Some one asks about sigma and instructor explains it without any reproach...this is rare humility. Instructor is no doubt well experienced is immensely knowledgeable so as to be able to explain the concepts in lucid terms.
Awesome lectures! The most didactic and coolest material in that topic I found until now. Thank you professor Mubarak, and everyone involved. PS: On the example showed in 10:42, at row 3, column 6 I think there is a mistype on the slide, it needs a "-" signal (-20 instead of 20).
At 13:33, there is an error in the slide (but not in the explanations). The gradient direction is atan of f_y over f_x, not f_x over f_y. Around 21', the definitions of correlation and convolution are fishy; it's not f * h that's defined, but f * h [i, j], and the indices i and j should be used to select cells in f, not in h.
good point. I noticed these but since i'm not confident i thought maybe there's something unearthly about it that's beyond my scope. thanks for sharing
Awesome videos Prof Shah! I believe in 1:02:30 the x derivative filter should be [-1 1] instead of [1 -1] if you follow the same coordinate system as earlier in the video.
For the correlation and convolution equations (19:31) do the i and j values represent the x and y dimensions of the kernel? so for a 3x3 kernel i and j would be 3 and 3?
+Benjamin Montz I believe they are related to the pixel of the resulting image.. For the convolution, at least, if you put the indices of the kernel as (-1, -1) to (1,1) (or generally, centralize the kernel, so for a 5x5 kernel, indices should go from -2 to 2), then f*h(i,j) fulfills the equation. For the correlation I found a formula that is basically the same, but f is the kernel and h is the image. Alternatively, one can make an index shift and sum over f(k,l)h(k-i,l-i) with f being the image and h being the kernel. But I'm not 100% sure I understood it, either.
gradient direction is fx/fy or fy/fx? because i hv studied in school time that y component/ x component gives the angle or the direction. if u can clear it how u took gradient direction?
@10:25 whle discussing forward / backward differences...... it would have been great to see the "Advantage" of doing such an operation ........ Else, lectures will be very dry.
It seems weird to me that x determines the row and y determines what column it is (0:03:38). When i think of x, it always spans from left to right and y (in 2D) spans from bottom to top when i think of math and top to bottom when i think of computer science. I realize that the assignment is completely arbitrary, but this declaration seems to add unnecessary confusion.
Does anyone understand what i and j represent in the convolution/correlation equations? There are summations over k and l but not over i and j. Thanks!
Brandon Pelfrey You say the central derivative should be divided by 2, but why is the mask still [-1 0 1]? It is stepping over 2 pixels, is delta x not 2 then?
And if you take a look here www.holoborodko.com/pavel/numerical-methods/numerical-derivative/central-differences/ the formula he derives ends up being divided by 2h, so I'm a bit confused how we can just find the difference of the previous and next pixel in our masks, without halving the result. Thanks!
Logan Dunbar i think that he made a mistake because by logic and even on wikipedia in the central difference there is a division by 2*delta x which in our case is one...
+theCguy L Yes, agree with you. The general form should be f(x+h1) - f(x-h2)/(h1+h2). Or the equation change to: f(x+1/2) - f(x-1/2) to be correct at 10:19, where f(x+1/2) is the mid point of f(x+1) and f(x), f(x-1/2) is the mid point of f(x-1) and f(x). f(x+1/2) should equal to (f(x+1) - f(x))/2, f(x-1/2) = (f(x)-f(x-1))/2 => f(x+1/2) - f(x-1/2) = (f(x+1) - f(x))/2 - (f(x) - f(x-1))/2 = (f(x+1) - f(x-1) / 2
Indeed, if you need accuracy for differential equation system solving (ex: fluid simulations), you should use the true central difference mask is [-0.5 0 -0.5], but in detection applications, where you are searching for extremums or zero crossings, you can drop the division.
tien dh got it, you use this linear algebra convention to calculate derivatives, so, if we want to calculate delta I(x) as I(x+1)-I(x) we calculate it as I(x+1) + (-I(x)) where I(x) is intensity value of a pixel number x now let us assume we have a matrix nx2 and we want to get its partial derivative wrt x axis, if we multiply this matrix by the vector [-1,1] we will have a vector I' with n dimensions, having each element as I(x+1)-I(x) the use of this filter is by repeating the above operation for each column from 2 to m where m is number of columns, as if we are sliding this vector across x axis we can perform the same operation across y axis, or z-axis if this is a volume image
+Ahmad Rashad Was looking at it for probably 15 min, no joke. The weird thing about it is the masks aren't used as matrices being multiplied, but rather just as an overlay on the corresponding positions, summed up.. So at 16:00, he says "-1 to 10" three times because he's multiplying each mask position to the corresponding number in the image. He then sums up the column, and divides by 3 to average. I don't know why the operation is defined so confusing.
