The goal of this exercise is to implement the meet-in-the-middle attack on double DES. For this, you are given a starter code (see below), implemented using the Pycrypto library: https://pypi.python.org/pypi/pycrypto

Perform the following tasks:

• a) Install the pycrypto library
• b) Starting from the starter code (see below), write methods to encrypt and decrypt using double-DES (2DES), defined as follows:

<quote> 2DES( (k1,k2), m) = DES(k1, DES(k2, m))</quote>

• c) You are given the ciphertexts

c1 = 'cda98e4b247612e5b088a803b4277710f106beccf3d020ffcc577ddd889e2f32'

c2 = '54826ea0937a2c34d47f4595f3844445520c0995331e5d492f55abcf9d8dfadf'

as hex strings (i.e. you need to decode them to get the actual byte strings to use with DES, e.g. c1.decode('hex'))

Decrypt them using the following keys:

k1 = 'Smerenie'

k2 = 'Dragoste'

The plaintext corresponding to c1 is m1='Fericiti cei saraci cu duhul, ca'. Find the plaintext m2 corresponding to c2.

• d) Decrypt the entire ciphertext (c1 || c2) with k1 and k2 using 2DES and check it matches the messages m1||m2 above.
• e) You are given the following ciphertext/plaintext pairs:

m1 = 'Pocainta' (in byte string, i.e. can be used directly with Pycrypto DES)

c1 = '9f98dbd6fe5f785d' (in hex string, you need to hex-decode)

m2 = 'Iertarea'

c2 = '6e266642ef3069c2'

Due to a problem with the PRG, you also know the last 6-bytes of each key (note we now have a different key than for the previous exercises): k1 (last 6 bytes) = 'oIkvH5' k2 (last 6 bytes) = 'GK4EoU'

To build a table, I recommend using a list of tuples, where you add new (key,enc) pairs as follows:

tb = []
tb.append(('keyval', 'encva'))

To sort the table, you can do this:

tbs = sorted(tb, key=itemgetter(1))

To search with binary search, first select just the second column (to search the encryptions):

tenc = [value for _,value in tbs]

then use the bisect library (e.g. bisect.bisect_left) https://docs.python.org/2/library/bisect.html

The starter code is this:

import sys
import random
import string
from operator import itemgetter
import time
import bisect
from Crypto.Cipher import DES

def strxor(a, b): # xor two strings (trims the longer input)
  return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])

def hexxor(a, b): # xor two hex strings (trims the longer input)
  ha = a.decode('hex')
  hb = b.decode('hex')
  return "".join([chr(ord(x) ^ ord(y)).encode('hex') for (x, y) in zip(ha, hb)])

def bitxor(a, b): # xor two bit strings (trims the longer input)
  return "".join([str(int(x)^int(y)) for (x, y) in zip(a, b)])

def str2bin(ss):
  """
    Transform a string (e.g. 'Hello') into a string of bits
  """
  bs = ''
  for c in ss:
    bs = bs + bin(ord(c))[2:].zfill(8)
  return bs

def str2int(ss):
  """
    Transform a string (e.g. 'Hello') into a (long) integer by converting
    first to a bistream
  """
  bs = str2bin(ss)
  li = int(bs, 2)
  return li

def hex2bin(hs):
  """
    Transform a hex string (e.g. 'a2') into a string of bits (e.g.10100010)
  """
  bs = ''
  for c in hs:
    bs = bs + bin(int(c,16))[2:].zfill(4)
  return bs

def bin2hex(bs):
  """
    Transform a bit string into a hex string
  """
  bv = int(bs,2)
  return int2hexstring(bv)

def byte2bin(bval):
  """
    Transform a byte (8-bit) value into a bitstring
  """
  return bin(bval)[2:].zfill(8)

def int2hexstring(bval):
  """
    Transform an int value into a hexstring (even number of characters)
  """
  hs = hex(bval)[2:]
  lh = len(hs)
  return hs.zfill(lh + lh%2)

def get_index(a, x):
  'Locate the leftmost value exactly equal to x in list a'
  i = bisect.bisect_left(a, x)
  if i != len(a) and a[i] == x:
    return i
  else:
    return -1

def des_enc(k, m):
  """
  Encrypt a message m with a key k using DES as follows:
  c = DES(k, m)

  Args:
    m should be a bytestring (i.e. a sequence of characters such as 'Hello' or '\x02\x04')
    k should be a bytestring of length exactly 8 bytes.

  Note that for DES the key is given as 8 bytes, where the last bit of
  each byte is just a parity bit, giving the actual key of 56 bits, as expected for DES.
  The parity bits are ignored.

