Strongly symmetric self-orthogonal diagonal Latin ... - Semantic Scholar
Discrete Mathematics 328 (2014) 79–87
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Strongly symmetric self-orthogonal diagonal Latin squares and Yang Hui type magic squares✩ Yong Zhang a , Kejun Chen a,b,∗ , Nanyuan Cao a , Hantao Zhang c a
School of Mathematical Sciences, Yancheng Teachers University, Jiangsu 224002, China
b
Department of Mathematics, Taizhou University, Jiangsu 225300, China
c
Computer Science Department, The University of Iowa, Iowa City, IA 52242, USA
article
abstract
info
Article history: Received 8 September 2013 Received in revised form 22 March 2014 Accepted 1 April 2014
In this paper, a strongly symmetric self-orthogonal diagonal Latin square of order n with a special property (∗ SSSODLS(n)) is introduced. It is proved that a ∗ SSSODLS(n) exists if and only if n ≡ 0 (mod 4) and n ̸= 4. As an application, it is shown that there exists a Yang Hui type magic square YMS(n, 4) if and only if n ≡ 0 (mod 4) and n ̸= 4. © 2014 Elsevier B.V. All rights reserved.
Keywords: Self-orthogonal Latin square Strongly symmetric Magic square Yang Hui type
1. Introduction A Latin square of order n, denoted by LS(n), is an n × n array such that every row and every column is a permutation of an n-set S. A diagonal Latin square is a Latin square with the additional property that the main diagonal and back diagonal are both permutations of S. Two LS(n)s are called orthogonal if each symbol in the first square meets each symbol in the second square exactly once when they are superposed. A Latin square is called self-orthogonal if it is orthogonal to its transpose. Usually, the (i, j) entry of a matrix A is denoted by ai,j . Let In = {0, 1, . . . , n − 1}. An LS(n)L over In is called strongly symmetric if li,j + ln−1−i,n−1−j = n − 1 for all i, j ∈ In . A strongly symmetric self-orthogonal diagonal LS(n) is denoted by SSSODLS(n). As an example, an SSSODLS(4) is listed below.
0 3 U4 = 2 1
1 2 3 0
3 0 1 2
2 1 . 0 3
Du and Cao [11], Cao and Li [5] investigated the existence of SSSODLS(n)s, they proved the following. Lemma 1.1 ([11,5]). There exists an SSSODLS(n) if and only if n ≡ 0, 1, 3(mod 4) and n ̸= 3.
✩ The research is supported by the National Natural Science Foundations of China (No. 11371308 and No. 11301457).
∗
Corresponding author at: School of Mathematical Sciences, Yancheng Teachers University, Jiangsu 224002, China. E-mail addresses: [email protected], [email protected] (K. Chen).
http://dx.doi.org/10.1016/j.disc.2014.04.002 0012-365X/© 2014 Elsevier B.V. All rights reserved.
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Let n be an even integer, and let A = (ai,j ) be an SSSODLS(n) satisfying
(∗)
a¯ 2i,j a¯ j,i =
i∈I n j∈I n 2
a¯ 2i,j a¯ j,i = 0,
i∈I n j∈In \I n
2
2
2
where a¯ = a − Then we say that A is a ∗ SSSODLS(n). By Lemma 1.1, there exists a ∗ SSSODLS(n) only if n ≡ 0(mod 4). We shall show that this necessary condition is also sufficient. Specially, we shall prove the following. n−1 . 2
Theorem 1.2. There exists a ∗ SSSODLS (n) if and only if n ≡ 0(mod 4) and n ̸= 4. SSSODLS(n)s can be used to construct Yang Hui type magic squares. An n × n matrix A consisting of all given n2 consecutive integers is a magic square of order n, denoted by MS(n), if the sum of elements in each row, each column, and each of two diagonals is the same. The study of magic squares probably dates back to prehistoric times [3]. A lot of work had been done on magic squares [9, 6,1–3]. Yang Hui [14] gave an MS(8) as follows. ∗
Y8 = (yi,j ) =
61 12 20 37 29 44 52 5
3 54 46 27 35 22 14 59
2 55 47 26 34 23 15 58
64 9 17 40 32 41 49 8
57 16 24 33 25 48 56 1
7 50 42 31 39 18 10 63
6 51 43 30 38 19 11 62
60 13 21 36 28 45 53 4
.
As pointed out by S. Chikaraishi et al. [8] that the Y8 has the additional property that for each e = 2, 3, 4,
7 7 i=4
e j=0 yi,j , and
7 3
e j=0 yi,j
i =0
=
3 7
e i=0 j=0 yi,j e j=4 yi,j . A magic square having such a property is usually called a Yang Hui type
=
7 7 i=0
magic square. Generally, for an even integer n and an integer t ≥ 2, an MS(n)A is a Yang Hui type magic square, denoted by YMS(n, t ), if, for each e = 2, 3, . . . , t, the sum of the elements of the first 2n rows of A∗e is the same as that of the last 2n rows, and the sum of the elements of the left 2n columns of A∗e is the same as that of the right 2n columns, where A∗e = (aei,j ). A YMS(n, t ) is also called a magic square with t-powered sum by Chikaraishi in [8]. We should mention that a YMS(n, t ) is a kind of weak form of a t-multimagic square. For the details of t-multimagic squares, we refer the readers to [10,15,7] and the references therein. For the existence of YMS(n, t )s, Chikaraishi et al. [8] gave a family of YMS(2t , 2t − 2)s. Recently, Cao, Chen and Zhang [4] determined a necessary and sufficient condition for the existence of a YMS(n, 2) for n ̸= 2. These results are listed as follows. Lemma 1.3 ([8,4]). (1) There exists a YMS (2t , 2t − 2) for all integers t ≥ 2. (2) There exists a YMS (n, 2) if and only if n is even and n ̸= 2. In this paper, we will investigate the existence of YMS(n, 4)s by making use of ∗ SSSODLS(n)s and get the following. Theorem 1.4. There exists a YMS (n, 4) if and only if n ≡ 0(mod 4) and n ̸= 4. A construction of YMS(n, 4) based on ∗ SSSODLS(n) is presented in Section 2, and some constructions of ∗ SSSODLS(n)s are provided in Section 3. The proofs of Theorems 1.2 and 1.4 are both given in Section 4. 2. A construction of YMS(n, 4) based on ∗ SSSODLS(n) Let n be an even integer, denote S11 = {(i, j)|i, j ∈ I n }, 2
S12 = {(i, j)|i ∈ I n , j ∈ In \ I n }, 2
S21 = {(i, j)|i ∈ In \ I , j ∈ I n }, n 2
2
2
S22 = {(i, j)|i, j ∈ In \ I n }. 2
For a matrix A of order n, the sum (i,j)∈S ai,j is written as Let C be a YMS(n, t ). It is easy to see that the conditions
cie,j +
S11
S11
cie,j =
S12
cie,j +
S21
cie,j +
S21
cie,j =
S12
cie,j ,
S22
cie,j +
S22
cie,j ,
S ai,j in short.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
are equivalent to that
cie,j =
S11
cie,j ,
S22
cie,j =
S12
cie,j ,
S21
where e = 2, 3, . . . , t. Now we give a construction of YMS(n, 4) based on ∗ SSSODLS(n) in the following. Construction 2.1. If there is a ∗ SSSODLS (n), then there is a YMS (n, 4). Proof. Let A be a ∗ SSSODLS(n), and let C = (ci,j ),
ci,j = nai,j + aj,i , i, j ∈ In .
