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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.

/m80 You're right; for invariance to scale, a good choice would be to use SIFT.

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.

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.

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?

Edited by author 05-28-2009 06:38 PM
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?

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.

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.

I think Chris in /m57 is right (to get subpixel values x_0), if I understand the notes correctly, b is simply A*x, so x_0 = inv(A)*A*x = x. Are we both interpreting the notes wrong?

For homework 4 prob 4. If you implement the your code exactly like pg 91 in MaSKS, you will find that there will often be clusters of points that can pass a threshold, when what you really need are a point from each of these clusters. Those cluster of points are often found around some "strong" corners, such as a corner of wooden lattice that support the dark glass. I came up with a way to isolate the best a single point within a set of corner clusters.

The method is below:
1) declare a matrix A that is same size of the image.
2) Assign A(I, J) = 1, if image(I, J) passed sigma2 threshold, otherwise A(I, J) = 0;
3) Perform 8 point connected component analysis on matrix A, store the result in B, the matlab command is B = bwlabel(A, 8). After this, every cluster will have a different label.
4) Iterate through all the labels in B. For each cluster in B, find the one point whose sigma2 is the highest, and label that point as the corner point, and discard the rest of the cluster.

This technique isolates the strongest corner point, and it seem to work quite well in hw 4 problem 4 and problem 5.

A quick way to map image via homography to another set of coordinates:
Given a homography matrix H, I previously had some trouble transforming one image to its corresponding view, then I did some research on matlab function and there's a solution that uses matlab internal 2D transformation function.

Given an H, and that you know the range of transformed image will be within [1 130] on X direction, and [1 200] in the Y direction, and it runs fairly fast, you can do the following:

%you have use H' in this maketform function, as the format of tranformation matrix is the transpose of the actual transformation matrix in the X Y coordinate system.
T = maketform('projective', H');
transformedI = imtransform(original_image, T, 'XData', [1 130], 'YData', [1 200]);
imagesc(transfromedI);

%for an example you can use a translation matrix H = [1 0 0; 0 1 0; 10 0 1]; to test it out first.

/m73 I don't have the book handy, but the difference would just be the frame in which the corner coordinates are computed -- in one case, it's relative to the window center, in the other it's relative to the image origin.