python 曲线垂直斜率 python求直线斜率

基本图形生成算法

直线段

基础算法

计算斜率和截距，通过y = kx + b的直线表达式计算每一个x对应的y值

'''基础算法'''
def drawLine_Basic(grid, start, end):
  k = (end.y-start.y)/(end.x-start.x)
  b = start.y - k * start.x

  for xi in range(start.x, end.x):    # 栅格的性质
    yi = k * xi + b
    drawPixel(xi, int(yi+0.5), 1, grid)     # y坐标要进行近似

---

数值微分算法(DDA)

• 采用**“增量”**的思想

• 当|k|<=1时，x每增加1，y增加k
• 当|k|>1时，y每增加1，x增加1/k

• 证明: (这里只考虑|k|<=1当情况)
  由$x_{i+1} = x_{i} + 1$
  $y_{i+1} = k*x_{i+1} + b = k*(x_{i}+1) + b = k*x_{i} + b + k = y_{i} + k$

'''数值微分算法（DDA）'''
def drawLine_DDA(grid, start, end):
  k = (end.y - start.y) / (end.x - start.x)
  xi, yi = start.x, start.y

  if(abs(k<=1)):
    for xi in range(start.x, end.x):
      drawPixel(xi, int(yi+0.5), 1, grid)
      yi += k
  else:
    for yi in range(start.y, end.y):
      drawPixel(int(xi+0.5), yi, 1, grid)
      xi += 1/k

如果不对k进行分类讨论

不对k进行分类讨论

![在这里插入图片描述]() 对k进行分类讨论 ------

中点画线法

• 设直线方程为：ax + by + c =0

• a = y0 - y1
• b = x1 - x0
• c = x0y1 - x1y0

• **考核点：**(xp+1, yp+0.5)
• 判别式：$\Delta = F(x_p+1, y_p+0.5) = a*(x_p+1) + b*(y_p+0.5) + c$

• 如果$\Delta<0$ => Q点在M下方 => 选p2 (x+1, y+1)
• else， 选p1 (x+1, y)

'''中点画线法(k<=1)'''
def drwaLine_MidPoint(grid, start, end):
  a, b, c = start.y-end.y, end.x-start.x, start.x*end.y-end.x*start.y

  xp, yp = start.x, start.y
  for xp in range(start.x, end.x):
    drawPixel(xp, yp, 1, grid)

    delta = a*(xp+1) + b*(yp+0.5) + c   # 考核点(xp+1, yp+0.5)
    if delta<0:
      yp += 1
    else:
      # yp += 0
      pass

在中点画线法中添加增量的思想

• 若取p1，增量为a
• 若取p2，增量为a+b
• 初值：d0 = a + 0.5*b
• 由于只用d的符号来判断，可以用2d代替d，摆脱浮点数

'''中点画线法 with DDA'''
def drawLine_MidPoint_with_DDA(grid, start, end):
  a, b = start.y-end.y, end.x-start.x

  d = a + (b<<2)      # 用2d代替d， 摆脱小数
  d1, d2 = a<<2, (a+b)<<2

  xp, yp = start.x, start.y
  for xp in range(start.x, end.x):
    drawPixel(xp, yp, 1, grid)

    if d<0:
      yp += 1
      d += d2
    else:
      d += d1

---

Bresenham画线法

• 由误差项符号决定下一个像素选正右方还是右上方
• 判别式：$\varepsilon = y_{i+1} - y_{i,r} - 0.5$

• $\varepsilon > 0$, 取右上 (x+1, y+1)
• else，取正右 (x+1, y)

• 引入增量思想：

• $\varepsilon > 0$，增量为 k-1
• else，增量为 k
• 初始值：-0.5

'''Bresenham画线法(k<=1)'''
def drawLine_Bresenham(grid, start, end):
  k = (end.y - start.y) / (end.x - start.x)
  x, y = start.x, start.y
  e = -0.5

  for x in range(start.x, end.x):
    drawPixel(x, y, 1, grid)

    if e > 0:
      e += k - 1
      y += 1
    else:
      e += k
      # y += 0

去点浮

• 用$\varepsilon$代替$\varepsilon$
• 去掉k的计算
• 引入增量思想：

• $\varepsilon > 0$，增量为 2(dy - dx)
• else，增量为 2dy
• 初始值：-dx

'''Bresenham画线法(去点浮)(k<=1)'''
def drawLine_Bresenham_nonreal(grid, start, end):
    dx, dy = (end.x - start.x), (end.y - start.y)
    x, y = start.x, start.y
    e = -dx

    for x in range(start.x, end.x):
      drawPixel(x, y, 1, grid)

      if e > 0:
        e += (dy - dx) << 2
        y += 1
      else:
        e += (dy) << 2
        # y += 0

---

圆弧

暴力算法

• $y^2 = \sqrt{R^2 - x^2}$

---

中点画圆法

• 只需画1/8圆（第一象限 y>x 部分）
• 判别式：$F(x,y) = x^2 + y^2 - R^2$

• F>0，取正右方 (x+1, y)
• else，取右下方 (x+1, y-1)

• 增量思想：

• d<0, 增量为 2*x + 3
• else, 增量为 2*(x-y) + 5
• 初始值：1.25 - R

• **去点浮：**用e = d - 0.25代替 d

• 初始值：e = 1-R
• 循环条件：d < 0 <=> e < 0.25 <= e始终为整数 => e < 0

'''中点画圆法(DDA)'''
def drawArc_MidPoint_with_DDA(grid, R):
  d = 1 - R

  x, y = 0, R
  while x < y:
    drawPixel_symmetry8(x, y, 1, grid)

    if d < 0:
      x += 1
      d += 2*x + 3
    else:
      x += 1
      y -= 1
      d += ((x-y) << 1) + 5

对增量本身再次使用增量思想

• x递增1，d递增$\Delta x = 2$
• y递减1，d递增$\Delta\ y = 2$
• 初始值：

• $\Delta x = 3$
• $\Delta\ y = -2r + 2$

'''中点画圆法(DDA)(去点浮)'''
def drawArc_MidPoint_with_DDA_nonreal(grid, R):
  d = 1 - R
  deltax, deltay = 3, 2 - (R << 1)

  x, y = 0, R
  while x < y:
    drawPixel_symmetry8(x, y, 1, grid)

    if d < 0:
      x += 1
      d += deltax
      deltax += 2
    else:
      x += 1
      y -= 1
      d += (deltax + deltay)
      deltax += 2
      deltay += 2

---

Bresenham画圆法

• 选取距离真正的圆曲线近的点进行扩展



• 如果|OP1| > |OP2|，则选P2<=> $\sqrt{x_1^2 + y_1^2 - R^2} > \sqrt{R^2 - x_2^2 - y_2^2}$
  <=> $x_1^2 + y_1^2 + x_2^2 + y_2^2 - 2R^2 > 0$
• 当前像素的下一个扩展节点：正右方、右下方、正下方

1. 取H
2. 在H、D中选更近的
3. 取D
4. 在V、D中选更近的
5. 取V

• 令$\Delta H = |OH|$, $\Delta D = |OD|$, $\Delta V = |OV|$
  令$\delta HD = |\Delta H| - |\Delta D| = |OH| - |OD|$, $\delta DV = |\Delta D| - |\Delta V|$

1. $\delta HD = 2\Delta D + 2y - 1$
2. $\delta DV = 2\Delta D - 2x - 1$

• $\Delta D > 0$，若$\delta DV <= 0$，取D；else取V
• $\Delta D < 0$，$\delta HD <= 0$，取H；else，取D
• $\Delta D = 0$，取D

• 用增量法简化$\Delta D$，下一个像素取：

• H，增量取2x + 1
• D，增量取2x - 2y + 2
• V，增量取-2y + 1
• 初始值：-2r + 2

'''Bresenham画圆法'''
def drawArc_Bresenham(grid, R):
    delta = (1 - R) << 1

    x, y = 0, R
    while y >= 0:
        drawPixel_symmetry4(x, y, 1, grid)

        if delta < 0:
            delta1 = ((delta + y) << 1) - 1
            if delta1 <= 0:
                direction = 1
            else:
                direction = 2
        elif delta > 0:
            delta2 = ((delta - x) << 1) - 1
            if delta2 <= 0:
                direction = 2
            else:
                direction = 3
        else:
            direction = 2


        if direction == 1:      # 前进到 正右
            x += 1
            delta += (x << 1) + 1
        elif direction == 2:    # 前进到 右下
            x += 1
            y -= 1
            delta += ((x - y) << 1) + 2
        else:                   # 前进到 正下
            y -= 1
            delta += 1 - (y << 1)

---

正负法

• 圆方程：$F(x,y) = x^2 + y^2 - R^2$
• 设$P(x_i, y_i)$

• P在圆内 -> $D(x_i,y_i) <= 0$
• P在圆外 -> $D(x_i,y_i) > 0$

• 引入增量思想：

• 当F<=0，增量为2x + 1
• else，增量为-2y + 1
• 初始值：0

'''正负法'''
def drawArc_PositiveNegative(grid, R):
    F = 0

    x, y = 0, R
    while x <= y:
        drawPixel_symmetry8(x, y, 1, grid)
        print(F)
        if F <= 0:
            F += (x << 1) + 1
            x += 1
        else:
            F += 1 - (y << 1)
            y -= 1

---

圆内接正多边形逼近法

• $lim_{n->\infin}(n边正内接多边形) = 圆$

• 工程上，$n -> 2\pi R$就可以了，两个点的距离达到1pixel之后再小也没用了

• 圆的参数方程:

• $x_i = Rcos\theta_i$
• $y_i = Rsin\theta_i$

• 引入增量思想：
  $(\begin{matrix}x_{i+1} \\ y_{i+1} \end{matrix}) = (\begin{matrix}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{matrix})(\begin{matrix}x_{i} \\ y_{i} \end{matrix}) = (\begin{matrix}cos\alpha x_i - sin\alpha y_i \\ sin\alpha x_i + cos\alpha y_i \end{matrix})$

'''圆内接正多边形逼近法'''
def drawArc_InscribedRegularPolygonApproximate(grid, R):
    Alpha = 1/R
    cosAlpha, sinAlpha = cos(Alpha), sin(Alpha)

    x, y = R, 0
    while x >= y:
        drawPixel_symmetry8(int(x+0.5), int(y+0.5), 1, grid)

        x = cosAlpha * x - sinAlpha * y
        y = sinAlpha * x + cosAlpha * y