Let F be the fixed point of the IFS W: W(F)=F. Prove that the fixed points p_i of the affine transformations w_i of W always lie on the fractal F (so if w_i(p_i) = p_i, then p_i is on F.)

Write down, in 'plain english', the rules of the Chaos Game for the von Koch and square von Koch curves.

(3a*) Make sketches of the Cantor set and the von Koch curve and indicate on your sketch the regions with addresses;
112, 212, 211(11) for the Cantor set,
12, 321, 2(22) for the von Koch curve.
(Recall that (xy) means xyxyxy......)

(3b*) Find the coordinates (x,y) of the points 2(22), 21(11), and 21(21) on the von Koch curve. (Hints: For 21(21) consider the affine transformation w₂₁ = w₂w₁. )

(3c) Let's define rational points of the von Koch curve K to be those points on K that have addresses that are eventually repeating (eg., 24312342342(342) ). Prove that every point on the von Koch curve is an accumulation point of rational points of the von Koch curve. (Hint: Follow the idea of the proof that every point in the Cantor set is an accumulation point of the set of end points of the intervals removed in the connstruction, which in turn is similar to the proof that every real number is an accumulation point of the set of rational numbers. Recall that p is an accumulation point of a set A if there is a sequence of points {x₁, x₂, . . . } from A such that x_n tends to p as n tends to infinity.)

Consider the Sierpinski IFS, and set the probabilities to be p₁ = 0.5, p₂ = p₃ = 0.25. Suppose the chaos game is played with 10,000 points.
(a) How many points will lie in the small triangles with address 11 ? Address 12 ? Adress 22 ?
(b) Among all the small triangles with addresses of length 3 (the D_ijk), which will have the most points? The least points? Intermediate number of points?
(c) Based on your answers to parts (a) and (b), and by considering other small triangles, what do you expect the final image to look like? Use the Chaos Game applet to play this game with these probabilities and verify your answer.

This question uses the updated Fractal Movie applet to calculate fractal dimension. Note that you could use the (updated) Chaos Game applet in addition to get a more accurate estimate of the fractal dimension.

• (5a) Let F(r) be the fractal dimension of the fractal square(r) where the IFS of square(r) is obtained from the IFS of full square by changing the reduction factors of the matrices of the w_i to r (i.e., change the a and c entries of all the matrices in the w_i from 0.5 to r).
  • (i) Use the Fractal Movie applet to sketch the graph of F(r) for 0 <= r <= 0.5 (so enter as start pattern parameters the full square data from the drop down menu, and for end pattern parameters change all the 0.5^'s in the a and c^'s to 0. It's better to set # iterations to 200,000 and # steps to 300. Note that you can compute exactly what the value of r is at any time step M (current M).
  • (ii) Find a formula for F(r) (do this by calculating the fractal dimension of square(r)). Compare this to your sketch in part (i). (This question investigates how the fractal dimension of a fractal changes if the reduction rates of the (non-overlapping) lenses changes.)
• (5b) Let F(a) be the fractal dimension of the fractal Sier(a) where Sier(a) is the fractal produced with the IFS obtained from the Sierpinski IFS by changing the shift vector (e,f) of lens 3 (w₃) to (0.25, a). So for example, Sier(0.5) is the Sierpinski triangle and so F(1) = 1.585..., Sier(0) is a line and so F(0) = 1. Use the Fractal Movie applet to sketch the graph of F(a) as accurately as you can for 0 <= a <= 0.5 (So for start pattern parameters use Sierpinski data from the drop down menu, and for end pattern parameters change f to 0 in w₃.) It will be better to set # of iterations to 200,000 and use 300 or more time steps (# of steps). Note that you can tell exactly what the value of a is for a given time step (current M). (This question investigates how the fractal dimension of a fractal changes if the images produced by the lenses move and overlap.)
• (5c) Let Sier(t) be the fractal obtained by changing the contraction rate of w₃ in the Sierpinski IFS from 0.5 to t (so change the a and c entries of the matrix from 0.5 to t) and the shift vector to (e,f) = (0.25, 1-t). ( So Sier(0.5) is the Sierpinski triangle, and Sier(0) is a line). Let F(t) be the fractal dimension of Sier(t). Use the Fractal Movie applet to sketch the graph of F(t) for 0 <= t <= 0.5 . (This question investigates how the fractal dimension of a fractal changes if the reduction rates of the lenses change from being all equal to being not equal.)

In this question the chaos game is played with the Sierpinski IFS.

• (6a*) Suppose that the sequence {2, 1, 3, 3} are the first 4 random numbers chosen in the game. If the first 5 digits of the address of the game point z₄ is 33123, where was the initial game point z₀?
• (6b*) Suppose the game point z₄ has address 33123(22). What is the address of the initial game point z₀? What are the coordinates (x,y) of z₀? Sketch the positions of the points z₀, z₁, z₂, z₃ and z₄ on the Sierpinski triangle.

(See pages 311-312 in the text, Question #3 above, and the weekly summary for 11 Feb.)

Run the Modified Chaos Game applet with the Sierpinski IFS and remove all occurences of the string 13. Explain the image you see.

