1
Mathematical Modelling
Worksheet 15
Ranking A Tournament (1)
A tournament in tennis for example involving n players is a contest in which every
player plays every other. We can assume that the game involved is one in which a
draw is not possible.
1.
We can work out how many individual matches there are in a tournament.
The number of individual matches in a tournament can be modelled by considering
the set of complete graphs K n , where the vertices represent the number of matches.
 Number of edges of K n = Number of matches in a tournament
 n 1
=    nn  1
 2 2
The results could be illustrated graphically by the K n graphs.
e.g. K 5
P1
P2
P5
P4
2.
The scores can be placed
on the edges of the graph
and an arrow can be used
to indicate which player
won.
P3
We can show that when all the matches in a tournament have been played, the
players can be arranged in a sequence p1 ,....., p n such that p1 beat p2 , p2
beat p3 ,....., p n1 beat p n .
2
Suppose that all the matches involving all other players except player p j are played
first. Then player p j plays all his matches against all of the other players. There will
be two possible outcomes, either player p j wins or he loses.
So if player p j beats p1 we can place p j in front of p1 to get the sequence
p j , p1 , p 2 ,....., p n .
Or considering if player p1 beats p j we can place p1 in front of p j get the sequence
p1 , p j , p 2 ,....., p n .
This can carry on with p j beats p 2 or p 2 beats p j etc. until we arrive at the desired
sequence p1 , p2 ,....., pn .
3.
We can show by means of an example that the player p1 in question (2) need
not be the one with the most wins.
P1
P5
P2
P4
P3
p1
p2
p3
p4
p5
wins 3
wins 1
wins 1
wins 2
wins 3
Rank
1
4
4
3
1
So p1 need not be the one with the most wins, he can share the same number of wins
with another player and be ranked equivalent. The ardent fan of p1 may use this to
argue the unfairness of league tables because (say) p1 could be put in 3rd when his
rank is equivalent to 2nd and 1st, making p1 look worse than the 1st and 2nd positions.
4.
We can call the results of a tournament decisive if the players can split into
two non-empty subsets, the top dogs and the bottom dogs, such that in any
match between a top dog and a bottom dog the top dog won.
We can show that in a decisive tournament the sequence in question (2) must
have p1 ,....., pr top dogs and pr 1 ,...., pn bottom dogs. Also, that either the
results of a tournament are decisive or that any of the players can be p1 in
question (2).
p1 , p2 ,....., pi , p j pk ,....., pn1
3
If the results of a tournament are split into two non-empty subsets then you cannot
find a path around the graph from p n1 back to p1 . Therefore, not anyone can be p1 .
In a decisive tournament the top dogs are always ranked above the bottom dogs, so we
might as well restrict ourselves to a tournament, which is not decisive – such a
tournament is called strongly connected. The definition of a strongly connected
tournament is that for any two players p, q there is a sequence p1  p, p2 ,....., pr  q
of players such that p1 beat p 2 , p 2 beat p3 ,....., pr 1 beat pr . In the graphical
presentation of the results of a tournament given in question (1) this means that there
must be a path from p to q which follows the direction of the arrows at every step.
5.
(i)
We can show that a decisive tournament is not strongly connected.
A decisive tournament has p1 ,....., pr top dogs and pr 1 ,...., pn bottom dogs. Given
this you cannot connect a path from p to q, which follows the direction of the arrows
at every step  a decisive tournament is not strongly connected.
(ii)
We can show that “connected in each way by a path which follows the
direction of the arrows at every step” is an equivalence relation on the
participants in a tournament.
Top dogs
Vs
Bottom dogs
(iii)
We can show that there is only one strongly connected tournament with three
players, and only one with four.
p1
Only 1 strongly connected tournament
with 3 players because there is only 1
unique path
p3
p2
4
p1
p2
p3
p4
Only 1 strongly connected tournament
with 4 players because there is only 1
unique path.
i.e. only 1 Eulerian path.
We can also consider how many strongly connected tournaments there are with five
players.
p1
p1
p5
p2
p3
p4
Fig (1)
p5
p2
p3
p4
Fig (2)
There are 3 strongly connected tournaments with 5 players. This is illustrated on the
diagrams above. Fig (1) shows the first two paths indicated in red and black. Fig (2)
shows the third path.
(iv)
We can show that in a tournament all the players can be arranged
p1 , p2 ,....., pn
p1 beat p2 , p2 beat p3 ,....., pn1 beat pn and
such
that
pn beat p1 , then the tournament is strongly connected.
pn
p1
p2
p3
5
If in a tournament all the players p1 , p2 ,....., pn can be arranged such that
p1 beat p2 ..... pn beat p1 . We can see from the diagram that there is a path from
