Hexagon as the attractor for the $Z_3$ Sierpinski Triangle
The IFS for the Sierpinski triangle is
\(
f_1(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
\)
scale by 1/2
\(
f_2(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1/2\\
0
\end{array}
\right]
\)
scale by 1/2
\(
f_3(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1/4\\
\sqrt{3}/4
\end{array}
\right]
\)
scale by 1/2
If the elements in the \(Z_3\) cyclic group \(\{r_0, r_{120}, r_{240}\}\) consisting of rotations by \(0^\circ\), \(120^\circ\), and \(240^\circ\), respectively, are composed with the functions in the Sierpinski IFS, then the resulting IFS will have the following 9 functions, all of which involve a scaling by 1/2 and the following rotations.
\(
h_1(\mathbf{x})=r_0 f_1(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
\)
no rotation
\(
h_2(\mathbf{x})=r_{120}f_1(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & -\sqrt{3}/4\\
\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
\)
rotate by 120°
\(
h_3(\mathbf{x})=r_{240}f_1(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & \sqrt{3}/4\\
-\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
\)
rotate by 240°
\(
h_4(\mathbf{x})=r_0f_2(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1/2\\
0
\end{array}
\right]
\)
no rotation
\(
h_5(\mathbf{x})=r_{120}f_2(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & -\sqrt{3}/4\\
\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
-1/4\\
\sqrt{3}/4
\end{array}
\right]
\)
rotate by 120°
\(
h_6(\mathbf{x})=r_{240}f_2(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & \sqrt{3}/4\\
-\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
-1/4\\
-\sqrt{3}/4
\end{array}
\right]
\)
rotate by 240°
\(
h_7(\mathbf{x})=r_0f_3(\mathbf{x})=
\left[
\begin{array}{cc}
1/2 & 0\\
0 & 1/2
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1/4\\
\sqrt{3}/4
\end{array}
\right]
\)
no rotation
\(
h_8(\mathbf{x})=r_{120}f_3(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & -\sqrt{3}/4\\
\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
-1/2\\
0
\end{array}
\right]
\)
rotate by 120°
\(
h_9(\mathbf{x})=r_{240}f_3(\mathbf{x})=
\left[
\begin{array}{cc}
-1/4 & \sqrt{3}/4\\
-\sqrt{3}/4 & -1/4
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1/4\\
-\sqrt{3}/4
\end{array}
\right]
\)
rotate by 240°
The attractor \(A\) for this IFS of nine functions is the unique set satisfying
\[A = \bigcup_{n=1}^9 h_n(A)\]
So we just need to show that a regular hexagon with sides of length 1 and one vertex at (1,0) will satisfy this condition. This hexagon is shown below in black. Click on each of the buttons to the left to successively build up the union from function 1 to function 9. Each image set \(h_n(A)\) will be shown in color, with an "L" representing how the original hexagon was rotated. It takes the union of all 9 sets to completely cover the original hexagon.
Here is an image of the $Z_3$ Sierpinski triangle drawn with pixel coloring. The overlapping of the nine scaled images shown above contributes to how some pixels are hit much more often than others when the fractal is drawn using the random algorithm.
Click the following link for more details about symmetric fractals
