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
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.
Iterated Function System
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.
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 5k squares in B(k), the total perimeter of B(k) 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.