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I'm a computer science student and I have an assignment requiring me to implement the algorithm proposed by this paper but I'm hitting a roadblock trying to understand the implementation of one of the methods proposed in the paper. I'm not sure if I understand Section IIIB well. This is what I understand so far.

In order to separate the histograms into subregions, we must first obtain the histogram of the intensity levels. Once achieved, we calculate the density of the histogram (equation 7). Furthermore we have to sort the density function values in descending order. How is this implemented? I assume its only in the following paragraph in the paper that goes on to describe this but I'm having a hard time understanding what the authors are trying to convey.

Once we find the groups of high density areas, we extend a group area surrounding these values. The paper talks about momentary changes. I'm not sure what this means, anyone have an idea?

The page is located at http://www.waset.org/journals/ijeee/v3/v3-6-53.pdf

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  • $\begingroup$ Please provide a link to the paper! Since you refer to numbered sections and equations inside, it will help people in helping you. $\endgroup$ – sansuiso Nov 27 '13 at 7:43
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The paper is not really well written, so it's not surprising that you're having a hard time. These are a few hints to guide you:

  • the authors try to do some mode detection on the image brightness histogram in order to segment it between common values in the image (= high density in their terminology) and rare values in the image (= low density);
  • in order to segment the histogram, they fix a window width (i.e., a number of bin of the histogram), apply this window over all possible positions in the histogram and sum the populations of the bins inside the window ($Eq. (7)$). This gives the density of a histogram bin (or equivalently of a brightness level);
  • their segmentation process consists then in going from the peaks of the histogram and merging neighbouring bins into it as long as the difference between the border bin and its neighbour (= the momentary change) is less than the average of the momentary changes in the region.
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  • $\begingroup$ I wish I could give you an upvote but man, that helped a lot! $\endgroup$ – fryBender Nov 30 '13 at 0:04

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