Histogram equalization: Difference between revisions

Revision as of 15:54, 16 June 2025

Revision as of 23:33, 17 July 2025

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:<math>\ k^\prime = k \cdot(\max(i) - \min(i)) + \min(i)= k \cdot(L- 1)</math><ref>{{web archive |url=https://web.archive.org/web/20200601000000*/https://www.math.uci.edu/icamp/courses/math77c/demos/hist_eq.pdf |title=University of California, Irvine Math 77C - Histogram Equalization}}</ref><!--[[User:Kvng/RTH]]-->

<math>k </math> is a real value while<math>\ k^\prime </math> has to be an integer. An intuitive and popular method<ref>{{Cite book|last=Gonzalez|first=Rafael C.|url=https://www.worldcat.org/oclc/991765590|title=Digital image processing|date=2018|publisher=Pearson|others=Richard E. Woods|isbn=978-1-292-22304-9|edition=4th|___location=New York, NY|pages=138–140|oclc=991765590}}</ref> is applying the round operation:

:<math>\ k^\prime = \operatorname{round} (k \cdot(L- 1))</math>.

However, detailed analysis results in slightly different formulation. The mapped value <math>k^\prime </math> should be 0 for the range of <math>0<k \leq1/L</math>. And <math>k^\prime =1</math> for <math>1/L < k \leq 2/L</math>, <math>k^\prime = 2 </math> for <math>2/L < k \leq 3/L</math>, ...., and finally <math>k^\prime =L-1</math> for <math>(L-1)/L < k \leq 1</math>. Then the quantization formula from <math>k</math> to <math>k^\prime </math> should be