Estimate how many game points would be needed to draw the von Koch curve using the chaos game with equal probabilities (assume that the von Koch curve sits in a square of size 512 by 512 pixels).

We saw in class that if you play the chaos game with the Sierpinski IFS and use equal probabilities p₁ = p₂ = p₃ = 1/3, that about 20,000 game points are needed (if the triangle is of size 512 pixels wide). Estimate how many game points would be needed to draw the Sierpinski triangle if the unequal probabilities p₁ = p₂ = 0.2, p 3 = 0.6 are used. Run the Chaos Game applet and compare to your answer.

Using the Fractal Movie applet for the following experiment.
(a) Start Pattern: Full Square (via drop down menu). End Pattern: Full Square, but with probabilities p₁ = p₂ = p₃ = .3334, p₄ = 0. (Try with m = 200, # of iterations = 50,000.)
(b) Start Pattern: Full Square (via drop down menu). End Pattern: Full Square, but with probabilities p₁ = 0, p₂ = 0, p₃ = 0.5, p₄ = 0.5. (Try with m = 200, # of iterations = 50,000.)

Either stop the movie at an intermediate pattern, or run the Chaos Game applet with intermediate probabilities and explain as many features of the intermediate pattern as you can. You will use addresses and probabilities.

Consider the Barnsley Fern. Suppose we play the chaos game with 100,000 points.
(a) How tall is the fern produced by this game when (i) the probabilities are all 1/4?, (ii) when the probabilities are p₁=0.85, p₂=p₃ =p₄ = 0.05?
(b) How long is the first branch on the right side of the fern when (i) the probabilities are all 1/4?, (ii) when the probabilities are p₁=0.85, p₂=p₃=p₄ = 0.05?
(By 'height of the fern' we mean how many of the segments that are drawn by w₄ appear by the end of the chaos game. (Note that each iteration of the IFS - if you were drawing the fractal by iterating the IFS - produces another segment of the stem. If you are drawing the fractal with the chaos game, and if at least one point in the chaos game appears in a particular segment, then we say that segment appears in the game). By 'length of the first branch' we mean how many of the segments of its stem appear by the end of the chaos game.)

(You may want to do question 16 below first to understand the addressing of the Fern.)

Use the 'full triangle' IFS to answer the following questions. (The full triangle' IFS has the three transformations that make up the Sierpinski IFS plus a forth one that 'fills' the middle hole of the Sierpinski IFS, i.e., w₄ = (-0.5, 0, 0, -0.5) + (0.75, 0.5) )

• (a) What image is produced from the chaos game using the sequence {1,2,3,4,1,2,3,4,...} (i.e., 1,2,3,4, repeating)? (Hint: consider the addresses 43214321...., 14321432..., 21432143..., and 32143214... )
• (b) Suppose every occurence of the string 1,2 is removed from a random sequence of the integers 1,2,3,4 and this resulting sequence is used in the chaos game for this IFS. What is the image produced using this sequence for the chaos game? (Hint: think of addresses!) Is the image produced a fractal? Can you estimate the fractal dimension of the image?

Follow the procedure discussed in class (and described on page 328 of the text) in the case of the Fern, to find 'better' probabilities than the equal ones p_i = 1/4 for the Crystal 4 fractal. That is, determine the probabilities such that there are an equal density of game points in each region with address of length 1. Check your answer by running the chaos game applet first with equal probabilities and then with the 'better' probabilities you found. (The Crystal 4 fractal is Figure 5.14 - the parameters for the IFS are listed on page 295.)

Write down a sequence that contains all strings of length 3 of the integers 1,2,3 that is more 'efficient' than the 'naive' one of length 3*3³ = 81. (The 'best' such sequence has length 3³ +1 = 28). Verify that your sequence is more efficient by writing down the addresses of the game points.

Use the Chaos Game applet, or better yet the VB program Fractal Pattern, to investigate how 'efficient' a random sequence is when used in the chaos game to draw the Cantor Maze fractal. Follow the procedure described in class;

• Determine the level m; the smallest addresses that can be resolved on your computer screen (these addresses have length m). Here are the sizes (in pixels) of the canvases that contains the image produced by the applet:
  • for the 640x480 version of the applet, canvas size is 304 by 304
  • for the 800x600 version of the applet, canvas size is 380 by 380
  • for the 1024x768 version of the applet, canvas size is 500 by 500
  • for the 1400x1050 version of the applet, canvas size is 800 by 800
• calculate the length of the 'best' sequence and the worst, 'naive', sequence that will draw the fractal to this level m;
• show that a string of random numbers of the same length as the 'best' sequence does not draw the fractal well (to level m), while the fractal can be drawn well (to level m) by a string of random numbers that is shorter than the 'naive' sequence.

Give an explanation of your results above. Based on your results, how much 'redundancy' does a random sequence have, i.e., how many addresses at level m are repeated in a random sequence?

Find the addresses of all the 'subferns' on the first branch on the left side of the Barnsley Fern (Fig. 6.24). Note that there are infinitely many such subferns, but the pattern of their addresses will become apparent.

End of Homework # 4 !

Back to the MAT 335 Home Page