MAS334 Combinatorics

13. Tournaments

Definition 13.1.

Consider a finite set P. A tournament on P is a subset TP×P such that

  • (a)

    No pair of the form (a,a) lies in T.

  • (b)

    For any two players ab, either (a,b)T or (b,a)T but not both.

Remark 13.2.

We interpret this as follows. The set P could be a set of players, who play some game against each other in pairs, with each pair playing precisely once. Then T is the set of pairs (a,b) such that a beats b.

Example 13.3.

Here are three different ways to represent the result of a tournament with players A,,E:

The left hand table shows who wins each game. For example, in the (A,D) position (i.e. the row marked A and the column marked D) we see a W, indicating that A wins against D. Correspondingly, we have an L in the (D,A) position, indicating that D loses against A. 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.

Remark 13.4.

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.

Example 13.5.

We say that a tournament T is consistent if the players can be listed as a1,,an in such a way that ai beats aj whenever i<j. (In particular, a1 beats everyone else and so is the “number one player” in the usual sense. Similarly, a2 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.

Definition 13.6.

A winning line for a tournament TA×A is a list a1,,an containing each player exactly once, such that ai beats ai+1 for i=1,,n-1.

Remark 13.7.

Although a1 beats a2 and a2 beats a3, we are not assuming here that a1 beats a3. Thus, having a winning line is much weaker than having a consistent ranking as in Example 13.5.

Example 13.8.

In Example 13.3, the sequence C,B,D,E,A is a winning line. This is most easily seen using the arrow graph:

Note that the effect mentioned in Remark 13.7 appears here: C beats B and B beats D, but C does not beat D.

Proposition 13.9.

Every tournament has a winning line.

Proof.

This is clear if there are 0, 1 or 2 players. Suppose instead that the set of players is P, with |P|=n>2, and argue by induction on n. Choose a player a*, and put P=P{a*}. We can apply the induction hypothesis to P, and thus list the elements of P as a1,,an-1, in such a way that ai beats ai+1 for i=1,,n-2. If a* does not beat any of these players, then a1,,an-1,a* is a winning line for T. Suppose instead that a* does beat some of the players ai. Let ak be the first one that a* beats. If k=1, then a*,a1,,an-1 is a winning line for T. Suppose instead that k>1 (so ak-1 is meaningful). As ak is the first player that a* beats, we see that ak-1 must beat a*. It follows that the sequence a1,,ak-1,a*,ak,,an-1 is a winning line. ∎

Definition 13.10.

The score of a player in a tournament is the number of games that they win. (More formally, the score of aA with respect to a tournament TA×A is |{bA|(a,b)T}|.) The score sequence of T is the list of scores of all players, written in decreasing order. We write scores(T) for the score sequence of T.

Example 13.11.

In Example 13.11, the scores of A, B, C, D and E are 2,2,3,2 and 1. The score sequence is therefore 3,2,2,2,1. Consider instead a consistent tournament (as in Example 13.5) with n players. Then player 1 beats players 2,,n and so scores n-1, and player 2 beats players 3,,n and so scores n-2, and so on. The last player (number n) beats no one and so scores zero. The score sequence is therefore n-1,n-2,,1,0.

Example 13.12.

Given m>1, we can construct a tournament with 2m+1 players as follows. The set of players is /(2m+1), so all expressions with player numbers must be interpreted modulo 2m+1. For each i/(2m+1), player i beats players i+1,,i+m, and is beaten by players i-1,,i-m. We call this the odd modular tournament of size 2m+1. For example, when m=2 (so 2m+1=5) we have the following pattern:

Note that in this kind of tournament, each player has a score of m. 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.

Definition 13.13.

Consider a sequence s=(s1,,sn) of nonnegative integers with s1sn. We define

length(s) =n
firstk(s) =s1+s2++sk=1iksi
lastk(s) =sn-k+1+sn-k+2++sn=n-k<insi
total(s) =firstn(s)=lastn(s)=1insi.

More generally, given any subset U{1,,n}, we put sU=iUsi.

Definition 13.14.

We say that a sequence s is realisable if there is a tournament T with scores(T)=s. Any such tournament is a realisation of s.

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:

Lemma 13.15.

If s is a realisable sequence of length n, then total(s)=(n2).

Proof.

As s is realisable, we can find a tournament with score sequence s. Put m=isi. 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 (n2). ∎

Another basic observation is as follows:

Lemma 13.16.

If a sequence is realisable, then it cannot contain two zeros.

Proof.

