Agnes Scott College
Larry Riddle, Agnes Scott College

Sierpinski Carpet Relatives


The construction of these fractals is motivated by the web page "Square divided by nine" by Robert Fathauer, part of his website on Gasket Fractals. Start with a square rotated by 45°. Divide this into 9 smaller congruent squares. If the middle square is removed and the construction is iterated on the remaining squares, the result is the Sierpinski Carpet, but rotated 45°.

rotatedcarpetdesign   rotatedcarpet

The image on the right shows the basic design of the remaining initial squares considered to have an upward orientation. But suppose the top square is removed rather than the middle square. Iterating that construction would produce a different fractal.

removetopdesign   removetop

In general, we can choose to remove from 1 to 8 squares to form the basic design. In addition, however, we can also choose to rotate any of the remaining squares by 90° counterclockwise, by 180°, or by 270° counterclockwise (i.e. 90° clockwise) as that will maintain their relative positions, but change their orientation. For example, suppose only the top square is removed as before, but now the three squares in the middle row are rotated by 90°, 180°, and 270° counterclockwise respectively as illustrated in the design below along with the first four iterations. View the animation to see how the arrows are rotated during these four iterations, and view the video to see how the eight scaled hearts fill their respective squares while points are plotted using the chaos game algorithm with the IFS described below.


Chaos Game

heart1 heart2 heart3 heart4

Because we are only interested in symmetric fractals, we will always require that the basic design be symmetric with respect to a vertical line through the middle.


code Take the initial square to have diagonals of length 1 and bottom vertex at the origin. If the nine squares are labeled from 1 to 9 as shown to the right, then each of the functions corresponding to a particular square can be indexed by the same number as the triangle. This gives the nine functions below. The translation vectors are determined by how the bottom vertex of square 1 has to be moved to the bottom vertices of the other eight squares. Each gridline corresponds to 1/6.

Use the dropdown menus to chose 0°, 90°, 180°, or 270° depending on how a square is to be rotated. If a square is to be deleted, choose "Remove" from the menu. Finally, click on the "Create IFS" button. The IFS for that fractal will consist of the functions below whose indices match those of the remaining squares and for which the appropriate angle of rotation is specified.

IFS Video
For example, if the top square is removed (square 9) as shown in the "heart" above, and \(\theta_4=90\), \(\theta_5 = 180\), and \(\theta_6 = 270\), then the IFS for the resulting fractal is the set \(\{f_1,f_2,f_4,f_5,f_6,f_7,f_8\}\) with the corresponding rotation angles.


1.   \({f_1}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0} \\ {0} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_1\) =  
2.   \({f_2}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/6} \\ {1/6} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_2\) =  
3.   \({f_3}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/6} \\ {1/6} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_3\) =  
4.   \({f_4}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/3} \\ {1/3} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_4\) =  
5.   \({f_5}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0} \\ {1/3} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_5\) =  
6.   \({f_6}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/3} \\ {1/3} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_6\) =  
7.   \({f_7}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {-1/6} \\ {1/2} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_7\) =  
8.   \({f_8}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/6} \\ {1/2} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_8\) =  
9.   \({f_9}({\bf{x}}) =\displaystyle \left[ {\begin{array}{*{20}{c}} {1/3} & {0} \\ {0} & {1/3} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {0} \\ {2/3} \\ \end{array}} \right]\)    scale by 1/3,  rotate by \(\theta_9\) =  


Suppose \(m\) squares are removed at each iteration of the construction, where \(1 \le m \le 7\) (if you remove 8 squares, leaving only one, the limit of the iteration will be a single point.) The resulting fractal is self-similar with \((9-m)\) non-overlapping copies of itself, each scaled by the factor r = 1/3. Therefore the similarity dimension, d, of the attractor of the IFS is the solution to

$$\sum\limits_{k = 1}^{9-m} {{r^d}} = 1\quad \Rightarrow \quad d = \frac{{\log (1/(9-m))}}{{\log (r)}} = \frac{{\log (9-m)}}{{\log (3)}}$$


How many examples are possible? There are \(\displaystyle \binom{9}{k}\) possible choices for how to remove \(k\) squares. Each of the remaining \(9-k\) squares can be rotated by one of 4 different angles (0, 90, 180, or 270). Assuming that only between 1 and 7 squares are removed, the total number of possibilities is \[\sum_{k=1}^7 \binom{9}{k}4^{9-k} = 1,690,944.\] However, this counts many designs that are not symmetric with respect to a vertical line through the middle of the initial square. In addition, some designs produce the same fractal and therefore should only be counted once. For example, if only the middle square is removed (square 5 in the illustration above), then the resulting fractal will be the Sierpinski carpet regardless of how the other eight squares are rotated (and there are 48=65536 different ways to rotate those eight squares!). Finally, some of the fractals are just reflections across the horizontal line through the middle of the original square and therefore not really distinct images. For example, the following two fractals are such reflections. On the left, squares 7 and 8 were removed at each iteration and five other squares were rotated, while on the right, squares 2 and 3 were removed at each iteration and five other squares were rotated. Click on the fractal on the right to see its reflection. We would not want to count these as separate examples.

1    3
1    3

[Another view]
[Another view with squares]
[IFS video demonstration]

Click on "Another view" to see a version of the fractal above that was drawn using the chaos game algorithm to plot over 350,000 points which were colored using the idea of "color stealing" developed by Michael Barnsley. The fractal consists of seven self-similar pieces which are easy to spot when each piece is colored differently. Click on "Another view with squares" to see the same image but with the nine squares superimposed on top of the fractal, with each self-similar piece contained entirely within a square. Finally, click on "IFS video demonstration" to see the first five iterations of the IFS construction of the fractal.

IFS Video
Here is an example where the top three squares are removed and three other squares are rotated as illustrated in the diagram. Click on the fractal for a larger view done in cross stitch.

flower design flowers

The following chart shows the total number of distinct symmetric fractals when removing from 1 to 7 squares at each iteration for a total of 1086 possible fractals [Details].

squares removed1234567
number of fractals257326230162653511



Suppose that the area of the initial right square is equal to 1 and that m subsquares are removed at each iteration of the construction. How much of the area of the original square is removed in the limit? At the first iteration, each subsquare that is removed has area 1/9, so the total area removed is m/9. There are (9-m) squares left. Each of these have m subsquares of area (1/9)2 removed. So after two iterations, the total area removed is \[\frac{m}{9}+(9-m)\frac{m}{9^2}.\] Continuing the same way with the additional iterations would produce the geometric series \[\sum_{m=1}^{\infty}(9-m)^{m-1}\frac{m}{9^m} = \frac{\frac{m}{9}}{1-\frac{9-m}{9}} = \frac{\frac{m}{9}}{1-1+\frac{m}{9}}=1.\] All of the area has been removed, just as with the Sierpinski carpet!

The two variations on the Sierpinski carpet that come from dividing the rotated square into nine subsquare pieces and dividing a right isosceles triangle into nine triangular pieces have some similarities in that the right isosceles triangle is half of the rotated square, and pairs of triangular pieces make up a subsquare. So it is not surprising that the family of Sierpinski carpet relatives shares some common fractals with the family that comes from subdividing the right isosceles triangle. Here are two examples.

square3ex1   square3ex2
square3ex1design  square3ex2design
triangle3ex1design   triangle3ex2design

The second row shows the design as a Sierpinski carpet relative where the bottom three squares have been removed and squares 7, 8, and 9 have been rotated. The third row shows the design when starting with the isosceles right triangle divided into nine pieces with three being removed. The reason these give the same fractal is that each has an iterated function system with the same scaling and rotation matrices, but just with different translation vectors.


  1. For other examples of similar gasket fractals, see Robert Fathauer's website at