# Mathematical explanation for image edge detection and denoising

I am trying to understand why the convolution kernel, $$\left[\begin{array}{rrr} -1&-1&-1\\ 2&2&2\\ -1&-1&-1 \end{array}\right]$$ detects the edges in an image. If anyone has a mathematical reason for this, please post it. Thanks.

Also, why does the a $3$-by-$3$ matrix of $1/9$ work for denoising?

• For your second question, the 3-by-3 matrix of $\frac{1}{9}$ does averaging, and you may think that as a low-pass filter to filter out or average out image noises. – peterwhy Oct 3 '13 at 23:19
• Thank you for your response. I do understand the averaging; If possible could you elaborate though - why does that specific matrix work for virtually any noisy image? Why not say a 3-by-3 matrix of 1/3 for example? – eTothEipiPlus1 Oct 3 '13 at 23:24
• I presume the above only detects edges in one direction? – copper.hat Oct 3 '13 at 23:26
• @user98592 $\frac{1}{9}$ is better because this convolution does not increase overall magnitude/amplitude/value/intensity (I don't know the right word for this...). Think of performing this convolution on a large array of $1$'s, you would not wish to make the image brighter (for example, if we are talking about brightness) after denoising. – peterwhy Oct 3 '13 at 23:28

The convolution with matrix $\left[\begin{array}{rrr} -1&-1&-1\\ 2&2&2\\ -1&-1&-1 \end{array}\right]$ gives a low value on a more even region of an array, and a high value if there is a large difference between that point and the previous or next point in vertical direction. You can try to perform a convolution on an array with horizontal edge like

$$\left[\begin{array}{ccccc} 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 1&1&1&1&1\\ 1&1&1&1&1\\ 1&1&1&1&1\\ \end{array}\right]$$

Around the horizontal edge, the convolution gives a higher absolute value than in the more even area. As @copper.hat pointed out, this convolution detects better for horizontal or almost horizontal edges, but not vertical edges, as this convolution kernel does not calculate differences between columns.

For your second question, the 3-by-3 matrix of $\dfrac{1}{9}\left[\begin{array}{rrr} 1&1&1\\ 1&1&1\\ 1&1&1 \end{array}\right]$ does averaging, and you may think that as a low-pass filter to filter out or average out image noises. $\frac{1}{9}$ is better than other values like $\frac{1}{3}$ because this convolution does not increase overall magnitude/amplitude/value/intensity (I don't know the right word for this...). Think of performing this convolution on a large array of $1$'s, you would not wish to make the image brighter (for example, if we are talking about brightness) after denoising.

There is a number of different ways to detect (horizontal) edges. The simplest way is to use the kernel $\left[\begin{array}{r} 1\\ 0\\ -1 \end{array}\right]$, which approximates the y-derivative. High values in the result of the convolution with this kernel will correspond to horizontal edges.

The convolution with matrix $\left[\begin{array}{rrr} -1&-1&-1\\ 2&2&2\\ -1&-1&-1 \end{array}\right]$ is a different way to detect horizontal edges. It approximates the second derivative the y-direction. Convolution with this kernel will give you a positive value on one side of the edge and a negative value on the other side. Then you can look for zero-crossings, which allows you to localize edges with sub-pixel accuracy. You can also generalize this approach to detecting edges of any orientation using the Laplacian.

Here is some more information on using the zeros-crossing of the Laplacian for edge detection.