The publication appeared in Szostek R.: Effective management system sporting competitions, Operations Research and Decisions, Wroclaw University of Technology, Wroclaw 2011, t. 21, pp. 65-75

An Effective System of Sports Competition Management

Roman Szostek, Ph.D., Eng.

Rzeszów University of Technology Department of Quantitative Methods ul. Wincentego Pola 2, 35-959 Rzeszów email: [email protected]

The paper presents an innovative system of managing sports competitions. Its advantages over the currently applied systems were discussed. The final part of the paper contains a theorem concerning the system. The formalization of the management system of sports competitions, the system presented and the theorem are the results of the author’s research.

Key words: sports games, championships, system of sports competitions, probabilistic sorting, computational complexity

INTRODUCTION

The aim of sports competitions is to rank teams or competitors taking part in sports games. Thus, the problem is how to sort teams by their ability to win matches. However, it is not possible to directly measure this team characteristic because that kind of ability depends on many factors that are difficult to define. It is possible, however, to compare two randomly chosen teams provided they have played a match. All the rules applied when the competitions

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are on form a system of managing sports competitions. Thus a system of sports competitions consists of the rules which decide who plays against whom, in what sequence and where as well as the criteria that decide the place on the final ranking list. Sorting sports teams differs from standard sorting problems due to the fact that the result of comparing two teams is sometimes incorrect. It does happen that a team of a greater ability to win gets defeated by a team whose winning potential is undoubtedly lower. Thus the problem of sorting teams is a probabilistic one. That is the reason why the traditional methods of sorting do not work in the case of the management of sports competitions. The system the article presents is innovative. Since it has some advantages that other known and applied management systems of sports competitions do not have, one may call it effective system.

THE FORMULATION OF THE PROBLEM

There are n objects d1, d2, ..., dn∈D. Some numerical value called a quality is assigned to each object. An object di has a quality which has a value of xi∈R (i=1, 2, ..., n). The measurement of the quality of none of the objects is possible. However, the comparison of any two objects selected according to the quality is possible. As a result of comparing two objects di and dj the relation between di ∠ dj or dj ∠ di is established given that ∀ xi ≤ x j i, j

where:

⇒

[Pr (d

i

∠ d j ) = pij

≥ Pr (d j ∠ d i ) = q ij

]

(1)

- pij≥½ is the probability that the relation between the compared objects equals the values of their qualities, - qij≤½ is the probability that the relation between the compared objects does not equal the values of their qualities, - pij+qij=1. We assume that probability pij is a non-decreasing function of the distance of qualities

xi and xj . Consequently, there are pij = p ji = f (| xi − x j |)

(2)

0≤a≤b ⇒

(3)

f ( 0) =

f (a ) ≤ f (b)

1 2

(4)

2

Objects belonging to the set D ought to be sorted according to conditions stated above.

DISCUSSION OF THE PROBLEM

The management system of sports competitions is the problem of probabilistic sorting. It consists in sorting objects under conditions when the only tool that we possess is probabilistic pan scales. Their properties are presented in (1)-(4) equations. The probabilistic character of the scales results in the fact that the relation established between the same objects can differ in different weighing. Dependence (1) guarantees, that if two objects have different qualities, the probability pij that the relation between them will be consistent with the property of the quality is greater than the probability qij that the established relation will not be consistent with the properties of the quality. The dependence also contains an assumption that the result of the comparison of two objects never indicates equality relation between them (pij+qij=1). Dependence (2) guarantees, that the probability pij that the relation between two compared objects will be consistent with the properties of their qualities depends only on the extent of differences between these properties. Therefore, the order of the compared objects is not important so pij = pji. Dependence (3) guarantees, that the probability pij that the relation between two compared objects will be consistent with the properties of their qualities for the objects whose qualities differ much is not smaller than the probability pij for the objects whose qualities are similar. The result of weighing is more often true for objects that differ much. Dependence (4) guarantees, that the result of the comparison of two objects whose values of properties are the same does not favour any of the objects. An example function pij is presented in Figure 1.

pij = f (| xj – xi|)

1 1 −λ | x − x | 1− e j i 2

0.5 0

| xj – xi|

0 Fig. 1. An example of function pij 3

It is obvious that in probabilistic sorting no method guarantees obtaining exactly the same order that results from the properties of the qualities of the sorted objects. The searched for methods may merely guarantee high probability of obtaining such an order.

