MAGIC SQUARES AND ORTHOGONAL ARRAYS

Transcription

1 MAGIC SQUARES AND ORTHOGONAL ARRAYS Donald L. Kreher Michigan Technological University 3 November University of Minnesota-Duluth

2 REFERENCES Andrews, W.S. Magic Squares And Cubes. Dover, (1960). Franklin, B., Autobiography of Benjamin Franklin. (Leonard W. Labaree (Editor)) Yale Univ Press, (1964). Hilliard, J.N. Greater Magic. Kaufman and Greenberg, (1994). Laywine, C.F. and G.L. Mullen. Discrete Mathematics Using Latin Squares, (1998). Lindner, C.C. and C.A. Rodger, Design Theory, CRC press, (1997). Rouse Ball, W.W. and H. S. M. Coxeter, Mathematical Recreations and Essays. Dover (1987).

3 CHAPTER 1 FUNDAMENTALS OF COMBINATORIAL MATHEMATICS 1. What is combinatorial mathematics? Combinatorial mathematics also referred to as combinatorial analysis or combinatorics, is a mathematical discipline that began in ancient times. According to legend the Chinese Emperor Yu (C. 00 B.C.) observed the magic square on the shell of a divine turtle.... H.J. Ryser,Combinatorial Mathematics, C.M. 14 (1963).

4

5 Definition: A magic square is an n by n array of integers with the property that the sum of the numbers in each row, each column and the the main and back diagonals is the same. This sum is the magic sum. The square is n-th order if the integers 1,, 3,..., n are used n 4 magic sum 34 The magic sum of an n-th order magic square is 1 n n 1 npn 1q

6 THE FRANKLIN SQUARE OF ORDER 8: I was at length tired with sitting there to hear debates, in which, as clerk, I could take no part, and which were often so unentertaining that I was induc d to amuse myself with making magic squares or circles, or any thing to avoid weariness; Benjamin Franklin Autobiography

7 For what n does does an n-th order magics square exist? Let M be the set of all positive integers n such that an n-th order magic square exists. So far we know 3, 4, 8 P M and it is easy to see that R M.

8 SQUARES OF ODD ORDER, DE LA LOUBÉRE (1693) First place 1 in the middle cell of the first row. The numbers are placed consecutively 1,, 3,..., n in diagonal lines which slope upward to the right except 1 When the top row is reached the next number is written in the bottom row as if it were the next row after the top. when a right hand column is reached, the next number is written in the left hand column as if it followed the right-hand column; 3 if a cell is reached that is already filled or if the upper right corner is reached then the next cell to be used is the one directly below it. Example This shows that m 1 P M for all m.

9 A Latin square of order n n by n array symbols in t0, 1,,..., n 1u each row and column contains each symbol n n Each symbol once

10 Two Latin squares A and B of order n are said to be orthogonal Latin squares if the n ordered pairs parx, ys, Brx, ysq are all distinct. (x 0, 1,, 3,..., n 1, y 0, 1,,..., n 1) Every ordered pair occurs on is

11 M J A 5B Example Better ( ) ( ) 5( ) (++++) ( ) ( ) (++++)

12 Squares of order pm 1q STRACHEY (1918) Let A be a magic square of order u m 1, m 1. Step 1: Construct this non-magic square of order u: Step : Interchange the indicated cells: m m 1 A A+u Swap Swap A+3u A+u Result is a magic square of order pm So pm 1q 4m P M for all m. 1q

13 Example: m, u 5. Step 1: Step : A magic square of order 10. Magic sum=

14 THEOREM. IF u m 1 P M, THEN u P M. Proof: Use the Strachey construction. m m 1 A A+u Swap A+3u A+u Step 1: Step : Magic sum of A is u3 u In Step we need magic sum puq3 u 4u 3 u. Swap

15 THEOREM. IF u m 1 P M, THEN u P M. Proof: Use the Strachey construction. m m 1 A A+u Swap A+3u A+u Step 1: Step : Magic sum of A is u3 u In Step we need magic sum puq3 u 4u 3 u. Entries: Swap Top left t1,, 3,..., u u 0u t1,, 3,..., u u Bottom right t1,, 3,..., u u u tu 1, u, u 3,..., u u Top right t1,, 3,..., u u u tu 1, u, u 3,..., 3u u Bottom left t1,, 3,..., u u 3u t3u 1, 3u, 3u 3,..., 4u u

16 THEOREM. IF u m 1 P M, THEN u P M. Proof: Use the Strachey construction. m m 1 A A+u Swap A+3u A+u Step 1: Step : Magic sum of A is u3 u In Step we need magic sum puq3 u 4u 3 u. Columns: In Step 1 the sum of the entries of columns 1 through u is u 3 u p u3 u u 3u q 4u 3 u of the entries of columns u through u is p u3 u u u q p u3 u u u q 4u 3 u Swap Step 1 to Step does not change the entries in any column.