Then C is an MS(n). We shall prove that C is also a YMS(n, 4), i.e., C satisfies the following.
cie,j =
S11
cie,j ,
S22
cie,j =
S12
cie,j ,
e = 2, 3, 4.
c˜ie,j ,
e = 2, 3, 4,
S21
It suffices to prove that
c˜ie,j =
S11
c˜ie,j ,
S22
c˜ie,j =
S12
S21
2
where c˜i,j = ci,j − n 2−1 , i, j ∈ In . By the definition of C , we have c˜i,j = nai,j + aj,i −
n2 − 1 2
n−1 n−1 = n ai , j − + aj , i − = na¯ i,j + a¯ j,i , 2
2
1 where a¯ i,j = ai,j − n− , i , j ∈ In . 2 Noting that a¯ i,j = −¯an−1−i,n−1−j , i, j ∈ In since A is strongly symmetric, we have the following
c˜n−1−i,n−1−j = na¯ n−1−i,n−1−j + a¯ n−1−j,n−1−i = −na¯ i,j − a¯ j,i = −˜ci,j , Clearly,
(˜ci,j )2 , (˜ci,j )2 =
(˜ci,j )2 , (˜ci,j )2 =
(˜ci,j )4 . (˜ci,j )4 = S21
S12
S22
S11
S21
S12
S22
S11
(˜ci,j )4 , (˜ci,j )4 = It remains to prove that
c˜i3,j =
S11
c˜i3,j ,
S22
c˜i3,j =
S12
c˜i3,j .
S21
Since c˜i,j = −˜cn−1−i,n−1−j , i, j ∈ In , we need only to prove that
c˜i3,j =
S11
c˜i3,j = 0.
S12
In fact, c˜i3,j = n3 a¯ 3i,j + 3n2 a¯ 2i,j a¯ j,i + 3na¯ i,j a¯ 2j,i + a¯ 3j,i , Since A is a Latin square, for each e > 0 we have
a¯ ei,j +
S11
a¯ ei,j =
S12
a¯ ei,j +
S11
a¯ ei,j +
S21
a¯ ei,j =
S21
a¯ ei,j ,
S22
a¯ ei,j +
S12
a¯ ei,j .
S22
Therefore
S11
a¯ ei,j =
S22
a¯ ei,j ,
S12
a¯ ei,j =
S21
a¯ ei,j .
i , j ∈ In .
i, j ∈ In .
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Since a¯ i,j = −¯an−1−i,n−1−j , i, j ∈ In , we have
a¯ 3i,j =
S11
a¯ 3i,j =
S22
a¯ 3i,j =
S12
a¯ 3i,j = 0.
S21
From the condition (∗)
a¯ 2i,j a¯ j,i = 0,
S11
a¯ 2i,j a¯ j,i = 0.
S12
By commutating i, j in the above two equalities, we get the following.
a¯ i,j a¯ 2j,i = 0,
S11
a¯ i,j a¯ 2j,i = 0.
S21
Hence
a¯ i,j a¯ 2j,i =
(−¯ai,j )(−¯aj,i )2 = − a¯ i,j a¯ 2j,i = 0. S21
S12
S21
Thus
c˜i3,j =
S11
c˜i3,j = 0.
S12
This completes the proof.
3. Constructions for ∗ SSSODLS(n)s In this section, we provide some constructions of ∗ SSSODLS(n)s. Construction 3.1. If there is a ∗ SSSODLS (n), and an SSSODLS (m), then there is a ∗ SSSODLS (mn). Proof. Let U be a ∗ SSSODLS(n), and let V be an SSSODLS(m). Let W be an mn × mn matrix given by
wi,j = mur ,s + vx,y ,
i = mr + x, j = ms + y, r , s ∈ In , x, y ∈ Im .
It is readily checked that W is an SSSODLS(mn). We shall show that W satisfies the condition (∗), i.e.,
w ¯ i2,j w ¯ j ,i =
S11
w ¯ i2,j w ¯ j ,i = 0 ,
w ¯ =w−
mn − 1 2
S12
.
By the definition of W , we have mn − 1
w ¯ i,j = wi,j −
2
= mur ,s + vx,y −
1 where u¯ = u − n− , v¯ = v − 2
m−1 . 2
mn − 1 2
= mu¯ r ,s + v¯ x,y ,
So
w ¯ i2,j w ¯ j,i = (mu¯ r ,s + v¯ x,y )2 (mu¯ s,r + v¯ y,x ), = m3 u¯ 2r ,s u¯ s,r + 2m2 u¯ r ,s u¯ s,r v¯ x,y + mu¯ s,r v¯ x2,y + m2 u¯ 2r ,s v¯ y,x + 2mu¯ r ,s v¯ x,y v¯ y,x + v¯ x2,y v¯ y,x , ¯ 2r ,s u¯ s,r = 0. Therefore By the condition (∗) we have r ∈I n s∈I n u 2
2
u¯ 2r ,s u¯ s,r
x∈Im y∈Im r ∈I n s∈I n 2
= u¯ 2r ,s u¯ s,r = 0. x∈Im y∈Im
2
r ∈I n s∈I n 2
2
v¯ x,y = 0 since V is a Latin square, we have v¯ x,y = 0, u¯ r ,s u¯ s,r v¯ x,y = u¯ r ,s u¯ s,r
Note that
x∈Im
y∈Im
x∈Im y∈Im r ∈I n s∈I n 2
r ∈I n s∈I n
2
2
x∈Im y∈Im r ∈I n s∈I n 2
2
¯ v¯
u2r ,s y,x
x∈Im y∈Im
2
2 = u¯ r ,s v¯ y,x = 0. r ∈I n s∈I n 2
2
x∈Im y∈Im
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
83
Since U is ∗ SSSODLS(n), the following is clear.
u¯ r ,s = 0,
r ∈I n s∈In
u¯ r ,s = 0,
r ∈In s∈I n
2
r ∈I n
2
u¯ s,r v¯ x2,y =
x∈Im y∈Im r ∈I n s∈I n 2
x∈Im y∈Im
2
r ∈I n s∈I n 2
2
v¯ x,y v¯ y,x = 0. = u¯ r ,s
u¯ r ,s v¯ x,y v¯ y,x
x∈Im y∈Im r ∈I n s∈I n 2
v¯ x2,y u¯ s,r = 0,
2
2
2
2
u¯ r ,s = 0, and
s∈I n
2
u¯ r ,s .
r ∈In \I n s∈I n
r ∈I n s∈In \I n
2
From the above we have
u¯ r ,s = −
x∈Im y∈Im
r ∈I n s∈I n
2
2
2
Since V is self-orthogonal, we have
v¯ x2,y v¯ y,x =
x∈Im y∈Im
x−
m−1
2
2
x∈Im y∈Im
y−
m−1 2
=
x−
m−1
x∈Im
2
2
y−
m−1
y∈Im
2
= 0.
Therefore
v¯ v¯
2 x,y y,x
=
x∈Im y∈Im r ∈I n s∈I n 2
Hence
r ∈I n s∈I n
2
2
2
v¯ v¯
2 x,y y,x
= 0.
x∈Im y∈Im
w ¯ w ¯ j,i = 0. 2 i ,j
S11
Similarly, one can prove that
S12
w ¯ i2,j w ¯ j,i = 0. Thus W is a ∗ SSSODLS(mn).
An m × n magic rectangle, denoted by MR(m, n), is an m × n matrix consisting of all the numbers of Imn such that the sum of the entries of each row is a constant and each column sum is also a constant (different if m ̸= n). Harmuth [12,13] proved that for m, n > 1, there exists an MR(m, n) if and only if m ≡ n(mod 2) and (m, n) ̸= (2, 2). We shall now construct a ∗ SSSODLS(4m) by using an SSSODLS(m) and an MR(2, 2m). Construction 3.2. If there exists an SSSODLS (m), then there exists a ∗ SSSODLS (4m). Proof. Let U be the U4 given in Section 1. Let V be an SSSODLS(m). A matrix W is given by
wi,j = mur ,s + vx,y ,
i = mr + x, j = ms + y, r , s ∈ I4 , x, y ∈ Im .
It is readily checked that W is an SSSODLS(4m). Let H be an MR(2, 2m), and let L = (lj ), where h0,j , h1,4m−1−j ,
j ∈ I2m , j ∈ I4m \ I2m .
lj =
Then lj + l4m−1−j = h0,j + h1,j = 4m − 1, j ∈ I2m . Let A = (ai,j ),
ai,j = lwi,j , i, j ∈ I4m .
Clearly, A is also an SSSODLS(4m). We shall prove that A satisfies the condition (∗), i.e.,
a¯ 2i,j a¯ j,i =
S11
a¯ 2i,j a¯ j,i = 0,
S12
where a¯ i,j = ai,j − In fact,
a¯ 2i,j a¯ j,i =
S11
4m−1 2
= lwi,j −
4m−1 2
(¯lwi,j )2¯lwj,i =
S11
= ¯lwi,j .
