Settlers of Catan - Help Needed Creating A Tournament Layout

[bannedtim]

Hello,

I am trying to run a 8-man Settlers of Catan tournament. There are a few restrictions we have been trying to build our tournament within. I am coming up with some problems, so I am here asking for your help.

The idea we wanted is two, four player games going on at the same time. We wanted it so that all players had a chance to play from every starting position(placing first, second, third, and the wheel), and on top of that we wanted to make sure that everyone plays everyone at least once, but no more then twice. This is where we are running into problem, I can get it where everyone plays everyone at least once from different positions, but I can't get it to where they play people no more then twice. Anyone have time to help me out and try to figure out this puzzle? Thanks,

Bannedtim

[RoundTower]

this is interesting

you want to have 4 sets of 2 games?

[RoundTower]

OK if this is what you mean:

you want to have four sets of 2 games, one after the other. Each player plays with each opponent exactly twice.

Spoiler in white:
<font color="white">I have a simple proof that this cannot be done. While this isn't much help for you organising your tournament, it is just as satisfying from a mathematical point of view.

Suppose A and B play together in rounds 1 and 2. Then at least one opponent must be in the game that does not include A and B, in both round 1 and 2. Call this opponent C. Now clearly A must play with C in rounds 3 and 4. But similarly B must play with C in rounds 3 and 4. Therefore A must play with B again.</font>

Hope this helps.

[bannedtim]

Round Tower,

Makes sense. Thanks for the help.

BT

[Neil S]

Just roll dice. That's already going to determine the winner.

<font color="white">I know I know...</font>

[RoundTower]

[ QUOTE ]
Round Tower,

Makes sense. Thanks for the help.

BT

[/ QUOTE ]
unfortunately I realised last night I answered the wrong question. I solved it for each pair of players playing exactly twice, it should have been between once and twice. However I got the right answer, it is still impossible. Proof this time:

first see that no three players can play together twice, because they cannot all be separated in the other two rounds. So suppose A and B play together in rounds 1 and 2. They play with C and D in the first round, and E and F in the second round. So we have

ABCD EFGH
ABEF CDGH
A??? B???
A??? B???

G must get a chance to play with both A and B, suppose with A in round 3 and B in round 4. Similarly H, but H must be kept apart from G. So

ABCD EFGH
ABEF CDGH
AG?? BH??
AH?? BG??

E must get a chance to play with both C and D. So the last two rounds must involve CE/DF and DE/CF in some order. However, you can see that whichever way you slot these in, at least one of C,D,E or F plays with A in both of the last rounds, and therefore plays with A three times in total.