In this episode, Charlie uses fractal dimension to detect forged comics. He argues that forgeries take longer to draw, causing ink to seep out in a way that increases "wrinkliness", which can be measured by fractal dimension. How appropriate, then, that the show features renowned travellers through the dimensions of time (Christopher Lloyd, Back to the Future) and space (Wil Wheaton, Star Trek.) Let's take a closer look at fractal dimension. There are many ways of associating a number called a dimension to a fractal object. Here let's focus on the Hausdorff dimension. The main idea is to generalize 1-dimensional length, 2-dimensional area, and 3-dimensional volume to d dimensions, where d does not have to be an integer! First, we must determine some properties this d-dimensional Hausdorff measure should have.
(1)A basic property of length, area, and volume is that it is additive. If we break a figure up into (reasonable) separate pieces, then the length, area, or volume of the whole figure is equal to the result of adding the measures of the pieces. It is reasonable to think our d-dimensional measure should satisfy this same property.
(2)If we multiply the sides of a rectangle by r, the area is multiplied by r^{2} (r*r). If we multiply the sides of a box by r, the volume is multiplied by r^{3} (r*r*r). This reflects the fact that area is 2-dimensional, and volume is 3-dimensional. So d-dimensional measure "should" satisfy the property that when a figure is scaled by a factor of r, its d-dimensional measure is multiplied by r^{d}.
Hausdorff proved that there is a consistent way of doing the above. Now we determine dimension the same way Goldilocks chooses porridge. If we try to measure a filled-in rectangle with a 3-dimensional measure (volume), we find that it is too small (0 volume) because it has no thickness. On the other hand, if we split the rectangle into (infinitely many) horizontal line segments, we find that the length is too big (infinite). The area measure is "just right", providing a finite positive measure.
For the fractal objects we will consider on this webpage, there will be precisely one number d≥0 where the d-dimensional measure is positive, but not infinite. For e<d, the e-dimensional measure will be infinite. For e>d, the e-dimensional measure will be 0. Our task will be to find the Hausdorff dimension, the d that is "just right".
Remove the middle third, but include the endpoints so that you now have 2 closed intervals each of length one third.
Remove the middle thirds of each of these two intervals, to form 4 smaller intervals.
Continue this process indefinitely.
The points that are never removed at any step in this process form the Cantor Set. What is its fractal dimension?
The key observation here is that the Cantor Set can be broken up into two pieces, the left side and right side, which are scaled down versions of the Cantor Set itself! The scale factor is 1/3.
Let M_d(a figure) denote the d-dimensional measure of that figure.
By property (1) above, M_d(Cantor Set)=M_d(left piece)+M_d(right piece).
By property (2), M_d(left piece)=M_d(right piece)=M_d(Cantor Set) * (1/3)^{d}
Putting these equations together, M_d(Cantor Set)=M_d(Cantor Set) * 2 * (1/3)^{d}
If d is the Hausdorff dimension, then M_d(Cantor Set) is just right - a nonzero finite number. So we can divide both sides of the equation by it to obtain:
1=2*(1/3)^{d}, or in other words 3^{d}=2.
In high school algebra, one learns that equations like this are solved via logarithms: d=log(2)/log(3).
Compute the Haudorff dimensions of the following fractals:
Step 1) Start with an equilateral triangle. Divide it into 4 congruent subtriangles, and remove the middle one, as in the Zelda Triforce.
Power! Wisdom! Courage! |
Step 2) For each remaining triangle, perform a scaled down version of step 1.
Step 3) Repeat indefinitely.
The points never removed in this process form Sierpinski's triangle.
Bonus: If you know Pascal's triangle, see if you can find a connection between it and Sierpinski's.
This is a piece of the more familiar Koch snowflake (which is obtained by arranging three of these to form a "triangle".) Start with a closed unit interval [0,1]. Remove the middle third, as when forming the Cantor set, but this time replace it with the other two sides of an equilateral triangle that could be placed on top of this middle segment, as in the figure below. Now, with the four segments of length 1/3, perform the same procedure. Repeating this process and taking the "limit" yields the Koch curve.
Estimate the Hausdorff dimension of a real-world curve, by measuring its length at several different scales. Using your data, find the dimension that best fits Hausdorff's conjecture. This could be something small, like a pen stroke, or something quite large. (If your curve is small, you will need to use a microscope.) Try to find something you can break into pieces, so that a group of people can cooperate in taking the length measurements for the smaller scales (i.e., the scales with greater precision).