(¯lmur ,s +vx,y )2¯lmus,r +vy,x .
r ∈I2 s∈I2 x∈Im y∈Im
By the definition of U, we have {(ur ,s , us,r )|r , s ∈ I2 } = {(0, 0), (1, 3), (3, 1), (2, 2)}, and
S11
a¯ 2i,j a¯ j,i =
x∈Im y∈Im
[(¯lvx,y )2¯lvy,x + (¯lm+vx,y )2¯l3m+vy,x + (¯l3m+vx,y )2¯lm+vy,x + (¯l2m+vx,y )2¯l2m+vy,x ].
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Noting that {(vx,y , vy,x )|x, y ∈ Im } = {(i, j)|i, j ∈ Im } since V is self-orthogonal, we have
(¯lkm+vx,y )2¯lsm+vy,x =
¯lsm+j , k, s ∈ I2 . (¯lkm+i )2¯lsm+j = (¯lkm+i )2
x∈Im y∈Im
i∈Im j∈Im
i∈Im
j∈Im
Hence
a¯ 2i,j a¯ j,i =
¯lj + ¯lj+3m + ¯lj+m + ¯lj+2m . (¯li )2 (¯li+m )2 (¯li+3m )2 (¯li+2m )2 i∈Im
S11
j∈Im
i∈Im
j∈Im
i∈Im
j∈Im
i∈Im
j∈Im
It is easy to see that ¯lj = −¯l4m−1−j , j ∈ I4m . By this we have
(¯lj+3m )2 = (−¯lm−1−j )2 = (¯lj )2 , j∈Im
j∈Im
j∈Im
(¯lj+2m )2 = (−¯lm+m−1−j )2 = (¯lj+m )2 . j∈Im
j∈Im
j∈Im
Consequently,
a¯ 2i,j a¯ j,i =
S11
=
(¯li )2
¯lj +
i∈Im
j∈Im
(¯li )2
i∈Im
It is easy to show that
¯lj+3m + ¯lj+2m ¯lj+m + (¯li+m )2 (¯li+m )2 (¯li )2 i∈Im
¯lj +
¯ +
j∈Im lj
¯lj+m +
j∈Im
j∈Im
j∈Im
i∈Im
¯
j∈Im lj+m
j∈Im
¯lj+3m . ¯lj+2m + (¯li+m )2 i∈Im
2 S11 ai,j aj,i
i∈Im
j∈Im
= 0,
j∈Im
j∈Im
¯
j∈Im lj+2m
+
¯
j∈Im lj+3m
= 0 since H is an MR(2, 2m). So we have
¯ ¯ = 0.
In the same way one can prove that
S12
a¯ 2i,j a¯ j,i = 0. So A is a ∗ SSSODLS(4m).
To illustrate the proof of Construction 3.2, we give an example as follows. Example 2. An SSSODLS(4) U is listed in Section 1. An SSSODLS(5) is given below.
1 2 V = 3 4 0
3 4 0 1 2
0 1 2 3 4
2 3 4 0 1
4 0 1 . 2 3
Let W = (wi,j ),
wi,j = 5ur ,s + vx,y , i = 5r + x, j = 5s + y, r , s ∈ I4 , x, y ∈ I5 ,
i.e.,
W = (wi,j ) =
1 2 3 4 0 16 17 18 19 15 11 12 13 14 10 6 7 8 9 5
3 4 0 1 2 18 19 15 16 17 13 14 10 11 12 8 9 5 6 7
0 1 2 3 4 15 16 17 18 19 10 11 12 13 14 5 6 7 8 9
2 3 4 0 1 17 18 19 15 16 12 13 14 10 11 7 8 9 5 6
4 0 1 2 3 19 15 16 17 18 14 10 11 12 13 9 5 6 7 8
6 7 8 9 5 11 12 13 14 10 16 17 18 19 15 1 2 3 4 0
8 9 5 6 7 13 14 10 11 12 18 19 15 16 17 3 4 0 1 2
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0 1 2 3 4
7 8 9 5 6 12 13 14 10 11 17 18 19 15 16 2 3 4 0 1
9 5 6 7 8 14 10 11 12 13 19 15 16 17 18 4 0 1 2 3
16 17 18 19 15 1 2 3 4 0 6 7 8 9 5 11 12 13 14 10
18 19 15 16 17 3 4 0 1 2 8 9 5 6 7 13 14 10 11 12
15 16 17 18 19 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
17 18 19 15 16 2 3 4 0 1 7 8 9 5 6 12 13 14 10 11
19 15 16 17 18 4 0 1 2 3 9 5 6 7 8 14 10 11 12 13
11 12 13 14 10 6 7 8 9 5 1 2 3 4 0 16 17 18 19 15
13 14 10 11 12 8 9 5 6 7 3 4 0 1 2 18 19 15 16 17
10 11 12 13 14 5 6 7 8 9 0 1 2 3 4 15 16 17 18 19
12 13 14 10 11 7 8 9 5 6 2 3 4 0 1 17 18 19 15 16
14 10 11 12 13 9 5 6 7 8 4 0 1 2 3 19 15 16 17 18
.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
85
An MR(2, 10) is listed below.
0 19
H =
1 18
3 16
9 10
12 7
11 8
13 6
14 5
15 4
17 2
.
Let L = (lj ), where h 0 ,j , h1,4m−1−j ,
j ∈ I10 , j ∈ I20 \ I10 .
lj = i.e.,
L = (lj ) =
0
1
3
9
11
12
13
14
15
17
2
4
5
6
7
8
10
16
18
19
.
Let A = (ai,j ), ai,j = lwi,j , i, j ∈ I20 , i.e.,
1 A=
3 9 11 0 10 16 18 19 8 4 5 6 7 2 13 14 15 17 12
9 11 0 1 3 18 19 8 10 16 6 7 2 4 5 15 17 12 13 14
0 1 3 9 11 8 10 16 18 19 2 4 5 6 7 12 13 14 15 17
3 9 11 0 1 16 18 19 8 10 5 6 7 2 4 14 15 17 12 13
11 0 1 3 9 19 8 10 16 18 7 2 4 5 6 17 12 13 14 15
13 14 15 17 12 4 5 6 7 2 10 16 18 19 8 1 3 9 11 0
One can check that A is a ∗ SSSODLS(20).
15 17 12 13 14 6 7 2 4 5 18 19 8 10 16 9 11 0 1 3
12 13 14 15 17 2 4 5 6 7 8 10 16 18 19 0 1 3 9 11
14 15 17 12 13 5 6 7 2 4 16 18 19 8 10 3 9 11 0 1
17 12 13 14 15 7 2 4 5 6 19 8 10 16 18 11 0 1 3 9
10 16 18 19 8 1 3 9 11 0 13 14 15 17 12 4 5 6 7 2
18 19 8 10 16 9 11 0 1 3 15 17 12 13 14 6 7 2 4 5
8 10 16 18 19 0 1 3 9 11 12 13 14 15 17 2 4 5 6 7
16 18 19 8 10 3 9 11 0 1 14 15 17 12 13 5 6 7 2 4
19 8 10 16 18 11 0 1 3 9 17 12 13 14 15 7 2 4 5 6
4 5 6 7 2 13 14 15 17 12 1 3 9 11 0 10 16 18 19 8
6 7 2 4 5 15 17 12 13 14 9 11 0 1 3 18 19 8 10 16
2 4 5 6 7 12 13 14 15 17 0 1 3 9 11 8 10 16 18 19
5 6 7 2 4 14 15 17 12 13 3 9 11 0 1 16 18 19 8 10
7 2 4 5 6 17 12 13 14 15 11 0 1 3 9 19 8 10 16 18
.
