[ XOR Python Text Encryption/Decryption ]
I know there is a built in xor operator that can be imported in Python. I'm trying to execute the xor encryption/decryption. So far I have:
def xor_attmpt():
message = raw_input("Enter message to be ciphered: ")
cipher = []
for i in message:
cipher.append(bin(ord(i))[2::])#add the conversion of the letters/characters
#in your message from ascii to binary withoout the 0b in the front to your ciphered message list
cipher = "".join(cipher)
privvyKey = raw_input("Enter the private key: ")
keydecrypt = []
for j in privvyKey:
keydecrypt.append(bin(ord(j))[2::]) #same
keydecrypt = "".join(keydecrypt )#same
print "key is '{0}'" .format(keydecrypt) #substitute values in string
print "encrypted text is '{0}'" .format(cipher)
from operator import xor
for letter in message:
print xor(bool(cipher), bool(keydecrypt))
This:
> for letter in message:
print xor(bool(cipher), bool(keydecrypt))
is where my python starts to go wrong.
The ouput looks like this
Enter message to be ciphered: hello
Enter the private key: \@154>
key is '10111001000000110001110101110100111110'
encrypted text is '11010001100101110110011011001101111'
False
False
False
False
False
What I'm messing up on is trying to get these two binary(key and encrypted) to be compared and give true(1) or false(being 0). The xor should then give me a resulting 1 and 0 binary list from comparing the two. Any input?
Answer 1
Here's a variation of the code example from XOR Cipher Wikipedia article:
def xor(data, key):
return bytearray(a^b for a, b in zip(*map(bytearray, [data, key])))
Example (Python 2):
>>> one_time_pad = 'shared secret'
>>> plaintext = 'unencrypted'
>>> ciphertext = xor(plaintext, one_time_pad)
>>> ciphertext
bytearray(b'\x06\x06\x04\x1c\x06\x16Y\x03\x11\x06\x16')
>>> decrypted = xor(ciphertext, one_time_pad)
>>> decrypted
bytearray(b'unencrypted')
>>> plaintext == str(decrypted)
True
Answer 2
Code below works both ways and does not need length checking as cycle is used.
from itertools import cycle, izip
cryptedMessage = ''.join(chr(ord(c)^ord(k)) for c,k in izip(message, cycle(key)))
Answer 3
somecode = 'asdfln3j34tnonfdkjnflksdfnla'
message = 'this is my message'
def str_xor(s1, s2):
return "".join([chr(ord(c1) ^ ord(c2)) for (c1,c2) in zip(s1,s2)])
encoded = str_xor(message, somecode)
# encoded == '\x15\x1b\r\x15L\x07@J^MT\x03\n\x1d\x15\x05\x0c\x0f'
decoded = str_xor(encoded, somecode)
# decoded == 'this is my message'
This is a naive / minimalistic implementation without error checking. Here len(somecode) >= len(message) is required.