How to Solve Histogram Equalization Numerical Problem in MATLAB?

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. 

0	1	1	3	4
7	2	5	5	7
6	3	2	1	1
1	4	4	2	1

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:

Range	Frequency	PDF	CDF	7*CDF	Round-off
0	1	0.0500	0.0500	0.3500	0
1	6	0.3000	0.3500	2.4500	2
2	3	0.1500	0.5000	3.5000	4
3	2	0.1000	0.6000	4.2000	4
4	3	0.1500	0.7500	5.2500	5
5	2	0.1000	0.8500	5.9500	6
6	1	0.0500	0.9000	6.3000	6
7	2	0.1000	1.0000	7.0000	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:

0	2	2	4	5
7	4	6	6	7
6	4	4	2	2
2	5	5	4	2