Math problem

[Gryphin]

Has something to do with "Pigeon Holing
Geo I don't know if these help. I did a search at google on the formula:

http://www.cs.mtu.edu/~shene/COURSES...06/fact-2.html

This is a word doc:
http://www.cecs.csulb.edu/~lam/cecs228/ch4.doc

I think this is a powerpoint presentation:
file:///D:/Temporary%20Internet%20Files/Temporary%20Internet%20Files/Content.IE5/I56W66T4/275,10,r-permutations

This one seemed the simplest:
http://fclass.vaniercollege.qc.ca/we...ombs_Intro.htm

[Gryphin]

This site has
"Real Life Mathmatics" and includes calculators.
http://fclass.vaniercollege.qc.ca/we...real/2real.htm

Edit:
Oh well, too little too late but I did learn a lot and the above link has some interesting math described in laymens terms. It also has some math word puzzles under the "Teasers"

[ August 06, 2003, 22:51: Message edited by: Gryphin ]

[Fyron]

Originally posted by Suicide Junkie:
More than two players would get you problems with teaming up, I imagine.

Yes, and it is not meant for 2-player games.

[Captain Kwok]

You need to change the format!

What about setting it up like a sports league?

For example, 12 players and 2-player games.

Make 2 divisions of 6 players. Each player plays two matches against their division mates and 1 game against the opposite division. This makes 16 games. The top 2 players from each division play a best-of-3 series to determine a 'pennant' winner, and then the pennant winners from each division play for the grand championship in a best-of-3 series.

The games can be played concurrently:

Day 1 to 75 > 1st games against own division
Day 76 to 150 > games against other division
Day 151-225 > 2nd games against own division
Day 225-? > Playoffs

If this is too long, you can set up this style for any amount of players. You could set up a smaller league that runs faster, say like 8-player leagues. You could even run several small leagues at different intervals, giving opportunities for new players to join etc...

[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 ]

[geoschmo]

Kwok and SJ, the idea isn't to use 2 player games. I am trying to figure out a way to mathematically derive the total number of players needed for a round robin tourney of larger then two player games. And I want to be able to figure it out for any possible number, not jsut three.

Coming up with alternate formats isn't the idea. I am not really trying to make a tourney. I am just trying to figure out the math side of it.

It's a tanget.

Gryphin, those sites are interesting, but they are have the same problem as Slicks formula. They are calculating all possible combinations. That's not what I am trying to do.

Geoschmo

[geoschmo]

Originally posted by cybersol:
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?

I suspect his program is hitting a point where it can't find a valid remaining set and is then deciding that number n has no solution. Whil ein fact as we showed earlier sometimes you can get stuck down a "blind-alley" where there is a solution for n, but not for every possible set of 3. Like what happened to Bbgemont. To completely rule out a possible solution it would have to back track when it reaches these end points and change an earlier set and rework from that point. Sounds complicated.

Geoschmo

[Gryphin]

(I don't really know anything about the math involved)
I did another google search:
"round robin tournament"
Then I tried:
"round robin tournament" +software

Quite a few hits that might help including calcuators.
Here is one link:
http://www.devenezia.com/downloads/round-robin/

I guess what I'm driving at here is others must have wanted this and their answer must be on the web.
Good luck

[ August 07, 2003, 00:58: Message edited by: Gryphin ]

[Captain Kwok]

Geo:

You have a difficult problem. Most programs used to set up match schedules like this are based on 2 players/teams.

What you are proposing is not a very common format to schedule and will be very difficult to organize. Other than finding the total number of games required (i.e. number of all the different combinations of players as calculated by others), you'll have to manually arrange the games or find someone to make a program for you that can do this automatically. On a more positive note, I'm sure there is some sort of combinations calculator out there on the net that lists each of the combinations...

[ August 07, 2003, 01:07: Message edited by: Captain Kwok ]

[Jack Simth]

Let's see: Floor function for the numbers:

Pp = players per game
Np = Number of players (total)
Gp = Games per player
Tg = Total games

Gp = (Np - 1) / (Pp - 1)
Tg = (Np * Gp) / Pp
= (Np*((Np - 1) / (Pp - 1)))/Pp
= (Np * (Np - 1)) / (Pp * (Pp - 1))

Gp = (Np - 1) / (Pp - 1)
Tg = (Np * (Np - 1)) / (Pp * (Pp - 1))

If Gp and Tg come out as positive integers, it should be doable - I'm not sure about the arrangement, however.

Edit: Arrangement method:

1) List players
2) Variables
Pp = players per game
Np = Number of players (total)
Gp = Games per player
Tg = Total games
Sk = Skip (counting variable; internal use only)
3) Gp = (Np - 1) / (Pp - 1)
4) Tg = (Np * (Np - 1)) / (Pp * (Pp - 1))
5) Sk = 0
6) Group, skipping Sk
7) Sk = Sk + Pp
8) If Sk < Np, Goto 6

[ August 07, 2003, 01:44: Message edited by: Jack Simth ]