Fractal dimension computing

fractal-dimension.py

# -----------------------------------------------------------------------------
# From https://en.wikipedia.org/wiki/Minkowski–Bouligand_dimension:
#
# In fractal geometry, the Minkowski–Bouligand dimension, also known as
# Minkowski dimension or box-counting dimension, is a way of determining the
# fractal dimension of a set S in a Euclidean space Rn, or more generally in a
# metric space (X, d).
# -----------------------------------------------------------------------------
import scipy.misc
import numpy as np


def fractal_dimension(Z, threshold=0.9):

    # Only for 2d image
    assert(len(Z.shape) == 2)

    # From https://github.com/rougier/numpy-100 (#87)
    def boxcount(Z, k):
        S = np.add.reduceat(
            np.add.reduceat(Z, np.arange(0, Z.shape[0], k), axis=0),
                               np.arange(0, Z.shape[1], k), axis=1)

        # We count non-empty (0) and non-full boxes (k*k)
        return len(np.where((S > 0) & (S < k*k))[0])


    # Transform Z into a binary array
    Z = (Z < threshold)

    # Minimal dimension of image
    p = min(Z.shape)

    # Greatest power of 2 less than or equal to p
    n = 2**np.floor(np.log(p)/np.log(2))

    # Extract the exponent
    n = int(np.log(n)/np.log(2))

    # Build successive box sizes (from 2**n down to 2**1)
    sizes = 2**np.arange(n, 1, -1)

    # Actual box counting with decreasing size
    counts = []
    for size in sizes:
        counts.append(boxcount(Z, size))

    # Fit the successive log(sizes) with log (counts)
    coeffs = np.polyfit(np.log(sizes), np.log(counts), 1)
    return -coeffs[0]

I = scipy.misc.imread("sierpinski.png")/256.0
print("Minkowski–Bouligand dimension (computed): ", fractal_dimension(I))
print("Haussdorf dimension (theoretical):        ", (np.log(3)/np.log(2)))

Thank you for sharing this.

I believe we can reduce

    # Greatest power of 2 less than or equal to p
    n = 2**np.floor(np.log(p)/np.log(2))

    # Extract the exponent
    n = int(np.log(n)/np.log(2))

to

    # largest n for which 2^n is less than or equal p
    n = int(np.floor(np.log2(p)))

sorry, want to ask if this can be used on iris recognition or is there any other version?
btw the image is

thanks before and really appreciate it