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Lecture 6: Fractals from Iterated Function Systems He draweth also the mighty with his power: Job 24:
1. Generating Fractals by Iterating Transformations The Sierpinski gasket and the Koch snowflake can both be generated in LOGO using recursive turtle programs. But in CODO there is no recursion. How then can we generate fractals in CODO? Consider the first few levels of the Sierpinski gasket depicted in Figure 1. We begin with a triangle. The next level contains three smaller triangles. (Ignore the central upside-down triangle.) Each of these smaller triangles is a scaled down version of the original triangle. If we denote the vertices of the original triangle by € P 1 , P 2 , P 3 , then we can generate the three small triangles by scaling by € 1 / 2 around each of the points € P 1 , P 2 , P 3 -- that is, by applying the three transformations € Scale ( P 1 ,1/ 2) , € Scale ( P 2 , 1 / 2), € Scale ( P 3 ,1/ 2) to € Δ P 1 P 2 P 3. Now the key observation is that to go from level 1 to level 2, we can apply these same three transformations to the triangles at level 1. Scaling by € 1 / 2 at any corner of the original triangle maps the three triangles at level 1 to the three smaller triangles at the same corner of level 2. We can go from level 2 to level 3 in the same way, by applying the transformations € Scale ( P 1 ,1/ 2) , € Scale ( P 2 , 1 / 2), € Scale ( P 3 ,1/ 2) to the triangles in level 2. And so on and so on. Thus starting with € Δ P 1 P 2 P 3 and iterating the transformations € Scale ( P 1 ,1/ 2) , € Scale ( P 2 , 1 / 2), € Scale ( P 3 ,1/ 2) n times generates the nth level of the Sierpinski gasket. Figure 1. Levels zero through five of the Sierpinski gasket.
Let’s try another example. Consider the first few levels of the Koch curve illustrated in Figure
- Here we begin with a triangular bump. The next level contains a small triangular bump on each line of the original bump. Each of these smaller bumps is a scaled down version of the original bump. Thus there are four scaling transformations that map the original bump at level 0 to the four smaller bumps at level 1. Once again the key observation is that to go from level 1 to level 2, we can apply these same four transformations. Now we can go from level 2 to level 3 in the same way; and on and on. Thus starting with a triangular bump and iterating four scaling transformations n times generates the nth level of the Koch curve. Figure 2. Levels zero through five of the Koch curve. Evidently then, if we know the first few levels of a fractal, we can discover transformations that iteratively generate the fractal by finding transformations that generate level 1 from level 0 and then checking that these transformations also generate level 2 from level 1. But suppose we encounter a fractal with which we are not so familiar such as one of the fractals shown in Figure 3. If we have not seen these fractals before, we may not know what the first few levels should look like. We could, of course, write a turtle program to generate these fractals and then observe the first few levels of the turtle program, but this approach is rather tortuous. Fortunately, there is a better, more direct approach. Figure 3. Three fractals for which the first few levels are not evident: a fractal flower, a fractal bush, and a fractal hangman.
M 2 = Scale ( A , 1 / 3)∗ Trans ( Vector ( A , D )) ∗ Rot ( D ,π / 3 )
M 3 = Scale ( C , 1 / 3 ) ∗ Trans ( Vector ( C , E )) ∗ Rot ( E ,− π / 3 )
M 4 = Scale ( C , 1 / 3 ) Method 2: Building each transformation by specifying the image of three points. € M 1 = AffineTrans ( A , B , C ; A , F , D ) € M 2 = AffineTrans ( A , B , C ; D , H , B ) € M 3 = AffineTrans ( A , B , C ; B , I , E ) € M 4 = AffineTrans ( A , B , C ; E , G , C ) Notice that in this example, both methods generate the same set of transformations. The first method may seem more natural, but the second method is often much simpler to apply. For yet another approach to deriving these same transformations, see Exercise 6. Figure 4: The Koch curve. Key points along the curve are labeled to help specify the four transformations in the IFS that generate this curve.