Suppose we have a tournament T, and two distinct players a and b. Then a and b 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 scores(T) contains at most one zero. Thus, if we have a sequence with two or more zeros, it cannot be realisable. ∎

Video (Lemma 13.17 to Lemma 13.20)

To formulate Landau’s Theorem, it will be convenient to have the following result.

Lemma 13.17.

For 0kn we have (n2)=(k2)+k(n-k)+(n-k2).

Algebraic proof.
(k2)+k(n-k)+(n-k2) =12k(k-1)+k(n-k)+12(n-k)(n-k-1)
=12k2-12k+nk-k2+12n2-nk+12k2-12n+12k
=12n2-12n=12n(n-1)=(n2).

Bijective proof.

Put N={1,,n}, and let KN be any subset of size k, so NK has size n-k. Then

  • (n2) is the number of subsets TN with |T|=2.

  • (k2) is the number of such subsets T with both elements of T in K.

  • k(n-k) is the number of such subsets T with one element in K and the other in NK.

  • (n-k2) is the number of such subsets T with both elements of T in NK.

From this it is clear that (n2)=(k2)+k(n-k)+(n-k2). ∎

Proposition 13.18.

Let s=(s1,,sn) be a sequence of nonnegative integers in descending order, and suppose that total(s)=(n2). Put N={1,,n}. Then the following conditions are equivalent:

  • (a)

    For all UN with |U|=k we have sU(k2).

  • (b)

    For all UN with |U|=k we have sUk(n-k)+(k2).

  • (c)

    For all k we have lastk(s)(k2).

  • (d)

    For all k we have firstk(s)k(n-k)+(k2).

Proof.

First suppose that (a) holds; we will prove that (b) also holds. Indeed, if UN with |U|=k, then |Uc|=n-k, so (a) tells us that sUc(n-k2), so

(n2)-sUc(n2)-(n-k2).

On the other hand, we have sU+sUc=total(s)=(n2), so (n2)-sUc=sU. Moreover, Lemma 13.17 shows that (n2)-(n-k2)=k(n-k)+(k2). Thus, the above inequality can be rewritten as sUk(n-k)+(k2), 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 sU for subsets U with |U|=k. Note that firstk(s) is one of these numbers (for U={1,,k}) and lastk(s) is also one of these numbers (for U={n-k+1,,n}). In fact, since the numbers si are in decreasing order, it is clear that firstk(s) is the largest of these numbers sU, and lastk(s) is the smallest. Thus, if lastk(s)(k2), then sU(k2) for all U with |U|=k. Similarly, if firstk(s)k(n-k)+(k2), then sUk(n-k)+(k2) for all U with |U|=k. This shows that (a) is equivalent to (c) and (b) is equivalent to (d). ∎

Definition 13.19.

We say that a sequence s is plausible if it has the equivalent conditions described in Proposition 13.18.

The easy half of Landau’s result is as follows.

Lemma 13.20.

Every realisable sequence is plausible.

Proof.

Let s be a realisable sequence of length n. As s is realisable, we can find an n-player tournament T such that si is the score of player i. Let U be a set of k players, so that sU is the total of their scores. We then have sU=p+q, where p is the total of scores that people in U earn by playing each other, and q is the total of scores that they earn by playing other people. The members of U play (k2) games against each other, and each of these earns a score of 1 for one or the other of the players, so we get p=(k2). On the other hand, the members of U play k(n-k) games against people who are not in U. They might lose all of these games, or they might win all of them, or something in between; so 0qk(n-k). From this it follows that

(k2)sUk(n-k)+(k2),

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. ∎

Example 13.21.

For a consistent tournament of size n we have si=n-i for all i, and it is not hard to deduce that lastk(s)=(k2) and firstk(s)=k(n-k)+(k2), so the score sequence is plausible, as predicted by Lemma 13.20. For an odd modular tournament of size 2m+1, we have si=m for all i. Thus, for any set U with |U|=k, we have sU=mk. On the other hand, we have 2m+1=nk so m(k-1)/2 so mkk(k-1)/2=(k2). This means that sU(k2), so again the score sequence is plausible, as predicted by Lemma 13.20. Finally, the score sequence for Example 13.3 was s=(3,2,2,2,1). This has

last1(s)=1=10=(12)last2(s)=2+1=31=(22)last3(s)=2+2+1=53=(32)last4(s)=2+2+2+1=76=(42)last5(s)=3+2+2+2+1=10=10=(52).

Once again, we see that the score sequence is plausible.

The hard part is the converse:

Theorem 13.22 (Landau’s Tournament Theorem).

Every plausible sequence is realisable.

Proof.

