In this section, we go through some very simple geometric fractals to give you a taste of how recursion really works. If you are familiar with exponents and logarithms, note that many characteristics of these fractals grow either exponentially or logarithmically. I highly suggest you scour the web for fractal applets - they are really helpful!
Note: I gave names to some of these fractals for ease of description, but you may find a different name somewhere else.

Binary Line

This is the fractal that was described on the front page in brief. It is a really simple fractal to draw and it exhibits exactly how self-similarity forms. The rule is that you draw a line, then draws two other lines half its length perpindicular on each end. This continues to inifinity.

The reason you can say that this has a fractional dimension is that the length happens to be infinite, but the volume occupied is finite.

Blocks

This is another simple fractal that happens to have a a neat optical illusion to it. It appears to be a bunch of curly brackets around a square. What is really going on is that you start with a square, then draw four squares half that size on each corner. Continue this to infinity.

Van Koch Curve/Snowflake

This is the curve mentioned in The Practical Fractal. Yet another simple curve with some bizzarre properties. Here is how it is generated: start with a line and divide it into three equal sections. Remove the middle section, and add two lines in its place to make an equilateral triangle. Do this for every line on each iteration. The Van Koch Snowflake is this exact curve only you start with an equilateral triangle rather than a line segment.

Let's take a look at this fractal for a moment and observe its properties. If we iterate the equation to infinity (as you do with nearly every fractal), the length of this curve is infinite. Why? Because after every iteration, each line segment will be 4/3 the length of segment before it.
So what about the area of the snowflake - is that infinite? Since you iterate by adding triangles to the figure, one would think that the area is infinite as well. As it turns out, this figure covers a finite area.

Let the original side of the Koch Snowflake be 1.
Area[0] is the original equilateral triangle with a side of 1.

So the whole series is:

By taking the limit of the sum of that series, we get that sequence to be , so we find that the area under a Van Koch Snowflake with a side of 1 would be *.
So basically, if you had a true Van Koch Snowflake on a piece of paper, you could color it in a certain amount of time, but it would take you forever to outline it! How does that work?

Page last updated: February 04, 2008