Pardon my ignorance on the matter but,
Verifying passwords = Polynomial (linear)
Guessing passwords = Exponential
Since each guess has nothing to do with one another, exponential time is best possible time (but verifiable in linear time).
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Sign up to join this communityPardon my ignorance on the matter but,
Verifying passwords = Polynomial (linear)
Guessing passwords = Exponential
Since each guess has nothing to do with one another, exponential time is best possible time (but verifiable in linear time).
The main problem with your argument is that guessing a completely unknown password doesn't fit in the framework of P vs NP. P and NP are classes of decision problems. This means that you are given an input and you have to answer "yes" or "no". For example, in the Hamiltonian graph problem you are given a graph and you answer either "this graph is Hamiltonian" or "this graph is not Hamiltonian". There is also no hidden information. If you answer "no", I can't suddenly say "aha, there is an additional edge in the graph that I didn't tell you about, which makes it Hamiltonian, so you're wrong". Guessing a password doesn't have a yes-no answer and it involves hidden information, so it's not a decision problem.
Maybe you are thinking of some interpretation of password guessing that is a decision problem. If so, you should clarify what you mean.
You just need a mathematical proof that cracking a password cannot possibly done in polynomial time. The only evidence is “all these clever people tried, and they did not succeed”, which is the exact same situation as for P≠NP.
Problems in complexity theory have to scale with respect to some parameter. A 9x9 Sudoku for example can be solved by brute forcing all solutions - this in itself is not an exponential algorithm until we measure its growth with respect to $n$.
Back to your problem. Suppose we phrase the problem of guessing passwords as a decision problem. What is your input to describe a particular instance of the problem, if not the password itself?
Now suppose again we have the password is fixed with the problem statement. For example, one could have the password be digits of $\pi$, and the input is the length $n$. Well, there always "exists" a non deterministic Turing machine that can simply output that string without calculuation as well.
This just sheds some light on NP hardness. This link provides you with a definition of P, NP and NP-hardness.