If you really want a bruteforce algorithm, don't save any big list in the memory of your computer, unless you want a slow algorithm that crashes with a MemoryError.
You could try to use itertools.product like this :
from string import ascii_lowercase
from itertools import product
charset = ascii_lowercase # abcdefghijklmnopqrstuvwxyz
maxrange = 10
def solve_password(password, maxrange):
for i in range(maxrange+1):
for attempt in product(charset, repeat=i):
if ''.join(attempt) == password:
return ''.join(attempt)
solved = solve_password('solve', maxrange) # This worked for me in 2.51 sec
itertools.product(*iterables) returns the cartesian products of the iterables you entered.
[i for i in product('bar', (42,))] returns e.g. [('b', 42), ('a', 42), ('r', 42)]
The repeat parameter allows you to make exactly what you asked :
[i for i in product('abc', repeat=2)]
Returns
[('a', 'a'),
('a', 'b'),
('a', 'c'),
('b', 'a'),
('b', 'b'),
('b', 'c'),
('c', 'a'),
('c', 'b'),
('c', 'c')]
Note:
You wanted a brute-force algorithm so I gave it to you. Now, it is a very long method when the password starts to get bigger because it grows exponentially (it took 62 sec to find the word 'solved').