smallestCircle.py

#
# Smallest enclosing circle - Library (Python)
#
# Copyright (c) 2017 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 Lesser 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 Lesser General Public License for more details.
#
# You should have received a copy of the GNU Lesser General Public License
# along with this program (see COPYING.txt and COPYING.LESSER.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.
#
# Initially: No boundary points known
def make_circle(points):
    # Convert to float and randomize order
    shuffled = [(float(x), float(y)) for (x, y) 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[: 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[: i + 1], p, q)
    return c


# Two boundary points known
def _make_circle_two_points(points, p, q):
    circ = make_diameter(p, q)
    left = None
    right = None
    px, py = p
    qx, qy = q

    # For each point not in the two-point circle
    for r in points:
        if is_in_circle(circ, r):
            continue

        # Form a circumcircle and classify it on left or right side
        cross = _cross_product(px, py, qx, qy, 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(px, py, qx, qy, c[0], c[1]) > _cross_product(px, py, qx, qy, left[0],
                                                                                            left[1])):
            left = c
        elif cross < 0.0 and (
                right is None or _cross_product(px, py, qx, qy, c[0], c[1]) < _cross_product(px, py, qx, qy, right[0],
                                                                                             right[1])):
            right = c

    # Select which circle to return
    if left is None and right is None:
        return circ
    elif left is None:
        return right
    elif right is None:
        return left
    else:
        return left if (left[2] <= right[2]) else right


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


def make_diameter(p0, p1):
    cx = (p0[0] + p1[0]) / 2.0
    cy = (p0[1] + p1[1]) / 2.0
    r0 = math.hypot(cx - p0[0], cy - p0[1])
    r1 = math.hypot(cx - p1[0], cy - p1[1])
    return (cx, cy, max(r0, r1))


_MULTIPLICATIVE_EPSILON = 1 + 1e-14


def is_in_circle(c, p):
    return c is not None and math.hypot(p[0] - c[0], p[1] - c[1]) <= c[2] * _MULTIPLICATIVE_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)