p1 to pn and back to p1 which follows the direction of the arrows at every step (this is
called a cycle of length n), therefore the tournament is strongly connected.
Theorem (J W Moon 1966).
If p is a participant in a strongly connected tournament (with n players) there are
cycles of length k=3,4,…..,n containing p.
6.
(i)
We can show that there is a cycle of length 3 containing p.
p
r
q
Starting at p. p has beat q; therefore q belongs to T, the set of players whom p beat.
Then because of strong connectedness q beat r. Because r beat p due to it belonging to
S the set consisting of players who beat p. Therefore due to strong connectedness, q a
member of T defeated r a member of S and so we have a cycle of length 3 containing
p.
Now let us assume that p1  p, p 2 ,....., pk 1 are the players (in order) in a cycle C of
length k-1 containing p.
(ii)
Suppose that there is a player s not in this cycle that beat some of the players
in this cycle and was beaten by the others. We can use the argument in
question (2) to show that p1 , p 2 ,....., p j 1 , s, p j ,....., p k 1 is a cycle of length k
for a suitably chosen j.
Since we are told that player s has beat some of the players in this cycle and was
beaten by the others, then player s cannot beat p1 and be placed at the front of the
sequence. Also pk 1 cannot beat s, so that s is placed at the end of the sequence.
Therefore s must be placed within the sequence in between the players which have
6
beat s and the players which s has beat. Such a sequence is
p1 , p 2 ,....., p j 1 , s, p j ,....., p k 1 and due to s being inserted the sequence is of length k.
Now let us assume that every player not in this cycle C either beat all of p1 ,....., p k 1
or was beaten by all of them. Let S be the set consisting of the players who beat
p1 ,....., p k 1 and T the set of players who was beaten by p1 ,....., p k 1 .
(iii)
We can explain why strong connectedness requires that some player t  T
defeated some player s  S.
s
t
S
p1 ,....., p k 1
T
If s has beat p1 ,....., p k 1 , then s belongs to the set S. Then by strong connectedness
p1 ,....., p k 1 beats t , which belongs to the set T. Now strong connectedness requires
player t to beat player s because there must be a path from t to s which follows the
direction of the arrows at every step.
(iv) We can show that p1 , t , s, p3 ,....., pk 1 is a cycle of length k.
s
S
t
p1 ,....., p k 1
T
s beat p1 ,....., pk 1 beat t , t beat s  p1 , t , s, p2 ,....., pk 1  length k  1
This can be re-arranged so that we get a cycle of length k.
 p1, t , s, p3 ,....., pk 1  length k
7
Mathematical Modelling
Worksheet 16
Ranking a Tournament (2)
7.
The score vector s 1 of a tournament is a vector whose i’th entry is the number
of wins gained by player i.
1
2
6
3
5
(i)
4
We can find the score vector s 1 of the above tournament.
So by counting the number of wins of each player indicated by the arrows, we get the
score vector
s 1 =(4,3,3,2,2,1)
It can be noted from the score vector that players 2 and 3 have the same number of
wins. It is possible to distinguish between their victories by considering who they
have played and beat.
Firstly consider Player 2:
Player 2 has beaten Players 4, 5, 6 who are ranked below Player 2.
Secondly consider Player3:
Player 3 has beaten Players 1, 2, 4. Player 1 has the top rank, so it could be said that
Player 3 has beaten the top player making a more outstanding victory. Also Player 3
has beaten Player 2 which currently holds equal rank.
Therefore it could be said that Player 3 deserves 2nd place given the greater victories
Player 3 has achieved.
8
(ii)
In the i’th level score vector in a tournament, each player’s score is the sum of
the (i-1)’th level scores of players beaten, starting with s 1 , the ordinary score
vector. Therefore we can find s 2 , s 3 , s 4 , s 5 , s 6 score vectors for the above
tournament.
s 2 =(8,5,9,3,4,3)
s 3 =(15,10,16,17,12,9)
s 4 =(38,28,32,21,25,16)
(iii)
s 5 =(90,62,87,41,48,32)
s 6 =(183,121,193,80,119,87)
What do we need to know about the score vectors as i   in order to
produce a stable ranking by this procedure? Does lim s i exist?
i
The limit lim s i does not exist, as there is no limit because the scores are continuously
i
increasing.
The matrix theory required to rank a tournament concerns the adjacency matrix of the
tournament. This is the n  n matrix whose (i, j)’th entry is 1 if player i beat player j
and 0 otherwise. In particular, the diagonal elements are all zero. Let A be the
adjacency matrix of a strongly connected tournament.
A path of length k from p to q in a tournament is a sequence of players (not
necessarily all different) p0  p, p1 , p2 ,....., pk  q such that p 0 beat p1 , p1 beat
p2 ,....., pk 1 beat pk .
8 (i)
We can show that the (i, j)’th entry of A k is the number of paths of length k
from player i to j.
From question (7) we obtain the adjacency matrix
9
Player
 1

