A Latin Square of order n is an n n array of n symbols where each symbol occurs once in each row and column. For example,

Transcription

1 1 Latin Squares A Latin Square of order n is an n n array of n symbols where each symbol occurs once in each row and column. For example, A B C D E B C A E D C D E A B D E B C A E A D B C is a Latin square of order 5. An orthogonal array OA(n, 3) or order n and depth 3 is a 3 n 2 array with integers 1, 2,..., n as entries where for any two rows, the n 2 vertical pairs occurring are all different. There is a one to one correspondence between orthogonal arrays and Latin squares as illustrated below To create a Latin square from an orthogonal array we do the following: in each column of the array, if the first entry is i, the second entry is j, and the third entry is k, then the ij th entry in the Latin square will be k. See the table above. 1.1 Construction of Latin Squares We shall use modular arithmetic to construct Latin squares. Let Z m = {0, 1,..., m 1} be the integers modulo m. We define addition and multiplication on Z m where x y = r if x + y = qm + r, 0 r < m. x y = r if x y = qm + r, 0 r < m. 1

2 One can check that and are commutative and associative (and distributive). x y = y x, x y = y x commutative x (y z) = (x y) z, x (y z) = (x y) z, associative. Each element x Z m has an additive inverse; that is, an element y such that x y = y x = 0. We denote such an element y by x. If there is an element y such that x y = y x = 1, then y is the multiplicative inverse of x and we denote such an element by y 1. Note that not all elements have a multiplicative inverse, for example the element 2 Z 6 has no inverse. Why? Elements in Z m which have multiplicative inverses are said to be invertible. One can determine exactly when an element is invertible. 1.1 Theorem An element x Z m is invertible iff x and m are coprime in Z. In particular, when p is prime, every element of Z p is invertible except 0. Proof. ( ) Suppose x Z m is invertible. Then x 1 exists and x x 1 = 1. This means that in Z it holds that x x 1 = qm + 1, and x x 1 qm = 1. Thus if d = gcd(x, m), then d x x 1 qm and hence d 1. This implies that d = 1 and x and m are coprime. ( ) Suppose that x and m are coprime. Then gcd(x, m) = 1. It is a known fact that for any integers a, b where gcd(a, b) = d there are integers α and β where α a + β b = d. Thus there are integers α and β where α x + β m = 1. Here we can assume 1 α < m. Then α x = β m + 1. Thus α x = 1, and x 1 = α. We conclude that x is invertible. For a Latin square L we define L(i, j) to be the ij th entry of L. We can easily construct Latin squares using modulo arithmetic. 1.2 Theorem For each m 2 the m m array defined by L(i, j) = i j i, j Z m is a Latin square. Proof. Suppose that in L the entries in positions (i, j) and (i, j ) are the same; that is, L(i, j) = L(i, j ). Then i j = i j i (i j) = i (i j ) ( i i) j = ( i i) j 0 j = 0 j j = j. 2

3 We conclude that each row in L has different entries. Using similar reasoning, one can also show that the entries in each column are also different. The example below illustrates how one can use the above theorem to construct a Latin square of order 4 (using Z 4 ) Suppose that L 1 and L 2 are two Latin squares of order n. We say that L 1 and L 2 are orthogonal if for each ordered pair (k, k ) of symbols, there is exactly one position (i, j) where L 1 (i, j) = k, and L 2 (i, j) = k. The example below illustrates an orthogonal pair of Latin squares of order 4. A B C D B A D C C D A B D C B A A B C D C D A B D C B A B A D C Each of the pairs (A, A), (A, B),..., (D, D) occurs exactly once if we take the pairs of elements in the corresponding positions in each Latin square. 1.2 Constructing Orthogonal Latin Squares Constructing orthogonal can be deceptively hard. For example Euler struggled with the following problem (which he never solved). Problem(Euler) Can one construct an orthogonal pair of Latin squares of order 6? It has since been shown that the answer that the answer to the above problem is negative such a pair doesn t exist. However, there is a good method to construct orthogonal pairs of Latin squares using modulo arithmetic. 1.3 Theorem Let p be a prime, and let t Z p, t 0. Then we define a Latin square L t whose entries are elements in Z p where L t (i, j) = (t i) j. In addition, if u Z, u t, then L u and L t are an orthogonal pair of Latin squares. 3

