Talk:Histogram equalization

I have created an example. But it seems a little awkwardly placed. Feel free to re-arrange it or put it in a different location in the article.--Konstable 05:56, 11 June 2006 (UTC)[reply]

Added a section on back project with a citation and footnote reference, but couldn't figure out how to get the superscript to link properly. kostmo 20:40, 18 August 2006 (UTC)[reply]

Excellent! I love the example. Wilson Harron 21:55, 22 October 2006 (UTC)[reply]

Thanks!--Konst.able^Talk 22:45, 22 October 2006 (UTC)[reply]

Sorry, but I don't understand why is this page into philosophy's project. It seems better suited to optics, computer science or something like this, is not?--Patillotes (talk) 07:47, 31 January 2008 (UTC)[reply]

Incorrect image

Isn't the example pictogram of the histograms wrong !? Histogram equalization should transfrom an image such that it's histogram becomes (approximately) uniform, i.e. "flat" -- the shown pictogram looks more like it's just about contrast and brightness change, i.e. linear transformation . -Seb.Haase —Preceding unsigned comment added by Sebhaase (talk • contribs) 08:40, 5 September 2008 (UTC)[reply]

That is not the point at all. The point is is that the cumulative histogram is linear:

H.eq. spreads out the histogram which is a form of contrast adjustment. The more "probable" a given intensity is in the original image that intensity has more local contrast. The maximum in the above image (~52) has the largest step to the previous and next intensity.

Flattening the histogram would result in global contrast change and a drastic change in the image. What transformation are you thinking of that would yield a flat histogram? Cburnett (talk) 20:07, 5 September 2008 (UTC)[reply]

The histogram would be flat only if histograms were continuous functions. Since they are discrete, it's not possible to guarantee a flat histogram. But you can still make the cumulative histogram quasi-linear. 89.214.104.88 (talk) 11:38, 12 March 2009 (UTC)[reply]

For the cumulative histogram to be linear, each step must contribute the same number of occurrences as every other step of the same size, and the histogram must therefore be constant...unless the x-axis is non-linear or not fully populated. —Preceding unsigned comment added by 128.170.116.22 (talk) 21:48, 31 July 2009 (UTC)[reply]

Error in formula

The formula in the example is not correct: it says "cdf(v) = round[cdf(v)..." when it should be something like "v_new(v_old) = round[cdf(v_old)...", as the formula gives us the new value that a certain gray level in the original image will assume in the equalized image. Gazilion (talk) 12:10, 12 March 2009 (UTC)[reply]

Histogram Matching and Histogram Hyperbolization

Maybe it makes sense to mention Histogram Matching (match histogram to a distribution of a 2nd source). Even more important is Histogram Equalization in log-log-domain (Histogram Hyperbolization). Hyperbolization is achieved by using the power function for the cdf. This leads to more "natural" results, since many quantities in nature are roughly linear in log-log domain (including light as perceived by the human visual system). -- 92.225.71.216 (talk) 09:50, 30 November 2009 (UTC)[reply]

Overview is bad

The overview is not an overview if it starts giving mathematical jargon and talks about DNA stuff. Applications of this technique shouldn't be discussed in that much detail. There are also a few grammatical mistakes in the overview. The discussion of when the equalization should be used is also simplifying it too much. The section is just bad; the main person to blame is whoever added that stuff about DNA.Owen214 (talk) 04:38, 13 June 2010 (UTC)[reply]

Invertible?

This article talks about the equalization being invertible, yet the german wikipedia (and my image processing course notes) say that is is explicitly NOT invertible. Which also makes sense to me, since this process can map multiple (infrequent) grey levels in the source image to the same grey level in the resulting image.