Construction of magic rectangles of odd order

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1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 55 (013), Pages Construction of magic rectangles of odd order Feng Shun Chai Institute of Statistical Science Academia Sinica, Taipei Taiwan, ROC Ashish Das Department of Mathematics Indian Institute of Technology Bombay Mumbai India Chand Midha Department of Statistics The University of Akron Akron, OH 4435 USA Abstract Magic rectangles are well-known for their very interesting and entertaining combinatorics Such magic rectangles have also been used in designing experiments In a magic rectangle, the integers 1 to pq are arranged in an array of p rows and q columns so that each row adds to the same total P and each column to the same total Q In the present paper we provide a systematic method for constructing a p by q magic rectangle where p>1 and q>1 are any odd integers 1 Introduction Magic rectangles are well-known for their very interesting and entertaining combinatorics A magic rectangle is an arrangement of the integers 1 to pq in an array of p rows and q columns so that each row adds to the same total P and each column to the same total Q The totals P and Q are termed the magic constants Since the average value of the integers is A =(pq +1)/, we must have P = qa and Q = pa

2 13 FS CHAI, A DAS AND C MIDHA The total of all the integers in the array is pqa = pp = qq Ifpq is even pq +1is odd and so for P = q(pq +1)/ andq = p(pq +1)/ tobeintegersp and q must both be even On the other hand, if pq is odd then p and q must both be odd In this case also P and Q are integers since pq + 1 is even Therefore, an odd by even magic rectangle is impossible For an update on available literature on magic rectangles we refer to Hagedorn [5], and Bier and Kleinschmidt [1] Recently, Reyes, Das and Midha [3], and Reyes, Das, Midha and Vellaisamy [4], have provided complete solutions for constructing an even by even magic rectangle The method of construction provided there is not extendable to the odd case Thus the construction of odd by odd magic rectangles has been more challenging Das and Sarkar [] provide constructions for a few odd by odd magic rectangles by taking a rectangle with constant column sums and transforming it into a magic rectangle via transpositions within columns They provide tables for the constructed magic rectangles Following the ideas of Das and Sarkar [], in the present paper we provide complete solutions to the construction problem by providing a systematic method of constructing a p by q magic rectangle where both p and q are odd A p q rectangle H p, containing integers1,,,pq once, with constant column sums, is readily produced This rectangle can be transformed into a magic rectangle M by performing transpositions within columns to make the row sums constant The change in a particular row sum from H p to M is equal to a certain sum of differences of transposed entries The method has been shaped in the form of an algorithm that is very convenient for writing a computer program In Section we construct p by q magic rectangles The supporting proofs related to the construction are given in Section 4 In Section 3 we illustrate our construction method through some examples of magic rectangles The construction We construct magic rectangles with p rows and q columns (p >1andq>1both odd) In order to build our construction method, we first define some matrices In the sequel, J mn denotes a m n matrix having all its elements equal to 1, and J m1 is denoted by 1 m Also, denotes the Kronecker product symbol Definition 1 For q =q 1, q 1, define R = ( ) ( 1 q 1 q RU = q q 1 1 R L ), 1 q q +1 q S = q q +1 q 1 q 1, q q 1 q 1

3 MAGIC RECTANGLES OF ODD ORDER 133 and T = q q 1 q 1 q 3 1 q q +1 q + q 1 3 q 4 q 1 q q 1 q q 1 q 1 = ( TU T L where each of R U and R L are of order 1 q and each of T U and T L are of order q R U Definition Let p 1 Define G 3 = S, G 5 = S R L, and 1 p T U S 1 p T L if p =4p +3, G p = R U G p if p =4p +5 R L ), Also, define, H p =(h ij ), where h ij G p =(g ij ) =(g ij 1)p + i, 1 i p, 1 j q, and Theorem 1 Suppose p>1, q>1, n>1 are odd Let M 1 be a p n magic rectangle Then there exists a p qn magic rectangle M Proof For 1 i p, 1 j q, let M j = M 1 +(j 1)pnJ pn and m ij represent the i-th row of M j Then M is obtained from G p by replacing the element j in the i-th row of G p by m ij,1 i p, 1 j q Using the fact that M 1 is p n magic rectangle and using the properties of the matrix G p (Lemma 41), it is easy to see that each row sum of M equals ( n(pn+1) )q + npn + npn + + n(q 1)pn = qn(pqn+1) and each column sum of M equals p(pn+1) +( p(q+1) p)pn = p(pqn+1) Hence M is a p nq magic rectangle Theorem For p odd, consider the p p matrices 1 p 1 p p p p 1 1 p 3 N 1 = and N = p 1 p p 3 p p p 3 p p 1 p 1 p p 1 p 1 p 1 p Then a p p magic rectangle M is given by M = N 1 + p(n J pp ) Proof Let a ij and b ij be the (ij)-th element of N 1 and N,1 i, j p Notice that N 1 and N are orthogonal latin squares since both of them are latin squares

