how to extract a watermark in matlab?

the matlab code which i have used is:-

clear all; close all; clc;

bit=input('Please define which bit you want to extract from? (8 | 7 | 6 | 5)'); image=imread('C:\Users\user\Desktop\wm pictures\lena.jpg');

imshow(image);

[row,col]=size(image);

Hlength=row*col;

i=1;

j=1;

k=1; row=128;

col=384;

wmimage=imread('C:\Users\user\Desktop\wm pictures\cameraman.jpg');

wm=dec2bin(wmimage);

Wlength=row*col*8;

host=dec2bin(image);

cnt=0;

while i <= Hlength

      cnt=cnt+1;

      if cnt>Wlength

          break;

      end
      wm(j,k)=host(i,bit);
      k=k+1;
      if k>8
          k=1;
          j=j+1;
      end
     i=i+1;
  end

wm1=bin2dec(wm);

wm2=reshape(wm1,row,col);

wmimage(1:row,1:col)=wm2(1:row,1:col);

imwrite(wmimage,'extracted.jpg');

imshow(wmimage);

This code is not giving the desired result.

If anyone can give me some other codes for extraction in spatial domain then please help me out.

thank you

Answer 1

[Image Analyst]

If your watermark is just in one bitplane, then perhaps you'd be interested in my bitplane viewer, or in my LSB watermark demo, both attached.

[richa sharma]

Thanks sir for your help. I hv one more question to ask:

i have to divide an image into m*n blocks and then find mean and variance of each block.

cn u help in any way regarding this question? if any matlab code is dere, plz help me wid dat.

thanks in advance

Answer 2

[Walter Roberson]

Re-read the documentation about size() !

[d1,d2,d3,...,dn] = size(X), for n > 1, returns the sizes of the dimensions of the array X in the variables d1,d2,d3,...,dn, provided the number of output arguments n equals ndims(X). If n does not equal ndims(X), the following exceptions hold:

n < ndims(X)

di equals the size of the ith dimension of X for 0<i<n, but dn equals the product of the sizes of the remaining dimensions of X, that is, dimensions n through ndims(X).

n > ndims(X)

size returns ones in the "extra" variables, that is, those corresponding to ndims(X)+1 through n.

[Image Analyst]

http://blogs.mathworks.com/steve/2011/03/22/too-much-information-about-the-size-function/

Of course she's hard coding in the rows and columns after she calls size() so that blows the values that the size() function returned. Not sure why she does that.

And we don't know the algorithm used to insert the watermark so we don't know if that cryptic, uncommented algorithm used to extract the watermark, or repair the original image by removing the watermark, is correct or not, but that's what the debugger is for.