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

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!

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 $s \approx 1.393\ldots$.