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• Alternative ways to implement a triangular recursion
• How to speed up the plotting of B-spline curve?

• Alternative ways to implement a triangular recursion
• How to speed up the plotting of B-spline curve?

In addition, $u_0 \in [u_i,u_{i+1})$ knots = $\{u_0,u_1, \cdots, u_m\}$

Here is a procedural implementaion

$B_{n,i}(u)=\binom n i u^i(1-u)^{n-i}$, where $0 \leq u \leq 1$

So we can use the following triangular schematic digram to calculate $\color{blue}{B_{n,0},B_{n,0}, \cdots B_{n,n}}$

In addition, $u_0 \in [u_i,u_{i+1})$ knots = $\{u_0,u_1, \cdots, u_m\}, 0\leq u_i \leq u_j$

Here is a procedural implementaion calculate $\color{blue}{N_{i-p,p}(u_0),B_{i-p+1,p}(u_0), \cdots, N_{i,p}(u_0)}$

$$B_{n,i}(u)=\binom n i u^i(1-u)^{n-i}$$, where $0 \leq u \leq 1$

So we can use the following triangular schematic digram to calculate $\color{blue}{B_{n,0}(u),B_{n,0}(u), \cdots, B_{n,n}(u)}$

$$\left( \begin{array}{ccccc} & & & & N_{i-p,p}\left(u_0\right) \\ & & N_{i-2,2}\left(u_0\right) & & \\ & N_{i-1,1}\left(u_0\right) & & & \\ N_{i,0}\left(u_0\right) & & N_{i-1,2}\left(u_0\right) & \cdots & \vdots \\ & N_{i,1}\left(u_0\right) & & & \\ & & N_{i,2}\left(u_0\right) & & \\ & & & & N_{i,p}\left(u_0\right) \end{array} \right)$$

where $N_{m,n}$ was stored in the position $(p+1-i+m,n+1)$

So we can use the following triangular schematic digram to calculate $B_{n,0},B_{n,0}, \cdots B_{n,n}$

$$\left( \begin{array}{ccccc} \text{} & \text{} & \text{} & \text{} & B_{n,0} (u) \\ \text{} & \text{} & \text{} & .\cdot{}^{\cdot} & \text{} \\ \text{} & \text{} & B_{2,0}(u)& \text{} & \text{} \\ \text{} & B_{1,0} (u) & \text{} & \text{} & \text{} \\ B_{0,0} (u)=1 & \text{} & B_{2,1}(u)& \vdots & \vdots \\ \text{} & B_{1,0}(u) & \text{} & \text{} & \text{} \\ \text{} & \text{} & B_{2,2}(u)& \text{ } & \text{} \\ \text{} & \text{} & \text{} & \ddots & \text{} \\ \text{} & \text{} & \text{} & \text{} & B_{n,n} (u) \\ \end{array} \right)$$

$$\left( \begin{array}{ccccc} & & & & N_{i-p,p}\left(u_0\right) \\ & & N_{i-2,2}\left(u_0\right) & & \\ & N_{i-1,1}\left(u_0\right) & & & \\ \color{red}{N_{i,0}\left(u_0\right)=1} & & N_{i-1,2}\left(u_0\right) & \cdots & \vdots \\ & N_{i,1}\left(u_0\right) & & & \\ & & N_{i,2}\left(u_0\right) & & \\ & & & & N_{i,p}\left(u_0\right) \end{array} \right)$$

where $N_{m,n}$ was stored in the position $(p+1-i+m,n+1)$ of local array

So we can use the following triangular schematic digram to calculate $\color{blue}{B_{n,0},B_{n,0}, \cdots B_{n,n}}$

$$\left( \begin{array}{ccccc} \text{} & \text{} & \text{} & \text{} & B_{n,0} (u) \\ \text{} & \text{} & \text{} & .\cdot{}^{\cdot} & \text{} \\ \text{} & \text{} & B_{2,0}(u)& \text{} & \text{} \\ \text{} & B_{1,0} (u) & \text{} & \text{} & \text{} \\ \color{red}{B_{0,0} (u)=1} & \text{} & B_{2,1}(u)& \vdots & \vdots \\ \text{} & B_{1,0}(u) & \text{} & \text{} & \text{} \\ \text{} & \text{} & B_{2,2}(u)& \text{ } & \text{} \\ \text{} & \text{} & \text{} & \ddots & \text{} \\ \text{} & \text{} & \text{} & \text{} & B_{n,n} (u) \\ \end{array} \right)$$