Consider a finite set . A tournament on is a subset such that
No pair of the form lies in .
For any two players , either or but not both.
We interpret this as follows. The set could be a set of players, who play some game against each other in pairs, with each pair playing precisely once. Then is the set of pairs such that beats .
Here are three different ways to represent the result of a tournament with players :
The left hand table shows who wins each game. For example, in the position (i.e. the row marked and the column marked ) we see a , indicating that wins against . Correspondingly, we have an in the position, indicating that loses against . In the middle, we list all ten possible pairs of players, with the winner marked bold and in red. On the right, we have an arrow between each pair of players, pointing from the winner to the loser.
Here is yet another way of representing the same information. We imagine making a medal for each pair of players, and giving it to the winner of the corresponding match. If we know each player’s collection of medals, then we know all the results of the tournament. The tournament in Example 13.3 gives the following medal collections.
We say that a tournament is consistent if the players can be listed as in such a way that beats whenever . (In particular, beats everyone else and so is the “number one player” in the usual sense. Similarly, is the “number two player” and so on.) This is the kind of tournament that we expect when the players have consistently different levels of skill, and the better player always wins, with no effect of randomness or anything else. In a realistic tournament, things will usually be more complicated than this.
A winning line for a tournament is a list containing each player exactly once, such that beats for .
Although beats and beats , we are not assuming here that beats . Thus, having a winning line is much weaker than having a consistent ranking as in Example 13.5.
Every tournament has a winning line.
This is clear if there are , or players. Suppose instead that the set of players is , with , and argue by induction on . Choose a player , and put . We can apply the induction hypothesis to , and thus list the elements of as , in such a way that beats for . If does not beat any of these players, then is a winning line for . Suppose instead that does beat some of the players . Let be the first one that beats. If , then is a winning line for . Suppose instead that (so is meaningful). As is the first player that beats, we see that must beat . It follows that the sequence is a winning line. ∎
The score of a player in a tournament is the number of games that they win. (More formally, the score of with respect to a tournament is .) The score sequence of is the list of scores of all players, written in decreasing order. We write for the score sequence of .
In Example 13.11, the scores of , , , and are and . The score sequence is therefore . Consider instead a consistent tournament (as in Example 13.5) with players. Then player beats players and so scores , and player beats players and so scores , and so on. The last player (number ) beats no one and so scores zero. The score sequence is therefore .
Given , we can construct a tournament with players as follows. The set of players is , so all expressions with player numbers must be interpreted modulo . For each , player beats players , and is beaten by players . We call this the odd modular tournament of size . For example, when (so ) we have the following pattern:
Note that in this kind of tournament, each player has a score of . This is in some sense opposite to the case of a consistent tournament: the scores give no reason to think that any player is better than any other player.
We next want to investigate some properties of score sequences.
Consider a sequence of nonnegative integers with . We define
More generally, given any subset , we put .
We say that a sequence is realisable if there is a tournament with . Any such tournament is a realisation of .
Our main aim in this section is to prove a theorem of Landau, which will tell us exactly which sequences are realisable. The following simple example illustrates the key ingredients:
The most basic property is as follows:
If is a realisable sequence of length , then .
As is realisable, we can find a tournament with score sequence . Put . This is the sum of the scores of all players. Each game contributes a score of one to one or the other player, so the sum of all scores is just equal to the number of games. We have one game between each pair of players, so the number of games is the number of pairs, which is . ∎
Another basic observation is as follows:
If a sequence is realisable, then it cannot contain two zeros.
Suppose we have a tournament , and two distinct players and . Then and play each other, and one of them must win, thereby gaining a score of one; so they cannot both have a score of zero. Thus, the score sequence contains at most one zero. Thus, if we have a sequence with two or more zeros, it cannot be realisable. ∎
To formulate Landau’s Theorem, it will be convenient to have the following result.
For we have .
∎
Put , and let be any subset of size , so has size . Then
is the number of subsets with .
is the number of such subsets with both elements of in .
is the number of such subsets with one element in and the other in .
is the number of such subsets with both elements of in .
From this it is clear that . ∎
Let be a sequence of nonnegative integers in descending order, and suppose that . Put . Then the following conditions are equivalent:
For all with we have .
For all with we have .
For all we have .
For all we have .
First suppose that (a) holds; we will prove that (b) also holds. Indeed, if with , then , so (a) tells us that , so
On the other hand, we have , so . Moreover, Lemma 13.17 shows that . Thus, the above inequality can be rewritten as , as required. This completes the proof that (a) implies (b), and the whole argument can be reversed in a straightforward way to prove that (b) implies (a), so (a) and (b) are equivalent.
Now consider all the numbers for subsets with . Note that is one of these numbers (for ) and is also one of these numbers (for ). In fact, since the numbers are in decreasing order, it is clear that is the largest of these numbers , and is the smallest. Thus, if , then for all with . Similarly, if , then for all with . This shows that (a) is equivalent to (c) and (b) is equivalent to (d). ∎
We say that a sequence is plausible if it has the equivalent conditions described in Proposition 13.18.
