Color Histogram Equalization - MATLAB CODE

                          Histogram Equalization can be considered as redistribution of the intensity of the image. Color histogram equalization can be achieved by converting a color image into HSV/HSI image and enhancing the Intensity while preserving hue and saturation components. 

                          However, performing histogram equalization on components of R,G and B independently will not  enhance the image. At the end of this post, check the histogram of before and after histogram equalization of an image which is obtained by performing histogram equalization on the components(R,G and B) independently.

Steps to be performed:

1.      Convert RGB image into HSI Image.
https://www.imageeprocessing.com/2013/05/converting-rgb-image-to-hsi.html

2.      Obtain the ‘Intensity Matrix’ from the HSI Image matrix

3.      Perform Histogram Equalization on the intensity Matrix
https://www.imageeprocessing.com/2011/04/matlab-code-histogram-equalization.html

4.      Update the Intensity Matrix from the HSI Image matrix with the histogram equalized Intensity matrix

5.      Convert HSI Image back to RGB Image
https://www.imageeprocessing.com/2013/06/convert-hsi-image-to-rgb-image.html

        

MATLAB CODE:

%COLOR HISTOGRAM EQUALIZATION

%READ THE INPUT IMAGE

I = imread('football.jpg');

%CONVERT THE RGB IMAGE INTO HSV IMAGE FORMAT

HSV = rgb2hsv(I);

%https://www.imageeprocessing.com/2013/05/converting-rgb-image-to-hsi.html

%PERFORM HISTOGRAM EQUALIZATION ON INTENSITY COMPONENT

Heq = histeq(HSV(:,:,3));

%https://www.imageeprocessing.com/2011/04/matlab-code-histogram-equalization.html

HSV_mod = HSV;

HSV_mod(:,:,3) = Heq;

RGB = hsv2rgb(HSV_mod);

%https://www.imageeprocessing.com/2013/06/convert-hsi-image-to-rgb-image.html

figure,subplot(1,2,1),imshow(I);title('Before Histogram Equalization');

       subplot(1,2,2),imshow(RGB);title('After Histogram Equalization');

EXPLANATION:

RGB image matrix is converted into HSI(Hue ,Saturation and Intensity) format and histogram equalization is applied only on the Intensity matrix . The Hue and Saturation matrix remains the same. The updated HSI image matrix is converted back to RGB image matrix.

%DISPLAY THE HISTOGRAM OF THE ORIGINAL AND THE EQUALIZED IMAGE

HIST_IN = zeros([256 3]);

HIST_OUT = zeros([256 3]);

%http://angeljohnsy.blogspot.com/2011/06/histogram-of-image.html

%HISTOGRAM OF THE RED,GREEN AND BLUE COMPONENTS

HIST_IN(:,1) = imhist(I(:,:,1),256); %RED

HIST_IN(:,2) = imhist(I(:,:,2),256); %GREEN

HIST_IN(:,3) = imhist(I(:,:,3),256); %BLUE

HIST_OUT(:,1) = imhist(RGB(:,:,1),256); %RED

HIST_OUT(:,2) = imhist(RGB(:,:,2),256); %GREEN

HIST_OUT(:,3) = imhist(RGB(:,:,3),256); %BLUE

mymap=[1 0 0; 0.2 1 0; 0 0.2 1];

figure,subplot(1,2,1),bar(HIST_IN);colormap(mymap);legend('RED CHANNEL','GREEN CHANNEL','BLUE CHANNEL');title('Before Applying Histogram Equalization');

       subplot(1,2,2),bar(HIST_OUT);colormap(mymap);legend('RED CHANNEL','GREEN CHANNEL','BLUE CHANNEL');title('After Applying Histogram Equalization');

EXPLANATION:

Obtain the histogram of each component (Red,Green and Blue) independently.

Define the colormap ‘mymap’ with three colors namely Red, Green and Blue.

Display the histograms of the components before and after histogram equalization.

