0
$\begingroup$

I am taking a computer vision class and I have just learnt about the Harris corner detection concept. A corner is detected when a small shift in a window function defined around the corner results in a large $E(u,v)$ term, which is the sum of squared difference of pixels between the previous window and the next window. There is a post on this forum that has a very nice conceptual explanation about harris corner detectors.

However, given that $E(u,v)$ should be large for small changes in $(u,v)$, why is $E(u,v)$ function usually depicted as a function that has a negative maxima ? Shouldn't $E(u,v)$ value be high when $(u,v)$ is close to $(0,0)$

enter image description here

enter image description here

$\endgroup$

1 Answer 1

1
$\begingroup$

$E(u,v)$ is zero for zero shift ($u=v=0$), where the two patches being compared are equal. As you increase the distance, the difference tends to increase. Depending on the local shape of the image, the $E(u,v)$ surface will have a different shape: at a corner or a small point, it will steeply increase from 0 in all directions. Along an edge, it will remain close to 0 along one direction, and increase steeply in the perpendicular direction. At flat image regions, the function will remain closer to zero in all directions.

The Harris corner detector (and other corner detectors too) look at the shape of this patch, if it increases steeply in all directions, the detector will produce a stronger response.

$\endgroup$
1
  • $\begingroup$ Taking your advise of taking advantage of all +1 in a day :-). $\endgroup$
    – Royi
    Jul 1, 2023 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.