Hello Martin,
again thanks for your endeavour, ist tought me to really think deeply how to
specify my task fort he Python language.
Before I start to work with your "heavy" piece of code for a beginner below
I like to make the following statements:
1. my task is to create lists or tupleS (whichever works best) containing 3
letters, later to be assigned names, in unique sequences.
2. My first approach with pairs like 'a b', 'a c',..... does not work with
itertools.combinations(s, 3): Although, it produces lists/tuples with 3
pairs
there are 4 letters in each list/tuple whereas I need only 3.
3. Therfore, I'am working with single letters creating lists/tuples with 3
letters: ('a', 'b', 'c'), ('a','b','d')........
Using all 26 letters of the alphabet would correspond to Problem A:
Impossible to handle.
Using only 15 letters would already create 455 lists/tuples.
So I am using 9 letters which is practical for one letter or name can be
combined with 2 others 5 times or on 5 days, each letter/name can occur
only once a day.
The 9 letters produce 84 lists/tuples. But if I am isolating lists/tuple
with unique sequences by hand I am left with 15 lists/tuples but with a
uneven distrubution of the 9 letters:
a 8
b 5
c 6
d 5
e 6
f 6
g 4
h 4
i 4
This variance gets the smaller the more letters are used.
4. I have come to the conclusion that my task is too mathematical for
itertools. I hope I can be successfull with your code below
although it will me take time to comprehend it.
Sorry for this long text, regards, Marcus.
-----Ursprüngliche Nachricht-----
Von: Martin A. Brown [mailto:mar...@linux-ip.net]
Gesendet: Dienstag, 22. September 2015 03:10
An: marcus lütolf <marcus.luet...@bluewin.ch>
Cc: tutor@python.org
Betreff: Re: [Tutor] Creating lists with 3 (later4) items occuring only once
Marcus,
I have more questions for you, as well as a possible solution (though it's a
bit more verbose than I would have liked to offer).
Question:
Problem A: Are you looking to identify the complete set of
possible options for constructing the triples of pairs?
Problem B: Are you looking only to construct one set that
satisfies the problem? [see my lousy solution]
You may observe that the question I ask is quite similar to the question
asked by Francesco [0].
If you are asking about the complete set of possible options (Problem A),
then I submit to you that you are asking a mathematical question, not a
Python question. If that's the case, perhaps you should look further into
the Steiner system and/or ask again on the list.
If you are asking about finding an individual solution satisfying the
constraints, I submit to you that either my approach or Francesco's approach
could work for you. If that's the case, then using random.sample may offer
you some help. See my sample, below--it should work on Python 2.x or Python
3.x.
Comments:
* There are rules in your system about when a player can play
again. The rules were not fully clear to me, so I allowed
players to be selecteda second time, if there were no more
players who had not already been chosen.
* This program produces only one possible scenario for 7
rounds of 3 distinct, simultaneously played 2-player games.
No player can play twice in the same round. (Simple arithmetic
determines the minimum number of players.)
If your question is Problem A, then I wonder if you know anybody who knows
combinatorics? I do not.
-Martin
[0] https://mail.python.org/pipermail/tutor/2015-September/106820.html
#! /usr/bin/python
from __future__ import print_function
import sys
import string
import random
import logging
logging.basicConfig(stream=sys.stderr, level=logging.INFO) logger =
logging.getLogger()
class NotEnoughPlayers(Exception):
pass
def choose_game_participants(players, pcount=2):
'''returns a tuple of players for a single game
'''
if len(players) < pcount:
raise NotEnoughPlayers("not enough players, need %d, have only %d:
%r" %
(pcount, len(players), players))
game = tuple(sorted(random.sample(players, pcount)))
return game
def choose_match_games(players, pcount=2, gcount=3):
'''returns a list of games where no player is duplicated
'''
mcount = pcount * gcount
if len(players) < mcount:
raise NotEnoughPlayers("not enough players, need %d, have only %d:
%r" %
(mcount, len(players), players))
eligible_players = random.sample(players, mcount)
match = list()
while eligible_players:
m = choose_game_participants(eligible_players, pcount)
for x in m:
eligible_players.remove(x)
match.append(m)
match.sort() # optional
return match
def generate_rounds(players, pcount, gcount, rcount):
games = set()
matches = list()
mcount = pcount * gcount
eligible_players = list(players)
if mcount > len(eligible_players):
raise NotEnoughPlayers("not enough players (%d) to guarantee %d
%d-player games per match" %
(len(eligible_players), gcount, pcount))
r = 1
while r <= rcount:
try:
proposed_match = choose_match_games(eligible_players, pcount,
gcount)
except NotEnoughPlayers:
logger.info("adding %d additional players in round %d to meet
minimum pool requirements",
mcount, r)
how_many = mcount - len(eligible_players)
eligible_players.extend(random.sample(players, how_many))
continue
already_played = games.intersection(set(proposed_match))
if already_played:
logger.info("discarding %d %r because %r have already played",
r, proposed_match, list(already_played))
continue
else:
games.update(proposed_match)
matches.append(proposed_match)
logger.info('Proposed match %r', proposed_match)
for game in proposed_match:
for player in game:
eligible_players.remove(player)
r = r + 1
return matches
def log_match_info(matches, detail=False):
for mnum, match in enumerate(matches, 1):
logger.info("match summary %d: %r", mnum, match)
for gnum, game in enumerate(match, 1):
if not detail:
continue
logger.info("match detail %d, game %d: players %r",
mnum, gnum, game)
def log_match_summary(matches):
log_match_info(matches, detail=False)
def log_match_detail(matches):
log_match_info(matches, detail=True)
if __name__ == '__main__':
players = list(string.ascii_uppercase)
random.shuffle(players)
# players = set('ABCDEFGHIJ')
pcount = 2 # players per game
gcount = 3 # games per match
rcount = 7 # rounds (of matches)
matches = generate_rounds(players, pcount, gcount, rcount)
log_match_detail(matches)
# -- end of file
--
Martin A. Brown
http://linux-ip.net/
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