-tournaments (multitournaments) are directed graphs in which each pair of vertices is connected with exactly arcs.
Reid formulated the following conjecture in [ 233 ].
Using Landau's theorem this conjecture can be formulated in the following arithmetic form too.
In this case we say that the sequence realizes the sequence or is a solution for .
Reid gave a constructive proof of his conjecture for sets containing one, two or three elements [ 233 ].
A score set is called k-unique, if there exists exactly 1 score sequence of -tournaments generating the given set. In the talk we investigate the following questions:
characterization of the unique score sets of 1-tournaments;
extension of the Reid's conjecture to 2-tournaments.
At first we formulate a useful necessary condition.
and in the case we have
The proof is left an exercise (see Exercise 30.3-1). This lemma implies the exact answer for the case .
Proof. Lemma 30.10 implies that only this solution is acceptable. One can check that it satisfies the required inequality and equality.
Now we present a useful method of the investigation of the uniqueness. Let . Then according to the Reid-equality we get
But here only the values are permitted where
By substitution from (30.12) we get
Here must be an integer, so transform this formula into
In the following cases there is only one solution:
In the following case there are at least two solutions:
is odd and .
b) Uniqueness. If , then , and is integer only if . So we get the unique solution.
If , then only is permitted, implying the unique solution .
If or is odd, then we also can analyse formula (30.15).
This theorem left open the case when the difference is odd and the investigated set is sparse and also the case when the difference is an even number greater then 2.
Now we present a new form of Reid-problem for 2-tournaments.
For a fixed sequence with of positive integers, we shall denote by the set of sequences such that
Here we also say that g is a solution for q.
We wish to give necessary and sufficient conditions for to have a solution that is a nonempty .)
Let be integers for which , where . If , then and let and . Hence we have
Now assume that and there is a prime such that . In this case and we choose as follows:
It is obvious that
Finally, assume that or and is the product of distinct primes. If there are positive integers such that and , then we have and
This is impossible.
30.3-1 Prove Lemma 30.10.
30.3-2 Find the score sequence of the 1-tournament with a score set [5} and design a schedule for the corresponding tournament.