Classic Iterated Function Systems

Lévy Diamonds

Instead of starting the construction with a isosceles right triangle, we start with an isosceles trapezoid $C_0$ with base of length 1 and the other three sides of length 1/2. Replace this trapezoid with three scaled copies whose bases lie on the equal sides of the old trapezoid. Place each new trapezoid so that it points out from the original trapezoid to get $C_1$. For the next step, repeat the process on each of the trapezoids that make up $C_1$ to get $C_2$, as illustrated in the figure below. Continue to repeat this construction on each of the scaled trapezoids.

Suppose that the original trapezoid is placed so that its base is the unit interval on the x-axis. The three scaled copies of this trapezoid are positioned as shown in the figure below.

The first scaled trapezoid must be rotated by 60°. The second scaled trapezoid is just translated so that the point at the origin is moved to the point \((1/4,\sqrt 3 /4)\). The third scaled trapezoid is rotated by −60° and then translated so that the point at the origin moves to the point \((3/4,\sqrt 3 /4)\). This yields the following IFS.

The Lévy diamonds consists of three self-similar pieces corresponding to the three functions in the iterated function system.

[Enlarge]

Angle 30
Axiom F
F —> ++F−−F−−F++

The Lévy diamonds is self-similar with 3 non-overlapping copies of itself, each scaled by the factor $r$ = 1/2. Therefore the similarity dimension, $d$, of the attractor of the IFS is the solution to

\[\sum\limits_{k = 1}^3 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/3)}}{{\log (1/2 )}} = \frac{\log 3}{\log 2} = 1.58496\]

The L-system construction is the same as iterating the IFS starting with the unit line segment as the initial set. It is instructive to observe where the endpoints of the line segments in each iteration touch the fractal. Click on the bottoms to the left to see what happens during the first 5 iterations.