| import numpy | |
| def norm2(X): | |
| return numpy.sqrt(numpy.sum(X ** 2)) | |
| def normalized(X): | |
| return X / norm2(X) | |
| ''' | |
| Returns the intersection, a line, between the plane A and B |
MIT License
Copyright (c) 2021 Devert Alexandre
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:
| import numpy | |
| ''' | |
| Pick a point uniformly from the unit circle | |
| ''' | |
| def circle_uniform_pick(size, out = None): | |
| if out is None: | |
| out = numpy.empty((size, 2)) | |
| angle = 2 * numpy.pi * numpy.random.random(size) |
This code sample demonstrates how to generates a set of points that covers a disc almost evenly, for any number of points. The trick is to use a Fermat's spiral and the golden angle.
| 42 points | 256 points | 999 points |
|---|---|---|
 |
 |
 |
This code sample demonstrates how to check if a box and a disc (in 3d, a sphere) intersects each other. The code works for 2d and 3d, and beyond, without any modifications.
The actual test is the function box_intersect_disc, the rest of the script is just to provide a visual demonstration of the test.
| import numpy | |
| def get_unit_simplex(N, dtype = float): | |
| # Original algorithm from John Burkardt | |
| A = numpy.full(N, (1. - numpy.sqrt(N + 1)) / N, dtype = dtype) | |
| S = numpy.concatenate([A[:,None].T, numpy.identity(N, dtype = dtype)]) | |
| S -= numpy.mean(S, axis = 0) | |
| S /= numpy.sqrt(2.) | |
| return S |
| import numpy | |
| def get_circumsphere(S): | |
| U = S[1:] - S[0] | |
| B = numpy.sqrt(numpy.sum(U ** 2, axis = 1)) | |
| U /= B[:,None] | |
| C = numpy.dot(numpy.linalg.solve(numpy.inner(U, U), .5 * B), U) | |
| return C + S[0], numpy.sqrt(numpy.sum(C ** 2)) |
| import numpy | |
| def quickhull_2d(S): | |
| def process(P, a, b): | |
| signed_dist = numpy.cross(S[P] - S[a], S[b] - S[a]) | |
| K = [i for s, i in zip(signed_dist, P) if s > 0 and i != a and i != b] | |
| if len(K) == 0: |
| import numpy | |
| def sample_annulus(r1, r2, k): | |
| theta = numpy.random.rand(k) | |
| theta *= 2 * numpy.pi | |
| R = numpy.sqrt(numpy.random.rand(k)) | |
| R *= r2 - r1 | |
| R += r1 | |
| S = numpy.array([numpy.cos(theta), numpy.sin(theta)]).T | |
| S *= R[:,None] |
| import itertools | |
| from sympy.combinatorics import Permutation | |
| def get_hexacosichoron_vertices(): | |
| phi = (1 + 5 ** .5) / 2 | |
| U = [.5 * phi, .5, .5 / phi] | |
| permutation_list = [P for P in itertools.permutations(range(4)) if Permutation(P).is_even] | |
| for i in range(16): |
