Larry Riddle, Agnes Scott College

## Hexagon as the attractor for the Z3 Sierpinski Triangle

The IFS for the Sierpinski triangle is
 $${f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}}$$ scale by 1/2 $${f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ 0 \\ \end{array}} \right]$$ scale by 1/2 $${f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/4} \\ {\sqrt{3}/4} \\ \end{array}} \right]$$ 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.

 $${h_1}({\bf{x}}) = {r_0}\,{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}}$$ no rotation $${h_2}({\bf{x}}) = {r_{120}}\,{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & -\sqrt{3}/4 \\ \sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}}$$ rotate by $$120^\circ$$ $${h_3}({\bf{x}}) = {r_{240}}\,{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & \sqrt{3}/4 \\ -\sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}}$$ rotate by $$240^\circ$$ $${h_4}({\bf{x}}) = {r_0}\,{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ 0 \\ \end{array}} \right]$$ no rotation $${h_5}({\bf{x}}) = {r_{120}}\,{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & -\sqrt{3}/4 \\ \sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/4} \\ \sqrt{3}/4 \\ \end{array}} \right]$$ rotate by $$120^\circ$$ $${h_6}({\bf{x}}) = {r_{240}}\,{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & \sqrt{3}/4 \\ -\sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/4} \\ -\sqrt{3}/4 \\ \end{array}} \right]$$ rotate by $$240^\circ$$ $${h_7}({\bf{x}}) = {r_0}\,{f_3}\,({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/4} \\ \sqrt{3}/4 \\ \end{array}} \right]$$ no rotation $${h_8}({\bf{x}}) = {r_{120}}\,{f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & -\sqrt{3}/4 \\ \sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/2} \\ 0 \\ \end{array}} \right]$$ rotate by $$120^\circ$$ $${h_9}({\bf{x}}) = {r_{240}}\,{f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {-1/4} & \sqrt{3}/4 \\ -\sqrt{3}/4 & {-1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/4} \\ -\sqrt{3}/4 \\ \end{array}} \right]$$ rotate by $$240^\circ$$
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 Z3 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.

More details about symmetric fractals can be found here.