In the present paper，we study square matrices in which the
sum of elements in any row，in any column ，in any extended diagonal
add up to a constant. We call such a matrix a pandiagonal
constant sum matrix. We will show that the number of independents
elements in a pandiagonal constant Sum matrix of order n is
n2 - 4n + 3 if n is odd or n2 - 4n + 4 if n is even

Ishihara, Toru

Tokushima University Institutional Repository

J. Math. Univ. Tokushima Vol. 41 (2007), 1-111Pandiagonal' Constant Sum Matrices By Toru ISHIHARA Fαculty of Integnαted Arts and Sciences， The University of Tokushima， Minαmijosαnjimα， Tokushimα770-8502， JAPAN e-mαilαddress: ishihαra@ias. tokushima-u.αc.jp (Received September 28， 2007) Abstract In the present paper， we study sqllare matrices in which the sum of elements in any row， inany colllmn ， inany extended di-agonal add up to a constant. We call sl1ch a matrix a pandiagona1 constant Sl1m matrix. We will show that the nllmber of indepen-dents elements in a pandiagonal constant Sl1m matrix of order ηis n2 - 4n + 3 ifn isodd or n2 -4n十4if n is even. 2000 Mathematics Sl1bject Classification. 05B20 Introduction Let .E be a a setof n di旺erentelements. A latin square of order n isa sqllare matrix with n entries of elements in .E， none of them occllrring twice within any row or colllmn of the matrix. A mαtrix of the sαme number n is defined to be a square matrix with n2 entries of n di百erentelements， each appeared exactly ηtimes. A latin square of order n isa matrix of the same number n. A magic square of order n isan arrangement of n2 consecutive integers in a square， such that the sums of each row each colllmn and each of the main diagonal are the same. If also the sum of each extended diagonal is the same， the magic square is called pandiagonal. Two latin Sql町田 A=(向jand B = (bij) of order n are said to be orthogonal if every ordered pair of symbols occurs exactly once among the ポ pairs(向j，bij) . We can define that two matrices of the same number n are orthogonal similarly. Let A = (αij )， α2)ξ R@be a square matrix of order n. It is ca11ed a constαnt sum mαtrix if the sums of each row and each column are the same. If moreover the sum of each main diagonal is the same ， it is called adiαgonal constαnt sum mαtr紅白ldif the Sl1m of each extended diagonal is the same， itis called a pαndiα，gonal constαnt sum mat向x.In the present paper， we take 0，1，" " n -1部 nconsecutive integers and put Toru Ishihara2 Pandiagonal Constant Sum Matrices 3L: = {O， 1，'・.