Hong-Yeop Song hysong@yonsei ac kr hysong@yonsei.ac.kr - - PowerPoint PPT Presentation

Ch i' th l L ti S Choi's orthogonal Latin Squares is at least 67 years earlier than y Euler's

2008 Global KMS International Conference

• 2008. 10. 23.~10. 25.

JEJU ICC, KOREA

Hong-Yeop Song

hysong@yonsei ac kr hysong@yonsei.ac.kr http://coding.yonsei.ac.kr School of Electrical and Electronic Engineering g g Yonsei University, Seoul, KOREA

Content l d Content ☞ Prelude M i S ☞ Magic Squares L i S d O h li ☞ Latin Squares and Orthogonality O h l L i S d M i S ☞ Orthogonal Latin Squares and Magic Squares S ☞ Summary

l d Prelude

Seok-Jeong Choi Seok-Jeong Choi Leonhard Paul Euler Sang Geun Hahn Hong-Yeop Song Hong Yeop Song

Leonhard Paul Euler (1707 – 1783) was a pioneering Swiss 최석정(崔錫鼎, 1646년~1715년)은 조선 후기 의 문신이다. 여러 요직을 거쳤고 1701년 영 의정이 되었다 그는 수학서 《구수략》을 저술 was a pioneering Swiss mathematician and physicist who spent most of his life in Russia and Germany Euler is considered to be 의정이 되었다. 그는 수학서 《구수략》을 저술 했는데 이는 그가 1710년 영의정을 그만둔 이 후 작성했을 것으로 추측된다. Choi, Seok-Jeong (1646-1715) has served Yi

• Germany. Euler is considered to be

the preeminent mathematician of the 18th century and one of the greatest of all time C o , Seo Jeo g ( 6 6 5) as se ed Dynasty in various positions, including the prime minister in 1701. He wrote Koo-Soo- Ryak which was believed to be written greatest of all time. y after he retired in 1710.

Pair of

• rthogonal
• rthogonal

Latin squares

1782 1715 67 years?? 67 years??

From Mathematics Genealogy Project North Dakota State University, http://genealogy.math.ndsu.nodak.edu

Hong-Yeop Song, Ph.D. University of Southern California 1991, USA,

Advisor: Solomon W Golomb Advisor: Solomon W. Golomb Solomon Wolf Golomb, Ph.D. Harvard University 1957, USA, Advisor: David Widder David Vernon Widder, Ph.D. Harvard University 1924, USA, Advisor: George Birkhoff George David Birkhoff, Ph.D. University of Chicago 1907, USA, Advisor: E. H. Moore

• E. H. (Eliakim Hastings) Moore, Ph.D. Yale University 1885, USA, Advisor: H. A. Newton
• H. A. (Hubert Anson) Newton, B.S. Yale University 1850, USA, Advisor: Michel Chasles

Michel Chasles, Ph.D. Ecole Polytechnique 1814, France, Advisor: Simeon Poisson Simeon Denis Poisson France Advisor: Joseph Lagrange Simeon Denis Poisson, France, Advisor: Joseph Lagrange Joseph Louis Lagrange, Advisor: Leonhard Euler

PART 1: Pre-history H YSong (Grad student at USC) has met Dr Denes (co-author of “Latin ~1991 H.Y.Song (Grad student at USC) has met Dr. Denes (co author of Latin Squares and Their Applications” 1974) who was a good friend of his advisor Dr. Golomb. ~1991 S G Hahn (Prof at KAIST Math dept ) has seen the original printing of ~1993 S.G.Hahn (Prof. at KAIST Math dept.) has seen the original printing of Koo-Soo-Ryak by Choi Seok-Jeong, and published a (Korean) paper on KSESM journal about the original Pair of Orthogonal Latin Squares of

• rder 9 He also had some correspondents with Dr Denes about this

~1993

• rder 9. He also had some correspondents with Dr. Denes about this.