4. Proofs of Theorems 1.2 and 1.4 In this section we shall give the proofs of Theorems 1.2 and 1.4. Lemma 4.1. There is no ∗ SSSODLS (4). Proof. Suppose on the contrary that A is a ∗ SSSODLS(4). Let a¯ i,j = ai,j − 23 , and let e = 2a¯ 0,0 , f = 2a¯ 0,1 , g = 2a¯ 1,0 , h = 2a¯ 1,1 . Then e, f , g , h ∈ {−3, −1, 1, 3}. Clearly, e, f , h are pairwise distinct, g ̸= e, g ̸= h and e3 + h3 ∈ {±28, ±26, 0}. By the condition (∗) we have e3 + h3 + fg (f + g ) = 0. We shall show that g ̸= f . In fact, If g = f , then e3 + h3 + 2f 3 = 0. If e3 + h3 = 0, then f = 0, a contradiction. So e3 + h3 ∈ {±28, ±26} and f 3 ∈ {±14, ±13}, which is impossible. Hence g ̸= f . From the above we have {e, f , g , h} = {−3, −1, 1, 3}. Hence f + g ∈ {±2, ±4, 0}, fg ∈ {−1, ±3, −9} and fg (f + g ) ∈ {0, ±2, ±4, ±6, ±12, ±18, ±36}. Therefore e3 + h3 = −fg (f + g ) = 0. It follows that h = −e, g = −f and a¯ 2,2 = −¯a1,1 = e = a¯ 0,0 , which contradicts the assumption that A is diagonal. So there is no ∗ SSSODLS(4). Lemma 4.2. There exists a ∗ SSSODLS (n) for each n ∈ {8, 12, 24}. Proof. Let
A8 =
0 6 1 7 3 5 2 4
2 4 3 5 1 7 0 6
4 2 5 3 7 1 6 0
6 0 7 1 5 3 4 2
5 3 4 2 6 0 7 1
7 1 6 0 4 2 5 3
1 7 0 6 2 4 3 5
3 5 2 4 0 6 1 7
.
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
It is readily checked that A8 is a ∗ SSSODLS(8). Let
A12
0 5 8 3 6 11 = 9 10 4 7 1 2
1 3 2 9 8 10 11 7 6 5 4 0
17 14 9 23 22 7 2 5 20 12 19 8 1 13 0 16 11 3 10 4 21 15 6 18
3 16 23 15 6 9 4 13 12 2 21 18 17 5 20 14 7 1 8 0 19 11 22 10
10 0 7 4 11 1 3 9 5 6 2 8
4 6 1 10 5 7 2 8 11 0 3 9
7 2 3 8 9 4 5 1 0 11 10 6
8 4 10 1 7 3 6 11 9 2 0 5
6 11 9 2 0 5 8 4 10 1 7 3
5 1 0 11 10 6 7 2 3 8 9 4
2 8 11 0 3 9 4 6 1 10 5 7
3 9 5 6 2 8 10 0 7 4 11 1
11 7 6 5 4 0 1 3 2 9 8 10
9 10 4 7 1 2 0 5 8 3 6 11
and
A24
=
22 7 6 20 15 12 11 23 14 8 1 19 2 18 13 5 16 4 21 10 17 9 0 3
18 3 22 1 16 8 15 4 21 9 20 12 7 23 6 0 17 11 2 14 13 10 5 19
2 5 0 9 8 22 17 14 13 1 16 10 3 21 4 20 15 7 18 6 23 12 19 11
7 21 20 5 0 3 18 10 17 6 23 11 22 14 9 1 8 2 19 13 4 16 15 12
21 13 4 7 2 11 10 0 19 16 15 3 18 12 5 23 6 22 17 9 20 8 1 14
8 20 19 16 11 1 0 9 4 7 22 14 21 10 3 15 2 18 13 5 12 6 23 17
0 12 5 18 1 17 8 11 6 15 14 4 23 20 19 7 22 16 9 3 10 2 21 13
1 19 10 22 21 18 13 3 2 11 6 9 0 16 23 12 5 17 4 20 15 7 14 8
It is routine to check that A12 and A24 are ∗ SSSODLSs.
23 15 2 14 7 20 3 19 10 13 8 17 16 6 1 22 21 9 0 18 11 5 12 4
16 10 3 21 12 0 23 20 15 5 4 13 8 11 2 18 1 14 7 19 6 22 17 9
14 6 1 17 4 16 9 22 5 21 12 15 10 19 18 8 3 0 23 11 2 20 13 7
19 11 18 12 5 23 14 2 1 22 17 7 6 15 10 13 4 20 3 16 9 21 8 0
15 9 16 8 3 19 6 18 11 0 7 23 14 17 12 21 20 10 5 2 1 13 4 22
10 2 21 13 20 14 7 1 16 4 3 0 19 9 8 17 12 15 6 22 5 18 11 23
6 0 17 11 18 10 5 21 8 20 13 2 9 1 16 19 14 23 22 12 7 4 3 15
13 1 12 4 23 15 22 16 9 3 18 6 5 2 21 11 10 19 14 17 8 0 7 20
5 17 8 2 19 13 20 12 7 23 10 22 15 4 11 3 18 21 16 1 0 14 9 6
9 22 15 3 14 6 1 17 0 18 11 5 20 8 7 4 23 13 12 21 16 19 10 2
11 8 7 19 10 4 21 15 22 14 9 1 12 0 17 6 13 5 20 23 18 3 2 16
12 4 11 0 17 5 16 8 3 19 2 20 13 7 22 10 9 6 1 15 14 23 18 21
4 18 13 10 9 21 12 6 23 17 0 16 11 3 14 2 19 8 15 7 22 1 20 5
20 23 14 6 13 2 19 7 18 10 5 21 4 22 15 9 0 12 11 8 3 17 16 1
.
Proof of Theorem 1.2. Let n = 4m. For m = 1, there is no ∗ SSSODLS(4) by Lemma 4.1. For m = 2, 3, 6, the ∗ SSSODLS(4m)s are given in Lemma 4.2. For m ≡ 0, 1, 3(mod 4), m > 3, there is an SSSODLS(m) by Lemma 1.1, so there is a ∗ SSSODLS(4m) by Construction 3.2. For m ≡ 2(mod 4), m ≥ 10, let m = 4k + 2, k ≥ 2, then 4m = 8(2k + 1). Since there is an SSSODLS(2k + 1) by Lemma 1.1 and there is a ∗ SSSODLS(8) by Lemma 4.2, there is a ∗ SSSODLS(8(2k + 1)) by Construction 3.1. The proof is now complete. To give the proof of Theorem 1.4 we first prove the following. Lemma 4.3. For n ≡ 2(mod 4), there is no YMS (n, t ) whenever t ≥ 3. Proof. Let n = 34k + 2. We need only to prove that there is no YMS(n, 3). Suppose on the contrary that C is a YMS(n, 3). Let T = S11 ∪S12 (ci,j − ci,j ). Let St (n) =
(
1 n
i∈I 2 n
it . We have S1 (n) =
1 n 2
(n2 − 1), S3 (n) =
1 3 n 4
(n2 − 1)2 . Noting that
S11 ∪S12
(ci3,j − ci,j ) =
− ci,j ) since C is a YMS(n, 3), we have n n 1 3 2 1 T = (S3 (n) − S1 (n)) = n (n − 1)2 − n(n2 − 1)
3 S21 ∪S22 ci,j
2
2
4
2
= (2k + 1)(2k + 1)(4k + 1)(4k + 3)(16k + 16k + 5)(8k2 + 8k + 1), 2
which is an odd integer. On the other hand, ci3,j − ci,j = ci,j (ci,j − 1)(ci,j + 1), which is even. It implies that T is also even, a contradiction. So there is no YMS(4k + 2, 3). Proof of Theorem 1.4. For n ≡ 2(mod 4), there does not exist a YMS(n, 4) by Lemma 4.3. For n ≡ 0(mod 4), let n = 4m. If m = 1, by using computer search, we know that there is no YMS(4, 3), hence there is no YMS(4, 4). For m ≥ 2, there is a ∗ SSSODLS(4m) by Theorem 1.2 and hence there is a YMS(4m, 4) by Construction 2.1. Remark. An MS(n) M is symmetric if mi,j + mn−1−i,n−1−j = n2 − 1, i, j ∈ In . Let M = nA + B, M is elementary if A, B are both Latin squares. Otherwise, it is nonelementary. If A, B are strongly symmetric, then M is symmetric. The YMS(n, 4)s obtained in this paper are all symmetric and elementary. So we have the following. Corollary 4.4. There exists a symmetric elementary YMS (n, 4) if and only if n ≡ 0(mod 4) and n ̸= 4.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Acknowledgment The authors would like to thank Professor L. Zhu of Suzhou University for helpful discussions and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
G. Abe, Unsolved problems on magic squares, Discrete Math. 127 (1994) 3–13. M. Ahmed, Algebric combinatorics of magic squares. Ph.D. Dissertation, University of California, Davis, 2004. W.S. Andrews, Magic Squares and Cubes, second ed., Dover, New York, 1960. N. Cao, K. Chen, Y. Zhang, Existence of Yang Hui type magic squares, preprint. H. Cao, W. Li, Existence of strongly symmetric self-orthogonal diagonal Latin squares, Discrete Math. 311 (2011) 841–843. S.v.R. Cammann, Magic squares, in: Encyclopæ dia Britannica, fourteenth ed., Chicago, 1973. K. Chen, W. Li, Existence of normal bimagic squares, Discrete Math. 312 (2012) 3077–3086. S. Chikaraishi, M. Kobayashi, N. Mutoh, G. Nakamura, Magic squares with powered sum, http://usr.u-shizuoka-ken.ac.jp/kn/AN10118525201111001030.pdf. C.J. Colbourn, J.H. Dinitz (Eds.), Handbook of Combinatorial Designs, second ed., Chapman and Hall/CRC, Boca Raton FL, 2007. H. Derksen, C. Eggermont, A.v.D. Essen, Multimagic squares, Amer. Math. Monthly 114 (2007) 703–713. B. Du, H. Cao, Existence of strongly symmetric self-orthogonal diagonal Latin squares, Acta Math. Appl. Sin. (Chin. Ser.) 25 (2002) 187–189. T. Harmuth, Über magische Quadrate und ähniche Zahlenfiguren, Arch. Math. Phys. 66 (1881) 286–313. T. Harmuth, Über magische Rechtecke mit ungeraden Seitenzahlen, Arch. Math. Phys. 66 (1881) 413–447. H. Yang, Yang Hun suan fa, 1274, http://www.tulips.tsuknba.acjp/limedio/dlam/B14B1407158/1.pdf, reprint, 1433. Y. Zhang, K. Chen, J. Lei, Large sets of orthogonal arrays and multimagic squares, J. Combin. Des. 21 (2013) 390–403.