 NOTE:

Histogram of the above image by processing the components independently gives bad result.

 

Histogram Equalization applied on individual components

Fast Fourier Transform on 2 Dimensional matrix using MATLAB

Fast Fourier transformation on a 2D matrix can be performed using the MATLAB built in function ‘fft2()’.

Fourier transform is one of the various mathematical transformations known which is used to transform signals from time domain to frequency domain.
The main advantage of this transformation is it makes life easier for many problems when we deal a signal in frequency domain rather than time domain.

Example:

%FOURIER TRANSFORM ON A MATRIX

A  = zeros(5);

A ( : ) = 1:25;

display(A);

F_FFT  = fft2(A);

display(F_FFT);

%INVERSE FOURIER TRANSFORM

I_FFT = ifft2(F_FFT);

display(abs(I_FFT));

NOTE on ABSOLUTE VALUE:

When we use FFT2() or FFT(),the result we obtain in the frequency domain is of complex data type.

i.e It contains both the real  as well as the imaginary part.

Let   A=10+5i

A is a complex number as it contains both real and imaginary part.In this particular case ‘10’ is the real part and ‘5’ is the imaginary part.

abs(A) = 11.1803 is the absolute (also called modulus in few books or notations) value of A which is nothing but the magnitude. It can be arrived by using the below mentioned formula:

abs(A) = sqrt(real part^2+imaginary part^2).

              = sqrt(10^2+5^2)

              = sqrt(125)

             = 11.1803 (approx)

Let’s try to understand how the Fourier transform on 2 dimensional data works with a simple example.

This method will be helpful to understand the up sampling and down sampling in both spatial and frequency domain.

1.       Consider a matrix A

2.       Perform 1 D Fast Fourier transform(FFT) on each row

1D FFT on first row (Note that the absolute value is only displayed and not the actual imaginary number):

Similarly, perform  1D FFT on each row:

NOTE: The figure represents the 1 D FFT on each row and the result is the absolute value of the complex data obtained using FFT.

3.       Perform 1 D Fast Fourier transform on each column.
On the matrix obtained from the previous step, compute 1D FFT column wise.

4.       Display the results obtained.

Flow Chart for Fast Fourier Transform on 2D :

INVERSE FOURIER TRANSFORM:

1.       Perform Inverse Fourier Transform on each column

2.       Perform IFFT on each row

3.       Display the original data

MATLAB CODE:

A=[110 20 140 0 220;

   60 34 23 198 20;

   15 12 126 230 15;

   140 28 10 28 10;

   11 12 19 85 100];

FFT_row = zeros(size(A));

FFT_col = zeros(size(A));

%Perform FFT on each row

for i=1:size(A,1)

FFT_row(i,:) = fft(A(i,:));

end

display(FFT_row);

%display(abs(FFT_row));

%Perform FFT on each column

for i=1:size(A,2)

FFT_col(:,i) = fft(FFT_row(:,i));

end

display(FFT_col);

%display(abs(FFT_col));

%INVERSE FOURIER TRANSFORM

IFFT_row = zeros(size(A));

IFFT_col = zeros(size(A));

%Perform Inverse Fourier Transform on each column

for i=1:size(A,2)

IFFT_col(:,i) = ifft(FFT_col(:,i));

end

%Perform IFFT on each row

for i=1:size(A,2)

IFFT_row(:,i) = ifft(IFFT_col(:,i));

end

display(abs(A))

ALTERNATE METHOD FOR INVERSE FOURIER TRANSFORM:

Instead of using ifft2() or ifft(), we can also use the following method to obtain the original data from the Fast Fourier transformed result :

1.       Obtain the conjugate of the Forward FFT

2.       Perform Forward fast Fourier transform

3.       Obtain the conjugate of the result from step 2.

4.       Divide it by the number of elements present in the matrix

5.       Obtain the original matrix

MATLAB CODE:

Conj_F = conj(F_FFT);

Conj_FFT = fft2(Conj_F);

IFFT_conj = conj(Conj_FFT)/numel(Conj_FFT)

display(abs(IFFT_conj));

Reference: Digital Image Processing  by Rafael C.Gonzalez, fourth Chapter.

