LATIN SQUARES AND THEIR APPLICATION TO THE FEASIBLE SET FOR ASSIGNMENT PROBLEMS TIMOTHY L. VIS Abstract. A significant problem in finite optimization is the assignment problem. In essence, the assignment problem asks for a minimum weight assignment of elements of distinct sets to one another. Assignment problems have many applications. We give descriptions of several of the most common assignment problems. We relate the planar assignment problems to Latin squares and orthogonal Latin squares and discuss many known results on the existence of Latin squares and orthogonal Latin squares as the feasible sets for these assignment problems. Finally, we give a mathematical formulation for planar assignment problems and look in some depth at the intersection graph induced by the four index planar assignment problem. This intersection graph has application to the structure of the solution polytope; however, we do not directly explore this application.. Introduction A common class of combinatorial optimization problems is the class of assignment problems. An assignment problem is any problem in which elements of one set are assigned to elements of another set (or sets). An assignment problem can generally be broken down into two parts: that of determining feasible assignments, and that of determining the optimal assignment. For some classes of assignment problems, elegant combinatorial constructions exist that exhibit the feasiblity of the problem. For some very limited subclasses of assignment problems, elegant and efficient algorithms exist for optimizing the assignment. However, in general, once the problems become more complex than the most basic assignment problems, both determining the set of feasible assignments and determining which of these assignments is optimal become very difficult problems. We will introduce several classes of assignment problems, with a particular focus on planar assignment problems, wherein the feasible solutions are closely related to Latin squares. The simplest type of assignment problem is the -index assignment problem. Mathematically, given sets G and H, with G = H = n, and a cost (benefit) associated to each ordered pair (g, h), where g G, h H, determine the bijection between G and H that minimizes (maximizes) the total cost (benefit). For example, suppose an office has five positions to fill and five staff members with which to fill the positions. What assignment of staff members to positions will minimize the cost (maximize the benefit) to this office? This sort of problem can be easily modeled by way of a weighted complete bipartite graph. Such a model easily adapts to the Date: February 6, 007.
TIMOTHY L. VIS more complex situation in which the sets may have differing sizes, or the situation where not every pairing is possible by the addition or removal of vertices and edges.. Multi-index Assignment Problems This notion of an assignment problem can be generalized to the case with a larger number of sets and with different means of assigning the sets. In the -index assignment problem, each assignment is an ordered pair, and each element of each of the sets is required to appear in exactly one ordered pair. In general, given k sets of n elements each, an assignment will be an ordered k-tuple. Note, however, that the requirement need no longer be that every element of each set appear in exactly one ordered k-tuple. The requirement could easily be amended to require that every pair of elements from distinct sets appear in exactly one ordered k-tuple, or likewise, every set of s elements from distinct sets appear in exactly one ordered k-tuple, or that they appear at least once. We will refer to the number k of sets as the index of the problem. Two varieties of multi-index assignment problems are of particular interest: the axial assignment problem and the planar assignment problem. The axial assignment problem asks for an optimal set of n k-tuples, such that each element appears in exactly one such k-tuple. The following far-fetched, but helpful example of an axial -index assignment problem is due to Spieksma [8]. A certain chaotic village has exactly 0 cats, 0 houses, 0 men, and 0 women. For each household unit consisting of one house, one cat, one man, and one woman, there is some number measuring the happiness of this household unit. What household units should be created in order to maximize the happiness of the village and to bring peace and quiet to this part of the world? In this situation, there are a possible (0!), or about.866 0 97, partitions of the village into household units. Of course, the problem is easily extended to higher indices by the addition of other sets to the system. It should be easy to see that the axial assignment problem, as described here, will always have a feasible solution. In fact, a simple counting argument will establish the existence of (n!) k distinct solutions, where n is the order of each of the sets and k the index of the problem. On the other hand, the planar assignment problem, which will be described shortly, given sufficiently high index, may not have any solution. The planar assignment problem asks for an optimal system of n k tuples, where every ordered pair of elements from two distinct sets appears exactly once. In 78, Euler considered the following problem. Thirty-six officers, six from each of six regiments, and six of each of six ranks are selected, so that any two officers from the same regiment are of different ranks. Is it possible to position these officers at the thirty-six intersections of a six-by-six street grid so that no two officers in the same street are of the same rank or are from the same regiment? The Euler Officer Problem (taken here from [6]) is a classic example of a -index planar assignment problem. In particular, Euler s Officer Problem is an example of an infeasible planar assignment problem, as was shown by Tarry in 900 [9]. For the balance of this paper, we shall concentrate only on the planar assignment problem, and in particular, on the combinatorial structures which give rise to its feasible solutions.