SOLUTIONS APPLIED IN SPORT

Nowadays in sport, there are two systems of managing sports games all over the world. The first of them is the round-robin system (the peer-to-peer system) known as an English system. The other one is a knockout system. Also various variants of joining these systems are applied. In the round-robin system all possible pairs of participating teams are made. Teams of each pair are compared in accordance with the number of times set in advance. One of the disadvantages of the round-robin system is the need for making numerous comparisons. The number of comparisons is the multiple of ½n(n–1), where n is the number of all teams. Thus, the computational complexity of this method is polynomial. Another disadvantage of this method is the fact that one cannot freely arrange the number of all planned comparisons during the course of games. This method is also ineffective because teams that differ considerably are compared unnecessarily. Another drawback concerns the impossibility of a precise, that is, frequent comparison of teams that are similar. It is the result of the comparison of such teams that is often incorrect and needs checking several times. The knockout system appears in several variants. In the basic variant, a team that has lost a match is eliminated from games. Thus, after each sequence, only half of the teams remain. A team that does not lose any match becomes a winner. The number of comparisons made equals n–1. The various combinations of the round-robin system and the knockout system are used. One disadvantage of these solutions is the fact that they do not allow arranging all the teams yet merely determining a winner. Besides, they assume that the probability pij is constant and equals 1. In chess games, the Swiss system is used. The drawback of this system of the arrangement of chess players concerns the fact that it does not guarantee the possibility of the more precise, that is, more frequent making of comparisons of players that are similar, since two players can be compared only once.

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THE ALGORITHM OF PROBABILISTIC SORTING

The proposed algorithm of probabilistic sorting is based on 5 properties. 1. Sorting is done in stages. In each stage objects are placed in a sequence. The neighbouring objects are compared, i.e. the first one with the second one, the third one with the fourth one. The first arrangement of objects, so-called starting arrangement, may be made at random. 2. After each comparison is made, the new arrangement of objects in a sequence is made. It is established in such a way that even if di ∠ dj, the object dj stays in its place or moves to the left side of the sequence, whereas the object di stays in its place or moves to the right side of the sequence. 3. Every object has its level. Thanks to this fact objects can obtain a diversified number of scores for the results of the comparison depending on the level. At the beginning, all objects are placed on the same level which is the lowest level. 4. After each comparison is made, objects move to new levels. They move in such a way that if di ∠ dj, then object dj moves to a level not lower than its current level, whereas the object di moves to a level not higher than its current level (according to this fact, the objects will change occupied levels in the further stages of the algorithm). 5. After each comparison is made, objects gain scores. If di ∠ dj, then dj gains more scores than di. Additionally, more scores are gained by the objects that are on higher levels than those on lower levels.

Many variants of the algorithm can be constructed in accordance with the above properties. If the number of objects is odd, it should be completed with a virtual object with a property that is smaller than any other properties.

Below there is the formal notation of the algorithm for the even number of n-objects. In a sequence objects are put in pairs which are numbered with values 1, 2, …, n/2. Let aik denote the number of a pair, i.e. the places of the di object in the k-nth sequence. aik ∈ {1, 2, 3, ..., n / 2} - number of a pair in the sequence with object di k ∈ {1, 2, 3, ..., K }

- number of the sequence

i ∈ {1, 2, 3, ..., n}

- number of the object

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The numbers of pairs (a1k , a2k , ..., ank ) are, for each sequence k, permutations of the elements of set {1, 1, 2, 2, 3, 3, ..., n / 2, n / 2} . If in a k-nth sequence two objects di and dj were compared (then aik = a kj ) and it was established that di ∠ dj, then the position of these objects in a next k+1 sequence satisfies the dependence

1 ≤ a kj +1 ≤ a kj = aik ≤ aik +1 ≤ n / 2 One of M levels is connected with each pair of the objects. Let bik denote the level of the pair with object di in k-nth sequence. bik ∈{1, 2, 3, ..., M }

- level of the pair with object di in k-nth sequence

All the pairs of the first sequence have the lowest 1st level. In all the next sequences the levels are assigned to the pairs in such a way that if, in an k-nth sequence, it turns out that di ∠ dj for two comparable objects, then

1 ≤ bik +1 ≤ bik = b kj ≤ b kj +1 ≤ M Beginning from a sequence, after each comparison of the objects, they obtain points. Their value depends on the pair’s level and the result of the comparison conducted in the above described way.