，n -1}. A pandiagonal latin square on L: is a pa吋 aiagonalconstant sum matrix of the same number n. Let A and B are orthogonal matrices of the same number n. Put C = nA十B.Then it is known [3] that if A and B are diagonal(resp. pandiagonal) constant sum matrices， C is a magic (resp. pandiagonal magic) square. 1. Pandiagonal constant sum matrices Let A = (αij) ，α2)ε R be a pandiagonal constant sum matrix of order n. In the present paper， subscripts have the range 0， 1，・，n-1(mod n). Then we have the following equations. n-l 2二α包j = C， 0 ~ j ~η-1 ， (1) i=O 九一1L αij ニ C，1 ~ミ i ~ n -1， (2) jニO叫 -1L αin+j-i C， 1壬j~ n -1， (3) i=O 九一1L aii+j C， 1 ~ j ~ n-1， (4) i=O where C is a constant. Notice that from (1) and (2)， (3)，(4) it follows n 乞αOj c. j=O n Lain-i C 包ニOγz 乞αii C. i=O When n iseven， that is， n = 2m， there is a redundant equation in (3) and (4). In fact， ifwe set 2j = 2k十 2i (mod 2r仏wehave ?????? ? ?? ??? ? ??????????? ? ? ?Hence， we can consider the equation Toru Ishihara2 Pandiagonal Constant Sum Matrices 3乞 αi1+i= c is redundant. Now， when n = 2m， we set 2m-l L αij C， 0 S;j S; 2m -1， (5) zニ O2m-l 玄 αり C， 1 S;i S;2m -1， (6) 3ニO2m-l 乞 αi2m十1ー 包 C， 1 S;j S; 2m -1， (7) 包ニO2m-l 玄白+j ニ C，2 S;j手2m-1. (8) i=O Theorem 1 When n is an odd number， the equations (1)，(2)，(3) and (4) are independent， and whenη = 2m， the equations (5)，(6)，(7) and (8) are independent. Proof. Set Xi = α。包，Yi = αli， Zi = α2i， 0 S; i三n-l.Then we have Aj :町 +Yj + kj 2二αij C， 0 S;j三n-1，i=3 九一1鳥 :Xj十 Yj-l十 Zj-2十 Laij-i C， 1 S;j三n-1，i=O D1: LYj = C， j=O D2: LZj = c， jニOn-l Dj :乞α1包 C， 3 ~ミ j 三 n -1. Toru Ishihara4 Pandiagonal Constant Sum Matrices 5(1) Now SllppOse that n isodd， that is n = 2m + 1.Then it holds Cj: Xj十Yj+l十Zj+2+玄白+j= C， 1 ~ j臼 -1Now we represent simply the above eqllations as Pllt Bj(l) Bπ-1(2) Bj(2) l~j 三; n-2. Aj Bj Cj D1 D2 Dj Especially， we have Next， we get (Xj'Yj，Zj，牢)， 052j壬n-1，(Xj，Yj-1，Zj-2，*) 1三j三η1，(Xj，Yj+l，Zj+2， *)， 1三j三n-1n-1 (0， LYj，O，牢)，jニO九-1(O，O'L引)，j=O (0，0，0，牢)， 3三j三η-1.Bj -Aj = (0， Yj-1 -Yj， Zj-2 勺，牢)， 1三j三n-1(9) Bn-1 (1) = (0， Yn-2 -Y.九一1，Znー3- Zn-I， *)， (10) Bj(l) + Bj+l (2) (0， Yj-1 -Yn-1， Zj-2 +勺-1- Zn-2 -Zn-I，牢)， (11) B1 (2) = (0， YO-Yn-1， Zo-Zn-2.*) Cj(l) Cj(2) Cηー1(2)匂-Aj = (0， -Yj + Yj+1， -Zj +勺+2，キ)， 1三j三η-1，Now， set Cj(3) Cnー 1(3)Cj(l) + Bj+l(l) = (0，0，勺一勺+1一勺+2十Zj+3，本)， 1三j三n-2，Cn-1(1) -B1(2) = (0，0，Zn-2 -Zn-1 -Zo + Z1，キ). Cj(2)十Cj十1(2) = (0，0， Zj-1 -2Zj+1十Zj十3，キ)， 1三j三η-2， Cηー1(2)= (0，0， Zηー2-Z礼一1-Zo十Z1，*). Toru Ishihara4 Pandiagonal Constant Sum Matrices 5PllもC2m-1(4) = C2m-1(3)， C2m-3(4) = C2m-3(3)十2C2mー1(4)， C2(m-k)+1 (4) = C2(m-k)+l (3)+2C2(m-k+l)+1 (4)-C2(m-k+2)+l (4)， 3三k三作品.Then we have C2(m-k)+1 (4) = (0，0， Z2(m-k)一(k十 1)z2m+ kz1，牢)， 1三k三m.Next， we pllt C2m(4)ニ C2m(3)，C2m-2(4) = C2m-2(3) + 2C2m(4)， C2(m-k)(4) = C2(m-k) (3) + 2C2(m-k+l) (4) -C2(m-k十2)(4)，2壬k三m-l.Then， we get C2(m-k)(4) = (0，0， z2(m-k)-1一(k+ 1)(z2m -zI) -zo，キ)， 0三k三m-l.Set C2(5) C2(m-k)+l (5) C2(m-k) (5) -i-(C2(4)+C1(4))=(o，OA-Z2m牢)，2m+ 1 C2(m-k)+l (4) -kC2(5)， 1三k三m，C2(m-k) (4) -(k十 1)C2(5)十C1(5)，0::; k三m-2.Thlls we obtain Cj(5) = (0，0，勺ー 1- z2m，牢)， 1三j三n-1ニ 2m.Now the eqllations Aj Bj(2) B礼一1(2)D1 Cj(5) D2 Dj (Xj， Yj，勺，*)， 0三j三η1，( ，約一1-Yn-1，Zj-2 +一勺 1ー ら 2ー ら 1，*)， 1壬j三n-2，(0， Yn-2 -Yn-1ヲzn-3-Zn-l， *)， 九一1(0，乞Yj，Oぅキ)，j=O (0，0， Zj-l -Z2m， *)， 1三j三n-1，(0，0， I:引)， (0，0，0，牢)， 3三j三n-1are eqllivalent to the eqllations given at白rst.It is evident that the rank of the coe伍cientmatrix of the equations is 4n -3. Hence， these eqllations are independent. Toru Ishihara6 Pandiagonal Constant Sum Matrices 7(2) S叩 posethat n is even， t出ha叫tis，η 二 2rr凡 By凶 n略gthe s討im凶il耐a訂rnotations， we consider the following 4n -4 eqllations Aj (Xj，Yj，Zj，牢)， 0三j三η-1，Bj ニ (Xj，Yj-1， Zj-2，ヰ)， 1三j三n-1，(7j (Xj+1，Yj+2，Zj+3，*) 1壬j三n-2D1 = (0，2::: Yj，仰)，j=O D2 = (川乞引)， j=O Dj 二 (0，0，0，本) 3三j三η-1.We define Bj(1)， Bj(2)， 1三j三n-1回 similarly凶 in(9)，(10)，(1). We pllt (7j(1) (7j(2) (7n-2(2) Now， set 匂(3)(7n-2(3) Pllt (7j -Aj+1 = (0， -Yj+1十約+2，-Zj+1十勺+3，*)， 1三j:壬η-2， (7j(1) + Bj+2(1) = (0，0，勺一勺+1一勺+2十勺+3ヲ牢)， 1三j三n-3，(7n-2(1) -B1(2) = (0， 0， Zn-2 -Zn-1 -Zo十 Zl，牢). (7j(2) + (7j+1(2) = (O，O，Zj -2Zj+2 + Zj+4バ)， 1三j壬n-2，(7n-2(2) = (0，0， Zn-2 -Zn-1 -Zo十九，*). (72m-2(4) = (72m-2(3)， (72m-4(4) = (72m-4(3) + 2(72m-2(4) (72(m-k)(4) = (72(m-k) (3) + 2(72(m-k+1) (4) -(72(m-k+2)， 3三j三m-1.Then we have (72(m-k)(4) = (0ぅ0，Z2(m-k) -k(Z2m-1 -Zl) -ZO， *)， 1三k三m-1.Next， we plt (72m-3(4) = (72m-3(3)， (72m-5(4) = (72m-5(3) + 2(72m-3(4)， (72(m-k)ー 1(4) = (72(m-k)ー 1(3)十2(72(m-k+1)ー 1(4)-(72(m-k+2)-b 3三k三m-1.Toru Ishihara6 Pandiagonal Constant Sum Matrices 7Then we get C2(m-k)-1 (4) = (0，0， Z2(m-k)-1 - (k + 1)Z2m-1十 kz1，*) 1三k三m-l.Set α(5)=tC1(4)=川 Z}-Z2m-1バ)， C2(m-k)(5) = C2(m-k) (4) -kC1(5)ニ (0，0，Z2(m-k) - ZO，キ)， 1:S k三m-1，C2(m-k)一1(5)ニ C2(m-k)-1(4)-kC1(5)= (0， 0， Z2(m-k)-1-Z2m-1，*) 1三k:S m-l. Now the equations Aj Bj(2) Bn-1(2) D1 C1(5) C2(m-k) (5) C2(m-k)ー 1(5)D2 Dj (Xj，Yj，Zj，牢)， 0三j三n-1，(， Yj-1 -Yn-I，勺-2+ -Zj-1 -Z丸一2- Zn-1， * )， 1三j三n-2， (O，y叫ー2-Ynー 1，Zn-3 - Zn-I，本)， n-1 (0，2ン'j，0， *)， j=O (0，0， Zl- Z2m-I， *)， (0，0，Z2(m-k)-ZO，*) 1三k三m-1，(0，0， Z2(叶み)ー1- Z2m-1，牢)， 1三k壬m-1，n-1 (0，0，2: Zjバ)，j=O (0，0，0，*)， 3三j三n-1are equivalent to the equations given at first. It is evident that the rank of the coe血cientmatrix of the equations is 4n -4. Hence， these equations are independent. 