87
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Strongly symmetric self-orthogonal diagonal Latin squares and Yang Hui type magic squares✩ Yong Zhang a , Kejun Chen a,b,∗ , Nanyuan Cao a , Hantao Zhang c a
School of Mathematical Sciences, Yancheng Teachers University, Jiangsu 224002, China
b
Department of Mathematics, Taizhou University, Jiangsu 225300, China
c
Computer Science Department, The University of Iowa, Iowa City, IA 52242, USA
article
abstract
info
Article history: Received 8 September 2013 Received in revised form 22 March 2014 Accepted 1 April 2014
In this paper, a strongly symmetric self-orthogonal diagonal Latin square of order n with a special property (∗ SSSODLS(n)) is introduced. It is proved that a ∗ SSSODLS(n) exists if and only if n ≡ 0 (mod 4) and n ̸= 4. As an application, it is shown that there exists a Yang Hui type magic square YMS(n, 4) if and only if n ≡ 0 (mod 4) and n ̸= 4. © 2014 Elsevier B.V. All rights reserved.
Keywords: Self-orthogonal Latin square Strongly symmetric Magic square Yang Hui type
1. Introduction A Latin square of order n, denoted by LS(n), is an n × n array such that every row and every column is a permutation of an n-set S. A diagonal Latin square is a Latin square with the additional property that the main diagonal and back diagonal are both permutations of S. Two LS(n)s are called orthogonal if each symbol in the first square meets each symbol in the second square exactly once when they are superposed. A Latin square is called self-orthogonal if it is orthogonal to its transpose. Usually, the (i, j) entry of a matrix A is denoted by ai,j . Let In = {0, 1, . . . , n − 1}. An LS(n)L over In is called strongly symmetric if li,j + ln−1−i,n−1−j = n − 1 for all i, j ∈ In . A strongly symmetric self-orthogonal diagonal LS(n) is denoted by SSSODLS(n). As an example, an SSSODLS(4) is listed below.
0 3 U4 = 2 1
1 2 3 0
3 0 1 2
2 1 . 0 3
Du and Cao [11], Cao and Li [5] investigated the existence of SSSODLS(n)s, they proved the following. Lemma 1.1 ([11,5]). There exists an SSSODLS(n) if and only if n ≡ 0, 1, 3(mod 4) and n ̸= 3.
✩ The research is supported by the National Natural Science Foundations of China (No. 11371308 and No. 11301457).
∗
Corresponding author at: School of Mathematical Sciences, Yancheng Teachers University, Jiangsu 224002, China. E-mail addresses: [email protected], [email protected] (K. Chen).
http://dx.doi.org/10.1016/j.disc.2014.04.002 0012-365X/© 2014 Elsevier B.V. All rights reserved.
80
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Let n be an even integer, and let A = (ai,j ) be an SSSODLS(n) satisfying
(∗)
a¯ 2i,j a¯ j,i =
i∈I n j∈I n 2
a¯ 2i,j a¯ j,i = 0,
i∈I n j∈In \I n
2
2
2
where a¯ = a − Then we say that A is a ∗ SSSODLS(n). By Lemma 1.1, there exists a ∗ SSSODLS(n) only if n ≡ 0(mod 4). We shall show that this necessary condition is also sufficient. Specially, we shall prove the following. n−1 . 2
Theorem 1.2. There exists a ∗ SSSODLS (n) if and only if n ≡ 0(mod 4) and n ̸= 4. SSSODLS(n)s can be used to construct Yang Hui type magic squares. An n × n matrix A consisting of all given n2 consecutive integers is a magic square of order n, denoted by MS(n), if the sum of elements in each row, each column, and each of two diagonals is the same. The study of magic squares probably dates back to prehistoric times [3]. A lot of work had been done on magic squares [9, 6,1–3]. Yang Hui [14] gave an MS(8) as follows. ∗
Y8 = (yi,j ) =
61 12 20 37 29 44 52 5
3 54 46 27 35 22 14 59
2 55 47 26 34 23 15 58
64 9 17 40 32 41 49 8
57 16 24 33 25 48 56 1
7 50 42 31 39 18 10 63
6 51 43 30 38 19 11 62
60 13 21 36 28 45 53 4
.
As pointed out by S. Chikaraishi et al. [8] that the Y8 has the additional property that for each e = 2, 3, 4,
7 7 i=4
e j=0 yi,j , and
7 3
e j=0 yi,j
i =0
=
3 7
e i=0 j=0 yi,j e j=4 yi,j . A magic square having such a property is usually called a Yang Hui type
=
7 7 i=0
magic square. Generally, for an even integer n and an integer t ≥ 2, an MS(n)A is a Yang Hui type magic square, denoted by YMS(n, t ), if, for each e = 2, 3, . . . , t, the sum of the elements of the first 2n rows of A∗e is the same as that of the last 2n rows, and the sum of the elements of the left 2n columns of A∗e is the same as that of the right 2n columns, where A∗e = (aei,j ). A YMS(n, t ) is also called a magic square with t-powered sum by Chikaraishi in [8]. We should mention that a YMS(n, t ) is a kind of weak form of a t-multimagic square. For the details of t-multimagic squares, we refer the readers to [10,15,7] and the references therein. For the existence of YMS(n, t )s, Chikaraishi et al. [8] gave a family of YMS(2t , 2t − 2)s. Recently, Cao, Chen and Zhang [4] determined a necessary and sufficient condition for the existence of a YMS(n, 2) for n ̸= 2. These results are listed as follows. Lemma 1.3 ([8,4]). (1) There exists a YMS (2t , 2t − 2) for all integers t ≥ 2. (2) There exists a YMS (n, 2) if and only if n is even and n ̸= 2. In this paper, we will investigate the existence of YMS(n, 4)s by making use of ∗ SSSODLS(n)s and get the following. Theorem 1.4. There exists a YMS (n, 4) if and only if n ≡ 0(mod 4) and n ̸= 4. A construction of YMS(n, 4) based on ∗ SSSODLS(n) is presented in Section 2, and some constructions of ∗ SSSODLS(n)s are provided in Section 3. The proofs of Theorems 1.2 and 1.4 are both given in Section 4. 2. A construction of YMS(n, 4) based on ∗ SSSODLS(n) Let n be an even integer, denote S11 = {(i, j)|i, j ∈ I n }, 2
S12 = {(i, j)|i ∈ I n , j ∈ In \ I n }, 2
S21 = {(i, j)|i ∈ In \ I , j ∈ I n }, n 2
2
2
S22 = {(i, j)|i, j ∈ In \ I n }. 2
For a matrix A of order n, the sum (i,j)∈S ai,j is written as Let C be a YMS(n, t ). It is easy to see that the conditions
cie,j +
S11
S11
cie,j =
S12
cie,j +
S21
cie,j +
S21
cie,j =
S12
cie,j ,
S22
cie,j +
S22
cie,j ,
S ai,j in short.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
are equivalent to that
cie,j =
S11
cie,j ,
S22
cie,j =
S12
cie,j ,
S21
where e = 2, 3, . . . , t. Now we give a construction of YMS(n, 4) based on ∗ SSSODLS(n) in the following. Construction 2.1. If there is a ∗ SSSODLS (n), then there is a YMS (n, 4). Proof. Let A be a ∗ SSSODLS(n), and let C = (ci,j ),
ci,j = nai,j + aj,i , i, j ∈ In .
