# Image Steganography using LSB?

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KARTHICK on 3 Mar 2013
Answered: Hassan Vakani on 15 Mar 2021
I have coded a LSB algorithm for Image Steganography. During retrieval process i'm getting different msg. Can anyone correct this code please!
Embedding code
message = 'hellokarthick'
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, 'hiddenmsgimage.bmp');
Retriever coding
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);
Tommy Halim on 5 Apr 2019
Very helpful, thank you Walter Roberson

Walter Roberson on 3 Mar 2013

Maged Rawash on 25 Apr 2015
it gives you different msg because you are using JPG and jpg using lossy mode to compress .. which change the pixel value and return different msg ..
you can use that code in imwrite();
imwrite(s,'img.jpg', 'Mode','lossless' );
% not all image application will read JPG lossless
so just work on PNG, TIFF it works with me ...
to retrieve the correct massage without guessing the char number of the massage ... just store the msg length in first pixel and get it in retrieving code like this ....
c(1:1:1)= length(msg) ; %to count massage Char to easly retrive all the massage
c=imresize(c,[size(c,1) size(c,2)],'nearest');
message = msg ; %add ' .' to prevint lossing one char
message = strtrim(message);
m = length(message) * 8;
AsciiCode = uint8(message);
binaryString = transpose(dec2bin(AsciiCode,8));
binaryString = binaryString(:);
N = length(binaryString);
b = zeros(N,1);
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);
elseif(LSB == 1)
s(i,j) = (c(i,j) - 1);
elseif(LSB == 0)
s(i,j) = (c(i,j) + 1);
end
k = k + 1;
end
end
imgWTxt = 'msgimage.png';
imwrite(s,imgWTxt);
----------------------------------- Retriever coding
height = size(s,1);
width = size(s,2);
m = double( s(1:1:1) ) * 8 ;
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,[]);
textString = char(binValues*binMatrix);
disp(textString);
thanks For the Code ...
Walter Roberson on 13 Mar 2019
That sounds like a university assignment that you would be expected to do your own work for.

supriya on 10 Mar 2013
Edited: supriya on 10 Mar 2013
Actually wat i did is..made an array of the the positions of the pixels whose lsb wud be changed due to the encryption method and then did some changes in ur decryption method..see
c = imread('C:\Documents and Settings\All Users\Documents\My Pictures\Sample Pictures\Sunset.jpg');
message = 'hellokarthick'
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; Array=[];l=1;my=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);
l=k+1;
else
if(LSB == 1)
s(i,j) = c(i,j) - 1;
else
s(i,j) = c(i,j) + 1;
Array(my)=l;
l=l+1;
my= my + 1;
end
k = k + 1;
end
end
end
imwrite(s, 'hiddenmsgimage.bmp');
Retriever code changes:
k = 1;my=1;ur=1;
for i = 1 : height
for j = 1 : width
if( k<=m )
if (my<numel(Array) && Array(my)==ur)
b(k)=~(mod(double(s(i,j)),2));
else
b(k) = mod(double(s(i,j)),2);
end
k = k + 1;
my= my + 1;
end
ur=ur+1;
end
end
Walter Roberson on 12 Mar 2018
Which MATLAB version are you using? What shows up for
which -all dec2bin

ARJUN K P on 16 May 2015
this code is not actally work correctly.. the extraction text is wrong...pls help me
the output is:
Image Analyst on 16 May 2015
What's that garbage at the end? Don't paste all that stuff into the command window. Learn how to use a script.

Mariam Chatha on 18 Sep 2016
how would the code change if we were doing MSB?
Image Analyst on 18 Sep 2017
See my code for hiding in the LSB http://www.mathworks.com/matlabcentral/fileexchange/54155-encoding-text-into-image-gray-levels Of course if you hid it in the most significant bit, that would corrupt the image quite a bit. But nonetheless, you can use it to do that if you want.

Aishwarya Rajan on 30 Nov 2016
Can anyone please explain how the LSB embedding part of the above code works?

Swati Nagpal on 21 Jul 2018
Their is problem with the retrieval code it is showing different unexpected results. So if anyone got the correct result please do share the code. swatinagpal087@gmail.com

please someone send correct code of image stegnography i will be thankfull
Walter Roberson on 9 Aug 2018
"ok correct my code and send it back."
No. I posted the link that will get you to a list of over 300 steganography postings. Some of them have complete code, and others have discussions of how you would need to deal with situations such as yours. I have no reason to do your work.

Edited: Walter Roberson on 19 Jan 2021
% Clear the existing workspace
clear all;
% Clear the command window
clc;
% Convert image to greyscale
input=rgb2gray(input);
% Resize the image to required size
input=imresize(input, [512 512]);
% Message to be embedded
message='geeksforgeeks';
% Length of the message where each character is 8 bits
len = length(message) * 8;
% Get all the ASCII values of the characters of the message
ascii_value = uint8(message);
% Convert the decimal values to binary
bin_message = transpose(dec2bin(ascii_value, 8));
% Get all the binary digits in separate row
bin_message = bin_message(:);
% Length of the binary message
N = length(bin_message);
% Converting the char array to numeric array
bin_num_message=str2num(bin_message);
% Initialize output as input
output = input;
% Get height and width for traversing through the image
height = size(input, 1);
width = size(input, 2);
% Counter for number of embedded bits
embed_counter = 1;
% Traverse through the image
for i = 1 : height
for j = 1 : width
% If more bits are remaining to embed
if(embed_counter <= len)
% Finding the Least Significant Bit of the current pixel
LSB = mod(double(input(i, j)), 2);
% Find whether the bit is same or needs to change
temp = double(xor(LSB, bin_num_message(embed_counter)));
% Updating the output to input + temp
output(i, j) = input(i, j)+temp;
% Increment the embed counter
embed_counter = embed_counter+1;
end
end
end
% Write both the input and output images to local storage
% Mention the path to a folder here.
imwrite(input, 'path_to_folder\originalImage.png');
imwrite(output, 'path_to_folder\stegoImage.png');
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 19 Jan 2021
There is a legal distinction between steganography, which "hides" data, and "encryption". If you need to decrypt data then you need to have encrypted it... and we cannot talk about encryption here for legal reasons.

Hassan Vakani on 15 Mar 2021
I know it is a quite late. First of all thank you for the code. Your logic doesn't have any problem. It is just the placement of the counter which is k=k+1. It should outside the outer if statement and it works.
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
end
k = k + 1;
end
end
imwrite(s, 'hiddenmsgimage.bmp');
No need for adding the b(k) or anything else