The last problem seemed pretty straightforward, but now consider the following scenario

given and , find corresponding such that a collision between the functions occurs.

The work for this problem is very similar to the last, and we find the second key with a series of XORs. The code should make this clear if you need a hint ;)

``````from Crypto.Cipher import AES
from Crypto.Util.strxor import strxor as xor
from Crypto import Random
from binascii import hexlify as hexify

# find a, b, x, and y such that
# AES(x, x) XOR y == AES(a, a) XOR b
# Note that for clarity, these names are not used
# each variable is referred to as a m (message) or a k(key)
# and AES takes arguments (key, message)
def collision2():
random = Random.new()

key1 = AES.new(m1, AES.MODE_ECB)
key2 = AES.new(m2, AES.MODE_ECB)

f = xor(key1.encrypt(m1), k1)

k2 = xor(xor(key1.encrypt(m1), key2.encrypt(m2)), k1)

g = xor(key2.encrypt(m2), k2)

print("f: %s\ng: %s" %(hexify(f), hexify(g)))

if __name__ == '__main__':
collision2()
``````

Source of this and previous question: Stanford Crypto Class HW from winter 2010 (problem #2).