直线段与圆弧光栅化的计算方法

直线段光栅化

数值微分法（DDA算法）

计算方法：

$\Delta$x = $x_2-x_1$，$\Delta y=y_2-y_1$ ，$k=\frac{\Delta y}{\Delta x}$

当$-1≤k≤1$ 时：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1 \quad \\ y_{i+1} = y_i + k \quad \\ \end{matrix}\right. \end{array}$$

当 $k>1$ 时：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + \frac{1}{k} \quad \\ y_{i+1} = y_i + 1 \quad \\ \end{matrix}\right. \end{array}$$

当$k<-1$ 时：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i - \frac{1}{k} \quad \\ y_{i+1} = y_i - 1 \quad \\ \end{matrix}\right. \end{array}$$

算法评价：

• 比直接使用$y=kx+b$更快
• 耗时（增加了浮点数运算，除法运算，取整操作等，不利于硬件实现）

$Bresenham$划线算法（重点掌握）

计算方法：

$\Delta$x = $x_2-x_1$，$\Delta y=y_2-y_1$ ，$k=\frac{\Delta y}{\Delta x}$ $d_i=\Delta x(s_i-t_i)$

• 当$0<k≤1$ 时：

​ $d_0 = 2dy -dx$
​ 绘制点的递推公式：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1 \quad \\ y_{i+1} = y_i + \Delta y \quad \begin{array}{l} \left\{\begin{matrix} \Delta y = 0 \quad d_i<0 \\ \Delta y = 1 \quad d_i≥0 \\ \end{matrix}\right. \end{array} \\ \end{matrix}\right. \end{array} \\ $$

​ 误差项递推公式：

$$d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2(dy-dx) \quad d_i≥0\\ d_i + 2dy \quad d_i<0 \end{matrix}\right. \end{array}$$

• 当 $1<k$ 时：

  ​ $d_0 = 2dx -dy$
  ​ 绘制点的递推公式：

  $$\begin{array}{l} \left\{\begin{matrix} y_{i+1} = y_i + 1 \quad \\ x_{i+1} = x_i + \Delta x \quad \begin{array}{l} \left\{\begin{matrix} \Delta y = 0 \quad d_i<0 \\ \Delta y = 1 \quad d_i≥0 \\ \end{matrix}\right. \end{array} \\ \end{matrix}\right. \end{array}$$

  ​ 误差项递推公式：

  $$d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2(dx-dy) \quad d_i≥0\\ d_i + 2dx \quad d_i<0 \end{matrix}\right. \end{array}$$

  • 当 $1<k$ 时：

    ​ $d_0 = 2dx -dy$
    ​ 绘制点的递推公式：

    $$\begin{array}{l} \left\{\begin{matrix} y_{i+1} = y_i + 1 \quad \\ x_{i+1} = x_i + \Delta x \quad \begin{array}{l} \left\{\begin{matrix} \Delta y = 0 \quad d_i<0 \\ \Delta y = 1 \quad d_i≥0 \\ \end{matrix}\right. \end{array} \\ \end{matrix}\right. \end{array}$$

    ​ 误差项递推公式：

    $$d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2(dx-dy) \quad d_i≥0\\ d_i + 2dx \quad d_i<0 \end{matrix}\right. \end{array}$$

注意： 感觉期末只可能考察斜率在 $0$~$1$之间且起点从原点开始的

若dx > 0, dy > 0, 0< m < 1:

xi = x1, yi = y1

第一项: pi = 2dy -dx

若pi < 0: pi = pi + 2dy, yi = yi

若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1

xi = xi + 1

输出: (xi, yi)

若dx > 0, dy > 0, m > 1:

xi = x1, yi = y1

第一项: pi = 2dx -dy

若pi < 0: pi = pi + 2dx, xi = xi

若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1

yi = yi + 1

输出: (xi, yi)

