def cubic_bezier_interpolation(
points: Iterable['Vertex']) -> Iterable[Bezier4P]:
""" Returns an interpolation curve for given data `points` as multiple cubic
Bézier-curves. Returns n-1 cubic Bézier-curves for n given data points,
curve i goes from point[i] to point[i+1].
Args:
points: data points
.. versionadded:: 0.13
"""
from ezdxf.math import tridiagonal_matrix_solver
# Source: https://towardsdatascience.com/b%C3%A9zier-interpolation-8033e9a262c2
points = Vec3.tuple(points)
if len(points) < 3:
raise ValueError('At least 3 points required.')
num = len(points) - 1
# setup tri-diagonal matrix (a, b, c)
b = [4.0] * num
a = [1.0] * num
c = [1.0] * num
b[0] = 2.0
b[num - 1] = 7.0
a[num - 1] = 2.0
# setup right-hand side quantities
points_vector = [points[0] + 2.0 * points[1]]
points_vector.extend(2.0 * (2.0 * points[i] + points[i + 1])
for i in range(1, num - 1))
points_vector.append(8.0 * points[num - 1] + points[num])
# solve tri-diagonal linear equation system
solution = tridiagonal_matrix_solver((a, b, c), points_vector)
control_points_1 = Vec3.list(solution.rows())
control_points_2 = [
p * 2.0 - cp for p, cp in zip(points[1:], control_points_1[1:])
]
control_points_2.append((control_points_1[num - 1] + points[num]) / 2.0)
for defpoints in zip(points, control_points_1, control_points_2,
points[1:]):
yield Bezier4P(defpoints)
def test_tridiagonal_matrix_solver(tridiag):
result = tridiagonal_matrix_solver(tridiag, zip(B1, B2, B3))
assert result == TRI_SOLUTION