I have a very fundamental question. An image (original) comes with inherent NOISE. In all these discussions .. I observe that a model for NOISE is added (superimposed) on the original image. Why should you voluntarily ADD NOISE and do all these unnecessary unwanted operations (filtering, averaging etc) instead of removing the INHERENT NOISE in the original image ? All these operations suggested are only trying to remove the NOISE voluntarily added to the original image. To make it simple .. Why add Noise and try to remove it ?
We are not adding noise to an image.The noisy image is represented as pure image + noise. Its just modeling the input (assuming all noise i the image are just additions over pure image ). In some scenarios it is not that straight forward. I guess that answers ur question :)
You are mixing two different scenarios. In reality, when you take a picture, usually the snapshot has imperfections. You can observe that in cheap cameras when the image contains imperfections or noise for short. Modern cameras like expensive cameras, you barely notice any thing wrong with the image and you will be happy with the high quality (i.e. a lot of filtering under the hood). The second scenario in which researchers need first to generate noise which should resemble the true noise (i.e. at an approximated level otherwise if we know it then we can remove it) and then add it to the images. They need to do this in order to test their filters and conclude if the filters are capable of handling real scenarios. Modeling noise is extremely important in all engineering fields alike.
I have problem on understanding how the equation of correlation and convolution works (especially the i and j, where they come from?). As for k and l, it is just the coordinate of the pixel right? Hope someone can explain this to me.
+Hafiz Sofian (i, j) is the (row, column) coordinate in image f. (k,l) are the coordinates in kernel h. Correlation or convolution is applied at every pixel (i,j) in image f by iterating over every element (k,l) in kernel h.
This lecture series is excellent, thank you for sharing them! I found the jump from the image derivative to correlation confusing--it made more sense, though, once the discussion got around to the topic of image filters.
i know Im asking the wrong place but does someone know a trick to log back into an instagram account..? I was stupid lost the login password. I would love any help you can offer me.
@Emerson Saint i really appreciate your reply. I got to the site through google and im in the hacking process atm. I see it takes a while so I will get back to you later when my account password hopefully is recovered.
This is a fair point , which is what I might have done as well, but I believe here he is trying to provide an intuition of the derivative process in images. An alternative way of thinking can be - Since you want to detect edges , you want to look at the two sides of a pixel , hence you ignore it ( by placing a zero at the pixel location ,like in centre derivate mask ) and hence you would have delX = 1 ...
Some one asks about sigma and instructor explains it without any reproach...this is rare humility. Instructor is no doubt well experienced is immensely knowledgeable so as to be able to explain the concepts in lucid terms.
Awesome lectures! The most didactic and coolest material in that topic I found until now. Thank you professor Mubarak, and everyone involved.
PS: On the example showed in 10:42, at row 3, column 6 I think there is a mistype on the slide, it needs a "-" signal (-20 instead of 20).
+Charles Prado thanks for saying that now I know I'm not alone.... :-)
Best filtering course I've seen :)
At 13:33, there is an error in the slide (but not in the explanations). The gradient direction is atan of f_y over f_x, not f_x over f_y.
Around 21', the definitions of correlation and convolution are fishy; it's not f * h that's defined, but f * h [i, j], and the indices i and j should be used to select cells in f, not in h.
good point. I noticed these but since i'm not confident i thought maybe there's something unearthly about it that's beyond my scope. thanks for sharing
Also at 10:16, the central difference is the given expression divided by 2
Probably the best lecture series on computer vision out there!
Watch the 2014 series if you're going to go ahead with watching these videos though. It's a lot clearer!
Why the gray level is from 0 to 250? Shouldn't it be from 0 to 255 (i.e. 2^8 = 256 ) at 1:13 min.
