Creating Fixtures in a Scrabble League

[wordsblade]

I run a Scrabble league in London

The number of players each "season" varies. Assume that we have y players. Individual fixtures take place as a round robin amongst 4 players (all play all 6 games in total). Fixtures take place regularly (currently monthly) with each player involved in a maximum of one fixture per month. Clearly if y is not divisible by 4 then some players will not be included in fixtures for a particular month. The maximum number of players who are not involved in a monthly round of fixtures can be no more than y mod 4 I am looking to create a set of fixtures such that over the course of a season all players in the league (ie y) have played all the other players in the league an agreed number of times (n). My questions are

1 Is it possible to show mathematically that there is a solution to this problem for given y and n
2. How can the minimum value of n be determined
3. If there is a solution what would the month by month fixtures be?

As an example I know that there is a solution with y = 16 and n=2 over 10 months. This year y=13 which probably complicates the situation gievn it is not divisible by 4.

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[Guest]

In general this is a very hard problem. There are obviously some constraints y and n have to satisfy, for example a player always plays against 3 other players in a fixture so n(y-1) (the total matches they've played) must be a multiple of 3. However determining whether a solution exists is very difficult and it looks like you'll have to use a computer to search through all the possibilities somehow. The best I can do is point you to the following:
http://en.wikipedia.org/wiki/Block_design
http://www.logic.at/prolog/mst.pdf
You really need to contact someone who is an expert in combinatorial block designs.

Hope this helped,

R. Baber.

[Guest]

Perhaps you can try contacting Professor Rosemary Bailey of Queen Mary University London. She's one of the only people I know of that is working in this area in London.

Hope this helped,

R. Baber.