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]

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]