- Email: [email protected]
Feasibility of home–away-pattern sets for round robin tournaments
Operations Research Letters 36 (2008) 283–284 www.elsevier.com/locate/orl
Feasibility of home–away-pattern sets for round robin tournaments Dirk Briskorn ∗ University of Kiel, Germany Received 26 January 2007; accepted 30 September 2007 Available online 23 December 2007
Abstract A Home–Away-Pattern (HAP) set defines each team’s venue in each period. We consider the decision problem of whether a round robin tournament can be arranged on the basis of a given HAP set. We give a necessary condition which can be checked in polynomial time and conjecture it to be sufficient. c 2008 Elsevier B.V. All rights reserved.
Keywords: Round robin tournaments; Home–away-pattern set feasibility problem
1. Problem definition We consider sports leagues having a set T of an even number n = |T | of teams. A single round robin tournament (RRT) structure is defined such that each pair of teams meets exactly once and such that each team plays exactly once per period p ∈ P, |P| = n − 1. Scheduling an RRT is a hard combinatorial problem as soon as additional constraints such as stadium availability are to be considered. There are several decomposition schemes based on the idea of separating the decision about the period when two teams compete and the decision about the venue where a match of two teams is carried out; see [1,5,6], for example. A Home–Away-Pattern (HAP) h i for team i is a string of length n − 1 containing 0 in slot p if team i plays at home in period p (1 otherwise). The collection of HAPs of all teams is called an HAP set. An example of an HAP set for n = 14 teams is provided in Table 1. An HAP set restricts the set of possible matches such that no two teams having the same entry in slot p can compete in period p. Obviously, a given HAP set h ∈ {0, 1}n×(n−1) may have no feasible solution, that is there is no single RRT corresponding to h. In the following, an HAP set h is called feasible if and only if h has a feasible solution. Then, the HAP set feasibility problem is to determine whether a given HAP set is feasible.
Miyashiro et al. [4] introduce (1) as a necessary condition for HAP sets to be feasible. 0 X |T | 0 0 min c0 (T , p), c1 (T , p) − ≥0 2 p∈P ∀T 0 ⊆ T.
(1)
Here, c0 (T 0 , p) and c1 (T 0 , p) denote the numbers of zeros and ones, respectively, corresponding to a subset T 0 ⊆ T of teams and period p. For each period p the number of matches between teams in T 0 cannot exceed min c0 (T 0 , p), c1 (T 0 , p) . Then, condition (1) states that the overall number 0 of possible |T | matches during the tournament must be at least 2 which is the exact number of matches between teams of T 0 . A break for team i in period p occurs if i plays twice at home or twice away, respectively, in periods p − 1 and p. As shown in [4], for HAP sets having the minimum number of breaks, necessary condition (1) can be checked in polynomial time and, moreover, it is conjectured to be sufficient. However, sufficiency is not proven and, therefore, complexity of the HAP set feasibility problem remains open so far for arbitrary HAP sets as well as for HAP sets having the minimum number of breaks. 2. A new necessary condition
∗ Corresponding address: Christian-Albrechts-Universit¨at zu Kiel, Institut f¨ur Betriebswirtschaftslehre, Olshausenstr. 40, D-24098, Kiel, Germany. E-mail address: [email protected].
c 2008 Elsevier B.V. All rights reserved. 0167-6377/$ - see front matter doi:10.1016/j.orl.2007.09.009
We formulate an integer linear programming model in order to check feasibility of a given HAP set h.
284
D. Briskorn / Operations Research Letters 36 (2008) 283–284
Table 1 Infeasible HAP set for n = 14 Team
1
2
3
4
5
6
7
8
9
10
11
12
13
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 0 0 0 0 0 1 1 1 1 1 0 1 1
0 0 0 0 0 0 1 1 1 1 1 0 1 1
0 0 0 0 0 0 1 1 1 1 1 1 0 1
0 0 0 0 0 0 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0 0 1 1
1 1 1 1 1 0 0 0 0 0 0 1 0 1
1 1 1 1 1 0 0 0 0 0 0 1 1 0
1 1 1 1 1 0 0 0 0 0 0 1 1 0
0 0 1 1 1 0 0 0 1 1 1 0 0 0
1 0 0 1 1 1 1 0 0 1 1 0 0 0
1 1 0 0 1 1 1 1 0 0 1 0 0 0
1 1 1 0 0 1 1 1 1 0 0 0 0 0
0 1 1 1 0 0 0 1 1 1 0 0 1 0
Let xi, j, p , i, j ∈ T , j < i, p ∈ P, be a binary variable that equals 1 if and only if teams i and j compete in period p. HAP set feasibility X X max z h =
X
(2)
xi, j, p
i∈T j∈T, j
s.t.
