Python Brute Force algorithm [closed]

Question

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

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 password 'solved'). You could also use an already existing password-dictionnary or generate it with tools like cupp.

Answer 6

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 7

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 8

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 9

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 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()