boxcount

Fractal dimension using the 'box-counting' method for 1D, 2D and 3D sets

Description

BOXCOUNT Box-Counting of a D-dimensional array (with D=1,2,3).

The Box-counting method is useful to determine fractal properties of a
1D segment, a 2D image or a 3D array. If C is a fractal set, with fractal dimension DF < D, then the number N of boxes of size R needed to cover the set scales as R^(-DF). DF is known as the Minkowski-Bouligand dimension, or Kolmogorov capacity, or Kolmogorov dimension, or simply box-counting dimension.

[N, R] = BOXCOUNT(C), where C is a D-dimensional array (with D=1,2,3), counts the number N of D-dimensional boxes of size R needed to cover the nonzero elements of C. The box sizes are powers of two, i.e., R = 1, 2, 4 ... 2^P, where P is the smallest integer such that MAX(SIZE(C)) <= 2^P. If the sizes of C over each dimension are smaller than 2^P, C is padded with zeros to size 2^P over each dimension (e.g., a 320-by-200 image is padded to 512-by-512). The output vectors N and R are of size P+1. For a RGB color image (m-by-n-by-3 array), a summation over the 3 RGB planes is done first.
 
BOXCOUNT(C,'plot') also shows the log-log plot of N as a function of R
(if no output argument, this option is selected by default).
 
BOXCOUNT(C,'slope') also shows the semi-log plot of the local slope DF = - dlnN/dlnR as a function of R. If DF is contant in a certain range of R, then DF is the fractal dimension of the set C.
 
The execution time depends on the sizes of C. It is fastest for powers
of two over each dimension.
 
Examples:
 
c = (rand(1,2048)<0.2);
boxcount(c);

c = randcantor(0.8, 512, 2);
boxcount(c);
figure, boxcount(c, 'slope');

• MATLAB release: MATLAB 7.2 (R2006a)
• Published M Files: Computing a fractal dimension with Matlab: 1D, 2D and 3D Box-counting

Other Files

boxcount/Apollonian_gasket.gif,
boxcount/boxcount.m,
boxcount/Contents.m,
boxcount/demo.m,
boxcount/dla.gif,
boxcount/fractal_tree.jpg,
boxcount/Thumbs.db,
boxcount/html/demo.png,
boxcount/html/demo_01.png,
boxcount/html/demo_02.png,
boxcount/html/demo_03.png,
boxcount/html/demo_04.png,
boxcount/html/demo_05.png,
boxcount/randcantor.m,
boxcount/html/demo_07.png,
boxcount/html/demo_08.png,
boxcount/html/demo_09.png,
boxcount/html/demo_10.png,
boxcount/html/demo_11.png,
boxcount/html/demo_12.png,
boxcount/html/demo_13.png,
boxcount/html/demo_14.png,
boxcount/html/demo_15.png,
boxcount/html/demo_16.png,
boxcount/html/Thumbs.db,
boxcount/html/demo_06.png

[SIMONA SSSS]

SEMBRA UN PROGRAMMA CHE FUNZIONI!

[Martin Lopez]

The program works.
It would be very useful more documentation.

[Guna Sekaran]

Very good tool. It works. Thanks to Author. May be more docs will help for better understanding.

[Fabien MONTEL]

clair et efficace ! merci pour le programme, il m'a fait gagné du temps

[ton malaka]

 

[Carissa Pan]

Thanks for sharing.

[Carissa Pan]

Thanks for sharing

[Owen Horsfield]

Consider:
[n,r]=boxcount([0 1 1 0])
I can put a single 1x2 box around the nonzero elements. So output should be:
n=[2 1 1]
r=[1 2 4]
However boxcount returns:
n=[2 2 1]
r=[1 2 4]
Algorithm problem? or am i missing something?

[Frederic Moisy]

Dear Owen Horsfield,
This is not really a bug: boxcount does not give the *minimum* number of boxes needed to cover the set, but only the number of boxes starting from the 1st element. However, it is true that starting the box covering with the 1st element is arbitrary, and other choice may lead to slightly different results.

[Ashwin Sundar]

 

[Marian Axente]

Hi Frederic

Could you point me in the right direction about your boxcounting algorithm? If I put a cube as input the program crashes; I thought that it was an easy way to check for consistency of the results.

[Frederic Moisy]

Dear Marian,
What does it mean "the program crashes"? Please provide an example. If I make the simplefollowing test to compute the "fractal" dimension of a square,
   a=zeros(4096,4096);
   a(600:800, 1200:1400)=1;
   boxcount(a,'slope')
it works properly (dimension close to 2 for small r, and close to 0 for large r, as expected)
F.

[Marian Axente]

Sorry...you are right. One more question in light of my fractal inexperience: why is the algorithm not give a full topological dimension whenever we put a square or a cube as input? Rather it gives 1.96 or 2.9. I thought that for smooth manifolds, the fractal dimension is equal to the topological dimension. Is there an approximation somewhere, or how could I modify your code to get a full 3 for a cube input?

[Frederic Moisy]

Dear Marian
Please use my email address for questions or discussions about this submission. Algorithms to compute a fractal dimension (box-counting or other methods) are all approximations, which converge towards the "true" topological dimension in the limit of large scale separation. I hope this helps. - F.

[Marian Axente]

[e-mail] bounces me every time...