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Smallest enclosing circle, error in the code

I have a set of points that represent the vertices (x, y) of a polygon.

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

I need to find the Smallest Enclosing Circle (area, x and y of center, and radius)

enter image description here

I am using the python code derived from this page: Smallest enclosing circle of Nayuki

when I run the code the results change every time, for example:

>>> 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)

where return is x, y (of the center of the circle), and radius

using a commercial software (i.e. ArcGiS) whin the some set of points the correct result is:

421646.74552, 4596389.82475, 24.323246

enter image description here

code used:

# 
# 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)
like image 980
Gianni Spear Avatar asked Dec 28 '14 03:12

Gianni Spear


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2 Answers

I am the author of the smallest enclosing circle implementation that you used. Please accept my apologies for the faulty code and the 2-years-late response.

The current version of the library fixes the issue that you experienced. Please update your copy to the latest version, and I assure you that the algorithm generates stable and sane results for the numerical case that you gave.


Another user came to me with a similar problem to you, where the algorithm outputs wildly different circles depending on how the list of points is randomized. His input data has a similar characteristic - the variation in the numbers is smaller than the magnitude of the numbers; for example 10.000001 versus 10.000002. I managed to thoroughly debug his test case because it contained only 5 points whereas yours has 32.

The root cause is that _make_circle() and _make_circumcircle() blindly calculate a radius that is mathematically correct, but fail to account for the distortion when coordinates of the center point is rounded. The correct way is to calculate the center point of the proposed circle, and then calculate the radius based on the maximum of how far each circumferential point is from the center.

For example, suppose that we want to find a circle that encloses (1,0) and (6,0), but every point is rounded to an integer. The true circle is of course (3.5, 0, 2.5). We calculate the x center as (1+6)÷2 = 3.5 → 4 (round to half even). If we calculate the radius separately and blindly, then it is the distance between 1 and 6, divided by 2, which is (6−1)÷2 = 2.5 → 2 (round to half even). But if we calculate the distance from the distorted center of (4,0) to (1,0) and (6,0), then we can see that we actually need a radius of 3. When the circle's radius is too small, it won't contain points that it was required to contain by design, and so the algorithm gets confused and tries to calculate new circles based on dubious data in dubious ways.

like image 159
Nayuki Avatar answered Oct 23 '22 04:10

Nayuki


Without understanding anything about your algorithm, I noticed one thing: the ratio between your coordinates and your radius is very large, about 2e5. Maybe, your algorithm is ill conditioned when trying to find a circle around points which are so far away from the origin. Especially in your _make_circumcircle function, this leads to the subtraction of large numbers, which is usually a bad thing for numerical errors.

Since fitting the radius and the center of the circle with respect to the points should be independent of a translation, you could simply subtract the mean of all points (the center of mass of your cloud of points), do the fitting, and then add the mean back to obtain the final result:

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])

Result:

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)

These numbers do not change a single digit from one run to the next one, and differ from your 'correct result' by only a tiny amount (probably different numerical errors due to a different implementation).

like image 25
Bas Swinckels Avatar answered Oct 23 '22 04:10

Bas Swinckels