Canny Edge Detection

This is a multi-step edge detection procedure by Canny. The purpose of the following two methods is to detect edges with noise supressed at the same time.

1. Smooth the image with a Gaussian filter to reduce noise and unwanted details and textures.
   
   $$g(m,n)=G_{\sigma}(m,n)*f(m,n)$$
   
   
   where
   
   $$G_{\sigma}=\frac{1}{\sqrt{2\pi\sigma^2}}exp[-\frac{m^2+n^2}{2\sigma^2}]$$
   
   

2. Compute gradient of $g(m,n)$ using any of the gradient operatiors (Roberts, Sobel, Prewitt, etc) to get:
   
   $$M(n,n)=\sqrt{g_m^2(m,n)+g_n^2(m,n)}$$
   
   
   and
   
   $$\theta(m,n)=tan^{-1}[g_n(m,n)/g_m(m,n)]$$
   
   

3. Threshold M:
   
   $$M_T(m,n)=\left\{ \begin{array}{ll} M(m,n) & \mbox{if $M(m,n)>T$} 0 & \mbox{otherwise} \end{array} \right.$$
   
   
   where $T$ is so chosen that all edge elements are kept while most of the noise is suppressed.

4. Suppress non-maxima pixels in the edges in $M_T$ obtained above to thin the edge ridges (as the edges might have been broadened in step 1). To do so, check to see whether each non-zero $M_T(m,n)$ is greater than its two neighbors along the gradient direction $\theta(m,n)$. If so, keep $M_T(m,n)$ unchanged, otherwise, set it to 0.

5. Threshold the previous result by two different thresholds $\tau_1$ and $\tau_2$ (where $\tau_1<\tau_2$) to obtain two binary images $T_1$ and $T_2$. Note that compared to $T_1$, $T_2$ has less noise and fewer false edges but larger gaps between edge segments.

6. Link edges segments in $T_2$ to form continuous edges. To do so, trace each segment in $T_2$ to its end and then search its neighbors in $T_1$ to find any edge segment in $T_1$ to bridge the gap untill reaching another edge segment in $T_2$.

Example 1:

Edge detection by gradient operators (Roberts, Sobel and Prewitt):

Edge detection by LoG and DoG:

Edge detection by Canny method ($\sigma=1,2,3$, $\tau_1=0.3$, $\tau_2=0.7$):

Example 2:

Edge detection results by Sobel, Prewitt gradient operators, by DoG method and by Canny's method ($\sigma=5$, $\tau_1=0.8$, $\tau_2=0.95$):