# A Tournament design problem

There is a need for a new tournament form both for training and teaching, where one wants to validate the good team player rather than focusing on the one scoring the goals. One may use the following scores:

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.

## Odd number of players

In case of an odd number, 2n + 1, of players, one needs a tournament plan with 2n + 1 rounds, so that each player only sits out one round. Here is the plan for 3, 5, 7 and 9 players.

### 3 players, n = 1

 round 1 2 3 side 1 B C A side 2 C A B sitting out A B C

### 5 players, n = 2

 round 1 2 3 4 5 side 1 AB DE AD BC CE side 2 CD AC BE AE BD sitting out E B C D A

### 7 players, n = 3

 round 1 2 3 4 5 6 7 side 1 ABC BEG ADE CDG AFG CEF BDF side 2 DFG ACD BCF AEF BDE ABG CEG sitting out E F G B C D A

### 9 players, n = 4

 round out team 1 - team 2 1 A B C E I - D F G H 2 B A D F I - C E G H 3 C A D E G - B F H I 4 D B C F G - A E H I 5 E A C F H - B D G I 6 F B D E H - A C G I 7 G C D H I - A B E F 8 H E F G I - A B C D 9 I A B G H - C D E F

### Theory

Presently, such a tournament plan can be made for all prime powers, 3, 5, 7, 9 = 3*3, 11 and 13, but not 15 = 3 * 5. This is achieved with a generalisation of corollary 2.2.4 of [1].

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.

## Even number of players

This problem appears more easy, because the number of rounds may be varied. But appearances deceive. In the language of [1] and the link above, the tournament plan holds a resolvable (2n,n,l) BIBD. One wants to minimise the number of rounds. With the usual notation,

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.

### 4 players, n = 2

In thise case, there exists a resolvable (4,2,1) BIBD. It leads to the tournament plan
 1 AB - CD 2 AC - BD 3 AD - BC

### 6 players, n = 3

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

(2n-1) + (2n-1) - (n-1) = 3n - 1

of the sets. Thus none of them occurs in
b - (3n-1) = 2(2n-1) - (3n-1) = n - 1

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

### 8 players

[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."

### references

1. Anderson, Ian: Combinatorial Designs and Tournaments, Oxford University Press, 1997