Thread Subject:

Convert 4096 (8 X 8) image blocks into an image

Dear all, I'm working on watermarking of digital image. A portion of the code is shown here:

clc;
clear all;
close all;

start_time = cputime; %Starting Time of Execution

I = imread('lena.jpg');

[r, c]=size(I);
bs=8; % Block Size (8x8)
nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks

% Dividing the Cover image into 8x8 Blocks
for i=1:(r/bs)
    for j=1:(c/bs)
        dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
    end
kk=kk+(r/bs);
end
size(dctimg) = 8 8 4096.
How can I get size(dctimg) = 512 X 512;
Please Help. thanks in advance.

"Krishna Prasad Satamraju" wrote in message <ietg1j$jf5$1@fred.mathworks.com>...
> Dear all, I'm working on watermarking of digital image. A portion of the code is shown here:
>
> clc;
> clear all;
> close all;
>
> start_time = cputime; %Starting Time of Execution
>
> I = imread('lena.jpg');
>
> [r, c]=size(I);
> bs=8; % Block Size (8x8)
> nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks
>
> % Dividing the Cover image into 8x8 Blocks
> for i=1:(r/bs)
> for j=1:(c/bs)
> dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
> end
> kk=kk+(r/bs);
> end
> size(dctimg) = 8 8 4096.
> How can I get size(dctimg) = 512 X 512;
> Please Help. thanks in advance.

doc reshape

but why not just use blkproc to do the whole thing for you?
doc blkproc

or even cell2mat->cellfun->mat2cell
?

On Dec 22, 1:29 pm, "Krishna Prasad Satamraju"
> size(dctimg) = 8  8 4096.
> How can I get size(dctimg) = 512 X 512;
> Please Help. thanks in advance.
--------------------------------
I don't know. How did you *want* to turn your 3D image into a 2D
image? I have no idea. Maybe summing, maybe reshaping, maybe just
building it up differently in the first place, I have no idea. I
don't even know why you created this 3D image in the first place
instead of doing what Sean suggested. Are you SURE you really need a
3D (8x8x4096) or 2D (512x512) image at all???

Dear Krishna,

Let me add my wild guessing also:

> > size(dctimg) = 8  8 4096.
> > How can I get size(dctimg) = 512 X 512;

  % D = [8 x 8 x 4096]
  D = reshape(D, 8, 8, 64, 64);
  E = permute(D, [1, 3, 2, 4]);
  F = reshape(D, 512, 512);

Perhaps, or perhaps not. Jan

"Sean de " <sean.dewolski@nospamplease.umit.maine.edu> wrote in message <ietgro$ddj$1@fred.mathworks.com>...
> "Krishna Prasad Satamraju" wrote in message <ietg1j$jf5$1@fred.mathworks.com>...
> > Dear all, I'm working on watermarking of digital image. A portion of the code is shown here:
> >
> > clc;
> > clear all;
> > close all;
> >
> > start_time = cputime; %Starting Time of Execution
> >
> > I = imread('lena.jpg');
> >
> > [r, c]=size(I);
> > bs=8; % Block Size (8x8)
> > nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks
> >
> > % Dividing the Cover image into 8x8 Blocks
> > for i=1:(r/bs)
> > for j=1:(c/bs)
> > dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
> > end
> > kk=kk+(r/bs);
> > end
> > size(dctimg) = 8 8 4096.
> > How can I get size(dctimg) = 512 X 512;
> > Please Help. thanks in advance.
>
> doc reshape
>
> but why not just use blkproc to do the whole thing for you?
> doc blkproc
>
> or even cell2mat->cellfun->mat2cell
> Thanks for your response to my recent post related to Image watermark. You quoted use of blkproc. I request you to please go throough the code once and suggest me to come out of the problem. I attached matlab file of the program
My aim is to
1. divide image(512 X 512 lena.jpg) into blocks of 8X8,
2. then perform DCT on each block.
3. In the DCT, replace the midfrequency coefficients (we get 64 coefficients out of which I pick 9) with watermark image. (my watermark image is divided into 3X3 block)
4. After replacing, perform IDCT to get the watermarked image. If I use blkproc I get dct for the entire image and cannot access individual 8 X 8 blocks
That's why I followed the above method, where I can access each block individually
The dctimg is an array of 8X8 dct coefficients.
size(dctimg) = 8 8 4096
Now I have to combine the 4096 blocks of image into a single image and calculate inverse DCT on it.
Please suggest a solution thank you

i am doing steganography in the similar way,
My aim is to
1. divide image(512 X 512 lena.jpg) into blocks of 8X8,

2. then perform DCT on each block and find quantized DCT values..

