PyCrypto problem using AES+CTR

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The counter must return the same on decryption as it did on encryption, as you intuit, so, one (NOT SECURE AT ALL) way to do it is:

>>> secret = os.urandom(16)
>>> crypto = AES.new(os.urandom(32), AES.MODE_CTR, counter=lambda: secret)
>>> encrypted = crypto.encrypt("aaaaaaaaaaaaaaaa")
>>> print crypto.decrypt(encrypted)
aaaaaaaaaaaaaaaa

CTR is a block cipher, so the “16-at-a-time” constraint that seems to surprise you is a pretty natural one.

Of course, a so-called “counter” returning the same value at each call is grossly insecure. Doesn’t take much to do better, e.g….:

import array

class Secret(object):
  def __init__(self, secret=None):
    if secret is None: secret = os.urandom(16)
    self.secret = secret
    self.reset()
  def counter(self):
    for i, c in enumerate(self.current):
      self.current[i] = c + 1
      if self.current: break
    return self.current.tostring()
  def reset(self):
    self.current = array.array('B', self.secret)

secret = Secret()
crypto = AES.new(os.urandom(32), AES.MODE_CTR, counter=secret.counter)
encrypted = crypto.encrypt(16*'a' + 16*'b' + 16*'c')
secret.reset()
print crypto.decrypt(encrypted)

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