Larry Riddle, Agnes Scott College

Construction

Animation

Begin with a line segment. For the first iteration, replace this
segment with two segments, the first scaled by **r** and the second scaled by **r**^{2}, where **r** is equal to (1/φ)^{(1/φ)} with φ being the golden ratio, i.e. \(\varphi = \frac{{1 + \sqrt 5 }}{2}\).
The new segments are positioned in such a way that
the original segment would have been the third side of the triangle formed by the new segments. Following along the original segment, we place the
two new segments to the left. For the second iteration, replace the first segment with two new segments scaled in order by **r** and **r**^{2} and placed to the left. Replace the second segment from the first iteration with two new segments scaled in order by **r**^{2} and **r** and placed to the right. Continue this
construction, always alternating the new segments between left and
right along the segments of the previous iteration, and alternating the scaling order between (**r**, **r**^{2}) and (**r**^{2}, **r**). This generates
the "golden dragon curve". The following figure shows the first three
iterations for this construction.

Function

System

To think of this construction as a result of affine
transformations, consider the first iteration of the line segment
from (0,0) to (1,0) as shown by the two red line segments in the
following figure.

Remember that \(r = {\left( {\dfrac{1}{\varphi }} \right)^{\frac{1}{\varphi }}} = 0.74274\) where \(\varphi = \dfrac{{1 + \sqrt 5 }}{2} = 1.618033988\).

We can use the Law of Cosines to determine the value of the angle A.

\[{r^4} = 1 + r^2 - 2r\cos A \Rightarrow \cos A = \frac{{1 + r^2 - {r^4}}}{{2r}} = 0.839678466\]
Therefore A = cos^{−1}(0.839678466) = 32.893818°. A similar calculation with the Law of Cosines for angle B shows that

Because of the need to alternate left and right when constructing
the dragon, we need to think of constructing the second line segment
by a rotation through 180°−B = 133.0140178° followed by a translation of the origin
to the point (1,0). This yields the following IFS

\({f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{0.62367} & { - 0.40337} \\
{0.40337} & {0.62367} \\
\end{array}} \right]{\bf{x}}\) |
scale by r, rotate by 32.893818° |

\({f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ - 0.37633} & { -0.40337} \\
{0.40337} & { -0.37633} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
1 \\
0 \\
\end{array}} \right]\) |
scale by r^{2}, rotate by 133.0140178° |

The attractor of this IFS will be the golden dragon curve. The golden dragon consists of two self-similar pieces corresponding to the two functions in the IFS.

Dimension

The golden dragon is self-similar with 2 non-overlapping copies of itself, one scaled by the factor
**r** and the other by the factor **r**^{2}. Therefore the similarity dimension, **d**, of the
attractor of the IFS is the unique solution to \({r^d} + {r^{2d}} = 1\). To see that **d** is the golden ratio φ, you need to first recall that φ satisfies φ^{2}−φ−1=0, and hence φ^{2}=φ+1. Therefore
\[{r^\varphi } + {r^{2\varphi }} = \frac{1}{\varphi } + \frac{1}{{{\varphi ^2}}} = \frac{{\varphi + 1}}{{{\varphi ^2}}} = 1\]

Properties

If two copies of the golden dragon are drawn vertically, with one reflected horizontally across a vertical line, they fit together nicely to form the following picture.

- Golden Ratio at Wikipedia.