Safe password check / Lost password
"I can just look in the source code" wrong
Firstly, all passwords and their messages are either hashed with SHA256 or encrypted with Fernet.
But to make it somehow possible in time, there are 2 Hints that perhabs even be in a rainbow table.
With them (yes this is also a clue) the original possiblities
95 ^ 20 = 3584859224085422343574104404449462890625
are shorten up so that it should be hopefully possible.
Furthermore, it should be fairly easy to implement it some kind of brute force algorithm.
Nevertheless, feel free to ask me if it would take too much time.
encryption.py, there need to be more than 100,000 iterations for secure derivatization. The official cryptology example has 390,000, but recommends as many as the computer can handle, which Is probably more than 1 million.