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Let $f$ be an $N \times M$-image and suppose that $g$ is its equalization. Question - Why the equalization does not change the objects, why a car on $f$ does not convert into a bus on $g$?. It is not obious that after the pixel intensity $r_k \mapsto s_k$ recalculation for $g$ $$ s_k =\frac{L-1}{NM}\sum_{j=0}^k n_k, k=0,1,\ldots,L-1, $$ we obtain the same scene as was for $f$! Here $n_k$ denotes the number of pixels that have intensity $r_k.$

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It probably has the potential to do so, but the probability is marginal. Result of equalization depends on the initial pixel intensity distribution.

Image histogram equalization can have different implementations. In one possible implementation, you would re-map a whole block of a given intensity level into a new calculated intensity. In such a case, objects will not change shape; boundaries will remain the same.

The aim of equalization is not necessarily the creation of a new image with strictly flat histogram will all bins equally occupied. That won't create a good looking image in any sense. Instead, the aim is to expand the pixel intensities to fully utilize the available dynamic range in the hope of getting a higher constrast with more visible dark regions and unsaturated highlights.

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  • $\begingroup$ I do not understand why the recalculation of the intensity of each pixel does not change the whole scene in the image. It's looks as a miracle. $\endgroup$ – Leox Sep 22 '19 at 7:32
  • $\begingroup$ @Leox: have you tried it? Imagine a contrast adjustment tool in Photoshop and you have a notion what image [histogram] equalization can do. In general, if you’re learning about some image processing method, first see the technique being used on a real image, then look at the math. $\endgroup$ – Rethunk Sep 23 '19 at 16:24

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