 2

A 3
 4

 5
 6

1 2 3 4 5 6
0 1 0 1 1 1
0 0 0 1 1 1

1 1 0 1 0 0
0 0 0 0 1 1

0 0 1 0 0 1
0 0 1 0 0 0
Therefore for k = 2 we get
0

0
1
A A  
0
0

0

0

0
0
A2  
0
1

1

1 0 1 1 1 0
 
0 0 1 1 1 0
1 0 1 0 0 1

0 0 0 1 1 0
0 1 0 0 1   0
0 1 0 0 0   0
1 0 1 1 1

0 0 1 1 1
1 0 1 0 0

0 0 0 1 1
0 1 0 0 1 
0 1 0 0 0 
0 2 1 2 3

0 2 0 1 2
1 0 2 3 3

0 2 0 0 1
1 1 1 0 0 
1 0 1 0 0 
A k  A  Every time we multiply the matrix we add an extra path i.e. k + 1.
Therefore the (i, j)’th entry of A k is the number of paths of length k from player i to j.
In a strongly connected tournament there is a path from player i to player j for every i
and j so there is a shortest path from player i to player j. Let d be the maximum over
all i, j of the length of the shortest path from player i to player j.
(ii)
We can show that in a strongly connected tournament with five or more
players Ad 3 is positive (i.e. has every entry >0).
10
For the (i, j)’th element the maximum number of steps from player i to player j is
 d . Then if we add a cycle at one end i.e. j  j cycle of length 3 to make the
length of the path d + 3.
Therefore since i  j  d and j  j is a cycle of length 3 we have a path of length
d + 3.
(iii)
We can show that the vector s k in question (7) is given by s k  A k j where
j is the column vector with each entry equal to 1.
The vector s 1 in question (7) can be written as a matrix:
0

0
1
A
0
0

0

1 0 1 1 1

0 0 1 1 1
1 0 1 0 0

0 0 0 1 1
0 1 0 0 1 
0 1 0 0 0 
For the other vectors s 2 , s 3 ,.....s k , by performing the operation A k , the matrix A is
multiplied by itself k times. Then by performing A k j where j is the column vector
with each entry equal to 1. This basically allows us to calculate the infinity norm i.e.
the row sum, which gives the players score to be the sum of the (i-1)’th level scores of
players defeated.
i.e.
(iv)
2

2
0
s3  
2
1

0

2 5 2 1 3 1 15 
   
2 3 2 0 1 1 10 
0 6 1 3 6 1 16 
    
2 1 2 0 0 1  7 
2 0 3 3 3 1 12 
1 0 2 3 3 1  9 
The behaviour of s k will depend on the dominant eigenvalue of A.
s k  A k j  k x
dominant
eigenvalue
if   1, s k is stable.
dominant
eigenvector
11
Theorem (Perron-Frobenius)
If M is a square matrix such that for some integer r every element of M r is positive,
then the eigenvalue of M with largest modulus is a real positive number, having a
corresponding eigenvector with every component positive. Furthermore, every other
eigenvalue of M has strictly smaller modulus.
9 (i)
The adjacency matrix A of a strongly connected tournament can be modified
in order to produce a matrix B whose eigenvalue of largest modulus is 1.
Ax  x
1
Ax  x

