As far as I know, the following fractal has a self-similar fractal dimension of
$D = -\log(3) / \log(1/2) = 1.5850$
But what is the fractal dimension of the following fractal (4 times the fractal of above)
Does the fractal dimension change when I include the fractal in each sqare, like partially drawn in the following graphic?
(The other squares, e.g. the yellow ones will also get the fractal inserted. I didn't draw it yet, since I only use Photoshop for the self-similar inclusion)
This inclusion of the fractal in each squares seems to be self-similar, but it cannot described with the self-similar fractal dimension formula, since the stretch-constant is not the same, since the squares, where the fractal is included, have different sizes.
If the self-similar fractal dimension cannot be applied, which method should I use to determinate the fractal dimension? Box counting?
UPDATE
Here is my non-formal construction description:
The central square has a size of $1^2$.
All subsequent squares have the half of the edge length. The distance is also half of the edge length of the previous square.
Using the distance scale down factor of 0.5 and the square scale down factor of 0.5, the resulting fractal has a perimeter of $4v$ where $v$ is $5*\sqrt{2}/2$ . Consequently, when each square is replaced by the whole set (like it is done with graphic #3), then the scale down factor of the whole fractal is $1/v$.
Here is a graphic for better explanation of the construction:
In the next level, each soldit square is replaced by the whole set itself, so that a true fractal is created.
It would be nice to have a precise description of the set that you are dealing with, rather than just a picture. Judging from the picture, though, I'd bet you're working with an IFS that is geometrically similar to the IFS consisting of the following three functions: \begin{align} f_1(x,y) &= \frac{1}{2}R\left(\frac{\pi}{2}\right) \left( \begin{array}{c} x \\ y \end{array} \right) + \left( \begin{array}{c} 1/2 \\ 0 \end{array} \right) \\ f_2(x,y) &= \frac{1}{2} \left( \begin{array}{c} x \\ y \end{array} \right)+ \left( \begin{array}{c} 0 \\ 1/2 \end{array} \right) \\ f_2(x,y) &= \frac{1}{2}R\left(-\frac{\pi}{2}\right) \left( \begin{array}{c} x \\ y \end{array} \right)+ \left( \begin{array}{c} 1/2 \\ 1 \end{array} \right), \\ \end{align} where $R(\theta)$ is the $2\times2$ rotation matrix through the angle $\theta$. The image of a set $A$ under this IFS is simply $$ \bigcup_{i=1}^3 f_i(A). $$ An oriented unit square, together with it's image under this IFS, looks like so:
If you start with a solid unit square and apply the IFS iteratively, you generate a sequence of images that looks like so:
If we iterate this 10 times and color the pieces, we get an image like so:
Let's call this set $S$, which is, indeed, a self-similar set, without overlap, consisting of 3 copies of itself scaled by the factor $1/2$. As such, its similarity dimension is, as you say, $\log(3)/\log(2)$.
However, it appears to me that your images are generated using a technique called condensation. This simply means that, rather than just applying the IFS to generate a new set, we retain the previous set as well. If we do this starting with a small, rotated square centered at $(1/2,1/2)$, we generate a sequence of images like so:
And, after a few more iterates:
Lets' call this set $T$, which I guess is your picture??
Now, some answers. Though, $\dim(S)=\log(3)/\log(2)$, this is not the case for $T$. $T$ contains solid squares and, as such, it has full dimension 2. Your 4-fold versions should have the same dimension as the set you start with, as they are just the union of the original set with a rotated copy of that set.
Finally, computation of the fractal dimension of your last set is a non-trivial matter. I believe that it is most easily described using a digraph iterated function system. Assuming so, I generated the following image:
Note that the central square is a copy of the whole scaled by the factor $1/4$. There is overlap between the pieces, which makes the computation of the dimension difficult. Assuming the central scaling factor is a power of $1/2$, the dimension can be computed using the techniques defined in this paper though, again, this is not particularly easy. Using those techniques, though, I computed the dimension of the last picture to be $$\frac{\log{\lambda}}{\log{4}} \approx 1.75829,$$ where $\lambda\approx11.44$ is the largest, positive root of $x^3-11 x^2-5 x-1$.