# How to Solve Histogram Equalization Numerical Problem in MATLAB?

• 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.

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:

```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:

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