Python Brute Force algorithm

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I need to generate every possible combination from a given charset to a given range. Like,

charset=list(map(str,"abcdefghijklmnopqrstuvwxyz"))
range=10

And the out put should be,

[a,b,c,d..................,zzzzzzzzzy,zzzzzzzzzz]

I know I can do this using already in use libraries.But I need to know how they really works.If anyone can give me a commented code of this kind of algorithm in Python or any programming language readable,I would be very grateful.

Answer 1

If you REALLY want to brute force it, try this, but it will take you a ridiculous amount of time:

your_list = 'abcdefghijklmnopqrstuvwxyz'
complete_list = []
for current in xrange(10):
    a = [i for i in your_list]
    for y in xrange(current):
        a = [x+i for i in your_list for x in a]
    complete_list = complete_list+a

On a smaller example, where list = 'ab' and we only go up to 5, this prints the following:

['a', 'b', 'aa', 'ba', 'ab', 'bb', 'aaa', 'baa', 'aba', 'bba', 'aab', 'bab', 'abb', 'bbb', 'aaaa', 'baaa', 'abaa', 'bbaa', 'aaba', 'baba', 'abba', 'bbba', 'aaab', 'baab', 'abab', 'bbab', 'aabb', 'babb', 'abbb', 'bbbb', 'aaaaa', 'baaaa', 'abaaa', 'bbaaa', 'aabaa', 'babaa', 'abbaa', 'bbbaa', 'aaaba','baaba', 'ababa', 'bbaba', 'aabba', 'babba', 'abbba', 'bbbba', 'aaaab', 'baaab', 'abaab', 'bbaab', 'aabab', 'babab', 'abbab', 'bbbab', 'aaabb', 'baabb', 'ababb', 'bbabb', 'aabbb', 'babbb', 'abbbb', 'bbbbb']

Answer 2

Use itertools.product, combined with itertools.chain to put the various lengths together:

from itertools import chain, product
def bruteforce(charset, maxlength):
    return (''.join(candidate)
        for candidate in chain.from_iterable(product(charset, repeat=i)
        for i in range(1, maxlength + 1)))

Demonstration:

>>> list(bruteforce('abcde', 2))
['a', 'b', 'c', 'd', 'e', 'aa', 'ab', 'ac', 'ad', 'ae', 'ba', 'bb', 'bc', 'bd', 'be', 'ca', 'cb', 'cc', 'cd', 'ce', 'da', 'db', 'dc', 'dd', 'de', 'ea', 'eb', 'ec', 'ed', 'ee']

This will efficiently produce progressively larger words with the input sets, up to length maxlength.

Do not attempt to produce an in-memory list of 26 characters up to length 10; instead, iterate over the results produced:

for attempt in bruteforce(string.ascii_lowercase, 10):
    # match it against your password, or whatever
    if matched:
        break

Answer 3

I found another very easy way to create dictionaries using itertools.

generator=itertools.combinations_with_replacement('abcd', 4 )

This will iterate through all combinations of 'a','b','c' and 'd' and create combinations with a total length of 1 to 4. ie. a,b,c,d,aa,ab.........,dddc,dddd. generator is an itertool object and you can loop through normally like this,

for password in generator:
        ''.join(password)

Each password is infact of type tuple and you can work on them as you normally do.

Answer 4

itertools is ideally suited for this:

itertools.chain.from_iterable((''.join(l)
                               for l in itertools.product(charset, repeat=i))
                              for i in range(1, maxlen + 1))

Answer 5

A solution using recursion:

def brute(string, length, charset):
    if len(string) == length:
        return
    for char in charset:
        temp = string + char
        print(temp)
        brute(temp, length, charset)

Usage:

brute("", 4, "rce")

Answer 6

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').

Answer 7

import string, itertools

    #password = input("Enter password: ")

    password = "abc"

    characters = string.printable

    def iter_all_strings():
        length = 1
        while True:
            for s in itertools.product(characters, repeat=length):
                yield "".join(s)
            length +=1

    for s in iter_all_strings():
        print(s)
        if s == password:
            print('Password is {}'.format(s))
            break

Answer 8

from random import choice

sl = 4  #start length
ml = 8 #max length 
ls = '9876543210qwertyuiopasdfghjklzxcvbnm' # list
g = 0
tries = 0

file = open("file.txt",'w') #your file

for j in range(0,len(ls)**4):
    while sl <= ml:
        i = 0
        while i < sl:
            file.write(choice(ls))
            i += 1
        sl += 1
        file.write('\n')
        g += 1
    sl -= g
    g = 0
    print(tries)
    tries += 1


file.close()

Answer 9

Try this:

import os
import sys

Zeichen=["a","b","c","d","e","f","g","h"­,"i","j","k","l","m","n","o","p","q­","r","s","­;t","u","v","w","x","y","z"]
def start(): input("Enter to start")
def Gen(stellen): if stellen==1: for i in Zeichen: print(i) elif stellen==2: for i in Zeichen:    for r in Zeichen: print(i+r) elif stellen==3: for i in Zeichen: for r in Zeichen: for t in Zeichen:     print(i+r+t) elif stellen==4: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in Zeichen:    print(i+r+t+u) elif stellen==5: for i in Zeichen: for r in Zeichen: for t in Zeichen: for u in    Zeichen: for o in Zeichen: print(i+r+t+u+o) else: print("done")

#*********************
start()
Gen(1)
Gen(2)
Gen(3)
Gen(4)
Gen(5)

Answer 10

Simple solution using the itertools and string modules

# modules to easily set characters and iterate over them
import itertools, string 

# character limit so you don't run out of ram
maxChar = int(input('Character limit for password: '))  

# file to save output to, so you can look over the output without using so much ram
output_file = open('insert filepath here', 'a+') 

# this is the part that actually iterates over the valid characters, and stops at the 
# character limit.
x = list(map(''.join, itertools.permutations(string.ascii_lowercase, maxChar))) 

# writes the output of the above line to a file 
output_file.write(str(x)) 

# saves the output to the file and closes it to preserve ram
output_file.close() 

I piped the output to a file to save ram, and used the input function so you can set the character limit to something like "hiiworld". Below is the same script but with a more fluid character set using letters, numbers, symbols, and spaces.

import itertools, string

maxChar = int(input('Character limit for password: '))
output_file = open('insert filepath here', 'a+')

x = list(map(''.join, itertools.permutations(string.printable, maxChar)))
x.write(str(x))
x.close()