Larry Riddle, Agnes Scott College

Terdragon Boundary

The boundary of the terdragon is the union of two self-similar pieces. The red piece is the top boundary and the blue piece (a rotation of the red piece by 180°) is the bottom boundary.

Put them together by clicking on the buttons to the left in sequence

The figure on the left below shows the first level in the construction of the terdragon. The figure to the right shows a red envelope on top and a blue envelope on the bottom.

If we go to the next level (shown in the figure on the left below), we can connect endpoints of the line segments that form level 2 of the terdragon to form an envelope around the terdragon (red on top, blue on bottom).

Convergence
Animation

Below we show the same construction for level 3 and level 4.

When this construction is continued ad infinitum, the red and blue envelopes will converge to the boundary of the terdragon.

Convergence
Animation

These envelopes are also what we get if we apply the IFS for the terdragon starting with the level 1 parallelogram shown above. You can see in the figure below that the level 2 envelope consists of three copies of the level 1 parallelogram, two of which (red and green) have been rotated by 30° and one (blue) by −90° (clockwise rotation). In the same way, the level 3 envelope is obtained by applying the terdragon IFS to the level 2 envelope.

IteratedFunctionSystem

The IFS for the top boundary is based on the scaling, rotations, and translations shown in the following figure where the scaling factor $$r = 1/ \sqrt 3$$. Each iteration replaces a line segment with a copy of this motif, alternating between placing the motif to the left and then to the right of the line segment it replaces.

This yields the following IFS for the top boundary:

 $${T_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & { - \sqrt 3 /6} \\ {\sqrt 3 / 6} & {1/2} \\ \end{array}} \right]{\bf{x}}$$ scale by $$\frac{1}{\sqrt 3}$$, rotate by 30° $${T_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/2} & { - \sqrt 3 / 6} \\ {\sqrt 3 / 6} & { -1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ \end{array}} \right]$$ scale by $$\frac{1}{\sqrt 3}$$, rotate by 150°

The IFS for the bottom boundary is based on the following figure. Again, starting at the origin, each segment is replaced by a copy of the motif alternating between left and right.

This gives the following IFS for the bottom boundary:

 $${B_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/2} & { - \sqrt 3 / 6} \\ {\sqrt 3 / 6} & { -1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ -\sqrt 3 / 6 \\ \end{array}} \right]$$ scale by $$\frac{1}{\sqrt 3}$$, rotate by 150° $${B_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/2} & { - \sqrt 3 / 6} \\ {\sqrt 3 / 6} & { 1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ -\sqrt 3 / 6 \\ \end{array}} \right]$$ scale by $$\frac{1}{\sqrt 3}$$, rotate by 30°

L-System

Angle 30
Axiom FX−−−−−−FX
F —> Z
X —> +FX−−FY+
Y —> −FX++FY−

The axiom creates two segments going in opposite directions. This will then produce both the top and bottom boundaries at the same time.

SimilarityDimension

The boundary of the terdragon consists of 2 parts, each of which is self-similar with 2 non-overlapping pieces scaled by $$r = \frac{1}{\sqrt 3}$$. Therefore the similarity dimension, d, of the attractor of the IFS is the solution to

$\sum\limits_{k = 1}^2 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/2)}}{{\log (1/\sqrt 3 )}} = \frac{\log 4}{\log 3} = 1.26186$