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# setup: N=10
# run: julia(1., 1., N, 1.5, 10., 1e4)
# pythran export julia(float, float, int, float, float, float)
import numpy as np
def kernel(zr, zi, cr, ci, lim, cutoff):
''' Computes the number of iterations `n` such that
|z_n| > `lim`, where `z_n = z_{n-1}**2 + c`.
'''
count = 0
while ((zr * zr + zi * zi) < (lim * lim)) and count < cutoff:
zr, zi = zr * zr - zi * zi + cr, 2 * zr * zi + ci
count += 1
return count
def julia(cr, ci, N, bound=1.5, lim=1000., cutoff=1e6):
''' Pure Python calculation of the Julia set for a given `c`. No NumPy
array operations are used.
'''
julia = np.empty((N, N), np.uint32)
grid_x = np.linspace(-bound, bound, N)
for i, x in enumerate(grid_x):
for j, y in enumerate(grid_x):
julia[i, j] = kernel(x, y, cr, ci, lim, cutoff)
return julia