3 ms·
I don't follow. If the algorithm is SHA1(SHA256(X)) all an attacker can modify is X. Yes it's possible to find a SHA1() collision, but finding X where the SHA25
by gregmac 4y ago
I don't follow. If the algorithm is SHA1(SHA256(X)) all an attacker can modify is X. Yes it's possible to find a SHA1() collision, but finding X where the SHA256() will generate a collision -- that is SHA1(SHA256(X)) == SHA1(SHA256(Y)) -- is still required.
The question is does the SHA1 step make this any easier?
Don't you still have to either break SHA256 (predicting the hash it will generate) or do this by brute force?
- mlindner 4y agoI was assuming that it was optionally SHA1(X) or SHA1(SHA256(X)) with the determination of which happening being something attacker controllable in X.
- TedDoesntTalk 4y agoI’m OP. I mean SHA1(SHA256(X)) but I have no idea if that makes a collision more difficult than SHA1(X) or any other implications. It was a way to reduce to hash length without truncation.