Sierpinski n-gons

By changing the initial polygon and scale factor, it is possible to generate other types of fractals. The idea is to start with a regular n-sided polygon, then scale the polygon by a factor r so that n copies of the scaled polygon exactly fit inside the original polygon. The Sierpinski pentagon is such an example with n = 5 and r = 0.381966. The figure below shows the first iteration for a hexagon and the limit set.

Here is the limit set starting with an octagon (n = 8 and r = 0.292893)

and the limit set starting with a regular polygon with 15 sides (n = 15 and r = 0.172909).

The scale factor \(r_n\) for an n-gon is

\[r_n = \frac{1}{{2 \; \left( {1 + \sum\limits_{k = 1}^{\left\lfloor {n/4} \right\rfloor } {\cos \left( {\frac{{2\pi k}}{n}} \right)} } \right)}}\]

where \({\left\lfloor {n/4} \right\rfloor }\) denotes the greatest integer less than or equal to n/4. As n increases, the scaling factor \(r_n\) converges to 0 and the Sierpinski n-gons converge to a circle.

The only affine transformations needed for the iterated function system is scaling by \(r_n\) and translations. The translations can be taken to be

\[\left[ {\begin{array}{*{20}{c}} {(1 - r_n)w\cos \left( {\frac{{2\pi k}}{n}} \right)} \\ {(1 - r_n)w\sin \left( {\frac{{2\pi k}}{n}} \right)} \\ \end{array}} \right]\]

for k = 1 to n, where w is the radius of the initial polygon, that is, the distance from the center of the polygon to each vertex.

Click here for the details.

The Sierpinski n-gon is self-similar with n non-overlapping copies of itself, each scaled by the factor \(r_n \lt 1\). Therefore the similarity dimension, d, of the attractor of the IFS is the solution to

\[\sum\limits_{k = 1}^n {{r^d}} = 1 \quad \Rightarrow \quad d = \frac{{\log (1/n)}}{{\log (r_n)}} \]

Here is a graph of the dimension of the n-gon for n from 3 to 30.

As n goes to infinity, the dimension converges to 1. [Details]

Suppose the area of the original n-gon G(0) is equal to 1, where \(n>4\). To get the (m+1)th iteration G(m+1), we scale the mth iteration G(m) by \(r_n\), which reduces the area by \(r_n^2\). But we make n copies of this scaled version to form G(m+1). Therefore the area of G(m+1) must be \(n\cdot r_n^2\) of the area of G(m). This means that the area of G(m) is \( (n \cdot r_n^2)^m\) for all m.

Here is a graph of \(n \cdot r_n^2\) for n from 3 to 16.

Notice that \(n \cdot r_n^2 \lt 1\) for \(n > 4\). Therefore the area of G(m) will go to 0 as m goes to infinity. This means that we have removed "all" of the area of the original n-gon in constructing the Sierpinski n-gon, for \(n > 4\). But of course there are many points still left in the fractal. That is one reason why area is not a useful dimension for this set.

The exception is n = 4. This value of n corresponds to starting with a square and dividing the square into 4 pieces that exactly fit within the original square. In this case, \(4 \cdot r_4^2 = 1\) and no area is removed at any stage in the iteration because each iteration is still the original square.

Rather than use the scaling ratio \(r(n)\) given above for an n-gon, one could continue to use the ratio r = 1/2 as with the original Sierpinski gasket. In this case the scaled copies will overlap. This makes them ideal candidates to color using pixel counting. Here are two examples with a pentagon and a hexagon. Click on each for a larger view.

  

 

1. Schlicker, Steven and Kevin Dennis. "Sierpinski n-gons," Pi Mu Epsilon Journal, 10 (1995), No. 2, 81-89. [Available at Pi Mu Epsilon Journal Past Issues]