Prob his typos
Awesome videos Prof Shah! I believe in 1:02:30 the x derivative filter should be [-1 1] instead of [1 -1] if you follow the same coordinate system as earlier in the video.
For the correlation and convolution equations (19:31) do the i and j values represent the x and y dimensions of the kernel? so for a 3x3 kernel i and j would be 3 and 3?
+Benjamin Montz I believe they are related to the pixel of the resulting image.. For the convolution, at least, if you put the indices of the kernel as (-1, -1) to (1,1) (or generally, centralize the kernel, so for a 5x5 kernel, indices should go from -2 to 2), then f*h(i,j) fulfills the equation. For the correlation I found a formula that is basically the same, but f is the kernel and h is the image. Alternatively, one can make an index shift and sum over f(k,l)h(k-i,l-i) with f being the image and h being the kernel. But I'm not 100% sure I understood it, either.
Thanks Dr. Mubarak your lecture was very helpful
Hi, There should be a division of 2 at 10:11 for central difference?
gradient direction is fx/fy or fy/fx? because i hv studied in school time that y component/ x component gives the angle or the direction. if u can clear it how u took gradient direction?
what is the difference between operator and filter ... i'm some confused ???
@10:25 whle discussing forward / backward differences...... it would have been great to see the "Advantage" of doing such an operation ........ Else, lectures will be very dry.
It seems weird to me that x determines the row and y determines what column it is (0:03:38). When i think of x, it always spans from left to right and y (in 2D) spans from bottom to top when i think of math and top to bottom when i think of computer science. I realize that the assignment is completely arbitrary, but this declaration seems to add unnecessary confusion.
he starts the next lecture on edge detection at 53:18
do we have lecture slides for this that I can print out?
Does anyone understand what i and j represent in the convolution/correlation equations? There are summations over k and l but not over i and j. Thanks!
I really wondered about those two symbols !!
Brandon Pelfrey You say the central derivative should be divided by 2, but why is the mask still [-1 0 1]? It is stepping over 2 pixels, is delta x not 2 then?
And if you take a look here www.holoborodko.com/pavel/numerical-methods/numerical-derivative/central-differences/ the formula he derives ends up being divided by 2h, so I'm a bit confused how we can just find the difference of the previous and next pixel in our masks, without halving the result. Thanks!
Logan Dunbar i think that he made a mistake because by logic and even on wikipedia in the central difference there is a division by 2*delta x which in our case is one...
+theCguy L Yes, agree with you. The general form should be f(x+h1) - f(x-h2)/(h1+h2). Or the equation change to:
f(x+1/2) - f(x-1/2) to be correct at 10:19, where f(x+1/2) is the mid point of f(x+1) and f(x), f(x-1/2) is the mid point of f(x-1) and f(x).
f(x+1/2) should equal to (f(x+1) - f(x))/2, f(x-1/2) = (f(x)-f(x-1))/2
=> f(x+1/2) - f(x-1/2) = (f(x+1) - f(x))/2 - (f(x) - f(x-1))/2 = (f(x+1) - f(x-1) / 2
Indeed, if you need accuracy for differential equation system solving (ex: fluid simulations), you should use the true central difference mask is [-0.5 0 -0.5], but in detection applications, where you are searching for extremums or zero crossings, you can drop the division.
I don't get the derivative mask part! how does it work?
Ahmad Rashad Same to me, it's confused
tien dh got it, you use this linear algebra convention to calculate derivatives, so, if we want to calculate delta I(x) as I(x+1)-I(x) we calculate it as I(x+1) + (-I(x)) where I(x) is intensity value of a pixel number x
now let us assume we have a matrix nx2 and we want to get its partial derivative wrt x axis, if we multiply this matrix by the vector [-1,1] we will have a vector I' with n dimensions, having each element as I(x+1)-I(x)
the use of this filter is by repeating the above operation for each column from 2 to m where m is number of columns, as if we are sliding this vector across x axis
we can perform the same operation across y axis, or z-axis if this is a volume image
+Ahmad Rashad Was looking at it for probably 15 min, no joke. The weird thing about it is the masks aren't used as matrices being multiplied, but rather just as an overlay on the corresponding positions, summed up.. So at 16:00, he says "-1 to 10" three times because he's multiplying each mask position to the corresponding number in the image. He then sums up the column, and divides by 3 to average. I don't know why the operation is defined so confusing.
