CSCE 441 Lecture 11

« previous | Friday, February 7, 2014 | next »

Condensation Sets

$X_{0}=C$ $X_{i+1}=\left(\bigcup _{j}F_{j}(X_{i})\right)\cup C$

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

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

• 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 $\infty$ as $i\to \infty$.
• 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 ${\frac {1}{1-{\frac {4}{9}}}}={\frac {9}{5}}$