Then C is an MS(n). We shall prove that C is also a YMS(n, 4), i.e., C satisfies the following.
cie,j =
S11
cie,j ,
S22
cie,j =
S12
cie,j ,
e = 2, 3, 4.
c˜ie,j ,
e = 2, 3, 4,
S21
It suffices to prove that
c˜ie,j =
S11
c˜ie,j ,
S22
c˜ie,j =
S12
S21
2
where c˜i,j = ci,j − n 2−1 , i, j ∈ In . By the definition of C , we have c˜i,j = nai,j + aj,i −
n2 − 1 2
n−1 n−1 = n ai , j − + aj , i − = na¯ i,j + a¯ j,i , 2
2
1 where a¯ i,j = ai,j − n− , i , j ∈ In . 2 Noting that a¯ i,j = −¯an−1−i,n−1−j , i, j ∈ In since A is strongly symmetric, we have the following
c˜n−1−i,n−1−j = na¯ n−1−i,n−1−j + a¯ n−1−j,n−1−i = −na¯ i,j − a¯ j,i = −˜ci,j , Clearly,
(˜ci,j )2 , (˜ci,j )2 =
(˜ci,j )2 , (˜ci,j )2 =
(˜ci,j )4 . (˜ci,j )4 = S21
S12
S22
S11
S21
S12
S22
S11
(˜ci,j )4 , (˜ci,j )4 = It remains to prove that
c˜i3,j =
S11
c˜i3,j ,
S22
c˜i3,j =
S12
c˜i3,j .
S21
Since c˜i,j = −˜cn−1−i,n−1−j , i, j ∈ In , we need only to prove that
c˜i3,j =
S11
c˜i3,j = 0.
S12
In fact, c˜i3,j = n3 a¯ 3i,j + 3n2 a¯ 2i,j a¯ j,i + 3na¯ i,j a¯ 2j,i + a¯ 3j,i , Since A is a Latin square, for each e > 0 we have
a¯ ei,j +
S11
a¯ ei,j =
S12
a¯ ei,j +
S11
a¯ ei,j +
S21
a¯ ei,j =
S21
a¯ ei,j ,
S22
a¯ ei,j +
S12
a¯ ei,j .
S22
Therefore
S11
a¯ ei,j =
S22
a¯ ei,j ,
S12
a¯ ei,j =
S21
a¯ ei,j .
i , j ∈ In .
i, j ∈ In .
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Since a¯ i,j = −¯an−1−i,n−1−j , i, j ∈ In , we have
a¯ 3i,j =
S11
a¯ 3i,j =
S22
a¯ 3i,j =
S12
a¯ 3i,j = 0.
S21
From the condition (∗)
a¯ 2i,j a¯ j,i = 0,
S11
a¯ 2i,j a¯ j,i = 0.
S12
By commutating i, j in the above two equalities, we get the following.
a¯ i,j a¯ 2j,i = 0,
S11
a¯ i,j a¯ 2j,i = 0.
S21
Hence
a¯ i,j a¯ 2j,i =
(−¯ai,j )(−¯aj,i )2 = − a¯ i,j a¯ 2j,i = 0. S21
S12
S21
Thus
c˜i3,j =
S11
c˜i3,j = 0.
S12
This completes the proof.
3. Constructions for ∗ SSSODLS(n)s In this section, we provide some constructions of ∗ SSSODLS(n)s. Construction 3.1. If there is a ∗ SSSODLS (n), and an SSSODLS (m), then there is a ∗ SSSODLS (mn). Proof. Let U be a ∗ SSSODLS(n), and let V be an SSSODLS(m). Let W be an mn × mn matrix given by
wi,j = mur ,s + vx,y ,
i = mr + x, j = ms + y, r , s ∈ In , x, y ∈ Im .
It is readily checked that W is an SSSODLS(mn). We shall show that W satisfies the condition (∗), i.e.,
w ¯ i2,j w ¯ j ,i =
S11
w ¯ i2,j w ¯ j ,i = 0 ,
w ¯ =w−
mn − 1 2
S12
.
By the definition of W , we have mn − 1
w ¯ i,j = wi,j −
2
= mur ,s + vx,y −
1 where u¯ = u − n− , v¯ = v − 2
m−1 . 2
mn − 1 2
= mu¯ r ,s + v¯ x,y ,
So
w ¯ i2,j w ¯ j,i = (mu¯ r ,s + v¯ x,y )2 (mu¯ s,r + v¯ y,x ), = m3 u¯ 2r ,s u¯ s,r + 2m2 u¯ r ,s u¯ s,r v¯ x,y + mu¯ s,r v¯ x2,y + m2 u¯ 2r ,s v¯ y,x + 2mu¯ r ,s v¯ x,y v¯ y,x + v¯ x2,y v¯ y,x , ¯ 2r ,s u¯ s,r = 0. Therefore By the condition (∗) we have r ∈I n s∈I n u 2
2
u¯ 2r ,s u¯ s,r
x∈Im y∈Im r ∈I n s∈I n 2
= u¯ 2r ,s u¯ s,r = 0. x∈Im y∈Im
2
r ∈I n s∈I n 2
2
v¯ x,y = 0 since V is a Latin square, we have v¯ x,y = 0, u¯ r ,s u¯ s,r v¯ x,y = u¯ r ,s u¯ s,r
Note that
x∈Im
y∈Im
x∈Im y∈Im r ∈I n s∈I n 2
r ∈I n s∈I n
2
2
x∈Im y∈Im r ∈I n s∈I n 2
2
¯ v¯
u2r ,s y,x
x∈Im y∈Im
2
2 = u¯ r ,s v¯ y,x = 0. r ∈I n s∈I n 2
2
x∈Im y∈Im
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
83
Since U is ∗ SSSODLS(n), the following is clear.
u¯ r ,s = 0,
r ∈I n s∈In
u¯ r ,s = 0,
r ∈In s∈I n
2
r ∈I n
2
u¯ s,r v¯ x2,y =
x∈Im y∈Im r ∈I n s∈I n 2
x∈Im y∈Im
2
r ∈I n s∈I n 2
2
v¯ x,y v¯ y,x = 0. = u¯ r ,s
u¯ r ,s v¯ x,y v¯ y,x
x∈Im y∈Im r ∈I n s∈I n 2
v¯ x2,y u¯ s,r = 0,
2
2
2
2
u¯ r ,s = 0, and
s∈I n
2
u¯ r ,s .
r ∈In \I n s∈I n
r ∈I n s∈In \I n
2
From the above we have
u¯ r ,s = −
x∈Im y∈Im
r ∈I n s∈I n
2
2
2
Since V is self-orthogonal, we have
v¯ x2,y v¯ y,x =
x∈Im y∈Im
x−
m−1
2
2
x∈Im y∈Im
y−
m−1 2
=
x−
m−1
x∈Im
2
2
y−
m−1
y∈Im
2
= 0.