I don't think he was following the rules of matrix multiplication. He was just multiplying corresponding elements.
I have a very fundamental question.
An image (original) comes with inherent NOISE. In all these discussions .. I observe that a model for NOISE is added (superimposed) on the original image. Why should you voluntarily ADD NOISE and do all these unnecessary unwanted operations (filtering, averaging etc) instead of removing the INHERENT NOISE in the original image ? All these operations suggested are only trying to remove the NOISE voluntarily added to the original image. To make it simple .. Why add Noise and try to remove it ?
We are not adding noise to an image.The noisy image is represented as pure image + noise. Its just modeling the input (assuming all noise i the image are just additions over pure image ). In some scenarios it is not that straight forward. I guess that answers ur question :)
You are mixing two different scenarios. In reality, when you take a picture, usually the snapshot has imperfections. You can observe that in cheap cameras when the image contains imperfections or noise for short. Modern cameras like expensive cameras, you barely notice any thing wrong with the image and you will be happy with the high quality (i.e. a lot of filtering under the hood). The second scenario in which researchers need first to generate noise which should resemble the true noise (i.e. at an approximated level otherwise if we know it then we can remove it) and then add it to the images. They need to do this in order to test their filters and conclude if the filters are capable of handling real scenarios. Modeling noise is extremely important in all engineering fields alike.
I have problem on understanding how the equation of correlation and convolution works (especially the i and j, where they come from?). As for k and l, it is just the coordinate of the pixel right? Hope someone can explain this to me.
+Hafiz Sofian
(i, j) is the (row, column) coordinate in image f.
(k,l) are the coordinates in kernel h.
Correlation or convolution is applied at every pixel (i,j) in image f by iterating over every element (k,l) in kernel h.
this is really good, it puts Things in perspective.
In the video. th-cam.com/video/1THuCOKNn6U/w-d-xo.htmlm14s. The slide indicates .... function f = image ... f = kernel . Is that correct?
no it was a mistake. The prof. didn't notice it. It should have been h = kernel.
very nice content
better than many overrated course..
This lecture series is excellent, thank you for sharing them!
I found the jump from the image derivative to correlation confusing--it made more sense, though, once the discussion got around to the topic of image filters.
i know Im asking the wrong place but does someone know a trick to log back into an instagram account..?
I was stupid lost the login password. I would love any help you can offer me.
@Kian Gus Instablaster :)
@Emerson Saint i really appreciate your reply. I got to the site through google and im in the hacking process atm.
I see it takes a while so I will get back to you later when my account password hopefully is recovered.
@Emerson Saint It did the trick and I finally got access to my account again. Im so happy:D
Thank you so much you saved my ass!
@Kian Gus Happy to help :D
wow ! the way he explained ! really good lecture !
Thanks a lot man :) , it helped me through my MV problems
can any one help me with gaussian filter with kernel convulation
Thank you Dr. it is helpful
Who the hell is Pavel Babenko...
in center derivative , I think, the value should be (f(x+1) - f(x-1))/2, because delX = 2 here. can anyone comment on it
This is a fair point , which is what I might have done as well,
but I believe here he is trying to provide an intuition of the derivative process in images.
An alternative way of thinking can be -
Since you want to detect edges , you want to look at the two sides of a pixel , hence you ignore it ( by placing a zero at the pixel location ,like in centre derivate mask ) and hence you would have delX = 1 ...
I think it too
Why there is not English Subtitle . My English Listening is so poor. I really admire those who are speak English.
What a nice lecture!!!!!!!
Thank You sir. Great lecture
thanks alot. it is too useful
Really helpful
Really nice, thank you!
really helpful!
like this, very helpful
In 19:10 f = Image and h = Kernel
thanks sir
Yes.
Watch on 1.5x, it's a lot better.
So... every teacher uses the same ppt.
good!
The explanation is quite good.
dont like the ads..
www.cs.cornell.edu/courses/cs6670/2011sp/lectures/lec02_filter.pdf for better understanding the video
most confused prof ever!! DO NOT WATCH!!
explain yourself.
seems extremely clear to me.
@amarug, I agreed.
can any any prof from anywhere just do this stuff as in uploading videos to teach???
Knowledge needs patience. I think this tutorial is amazing and upto the point.
very clear and helpful, Thx!