Math problem
 

Ok, I like math but my brain doesn't work well when you get past two dimensions. http://forum.shrapnelgames.com/images/icons/icon10.gif

I am trying to figure something out. Assume a hypothetical tournament which are all 3 player games. In the tournament everybody has to play everybody else once, but only once.

If there are three players it's easy, A,B,C. What's the next number of players that this format will work for? Can it be done with 9 or do you need 27, or is it some other number completely?

Is there an easy way to figure this out including who has to play who?

Re: Math problem
 

Ok, I figured it out the long way using graph paper. http://forum.shrapnelgames.com/images/icons/blush.gif

So I guess then it's x squared, where x is the number of players per game. Not x to the power of x. So with 4 player games it would take 16 people in the tourney and 5 player games would take 25? Does that sound right? Is there a way to verify that without using graph paper? http://forum.shrapnelgames.com/images/icons/shock.gif

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution http://forum.shrapnelgames.com/images/icons/icon7.gif

Re: Math problem
 

hey geo i use that for figuring out how many tickets i need to do in betting on sport games

3 of 3 rotations ( 3 teams to 25) right up to 12 of 12 rotations

3 of 4 is 4
3 of 5 is 10
3 of 6 is 20
3 of 7 is 35
3 of 8 is 56
3 of 9 is 84
3 of 10 is 120
4 of 5 is 5
4 of 6 is 15
4 of 7 is 35
4 of 8 is 70
4 of 9 is 126
4 of 10 is 210
5 of 6 is 6
5 of 7 is 21
5 of 8 is 56
5 of 9 is 126
5 of 10 is 252
6 of 7 is 7
6 of 8 is 28
6 of 9 is 84
6 of 10 is 210
7 of 8 is 8
7 of 9 is 36
7 of 10 is 120
8 of 9 is 9
8 of 10 is 45

many different rotations...

Re: Math problem
 

The number of matches you need is n!
where n! = 1*2*3*4*...*(n-1)*(n)
and is called factorial and can be used to determine the numbers of permutations of n tokens.

In a 3 players match where anyone is playing a role only once is 3! (1*2*3 = 6)

Empirical solution:
123
213
312
321
132
231

If I'm not right you can spank me with a crowbar

Re: Math problem
 

umm mine was total games that would be needed to play each other in every combination...

ignore

Re: Math problem
 

Spank me.

Is not right, every player plays two times in the same role.

But I'm sure that the number will be a subset of n!, but I also believe that you need an even number of players to not have conflicts.

Re: Math problem
 

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

Re: Math problem
 

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.

[ August 06, 2003, 18:41: Message edited by: Erax ]

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">The problem is not every legal combination will work. For example, by using
1) 123
2) 145
3) 167
4) 189
you prevent the option of trying 146. This is fine for 1, 4, and 6 as long as they all play each other once, but might prevent others from being able to play each other, as you found out.

Instead if you do something like...
1) 123
2) 456
3) 789
4) 147
5) 158
6) 169
7) 249
8) 275
9) 268
10) 348
11) 359
12) 367

then it works.

Re: Math problem
 

When in doubt, ask a programmer (if you have one handy...) Here's the reply I got:
-----
If you want to do all x*x sized games, then any power of x will support your goal. However, be warned that in a 27-man tournament of 3x3 matchings, you are talking about 9 games apiece (3^n-1).

Scheduling is a bit of a pain but manageable. The way I would do it is to schedule x^n-1 guys for the first round, and then everything sort of falls out of that just by pairing that guy off with the next guy down the list, then 2 down the list, etc.. For 27 guys (a-z and 1) at x=3. You essentially match up randomly in the first round, and then down diagonals for each subsequent round based on the first round.

ABC
DEF
GHI
JKL
MNO
PQR
STU
VWX
YZ1

AEI
DHL
GKO
JNR
MQU
PTX
SW1
VZC
YBF

AHO
DKR
GNU
JQX
MT1
PWC
SZF
VBI
YEL

etc.

Re: Math problem
 

I used a program. Here are two solutions:

N=7
3,5,6
3,4,7
2,5,7
2,4,6
1,6,7
1,4,5
1,2,3

N=15
7,11,12
7,10,13
7,9,14
7,8,15
6,11,13
6,10,12
6,9,15
6,8,14
5,11,14
5,10,15
5,9,12
5,8,13
4,11,15
4,10,14
4,9,13
4,8,12
3,13,14
3,12,15
3,9,10
3,8,11
3,5,6
3,4,7
2,13,15
2,12,14
2,9,11
2,8,10
2,5,7
2,4,6
1,14,15
1,12,13
1,10,11
1,8,9
1,6,7
1,4,5
1,2,3

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">This isn't what I was looking for. This is everybody playing 3 three man games. I wanted everybody to play EVERYBODY else once and only once in a three man game. 7 players won't work for that. You need a power of 3 as Spoon says so 9, or 27, etc.

