Set Datang di SMAN 8 Batam Fractal geometry
Fractal Geometry
Isa S. Jubran – jubrani@cortland.edu – http://web.cortland.edu/jubrani
A Fractal is a geometric object whose dimension is fractional. Most fractals are self
similar, that is when any small part of a fractal is magnified the result resembles
the original fractal. Examples of fractals are the Koch curve, the Seirpinski gasket,
and the Menger sponge.
Similarity Dimension:
• Take a line segment of length one unit and divide it into N equal parts. let
old length
S = new
, then N = S 1 .
length
N = 3, S = 3
unit and divide it into N
• Take a square (dimension 2)
√ of area one square
2
congruent parts. Then S = N and N = S .
N = 4, S = 2
• Take a cube (dimension 3) of volume one cubic unit and divide it into N
1
congruent parts. Then S = N 3 and N = S 3 .
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....
N = 8, S = 2
So given set of dimension d consisting of N parts congruent part then N = S d ,
old length
. Solving for d yields
where S = new
length
d=
ln N
ln S
1
Exercise 1: T∞ = limn→∞ Tn is called the Seirpinski Triangle or Gasket.
1. What is the perimeter of Tn ?
2. What is the area of Tn ?
3. Compute the fractal dimension of T∞ .
T1
T0
T2
Solution:
1. Pn = ( 32 )n P0 .
2. An = ( 34 )n A0 .
3. N = 3 and S = 2, hence d =
ln3
.
ln2
Exercise 2: C∞ = limn→∞ Cn is called the Seirpinski Carpet.
C0
C1
C2
1. What is the perimeter of Cn ?
2. What is the area of Cn ?
3. Compute the fractal dimension of C∞ .
Solution:
1. Pn = P0 + 13 P0 +
8
P
32 0
+ ··· +
8n−1
P0 .
3n
2. An = ( 89 )n A0 .
3. N = 8 and S = 3, hence d =
ln8
.
ln3
2
Exercise 3: M∞ = limn→∞ Mn is called the Menger Sponge.
2
0
M0
2
M1
M2
1. What is the volume of Mn .
2. What is the surface area of Mn ?
3. Compute the fractal dimension of M∞ .
Solution:
1. V0 = 23 = 8 and Vn = 20 · (1/3)3 · (Vn−1 ).
2. A0 = 6 · 22 = 24 and An = 20 · (1/3)2 · (An−1 ) − 24 · (1/3)2 · (At n−1 ) · 2
= (20/9) · Area(Cn−1 ) − (64/3) · (8/9)n−1 where Area(Cn−1 ) is the area of
the n − 1 stage in the construction of the Sierpinski carpet. Hence An =
(8/9)n · 16 + (20/9)n · 8.
3. N = 20 and S = 3, hence d =
ln20
.
ln3
Exercise 4: Draw a fractal with dimension 23 .
Solution:
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Demonstrate the applet ”Fractal Dimension” at:
http://www.shodor.org/interactivate/activities/fracdim/index.html
3
Box Dimension:
Box dimension is another way to measure fractal dimension, it is defined as follows:
If X is a bounded subset of the Euclidean space, and ǫ ≥ 0. Let N (X, ǫ) be the
minimal number of boxes in the grid of side length ǫ which are required to cover
X. We say that X has box dimension D if the following limit exists and has value
D.
lim+
ǫ→0
ln(N (X, ǫ))
ln( 1ǫ )
Consider for example the Seirpinski gasket:
✻
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1 .................................................
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. . .
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0
1
ǫ
1
1
2
1
4
..
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1
2n
N (X, ǫ)
1
3
9
..
.
3n
ln(4 · 3n )
ln 3
Hence D = lim
=
n→∞ ln(2n )
ln 2
Demonstrate the applet ”Box Counting Dimension” at:
http://classes.yale.edu/fractals/Software/boxdim.html
4
Geometrically, if you plot the results on a graph with ln N (X, ǫ) on the vertical
axis and ln(1/ǫ) on the horizontal axis, then the slope of the best fit line of the
data will be an approximation of Box dimension of the fractal.
✻
✛ (ln(4), ln(9))
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...........ln(3))
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(ln(1), ln(1))
Discuss computing the fractal dimension of a bunch of broccoli.
http://hypertextbook.com/facts/2002/broccoli.shtml
My Results:
1
ǫ
Ball Size
N
ln( 1ǫ )
ln(N )
6x5x2.5
4x4x2.5
3.5x3x2
3x2.5x2
2.5x2.5x2
10 1/4.5 -1.504
19 1/3.5 -1.253
28 1/2.83 -1.04
41 1/2.5 -.916
64 1/2.33 -.847
2.303
2.94
3.33
3.71
4.158
Use the linear regression applet at http://math.hws.edu/javamath/ryan/Regression.html
to plot the data points and find the slope of the best fit line.
