Starting with a square of size 500 by 500 pixels, and assuming your computer can draw 10,000 squares per second, estimate how long your computer will take to draw the von Koch and the square von Koch curves using an IFS (see page 260 in the text).

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.

(4a*) 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......)

(4b*) 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. )

(4c) 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.)

What is the fractal dimension of the set of numbers in [0,1] that have no occurence of the string '73' in their decimal expansion? (This a hard one - I think, so even just and estimate (like "the fractal dimension is greater than y but less than x") would be something.) Note that this is similar to asking what is the factal dimension of the image produced by the full triangle IFS when the occurence of the string '12' is removed from the random sequence before using the chaos game to draw the image (this is displayed in the Modified Chaos Game applet).

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, for example, D₁₃₃ denotes the small triangle with address 133), 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.

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

• (7a*) 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?
• (7b*) 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 #4 above, and the weekly summary for 10 Feb.)

Run the Modified Chaos Game applet with the Sierpinski IFS and remove all occurences of the string 13. Explain the image you see. Do the same but remove the sequences 123, and 3. Before running the applet try to make a sketch of the image.

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. (See the Chaos Game with Varying Probabilities page for images.)

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 14 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... . What transformations have these as their fixed points?)
• (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?

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.

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. (See the handout, The Chaos Game with Non-random Sequences.)

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?