若dx > 0, dy < 0, 0< m < 1:

xi = x1, yi = -y1

第一项: pi = 2dy -dx

若pi < 0: pi = pi + 2dy, yi = yi

若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1

xi = xi + 1

输出: (xi, -yi)

若dx > 0, dy < 0, m > 1:

xi = x1, yi = -y1

第一项: pi = 2dx -dy

若pi < 0: pi = pi + 2dx, xi = xi

若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1

yi = yi + 1

输出: (xi, -yi)

若dx < 0, dy > 0, 0< m < 1:

xi = -x1, yi = y1

第一项: pi = 2dy -dx

若pi < 0: pi = pi + 2dy, yi = yi

若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1

xi = xi + 1

输出: (-xi, yi)

若dx < 0, dy > 0, m > 1:

xi = -x1, yi = y1

第一项: pi = 2dx -dy

若pi < 0: pi = pi + 2dx, xi = xi

若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1

yi = yi + 1

输出: (-xi, yi)

若dx < 0, dy < 0, 0< m < 1:

xi = -x1, yi = -y1

第一项: pi = 2dy -dx

若pi < 0: pi = pi + 2dy, yi = yi

若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1

xi = xi + 1

输出: (-xi, -yi)

若dx < 0, dy < 0, m > 1:

xi = -x1, yi = -y1

第一项: pi = 2dx -dy

若pi < 0: pi = pi + 2dx, xi = xi

若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1

yi = yi + 1

输出: (-xi, -yi)

评价方法：

• 只有整数运算，不含乘除法
• 只有加法和乘2运算，效率高

中点划线算法（重点掌握）

计算方法：（假定$0≤k≤1$ ，$x$是最大位移方向）

$d_i = F(M) = y_i+0.5-k(x_i+1)-b$

​ 绘制点的递推公式：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1\\ y_{i+1} = \begin{array}{l} \left\{\begin{matrix} y_{i} + 1 \quad d<0 \\ y_{i} \quad d≥0 \end{matrix}\right. \end{array} \end{matrix}\right. \end{array}$$

​ 误差项递推公式：

​ $d_1 = 0.5 - k$

$$d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 1 - k \quad d_i<0\\ d_i - k \quad d_i≥0 \end{matrix}\right. \end{array}$$

改进的计算方法：（假定$0≤k≤1$ ，$x$是最大位移方向）

用 $2d\Delta x$ 代替 $d$ ，令$D=2d\Delta x$

​ 绘制点的递推公式：

$$\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1\\ y_{i+1} = \begin{array}{l} \left\{\begin{matrix} y_{i} + 1 \quad d<0 \\ y_{i} \quad d≥0 \end{matrix}\right. \end{array} \end{matrix}\right. \end{array}$$

​ 误差项递推公式：

​ $D_1 = \Delta x - 2\Delta y$

$$D_{i+1}= \begin{array}{l} \left\{\begin{matrix} D_i + 2\Delta x - 2\Delta y \quad D_i<0\\ D_i- 2\Delta y \quad D_i≥0 \end{matrix}\right. \end{array}$$

圆弧光栅化

八分法画圆

中点画圆算法

计算方法：

• 误差项

$$d = F(x_M,y_M)=F(x_i+1，y_i-0.5)=(x+1)^2+(y_i-0.5)^2-R^2$$

• $d<0$时，下一点取$Pu(x_i+1,y_i)$
• $d_i≥0$时，下一点取$Pd(x_i+1,y_i-1)$

$初项：d_0 =1.25-R$

$$d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2x_i + 3 \quad d_i<0\\ d_i + 2(x_i-y_i) + 5 \quad d_i≥0 \end{matrix}\right. \end{array}$$

改进计算方法：

用$e=d-0.25代替d$

$初项：e_0 =1-R$

$$e_{i+1}= \begin{array}{l} \left\{\begin{matrix} e_i + 2x_i + 3 \quad e_i<0\\ e_i + 2(x_i-y_i) + 5 \quad e_i≥0 \end{matrix}\right. \end{array}$$