17 THEOREM. IF u m 1 P M, THEN u P M. Proof: Use the Strachey construction. m m 1 A A+u Swap A+3u A+u Step 1: Step : Magic sum of A is u3 u In Step we need magic sum puq3 u 4u 3 u. Rows: In Step the sum Swap of the entries of rows 1 through u is u 3 u m 3u u 3 u pm q u pm 1q u 4u 3 u of the entries of rows u through u is u 3 u pm 1q 3u u 3 u pm q u pm 1q u 4u 3 u

18 THEOREM. IF u m 1 P M, THEN u P M. Proof: Use the Strachey construction. m m 1 A A+u Swap A+3u A+u Step 1: Step : Magic sum of A is u3 u In Step we need magic sum puq3 u 4u 3 u. Diagonals: In Step the sum of the entries on the forward diagonal is u 3 u pm 1q 3u u 3 u pm q u pm 1q u 4u 3 u of the entries on the backward diagonal is u 3 u m 3u u 3 u Swap pm q u pm 1q u 4u 3 u

19 PRODUCT CONSTRUCTION Let A ra ij s be a magic square of order m and magic sum α. Let B rb ij s be a magic square of order n and magic sum β. Then the mn by mn square A b B pa 11 1qn B pa 1 1qn B pa 1m 1qn B pa 1 1qn B pa 1qn B pa m 1qn B pa m1 1qn B pa m 1qn B pa mm 1qn B is a magic square of order mn and magic sum αn mn Thus: if m, n P M, then mn P M. mβ.

20 EXAMPLE: b

21 THEOREM. THERE EXISTS A MAGIC SQUARE OF EVERY ORDER n PROOF. Let M be the set of n, for which there exists a magic square of order n. We know: (A) 1, 3, 4, 8 P M, (B) n 1 P M, for all n, (C) 4n P M, for all n 1, (D) if m, n P M, then mn P M. (b) and (c) show that n P M for all n 1,, 3 pmod 4q, n. Use, (a) and (d) to get 4 l, P M, for all l 1,, 3,.... Let n 4k 4 l n 1, where n 1 0 pmod 4q, n 1. Then, n 1 P M and 4 l P M Use (d) to get n P M. Therefor M tn P Z : n 0, n u.

22 Recall: Two Latin squares A and B of order n are said to be orthogonal Latin squaresif the n ordered pairs ( A[x,y], B[x,y]) are all distinct. (x 0, 1,, 3,..., n 1, y 0, 1,,..., n 1 ) Every ordered pair occurs on is

23 A set of k orthogonal Latin squares A 1, A,..., A k are said to be k mutually orthogonal Latin squares of order n if they are pairwise orthogonal. k MOLSpnq Example: 3 MOLSp4q , ,

24 OApk, nq ORTHOGONAL ARRAY k a b i i all ordered pairs n Example: OAp3, 3q Example: OAp5, 4q $ & % $ '& '% Theorem: k MOLSpnq is equivalent to an OApk, nq.

25 THE PAIR OF ORTHOGONAL LATIN SQUARES TIME LINE. 178 Euler conjectures that MOLSpnq exists ô n 0, 1, 3 pmod 4q. He couldn t construct any pair of MOLS of order pmod 4q G. Tarray In a 33 page paper showed: E MOLSp6q. 19 MacNeise conjectures that if n p r1 1 pr prx x, where p 1, p,..., p x are distinct primes, then the maximum number of MOLSpnq is Mpnq mintp r1 1, pr,..., prx x u E.T. Parker constructs 3 MOLSp1q. This disproves MacNeise s conjecture that the maximum number would be Mp1q mint3, 7u R.S. Bose and S.S. Shrikande construct MOLSpq disproving Euler s conjecture. This result made the New York times E.T. Parker constructs MOLSp10q Bose, Shrikande, and Parker proves that there exists MOLSpnq for every n except for n or n 6 when they cannot exist Stinson gives a clever 3 page proof that there do not exist MOLSp6q.

26 THE PAIR OF ORTHOGONAL LATIN SQUARES TIME LINE. 178 Euler conjectures that MOLSpnq exists ô n 0, 1, 3 pmod 4q. He couldn t construct any pair of MOLS of order pmod 4q G. Tarray In a 33 page paper showed: E MOLSp6q. 19 MacNeise conjectures that if n p r1 1 pr prx x, where p 1, p,..., p x are distinct primes, then the maximum number of MOLSpnq is Mpnq mintp r1 1, pr,..., prx x u E.T. Parker constructs 3 MOLSp1q. This disproves MacNeise s conjecture that the maximum number would be Mp1q mint3, 7u R.S. Bose and S.S. Shrikande construct MOLSpq disproving Euler s conjecture. This result made the New York times E.T. Parker constructs MOLSp10q Bose, Shrikande, and Parker proves that there exists MOLSpnq for every n except for n or n 6 when they cannot exist Stinson gives a clever 3 page proof that there do not exist MOLSp6q.