ASSIGNMENT PROBLEMS.. Orthogonal Arrays.. Orthogonal Arrays and Latin Squares Definition.. An orthogonal array of size N, k constraints, n levels, strength t, and index λ is a k N matrix M having n different elements and with the property that each different ordered t-tuple of elements occurs exactly λ times as a column in any t-rowed submatrix of M [5]. It should be clear from this definition that there is a fairly direct correlation between orthogonal arrays and the feasible sets to general assignment problems. We are interested in particular in the planar assignment problem. With this in mind, we refine the definition. Definition.. An orthogonal array corresponding to a k-index planar assignment problem on sets of n elements is a k n matrix having n different elements with the property that each different ordered pair of elements occurs exactly once as a column in any n submatrix of M.. Latin Squares. Definition.. A Latin Square of order n is an n n matrix L with n different elements and the property that every element occurs exactly once in each row and column of the matrix. The following theorem establishes the existence of Latin squares of every possible order. Theorem.. For every positive integer n, there exists a Latin square of order n. Proof. Given n, define L in the following manner: set L i,j = i + j and reduce modulo n. Consider L i,j and L i,k, where j k. Then j and k differ by at most n, so that i+j and i+k are distinct when reduced modulo n. That is L i,j L i,k. Likewise, L i,k and L j,k are distinct when i j. As such, L is a Latin square of order n. If we assign to each entry in a Latin square a column vector, such that the first entry gives its row, the second entry its column, and the third its symbol, we create n entry column vectors. If we let these be the columns of a n matrix M, it is easily seen that this matrix is an orthogonal array. Conversely, given a n orthogonal array, we can construct a Latin square by letting the third entry of any column denote the symbol placed in the position whose row and column are determined by the first and second entry of that same column. As such, a n orthogonal array is equivalent to a Latin square of order n... Mutually Orthogonal Latin Squares. Definition.5. Two Latin squares on the integers,,...,n L = a i,j and L = b i,j of order n are orthogonal if and only if every ordered pair of symbols occurs exactly once among the n pairs (a i,j, b i,j ), where i, j =,,..., n. The following figure illustrates a pair of orthogonal Latin squares of order.
TIMOTHY L. VIS By a simple relabelling, we may assume that any Latin square of order n uses as its symbols the integers,,..., n. so that the definition is easily extended to any pair of Latin squares. This definition is easily extended to a set of k Latin squares in the following manner: Definition.6. A set of k Latin squares is mutually orthogonally if and only if any two distinct Latin squares in the set are orthogonal. We can easily extend the earlier relation between Latin squares and orthogonal arrays to relate sets of mutually orthogonal Latin squares (or MOLS) and orthogonal arrays. We simply add a row to our orthogonal array for each additional Latin square. We simply replace the third row in the earlier argument with the row corresponding to a given Latin square in the set of MOLS. As such, an k + n orthogonal array is equivalent to a set of k MOLS of order n. Thus, there is a direct correspondence between sets of k MOLS of order n and feasible solutions to the k + -index planar assignment problem with sets of size n, where a set of one MOLS is simply any Latin square. Lemma.7. The maximum size k of a set of mutually orthogonal Latin squares of order n is n. Proof. Without loss of generality, we can assume that the Latin squares of order n use the symbols,,...,n, and that the entry L,k = k by way of some relabelling or permutation of the symbols. As such, the ordered pair (k, k) occurs in the position L,k for any pair of the Latin squares. Consider now the possible entries in position L,. Since occurs in L, for each of the squares, cannot ever occur as the entry L,. Further, if two of the squares have L, = k, then those squares are not orthogonal, since L,k = k for both of those squares. Thus, only n squares are possible, each having a distinct entry L,. Definition.8. A set of n mutually orthogonal Latin squares of order n is called complete. We now present a construction for Latin squares which has the property that it is easily extended to produce sets of mutually orthogonal Latin squares. Let GF (q) be the finite field of order q with set of elements F and operations addition and multiplication. For each k GF (q), k 0, define a binary operation k on F by x k y = xk + y. Then the operation table for (F, k ) is a Latin square. As a proof, suppose that x k y = x k z. Then xk + y = xk + z, so that y = z. Suppose that x k z = y k z. Then xk + z = yk + z, so that xk = yk, and since k 0, x = y. Thus, a Latin square is produced. Theorem.9. The Latin squares determined by {(F, k )} where k ranges over the nonzero elements of GF (n) form a complete set of mutually orthogonal Latin squares of order n.
ASSIGNMENT PROBLEMS 5 Proof. Suppose k l and consider the Latin squares determined by (F, k ) (L k ) and (F, l ) (L l ). It suffices to show that if L k i,j = Lk u,v, then L l i,j Ll u,v. So suppose that L k i,j = Lk u,v and Ll i,j = Ll u,v. Then we have i k + j = u k + v and i l + j = u l + v. Subtracting the two equations, we have i (k l) = u (k l). Since k and l are assumed to be distinct, this implies that i = u. But then L k i,j and L k u,v must represent the same entry in each of the Latin squares. Thus, for distinct (i, j) and (u, v), L k i,j = Lk u,v implies that Ll i,j Ll u,v, so that Lk and L l are orthogonal. Since finite fields of order p m exist for every prime power p m, a complete set of MOLS exists for every prime power p m. These sets of MOLS can be used to construct sets of MOLS of higher order. Definition.0. The direct product of an m m Latin square L and an n n Latin square L is the mn mn matrix formed by replacing each entry of L with a copy of L whose entries are changed to ordered pairs with first coordinate the entry in L replaced by the copy of L and with second coordinate the entry in L. Lemma.. The direct product of two Latin squares is a Latin square. Proof. Consider two distinct entries in the same row (column). If the two distinct entries have the first coordinate in common, they sit in the same copy of L and have distinct second coordinates. Thus, no two entries in the same row (column) have both the first and second coordinates in common. Lemma.. If K and K are orthogonal Latin squares of order m and L and L are orthogonal Latin squares of order n, then the direct product of K and L and the direct product of K and L are orthogonal. Proof. We will again show that if two cells contain the same entry in the first direct product, they cannot contain the same entry in the second direct product. Suppose that two cells contain the same entry in the first direct product, and contain the same first coordinate in the entry in the second direct product. Then these cells must be in the same copy of L and L in each direct product (otherwise the orthogonality of K and K is violated. Since the cells are distinct, however, these cells must correspond to different cells in L and L. Since L and L are orthogonal, the second coordinates must differ. That is, two cells containing the same entry in the direct product of K and L cannot contain the same entry in the direct product of K and L. The preceding Lemma and the following theorem are due to MacNeish [7]. Theorem.. If there exist r MOLS of order m and r MOLS of order n, then there exist r MOLS of order mn. Proof. Taking the direct product of each of the MOLS of order m with a different Latin square of order n yields pairwise orthogonal Latin squares of order mn by the preceding two Lemmas. Thus, the r Latin squares of order mn thus produced are a set of r MOLS of order mn. Finally, we have the following important corollary. Corollary.. If n = p r pr pr k k, where each p i is a distinct prime, then there exist min {p r, pr,..., pr k k } MOLS of order n.
6 TIMOTHY L. VIS Proof. For each p ri i there exists a finite field, and thus a set of p ri i MOLS by the finite field construction. Repeated application of the direct product construction will then produce the desired set of MOLS. Further constructions exist that yield larger sets than are given by this result, and in fact, at least a pair of orthogonal Latin squares of order n exists for all n except n = and n = 6. A more complicated construction due to Richard Wilson (in [6]) establishes existence for all cases other than 0 and where existence does not follow from the preceding work. However, this result is extremely important, as it gives a lower bound on the number of MOLS of a given order that exist.. Mathematical Formulation In order to generally formulate the planar assignment problem, we first consider the -index planar assignment problem. The formulation that follows is taken from Appa, Magos, and Mourtos []. Let I, J, and K, with I = J = K = n be the sets under consideration. We define x ijk (i I, j J, k K) as one or zero, according as (i, j, k) appears as a triple. We further define w ijk as the cost of the triple (i, j, k). Then, given a valid assignment, and a pair of elements (i, j), i I, j J, exactly one k K occurs in a triple with i and j in this assignment. We formulate this as follows: x ijk = for all i I j J k K This argument is symmetric in all three variables, so that we also obtain the constraints: x ijk = for all j J k K i I x ijk = for all i I k K j J Further, with the costs of each triple defined, and the variables x determining whether or not a triple is in the assignment, we obtain the following objective function to be minimized: w ijk x ijk. i I j J k K Thus, the mathematical formulation of the problem is minimize w ijk x ijk subject to i I j J k K x ijk = for all j J k K i I x ijk = for all i I k K j J x ijk = for all i I j J k K If we extend this to the -index planar assignment problem, the constraint desired is that each ordered pair in each set appears exactly twice. Let our sets be I, J, K,
ASSIGNMENT PROBLEMS 7 and L, with I = J = K = L = n and the variable x ijkl (i I, j J, k K, l L) equal to one or zero, according as (i, j, k, l) appears as a quadruple, and let w ijkl be the cost of the quadruple (i, j, k, l). Then we consider a pair of elements from each of two sets, i I and j J, for instance. This pair must occur only once in all of the possible combinations of values for k and l. That is, x ijkl = for all i I k K k K l L As with the -index assignment problem, the costs of each quadruple are defined, and the variables x determine whether or not a quadruple is in a given assignment. Thus, we obtain the following objective function to be minimized: w ijkl x ijkl i I j J k K l L To obtain the final formulation of the problem, we add the remaining constraints, corresponding to each of the other pairs of sets. The final formulation (due to Gale in []) of the problem is the following: minimize w ijkl x ijkl subject to i I j J k K l L x ijkl = for all k K l L i I j J x ijkl = for all j J l L i I k K x ijkl = for all j J k K i I l L x ijkl = for all i I l L j J k K x ijkl = for all i I k K j J l L x ijkl = for all i I j J k K l L In general, the objective function will be the sum over all elements in all sets of w ii i k x ii i k. There will be a set of constraints for every pair of sets, such that the sum over all element in all other sets of x is. We have a total of n k binary variables, and ( k ) n k equality constraints. We now consider some of the work concerning the polytope derived from the constraints of the -index assignment problem. Although, we will not delve into the details of the polytope, we will discuss the intersection graph, a graph which plays a key role in the discussion of the polytope, at some length. The work that follows is taken from []. 5. The Intersection Graph Recall that in the formulation of the k-index planar assignment problem, we had n k binary variables and ( k ) n k equality constraints. In particular, the - index planar assignment problem involves n binary variables and ( ) n = 6n constraints. If we arrange these entries in a matrix A, we find that A has n columns