2. Pandiagonal zero sum matrices Let A = (αり)，αij E R be a pandiagonal constant sum matrix of order n with constant C. In this section， we have from here on ， s山 tractedSjn from every elements in the matrix so that the sum of the elements in any row， column or diagonal will be zero， and we call such modified a zero-sum mαtrix. The results in this section mainly owe to W. R. Andress [1]. We introduce operators R， Csuch that Toru Ishihara8 Pandiagonal Constant Sum Matrices 9Set Then we have Rα丸j=αi+l，j Cαi，j -αω+1・ん(R)=LRi，Dn(R，C)=乞Rn-1-iCi包ニO $ニOcollIllIll: Ln(R)何 0，row: Ln(C)αω0， diagonal : Ln(RC)αω ニ Oヲdiagonal: Dn(R， C)αω= O. Lemma Let Qi，j be elements of a sqllare matrix of order n. If any one of the three conditions (1) (C -l)Qi，j = 0ε;;::~ Qi，j = 0， (2) (R -l)Qi，j = o，ZUQLjニ 0，(3) (R -C)Qi，j = 0， L::01 Qi，i二 oholds， itfollows that Qi，j = O. Proof. Assllme that the condition (1) holds. Then Qi，j = 0 isindependent of colllmn. Hence llsing the second eqllation of (1)， we get Qi，j = O. The other res1l1ts follow similarly. From (Ln(RC) -Ln(R))向、j= R(C -1) L:~:11 Ri-l Li-1 (C)αω= 0， llsing Lemma， we get L:ご~1Ri-1 Li-1 (C)仇 j= O. Since this is true for al val附 ofi，j， itis convenient to Sllppress the operand向，jso that L:RiLi(C) = 0 (12) 包ニOThis is a triangle-invariant. We may interchange the operations R， Cin the above formula 80 that 乞CtLi(R)二。 (13) 包ニOThe tria時le-invariant(12) remains invariant if we replace R by 1/ R and multi ple Rn-l儲 thismerely represents a reflection in a horizontal line， and glve LRiLn一日(C)=。 (14) Toru Ishihara8 Pandiagonal Constant Sum Matrices 9Put Sη = Ln(R)Ln(C). Then， this presents a s叩1訂 eof order η. Sl1btracting (12)-(13)， and jlstiかingthe removal of the factor R -C by Lemma 1， we obtain the invariant 乞 (RC)tSn-2-2i = 0 (15) i=O Sl1btracti時 (14)from (15)， addi時 C-1Dn and mll1tiplyi時 (RC)-l， we get the invariant l (π-5)/2J 玄 (RC)包Sηー 3-2i十Rn-2cn-2ニ O. (16) 包ニOSllbtracting 2:コ~03RiLn(R) from (15) and adding (cn-3 + cn-2)Ln(R) gives the invariant l(n-5)/2J 乞 (RC)弘一4-2i+ (RC)叫ー3S2= 0 (17) i=O From now on， we assllme that n 1Sodd， that is， nニ 2m+1. The triangle-invariant (12) remains invariant if we replace C by l/C槌 thismerely represents a refiection in a vertical line， and give LRiLi(C-1)二 O (18) $ニOS由tractingthis from (12) and removing R(C -C 1)， we get 玄RiLl(i+2-j)/2j(Cj+C-j)+玄Ri乞 l(m-i-j)/2j(ぴ+C-J)= 0 包ニo j=O 包ニO j=O (19) Theorem 2 Let A = (αi，j) be a pandiagonal constant Sllm matrix of order n. For an odd n， ifn2-4n+3 elements向小 O::;i三η-4，O::;j三n-2are given， the other elements decided llniqllely. For an even n， if n2 -4n+4 elements α包小 O::;i 三n-4， 0::; j 三n -2 and any any one of αn-3，j， 0 ~ミ j::;n-1are given， the other elements decided uniquely. Proof. Using (12)， we can getαi，n-1， 0三i三n-4. Assllme that n is an odd number. Using (19)， we obtainαn-3山 O三j三n-1 and then αn-2，j，α九一1小 O::;j三n-1. Next， Let n be an even nllmber. Using (16)， we can determineαn-2，j， 0三j壬n-1. Then using (17)，企omany one of α九一3小 。