraction factor 1

3, then rotating it by angle

−π3and finally translating it to (1

2,√

3

6). all of these actions are accomplished with the following affine

map,

f3(x, y) =1

3

1

2

√

3

2

−

√

3

2

1

2

x

y

+

1

2√

3

6

. (112)

The fourth copy, which ends at (1,0) may be obtained by once again shrinking the von Koch curve

toward (0,0) with contraction factor 1

3and then simply translating it to (2

3, 0). This is accomplished

with the following affine map,

f4(x, y) =

1

30

0 1

3

x

y

+

2

3

0

. (113)

We have achieved our goal: The four contraction maps fi, 1 ≤ i ≤ 4, produce four copies f̂i(C) the

von Koch curve C such that the union of these copies reproduces C, i.e.,

C =

4⋃

i=1

f̂1(C) . (114)

The results are summarized on the next page.

443

von Koch curve

f1(x, y) =

1

30

0 1

3

x

y

+

0

0

f2(x, y) =

1

6−

√

3

6√

3

6

1

6

x

y

+

1

3

0

Contraction factor r = 1

3, rotation π/3, translation.

f3(x, y) =

1

6

√

3

6

−

√

3

6

1

6

x

y

+

1

2√

3

6

Contraction factor r = 1

3, rotation −π/3, translation.

f4(x, y) =

1

30

0 1

3

x

y

+

2

3

0

.

Contraction factor r = 1

3, translation.

444

Sierpinski gasket

The Sierpinski gasket, seen before and shown below, is a union of three copies of itself, so we shall

need three maps in the IFS. Each IFS map fi will be a 1

2contraction with no rotation. The affine

maps are quite easy to determine and are listed below.

Sierpinski gasket

f1(x, y) =

1

20

0 1

2

x

y

+

0

0

Contraction factor r = 1

2, rotation 0.

f2(x, y) =

1

20

0 1

2

x

y

+

1

4√

3

4

Contraction factor r = 1

2, rotation 0, translation.

f3(x, y) =

1

20

0 1

2

x

y

+

1

2

0

Contraction factor r = 1

2, rotation 0, translation.

445

Modified Sierpinski gasket - “Cantor tree”

If we alter the contraction factors of the three affine maps so that the first and third copies touch

the second copy, but do not touch each other, we produce the following fractal set. A horizontal line

intersecting the set will generally intersect it over a Cantor-like set.

Modified Sierpinski gasket - “Cantor tree”

f1(x, y) =

2

50

0 2

5

x

y

+

0

0

f2(x, y) =

3

50

0 3

5

x

y

+

1

5√

3

5

f3(x, y) =

2

50

0 2

5

x

y

+

3

5

0

446

Another modified Sierpinski gasket - “Cantor dust”

If the contraction factors of the three affine maps are adjusted so that there is no intersection of copies,

then the result is a “2D Cantor dust”, a set which is totally disconnected in R2.

Modified Sierpinski gasket - “2D Cantor dust”

f1(x, y) =

2

50

0 2

5

x

y

+

0

0

f2(x, y) =

2

50

0 2

5

x

y

+

3

10

3√

3

10

f3(x, y) =

2

50

0 2

5

x

y

+

3

5

0

447

And yet another modified Sierpinski gasket

The contraction factors of the three maps have now been increased beyond 1

2so that there is overlap-

ping between the copies f̂i(S). This overlapping will occur in a self-similar manner throughout the

set.

Modified Sierpinski gasket with overlap

f1(x, y) =

3

50

0 3

5

x

y

+

0

0

f2(x, y) =

3

50

0 3

5

x

y

+

1

5√

3

5

f3(x, y) =

3

50

0 3

5

x

y

+

2

5

0

448

Modified Sierpinski gasket with rotations

In the following examples, the maps have the general form,

fi(x, y) = r

cos θ − sin θ

sin θ cos θ

x

y

+

b1

b2

Modified Sierpinski gasket with one rotation map

f1(x, y) : r = 0.5 , θ =π

20.

f2(x, y) : r = 0.5 , θ = 0 .

f3(x, y) : r = 0.5 , θ = 0 .

449

Modified Sierpinski gasket with rotations

fi(x, y) = r

cos θ − sin θ

sin θ cos θ

x

y

+

b1

b2

Modified Sierpinski gasket with two rotation maps

f1(x, y) : r = 0.5 , θ =π

20.

f2(x, y) : r = 0.5 , θ = 0 .

f3(x, y) : r = 0.5 , θ =π

20.

