Brute force script

Question

In Python 3.2 I wanted to brute-force a random password like:

PassWord = random.randint(0,9999)

I made something that could try random passwords:

import random
PassWord = str(random.randint(0,9999))
Trial = ' '
while Trial != PassWord:
    Trial = str(random.randint(0,9999))
    print(Trial)
    if Trial == PassWord:
        print('The password is: '+PassWord)
        input()

It's trying to randomly guess a password but a brute-force attack tries all possibilities. How do I first checks all possibilities with 1 digit, in the right order (0, 1, 2, 3 and so on), then 2, 3 and 4 digits?

Answer 1

Code first:

from itertools import product

chars = '0123456789' # chars to look for

for length in range(1, 3): # only do lengths of 1 + 2
    to_attempt = product(chars, repeat=length)
    for attempt in to_attempt:
        print(''.join(attempt))

itertools.product produces a Cartesian join of its input(s) - in this case, it's being 'joined' to itself. So in the first iteration, each single character is printed. Then in the next iteration, because of repeat=length (and length is now == 2), generates '00', '01', etc... It's worth trying it and seeing the output to understand it better.

This also means you can throw in letters (uppercase/lowercase), and change the upperbound in the range function.

It's certainly not going to break the world of code-breaking, but should give you an idea of the flexibility of Python and the tools available to you.

I'll leave you to check the passwords match and break out the loop.