I am getting wrong result while extracting the text

%texthide   
    c = imread('Ramya.jpg');
    message = 'Ramyashree'
    message = strtrim(message);
    m = length(message) * 8;
    AsciiCode = uint8(message); 
    binaryString = transpose(dec2bin(AsciiCode,8));
    binaryString = binaryString(:);
    N = length(binaryString);
    b = zeros(N,1); %b is a vector of bits
    for k = 1:N
        if(binaryString(k) == '1')
            b(k) = 1;
        else
            b(k) = 0;
        end
    end
    s = c;
    height = size(c,1);
    width = size(c,2);
    k = 1;
    for i = 1 : height
        for j = 1 : width
             LSB = mod(double(c(i,j)), 2);
            if (k>m || LSB == b(k))
                s(i,j) = c(i,j);
            else
                if(LSB == 1)
                    s(i,j) = c(i,j) - 1;
                else
                    s(i,j) = c(i,j) + 1;
                end
            k = k + 1;
        end
        end
    end
     imwrite(s, 'Ramya_hiddenmsgimage.jpg');
     
     
     
     %extract.m
     
    s = imread('Ramya_hiddenmsgimage.jpg');
    height = size(s,1);
    width = size(s,2);
    %For this example the max size is 100 bytes, or 800 bits, (bytes * = bits
    m = 800;
    k = 1;
    for i = 1 : height
        for j = 1 : width
            if (k <= m)
                b(k) = mod(double(s(i,j)),2);
                k = k + 1;
            end
        end
    end
    binaryVector = b;
    binValues = [ 128 64 32 16 8 4 2 1 ];
    binaryVector = binaryVector(:);
    if mod(length(binaryVector),8) ~= 0
        error('Length of binary vector must be a multiple of 8.');
    end
    binMatrix = reshape(binaryVector,8,100);
    display(binMatrix);
    textString = char(binValues*binMatrix);    
    disp(textString);
    
    
    Output is:À                           òꪪªªªÿÿ°   ÿÿÕRÿÿ÷Ô      ` Àyÿá?ÿÀ *?ÿ½ÒÅJ      ÿáØ 1þ

[Chirag Sarda]

Can you plese descibe what result you were expecting?

[Jan]

    N = length(binaryString);
    b = zeros(N,1); %b is a vector of bits
    for k = 1:N
        if(binaryString(k) == '1')
            b(k) = 1;
        else
            b(k) = 0;
        end
    end
    
% Simplification:
b = (binaryString == '1');

Answer 1

[DGM]

You're writing to non-consecutive pixels based on their original values and the values of the embedded message. When you read the marked image, you're reading a fixed number of consecutive pixels with no knowledge of which pixels were altered or what the original values were (either the image or the message).

% inputs
testfilename = 'test.png';
cleanpict = imread('cameraman.tif'); % you're presuming that the input is single-channel
message = 'Potato';

% simplify things for testing
cleanpict = cleanpict(1,:);

% convert message to a logical column vector (binary representation at 8b/char)
message = strtrim(message);
b = dec2bin(message,8) == '1'; % convert the whole thing first; don't need to cast uint8
b = reshape(b.',[],1); % then do all the reshaping
m = numel(b); % this just seems safer

markedpict = cleanpict;
height = size(cleanpict,1);
width = size(cleanpict,2);
k = 1;
idx = []; % just for testing
for i = 1 : height
    for j = 1 : width
        LSB = mod(double(cleanpict(i,j)), 2);
        if (k>m || LSB == b(k))
            % skip if remainder matches message
            markedpict(i,j) = cleanpict(i,j);
        else
            % if remainder differs from message
            if(LSB == 1)
                % if odd, make even
                markedpict(i,j) = cleanpict(i,j) - 1;
            else
                % if even, make odd
                markedpict(i,j) = cleanpict(i,j) + 1;
            end
            k = k + 1;
            idx = [idx; j]; % keep track of which indices are altered
        end
    end
end
imwrite(markedpict, testfilename);

% the indices that are altered are not consecutive
idx.'

ans = 1×48

2 3 4 5 7 9 12 17 18 19 20 23 28 30 34 35 36 37 38 42 45 47 49 51 52 53 61 62 63 64

% demonstrate that the payload can be recovered 
b = mod(markedpict(idx),2); % with knowledge of altered indices
key = '01';
b = reshape(key(b+1),8,[]).'; % convert to char representation of binary
b = char(bin2dec(b).') % convert to numeric -> char

b = 'Potato'

Also, there's no reason to assume that any such watermarking would survive being written to a JPG anyway.

[Ramya]

thank you so much