bspline的渐进迭代逼近（PIA）方法无法工作？

如果bspline的PIA方法无法工作，有一种可能的解决方法是检查bspline的控制点是否被正确输入。以下是一个示例代码，它可以使用PIA方法计算bspline曲线：

import numpy as np
import matplotlib.pyplot as plt

def bspline(control_points, degree, n_samples):
    n_control_points = len(control_points)
    knot_vector = generate_knot_vector(n_control_points, degree)

    # Evaluate bspline curve at given parameter values
    parameter_values = np.linspace(0, n_control_points - 1, n_samples)
    curve_points = np.zeros((n_samples, 2))
    for i, t in enumerate(parameter_values):
        curve_points[i] = evaluate_bspline_at_parameter_value(control_points, degree, knot_vector, t)

    return curve_points

def generate_knot_vector(n_control_points, degree):
    # Construct a clamped knot vector
    n_knots = n_control_points + degree + 1
    knot_vector = np.zeros(n_knots)
    knot_vector[degree + 1:n_control_points + 1] = np.arange(1, n_control_points - degree)
    knot_vector[n_control_points + 1:] = n_control_points - degree

    return knot_vector

def evaluate_bspline_at_parameter_value(control_points, degree, knot_vector, t):
    n_control_points = len(control_points)
    i = find_span(n_control_points, degree, t, knot_vector)
    basis_functions = compute_basis_functions(i, t, knot_vector, degree)

    curve_point = np.zeros(2)
    for j in range(degree + 1):
        curve_point += basis_functions[j] * control_points[i - degree + j]

    return curve_point

def find_span(n_control_points, degree, t, knot_vector):
    if t == knot_vector[-1]:
        return n_control_points - 1

    low = degree
    high = n_control_points
    mid = (low + high) // 2

    while t < knot_vector[mid] or t >= knot_vector[mid + 1]:
        if t < knot_vector[mid]:
            high = mid