#
# Smallest enclosing circle - Library (Python)
#
# Copyright (c) 2020 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
from numba.pycc import CC

cc = CC('smallest_circle')
cc.verbose = True

# 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


@cc.export('is_in_circle', 'b1(f8[:], f8[:])')
def is_in_circle(c, p):
    if (c is not None) and (math.hypot(p[0] - c[0], p[1] - c[1]) <= c[2] * _MULTIPLICATIVE_EPSILON):
        res = True
    else:
        res = False
    return res


@cc.export('make_circle', '(f8[:,:],)')
def make_circle(points):
    # points : numpy.ndarray (3,2) float64
    # Convert to float and randomize order
    shuffled = points[np.random.permutation(len(points))]

    # Progressively add points to circle or recompute circle
    first_pass = True
    c = (0,0,0)
    for (i, p) in enumerate(shuffled):
        if first_pass is True or not is_in_circle(c, p):
            first_pass = False
            c = _make_circle_one_point(shuffled[: i + 1], p)
    return c


# One boundary point known
@cc.export('make_circle_one_point', '(f8[:,:], f8[:])')
def _make_circle_one_point(points, p):
    # points : numpy.ndarray (n,2) float64
    # p : numpy.ndarray (2,) float64
    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
@cc.export('make_circle_two_points', '(f8[:,:],f8[:],f8[:])')
def _make_circle_two_points(points, p, q):
    # points : numpy.ndarray (n,2) float64
    # p : numpy.ndarray (2,) float64
    # q : numpy.ndarray (2,) float64
    circ = make_diameter(p, q)
    left = (0, 0, 0)
    left_valid = False
    right = (0, 0, 0)
    right_valid = False
    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, c_valid = make_circumcircle(p, q, r)
        if c_valid is False:
            continue
        elif cross > 0.0 and (
                left_valid is False or
                _cross_product(px, py, qx, qy, c[0], c[1]) > _cross_product(px, py, qx, qy, left[0], left[1])):
            left = c
            left_valid = True
        elif cross < 0.0 and (right_valid is False or _cross_product(px, py, qx, qy, c[0], c[1]) < _cross_product(px, py, qx, qy, right[0], right[1])):
            right = c
            right_valid = True
    # Select which circle to return
    if left_valid is False and right_valid is False:
        return circ
    elif left_valid is False:
        return right
    elif right_valid is False:
        return left
    else:
        return left if (left[2] <= right[2]) else right


@cc.export('make_diameter', '(f8[:],f8[:])')
def make_diameter(a, b):
    # a : numpy.ndarray (2,) float64
    # b : numpy.ndarray (2,) float64
    cx = (a[0] + b[0]) / 2
    cy = (a[1] + b[1]) / 2
    r0 = math.hypot(cx - a[0], cy - a[1])
    r1 = math.hypot(cx - b[0], cy - b[1])
    return (cx, cy, max(r0, r1))


@cc.export('make_circumcircle', '(f8[:], f8[:], f8[:])')
def make_circumcircle(a, b, c):
    # Mathematical algorithm from Wikipedia: Circumscribed circle
    # a : numpy.ndarray (2,) float64
    # b : numpy.ndarray (2,) float64
    # c : numpy.ndarray (2,) float64
    ox = (min(a[0], b[0], c[0]) + max(a[0], b[0], c[0])) / 2
    oy = (min(a[1], b[1], c[1]) + max(a[1], b[1], c[1])) / 2
    ax = a[0] - ox
    ay = a[1] - oy
    bx = b[0] - ox
    by = b[1] - oy
    cx = c[0] - ox
    cy = c[1] - oy
    d = (ax * (by - cy) + bx * (cy - ay) + cx * (ay - by)) * 2.0
    if d == 0.0:
        return (0,0,0), False
    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 - a[0], y - a[1])
    rb = math.hypot(x - b[0], y - b[1])
    rc = math.hypot(x - c[0], y - c[1])
    return (x, y, max(ra, rb, rc)), True


_MULTIPLICATIVE_EPSILON = 1 + 1e-14


# Returns twice the signed area of the triangle defined by (x0, y0), (x1, y1), (x2, y2).
@cc.export('cross_product', 'f8(f8,f8,f8,f8,f8,f8)')
def _cross_product(x0, y0, x1, y1, x2, y2):
    return (x1 - x0) * (y2 - y0) - (y1 - y0) * (x2 - x0)

if __name__ == '__main__':
    cc.compile()

    import matplotlib.pyplot as plt
    import matplotlib.patches as patches
    import numpy as np

    # Create 30 random points
    points = np.random.rand(30, 2)

    # Compute the smallest enclosing circle
    c = make_circle(points)

    # Plot the points and the enclosing circle
    fig, ax = plt.subplots()
    ax.set_aspect('equal')
    ax.add_patch(patches.Circle(c[:2], c[2], fill=False))
    ax.scatter(points[:, 0], points[:, 1])
    plt.show()