3. Fractals as Attractors We have explained how to find an IFS to generate a given fractal curve, but we still do not know on what shape to start the iteration. That is, we still do not know how, in general, to find level 0 of a given fractal curve. For the Sierpinski triangle, we know to begin with an ordinary triangle, and for the Koch curve we know to begin with a triangular bump. But how do we begin for fractals with which we are not so familiar such as the fractals shown in Figure 3? Fractals generated by recursive turtle programs are independent of the base case. Could the same independence property hold for fractals generated by iterated function systems? Yes! Figure 5 illustrates this independence for the Sierpinski gasket. The same IFS is applied to three different base cases: a square, a horizontal line, and a single point. In each case after a few iterations the figures all converge to the Sierpinski gasket.
Figure 5: Levels 1,3,5 of the Sierpinski gasket. On the top, the base case is a square; in the middle, the base case is a horizontal line; on the bottom, the base case is a single point. Fractals are attractors. No matter what geometry we choose for the initial shape, the subsequent shapes are inexorably attracted to the fractal generated by the IFS. We shall explain the reason for this attraction in the next lecture. For now it is enough to know that it does not matter how we choose the initial shape. The base case is irrelevant. If we choose the right IFS, the fractal we want will magically emerge no matter what geometry we choose at the start.
Figure 7: Levels 1,3,5 of the fractal staircase. The base case is the large square, which is also the condensation set. Here the identity transformation is included in the IFS. Figure 8: Levels 1,3,5 of the fractal staircase. The base case is a pentagon and the condensation set is the large square. Notice that the pentagons converge to the diagonal of the staircase, and that the square generates the main part of the fractal. Here the identity transformation is not included in the IFS, but at each level after applying the transformations in the IFS, the large square is added back into the curve.
5. Summary Fractals are fixed points of iterated functions systems. This observation allows us to find transformations that generate a given fractal curve by finding a set of transformations that map the fractal onto itself as the union of a collection of smaller self-similar copies. Fractals are attractors. No matter what base case we choose to initiate the iteration, for a fixed IFS the shapes we generate will automatically converge to the identical fractal. Fractals may have condensation sets. We can incorporate these condensation sets into our scheme for generating fractals from an IFS by adding the condensation set back into the curve at every stage of the iteration after applying the transformations in the IFS.
Exercises
- Find an IFS to generate a polygonal gasket with n sides.
- Find an IFS to generate each of the fractals in Figure 3 by: a. composing translations, rotations, and uniform scalings b. specifying the images of three points in the plane.
- Find an IFS to generate each of the fractals in Lecture 2, Figure 8.
- Find an IFS to generate each of the fractals in Lecture 2, Figure 16.
- Find explicit expressions for the IFS and the condensation set of the fractal tree in Figure 6.
- Consider the Koch curve in Figure 4. Let M denote the point midway between D and E , and let € v = E − D. a. Show that the Koch curve is generated by the IFS containing the four transformations: € T 1 = Scale ( M , 1 / 3)∗ Trans ( v ) , € T 2 = Scale ( M ,1/ 3 ) ∗ Trans (− v ) , € T 3 = Scale ( M , 1 / 3)∗ Rot ( D ,π / 3), € T 4 = Scale ( M , 1 / 3)∗ Rot ( E , − π / 3). b. Verify that these transformations are identical to the transformations in Example 1.
- Prove that for any self-similar fractal generated by an iterated function system, € Fractal Dimension = −
Log ( # Transformations in the IFS )
Log ( ScaleFactor )
- Using the result of Exercise 7, compute the fractal dimension of the fractals in Figure 3.
- The Cantor set is generated by starting with the unit interval and recursively removing the middle third of every line segment. a. Show that the Cantor set is the fractal generated by the IFS consisting of the two conformal transformations € Scale ( Origin , 1 / 3 ) and € Scale ( P 1 , 1 / 3 ) , where €
P 1 = ( 1 ,0).
b. Using part a and the result in Exercise 7, show that € 0 < Dimension ( Cantor Set ) < 1. c. Show that the length of the Cantor set is zero. d. Consider the set generated by starting with the unit interval and recursively removing the middle half of every line segment. i. Find the transformations in the iterated function system that generate this fractal. ii. Using the result in Exercise 7, show that the dimension of this set is €
Programming Project:
- CODO Implement CODO in your favorite programming language using your favorite API. a. Write CODO programs for the curves discussed in the text and in the exercises for Lectures 5 and 6. b. Write a CODO program to design a novel flag for a new country. c. Develop some iterated function systems to create your own novel fractals with and without condensation sets.