H.Y.Song (Senior Engineer at Qualcomm) published a contribution to CRC Handbook of Combinatorial Designs, co-authored with one of co- editors, J. Dinitz (Prof. at U. Vermont). 1994

PART 2: how it begins H Y S (P f t Y i U i ) 1997

• H. Y. Song (Prof. at Yonsei Univ.)
• met Dr. Denes at Ulm, Germany, attending IEEE ISIT, and he was

asked to find some Korean journal paper;

• was SO SURPRISED to see the paper by S G Hahn sent the paper

1997

• was SO SURPRISED to see the paper by S. G. Hahn, sent the paper

and its English translation to Dr. Denes;

• made a (first) call to Prof. Hahn asking a copy of the original printing.
• Prof. Hahn published a newsletter contribution to KSESM about this.

1998

PART 3: how it ends

• Dr. Denes has passed away. He was not able to publish the 2nd edition
• f his book, Latin Squares and Their Applications. We guess he

was planning to say something about Choi’s POLS there 2001(?) was planning to say something about Choi s POLS there.

• H. Y. Song
• was asked to revised his contribution to CRC Handbook of

2005

• was asked to revised his contribution to CRC Handbook of

Combinatorial Designs for 2nd edition;

• has asked to see the galley version of a newly added section on

“History of Combinatorial mathematics” and decided it is the time; History of Combinatorial mathematics , and decided it is the time;

• sent all he had so far to Dr. Dinitz.

The page of the book discussing Choi’s POLS of order 9

The page of the book showing Choi’s birth place and year.

Pair of orthogonal Latin squares of order 9

Koo-Soo-Ryak, 1715(?), written by Seok-Jeong Choi (최석정).

Koo-Soo-Ryak, 2006, translated by Hae-Nam Chung and Min Heo (정해남∙허민, 교우사)

Choi has constructed the following POLS of order 9 to form a magic square.

Pictures from Monthly Mag. of Dong-Ah Sciences, Aug. 2008, Article by S.K. Kang

Handwritten copy (early 1700?) – Library of Yonsei University C.H. Lee (Prof. Math, Yonsei University) has donated this in the 1950s.

Handwritten copy (early 1700?) – Library of Yonsei University

Magic Squares Magic Squares

A magic square of order n with magic constant d is an n ⅹ n array of n2 g q g y integers such that

• 1. Every row-sum is d
• 2. Every column-sum is d

y

• 3. LR diagonal-sum is d
• 4. RL diagonal-sum is d.

A magic square of order n is called normal if the entries are 1,2,…,n2. A semi-magic square satisfies only the conditions 1 and 2 above. There are numerous ways to generalize the magic squares: magic circles, magic cubes, magic graphs, magic series, magic tesseracts, magic hexagons magic diamonds magic domino tilings magic stars magic hexagons, magic diamonds, magic domino tilings, magic stars, bimagic squares, trimagic squares, etc. See http://mathworld.wolfram.com/topics/MagicFigures.html See ttp:// at

• d. o

a .co /top cs/ ag c gu es. t Or http://www.multimagie.com/indexengl.htm

Picture from Mathematics Concert (수학 콘서트, 박경미, 동아시아 2006)

Lo Shu [洛書] (낙서)

Some three thousands years ago in China, when King y g , g tried to appease the river god, a turtle with the following pattern of numbers on its back had appeared…..

For an odd integer n≥1, there is an “easy” construction for a (normal) magic square of order n. Example of order 5 Example of order 5

RU corner of Melencolia I Melencolia I Engraving (1514) Albrecht Dürer (1471∼1528) Albrecht Dürer (1471 1528)

It h th dditi l t th t It has the additional property that the sums in any of the four quadrants (composite magic square), as well as the sum of the middle four numbers, are all 34 (Hunter and Madachy 1975, p. 24). It is thus a gnomon magic square. In addition, any pair of numbers symmetrically placed about the cen y p y y p ter of the square sums to 17, a property making the square even m

• re magical.

Picture from Mathematics Concert (수학 콘서트, 박경미, 동아시아 2006)

Magic square of order 4 (non-normal) by Srinivasa Ramanujan (1887 – 1920, India)

22 12 18 87 22 12 18 87 21 2 84 32 21 2 84 32 16 7 92 24 16 7 92 24 26 4 82 27

Magic constant = 139 Magic constant = 139

Latin Squares Latin Squares

A latin square of order n is an n ⅹ n array of n different symbols such that each row and each column is a permutation of these n symbols. 1 2 3 3 1 2 2 3 1 1 2 3 2 3 1 3 1 2 1 2 3 4 2 1 4 3 3 4 1 2 1 2 2 1 1 1 2 3 4 3 4 1 2 4 3 2 1 4 3 2 1 2 1 4 3 A transversal in a latin square of order n is a set of n cells (or positions), Two latin squares A=(aij) and B=(bij) of order n are called a pair of

• rthogonal latin squares (POLS or a graeco-latin square) of order n
• ne from each row and column, containing each of n symbols exactly once.
• rthogonal latin squares (POLS, or a graeco latin square) of order n

if the n2 ordered pairs (aij,bij) are all distinct for 1≤i,j≤n. In this case, we say that A is orthogonal to B or vice versa.

α β γ γ α β a b c b c a αa βb γc γb αc βa γ α β β γ α b c a c a b γb αc βa β c γa αb

Theorem: A Latin square of order n has an orthogonal mate if and only if it can be decomposed into n disjoint transversals.

Euler’s 36 officers Problem and the conjecture

How can a delegation of six regiments, each of which sends a colonel, a lieutenant-colonel, a major, a captain, a lieutenant, and a sub-lieutenant be arranged in a regular 6 ⅹ 6 array such that no row or column g g y duplicates a rank or a regiment? Does there exist a pair of orthogonal latin squares of order 6? Euler conjectured that there do not

Euler, L., Recherches sur une nouvelle espece de quarr

j exist Graeco-Latin squares of order 4k+2 for k=1, 2, ....

es magiques (1782). Tarry, G. "Le problème de 36 officiers." Compte Rendu de l'Assoc. Français Avanc. Sci. Naturel 1, 122-123, 1900.

While it is true that no such square of

• rder six exists,

Parker, E. T. "Orthogonal Latin Squares."

• Not. Amer. Math. Soc. 6, 276, 1959.

Bose, R. C.; Shrikhande, S. S.; and Parker, E. T. "Further Results on the Construction of Mutually Orthogonal L

such squares were found to exist for all other orders of the form .

Results on the Construction of Mutually Orthogonal L atin Squares and the Falsity of Euler's Conjecture."

• Canad. J. Math. 12, 189, 1960.

Picture from Mathematics Concert (수학 콘서트, 박경미, 동아시아 2006)

Pair of orthogonal Latin squares of order 10

Some Applications of Latin Squares I Sudoku Sudoku

Some Applications of Latin Squares II - parallel access

Kichul Kim and Viktor K. Prasanna, “Latin Squares for Parallel Array Access," IEEE Transactions and Parallel and Distributed Systems, y

• vol. 4, Issue 4, pp. 361-370, April 1993.

Some Applications of Latin Squares III -- parallel access

Dae-Son Kim, Hyun-Young Oh, and Hong-Yeop Song, "Collision-free Interleaver composed of a Latin Square Matrix for Parallel-architecture Turbo Codes," IEEE Communications Letters,

• vol. 12, Issue 3, pp. 203-205, March 2008.

INTERLEAVER – PARALLELIZE??

Some Applications of Latin Squares IV - more than a pair?