4 134 FS CHAI, A DAS AND C MIDHA and the collection {(a ij,b ij ), 1 i, j p} collects all p ordered pairs Based on the orthogonality of N 1 and N, the collection of all elements of M is the set {i +(j 1)p, 1 i, j p} which contains p consecutive integers from 1 to p Now, it is clear to see each column and row sums of M equals p(1 + p)/+p(p(1 + p)/ p) = p(1 + p )/ Hence M is a p p magic rectangle Magic rectangle construction method: step by step descriptions Suppose p>1andq>1 are odd The p q magic rectangle M is constructed as follows Step I Construct G p and H p ( both matrices are defined in Definition ) Each row of G p is a permutation of (1,,,q) and its column sums are all equal H p has the following properties (i) each of the numbers 1 through pq appear once; (ii) each column sum equals p(pq +1)/; (iii) the i-th row sum is pq(q 1)/ +qi; (iv)for 1 i (p 1)/, the difference between the (p +1 i)-th row sum and the i-th row sum is q(p +1 i); (v) the (p +1)/-th row sum is q(pq +1)/ Step II Each column sum of H p are equal, that matches the column sum requirements of a magic rectangle Hence, in this step, we will rearrange the element orders within columns such that the resulting H p has the equal row sums, therefore, it is the required magic rectangle Observe that (iii), (iv) and (v) properties of H p in Step I, we wish to exchange some elements of the i-th row with the corresponding elements of the (p +1 i)-th row, such that the resulting row sum of the i-th row gains q(p +1 i)/ and the resulting row sum of the (p +1 i)-th row loses q(p +1 i)/ Then, the resulting row sum of i-th equals to pq(q 1)/ +qi + q(p +1 i)/= q(pq +1)/ and the resulting row sum of the (p +1 i)-th row equals to pq(q 1)/+q(p +1 i) -q(p +1 i)/ =q(pq +1)/ We now explain how this row element exchanges procedure works For 1 i p, let g i = (g i1,g i,,g iq ), g (p+1 i) = (g (p+1 i),1, g (p+1 i),,,g (p+1 i),q ), h i = (h i1,h i,,h iq )andh (p+1 i) =(h (p+1 i),1,h (p+1 i),,,h (p+1 i),q )denotethei-th and (p +1 i)th rows of G p and H p Based on h i = pg i + i1 q, the above row element exchanges on rows of H p can be done through the row element exchanges on the rows of G p We prefer to do the row element exchanges on rows of G p, since the row differences of G p are easier to be characterized Let the difference row vector d i = g (p+1 i) g i =(d i1,d i,,d iq ), 1 i (p 1)/ Assume, in d i,thel elements, d ij1, d ij,,d ijl,arechosen,suchthatp l u=1 d ij u + l(p+1 i) =q(p+1 i)/ This suggests that, in G p,thelelements g ij1,g ij,,g ijl, of the i-th row should be exchanged with the l elements, g (p+i 1),j1,g (p+i 1),j,, g (p+i 1),jl of the (p + i 1)-th row Also, it indicates, in H p, the l elements, h ij1,h ij,,h ijl,ofthei-th row should be exchanged with the l elements, h (p+i 1),j1, h (p+i 1),j,,h (p+i 1),jl of the (p + i 1)-th row Then the i-th and (p +1 i)-th row sums of the resulting H p are both equal to q(pq +1)/ Step III Do the above row element exchanges of G p and H p,for1 i (p 1)/ Keep the (p +1)/-th row of H p intact, since its row sum equals to q(pq +1)/