Fractal dimension = slope = 2.62.
If you throw the 1st point out you get fractal dimension=slope =2.82.
5
Divider Dimension:
A fractal curve has fractal (or divider) dimension D if its length L when measured
with rods of length l is given by
L = C · l1−D
C is a constant that is a certain measure of the apparent length, and the equation
above must be true for several different values of l. Consider the Koch curve:
K2
l
1
L
1
1
3
1
9
4
3
16
9
..
.
1
3n
..
.
4 n
(3)
It appears to have length L = ( 43 )n when measured with rods of length l = ( 31 )n .
1
4
Hence, ( )n = C · ( )n(1−D)
3
3
4
1
This implies that n ln 3 = ln(C) + n(1 − D) ln 3 .
In order for the equation to be satisfied for all n, ln C must be zero, hence C = 1
ln 4
and so D = ln
.
3
Geometrically, if you graph ln(L) versus ln(l) the D = 1 − Slope of line.
Exercise: The west coast of Britain when measured with rods of length 10 km is
3020 km. But when measured with rods of length 100 km it is 1700 km. What is
the fractal dimension of the west coast of Britain?
Demonstrate the applet ”Coastline of Hong Kong” at:
http://www.eie.polyu.edu.hk/∼cktse/movies/pheno/map.html
6
Affine transformations:
An affine transformation is of the form
Ã
f(
x
y
!
)=
Ã
a b
c d
! Ã
·
!
+
Ã
e
f
! Ã
x
y
!
x
y
!
or
Ã
f(
!
x
y
)=
Ã
r cos(θ) −s sin(φ)
r sin(θ)
s cos(φ)
·
+
Ã
e
f
!
Where r and s are the scaling factors in the x and y directions respectively. θ and φ
measure rotation of horizontal and vertical lines resp. e and f measure horizontal
and vertical translations resp.
It can be easily shown that r2 = a2 + c2 , s2 = b2 + d2 , θ = arctan(c/a) and
φ = arctan(−b/d).
✻
b ✻
c
✛
θ
a
✲
✛
❄
d
φ
✲
❄
✻
B..
C........................
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Similarity
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B..
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C ′.........................
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D
A
A′
1
′
x = 2x
y ′ = 12 y
✻
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C..........................................B
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Af f inity
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B.. ′
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C ′................................
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D
A′ A
x′ = 43 x
y ′ = 21 y
A reflection about the y-axis would be given by
′
x = −x
′
and
y =y
−−−−−−−
matrix f orm
Ã
−1 0
0 1
!
matrix f orm
Ã
1
0
0 −1
!
A reflection about the x-axis would be given by
′
x =x
and
′
y = −y
−−−−−−−
7
Rotation is represented as follows:
✻
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B
C ...
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B..′
C ′..............................
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θ=0
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φ = 45
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D
A
′
B
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θ = 45
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φ=0
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D
A
x′ = √12 x
y ′ = √12 x + y
x′ = x − √12 y
y ′ = √12 y
M atrixf orm :
Ã
cos(θ) − sin(φ)
sin(θ)
cos(φ)
!
=
✻
...′
.
B
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B
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C
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C ′.....
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...A′
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θ = 45
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φ
= 45
... .. ...
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A
D
x′ = √12 x − √12 y
y ′ = √12 x + √12 y
8
Ã
a b
c d
!
Shears are represented as follows:
B′
✻
.......................A
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D
A
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..C ′
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C ..
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D
A
x′ = x + 2y
y′ = y
x′ = x
y = 2x + y
′
Shear in the y direction
Shear in the x direction
9
B′
Iterated Function Systems:
An Iterated Function System in R2 is a collection {F1 , F2 , . . . , Fn } of contraction
mappings on R2 . (Fi is a contraction mapping if given x, y then d(Fi (x), Fi (y)) ≤
sd(x, y) where 0 ≤ s < 1)
The Collage Theorem says that there is a unique nonempty compact subset A ⊂ R2
called the attractor such that
A = F1 (A) ∪ F2 (A) ∪ . . . ∪ Fn (A)
Sketch of Proof:
Given a complete metric space X, let H(X) be the collection of nonempty compact
subsets of X. Given A and B in H(X) define d(A, B) to be the smallest number
r such that each point of in A is within r of some point in B and vice versa. If X
is complete so is H(X).
