I've asked this question once before on here and didn't get the answer I was looking for, but I got another method of doing it. so here I go again. (The problem with the previous found answer was that it was extremely efficient, a bit too efficient. I couldn't count comparisons, and I wanted it to be the most bare bone way of finding the password, so that I can implement a rating system to it.) I am looking to make a password rating program that rates a password based off of the length of time, and the amount of comparisons the program had to make in order to find the correct password. I'm looking to use multiple different methods in order to crack the input, ex : comparing the input to a database of common words, and generating all possibilities for a password starting with the first character. Problem : I can't find a way to start with one element in a list, we will call it A. Running A through chr(32) - chr(127) ('space' - '~'), and then adding a second element to a list called B. For the second loop, set A = chr(32) and B would then run through chr(32) - chr(127). A would then turn into chr(33) and B would run through all characters and so forth. until all possible options have been compared, then for the next loop it would add another element onto the list and continue the search, starting with chr(32), chr(32), chr(32-127). Continuing this pattern until it finds the correct password. This is the closest I could get to something that would work (I know it's terrible). while ''.join(passCheck) != ''.join(usrPassword) : for i in range(0, len(usrPassword)) : if ''.join(passCheck) != ''.join(usrPassword) : passCheck.append(' ') for j in range(32, 127) : if ''.join(passCheck) != ''.join(usrPassword) : passCheck = chr(j) for k in range(0, len(passCheck)) : for l in range(32, 127) : if ''.join(passCheck) != ''.join(usrPassword) : passCheck[k] = chr(l) print(passCheck) Continue reading...