3. modify the non zero DCT quantized values coefficients (we get only 6-9 coeff out of 64 coefficients) with secret message.

4. After modifying, perform IDCT to get the stego image.

If I use blkproc I get dct for the entire image and cannot access individual 8 X 8 blocks

please help me how to proceed with the 8x8 block.








> > > clc;
> > > clear all;
> > > close all;
> > >
> > > start_time = cputime; %Starting Time of Execution
> > >
> > > I = imread('lena.jpg');
> > >
> > > [r, c]=size(I);
> > > bs=8; % Block Size (8x8)
> > > nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks
> > >
> > > % Dividing the Cover image into 8x8 Blocks
> > > for i=1:(r/bs)
> > > for j=1:(c/bs)
> > > dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
> > > end
> > > kk=kk+(r/bs);
> > > end
> > > size(dctimg) = 8 8 4096.
> > > How can I get size(dctimg) = 512 X 512;

> My aim is to
> 1. divide image(512 X 512 lena.jpg) into blocks of 8X8,
> 2. then perform DCT on each block.
> 3. In the DCT, replace the midfrequency coefficients (we get 64 coefficients out of which I pick 9) with watermark image. (my watermark image is divided into 3X3 block)
> 4. After replacing, perform IDCT to get the watermarked image. If I use blkproc I get dct for the entire image and cannot access individual 8 X 8 blocks
> That's why I followed the above method, where I can access each block individually
> The dctimg is an array of 8X8 dct coefficients.
> size(dctimg) = 8 8 4096
> Now I have to combine the 4096 blocks of image into a single image and calculate inverse DCT on it.
> Please suggest a solution thank you

"Krishna Prasad Satamraju" wrote in message <ietg1j$jf5$1@fred.mathworks.com>...
> Dear all, I'm working on watermarking of digital image. A portion of the code is shown here:
>
> clc;
> clear all;
> close all;
>
> start_time = cputime; %Starting Time of Execution
>
> I = imread('lena.jpg');
>
> [r, c]=size(I);
> bs=8; % Block Size (8x8)
> nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks
>
> % Dividing the Cover image into 8x8 Blocks
> for i=1:(r/bs)
> for j=1:(c/bs)
> dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
> end
> kk=kk+(r/bs);
> end
> size(dctimg) = 8 8 4096.
> How can I get size(dctimg) = 512 X 512;
> Please Help. thanks in advance.


HI, CAN I GET THE SOLUTION FOR DIVIDING THE 512X512 IMAGE INTO 8X8 BLOCK

"Krishna Prasad Satamraju" wrote in message <ietg1j$jf5$1@fred.mathworks.com>...
> Dear all, I'm working on watermarking of digital image. A portion of the code is shown here:
>
> clc;
> clear all;
> close all;
>
> start_time = cputime; %Starting Time of Execution
>
> I = imread('lena.jpg');
>
> [r, c]=size(I);
> bs=8; % Block Size (8x8)
> nob=(r/bs)*(c/bs); % Total number of 8x8 Blocks
>
> % Dividing the Cover image into 8x8 Blocks
> for i=1:(r/bs)
> for j=1:(c/bs)
> dctimg(:,:,kk+j)=dct2(I((bs*(i-1)+1:bs*(i-1)+bs),(bs*(j-1)+1:bs*(j-1)+bs)));
> end
> kk=kk+(r/bs);
> end
> size(dctimg) = 8 8 4096.
> How can I get size(dctimg) = 512 X 512;
> Please Help. thanks in advance.


HI, CAN I GET THE SOLUTION FOR DIVIDING THE 512X512 IMAGE INTO 8X8 BLOCK

ImageAnalyst <imageanalyst@mailinator.com> wrote in message <e249575d-0b6e-462e-9a6b-ea9d090168e1@g26g2000vbi.googlegroups.com>...
> On Dec 22, 1:29 pm, "Krishna Prasad Satamraju"
> > size(dctimg) = 8  8 4096.
> > How can I get size(dctimg) = 512 X 512;
> > Please Help. thanks in advance.
> --------------------------------
> I don't know. How did you *want* to turn your 3D image into a 2D
> image? I have no idea. Maybe summing, maybe reshaping, maybe just
> building it up differently in the first place, I have no idea. I
> don't even know why you created this 3D image in the first place
> instead of doing what Sean suggested. Are you SURE you really need a
> 3D (8x8x4096) or 2D (512x512) image at all???

Can any one tel that how we can use this code for Color images.