X
xi, j, p ≤ 1
∀i, j ∈ T, j < i
(3)
p∈P
X
xi, j, p +
j∈T, j
X
xi, j, p ≤ |h i, p − h j, p | xi, j, p ∈ {0, 1}
x j,i, p ≤ 1
∀i ∈ T, p ∈ P
(4)
j∈T, j>i
∀i, j ∈ T, j < i, p ∈ P
∀i, j ∈ T, j < i, p ∈ P.
(5) (6)
The objective function (2) represents the goal of maximizing the number of matches while constraints (3) and (4) assure a single RRT. More precisely, constraint (3) forces each pair of teams to meet at most once while constraint (4) restricts the number of matches per team and period to be less than or equal to one. The entry of HAP set h corresponding to team i and period p is denoted by h i, p . Then, (5) takes care of the HAP set h such that no pair of teams can meet in a period where both of them play at home or away, respectively. Clearly, h is feasible if and only if z h = n(n−1) 2 . Let z¯ h be the objective value of the LP relaxation of HAP set feasibility which obviously can be computed in polynomial time. Then, z¯ h ≥ z h and, hence, z¯ h = n(n−1) is a necessary condition for h 2 to be feasible. Theorem 1. Condition z¯ h than (1).
=
n(n−1) 2
is strictly stronger
Proof. First, we show that each HAP set h that violates condition (1) violates condition z¯ h = n(n−1) 2 , as well. Suppose T 0 be the number of matches between z¯ h = n(n−1) . Let z ¯ h, p 2 P T0 teams in T 0 ⊆ T in period p. Obviously, p∈P z¯ h, p = |T 0 |(|T 0 |−1) T 0 and, . Moreover, min c0 (T 0 , p), c1 (T 0 , p) ≥ z¯ h, p 2 P 0 thus, p∈P min c0 (T 0 , p), c1 (T 0 , p) ≥ |T2 | holds.
Second, we consider an HAP set h provided by [3] with n = 14 shown in Table 1. HAP set h fulfills (1) but has z¯ h = 90 < 91 = n(n−1) 2 . Consequently, h is infeasible which is detected by z¯ h < n(n−1) but not by (1). 2 Moreover, we conjecture that z¯ h = z h for each h ∈ {0, 1}n×(n−1) . We failed to prove sufficiency of condition z¯ h = n(n−1) 2 but, on the basis of comprehensive computational results, we conjecture it. In order to support the conjecture we have carried out computational tests for two different classes of HAP sets solving HAP set feasibility and the corresponding LP relaxation (using Ilog Cplex [2]). For given n these classes of HAP sets were generated as follows: I Each entry was randomly chosen from {0, 1}. II For each slot p we randomly chose n2 teams having 0 in p. The remaining teams had 1 in p. Obviously, instances with class II HAP sets have much higher probability of being feasible since identical amounts of zeros and ones in each slot are necessary for an HAP set to be feasible. We carried out test runs for 10,000 instances for each class and each n ∈ {6, . . . , 30}. For each of the 260,000 considered instances, we had that z¯ h = z h for the HAP set h at hand and, therefore, h was feasible if z¯ h = n(n−1) 2 . References [1] K. Easton, G. Nemhauser, M. Trick, Sports Scheduling, in: J. Leung (Ed.), Handbook of Scheduling, CRC Press, 2004, pp. 52.1–52.19. [2] Ilog Cplex, URL: http://www.ilog.com/products/cplex/ visited at 12, September 2007. [3] K. Kashiwabara, Personal communication, 2005. [4] R. Miyashiro, H. Iwasaki, T. Matsui, Characterizing feasible pattern sets with a minimum number of breaks, in: E. Burke, P. de Causmaecker (Eds.), Proceedings of the 4th International Conference on the Practice and Theory of Automated Timetabling, in: Lecture Notes in Computer Science, vol. 2740, Springer, Berlin, Germany, 2003, pp. 78–99. [5] G.L. Nemhauser, M.A. Trick, Scheduling a Major College Basketball Conference, Operations Research 46 (1998) 1–8. [6] G.F. Post, G.J. Woeginger, Sports tournaments, Home–Away-Assignments, and the break minimization problem, Discrete Optimization 3 (2006) 165–173.
