如果已知 NURBS 曲线的控制点，如何找到结向量？

Question

我有一组控制点

pts = [[849, 1181],
       [916, 1257],
       [993, 1305],
       [1082,1270], 
       [1137,1181],
       [1118,1055], 
       [993,1034], 
       [873,1061], 
       [849, 1181]]

我有生成开结向量的逻辑：

/*
Subroutine to generate a B-spline open knot vector with multiplicity
equal to the order at the ends.

c            = order of the basis function
n            = the number of defining polygon vertices
nplus2       = index of x() for the first occurence of the maximum knot       vector value
nplusc       = maximum value of the knot vector -- $n + c$
x()          = array containing the knot vector
*/

knot(n,c,x)

int n,c;
int x[];

{
    int nplusc,nplus2,i;
nplusc = n + c;
nplus2 = n + 2;

x[1] = 0;
    for (i = 2; i <= nplusc; i++){
        if ( (i > c) && (i < nplus2) )
            x[i] = x[i-1] + 1;
    else
            x[i] = x[i-1];


    }
}

还有一个用于生成周期性节点向量：

/*  Subroutine to generate a B-spline uniform (periodic) knot vector.

c            = order of the basis function
n            = the number of defining polygon vertices
nplus2       = index of x() for the first occurence of the maximum knot vector value
nplusc       = maximum value of the knot vector -- $n + c$
x[]          = array containing the knot vector
*/

#include    <stdio.h>

knotu(n,c,x)

int n,c;
int x[];

{
    int nplusc,nplus2,i;

nplusc = n + c;
nplus2 = n + 2;

x[1] = 0;
for (i = 2; i <= nplusc; i++){
    x[i] = i-1;
}
}

然而，我需要在 [0,1]

范围内生成一个非均匀节点矢量

以上算法产生统一的结向量。

如果有什么办法，请提出建议。如果代码在 python

中会更可取

Answer 1

结向量（均匀或不均匀）是 NURBS 曲线定义的一部分。所以，你实际上可以定义自己的 non-uniform 结向量，只要结向量遵循基本规则：

1) 节点值的数量 = 控制点的数量 + 顺序

2) 所有节点值必须是non-decreasing。即，k[i] <= k[i+1].

对于具有 9 个控制点的示例，您可以使用 non-uniform 节点向量，例如 [0, 0, 0, 0, a, b, c, d, e, 1, 1, 1, 1 ]，其中 0.0 < a <= b <= c <=d <=e < 1.0 对于 3 B-spline 曲线。当然，为a、b、c、d和e选择不同的值会导致不同形状的曲线。

如果已知 NURBS 曲线的控制点，如何找到结向量？

How to find knot vector if control points are known for a NURBS curve?

python

curve-fitting

nurbs

bspline

basis