Stav 23.06.2026

2026-06-23 15:20:56 +02:00
commit 6d91e83e8c
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+'''
+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)