Given N teams, how can they play a round-robin tournament, using N-1 prepared games, so that each team...

Suppose p=N-1 is a prime number, at least 7, and N is a multiple of 4.
The following strategy will work for p+1 teams, each playing p games.

Label the teams 0,1,...,p, and the games A,B,...,G.  The following
tables illustrate the process for p=7.  The left-hand table shows who
plays who in each round, and the right-hand table shows which prepared
game each pair of opponents plays.

Round                                   Game
  1    0v7 1v6 2v5 3v4                    A    0v7 1v3 4v5 2v6
  2    0v1 2v7 3v6 4v5                    B    0v1 2v4 5v6 3v7
  3    0v2 3v1 4v7 5v6                    C    0v2 3v5 6v7 4v1
  4    0v3 4v2 5v1 6v7                    D    0v3 4v6 7v1 5v2
  5    0v4 5v3 6v2 7v1                    E    0v4 5v7 1v2 6v3
  6    0v5 6v4 7v3 1v2                    F    0v5 6v1 2v3 7v4
  7    0v6 7v5 1v4 2v3                    G    0v6 7v2 3v4 1v5

By reading down the columns, it is clear how the entire table is built
up from the first row.  So we only need to say who plays who in Round
1, and which opposing teams play Game A.

In round 1, the opposing teams all add up to p (=7 here).

Let h=(p-1)/2.  (=3 here)
Let S be the set of remainders when you divide the first h square
numbers by p.  (Here, S = {1,4,9} mod 7 = {1,4,2}.)

Let m be any number from 2 to p-2 which is not in S.  (Here, m=3 or 5;
set m=3)

Let Game A be played between 0 and p, and between s and m*s (mod p) for
every s in S.

Build up the other rounds and prepared games by cycling the numbers
1,2,...,p.

There follows the skeleton of a solution for other numbers of players:

12 players
p =  11
h =  5
S =  {1,4,9,5,3}
m =  2,6,7 or 8
Round 1 : 0v11 1v10 2v9 3v8 4v7 5v6
Game A : 0v11 1v2 3v6 4v8 5v10 9v7

20 players
p =  19
h =  9
S =  {1,4,9,16,6,17,11,7,5}
m =  2,3,8,10,12,13,14 or 15
Game A : 0v19 1v2 4v8 5v10 6v12 7v14 9v18 11v3 16v13 17v15

24 players
p =  23
h =  11
S =  {1,2,3,4,6,8,9,12,13,16,18}
m =  5,7,10,11,14,15,17,19,20 or 21
Game A : 0v23 1v5 2v10 3v15 4v20 6v7 8v17 9v22 12v14 13v19 16v11 18v21

32 players
p =  31
h =  15
S =  {1,2,4,5,7,8,9,10,14,16,18,19,20,25,28}
m =  3,6,11,12,13,15,17,21,22,23,24,26,27 or 29
Game A : 0v31 1v3 2v6 4v12 5v15 7v21 8v24 9v27 10v30 14v11 16v17 18v23
         19v26 20v29 25v13 28v22

44 players
p =  43
h =  21
S =  {1,4,6,9,10,11,13,14,15,16,17,21,23,24,25,31,35,36,38,40,41}
m =  2,3,5,7,8,12,18,19,20,22,26,27,28,29,30,32,33,34,37 or 39
Game A : 0v43 1v2 4v8 6v12 9v18 10v20 11v22 13v26 14v28 15v30 16v32
         17v34 21v42 23v3 24v5 25v7 31v19 35v27 36v29 38v33 40v37 41v39

-- Michael Peake <mrp@resumix.portal.com>