1
AB
λ
Bx  x
(ii)
for largest eigenvalue
Suppose for simplicity that this n  n matrix B has n linearly independent
eigenvectors. By expressing the vector j as a linear combination of these
eigenvectors we can show that B k j  x 1 as k   , where x 1 is an
eigenvector of B corresponding to eigenvalue 1 with all its components
positive.
j  x1  a 2 x 2  .....  a n x n
B k j  B k x1  a 2 B k x 2  .....  a n x n
 B k 1  Bx1  a 2 B k 1  Bx 2  .....  a n B k 1  Bx n
 B k 1 x1  a 2 B k 1 2 x 2  .....  a n B k 1 n x n
 x1  a 2  2  x 2  .....  a n  n  x n
k
(iii)
k
What ranking procedure for a strongly connected tournament does this give?
Find the largest of the eigenvalues of A and use the result of question (9 (ii)) to obtain
the ranking procedure.
12
10.
We can rank the tournament with adjacency matrix:
j
i
A
A B C
0 1 1
D E
1 1
F
1
G
0
H
0
I J
1 1
B
0
0
1
0
0
1
0
0
0 0
C
D
0
0
0
1
0
1
0
0
0
1
0
1
0
0
0
0
0 0
1 0
E
F
0
0
1
0
1
1
0
0
0
1
0
0
0
0
0
0
0 0
1 0
G
1
1
1
1
1
1
0
0
1
H
I
1
0
1
1
1
1
1
0
1
1
1
0
1
0
0
0
1 1
0 0
J
0
1
1
1
1
1
1
0
1
0
0
The adjacency matrix can be ranked by deleting rows and columns by considering
how many matches the players have won.
Firstly looking for the player with the most wins.
The player with the most wins is H who has not been beaten by any other player.
Therefore player H can be given rank 1.
Now deleting row and column H we are left with a 9 9 matrix.
j
A B C
D
E
F
G
I
J
A
B
0
0
1
0
1
1
1
0
1
0
1
1
0
0
1 1
0 0
C
0
0
0
0
0
0
0
0 0
D
i E
0
0
1
1
1
1
0
0
1
0
1
0
0
0
1 0
0 0
F
0
0
1
0
1
0
0
1
G
I
1
0
1
1
1
1
1
0
1
1
1
0
0
0
1 0
0 0
J
0
1
1
1
1
1
1
1
0
0
13
Again looking for the players with the most wins we find that players A, G and J have
won the same number of matches.
Since a beats g , g beats j , j beats a forms a cycle we can rank them together with a
rank of 2.
Having determined the rank of players A, G and J, we can delete their rows and
columns from the adjacency matrix. This then forms a 6 6 matrix.
j
B C
0 1
B
C 0
D 1
i
E 1
F 0
I 1
Here the player with the most wins
rank of 5.
D E
0 0
F
1
0 0 0
1 0 1
1 0 0
1 0 1
1 0 1
is player D.
I
0
0 0
1 1
0 0
0 1
0 0
Therefore player D can be given a
Once again deleting the rows and columns of D to reduce the matrix to a 5 5 .
j
B C
E
F
I
B
C
0
0
1
0
0
0
1
0
0
0
iE
F
I
1
0
1
1
0
1
0
0
0
1
1
1
1
0
0
From this matrix the players with the most wins are F and I. Therefore we can rank
them together with a rank of 6.
Deleting the rows and columns of F and I reduces the matrix to a 3 3 .
j
B
i C
E
B C
0 1
0 0
1 1
E
0
0
0
14
From this simple matrix it is clear that player E has the highest number of wins and
can be given a rank of 8.
Player B has the second highest number of wins and can be given a rank of 9.
Player C is the only remaining player and has no wins. Therefore player C can be
ranked in last place with a rank of 10.
The ranking of the tournament can be summarised as follows:
Player
H
Rank
1
A
2
G
J
2
2
D
F
5
6
I
6
E
B
8
9
C
10