Filter that gives the max change in matrix value

Question

I have a 3x3 matrix, I want to find the maximum change in value that any one element in the vector has with the centre element (given that this element has the opposite sign to the centre element). Out of curiosity, would there be a filter that I could convolve with the matrix, which would give me this output?

say I have this matrix: \begin{bmatrix} 1 & 2 & 3\\ 2 &-1 & 3\\ 1 & 5& 2 \end{bmatrix} would there be a filter that could give the output $6$?

Answer 1

No, because the function $f$ that you are interested in is a nonlinear function of the input matrix, while convolving matrices $A, B$ with a filter $F$ would always satisfy $$convolve(F, A+B) = convolve(F, A) + convolve(F, B).$$

To see that the function you are interested in is nonlinear, consider the sum of two matrices $$ A+ B = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 5 & 0 \\ \end{array}\right] + \left[\begin{array}{ccc} 0 & 4 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \\ \end{array}\right] = \left[\begin{array}{ccc} 0 & 4 & 0 \\ 0 & -3 & 0 \\ 0 & 5 & 0 \\ \end{array}\right] $$ Then $f(A)=6$, $f(B)=6$, but $f(A+B) = 8 \neq 6+6$.

Answer 2

There is a sub-field if image processing called Mathematical Morphology. In this field, we focus on non-linear filters based on max and min operators.

One of the basic filters is thus the dilation, a local maximum filter. Think of a convolution, but instead of multiplying kernel values to image values and adding up the results, we add kernel values to the image values and take the maximum of the result.

With a kernel that is composed of only zeros, this turns into finding the maximum pixel value in a neighborhood.

Using a 3x3 square neighborhood, we obtain the maximum value over the neighboring elements. Subtracting the input, we obtain the largest difference of a pixel with its neighbors.

This difference is not symmetric. That is, if you invert the image by negating each pixel, you would get a different result. By adding in the opposite operation (erosion, a local minimum), it is possible to make the difference symmetric:

$$ \left( \delta(f)-f \right) \vee \left( f-\epsilon(f) \right)$$

($\delta$ is the dilation, $\epsilon$ is the erosion, $f$ is the image, and $\vee$ is the supremum/maximum)