  Return:
    The bytestring ciphertext c
  """
  d = DES.new(k)
  c = d.encrypt(m)

  return c

def des_dec(k, c):
  """
  Decrypt a message c with a key k using DES as follows:
  m = DES(k, c)

  Args:
    c should be a bytestring (i.e. a sequence of characters such as 'Hello' or '\x02\x04')
    k should be a bytestring of length exactly 8 bytes.

  Note that for DES the key is given as 8 bytes, where the last bit of
  each byte is just a parity bit, giving the actual key of 56 bits, as expected for DES.
  The parity bits are ignored.

  Return:
    The bytestring plaintext m
  """
  d = DES.new(k)
  m = d.decrypt(c)

  return m
  
def main():

  # Exercitiu pentru test des2_enc
  key1 = 'Smerenie'
  key2 = 'Dragoste'
  m1_given = 'Fericiti cei saraci cu duhul, ca'
  c1 = 'cda98e4b247612e5b088a803b4277710f106beccf3d020ffcc577ddd889e2f32'
  # TODO: implement des2_enc and des2_dec
  m1 = des2_dec(key1, key2, c1.decode('hex'))

  print 'ciphertext: ' + c1
  print 'plaintext: ' + m1
  print 'plaintext in hexa: ' + m1.encode('hex')

  # TODO: run meet-in-the-middle attack for the following plaintext/ciphertext
  m1 = 'Pocainta'
  c1 = '9f98dbd6fe5f785d' # in hex string
  m2 = 'Iertarea'
  c2 = '6e266642ef3069c2'
  
  # Note: you only need to search for the first 2 bytes of the each key:
  k1 = '??oIkvH5'
  k2 = '??GK4EoU'



if __name__ == "__main__":
  main()

In this exercise you'll try to implement the CBC-padding attack of Serge Vaudenay: http://link.springer.com/content/pdf/10.1007%2F3-540-46035-7_35.pdf

See sections 3.1 to 3.3 of chapter 3 in the paper above for more details.

Another useful link: https://robertheaton.com/2013/07/29/padding-oracle-attack/

Given a message of length L bytes, we need to pad with b more bytes such that L+b is a multiple of the cipher's block size (e.g. 16 bytes in the case of AES). Each of the last b bytes will have the value b. For example, for a message that has 30 bytes, we shall add two bytes with the value 0x02 at the end, obtaining m = [b1 | b2 | .. | b30 | 0x02 | 0x02], where bi are the 30 original bytes.

The idea in short is the following: given a 1-block ciphertext c = [c1 | c2 | … | c16], you can find its last byte by crafting a 2-block ciphetext c' = [r | c], where r = [r1 | r2 | … | r16] contains random bytes. Then, by using an oracle that tells you whether the padding of the decryption of c' is correct or not, you can determine the value of c16. For this you just try all possible values of r16, until for one of them you will find that the oracle returns a successful result (it will do that for one value). In that case, due to the workings of CBC, you found out that r16 ⊕ LSB(Dec(k, c)) = 1. Hence you find that the last byte of the decryption of c is r16 ⊕ 1.

Once you found the last byte c16 of the decryption of c, you can work your way towards the following byte, c15. This time you need to target the two-byte padding [0x02 | 0x02]. For this you simply use for r16 the value LSB(Dec(k, c)) ⊕ 0x02, and try all values for r15 just as we did above for r16. In the same manner you can decrypt all the ciphertext blocks c1 … c16.

You can then repeat the above process for any ciphertext of arbitrary length. Simply apply the attack on all blocks.

You are given the following ciphertext: 553b43d4b821332868fece8149eea14a2b0a98c7bed43cc1cf75f4e778cb315dc1d92 8d0340e0aab4900ca8af9adaee761e2affa3e9996d81483e950b913492b

and a padding oracle (see the method “check_cbcpad” below), which knows the key that was used to encrypt the ciphertext and which will answer 1 if the padding of the ciphertext is correct and 0 otherwise.

Your task is to decrypt the message by using the CBC-padding attack.

Here is a starting python script, which contains your padding oracle:

cbc_padding.py

import sys
import random
import string
import time
from Crypto.Cipher import AES
 
IV = 'Hristos a inviat'
 
def strxor(a, b): # xor two strings (trims the longer input)
  return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b)])
 
def hexxor(a, b): # xor two hex strings (trims the longer input)
  ha = a.decode('hex')
  hb = b.decode('hex')
  return "".join([chr(ord(x) ^ ord(y)).encode('hex') for (x, y) in zip(ha, hb)])
 
def bitxor(a, b): # xor two bit strings (trims the longer input)
  return "".join([str(int(x)^int(y)) for (x, y) in zip(a, b)])
 
def str2bin(ss):
  """
    Transform a string (e.g. 'Hello') into a string of bits
  """
  bs = ''
  for c in ss:
    bs = bs + bin(ord(c))[2:].zfill(8)
  return bs
 
def str2int(ss):
  """
    Transform a string (e.g. 'Hello') into a (long) integer by converting
    first to a bistream
  """
  bs = str2bin(ss)
  li = int(bs, 2)
  return li
 
def hex2bin(hs):
  """
    Transform a hex string (e.g. 'a2') into a string of bits (e.g.10100010)
  """
  bs = ''
  for c in hs:
    bs = bs + bin(int(c,16))[2:].zfill(4)
  return bs
 
def bin2hex(bs):
  """
    Transform a bit string into a hex string
  """
  bv = int(bs,2)
  return int2hexstring(bv)
 
def byte2bin(bval):
  """
    Transform a byte (8-bit) value into a bitstring
  """
  return bin(bval)[2:].zfill(8)
 
def int2hexstring(bval):
  """
    Transform an int value into a hexstring (even number of characters)
  """
  hs = hex(bval)[2:]
  lh = len(hs)
  return hs.zfill(lh + lh%2)
 
def check_cbcpad(c):
  """
  Oracle for checking if a given ciphertext has correct CBC-padding.
  That is, it checks that the last n bytes all have the value n.
 
  Args:
    c is the ciphertext to be checked. Note: the key is supposed to be
    known just by the oracle.
 
  Return 1 if the pad is correct, 0 otherwise.
  """
  ko = 'Sfantul Gheorghe'
  m = aes_dec_cbc(ko, c, IV)
  lm = len(m)
  lb = ord(m[lm-1])
 
  if lb > lm:
    return 0
 
  for k in range(lb):
    if ord(m[lm-1-k]) != lb:
      return 0
 
  return 1
 
def aes_enc(k, m):
  """
  Encrypt a message m with a key k in ECB mode using AES as follows:
  c = AES(k, m)
 
  Args:
    m should be a bytestring multiple of 16 bytes (i.e. a sequence of characters such as 'Hello...' or '\x02\x04...')
    k should be a bytestring of length exactly 16 bytes.
 
  Return:
    The bytestring ciphertext c
  """
  aes = AES.new(k)
  c = aes.encrypt(m)
 
  return c
 
def aes_dec(k, c):
  """
  Decrypt a ciphertext c with a key k in ECB mode using AES as follows:
  m = AES(k, c)
 
  Args:
    c should be a bytestring multiple of 16 bytes (i.e. a sequence of characters such as 'Hello...' or '\x02\x04...')
    k should be a bytestring of length exactly 16 bytes.
 
  Return:
    The bytestring message m
  """
  aes = AES.new(k)
  m = aes.decrypt(c)
 
  return m
 
def aes_enc_cbc(k, m, iv):
  """
  Encrypt a message m with a key k in CBC mode using AES as follows:
  c = AES(k, m)
 
  Args:
    m should be a bytestring multiple of 16 bytes (i.e. a sequence of characters such as 'Hello...' or '\x02\x04...')
    k should be a bytestring of length exactly 16 bytes.
    iv should be a bytestring of length exactly 16 bytes.
 
  Return:
    The bytestring ciphertext c
  """
  aes = AES.new(k, AES.MODE_CBC, iv)
  c = aes.encrypt(m)
 
  return c
 
def aes_dec_cbc(k, c, iv):
  """
  Decrypt a ciphertext c with a key k in CBC mode using AES as follows:
  m = AES(k, c)
 
  Args:
    c should be a bytestring multiple of 16 bytes (i.e. a sequence of characters such as 'Hello...' or '\x02\x04...')
    k should be a bytestring of length exactly 16 bytes.
    iv should be a bytestring of length exactly 16 bytes.
 
  Return:
    The bytestring message m
  """
  aes = AES.new(k, AES.MODE_CBC, iv)
  m = aes.decrypt(c)
 
  return m
 
def main():
 
  # Find the message corresponding to this ciphertext by using the cbc-padding attack
  c = '553b43d4b821332868fece8149eea14a2b0a98c7bed43cc1cf75f4e778cb315dc1d928d0340e0aab4900ca8af9adaee761e2affa3e9996d81483e950b913492b'
  ct = c.decode('hex')
 
  #Check correct padding
  print 'Oracle check of pad = ' + str(check_cbcpad(ct))
 
  # TODO: implement the CBC-padding attack to find the message corresponding to the above ciphertext
  # Note: you cannot use the key known by the oracle
  # You can use the known IV in order to recover the full message
 
 
if __name__ == "__main__":
  main()