Math problem

geoschmo · #1 August 6th, 2003, 10:06 PM

By the way LGM, what is the reason for using the prime numbers in the program? I'm not a programmer so I can't really make heads or tails of your code there, not that it wasn't apprecieated.

LGM · #2 August 6th, 2003, 10:07 PM

Try X raised to the Yth power and subtract one. Just a theory. I am not sure if that works for 4 or 5 man games yet. X is Number of players per game minus 1. Incidentally, that does not work for 2 players games as any number works for 'round robin' leagues.

LGM · #3 August 6th, 2003, 10:09 PM

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 · #4 August 6th, 2003, 10:12 PM

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 · #5 August 6th, 2003, 10:16 PM

Quote:

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 · #6 August 6th, 2003, 10:21 PM

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

Slick.

Ack · #7 August 6th, 2003, 10:35 PM

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.