LSB new twist : text based Stegnograph

Today i am going to discuss about a simple twist in existing Stegnography and text base data hiding techniques. While creating challenges for Preconference hacking challenge “HACKIM” for nullcon conference. we cameup with text based LSB hiding challenge as crypto challenge part 5. today i am releasing tools i used to encoding and decoding such values.

The basic prremise behind the tool is the simplest Image based Stegnographic technique where data is hidden in the least significant bit of the Pixel. Here we implemented the same principle on a normal text. The difference in both is in case of the image there is very less distortion and original image is looks pretty much the same however in text implementation the text looks a lot different.

Example

text to hide = a

binary of text to hide = 01000001

So for this we need a cover of size = 8char

lets assume the cover is abcdefgh

We will replace the LSB of each character with one bit of the text which we need to hide. starting from HSB to LSB of text.

S.NoTextbinaryconverted BinaryFinal Text
1a0110000101100000`
2b0110001001100011c
3c0110001101100011c
4d0110010001100100d
5e0110010101100100d
6f0110011001100110f
7g0110011101100110f
8h0110100001101001i

now this new text could be send across

final text becomes `ccddffi

Now as we can see the text is changed a lot however the important thing to remember is first and formost the cover is extracted and is most easily available.

by just random variation of 1 and 0 you can get the actual value.

Here is an funny tip

Make your envelop text something interesting so that focus shifts on the envelope rather then the variation in data.

For the above task to be simplified i wrote simple encode and decode scripts as listed below. (Scripts are at there crude level’s right now may improve on them if the need arises.)

Encode

#License : GPL2
#License File : http://www.gnu.org/licenses/gpl-2.0.html
import sys
Key="CoDe"
Cover="Secret is hidden at /home/anant/"
km=""
if (len(Key) * 8) == len(Cover):
	print "Original : " + Cover
	for k in Key:
		km += bin(ord(k))[2:].zfill(8)
	sys.stdout.write("Conceled : ")
	i=0
	for c in Cover:
		y=bin(ord(c))[2:].zfill(8)
		s = list(y)
		s[7]=km[i]
		i+=1
		sys.stdout.write(chr(int('0b' + "".join(s),2)))
	print ""
else:
	print "Invalid Cover Size : " + str(len(Cover)) + " : " + str((len(Key) * 8))

Decoding

#License : GPLv2
#License URL : http://www.gnu.org/licenses/gpl-2.0.html
import sys
crypt="Rebrdt!ir!iheeeo at .inld/an`ot/"
cnt=0
decode=list()
#print len(crypt)
if (len(crypt) % 8) == 0:
	for c in crypt:
		y=bin(ord(c))
		decode.append(y[len(y)-1])
		cnt = cnt + 1
		if (cnt % 8 == 0):
			sys.stdout.write(chr(int('0b' + "".join(decode),2)))
			decode = []
else:
	print "Invalid Input"
print " "

Hope people might find some use of this somewhere..

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