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I have applied SIFT on one image but two times ,example:

[image1, descript1, location1] = sift('book.pgm'); [image2, descript2, location2] = sift('book.pgm');

after matching function it will show all points are matching, I want to eliminate all the auto matches, i.e. the match of a point with itself.

In the image some areas are copy-pasted, I want to show only that matched points.

Anybody have suggestions how to do it? Thanks

user615864
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  • This is counter-intuitive. The whole purpose of feature matching is to find correspondences between same patterns in a pair of images. Why would you want to defeat this purpose? – Zaphod Apr 19 '13 at 09:32
  • I want to detect if there is any copy -paste in the image, if copy some part of the image and paste it to another part ,I want to detect these two part, because I used same image two times then it will detect all points in both images are connected, I want to remove the connected points rather than copy-pasted areas. – user615864 Apr 19 '13 at 11:38

2 Answers2

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The following steps might work, but I can't be sure.

Get a set of matches such that you extract the 2 best matches for every descriptor in the first image. Then compute a homography using RANSAC. You can find a tutorial here, but you can replace SURF with SIFT or any other descriptor quite easily. Find the inlier set of matches, which should consist of matches between corresponding points in the identical images. Delete all these matches.

Now compute a RANSAC-based homography on the remaining matches and find the inlier set. This set may correspond to the copy-pasted region.

Zaphod
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Another method that you may want to try out is:

  • Get the pixel-wise difference of the two images. The difference image will then only contain the copy-pasted region.
  • Create a mask from the difference image i.e. any non-zero pixel in the difference image becomes 1 on the mask.
  • Do feature matching between image 1 and the masked version of image 2.

Remember this will only work if the duplicate image is not a rotated version of the first one.

Zaphod
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