Question

我有一组代表多边形顶点（x，y）的点。

points= [(421640.3639270504, 4596366.353552659), (421635.79361391126, 4596369.054192241), (421632.6774913164, 4596371.131607305), (421629.14588570886, 4596374.870954419), (421625.6142801013, 4596377.779335507), (421624.99105558236, 4596382.14190714), (421630.1845932406, 4596388.062540068), (421633.3007158355, 4596388.270281575), (421637.87102897465, 4596391.8018871825), (421642.4413421138, 4596394.918009778), (421646.5961722403, 4596399.903805929), (421649.71229483513, 4596403.850894549), (421653.8940752105, 4596409.600842565), (421654.69809098693, 4596410.706364258), (421657.60647207545, 4596411.329588776), (421660.514853164, 4596409.875398233), (421661.3458191893, 4596406.136051118), (421661.5535606956, 4596403.22767003), (421658.85292111343, 4596400.94251346), (421656.5677645438, 4596399.696064423), (421655.52905701223, 4596396.164458815), (421652.82841743, 4596394.502526765), (421648.46584579715, 4596391.8018871825), (421646.38843073393, 4596388.270281575), (421645.55746470863, 4596386.400608018), (421647.21939675923, 4596384.115451449), (421649.5045533288, 4596382.661260904), (421650.7510023668, 4596378.714172284), (421647.8426212782, 4596375.8057911955), (421644.9342401897, 4596372.897410107), (421643.6877911517, 4596370.404512031), (421640.3639270504, 4596366.353552659)]

enter image description here

我需要找到最小的封闭圆（区域，中心的x和y，以及半径）

enter image description here

我正在使用从此页面派生的python代码：Smallest enclosing circle of Nayuki

当我运行代码时，结果每次都会改变，例如：

>>> make_circle(points)
(421643.0645666326, 4596393.82736687, 23.70763190712525)
>>> make_circle(points)
(421647.8426212782, 4596375.8057911955, 0.0)
>>> make_circle(points)
(421648.9851995629, 4596388.841570718, 24.083963460031157)

其中return是x，y（圆心）和radius

使用商业软件（即ArcGiS），在某些点上正确的结果是：

421646.74552, 4596389.82475, 24.323246

enter image description here

使用的代码：

# 
# Smallest enclosing circle
# 
# Copyright (c) 2014 Project Nayuki
# https://www.nayuki.io/page/smallest-enclosing-circle
# 
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
# 
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
# 
# You should have received a copy of the GNU General Public License
# along with this program (see COPYING.txt).
# If not, see <http://www.gnu.org/licenses/>.
# 

import math, random


# Data conventions: A point is a pair of floats (x, y). A circle is a triple of floats (center x, center y, radius).

# 
# Returns the smallest circle that encloses all the given points. Runs in expected O(n) time, randomized.
# Input: A sequence of pairs of floats or ints, e.g. [(0,5), (3.1,-2.7)].
# Output: A triple of floats representing a circle.
# Note: If 0 points are given, None is returned. If 1 point is given, a circle of radius 0 is returned.
# 
def make_circle(points):
    # Convert to float and randomize order
    shuffled = [(float(p[0]), float(p[1])) for p in points]
    random.shuffle(shuffled)

    # Progressively add points to circle or recompute circle
    c = None
    for (i, p) in enumerate(shuffled):
        if c is None or not _is_in_circle(c, p):
            c = _make_circle_one_point(shuffled[0 : i + 1], p)
    return c


# One boundary point known
def _make_circle_one_point(points, p):
    c = (p[0], p[1], 0.0)
    for (i, q) in enumerate(points):
        if not _is_in_circle(c, q):
            if c[2] == 0.0:
                c = _make_diameter(p, q)
            else:
                c = _make_circle_two_points(points[0 : i + 1], p, q)
    return c


# Two boundary points known
def _make_circle_two_points(points, p, q):
    diameter = _make_diameter(p, q)
    if all(_is_in_circle(diameter, r) for r in points):
        return diameter

    left = None
    right = None
    for r in points:
        cross = _cross_product(p[0], p[1], q[0], q[1], r[0], r[1])
        c = _make_circumcircle(p, q, r)
        if c is None:
            continue
        elif cross > 0.0 and (left is None or _cross_product(p[0], p[1], q[0], q[1], c[0], c[1]) > _cross_product(p[0], p[1], q[0], q[1], left[0], left[1])):
            left = c
        elif cross < 0.0 and (right is None or _cross_product(p[0], p[1], q[0], q[1], c[0], c[1]) < _cross_product(p[0], p[1], q[0], q[1], right[0], right[1])):
            right = c
    return left if (right is None or (left is not None and left[2] <= right[2])) else right


def _make_circumcircle(p0, p1, p2):
    # Mathematical algorithm from Wikipedia: Circumscribed circle
    ax = p0[0]; ay = p0[1]
    bx = p1[0]; by = p1[1]
    cx = p2[0]; cy = p2[1]
    d = (ax * (by - cy) + bx * (cy - ay) + cx * (ay - by)) * 2.0
    if d == 0.0:
        return None
    x = ((ax * ax + ay * ay) * (by - cy) + (bx * bx + by * by) * (cy - ay) + (cx * cx + cy * cy) * (ay - by)) / d
    y = ((ax * ax + ay * ay) * (cx - bx) + (bx * bx + by * by) * (ax - cx) + (cx * cx + cy * cy) * (bx - ax)) / d
    return (x, y, math.hypot(x - ax, y - ay))


def _make_diameter(p0, p1):
    return ((p0[0] + p1[0]) / 2.0, (p0[1] + p1[1]) / 2.0, math.hypot(p0[0] - p1[0], p0[1] - p1[1]) / 2.0)


_EPSILON = 1e-12

def _is_in_circle(c, p):
    return c is not None and math.hypot(p[0] - c[0], p[1] - c[1]) < c[2] + _EPSILON


# Returns twice the signed area of the triangle defined by (x0, y0), (x1, y1), (x2, y2)
def _cross_product(x0, y0, x1, y1, x2, y2):
    return (x1 - x0) * (y2 - y0) - (y1 - y0) * (x2 - x0)

Answer 1

如果不了解您的算法，我注意到一件事：坐标和半径之间的比例非常大，约为2e5。也许，当你试图在远离原点的点周围找到一个圆时，你的算法会受到限制。特别是在你的_make_circumcircle函数中，这会导致大数的减法，这对于数值误差通常是一件坏事。

由于相对于点拟合圆的半径和圆心应该与平移无关，你可以简单地减去所有点的平均值（你的点云的质心），做拟合，然后再添加均值以获得最终结果：

def numerical_stable_circle(points):
    pts = np.array(points)
    mean_pts = np.mean(pts, 0)
    print 'mean of points:', mean_pts
    pts -= mean_pts  # translate towards origin
    result = make_circle(pts)
    print 'result without mean:', result
    print 'result with mean:', (result[0] + mean_pts[0], 
    result[1] + mean_pts[1], result[2])

结果：

mean of points: [  421645.83745955  4596388.99204294]
result without mean: (0.9080813432488977, 0.8327111343034483, 24.323287017466253)
result with mean: (421646.74554089626, 4596389.8247540779, 24.323287017466253)

这些数字不会将一个数字从一个数字更改为下一个数字，并且与您的“正确结果”不同。只有很小的数量（由于不同的实现可能会产生不同的数值误差）。

Answer 2

我是您使用的smallest enclosing circle implementation的作者。请接受我对错误代码和2年迟到回复的道歉。

该库的当前版本修复了您遇到的问题。请将您的副本更新到最新版本，我向您保证，算法会为您提供的数字案例生成稳定且合理的结果。

另一位用户带着similar problem来找我，算法会输出截然不同的圆圈，具体取决于点列表的随机化方式。他的输入数据具有类似的特征 - 数字的变化小于数字的大小;例如10.000001与10.000002。我设法彻底调试他的测试用例，因为它只包含5个点，而你的测试用例只有32个。

根本原因是_make_circle()和_make_circumcircle()盲目地计算数学上正确的半径，但是当中心点的坐标被舍入时无法解释失真。正确的方法是计算拟议圆的中心点，然后根据每个圆周点离中心的距离的最大值计算半径。

例如，假设我们想要找到一个包围（1,0）和（6,0）的圆，但每个点都舍入为整数。真正的圆圈当然是（3.5,0,2.5）。我们将 x 中心计算为（1 + 6）÷2 = 3.5→4（舍入到半偶数）。如果我们单独和盲目地计算半径，则它是1和6之间的距离除以2，即（6-1）÷2 = 2.5→2（圆到半偶）。但是如果我们计算从（4,0）到（1,0）和（6,0）的扭曲中心的距离，那么我们可以看到我们实际上需要半径为3.当圆的半径太小时，它不会包含设计中需要包含的点，因此算法会混淆并尝试以可疑的方式基于可疑数据计算新的圈子。

Python中最小的封闭圆，代码中的错误

2 个答案: