如何测量控制多边形与第n条细分曲线之间的距离？

Question

• 我有一个 m 个点的控制多边形，比如 m = 6，我正在应用 n 个细化，比如 n = 5，使用不同的细分方案：

  Chaikin，拐角切割，4 点方案，广义 4 点方案，3 次均匀样条。

我的问题是：

如何测量得到的具有不同点数的细分曲线与原始控制多边形之间的距离？

#========================
# Import Libraries
#========================
import numpy as np
import matplotlib.pyplot as plt
# !pip install shapely
from shapely.geometry import Point, Polygon 

#========================
# Define the Polygon
#========================
# Define Control Points
P1 = Point(0.0, 0.0)
P2 = Point(1.0, 2.0)
P3 = Point(2.0, 2.0)
P4 = Point(4.0, 4.0)
P5 = Point(5.0, 9.0)
P6 = Point(1.0, 7.0)
# control Points List
PointList = [P1, P2, P3, P4, P5, P6]
# Define a Polygon from the list of the control points
Polygn = Polygon(PointList)
#==========================================
# Uniform Spline of Degree k = 3
#==========================================
plt.figure(num='Uniform Spline of degree k =3 Subdivision Scheme')

for n in range(5):
    PointList2 = []
    for idx in range(len(PointList)):
        # Get Three of adjacent points
        P1n = PointList[idx-2]
        P2n = PointList[idx-1]
        P3n = PointList[idx]
        # Get new First Control Point
        x21n = 0.125 * (P1n.x + 6*P2n.x + P3n.x)
        y21n = 0.125 * (P1n.y + 6*P2n.y + P3n.y)
        P21n = Point(x21n, y21n)
        PointList2.append(P21n)
        # Get new Second Control Point
        x22n = 0.5 * (P2n.x+P3n.x)
        y22n = 0.5 * (P2n.y+P3n.y)
        P22n = Point(x22n, y22n)
        PointList2.append(P22n)
    # Assign the New list of points to the old one
    PointList = PointList2
    # print(len(PointList2))
    Polygn2 = Polygon(PointList2)
    
    # 
    plt.plot(*Polygn.exterior.xy, 'ro--', linewidth=1, markersize=4, label='Original')
    plt.plot(*Polygn2.exterior.xy, 'go-', linewidth=1, markersize=2, label='Subdivision ({})'.format(n+1))
    plt.legend(bbox_to_anchor=(.68, .14), loc='upper left', borderaxespad=0.)
    plt.show(block=False)
    plt.pause(1)
    plt.clf()
#=========================
# Distance Finction
#=========================
def dist(Polygn, Polygn2):
 pass
 # return max_dist

Answer 1

解决方案在 Shapely 库中，Hausdorff Distance 已经在那里实现：

polygon2.hausdorff_distance(polygon)