Figs 2–5 show an example algorithm for 16 objects. In each sequence the objects form eight pairs- match bases. Each pair is marked with a circle. The numbers of the circles stand for the place in the sequence and not for a particular pair of objects. First, the starting arrangement of the objects is made as well as comparisons of the first sequence. After eight comparisons have been made in the first sequence, eight objects that won and eight objects defeated are obtained. Then all the objects are given a new arrangement in the next sequence in the way shown in the figures. The curves going upwards indicate the arrangement of the objects which won in the next sequence while those going downwards point to the location of the objects which lost in the next sequence. As it can be seen in the figures, the winning objects move to the left side of the sequence, whereas the objects that lost move to the right side of the sequence. Thanks to this objects of similar characteristics are compared in the next sequences. It was originally assumed that a draw is out of the question. That is why, if a match ends in a draw, its result is decided by an extra time or randomly. 6

Each object is on one of the levels: A, B, C or D. The lowest level is A and the highest one is D. At first all the objects are on level A. It is shown in Fig.2, which represents the first sequence. After the comparisons of the first sequence are made, the objects which won move to higher level B while the objects which lost stay on level A. The new levels are shown in Fig. 3, which represents the second sequence. Now, the same as before, objects which won move to a higher level, i.e. from A to B, from B to C. The objects which lost remain on their old levels. New levels are shown in Fig. 4, which represents the third sequence. In the case of sequence three the procedure is the same. A change begins from the fourth sequence. Now the objects will only change the occupied levels. It occurs in the way shown in Fig. 5. For instance, if one looks at the objects from base 1, they are on level D. After the objects are compared, the object that won moves to base 1, whereas the one that lost moves to base 3. Thus, the winning objects stay on their levels or advance to higher ones. On the other hand, the objects that lost remain on their levels or fall dawn to lower levels. Other sequences are held in the same manner as in the fourth round. “Stop the execution” condition is established at the beginning of the number of queues. After each sequence objects get scores for achieved results. The number of obtained scores depends on a level on which a given object is placed. An example score is presented in Table 1. The objects placed on a higher level obtain more scores than objects from lower levels. In the final ranking points obtained by each object in all sequences are added. The object that gets the most scores is given the 1st place. In the conducted simulations very good results were obtained when the objects received the scores after the fourth sequence. In the first three sequences the objects competed just for the level.

1

2

3

4

5

6

7

A Fig. 2. The algorithm of sorting – the first sequence

7

8

1

2

3

4

5

7

6

B

8

A

Fig. 3. The algorithm of sorting – the second sequence

1

2

3

4

C

5

6

7

B

8

A

Fig. 4. The algorithm of sorting – the third sequence

1

D

2

3

4

5

C

6

7

B

8

A

Figure. 5. The algorithm of sorting – the fourth, fifth, … sequence

The level of an object the highest – D C B the lowest – A

Scores for an object Win 4 the highest – D 3 C 2 B 1 the lowest – A

Table 1. An example scoring

The number of required comparisons in the proposed algorithm of probabilistic sorting equals ½⋅n⋅(t+log2n–1), where t is the number of last sequence repetitions, which is shown in

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Figure 5. It seems that it is enough that t is log2n thus the complexity of the algorithm equals Ө(n⋅log2n) and it is the same as the mean complexity of a quicksort algorithm.

THEOREM OF NEAT SYSTEMS

Let X and Y be match bases of a sequence. Notation X>Y will be adopted when base X is located higher than base Y, i.e. following the notation used in the example the number of base X is lower than that of base Y.

Definition of the neat system The neat system is a system where, for each X and Y pair of match bases, if X>Y, then in the next sequence the winner from base X cannot be in a base lower than the winner from base Y. Similarly, a loser from base X cannot be in a base lower than a loser from base Y. ■ The notations accepted are as follows: S1 – set of systems where, in each sequence, winners (of previous sequence matches) can compete only with winners and losers only with losers. S2 – set of systems where, for each pair of X and X match bases, if winners from the bases play with each other in the next sequence, then losers from the bases also play with each other in the next sequence.

Theorem on neat systems (winners – losers) Premise:

x∈S1

Thesis:

x∈S2 ■

Proof (indirect) Figure 6 presents three different fragments of the system where moves from any selected sequence (bases X, Y, M) to the next sequence (bases A, B, N) were marked. In the previous sequence (enclosed with a broken line), the bases situated on the left are higher than those on the right. According to the figure, winners from bases X and Y meet in the next sequence, but the losers from the bases do not. A loser from base X meets with a loser from any other base M. The sections presented differ in the location of base M with regard to bases

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X and Y. It can be seen that each x ∉ S2 (and x∈S1 as well) presented as a system of graphs has to include at least one of the three fragments.

