Question

I am reading lot of post for object detection using feature extraction (sift ecc).

After having calculate descriptors on both images, to get good matches they are using crossCheckMatching. (found on sample/cpp/descritpor_extractor_matcher.cpp)

Coudl I understand why this choice?

Why we need to evalute both

descriptorMatcher->knnMatch( descriptors1, descriptors2, matches12, knn );
descriptorMatcher->knnMatch( descriptors2, descriptors1, matches21, knn );

I don’t understand it.

Computing the Euclian distance for example doesn’t return the same result in both direction ?

Answer 1

You can’t generally assume that the Eucludian distance will be used by your matcher. For instance, the BFMatcher supports different norms : L1, L2, Hamming…

You can check the documentation here for more details : http://docs.opencv.org/modules/features2d/doc/common_interfaces_of_descriptor_matchers.html

Anyway, all these distance measures are symmetric and it doesn’t matter which one you use to answer your question.

And the answer is : calling knnMatch(A,B) is not the same as calling knnMatch(B,A).

If you don’t trust me, I’ll try to give you a graphical and intuitive explanation. I assume for the sake of simplicity that knn==1, so that for each queried descriptor, the algorithm will only find 1 correspondence (much easier to plot 🙂

I randomly picked few 2D samples and created two data-sets (red & green). In the first plot, the greens are in the query data-set, meaning that for each green point, we try to find the closest red point (each arrow represents a correspondence).

In the second plot, the query & train data-sets has been swapped.

Finally, I also plotted the result of the crossCheckMatching() function which only conserve the bi-directional matches.

And as you can see, the crossCheckMatching()‘s output is much better than each single knnMatch(X,Y) / knnMatch(Y,X) since only the strongest correspondence have been kept.