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Advanced Computer Vision

This lecture by Roger S. Gaborski delves into advanced concepts of histograms in computer vision. It explains how a histogram maps pixel values from a 2D matrix into a vector, highlighting the significance of each bin representing a range of gray levels. The discussion includes methods to assess the similarity between histograms, utilizing the histogram intersection method. Examples are provided to illustrate cases of identical and different histograms, emphasizing the limitations of histograms in capturing spatial information in image analysis.

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Advanced Computer Vision

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  1. Lecture 08 Roger S. Gaborski Advanced Computer Vision Roger S. Gaborski

  2. Histograms • Histogram is nothing more than mapping the pixels in a 2 dimensional matrix into a vector • Each component in the vector is a bin (range of gray level values) and the corresponding value is the number of pixels with that gray level value

  3. Similarity between Histograms • Similarity between histogram bins: • Assuming both histograms have ∑nj j=1…B pixels M. Swain and D. Ballard. “Color indexing,”International Journal of Computer Vision, 7(1):11–32, 1991.

  4. Histogram Intersection • A simple example: • g = [ 17, 23, 45, 61, 15]; (histogram bins) • h = [ 15, 21, 42, 51, 17]; • in=sum(min(h,g))/min( sum(h),sum(g)) • in = 0.9863

  5. >> g = [17,23,45,61,15]; >> h = [15,21,42,51,17]; >> min(g,h) ans= 15 21 42 51 15 >> N = sum(min(g,h)) N = 144 >> D=min(sum(h),sum(g)) D = 146 >> intersection = N/D intersection = 0.9863 Roger S. Gaborski

  6. If Histograms Identical • g = 15 21 42 51 17 • h = 15 21 42 51 17 • >> in=sum(min(h,g))/min( sum(h),sum(g)) • in = 1

  7. Different Histograms • h = 15 21 42 51 17 • g = 57 83 15 11 1 • >> in=sum(min(h,g))/min( sum(h),sum(g)) • in = 0.4315

  8. Use Gray Scale for Example

  9. Region and Histogram Similarity with itself: >>h = hist(q(:),256); >> g=h; >> in=sum(min(h,g))/min( sum(h),sum(g)) in = 1

  10. >> r=236;c=236; >> g=im(1:r,1:c); >> g= hist(g(:),256); >> in=sum(min(h,g))/min( sum(h),sum(g)) in = 0.5474

  11. Partial Matches >> g= hist(g(:),256); >> in=sum(min(h,g))/min( sum(h),sum(g)) in = 0.8014 in=sum(min(h,g))/min( sum(h),sum(g)) in = 0.8566

  12. Lack of Spatial Information • Different patches may have similar histograms

  13. in=sum(min(h,g))/min( sum(h),sum(g)) in = 1

  14. For Howard Ross Roger S. Gaborski

  15. Roger S. Gaborski

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