Given an IFS on X, define G : H(X) → H(X) by G(K) = F1 (K) ∪ . . . ∪ Fn (K). If
each Fi is a contraction then so is G. The contraction mapping theorem says that
a contraction mapping on a complete space has a unique fixed point. Hence there
is A ∈ H(X) such that G(A) = A = F1 (A) ∪ F2 (A) ∪ . . . ∪ Fn (A).
Given K0 ⊂ R2 , let Kn = Gn (K0 ). Then
d(A, Ki ) ≤
si d(K0 , G(K0 ))
1−s
Hence Ki is a very good approximation of A for large enough i.
Deterministic Algorithm: (Slow!)
Given K0 , let Kn = Gn (K0 ). Now Kn → A.
Random Algorithm: (Fast!)
Choose P0 ∈ K0 and let Pi+1 = Fj (Pi ) where Fj is chosen at random from all of
the Fi . Clearly Pi ∈ Ki and for sufficiently large i,Pi is arbitrarily close to some
point in A.
10
Finding IFS Rules from Images of Points
Given three non-collinear initial points p1 = (x1 , y1 ), p2 = (x2 , y2 ), p3 = (x3 , y3 )
and three image points q1 = (u1 , v1 ), q2 =
, v2 ), qÃ3 = (u3!, v3Ã) respectively.
à (u2!
! Ã
! Find
x
a b
x
e
an affine transformation T defined by T (
)=
·
+
such
y
c d
y
f
that T (p1 ) = q1 , T (p2 ) = q2 , and T (P3 ) = q3 .
This gives the following six equations in six unknowns:
ax1 + by1 + e = u1
cx1 + dy1 + f = v1
ax2 + by2 + e = u2
cx2 + dy2 + f = v2
ax3 + by3 + e = u3
cx3 + dy3 + f = v3
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.. 0, .5
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.5,
0,
0,
.5,
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.5,
0,
0,
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.5, 0, 0, .5, 0, 0
The IFS for the Seirpinski triangle is {T1 , T2 , T2 } where
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
·
x
y
!
T2 (
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
·
x
y
!
T3 (
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
x
y
!
T1 (
11
·
+
Ã
0
0
!
+
Ã
.5
0
!
+
Ã
0
.5
!
Example:
✻
1 .....................................................................
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V
I
V
II
V
III
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2 ......................
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IV
V
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1 ......................
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3 ...
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I
II
III
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0
1
2
0
1
3
3
The IFS for the Seirpinski carpet is {T1 , T2 , T3 , T4 , T5 , T6 , T7 , T8 } where
Ã
T1 (
T2 (
x
y
Ã
x
y
!
Ã
x
y
Ã
!
)=
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
..333
0
0 .333
! Ã
·
x
y
!
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
x
y
!
T3 (
T4 (
T5 (
T6 (
T7 (
T8 (
)=
)=
)=
)=
)=
)=
)=
12
·
+
Ã
0
0
!
+
Ã
.333
0
!
+
Ã
.666
0
!
+
Ã
0
.333
!
+
Ã
.666
.333
!
+
Ã
0
.666
!
+
Ã
.333
.666
!
+
Ã
.666
.666
!
Example:
✻
20)
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.....(15,
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......(5,
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.20)
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.5, −.5, 5, .5,
.5,
.5,
5,
...
... .5, 5
.
... −.5, .5, 15
.
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.15)
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. 15)
(5,
(15,
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(0, 5) ..
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(10,.... 10)
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...
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... −10 < x < 30
..
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0 < y < 40
...
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.. T runk 0, 0,... 10, 0, .333, 0 ...
..
...
...
...
...
...
..
..
..
..
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..
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..
..
.
.................................................................................. ✲
(10, 0)
The IFS for the tree is {T1 , T2 , T2 } where
Ã
x
y
!
Ã
x
y
Ã
x
y
T1 (
T2 (
T3 (
Ã
.5 .5
−.5 .5
! Ã
·
x
y
!
!
)=
Ã
.5 −.5
.5
.5
! Ã
x
y
!
!
)=
Ã
0
0
0 .333
)=
·
! Ã
·
x
y
!
+
Ã
5
15
!
Ã
5
5
!
10
0
!
+
+
Ã
Download and experiment with the freeware ”IFS Construction Kit” which can be
used to design and draw fractals based on iterated function systems.
You can find it at:
http://ecademy.agnesscott.edu/∼lriddle/ifskit/
• Discuss the IFS for the Fern.