4 Proof. Let L t be as defined in the theorem. We shall first show that L t is a Latin square. Suppose two different entries in a row are equal; that is L t (i, j) = L t (i, j ). We note first that t is invertible (since p is prime, and t 0). Then (t i) j = (t i) j t 1 (t i) j = t 1 (t i) j ((t 1 t) i) j = ((t 1 t) i) j (1 i) j = (1 i) j i j = i j j = j. Thus the entries in any row must be different. Similar reasoning shows that the same is true for the columns. Thus L t is indeed a Latin square. To show that L t and L u are orthogonal, suppose for some ordered pair of elements (k, k ) in Z m it holds L t (i 1, j 1 ) = k, L u (i 1, j 1 ) = k L t (i 2, j 2 ) = k, L u (i 2, j 2 ) = k where (i 1, j 1 ) (i 2, j 2 ). Subtracting equations, we obtain t (i 1 i 2 ) = j 2 j 1, u (i 1 i 2 ) = j 2 j 1. If i 1 = i 2, then j 2 j 1 = 0, and j 2 = j 1. This gives a contradiction. So i 1 i 2, and i 1 i 2 0. This means that (i 1 i 2 ) 1 exists (since p is prime and all nonzero elements of Z p are invertible). Now multiply both equations above by (i i i 2 ) 1, to get (t (i 1 i 2 )) (i 1 i 2 ) 1 = (j 2 j 1 ) (i 1 i 2 ) 1, (u (i 1 i 2 )) (i 1 i 2 ) 1 = (j 2 j 1 ) (i 1 i 2 ) 1. Thus u = t = (j 2 j 1 ) (i 1 i 2 ) 1. And we reach a contradiction. We conclude that for each pair of elements (k, k ) there is exactly one position (i, j) such that L t (i, j) = k and L u (i, j) = k. Thus L u and L t are orthogonal. Taking p = 5 and t = 2, and u = 3 we can construct a pair of orthogonal Latin squares as illustrated below. 4

5 Completing Latin Squares. If the first k rows (k n) of a Latin square of order n are filled, then we call the k n array a Latin rectangle When can we complete a Latin rectangle to a Latin square? The next theorem says that we can complete a Latin rectangle to a Latin square by filling in the empty cells row-by-row. 2.1 Theorem A k n Latin rectangle (k < n) with entries 1, 2,..., n can be extended to a (k + 1) n Latin rectangle. Proof. For i = 1, 2,..., n let B j be the set of numbers which do not occur in the j th column. Since the numbers 1, 2,..., n occur exactly k times in the Latin rectangle, each number does not appear in exactly n k columns, and thus each number belongs to exactly n k of the sets B j, j = 1,..., n. To extend the rectangle to the k + 1 th row, we need only show that the sets B j, j = 1,..., n have an SDR (system of distinct representatives). It suffices to show that Hall s condition holds; that is, for each choice of l sets, say B i1, B i2,..., B il it holds that B i1 B il l. We have that B ij = n k, j = 1,..., l. Thus B i1 + B i2 + + B il = l (n k). 5

6 In the above sum, each number is counted exactly n k times, and as such there must be l (n k) n k distinct numbers in the union B i1 B il. Thus Hall s condition holds, and the sets have an SDR. Let b 1 B 1, b 2 B 2,..., b n B n be an SDR. Then if we make b 1 b 2 b n the k + 1 th row, then it is seen that we get (k + 1) n Latin rectangle. An n n array A with cells which are either empty of contain one of n symbols is called a partial Latin square if no symbol appears more than once in any row or column. A D B E A C D When can we complete a partial Latin square to a Latin square? The examples below show that this cannot always be done, even when relatively few cells are filled in Conjecture ( Evans ) If an n n partial Latin square has fewer than n nonempty cells, then it can be completed to a Latin square. This conjecture was proven by Smetaniuk in 1981, but the proof of it is beyond the scope of this course. We shall instead look at the problem of completing certain kinds of partial Latin squares. To do this, we shall first look at edge-coloring of graphs. 6

