XOR Properties

In the last challenge, you saw how XOR worked at the level of bits. In this one, we're going to cover the properties of the XOR operation and then use them to undo a chain of operations that have encrypted a flag. Gaining an intuition for how this works will help greatly when you come to attacking real cryptosystems later, especially in the block ciphers category. There are four main properties we should consider when we solve challenges using the XOR operator Commutative: A ⊕ B = B ⊕ A Associative: A ⊕ (B ⊕ C) = (A ⊕ B) ⊕ C Identity: A ⊕ 0 = A Self-Inverse: A ⊕ A = 0 Let's break this down. Commutative means that the order of the XOR operations is not important. Associative means that a chain of operations can be carried out without order (we do not need to worry about brackets). The identity is 0, so XOR with 0 "does nothing", and lastly something XOR'd with itself returns zero. Let's put this into practice! Below is a series of outputs where three random keys have been XOR'd together and with the flag. Use the above properties to undo the encryption in the final line to obtain the flag. KEY1 = a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313 KEY2 ^ KEY1 = 37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e KEY2 ^ KEY3 = c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1 FLAG ^ KEY1 ^ KEY3 ^ KEY2 = 04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf Before you XOR these objects, be sure to decode from hex to bytes.

Goodbye XOR

The most important property is the self-inverse property since it gives us an easy way to undo the keys

The critical thing to note is (A ⊕ B) ⊕ A = B so we can remove the effect of A XOR on B

What we want is FLAG, so we need to remove KEY1, KEY2, and KEY3 by XORing the respective values

• We have FLAG ^ KEY1 ^ KEY3 ^ KEY2

• We are also given KEY1 and KEY2 ^ KEY3 and these 2 strings give us one instance of each key

• Thus if we XOR all of these strings together, we can isolate the FLAG

-> Due to associativity, the order in which things are XORed does not matter

from pwn import xor

KEY1 = "a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313"
KEY2_KEY1 = "37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e"
KEY2_KEY3 = "c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1"
FLAG_KEY1_KEY3_KEY2 = "04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf"

# Convert to bytes
k1 = bytes.fromhex(KEY1)
k2k3 = bytes.fromhex(KEY2_KEY3)
fk1k3k2 = bytes.fromhex(FLAG_KEY1_KEY3_KEY2)

# XOR everything
flag = xor(k1, xor(k2k3, fk1k3k2))

print(flag)

Flag

crypto{x0r_i5_ass0c1at1v3}