Computing the difference between images

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傲寒
傲寒 2020-12-04 07:00

Do you guys know of any algorithms that can be used to compute difference between images?

Take this webpage for example http://tineye.com/ You give it a link or uplo

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  • 2020-12-04 07:13

    One technique is to use color histograms. You can use machine learning algorithms to find similar images based on the repesentation you use. For example, the commonly used k-means algorithm. I have seen other solutions trying to analyze the vertical and horizontal lines in the image after using edge detection. Texture analysis is also used.

    A recent paper clustered images from picasa web. You can also try the clustering algorithm that I am working on.

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  • 2020-12-04 07:14

    What TinEye does is a sort of hashing over the image or parts of it (see their FAQ). It's probably not a real hash function since they want similar "hashes" for similar (or nearly identical) images. But all they need to do is comparing that hash and probably substrings of it, to know whether the images are similar/identical or whether one is contained in another.

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  • 2020-12-04 07:22

    Consider using lossy wavelet compression and comparing the highest relevance elements of the images.

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  • 2020-12-04 07:27

    Correlation techniques will make a match jump out. If they're JPEGs you could compare the dominant coefficients for each 8x8 block and get a decent match. This isn't exactly correlation but it's based on a cosine transfore, so it's a first cousin.

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  • 2020-12-04 07:31

    here is some code i wrote, 4 years ago in java yikes that does image comparisons using histograms. dont look at any part of it other than buildHistograms()

    https://jpicsort.dev.java.net/source/browse/jpicsort/ImageComparator.java?rev=1.7&view=markup

    maybe its helpful, atleast if you are using java

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  • 2020-12-04 07:33

    Heres an image similarity page, but its for polygons. You could convert your image into a finite number of polygons based on color and shape, and run these algorithm on each of them.

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