12 players, 10 venues tricky

[hassehulabeck]

Dear greater thinker,

I'm organizing a pinball competition. 12 players will face each other, playing on 10 pinball machines. To get a scheme for all 12 to play against each other isn't hard, but to let them play only once on each machine is a bit trickier. If anyone can help, I'll be delighted.

Twelve players: A-M ("I" not used because of the audible similarity with "J" in swedish)
Ten pinball machines: 1-10

And actually, for special reasons, each player will only have ten games, leaving one out for a full round robin, but don't mind about that.

/Hans.

[Ian Wakeling]

Dear Hans,

I think you may be looking for a schedule like this:

(4 5)  (- -)  (0 x)  (- -)  (- -)  (8 2)  (7 y)  (3 6)  (- -)  (9 1)
(0 2)  (5 6)  (- -)  (1 x)  (- -)  (- -)  (9 3)  (8 y)  (4 7)  (- -)
(- -)  (1 3)  (6 7)  (- -)  (2 x)  (- -)  (- -)  (0 4)  (9 y)  (5 8)
(6 9)  (- -)  (2 4)  (7 8)  (- -)  (3 x)  (- -)  (- -)  (1 5)  (0 y)
(1 y)  (7 0)  (- -)  (3 5)  (8 9)  (- -)  (4 x)  (- -)  (- -)  (2 6)
(3 7)  (2 y)  (8 1)  (- -)  (4 6)  (9 0)  (- -)  (5 x)  (- -)  (- -)
(- -)  (4 8)  (3 y)  (9 2)  (- -)  (5 7)  (0 1)  (- -)  (6 x)  (- -)
(- -)  (- -)  (5 9)  (4 y)  (0 3)  (- -)  (6 8)  (1 2)  (- -)  (7 x)
(8 x)  (- -)  (- -)  (6 0)  (5 y)  (1 4)  (- -)  (7 9)  (2 3)  (- -)
(- -)  (9 x)  (- -)  (- -)  (7 1)  (6 y)  (2 5)  (- -)  (8 0)  (3 4)

I have used 0 to 9 and x and y as the symbols for the 12 players so you will have to convert these back to A-M.  All 12 players should appear exactly once in each row and in each column, so rows can be rounds, and columns can be machines, or vice versa.  If you look down the diagonals, the cyclic construction of the schedule should be visible.

Ian.

[hassehulabeck]

Ah, very good. Unfortunately, I was a bit sloppy in my information, as I left out an important bit. I noticed this just now, when I used your solution.

The problem is, 12 players will face each other on ten machines. In each round, six matches will be played on six of the ten machines. What I would like to do, is to use a schedule that doesn't put a player on each machine more than once (and just now I was aware that I need 11 machines for that. Stupid me.)
I believe that this would be possible, but the really tricky thing is to make sure that machine x isn't used twice in any round, as this would cause a little disturbance.

So, to clear it. A single round robin for 12 players, 11 machines and 11 rounds. No player should play on the same machine twice, and no machine should be used twice in any round. Is this possible?

And of course, thanks for your effort Ian. Great thinking for some dull instructions.

[hassehulabeck]

Oh, wait a minute. I was stubborn enough to try and mix your solution with my already plotted schedule. And that didn't work, but yours actually does very well.

May I ask if there's a *simple* method that you've used for this?

[Ian Wakeling]

Hans,

Well it's simple if you have the first row of the schedule, then the rest of it can be generated fairly easily.  These schedules are called Howell designs and the mathematics that you need to find a suitable first row for a cyclic schedule is not so simple.

I used the attached Excel workbook that I put together a while ago,  it has a number of constructions that you can scroll through by clicking the arrow button.  Where it says H(a,b) for a schedule, a is the number of venues (machines) and b is the number of players.

Hope that helps.

[hassehulabeck]

Thank you. Really nice file, it will be useful for years.
I tried to search for Howell design, but couldn't find anything at first. Would be fun to see if I could understand the maths beneath the schedules.

[Ian Wakeling]

If you look here on Wolfram MathWorld then there is a definition of a Howell design and also a link to the reference that I used.