5 MAGIC RECTANGLES OF ODD ORDER 135 Now, the resulting H p has equal column sums and row sums, and is a p q magic rectangle 3 Some illustrative examples The method provided in this paper allows one to write a simple computer program for obtaining any odd by odd magic rectangle We provide two examples of magic rectangle of sides 7 by 11 and sides 13 by 19 Magic rectangle of order 7 by 11 p =7=4+3,q =11= 6 1 Thus p =1andq = G 7 = From G 7 above, we obtain H 7 as below: H 7 = For rows i and 7 i, 1 i, we carry out the corresponding interchanges as in Case II of Theorem 41 Here, l =(11 7)/ == 0+ Thus y =0 Sowe interchange (i) between rows 1 and 7 at positions 1 and (7+3 )/ = 4, (ii) between rows and 6 at positions 1 and ( )/ = 3 For rows 3 and 5, we carry out the corresponding interchanges as in Case I of Theorem 41 Here, q = 11 = 6 1 (ie, k=), l ==4 0 + Thus y = 0 Hence, we interchange between rows 3 and 5 at positions =4and4 =8 From H 7 above, on carrying out the interchanges, we obtain 7 11 magic rectangle M as below:

6 136 FS CHAI, A DAS AND C MIDHA M = Magic rectangle of order 13 by 19 p =13=4 +5,q =19= 10 1 Thus p =andq = G 13 = From G 13 above, we obtain H 13 =(H13 1,H 13 ), where H13 1 = and

7 MAGIC RECTANGLES OF ODD ORDER H13 = For rows 1 and 13, we carry out the corresponding interchanges as in Case III of Theorem 41 Here, l =(19 13)/ =3= Thus y =1 Soforrow 1 we carry out the corresponding interchanges which is between rows 1 and 13 at positions 1, 19 and 7 For rows i and 14 i, i 5, we carry out the corresponding interchanges as in Case II of Theorem 41 Again, l =y +1 with y =1 Sowe interchange (i) between rows and 1 at positions 10, 11 and 6, (ii) between rows 3 and 11 at positions 10, 11 and 5, (iii) between rows 4 and 10 at positions 10, 11 and 4, (iv) between rows 5 and 9 at positions 10, 11 and 3 Finally, for rows 6 and 8, we carry out the corresponding interchanges as in Case I of Theorem 41 Here, q =19= (ie, k=3), l =3=4 0+3 Thus y = 0 Hence we interchange between rows 6 and 8 at positions 13, 6 and 8 From H 13 above, on carrying out the interchanges, we obtain magic rectangle M =(M 1,M ), where M 1 = and

8 138 FS CHAI, A DAS AND C MIDHA M = Proof of the validity of the construction We first establish some preliminary results Recall that matrices R, S, T, G p and H p are defined in Section Lemma 41 The rows of R, S, T and G p are permutations of (1,, 3,,q) and the column sums of R, S, T and G p are q +1, 3(q +1)/, (q +1) and p(q +1)/ respectively Proof For R, S and T, it is easy to see that each row being a permutation of (1,,,q)hasrowsumq(q+1)/ and each column has column sum q+1, 3(q+1)/, (q + 1), respectively G p is formed as the combination of copies of R, S and T,has the above row properties and equal column sums p(q +1)/ The following quantities are needed ( in Lemma 4 ) For a positive integer k, let e =( 1 1),e 3 =( 1 01),e 4 =, v =(6k 6k 5 3k+10), v =(3k 3k 5 1), v 3 =( 5 3k +1), v 4 =(3k 33k 6 3), v 5 =( 3 6 3k +3),v 6 =( 3k 3k 3 6k +3), w 1 =(6k 6k 3 3k +3), w =(3k 3k 3 3), w 3 =( 3 6 3k), w 4 =(3k 3k 5 1), w 5 =( 5 3k +1), w 6 =( 3k 3k 5 6k +1) Observe that v 4 and v 5 are of length k 1, while the others are of length k Lemma 4 Let q =q 1 Then we have (i) e R = ( q 1 q q +3 q +1 ) ( ), v1 v v 3 v 4 0 v 5 v 6 if q =6k 1 (ii) e 3 S = ( ) w1 w 0 w 3 w 4 w 5 w 6 if q =6k +1,

