1. Latin squares Def: A Latin square of order n is a quadruple (R, C, S; L) where R, C and S are sets of cardinality n and L is a mapping L: R × C → S such that for any i ∈ R and x ∈ S, the equation L(i, j) = x has a unique solution j ∈ C, and for any j ∈ C, x ∈ S, the same equation has a unique solution i ∈ R. R: rows, C: columns, S: symbols. Permuting R, C, S of a Latin square gets an equivalent Latin square. Eg. abcde baecd cdbea deabc ecdab

2. Two Latin square are called conjugate, if they have the same three rows of OA (may in different order). Def: An orthogonal array OA(n, 3) of order n and depth 3 is a 3 by n 2 array with the integers 1 to n as entries, such that for any 2 rows of the array, the n 2 vertical pairs occurring in these rows are different. rows 1111222233334444 Columns 1234123412341234 symbols 3241142343122134 3241 1423 4312 2134 1111222233334444 Columns 1234123412341234 rows3241142343122134 2431 4123 1342 3214

3. Two Latin squares for which the corresponding orthogonal arrays have the same three rows (possibly in different order) are called conjugates. Def: An n  n array A with cells empty and no symbol occurs more than once in any row or column is called a partial Latin square. When and how can we complete a partial Latin square to a Latin square of order n? Eg. Can the following partial Latin squares be completed? 1234 5 1 1 1 1 2 Conjugate?

4. Thm 1: A k  n Latin rectangle, k<n, can be extended to a (k+1)  n Latin rectangle. Pf: Let B j denote the set of positive integers that do not occur in column j of A. Each of 1, …, n occurs k times in A and therefore n-k times in B j ’ s. Any ℓ of B j ’ s contains ℓ(n-k) elements and thus at least ℓ different ones. (why?) B j ’s have the property H, and therefore have an SDR. This SDR can be adjoined as the (k+1)-st row. ▧

5. Def: L(n): the total number of different Latin squares of L(n) is known for only n=1, …,9. Thm 2: L(n) ≥ (n!) 2n /n n 2. Pf: The first row of a Latin square can be any permutation of 1,…,n. Suppose we have a Latin rectangle with k rows. By Thm 1, the number of choices for the next row is per B, where b ij =1 if i ∈ B j. By Thm 12.8, per J n = n!/n n, per A > per J n for any A≠J n. Thus per B ≥ (n-k) n n!/n n So we have ▧

6. Thm 4: Let A be a partial Latin square of order n in which cell (i, j) is filled iff i ≤ r and j ≤ s. Then A can be completed iff N(i) ≥ r+s-n for i=1, …,n, where N(i) denotes the number of elements of A that are equal to i. Pf: “ ⇒ ” s r A Latin square. For any i, it must appear r times in the first r rows and at most n-s of which appear in the last n-s columns. Thus N(i) ≥ r-(n-s) = r+s-n.

7. “ ⇐ ” Let B be the (0,1)-matrix of size r  n with b ij =1 iff the element j does not occur in row i of A. Every row of B has sum n-s. The j-th column of B has sum r-N(j) ≤ n-s. By Thm 7.5 (d=n-s) If A is an integral matrix and d any positive integer, then A=B 1 +B 2 + … +B d where each B i is an integral matrix whose entries, row-sums, col-sums, and sum of all entries are those of (1/d)·A, rounded up or down. s r n-s r 1 0 0 1 j

8. We have B=L (s+1) + … +L (n), where each L (t) is an r  n (0,1)-matrix with one 1 in each row and at most one 1 in each col. Eg. Let r=s=4, n=7 1234 5316 3152 7425 A:

9. Say L (t) =. For i=1, …,r, j=s+1, …,n by k if The example will add in the last column. Then by Thm 1, A can be completed to a Latin square. ▧

10. Thm 5. Let A be a Latin square of order n. Let B be an (n+1)  (n+1) array whose (i, j)-entry is the (i, j)- entry of A for i, j ≥ 1, i+j ≤ n+1, which has a new symbol α on the back-diagonal; and whose cells below the back-diagonal are empty. Then B can be completed to a Latin square of order n+1. Eg. 12345 43512 25134 51423 34251 A 12345 α 4351 α 251 α 51 α 3 α α B 12345 4351α 251α 51α 3α C

