I was doing the cryptopals challenge 29, and I found something weird. When I try to create a forged hash for a forged message I get always the same hash for different forged messages.

    def attack(self, hash, message, payload, oracle):
        for k in range(100):
            forged = self.glue_padding(b'a' * k + message)[k:] + payload
            ml = (k + len(forged)) * 8
            r = self.get_register(hash)
            sha1 = SHA1(payload, r, ml)
            forged_hash = sha1.digest()
            # SAME VALUE
            if oracle.validate(forged, forged_hash):
                print("[*] forged message : " + str(forged))
                return True
        return False

is this the normal behaviour?

  • $\begingroup$ There are several things you do to the input parameters before feeding it to SHA-1 algorithm in your code, and even the function itself seems to accept multiple arguments (where SHA-1 only has one: the message). I'd try and print out the direct input to SHA-1 (in case there is a collision anyway), and you'll likely find that it is identical as well. $\endgroup$ – Maarten Bodewes Nov 26 '19 at 20:12

Is this the normal behaviour?

No, it is not. While it is known that SHA-1 collisions can be found, it's a lot more work than what you do.

It's likely to be a bug in your code. I went through your code, and I'm not sure whether you might be computing the hash of the same preimage each iteration - that would certainly cause each hash to be the same.

| improve this answer | |
  • $\begingroup$ I found out that in fact I pass to the SHA1() always the same values, so it's always the same the digest() $\endgroup$ – meowmeowxw Nov 26 '19 at 17:18

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