Classic Iterated Function Systems

Box Fractal

Start with a solid (filled) square $B_0$. Divide this into 9 smaller congruent squares. Keep the four corner squares and the middle squares but remove the other four squares to get $B_1$. Now subdivide each of the five remaining solid squares into 9 congruent squares and remove the corresponding set of 4 squares to get $B_2$. Continue to repeat the construction to obtain a decreasing sequence of sets

$$ B_0 \supset B_1 \supset B_2 \supset B_3 \supset \cdots $$

The box fractal is the intersection of all the sets in this sequence, that is, the set of points that remain after this construction is repeated infinitely often. The figures below show the first three iterations.

The original square is scaled by a factor $r=1/3$. This is done 5 times followed by the necessary translations to arrange the eight squares as depicted for $B_1$ If we take the original square to be a unit square with opposite corners at (0,0) and (1,1), then the IFS would be given by the following functions.

\[ \begin{align} f_1 ({\bf{x}}) &= \left[ {\begin{array}{*{20}c} {1/3} & 0 \\ 0 & {1/3} \\ \end{array}} \right]{\bf{x}} \\ f_2 ({\bf{x}}) &= \left[ {\begin{array}{*{20}c} {1/3} & 0 \\ 0 & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}c} 0 \\ {2/3} \\ \end{array}} \right] \\ f_3 ({\bf{x}}) &= \left[ {\begin{array}{*{20}c} {1/3} & 0 \\ 0 & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}c} {1/3} \\ {1/3} \\ \end{array}} \right] \\ f_4 ({\bf{x}}) &= \left[ {\begin{array}{*{20}c} {1/3} & 0 \\ 0 & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}c} {2/3} \\ 0 \\ \end{array}} \right] \\ f_5 ({\bf{x}}) &= \left[ {\begin{array}{*{20}c} {1/3} & 0 \\ 0 & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}c} {2/3} \\ {2/3} \\ \end{array}} \right] \\ \end{align} \]

The box fractal consists of 5 self-similar pieces corresponding to the five functions in the iterated function system.

[Enlarge]

Angle 90
Axiom F+F+F+F
F —> F+F−F−F+F

This will produce the outline of the box fractal.

     

The box fractal is self-similar with 5 non-overlapping copies of itself, each scaled by the factor $r = 1/3$. Therefore the similarity dimension, $d$, of the unique attractor of the IFS is the solution to

$$\sum\limits_{k = 1}^5 {{r^d}} = 1\quad \Rightarrow \quad d = \frac{{\log (1/5)}}{{\log (r)}} = \frac{{\log (1/5)}}{{\log (1/3)}} = \frac{{\log (5)}}{{\log (3)}} = 1.46497$$

Suppose the original square $B_0$ has sides of length 1. Each square in $B_k$ has side length $(1/3)^k$ and thus perimeter $4(1/3)^k$. Since there are $5^k$ squares in $B_k$, the total perimeter of $B_0$ is $4(5/3)^k$. In addition, each square in $B_k$ has area $(1/3)^{2k}$ and thus the total area of $B_k$ is $(5/9)^k$. Thus in the limit as $k$ goes to infinity, the perimeter of $B_k$ goes to infinity while the area of $B_k$ goes to 0. In this sense, the box fractal has infinity perimeter but no area.

The box fractal is one example of a fractal antenna that is used in cell phones to maximize reception in a minimum amount of space.

At each iteration remove the 4 corner squares rather than the four squares along the middle of each side. This will yield the box fractal rotated by 45 degrees.