CSCE 441 Lecture 11

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Condensation Sets

$X_0=C$ $X_{i+1}=(\bigcup _j{F}_j(X_i))\cup C$

where $C$ is a condensation set, a "starting set" that is common to all levels of rendering.

1. Apply transformations
2. Add in condensation set

Cannot be rendered by fractal tennis; only a depth-first traversal with the condensation set added at the end of each iteration

Fractal Dimension

$\text{fractal dimension}=-\frac{\log (\text{number of transformations})}{\log (\text{scale factor})}$

• A line is a fractal like a Serpinski triangle with only two points(scale about 0.5 at each endpoint). $-\frac{\log 2}{\log \frac{1}{2}}=1$
• A square can similarly be constructed with 4 transformations $-\frac{\log 4}{\log \frac{1}{2}}=2$
• The Koch curve has dimensionality $-\frac{\log 4}{\log \frac{1}{3}}\approx 1.26$
• The Dragon curve has dimensionality $-\frac{\log 2}{\log \frac{1}{\sqrt{2}}}=2$

Fractal curves can have infinite length, but enclose a finite area:

• The Koch curve has length ${\left(\frac{4}{3}\right)}^{i}4$ at the $i$-th iteration. The length approaches $\mathrm{\infty }$ as $i\to \mathrm{\infty }$.
• The enclosed area is $\frac{\sqrt{3}}{4}\sum _{j=0}^i{\left(\frac{4}{9}\right)}^j$ at the $i$-th iteration. This is a geometric series, and its limit is $\frac{1}{1-\frac{4}{9}}=\frac{9}{5}$