Take a line segment of length one unit and divide it into N equal old length. Take a square (dimension 2) of area one square unit and divide

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1 Fractal Geometr A Fractal is a geometric object whose dimension is fractional Most fractals are self similar, that is when an 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 Similarit Dimension: Take a line segment of length one unit and divide it into N equal old length parts let S = new length, then N = S N = 3, S = 3 Take a square (dimension 2) of area one square unit and divide it into N congruent parts Then S = N and N = S 2 N = 4, S = 2 Take a cube (dimension 3) of volume one cubic unit and divide it into N congruent parts Then S = N 3 and N = S 3 N = 8, S = 2

2 So given set of dimension d consisting of N parts congruent part then N = S d, where S = Solving for d ields old length new length d = ln N ln S Exercise : T = lim n T n is called the Seirpinski Triangle or Gasket What is the perimeter of T n? 2 What is the area of T n? 3 Compute the fractal dimension of T T 0 T T 2 Solution: P n = ( 3 2 )n P 0 2 A n = ( 3 4 )n A 0 3 N = 3 and S = 2, hence d = ln3 ln2 2

3 Exercise 2: C = lim n C n is called the Seirpinski Carpet C 0 C C 2 What is the perimeter of C n? 2 What is the area of C n? 3 Compute the fractal dimension of C Solution: P n = P P P n 3 n P 0 2 A n = ( 8 9 )n A 0 3 N = 8 and S = 3, hence d = ln8 ln3 3

4 Exercise 3: M = lim n M n is called the Menger Sponge 2 0 M 0 2 M M 2 What is the volume of M n 2 What is the surface area of M n? 3 Compute the fractal dimension of M Solution: V 0 = 2 3 = 8 and V n = 20 (/3) 3 (V n ) 2 A 0 = = 24 and A n = 20 (/3) 2 (A n ) 24 (/3) 2 (A tn ) 2 = (20/9) Area(C n ) (64/3) (8/9) n where Area(C n ) is the area of the n stage in the construction of the Sierpinski carpet Hence A n = (8/9) n 6 + (20/9) n 8 3 N = 20 and S = 3, hence d = ln20 ln3 4

5 Exercise 4: Draw a fractal with dimension 3 2 Solution: Demonstrate the applet Fractal Dimension at: 5

6 Box Dimension: Box dimension is another wa to measure fractal dimension, it is defined as follows: If X is a bounded subset of the Euclidean space, and ɛ 0 Let N(X, ɛ) be the minimal number of boxes in the grid of side length ɛ which are required to cover X We sa that X has box dimension D if the following limit exists and has value D ln(n(x, ɛ)) lim ɛ 0 + ln( ɛ ) Consider for example the Seirpinski gasket: 0 ɛ N(X, ɛ) n 2 n Hence D = lim n ln(4 3 n ) ln(2 n ) = ln 3 ln 2 6

7 Demonstrate the applet Box Counting Dimension at: Geometricall, if ou plot the results on a graph with ln N(X, ɛ) on the vertical axis and ln(/ɛ) 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(), ln()) 7

8 Discuss computing the fractal dimension of a bunch of broccoli M Results: Ball Size N ɛ ln( ɛ ) ln(n) 6x5x25 0 / x4x25 9 / x3x2 28 / x25x2 4 / x25x2 64 / Use the linear regression applet at to plot the data points and find the slope of the best fit line Fractal dimension = slope = 262 If ou throw the st point out ou get fractal dimension=slope =282 8

9 Divider Dimension: A fractal curve has fractal (or divider) dimension D if its length L when measured with rods of length l is given b L = C l 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: K 2 l L n ( 4 3 )n It appears to have length L = ( 4 3 )n when measured with rods of length l = ( 3 )n Hence, ( 4 3 )n = C ( 3 )n( D) This implies that n ln 4 3 = ln(c) + n( D) ln 3 In order for the equation to be satisfied for all n, ln C must be zero, hence C = and so D = ln 4 ln 3 Geometricall, if ou graph ln(l) versus ln(l) the D = Slope of line 9

10 Exercise 5: The west coast of Britain when measured with rods of length 0 km is 3020 km But when measured with rods of length 00 km it is 700 km What is the fractal dimension of the west coast of Britain? Demonstrate the applet Coastline of Hong Kong at: cktse/movies/pheno/maphtml 0

11 Affine transformations: An affine transformation is of the form or f( f( ) = ) = a b c d x r cos(θ) s sin(φ) r sin(θ) s cos(φ) + e f x + e f Where r and s are the scaling factors in the x and directions respectivel θ and φ measure rotation of horizontal and vertical lines resp e and f measure horizontal and vertical translations resp It can be easil shown that r 2 = a 2 +c 2, s 2 = b 2 +d 2, θ = arctan(c/a) and φ = arctan( b/d) b θ a c d φ C B B C Similarit Affinit B C B C D A A D A A x = = 2 x 2 x = 3 4 x = 2

