Fractals and iteration function systems II Complex systems simulation

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1 Fractals and iteration function systems II Complex systems simulation Facultad de Informática (UPM) Sonia Sastre November 24, 2011 Sonia Sastre () Fractals and iteration function systems November 24, / 26

2 The metric space (R n, d u ) A metric space is a set X with a function d : X X [0, ] called distance function or metric if it is verifies the following properties 1 d(x, y) 0 for all x, y X, y d(x, y) = 0 x = y. 2 Symmetric: d(x, y) = d(y, x) for all x, y X. 3 Triangle inequality: d(x, y) + d(y, z) d(x, z) for all x, y, z X. Consider in X = R n euclidean metric d = d u. d(x, y) = (x 1 y 1 ) 2 + (x 2 y 2 ) (x n y n ) 2 wehere x = (x 1, x 2,..., x n ), y = (y 1, y 2,..., y n ) R n. Sonia Sastre () Fractals and iteration function systems November 24, / 26

3 The Contraction Mapping Principle The Contraction Mapping Principle The Contraction Mapping Principle Let X be a complete metric space, and let f : X X be a contraction map. Then,there is a unique point p X such that f (p) = p. 1 Complete metric space. 2 Contractive map. Proof. Let r be the reduction factor of f. We will the proof in three steps. 1. If there is a fixed point for f then, it is unique. (Reductio ad absurdum) 2. For any x 0 X, the sequence x n = f n (x 0 ) converges to a point p X, where f n is the n-times composition of f. (Due to the fact that f is a contraction map) 3. The above point P is the fixed point of f, i. e. f (p) = p. (Due to the continuity of f ) Sonia Sastre () Fractals and iteration function systems November 24, / 26

4 The Contraction Mapping Principle The Contraction Mapping Principle. Proof 2 2. Applying the triangle inequality iteratively, we have d(x, f k (x)) d(x, f (x)) + d(f (x), f 2 (x)) + + d(f k 1 (x), f k (x)) (1 + r + r r k 1 )d(x, f (x)) d(x, f (x)) r k = 1 d(x, f (x)). 1 r k=0 Then, for all n, k N, we have d(f n (x), f n+k (x)) 1 1 r d(f n (x), f n+1 (x)) 1 1 r r n d(x, f (x)), Due to the fact that X is complete, there exists the limit of the sequence x n = f n (x 0 ). Observation. The Contraction Mapping Principle, give us a method to approximate the fixed point iterating the map starting at any point. Think about the MRCM algorithm. Enumerate three similarities with the Contraction Mapping Principle. Sonia Sastre () Fractals and iteration function systems November 24, / 26

5 The Contraction Mapping Principle MRCM 1 Iterate the function make a copy. 2 The limit always is the Sierpinski gasket, no matter what the starting point is. 3 The Sierpinski gasket is invariant for the MRCM. Now, we will show that the MRCM is just the Contraction Theorem. We will need 1 A space where the elements are sets of points of R n : Let H(R n ) the set of all compact sets in R n. 2 A metric in H(R n ) such that this space will be complete: The Hausdorff distance d H. 3 The function make a copy with the MRCM will be a contraction map in the complete metric space (H(R n ), d H ): Iterated Function Systems(IFS) Sonia Sastre () Fractals and iteration function systems November 24, / 26

6 The metric space of compact sets Basisc definitions 1 The δ-collar of a compact set A is A δ = {x R n d(x, a) < δ for some a A} 2 the diameter of a compact set A: diam(a) = sup {d(x, y) x, y A} 3 Hausdorff distance between compact sets is d H (A, B) = max {δ > 0 A B δ y B A δ} A δ δ 1 A B δ 2 δ-collar of A Hausdorff distance d H (A, B) = δ 2 Sonia Sastre () Fractals and iteration function systems November 24, / 26

7 The metric space of compact sets Hausdorff distance. Exercises How long is the Hausdorff distance between the following sets 1 B r+a (p) and B r (p). 2 Two unitary sets (two points). 3 The square [0, 1] [0, 1] and the point (0, 0). 4 The Cantor set and theinterval [0, 1]. 5 The Cantor dust C 2 1/4 and the square [0, 1] [0, 1]. Theorem The space H(R n ) with the Hausdorff distance d H is a complete metric space Sonia Sastre () Fractals and iteration function systems November 24, / 26

