I'm currently reading up on lookup tables and efficiency. In my uni script it says the following:
For Brute Force:
Preparation time: $O(1)$
Disk space requirement: $O(1)$
Time required to crack the password: $O(2^n)$
Full lookup table:
Preparation time: $O(2^n)$
Disk space requirement: $O(2^n)$
Time required to crack the password: $O(1)$
I'm not sure I'm understanding this correctly.
As far as I understand, for Brute Force:
$O(1)$ is the time required to look up the password in my table of all possible passwords. The disk space requirement is simply as large as needed for a list of all possible passwords, so $O(1)$ as well
Why is the time needed to crack the password $O(2^n)$? How is it determined?
It seems I don't understand the concept of a lookup table either. As far as I see, a lookup table will simply hash the full list of possible passwords (so required disk space is larger) but what then? Why is the cracking of the password per se faster?
I think I'm missing something crucial here. Any help will be appreciated.