7 2.1 Edge-Colorings An edge-coloring of a graph is an assignments of colors to the edges so that no two edges of the same color meet at a vertex. That is, edges of the same color form a matching in the graph. For a graph G, we define the chromatic index χ (G) to be the minimum number of colors needed in an edge-coloring of G. Given that (G) is the maximum degree of G, it clearly holds that χ (G) (G). For bipartite graphs, they are in fact equal. 2.3 Theorem Let G be a bipartite graph. Then χ (G) = (G). Proof. Since G is bipartite, we can assume that the vertices of G are divided into two sets X and Y, where all the edges of G have one end in X and the other in Y. We shall show that the edges of G can be divided into exactly = (G) matchings. Case 1: Suppose that G is regular. Since G is regular, it is -regular. We shall show that G has a complete matching. To do this, we need only show that Hall s condition is satisfied; that is, N G (X ) X, X X. Here N G (X ) denotes the set of neighbors of vertices of X. We observe that the number of edges incident with vertices of X equals X, which is at most the number of edges incident with N G (X ) which equals N G (X ). Consequently, X N G (X ), N G (X ) X. Thus Hall s condition holds, and there is a matching from X to Y, which must also be perfect. If we now delete the edges of this matching, then we obtain a 1-regular graph. Again, we could find another perfect matching, and continuing in this fashion, we could find perfect matchings. Assigning a different color to the edges of each matching, we obtain a edge-coloring of G with colors. Thus χ (G) =. Case 2: Suppose G is not regular. We can assume that X = Y ; otherwise, just add some isolated vertices 7

8 to the graph so that this becomes so. Since G is not regular, there must be a vertex x X with deg(x) < and a vertex y Y with deg(y) <. Join x and y with a new edge. If the resulting graph is not -regular, then continue adding edges. Eventually we obtain a -regular graph G. According to case 1, χ (G ) =, and since G is a subgraph of G, it also holds that χ (G) =. The proof follows from cases 1 and 2 above. 2.2 ATheorem about Partial Latin Squares 2.4 Theorem Let A be a partial Latin square of order n in which a cell (i, j) is filled iff i r and j s. Then A can be completed to a Latin square iff N(i) r + s n, i = 1, 2,..., n. Here N(i) equals the number of cells in A containing i. Proof. To show that the above condition N(i) r + s n, i = 1, 2,..., n is necessary, suppose that we have completed A to a Latin square. Then in rows 1,..., r, each element i occurs exactly r times in a completed Latin square. However, i can occur at most n s times in the last n s columns, and as such, it must hold N(i) r (n s) = r + s n. So this proves necessity. To prove sufficiency, suppose that N(i) r + s n, i = 1, 2,..., n. We construct a bipartite graph G with vertices X = {u 1, u 2,..., u r } and Y = {v 1, v 2,..., v n } where u i is joined to v j if j does not occur in row i. Then we have deg(u i ) = n s, i = 1, 2,..., s. Furthermore, each j (j = 1, 2,..., n) does not occur in r N(j) rows of A, and r N(j) r (r + s n) = n s. Thus deg(v j ) = s N(j) n s, j = 1, 2,..., n. 8

9 We conclude that (G) = n s, and given that G is bipartite, the previous theorem implies that χ (G) = n s.. Thus we can color the edges with n-t colors, say c s+1, c s+2,..., c n. Now we can complete A to an s n Latin rectangle by the following rule: if edge u i v j is colored with color c k, then put j in the i th row and k th column of the rectangle. One can easily check that this results in a s n Latin rectangle. Now using our theorem about Latin rectangles, we can complete A to a Latin square. 3 Exercises 1. An ordinary deck of cards contains four Jacks, four Queens, four Kings, and four Aces. Explain how to arrange these 16 cards in a 4 4 square is such a way that each row contains one card from each suit and one card from each denomination. Interpret your results in terms of Latin squares. 2. Construct a pair of orthogonal Latin squares of order Find all values of Q for which the rectangle R 1 can be extended to a 6 6 Latin square. Show that R 2 cannot be so extended, whatever value Q has. R 1 = A B C D F E A B C D F A D A B Q R 2 = A B C D F E A B B D F A D A B Q 4. Show that any n n Latin square can be used as the top left quarter of a 2n 2n Latin square. 5. Let 1, 2, 3, 4, 5 be the first row of a Latin square of order 5, and let 2, 3, 4, 5, 1 be the second row. How many different candidates are there for the third row? 6. Suppose the first row of an n n array is x 1 x 2 x 3 x n 1 x n and suppose also that each successive row is obtained from the previous one be a cyclic shift of r places so that the second row is x r+1 x r+2 x r+3 x r 1 x r 9

10 and so on. If n is given, for which values of r does this construction yield a Latin square? 7. Complete the Sudoko puzzle