12 A reflection about the -axis would be given b x = x and = matrix form A reflection about the x-axis would be given b x = x and = matrix form Rotation is represented as follows: B C B C B θ = 0 φ = 45 D A x = x = 2 2 B C A θ = 45 φ = 0 D A x = = 2 x 2 x + Matrixform : cos(θ) sin(φ) sin(θ) cos(φ) = a b c d C B B C A θ = 45 φ = 45 D A x = = 2 x 2 x

13 Shears are represented as follows: B A B B C C C D A D A x = x = 2x + Shear in the direction x = x + 2 = Shear in the x direction B 3

14 Iterated Function Sstems: An Iterated Function Sstem in R 2 is a collection {F, F 2,, F n } of contraction mappings on R 2 (F i is a contraction mapping if given x, then d(f i (x), F i ()) sd(x, ) where 0 s < ) The Collage Theorem sas that there is a unique nonempt compact subset A R 2 called the attractor such that A = F (A) F 2 (A) F n (A) Sketch of Proof: Given a complete metric space X, let H(X) be the collection of nonempt 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) b G(K) = F (K) F n (K) If each F i is a contraction then so is G The contraction mapping theorem sas that a contraction mapping on a complete space has a unique fixed point Hence there is A H(X) such that G(A) = A = F (A) F 2 (A) F n (A) Given K 0 R 2, let K n = G n (K 0 ) Then d(a, K i ) si d(k 0, G(K 0 )) s Hence K i is a ver good approximation of A for large enough i Deterministic Algorithm: (Slow!) Given K 0, let K n = G n (K 0 ) Now K n A Random Algorithm: (Fast!) Choose P 0 K 0 and let P i+ = F j (P i ) where F j is chosen at random from all of the F i Clearl P i K i and for sufficientl large i,p i is arbitraril close to some point in A 4

15 Finding IFS Rules from Images of Points Given three non-collinear initial points p = (x, ), p 2 = (x 2, 2 ), p 3 = (x 3, 3 ) and three image points q = (u, v ), q 2 = (u 2, v 2 ), q 3 = (u 3, v 3 ) respectivel Find an affine transformation T defined b T ( ) = a b c d x + e f such that T (p ) = q, T (p 2 ) = q 2, and T (P 3 ) = q 3 This gives the following six equations in six unknowns: ax + b + e = u cx + d + f = v ax 2 + b 2 + e = u 2 cx 2 + d 2 + f = v 2 ax 3 + b 3 + e = u 3 cx 3 + d 3 + f = v 3 This sstem has a unique solution if and onl if the points p, p 2, and p 3 are noncollinear 5

16 Example: 5, 0, 0, 5, 0, 5 5, 0, 0, 5, 5, 0 5, 0, 0, 5, 0, 0 The IFS for the Seirpinski triangle is {T, T 2, T 2 } where T ( T 2 ( T 3 ( ) = ) = ) = x x x

17 Example: V I V II V III IV V I II III The IFS for the Seirpinski carpet is {T, T 2, T 3, T 4, T 5, T 6, T 7, T 8 } where T ( T 2 ( T 3 ( T 4 ( T 5 ( T 6 ( T 7 ( T 8 ( ) = ) = ) = ) = ) = ) = ) = ) = x x x x x x x x

18 Example: (5, 20) (5, 20) 5, 5, 5, 5, 5, 5 5, 5, 5, 5, 5, 5 (5, 5) (5, 5) (0, 5) (20, 5) (0, 0) 0 < x < 30 0 < < 40 T runk 0, 0, 0, 0, 333, 0 (0, 0) The IFS for the tree is {T, T 2, T 2 } where T ( T 2 ( T 3 ( ) = ) = ) = x x x

19 Download and experiment with the freeware IFS Construction Kit which can be used to design and draw fractals based on iterated function sstems You can find it at: 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 9

20 Mandelbrot and Julia Sets: Consider the complex function F c (z) = z 2 + c Given an initial point z 0 define its orbit under F c to be z 0, z, z 2, where z n+ = z 2 n + c Define the escape set, E c, and the prisoner set, P c, for for the parameter c to be: E c = {z 0 : z n as n }, and E c = {z 0 z 0 / E c } Now the Julia set for the parameter c, J c, is the boundar of the escape set E c For example, if c = 0 then F c (z) = z 2 Now if z = re iθ then F (z) = r 2 e 2iθ z 0 θ z 0 θ z 0 θ z z z z z z z 3 z 2 z 3 z 4 z 2 z z 0 z z 0 z 4 z 0 < z 0 = 20

21 Hence E c = {z : z > }, P c = {z : z }, and therefore J 0 is the unit circle The Julia set is either connected (one piece) or totall disconnected(dust) Now we define the Mandelbrot set to be M = {c C : J c is connected} Experiment with the Java applet Mandelbrot and Julia Sets at: Download winfeed at the address below and use it to plot the Julia sets for the constants c = 5 + 5i, c = i, and others 2