c++ - 如何在给定 dims+1 点的情况下找到任意维球体的中心和半径

Question

给定一个 N 维点 vector 。 vector 的大小为 N+1。

是否有一种通用算法可以使用球体与每个点相交的点来查找 ND 球体的中心和半径？

Answer 1

同样的问题已在 math stackexchange 上提出，并得到了建设性的答案:

• Does a set of n+1 points that affinely span R^n lie on a unique (n-1)-sphere?

这是该答案中描述的算法的 python/numpy 实现。

import numpy as np

def find_sphere_through_points(points):
    n_points, n_dim = points.shape
    if (n_points != n_dim + 1):
        raise ValueError('Number of points must be equal to 1 + dimension')
    a = np.concatenate((points, np.ones((n_points, 1))), axis=1)
    b = (points**2).sum(axis=1)
    x = np.linalg.solve(a, b)
    center = x[:-1] / 2
    radius = x[-1] + center@center
    return center, radius

为了测试此方法，我们可以使用此相关问题中描述的方法在球体表面生成随机点:

• Generate a random sample of points distributed on the surface of a unit spher

import numpy as np

def sample_spherical(npoints, ndim=3, center=None):
    vec = np.random.randn(npoints, ndim)
    vec /= np.linalg.norm(vec, axis=1).reshape(npoints,1)
    if center is None:
        return vec
    else:
        return vec + center

n = 5
center = np.random.rand(n)
points = sample_spherical(n+1, ndim=n, center=center)

guessed_center, guessed_radius = find_sphere_through_points(points)

print('True center:\n    ', center)
print('Calc center:\n    ', guessed_center)
print('True radius:\n    ', 1.0)
print('Calc radius:\n    ', guessed_radius)

# True center:
#      [0.18150032 0.94979547 0.07719378 0.26561175 0.37509931]
# Calc center:
#      [0.18150032 0.94979547 0.07719378 0.26561175 0.37509931]
# True radius:
#      1.0
# Calc radius:
#      0.9999999999999997