Asked by mcc mscelec
on 14 Dec 2011

*Dear Sir,
I have an issue in extracting the watermark..*

**PROPOSED APPROACH :**

1.Bring the image to frequency domain by applying the DCT

2.generate a watermark signal

3.Use the thousand largest coefficients of the original image to embed a watermark sequence of length 1000.

4.Coefficients are modified according to the stream bits of the message using the equation below,

|CAW = CA (1 + α Wi)|

5.Extraction process – simply subtracting the original DCT coefficients from the watermarked image coefficients. --------------------------------------------------------------------------------------------------------------------------------

***The code for embedding and extraction is as follows :*** *______________________________________________________________*

[fname pthname]=uigetfile('*.jpg;*.png;*.tif;*bmp','Select the Asset Image'); %select image

I=imread([pthname fname]);

wmsz=1000; %watermark size

I=I(:,:,1);%get the first color in case of RGB image

[r,c]=size(I);

D=dct2(I);%get DCT of the Asset

D_vec=reshape(D,1,r*c);%putting all DCT values in a vector

[D_vec_srt,Idx]=sort(abs(D_vec),'descend');%re-ordering all the absolute values

W=randn(1,wmsz);%generate a Gaussian spread spectrum noise to use as watermark signal

Idx2=Idx(2:wmsz+1);%choosing 1000 biggest values other than the DC value

%finding associated row-column order for vector values

IND=zeros(wmsz,2);

for k=1:wmsz

x=floor(Idx2(k)/r)+1;%associated culomn in the image

y=mod(Idx2(k),r);%associated row in the image

IND(k,1)=y;

IND(k,2)=x;

end

D_w=D;

for k=1:wmsz

%insert the WM signal into the DCT values

D_w(IND(k,1),IND(k,2))=D_w(IND(k,1),IND(k,2))+.1*D_w(IND(k,1),IND(k,2)).*W(k);

end

I2=idct2(D_w);%inverse DCT to produce the watermarked asset

%The extraction process is simply subtracting the original DCT %coefficients from the %watermarked image ones. The code can be written like below:

W2=[];%will contain watermark signal extracted from the image for k=1:wmsz W2=[W2(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];%watermark extraction

end | |MONOSPACED TEXT|

*_____________________________________________________________*

The ERROR appears as follows in the 25th line :

??? Attempted to access W2(-0.0432565); index must be a positive integer or logical.

Error in ==> pooya at 29

W2=[W2(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];%watermark extraction

-----------------------------------------------------------

Please help

Thank you.

*No products are associated with this question.*

Answer by Eric Pahlke
on 15 Dec 2011

Accepted answer

There's a lot of code to sort through to figure out exactly what you need to fix, but what stands out is this line:

D_w(IND(k,1),IND(k,2))=D_w(IND(k,1),IND(k,2))+.1*D_w(IND(k,1),IND(k,2)).*W(k);

You're using D_w as an index on line 25(where you get the error), which means it has to be an integer.

However in the line above you're assigning fractional values to D_w:

(D_w(stuff) = D_w(stuff) + .1*D_w(stuff)

Was the .1* intended to be a .* ?

mcc mscelec
on 16 Dec 2011

No.. the value is (0.1)

And i'm extremely sorry to say that it was my mistake to mention as to where the error appears. The error actually appears in the 29th line:

W2=[W2(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];

The error says

"??? Attempted to access W2(-0.0432565); index must be a positive integer or logical."

Could you help me in debugging the error.

Thank you :)

Walter Roberson
on 17 Mar 2012

Answer by mcc mscelec
on 18 Mar 2012

I figured out the flaw in the code line:

W2=[W2(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];%watermark extraction

All this while i haven't been storing the new values of W2. The code worked successfully when i made a small modification as follows:

W2(k)=[(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];%watermark extraction

**------------------Here is the complete working program:-----------------------------**

I=imread('girl512.bmp'); subplot(2,3,1),imshow(I,[]),title('Original Image'); wmsz=1000; I=I(:,:,1); [r,c]=size(I); D=dct2(I); D_vec=reshape(D,1,r*c); [D_vec_srt,Idx]=sort(abs(D_vec),'descend'); W=randn(1,wmsz); subplot(2,3,2),plot(W),title('Watermark'); Idx2=Idx(2:wmsz+1);%choosing 1000 biggest values other than the DC %finding associated row-column order for vector values IND=zeros(wmsz,2); for k=1:wmsz x=floor(Idx2(k)/r)+1;%associated culomn in the image y=mod(Idx2(k),r);%associated row in the image IND(k,1)=y; IND(k,2)=x; end D_w=D;

%WATERMARK EMBEDDING for k=1:wmsz fw=D_w((IND(k,1),IND(k,2)); fw=fw+0.1*fw.*W(k); end I2=idct2(D_w);%inverse DCT to produce the watermarked asset I2_int=uint8(I2); imwrite(I2_int,'I2_watermarkedn.bmp','bmp'); subplot(2,3,3),imshow('I2_watermarkedn.bmp'),title('Watermarked Image');

%WATERMARK EXTRACTION W2=[]; for k=1:wmsz W2(k)=[(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10]; end subplot(2,3,4),plot(W2),title('Extracted Watermark');

Thank you for all the comments and suggestions :)

Show 2 older comments

Pardeep Kumar
on 15 Apr 2013

pls correct these errors

fw=D_w(IND(k,1),IND(k,2));

and

%WATERMARK EXTRACTION W2=(2500000); for k=1:wmsz W2(k)=((D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10); end subplot(2,3,4),plot(W2),title('Extracted Watermark');

Answer by mcc mscelec
on 14 Apr 2012

Code for WATERMARK DETECTOR RESPONSE FUNCTION

function WM_plot(r,c,ext_wm,orig_wm) for k=1:1000 wm=randn(r,c);%depending on the size of the watermark wm=uint8(wm);%if necessary store(k)=WM_detect(ext_wm,wm);%wrong watermarks if k == 400 store(k)=WM_detect(ext_wm,orig_wm);%original watermark detection end end figure(1),plot(1:k,[store]),ylabel('Watermark detector response'),xlabel('random watermarks'); hold on %threshold calculation [peak,ind]=sort(store,'descend'); threshold=peak(2)+(peak(2)*0.1);%T=second highest peak+10percentof the same figure(1),plot(1:1000,[threshold],'red'); hold on figure(1),plot(1:1000,peak(2),'green');

Regards, mcc :)

Pardeep Kumar
on 26 May 2013

sir ,there is error showing in the line store(k)=WM_detect(ext_wm,orig_wm);

ie.value assigned to variable store might be unused .

Answer by Lester
on 2 Mar 2013

sir ,

may i know what are the steps followed for embedding and extracting the image in DCT domain in the above code....???

Answer by Alaa Eleyan
on 22 Nov 2013

the code is working properly..

You just need to add space between W2 and (D_w(....... in the for loop :

Wrong : W2=[W2(D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];

Correct : W2=[W2 (D_w(IND(k,1),IND(k,2))/D(IND(k,1),IND(k,2))-1)*10];

Opportunities for recent engineering grads.

## 1 Comment

## Walter Roberson (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/23939#comment_52335

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup