#!/usr/bin/env python # file : every.py print 'Running "every.py"'
possible = "234567abcdefghijklmnopqrstuvwxyz" word_length = 16 word_list = [] def add_word(word): if len(word)==word_length: word_list.append(word) print word # There may come a time you won't want this line. else: for c in possible: new_word = word + c add_word(new_word) add_word("") # print word_list > That solution is really nice, thanks, I doubt I would've come up with > something as nice. What I'll do is have that write to a file then use > regex > to strip out any characters like whitespace or commas from the list and > that should give me a good solid list, no? > > I'm using this to write a program I'm calling TORdialer with the goal of > attempting to basically ping all possible TOR hidden service pages with > the > goal of having an absolutely complete list of all TOR hidden services on > the net right now. Unfortunately it will probably be a lot of fucked up CP > sites and drug related marketplaces. I'm mostly interested in the sites > that aren't publicized (ie- government related sites, etc). > Thanks for your help and I'll try to keep everyone posted. > There won't be any whitespace or commas in any of the derived output. For your purposes, you might want to make this into a generator although that would be getting too sophisticated for me to help you. Since in its present form the algorithm is a recursive one, I'm guessing it'll run out of memory long before it comes to completion although I haven't let it run for more than a few seconds, just enough to see that it seemed to be doing what I wanted. >From what I know about computer science, I believe that most if not all recursive algorithms can be re-written to be non recursive. If you could do that, you could then make a generator and presumably make use of it without overwhelming resources. I look forward to hearing what those with more expertise might have to say about that. _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor