I've implemented CTR mode by myself (only decryption for now), using only AES built-in functions from pycrypto. It means that I'm not supposed to use mode=AES.MODE_CTR. However, I know that using AES.MODE_CTR would be more simple, but I'm doing this as a learning experience.
Is it a safe AES CTR??
(non-parallel version)
from Crypto.Cipher import AES
from Crypto.Cipher import AES
ciphers = ["69dda8455c7dd4254bf353b773304eec0ec7702330098ce7f7520d1cbbb20fc3" + \
"88d1b0adb5054dbd7370849dbf0b88d393f252e764f1f5f7ad97ef79d59ce29f5f51eeca32eabedd9afa9329", \
"770b80259ec33beb2561358a9f2dc617e46218c0a53cbeca695ae45faa8952aa" + \
"0e311bde9d4e01726d3184c34451"]
key = "36f18357be4dbd77f050515c73fcf9f2"
class IVCounter(object):
def __init__(self, value):
self.value = value
def increment(self):
# Add the counter value to IV
newIV = hex(int(self.value.encode('hex'), 16) + 1)
# Cut the negligible part of the string
self.value = newIV[2:len(newIV) - 1].decode('hex') # for not L strings remove $ - 1 $
return self.value
def __repr__(self):
self.increment()
return self.value
def string(self):
return self.value
class CTR():
def __init__(self, k):
self.key = k.decode('hex')
def __strxor(self, a, b): # xor two strings of different lengths
if len(a) > len(b):
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a[:len(b)], b)])
else:
return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b[:len(a)])])
def __split_len(self, seq, lenght):
return [seq[i:i+lenght] for i in range(0, len(seq), lenght)]
def __AESencryptor(self, cipher):
encryptor = AES.new(self.key, AES.MODE_ECB)
return encryptor.encrypt(cipher)
def decrypt(self, cipher):
# Split the CT into blocks of 16 bytes
blocks = self.__split_len(cipher.decode('hex'), 16)
# Takes the initiator vector
self.IV = IVCounter(blocks[0])
blocks.remove(blocks[0])
# Message block
msg = []
# Decrypt
for b in blocks:
aes = self.__AESencryptor(self.IV.string())
msg.append(self.__strxor(b, aes))
self.IV.increment()
return ''.join(msg)
def main():
decryptor = CTR(key)
for c in ciphers:
print 'msg = ' + decryptor.decrypt(c)
if __name__ == '__main__':
main()
This code was supposed to do the same that the code below:
import Crypto.Util.Counter
ctr_e = Crypto.Util.Counter.new(128, initial_value=long(IV.encode('hex'), 16))
decryptor = AES.new(key.decode('hex'), AES.MODE_CTR, counter=ctr_e)
print decryptor.decrypt(''.join(blocks))
2 Answers 2
I couldn't spot any problems with your implementations, but caveat lector: I'm not a cryptography expert, so take that with an appropriate pinch of salt.
(削除) I'm also concerned about the output:
$ python reinvent.py
msg = CTR mode lets yo▒B▒▒▒c▒?▒N▒▒w▒▒ap▒9▒uz▒▒▒3▒▒▒o▒V_M▒▒A
msg = Always avoid the▒?A▒▒r ▒ά
It looks as if the first block has been decrypted correctly, and then it all falls over. Probably a bug, but I didn't dig into what's causing it. (削除ここまで)
Edit: Turns out this was an indentation problem – the code as posted had lost some indentation, and I guessed wrong when I tried to add it back.
The biggest problem with this code is the lack of docstrings and comments. Especially when you're dealing with cryptography – it's important to explain why you wrote the code you did, and how it relates to the original problem. This makes code easier to read, review and maintain – and would probably make it much easier to spot the bug above.
A variety of comments below.
IVCounter class
There should be a docstring explaining what this class represents, and what sort of values it expects as input. I had lots of
(削除) fun (削除ここまで)hassle trying different inputs and hitting a variety of errors before I got it working.Your __repr__ method is mutating the internal state of the object – that seems like a really bad idea. The usual use case for repr() is as a debugging tool; changing the instance you're trying to debug will be a bundle of laughs.
Here's an alternative repr you could use:
def __repr__(self): return '%s(%r)' % (self.__class__.__name__, self.value)
This returns a string that could be eval'd to get something equivalent to the current object, which is a much more conventional use of this function.
I'm not sure why you have a string method. It would be better for callers to access self.value directly, and if you want a string representation of an object, you should define __str__ instead.
What happens when the counter overflows? You get a nasty TypeError:
>>> x = IVCounter('\xff\xff\xff\xff\xff\xff\xff\xff\xff\xfe') >>> x.increment() '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff' >>> x.increment() Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 8, in increment File "/usr/lib/python2.7/encodings/hex_codec.py", line 42, in hex_decode output = binascii.a2b_hex(input) TypeError: Odd-length string
You should think about how you want to handle an overflow like this – at the very least, you should wrap the exception and provide a more intelligible error message.
CTR class.
- Should subclass from object.
- Avoid single letter variable names where possible; try to prefer descriptive and expressive names. For example,
__init__(self, key)
instead of__init__(self, key)
. Orfor block in blocks
. - In your
__strxor
method, you have repeated code – the only thing that changes is the arguments tozip()
. Would be good to cut down that repetition. - The variable name
lenght
is misspelt in the definition of the__split_len
method. - When I hear the word
cipher
, I think of things like AES, whereas you seem to be using it to describe a message text. That's a little confusing – docstrings would help. In the
decrypt()
method, rather than:self.IV = IVCounter(blocks[0]) blocks.remove(blocks[0])
you could just use:
self.IV = IVCounter(blocks.pop(0))
which will remove the item and extract it.
Misc
- You appear to be importing Crypto.Cipher.AES twice. Why?
The Pycrpto library has both encrpypt and decrypt functions for the class AES
, but you have used the encrypt function for decryption:
For b in blocks:
aes = self.__AESencryptor(self.IV.string())"
It can be corrected by first defining an AES decryptor function as:
def __AESdecryptor(self, cipher):
dec = AES.new(self.key, AES.MODE_ECB)
return dec.decrypt(cipher)
and calling it inside your CTR decryptor.
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