Python Encrypting with PyCrypto AES

Question

I just found pycrypto today, and I've been working on my AES encryption class. Unfortunately it only half-works. self.h.md5 outputs md5 hash in hex format, and is 32byte. This is the output. It seems to decrypt the message, but it puts random characters after decryption, in this case \n\n\n... I think I have a problem with block size of self.data, anyone know how to fix this?

Jans-MacBook-Pro:test2 jan$ ../../bin/python3 data.py b'RLfGmn5jf5WTJphnmW0hXG7IaIYcCRpjaTTqwXR6yiJCUytnDib+GQYlFORm+jIctest 1 2 3 4 5 endtest\n\n\n\n\n\n\n\n\n\n'

from Crypto.Cipher import AES
from base64 import b64encode, b64decode
from os import urandom

class Encryption():
    def __init__(self):
        self.h = Hash()

    def values(self, data, key):
        self.data = data
        self.key = key
        self.mode = AES.MODE_CBC
        self.iv = urandom(16)
        if not self.key:
            self.key = Cfg_Encrypt_Key
        self.key = self.h.md5(self.key, True)

    def encrypt(self, data, key):
        self.values(data, key)
        return b64encode(self.iv + AES.new(self.key, self.mode, self.iv).encrypt(self.data))

    def decrypt(self, data, key):
        self.values(data, key)
        self.iv = b64decode(self.data)[:16]
        return AES.new(self.key, self.mode, self.iv).decrypt(b64decode(self.data)[16:])

Answer 1

To be honest, the characters "\n\n\n\n\n\n\n\n\n\n" don't look that random to me. ;-)

You are using AES in CBC mode. That requires length of plaintext and ciphertext to be always a multiple of 16 bytes. With the code you show, you should actually see an exception being raised when data passed to encrypt() does not fulfill such condition. It looks like you added enough new line characters ('\n' to whatever the input is until the plaintext happened to be aligned.

Apart from that, there are two common ways to solve the alignment issue:

1. Switch from CBC (AES.MODE_CBC) to CFB (AES.MODE_CFB). With the default segment_size used by PyCrypto, you will not have any restriction on plaintext and ciphertext lengths.

2. Keep CBC and use a padding scheme like PKCS#7, that is:

   • before encrypting a plaintext of X bytes, append to the back as many bytes you need to to reach the next 16 byte boundary. All padding bytes have the same value: the number of bytes that you are adding:

     length = 16 - (len(data) % 16)
     data += bytes([length])*length
     

     That's Python 3 style. In Python 2, you would have:

     length = 16 - (len(data) % 16)
     data += chr(length)*length
     

   • after decrypting, remove from the back of the plaintext as many bytes as indicated by padding:

     data = data[:-data[-1]]
     

Even though I understand in your case it is just a class exercise, I would like to point out that it is insecure to send data without any form of authentication (e.g. a MAC).

Answer 2

from hashlib import md5
from Crypto.Cipher import AES
from Crypto import Random
import base64

def derive_key_and_iv(password, salt, key_length, iv_length):
    d = d_i = ''
    while len(d) < key_length + iv_length:
        d_i = md5(d_i + password + salt).digest()
        d += d_i
    return d[:key_length], d[key_length:key_length+iv_length]

def encrypt(in_file, out_file, password, key_length=32):
    bs = AES.block_size
    salt = Random.new().read(bs - len('Salted__'))
    key, iv = derive_key_and_iv(password, salt, key_length, bs)
    cipher = AES.new(key, AES.MODE_CBC, iv)
    #print in_file
    in_file = file(in_file, 'rb')
    out_file = file(out_file, 'wb')
    out_file.write('Salted__' + salt)
    finished = False
    while not finished:
        chunk = in_file.read(1024 * bs)
        if len(chunk) == 0 or len(chunk) % bs != 0:
            padding_length = bs - (len(chunk) % bs)
            chunk += padding_length * chr(padding_length)
            finished = True
        out_file.write(cipher.encrypt(chunk))
    in_file.close()
    out_file.close()

def decrypt(in_file, out_file, password, key_length=32):
    bs = AES.block_size

    in_file = file(in_file, 'rb')
    out_file = file(out_file, 'wb')
    salt = in_file.read(bs)[len('Salted__'):]
    key, iv = derive_key_and_iv(password, salt, key_length, bs)
    cipher = AES.new(key, AES.MODE_CBC, iv)
    next_chunk = ''
    finished = False
    while not finished:
        chunk, next_chunk = next_chunk, cipher.decrypt(in_file.read(1024 * bs))
        if len(next_chunk) == 0:
            padding_length = ord(chunk[-1])
            if padding_length < 1 or padding_length > bs:
               raise ValueError("bad decrypt pad (%d)" % padding_length)
            # all the pad-bytes must be the same
            if chunk[-padding_length:] != (padding_length * chr(padding_length)):
               # this is similar to the bad decrypt:evp_enc.c from openssl program
               raise ValueError("bad decrypt")
            chunk = chunk[:-padding_length]
            finished = True
        out_file.write(chunk)    
    in_file.close()
    out_file.close()

def encode(in_file, out_file):
    in_file = file(in_file, 'rb')
    out_file = file(out_file, 'wb')
    data = in_file.read()
    out_file.write(base64.b64encode(data))    
    in_file.close()
    out_file.close()

def decode(in_file, out_file):
    in_file = file(in_file, 'rb')
    out_file = file(out_file, 'wb')
    data = in_file.read()
    out_file.write(base64.b64decode(data))    
    in_file.close()
    out_file.close()

Answer 3

You can use a fix character as long as you remember the length of your initial payload, so you don't "throw" useful end bytes away. Try this:

import base64

from Crypto.Cipher import AES

def encrypt(payload, salt, key):
    return AES.new(key, AES.MODE_CBC, salt).encrypt(r_pad(payload))


def decrypt(payload, salt, key, length):
    return AES.new(key, AES.MODE_CBC, salt).decrypt(payload)[:length]


def r_pad(payload, block_size=16):
    length = block_size - (len(payload) % block_size)
    return payload + chr(length) * length


print(decrypt(encrypt("some cyphertext", "b" * 16, "b" * 16), "b" * 16, "b" * 16, len("some cyphertext")))

Answer 4

AES.new().encrypt() and .decrypt() take as both input and output strings whose length is a multiple of 16. You have to fix it in one way or another. For example you can store the real length at the start, and use that to truncate the decrypted string.

Note also that while it's the only restriction for AES, other modules (notably in Crypto.PublicKey) have additional restrictions that comes from their mathematical implementation and that shouldn't (in my opinion) be visible to the end user, but are. For example Crypto.PublicKey.ElGamal will encrypt any short string, but if it starts with null characters, they are lost upon decryption.