• Discuss the IFS for the Tree.
• Discuss the IFS for the mountain range.
• Discuss the IFS for the Wave.
13
Mandelbrot and Julia Sets:
Consider the complex function Fc (z) = z 2 + c. Given an initial point z0 define its
orbit under Fc to be z0 , z1 , z2 , . . . where zn+1 = zn2 + c.
Define the escape set, Ec , and the prisoner set, Pc , for for the parameter c to be:
Ec = {z0 : |zn | → ∞ as n → ∞},
and
Ec = {z0 |z0 ∈
/ Ec }
Now the Julia set for the parameter c, Jc , is the boundary of the escape set Ec .
For example, if c = 0 then Fc (z) = z 2 . Now if z = reiθ then F (z) = r2 e2iθ .
z
z2
z4
z8
z16
z 32
|z0 |
θ
.8
15
.64
30
.4096 60
.1678 120
.0281 240
.0008 120
|z0 | θ
1
15
1
30
1
60
1 120
1 240
1 120
|z0 |
θ
1.5
15
2.25
30
5.0625
60
25.629
120
656.84
240
431439.88 120
✻
✻
z3
z2
✛
z3
z4
z2
z1
z0
✲
z1
z0
✲
✛
z4
❄
|z0 | < 1
❄
|z0 | = 1
Hence Ec = {z : |z| > 1}, Pc = {z : |z| ≤ 1}, and therefore J0 is the unit circle.
14
The Julia set is either connected (one piece) or totally disconnected(dust).
Now we define the Mandelbrot set to be
M = {c ∈ C : Jc is connected}
Experiment
with
the
Java
applet
”Mandelbrot
and
Julia
Sets”
at:
http://classes.yale.edu/fractals/
Download ”winfeed” at the address below and use it to plot the Julia sets for the
constants c = −.5 + .5i, c = −.7454285 + 0.1130089i, and others.
http://math.exeter.edu/rparris/winfeed.html
15
Isa S. Jubran – jubrani@cortland.edu – http://web.cortland.edu/jubrani
A Fractal is a geometric object whose dimension is fractional. Most fractals are self
similar, that is when any small part of a fractal is magnified the result resembles
the original fractal. Examples of fractals are the Koch curve, the Seirpinski gasket,
and the Menger sponge.
Similarity Dimension:
• Take a line segment of length one unit and divide it into N equal parts. let
old length
S = new
, then N = S 1 .
length
N = 3, S = 3
unit and divide it into N
• Take a square (dimension 2)
√ of area one square
2
congruent parts. Then S = N and N = S .
N = 4, S = 2
• Take a cube (dimension 3) of volume one cubic unit and divide it into N
1
congruent parts. Then S = N 3 and N = S 3 .
..
......................
..
.... ..
..... ... ..
..
.
.
.
.
.
.
..
.
.
.
.. ..
... ...
..
..
... ......................................
..............................
.... .. ..... ..
..
.... ... .... ... ...
..
.. ........................
.
....... ...........
.. ......... .........
.... .......
.
....
N = 8, S = 2
So given set of dimension d consisting of N parts congruent part then N = S d ,
old length
. Solving for d yields
where S = new
length
d=
ln N
ln S
1
Exercise 1: T∞ = limn→∞ Tn is called the Seirpinski Triangle or Gasket.
1. What is the perimeter of Tn ?
2. What is the area of Tn ?
3. Compute the fractal dimension of T∞ .
T1
T0
T2
Solution:
1. Pn = ( 32 )n P0 .
2. An = ( 34 )n A0 .
3. N = 3 and S = 2, hence d =
ln3
.
ln2
Exercise 2: C∞ = limn→∞ Cn is called the Seirpinski Carpet.
C0
C1
C2
1. What is the perimeter of Cn ?
2. What is the area of Cn ?
3. Compute the fractal dimension of C∞ .
Solution:
1. Pn = P0 + 13 P0 +
8
P
32 0
+ ··· +
8n−1
P0 .
3n
2. An = ( 89 )n A0 .
3. N = 8 and S = 3, hence d =
ln8
.
ln3
2
Exercise 3: M∞ = limn→∞ Mn is called the Menger Sponge.
2
0
M0
2
M1
M2
1. What is the volume of Mn .
2. What is the surface area of Mn ?
3. Compute the fractal dimension of M∞ .
Solution:
1. V0 = 23 = 8 and Vn = 20 · (1/3)3 · (Vn−1 ).
2. A0 = 6 · 22 = 24 and An = 20 · (1/3)2 · (An−1 ) − 24 · (1/3)2 · (At n−1 ) · 2
= (20/9) · Area(Cn−1 ) − (64/3) · (8/9)n−1 where Area(Cn−1 ) is the area of
the n − 1 stage in the construction of the Sierpinski carpet. Hence An =
(8/9)n · 16 + (20/9)n · 8.