8 TIMOTHY L. VIS and 6n rows. Each row corresponds to one constraint. Since each constraint fixes the element in each of two sets and sums over the remaining constraints, a bijective correspondance is induced between the constraints and elements of the set R = (K L) (J L) (J K) (I L) (I K) (I J). We thus index each constraint and each row of A with its corresponding element in R. Similarly, each column corresponds to one binary variable. Since each variable corresponds to a set of one element from each of the sets I, J, K, and L, we can index each variable and each column of A with its corresponding element in the set C = I J K L. Note that each element of C can be denoted as an ordered quadruple (i, j, k, l), where i I, j J, k K, and l L. Any pair of distinct quadruples will then share zero, one, two, or three indices in common. We define the intersection graph in the following manner: Definition 5.. The intersection graph for a pair of orthogonal Latin squares is the graph with a vertex for every element of C and an edge between two elements of C if and only if they share two or three indices in common. Proposition 5.. The intersection graph thus defined is (n )(n )-regular. Proof. The proof is a simple counting argument. Let v be a vertex in the intersection graph. There are a total of n possible vertices, of which v is one. Since v corresponds to a particular quadruple, there are n choices for each entry in the quadruple that will be different from the same entry in v. Thus, there are (n ) vertices corresponding to elements of C sharing no common index with v. Further, there are four choices for a shared index with v and (n ) ways to pick the remaining indices to be different from v, for a total of (n ) vertices corresponding to elements of C sharing one index with v. The remaining vertices must share two or three indices in common with v and must thus be adjacent to v. This total is d = n (n ) (n ) = n ( n n + 6n n + ) ( n n + n ) = n n + n 6n + n n + n n + = 6n 8n + = (n )(n ) Since v was chosen arbitrarily, every vertex is adjacent to (n )(n ) other vertices; the intersection graph is (n )(n )-regular. Corollary 5.. The graph has n (n )(n ) edges. Proof. Each vertex is an end-vertex for (n )(n ) edges. Multiplying this by n vertices and dividing by vertices for each edge gives the desired result. Of particular interest is the clique structure of the intersection graph. We wish to determine all of the cliques in the intersection graph. Consider first a row r of the matrix A, and denote by R (r) the set of entries in C having a non-zero entry in row r. Recall that each row corresponds to a unique constraint and that each constraint sums over all assignments fixing two constraints. We have the following:
ASSIGNMENT PROBLEMS 9 Proposition 5.. For each row r, the set of vertices corresponding to the elements of R (r) induces a maximal clique in the intersection graph of order n, and there are 6n such cliques. Proof. That the set of vertices is a clique is a direct consequence of the fact that the row is a constraint summing over terms with two indices in common. That is, each pair of vertices is adjacent. That the clique is maximal follows from the fact that the constraint sums over all terms with two indices in common. In order for another vertex to have at least two indices in common with all other vertices in the clique, it would need to share those two indices, and is thus already included. Finally, since there are 6n rows, there are 6n such cliques. We will refer to these cliques as Type I cliques. Now consider a vertex v and all vertices that share three indices with v. We denote this set by Q (v). Proposition 5.5. For each vertex v, the set of vertices Q (v) induces a maximal clique in the intersection graph of order n and there are n such cliques. Proof. Consider two elements x and y of the set Q (v). Each of these elements differs from v in exactly one index. Then x and y differ from one another in at most two indices (those in which they differ from v). But then x and y share at least two indices in common and are thus adjacent. So Q (v) induces a clique. To see that the clique is maximal, consider some other element z adjacent to v. This element agrees with v in exactly two indices. But then there exists an element of Q (v) that agrees with v in only one of those two indices and differs from v elsewhere. In particular, this element is not adjacent to z, so the clique is maximal as it is. Finally, each element induces a different clique, so as there are n elements, there are n cliques of this type. We will refer to these cliques as Type II cliques. We now consider one further type of clique. Consider a pair of vertices u and v, such that their elements of C share exactly one index in common. Then consider the set Q (u, v) of vertices that share three indices with v and two indices with u along with u. Proposition 5.6. For each pair of vertices u and v sharing one common index, the set Q (u, v) induces a maximal clique in the intersection graph of order. There are n (n ) such cliques. Proof. Any element other than u in the set Q (u, v) shares, by definition, two indices with u and is adjacent to u. Further, any two other elements share three indices with v. Since