report_draft.md

1

题

给定以下三次多项式曲线 $$ P ( u ) = - \left( \begin{array} { c } { 7 / 8 } \ { 5 / 8 } \end{array} \right) u ^ { 3 } + \left( \begin{array} { c } { 9 } \ { 15 / 4 } \end{array} \right) u ^ { 2 } - \left( \begin{array} { c } { 57 / 2 } \ { 9 / 2 } \end{array} \right) u + \left( \begin{array} { c } { 30 } \ { - 1 } \end{array} \right) $$

1. 计算 $P(u)$ 的极形式及其在区间 $[2,4]$ 内的 Bezier 控制多边形的顶点 $P_0,P_1,P_2,P_3$，并大致勾勒出该控制多边形；
2. 用 de Casteljau 算法计算在采样点 $u={4/2,3,7/2}$ 处的多项式曲线 $P(u)$，并在 1. 图中画出；
3. 用 2. 中结果将曲线在 $u=3$ 处细分，再将右边部分曲线在中点 $u=7/2$ 处细分。将控制多边形在 1. 图中画出，并画出 $P(u)$ 表示的曲线。

解

• 红色的是 1. 的控制多边形
• 蓝色圆点是 2. 的三个采样点
• 绿色的多边形是 $u=3$ 处细分曲线的左部分的控制多边形，黄色和洋红色的多边形是右部分在中点细分的两曲线的控制多边形
• 青色的是曲线 $P(u)$

1

极形式为 $$ \begin{aligned} q(u_1,u_2,u_3)= &- \left( \begin{array} { c } { 7 / 8 } \ { 5 / 8 } \end{array} \right) u_1 u_2 u_3

• \left( \begin{array} { c } { 9 } \ { 15 / 4 } \end{array} \right) \frac{u_1u_2+u_2u_3+u_1u_3}{3}\ &- \left( \begin{array} { c } { 57 / 2 } \ { 9 / 2 } \end{array} \right) \frac{u_1+u_2+u_3}{3}
• \left( \begin{array} { c } { 30 } \ { - 1 } \end{array} \right) \end{aligned} $$

则控制多变性顶点为 $$ \begin{aligned} P_0&=q(2,2,2)=\left( \begin{array} { c } { 2 } \ { 0 } \end{array} \right)\ P_1&=q(2,2,4)=\left( \begin{array} { c } { 0 } \ { 2 } \end{array} \right)\ P_2&=q(2,4,4)=\left( \begin{array} { c } { 3 } \ { 4 } \end{array} \right)\ P_3&=q(4,4,4)=\left( \begin{array} { c } { 4 } \ { 1 } \end{array} \right)\ \end{aligned} $$

2

t = 0.75
P00=[2,0];
P01=[0,2];
P02=[3,4];
P03=[4,1];

P10=(1-t)*P00 + t*P01
P11=(1-t)*P01 + t*P02
P12=(1-t)*P02 + t*P03

P20=(1-t)*P10 + t*P11
P21=(1-t)*P11 + t*P12

P30=(1-t)*P20 + t*P21

$$ P^{(k)}i(u)=\frac{4-u}{4-2}P_i^{(k-1)}+\frac{u}{4-2}P^{(k-1)}{i+1}\quad(k=1,\dots,3,\ i=0,\dots3-k) $$

u = 4/2

$$ \begin{aligned} P^{(1)}0&=1\cdot P_0^{(0)}+0\cdot P^{(0)}{1}=\left( \begin{array} { c } { 2 } \ { 0 } \end{array} \right)\ P^{(1)}1&=1\cdot P_1^{(0)}+0\cdot P^{(0)}{2}=\left( \begin{array} { c } { 0 } \ { 2 } \end{array} \right)\ P^{(1)}2&=1\cdot P_2^{(0)}+0\cdot P^{(0)}{3}=\left( \begin{array} { c } { 3 } \ { 4 } \end{array} \right)\ P^{(2)}0&=1\cdot P_0^{(1)}+0\cdot P^{(1)}{1}=\left( \begin{array} { c } { 2 } \ { 0 } \end{array} \right)\ P^{(2)}1&=1\cdot P_1^{(1)}+0\cdot P^{(1)}{2}=\left( \begin{array} { c } { 0 } \ { 2 } \end{array} \right)\ P^{(3)}0&=1\cdot P_0^{(2)}+0\cdot P^{(2)}{1}=\left( \begin{array} { c } { 2 } \ { 0 } \end{array} \right)\ \end{aligned} $$

