Larry Riddle, Agnes Scott College

## Sierpinski n-gons

#### Description

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).

#### IteratedFunctionSystem

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.

#### SimilarityDimension

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]

#### Area

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.

#### Variations

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.

#### References

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]