Proposition: If there exists a set S of mutually orthogonal Latin squares Definition: A set of Latin squares of order n is called mutually orthogonal if any pair in the set is orthogonal. p y g q (MOLS) of order n, then the cardinality of S is at most n-1. When |S|=n-1, it is called a complete set of MOLS of order n.

Th Th i t l t t f MOLS f d if d Theorem: There exists a complete set of MOLS of order n if and

• nly if there exists a finite projective plane of order n.

Theorem: There exists a finite projective plane of order n if n is a power

• f a prime.

Proof: n is a power of a prime → there exists a finite field Fn of size n. p p

n

Let e0=0, e1, e2, … , en-1 be some ordering of the elements of Fn, and Put A(k) = (a(k)

ij) be the k-th latin square of order n. Then

a(k)

ij =ekei+ej for k=1,2,…,n-1, and i,j=0,1,2,…,n works. ■

Conjecture (PPC): The above is also a necessary condition. That is, a finite projective plane of order n exists if and only if n is a power of a

ij k i j

j p j p y p prime.

• 1994. D. Gordon. PPC is true for n up to 2,000,000.

Pair of Pair of Orthogonal Latin Orthogonal Latin S d Squares and Magic Squares g q

Pictures from Monthly Mag. of Dong-Ah Sciences, Aug. 2008, Article by S.K. Kang

Proof of constant row sum and column sum:

sum column any ) 1 ( 3 sum row any = + − =

∑ ∑

q p

q p

Question: What about diagonal sums? Will it work for any pair of orthogonal Latin squares of order n ?

Consider, for example, the earlier example of POLS of order 4. 1 2 3 4

1 6 11 16 7 4 13 10

1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 1 2 3 4 1 2 3 4 2 1 4 3

7 4 13 10 12 15 2 5

1 2 3 4 3 4 1 2 4 3 2 1 3 4 1 2 4 3 2 1 4 2 1 4 3 3 4 1 2 4 3 2 1

14 9 8 3

column-sum = row-sum = 34 = magic constant 4 3 2 1 2 1 4 3 2 1 4 3 4 3 2 1 g 1diagonal-sum = 1+4+2+3 = 10 2diagonal-sum = 13+14+15+16 = 58

It is only a semi magic square!!! It is only a semi-magic square!!!

Theorem: For any given pair of orthogonal Latin squares of order n, one can construct a (normal) semi-magic square of order n It will be magic if can construct a (normal) semi magic square of order n. It will be magic if

• 1. Both diagonals are transversals in both Latin squares, or
• 2. n=odd and two of the four diagonals of squares are constant and

equal to (n+1)/2. equal to (n 1)/2.

Examples 1 4 1 2 1 3 5 2 4 1 5 4 3 2 1 4 2 1 3 5 4 1 2 2 5 3 1 4 and 1 3 5 2 4 5 2 4 1 3 4 1 3 5 2 3 5 2 4 1 1 5 4 3 2 3 2 1 5 4 5 4 3 2 1 2 1 5 4 3 and 5 4 2 5 1 4 4 5 3 5 2 4 1 2 4 1 3 5 4 3 2 1 5 3 1 2 3

• r

3 5 2 4 1 1 3 5 2 4 2 1 5 4 3 1 5 4 3 2 3 2 3 4 3 5 3 3 3 1 3 4 and 1 3 5 2 4 4 1 3 5 2 2 4 1 3 5 5 2 4 1 3 1 5 4 3 2 5 4 3 2 1 4 3 2 1 5 3 2 1 5 4 and 5 3 3 4 5 2 4 1 3 3 2 1 5 4

Some observations: Any row permutation or column permutation or symbol permutation will not change the row sum or the column sum of a semi-magic square. I th bi ti f / l / b l t ti th t t Is there a combination of row/column/symbol permutations that convert the semi-magic square from any POLS into a magic square? E l

1 6 11 16

Example:

14 9 3 8

Sum=34

7 4 13 10 12 15 2 5 7 4 10 13 12 15 5 2 12 15 2 5 14 9 8 3 12 15 5 2 1 6 16 11

Sum=34

Three Questions:

• 1. Is it true that the semi-magic square from a POLS can be

transformed into a magic square by a combination of row/column/symbol permutations?

We need both the diagonal sums to be the magic constant.

• 2. Is it true that the semi-magic square from a POLS can be

transformed into a semi-magic square by a combination g q y

• f row/column/symbol permutations such that at least
• ne diagonal-sum is the magic constant?
• 3. Does there exists a semi-magic square (from a POLS)

that can never be transformed into a form in which that can never be transformed into a form in which none of the two diagonal sums is the magic constant?

Some observation

1 6 11 16 7 4 13 10 14 9 3 8 7 4 10 13 12 15 2 5 7 4 10 13 12 15 5 2 14 9 8 3 1 6 16 11

These 4 positions are symmetric pattern of two transversals

• f both RED and BLUE latin squares at the same set of positions.

1 6 11 16 7 4 13 10 14 9 3 8 7 4 10 13 12 15 2 5 7 4 10 13 12 15 5 2 14 9 8 3 1 6 16 11

Three Questions Again:

• 1. Is it true that, for any POLS, either (1) they have a

symmetric pattern of two transversals in both squares at the same set of n positions or (2) n=odd and two of such four transversal positions contain a constant?

• pen (I think it is false)
• - open. (I think it is false)
• 2. Is it true that, for any POLS, either (1) they have a
• 2. Is it true that, for any POLS, either (1) they have a

transversal in both squares at the same set of n positions or (2) n=odd and the set of n positions containing a constant in one square is a transversal in the other square?

• pen (I think it is true)
• - open. (I think it is true)
• 3. Find a pair of orthogonal Latin squares of order, say n,
• 3. Find a pair of orthogonal Latin squares of order, say n,

such that none of the above is satisfied.

Observation 1:

Given ANY POLS, A and B, of order n (even or odd) and any n cells of A containing a constant, the n positions in B corresponding to these n cells in A must be a transversal in B. corresponding to these n cells in A must be a transversal in B.

constant transversal A= =B Remark: From the above observation, when n=odd, any given POLS , , y g can be transformed by a combination of symbol/row/column permutations into a form that leads to a semi-magic square in which at least one of LR or RL diagonal sums is the magic constant. g g

Observation 2 (I do not have a proof but …)

Any POLS has a transversal in both squares at the same set of n positions.

transversal transversal A= =B Remark: From the above observation, when n=even, any given POLS , , y g can be transformed by a combination of symbol/row/column permutations into a form that leads to a semi-magic square in which at least one of LR or RL diagonal sums is the magic constant. g g

Pictures from Monthly Mag. of Dong-Ah Sciences, Aug. 2008, Article by S.K. Kang

It is surprising that both diagonal- sums equal to the magic constant sums equal to the magic constant 369.

Row sum = 42 Column sum = 42 RL diagonal sum = 42 LR diagonal sum = 42 114 213 42 51 123 195 Row sum = 369 Column sum = 369 51 123 195 204 33 132 Column sum = 369 RL diagonal sum = 369 LR diagonal sum = 369

Choi’s POLS of order 9 (1715, KOO-SOO-RYAK) for magic square

5 6 4 8 9 7 2 3 1 4 5 6 7 8 9 1 2 3 1 3 2 7 9 8 4 6 5 3 2 1 9 8 7 6 5 4

for magic square Rows are palindromes.

4 5 6 7 8 9 1 2 3 6 4 5 9 7 8 3 1 2 2 3 1 5 6 4 8 9 7 3 2 1 9 8 7 6 5 4 2 1 3 8 7 9 5 4 6 7 9 8 4 6 5 1 3 2

As 3 x 3 arrays, they

2 3 1 5 6 4 8 9 7 1 2 3 4 5 6 7 8 9 3 1 2 6 4 5 9 7 8 7 9 8 4 6 5 1 3 2 9 8 7 6 5 4 3 2 1 8 7 9 5 4 6 2 1 3

are POLS of order 3. And each 3 x 3 array is a latin square.

8 9 7 2 3 1 5 6 4 7 8 9 1 2 3 4 5 6 9 7 8 3 1 2 6 4 5 4 6 5 1 3 2 7 9 8 6 5 4 3 2 1 9 8 7 5 4 6 2 1 3 8 7 9

They are NOT Sudoku, but both can be

9 7 8 3 1 2 6 4 5 5 4 6 2 1 3 8 7 9

Two of the four diagonals contain the constant 5 transformed simultaneously to Sudoku. Two of the four diagonals contain the constant 5 and the remaining two are transversals.

Choi’s POLS of order 9 (1715, KOO-SOO-RYAK) Kim and Prasanna’s POLS of order 9 (1993 IEEE Trans PD S) 5 6 4 8 9 7 2 3 1 1 3 2 7 9 8 4 6 5 ( , ) for magic square (1993, IEEE Trans P .D.S) for parallel access 0 1 2 3 4 5 6 7 8 0 3 6 2 5 8 1 4 7 4 5 6 7 8 9 1 2 3 6 4 5 9 7 8 3 1 2 2 3 1 5 6 4 8 9 7 1 2 3 4 5 6 7 8 9 3 2 1 9 8 7 6 5 4 2 1 3 8 7 9 5 4 6 7 9 8 4 6 5 1 3 2 9 8 7 6 5 4 3 2 1 3 4 5 6 7 8 0 1 2 6 7 8 0 1 2 3 4 5 2 0 1 5 3 4 8 6 7 5 3 4 8 6 7 2 0 1 1 4 7 0 3 6 2 5 8 2 5 8 1 4 7 0 3 6 3 6 0 5 8 2 4 7 1 4 7 1 3 6 0 5 8 2 1 2 3 4 5 6 7 8 9 3 1 2 6 4 5 9 7 8 8 9 7 2 3 1 5 6 4 7 8 9 1 2 3 4 5 6 9 8 7 6 5 4 3 2 1 8 7 9 5 4 6 2 1 3 4 6 5 1 3 2 7 9 8 6 5 4 3 2 1 9 8 7 5 3 4 8 6 7 2 0 1 8 6 7 2 0 1 5 3 4 1 2 0 4 5 3 7 8 6 4 5 3 7 8 6 1 2 0 4 7 1 3 6 0 5 8 2 5 8 2 4 7 1 3 6 0 6 0 3 8 2 5 7 1 4 7 1 4 6 0 3 8 2 5 7 8 9 1 2 3 4 5 6 9 7 8 3 1 2 6 4 5 6 5 4 3 2 1 9 8 7 5 4 6 2 1 3 8 7 9 palindromic symmetric 4 5 3 7 8 6 1 2 0 7 8 6 1 2 0 4 5 3 7 1 4 6 0 3 8 2 5 8 2 5 7 1 4 6 0 3 palindromic symmetric Sudoku Leads to a magic square All the 3 x 3 window sum is l h i NOT Sudoku Leads to a magic square equal to the magic constant.

Start with this (CHOI) Row permutation: 5 6 4 8 9 7 2 3 1 4 5 6 7 8 9 1 2 3 6 4 5 9 7 8 3 1 2 4 5 3 7 8 6 1 2 0 3 4 5 6 7 8 0 1 2 5 3 4 8 6 7 2 0 1 0 1 2 3 4 5 6 7 8 3 4 5 6 7 8 0 1 2 6 7 8 0 1 2 3 4 5 6 4 5 9 7 8 3 1 2 2 3 1 5 6 4 8 9 7 1 2 3 4 5 6 7 8 9 3 1 2 6 4 5 9 7 8 5 3 4 8 6 7 2 0 1 1 2 0 4 5 3 7 8 6 0 1 2 3 4 5 6 7 8 2 0 1 5 3 4 8 6 7 6 7 8 0 1 2 3 4 5 2 0 1 5 3 4 8 6 7 5 3 4 8 6 7 2 0 1 8 6 7 2 0 1 5 3 4 3 6 5 9 8 8 9 7 2 3 1 5 6 4 7 8 9 1 2 3 4 5 6 9 7 8 3 1 2 6 4 5 5 3 8 6 7 8 6 1 2 0 4 5 3 6 7 8 0 1 2 3 4 5 8 6 7 2 0 1 5 3 4 8 6 5 3 1 2 0 4 5 3 7 8 6 4 5 3 7 8 6 1 2 0 7 8 6 1 2 0 4 5 3 Symbol substitution: 1 → 0 End up with this (K&P) 2 → 1 3 → 2 4 → 3 5 4

Thus, they are NOT essentially distinct!!!

5 → 4 6 → 5 7 → 6 8 → 7 8 → 7 9 → 8

New Generalization of Magic Squares

Magic squares over Finite Fields (NEW)

What would happen to a magic square of order n over F ? What would happen to a magic square of order n over Fq ? When q<n2, obviously, impossible to exist. When q=n2, or q<n2, something interesting is happening… q , q , g g pp g The case where q=n2 and n=prime seems to be a bit trivial.

Theorem(Song 2008): Let q=n2 and n=prime. Construct Latin squares

A=(aij) and B=(bij) both of order n as follows:

ij ij

aij = i-j (mod n), bij = i+j (mod n)

Let α be a primitive element of F where q=n2 Then Let α be a primitive element of Fq where q=n . Then

M=(mij) where mij= aij α+ bij for i,j=0,1,…,n-1

i ( l) i f d F i h i is a (normal) magic square of order n over Fq with magic constant 0. (NOT semi-magic, but magic.)

Example

4 3 2 1 1 2 3 4 aij = i-j (mod 5), bij = i+j (mod 5) 0 4 3 2 1 1 0 4 3 2 2 1 0 4 3 3 2 1 4 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 1 2 3 2 1 0 4 4 3 2 1 0 3 4 0 1 2 4 0 1 2 3

0α+ 4α+ 3α+ 2α+ 1α+ 1α+ 0α+ 4α+ 3α+ 2α+ 0 1 2 3 4 1 2 3 4 1α+ 0α+ 4α+ 3α+ 2α+ 2α+ 1α+ 0α+ 4α+ 3α+ 1 2 3 4 0 2 3 4 0 1 3α+ 2α+ 1α+ 0α+ 4α+ 4 3 2 1 3 4 0 1 2 4α+ 3α+ 2α+ 1α+ 0α+ 4 0 1 2 3

Corollary 1: The above construction works even when n is a power of a prime. You need to order the elements of Fq and use ai instead of i in the theorem.

What happens when n is not (a power of) the characteristic of the field?

Theorem(Song 2008): There does not exist a magic square of order 3 over a finite field of characteristic 2.

b e Necessarily, a+d=b+c and b+e=a+c Proof: a d c → a+b+d+e = a+b+c+c → d+e=0 → d=e, which is impossible.

What else can we say? I would like to invite all of you to this interesting world of combinatorics. I have never seen so far this type of generalization of i fi it fi ld magic squares over finite fields.

Summary Summary

Summary

• 1. Choi Seok-Jeong in Korea has constructed a pair of orthogonal latin squares

(POLS) of order 9 in 1715 or earlier, which is at least 67 years earlier than that (POLS) of order 9 in 1715 or earlier, which is at least 67 years earlier than that

• f Euler’s.

This has appeared in “Handbook of Combinatorial Designs” 2nd ed., edited by C Colbourn and J Dinitz 2007 published by Chapman Hall & CRC by C. Colbourn and J. Dinitz, 2007, published by Chapman Hall & CRC. This POLS is so special that they construct a magic square of order 9.

• 2. Given ANY POLS A and B of order n, there is a transversal in A such that the n

iti i B di t th t l i A t i ith t t positions in B corresponding to the transversal in A contain either a constant or n distinct symbols exactly once.

• 3. A sufficient condition for a POLS of order n to construct a magic square, some

h bl

• ther open problems.
• 4. Kim & Prasanna’s POLS of order n2 constructs a (normal) magic square whose

n2 window sums (each of size n x n) are all equal to the magic constant.

• 5. Theorem: Existence of magic squares over finite fields:

Construction of magic squares of order n over a finite field of size q where p is an odd prime and n2=q=p2t for any positive integer t. p is an odd prime and n q p for any positive integer t. No 3 x 3 magic squares over finite fields of characteristic 2.

References (Korean/Chinese)

1 S k J Ch i (1646 1715) K S R k 1710 1715(?)

• 1. Seok-Jeong Choi (1646-1715), Koo-Soo-Ryak, 1710-1715(?).
• 2. H. N. Chung and M. Heo, Korean Translation of Koo-Soo-Ryak, 2006.

3 S K Hahn and Y Y Oh “Choi Seok Jeong and his Magic Squares” Journal

• 3. S. K. Hahn and Y. Y. Oh, Choi Seok-Jeong and his Magic Squares, Journal
• f KSESM, vol. 32, pp. 205-219, 1993.

4 S K Hahn “Choi Seok Jeong and Koo Soo Ryak ” Newsletter of KSESM vol

• 4. S. K. Hahn, Choi Seok-Jeong and Koo-Soo-Ryak, Newsletter of KSESM, vol.

14, April, 1998. 5 S K Kang “Choi was earlier than Euler” Monthly Magazine of Dong-Ah

• 5. S. K. Kang, Choi was earlier than Euler, Monthly Magazine of Dong Ah

Sciences, Aug. 2008.

• 6. H. W. Chang, Mathematics in Choseon (Lee Dynasty), KM Press, 2007.
• 6. H. W. Chang, Mathematics in Choseon (Lee Dynasty), KM Press, 2007.
• 7. K. M. Park, Mathematics Concert, Dong Asia Press, 2006.

References (English)

1 L E l R h h ll d i 1782 1.

• L. Euler, Recherches sur une nouvelle espece de quarres magiques, 1782.

2.

• G. Tarry, G. "Le problème de 36 officiers.“ Compte Rendu de l'Assoc. Français Avanc. Sci. Naturel vol.

1, pp. 122-123, 1900. 3.

• E. T. Parker, "Orthogonal Latin Squares.“ Not. Amer. Math. Soc. vol. 6, p. 276, 1959.

4 R C B S S Sh ikh d d E T P k "F h R l h C i f M ll O h 4.

• R. C. Bose, S. S. Shrikhande, and E. T. Parker, "Further Results on the Construction of Mutually Ortho

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16 V i titl i Wiki di t htt // iki di / iki/M i P

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http://coding.yonsei.ac.kr/~hysong 1995.9 - current: Professor, Electrical & Electronics Engineering, Yonsei University. 1994.1 - 1995.8: Senior Engineer, g Qualcomm Inc., San Diego, California. 1992.1 - 1993.12: Post-Doc Research Associate, USC. 1986.9 - 1991.12: PHD at USC, Dept of EE-Systems 1984.9 - 1986.5: MSEE at USC, Dept of EE-Systems. 1980.3 - 1984.2: BS at Yonsei University, Dept of Electronics. Research Area Communication & Coding Theory, PN Sequences, Cryptography, Discrete Mathematics, etc. Send me an email at hysong@yonsei.ac.kr for MS or PHD PROGRAM in EE! y g@y You will have a chance to become a 10th generation PHD from L. Euler, and to have Erdos number 3 (because I have Erdos number 2).