there are only four indices, any pair may differ in at most two indices and must then share at least two indices. As such, every vertex in Q (u, v) is adjacent to every other vertex in Q (u, v). To see that the clique is maximal, consider some other vertex. This vertex must share at least two indices in common with u and at most two indices in common with v. We have two cases: either this vertex w shares the same index with u as does v or it does not. In the former case, w shares only one of the other three indices with v at most. But then one of the vertices of Q (u, v) shares only the other two indices and the common index to u and v with v, and thus is not adjacent to w. In the latter case, w can only share one index with v if it is to share two with u. But then the vertex from Q (u, v) that shares the other three indices with v is not adjacent to w. To see that the clique has size four, recognize that the index shared by u and v must be shared by any vertex of
0 TIMOTHY L. VIS Q (u, v) in order to allow it to both share two indices with u and three with v. Since every such clique is induced by an ordered pair sharing one index, and there are n (n ) choices of these ordered pairs (by the argument in Proposition 5.), there are at most n (n ) such cliques. However, each clique can be induced by four different ordered pairs (by letting each of the vertices play the role of u and altering v accordingly. Further, no other ordered pairs can induce the same clique. Thus, there are n (n ) such cliques. We will refer to these cliques as Type III cliques. As such, we have now defined three distinct classes of cliques. The final theorem in this paper asserts that there are no other cliques. Theorem 5.7. The only maximal cliques in the intersection graph are the Type I, Type II, and Type III cliques. Proof. Suppose Q is a set of vertices inducing a maximal clique in the intersection graph. Fix some vertex u in Q. Then if v is some other vertex in Q, v must share at least two indices with u. If no vertex in Q shares only two indices with u, then Q is a Type II clique. Further, if every vertex in Q shares the same two indices with u, then Q is a Type I clique. So assume that some vertex in Q shares only two indices with u and that no two vertices are common to all vertices in Q. Suppose then that v has only two indices in common with u and that w does not share those same two vertices in common with u. Now w must share at least one of the two indices common to both u and v. If not, it shares the other two indices with u and no indices with v. Further, of the two indices not common between u and v, w shares one with u and the other with v. Again, this is necessary to ensure that w is adjacent to both u and v. But in this case the indices of w not agreeing with u are determined, so that each element of Q shares exactly two indices with u, and three indices with some element not in the clique. That is, the clique is a Type III clique. 6. Conclusion Although we do not discuss them in this paper, the classification of all cliques in the intersection graph allows the definition of inequalities which define facets of the polytope necessary for the use of various programming techniques in the solution of problems based on Latin squares. Again, while we do not discuss those here, the work put forth provides a solid foundation for further reading into the usage of this work. A great deal of further discussion is provided by Appa, Magos, Mourtos and Janssen [], who give a general description of many properties of the polytope, and by Appa, Magos, and Mourtos [], who exploit further structures within the intersection graph. References [] G. Appa, D. Magos, and I. Mourtos. The wheels of the orthogonal latin squares polytope: Classification and valid inequalities. J. Combin. Optimization, 0:65 89, 005. [] G. Appa, D. Magos, and I. Mourtos. A new class of facets for the latin square polytope. Disc. App. Math., 5:900 9, 006. [] G. Appa, D. Magos, I. Mourtos, and J. Janssen. On the orthogonal latin squares polytope. Disc. Math., 06:7 87, 006. [] G. Dantzig. Linear Programming and Extensions. Princeton University Press, 96.
ASSIGNMENT PROBLEMS [5] J. Dénes and A. Keedwell. Latin Squares and their Applications. Academic Press, New York, 97. [6] C. Lindner and C. Rodger. Design Theory. CRC Press, Boca Raton, 997. [7] H. MacNeish. Euler squares. Ann. Math., : 7, 9. [8] F. Spieksma. Multi-index assignment problems: complexity, approximation, applications. In L. Pitsoulis and P. Pardalos, editors, Nonlinear Assignment Problems: Algorithms and Applications, pages. Kluwer Academic Publishers, Amsterdam, 000. [9] G. Tarry. Le problème des 6 officers. C. R. Assoc. Fr. Av. Sci., 9:70 0, 900.