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How to Solve Histogram Equalization Numerical Problem in MATLAB?

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  • Difficulty Level : Expert
  • Last Updated : 22 Nov, 2021

Histogram Equalization is a mathematical technique to widen the dynamic range of the histogram. Sometimes the histogram is spanned over a short range, by equalization the span of the histogram is widened. In digital image processing, the contrast of an image is enhanced using this very technique. 

Use of Histogram Equalization:

It is used to increase the spread of the histogram. If the histogram represents the digital image, then by spreading the intensity values over a large dynamic range we can improve the contrast of the image. 

Algorithm:

  • Find the frequency of each value represented on the horizontal axis of the histogram i.e. intensity in the case of an image.
  • Calculate the probability density function for each intensity value.
  • After finding the PDF, calculate the cumulative density function for each intensity’s frequency.
  • The CDF value is in the range 0-1, so we multiply all CDF values by the largest value of intensity i.e. 255.
  • Round off the final values to integer values.

Example:

Matlab




% MATLAB code for Histogram equalisation
% function to return resultant
% matrix and summary
function [f,r]=HistEq(k,max)
Freq=zeros(1,max);
[x,y]=size(k);
  
% Calculate frequency of each
% intensity value.
for i=1:x
    for j=1:y
        Freq(k(i,j)+1)=Freq(k(i,j)+1)+1;
    end
end
  
% Calculate PDF for each intensity value.
PDF=zeros(1,max);
Total=x*y;
for i=1:max
    PDF(i)=Freq(i)/Total;
end
  
% Calculate the CDF for each intensity value.
CDF=zeros(1,max);
CDF(1)=PDF(1);
for i=2:max
    CDF(i)=CDF(i-1)+PDF(i);
end
  
% Multiply by Maximum intensity value
% and round off the result.
Result=zeros(1,max);
for i=1:max
    Result(i)=uint8(CDF(i)*(max-1));
end
  
% Compute the Equalized image/matrix.
mat=zeros(size(k));
for i=1:x
    for j=1:y
        mat(i,j)=Result(k(i,j)+1);
    end
end
    
f=mat;
r=Result;
end
  
% Utility code here.
k=[0, 1, 1, 3, 4; 
   7, 2, 5, 5, 7; 
   6, 3, 2, 1, 1; 
   1, 4, 4, 2, 1];
     
[new_matrix, summary]=HistEq(k,8);

Output:

A 3-bit image of size 4×5 is shown below. Compute the histogram equalized image. 

01134
72557
63211
14421

Steps:

  • Find the range of intensity values.
  • Find the frequency of each intensity value.
  • Calculate the probability density function for each frequency.
  • Calculate the cumulative density function for each frequency.
  • Multiply CDF with the highest intensity value possible.
  • Round off the values obtained in step-5.
Overview of calculation:
Range of intensity values = [0, 1, 2, 3, 4, 5, 6, 7]
Frequency of values = [1,  6, 3, 2, 3, 2, 1, 2] 
total = 20 = 4*5
Calculate PDF = frequency of each intensity/Total sum of
all frequencies, for each i value of intensity
Calculate CDF =cumulative frequency of each intensity 
value = sum of all PDF value (<=i)
Multiply CDF with 7.
Round off the final value of intensity.

The tabular form of the calculation is given here:

RangeFrequencyPDFCDF7*CDFRound-off
010.05000.05000.3500 0
16 0.3000            0.35002.45002
230.1500            0.50003.50004
320.1000           0.60004.20004
430.1500            0.75005.25005
520.1000           0.85005.95006
61  0.0500            0.90006.30006
720.1000           1.00007.0000 7
Interpretation: 
The pixel intensity in the image has modified.
0 intensity is replaced by 0.
1 intensity is replaced by 2.
2 intensity is replaced by 4.
3 intensity is replaced by 4.
4 intensity is replaced by 5.
5 intensity is replaced by 6.
6 intensity is replaced by 6.
7 intensity is replaced by 7.

Output: The new image is as follow:

 

0245
74667
64422
25542

 


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