450

Modified Sierpinski gasket with rotations

fi(x, y) = r

cos θ − sin θ

sin θ cos θ

x

y

+

b1

b2

Modified Sierpinski gasket with two rotation maps

f1(x, y) : r = 0.5 , θ =π

20.

f2(x, y) : r = 0.5 , θ = 0 .

f3(x, y) : r = 0.5 , θ = −π

20.

451

Sierpinski gasket with an additional map

We now add a fourth map to the Sierpinski gasket IFS. This fourth map is a contraction map with

fixed point at the center of the gasket. The contraction factor is 1

5– small enough to map the entire

triangle into the formerly empty space. Note that these copies are propagated to all “formerly empty”

spots.

Sierpinski gasket with extra map in middle

f1(x, y) =

1

20

0 1

2

x

y

+

0

0

f2(x, y) =

1

20

0 1

2

x

y

+

1

4√

3

4

f3(x, y) =

1

20

0 1

2

x

y

+

1

2

0

f4(x, y) =

1

50

0 1

5

x

y

+

2

5√

3

8

452

“Twin dragon” attractor in R2

This is a well-known set that is the attractor of a two-map IFS in the plane. The fixed points of the

two maps lie on the real line.

f1(x, y) =1√2

1√

2− 1

√

2

1√

2

1√

2

x

y

+

1

0

f2(x, y) =1√2

1√

2− 1

√

2

1√

2

1√

2

x

y

−

1

0

Both affine maps involve (i) a contraction of1√2and (ii) a rotation by

π

4followed by a translation.

The twin-dragon attractor tiles the plane R2. There are no holes in this set. The two copies of this set

f1(A) and f2(A) have been shaded differently so that they can be viewed more easily as contracted

(by 1/√2) and rotated (by π/4) copies of A.

453

Tree-like sets

The following 3-map IFS shows how tree-like objects can be generated. The map f3 generates the

main stem. The other two maps take the stem, translate it upwards and then rotate it in opposite

directions. Of course, these maps operate not only on the stem but on the entire set. For this reason,

the self-similar tree-like attractor set is produced.

Simple tree on [−1, 1] × [0, 2]

f1(x, y) =

0.353 −0.353

0.353 0.353

x

y

+

0.0

0.5

Contraction factor r = 0.5, rotation π/4.

f2(x, y) =

0.353 0.353

−0.353 0.353

x

y

+

0.0

0.5

Contraction factor r = 0.5, rotation −π/4.

f3(x, y) =

0.0 0.0

0.0 0.55

x

y

+

0.0

0.0

“Terminator”: Squash onto y-axis.

454

Tree-like sets (cont’d)

A slight modification of the previous IFS – the maps f1 and f2 no longer have pure rotations, but

shear transformations.

Less simple tree on [−1, 1]× [0, 2]

f1(x, y) =

0.4 −0.433

0.433 0.4

x

y

+

0.0

0.5

Shear transformation

f2(x, y) =

0.4 0.433

−0.433 0.4

x

y

+

0.0

0.5

Shear transformation

f3(x, y) =

0.0 0.0

0.0 0.47

x

y

+

0.0

0.0

“Terminator”: Squash onto y-axis.

455

The epitome of fractal attractors – at least in 1986

This is Prof. Michael Barnsley’s celebrated “spleenwort fern,” the attractor of a four-map IFS.

Barnsley’s Spleenwort Fern

f1(x, y) =

0.0 0.0

0.0 0.16

x

y

+

0.50

0.0

“Terminator”: Squash onto y-axis to make stem.

f2(x, y) =

0.20 −0.26

0.23 0.22

x

y

+

0.40

0.05

f3(x, y) =

−0.15 0.28

0.26 0.24

x

y

+

0.57

−0.12

f4(x, y) =

0.85 0.04

−0.04 0.85

x

y

+

0.08

0.18

456

In the figure below, the action of the IFS maps, denoted as wi, is shown, in order to give the reader an

idea of how the smaller and smaller copies are produced in a geometric cascade as we move upwards

toward the peak of the fern. This cascade is produced by map w4. Map w1, the “terminator”, produces

the stem. Map w2 produces the lower right copy of the fern and map w3 flips the fern and produces

the lower left copy. Some playing around of the map parameters is necessary in order to ensure that

the bottoms of the stems of the two copies produced by w2 and w3 touch the main stem – otherwise,

there would be gaps that propagate throughout the fern. For the same reason, it is important that

the upward translation produced by w4 also touch the main stem.

