Classic Bleichenbacher RSA Padding Oracle Attack

bleichenbacher.py

#! /usr/bin/env python3

# Copyright (C) 2019 Karim Kanso. All Rights Reserved.
#
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program.  If not, see <http://www.gnu.org/licenses/>.




#
# The program in this file is as close as possible to the classic
# Bleichenbacher RSA padding algorithm described in Daniel
# Bleichenbacher's 1998 paper: "Chosen Ciphertext Attacks Against
# Protocols Based on the RSA Encryption Standard PKCS #1". It is
# suggested to read the paper for information about how it works.
#
# In most cases, the implementation here uses standard python
# definitions and does not attempt to do anything fancy that would
# otherwise obfuscate the implementation or add dependencies.
#
# This script was written to find the flag during a crypto challenge.
#
# To use the program, provide an alternative version of the
# local_setup function that returns a 4-tuple of the needed parameters
# for the algorithm.
#  1. ciphertext encoded as a byte object
#  2. an oracle function that goes from a byte object to boolean value
#      (if oracle throws an exception the program will crash, ensure
#         all needed error handling is provided)
#  3. public exponent
#  4. modulus
#
# The default local_setup function will either generate a new key pair
# or use a pre-generated pair. This is then fed into the oracle to
# decrypt the content. The oracle directly uses a decryption library
# so it will check the full format of the message is PKCS1.5 compliant
# (i.e. the padding aswell) and not just the first two bytes.
#

import cryptography.hazmat.primitives.asymmetric.rsa as rsa
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives.asymmetric import padding
from cryptography.hazmat.primitives import serialization
import binascii
import math
import textwrap

def local_setup(newkey=False):
    'generates a key pair for local testing'
    print('Using local loop back oracle for testing')

    if newkey:
        print('* Generating a new key pair')
        priv_key = rsa.generate_private_key(
            public_exponent=65537,
            key_size=1024,
            backend=default_backend()
        )
    else:
        print('* Using hard coded key pair')
        priv_key = '''\
            -----BEGIN RSA PRIVATE KEY-----
            MIIBOQIBAAJBALSh8S5eL8uEskw1L7QvoVOuLwK7rpKmdu0IQvu4e1gYf/HdsMp6
            P7V7h+5D8siAbBigk2lphU60skVYK6OXCmUCAwEAAQJACJ5dep/l2ekX9Mjo4MkR
            AoQiHBhGaRrmO8MUJxyTTg2wptfKN/4qWPMtssACA6V3db4x4fyKeZi0MMAeEzAt
            QQIhAOP2J3Uggj8y6RxqWdS21z+qpdznwl5tzxK+pOEiW8+1AiEAytmJckmUQmdu
            Q2fwmrMlcJuwkk5ef126/NFrlty9HfECIF1IY0kYrnOyH5YTJwNWdqqE6C6HYBBI
            Gw5umQXPi4ZpAiBJXL9+2+mI0otoSXEVIfFKdqQ3Zax7d9Smlr7IgvDKoQIgO5YK
            bzoHpEb1T/UZ1MZFWNhoMvPG2MQs9B6KGVz0gGM=
            -----END RSA PRIVATE KEY-----'''

        priv_key = serialization.load_pem_private_key(
            textwrap.dedent(priv_key).encode('utf-8'),
            password=None,
            backend=default_backend()
        )

    pub_key = priv_key.public_key()
    pn = pub_key.public_numbers()

    print('  keysize: {}'.format(priv_key.key_size))
    print('  e: {}'.format(pn.e))
    print('  n: {}'.format(pn.n))
    print('  p: {}'.format(priv_key.private_numbers().p))
    print('  q: {}'.format(priv_key.private_numbers().q))
    print('  d: {}'.format(priv_key.private_numbers().d))

    if newkey:
        ciphertext = pub_key.encrypt(
            b'hello world!!',
            padding.PKCS1v15()
        )
    else:
        ciphertext = binascii.unhexlify(
            '1e2fb249ddd03554d0a7c27cc276ded8' +
            'cbf5e4daf1b84c28eccd37118adb7d46' +
            '9c29eab603220057df68b84d9fdd40b8' +
            'b5835c16e09ecbcf8ee7ec634e534f32'
        )

    print('  c: {}'.format(binascii.hexlify(ciphertext)))
    print()

    def oracle(ct):
        c = int.from_bytes(ct, 'big')

        try:
            priv_key.decrypt(
                ct,
                padding.PKCS1v15()
            )
            return True
        except ValueError:
            return False

    return ciphertext, oracle, pn.e, pn.n

# these two defs avoid rounding issues with floating point during
# division (especially with large numbers)
def ceildiv(a, b):
    return -(-a // b)

def floordiv(a, b):
    return (a // b)

oracle_ctr = 0
def main():
    print('Bleichenbacher RSA padding algorithm')
    print('  for more info see 1998 paper.')
    print()

