Math problem

[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.

[spoon]

When in doubt, ask a programmer (if you have one handy...) Here's the reply I got:
-----
If you want to do all x*x sized games, then any power of x will support your goal. However, be warned that in a 27-man tournament of 3x3 matchings, you are talking about 9 games apiece (3^n-1).

Scheduling is a bit of a pain but manageable. The way I would do it is to schedule x^n-1 guys for the first round, and then everything sort of falls out of that just by pairing that guy off with the next guy down the list, then 2 down the list, etc.. For 27 guys (a-z and 1) at x=3. You essentially match up randomly in the first round, and then down diagonals for each subsequent round based on the first round.

ABC
DEF
GHI
JKL
MNO
PQR
STU
VWX
YZ1

AEI
DHL
GKO
JNR
MQU
PTX
SW1
VZC
YBF

AHO
DKR
GNU
JQX
MT1
PWC
SZF
VBI
YEL

etc.

[LGM]

I used a program. Here are two solutions:

N=7
3,5,6
3,4,7
2,5,7
2,4,6
1,6,7
1,4,5
1,2,3

N=15
7,11,12
7,10,13
7,9,14
7,8,15
6,11,13
6,10,12
6,9,15
6,8,14
5,11,14
5,10,15
5,9,12
5,8,13
4,11,15
4,10,14
4,9,13
4,8,12
3,13,14
3,12,15
3,9,10
3,8,11
3,5,6
3,4,7
2,13,15
2,12,14
2,9,11
2,8,10
2,5,7
2,4,6
1,14,15
1,12,13
1,10,11
1,8,9
1,6,7
1,4,5
1,2,3