[ August 06, 2003, 19:48: Message edited by: geoschmo ]

Re: Math problem
 

Here is another solution. The patern appears to be 2^x-1. I would suspect that the next solution is at 63, but 31 took long enough to prove and my algorythm uses prime numbers with the mod function to find matches, so I don't want to look up another 32 prime numbers to make it that far.

N=31
15,23,24
15,22,25
15,21,26
15,20,27
15,19,28
15,18,29
15,17,30
15,16,31
14,23,25
14,22,24
14,21,27
14,20,26
14,19,29
14,18,28
14,17,31
14,16,30
13,23,26
13,22,27
13,21,24
13,20,25
13,19,30
13,18,31
13,17,28
13,16,29
12,23,27
12,22,26
12,21,25
12,20,24
12,19,31
12,18,30
12,17,29
12,16,28
11,23,28
11,22,29
11,21,30
11,20,31
11,19,24
11,18,25
11,17,26
11,16,27
10,23,29
10,22,28
10,21,31
10,20,30
10,19,25
10,18,24
10,17,27
10,16,26
9,23,30
9,22,31
9,21,28
9,20,29
9,19,26
9,18,27
9,17,24
9,16,25
8,23,31
8,22,30
8,21,29
8,20,28
8,19,27
8,18,26
8,17,25
8,16,24
7,27,28
7,26,29
7,25,30
7,24,31
7,19,20
7,18,21
7,17,22
7,16,23
7,11,12
7,10,13
7,9,14
7,8,15
6,27,29
6,26,28
6,25,31
6,24,30
6,19,21
6,18,20
6,17,23
6,16,22
6,11,13
6,10,12
6,9,15
6,8,14
5,27,30
5,26,31
5,25,28
5,24,29
5,19,22
5,18,23
5,17,20
5,16,21
5,11,14
5,10,15
5,9,12
5,8,13
4,27,31
4,26,30
4,25,29
4,24,28
4,19,23
4,18,22
4,17,21
4,16,20
4,11,15
4,10,14
4,9,13
4,8,12
3,29,30
3,28,31
3,25,26
3,24,27
3,21,22
3,20,23
3,17,18
3,16,19
3,13,14
3,12,15
3,9,10
3,8,11
3,5,6
3,4,7
2,29,31
2,28,30
2,25,27
2,24,26
2,21,23
2,20,22
2,17,19
2,16,18
2,13,15
2,12,14
2,9,11
2,8,10
2,5,7
2,4,6
1,30,31
1,28,29
1,26,27
1,24,25
1,22,23
1,20,21
1,18,19
1,16,17
1,14,15
1,12,13
1,10,11
1,8,9
1,6,7
1,4,5
1,2,3

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">Ok, I see how I need a power of x, so for 4 man games I'd need at least 16 players, and for 5 man I'd need 25 players. How did you calculate nine games each for 27 players in a 3 on 3 tourney though? With 27 player field if you play everybody once and only once and do it with 3 man games that would be (27-1)/2=13 games. With a nine player player field it would be (9-1)/2=4 games each.

Geoschmo

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">This isn't what I was looking for. This is everybody playing 3 three man games. I wanted everybody to play EVERYBODY else once and only once in a three man game. 7 players won't work for that. You need a power of 3 as Spoon says so 9, or 27, etc.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I am not sure I follow. Player 1 places all 3 player games and plays every player once and only once. The same applies for player 2, 3, etc.

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">Oh that? Must be programmer error... heh. That's why you should never listen to them... http://forum.shrapnelgames.com/images/icons/icon12.gif

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">What part of the requirement did I neglect?

Re: Math problem
 

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 ]

Re: Math problem
 

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

Re: Math problem
 

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.

Re: Math problem
 

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.

Re: Math problem
 

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? http://forum.shrapnelgames.com/image...s/confused.gif

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">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

Re: Math problem
 

Here is the code. Crudely done by brute force in a rather esoteric flavor of basic:

