# python – How to create a brute-force password cracker for alphabetical and alphanumerical passwords?

## python – How to create a brute-force password cracker for alphabetical and alphanumerical passwords?

Heres a naiive brute force method that will guess numbers (`string.digits`) and lower case letters (`string.ascii_lowercase`). You can use `itertools.product` with `repeat` set to the current password length guessed. You can start at `1` character passwords (or whatever your lower bound is) then cap it at a maximum length too. Then just `return` when you find the match.

``````import itertools
import string

chars = string.ascii_lowercase + string.digits
attempts = 0
attempts += 1
guess = .join(guess)
if guess == real:
return password is {}. found in {} guesses..format(guess, attempts)
# uncomment to display attempts, though will be slower
#print(guess, attempts)

``````

Output

``````a 1
b 2
c 3
d 4
...
aba 1369
abb 1370
password is abc. found in 1371 guesses.
``````

One possible option which would preserve almost exactly your current code is to convert to base 36 with the following digits: `0-9a-z`. This will give you every possible alpha-numeric combination for n characters if you search in `range(36**n)`.

Using a simplified function from How to convert an integer in any base to a string?:

``````def baseN(num, b=36, numerals=0123456789abcdefghijklmnopqrstuvwxyz):
return ((num == 0) and numerals) or (baseN(num // b, b, numerals).lstrip(numerals) + numerals[num % b])
``````

You can then loop through numbers as in your example:

``````>>> for i in range(10000, 10005):
...     print(baseN(i).zfill(5))
...
007ps
007pt
007pu
007pv
007pw
``````

To get all 3-letter possibilities, you can loop as follows:

``````for i in range(36**3):
possible = baseN(i).zfill(3)
``````