1. 零次 B 样条

$${\mathbf{F}}_{i，0}(t)=\{\begin{array}{ll}1& t_i\le t<t_{i+1}\\ 0& others\end{array}$$

2. 一次 B 样条，
$${\mathbf{F}}_{i,1}(t)=\frac{t-t_i}{t_{i+1}-t_i}{\mathbf{F}}_{i,0}(t)+\frac{t_{i+2}-t}{t_{i+2}-t_{i+1}}{\mathbf{F}}_{i+1,0}(t)$$

通过

$${\mathbf{F}}_{i,0}(t)=\{\begin{array}{ll}1& t_i\le t<t_{i+1}\\ 0& others\end{array}{\mathbf{F}}_{i+1,0}(t)=\{\begin{array}{ll}1& t_{i+1}\le t<t_{i+2}\\ 0& others\end{array}$$

可以得到

$${\mathbf{F}}_{i,1}(t)=\{\begin{array}{ll}\frac{t-t_i}{t_{i+1}-t_i}& t-i\le t<t_{i=1}\\ & \\ \frac{t_{i+2}-t}{t_{i+2}-t_{i+1}}& t_{i+1}\le t<t_{i+2}\\ & \\ 0& others\end{array}$$

3. 二次 B 样条

$${\mathbf{F}}_{i,2}(t)=\frac{t-t_i}{t_{i+2}-t_i}{\mathbf{F}}_{i,1}(t)+\frac{t_{i+3}-t}{t_{i+3}-t_{i+1}}{\mathbf{F}}_{i+1,1}(t)$$

由 $${\mathbf{F}}_{i,1}(t)=\{\begin{array}{ll}\frac{t-t_i}{t_{i+1}-t_i}& t_i\le t<t_{i+1}\\ & \\ \frac{t_{i+2}-t}{t_{i+2}-t_{i+1}}& t_{i+1}\le t<t_{i+2}\\ & \\ 0& others\end{array}{\mathbf{F}}_{i+1,1}(t)=\{\begin{array}{ll}\frac{t-t_{i+1}}{t_{i+2}-t_{i+1}}& t_{i+1}\le t<t_{i+2}\\ & \\ \frac{t_{i+3}-t}{t_{i+3}-t_{i+2}}& t_{i+2}\le t<t_{i+3}\\ & \\ 0& others\end{array}$$

得到

$${\mathbf{F}}_{i,1}(t)=\{\begin{array}{llc}\frac{t-t_i}{t_{i+2}-t_i}\frac{t-t_i}{t_{i+1}-t_i}& t_i\le t<t_{i+1}& \\ & & \\ \frac{t-t_i}{t_{i+2}-t_i}\frac{t_{i+2}-t}{t_{i+2}-t_{i+1}}+\frac{t_{i+3}-t_{i+1}}{t_{i+3}-t_{i+1}}\frac{t-t_{i+1}}{t_{i+2}-t_{i+1}}& t_{i+1}\le t<t_{i+2}& \\ & & \\ \frac{(t_{i=3}-t{)}^2}{(t_{i=3}-t_{i+1})(t_{i+3}-t_{i+2})}& t_{i+2}\le t<t_{i+3}& \\ & & \\ 0& others& \end{array}$$

根据参数 $t$ 的值和次数 $k$ 与节点矢量双精度数组 $Knot$ 计算第 $i$ 个 $k$ 次的 B 样条基函数的 $F_{i,k}(t)$

分析

$$F_{i,k}(t)=\frac{t-t_i}{t_{i+k}-t_i}{F}_{i,k-1}(t)+\frac{t_{i+k+1}-t}{t_{i+k+1}-t_{i+1}}{F}_{i+1,k-1}(t)$$

$$\begin{gathered} numberato{r}_1=t-t_i \\ numberato{r}_2=t_{i+k+1}-t \\ denominato{r}_1=t_{i+k}-t_i \\ denominato{r}_2=t_{i+k+1}-t_{i+1} \\ cofficien{t}_1=\frac{numberato{r}_1}{denominato{r}_1} \\ cofficien{t}_1=\frac{numberato{r}_2}{denominato{r}_2} \\ valu{e}_1=cofficien{t}_1\times F_{i,k-1}(t) \\ valu{e}_2=cofficien{t}_2\times F_{i+1,k-1}(t) \end{gathered}$$

double BasisFunctionValue(double t, int i, int k)
{double val1, val2, val;if (k == 0){if ((t >= knot[i]) && t < knot[i + 1]){return 1.0;}else{	// 其它return 0.0;}}if (k > 0) {if (t < knot[i] || t > knot[i + k + 1]) {return 0.0;		// 其它}else{double coffcient1, coffcient2;	// 凸组合系数1 凸组合系数 2double denominator = 0.0;		// 分母denominator = knot[i + k] - knot[i];	if (denominator == 0.0){// 约定 0/0 = 0coffcient1 = 0.0;}else{coffcient1 = (t - knot[i]) / denominator;	// 计算的第一项}denominator = knot[i + k + 1] - knot[i + 1];	// 递推公式第二项分母if (denominator == 0.0){// 约定 0/0 = 0coffcient2 = 0.0;}else{coffcient2 = (knot[i + k + 1] - t) / denominator;	// 递推公式第二项}val1 = coffcient1 * BasisFunctionValue(t, i, k - 1);	// 递推公式第一项的只val2 = coffcient2 * BasisFunctionValue(t, i+1, k - 1);	// 递推公式第二项的只val = val1 + val2;	// 基函数的值}}return val;
}