Fractals - when the number of seed shapes that can fit into the scaled-up copy is non-integer.

Question

I've heard people say (for eg. here) that the dimension of fractal patterns (particularly, in this question, Lindenmayer fractals) can be formulated as follows:

$$D=\frac{\ln N}{\ln S}$$

Where $N$ is the number of copies of the seed shape that fit into a copy scaled by the scaling factor $S$.

The logic is simple and works for simple shapes of integer dimension. Take a line of length $1$ and scale it by $S=2$, and you can fit 2 original lines in the scaled up copy, giving $N=2$ and thus $D=1$, as you would expect. Take a unit square and scale it by $S=2$ and you get $N=4$ and thus $D=2$. Do same process for cube and you get $D=3$.

This logic is then extended, inductively, to patterns with non-integer dimension like the Cantor Dust, the Sierpinski gasket, the Sierpinski carpet.

The seed shape of the Sierpinski gasket for example is:

Scale it up by $S=2$ and you get:

You can easily see that 3 seed shapes fit into the scaled up copy. Thus,

$$D=\frac{\ln(3)}{\ln(2)}$$

as is well known.

Now my question: What about seed shapes which do not fit neatly into scaled up copies? For example, what if I add a little triangle in the center of the Sierpinski gasket seed shape...

And then scale up by $S=2$ again:

How many seed shapes fit into this scaled up copy? Clearly, three whole seed shapes fit, but there is also a residual triangle in the center of the copy. So $N=3+r$ and $$D=\frac{\ln(3+r)}{\ln(2)}$$

where $r$ is "some residual" of the seed shape. How to calculate $r$? This raises other questions like what exactly constitutes a "copy." What does it mean to talk about a fraction of a copy, etc.?

The logic of the first equation in this post is fun and intriguing but it is (I think) purely inductive. There is no theorem proving it. (Correct me if I'm wrong). Does the example I provide here point to a snare in the logic?

Answer 1

Standard self-similarity

Fractals are often constructed using a recursive procedure and self-similar sets in particular, are always constructed this way. I think that most folks working in fractal geometry would guess that your picture implies a recursive construction like so:

Note that the first set is an initial seed. The second set is composed of three copies of the first scaled by the factor $1/2$ (the three outer blue triangles) and one copy of the first scaled by the factor $1/4$ (the inner red-ish piece). This pattern can be analyzed as a self-similar set. In general, we have a single set composed of copies of itself scaled by the factors $r_1,r_2,\ldots,r_n$. Assuming the copies don't overlap, the dimension of the set is the unique number $s$ such that $$\sum_{i=1}^n r_i^s = 1.$$ I actually answered a question today indicating why there is a unique such $s$ and how to solve for it in general.

The dimension is the unique number $s$ such that $$\frac{3}{2^s} + \frac{1}{4^s} = 1.$$ This can be written $$\left(\left(\frac{1}{2}\right)^s\right)^2 + 3\left(\frac{1}{2}\right)^s - 1 = 0.$$ Substituting $q=(1/2)^s$, we get $$q^2+3q-1=0,$$ whose unique positive solution is $$q=\frac{-3+\sqrt{13}}{2}.$$ The dimension of your set is therefore $$\frac{\log(\frac{-3+\sqrt{13}}{2})}{\log(1/2)} \approx 1.72368.$$ As pointed out in the comments, we expect this to be larger than the fractal dimension of the Sierpinski gasket or $$\frac{\log(3)}{\log(2)} \approx 1.58496.$$

Condensation

As you mention, the central triangle in your example stays fixed throughout from the first to the second step. While I don't think you've given a precise definition of exactly what's going on here, this is reminiscent of an Iterated Function System with condensation - a term coined by Michael Barnsley in his book Fractals Everywhere. In this scheme, the condensation set proliferates throughout the fractal further with each set. More precisely, we have a list of functions $(f_1,f_2,\ldots,f_m)$ and we define an operation $T$ on sets by $$T(E) = \bigcup_{i=1}^m f_i(E).$$ At the $n^{\text{th}}$ stage of our construction, we include $$\bigcup_{k=1}^n T^k(E)$$ in our picture. The sequence of approximations now looks more like so:

Note that the black triangles are exactly those in the standard approximation to the Sierpinski gasket. That is, if \begin{align} f_1(x) &= x/3 \\ f_2(x) &= x/3 + \langle 1/2,0 \rangle \\ f_3(x) &= x/3 + \langle 1/4, \sqrt{3}/4 \rangle \end{align} and $D$ is the equilateral triangle whose base is the unit interval, then the black portion at the $n^{\text{th}}$ stage is exactly $T^n(D)$.

The initial gray portion is the condensation set $C$ and all the gray throughout the $n^{\text{th}}$ approximation is $$\bigcup_{i=1}^n T^k(C).$$

Note that the only difference between the generation of the attractor and the generator of the condensation portion is the union in the condensation is absent for the attractor and this union is why the condensation set persists throughout the construction.

Barnsley, it turns out, was primarily interested applications involving image compression so his introduction of condensation in iterated function systems was motivated largely by that application.

The dimensional analysis of such sets, by contrast, is not terribly exciting. Your example (if I've got it right), must have dimension two. It certainly can't have dimension greater than two, since it is contained in the plane, but it must have dimension at least two, since it contains a solid triangle. There's not much fancy work to be done.

More generally, the dimension of a self-similar set with condensation is the larger of the dimension of the condensation set and the dimension of the attractor.