11. Pf: “ Strategy ” : Fill in the empty cells of C row by row. After the first r rows are filled in, we will have r  n a Latin rectangle in n+1 symbols 1, 2, …, n, α so that j-th column contains the same symbols, in some order, as they were in the first r rows of A; except the missing symbol α. Also we have that the r-1 missing symbols are all distinct. If is trivial for r=1 and 2. Assume it is done for some r, 2 ≤ r < n and that the missing symbols at this point are x n-r+2, x n-r+3, … x n.

12. To fill in the (r+1)-st row, we proceed as follows. Let the last removed r symbols in row r+1 of A be y n-r+1, y n-r+2, …, y n-1, y n. y n-r+1 is the symbol displaced by α and must be the new missing element in that column. If y n-r+1, x n-r+2, …, x n are distinct, we fill in row r+1 with y n-r+2, y n-r+3, …, y n. If not all distinct, consider y n-r+1 = x k 1 y k 1 = x k 2 y k m-1 = x k m so that y k m is not equal to any of the x j ’s. … y n-r+1, x n-r+2, …, x k 1,...,x k 2,...,x n y n-r+2, …, y k 1,..., y k 2,...,y n yk1xk1yk1xk1 yk2xk2yk2xk2

13. Then fill in row r+1 with y n-r+2, y n-r+3, …, y n except that y k 1, y k 2 …, y k m are omitted and replaced by x k 1, x k 2,…, x k m. Need to check that row r+1 contains distinct symbols. The new missing symbols will be y n-r+1 and x n-r+2, …, x n, except that x k 1, …, x k m are omitted and replaced by y k 1, …, y k m. Once all the cells of C are filled, the first n cells of the last row of B are to be the missing symbols x, thus an (n+1) x n Latin rectangle, and so obtain an Latin square of order n+1. ▧

14. Eg. 12345 4351α 251α 51α 3α 12345 4351α 251α4 51α 3α 32 12345 4351α 251α4 51α23 3α 432 missing 12345 4351α 251α4 51α23 3α452 x443242324231 y251451452 12345 α 4351α 251α 4 51α 23 3α 452 α 4231 B: 12345 43512 25134 51423 34251 A:

15. Thm 6. A Latin square of order n with at most n-1 filled cells can be completed to a Latin square of order n. Eg. 3 5 51 12345 43512 25134 51423 34251

16. Proof: (By induction on n). Suppose it holds up to n. Let L be a partial Latin square of order n+1 with at most n filled cells. First suppose there is a symbol x appears only once in L. Then we can permute rows and columns to move the filled cells above the back-diagonal, except x will be on the back- diagonal. Let the rows with filled cells have f 1,..., f k filled cells and x is in the row with f 1 filled cells, where f 1 +...+ f k ≤ n. x n+1- f 1 n+1- f 1 - f 2 If x is in column j, interchange with column f 1 +1.

17. By the induction hypothesis, the part above the back diagonal (with at most n-1 filled cells) can be completed to a Latin square A of order n based on symbols other than x. By Thm 5, it can be completed to a Latin square A of order n+1. If L has a row (or column) with exactly one filled cell, then the above also holds, because a conjugate of L will have the property that some symbol appears exactly once and a conjugate of L can be completed iff L can be completed. x 1 1 1 2 123 4 R 1 2 3 4 C 1 2 3 4 S 1 1 1 2 R 1 2 3 4 C 1 2 3 4

18. If no row or column contains exactly one filled cell, then the filled cells are contained in at most m rows and m columns, where m = [n/2]. We may permute rows and columns so that filled cells lie in the upper left sub-array of order m. And it is easy fill every empty cell in the m  m, since m = [n/2] and we have n+1 symbols available. By Thm 4, every m  m Latin rectangle can be completed to a Latin square of order n+1. 1 1 1 2 m=4, n= 2 m+1= 9