There are some errors in the notations, aren't there? at 32:17 the magnitude of the gradient is compared to the absolute value of the Laplacian, at 32:18 there is also the Laplacian sign instead of the gradiant..
on 15:00 why is the extra g(x) and g(y) introduced? The Laplace operator on the gaussian is (dg/dx) + (dg/dy) (apologize for d, I don't know where this partial derivative sign is on the keyboard...). So due to the distributivity of the convolution (which is f * (g + x) = (f*g) + (f*x)) it should become (D^2 g) * I = (dg/dx + dg/dy) * I = (dg/dx) * I + (dg/dy) * I. Where did these extra terms came from? While they only filter/blur (extra gaussian filtering) even more, I don't think the derivation is correct... :|
+Achilles Kappis Here gxx(x) is the laplacian operator (1D laplacian) and g(x) is the gaussian operator.(1D again). As he mentioned in the previous slides you can first perform laplacian and then use the gaussian operator. He is doing that here.
+Vaibhav Deshu Well, thanks for the reply. I actually found the solution of that, which is based on multiplicatevily separable functions. So the statement is that if f(x,y) = g(x) * z(y) (the product of two functions each of which is a function of each independent variable alone), then the second derivative in respect to x of f(x,y) = (d^2 g(x)/dx^2 ) * z(y) and the second derivative in respect to y of f(x,y) = (d^2 z(y)/dy^2 ) * g(x) (as we can see here the second derivative g(x) is multiplied with the z(y), and this is the reason later on the video we can see it like this... so in this equation the gx and gy should change places), so the Laplacian operator is actually the sum of the above two which equals [(d^2 g(x)/dx^2 ) * z(y)] + [(d^2 z(y)/dy^2 ) * g(x)]. And all this thing is convolved with the image, but because convolution is distributive we have the result he is showing. What I haven't found a proof yet is why if a function is multiplicatively separable (like the Gaussian) f(x,y) = g(x) * z(y) and you wanna convolve it with another function (which is the image) you can convolve the two functions (g(x) and z(y)) separately... I mean..., why [g(x)*z(y)]**w(x,y) = [w(x,y)**g(x)]**z(y) (here ** means convolution 'cause I use * for multiplication :D). If you have any clue about it or you can point me to some direction your comments would be most welcome...
I currently being tortured with a combination of neuroscience and doppler radar, someone, please help. I have been undergoing this tortures for at least 3 years. I started in december 2012, when a socialworker student marked me for torture.
Most intuitive series I have found so far on computer vision. Really puts things into perspective. Thanks Prof!
Who would downvote this kind of lecture?
i guess some other jealous professors ;)
Very well explained sir, may Allah bless you
allah is a false god.
who is your god ??? cow ??? rat ??idol ??darwen??
continued from L2 at 53:08
Thanks a lot. best lecture for canny.. :)
May Allah bless u sir,,wonderful work
23:09 Canny Edge Detector
tq
There are some errors in the notations, aren't there? at 32:17 the magnitude of the gradient is compared to the absolute value of the Laplacian, at 32:18 there is also the Laplacian sign instead of the gradiant..
Prof. Shah thanks a lot.
Wonderful elaboration .. Thank you very much.
18:33 gxx(x) then g(y) is still unclear to me. It would be great if anyone can explain in great detail. Thanks!
on 15:00 why is the extra g(x) and g(y) introduced? The Laplace operator on the gaussian is (dg/dx) + (dg/dy) (apologize for d, I don't know where this partial derivative sign is on the keyboard...). So due to the distributivity of the convolution (which is f * (g + x) = (f*g) + (f*x)) it should become (D^2 g) * I = (dg/dx + dg/dy) * I = (dg/dx) * I + (dg/dy) * I. Where did these extra terms came from? While they only filter/blur (extra gaussian filtering) even more, I don't think the derivation is correct... :|
+Achilles Kappis Here gxx(x) is the laplacian operator (1D laplacian) and g(x) is the gaussian operator.(1D again). As he mentioned in the previous slides you can first perform laplacian and then use the gaussian operator. He is doing that here.
+Vaibhav Deshu Well, thanks for the reply. I actually found the solution of that, which is based on multiplicatevily separable functions.
So the statement is that if f(x,y) = g(x) * z(y) (the product of two functions each of which is a function of each independent variable alone), then the second derivative in respect to x of f(x,y) = (d^2 g(x)/dx^2 ) * z(y) and the second derivative in respect to y of f(x,y) = (d^2 z(y)/dy^2 ) * g(x) (as we can see here the second derivative g(x) is multiplied with the z(y), and this is the reason later on the video we can see it like this... so in this equation the gx and gy should change places), so the Laplacian operator is actually the sum of the above two which equals [(d^2 g(x)/dx^2 ) * z(y)] + [(d^2 z(y)/dy^2 ) * g(x)]. And all this thing is convolved with the image, but because convolution is distributive we have the result he is showing.
What I haven't found a proof yet is why if a function is multiplicatively separable (like the Gaussian) f(x,y) = g(x) * z(y) and you wanna convolve it with another function (which is the image) you can convolve the two functions (g(x) and z(y)) separately... I mean..., why [g(x)*z(y)]**w(x,y) = [w(x,y)**g(x)]**z(y) (here ** means convolution 'cause I use * for multiplication :D). If you have any clue about it or you can point me to some direction your comments would be most welcome...
do anyone have link for download book introductory technics of 3d computer vision
Thanks a lor for the excellent lecture
thank you
I currently being tortured with a combination of neuroscience and doppler radar, someone, please help. I have been undergoing this tortures for at least 3 years. I started in december 2012, when a socialworker student marked me for torture.