Bit Plane Compression

                               This tutorial  explains in brief the compression of an image using bit plane slicing technique. This is a lossy compression technique. In other words,  part of the data is lost in the compression process compared to the original image.

Check out the post on bit plane slicing to understand better.

Let’s explain with a simple example how encoding and decoding is carried out in Bit plane compression. In this example, the Most Significant Bit(MSB) alone is considered and encoded. For better quality image retrieval, combination of various bit planes such as [8 7] ,[[8 7 6], [8 7 6 5] ..etc can be encoded and decoded. The numbers 8,7,6 , 5 ,..etc represent bit positions.

Compression:

Step 1: Consider a matrix A of size 3 x 5

Step 2: Obtain the binary equivalent of the values in the matrix A. The MATLAB function ‘dec2bin’ can be used for conversion

Step 3: Extract the Most Significant Bit(MSB)  for each value in the matrix from the binary representation. The MATLAB function ‘bitget’ can be used for the same.

Step 4:Rearrange the above MSB values such that each row  contains 8 columns or 8 bits.

In the above example, we  have 3x5=15 values but we need 8 columns in each row.  It can be achieved by padding the matrix with zeros in the end in order to form a matrix which has 8 columns in each row.

Step 5: Convert the binary representation in each row to a decimal number and store it.

Use ‘bin2dec’ MATLAB function to convert binary values to decimal values.

In our example, decimal equivalent of [1 0 0 0 1 0 0 0] = 136 and [1 0 1 0 1 0 1 0] = 170

MATLAB CODE:

clear all

clc

A = [180 4 80 33 201; 120 27 11 160 28; 224 1 133 67 144];

A = uint8(A);

%Encoding

%Check whether zeros has to be appended to the matrix

rem = mod(numel(A),8);

if rem~=0

rem = 8-rem;

end

%Extract the MSB

bit8 = bitget(A,8);

b8 = bit8';

b8 = b8(:);

b8 = [b8;zeros(rem,1)];

%Reshape the matrix as such each row contains 8 columns

mat8 = reshape(b8,8,[])';

str8 = num2str(mat8);

str8 = str8(:,1:3:end);

%Convert the binary to decimal

compressedbit8 = uint8(bin2dec(str8));

Verify the compressed data and original data size for comparison of size used for storage.

MATLAB CODE:

whos A compressedbit8

Decompression:

For Decoding an image/matrix, the compressed / encoded data has to be provided as the input along with size of the original image/matrix and  the vector containing the position of the bits used for encoding in order.

Step 1: Convert the decimal value of the compressed data into binary format.

Step 2: Remove the extra zeros appended to the matrix, if needed.

Step 3: Reshape the matrix to size of the original matrix A using the MATLAB function ‘reshape’.

Step 4: Preallocate a matrix of same size of original matrix A and replace the MSB(Most Significant Bit) of each value in the matrix with the bit we decompressed in the previous step.

Use the MATLAB function ‘bitset’

Step 5: Display the final data.

MATLAB CODE:

%Decoding

%Convert Decimal to Binary

decompressedbit8 = dec2bin(compressedbit8,8);

%Reshape the matrix to the size of original matrix size

%And remove extra zeros appended to the matrix

dbit8 = decompressedbit8';

dbit8 = dbit8(:);

dbit8 = dbit8(1:end-rem);

dbit8 = reshape(dbit8,size(A,2),size(A,1))';

%Preallocate a matrix

Image = zeros([size(A,1) size(A,2)]);

slice8 = zeros([size(A,1) size(A,2)]);

%Set the MSB with the binary values obtained from decompressed matrix

ind_bit8 = find(dbit8=='1');

slice8(ind_bit8) = 1;

Image = bitset(Image,8,slice8);

Image = uint8(Image);

%Display data

display(Image);

The above method can be extended for images by extracting combination of bits.

