Bezier曲线——de Casteljau递推算法实现

1. 定义

给定 n + 1 n+1 n+1个点的位置矢量 P i ( i = 0 , 1 , … , n ) P_i(i=0,1,\dots,n) Pi(i=0,1,…,n)，则Bezier曲线可以定义为
P ( t ) = ∑ i = 0 n P i B i , n ( t ) , t ∈ [ 0 , 1 ] P(t)=\sum_{i=0}^nP_iB_{i,n}(t),\quad t \in [0,1] P(t)=i=0∑nPiBi,n(t),t∈[0,1]
其中 P i P_i Pi(i=0,1,\dots,n)构成该Bezier曲线的特征多边形， B i , n ( t ) B_{i,n}(t) Bi,n(t)是 n n n次Bernstein基函数
B i , n ( t ) = C n i t i ( 1 − t ) n − i = n ! i ! ( n − i ) ! t i ⋅ ( 1 − t ) n − i , ( i = 0 , 1 , … , n ) B_{i,n}(t) = C_n^it^i(1-t)^{n-i}=\frac{n!}{i!(n-i)!}t^i\cdot (1-t)^{n-i},\quad (i=0,1,\dots,n) Bi,n(t)=Cniti(1−t)n−i=i!(n−i)!n!ti⋅(1−t)n−i,(i=0,1,…,n)
其中 0 0 = 1 , 0 ! = 1 0^0=1,0!=1 00=1,0!=1。

2. Bezier曲线的递推算法

计算Bez曲线上的点，可用Bezier曲线方程直接计算，但使用de Casteljau提出的递推算法则简单得多。
由 n + 1 n+1 n+1个控制点 P i ( i = 0 , 1 , … , n ) P_i(i=0,1,\dots,n) Pi(i=0,1,…,n)定义的 n n n次Bezier曲线 P 0 n P_0^n P0n可被定义为分别由前、后 n n n个控制点定义的两条 n − 1 n-1 n−1次Bezier曲线 P 0 n − 1 P_0^{n-1} P0n−1与 P 1 n − 1 P_1^{n-1} P1n−1的线性组合
P 0 n = ( 1 − t ) P 0 n − 1 + t P 1 n − 1 , t ∈ [ 0 , 1 ] P_0^n = (1-t)P_0^{n-1}+tP_1^{n-1},\quad t \in[0,1] P0n=(1−t)P0n−1+tP1n−1,t∈[0,1]
由此得到Bezier曲线的递推计算公式为
P i k = { P i , k = 0 ( 1 − t ) P i k − 1 + t P i + 1 k − 1 , k = 1 , 2 , … , n , i = 0 , 1 , … , n − k P_i^k= \begin{cases} P_i, & k=0 \\ (1-t)P_i^{k-1}+tP_{i+1}^{k-1}, & k=1,2,\dots,n,i=0,1,\dots,n-k \end{cases} Pik={Pi,(1−t)Pik−1+tPi+1k−1,k=0k=1,2,…,n,i=0,1,…,n−k

3. 代码实现(python)

如果直接采用递归的方法会有大量的重复计算，这样的求解和斐波那契数列的递归方法类似，参考网上做法，可以使用三个for循环可以简化该方法。

# -*- coding: UTF-8 -*-
import numpy as np
from scipy.special import comb, perm
from matplotlib import pyplot as pltclass MyBezier:def __init__(self, line):self.line = lineself.index_02 = None  # 保存拖动的这个点的索引self.press = None     # 状态标识，1为按下，None为没按下self.pick = None      # 状态标识，1为选中点并按下,None为没选中self.motion = None    # 状态标识，1为进入拖动,None为不拖动self.xs = list()      # 保存点的x坐标self.ys = list()      # 保存点的y坐标self.cidpress = line.figure.canvas.mpl_connect('button_press_event', self.on_press)        # 鼠标按下事件self.cidrelease = line.figure.canvas.mpl_connect('button_release_event', self.on_release)  # 鼠标放开事件self.cidmotion = line.figure.canvas.mpl_connect('motion_notify_event', self.on_motion)     # 鼠标拖动事件self.cidpick = line.figure.canvas.mpl_connect('pick_event', self.on_picker)                # 鼠标选中事件def on_press(self, event): # 鼠标按下调用if event.inaxes!=self.line.axes: returnself.press = 1def on_motion(self, event): # 鼠标拖动调用if event.inaxes!=self.line.axes: returnif self.press is None: returnif self.pick is None: returnif self.motion is None: # 整个if获取鼠标选中的点是哪个点self.motion = 1x = self.xsxdata = event.xdataydata = event.ydataindex_01 = 0for i in x:if abs(i - xdata) < 0.02: # 0.02 为点的半径if abs(self.ys[index_01] - ydata) < 0.02:breakindex_01 = index_01 + 1self.index_02 = index_01if self.index_02 is None: returnself.xs[self.index_02] = event.xdata # 鼠标的坐标覆盖选中的点的坐标self.ys[self.index_02] = event.ydataself.draw_01()def on_release(self, event): # 鼠标放开调用if event.inaxes!=self.line.axes: returnif self.pick == None: # 如果不是选中点，那就添加点self.xs.append(event.xdata)self.ys.append(event.ydata)if self.pick == 1 and self.motion != 1: # 如果是选中点，但不是拖动点，那就降阶x = self.xsxdata = event.xdataydata = event.ydataindex_01 = 0for i in x:if abs(i - xdata) < 0.02:if abs(self.ys[index_01] - ydata) < 0.02:breakindex_01 = index_01 + 1self.xs.pop(index_01)self.ys.pop(index_01)self.draw_01()self.pick = None # 所有状态恢复，鼠标按下到释放为一个周期self.motion = Noneself.press = Noneself.index_02 = Nonedef on_picker(self, event): # 选中调用self.pick = 1def draw_01(self): # 绘图函数self.line.clear()            # 不清除的话会保留原有的图self.line.axis([0,1,0,1])   # x和y范围0到1self.bezier(self.xs,self.ys) # Bezier曲线self.line.scatter(self.xs, self.ys,color='b',s=200, marker="o",picker=5) # 画点self.line.plot(self.xs, self.ys,color='r') # 画线self.line.figure.canvas.draw() # 重构子图def bezier(self,*args): # Bezier曲线公式转换，获取x和yn = len(args[0])  # 点的个数xarray,yarray = [],[]x,y = [],[]index = 0for t in np.linspace(0,1):for i in range(1,n):for j in range(0,n-i):if i == 1:xarray.insert(j,args[0][j]*(1-t) + args[0][j+1]*t)yarray.insert(j,args[1][j]*(1-t) + args[1][j+1]*t)continue# i != 1时,通过上一次迭代的结果计算xarray[j] = xarray[j]*(1 - t) + xarray[j+1]*tyarray[j] = yarray[j]*(1 - t) + yarray[j+1]*tif n == 1:x.insert(index,args[0][0])y.insert(index,args[1][0])else:x.insert(index,xarray[0])y.insert(index,yarray[0])xarray = []yarray = []index = index+1self.line.plot(x,y)fig = plt.figure(2,figsize=(12,6)) #创建第2个绘图对象,1200*600像素
ax = fig.add_subplot(111) #一行一列第一个子图
ax.set_title('My Bezier')
myBezier = MyBezier(ax)
plt.xlabel('X')
plt.ylabel('Y')
plt.show()

结果演示：

参考博客：
https://www.cnblogs.com/GH-123/p/7898774.html
https://blog.csdn.net/lafengxiaoyu/article/details/51296411
https://blog.csdn.net/lafengxiaoyu/article/details/56294678