Thank you again. I now have enough to keep me happily busy for days.
Robert Michael Langford wrote:
I understand that each response is unique Robert and no caching is required to solve the problem at hand. However in a real program, the chance you're brute forcing just one password is small (usually you would brute force many); additionally, the question posted specifically asked that the trials be cached, which is why I suggest you do so, not out of any real need for just testing on the "loner" testcase. Additionally, the reason the aaaaa->zzzzz method is suggested is that it a much faster algorithm to test every available combination than the random method. The random method may never find the data. I've created a frame work for you to play around with on this problem. It shows how long different methods take in a more realistic scenario. Try some other methods if you want, as well play around with password length, the aaaaa->zzzzz method, etc. I've shorted then password length to 3 to make it run faster as well. Enough procrastination, back to google app engine stuff. --Michael #!/usr/bin/env python import string import random # test code to simulate a computer system under # attack, no need to understand now class Computer(object): def __init__(self): self.reset_password() def reset_password(self,length=3): self.secret_password=''.join(random.sample(string.ascii_lowercase,length)) print('The new password is: ' + self.secret_password) def enter_password(self,teststr): """returns true if the password was correct""" return teststr==self.secret_password # end test code def rand_string(length = 3): """generates a random string of the given length""" return ''.join(random.sample(string.ascii_lowercase,length)) passcache = list() def crack_password(comp): """repeatedly tries passwords on the given computer""" tries = 0 while True: tries += 1 if comp.enter_password(rand_string()): print( 'Password found in ' + str(tries) + ' tries.') return def crack_password_cached(comp): """repeatedly tries passwords on the given computer""" for password in passcache: if comp.enter_password(password): print 'Password found in cache' return tries = 0 while True: tries += 1 s = rand_string() passcache.append(s) if comp.enter_password(s): print( 'Password found in ' + str(tries) + ' tries.') return def hack(): """hacks a computer via a brute force method""" my_computer = Computer() crack_password(my_computer) def hack_cached(): """hacks a computer via a brute force method with cached results""" my_computer = Computer() crack_password_cached(my_computer) def mean(ls): """finds the mean of a list of numbers""" return sum(ls)/len(ls) def test_case(funcname, repeats): """ timeit is a python library for timing code. No need to understand how this works, just that its repeating the command given the number times specified by repeats. It's returning a list of the execution times of each of the trials """ import timeit timer = timeit.Timer("%s()"%funcname,"from __main__ import %s"%funcname) return timer.repeat(number=repeats) if __name__ == '__main__': times_to_repeat = 100 uc_results = test_case("hack",times_to_repeat) c_results = test_case("hack_cached", times_to_repeat) print( "Cached mean time: " + repr(mean(c_results)) ) print( "Cached min. time: " + repr(min(c_results)) ) print( "Uncached mean time: " + repr(mean(uc_results)) ) print( "Uncached min. time: " + repr(min(uc_results)) ) On Thu, Jan 8, 2009 at 1:31 PM, Robert Berman <berma...@cfl.rr.com> wrote:Michael, Thank you for your code and your commentary. The code tells me this is really an ongoing learning process; almost as convoluted as linguistics and certainly every bit as interesting. Your concept of brute force in this example is intriguing. It is as if I have five cogs spinning, 'a' the slowest and 'e' flying as fast as its spokes can engage, and every click of every wheel produces a viable but possibly wrong response. However, each response is unique and in the nth iteration where password is found, all previous iterations are unique so there is certainly no need to carry the previous responses in either a list or a dictionary. They do not duplicate since they cannot repeat. Other than to include extraneous code, why include repetitive checking at all? In spite of that, your code and your remarks are most appreciated. Robert Michael Langford wrote: Here is your algorithm made more pythonic. Notice the use of default parameters, doc strings, not abbreviated variable names, unix C style capitolization (very subjective, but often the one found in python libs), the avoidance of control variables when possible, the use of ascii_lowercase instead of your own string and the not calling functions main that aren't __main__. #!/usr/bin/env python import string import random def rand_string(length = 5): """returns a random string of numbers""" return ''.join(random.sample(string.ascii_lowercase,length)) def try_password(match="loner"): """randomly tries 5 letter long passwords until the user finds the correct one""" tries = 0 while True: tries += 1 if rand_string() == match: print 'Password found in ' + str(tries) + ' tries.' return if __name__ == '__main__': try_password() Note: I do not think you're doing the problem the way the author intended. As it wants you to try all combinations, and it's about brute force, not finding it quickly, you most likely should start at "aaaaa" and move to "zzzzz", and cache the results in a list like they ask you too. At the very least, you should be saving these combinations you are generating randomly in a list (as it asks you to) and not testing the password if the generated string was already in the list (as in a real application, that is the time consuming or dangerous operation). --Michael PS: If you wish to assure direct replies reach me On Thu, Jan 8, 2009 at 10:49 AM, Robert Berman <berma...@cfl.rr.com> wrote: Hi, One of the challenges on the challenge you web page appropriately titled 'Brute force' reads as follows: "The password you have to guess is 'loner' . Try all combinations of lowercase letters until you guess it. Try not to loop much for example, save all used combinations in an array so you don't repeat." My code is as follows: #!/usr/bin/env python import random def GetString(): alphabet='abcdefghijklmnopqrstuvwxyz' return ''.join(random.sample(alphabet,5)) def main(): password='loner' errknt = 0 control = True while control is True: if GetString() != password: errknt +=1 else: print 'Password found in ',errknt,' tries.' control = False if __name__ == '__main__': main() The code does work. I am looking for suggestions to learn more about both efficiency and optimization of code. Since the challenge revolves around the use of randomized retrieval, I'm not too sure how to optimize the process. The authors concept of using arrays seem a bit superfluous as I think it takes longer to add an item to a dictionary and retrieve an item from a dictionary than it does to do an if compare of two 5 character strings. So, I left that code out of the program entirely. If that was wrong, or there is a better way to avoid duplication, please point me in the right direction. I think, perhaps, I could make it a tad more efficient if I changed 'alphabet' from a string to a list as I remember reading that lists are significantly faster to manipulate than are strings. Is that true and is it a viable change. I realize my code looks like modified C++ structured code. I am trying to become more Python concise but I think that is a matter of writing more and more python code. All suggestions, ideas, critiques are most welcome. Thank you, Robert _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
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