457

Using IFS to model the real world?

Professor Barnsley’s beautiful discovery gave rise to the question – and the tremendous amount of

research it inspired – of how well IFS could be used to generate sets that approximated real-world

objects. This eventually led to the idea of fractal image coding – a way of representing photos

and videos as attractors of special types of IFS. Early in this research (mid 1980’s and early 1990’s),

fractal image coding was shown to be an effective method of image compression, i.e., reducing the

amount of computer memory required to store a representation of an image, from which it could be

regenerated. (JPEG compression has, and continues to be, a standard method of image compression,

although much more powerful methods of compression exist today. JPEG compression, by the way,

is based on the method of Fourier series representation of functions. It is actually a version of the

discrete cosine transform for discrete – in this case, digital – data sets.)

Methods to generate fractal attractor sets of IFS

We have not yet addressed one important matter regarding IFS and their attractor sets: Once we have

a set of N contraction maps f = {f1, f2, · · · , fn}, how do we produce a picture of the attractor A of

the parallel IFS operator f̂ which they define, i.e., the set which possesses the following self-similarity

property,

A =N⋃

i=1

f̂i(A) ? (115)

In what follows, we describe briefly two principal methods that can be used to construct the fractal

attractor set A of an N -map IFS in Rn.

Random iteration algorithm: In the research literature, this method is known as the “Chaos

Game.” (See the book, Fractals Everywhere, by M.F. Barnsley.) This method, the main ideas of

which were described in a question in Problem Set No. 5, was used to generate the pictures of IFS

attractors presented earlier. We start with a seed point x0 ∈ Rn and perform the following random

iteration procedure:

xn+1 = fσn(xn) , n ≥ 0 . (116)

At each step n, the index σn is chosen randomly and independently from the set of indices {1, 2, · · · , N}.

If the seed point x0 happens to lie in A, then all future xn will also be in A. However, just to be

safe, one should wait for a sufficient number of iterations, say 10 or, better yet, 50, before plotting

458

the points xn. If x0 ∈ A, then, because of the contractivity of the fi maps, future iterates xn will be

attracted to A.

Because the maps fi are chosen at random, the iterates xn will be travelling or (or at least near)

the attractor A in a random way. It is necessary to plot a sufficiently large number of xn so that all

regions of the attractor A are visited.

There still remains the question of what probabilities to employ in the selection of the maps fi.

For many of the “standard” IFS attractors, e.g., von Koch curve, Sierpinski gasket, good results will

be obtained if the maps are chosen with equal probability, i.e., the probability pi of choosing index

i ∈ {1, 2, · · · , N} is pi = 1

N, 1 ≤ i ≤ N . For more complicated sets such as the Spleenwort fern,

however, it is necessary to employ nonuniform probabilities. To see this, note that the “terminator”

map f1 (or w1) for the Spleenwort fern maps the entire attractor A onto the tiny stem situated at the

base of the fern. On the other hand, map f4 (or w4) is responsible for generating about 90% of the

fern in terms of the geometric cascade. As such, it is important that a much higher probability be

assigned to map w4. A rough rule of thumb is that the probability pi of choosing map fi should be in

some way related to the ratio of the area of the copy fi(A) to the area of A.

The idea of probabilities pi being associated with the IFS maps leads to another important con-

cept, that of invariant measures that “live” on their attractors A. Unfortunately, there is no time

to discuss this topic in this course.

On the next page is presented a very simple MATLAB program, “ifs2d.m”, to plot the attractor

of an N -map IFS in R2. The program reads the IFS parameters, i.e., the elements of the matrix A

and vector b defining each affine IFS map, from a file – in this case, the file is called “ifs2d.dat”. A

sample data file is given below the program. Its three lines contain the components of the three maps

which define the IFS with Sierpinski gasket as attractor, given earlier. The output generated from

this data is shown on the next page.

The files ifs2d.m and ifs2d.dat have been posted at the course site on LEARN.