Let s=(s1,,sn) be a plausible sequence, so s1sn and s1++sk(k2) for all k, and s1++sn=(n2). We will set up a kind of team allocation problem, to which we can apply Proposition 12.13. Consider a pair of players, say {Alice,Bob}. 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 x1,,xN, where N=(n2). For the medal ceremony, we recruit children c1,,cN, and give medal xj to child cj. 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 i must win si games, so they must receive si medals, so they need a team of si children to present medals to them. These children must be qualified, in the sense that they must carry a medal with player i’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 U of players, and put k=|U|. Let CU be the set of candidates for the corresponding jobs. Equivalently, CU is the set of children who carry a medal that could be presented to one of the players in U. There are (k2) medals where both names are in U, and k(n-k) medals where one name is in U and the other is not; thus, we have |CU|=(k2)+k(n-k). The total number of children needed to present medals to players in U is iUsi=sU. Thus, the numerical condition in Proposition 12.13 is that (nk)+k(n-k)sU for all U. 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. ∎

Example 13.23.

Consider the sequence s=(5,3,2,2,2,1), of length 6. We have

1=10=(12)2+1=31=(22)2+2+1=53=(32)2+2+2+1=76=(42)3+2+2+2+1=1010=(52)5+3+2+2+2+1=15=15=(62)

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:

Example 13.24.

For the sequence s=(4,4,4,2,1,1,1) we have last4(s)=5, which is strictly less than (42)=6, so the sequence is not plausible, so it cannot be the score sequence for a tournament.

We next discuss two ways of combining tournaments.

Video (Definition 13.25 to Example 13.31)

Definition 13.25.

Let T be a tournament with n players and scores t1tn. Let U be another tournament, with m players and scores u1um. We then let T:U denote the combined tournament with n+m players given by the following rules:

  • (a)

    For a game between two players in T, the result is the same as it was in T.

  • (b)

    For a game between two players in U, the result is the same as it was in U.

  • (c)

    For a game with one player from T and one player from U, the player from T always wins.

Note that the players from U have the same score in the combined tournament as they did in U. However, the players from T have their original score plus an extra m points for beating all the players from U. In particular, the players from T all have at least m points, but the players from U all have less than m points. Thus, the score sequence for the combined tournament is

t1+mt2+mtn+mu1um.
Remark 13.26.

You can think of T as the Sheffield primary school football league, and U 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 T:U.

Example 13.27.

Suppose we want to construct a tournament with score sequence (7,7,7,7,7,2,2,2,2,2). We start with an odd modular tournament T with players (1,2,3,4,5) and scores (2,2,2,2,2). We then let U be another copy of T, with players (6,7,8,9,10) and scores (2,2,2,2,2) again. The combined tournament T:U then has scores (7,7,7,7,7,2,2,2,2,2) as required.

Definition 13.28.

Again let T and U be two tournaments, with n and m players respectively. We can combine them in a different way to produce a tournament T*U as follows. The players are pairs (i,j), where i is a player from T and j is a player from U, so there are nm players altogether. Consider two players (i,j) and (i,j) with (i,j)(i,j), so either ii or (i=i and jj).

  • (a)

    If ii then the result of (i,j) playing (i,j) in T*U is the same as the result of i playing i in T.

  • (b)

    If i=i and jj then the result of (i,j) playing (i,j) in T*U is the same as the result of j playing j in U.

Remark 13.29.

You can think of T as a popularity contest between various potential birthday presents, and U as a popularity contest between various kinds of fancy bags. Then T*U 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.

Lemma 13.30.

Suppose that the scores in T are t1tn and the scores in U are u1um. Then the score for player (i,j) in T*U is mti+uj.

Proof.

Let Pi be the set of players beaten by i in T, and let Qj be the set of players beaten by j in U. We then have |Pi|=ti and |Qj|=uj. The players beaten by (i,j) are then the players (i,j) with either iPi and j arbitrary, or i=i and jQj. There are mti possibilities of the first type, and uj possibilities of the second type. This shows that the score for (i,j) is mti+uj as claimed. ∎

Example 13.31.

Let T be an odd modular tournament of size n=3, so the scores are (1,1,1), or in other words ti=1. Let U be a consistent tournament of size m=5, so the scores are (4,3,2,1,0), or in other words uj=5-j. In T*U, the score for (i,j) is mti+uj=5+5-j=10-j. Each number appears three times, because there are three possible choices for i, so the full score sequence is

(9,9,9,8,8,8,7,7,7,6,6,6,5,5,5).

We could instead consider the tournament U*T. Here the score for (j,i) is nuj+ti=3(5-j)+1=16-3j. The full score sequence is

(13,13,13,10,10,10,7,7,7,4,4,4,1,1,1).