Feasibility of home–away-pattern sets for round robin tournaments Dirk Briskorn ∗ University of Kiel, Germany Received 26 January 2007; accepted 30 September 2007 Available online 23 December 2007
Abstract A Home–Away-Pattern (HAP) set defines each team’s venue in each period. We consider the decision problem of whether a round robin tournament can be arranged on the basis of a given HAP set. We give a necessary condition which can be checked in polynomial time and conjecture it to be sufficient. c 2008 Elsevier B.V. All rights reserved.
Keywords: Round robin tournaments; Home–away-pattern set feasibility problem
1. Problem definition We consider sports leagues having a set T of an even number n = |T | of teams. A single round robin tournament (RRT) structure is defined such that each pair of teams meets exactly once and such that each team plays exactly once per period p ∈ P, |P| = n − 1. Scheduling an RRT is a hard combinatorial problem as soon as additional constraints such as stadium availability are to be considered. There are several decomposition schemes based on the idea of separating the decision about the period when two teams compete and the decision about the venue where a match of two teams is carried out; see [1,5,6], for example. A Home–Away-Pattern (HAP) h i for team i is a string of length n − 1 containing 0 in slot p if team i plays at home in period p (1 otherwise). The collection of HAPs of all teams is called an HAP set. An example of an HAP set for n = 14 teams is provided in Table 1. An HAP set restricts the set of possible matches such that no two teams having the same entry in slot p can compete in period p. Obviously, a given HAP set h ∈ {0, 1}n×(n−1) may have no feasible solution, that is there is no single RRT corresponding to h. In the following, an HAP set h is called feasible if and only if h has a feasible solution. Then, the HAP set feasibility problem is to determine whether a given HAP set is feasible.
Miyashiro et al. [4] introduce (1) as a necessary condition for HAP sets to be feasible. 0 X |T | 0 0 min c0 (T , p), c1 (T , p) − ≥0 2 p∈P ∀T 0 ⊆ T.
(1)
Here, c0 (T 0 , p) and c1 (T 0 , p) denote the numbers of zeros and ones, respectively, corresponding to a subset T 0 ⊆ T of teams and period p. For each period p the number of matches between teams in T 0 cannot exceed min c0 (T 0 , p), c1 (T 0 , p) . Then, condition (1) states that the overall number 0 of possible |T | matches during the tournament must be at least 2 which is the exact number of matches between teams of T 0 . A break for team i in period p occurs if i plays twice at home or twice away, respectively, in periods p − 1 and p. As shown in [4], for HAP sets having the minimum number of breaks, necessary condition (1) can be checked in polynomial time and, moreover, it is conjectured to be sufficient. However, sufficiency is not proven and, therefore, complexity of the HAP set feasibility problem remains open so far for arbitrary HAP sets as well as for HAP sets having the minimum number of breaks. 2. A new necessary condition
∗ Corresponding address: Christian-Albrechts-Universit¨at zu Kiel, Institut f¨ur Betriebswirtschaftslehre, Olshausenstr. 40, D-24098, Kiel, Germany. E-mail address: [email protected].
c 2008 Elsevier B.V. All rights reserved. 0167-6377/$ - see front matter doi:10.1016/j.orl.2007.09.009
We formulate an integer linear programming model in order to check feasibility of a given HAP set h.
284
D. Briskorn / Operations Research Letters 36 (2008) 283–284
Table 1 Infeasible HAP set for n = 14 Team
1
2
3
4
5
6
7
8
9
10
11
12
13
1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 0 0 0 0 0 1 1 1 1 1 0 1 1
0 0 0 0 0 0 1 1 1 1 1 0 1 1
0 0 0 0 0 0 1 1 1 1 1 1 0 1
0 0 0 0 0 0 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0 0 1 1
1 1 1 1 1 0 0 0 0 0 0 1 0 1
1 1 1 1 1 0 0 0 0 0 0 1 1 0
1 1 1 1 1 0 0 0 0 0 0 1 1 0
0 0 1 1 1 0 0 0 1 1 1 0 0 0
1 0 0 1 1 1 1 0 0 1 1 0 0 0
1 1 0 0 1 1 1 1 0 0 1 0 0 0
1 1 1 0 0 1 1 1 1 0 0 0 0 0
0 1 1 1 0 0 0 1 1 1 0 0 1 0
Let xi, j, p , i, j ∈ T , j < i, p ∈ P, be a binary variable that equals 1 if and only if teams i and j compete in period p. HAP set feasibility X X max z h =
X
(2)
xi, j, p
i∈T j∈T, j
s.t.