PRIMES = '2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61 ,67,71,73,79,83,89,97,101,103,109,113,127,131,137'
SWAP ',' WITH @AM IN PRIMES
ILIMIT = DCOUNT(PRIMES,@AM)
PRINT ILIMIT
FOR I = 4 TO ILIMIT
NBR = I
COMBOS = ''
COMBOX = ''
IDX = 1
FOR J = 1 TO I
FOR K = J+1 TO I
FOR L = K+1 TO I
GOSUB FIND.DUPLICATE
IF NOT(FOUND.DUPLICATE) THEN
COMBOS<-1> = PRIMES<J>*PRIMES<K>*PRIMES<L>
COMBOX<-1> = J:',':K:',':L
END
NEXT L
NEXT K
NEXT J
GOSUB CHECK.SOLUTION
NEXT I
PRINT 'DONE'
STOP
*
SOLVED:*
PRINT '---------------------'
PRINT 'SOLVED'
PRINT NBR
FOR M = DCOUNT(COMBOX,@AM) TO 1 STEP -1
PRINT COMBOX<M>
NEXT M
INPUT CH,1_
RETURN
*
CHECK.SOLUTION: *
FOR J = 1 TO I
FOR K = J + 1 TO I
NBR.MATCHES = 0
FOR M = DCOUNT(COMBOS,@AM) TO 1 STEP -1
A.COMBO = COMBOS<M>
IF MOD(A.COMBO,PRIMES<J>)=0 AND MOD(A.COMBO,PRIMES<K>)=0 THEN
NBR.MATCHES = NBR.MATCHES + 1
IF NBR.MATCHES > 1 THEN RETURN
END
NEXT M
IF NBR.MATCHES = 0 THEN RETURN
NEXT K
NEXT J
GOSUB SOLVED
RETURN
*
FIND.DUPLICATE: *
FOUND.DUPLICATE = 0
FOR M = DCOUNT(COMBOS,@AM) TO 1 STEP -1
A.COMBO = COMBOS<M>
J.MOD = MOD(A.COMBO,PRIMES<J>)
K.MOD = MOD(A.COMBO,PRIMES<K>)
L.MOD = MOD(A.COMBO,PRIMES<L>)
IF J.MOD = 0 THEN
IF K.MOD=0 OR L.MOD=0 THEN
FOUND.DUPLICATE = 1
END
END ELSE IF K.MOD=0 AND L.MOD=0 THEN
FOUND.DUPLICATE = 1
END
NEXT M
RETURN

Re: Math problem
 

Ok, so bascially you are having the couputer run through all the possible options fast and see if a number works. Ok, that does the job but isn't really what I was looking for. Mayeb there is no simple mathematical formula for what I am looking for.

Basically I can take x players and group them into games of three players each and see fairly quickly that 7 players will work, but 5 will not. What I was hoping to find was a formula that I could plug the numbers in and be able to tell if a combination of x and y would work without having to go through all teh grouping by hand.

Geoschmo

Re: Math problem
 

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. http://forum.shrapnelgames.com/images/icons/icon10.gif

Re: Math problem
 

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.

Re: Math problem
 

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.

Re: Math problem
 

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 ]

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">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.

Re: Math problem
 

Nevermind. Forget everything I said. C(n,r) is the wrong formula for this. Sorry for the confusion. http://forum.shrapnelgames.com/image...s/rolleyes.gif

Slick.

Re: Math problem
 

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.

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">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

Re: Math problem
 

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.

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">That's allright. I got Fyron to explain to me offline how that formula worked and it is interesting. It's close, but as you say not exactly right. It looks like it's counting all possible combinations, instead of each player facing every other player once and only once.

Maybe there is no forumla and LGM's way is the solution. Or maybe there is one and we'll all get the Nobel prize for mathematics for working it out. http://forum.shrapnelgames.com/images/icons/icon7.gif

Re: Math problem
 

There is certainly a formula for it, we just do not know it. Any professional mathematicians out there? http://forum.shrapnelgames.com/images/icons/icon10.gif

Re: Math problem
 

Choose 2 from N, since order is unimportant.
On my calculator, that's the "nCr" button
For N=27, that's 351 games to be played round-robin.

<font size="2" face="Verdana, Helvetica, sans-serif">No; Player 1 goes up against all 26 other players. Similarily for the others, so that's 26 games total for each player. http://forum.shrapnelgames.com/images/icons/icon12.gif

26games*27people = 702 plays
That makes for 351 two-player games.

[ August 06, 2003, 22:11: Message edited by: Suicide Junkie ]

Re: Math problem
 

But, each player only plays the others once each, so it does not work for larger games than 2 player ones.

Re: Math problem
 

More than two players would get you problems with teaming up, I imagine.

Re: Math problem
 

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

Re: Math problem
 

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 ]

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">Yes, and it is not meant for 2-player games. http://forum.shrapnelgames.com/images/icons/icon12.gif

Re: Math problem
 

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...

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">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?

<font size="2" face="Verdana, Helvetica, sans-serif"> <font size="2" face="Verdana, Helvetica, sans-serif">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 ]

Re: Math problem
 

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. http://forum.shrapnelgames.com/images/icons/icon7.gif

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

Re: Math problem
 

<font size="2" face="Verdana, Helvetica, sans-serif">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. http://forum.shrapnelgames.com/images/icons/icon7.gif

Geoschmo

Re: Math problem
 

(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. http://forum.shrapnelgames.com/images/icons/icon12.gif
Good luck

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

Re: Math problem
 

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 ]

Re: Math problem
 

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 ]