Question

我有一组2D点，我想找到它的最小外接圆。点数绘制如下（我将它们作为Python中元组的集合）：

原因如下：每个红点都是＆＃34;种子＆＃34;查询在线地图方向服务以寻找可能的路线，以便我可以逐步增加道路网络。问题是：因为我在种子附近询问，内种子往往会得到重复的结果，所以我正在考虑修剪＆＃34;修剪＆＃34;他们。为此，我需要找到包含圆的中心和直径，以便我可以删除最里面的那些 - 例如，圆圈内的圆不到圆周半径的一半。

Answer 1

我找到了一个实时演示here的工作解决方案（使用Python和JavaScript）。

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)

在一组2D点周围找到外接圆的算法

1 个答案: