Classic Iterated Function Systems

Durer's Pentagons

"Tile Patterns Formed by Pentagons" by Albrecht Durer in The Painter's Manual, 1525.

"Fifth, you can combine pentagons in the following manner: First draw a pentagon and place pentagons of the same size on each side. Then place five pentagons on their sides, particularly along the two sides. This will result in the formation of five narrow lozenges between them. Then add pentagons in the angles which will have formed, so that these will touch the narrow lozenges with their corners. You can continue in this manner as long as you desire."

To convert this into an iterated function system, image starting with the initial pentagon and scaling by the same factor $r$ = 0.382 used for the Sierpinski pentagon. This time, however, translate six copies so that they would fit exactly inside the initial pentagon. The copy in the middle must first be rotated by 180°. Now repeat the construction on each of the six pentagons to get the second iteration. Continue indefinitely.

The IFS for this construction is the same as the one for the Sierpinski pentagon with the addition of a sixth function for the scaled polygon in the middle. The scale factor is \(r = \frac{{3 - \sqrt 5 }}{2}\) = 0.381966. The sixth function includes a rotation by 180°.

\[\begin{align} {f_1}({\bf{x}}) &= \left[ {\begin{array}{cc} {0.382} & 0 \\ 0 & {0.382} \\ \end{array}} \right]{\bf{x}} \\ {f_2}({\bf{x}}) &= \left[ {\begin{array}{cc} {0.382} & 0 \\ 0 & {0.382} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{c} {0.618} \\ 0 \\ \end{array}} \right] \\ {f_3}({\bf{x}}) &= \left[ {\begin{array}{cc} {0.382} & 0 \\ 0 & {0.382} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{c} {0.809} \\ {0.588} \\ \end{array}} \right] \\ {f_4}({\bf{x}}) &= \left[ {\begin{array}{cc} {0.382} & 0 \\ 0 & {0.382} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{c} {0.309} \\ {0.951} \\ \end{array}} \right] \\ {f_5}({\bf{x}}) &= \left[ {\begin{array}{cc} {0.382} & 0 \\ 0 & {0.382} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{r} { - 0.191} \\ {0.588} \\ \end{array}} \right] \\ {f_6}({\bf{x}}) &= \left[ {\begin{array}{cc} {-0.382} & 0 \\ 0 & {-0.382} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{c} { 0.691} \\ {0.951} \\ \end{array}} \right] \\ \end{align}\]

Scaling and IFS Details are the same as for the Sierpinski pentagon.

The Durer fractal is self-similar with 6 non-overlapping copies of itself, each scaled by the factor $r \lt 1$. Therefore the similarity dimension, $d$, of the attractor of the IFS is the solution to

\[\sum\limits_{k = 1}^6 {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/6)}}{{\log (r)}} = \frac{{\log (6)}}{{\log (2/(3 - \sqrt 5 )}} = 1.86172\]