Table Of Contents

Source code for kivy.geometry

'''
Geometry utilities
==================

This module contains some helper functions for geometric calculations.
'''

__all__ = ('circumcircle', 'minimum_bounding_circle')

from kivy.vector import Vector


def circumcircle(a, b, c):
    '''
    Computes the circumcircle of a triangle defined by a, b, c.
    See: http://en.wikipedia.org/wiki/Circumscribed_circle

    :Parameters:
        `a`: iterable containing at least 2 values (for x and y)
            The 1st point of the triangle.
        `b`: iterable containing at least 2 values (for x and y)
            The 2nd point of the triangle.
        `c`: iterable containing at least 2 values (for x and y)
            The 3rd point of the triangle.

    :Return:
        A tuple that defines the circle :
         * The first element in the returned tuple is the center as (x, y)
         * The second is the radius (float)
    '''
    P = Vector(a[0], a[1])
    Q = Vector(b[0], b[1])
    R = Vector(c[0], c[1])

    mPQ = (P + Q) * .5
    mQR = (Q + R) * .5

    numer = -(- mPQ.y * R.y + mPQ.y * Q.y + mQR.y * R.y - mQR.y * Q.y -
              mPQ.x * R.x + mPQ.x * Q.x + mQR.x * R.x - mQR.x * Q.x)
    denom = (-Q.x * R.y + P.x * R.y - P.x * Q.y +
             Q.y * R.x - P.y * R.x + P.y * Q.x)

    t = numer / denom

    cx = -t * (Q.y - P.y) + mPQ.x
    cy = t * (Q.x - P.x) + mPQ.y

    return ((cx, cy), (P - (cx, cy)).length())


def minimum_bounding_circle(points):
    '''
    Returns the minimum bounding circle for a set of points.

    For a description of the problem being solved, see the `Smallest Circle
    Problem <http://en.wikipedia.org/wiki/Smallest_circle_problem>`_.

    The function uses Applet's Algorithm, the runtime is O\(h^3, \*n\),
    where h is the number of points in the convex hull of the set of points.
    **But** it runs in linear time in almost all real world cases.
    See: http://tinyurl.com/6e4n5yb

    :Parameters:
        `points`: iterable
            A list of points (2 tuple with x,y coordinates)

    :Return:
        A tuple that defines the circle:
            * The first element in the returned tuple is the center (x, y)
            * The second the radius (float)

    '''
    points = [Vector(p[0], p[1]) for p in points]

    if len(points) == 1:
        return (points[0].x, points[0].y), 0.0

    if len(points) == 2:
        p1, p2 = points
        return (p1 + p2) * .5, ((p1 - p2) * .5).length()

    # determine a point P with the smallest y value
    P = min(points, key=lambda p: p.y)

    # find a point Q such that the angle of the line segment
    # PQ with the x axis is minimal
    def x_axis_angle(q):
        if q == P:
            return 1e10  # max val if the same, to skip
        return abs((q - P).angle((1, 0)))
    Q = min(points, key=x_axis_angle)

    for p in points:
        # find R such that angle PRQ is minimal
        def angle_pq(r):
            if r in (P, Q):
                return 1e10  # max val if the same, to skip
            return abs((r - P).angle(r - Q))
        R = min(points, key=angle_pq)

        # check for case 1 (angle PRQ is obtuse), the circle is determined
        # by two points, P and Q. radius = |(P-Q)/2|, center = (P+Q)/2
        if angle_pq(R) > 90.0:
            return (P + Q) * .5, ((P - Q) * .5).length()

        # if angle RPQ is obtuse, make P = R, and try again
        if abs((R - P).angle(Q - P)) > 90:
            P = R
            continue

        # if angle PQR is obtuse, make Q = R, and try again
        if abs((P - Q).angle(R - Q)) > 90:
            Q = R
            continue

        # all angles were acute..we just need the circle through the
        # two points furthest apart!
        break

    # find the circumcenter for triangle given by P,Q,R
    return circumcircle(P, Q, R)