Finding the fractal dimension of the Mandelbrot set using the box counting method So I'm trying to calculate the fractal dimension of the perimeter of the mandelbrot set using the box-counting or Minkowski–Bouligand definition of fractal dimension. According to this definition, my results should be greater than 2, but for some reason, I keep getting around 1.36 as my dimension value.
I was wondering if what I'm doing is incorrect, even though it seems to be the proper method.
Box side length 2, 45 boxes
Box side length 2, 45 boxes
Box side length 1, 122 boxes
Box side length 1, 122 boxes
Box side length 0.5, 314 boxes
Box side length 0.5, 314 boxes
As far as I understand it, with the box side length increasing in size by 2x, the number of perimeter boxes should be divided by 2^d, where although d should be 2, I'm getting a value of about 1.36. Any help?
 A: Your measure is correct!
For the boundary of the standard Mandelbrot set (https://github.com/sjhalayka/random_images/raw/master/mandel1.tga), I get a box-counting dimension of $d \approx 1.3$. You can save the Mandelbrot sets of arbitrary size to TGA file using my code (https://github.com/sjhalayka/mandelbrot_set_to_tga)
I go about obtaining this measure of $d \approx 1.3$ by using Marching Squares. Any time a marched square generates geometric primitives (line segments), the box count is incremented by 1 -- see https://github.com/sjhalayka/ms_curvature
It's all pretty straightforward, like you point out: you're counting the number of boxes that are needed to cover the boundary, given an arbitrary step size.
Contrary to some comments, the Mandelbrot's boundary will never be of box-counting dimension $d = 2.0$. The only way for this maximum to occur is if the entire input 2-plane is in the set. It doesn't matter how small the step sizes are -- a 2-plane is obviously not the exact same thing as the Mandelbrot's boundary. What I mean is that a straight line is of dimension $d = 1$. A wavy line is of dimension $d \in (1, 2)$. So, no, the dimension should not be 2 or greater. Perhaps you were thinking of generating 3D Mandelbrot sets using Marching Cubes? A flat plane has dimension $d = 2$. A wavy surface is of dimension $d \in (2, 3)$.  I discuss the usage of Marching Cubes in this paper: https://vixra.org/abs/1812.0423 -- I also discuss a new curvature-based dimension that uses the Marching Squares/Cubes/Hypercubes geometric primitive data as well.
A: Using my own box-counter (sig) software and a large (19200x10800) image rendered with interior and exterior distance estimation I get this table of box counts:




$\log_2 E$
$N(E)$
$\log_2 N(E)$




0
3228102



1
845624
19.689656797455456


2
230136
17.812127155343557


3
66178
16.014064071416254


4
20056



5
6432



6
2156



7
762



8
273



9
109



10
42



11
16



12
8



13
3



14
1



15
1





Linear regression of the first and third columns (over the range 1-3) in gnuplot gives:
> fit -D*x+k "boxes.dat" u 1:3 via D,k
...
Final set of parameters            Asymptotic Standard Error
=======================            ==========================
D               = 1.8378           +/- 0.02294      (1.248%)
k               = 21.5142          +/- 0.04956      (0.2303%)
...

which gives a box counting dimension estimate of $1.84$.  However as Glougloubarbaki states in the comments, the dimension has been proven to be $2$.  Using a larger image may give a higher box counting dimension estimate, as the log-log plot seems to be curving upwards towards the left:

