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Python: Making a Brute Force Hash Checker with Known Characters

I've created a rather simple program to brute force the specified type of hash, with the ability to append known characters to the beginning of it.

import datetime
import hashlib
import itertools

characters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'

def checkHashes(targetHash, unhashedStringLength, hashType, discovered):

print("Start = " + str(datetime.datetime.now()))
typeOfHash = getattr(hashlib, hashType)

for trialString in itertools.product(characters, repeat=unhashedStringLength-len(discovered)):

s = discovered + "".join(trialString)

if(typeOfHash(s.encode('utf-8')).hexdigest() == targetHash):
print("Finish = " + str(datetime.datetime.now()))
return s


That being said, I'm currently unable to solve a hash with known characters throughout it. For example, if I were to have "AB234***ANS*34" presented to me, with the "*" being unknown characters, I wouldn't be able to tell the program that I know more than the first five characters. Is there any way to brute force a string of characters with known and unknowns throughout it without slowing the process? Also, if there is anything I can do to make the program run more efficiently, I'm more than happy to take suggestions.

Sorry if my wording is confusing. If you need anything clarified, please ask!

1 Answer
1

You can intersperse fixed and variable parts in your arguments to product(). For example, using your "AB234***ANS*34" example, but with a much smaller character set to make the output short (2 possible characters for each of the 4 wildcards = 2**4 = 16):

product()

"AB234***ANS*34"

from itertools import product
characters = "xy"
template = [["AB234"],
characters, characters, characters,
["ANS"],
characters,
["34"]]
for trialString in product(*template):
s = "".join(trialString)
print(s)


displays the 16 possibilities:

AB234xxxANSx34
AB234xxxANSy34
AB234xxyANSx34
AB234xxyANSy34
AB234xyxANSx34
AB234xyxANSy34
AB234xyyANSx34
AB234xyyANSy34
AB234yxxANSx34
AB234yxxANSy34
AB234yxyANSx34
AB234yxyANSy34
AB234yyxANSx34
AB234yyxANSy34
AB234yyyANSx34
AB234yyyANSy34


As far as product() is concerned, something like ["AB234"] is a sequence with a single element, so that single element is included in every product it returns.

product()

["AB234"]

As to efficiency, the task is inherently exponential-time. Nothing will make it fantastically faster. Industrial-strength crackers resort to assembler for low-level speed, and spread the task over as many CPUs as they can afford to run ;-)

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