1
$\begingroup$

I've already asked a similar question in mathematics exchange without getting an answer, I read a bit more and I think the question might be more suitable for the signal processing exchange.

Basically I was reviewing my background in edge detection methods and I've found the following statement:

For many applications, however, we wish to think such a continuous gradient image to only return isolated edges, i.e., as single pixels at discrete locations along the edge contours. This can be achieved by looking for maxima in the edge strength (gradient magnitude) in a direction perpendicular to the edge orientation, i.e., along the gradient direction. Finding this maximum corresponds to taking a directional derivative of the strength field in the direction of the gradient and then looking for zero crossing. The desired directional derivative is equivalent to the dot product between a second gradient operator and the result of the first... The gradient dot product with the gradient is called the Laplacian.

I did the calculations but I'm not getting the laplacian, however I've accidentally found this Jim Little - CPSC 505 Example: Laplacian vs Second Directional Derivative, where basically the explanation seems essentially to be "directional derivative and laplacian are different but not that much, so given that taking the laplacian instead of the second directional derivative is less expensive (computationally speaking) we can use the laplacian"

My question is then... is this what Szelisky mean?

$\endgroup$

1 Answer 1

0
$\begingroup$

The 2nd directional is a linear combination of the 2nd derivative in each direction.
The Laplacian is a specific combination as such.
The source you linked indeed specify that in practice it coincide:

enter image description here

$\endgroup$
2
  • $\begingroup$ Yes but "in practice" i.e. there's no theoretical equality between the two. $\endgroup$ Jul 23 at 3:14
  • $\begingroup$ @user8469759, He went for intuition over formulation. $\endgroup$
    – Mark
    Jul 23 at 6:35

Your Answer

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

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