How to compute the Hausdorff dimension of a "semi" self-similar shape?

Question

The essence of self-similar fractals is that rescaling the original shape and gluing together a number of identical copies will produce the same overall shape. The quadratic type 2 curve is an example of such a fractal. You scale down the original shape by a factor of $1/4$ and piece together $8$ copies to get the same fractal:

This results in a fractal with dimension $\log_{4}{8} = 1.5$:

What I'm interested in, and what I mean by "semi" self-similar, is fractals where there are differently scaled copies of the shape building up the entire image. For example, if the vertical "middle" of the quadratic type 2 curve is treated as one iteration, rather than two, then the copy of the curve making up this piece is only scaled by $1/2$, while all of the other pieces are scaled by the original $1/4$. This ends up producing quite a different-looking fractal:

Basic Structure:

Limit Fractal:

Another example would be a "shark fin" fractal, similar to the Koch snowflake, but where the "middle third" has a right triangle with height $1/3$, and hypotenuse $\sqrt{2}/3$:

Basic Structure:

Limit Fractal:

Any ideas about how to calculate the Hausdorff dimensions of such shapes would be greatly appreciated!

Answer 1

The similarity dimension of $\log_4 8$ comes from solving $8\left(\frac{1}{4}\right)^s = 1$. Your modified fractal's dimension is the solution of $6\left(\frac{1}{4}\right)^s + \left(\frac{1}{2}\right)^s = 1$, which Wolfram Alpha tells me is $s = \log_2 3$. In general there may not be a nice closed-form solution, and indeed your "shark fin" fractal requires solving $3\left(\frac{1}{3}\right)^s + \left(\frac{\sqrt{2}}{3}\right)^s = 1$, which Wolfram Alpha gives numerically as $s \approx 1.393\ldots$.

I'm not sure what is necessary to prove that this similarity dimension is equal to the Hausdorff dimension, probably it involves showing that the shape satisfies an "open set condition" (essentially, that it doesn't self-overlap too much).