In the previous paragraph, we mentioned that the filtering process typically takes place on a specific neighborhood of pixels. When this neighborhood process is applied for all pixels, it is called sliding neighborhood operation. In it, we slide a rectangular neighborhood window through all possible positions of the image and modify its central pixel using a function of the pixels in the neighborhood.

Let's see how this is done, using a numeric example. We'll start with something simple, like a linear filtering process, that is, averaging. Let's suppose that we have a small image, sized 8x8 pixels and we want to modify its pixel values, so that they get assigned with the rounded average of the pixels' values in their 3x3 neighborhoods.

This will be easier to explain by using a real numeric example. Let's explain what happens in the step shown in the following image, in which the central pixel of the highlighted 3x3 neighborhood (in the fourth row and sixth column) will be replaced by the average value of all the pixels in the neighborhood (rounded to the nearest integer):

Let the image be called I, the result in pixel I(4,6) will be:

Substituting the values of the pixels, we can calculate the average value:

Hence, the value of the central pixel of the neighborhood will become 121 (the closest integer to 120.89).

By repeating the process described previously for all the pixels of the image, we get a result commonly known as mean filtering or average filtering. The final result of the entire process is shown in the following figure:

You may be wondering now; the choice of neighborhood, for the example, was very convenient, but what happens when we want to change the value of a pixel on the borders of the image such as let's say pixel I(1,4)? Why was it set to 77 as shown in the image?

This is indeed a valid and natural question, and you are very intuitive if you already thought about it. The answer is that the way to tackle this problem when you want your resulting image to be the same size as your original image is to involve only the neighboring pixels that exist in your calculations. However, since in our example, the calculation that has to be performed is averaging the neighborhood pixels, the denominator will still be 9, hence, it will be like we pad the rest of the neighborhood with zeros. Let's demonstrate this example as well:

As shown in the previous image, the central pixel value gets evaluated as follows:

Of course, since there is no 0th line, the first three operands of the addition are non-existent, hence set to zero:

Therefore, the result of the averaging process for the aforementioned neighborhood will be equal to 77 (as shown in the image). This approach is not the only one we have for the image borders. We could assign the maximum possible value (255 for our example) to the non-existent pixels, or assign them the mean value of the rest of the neighborhood, and so on. The choice we make affects the quality of the borders of the image, as we will see in real pictures later on.