    # setup parameters, change local_setup with alternative
    # implementation, such as an oracle that uses a real server
    ct, oracle, e, n = local_setup(newkey=True)

    # byte length of n
    k = int(ceildiv(math.log(n,2), 8))

    # convert ciphertext from bytes into integer
    c = int.from_bytes(ct, 'big')

    # lift oracle defition to take integers
    def oracle_int(x):
        global oracle_ctr
        oracle_ctr = oracle_ctr + 1
        if oracle_ctr % 100000 == 0:
            print("[{}K tries] ".format(oracle_ctr // 1000), end='', flush=True)
        return oracle(x.to_bytes(k, 'big'))

    # define B as size of ciphertext space
    #   as first two bytes are 00 02, use 2^(keysize - 16)
    B = pow(2, 8 * (k-2))

    # precompute constants
    _2B = 2 * B
    _3B = 3 * B

    multiply = lambda x, y: (x * pow(y, e, n)) % n

    # should be identity as c is valid cipher text
    c0 = multiply(c, 1)
    assert c0 == c
    i = 1
    M = [(_2B, _3B - 1)]
    s = 1

    # ensure everything is working as expected
    if oracle_int(c0):
        print('Oracle ok, implicit step 1 passed')
    else:
        print('Oracle fail sanity check')
        exit(1)

    while True:
        if i == 1:
            print('start case 2.a: ', end='', flush=True)
            ss = ceildiv(n, _3B)
            while not oracle_int(multiply(c0, ss)):
                ss = ss + 1
            print('done. found s1 in {} iterations: {}'.format(
                ss - ceildiv(n, _3B),ss))
        else:
            assert i > 1
            if len(M) > 1:
                print('start case 2.b: ', end='', flush=True)
                ss = s + 1
                while not oracle_int(multiply(c0, ss)):
                    ss = ss + 1
                print('done. found s{} in {} iterations: {}'.format(
                    i, ss-s, ss))
            else:
                print('start case 2.c: ', end='', flush=True)
                assert len(M) == 1
                a, b = M[0]
                r = ceildiv(2 * (b * s - _2B), n)
                ctr = 0
                while True:
                    # note: the floor function below needed +1 added
                    # to it, this is not clear from the paper (see
                    # equation 2 in paper where \lt is used instead of
                    # \lte).
                    for ss in range(
                            ceildiv(_2B + r * n, b),
                            floordiv(_3B + r * n, a) + 1):
                        ctr = ctr + 1
                        if oracle_int(multiply(c0, ss)):
                            break
                    else:
                        r = r + 1
                        continue
                    break
                print('done. found s{} in {} iterations: {}'.format(i, ctr, ss))
        # step 3, narrowing solutions
        MM = []
        for a,b in M:
            for r in range(ceildiv(a * ss - _3B + 1, n),
                           floordiv(b * ss - _2B, n) + 1):
                m = (
                    max(a, ceildiv(_2B + r * n, ss)),
                    min(b, floordiv(_3B - 1 + r * n, ss))
                )
                if m not in MM:
                    MM.append(m)
                    print('found interval [{},{}]'.format(m[0],m[1]))
        # step 4, compute solutions
        M = MM
        s = ss
        i = i + 1
        if len(M) == 1 and M[0][0] == M[0][1]:
            print()
            print('Completed!')
            print('used the oracle {} times'.format(oracle_ctr))
            # note, no need to find multiplicative inverse of s0 in n
            # as s0 = 1, so M[0][0] is directly the message.
            message = M[0][0].to_bytes(k, 'big')
            print('raw decryption: {}'.format(
                binascii.hexlify(message).decode('utf-8')))
            if message[0] != 0 or message[1] != 2:
                return
            message =  message[message.index(b'\x00',1) + 1:]
            print('unpadded message hex: {}'.format(
                binascii.hexlify(message).decode('utf-8')))
            try:
                print('unpadded message ascii: {}'.format(
                    message.decode('utf-8')))
            except UnicodeError:
                pass
            return

if __name__ == "__main__":
    main()