27 FINITE FIELD CONSTRUCTION THEOREM Let n p α, where p is a prime and α is a positive integer. Then for n 3, there exists n 1 MOLSpnq. PROOF. Let F n be a finite filed of order n. For each f P F n define the F n F n matrix A f by A f rx, ys fx y, for all x, y P F n. We claim that the n 1 squares A f, f P F q, f 0 are MOLSpnq. 1 (Row Latin) If fx y 1 fx y, then y 1 y. (Column Latin) If fx 1 y fx y, then fx 1 fx and so x 1 x. 3 (Orthogonal) Suppose pa f rx, ys, A l rx, ysq pa f rx 1, y 1 s, A l rx 1, y 1 sq for some f l. Then fx y fx 1 y 1 lx y lx 1 y 1 Subtracting we see that pf lqx pf lqx 1. Thus x x 1 and then y y 1.

28 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

29 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

30 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

31 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

32 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

33 EXAMPLE: n 5, F 5 Z A 1 rx, ys x y A rx, ys x y A 3 rx, ys 3x y A 4 rx, ys 4x y

34 A Latin square L is idempotent if Lri, is i for all i IMOLS idempotent mutually orthogonal Latin squares. Example: 3 IMOLS(5)

35 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT

36 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk 1, nq WITH n CONSTANT COLUMNS simultaneously permute rows

37 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk 1, nq WITH n CONSTANT COLUMNS simultaneously permute rows

38 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT

39 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT individually permute symbols

40 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT individually permute symbols

41 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT

42 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT MOLSp5q loooooooooooooooooooooooooooooooomoooooooooooooooooooooooooooooooon

43 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT MOLSp5q loooooooooooooooooooooooooooooooomoooooooooooooooooooooooooooooooon

44 LEMMA. k MOLSpnq ñ k 1 IMOLSpnq ñ OApk COLUMNS. 1, nq WITH n CONSTANT MOLSp5q loooooooooooooooooooooooooooooooomoooooooooooooooooooooooooooooooon looooomooooon C B ó OAp5, 5q with 5 constant columns looooooooooooooooooooooooooooooomooooooooooooooooooooooooooooooon OAp5,BqzC B, where Bt0,1,,3,4u

45 DIRECT PRODUCT CONSTRUCTION THEOREM Let A be a Latin square on X and let B be a Latin square on Y. Define the square A B on X Y by A Brpx 1, y 1 q, px, y qs parx 1, x s, Bry 1, y sq. Then A B is a Latin square. Example: 1 3 X O O X X1 X X3 O1 O O3 X3 X1 X O3 O1 O X X3 X1 O O3 O1 O1 O O3 X1 X X3 O3 O1 O X3 X1 X O O3 O1 X X3 X1 THEOREM If A 1, A are MOLSpmq and B 1, B are MOLSpnq, then A 1 B 1 and A 1 B 1 are MOLSpmnq,

46 MACNEISE THEOREM If n p r 1 1 pr prx x, where p 1, p,..., p x are distinct primes, then there are at least MOLSpnq. PROOF. Mpnq mintp r 1 1, pr,..., prx x u 1 Finite field construction ñ p r i i 1 MOLSpp r i i q for each i. So there exists Mpnq MOLSpp r i i q, for each i. Apply the direct product construction to get Mpnq MOLSpnq. Recall: MacLeish conjectures that there are at most Mpnq MOLSpnq.

47 A pairwise balanced design PBD of type -pv, K, λq is a pair px, Bq where 1 X is a v-element set of points, B is a collection of subsets of X called blocks, 3 B P K for every block B P B, and 4 every pair of points is in λ blocks. Example 1: A PBD of type -(7,3,1) is given by: X t0, 1,, 3, 4, 5, 6u B t130, 14, 35, 346, 450, 156, 60u

48 EXAMPLE: A -(1,5,1) PBD. Recall: 3 MOLSp4q , , Gave a OAp5, 4q Take as points X t8, 0, 1,, 3, 0, 1,, 3, 0, 1,, 3, 0, 1,, 3, 0, 1,, 3u and blocks the columns of THEOREM n 1 MOLSpnq ô OApn 1, nq ô PBD of type -pn n 1, n 1, 1q

49 PBD CONSTRUCTIONS FOR MOLS THEOREM Let px, Bq be a PBD of type -pv, K, 1q Suppose for each B P B, there exists an k IMOLSp B q Then there exists k IMOLSp X q. PROOF. Then Let B tb 1, B,..., B b u be a -pv, K, 1q. k IMOLSpB i q ô OApk, B i q C Bi Y OApk, B i qzc Bi, C Bi the set of constant columns. is an OApk CX, OApk, B 1 qzc B1, OApk, B qzc B,..., OApk, B b qzc Bb, Xq with X constant columns, i.e. k MOLSp X q. Example: Three MOLS of order 1. 4, is a prime power ñ 3 MOLSp4q ñ -p1, 5, 1q PBD. 5 is a prime power ñ 4 MOLSp5q ñ 3 IMOLSp5q Apply the PBD construction to get 3 MOLSp1q. This disproves the MacNeise conjecture which claimed that there would be at most Mp1q mint3, 7u 1 of them