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Hello. A distance of 2 is fine, but I think there should be a lot more inliers. So either you have too few inliers from the start or you do not find all of them. When visualizing your best pairs, how many inliers do you count ? With my parameter setting I have a rather large fraction of inliers. I g...

Your result looks almost correct.

But it seems that you do not find the best homography.
If you visualize your result without blending, that should be clear.

Your homography points are very localized, therefore small errors in the estimation might become more severe.


Best regards,
Christoph

Hello everyone,

maybe I did not make myself clear enough.

All the functions you need to call from exercise 3 are already added to the startingKit.
Please take a look at the utils folder.

Best regards,
Christoph

You should return the orientation matrix, not the rotation. But that does not solve your problem here. We have the knowledge that we have a right handed system. Therefore we can verify that this is the case in our decomposition. The rq-decomposition does not guarantee that the second matrix is a rot...

You do not have to do that before, you can also do the re-scaling after the rq-decomposition. I will shortly explain why that re-scaling is equivalent to the early re-scaling of P and why the re-scaling is helpful to get the principal point. Rescaling is equivalent: The reason is that the third row ...

This is an answer to both questions. Let me start with the answer for Marc Schmitt. At first the rq-decomposition is not uniquely defined. As in the slides, if M = K*R, the following is also a valid decomposition: K * D * D^-1 R with D = (+- 1 0 0; 0 +-1 0; 0 0 +-1). Since if K is an upper triangle,...

No do not use the original squares, but build a complete matrix by concatenating the previous matrices.
This also applies for the calibration matrices.

Best regards,
Christoph

At first sight I am not sure what your problem is, therefore I ask some questions first. 1) When only the scale of the points is wrong here, you are probably very close to the solution. Is your conversion from homogeneous to cartesian coordinates doing the right thing? Your cartesian coordinates are...

Yes indeed, since the matrix squares stores cartesian coordinates, the call is the wrong way round.
Nevertheless the result of the line is nowhere used.

I am sorry for this inconvenience. I hope the line did not confuse somebody.

Dear all, the oral exam of the computer vision lecture will take place on the 22nd and 23rd of July. You need to register to us if you want to participate, and you will be notified about your exact date and time after we received all the emails. The DEADLINE FOR REGISTRATION is the 23rd of June. Ple...

The parameter PARAMS.svmThreshold is used to reduce the amount of bounding boxes returned by the detector, while the parameter PARAMS.prec_recall_threshold is used for the evaluation. Both are used in Part 3 of the project. More precisely, PARAMS.svmThreshold is considered for the non-max suppressio...

This is probably because your evaluation is currently dominated by the computation of the descriptor. The question was aiming at the evaluation of the SVM and not at the whole process of descriptor extraction. But I think that is not completely clear from the task description. So you could just ment...

The larger the training set the more the results will be unaffected. I actually didn' t and will not do a study on how much training with border artefacts will hurt your performance here. You might loose a percent in ERR or more or maybe nothing at all. In general you might introduce a pattern that ...

Thank you for the pictures. What you are seeing here is exactly the case of border effects I shortly described in my last reply. The red areas are different, which are the borders of your sliding window. These artefacts occur because of the behaviour of the derivative filter you are using. The borde...

There should be no difference in the gradients, regradless whether it is computed over the whole image or just over a small patch. The values at the pixel coverd by the patch should be equal to the corresponding values in the image. Regarding your comment 1.1 this seems to be the case for you. Did y...