Math problem

[LGM]

Originally posted by geoschmo:
Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

What part of the requirement did I neglect?

[geoschmo]

Oh wait LGM, I think I see what you are saying now. Did your program find solutions for anything between 7 and 15. The way I am looking at this now it should be possible to do for many numbers, not just powers of x.

Approaching the problem from the other end was the key if I am correct. For example, in a tournament of three man games the minimum number of players would be 3, and the number of games would be 1. The forumula should be:

x, game size
y, number of tourney games
n, total tourney field

EDIT: IGNORE EVERYTHING IN THIS POST AFTER THIS POINT. I haven't figured out what I did wrong, but this isn't right at all.

(y(x-1))+1=n

plugging in the ones we know already
(1(3-1))+1=3
(3(3-1))+1=7
(7(3-1))+1=15
(13(3-1))+1=27

We should be able to work out an n for any positive real integer X and Y fairly easy.
If x=3
y=1, n=3
y=2, n=5
y=3, n=7
y=4, n=9
.
.
y=15, n=31

So for x=3 n can be any odd number except 1.

The hard part would be actually coming up with the y number of combinations so that each player doesn't play anyone else more then once.

[ August 06, 2003, 20:44: Message edited by: geoschmo ]

[geoschmo]

No, that's not right.

LGM, I didn't mean any offense. I looked at your initial post and didn't think it solved teh problme, but after looking at it again I see it does. Can you explain the math your program is doing to find the solutions? That is what I am trying to figure out.

Geoschmo

[Slick]

I didn't read all the replies, so pardon this if already stated, but aren't you talking about the statistical formula for combinations? Mathematically, it means: how many ways can I group n things in groupings of r where the order of grouping is not important. (There is a separate formula for permutations where order is important)

The general formula for C(n,r) = n!/[r!(n-r)!]

Where the number of combinations C(n,r) is a function of the total number of people n taken r at a time.

You don't need any particular number of players to start with. If n = 10 and r = 3 then the required number of games is 10!/[3!x7!] = 120

Exactly how to arrange the people in each game is most easily worked out by hand although I suppose it could be done with a program.

Slick.

[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.

[geoschmo]

Sorry slick, this is way over my head.

"The general formula for C(n,r) = n!/[r!(n-r)!]"

What do n and r represent in the formula, and what are the exclamation points for?

And is C(n,r) the total numebr of players needed in the tourney, or the number of games played by each person in the tourney, or something else alltogether?

[geoschmo]

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.

Right, but other then working them out by hand, or having you plug it into yoru black box and give me the result, can you explain how you know this? What I am trying to learn is not the answer, but how the answer is derived.

Geoschmo