Math problem

[cybersol]

Originally posted by LGM:
Only solutions are 3, 7, 15, 31,...

Other numbers of players leave a three some with two players and no one to match them with. Someone demonstrated this earlier with 9 players.

I have verified the following solution by Erax for 9 players. LGM, why do you think it does not work and why does your program fail to find it?

Originally posted by Erax:
It will work for nine players. Here's how :

1-2-3 | 1-4-7 | 1-5-9 | 1-8-6
4-5-6 | 2-5-8 | 2-6-7 | 4-2-9
7-8-9 | 3-6-9 | 3-4-8 | 7-5-3

Nope, used no math, did it empirically.

Originally posted by geoschmo:
Is there an easy way to figure this out including who has to play who?

Here is what I know so far:
n = total number of players in tournament
x = number of player in each game

(n-1)/(x-1) = number of games played in the torunament by each player (#gpp)
This must be an integer for the tournament to work, so for x=3 player games there must be an odd number of players.

(#gpp*n)/x = total number of games in tournament
This must be an integer to work also, so for x=3 that leaves valid n=3,7,9,13,15,19,21,25,etc. This is really two series n=3,9,15,21,etc. and n=7,13,19,25,etc. deriving from whether n or #gpp is divisible by 3 respectively.

We have seen solutions for 3,7,9, and 15. If anyone wants to look for another empirical solution, I suggest seeing if 13 works.

Hope this helps,
cybersol

[ August 07, 2003, 00:17: Message edited by: cybersol ]