Formula derivation and code realization of B-spline curve

This article only briefly describes the derivation of the formula of the B-spline curve, and gives a code implementation. Before reading this article, make sure you have some background knowledge of B-spline curves. Relevant knowledge can be learned through the following courses: MOOC-Computer Graphics-China Agricultural University-Zhao Ming or watch the port version of Station B.

formula definition

Given the following parameters:

• $n+1$ control point$P_i(i=0,1,\cdots ,n)$
• $n+k+1(2⩽k⩽n+1)$ Numbers:{ ui ∣ i = 0 , 1 , ⋯ , n + k } \{u_i|i=0,1,\cdots,n+k\}{ ui​∣i=0,1,⋯,n+k}

Structure polynomial function
$$\mathcal{P}(u)=\sum _{i=0}^n{P}_i{B}_{i,k}(u),u\in [u_{k-1},u_{n+1}]\text{\hspace{1em}( 1 )}$$ to approximate the curve. where$B_{i,k}\left(u\right)$ is called$k$ order ($k-1$次）B样条基函数，它是由DeBoor-Cox递推公式来定义的：
$$\begin{array}{rl}& B_{i,1}(u)=\{\begin{array}{ll}1,& u_i⩽u<u_{i+1}\\ 0,& otherwise\end{array}，\\ & B_{i,k}(u)=\frac{u-u_i}{u_{i+k-1}-u_i}{B}_{i,k-1}\left(u\right)+\frac{u_{i+k}-u}{u_{i+k}-u_{i+1}}{B}_{i+1,k-1}\left(u\right)。\end{array}\text{\hspace{1em}( 2 )}$$ If 0 0is encountered in the recursive formula$0$ divided by$0$ , the result of the division is$0$。

$B_{i,k}\left(u\right)$ can be regarded as the control point$P_i$in $The\; weight\; under\; u$ , because the B-spline basis function has weight:
$$\sum _{i=0}^n{B}_{i,k}(u)\equiv 1,\forall u。\text{\hspace{1em}(3)}$$

对于某个$u\in [u_j,u_{j+1})$ , use formula (2) to deduce$B_{i,k}\left(u\right)$ , so that the final result only contains$B_{x,1}\left(u\right)$ , we can derive$x=i,i+1,\cdots ,i+k-1$ . And according to$B_{x,1}\left(u\right)$ definition (2), we find that only whenj ∈ { i , i + 1 , ⋯ i + k − 1 } j\in\{i,i+1,\cdots i+k-1\}j∈{ i,i+1,⋯i+k−1}，即 i ∈ { j − ( k − 1 ) , ⋯   , j } i\in\{j-(k-1),\cdots,j\} i∈{ j−(k−1),⋯,j}时，$B_{i,k}\left(u\right)$ may not be 0.

Let me explain with an example. With $i=7,k=3$ as an example, use formula (2) to deduce$B_{7,3}(u)$：
$$\begin{array}{rl}{B}_{7,3}(u)& =c_0{B}_{7,2}\left(u\right)+c_1{B}_{8,2}\left(u\right)\\ & =c_2{B}_{7,1}\left(u\right)+c_3{B}_{8,1}\left(u\right)+c_4{B}_{9,1}(u)\end{array}$$(where $c_0,c_1,\cdots ,c_4$are coefficients), we found that the final result only contains $B_{7,1}(u),B_{8,1}\left(u\right),B_{9,1}\left(u\right)$。对于某个$u\in [u_j,u_{j+1})$ , only whenj ∈ { 7 , 8 , 9 } j\in\{7,8,9\}j∈{ 7,8,9}时$B_{7,1}(u),B_{8,1}\left(u\right),B_{9,1}\left(u\right)$ are not all 0, so that$B_{7,3}\left(u\right)$ may not be 0.

Algorithm principle

对于某个$u\in [u_j,u_{j+1}]$ , we define
$$d_{i,0}=P_i,i=j-(k-1),\cdots ,j，\text{\hspace{1em}(4)}$$ $$\begin{array}{rl}{a}_{i,r}& =\frac{u-u_i}{u_{i+k-r}-u_i},\\ d_{i,r}& =(1-a_{i,r})d_{i-1,r-1}+a_{i,r}{d}_{i,r-1}\end{array}\left(\begin{array}{l}r=1,2,\cdots ,k-1\\ i=j-(k-1)+r,\cdots ,j\end{array}\right)。\text{\hspace{1em}(5)}$$

Here, for i = j − ( k − 1 ) + r in formula (5) , ⋯ , ji = j-(k-1)+r,\cdots,j$i=j-(k-1)+r,\cdots ,j$ makes an explanation: recursively$d_{i,r}$Make the final result only contain $d_{x,0}$，得到$x=i-r,\cdots ,i$ . by$d_{x,0}$The definition formula (4) knows, ∀ x , x ∈ { j − ( k − 1 ) , ⋯ , j } \forall x, x\in\{j-(k-1),\cdots,j\}∀x,x∈{ j−(k−1),⋯,j}。因此有$i-r\in \{j-(k-1),\cdots ,j\},\cdots ,i\in \{j-(k-1),\cdots ,j\}$ , so that$i\in \{j-(k-1)+r,\cdots ,j\}$。

