Math problem

[LGM]

Prime numbers are a nice way to test if something is a member of a set. Products of prime numbers is a common way of encoding sets because you can divide by a given prime. If the remainder is zero, it is an element of the set.

[LGM]

You could use 21 players (3 sets of 7) with a championship game from the winners of each set.

Or you could use 49 players (7 sets of 7) and have a championship round of the top 7 playing each of the top 7.

[ August 06, 2003, 21:13: Message edited by: LGM ]

[Slick]

Originally posted by 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?

Sorry, no offense intended...

n are the total number of players in the tourney
r are the number of players in the group (i.e. 3 for your case)

x! (spoken "x factorial") is defined as x! = x(x-1)(x-2)(x-3)...(3)(2)(1)

So 10! = 10x9x8x7x6x5x4x3x2x1

C(n,r) means C is a function of the variables n & r.

Bottom line:

The number of ways to group n people in Groups of r (i.e. the total number of games required to have everyone play everyone else in one and only one game) is C(n,r) = n!/[r!(n-r)!]

I seriously recommend you think up a better tourney. For 10 people, you need 120 games.

Slick.

[Slick]

Nevermind. Forget everything I said. C(n,r) is the wrong formula for this. Sorry for the confusion.

Slick.

[Ack]

Outside of being an interesting mathematical series problem, the general question is void. It would be incredibly difficult to get 27 players in a round-robin style game. It would take months if not years for every combination to be played out. I'd recommend a bracket style where only the top one or two players advance to the next round. Which is how the NCAA tournament decides a winner in just 6 rounds out of a pool of 64 teams.

[geoschmo]

Originally posted by Ack:
Outside of being an interesting mathematical series problem, the general question is void. It would be incredibly difficult to get 27 players in a round-robin style game. It would take months if not years for every combination to be played out.

Not really. I suppose it depends on the players in the tourney though. Months I can see, but it should not take years.

27 players in games of three, where each player must play every other player once and only once. Each player only has to play 13 games total. That's a lot, but you can play them simultaneously, or at least 3 or 4 at a time. And 3 man games of SE4 go pretty quick.

But that would be a lot of games nonetheless. It would be hard to find 27 people willing to do that many probably.

Geoschmo

[Slynky]

One of the good things about chess and its rating system (see my remark about one possible way to have a rating system in SE4 in the "New League" thread) is the wonderful way tournements can be run. Using the rating of the player, you just use Swiss-System pairing and get a tourney done in 4-5 rounds...considering anywhere from 15-30 people.