3. N = 20 and S = 3, hence d =
ln20
.
ln3
Exercise 4: Draw a fractal with dimension 23 .
Solution:
✻
..........
....................
..
.........
..
..
..
..
..
...
...
...
..
..
..
..
..
.
.
.
.
..✲
✛...........
.
.
.........
.........
.................
.
.
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..
...
...
...
..
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..
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..
..
..
..................................
❄
Demonstrate the applet ”Fractal Dimension” at:
http://www.shodor.org/interactivate/activities/fracdim/index.html
3
Box Dimension:
Box dimension is another way to measure fractal dimension, it is defined as follows:
If X is a bounded subset of the Euclidean space, and ǫ ≥ 0. Let N (X, ǫ) be the
minimal number of boxes in the grid of side length ǫ which are required to cover
X. We say that X has box dimension D if the following limit exists and has value
D.
lim+
ǫ→0
ln(N (X, ǫ))
ln( 1ǫ )
Consider for example the Seirpinski gasket:
✻
.. ..
..
.. .. ..
1 .................................................
... ...
...
... ... ...
.. ..
.. .. ..
..
.................................................
.. ..
.. .. ..
..
... ...
...
... ... ...
... ...
...
... ... ...
.. ..
..
.. .. ..
.. ..
.
.
.
.
.................................................
..
.. .. ..
... ...
.. .. ..
..
.. ..
.. .. ..
..
.. ..
.................................................
..
.. .. ..
... ...
...
... ... ...
.. ..
.. .. ..
..
.................................................
.. ..
.. .. ..
..
... ...
...
... ... ...
... ...
...
... ... ...
.. ..
..
.. .. ..
.. ..
.
.
.
.
.................................................
..
.. .. ..
... ...
..
.. .. ..
.. ..
.. .. ..
..
. .✲
. . .
.
0
1
ǫ
1
1
2
1
4
..
.
1
2n
N (X, ǫ)
1
3
9
..
.
3n
ln(4 · 3n )
ln 3
Hence D = lim
=
n→∞ ln(2n )
ln 2
Demonstrate the applet ”Box Counting Dimension” at:
http://classes.yale.edu/fractals/Software/boxdim.html
4
Geometrically, if you plot the results on a graph with ln N (X, ǫ) on the vertical
axis and ln(1/ǫ) on the horizontal axis, then the slope of the best fit line of the
data will be an approximation of Box dimension of the fractal.
✻
✛ (ln(4), ln(9))
...................... ......................
..
..
..
..
...
...
...
...
..
..
..
..
.
...
.
✛..
....................... .....(ln(2),
...........ln(3))
.......
..
..
..
..
..
..
..
..
..
..
..
..
..
.
...
.
..
..
...
..
..
..
..
..
..
..
..
✲
.
......................
(ln(1), ln(1))
Discuss computing the fractal dimension of a bunch of broccoli.
http://hypertextbook.com/facts/2002/broccoli.shtml
My Results:
1
ǫ
Ball Size
N
ln( 1ǫ )
ln(N )
6x5x2.5
4x4x2.5
3.5x3x2
3x2.5x2
2.5x2.5x2
10 1/4.5 -1.504
19 1/3.5 -1.253
28 1/2.83 -1.04
41 1/2.5 -.916
64 1/2.33 -.847
2.303
2.94
3.33
3.71
4.158
Use the linear regression applet at http://math.hws.edu/javamath/ryan/Regression.html
to plot the data points and find the slope of the best fit line.
Fractal dimension = slope = 2.62.
If you throw the 1st point out you get fractal dimension=slope =2.82.
5
Divider Dimension:
A fractal curve has fractal (or divider) dimension D if its length L when measured
with rods of length l is given by
L = C · l1−D
C is a constant that is a certain measure of the apparent length, and the equation
above must be true for several different values of l. Consider the Koch curve:
K2
l
1
L
1
1
3
1
9
4
3
16
9
..
.
1
3n
..
.
4 n
(3)
It appears to have length L = ( 43 )n when measured with rods of length l = ( 31 )n .
1
4
Hence, ( )n = C · ( )n(1−D)
3
3
4
1
This implies that n ln 3 = ln(C) + n(1 − D) ln 3 .
