Fractal Geometry Isa S. Jubran – [email protected] – 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: ² Take a line segment of length one unit and divide it into N equal parts. let old length 1 S = new length , then N = S . N = 3;S = 3 ² Take a square (dimension 2)p of area one square unit and divide it into N congruent parts. Then S = N and N = S2. N = 4;S = 2 ² Take a cube (dimension 3) of volume one cubic unit and divide it into N 1 3 congruent parts. Then S = N 3 and N = S . ......................... .... .... .... .... ...................... ..... ..... N = 8;S = 2 ..... ..... .... ..... ..................... .... .... ... ......................... .... .... .... .... .... ...................... .... .... .... .... So given set of dimension d consisting of N parts congruent part then N = Sd, old length where S = new length . Solving for d yields ln N d = ln S 1 Exercise 1: T1 = limn!1 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 T1. T0 T1 T2 Solution: 3 n 1. Pn = ( 2 ) P0. 3 n 2. An = ( 4 ) A0. ln3 3. N = 3 and S = 2, hence d = ln2 . Exercise 2: C1 = limn!1 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 C1. Solution: 1 8 8n¡1 1. Pn = P0 + 3 P0 + 32 P0 + ¢ ¢ ¢ + 3n P0. 8 n 2. An = ( 9 ) A0. ln8 3. N = 8 and S = 3, hence d = ln3 . 2 Exercise 3: M1 = limn!1 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 M1. Solution: 3 3 1. V0 = 2 = 8 and Vn = 20 ¢ (1=3) ¢ (Vn¡1). 2 2 2 2. A0 = 6 ¢ 2 = 24 and An = 20 ¢ (1=3) ¢ (An¡1) ¡ 24 ¢ (1=3) ¢ (Atn¡1) ¢ 2 n¡1 = (20=9) ¢ Area(Cn¡1) ¡ (64=3) ¢ (8=9) 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. ln20 3. N = 20 and S = 3, hence d = ln3 . 3 Exercise 4: Draw a fractal with dimension 2 . Solution: 6 .................... ........... .......... .................... .................... ................... ........... - . ......... .......... .......... ........... ? 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 ² ¸ 0. Let N(X; ²) be the minimal number of boxes in the grid of side length ² which are required to cover X. We say that X has box dimension D if the following limit exists and has value D. ln(N(X; ²)) lim 1 ²!0+ ln( ² ) Consider for example the Seirpinski gasket: 6 .................................................. 1 . .................................................. .................................................. .................................................. .................................................. .................................................. .- 0 1 ² N(X; ²) 1 1 1 2 3 1 4 9 . 1 3n 2n ln(4 ¢ 3n) ln 3 Hence D = lim = n!1 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; ²) on the vertical axis and ln(1=²) 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. 6 (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 1 Ball Size N ² ln( ² ) ln(N) 6x5x2.5 10 1/4.5 -1.504 2.303 4x4x2.5 19 1/3.5 -1.253 2.94 3.5x3x2 28 1/2.83 -1.04 3.33 3x2.5x2 41 1/2.5 -.916 3.71 2.5x2.5x2 64 1/2.33 -.847 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 L 1 1 1 4 3 3 1 16 9 9 . 1 4 n 3n ( 3 ) 4 n 1 n It appears to have length L = ( 3 ) when measured with rods of length l = ( 3 ) . 4 1 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 à ! à ! à ! à ! x a b x e f( ) = ¢ + y c d y f or à ! à ! à ! à ! x r cos(θ) ¡s sin(Á) x e f( ) = ¢ + y r sin(θ) s cos(Á) y f Where r and s are the scaling factors in the x and y directions respectively. θ and Á 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 Á = arctan(¡b=d). 6 b 6 c d Á θ - - a ? ? 6 B 6 B C. ..................... C. ..................... ..................... Similarity . Affinity . B0 . B0 C0. ....................... C0................................. ....................... ........................ - ....................... ....................... - D A0 A D A0 A x0 = 1 x x0 = 3 x 0 12 0 14 y = 2 y y = 2 y A reflection about the y-axis would be given by à ! ¡1 0 x0 = ¡x and y0 = y ¡¡¡¡¡¡¡ matrix form 0 1 A reflection about the x-axis would be given by à ! 1 0 x0 = x and y0 = ¡y ¡¡¡¡¡¡¡ matrix form 0 ¡1 7 Rotation is represented as follows: 0 B.. .... ... 6 6 ... .. ... ... ... ... ... B ... B C. C. 0 . 0 . C ............................... B. 0 ... ... ...A ... ... ... ... ... ... .. .. .. ... ... ... ... ... θ = 0 . ... θ = 45 ... ... ... ... ... Á = 45 . ... Á = 0 ... ... ... ................................................. - ......................... ..................... - D A D A x0 = x ¡ p1 y x0 = p1 x 2 2 y0 = p1 y y0 = p1 x + y 2 2 à ! à ! cos(θ) ¡ sin(Á) a b Matrixform : = sin(θ) cos(Á) c d 6 . 0 .B..... ... ... ... ... ... ... ... ... ... ... B ... C ... ... ... 0 .. .. C ..... ... 0 ... ...A ... ... ... ... .. .. ... ... ... ... θ = 45 ... ... ... ... Á = 45 ... ... ........................... ..................... - D A x0 = p1 x ¡ p1 y 2 2 y0 = p1 x + p1 y 2 2 8 Shears are represented as follows: B0 6 6 . ..................... 0 ...................... ..................... ...................... .A . 0 ...............................................B ... ....................... ........................B .C 0 C. C. B . ....................... ........................ - ...................... ........................ - D A D A x0 = x x0 = x + 2y y0 = 2x + y y0 = y Shear in the y direction Shear in the x direction 9 Iterated Function Systems: 2 An Iterated Function System in R is a collection fF1;F2;:::;Fng of contraction 2 mappings on R .(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 ½ 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 2 H(X) such that G(A) = A = F1(A) [ F2(A) [ ::: [ Fn(A). 2 n Given K0 ½ R , let Kn = G (K0). Then sid(K ;G(K )) d(A; K ) · 0 0 i 1 ¡ s Hence Ki is a very good approximation of A for large enough i. Deterministic Algorithm: (Slow!) n Given K0, let Kn = G (K0). Now Kn ! A. Random Algorithm: (Fast!) Choose P0 2 K0 and let Pi+1 = Fj(Pi) where Fj is chosen at random from all of the Fi. Clearly Pi 2 Ki and for sufficiently large i,Pi is arbitrarily close to some point in A.