4 ms·
Yes this is possible. It's called a preimage attack: https://en.wikipedia.org/wiki/Preimage_attack https://en.wikipedia.org/wiki/Preimage_attack It's computat
by bascule 10y ago
Yes this is possible. It's called a preimage attack:
https://en.wikipedia.org/wiki/Preimage_attack https://en.wikipedia.org/wiki/Preimage_attack
It's computationally infeasible against most cryptographically secure hash functions. However, Grover's algorithm results in a sqrt(keyspace) reduction in security levels (effectively halving bit-sized security levels):
https://en.wikipedia.org/wiki/Grover's_algorithm https://en.wikipedia.org/wiki/Grover's_algorithm
- AgentME 10y agoA successful preimage attack against a cryptographic hash would most likely find you a different message that gets you the same hash. You wouldn't do a preimage attack to find what the original input to create a hash was. You would do a preimage attack to find a new input with the same hash so that you could pass this new input around claiming it was the original (and have any signatures for the old input be valid for your new input, etc).
- oleganza 10y agoWhat you described is called "second preimage attack", that's != "first preimage attack".
- AgentME 10y agoIn a second preimage attack, you start with an input that comes out to a hash, and you're tasked with making a distinct input that comes out to a hash. In a first preimage attack, you start with only the hash, and have to come up with an input from scratch that comes out to that hash. Even in a first preimage attack, you're extremely unlikely to craft the same input that someone else used to create the hash you were given. By pigeonhole principle, the number of inputs that come out to a shorter hash is extreme.