Won match: 10 points

Draw:
5 points

Lost match: 0 points

Both teams get 1 point for each goal they score.

A system of scores and a fair tournament plan is essential to stimulate the trainees by the competitive element. The most common number of players will be 5, 6, 7 and 8.

The following sections contain tournament plans for n = 3, 4,..., 9. A little theory is given as well.

round | out | |

B C E I - D F G H | ||

A D F I - C E G H | ||

A D E G - B F H I | ||

B C F G - A E H I | ||

A C F H - B D G I | ||

B D E H - A C G I | ||

C D H I - A B E F | ||

E F G I - A B C D | ||

A B G H - C D E F |

**Theorem:** If q = 2n+1 is an odd prime power, there exists a
(2n+1,n,n-1)
BIBD with pair-wise disjunct sets.

**Proof:** Let c be the non-trivial character
on the multiplicative group (F_q)* of order two on the Galois field F_q. It takes the
value -1 n times, and the value +1 n times. In The members of (F_q)* are
used to identify the rounds. In round a, the teams consist of the elements
of F_q having the same value of c(x-a). The
value c(x-a) = 0 means that x=a is sitting out
in round a.

One must just show that each pair occurs n - 1 times. Choose an arbitrary pair, a,b Î F_q. They are on the same team in round x, iff c(x-a) = c(x-b). Using the multiplicativity of the character, this may be written

c((x-a)/(x-b)) = c(x-a) / c(x-b) = 1.

Consider the map: x mapsto (x-a)/(x-b). It is 1-1 and maps F_q - {b} onto F_q - {1}. Because c(1) = 1, there are n-1 rounds with a and b playing on the same side.

l = r(n-1)/(2n-1).

The greatest common factor (n-1,2n-1) = 1, so r must be a multiple of 2n-1. In this case, r is equal to the number of rounds. This means that lambda must be a multiple of n - 1.

1 | AB - CD |

2 | AC - BD |

3 | AD - BC |

There is no resolvable (6,3,2) BIBD. But in [1], one may find a non-resolvable (6,3,2) BIBD. It leads readily to a resolvable (2n,n,2n-2) BIBD.

**Proposition**: Assume that a (2n,n,n-1) BIBD exists. Then a resolvable
(2n,n,2n-2) BIBD exists.

**Proof:** The sets together with
their complements form a BIBD of the required form. Choose two
elements, a and b. We will compute the number of times none of
them occurs in the original BIBD. They each occur r = 2n - 1
times, but l = n - 1 times together. At
least one of them occurs in

of the sets. Thus none of them occurs in

of the sets. Thus a and b occur together in n-1 of the complements.

That way one obtains the following tournament plan.

1 | ABC - DEF |

2 | ABD - CEF |

3 | ACE - BDF |

4 | ADF - BCE |

5 | AEF - BCD |

6 | BCF - ADE |

7 | BDE - ACD |

8 | BEF - ACD |

9 | CDE - ABF |

10 | CDF - ABE |

[1] gives an non-resolvable (8,4,3) BIBD, which would lead to a tournament with 14 rounds. In the search of a solution with only 7 rounds, it was discovered that theorem 2.2.7 of [1] solves the problem, because the construction leads to an obviously resolvable design. The theorem is stated here in the form, in which we need it.

**Theorem:** If n is even, and 2n - 1 is a prime power, then a resolvable
(2n,n,n.-1) BIBD exists.

1 | ABCE - DFGH |

2 | BCDF - AEGH |

3 | CDEG - ABFH |

4 | DEFA - BCGH |

5 | DEFB - CDHA |

6 | FGAC - BDEH |

7 | GABD - CEFH |

Ian Anderson has commented that the result above may be improved upon. He wrote:

The "theorem about resolvable (2n,n,n-1) BIBD whenever n is even and 2n-1 prime can have this last condition replaced by the existence of a Hadamard matrix of order 2n. If you look at Corollary 6.3.5 in my book you will see that we take the complements of the blocks of a (4m-1,2m-1,m-1) design and adjoin a new element to the original blocks to obtain a design as required which actually has the extra property that every TRIPLE of elements occurs in the same number of blocks. This is a more general result than your condition about prime powers gives, since Hadamard designs exist in the prime power cases but in lots of other cases as well. In fact, a resolvable (4n,2n,2n-1) design is eqivalent to a Hadamard matrix of order 4n."