Math problem

[geoschmo]

Ok, I like math but my brain doesn't work well when you get past two dimensions.

I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

If there are three players it's easy, A,B,C. What's the next number of players that this format will work for? Can it be done with 9 or do you need 27, or is it some other number completely?

Is there an easy way to figure this out including who has to play who?

[geoschmo]

Ok, I figured it out the long way using graph paper.

So I guess then it's x squared, where x is the number of players per game. Not x to the power of x. So with 4 player games it would take 16 people in the tourney and 5 player games would take 25? Does that sound right? Is there a way to verify that without using graph paper?

[BBegemott]

Originally posted by Geoschmo:
I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution

[tesco samoa]

hey geo i use that for figuring out how many tickets i need to do in betting on sport games

3 of 3 rotations ( 3 teams to 25) right up to 12 of 12 rotations

3 of 4 is 4
3 of 5 is 10
3 of 6 is 20
3 of 7 is 35
3 of 8 is 56
3 of 9 is 84
3 of 10 is 120
4 of 5 is 5
4 of 6 is 15
4 of 7 is 35
4 of 8 is 70
4 of 9 is 126
4 of 10 is 210
5 of 6 is 6
5 of 7 is 21
5 of 8 is 56
5 of 9 is 126
5 of 10 is 252
6 of 7 is 7
6 of 8 is 28
6 of 9 is 84
6 of 10 is 210
7 of 8 is 8
7 of 9 is 36
7 of 10 is 120
8 of 9 is 9
8 of 10 is 45

many different rotations...

[Z.e.r.o]

The number of matches you need is n!
where n! = 1*2*3*4*...*(n-1)*(n)
and is called factorial and can be used to determine the numbers of permutations of n tokens.

In a 3 players match where anyone is playing a role only once is 3! (1*2*3 = 6)

Empirical solution:
123
213
312
321
132
231

If I'm not right you can spank me with a crowbar

[tesco samoa]

umm mine was total games that would be needed to play each other in every combination...

ignore

[Z.e.r.o]

Spank me.

Is not right, every player plays two times in the same role.

But I'm sure that the number will be a subset of n!, but I also believe that you need an even number of players to not have conflicts.

[tesco samoa]

i have a question along this line

say you have 24 numbers and you want to sort them in combinations of 4 where each number only appears once with each other how many combinations would that be ???

now if some one wanted to whip up a little program that does that ( but i can select the numbers (say up to 50), combinations ( 2 to 12 ) and uniqueness ( say once to 6 times ) and can produce a text file output i would be forever thankful

[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.

[ August 06, 2003, 18:41: Message edited by: Erax ]

[geoschmo]

Originally posted by BBegemott:
IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution

The problem is not every legal combination will work. For example, by using
1) 123
2) 145
3) 167
4) 189
you prevent the option of trying 146. This is fine for 1, 4, and 6 as long as they all play each other once, but might prevent others from being able to play each other, as you found out.

Instead if you do something like...
1) 123
2) 456
3) 789
4) 147
5) 158
6) 169
7) 249
8) 275
9) 268
10) 348
11) 359
12) 367

then it works.