u = 3

$$ \begin{aligned} P^{(1)}0&=0.5\cdot P_0^{(0)}+0.5\cdot P^{(0)}{1}=\left( \begin{array} { c } { 1 } \ { 1 } \end{array} \right)\ P^{(1)}1&=0.5\cdot P_1^{(0)}+0.5\cdot P^{(0)}{2}=\left( \begin{array} { c } { 1.5 } \ { 3 } \end{array} \right)\ P^{(1)}2&=0.5\cdot P_2^{(0)}+0.5\cdot P^{(0)}{3}=\left( \begin{array} { c } { 3.5 } \ { 2.5 } \end{array} \right)\ P^{(2)}0&=0.5\cdot P_0^{(1)}+0.5\cdot P^{(1)}{1}=\left( \begin{array} { c } { 1.25 } \ { 2 } \end{array} \right)\ P^{(2)}1&=0.5\cdot P_1^{(1)}+0.5\cdot P^{(1)}{2}=\left( \begin{array} { c } { 2.5 } \ { 2.75 } \end{array} \right)\ P^{(3)}0&=0.5\cdot P_0^{(2)}+0.5\cdot P^{(2)}{1}=\left( \begin{array} { c } { 1.875 } \ { 2.375 } \end{array} \right)\ \end{aligned} $$

u = 7/2

$$ \begin{aligned} P^{(1)}0&=0.25\cdot P_0^{(0)}+0.75\cdot P^{(0)}{1}=\left( \begin{array} { c } { 0.5 } \ { 1.5 } \end{array} \right)\ P^{(1)}1&=0.25\cdot P_1^{(0)}+0.75\cdot P^{(0)}{2}=\left( \begin{array} { c } { 2.25 } \ { 3.5 } \end{array} \right)\ P^{(1)}2&=0.25\cdot P_2^{(0)}+0.75\cdot P^{(0)}{3}=\left( \begin{array} { c } { 3.75 } \ { 1.75 } \end{array} \right)\ P^{(2)}0&=0.25\cdot P_0^{(1)}+0.75\cdot P^{(1)}{1}=\left( \begin{array} { c } { 1.8125 } \ { 3 } \end{array} \right)\ P^{(2)}1&=0.25\cdot P_1^{(1)}+0.75\cdot P^{(1)}{2}=\left( \begin{array} { c } { 3.375 } \ { 2.1875 } \end{array} \right)\ P^{(3)}0&=0.25\cdot P_0^{(2)}+0.75\cdot P^{(2)}{1}=\left( \begin{array} { c } { 2.984375 } \ { 2.390625 } \end{array} \right)\ \end{aligned} $$

3

在 $u=3$ 处细分的两条 Bezier 曲线的控制点分别为 $\left{P_0^{(0)},P_0^{(1)},P_0^{(2)},P_0^{(3)}\right}$ 和 $\left{P_0^{(3)},P_1^{(2)},P_2^{(1)}, P_3^{(0)}\right}$

右部曲线细分后的两条 Bezier 曲线的控制点分别为 $$ \left{ \left(\begin{matrix} 1.875\ 2.375 \end{matrix}\right),

\left(\begin{matrix} 2.1875\ 2.5625 \end{matrix}\right),

\left(\begin{matrix} 2.59375\ 2.59375 \end{matrix}\right),

\left(\begin{matrix} 2.984375\ 2.390625 \end{matrix}\right) \right} $$

$$ \left{ \left(\begin{matrix} 2.984375\ 2.390625 \end{matrix}\right),

\left(\begin{matrix} 3.375\ 2.1875 \end{matrix}\right),

\left(\begin{matrix} 3.75\ 1.75\ \end{matrix}\right),

\left(\begin{matrix} 4\ 1 \end{matrix}\right) \right} $$

2

题

给定以下三次多项式曲线 $$ F ( u ) = \left( \begin{array} { c } { 15 } \ { - 6 } \end{array} \right) u ^ { 3 }

• \left( \begin{array} { c } { 27 } \ { 10 } \end{array} \right) u ^ { 2 }

• \left( \begin{array} { l } { 9 } \ { 9 } \end{array} \right) u $$ 及参数区间 $[0,1]$

1. 计算 $F$ 的一阶和二阶导数；

2. 计算 $F$ 的极形式 $f(u_1,u_2,u_3)$ 及导数 $F^\prime$ 和 $F^{\prime\prime}$ 的极形式，证明它们分别等于 $3f(u_1,u_2,\hat{1})$ 和 $6f(u_1,\hat{1},\hat{1})$。

   注：$f \left( u _ { 1 } , u _ { 2 } , \hat { 1 } \right) = f \left( u _ { 1 } , u _ { 2 } , 1 \right) - f \left( u _ { 1 } , u _ { 2 } , 0 \right)$

解

1

$$ \begin{aligned} F^\prime(u) &= \left( \begin{array} { c } { 45 } \ { - 18 } \end{array} \right) u ^ { 2 }

• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right) u

• \left( \begin{array} { l } { 9 } \ { 9 } \end{array} \right)\ F^{\prime\prime}(u) &= \left( \begin{array} { c } { 90 } \ { - 36 } \end{array} \right) u

• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right)\ \end{aligned} $$

2

$$ f(u_1,u_2,u_3)= \left( \begin{array} { c } { 15 } \ { -6 } \end{array} \right) u_1 u_2 u_3

• \left( \begin{array} { c } { 27 } \ { 10 } \end{array} \right) \frac{u_1u_2+u_2u_3+u_1u_3}{3}

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right) \frac{u_1+u_2+u_3}{3} $$

记 $f^{(1)}$ 和 $f^{(2)}$ 分别为 $F^\prime$ 和 $F^{\prime\prime}$ 的极形式，有 $$ \begin{aligned} f^{(1)}(u_1,u_2)&= \left( \begin{array} { c } { 45 } \ { -18 } \end{array} \right) u_1u_2

• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right) \frac{u_1+u_2}{2}

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right)\

f^{(2)}(u_1)&= \left( \begin{array} { c } { 90 } \ { -36 } \end{array} \right) u_1

• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right)\ \end{aligned} $$ 由于 $$ \begin{aligned} 3f(u_1,u_2,\hat{1}) =&3\left(f(u_1,u_2,1)-f(u_1,u_2,0)\right)\ =&\left( \begin{array} { c } { 45 } \ { -18 } \end{array} \right) u_1 u_2
• \left( \begin{array} { c } { 27 } \ { 10 } \end{array} \right) (u_1u_2+u_2+u_1)

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right) (u_1+u_2+1)\ &- \left( \begin{array} { c } { 27 } \ { 10 } \end{array} \right) u_1u_2

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right) (u_1+u_2)\ =& \left( \begin{array} { c } { 45 } \ { -18 } \end{array} \right) u_1 u_2
• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right) \frac{u_2+u_1}{2}

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right)\ \end{aligned} $$ 故 $f^{(1)}(u_1,u_2)=3f(u_1,u_2,\hat{1})$

由于 $$ \begin{aligned} 6f(u_1,\hat{1},\hat{1})&=2\left(3f(u_1,1,\hat{1})-3f(u_1,0,\hat{1})\right)\ &=2\left(f^{(1)}(u_1,1)-f^{(1)}(u_1,0)\right)\ &=2\left( \left( \begin{array} { c } { 45 } \ { -18 } \end{array} \right) u_1

• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right) \frac{u_1+1}{2}

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right)
• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right) \frac{u_1}{2}

• \left( \begin{array} { c } { 9 } \ { 9 } \end{array} \right) \right)\ &= \left( \begin{array} { c } { 90 } \ { -36 } \end{array} \right) u_1
• \left( \begin{array} { c } { 54 } \ { 20 } \end{array} \right)\ \end{aligned} $$ 故 $f^{(2)}(u_1)=6f(u_1,\hat{1},\hat{1})$

3

题

给定由一下四点及结点向量 $[0,1,2,3,4,5]$ 定义的均匀 B 样条 $$ P _ { 0 } = \left( \begin{array} { c } { - 2 } \ { - 10 } \end{array} \right) , \quad P _ { 1 } = \left( \begin{array} { c } { - 4 } \ { 2 } \end{array} \right) , \quad P _ { 2 } = \left( \begin{array} { c } { 6 } \ { 5 } \end{array} \right) , \quad P _ { 3 } = \left( \begin{array} { c } { 4 } \ { - 7 } \end{array} \right) $$

1. 用 de Boor 算法计算曲线在 $t=2.5$ 处的位置。勾勒出控制多边形和此算法构造出的相关点；
2. 对于 1. 中的 B 样条，计算能表示同一曲线的相应 Bezier 控制顶点。在 1. 图中画出控制顶点和 Bezier 曲线

