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cse252c sp09 discussion board

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93
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85
Peter
05-31-2009
04:14 PM ET (US)
Regarding /m81. I'm also having trouble with getting a homography from my K matrix.
It is not clear to me how /m83 helps here.
84
Serge BelongiePerson was signed in when posted
05-29-2009
07:26 PM ET (US)
/m80 You're right; for invariance to scale, a good choice would be to use SIFT.
83
Steve
05-29-2009
05:42 PM ET (US)
Regarding /81, I think that the matrix K directly gives you a homography matrix, but it is a homography relating lines. So you need to do a covariant/contravariant conversion (lecture notes 4) to get a homography you can use to rectify the image.
82
Steve
05-29-2009
01:21 PM ET (US)
On homework 2 prob 2 parts 4-5, I'm having trouble coming up with an example and am wondering if the problem might be incorrect. The rotation R'=-I+2aa^T is a rotation of 180 degrees about the vector a (this can be verified by evaluating rodrigues formula for e^{\hat{a} \pi}=-I+2aa^T).

We are trying to show that translating a camera with no rotation can somehow produce the same effect as rotating 180 degrees. The two solutions seem to be twisted pairs, where in one solution one camera is at the wrong side of the object plane and imaging it backward.

It doesn't seem consistent with the Weng paper on the class website, which talks about the 2 physically possible solutions to the homography decomposition. The paper is over my head, but Theorem 3 seems to indicate that the two homographies should have the same sign, whereas our homographies are negations of one another. In their example, both solutions have a slight rotation, whereas in our case we have no rotation in one and a 180 degree rotation the the other.
81
Ben
05-29-2009
02:42 AM ET (US)
For homework 3, problem 3 (the affine_tile.gif problem), it is unclear to me what vector v to use. I think I found a good K, using the H&Z p. 56 trick, but putting that back into C_inf^* ' is eluding me. My first guess is to just use v = 0, but this does not give the correct homography. Any ideas?
80
Tingfan
05-28-2009
06:35 PM ET (US)
I tried RANSAC on some additional image sets.
It seems that normalized cross correlation is not robust for two matching interest points but in different scale. Any idea?
Edited 05-28-2009 06:38 PM
79
Kai
05-28-2009
12:21 AM ET (US)
regarding /m77, i think the x' values in the b column vector are the coordinates of the pixels being integrated -- or in our case, summed -- over, and not the centers of the windows. if you write out the entries of b by multiplying out the A matrix and x' coords, you can see how to construct the vector. that being said, i didn't notice any gain by using subpixel coordinates in practice. MaSKS also seems to ignore it in their book algorithm, so, oh well.
78
Emmett
05-27-2009
03:34 PM ET (US)
If people are having trouble getting the same homographies for HW3.4 in /m38, they are normalized by the 2nd singular value of the computed homography 3x3.
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