In order for the equation to be satisfied for all n, ln C must be zero, hence C = 1
ln 4
and so D = ln
.
3
Geometrically, if you graph ln(L) versus ln(l) the D = 1 − Slope of line.
Exercise: The west coast of Britain when measured with rods of length 10 km is
3020 km. But when measured with rods of length 100 km it is 1700 km. What is
the fractal dimension of the west coast of Britain?
Demonstrate the applet ”Coastline of Hong Kong” at:
http://www.eie.polyu.edu.hk/∼cktse/movies/pheno/map.html
6
Affine transformations:
An affine transformation is of the form
Ã
f(
x
y
!
)=
Ã
a b
c d
! Ã
·
!
+
Ã
e
f
! Ã
x
y
!
x
y
!
or
Ã
f(
!
x
y
)=
Ã
r cos(θ) −s sin(φ)
r sin(θ)
s cos(φ)
·
+
Ã
e
f
!
Where r and s are the scaling factors in the x and y directions respectively. θ and φ
measure rotation of horizontal and vertical lines resp. e and f measure horizontal
and vertical translations resp.
It can be easily shown that r2 = a2 + c2 , s2 = b2 + d2 , θ = arctan(c/a) and
φ = arctan(−b/d).
✻
b ✻
c
✛
θ
a
✲
✛
❄
d
φ
✲
❄
✻
B..
C........................
...
...
....
...
...
...
Similarity
....
...
...
..
′
.
B..
.....
C ′.........................
.
...
....
...
...
....
...
...
...
....
...
...
...
...
...
................................................ ✲
D
A
A′
1
′
x = 2x
y ′ = 12 y
✻
..
C..........................................B
...
...
...
Af f inity
...
...
..
B.. ′
...
C ′................................
...
...
....
...
...
...
....
....
....
...
...
...
...
...
.... ✲
............................................
...
D
A′ A
x′ = 43 x
y ′ = 21 y
A reflection about the y-axis would be given by
′
x = −x
′
and
y =y
−−−−−−−
matrix f orm
Ã
−1 0
0 1
!
matrix f orm
Ã
1
0
0 −1
!
A reflection about the x-axis would be given by
′
x =x
and
′
y = −y
−−−−−−−
7
Rotation is represented as follows:
✻
...
..
..
..
..
.
B
C ...
..
..
..
B..′
C ′..............................
.
...
...
.
..
...
...
..
...
...
..
...
...
..
...
...
θ=0
...
...
...
... ..
...
φ = 45
... ..
...
.....
...
✲
..........................................
.
D
A
′
B
....
.
.
... ..
✻
... ...
.
.
..
...
...
..
..
...
..
.
..
.
.. ...
..
.. ...
..
.....
..
B
.
..
C ..
..
..
..
..
..
...A′
..
.
.
..
...
..
...
.
..
.
..
..
..
θ = 45
...
.. .....
φ=0
.. ...
.. ...
..................... ✲
.......................
D
A
x′ = √12 x
y ′ = √12 x + y
x′ = x − √12 y
y ′ = √12 y
M atrixf orm :
Ã
cos(θ) − sin(φ)
sin(θ)
cos(φ)
!
=
✻
...′
.
B
.
... .....
...
...
.
.
...
..
.
.
...
B
..
.
...
.
C
.
.
...
.
.
...
...
...
C ′.....
...
...
...A′
.
.
..
...
.
.
..
...
...
..
...
...
.
..
...
.
..
θ = 45
...
...
... ... .....
φ
= 45
... .. ...
........
.......................................... ✲
A
D
x′ = √12 x − √12 y
y ′ = √12 x + √12 y
8
Ã
a b
c d
!
Shears are represented as follows:
B′
✻
.......................A
...′..................
..
..
...
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
..
...
..
..B
..
.
....................
....................
.
.
..
..
C ..
..
..
..
..
..
..
..
..
..
...
...
...
..
..
..
..
..
..
..
..
...
..
..
............................................ ✲
D
A
✻
............................................
..
..
...
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
..
...
..
..B
..C ′
.
..
....................
.
....................
..
C ..
..
...
..
..
..
..
..
..
..
...
...
...
..
..
..
..
..
..
..
..
...
..
..
.
.....................
