Break up the problem to reduce memory needed

brute-pin-divide.py

import time
import itertools

def brute_force_digits(pin):
    """ Uses itertools to create a cartesian product of the possible pins """
    for guess in list(itertools.product(list(range(10)),repeat=len(str(pin)))):
        if pin == int(len(str(pin))*'%d' % (guess)):
            return guess        

def brute_force_split(pin,n):
    """ Breaking up the problem space to minimize needed memory for brute_force_digits """
    split = len(pin)//n

    l = [ int(pin[i*n:(i+1)*n]) for i in range(split) ]

    return [ brute_force_digits(ppin) for ppin in l ]


if __name__ == '__main__':
    """ Actually works really well on single machine when dividing by 2. 
 
    Somewhat contrived here in order to illustrate parallel (lambda) processing advantages in
    certain situations.
    
    In this case, dividing by 6; breaks down with pin size of 360 (i7 16GB)

    """
    print('Provide a digit-only pin whose length is divisible by 6 to be brute-forced guessed')
    pin = input()
    t1 = time.time()
    guess = brute_force_split(pin,6)
    print('Found: ' + str(guess))
    t2 = time.time()
    print('Elapsed time was ' + str(t2-t1))