Comp 2673
Homework 1 Solutions
Fractals
  1. Draw the 4th iteration of a Koch's curve. Is this a fractal? To answer this question, see which of the 4 properties of fractals it has.


    This not a fractal:
  2. Showing that the length of the Koch's curve is infinite would indicate that its dimension should be more than 1. Show that the length of Koch's curve is indeed infinite (find a formula for the length of the 1st, 2nd, 3rd,...,nth iteration, and take the limit as n goes to ∞).

    Iteration:    Length:
    0 1
    1 4/3
    2 16/9 = (4/3)²
    n (4/3)n
    lim n → ∞ (4/3)n = ∞
    Length goes to infinity.

  3. Showing that the area of the Sierpinski gasket is 0 would indicate that its dimension should be less than 2. Show that the area of the Sierpinski gasket is 0. Do this by showing that the area of the removed triangles totals to the area of the original triangle. Do this by calculating the total area removed at the 1st, 2nd, 3rd, ..., nth step, then take the limit as n goes to infinity.

    Iteration:     Area remaining:
    0 1
    1 3/4
    2 9/16 = (3/4)2
    n (3/4)n
    lim n → ∞ (3/4)n = 0
    Area goes to zero.

    You can also use the area removed from the triangle, and sigma notation, to show that the area removed goes to 1, the same as the starting area.
    Iteration:     Area removed:
    0 0
    1 1/4
    2 1/4 + 3/16
    3 1/4 + 3/16 + 9/64
    n Σi=1 to n (3i-1)/(4i)
    lim n → ∞ ( Σi=1 to n (3i-1)/(4i) ) = 1
    Area removed goes to one, the original area, so all area is removed.

  4. Calculate the scaling dimension of the Sierpinski gasket.

    The formula for calculating dimension is: (magnification factor)dimension = size change

    From the image, we can see that if we magnify the fractal by two, we need three copies of the original fractal to make the new one.
    Thus, we have: 2d = 3 ⇒ d = ln3/ln2 ≈ 1.585

  5. Here's the description of an iterated function system. The iterated function sytem has 3 transformations:
    1. Start with ANY square non-blank image of your choice. Use a scaling photocopy machine, scissors and tape to create a 3rd iteration of the above iterated function system. Turn in the original image, the 1st, 2nd, and 3rd images.

      Sample transformation
      Iteration:       Image:       Iteration:       Image:
      0       3
      1       4
      2

    2. Before you begin, tell me how many photocopies you will have to make to get the final product (if you don't make any mistakes.)

      Each iteration takes 3 photocopies, so to get to the third iteration is a total of 9 photocopies. If you count the original as one, then 10 copies are required. Either answer is acceptable.

    3. Write down the matrices that describe these 3 linear transformations.

      There were several possible variations of these matrices. One big difference is where you consider the origin to be.
      If the origin is in the lower left of the original image, the transformations are: 
      1. 	| 1/2 1/2||x| + | 0 | = |x'|
	|-1/2 1/2||y|   |1/2|   |y'|
	
      2. 	|-1/3  0 ||x| + |2/3| = |x'|
	|  0 -1/3||y|   |1/3|   |y'|
	
      3. 	|1/3  0 ||x| + |0| = |x'|
	| 0  1/3||y|   |0|   |y'|
      If the origin is consider to be in the center of the square (for ease of rotations), then the transformations are: 
      1. 	| 1/2 1/2||x| + |0| = |x'|
	|-1/2 1/2||y|   |0|   |y'|
	
      2. 	|-1/3  0 ||x| + |  0 | = |x'|
	|  0 -1/3||y|   |-1/3|   |y'|
	
      3. 	|1/3  0 ||x| + |-1/3| = |x'|
	| 0  1/3||y|   |-1/3|   |y'|
      There may be other correct solutions in addition to these.

  6. Use the IFS from the previous problem. Starting with the point (0,0), find the image point after 3 iterations if you choose function 1, followed by function 3, followed by function 2.

    Origin lower left:     Origin centered:
    Starting point:   (0,0) (0,0)
    Function 1: (0,1/2) (0,0)
    Function 3: (0,1/6) (-1/3,-1/3)
    Function 2: (2/3,5/18) (1/9,-1/9)