MIN FILTER - MATLAB CODE

MIN FILTER

•      To find the darkest points in an image.
•       Finds the minimum value in the area encompassed by the filter.
•       Reduces the salt noise as a result of the min operation.
•       The 0^th percentile filter is min filter.

The MIN Filtering is similar to MAX filter. Check MAX filter post to know how the formula is used to filter the pepper noise.

MATLAB CODE:

%READ AN IMAGE

A = imread('board.tif');

A = rgb2gray(A(1:300,1:300,:));

figure,imshow(A),title('ORIGINAL IMAGE');

%PREALLOCATE THE OUTPUT MATRIX

B=zeros(size(A));

%PAD THE MATRIX A WITH ZEROS

modifyA=padarray(A,[1 1]);

        x=[1:3]';

        y=[1:3]';

       

for i= 1:size(modifyA,1)-2

    for j=1:size(modifyA,2)-2

      

       %VECTORIZED METHOD 

       window=reshape(modifyA(i+x-1,j+y-1),[],1);

       %FIND THE MINIMUM VALUE IN THE SELECTED WINDOW

        B(i,j)=min(window);

    end

end

%CONVERT THE OUTPUT MATRIX TO 0-255 RANGE IMAGE TYPE

B=uint8(B);

figure,imshow(B),title('IMAGE AFTER MIN FILTERING');

MAX FILTER - MATLAB CODE

• To find the brightest points in an image.
•  Finds the maximum value in the area encompassed by the filter.
•  Reduces the pepper noise as a result of the max operation.
•  The 100^th percentile filter is max filter. Check the 50^thpercentile filter i.e the median filter.

MATLAB CODE:

%READ AN IMAGE

A = imread('board.tif');

A = rgb2gray(A(1:300,1:300,:));

figure,imshow(A),title('ORIGINAL IMAGE');

%PREALLOCATE THE OUTPUT MATRIX

B=zeros(size(A));

%PAD THE MATRIX A WITH ZEROS

modifyA=padarray(A,[1 1]);

        x=[1:3]';

        y=[1:3]';

       

for i= 1:size(modifyA,1)-2

    for j=1:size(modifyA,2)-2

      

       %VECTORIZED METHOD

       window=reshape(modifyA(i+x-1,j+y-1),[],1);

       %FIND THE MAXIMUM VALUE IN THE SELECTED WINDOW

        

       B(i,j)=max(window);

   

    end

end

%CONVERT THE OUTPUT MATRIX TO 0-255 RANGE IMAGE TYPE

B=uint8(B);

figure,imshow(B),title('IMAGE AFTER MAX FILTERING');

2-D Inverse Discrete Cosine Transform

Consider the result obtained after DCT. (Check 2d-DCT )

Apply Inverse Discrete Cosine Transform to obtain the original Image.

MATLAB CODE:

%2-D INVERSE DISCRETE COSINE TRANSFORM

%PREALLOCATE THE MATRIX

A=zeros(size(B));

Temp=zeros(size(B));

[M N]=size(B);

x=1:M;

x=repmat(x',1,N);

y=repmat(1:N,M,1);

figure,

imshow(log(abs(B)),[]);colormap(jet);title('After DCT');

for i=1:M

    for j = 1: N

   

        if(i==1)

          AlphaP=sqrt(1/M);

        else

          AlphaP=sqrt(2/M);

        end

       

        if(j==1)

          AlphaQ=sqrt(1/N);

        else

          AlphaQ=sqrt(2/N);

        end

                 

        cs1=cos((pi*(2*x-1)*(i-1))/(2*M));

        cs2=cos((pi*(2*y-1)*(j-1))/(2*N));

        Temp=B.*cs1.*cs2*AlphaP*AlphaQ;

              

          A(i,j)=sum(sum(Temp));

    end

end

%OUTPUT

figure,

imshow(abs(A),[0 255]);title('Image after IDCT');

2-D Discrete Cosine Transform

DCT is a technique for converting a signal into elementary frequency components.

It is widely used in image compression. Check Inverse discrete cosine transform for the reverse process.

MATLAB CODE:

%CONSIDER A MATRIX

A=[4 2 9 60; 7 10 5 77;88 66 44 3];



%PREALLOCATE THE MATRICES

B=zeros(size(A));

Temp=zeros(size(A));

display(A);

%FIND THE SIZE OF THE
MATRIX

[M N]=size(A);



for i=1:M

    for j=1:N

        

        if(i==1)

          AlphaP=sqrt(1/M);

        else

          AlphaP=sqrt(2/M);

        end

        

        if(j==1)

          AlphaQ=sqrt(1/N);

        else

          AlphaQ=sqrt(2/N);

        end

For A(1,1),  m=1 and n=1. If m= 1 then AlphaP=sqrt(1/M)=0.5774

If n=1 then AlphaQ=sqrt(1/N)=0.5000

For A(2,3),m=1 and n=3. AlphaP=sqrt(2/M)=0.8165
 AlphaQ =  sqrt(2/N) = 0.7071

Temp=0;

       for x=1:M

            for y=1:N

                    cs1=cos((pi*(2*x-1)*(i-1))/(2*M));

                    cs2=cos((pi*(2*y-1)*(j-1))/(2*N));