B ... N ... B

N

... X ... Y ... M ...

... X ... M ... Y ...

B ... A ... B

B1

...

N

... M ... X ... Y ...

A

A

...

B2

Fig. 6. Three possible fragments of system x, such that x ∈ S1 ∧ x ∉ S2

It is enough to prove that the neat system x ∈ S1 cannot contain any of the three presented fragments, i.e. x ∈ S2. For the first fragment of the system from Fig. 6, base Y is higher than base M (Y>M) so, with regard to the neatness of the system, base B should be placed higher than base A (B>A). At the same time, as base X is higher than base Y (x>Y), base A should be higher than base B (A>B). We arrive at a contradiction. It means that the neat system x ∈ S1 cannot contain the first of the tree fragments. Similarly, one can prove that the neat system x ∈ S1 cannot contain the second of the tree fragments. It is also necessary to prove that an x ∈ S1 neat system with the third fragment is not possible. Even if the third fragment is extended by extra bases, it cannot contain either the first or the second fragment (it must look like a “zigzag”). Thus, the number of the bases in a previous sequence will always be smaller than that in the later one. It is impossible as the number of bases in each sequence is to be constant. It appears that the losers of bases X and Y have to meet in the next sequence, i.e. x ∈ S 2. ■ Thus, S1 ⊂ S2. Since one can prove that S2 ⊂ S1, then S1 = S2. There arises a question if there are neat y ∉ S1 = S2 systems. If a base A can be joined by a winner from base M and not a loser, as follows from the assumption of the theorem, then

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it is possible to create a system, e.g. the one shown in Fig. 7. In this system a winner meets a loser in basis A, so it is an example of a y ∉ S1= S2 system.

N

X

Y

A

B

M

Fig. 7. An example of a y ∉ S1 = S2 neat system.

In the example each of the bases: N, X, Y, A, B, M is different, but, generally, bases X, Y, M may coincide with some A, B, N bases.

CONCLUSION

The article presents an effective system of managing sports competitions. The problem of this kind of managing is that of probabilistic sorting. The proposed method of organizing competitions is modern and has assets that other known and currently applied methods of sports competition management are deprived of: 1. A league can consist of a practically unlimited number of teams so there is no need to play a greater number of sequences. 2. The number of sequences played in a season can be established in advance. 3. Every team plays the same number of matches in a season as none of them is eliminated from games. 4. Every team takes a specific, individual place in the ranking owing to the fact that a group is uniform after each sequence is completed. There are no ‘death groups’ which are unfair. 5. Teams compete with each other on various levels. On these levels a constant competition concerning promotion to a higher level and defence against a fall to a lower level is in progress. Thus, the games are the source of entertainment. 6. Strong teams play more frequently with other strong teams, whereas weak teams compete more frequently with other weak teams. Therefore, there are no unnecessary matches to play. Moreover, every team compete with a team they deserve. Therefore, no matches that are unnecessary and not so much spectacular are played in general.

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7. There are several matches that are spectacular in each sequence. Theses matches are played between top teams. 8. There is a chance that teams that have already played with each other can meet again. It is possible if their sports levels and scores are similar. In this way a match result between specific teams can be verified. 9. It makes corruption difficult as matches are played between teams that are classified according to similar levels. Furthermore, the result of each match is of great importance to the final ranking.

There are still numerous open questions. How does the probability of establishing a true ranking depend when taking into account: 1. a starting arrangement, 2. the variant of the algorithm presented, 3. the number of games to play, 4. the form of the function from Figure 1, 5. the number of levels on which comparisons take place, 6. received score?

The probability of establishing the true ranking regarding each of the above questions can be treated as the probability that all teams will be arranged according to the values of the properties (a global order) or as the probability that a specific team will be found in the final ranking in a position resulting from the value of its property (a local order).

REFERENCES

[1]

Illustrated encyclopaedia of sports, collective work, Publisher Muza S.A., Poland 2001

[2]

Harel Dawid: Algorythmics: the spirit of computing, Pearson Education, 1992

[3]

Mitzenmacher Michael, Upfal Eli: Probability and Computing. Randomized Algorithms

and Probabilistic Analysis, Cambridge University Press 2006 [4]

Wirth Niklaus: Algorithms + Data Structures = Programs, Prentice Hall, 1976

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