解

• 红色的圆点是 $P_0^{(0)},P_1^{(0)},P_2^{(0)},P_3 ^{(0)}$，红色虚线是相应控制多边形
• 绿色的圆点是 $P_0^{(1)},P_1^{(1)},P_2^{(1)}$
• 黄色的圆点是 $P_0^{(2)},P_1^{(2)}$
• 青色的圆点是 $P_0^{(3)}$
• 青色的实曲线是 Bezier 曲线

1

$$ \begin{aligned} P^{(1)}0&=\frac{0.5}{3}\cdot P_0^{(0)}+\frac{2.5}{3}\cdot P^{(0)}{1}=\left( \begin{array} { c } { -11/3 } \ { 0 } \end{array} \right)\ P^{(1)}1&=\frac{1.5}{3}\cdot P_1^{(0)}+\frac{1.5}{3}\cdot P^{(0)}{2}=\left( \begin{array} { c } { 1 } \ { 3.5 } \end{array} \right)\ P^{(1)}2&=\frac{2.5}{3}\cdot P_2^{(0)}+\frac{0.5}{3}\cdot P^{(0)}{3}=\left( \begin{array} { c } { 17/3 } \ { 3 } \end{array} \right)\ P^{(2)}0&=\frac{0.5}{2}\cdot P_0^{(1)}+\frac{1.5}{2}\cdot P^{(1)}{1}=\left( \begin{array} { c } { -1/6 } \ { 2.625 } \end{array} \right)\ P^{(2)}1&=\frac{1.5}{2}\cdot P_1^{(1)}+\frac{0.5}{2}\cdot P^{(1)}{2}=\left( \begin{array} { c } { 13/6 } \ { 3.375 } \end{array} \right)\ P^{(3)}0&=\frac{0.5}{1}\cdot P_0^{(2)}+\frac{0.5}{1}\cdot P^{(2)}{1}=\left( \begin{array} { c } { 1 } \ { 3 } \end{array} \right)\ \end{aligned} $$

2

设极形式为 $$ \pmb{p}(t_1,t_2,t_3)=\pmb{c}_0+\frac{t_1+t_2+t_3}{3}\pmb{c}_1+\frac{t_1t_2+t_2t_3+t_1t_3}{3}\pmb{c}_2+t_1t_2t_3\pmb{c}_3 $$ 则 $$ \begin{aligned} P_0&=\pmb{p}(0,1,2)=\pmb{c}_0+\pmb{c}_1+\frac{2}{3}\pmb{c}_2\ P_1&=\pmb{p}(1,2,3)=\pmb{c}_0+2\pmb{c}_1+\frac{11}{3}\pmb{c}_2+6\pmb{c}_3\ P_2&=\pmb{p}(2,3,4)=\pmb{c}_0+3\pmb{c}_1+\frac{26}{3}\pmb{c}_2+24\pmb{c}_3\ P_3&=\pmb{p}(3,4,5)=\pmb{c}_0+4\pmb{c}_1+\frac{47}{3}\pmb{c}_2+60\pmb{c}_3\ \end{aligned} $$ 令 $$ A=\left[\begin{matrix} 1 & 1 & 1 & 1\ 1 & 2 & 3 & 4\ \frac{ 2}{3} & \frac{ 11}{3} & \frac{26}{3} & \frac{47}{3}\ 0 & 6 & 24 & 60 \end{matrix}\right], P=\left[\begin{matrix} P_0 & P_1 & P_2 & P_3 \end{matrix}\right] $$ 则 $$ \left[\begin{matrix} \pmb{c}_0 & \pmb{c}_1 & \pmb{c}_2 & \pmb{c}_3 \end{matrix}\right]A=P $$ 即 $$ \left[\begin{matrix} \pmb{c}_0 & \pmb{c}_1 & \pmb{c}_2 & \pmb{c}_3 \end{matrix}\right]=PA^{-1}=

\left[\begin{matrix} 46& -68 & 30 & -4\ -24.5& 13.5 & 1.5 & -1\ \end{matrix}\right] $$ 故 Bezier 控制顶点为 $$ \begin{aligned} \pmb{b}_0&=\pmb{p}(2,2,2)=\left[\begin{matrix} -2\ 0.5 \end{matrix}\right]\ \pmb{b}_1&=\pmb{p}(2,2,3)=\left[\begin{matrix} -\frac{2}{3}\ 3 \end{matrix}\right]\ \pmb{b}_2&=\pmb{p}(2,3,3)=\left[\begin{matrix} \frac{8}{3}\ 4 \end{matrix}\right]\ \pmb{b}_3&=\pmb{p}(3,3,3)=\left[\begin{matrix} 4\ 2.5 \end{matrix}\right]\ \end{aligned} $$