9 (iii) e 4 T = MAGIC RECTANGLES OF ODD ORDER 139 ( ) 0 1 q 1 q +1 q q 1 q +1 q + 1 Proof This follows from straight observations on row differences in R, S and T The following two lemmas describe detailed element relations among v i s and w i s, 1 i 6, which appear in Lemma 4 (ii) Lemma 43 For u =1,, 3, 6, letv u = ( v u1 v u = ( ) v u1 v u v u,k 1 Then, v u ) v uk and for u =4, 5, let (i) v 1j =6k 3(j 1), 1 j k, (ii) v j =3k 3(j 1), 1 j k, (iii) v 3j = 3(j 1), 1 j k, (iv) v 4j =3k 3 3(j 1), 1 j k 1, (v) v 5j = 3 3(j 1), 1 j k 1, (vi) v 6j = 3k 3(j 1), 1 j k Also, the following hold: (a) For 1 j k, v 1j + v 6,k+1 j =1and v j + v 3,k+1 j = 1, (b) For 1 j k 1, v 4j + v 5,k j =0, (c) For 1 j k, v 1j + v j + v 3,k+1 j + v 6,k+1 j =0, (d) v k =1; v 4,k 1 + v 51 =0 Lemma 44 For 1 u 6, letw u = ( w u1 w u ) w uk Then1 j k, (i) w 1j =6k 3(j 1), (ii) w j =3k 3(j 1), (iii) w 3j = 3 3(j 1), (iv) w 4j =3k 3j +1, (v) w 5j = 3(j 1), (vi) w 6j = 3k 3(j 1) Also, the following hold: (a) w 1j + w 6,k+1 j =1; w j + w 3,k+1 j =0; w 4j + w 5,k+1 j = 1, (b) w 1j + w 4j + w 5,k+1 j + w 6,k+1 j =0, (c) w 4k =1; w k + w 31 =0 Lemma 45 The p q matrix H p has the following properties, (i) each of the numbers 1 through pq appear once, (ii) each column sum equals p(pq+1), (iii) the i-th row sum is pq(q 1) + qi, (iv) for 1 i p 1, the difference between the (p +1 i)-th row sum and the i-th row sum is q(p +1 i), (v) the ( p+1 )-th row sum is q(pq+1) Proof This follows from Lemma 41

10 140 FS CHAI, A DAS AND C MIDHA Theorem 41 For given odd integers p>1 and q>1, there exist a p q magic rectangle M Proof By the application of Theorems 1 and and without loss of generality, we can assume that p<qand q does not have a factor 3 First, construct G p and H p From Lemma 45, in H p, each column sum equals p(pq+1) and the ( p+1 )-th row sum equals q(pq+1) Therefore we will rearrange the elementorders within columns of H p such that the resulting H p has equal row sums More specific, our objective, for i p 1, is to interchange some elements of the i-th row of H p with the corresponding elements of (p +1 i)-th row such that it results in the two row sums being equal to q(pq+1) The row elements interchanges strategy in H p as follows Suppose we interchange l elements h ij1,h ij,,h ijl of the i-th row with corresponding elements h p+1 i,j1,h p+1 i,j,,h p+1 i,jl of the (p +1 i)-th row of H p Then, by Lemma 45 (ii), (iii) and (iv), we must have l u=1 [h p+1 i,j u h iju ]= q(p+1 i) in order to make the corresponding row sums equal to q(pq+1) This implies p l (g p+1 i,ju g iju )+(p +1 i)l = u=1 q(p +1 i), which has a solution l u=1 [g p+1 i,j u g iju ]= p+1 i and l = q p for 1 i p 1 Thus, for l =(q p)/ we seek j 1,j,,j l such that l u=1 [g p+1 i,j u g iju ]= (p+1 i) Based on the different row positions of H p, there are three cases of positions interchanges, Cases I, II and III, which are shown at the end of the proof Let the desired magic rectangle M be the matrix obtained from H p after carrying out interchanges at positions as below: (A) If p=3, then do the positions interchanges as in Case I (B) If p=5, then the first and the 5-th rows do the positions interchanges as in Case III and the rest of rows do the positions interchanges as in Case I (C) If p =4p +3and p 1, then the (p 1)/th and (p +1)/-th do the positions interchanges as in Case I and the rest rows do the positions interchanges as in Case II with i =1, 3, 5,,p 1 (D) If p =4p +5 and p 1, then the first and the last rows do the positions interchanges as in Case III, the (p 1)/th and (p +1)/)-th rows do the positions interchanges as in Case I and the rest of rows do the positions interchanges as in Case II with i =, 4, 6,,p Then, following the effects of interchange of elements in H p,thep q matrix M has the following properties, (i) each of the numbers 1 through pq appear once, (ii) each row sum equals q(pq+1),