ごミ j:; n -1， we obtain α礼一3，j，0::; j :;n -1. Now， it is easy to getαπ-1小 O::;j::;n-1.Toru Ishihara10 Pandiagonal Constant Sum Matrices 113. Orthogonal mattices of the same number AmαtTIx of the same number n is a sqll訂 ematrix with n2 entries of n different elements， each appeared exactly n times. Two matrices of the same nllmber n A = (向jand B = (bり)are defined to be orthogonal if every ordered pair of symbols OCCllrs exactly once among the n 2 pairs (αij， bij) . Itis well known that the largest value of r for which there exist r mutually orthogonal Latin squares of order ηis less than n. Now， we have Theorem 3. Denote by N(n) the largest value of r for which there exist r mutually orthogonal matrices of order n. it holds N(η)三η+1.Proof. Suppose Al'・，Atare mutually orthogonal matrices of order n on the symbols {O， 1，'・ヲn-1}. Take an n-square matrix S = (Si，j) whose η2 positions are labelled 0，1， . . . ，η2-1 as follows: Si，j = ni + j， 0::; i三j，O:; j :;n -1 . Then consider the collection of subsets Br，m defined by Br，m = {x : x isthe label inSof aposition in which Ar has entry m}， where 1 :;r :;t， 0::; m :;η 1. There are tn subsets of size n. It follows from the orthogonality of the Ar that no pair of elements can occur in more than one block. Sllppose for example .T and y both occur in Br1・mland Br2，m2' Then Ar1 h錨 thesame entry ml in x and y， while Ar2 has entry m2 in these positions. Hence the pair (mI， m2) occurs twice， contradicting to the orthogonality of A1 and A 2 • Note the number of pairs of elements in the sllbsets is 、 、 ? ? ? ????， ，?? 、? ??? ????? ???』????????，?This number must be more than (ゴ).Hence jd(n-山 jn2(η2ー 1)gives t :;n + 1.References [1] W. R. Andress， Basic properties ofpandigonal magic squraes， Amer. Math. :tvlonthly 67， 143-152， 1960. Toru Ishihara10 Pandiagonal Constant Sum Matrices 11[2] Y. C. Chen and G. 1¥1. Fu， Construction and enllmeration of pandiagonal magic sqllares of order n from step method， Ars Combinatoria 48， 233-244， 1998. [3] W. Proskllrowski and A. Proskllrowsk， Constrllction of pandiagonal magic squares from circulant pandiagonal matrices， Ars Combinatoria 15， 153-162，1983. 

Pandiagonal Constant Sum Matrices

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Pandiagonal Constant Sum Matrices

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