I had the correct idea, but unfortunately could not come up with the correct implementation to this problem. We can notice that shuffle is a function that uses the value of each byte in mixed_and as an index to select which byte in mixed_xor will be outputted. Hence, with the same value for every byte in mixed_and, the shuffle output will only be a repeat of some byte len(FLAG) times.

This is a challenge about length extension attack on hashes like MD5 and SHA1 that uses the Merkle-Damgard construction. More on that can be found in this link, and a good Youtube video about this topic, if you can understand Vietnamese, is this series by CyberJutsu. The following is written under the assumption that you have some experience with the attack. In MD5 hash extension attack, it is often the case that we are allowed to extend the secret with some arbitrary data that we decide.

The code is vulnerable to hash length extension attack. The hash for a message with a new block appended can be computed from the hash of the old message, and xor with the result of encrypting the old hash with the key as the new block. Denote $H$ as the old hash, $H’$ as the new hash, the newly appended block as $B$, and encryption as $E$, we have the relation:

There are two possible solutions to this problem. One involves the method of differential cryptanalysis (normally done on the substitution box SBOX like these), and another brute-force, or more specifically birthday attack solution. I attempt this using the birthday attack solution. The idea is that the space of state is too small. Indeed, the shuffling of the first stage of the hash: state = [16, 32, 48, 80, 80, 96, 112, 128] for i in range(0, len(data), 4): block = data[i:i+4] state[4] ^= block[0] state[5] ^= block[1] state[6] ^= block[2] state[7] ^= block[3] state = permute(state) state = substitute(state) Only state[4:] are modified from the block’s content.

In the two keys that we have to “insert”, one has to start with the prefix of “CryptoHack Secure Safe”, and the other must not have this prefix. The two keys have to pass this check: h1 = hashes[0] h2 = hashes[1] for i in range(2, 2**(random.randint(2, 10))): h1 = xor(self.magic1, xor(h2, xor(xor(h2, xor(h1, h2)), h2))) h2 = xor(xor(xor(h1, xor(xor(h2, h1), h1)), h1), self.magic2) ... if h1 == h2: return {"msg": f"The safe clicks and the door opens.