8 Iterated Function Systems Iterated Function Systems Lema Let Φ = {f 1, f 2,..., f k } be a set of contraction maps with reduction factors r 1, r 2,..., r k respectively. Then, consider the Hutchinson operator S Φ : H(R n ) H(R n ) given by S Φ(A) = k f i (A), i=1 which is a contraction map with the Hausdorff distance. reduction factor is r = max {r i i = 1, 2,..., k}. Therefore, Its Sonia Sastre () Fractals and iteration function systems November 24, / 26

9 Iterated Function Systems Iterated Function Systems In a complete metric space X, we call an Iterated Function system a set Φ of finite contraction maps. This IFS defines the Hutchinson operator by Φ = {f 1, f 2,..., f k } S Φ(A) = k i=1f i (A), in the metric space H(X ),which is a contraction map with reduction factor r = max {factores de contracción de f i }. Theorem. Let Φ = {f 1, f 2,..., f k } be an IFS with reduction factor r. Then, there exists a unique fixed set F H(R n ) such that F = S Φ(F ) = lim i S Φ i (F ). This set is called the attractor of the IFS. Sonia Sastre () Fractals and iteration function systems November 24, / 26

10 Iterated Function Systems Iterated Function System of similarities Let E be the attractor of the IFS of similarities Φ = {f 1, f 2,..., f k }, with reduction factors r 1, r 2,..., r k. The unique number verifying r s 1 + r s r s k = 1, is called the similarity dimension of the set E. The similarity dimension is greater or equal than the Hausdorff dimension, although they agree in most of the cases we had studied as classical fractals. dim H (C 1 ) = log 2 3 log 3 ; dim H(T ) = log 3 log 2 ; dim H(C ) = log 4 log 4 = 1; dim H (K) = log 4 log 3 ; dim H(SC)) = log 8 log 3 ; dim log 20 H(M) = log 3 Sonia Sastre () Fractals and iteration function systems November 24, / 26

11 The Inverse Problem The Inverse Problem In order to make use of IFSs for a real image, one first has to construct a suitable MRCM for the given image. This is the inverse problem.we cannot expect to be universally able to build an MRCM which produces exactly the given image. However, approximations should be possible. We can make these as close to the original as we desire. The collage theorem for iterated function systems, M.F.Barnsley. Let E H(R n ) be a given set (image) and consider a fixed ε > 0. Let Φ = {f 1, f 2,, f n } an IFS with reduction factor 0 r < 1, satisfying then, we have, d H (E, SΦ(E)) ε, d H (E, A) Where A is the attractor of the IFS Φ. This is the idea for fractal image encoding. ε 1 r, Sonia Sastre () Fractals and iteration function systems November 24, / 26

12 The Inverse Problem The collage theorem for IFS. Proof d(x, x f ) = d(x, lim f n (x)) = lim d(x, f n (x)) n n n lim d(f m 1 (x), f m (x)) n m=1 lim d(x, f (x))(1 + r + r r n 1 ) n d(x, f (x)). 1 r Sonia Sastre () Fractals and iteration function systems November 24, / 26

13 Fractal image encoding Fractal image encoding To save a binary image of pixels, we need bit = bits 9KB. How many KB we need to save the IFS corresponding to Sierpinski gasket? We have 3 similarities and every one of them has six parameters belonging to the set {0, 1/2, 1/4}, so we need 2 bits for every parameter bits = 36 bits = 4, 5 Bytes 0, 0045 KB. The problem now is that real images do not use to be self-similar. If we could find an IFS for every real image we would have a good ratio of compression. By the collage theorem and the last estimation the problem is: How can we generate an IFS such that the union of the transformed images cover the image as closely as possible? Sonia Sastre () Fractals and iteration function systems November 24, / 26

