1. 拉丁方陣 交大應數系 蔡奕正

2. Definition A Latin square of order n with entries from an n-set X is an n ＊ n array L in which every cell contains an element of X such that every row of L is a permutation of X and every column of L is a permutation of X.

3. Definition Let X be a finite set of cardinality n, and let ο be a binary operation defined on X. We say that the pair ( X, ο ) is a quasigroup of order n provided that the following two properties are satisfied: 1.For every x,y  X, the equation x ο z = y has a unique solution for z  X. 2.For every x,y  X, the equation z ο x = y has a unique solution for z  X.

4. Theorem: Suppose ο is a binary operation defined on a finite set X of cardinality n. Then ( X, ο ) is a quasigroup if and only if its operation table is a Latin square of order n. Note: We obtain Latin squares that are the multiplication tables of groups.

5. Example: ( c ⊕ e ) ⊕ d ≠ c ⊕ ( e ⊕ d ) a b c d e b a e c d c d b e a d e a b c e c d a b

6. Definition: 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 two rows of the array, the n^2 vertical pairs occurring in these rows are different. Definition: Two Latin squares for which the corresponding orthogonal arrays have the same three rows ( possibly in different order ) are called conjugates.

7. Example: 1111222233334444 1234123412341234 3241142343122134 3241142343122134 1234123412341234 1111222233334444 3241 1423 4312 2134 2431 4123 1342 3214

8. 1. 兩拉丁方陣被稱為等價 (equivalent) 代表此 兩拉丁方陣可利用換行換列或換元素而 得到相同的拉丁方陣. 2. 兩拉丁方陣為等價之例子 : ( 先作一二列交換, 再把元素 1 及元素 3 交換 ) 1 2 3 2 3 1 3 1 2 2 1 3 3 2 1 1 3 2

9. Note: Two orthogonal arrays are called isomorphic if one can be obtained from the other by permutations of the elements in each of the rows and permutations of the rows and columns of the array. Two Latin squares for which the corresponding orthogonal arrays are isomorphic are also called isomorphic. This means that one is equivalent to a conjugate of the other.

10. Example: 111222333 123123123 123231312 123231312 123123123 111222333 231312123 312312312 111222333 1 2 3 2 3 1 3 1 2 1 3 2 2 1 3 3 2 1 2 1 3 3 2 1 1 3 2

11. Definition: An n ＊ n array A with cells which are either empty or contain exactly one symbol is called a partial Latin square if no symbol occurs more than once in any row or column. Example: 以下兩個 partial Latin square 都不能 形成完整的拉丁方陣. 1 2 3 4 1 1 1 2

12. Note: 上述兩個 partial Latin square 並非真的在實 質上不同， 實際上為 conjugates 。 Definition: If the first k rows of a partial Latin square ( k ≦ n) are filled and the remaining cells are empty, then A is called a k ＊ n Latin rectangle. 1 1 1 2 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 1 2 2

13. Theorem: A k ＊ n Latin rectangle, k ＜ n, can be extended to a ( k+1 ) ＊ n Latin rectangle. Proof: Let B j ={ a | a 不出現在第 j 行 } j=1,2,……,n ∴只須證明 B 1,B 2, ……,B n 具 SDR 即可 W.L.O.G. 取 r 個集合為 B 1,B 2, ……,B r ∵ B 1,B 2, ……,B r 共 r(n-k) 個元素 ( 可能重複 ) 又每個元素至多出現 (n-k) 次 ∴ B 1,B 2, ……,B r 至少有 r 個不同元素 ∴ B 1,B 2, ……,B n 具 SDR

14. Theorem:L(n) ≧ (n  )^2n / n^(n^2) (We denote by L(n) the total number of different Latin squares of order n) Proof: to construct a Latin square of order n, we can take any permutation of 1 to n as the first row. If we have constructed a Latin rectangle with k rows, then the number of choices for the next row is perB where if b ij =1 if i  B j. By theorem 12.8, this permutation is at least (n-k)^n ． n!/n^n. so we find L(n) ≧ [n!][ (n-1)^n ． n!/n^n][ (n-2)^n ． n!/n^n] ……[1^n ． n!/n^n] = (n  )^2n / n^(n^2)

15. Theorem:(due to H. J. Ryser in 1951) Let A be a partial Latin square of order n in which cell ( i, j ) is filled if and only if i ≦ r and j ≦ s. Then A can be completed to a Latin square if and only if A( i )  r+s-n for i=1,……,n, where A( i ) denotes the number of elements of A that are equal to i. Proof: (  ) 把完成好的 Latin square 分成四部份如右 考慮 B+D, i 在 B+D 中恰好出現 n-s 次, i=1,……,n ∴ B( i ) ≦ n-s 又 A( i )+ B( i ) = r, ∴ B( i ) = r- A( i ) ∴ r- A( i ) ≦ n-s i.e A( i ) + n-s  r, 即 A( i )  r+s-n A B C D

16. (  ) Let B be the (0,1)-matrix of size r ＊ n with bij=1, if and only if the element j does not occur in row i of A. Clearly every row of B has sum n-s. The j-th column of B has sum r-A( j ) ≦ n-s. ( ∵ A( j )  r+s-n ) By Theorem7.5 ( with d:=n-s ) 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 column. Say L (t) = [l ij(t) ]. Then we fill the cell in position ( i,j ) of A, i = 1,……,r, j = s+1,……,n, by k if l ik(j) = 1. Thus A is changed into a partial Latin square of order n with r complete rows, i.e. a Latin rectangle. So this can be completed to a Latin square of order n.

17. Theorem : A partial Latin square of order n with at most n-1 filled cells can be completed to a Latin square of order n. ( 此為有名的 Evans conjecture, 且被 B. Smetaniuk 於 1981 年證出 ) Proof: case1: Assume that no row or column contains exactly one filled cell. Thus the filled cells are contained in at most m rows and columns, where m:=  n/2 . We may permute rows and columns so that all filled cells lie in the upper left subarray of order m. ∵ Every m ＊ m Latin rectangle can be completed to a Latin square of order n. So it will suffice to fill in the unfilled cells of the upper left subarray of order m to get a Latin rectangle. But this is easy, as we have n symbols available and we may proceed in any order, filled in a cell of the subsquare with a symbol not yet used in the row or column containing the cell.

18. case2: Assume there is a symbol x of L that occurs only once in L. 詳細證明頗複雜, 所以省略 case1 之 Example: 14 13 25 14 13 25 314 136 245