Source code for pcr.xtea

# Copyright (c) 2013 Stefano Palazzo <stefano.palazzo@gmail.com>
#
# 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/>.

''' XTEA block cipher (32 rounds) '''

import struct


def encrypt(block, key):
    s, d, m = 0, 0x9e3779b9, 0xffffffff
    key, (v0, v1) = struct.unpack(">4L", key), struct.unpack(">2L", block)
    for _ in range(32):
        v0 = (v0 + (((v1 << 4 ^ v1 >> 5) + v1) ^ (s + key[s & 3]))) & m
        s = (s + d) & m
        v1 = (v1 + (((v0 << 4 ^ v0 >> 5) + v0) ^ (s + key[s >> 11 & 3]))) & m
    return struct.pack(">2L", v0, v1)


def decrypt(block, key):
    s, d, m = 0xc6ef3720, 0x9e3779b9, 0xffffffff
    key, (v0, v1) = struct.unpack(">4L", key), struct.unpack(">2L", block)
    for _ in range(32):
        v1 = (v1 - (((v0 << 4 ^ v0 >> 5) + v0) ^ (s + key[s >> 11 & 3]))) & m
        s = (s - d) & m
        v0 = (v0 - (((v1 << 4 ^ v1 >> 5) + v1) ^ (s + key[s & 3]))) & m
    return struct.pack(">2L", v0, v1)