# Leaugue fixture puzzle

• Aug 20th 2010, 07:09 AM
Jevs
Leaugue fixture puzzle
Last season someone on this forum very kindly helped me with our 5-a-side a league fixtures. The fixture structure has stayed the same but the number of teams has changed so I wondered if anyone out there could help me work out the new fixtures according to the rules..

We now have 9 teams in the league, call them Team A, B C.. to team H.

We have available 24 weekly blocks (although judging by my rude calculations we should only need 18 of these?).

Each week block has 6 game slots of 20 minutes each (2hrs).

The rules for this puzzle are..

All teams have to play each other 3 times over the course of the league (the concept of home and away does not need to preserved).
On a week that a team does play, that team has to play exactly two games
The two games can be played consecutively, but no team should have to wait more than 20 minutes between games

We had two leagues of 6 teams last season and this worked fine, and I'm told it should be possible with 9 but my maths just isn't up to even knowing where to start.

Any help will be very gratefully received!!
• Aug 21st 2010, 08:36 AM
Wilmer
• Aug 22nd 2010, 01:18 PM
Jevs
Thanks Wilmer - those links are OK as a general guide (although most of the images appear to be missing in the top link?) however the generator isn't intelligent enough to observe the practical rules of our league, i.e we have to have exactly 6 matches in a round (i.e one night) and teams have to play exactly 2 games, and each team can not have more than a 20 minute (1 game) wait between games.

That's where the tricky bit comes in!
• Nov 22nd 2010, 01:28 PM
alan2
If you wish then skip ahead to the in summery part of this post.

Detail:

If I understand this correctly then the nuber of games in a league should be three times:
\$\displaystyle g_t = g_t_-_1 + (t - 1)\$

Where g is the number of games that will occur and t is the number of teams in the league. I'm much better at algorythms so I will write one of those as well.

Code:

```games = 0 for p = 1 to teams (         games += p ) games *= 3```
According to Wolfram Alpha:
\$\displaystyle g_t = ((t - 1) * t) * 0.5\$

For example if we have 4 teams then \$\displaystyle g_4 = ((4 - 1) * 4) * 0.5\$, therefore \$\displaystyle g_4 = 6\$.

In our case all this must occur three times so:
\$\displaystyle g_t = ((t - 1) * t) * 1.5\$
and
\$\displaystyle g_4 = 18\$

You play 6 matches per day and there are 365 days in a year therefore you could have 38 teams per yearly league so there is no trouble with 9 time wise. However there are other problems with your set-up.

Summary:

You need an odd number of teams to satisfy having teams play there matches in sets of 2.
You need a number of teams that gives an integer when divided by 4 to satisfy playing games in daily rounds of 6.
games that will be played = \$\displaystyle ((teams - 1) * teams) * 1.5\$

The first two conditions wont ever be true together.

For example:
3 teams creates 9 games, thats 2 per team. Exactly 6 games per round (not ok), teams play exactly 2 games per playing week (ok).
4 teams creates 18 games, thats 9 games per team. Exactly 6 games per round (ok), teams play exactly 2 games per playing week (not ok).
• Jan 8th 2011, 02:53 PM
alan2
Hope that was helpfull.