How long is the coast of Britain? The divider method

Walks a pair of dividers along a smooth circle and along the Koch curve with ever smaller openings. The circle’s measured length converges, but the Koch curve’s grows as a power of the ruler, the scaling Richardson observed for real coastlines and Mandelbrot interpreted as a fractal dimension.

import matplotlib.pyplot as plt
import numpy as np

from mathematicskit.fractals_chaos import divider_length, koch_curve

Measured length against ruler size

t = np.linspace(0, 2 * np.pi, 20001)
circle = np.column_stack([np.cos(t), np.sin(t)]) / (2 * np.pi)  # circumference 1
koch = koch_curve(8)
rulers = 3.0 ** -np.arange(1, 7)
circle_lengths = [divider_length(circle, r) for r in rulers]
koch_lengths = [divider_length(koch, r) for r in rulers]

slope = np.polyfit(np.log(rulers), np.log(koch_lengths), 1)[0]
print(f"Koch curve: L ~ ruler^{slope:.4f}, so D = 1 - slope = {1 - slope:.4f} (log 4 / log 3 = {np.log(4) / np.log(3):.4f})")

fig, ax = plt.subplots()
ax.loglog(rulers, circle_lengths, "o-", label="circle (smooth)")
ax.loglog(rulers, koch_lengths, "s-", label="Koch curve (fractal)")
ax.set_xlabel("ruler length")
ax.set_ylabel("measured length")
ax.invert_xaxis()
ax.legend()
ax.set_title("Richardson plot")

Koch curve: L ~ ruler^-0.2619, so D = 1 - slope = 1.2619 (log 4 / log 3 = 1.2619)

Text(0.5, 1.0, 'Richardson plot')