14 Fractal image encoding Fractal image encoding A single iteration starting from an initial point X 0 gives us an estimate of how far is X 0 from the attractor X f with respect to the Hausdorff metric d H : d H (X 0, X f ) d H(X 0, f (X 0 )) 1 r When the copy is identical to the original, then the corresponding IFS codes the target image perfectly. When the distance of the copy to the target is small, then the attractor of the IFS is not far from the initial image. Sonia Sastre () Fractals and iteration function systems November 24, / 26

15 Fractal image encoding Fractal image encoding In the fern, the part R (1) is a slightly smaller and rotated copy of the whole fern. The same procedure applies to the copies R (2) and R (3) in the figure. Even the bottom part of the stem is a copy of the whole. However, the fern is reduced to a line. Since the 4 portions completely cover the fern, they form an IFS for the fern. Sonia Sastre () Fractals and iteration function systems November 24, / 26

16 Fractal image encoding Fractal image encoding Design stages for a leaf: 1 Scanned image of a real leaf. 2 Take the polygon capturing its outline. 3 Collage by seven transformed images of the polygon. 4 Construct the IFS and the corresponding attractor. Sonia Sastre () Fractals and iteration function systems November 24, / 26

17 What happens when we change a little one function of the IFS for the Sierpinski gasket? Consider the IFS corresponding to the Sierpinski gasket, Φ 1 = {f 1, f 2, f 3 }, and consider the IFS Φ 2 = {f 1, f 2, f 3 } where f 1 is as f 1, changing the contraction factor r 1 = Figure: Deformed Sierpinski gasket Note that we get a set looking like the Sierpinski gasket. Sonia Sastre () Fractals and iteration function systems November 24, / 26

18 Next result allows us to get deformation or motion effects on the atractor of an IFS. Theorem Let f : R R n R n be a family of contraction mappings on R n, with contractivity factor 0 r < 1, such that 1 For every p R the function f p = f (p, ) is a contraction mapping on R n. 2 For every x R n the function f x = f (, x) is continuous on R. Then, the fixed point of f depends continuously on p. That is, the function F : R R n defined by F (p) = x fp, where x fp is the fixed point of f p, is continuous. Sonia Sastre () Fractals and iteration function systems November 24, / 26

19 Proof By the continuity of f x we have that ε > 0, δ > 0 such that Then, d u (f x (p), f x (q)) < ε q R with p q < δ. d u (F (p), F (q)) = d u (x f (p), x f (q)) Which implies = d u (f (p, x fp ), f (q, x fq )) d u (f (p, x fp ), f (q, x fp )) + d u (f (q, x fp ), f (q, x fq )) ε + rd u (x f (p), x f (q)). d u (F (p), F (q)) ε, : q with p q < δ. 1 r Sonia Sastre () Fractals and iteration function systems November 24, / 26

20 Remark Now, we have a problem: δ depends on x and on p, then we can not bound the distance between two fixed points, in terms of the distance between the two corresponding parameters. In order to get such a bound we have to require f x to be Lipschitz, that is that there exits K such that Then, we have d u (f (p, x), f (q, x)) k p q p, q R x R n. d u (x f (p), x f (q)) k p q. 1 r Sonia Sastre () Fractals and iteration function systems November 24, / 26

21 Applying these results to an IFS, we have the following Theorem Consider an IFS Φ = {f 1, f 2,... f N }, where the contraction mappings f 1,..., f N depend continuously on a parameter p R. Let F : R H(R n ) H(R n ) be a family of IFS on H(R n ), with contractivity factor 0 r < 1. Then, the attractor X p H(R n ) of the IFS F p depends continuously on p In addition, the following relation is satisfied d h (A p, A q ) k p q. 1 r Sonia Sastre () Fractals and iteration function systems November 24, / 26

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26 Exercices 1 Calculate the number of IFS to transform the Sierpinski gasket to the Cantor dust in order to have d h (A pi, A pi+1 ) Implement the algorithm to transform the Cantor dust into the Sierpinski gasket. 3 Implement an algorithm to simulate the movement of a fern. Sonia Sastre () Fractals and iteration function systems November 24, / 26