图像边缘检测拉普拉斯推导

图像点、线、边缘检测（1）——拉普拉斯

闲着无聊，整理一下图像中拉普拉斯算子的计算过程（离散差分方程）
公式来源：Digital Image Processing, 4th

关于$f(x)$的导数公式推导

• 将函数 $f(x+\Delta x)$ 关于$x$ 进行泰勒系数展开：
  $$\begin{array}{rl}f(x+\Delta x)& =f(x)+\Delta x\frac{\partial f(x)}{\partial x}+\frac{\Delta x^2}{2!}\frac{{\partial }^{2}f(x)}{\partial x^2}+\frac{\Delta x^3}{3!}\frac{{\partial }^{3}f(x)}{\partial x^3}+...\\ & =\sum _{n=0}^{\infty }\frac{(\Delta x{)}^n}{n!}\frac{{\partial }^{n}f(x)}{\partial x^n}\end{array}\text{\hspace{1em}(1)}$$

• 当 $\Delta x=1$(即像素后一位):
  $$\begin{array}{rl}f(x+1)& =f(x)+\frac{\partial f(x)}{\partial x}+\frac{1}{2!}\frac{{\partial }^{2}f(x)}{\partial x^2}+\frac{1}{3!}\frac{{\partial }^{3}f(x)}{\partial x^3}+\dots \\ & =\sum _{n=0}^{\infty }\frac{1}{n!}\frac{{\partial }^{n}f(x)}{\partial x^n}\end{array}\text{\hspace{1em}(2)}$$

• 当 $\Delta x=-1$(即像素前一位):
  $$\begin{array}{rl}f(x-1)& =f(x)-\frac{\partial f(x)}{\partial x}+\frac{1}{2!}\frac{{\partial }^{2}f(x)}{\partial x^2}-\frac{1}{3!}\frac{{\partial }^{3}f(x)}{\partial x^3}+...\\ & =\sum _{n=0}^{\infty }\frac{(-1{)}^n}{n!}\frac{{\partial }^{n}f(x)}{\partial x^n}\end{array}\text{\hspace{1em}(3)}$$

• 由$Eq.(2)$得到前向差分:
  $$\frac{\partial f(x)}{\partial x}=f(x+1)-f(x)\text{\hspace{1em}(4)}$$

• 同理, 由 $Eq.(3)$得到后向差分:
  $$\frac{\partial f(x)}{\partial x}=f(x)-f(x-1)\text{\hspace{1em}(5)}$$

• 由 $Eq.(3)$ - $Eq.(2)$得到中心差分:
  $$\frac{\partial f(x)}{\partial x}=f^{\prime }(x)=\frac{f(x+1)-f(x-1)}{2}\text{\hspace{1em}(6)}$$

• 二阶导数, ${\partial }^{2}f(x)/\partial x^2$ $Eq.(2)$ $Eq.(3)$:
  $$\frac{{\partial }^{2}f(x)}{\partial x^2}=f^{\prime \prime }(x)=f(x+1)-2f(x)+f(x-1)\text{\hspace{1em}(7)}$$

$f(x,y)$的导数推导

• 由$Eq.(7)$扩展至$f(x,y)$，对$x$求导可得：
  $$\frac{{\partial }^{2}f(x,y)}{\partial x^2}=f(x+1,y)-2f(x,y)+f(x-1,y)\text{\hspace{1em}(8)}$$

• 同理，对$y$求导得：
  $$\frac{{\partial }^{2}f(x,y)}{\partial y^2}=f(x,y+1)-2f(x,y)+f(x,y-1)\text{\hspace{1em}(9)}$$

拉普拉斯算子

• 拉普拉斯算子(Laplacian):
  $${\nabla }^{2}f(x,y)=\frac{{\partial }^{2}f(x,y)}{\partial x^2}+\frac{{\partial }^{2}f(x,y)}{\partial y^2}\text{\hspace{1em}(10)}$$

• 根据$Eq.(8)$$Eq.(9)$以及拉普拉斯算子可得二阶差分展开式：
  $$\begin{array}{rl}{\nabla }^{2}f(x,y)& =f(x+1,y)-2f(x,y)+f(x-1,y)+f(x,y+1)-2f(x,y)+f(x,y-1)\\ & =f(x+1,y)+f(x-1,y)+f(x,y+1)+f(x,y-1)-4f(x,y)\end{array}\text{\hspace{1em}(11)}$$