Steganography

The images that I hide a message are the 001.png, 003.png and 005.png. I used big images because it can make more difficult to spot the changes.


I created two codes, one to hide the message and another to show the message:

Hide message

I used the LSB algorithm to encrypt the message. Before, I created a list with the phrase in binary. Then I put this bits in the LSB but I started to put the bits in the image  from the middle of the image and I used a space for each word of 10 bits.


Code:

	#!/usr/bin/python
	import Image
	import sys
	from termcolor import colored, cprint
	import numpy as np
	import matplotlib.pyplot as plt
	import enchant
	print_color = lambda x: cprint(x,'cyan')
	def convertir_binario(texto):
	'''recibe una cadena de texto y regrasa su valor
	en binario, las pone en bloques de 10
	'''
	ant = ''
	numeros = []
	for i in texto:
	numero = ord(i)
	numeros.append(numero)
	binario = bin(numero)
	binario = str(binario)
	binario = binario.split('b')
	binario = binario[1]
	if binario == '100000':
	binario = '0100000'
	binario = ant +'000'+binario
	ant = binario
	binario = ant
	return binario, numeros
	def encripta(imagen, lista):
	'''esta funcion lee la imagen original y crea una aparir de ella
	cambiando el LSB de cada RGB de acuerdo a la lista
	que se recibe como argumento, solo se cambian los valores que son pares si no son pares se salta al siguiente byte, empieza a escribir apartir de el ultimo pixel de la imagen menos el tamano de la cadena menos 100 solo para despistar '''
	im = Image.open(imagen)
	im2 = Image.new("RGB",im.size)
	print im
	l = im.histogram()
	colores = list(im.getdata())
	plt.plot(l,'go')
	cont = 0
	nueva = []
	print_color('Se crea la nueva imagen')
	for i in colores[:len(colores)-(len(lista)+100)]:
	r = i[0]
	g = i[1]
	b = i[2]
	nueva.append((r,g,b))
	for i in colores[len(colores)-(len(lista)+100):]:
	if cont < len(lista):
	r = (i[0] & ~1) | lista[cont]
	cont = cont + 1
	else:
	r = i[0]
	if cont < len(lista):
	g = (i[1] & ~1) | lista[cont]
	cont = cont + 1
	else:
	g = i[1]
	if cont < len(lista):
	b = (i[2] & ~1) | lista[cont]
	cont = cont + 1
	else:
	b = i[2]
	nueva.append((r,g,b))
	im2.putdata(nueva)
	nueva = imagen.split('.')[0] + '_mod.png'
	im2.save(nueva)
	print im2
	z = im2.histogram()
	return nueva
	def main():
	texto = "steganography is the art and science of writing hidden messages in such a way that no one apart from the sender and intended recipient suspects the existence of the message a form of security through obscurity the word steganography is of greek origin and means concealed writing from the greek words steganos meaning covered or protected and graphei meaning writing the first recorded use of the term was in by johannes trithemius in his steganographia a treatise on cryptography and steganography disguised as a book on magic generally messages will appear to be something else images articles shopping lists or some other covertext and classically the hidden message may be in invisible between the visible lines of a private letter the advantage of steganography over cryptography alone is that messages do not attract attention to themselves plainly visible encrypted messages no matter how unbreakable will arouse suspicion and may in themselves be incriminating in countries where encryption is illegal therefore whereas cryptography protects the contents of a message steganography can be said to protect both messages and communicating parties "
	imagen = sys.argv[1]
	mensaje, numeros = convertir_binario(texto)
	print_color('Texto que se va a esconder:')
	print texto
	print_color('Se convierte a binario')
	s = []
	for i in mensaje:
	s.append(int(i))
	print_color('Se lee la imagen')
	nueva = encripta(imagen, s)
	print_color('Imagen nueva encriptada con nombre ' + nueva)
	if __name__ == "__main__":
	main()

view raw gistfile1.py hosted with ❤ by GitHub

Execution:

Recover the message

This is a little more complicated. I created an algorithm for look all LSB and then I search for differents sizes than can be the block of word. Once I find this, I start to check all the image and I get many data that can be the phrase that I want, but this doesn't make sense. Now I used a library that can help me to know if a word exist or not and automatically I get a phrase coherent.


Code:

	#!/usr/bin/python
	import Image
	import sys
	from termcolor import colored, cprint
	import numpy as np
	import matplotlib.pyplot as plt
	import enchant
	print_color = lambda x: cprint(x,'cyan')
	def abc():
	'''regresa una lista con todos los
	valores posibles del abecedario en ascci
	y el espacio'''
	abcd = []
	m = 97
	for i in range(25):
	abcd.append(m)
	m = m + 1
	abcd.append(32)
	return abcd
	def recuperar(nueva):
	d = enchant.Dict("en_US")
	print_color('Desencriptando...')
	im = Image.open(nueva)
	print im
	l = im.histogram()
	colores = list(im.getdata())
	lis = []
	letras = []
	for i in colores:
	a = str(bin(i[0])).split('b')[1]
	b = str(bin(i[1])).split('b')[1]
	c = str(bin(i[2])).split('b')[1]
	a = a[-1]
	b = b[-1]
	c = c[-1]
	lis.append(a)
	if len(lis)%10 == 0:
	letras.append(lis[len(lis)-10:len(lis)])
	lis.append(b)
	if len(lis)%10 == 0:
	letras.append(lis[len(lis)-10:len(lis)])
	lis.append(c)
	if len(lis)%10 == 0:
	letras.append(lis[len(lis)-10:len(lis)])
	otra = []
	abcd = abc()
	for i in letras:
	cadena = ''
	for j in i:
	cadena = cadena + str(j)
	s = int(cadena, 2)
	try:
	abcd.index(s)
	otra.append(chr(s))
	except:
	pass
	cadenota = ''
	for i in otra:
	cadenota = cadenota + i
	br = cadenota.split(' ')
	descubierto = ''
	for i in br:
	if len(i)>1:
	if d.check(i):
	descubierto = descubierto + i + ' '
	else:
	pass
	print_color("El mensaje es: ")
	print descubierto
	def main():
	imagen = sys.argv[1]
	recuperar(imagen)
	if __name__ == "__main__":
	main()

view raw gistfile1.py hosted with ❤ by GitHub

Execution:

Library pyenchant python