Dimensions of Magic Squares and Magic Square Codes

Transcription

1 Dimensions of Magic Squares and Magic Square Codes Jon-Lark Kim This is a joint work with Wooseok Jung, Yeonho Kim, Kisun Lee Department of Mathematics Sogang University, S. Korea The SIAM Conference on Applied Algebraic Geometry August 4, 2015

2 Outline Seok-Jeong Choi s orthogonal Latin squares MDS codes from Latin squares Dimension of the space of magic squares Open problems

3 History Seok-Jeong Choi is a Korean mathematician who invented orthogonal Latin squares 61 years before Euler found them. In this talk, we describe his work on Latin squares and associated magic squares. Then we describe the interaction among Latin squares, magic squares and codes, including some interesting research problems in this area.

4 Last Year Conference Poster

5 Seok-Jeong Choi Seok-Jeong Choi (or Seok Jung Choi) ( ) was a prime minister under King Suk Jong, Chosun Dynasty. As a mathematician, he wrote a book Koo-Soo-Ryak. In this book, he introduced two orthogonal Latin squares of order 9 to construct a magic square of order 9. His idea of Latin squares was 61 years before Euler presented them in This fact has been less known to many mathematicians. This is why I want to introduce him and other related works.

6 Handbook of Combinatorial Designs, 2nd Edition, 2007

7 Choi s two orthogonal Latin squares of order 9

8 Magic square of order 9 with magic sum 369 Transform (i, k) in the orthogonal Latin squares into 9(i 1) + j in the magic square of order 9.

9 Choi-Golomb-Posner s MDS code Theorem (Golomb-Posner, 1964) There is a q-ary MDS (n, q 2, n 1) code if and only if there are n 2 mutually orthogonal Latin squares of order q. In particular, if n = 4 and q = 9, then we obtain a 9-ary MDS (4, 9 2, 3) code from Choi s two orthogonal Latin squares of order 9. We call it Choi-Golomb-Posner s MDS code, CGP9!. CGP9 has 81 codewords of the form (1, 1, 5, 1), (1, 2, 6, 3), (1, 2, 4, 2),..., (1, 9, 1, 5) (2, 1, 4, 3), (2, 2, 5, 2), (2, 3, 6, 1),..., (2, 9, 3, 4)... (9, 1, 9, 5), (9, 2, 7, 4), (9, 3, 8, 6),..., (9, 9, 5, 9).

10 Definition of magic squares A magic square of size n and weight d over a field F is a matrix (a ij ) (1 i, j n) with entries in F (repetition allowed) satisfying the following conditions: n j=1 n i=1 i j=0 i+j=n+1 a ij = d for all 1 i n (row sums) a ij = d for all 1 j n (column sums) a ij = d (diagonal sum) a ij = d (antidiagonal sum) where n N and d F.

11 The vector of magic squares The set of all magic squares of size n over a field F forms a vector space. Ratliff (Monthly, 1959) and Small (Monthly, 1988) found the dimension of the vector space of all magic squares of size n.

12 Pandiagonal magic square This magic square has the additional property that all of its 8 diagonals also have the same sum. A magic square with this extra property is called a pandiagonal magic square.

13 Generalized magic squares Hou, Lecuona, Mullen, and Sellers (Linear Algebra and Its Applications 2013) generalized the concept of the magic square over a field in the following manner. Let F be a field, d F, n N, and 0 k n. Let M n,k (d) be the set of all n n matrices (a ij ) (1 i, j n) with entries in F satisfying the following conditions: n j=1 n i=1 i j= l+1 i+j= l+2 a ij = d for all 1 i n (row sums) a ij = d for all 1 j n (column sums) a ij = d for all 1 l k (diagonal sums) a ij = d for all 1 l k (antidiagonal sums) where the subscripts are taken modulo n.

14 Define M n,k = Dimension of M n,k over F d F M n,k (d). When k = 1, we have the space of magic squares over a field. When k = n, we have the space of pandiagonal magic squares over a field. Then one has dim F M n,k = dim F M n,k (0) + { 1 if M n,k (1) 0 if M n,k (1) =. Hou, et al. (2013) determined dim F M n,k (0) completely. They also proved that if the characteristic of a field F is neither 2 nor 3, then M n,k (1) for all n N, and 0 k n. Open question: Determine a necessary and sufficient condition that M n,k (1) when char F is 2 or 3.

15 Our results when charf is 2 With three undergraduate students at Sogang University, Wooseok Jung, Yeonho Kim, Kisun Lee, we completely solve this open question as follows. [Appeared in Linear Algebra and its Applications, The dimension of magic squares over fields of characteristics two and three, Vol. 472, 1 May 2015, pp ] charf n k dim F M n,k n 2 k n n 2 4n + 8 n 0 (mod 4) k = n 3 n 2 4n + 9 n 0 (mod 2) 0 k n 4 n 2 2n 2k + 2 = 2 n 1 k n n 2 4n + 5 n 2 (mod 4) k = n 2 n 2 4n k n 3 n 2 2n 2k + 2 n 1 (mod 2) n 1 k n n 2 4n k n 2 n 2 2n 2k + 2 Table : Values of dim F M n,k

16 Our results when charf is 3 charf n k dim F M n,k n 2 k n n 2 4n + 8 n 0 (mod 18) k = n 3 n 2 4n + 9 n 0 (mod 6) 0 k n 4 n 2 2n 2k + 2 n 2 k n n 2 4n + 7 n 6, 12 (mod 18) k = n 3 n 2 4n k n 4 n 2 2n 2k + 2 = 3 n 9 (mod 18) n 2 k n n 2 4n + 7 n 3 (mod 6) 0 k n 3 n 2 2n 2k + 2 n 3, 15 (mod 18) n 2 k n n 2 4n k n 3 n 2 2n 2k + 2 n 2, 4 (mod 6) n 1 k n n 2 4n k n 2 n 2 2n 2k + 2 n 1, 5 (mod 6) n 1 k n n 2 4n k n 2 n 2 2n 2k + 2 Table : Values of dim F M n,k

17 Observation Proposition If M n,k (1), then M n,k (1) for all 0 k k n. Equivalently, if M n,k (1) =, then M n,k (1) = for all 0 k k n.

18 charf = 2 case n 1 (mod 2) Lemma Let the characteristic of a field F be 2. If n 1 (mod 2), then M n,k (1) for all k with 0 k n. 1 n M n,n (1) 1 1

19 charf = 2 case n 0 (mod 4) Lemma Let the characteristic of a field F be 2. If n 0 (mod 4), then M n,k (1) for all k with 0 k n. Add the below matrix to another matrix in the next page

20 Proof: Continued r : 0 c : 0 d : 1 ad : 1 ( n 2 )th ( n 2 + 1)th

21 charf = 2 case n 2 (mod 4) I Lemma Let the characteristic of a field F be 2. If n 2 (mod 4), then M n,k (1) for all k with 0 k n 2.

22 Example We give a small size example which is in M 6,4 (1)

23 Example This is the sum of the followings : =

24 Generalization =

25 charf = 2 case n 2 (mod 4) II Lemma Let the characteristic of a field F be 2. If n 2 (mod 4), then M n,k (1) = for all k with n 1 k n.

26 Idea of the proof By proposition, it is enough to show that M n,n 1 (1) is empty. Assume that there is a matrix in M n,n 1 (1) for n 2 (mod 4). n/2 n/2 n/2 c 2k 1 + r 2k 1 + d 2k 1 = k=1 k=1 k=1 1 i,j n a ij A contradiction.

27 Summary for charf = 2 case Theorem Let the characteristic of a field F is 2. Then M n,k (1) = if and only if n 2 (mod 4) and n 1 k n.

28 charf = 3 case Now we assume that charf = 3. For convenience, we assume that F = {0,1,2}. We break n into four main cases as follows. n 0 (mod 6) n 0 (mod 18) n 6, 12 (mod 18) n 3 (mod 6) n 9 (mod 18) n 3, 15 (mod 18) n 2, 4 (mod 6) n 1, 5 (mod 6) We apply various arguments as well as the methods for charf = 2 case. Omit the details.

29 Summary for charf = 3 case Theorem Let the characteristic of a field F is 3. Then M n,k (1) = if and only if n 3, 15 (mod 18) and n 2 k n or n 6, 12 (mod 18) and n 3 k n.

30 Open problems Let M C = M n,k (0) and k = dim F M n,k (0). We call M C the magic square [n, k] code over F. Questions: 1. Find a (canonical) basis for M C. 2. Find the minimum distance of M C.

31 Open problems Let M C = M n,k (0) and k = dim F M n,k (0). We call M C the magic square [n, k] code over F. Questions: 1. Find a (canonical) basis for M C. 2. Find the minimum distance of M C. THANK YOU FOR YOUR ATTENTION!