Agnes Scott College
Larry Riddle, Agnes Scott College

Z2 Golden Dragon




Let K = {f1, f2} be the two functions in the Golden Dragon IFS. Let G = {g1, g2} be the Z2 cyclic group. Then g1 is the identity element and g2 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 = {g1f1, g2f1, g1f2, g2f2}. 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 \({\bf{r}} = \left( \dfrac{1}{\phi}\right)^{\frac{1}{\phi}} = 0.74274\) 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 Z2 Golden dragon.




(with squares)
The IFS for the Golden Z2 dragon will consist of four functions: the original two functions from the Golden dragon IFS and the rotation of those two functions through 180°.
\({h_1}({\bf{x}}) = {g_1}{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {0.623653} & { - 0.403372} \\ {0.403372} & {0.623653} \\ \end{array}} \right]{\bf{x}}\)
   scale by r, rotate by 32.894°
\({h_2}({\bf{x}}) ={g_2}{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-0.623653} & { 0.403372} \\ {-0.403372} & {-0.623653} \\ \end{array}} \right]{\bf{x}}\)
   scale by r, rotate by −147.106°
\({h_3}({\bf{x}}) ={g_1}{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 0.376347} & { - 0.403372} \\ {0.403372} & { - 0.376347} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ \end{array}} \right]\)
   scale by r2, rotate by 133.014°
\({h_4}({\bf{x}}) = {g_2}{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 0.376347} & { 0.403372} \\ {-0.403372} & { 0.376347} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} -1 \\ 0 \\ \end{array}} \right]\)
   scale by r2, 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 g2 from the cyclic group Z2 to the set H. Because g2g2 = g1 (identity) and g2g1 = g2, 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 h1(H) = h2(H). That is because h2 does the same scaling to H as h1 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.


More details on symmetric fractals can be found here and in the book by Field and Golubitsky.



  1. Michael Field and Martin Golubitsky, "Symmetric Fractals" (chapter 7) in Symmetry in Chaos: A Search for Pattern in Mathematics, Art and Nature, Oxford University Press (Edition 1, 1992/1995) and Siam (Edition 2, 2009) [see Preview at Google Books].