Example: The 7 and the 8 th bit can be extracted and stored or 2,4,6 and 8 bit can also be extracted.

COMPRESSION:

MATLAB CODE:

%ENCODING

clear all

clc

%INPUT IMAGE

A = imread('cameraman.tif');

%Encoding

bitnums = [6;7;8]; %BIT PLANES

%CHECK IF PADDING WITH ZEROS IS NEEDED OR NOT

rem = mod(numel(A),8);

if(rem~=0)

rem = 8-rem;

end

%EXTRACT EACH BIT AND STORE IT IN THE MATRIX

forinc =1:length(bitnums)

Ind = bitnums(inc);

%EXTRACT THE 'n'th BIT

bitval = bitget(A,Ind);

%PAD WITH ZEROS AND RESHAPE THE MATRIX

bval = bitval';

bval = bval(:);

bval = [bval;zeros(rem,1)];

matv = reshape(bval,8,[])';

strv = num2str(matv);

strv = strv(:,1:3:end);

%CONVERT BINARY TO DECIMAL

compressedbitv(:,inc) = uint8(bin2dec(strv));

end

%STORE THE COMPRESSED DATA IN A FILE

%OPTIONAL

fp = fopen('compressed_data678.data','wb');

fwrite(fp,compressedbitv','uint8');

fclose(fp);

EXPLANATION:

In the given example, 6,7 and 8 bit planes are extracted and compressed.

The compressed data can be stored in a file, if needed.

Original Image size       = 64 KB

Compressed Image size = 24 KB 

NOTE: bitnums = [6;7;8]; Modify this line to compress combination of bits.

Some examples: bitnums=[8] or bitnums=[2;4;6;8] or bitnums=[5;6;7;8]

DECOMPRESSION:

%DECOMPRESSION

clear all

clc

%INPUT FROM THE USER

M = 256; %SIZE OF THE ORIGINAL IMAGE

N = 256; %SIZE OF THE ORIGINAL IMAGE

bitnums = [6;7;8]; %BIT PLANES USED

rem = mod(M*N,8);

if(rem~=0)

rem = 8-rem;

end

%READ THE COMPRESSED DATA

fp = fopen('compressed_data678.data','rb');

compressedbitv = fread(fp,[length(bitnums),Inf],'uint8')';

fclose(fp);

%PREALLOCATE THE MATRIX

Image = zeros([M N]);

forinc = 1:length(bitnums)

Ind = bitnums(inc);

%CONVERT DECIMAL TO BINARY

decompressedbitv = dec2bin(compressedbitv(:,inc),8);

%REMOVE EXTRA ZEROS AND RESHAPE THE MATRIX

dbitv = decompressedbitv';

dbitv = dbitv(:);

dbitv = dbitv(1:end-rem);

dbitv = reshape(dbitv,N,M)';

%SET THE 'n'th BIT

slicev = zeros([M N]);

ind_bitv = find(dbitv == '1');

slicev(ind_bitv) = 1;

     Image = bitset(Image,Ind,slicev);

end

%DISPLAY THE IMAGE

Image = uint8(Image);

figure,imagesc(Image);colormap(gray);

Bit Plane 8

Bit Planes 7,8

Bit Planes 2,4,6 and 8

EXPLANATION:

The bit planes 6, 7 and 8 are extracted and the image is formed using those bits.

NOTE:

For decoding the following lines should be modified during each run.

%INPUT FROM THE USER

M = 256; %SIZE OF THE ORIGINAL IMAGE

N = 256; %SIZE OF THE ORIGINAL IMAGE

bitnums = [6;7;8]; %BIT PLANES USED

The size of the original image should be given as an input. Update M and N with the number of rows and columns of the original image.

The vector ‘bitnums’ should be exactly same as the one in the encoding procedure.

Whenever, ‘bitnums’ is modified during encoding then it should be modified during decoding process as well.