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August 7th, 2003, 02:18 AM
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Re: Math problem
geo you should grab one of those wheel systems for lotteries.
you need a 3 of n wheeler with a filter
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Hey GUTB where did you go...???
He is still driving his mighty armada at 3 miles per month along the interstellar highway bypass and will be arriving shortly
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August 7th, 2003, 02:19 AM
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Re: Math problem
Quote:
Originally posted by 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
...
If Gp and Tg come out as positive integers, it should be doable - I'm not sure about the arrangement, however.
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I think this is the same as what I had a few Posts back, or am I missing something?
Edit: Oh sure add more
Quote:
Originally posted by Jack Simth:
Edit: Arrangement method:
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
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Looks promising, but I don't understand exactly what you mean by group, skipping sk. Could you show the Np=13 and Pp=3 case I mentioned earlier as an example (since no one has shown it yet)?
[ August 07, 2003, 01:56: Message edited by: cybersol ]
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August 7th, 2003, 02:23 AM
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National Security Advisor
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Re: Math problem
Quote:
Originally posted by tesco samoa:
geo you should grab one of those wheel systems for lotteries.
you need a 3 of n wheeler with a filter
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This is exactly what he needs! Some sort of calculator that will list all the possible combinations of numbers (i.e. players) for n number of players, and r number of players per game!
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August 7th, 2003, 02:45 AM
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Re: Math problem
A few people posted while I was editing, so I'll put it back up:
Quote:
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
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I might have an off by one error in line 7.
[ August 07, 2003, 09:16: Message edited by: Jack Simth ]
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Of course, by the time I finish this post, it will already be obsolete. C'est la vie.
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August 7th, 2003, 02:46 AM
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Re: Math problem
Geoschmo is correct that you cannot use a combination equation for a round robin. This is from some website:
Example:
How many ways can we select three letters from the letters of RSTUV?
n = 5 r = 3
These are: RST, RSU, RSV, RTU, RTV, RUV, STU, STV, SUV and TUV.
From this you can see that, if RSTUV represented players, players would meet more than once (RS for example).
Permutation equations do not work for exactly the same reason.
Geoschmo,
I believe the only way you can avoid having players meet more than once is if the number of players is the square of the number of players per game. Actually another case is if the number of players is equal to the number of players per game (or 1 game). Anything other than this and you'll either having players facing each other multiple times or rounds where players do not play.
Next, the most possible games occurs if the number of players per game is equal to one. In this case the number of games is the summation of A (from A=1 to A= n-1). Where n is the total number of players. This is an important event.
So adjusting for the number of players, which should be a simple division, gives this:
Pg = players per game
n = total number of players
# games = (1/Pg) Summation (A=1 to A=n-1)
Solving a few cases:
Pg = 1, n = 1: # games = 1
Pg = 4, n = 2: # games = 3 (wrong, should be 6)
Pg = 9, n = 3: # games = 12
In summations, having variation between the odd and even entries is fairly common. So you need another equation for the even values, which I'm not going to work out tonight.
I suspect there is a more graceful solution by using some series expansions. Try looking at Taylor, Binomial, Geometric, etc. Series Expansions to find a better solution. To help you along, try determining the number of games required for 25-5 and 36-6.
One more quick thing. The number of games per round is simply n/Pg. So another option would be to use this and find a series that describes the number of rounds.
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August 7th, 2003, 02:48 AM
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Re: Math problem
i posted this earlier
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
And in VB and send me the source 
__________________
RRRRRRRRRRAAAAAGGGGGGGGGHHHHH
old avatar = http://www.shrapnelgames.com/cgi-bin...1051567998.jpg
Hey GUTB where did you go...???
He is still driving his mighty armada at 3 miles per month along the interstellar highway bypass and will be arriving shortly
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August 7th, 2003, 02:49 AM
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General
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Join Date: Jul 2001
Location: Canada
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Re: Math problem
i posted this earlier
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
And in VB and send me the source 
__________________
RRRRRRRRRRAAAAAGGGGGGGGGHHHHH
old avatar = http://www.shrapnelgames.com/cgi-bin...1051567998.jpg
Hey GUTB where did you go...???
He is still driving his mighty armada at 3 miles per month along the interstellar highway bypass and will be arriving shortly
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