Therefore
v¯ v¯
2 x,y y,x
=
x∈Im y∈Im r ∈I n s∈I n 2
Hence
r ∈I n s∈I n
2
2
2
v¯ v¯
2 x,y y,x
= 0.
x∈Im y∈Im
w ¯ w ¯ j,i = 0. 2 i ,j
S11
Similarly, one can prove that
S12
w ¯ i2,j w ¯ j,i = 0. Thus W is a ∗ SSSODLS(mn).
An m × n magic rectangle, denoted by MR(m, n), is an m × n matrix consisting of all the numbers of Imn such that the sum of the entries of each row is a constant and each column sum is also a constant (different if m ̸= n). Harmuth [12,13] proved that for m, n > 1, there exists an MR(m, n) if and only if m ≡ n(mod 2) and (m, n) ̸= (2, 2). We shall now construct a ∗ SSSODLS(4m) by using an SSSODLS(m) and an MR(2, 2m). Construction 3.2. If there exists an SSSODLS (m), then there exists a ∗ SSSODLS (4m). Proof. Let U be the U4 given in Section 1. Let V be an SSSODLS(m). A matrix W is given by
wi,j = mur ,s + vx,y ,
i = mr + x, j = ms + y, r , s ∈ I4 , x, y ∈ Im .
It is readily checked that W is an SSSODLS(4m). Let H be an MR(2, 2m), and let L = (lj ), where h0,j , h1,4m−1−j ,
j ∈ I2m , j ∈ I4m \ I2m .
lj =
Then lj + l4m−1−j = h0,j + h1,j = 4m − 1, j ∈ I2m . Let A = (ai,j ),
ai,j = lwi,j , i, j ∈ I4m .
Clearly, A is also an SSSODLS(4m). We shall prove that A satisfies the condition (∗), i.e.,
a¯ 2i,j a¯ j,i =
S11
a¯ 2i,j a¯ j,i = 0,
S12
where a¯ i,j = ai,j − In fact,
a¯ 2i,j a¯ j,i =
S11
4m−1 2
= lwi,j −
4m−1 2
(¯lwi,j )2¯lwj,i =
S11
= ¯lwi,j .
(¯lmur ,s +vx,y )2¯lmus,r +vy,x .
r ∈I2 s∈I2 x∈Im y∈Im
By the definition of U, we have {(ur ,s , us,r )|r , s ∈ I2 } = {(0, 0), (1, 3), (3, 1), (2, 2)}, and
S11
a¯ 2i,j a¯ j,i =
x∈Im y∈Im
[(¯lvx,y )2¯lvy,x + (¯lm+vx,y )2¯l3m+vy,x + (¯l3m+vx,y )2¯lm+vy,x + (¯l2m+vx,y )2¯l2m+vy,x ].
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Noting that {(vx,y , vy,x )|x, y ∈ Im } = {(i, j)|i, j ∈ Im } since V is self-orthogonal, we have
(¯lkm+vx,y )2¯lsm+vy,x =
¯lsm+j , k, s ∈ I2 . (¯lkm+i )2¯lsm+j = (¯lkm+i )2
x∈Im y∈Im
i∈Im j∈Im
i∈Im
j∈Im
Hence
a¯ 2i,j a¯ j,i =
¯lj + ¯lj+3m + ¯lj+m + ¯lj+2m . (¯li )2 (¯li+m )2 (¯li+3m )2 (¯li+2m )2 i∈Im
S11
j∈Im
i∈Im
j∈Im
i∈Im
j∈Im
i∈Im
j∈Im
It is easy to see that ¯lj = −¯l4m−1−j , j ∈ I4m . By this we have
(¯lj+3m )2 = (−¯lm−1−j )2 = (¯lj )2 , j∈Im
j∈Im
j∈Im
(¯lj+2m )2 = (−¯lm+m−1−j )2 = (¯lj+m )2 . j∈Im
j∈Im
j∈Im
Consequently,
a¯ 2i,j a¯ j,i =
S11
=
(¯li )2
¯lj +
i∈Im
j∈Im
(¯li )2
i∈Im
It is easy to show that
¯lj+3m + ¯lj+2m ¯lj+m + (¯li+m )2 (¯li+m )2 (¯li )2 i∈Im
¯lj +
¯ +
j∈Im lj
¯lj+m +
j∈Im
j∈Im
j∈Im
i∈Im
¯
j∈Im lj+m
j∈Im
¯lj+3m . ¯lj+2m + (¯li+m )2 i∈Im
2 S11 ai,j aj,i
i∈Im
j∈Im
= 0,
j∈Im
j∈Im
¯
j∈Im lj+2m
+
¯
j∈Im lj+3m
= 0 since H is an MR(2, 2m). So we have
¯ ¯ = 0.
In the same way one can prove that
S12
a¯ 2i,j a¯ j,i = 0. So A is a ∗ SSSODLS(4m).
To illustrate the proof of Construction 3.2, we give an example as follows. Example 2. An SSSODLS(4) U is listed in Section 1. An SSSODLS(5) is given below.
1 2 V = 3 4 0
3 4 0 1 2
0 1 2 3 4
2 3 4 0 1
4 0 1 . 2 3
Let W = (wi,j ),
wi,j = 5ur ,s + vx,y , i = 5r + x, j = 5s + y, r , s ∈ I4 , x, y ∈ I5 ,
i.e.,
W = (wi,j ) =
1 2 3 4 0 16 17 18 19 15 11 12 13 14 10 6 7 8 9 5
3 4 0 1 2 18 19 15 16 17 13 14 10 11 12 8 9 5 6 7
0 1 2 3 4 15 16 17 18 19 10 11 12 13 14 5 6 7 8 9
2 3 4 0 1 17 18 19 15 16 12 13 14 10 11 7 8 9 5 6
4 0 1 2 3 19 15 16 17 18 14 10 11 12 13 9 5 6 7 8
6 7 8 9 5 11 12 13 14 10 16 17 18 19 15 1 2 3 4 0
8 9 5 6 7 13 14 10 11 12 18 19 15 16 17 3 4 0 1 2
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0 1 2 3 4
7 8 9 5 6 12 13 14 10 11 17 18 19 15 16 2 3 4 0 1
9 5 6 7 8 14 10 11 12 13 19 15 16 17 18 4 0 1 2 3
16 17 18 19 15 1 2 3 4 0 6 7 8 9 5 11 12 13 14 10
18 19 15 16 17 3 4 0 1 2 8 9 5 6 7 13 14 10 11 12
15 16 17 18 19 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
17 18 19 15 16 2 3 4 0 1 7 8 9 5 6 12 13 14 10 11
19 15 16 17 18 4 0 1 2 3 9 5 6 7 8 14 10 11 12 13
11 12 13 14 10 6 7 8 9 5 1 2 3 4 0 16 17 18 19 15
13 14 10 11 12 8 9 5 6 7 3 4 0 1 2 18 19 15 16 17
10 11 12 13 14 5 6 7 8 9 0 1 2 3 4 15 16 17 18 19
12 13 14 10 11 7 8 9 5 6 2 3 4 0 1 17 18 19 15 16
14 10 11 12 13 9 5 6 7 8 4 0 1 2 3 19 15 16 17 18
.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
85
An MR(2, 10) is listed below.
0 19
H =
1 18
3 16
9 10
12 7
11 8
13 6
14 5
15 4
17 2
.
Let L = (lj ), where h 0 ,j , h1,4m−1−j ,
j ∈ I10 , j ∈ I20 \ I10 .
lj = i.e.,
L = (lj ) =
0
1
3
9
11
12
13
14
15
17
2
4
5
6
7
8
10
16
18
19
.
Let A = (ai,j ), ai,j = lwi,j , i, j ∈ I20 , i.e.,
1 A=
3 9 11 0 10 16 18 19 8 4 5 6 7 2 13 14 15 17 12
9 11 0 1 3 18 19 8 10 16 6 7 2 4 5 15 17 12 13 14
0 1 3 9 11 8 10 16 18 19 2 4 5 6 7 12 13 14 15 17
3 9 11 0 1 16 18 19 8 10 5 6 7 2 4 14 15 17 12 13
11 0 1 3 9 19 8 10 16 18 7 2 4 5 6 17 12 13 14 15
13 14 15 17 12 4 5 6 7 2 10 16 18 19 8 1 3 9 11 0
One can check that A is a ∗ SSSODLS(20).
