> The mention of "NP hard" in the first place just seems...unnecessary. Furthermore, trying all possible passwords of length n for a fixed alphabet can be done

> The mention of "NP hard" in the first place just seems...unnecessary. Furthermore, trying all possible passwords of length n for a fixed alphabet can be done in polynomial time, hence finding the password deterministically only takes polynomial time. Thus, it's only NP-hard if P=NP. Edit: You're right, I mixed up length and size of the alphabet, so disregard that comment. I probably should go to sleep.