包围给定点集的边界

Question

我目前正在使用的算法存在一些问题。我希望它成为一个界限。

这是当前行为的示例：

这是通缉行为的MSPaint示例：

C＃中凸包的当前代码：https : //hastebin.com/dudejesuja.cs

所以这是我的问题：

1）这有可能吗？

R：是的

2）这甚至被称为凸包吗？（我不这么认为）

R：不，这叫做边界，链接：https : //www.mathworks.com/help/matlab/ref/boundary.html

3）与传统的凸包相比，这是否会降低性能？

R：据我研究，它应该具有相同的性能

4）使用伪代码或类似方法的此算法示例？

R：尚未回答，或者我尚未找到解决方案

Answer 1

这是一些计算alpha形状（凹壳）并仅保留外部边界的Python代码。这可能是matlab的边界在内部执行的操作。

from scipy.spatial import Delaunay
import numpy as np


def alpha_shape(points, alpha, only_outer=True):
    """
    Compute the alpha shape (concave hull) of a set of points.
    :param points: np.array of shape (n,2) points.
    :param alpha: alpha value.
    :param only_outer: boolean value to specify if we keep only the outer border
    or also inner edges.
    :return: set of (i,j) pairs representing edges of the alpha-shape. (i,j) are
    the indices in the points array.
    """
    assert points.shape[0] > 3, "Need at least four points"

    def add_edge(edges, i, j):
        """
        Add an edge between the i-th and j-th points,
        if not in the list already
        """
        if (i, j) in edges or (j, i) in edges:
            # already added
            assert (j, i) in edges, "Can't go twice over same directed edge right?"
            if only_outer:
                # if both neighboring triangles are in shape, it's not a boundary edge
                edges.remove((j, i))
            return
        edges.add((i, j))

    tri = Delaunay(points)
    edges = set()
    # Loop over triangles:
    # ia, ib, ic = indices of corner points of the triangle
    for ia, ib, ic in tri.vertices:
        pa = points[ia]
        pb = points[ib]
        pc = points[ic]
        # Computing radius of triangle circumcircle
        # www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-formula-for-radius-of-circumcircle
        a = np.sqrt((pa[0] - pb[0]) ** 2 + (pa[1] - pb[1]) ** 2)
        b = np.sqrt((pb[0] - pc[0]) ** 2 + (pb[1] - pc[1]) ** 2)
        c = np.sqrt((pc[0] - pa[0]) ** 2 + (pc[1] - pa[1]) ** 2)
        s = (a + b + c) / 2.0
        area = np.sqrt(s * (s - a) * (s - b) * (s - c))
        circum_r = a * b * c / (4.0 * area)
        if circum_r < alpha:
            add_edge(edges, ia, ib)
            add_edge(edges, ib, ic)
            add_edge(edges, ic, ia)
    return edges

如果使用以下测试代码运行它，您将获得此图，该图看起来像您所需要的：

from matplotlib.pyplot import *

# Constructing the input point data
np.random.seed(0)
x = 3.0 * np.random.rand(2000)
y = 2.0 * np.random.rand(2000) - 1.0
inside = ((x ** 2 + y ** 2 > 1.0) & ((x - 3) ** 2 + y ** 2 > 1.0)
points = np.vstack([x[inside], y[inside]]).T

# Computing the alpha shape
edges = alpha_shape(points, alpha=0.25, only_outer=True)

# Plotting the output
figure()
axis('equal')
plot(points[:, 0], points[:, 1], '.')
for i, j in edges:
    plot(points[[i, j], 0], points[[i, j], 1])
show()

编辑：在注释中的请求之后，这是一些将输出边集“缝合”为连续边序列的代码。

def find_edges_with(i, edge_set):
    i_first = [j for (x,j) in edge_set if x==i]
    i_second = [j for (j,x) in edge_set if x==i]
    return i_first,i_second

def stitch_boundaries(edges):
    edge_set = edges.copy()
    boundary_lst = []
    while len(edge_set) > 0:
        boundary = []
        edge0 = edge_set.pop()
        boundary.append(edge0)
        last_edge = edge0
        while len(edge_set) > 0:
            i,j = last_edge
            j_first, j_second = find_edges_with(j, edge_set)
            if j_first:
                edge_set.remove((j, j_first[0]))
                edge_with_j = (j, j_first[0])
                boundary.append(edge_with_j)
                last_edge = edge_with_j
            elif j_second:
                edge_set.remove((j_second[0], j))
                edge_with_j = (j, j_second[0])  # flip edge rep
                boundary.append(edge_with_j)
                last_edge = edge_with_j

            if edge0[0] == last_edge[1]:
                break

        boundary_lst.append(boundary)
    return boundary_lst

然后，您可以越过边界列表的列表，并在每个边中附加与第一个索引对应的点，以获得边界多边形。