# How to create a Bruteforce password cracker for alphabetical and alphanumerical passwords?

Hello Guys, How are you all? Hope You all Are Fine. Today We Are Going To learn about How to create a Bruteforce password cracker for alphabetical and alphanumerical passwords in Python. So Here I am Explain to you all the possible Methods here.

## How to create a Bruteforce password cracker for alphabetical and alphanumerical passwords?

1. How to create a Bruteforce password cracker for alphabetical and alphanumerical passwords?

Here's a naiive brute force method that will guess numbers (`string.digits`) and lower case letters (`string.ascii_lowercase`).

Here's a naiive brute force method that will guess numbers (`string.digits`) and lower case letters (`string.ascii_lowercase`).

## Method 1

Here’s 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.```

## Method 2

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)`.

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

## Conclusion

It’s all About this issue. Hope all Methods helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which Method worked for you? Thank You.