11 MAGIC RECTANGLES OF ODD ORDER 141 (iii) each column sum equals p(pq+1) Thus M is a magic rectangle Case I: Interchanges between row elements of H p corresponding to S The three rows of S appear as ( p 1 p+1 p+3 )-th, ( )-th and ( )-th rows of G palso,the difference between the ( p+3 p 1 )-th and ( )-th row sums of H p is q Suppose l corresponding elements between the ( p+3 p 1 )-th and ( )-th rows of H p are interchanged Let the sum of the differences between these l elements in G p equals s (say) Then l and s satisfies ps +l = q with a solution s =1andl =(q p)/ Let l =(q p)/ =4y + z (say), y 0, 1 z 4 We have Case I (a) q =6k 1 The choice of the l interchanges, g (p+1)/,ju g (p 1)/,ju, 1 u l, means the choice of the l elements from (v 1,v,v 3,v 4, 0,v 5,v 6 ) Let V 0 = φ and V j = {v 1j,v j,v 3,k+1 j,v 6,k+1 j },1 j k 1 The l interchanges are as follows (1) If l =4y + 1, then choose V j,0 j y and v k () If l =4y +, then choose V j,0 j y and v k,0 (3) If l =4y + 3, then choose V j,0 j y and v k, v 4,k 1, v 51 (4) If l =4y + 4, then choose V j,0 j y and v k, v 4,k 1,0,v 51 By Lemma 43, we got l V j l =0,j =0, 1,,y k 1, v k =1andv 4,k 1 + v 51 =0, hence, l u=1 [g (p+1)/,j u g (p 1)/,ju ] = 1 for (1) to (4) Case I (b) q = 6k + 1 The choice of the l interchanges, g (p+1)/,ju g (p 1)/,ju, 1 u l, means the choice of the l elements from (w 1,w, 0,w 3,w 4,w 5,w 6 ) Let W 0 = φ and W j = {w 1j,w 4j,w 5,k+1 j,w 6,k+1 j },1 j k 1 The l interchanges are as follows (1) If l =4y + 1, then choose W j,0 j y and w 4k () If l =4y +, then choose W j,0 j y and w 4k,0 (3) If l =4y + 3, then choose W j,0 j y and w 4k, w k, w 31 (4) If l =4y + 4, then choose W j,0 j y and w 4k, w k,0,w 31 By Lemma 44, we obtained l U j l =0,j =0, 1,,y k 1, w 4k =1andv k +v 31 =0, hence, l u=1 [g (p+1)/,j u g (p 1)/,ju ] = 1 for (1) to (4) Let c j,1 j q denote the j-th column position of H p Then the above solution suggests interchange between ( p 1 p+3 )-th and ( )-th rows of H p at l positions as below: For q =6k 1, let C 1 0 = φ and C1 j = {c j,c k+j,c 3k+1 j,c 6k j },1 j (k 1) (1) If l =4y + 1, then choose Cj 1,0 j y and c k () If l =4y +, then choose Cj 1,0 j y and c k, c 4k

