Dryice Liu's Blog http://dryice.name/blog Tue, 31 Aug 2010 12:01:11 +0000 http://backend.userland.com/rss092 en challenge 6, continued A year has passed since I visited Python Challenge the last time. And it's not easy to start it again :) I've got a project that deals with Python image process, so I spend some time on the PIL handbook. And after that, I think trying this challenge is a ... http://dryice.name/blog/python/python_challenge/challenge-6-continued/ Challenge 5 Well this is not an easy one. The unpickling is as easy as it should: import pickle f = open("banner.p", "r") x = pickle.load(f) But after that I was lost. x is a list of lists, with 23 items, each item are constructed with one or more tuples, each tuple has two elements: one " ... http://dryice.name/blog/python/python_challenge/challenge-5/ Challenge 6 Well first I need to figure out what to zip. I took a look at the HTML source again and found something unusual: the hint looks like: <html> <!-- <-- zip --> There's an extra "<—" in the HTML comment. I believe this is pointing me what to zip. But after I ... http://dryice.name/blog/python/python_challenge/challenge-6/ Challenge 4 Following the idea yesterday, here is my first script: import urllib; import string; import re; def followNothing(url): page=urllib.urlopen(url) pagecontent = page.read() print pagecontent result =re.search("[0-9]{5}", pagecontent) if result: num = pagecontent[result.start():result.end()] nexturl = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=" + num followNothing(nexturl) else: print "fix the script!" and I got: In [12]: followNothing("http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345") and the next nothing is 92512 and the next nothing is 64505 <font color=red>Your hands are getting tired </font>and the next ... http://dryice.name/blog/python/python_challenge/challenge-4/ Challenge 3 Well, it said EXACTLY three big bodyguard, so the first try is like this: import re target = re.compile(r"[!A-Z][A-Z]{3}([a-z])[A-Z]{3}[!A-Z]") f = open("mess.txt", "r") myout = target.findall(f.read()) And I got this: In [7]: print myout ['e', 'i', 'z', 'r', 'v', 'g', 'b', 'm', 's', 'z', 'u', 'o', 'w', 'i', 'b', 'z', 'b', 'c', 'j', 'g', 'd', 'o', 'h', ... http://dryice.name/blog/python/python_challenge/challenge-3/ Challenge 2 First of all I checked the lib to see if there's an OCR module, looks no, at least not in the standard distribution :) OK, it says to find the rare chars, so here is my first try: f = open("mess.txt", "r") ss = {} for c in f.read(): if ss.has_key(c): ss[c] += 1 else: ss[c]=1 And here is ... http://dryice.name/blog/python/python_challenge/challenge-2/ Challenge 1 After staring at the mess sentence at the end for a while, I realize there's not too mucy "k,o,e" in it so the idea yesterday won't work. Look again at the pic and I think I found what's in it: int(k)+2 = m, int(o)+2 = q, int(e)+2 = g. I think ... http://dryice.name/blog/python/python_challenge/challenge-1/ Challenge 0 OK, this is marked as "warming up" so it should be easy. The hint is "look at the URL address.". OK, this is challenge 0 and the url is 0.html. So I changed it to 1.html. There is a page there! It shows "2**38 is much much larger.". OK I know what ... http://dryice.name/blog/python/python_challenge/challenge-0/