What is the fractal dimension of this tetramino shape?

Question

This tetramino is used to form a fractal by copying a smaller copy of the shape into each of the four square parts that form the original shape. The picture only has the first step, but assume the final shape contains additional steps that do the same thing. What is the fractal dimension of this shape?

Here is my proposed calculation, but I am not sure if it is correct.

The fractal dimension is given by the equation $D = log(N) / log(S)$ where N is the number of scaled down shapes that comprise the new shape, and S is how much smaller the scaled down shape used for the measurement of N is.

For my measurement, the scaled down shape is a copy of the original tetramino with 1/2 height and 1/3 width. Four of these scaled down shapes fit in the new shape, so $N = 4$. The scaled down shape has 1/6 the area, so $S = \sqrt6$. Thus, $D = log(4)/log(\sqrt6)$ or approximately 1.547.

Answer 1

No, your computation is not correct. It looks to me like you've taken a well known formula for similarity dimension, namely $$ D = \frac{\log(N)}{\log(1/r)}, $$ where $r$ is the linear scaling factor in all directions, and adjusted it to account for the fact that for a similarity transformation the area scales according to the square of $r$. That's a rather cool idea and natural to explore, but I'm afraid it doesn't quite work. That formula only works for self-similar sets; it's crucial that there be a single scaling factor in all-directions.

Your image is an example of a self-affine set, consisting of four pieces scaled by the factor $1/3$ in the horizontal direction and $1/4$ in the vertical direction. Computing the dimension of a self-affine set is a much more difficult problem in general. There's a good MathOverflow discussion on this very issue.

While there's no single formula that works to compute the dimension of all self-affine sets, there are some special cases where the fractal dimension can be computed. In particular, your set is a slight variation of a McMullen Carpet first studied in 1984 by Curt McMullen. To form a McMullen carpet, choose integers $m$ and $n$ such that $1<m<n$. Then, decompose the unit square into an $m\times n$ grid and select a subset of the resulting rectangles. If $m=3$ and $n=4$, then one such decomposition might look like so:

If we perform a similar decomposition to each of the resulting rectangles and color the pieces to clarify what's what, we get a picture that looks like so:

If we did this for a while, we'd get:

Note that this process is very similar to yours. If we had started with a $3\times4$ rectangle, divided it into $12$ squares, and proceeded recursively in the same manner, we'd get exactly your set. Thus, one is an affine image of the other and they have same dimension.

The best part about McMullen's analysis is that there is a formula for the box-counting dimension. In addition to $m$ and $n$ as defined above, let $N$ denote the number of rectangles chosen to form the decomposition and let $M$ denote the number of columns with at least one chosen rectangle. Then, the box-counting dimension of the set is $$ \frac{\log M}{\log m} + \frac{\log\left(N/M\right)}{\log n}. $$

In your example, we have $$ \frac{\log 3}{\log 3} + \frac{\log\left(4/3\right)}{\log 4} \approx 1.20752. $$

Note that a McMullen carpet can be easily specified by a zero-one matrix with zeros indicating empty regions and ones indicating occupied regions. I used this Observable notebook to do exactly that to generate the pictures here.