直線段與圓弧光柵化的計算方法
阿新 • • 發佈：2022-05-15
直線段光柵化
數值微分法（DDA演算法）
計算方法：
\(\Delta\)x = \(x_2-x_1\)，\(\Delta y=y_2-y_1\) ，\(k=\frac{\Delta y}{\Delta x}\)
當$ -1≤k≤1 $ 時：
\[\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1 \quad \\ y_{i+1} = y_i + k \quad \\ \end{matrix}\right. \end{array} \]當 $ k>1 $ 時：
\[\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + \frac{1}{k} \quad \\ y_{i+1} = y_i + 1 \quad \\ \end{matrix}\right. \end{array} \]當$ k<-1 $ 時：
演算法評價：
- 比直接使用\(y=kx+b\)更快
- 耗時（增加了浮點數運算，除法運算，取整操作等，不利於硬體實現）
\(Bresenham\)劃線演算法（重點掌握）
計算方法：
\(\Delta\)x = \(x_2-x_1\)，\(\Delta y=y_2-y_1\) ，\(k=\frac{\Delta y}{\Delta x}\)
- 當$ 0<k≤1 $ 時：
\(d_0 = 2dy -dx\)
繪製點的遞推公式：
誤差項遞推公式：
-
當 $ 1<k $ 時：
\(d_0 = 2dx -dy\)
\[\begin{array}{l} \left\{\begin{matrix} y_{i+1} = y_i + 1 \quad \\ x_{i+1} = x_i + \Delta x \quad \begin{array}{l} \left\{\begin{matrix} \Delta y = 0 \quad d_i<0 \\ \Delta y = 1 \quad d_i≥0 \\ \end{matrix}\right. \end{array} \\ \end{matrix}\right. \end{array} \]
繪製點的遞推公式： 誤差項遞推公式：
\[d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2(dx-dy) \quad d_i≥0\\ d_i + 2dx \quad d_i<0 \end{matrix}\right. \end{array} \]-
當 $ 1<k $ 時：
\(d_0 = 2dx -dy\)
\[\begin{array}{l} \left\{\begin{matrix} y_{i+1} = y_i + 1 \quad \\ x_{i+1} = x_i + \Delta x \quad \begin{array}{l} \left\{\begin{matrix} \Delta y = 0 \quad d_i<0 \\ \Delta y = 1 \quad d_i≥0 \\ \end{matrix}\right. \end{array} \\ \end{matrix}\right. \end{array} \]
繪製點的遞推公式： 誤差項遞推公式：
\[d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2(dx-dy) \quad d_i≥0\\ d_i + 2dx \quad d_i<0 \end{matrix}\right. \end{array} \]
-
注意： 感覺期末只可能考察斜率在 \(0\)~\(1\)之間且起點從原點開始的
若dx > 0, dy > 0, 0< m < 1:
xi = x1, yi = y1
第一項: pi = 2dy -dx
若pi < 0: pi = pi + 2dy, yi = yi
若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1
xi = xi + 1
輸出: (xi, yi)
若dx > 0, dy > 0, m > 1:
xi = x1, yi = y1
第一項: pi = 2dx -dy
若pi < 0: pi = pi + 2dx, xi = xi
若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1
yi = yi + 1
輸出: (xi, yi)
若dx > 0, dy < 0, 0< m < 1:
xi = x1, yi = -y1
第一項: pi = 2dy -dx
若pi < 0: pi = pi + 2dy, yi = yi
若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1
xi = xi + 1
輸出: (xi, -yi)
若dx > 0, dy < 0, m > 1:
xi = x1, yi = -y1
第一項: pi = 2dx -dy
若pi < 0: pi = pi + 2dx, xi = xi
若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1
yi = yi + 1
輸出: (xi, -yi)
若dx < 0, dy > 0, 0< m < 1:
xi = -x1, yi = y1
第一項: pi = 2dy -dx
若pi < 0: pi = pi + 2dy, yi = yi
若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1
xi = xi + 1
輸出: (-xi, yi)
若dx < 0, dy > 0, m > 1:
xi = -x1, yi = y1
第一項: pi = 2dx -dy
若pi < 0: pi = pi + 2dx, xi = xi
若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1
yi = yi + 1
輸出: (-xi, yi)
若dx < 0, dy < 0, 0< m < 1:
xi = -x1, yi = -y1
第一項: pi = 2dy -dx
若pi < 0: pi = pi + 2dy, yi = yi
若pi > 0: pi = pi + 2dy - 2dx, yi = yi + 1
xi = xi + 1
輸出: (-xi, -yi)
若dx < 0, dy < 0, m > 1:
xi = -x1, yi = -y1
第一項: pi = 2dx -dy
若pi < 0: pi = pi + 2dx, xi = xi
若pi > 0: pi = pi + 2dx - 2dy, xi = xi + 1
yi = yi + 1
輸出: (-xi, -yi)
評價方法：
- 只有整數運算，不含乘除法
- 只有加法和乘2運算，效率高
中點劃線演算法（重點掌握）
計算方法：（假定\(0≤k≤1\) ，\(x\)是最大位移方向）
\(d_i = F(M) = y_i+0.5-k(x_i+1)-b\)
繪製點的遞推公式：
\[\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1\\ y_{i+1} = \begin{array}{l} \left\{\begin{matrix} y_{i} + 1 \quad d<0 \\ y_{i} \quad d≥0 \end{matrix}\right. \end{array} \end{matrix}\right. \end{array} \] 誤差項遞推公式：
\(d_1 = 0.5 - k\)
\[d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 1 - k \quad d_i<0\\ d_i - k \quad d_i≥0 \end{matrix}\right. \end{array} \]改進的計算方法：（假定\(0≤k≤1\) ，\(x\)是最大位移方向）
用 \(2d\Delta x\) 代替 \(d\) ，令\(D=2d\Delta x\)
繪製點的遞推公式：
\[\begin{array}{l} \left\{\begin{matrix} x_{i+1} = x_i + 1\\ y_{i+1} = \begin{array}{l} \left\{\begin{matrix} y_{i} + 1 \quad d<0 \\ y_{i} \quad d≥0 \end{matrix}\right. \end{array} \end{matrix}\right. \end{array} \] 誤差項遞推公式：
\(D_1 = \Delta x - 2\Delta y\)
\[D_{i+1}= \begin{array}{l} \left\{\begin{matrix} D_i + 2\Delta x - 2\Delta y \quad D_i<0\\ D_i- 2\Delta y \quad D_i≥0 \end{matrix}\right. \end{array} \]圓弧光柵化
八分法畫圓
中點畫圓演算法
計算方法：
- 誤差項
- \(d<0\)時，下一點取\(Pu(x_i+1,y_i)\)
- \(d_i≥0\)時，下一點取\(Pd(x_i+1,y_i-1)\)
\(初項：d_0 =1.25-R\)
\[d_{i+1}= \begin{array}{l} \left\{\begin{matrix} d_i + 2x_i + 3 \quad d_i<0\\ d_i + 2(x_i-y_i) + 5 \quad d_i≥0 \end{matrix}\right. \end{array} \]改進計算方法：
用\(e=d-0.25代替d\)
\(初項：e_0 =1-R\)
\[e_{i+1}= \begin{array}{l} \left\{\begin{matrix} e_i + 2x_i + 3 \quad e_i<0\\ e_i + 2(x_i-y_i) + 5 \quad e_i≥0 \end{matrix}\right. \end{array} \]