12 14 FS CHAI, A DAS AND C MIDHA (3) If l =4y + 3, then choose Cj 1,0 j y and c k, c 4k 1, c 4k+1 (4) If l =4y + 4, then choose Cj 1,0 j y and c k, c 4k 1, c 4k, c 4k+1 For q =6k + 1, let C0 = φ and C j = {c j,c 3k+1+j,c 5k+ j,c 6k+ j },1 j k 1 (1) If l =4y +1,Cj,0 j y and c 4k+1 () If l =4y +,Cj,0 j y and c 4k+1, c k+1 (3) If l =4y +3,Cj,0 j y and c 4k+1, c k, c k+ (4) If l =4y +4,Cj,0 j y and c 4k+1, c k, c k+1, c k+ Case II: Interchanges ( ) between row ( elements ) of H p corresponding to T U and T L TU1 TL1 Let T U = and T T L = Let z denote the largest integer less U T L than or equal to z As rows of G p, the four rows of T appears as follows For p 1 p 3 p 3 i 1anditaking alternate values (ie, 1, 3, or, 4,), 4 T U1 appear as i-th row, T U appear as (i + 1)-th row, T L1 appear as (p +1 i 1)- th row and T L appear as (p +1 i)-th row of G p For p 1 p 3 p 3 i, 4 the difference between the (p +1 i)-th and i-th row sums of H p is q(p +1 i) Suppose l corresponding elements between the (p +1 i)-th and i-th rows of H p are interchanged Again, let the sum of the differences between these l elements in G p equals s Then l and s satisfies ps +(p +1 i)l = q(p +1 i)/ witha solution s =(p +1 i)/ andl =(q p)/ for carrying out interchanges between (p +1 i)-th and i-th rows of G p Let l =(q p)/ =y + z (say), y 0, z =1, The choice of the l interchanges, g (p+1 i),ju g i,ju,1 u l, means the choice of the l elements from (0, 1,,,q,q 1, q +1, q +,,, 1) Let V 0 = φ and V j = {q j, q + j}, 1 j q The l interchanges are as follow (1) If l =y + 1, then choose V j,0 j y and 1 () If l =y +, then choose V j,0 j y and 0, 1 It is clear to see l V j l =0,j =0, 1,,y q, hence, l u=1 [g (p+1 i),j u g i,ju ]=1 for (1) and () Let c j,1 j q denote the j-th column position of H p Then the above solution suggests interchange between (p +1 i)-th and i-th rows of H p at l positions as below: Let C 0 = φ and C j = {c q+3 j,cq+1 j}, 1 j q (1) If l =y + 1, then choose C j,0 j y and c p+3 i () If l =y +, then choose C j,0 j y and c 1, c p+3 i

13 MAGIC RECTANGLES OF ODD ORDER 143 Case III: Interchanges between row elements of H p corresponding to R U and R L The two rows of R appear as 1-st and p-th rows of G p Also, the difference between the p-th and 1-st row sums of H p is q(p 1) Suppose l corresponding elements between the p-th and 1-st rows of H p are interchanged As earlier, let the sum of the differences between these l elements in G p equals s Then l and s satisfies ps +(p 1)l = q(p 1)/ with a solution s =(p 1)/ andl =(q p)/ for carrying out interchanges between p-th and 1-st rows of G p Let l =(q p)/ =y + z (say), y 0, z =1, The choice of the l interchanges, g p,ju g 1,ju,1 u l, means the choice of the l elements from e R=(q 1,q 3,,4,, 0,, 4,, q +3, q +1) LetV 0 = φ and V j = {q j, q + j}, 1 j q The l interchanges are as follow (1) If l =y + 1, then choose V j,0 j y and (p 1)/ () If l =y +, then choose V j,0 j y and (p 1)/, 0 It can be checked that (p 1)/ isthe(q p +3)/4-th element of e R and l V j l = 0, j =0, 1,,y q ; hence l u=1 [g p,j u g 1,ju ]=(p 1)/ for (1) and () Let c j,1 j q, denotethej-th column position of H p Then the above solution suggests interchange between p-th and 1-th rows of H p at l positionsasbelow: Let C 0 = φ and C j ={c j, c q+1 j },1 j q (1) If l =y + 1, then choose C j,0 j y and c (q+3 p)/4 () If l =y +, then choose C j,0 j y and c (q+3 p)/4, c (q+1)/ Acknowledgements Thanks are due to a referee for some constructive suggestions Part of this work was carried out while the second author was visiting The University of Akron References [1] T Bier and A Kleinschmidt, Centrally symmetric and magic rectangles, Discrete Math 176 (1997), 9 4 [] M N Das and P Sarkar, Construction of magic rectangles with tables when row and column numbers are both odd, J Statist Comp Applic 4 (006), [3] J P De Los Reyes, A Das and C K Midha, A matrix approach to construct magic rectangles of even order, Australas J Combin 40 (008),

14 144 FS CHAI, A DAS AND C MIDHA [4] J P De Los Reyes, A Das, C K Midha and P Vellaisamy, On a method to construct magic rectangles of even order, Utilitas Math 80 (009), [5] T Hagedorn, Magic rectangles revisited, Discrete Math 07 (1999), 65 7 (Received 30 Aug 011; revised 3 Aug 01)

RESEARCH ARTICLE. Linear Magic Rectangles

RESEARCH ARTICLE. Linear Magic Rectangles Linear and Multilinear Algebra Vol 00, No 00, January 01, 1 7 RESEARCH ARTICLE Linear Magic Rectangles John Lorch (Received 00 Month 00x; in final form 00 Month 00x) We introduce a method for producing