Let $K = \{f_1,f_2\}$
be the two functions in the
Golden Dragon IFS. Let
$G = \{g_1,g_2\}$
be the $Z_2$ cyclic group where $g_1$ is the identity element and $g_2$ is rotation through 180°. We form a new IFS by applying each of the symmetries in G to each of the functions in K to form the set of affine transformations
$GK = \{g_1f_1,g_2f_1,g_1f_2,g_2f_2\}$.
The first iteration would transform the horizontal line segment into the following figure using the four functions in this set. The red line segments would be the results of the usual Golden dragon IFS and the blue line segments are the rotations by 180°.
Recall that \(r = \left( \dfrac{1}{\phi}\right)^{\frac{1}{\phi}} = 0.74274\) where \(\phi\) is the golden ratio, and that A = 32.893818° and B = 46.98598225°. Now continue to repeat the construction on each of the four new segments to get the $Z_2$ Golden dragon.
[Enlarge]
The IFS for the $Z_2$ Golden dragon will consist of four functions: the original two functions from the Golden dragon IFS and the rotation of those two functions through 180°.
\(
h_1(\mathbf{x})=
g_1f_1(\mathbf{x})=
\left[
\begin{array}{cc}
0.623653 & -0.403372\\
0.403372 & 0.623653
\end{array}
\right]\mathbf{x}
\)
scale by $r$, rotate by 32.894°
\(
h_2(\mathbf{x})=
g_2f_1(\mathbf{x})=
\left[
\begin{array}{cc}
-0.623653 & 0.403372\\
-0.403372 & -0.623653
\end{array}
\right]\mathbf{x}
\)
scale by $r$, rotate by −147.106°
\(
h_3(\mathbf{x})=
g_1f_2(\mathbf{x})=
\left[
\begin{array}{cc}
-0.376347 & -0.403372\\
0.403372 & -0.376347
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
1\\
0
\end{array}
\right]
\)
scale by $r^2$, rotate by 133.014°
\(
h_4(\mathbf{x})=
g_2f_2(\mathbf{x})=
\left[
\begin{array}{cc}
0.376347 & 0.403372\\
-0.403372 & 0.376347
\end{array}
\right]\mathbf{x}
+
\left[
\begin{array}{c}
-1\\
0
\end{array}
\right]
\)
scale by $r^2$, rotate by −46.986°
If H is the unique attractor for this IFS, then
\[H = h_1(H) \cup h_2(H) \cup h_3(H) \cup h_4(H) \]
Rotating H by 180° counterclockwise is the same as applying the function $g_2$ from the cyclic group $Z_2$ to the set H. Because $g_2g_2 = g_1$ (identity) and $g_2g_1 = g_2$, we get
\[\begin{align}
{g_2}(H) &= {g_2}\left( {{h_1}(H) \cup {h_2}(H) \cup {h_3}(H) \cup {h_4}(H)} \right)\\
\\
&= {g_2}{g_1}{f_1}(H) \cup {g_2}{g_2}{f_1}(H) \cup {g_2}{g_1}{f_2}(H) \cup {g_2}{g_2}{f_2}(H)\\
\\
&= {g_2}{f_1}(H) \cup {g_1}{f_1}(H) \cup {g_2}{f_2}(H) \cup {g_1}{f_2}(H)\\
\\
&= h_2(H) \cup h_1(H) \cup h_4(H) \cup h_3(H) = H
\end{align}\]
This shows that H has 180° rotational symmetry. Also notice that $h_1(H) = h_2(H)$. That is because $h_2$ does the same scaling to H as $h_1$ but the two rotations differ by exactly 180°. Because of the 180° rotational symmetry of H, however, the result will be the same in both cases. You can see what happens by clicking on each of the buttons to the left to cover H with the four scaled and rotated copies.
Click for more details on symmetric fractals or see the book by Field and Golubitsky.