459

% ifs2d.m

% routine to produce plots of attractors sets of 2D

% iterated function systems with probabilities

% using the random iteration algorithm, or "Chaos Game"

% nmaps IFS maps w(i) = A(i)*x + b(i), i=1..nmaps, with probabilities p(i)

% the attractor A of an IFS is presented as a scatter plot

% in the region [xmin,xmax] X [ymin,ymax]

% (the user may have to adjust these values accordingly)

nmaps=3;

niter=20000;

xmin=0.0;

xmax=1.0;

ymin=0.0;

ymax=1.0;

kndx=zeros(1,1,’uint8’);

% these vectors will store the x and y coordinates of the points

% generated on the attractor

xx=zeros(niter,1);

yy=zeros(niter,1);

% these vectors store the IFS parameters

a11=zeros(nmaps,1);

a12=zeros(nmaps,1);

a21=zeros(nmaps,1);

a22=zeros(nmaps,1);

b1=zeros(nmaps,1);

b2=zeros(nmaps,1);

p=zeros(nmaps,1);

% input IFS parameters from a data file

[a11,a12,a21,a22,b1,b2,p] = textread(’ifs2d.dat’,’%f %f %f %f %f %f %f’,nmaps);

% start with (0,0) and iterate first IFS map until points get close to its

% fixed point (which is a point on the attractor)

x=0.0;

y=0.0;

for i=1:50

x1=a11(1)*x + a12(1)*y + b1(1);

y1=a21(1)*x + a22(1)*y + b2(1);

460

x=x1;

y=y1;

end

% now continue random iteration of IFS over points on attractor

% here, the IFS maps are chosen with equal probabilities

% this program can be modified so that the ith IFS map is chosen

% with probability p(i)

for i=1:niter

xran=rand;

kndx=round(nmaps*xran+0.5);

kk(i)=kndx;

x1=a11(kndx)*x + a12(kndx)*y + b1(kndx);

y1=a21(kndx)*x + a22(kndx)*y + b2(kndx);

x=x1;

y=y1;

xx(i)=x;

yy(i)=y;

end

% plot the attractor

figure(1)

scatter(xx,yy,’.’),xlim([xmin xmax]),ylim([ymin ymax]);

SAMPLE FILE "ifs2d.dat" (Sierpinski gasket)

0.5 0.0 0.0 0.5 0.0 0.0 0.3333

0.5 0.0 0.0 0.5 0.25 0.433 0.3333

0.5 0.0 0.0 0.5 0.5 0.0 0.3333

461

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Output from MATLAB program ifs2d.m using data from ifs2d.dat from previous page: TheSierpinski gasket attractor.

462

Deterministic algorithm: Recall how the IFS “parallel operator” f̂ associated with a set of N

contraction maps fi acts on sets to produce sets: For a set S ∈ D (where D ⊂ Rk is our region of

interest),

f̂(S) =

N⋃

i=1

f̂i(S) . (117)

In other words, f̂ produces N contracted copies of S. Also recall that for any S ∈ D, the iterates of

S under the action of f̂ converge to the attractor/fixed point A of the IFS:

limn→∞

f̂n(S) = A = f̂(A) =N⋃

i=1

f̂i(A) . (118)

We can start the deterministic algorithm with a seed point x0 ∈ Rn, which defines our set S0. The

action of the IFS parallel operator on S0 is to produce a set S1 consisting of N (or possibly less)

points, i.e.,

S1 = f̂(S0) = f̂(x0) =

N⋃

i=1

fi(x0) . (119)

Applying the IFS parallel operator f̂ on S1 will produce a set S2 consisting of N2 (or possibly less)

points, i.e.,

S2 = f̂(S1) =

N⋃

j=1

fj

(

N⋃

i=1

fi(x0)

)

=N⋃

j=1

N⋃

i=1

(fj ◦ fi)(x0) . (120)

The reader should see the pattern. N applications of the IFS parallel operator f̂ will produce a set

Sn consisting of (at most) Nn points,

Sn = {(fi1 ◦ fi2 ◦ · · · ◦ fin)(x0) , i1, i2, · · · , in ∈ {1, 2, · · · , N} } . (121)

If the point x0 ∈ A, the attractor of the IFS, then all points in S1, S2, etc., will lie in A. So if

you happen to know a point in A, it’s good to use it. But even if x0 is not in A, the contractivity

of the fi will bring the points in Sn closer and closer to A as n increases. What is more important,

however, is to generate a sufficient number of points to ensure that most of the set A is visited by them.

The above algorithm corresponds to taking all possible Nn paths from x0 down an N -tree to the

Nn points which comprise the set Sn. This can be done by means of a recursive calling of a subroutine,

provided that the computer language you are using supports recursion.

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