Brute Force Password Program / Python 3.9

Question

I've been making a brute force program for fun. This program tries to guess a password using brute force. I've tried many things to help increase the efficiency. Here it is: (Sorry for the lack of comments)

from itertools import product
from time import time
def findamount(find):
 constValues = ["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",
 "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",
 "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
 dictValues = {constValues[n]:n+1 for n in range(62)}
 value = 0
 for n in range(len(str(find))):
 x = len(str(find)) - (n+1)
 value += dictValues[find[n]] * (62**x)
 return value
def fast(password):
 password = tuple(password)
 origtime = time()
 length = 1
 constValues = ["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",
 "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",
 "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
 for a in range(10000000): #More efficient than while True
 for n in product(constValues, repeat=length):
 if n == password:
 amount = findamount(''.join(list(n)))
 amountTime = time() - origtime
 if amountTime == 0:
 print(f"Length = {length}, Amounts = {amount}\nTime elapsed = 0 secs\nAttempts per sec = ? TIME TOO SHORT")
 else:
 print(f"Length = {length}, Amounts = {amount}\nTime elapsed = {amountTime} secs\n Attempts per sec (estimate) = {amount/amountTime}")
 print(f"DONE! PASSWORD: {password}")
 return
 length += 1
 if length > 3:
 amount = findamount(''.join(n))
 amountTime = time() - origtime
 print(f"Length = {length-1}, Amounts = {amount}, Password = {''.join(n)}\nTime elapsed = {amountTime} secs\nAttempts per sec = {amount/amountTime}\nEstimated Time: {findamount(''.join(password))/(amount/amountTime)} secs")
password = input(">>> ")
fast(password)

My current average is 6-16M per sec. However, I want to see how to increase it on my code. I tried using threading and this is the attempt:

from itertools import product
from time import time
from threading import Thread
from more_itertools import divide
def findamount(find):
 constValues = ["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",
 "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",
 "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
 dictValues = {constValues[n]:n+1 for n in range(62)}
 value = 0
 for n in range(len(str(find))):
 x = len(str(find)) - (n+1)
 value += dictValues[find[n]] * (62**x)
 return value
def fast(password, amountOfThreads, thread):
 password = tuple(password)
 origtime = time()
 length = 1
 constValues = ["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",
 "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",
 "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
 for a in range(10000000): #More efficient than while True (I think)
 for n in divide(amountOfThreads, product(constValues, repeat=length))[thread-1]:
 if n == password:
 amount = findamount(''.join(list(n)))
 amountTime = time() - origtime
 if amountTime == 0:
 print(f"Length = {length}, Amounts = {amount}\nTime elapsed = 0 secs\nAttempts per sec = ? TIME TOO SHORT")
 else:
 print(f"Length = {length}, Amounts = {amount}\nTime elapsed = {amountTime} secs\n Attempts per sec (estimate) = {amount/amountTime}")
 print(f"DONE! PASSWORD: {password}")
 raise SystemExit
 length += 1
 if length > 3:
 amount = findamount(''.join(n))
 amountTime = time() - origtime
 print(f"Length = {length-1}, Amounts = {amount}, Password = {''.join(n)}\nTime elapsed = {amountTime} secs\nAttempts per sec = {amount/amountTime}\nEstimated Time: {findamount(''.join(password))/(amount/amountTime)} secs")
password = input(">>> ")
if __name__ == "__main__":
 amountOfThreads = int(input("How much threads?"))
 for n in range(amountOfThreads):
 Thread(target=fast, args = (password, amountOfThreads, n+1)).start()

It gets much slower. If you input 1 thread it does 2-3M per sec and 2+ threads in total does about 200k per sec (totalling all threads). Is there any way to improve this code to allow threading to be useful, or any other efficient code I should add? Also, is threading the most efficient way to do this?

• FMc: The constraint in the program is computation, not IO. Using threads, as you discovered, won't help (search for the Python GIL). Instead see the multiprocessing library.

Answer 1

I've found a few ways to improve my code:

1. Instead of doing for a in range(10000000): and length += 1, you can just merge the two and do for length in range(1, 10000000):. This would help increase efficiency because in the previous case, a would be length-1, meaning if I use range(1, 10000000) (meaning a starts at 1 instead of 0), a would be equal to length.

2. The second improvement is in this code:

if length > 3: 
 amount = findamount(''.join(n))
 amountTime = time() - origtime
 print(f"Length = {length}, Amounts = {amount}, Password = {''.join(n)}\nTime elapsed = {amountTime} secs\nAttempts per sec = {amount/amountTime}\nEstimated Time: {findamount(''.join(password))/(amount/amountTime)} secs")

''.join(n) is calculated twice, I could just do currentPassword = ''.join(n) so ''.join(n) would only be calculated once, and use currentPassword for the two times I used ''.join(n).

3. This improvement is also on the code above. the print part can be improved. This isnt an efficiency improvement, but a readability improvement. Changing the print function to:

print(f"Length = {length}, Amounts = {amount}, Password = {currentAmount}\nTime elapsed = {amountTime} secs",
"\nAttempts per sec = {amount/amountTime}\nEstimated Time: {findamount(''.join(password))/(amount/amountTime)} secs")

I'll update this every time I find something new.