....................... ✲
D
A
x′ = x + 2y
y′ = y
x′ = x
y = 2x + y
′
Shear in the y direction
Shear in the x direction
9
B′
Iterated Function Systems:
An Iterated Function System in R2 is a collection {F1 , F2 , . . . , Fn } of contraction
mappings on R2 . (Fi is a contraction mapping if given x, y then d(Fi (x), Fi (y)) ≤
sd(x, y) where 0 ≤ s < 1)
The Collage Theorem says that there is a unique nonempty compact subset A ⊂ R2
called the attractor such that
A = F1 (A) ∪ F2 (A) ∪ . . . ∪ Fn (A)
Sketch of Proof:
Given a complete metric space X, let H(X) be the collection of nonempty compact
subsets of X. Given A and B in H(X) define d(A, B) to be the smallest number
r such that each point of in A is within r of some point in B and vice versa. If X
is complete so is H(X).
Given an IFS on X, define G : H(X) → H(X) by G(K) = F1 (K) ∪ . . . ∪ Fn (K). If
each Fi is a contraction then so is G. The contraction mapping theorem says that
a contraction mapping on a complete space has a unique fixed point. Hence there
is A ∈ H(X) such that G(A) = A = F1 (A) ∪ F2 (A) ∪ . . . ∪ Fn (A).
Given K0 ⊂ R2 , let Kn = Gn (K0 ). Then
d(A, Ki ) ≤
si d(K0 , G(K0 ))
1−s
Hence Ki is a very good approximation of A for large enough i.
Deterministic Algorithm: (Slow!)
Given K0 , let Kn = Gn (K0 ). Now Kn → A.
Random Algorithm: (Fast!)
Choose P0 ∈ K0 and let Pi+1 = Fj (Pi ) where Fj is chosen at random from all of
the Fi . Clearly Pi ∈ Ki and for sufficiently large i,Pi is arbitrarily close to some
point in A.
10
Finding IFS Rules from Images of Points
Given three non-collinear initial points p1 = (x1 , y1 ), p2 = (x2 , y2 ), p3 = (x3 , y3 )
and three image points q1 = (u1 , v1 ), q2 =
, v2 ), qÃ3 = (u3!, v3Ã) respectively.
à (u2!
! Ã
! Find
x
a b
x
e
an affine transformation T defined by T (
)=
·
+
such
y
c d
y
f
that T (p1 ) = q1 , T (p2 ) = q2 , and T (P3 ) = q3 .
This gives the following six equations in six unknowns:
ax1 + by1 + e = u1
cx1 + dy1 + f = v1
ax2 + by2 + e = u2
cx2 + dy2 + f = v2
ax3 + by3 + e = u3
cx3 + dy3 + f = v3
.....................
....................................................................
...
..
..
..
..
..
..
..
..
..
...
.
.
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
..
..
..
..
.. 0, .5
..
.5,
0,
0,
.5,
.......................
.
.
.
..
.....................
....................
.....................
.
.
....................
.
....................
.
....................
..
..
..
..
..
..
..
..
..
..
..
..
..
...
...
..
..
..
..
...
..
..
...
..
..
..
..
..
..
..
..
..
..
..
.
..
..
.
...
..
..
..
..
.......................................................................................
.........................
.....................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.5,
0,
0,
.
.
.
..........................5, .5, 0
.....................
..........................................................................................
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...
..
..
..
..
..
..
..
..
..
...........................................................................................
.5, 0, 0, .5, 0, 0
The IFS for the Seirpinski triangle is {T1 , T2 , T2 } where
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
·
x
y
!
T2 (
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
·
x
y
!
T3 (
Ã
x
y
!
)=
Ã
.5 0
0 .5
! Ã
x
y
!
T1 (
11
·
+
Ã
0
0
!
+
Ã
.5
0
!
+
Ã
0
.5
!
Example:
✻
1 .....................................................................
..
..
..
...
..
..
..
..
..
...
...
...
..
..
..
...
V
I
V
II
V
III
..
..
..
..
..
..
..
..
..
..
..
.
.
.
..
.
.
2 ......................
.........................................
.........................................
......................
3 ...
..
..
..
...
...
...
...
..
..
..
...
..
..
..
..
..
..
..
IV
V
..
..
..
..
..
..
..
...
..
..
..
..
.
.
1 ......................
......................
.........................................
.........................................
3 ...
..
..
..
..
..
..
..
..
..
..
..
...
...
...
...
..
..
..
I
II
III
...
..
..
..
..
..
..
..
..
.
.
..
....................................................................... ✲
0
1
2
0
1
3
3
The IFS for the Seirpinski carpet is {T1 , T2 , T3 , T4 , T5 , T6 , T7 , T8 } where
Ã
T1 (
T2 (
x
y
Ã
x
y
!
Ã
x
y
Ã
!
)=
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
..333
0
0 .333
! Ã
·
x
y
!
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
·
x
y
!
