The first of a few problems is:

Given a list of words on stdin (one per line), find the words that have the largest number of anagrams in that list.

Print all of the words that have the meet the criteria of having the largest number anagrams (One per line, in alphabetical order).

This competition has now finished. The winner was Nick Cooper at 103 bytes, with the following awesome solution:

import sys
s=sorted
o=s(sys.stdin)
r=map(s,o)
d=map(r.count,r)
for e,t in zip(o,d):print e*(max(d)==t),

Input:
caret
crate
react
trace
ester
reset
steer
terse
organ
groan

Output:
caret
crate
ester
react
reset
steer
terse
trace

The main difference came down to the fact that Katie decided to avoid using the regular expression that was given in the sample, and just parse the lines with str.split.
So without much more talking, here is the winning solution.

import sys
s=sum(([l[0]]*int(l[-1])for l in map(str.split,sys.stdin)if l[-1]!='-'),[])
for x in set(s):print x,s.count(x)

It is also interesting to note how she uses the sum function to append a set of lists together.

Solutions must be emailed to me. In the case where two people have the same character count, the solution that arrived in my inbox first will be declared the winner (this is to stop people from playing with the date header in the email ^^). The winner will receive a chocolate bar or coffee — their choice.

The problem this week is one of parsing log files. You must parse a log file in the common log format that will be given on to your program on stdin. You must then print on stdout the amount of data that was sent to each ip address (and the ip address). A sample program has been provided:

#!/usr/bin/env python2.5
# Copyright 2008 Greg Darke <greg+laptop@tsukasa.net.au>
# Licensed for distribution under the GPL version 2, check COPYING for details
# Sample program to show the output format for the python golf problem posed at
# http://blag.tsukasa.net.au/2008/11/23/python-golf-competitionpython-golf-competition/

import re, sys

def main():
    clf_regexp = re.compile(r'''^(\S+)\s(\S+)\s(\S+)\s\[([^\]]*)\]\s"([^"]*)"\s(\d*)\s(\d*)$''')
    mapping = {}

    for line in sys.stdin:
        m = clf_regexp.match(line)
        if not m:
            continue
        ip, _, _, _, _, _, size = m.groups()
        size = int(size)

        if ip not in mapping:
            mapping[ip] = 0

        mapping[ip] += size

    for ip in mapping:
        print '%s %d' % (ip, mapping[ip])

if __name__ == "__main__":
    main()

Solutions will be accepted until 11:59:59pm(EDT) on Friday 28th November 2008.

What you see below is actually a prime sieve that is written in a single statement. I have used as many tricks as I could think of to lower the number of characters. Note, this code only works on python 2.5 (as I am using the new ternary operator). I also know that this code give a deprecation warning, that is because I am using 1e3 instead of the longer 1000.

Please feel free to leave a comment if you either have a shorter version, or an idea on how to make it shorter.

print[n for(n,l)in[((i,locals()['_[2]']) if i>1 else(0,[]))for i in range(1e3)]if n and(not l.__setitem__(slice(n*2,None,n),[(0,None)]*len(l[n*2::n])))]

Edit: After a little trip home on the bus, I have it down to 132 characters ^^.

print[n for(n,l)in[(i,vars()['_[2]'])for i in range(2,1e3)]if n and(l.__setitem__(slice(n*2-2,None,n),[(0,0)]*len(l[n*2-2::n]))!=0)]

Edit: Make that 128 characters

print[n for(n,l)in[(i,vars()['_[2]'])for i in range(1e3)]if n>1 and(l.__setitem__(slice(n*n,None,n),[(0,0)]*len(l[n*n::n]))!=0)]

Edit: 84 characters, though it is no longer a one line/one statement thing

z=range(1e3);p=(n for n in z if n>1)for n in p:z[n*n::n]=[0]*len(z[n*n::n]);print n

Why? Well, because I can’t be bother to do it anywhere else on the internet, so I thought I would use it here.

Welcome to those of you who have managed to find my blog, I hope to treat your all with some interesting tid-bits of programming, geeky stuff and a little bit of anything else. My present to people reading the first post here is a little bit of python that we came up with after a game of golf (python golf that is).

print[x for x in range(2,1e3)if all(x%_ for _ in range(2,x))]

What does this do? Well you run it for yourself (or are feeling adventurous, you could work it out by hand).