[pycrypto] Bug in Crypto.PublicKey.RSA?
Dwayne C. Litzenberger
dlitz at dlitz.net
Mon Sep 8 08:42:01 CST 2008
On Sun, Sep 07, 2008 at 10:48:49PM -0700, Bill Broadley wrote:
>Shouldn't RSA.generate(keysize, rpool.get_bytes) generate a key that is
>keysize bits and that can encrypt that many bits?
No. RSA with an N-bit modulus can only guarantee encryption of up to N-1
bits.
When you generate an RSA key, you get three values:
- n: the modulus (when you talk about a 368-bit key, you are actually
talking about a 368-bit modulus)
- e: the public exponent (PyCrypto defaults to 65537)
- d: the private exponent
The public key is the pair (n, e), and the private key is the pair (n, d).
To encrypt, you perform the following operation:
C = M**e (mod n)
= M**e % n # Python notation
= pow(M, e, n) # Fast Python notation
To decrypt, you perform the following operation:
M = C**d (mod n)
= C**d % n # Python notation
= pow(C, d, n) # Fast Python notation
Notice how in the decryption, we do a "% n" operation. This means that
whatever the value we decrypt, we will only ever get a value between 0 and
n-1. Thus, PyCrypto only lets you encrypt a value between 0 and n-1.
This program should demonstrate:
# ==== BEGIN ====
>>> from Crypto.PublicKey import RSA
>>> from binascii import a2b_hex, b2a_hex
>>>
>>> # Generate a random RSA key (see my note below about RandomPool)
>>> rsaobj = RSA.generate(368, open("/dev/urandom", "rb").read)
>>>
>>> # Generate two messages
>>> M0 = a2b_hex("%02X" % rsaobj.key.n) # M0 = n
>>> M1 = a2b_hex("%02X" % (rsaobj.key.n-1)) # M1 = n-1
>>>
>>> # Try to encrypt n
>>> rsaobj.encrypt(M0, 0)
Traceback (most recent call last):
File "<stdin>", line 2, in ?
File "/usr/lib/python2.4/site-packages/Crypto/PublicKey/pubkey.py", line 50, in encrypt
ciphertext=self._encrypt(plaintext, K)
File "/usr/lib/python2.4/site-packages/Crypto/PublicKey/RSA.py", line 181, in _encrypt
return (self.key._encrypt(plain),)
_fastmath.error: Plaintext too large
>>>
>>> # Try to encrypt n-1
>>> rsaobj.encrypt(M1, 0)
('\x82\x9cp4\x0c@\xcd\t\x1f\x10\xed\x06Z*\x00^\xcf\xf1\xb2\xc0h\'%M+\x92\x91\xf8\xc3TD\xf5\xd4\xa3\r\xf7\x11\xa2\xf9\xec\x01"\x05\xd3\x89@',)
>>> # Success
>>>
# ==== END ====
>Then used this piece of code:
>
>from Crypto.PublicKey import RSA
>from Crypto.Util.randpool import RandomPool
>
>rpool = RandomPool()
>keysize=368
>privkeyA = RSA.generate(keysize, rpool.get_bytes)
As an aside, don't use RandomPool to generate random numbers. It doesn't
do what you think it does. See this thread:
http://lists.dlitz.net/pipermail/pycrypto/2008q3/000000.html
The next release of PyCrypto will provide an API that does what you want.
Until then, I would do something like this (Python 2.5 only):
import os
privkeyA = RSA.generate(keysize, os.urandom)
Or this:
privkeyA = RSA.generate(keysize, open("/dev/urandom", "rb").read)
Cheers,
- Dwayne
--
Dwayne C. Litzenberger <dlitz at dlitz.net>
Key-signing key - 19E1 1FE8 B3CF F273 ED17 4A24 928C EC13 39C2 5CF7
Annual key (2008) - 4B2A FD82 FC7D 9E38 38D9 179F 1C11 B877 E780 4B45
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