Ã
x
y
!
Ã
.333
0
0 .333
! Ã
x
y
!
T3 (
T4 (
T5 (
T6 (
T7 (
T8 (
)=
)=
)=
)=
)=
)=
)=
12
·
+
Ã
0
0
!
+
Ã
.333
0
!
+
Ã
.666
0
!
+
Ã
0
.333
!
+
Ã
.666
.333
!
+
Ã
0
.666
!
+
Ã
.333
.666
!
+
Ã
.666
.666
!
Example:
✻
20)
....................
.....(15,
..................
.............
....................
......(5,
................
....................
.......................
.20)
...................
............
...
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
.5, −.5, 5, .5,
.5,
.5,
5,
...
... .5, 5
.
... −.5, .5, 15
.
.
...
...
...
...
...
..
..
..
..
..
..
..
..
..
..
.
.15)
.
. 15)
(5,
(15,
.
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...........
...........
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.
...........
...........
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.
...........
.
...........
...........
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.
.
...........
(0, 5) ..
.. (20, 15)
..
..
..
..
..
..
..
..
...
...
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...
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..
..
..
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..
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...
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...........
..............
....................
......................
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.
...........
...........
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.
...........
...........
.
.
..
..
..
(10,.... 10)
..
...
..
..
..
..
..
..
..
..
..
..
..
..
...
...
...
...
... −10 < x < 30
..
..
..
..
..
..
..
..
..
..
0 < y < 40
...
...
...
...
...
.....................
............
............
.
.
...........
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.
...........
.
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..............
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....................
......................
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.
..
..
..
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...
...
...
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.
..
...
...
...
...
.
.
..
.
..
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..
.. T runk 0, 0,... 10, 0, .333, 0 ...
..
...
...
...
...
...
..
..
..
..
..
..
..
..
..
.
.................................................................................. ✲
(10, 0)
The IFS for the tree is {T1 , T2 , T2 } where
Ã
x
y
!
Ã
x
y
Ã
x
y
T1 (
T2 (
T3 (
Ã
.5 .5
−.5 .5
! Ã
·
x
y
!
!
)=
Ã
.5 −.5
.5
.5
! Ã
x
y
!
!
)=
Ã
0
0
0 .333
)=
·
! Ã
·
x
y
!
+
Ã
5
15
!
Ã
5
5
!
10
0
!
+
+
Ã
Download and experiment with the freeware ”IFS Construction Kit” which can be
used to design and draw fractals based on iterated function systems.
You can find it at:
http://ecademy.agnesscott.edu/∼lriddle/ifskit/
• Discuss the IFS for the Fern.
• Discuss the IFS for the Tree.
• Discuss the IFS for the mountain range.
• Discuss the IFS for the Wave.
13
Mandelbrot and Julia Sets:
Consider the complex function Fc (z) = z 2 + c. Given an initial point z0 define its
orbit under Fc to be z0 , z1 , z2 , . . . where zn+1 = zn2 + c.
Define the escape set, Ec , and the prisoner set, Pc , for for the parameter c to be:
Ec = {z0 : |zn | → ∞ as n → ∞},
and
Ec = {z0 |z0 ∈
/ Ec }
Now the Julia set for the parameter c, Jc , is the boundary of the escape set Ec .
For example, if c = 0 then Fc (z) = z 2 . Now if z = reiθ then F (z) = r2 e2iθ .
z
z2
z4
z8
z16
z 32
|z0 |
θ
.8
15
.64
30
.4096 60
.1678 120
.0281 240
.0008 120
|z0 | θ
1
15
1
30
1
60
1 120
1 240
1 120
|z0 |
θ
1.5
15
2.25
30
5.0625
60
25.629
120
656.84
240
431439.88 120
✻
✻
z3
z2
✛
z3
z4
z2
z1
z0
✲
z1
z0
✲
✛
z4
❄
|z0 | < 1
❄
|z0 | = 1
Hence Ec = {z : |z| > 1}, Pc = {z : |z| ≤ 1}, and therefore J0 is the unit circle.
14
The Julia set is either connected (one piece) or totally disconnected(dust).
Now we define the Mandelbrot set to be
M = {c ∈ C : Jc is connected}
Experiment
with
the
Java
applet
”Mandelbrot
and
Julia
Sets”
at:
http://classes.yale.edu/fractals/
Download ”winfeed” at the address below and use it to plot the Julia sets for the
constants c = −.5 + .5i, c = −.7454285 + 0.1130089i, and others.
http://math.exeter.edu/rparris/winfeed.html
15