Larry Riddle, Agnes Scott College

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).

Function

System

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.

Dimension

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.

- 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]