                    Temp=Temp+(A(x,y)*cs1*cs2);

             end

        end

        B(i,j)=AlphaP*AlphaQ*Temp;

    end

end

%2-D discrete Cosine Transform



display(B);

EXAMPLE 2

%2-D DISCRETE COSINE TRANSFORM FOR AN IMAGE

%READ THE IMAGE

I=imread('coins.png');

I=I(90:150,140:210);

A=double(I);

%PREALLOCATE THE MATRIX

B=zeros(size(A));

Temp=zeros(size(A));

[M N]=size(A);

%

x=1:M;

x=repmat(x',1,N);

y=repmat(1:N,M,1);

for i=1:M

    for j = 1: N

   

        if(i==1)

          AlphaP=sqrt(1/M);

        else

          AlphaP=sqrt(2/M);

        end

       

        if(j==1)

          AlphaQ=sqrt(1/N);

        else

          AlphaQ=sqrt(2/N);

        end

       

             

               

        cs1=cos((pi*(2*x-1)*(i-1))/(2*M));

        cs2=cos((pi*(2*y-1)*(j-1))/(2*N));

        Temp=A.*cs1.*cs2;

              

      

        B(i,j)=AlphaP*AlphaQ*sum(sum(Temp));

    end

end

%OUTPUT

figure,

subplot(2,1,1);imshow(I);title('Original Image');

subplot(2,1,2),imshow(log(abs(B)),[]);
colormap(jet);title('After DCT');

Bit-Plane Slicing

            Digitally, an image is represented in terms of pixels.

These pixels can be expressed further in terms of bits.

Consider the image ‘coins.png’ and the pixel representation of the image.

Consider the pixels that are bounded within the yellow line. The binary formats for those values are (8-bit representation)

The binary format for the pixel value 167 is 10100111

Similarly, for 144 it is 10010000

This 8-bit image is composed of eight 1-bit planes.

Plane 1 contains the lowest order bit of all the pixels in the image.

And plane 8 contains the highest order bit of all the pixels in the image.

Let’s see how we can do this using MATLAB

A=[167 133 111

      144 140 135

      159 154 148]

B=bitget(A,1);  %Lowest order bit of all pixels

‘bitget’ is a MATLAB function used to fetch  a bit from the specified position from all the pixels.

B=[1 1 1

      0 0 1

      1 0 0]

B=bitget(A,8);%Highest order bit of all pixels

B=[1 1 0

      1 1 1 

      1 1 1]

MATLAB CODE:

%Bit Planes from 1 to 8. Output Format: Binary

A=imread('coins.png');

B=bitget(A,1);

figure,

subplot(2,2,1);imshow(logical(B));title('Bit plane 1');

B=bitget(A,2);

subplot(2,2,2);imshow(logical(B));title('Bit plane 2');

B=bitget(A,3);

subplot(2,2,3);imshow(logical(B));title('Bit plane 3');

B=bitget(A,4);

subplot(2,2,4);imshow(logical(B));title('Bit plane 4');

 

B=bitget(A,5);

figure,

subplot(2,2,1);imshow(logical(B));title('Bit plane 5');

B=bitget(A,6);

subplot(2,2,2);imshow(logical(B));title('Bit plane 6');

B=bitget(A,7);

subplot(2,2,3);imshow(logical(B));title('Bit plane 7');

B=bitget(A,8);

subplot(2,2,4);imshow(logical(B));title('Bit plane 8');

 

Image reconstruction using n bit planes.

1.     The nth plane in the pixels are multiplied by the constant 2^n-1

2.     For instance, consider the matrix

A= A=[167 133 111

      144 140 135

      159 154 148] and the respective bit format

         

3.     Combine the 8 bit plane and 7 bit plane.

For 10100111, multiply the 8 bit plane with 128 and 7 bit plane with 64.

(1x128)+(0x64)+(1x0)+(0x0)+(0x0)+(1x0)+(1x0)+(1x0)=128

4.     Repeat this process for all the values in the matrix and the final result will be

[128 128 64

 128 128 128

128 128 128]

MATLAB CODE:

%Image reconstruction by combining 8 bit plane and 7 bit plane

A=imread('coins.png');

B=zeros(size(A));

B=bitset(B,7,bitget(A,7));

B=bitset(B,8,bitget(A,8));

B=uint8(B);

figure,imshow(B);

Image reconstructed using 8 and 7 bit planes

Explanation:

‘bitset’ is used to set  a bit at a specified position. Use ‘bitget’ to get the bit at the positions 7 and 8 from all the pixels in matrix A and use ‘bitset’ to set these bit values at the positions 7 and 8 in the matrix B.

%Image reconstruction by combining 8,7,6 and 5 bit planes

A=imread('coins.png');

B=zeros(size(A));

B=bitset(B,8,bitget(A,8));

B=bitset(B,7,bitget(A,7));

B=bitset(B,6,bitget(A,6));

B=bitset(B,5,bitget(A,5));

B=uint8(B);

figure,imshow(B);

Image reconstructed using 5,6,7 and 8 bit planes

              Also check  bit plane compression.