It can be proved that
$$\mathcal{P}(u)=d_{j,k-1}。\text{\hspace{1em}(6)}$$

So the logic of the algorithm is as follows (assuming that the input only provides $n+1$ control point, parameterui ( i ∈ { 0 , 1 , ⋯ , n + k } ) u_i(i\in\{0,1,\cdots,n+k\})ui​(i∈{ 0,1,⋯,n+k } ) is selected by us):

• Take any interval in the real number field, you might as well take $[0,1]$。
• will $[0,1]$ evenly spaced as$n+k$ segment (number of control points =$n+1$ , curve order =$k$）。
• After separation, n + k + 1 n+k+1 is generated$n+k+1$ separation point, recorded as$u_0,u_1,\cdots ,u_{n+k}$. So $u_j=d/(n+k)$。
• interval $[u_{k-1},u_{n+1}]$ on a large number of samples. For each sample$u$ , computation$\mathcal{P}(u)$。
• According to formula (6), we can calculate $d_{j,k-1}$To calculate $\mathcal{P}(u)$。
• $d_{j,k-1}$It can be calculated recursively by formula (5).

Algorithm implementation

def draw_curve(p_list, k: int = 4):
	"""
    :param p_list: (list of list of int: [[x0, y0], [x1, y1], [x2, y2], ...]) 曲线的控制点坐标列表
    :param k: B样条曲线的阶数，默认为4阶，即3次均匀B样条曲线
    :return: (list of list of int: [[x_0, y_0], [x_1, y_1], [x_2, y_2], ...]) 绘制结果的像素点坐标列表
    """
    n = len(p_list) - 1
    if n == 0:
        return p_list
    elif n == 1:
        return draw_line(p_list) # 调用绘制直线的算法
        
    assert 2 <= k <= n + 1
    freq = 500
    result = []
    
    for j in range(k - 1, n + 1):
        seg = n + k
        u_j = j / seg
        for u in range(freq + 1):
            u = u / (freq * seg) + u_j
            tmp = [
                [p_list[i + j - (k-1)][0],
                 p_list[i + j - (k-1)][1]
                ] for i in range(k)
            ]
            for r in range(1, k):
                for i in range(k - 1, r - 1, -1):
                    ii = i + j - (k - 1)
                    u_i = ii / seg
                    alpha = (u - u_i) / ((ii+k-r)/seg - u_i)
                    tmp[i] = [
                        (1-alpha)*tmp[i-1][0] + alpha*tmp[i][0],
                        (1-alpha)*tmp[i-1][1] + alpha*tmp[i][1]
                    ]
            result.append(
                [round(tmp[k-1][0]), round(tmp[k-1][1])]
            )
    return result

Supplement: $Derivation\; of\; P\left(u\right)$ Domain

In formula (1), we directly define the variable uuThe value range of$u$$[u_{k-1},u_{n+1}]$ . In this section we will explain its principle in detail.

According to the recursive formula (2), we can infer:

• $B_{i,1}$and 1 interval ( $[u_i,u_{i+1}]$ ) 2 control points ($u_i$sum $u_{i+1}$）
• $B_{i,2}$Reason $B_{i,1}$sum $B_{i+1,1}$Composition, involving 2 intervals and 3 control points
• $B_{i,3}$Reason $B_{i,2}$Sum $B_{i+1,2}$Composition, involving 3 intervals and 4 control points
• $\cdots$
• $B_{i,k}$involving $k$ intervals$k+1$ control point

Thus it is possible to define $B_{i,k}$support interval is $[u_i,u_{i+k}]$。

For example, as shown above , where $n=4,k=4$ . According to the definition of the support interval, we have:

• $P_0{B}_{0,4}$Involves node $u_0$to $u_4$
• $P_1{B}_{1,4}$Involving node $u_1$to $u_5$
• $\cdots$
• $P_4{B}_{4,4}$Involving node $u_4$to $u_8$

For each interval $[u_i,u_{i+1}]$ , which contains at most$In\; k$ different support intervals, that is, it is at most$K$ control points are related. The figure belowgraphically demonstrates this relationship.

Our goal is, forP ( u ) \mathcal{P}(u)Each interval in the domain of$P$$($$u$$)\; should\; be\; related\; to\; as\; many\; control\; points\; as\; possible.$So we finally chose$[u_{k-1},u_{n+1}]$ . possible,$\forall [u_i,u_{i+1}]\subseteq [u_{k-1},u_{n+1}]$，$[u_i,u_{i+1}]$ both with$K$ control points are related.

more details

Due to space limitations, some proofs are not described in detail in this paper. More proof details can be found in the related paper. For example C. DE BOOR, On calculating with B-splines , 1972 , etc.