The easy half of Landau’s result is as follows.
Every realisable sequence is plausible.
Let be a realisable sequence of length . As is realisable, we can find an -player tournament such that is the score of player . Let be a set of players, so that is the total of their scores. We then have , where is the total of scores that people in earn by playing each other, and is the total of scores that they earn by playing other people. The members of play games against each other, and each of these earns a score of for one or the other of the players, so we get . On the other hand, the members of play games against people who are not in . They might lose all of these games, or they might win all of them, or something in between; so . From this it follows that
so conditions (a) and (b) in Proposition 13.18 are satisfied. We have shown that conditions (a) to (d) are all equivalent, so in fact they are all satisfied. ∎
For a consistent tournament of size we have for all , and it is not hard to deduce that and , so the score sequence is plausible, as predicted by Lemma 13.20. For an odd modular tournament of size , we have for all . Thus, for any set with , we have . On the other hand, we have so so . This means that , so again the score sequence is plausible, as predicted by Lemma 13.20. Finally, the score sequence for Example 13.3 was . This has
Once again, we see that the score sequence is plausible.
The hard part is the converse:
Every plausible sequence is realisable.
Let be a plausible sequence, so and for all , and . We will set up a kind of team allocation problem, to which we can apply Proposition 12.13. Consider a pair of players, say . As in Remark 13.4, we can imagine making a medal which says “Alice vs Bob” (but does not say who won). Alice or Bob could receive this medal, but no one else could. We can repeat this for every pair of players, giving a set of medals , where . For the medal ceremony, we recruit children , and give medal to child . To specify the tournament, we just need to specify which children present medals to each player (because this determines who gets each medal, and thus who wins each game). Player must win games, so they must receive medals, so they need a team of children to present medals to them. These children must be qualified, in the sense that they must carry a medal with player ’s name on it. If we can show that this team allocation problem is solvable, this will give a tournament with the required scores. For this, we use the numerical condition in Proposition 12.13. Consider a set of players, and put . Let be the set of candidates for the corresponding jobs. Equivalently, is the set of children who carry a medal that could be presented to one of the players in . There are medals where both names are in , and medals where one name is in and the other is not; thus, we have . The total number of children needed to present medals to players in is . Thus, the numerical condition in Proposition 12.13 is that for all . This is is precisely the same as the plausibility condition from Proposition 13.18, and we are assuming that that condition is satisfied. Thus, Proposition 12.13 tells us that the team allocation problem is solvable, as required. ∎
Consider the sequence , of length . We have
This shows that the sequence is plausible, so Landau’s theorem tells us that it is possible to find a corresponding tournament. In fact, the following tournament works:
For the sequence we have , which is strictly less than , so the sequence is not plausible, so it cannot be the score sequence for a tournament.
We next discuss two ways of combining tournaments.
Let be a tournament with players and scores . Let be another tournament, with players and scores . We then let denote the combined tournament with players given by the following rules:
For a game between two players in , the result is the same as it was in .
For a game between two players in , the result is the same as it was in .
For a game with one player from and one player from , the player from always wins.
Note that the players from have the same score in the combined tournament as they did in . However, the players from have their original score plus an extra points for beating all the players from . In particular, the players from all have at least points, but the players from all have less than points. Thus, the score sequence for the combined tournament is
You can think of as the Sheffield primary school football league, and as the premier league. One year they decide to have a joint tournament for charity, and of course the professionals always let the children win; the resulting tournament is .
Suppose we want to construct a tournament with score sequence . We start with an odd modular tournament with players and scores . We then let be another copy of , with players and scores again. The combined tournament then has scores as required.
Again let and be two tournaments, with and players respectively. We can combine them in a different way to produce a tournament as follows. The players are pairs , where is a player from and is a player from , so there are players altogether. Consider two players and with , so either or ( and ).
If then the result of playing in is the same as the result of playing in .
If and then the result of playing in is the same as the result of playing in .
You can think of as a popularity contest between various potential birthday presents, and as a popularity contest between various kinds of fancy bags. Then is the corresponding popularity contest between presents-in-bags, where we assume that the recipient mostly cares about the present, and only compares the bags if the presents are the same.
Suppose that the scores in are and the scores in are . Then the score for player in is .
Let be the set of players beaten by in , and let be the set of players beaten by in . We then have and . The players beaten by are then the players with either and arbitrary, or and . There are possibilities of the first type, and possibilities of the second type. This shows that the score for is as claimed. ∎
Let be an odd modular tournament of size , so the scores are , or in other words . Let be a consistent tournament of size , so the scores are , or in other words . In , the score for is . Each number appears three times, because there are three possible choices for , so the full score sequence is
We could instead consider the tournament . Here the score for is . The full score sequence is