X
xi, j, p ≤ 1
∀i, j ∈ T, j < i
(3)
p∈P
X
xi, j, p +
j∈T, j
X
xi, j, p ≤ |h i, p − h j, p | xi, j, p ∈ {0, 1}
x j,i, p ≤ 1
∀i ∈ T, p ∈ P
(4)
j∈T, j>i
∀i, j ∈ T, j < i, p ∈ P
∀i, j ∈ T, j < i, p ∈ P.
(5) (6)
The objective function (2) represents the goal of maximizing the number of matches while constraints (3) and (4) assure a single RRT. More precisely, constraint (3) forces each pair of teams to meet at most once while constraint (4) restricts the number of matches per team and period to be less than or equal to one. The entry of HAP set h corresponding to team i and period p is denoted by h i, p . Then, (5) takes care of the HAP set h such that no pair of teams can meet in a period where both of them play at home or away, respectively. Clearly, h is feasible if and only if z h = n(n−1) 2 . Let z¯ h be the objective value of the LP relaxation of HAP set feasibility which obviously can be computed in polynomial time. Then, z¯ h ≥ z h and, hence, z¯ h = n(n−1) is a necessary condition for h 2 to be feasible. Theorem 1. Condition z¯ h than (1).
=
n(n−1) 2
is strictly stronger
Proof. First, we show that each HAP set h that violates condition (1) violates condition z¯ h = n(n−1) 2 , as well. Suppose T 0 be the number of matches between z¯ h = n(n−1) . Let z ¯ h, p 2 P T0 teams in T 0 ⊆ T in period p. Obviously, p∈P z¯ h, p = |T 0 |(|T 0 |−1) T 0 and, . Moreover, min c0 (T 0 , p), c1 (T 0 , p) ≥ z¯ h, p 2 P 0 thus, p∈P min c0 (T 0 , p), c1 (T 0 , p) ≥ |T2 | holds.
Second, we consider an HAP set h provided by [3] with n = 14 shown in Table 1. HAP set h fulfills (1) but has z¯ h = 90 < 91 = n(n−1) 2 . Consequently, h is infeasible which is detected by z¯ h < n(n−1) but not by (1). 2 Moreover, we conjecture that z¯ h = z h for each h ∈ {0, 1}n×(n−1) . We failed to prove sufficiency of condition z¯ h = n(n−1) 2 but, on the basis of comprehensive computational results, we conjecture it. In order to support the conjecture we have carried out computational tests for two different classes of HAP sets solving HAP set feasibility and the corresponding LP relaxation (using Ilog Cplex [2]). For given n these classes of HAP sets were generated as follows: I Each entry was randomly chosen from {0, 1}. II For each slot p we randomly chose n2 teams having 0 in p. The remaining teams had 1 in p. Obviously, instances with class II HAP sets have much higher probability of being feasible since identical amounts of zeros and ones in each slot are necessary for an HAP set to be feasible. We carried out test runs for 10,000 instances for each class and each n ∈ {6, . . . , 30}. For each of the 260,000 considered instances, we had that z¯ h = z h for the HAP set h at hand and, therefore, h was feasible if z¯ h = n(n−1) 2 . References [1] K. Easton, G. Nemhauser, M. Trick, Sports Scheduling, in: J. Leung (Ed.), Handbook of Scheduling, CRC Press, 2004, pp. 52.1–52.19. [2] Ilog Cplex, URL: http://www.ilog.com/products/cplex/ visited at 12, September 2007. [3] K. Kashiwabara, Personal communication, 2005. [4] R. Miyashiro, H. Iwasaki, T. Matsui, Characterizing feasible pattern sets with a minimum number of breaks, in: E. Burke, P. de Causmaecker (Eds.), Proceedings of the 4th International Conference on the Practice and Theory of Automated Timetabling, in: Lecture Notes in Computer Science, vol. 2740, Springer, Berlin, Germany, 2003, pp. 78–99. [5] G.L. Nemhauser, M.A. Trick, Scheduling a Major College Basketball Conference, Operations Research 46 (1998) 1–8. [6] G.F. Post, G.J. Woeginger, Sports tournaments, Home–Away-Assignments, and the break minimization problem, Discrete Optimization 3 (2006) 165–173.