# Exercises for Chaos Under Control

## Chapter 2: Deterministic Iterated Function Systems

In this set of exercises "DET" and "RAND" refer to Options in the TreenessEmerging Program.

9.[C] Looking at the Koch curve, it isn't clear whether the transformation that codes the right-hand side of the tent shouldn't be: scale by 0.333 toward the origin, rotate by +120¡, then translate by 0.667 in the positive x-direction. Change the Koch parameter table accordingly and run DET. See that this doesn't produce the correct picture. Explain why not.

10.[C] In the Koch Curve parameters try some other rotation values, such as +30¡ instead of +60¡ and -30¡ instead of -60¡. In each case, predict what you think will happen, run DET then explain why your guess was correct or incorrect.

11. Consider the IFS defined by these rules

scale by 0.5;

scale by 0.5 and translate by 0.5 in the y-direction;

scale by 0.5, rotate (counterclockwise) by 90¡, translate by 1 in the x-direction;

collage the pieces.

(a) Sketch the result of one application of these three rules to the picture shown in the picture. Answer

(b) Now sketch the result of one application of these rules to the picture you got in part (a). Answer

(c)[C] Run DET to find the limiting shape. Does this result surprise you? Answer

12. Draw the first iteration of the Deterministic IFS
R S Theta Phi E F
.5 .5 0 0 0 0
-.5 .5 0 0 .5 .5
.5 .5 90 90 0 .5
.5 .5 0 0 .5 .5

starting with a unit square (square of side equal to 1, on the left); the lower left hand corner of the square is the origin.

13. Draw the first iteration of the Deterministic IFS
R S Theta Phi E F
-.5 .5 0 0 0 0
.5 .5 0 0 .5 .5
.5 .5 90 90 0 .5
-.5 .5 0 0 .5 .5

starting with a unit square (square of side equal to 1, on the left); the lower left hand corner of the square is the origin. Answer

14. Draw the first iteration of the Deterministic IFS
R S Theta Phi E F
-.5 .5 0 0 0 1
.5 .5 45 45 1 1

starting with a unit square (square of side equal to 1, on the left); the lower left hand corner of the square is the origin.

15. Draw the first two iterations of the Deterministic IFS
R S Theta Phi E F
.5 .5 0 0 0 0
.5 .5 0 0 0 .5
.333 .333 0 0 .5 0

starting with a unit square (square of side equal to 1, on the left); the lower left hand corner of the square is the origin. Answer

16. In the first line of the Eq. Gasket parameters, change R from 0.500 to -0.500.

(a) Guess what picture will occur.

(b)[C] Using these parameters, run DET starting from a single point. Can you explain what results?

(c)[C] Try other changes in R and S in DET.

17. In the Koch Curve parameters, change all R and S values from 0.333 to -0.333.

(a) Guess what picture will occur.

(b)[C] Using these parameters, run DET starting from a single point. Can you explain what results?

(c)[C] Explore the impact of some other reflections.

18. One application of the deterministic algorithm for a particular three-function IFS takes the square on the left to the three squares on the right in the picture. Sketch the result of applying the same deterministic algorithm to the shape on the right. (Note: there might be more than one answer, so be sure to describe the IFS rules you use.)

19. [C] What happens if, in an IFS construction, we don't shrink the previous picture each time? Start with the original Eq. Gasket parameters. Change the "R" and "S" values all to 1.000. Try to figure out what will happen before running the program. You will want to change the size box before running. Make the value there really small, like 1, say. Run DET. Do you get what you expected? Contrast the results you just observed with what happens if you reset the "R" and "S" values to 0.900, for example. Run with size box = 1; then again with size box = 30, say. Can you characterize the differences between "no shrink" and "shrink?" (Answer: If you don't shrink, the picture always runs off the screen, whereas, any amount of shrink produces a bounded picture.)