15 17 12 13 14 6 7 2 4 5 18 19 8 10 16 9 11 0 1 3
12 13 14 15 17 2 4 5 6 7 8 10 16 18 19 0 1 3 9 11
14 15 17 12 13 5 6 7 2 4 16 18 19 8 10 3 9 11 0 1
17 12 13 14 15 7 2 4 5 6 19 8 10 16 18 11 0 1 3 9
10 16 18 19 8 1 3 9 11 0 13 14 15 17 12 4 5 6 7 2
18 19 8 10 16 9 11 0 1 3 15 17 12 13 14 6 7 2 4 5
8 10 16 18 19 0 1 3 9 11 12 13 14 15 17 2 4 5 6 7
16 18 19 8 10 3 9 11 0 1 14 15 17 12 13 5 6 7 2 4
19 8 10 16 18 11 0 1 3 9 17 12 13 14 15 7 2 4 5 6
4 5 6 7 2 13 14 15 17 12 1 3 9 11 0 10 16 18 19 8
6 7 2 4 5 15 17 12 13 14 9 11 0 1 3 18 19 8 10 16
2 4 5 6 7 12 13 14 15 17 0 1 3 9 11 8 10 16 18 19
5 6 7 2 4 14 15 17 12 13 3 9 11 0 1 16 18 19 8 10
7 2 4 5 6 17 12 13 14 15 11 0 1 3 9 19 8 10 16 18
.
4. Proofs of Theorems 1.2 and 1.4 In this section we shall give the proofs of Theorems 1.2 and 1.4. Lemma 4.1. There is no ∗ SSSODLS (4). Proof. Suppose on the contrary that A is a ∗ SSSODLS(4). Let a¯ i,j = ai,j − 23 , and let e = 2a¯ 0,0 , f = 2a¯ 0,1 , g = 2a¯ 1,0 , h = 2a¯ 1,1 . Then e, f , g , h ∈ {−3, −1, 1, 3}. Clearly, e, f , h are pairwise distinct, g ̸= e, g ̸= h and e3 + h3 ∈ {±28, ±26, 0}. By the condition (∗) we have e3 + h3 + fg (f + g ) = 0. We shall show that g ̸= f . In fact, If g = f , then e3 + h3 + 2f 3 = 0. If e3 + h3 = 0, then f = 0, a contradiction. So e3 + h3 ∈ {±28, ±26} and f 3 ∈ {±14, ±13}, which is impossible. Hence g ̸= f . From the above we have {e, f , g , h} = {−3, −1, 1, 3}. Hence f + g ∈ {±2, ±4, 0}, fg ∈ {−1, ±3, −9} and fg (f + g ) ∈ {0, ±2, ±4, ±6, ±12, ±18, ±36}. Therefore e3 + h3 = −fg (f + g ) = 0. It follows that h = −e, g = −f and a¯ 2,2 = −¯a1,1 = e = a¯ 0,0 , which contradicts the assumption that A is diagonal. So there is no ∗ SSSODLS(4). Lemma 4.2. There exists a ∗ SSSODLS (n) for each n ∈ {8, 12, 24}. Proof. Let
A8 =
0 6 1 7 3 5 2 4
2 4 3 5 1 7 0 6
4 2 5 3 7 1 6 0
6 0 7 1 5 3 4 2
5 3 4 2 6 0 7 1
7 1 6 0 4 2 5 3
1 7 0 6 2 4 3 5
3 5 2 4 0 6 1 7
.
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Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
It is readily checked that A8 is a ∗ SSSODLS(8). Let
A12
0 5 8 3 6 11 = 9 10 4 7 1 2
1 3 2 9 8 10 11 7 6 5 4 0
17 14 9 23 22 7 2 5 20 12 19 8 1 13 0 16 11 3 10 4 21 15 6 18
3 16 23 15 6 9 4 13 12 2 21 18 17 5 20 14 7 1 8 0 19 11 22 10
10 0 7 4 11 1 3 9 5 6 2 8
4 6 1 10 5 7 2 8 11 0 3 9
7 2 3 8 9 4 5 1 0 11 10 6
8 4 10 1 7 3 6 11 9 2 0 5
6 11 9 2 0 5 8 4 10 1 7 3
5 1 0 11 10 6 7 2 3 8 9 4
2 8 11 0 3 9 4 6 1 10 5 7
3 9 5 6 2 8 10 0 7 4 11 1
11 7 6 5 4 0 1 3 2 9 8 10
9 10 4 7 1 2 0 5 8 3 6 11
and
A24
=
22 7 6 20 15 12 11 23 14 8 1 19 2 18 13 5 16 4 21 10 17 9 0 3
18 3 22 1 16 8 15 4 21 9 20 12 7 23 6 0 17 11 2 14 13 10 5 19
2 5 0 9 8 22 17 14 13 1 16 10 3 21 4 20 15 7 18 6 23 12 19 11
7 21 20 5 0 3 18 10 17 6 23 11 22 14 9 1 8 2 19 13 4 16 15 12
21 13 4 7 2 11 10 0 19 16 15 3 18 12 5 23 6 22 17 9 20 8 1 14
8 20 19 16 11 1 0 9 4 7 22 14 21 10 3 15 2 18 13 5 12 6 23 17
0 12 5 18 1 17 8 11 6 15 14 4 23 20 19 7 22 16 9 3 10 2 21 13
1 19 10 22 21 18 13 3 2 11 6 9 0 16 23 12 5 17 4 20 15 7 14 8
It is routine to check that A12 and A24 are ∗ SSSODLSs.
23 15 2 14 7 20 3 19 10 13 8 17 16 6 1 22 21 9 0 18 11 5 12 4
16 10 3 21 12 0 23 20 15 5 4 13 8 11 2 18 1 14 7 19 6 22 17 9
14 6 1 17 4 16 9 22 5 21 12 15 10 19 18 8 3 0 23 11 2 20 13 7
19 11 18 12 5 23 14 2 1 22 17 7 6 15 10 13 4 20 3 16 9 21 8 0
15 9 16 8 3 19 6 18 11 0 7 23 14 17 12 21 20 10 5 2 1 13 4 22
10 2 21 13 20 14 7 1 16 4 3 0 19 9 8 17 12 15 6 22 5 18 11 23
6 0 17 11 18 10 5 21 8 20 13 2 9 1 16 19 14 23 22 12 7 4 3 15
13 1 12 4 23 15 22 16 9 3 18 6 5 2 21 11 10 19 14 17 8 0 7 20
5 17 8 2 19 13 20 12 7 23 10 22 15 4 11 3 18 21 16 1 0 14 9 6
9 22 15 3 14 6 1 17 0 18 11 5 20 8 7 4 23 13 12 21 16 19 10 2
11 8 7 19 10 4 21 15 22 14 9 1 12 0 17 6 13 5 20 23 18 3 2 16
12 4 11 0 17 5 16 8 3 19 2 20 13 7 22 10 9 6 1 15 14 23 18 21
4 18 13 10 9 21 12 6 23 17 0 16 11 3 14 2 19 8 15 7 22 1 20 5
20 23 14 6 13 2 19 7 18 10 5 21 4 22 15 9 0 12 11 8 3 17 16 1
.
Proof of Theorem 1.2. Let n = 4m. For m = 1, there is no ∗ SSSODLS(4) by Lemma 4.1. For m = 2, 3, 6, the ∗ SSSODLS(4m)s are given in Lemma 4.2. For m ≡ 0, 1, 3(mod 4), m > 3, there is an SSSODLS(m) by Lemma 1.1, so there is a ∗ SSSODLS(4m) by Construction 3.2. For m ≡ 2(mod 4), m ≥ 10, let m = 4k + 2, k ≥ 2, then 4m = 8(2k + 1). Since there is an SSSODLS(2k + 1) by Lemma 1.1 and there is a ∗ SSSODLS(8) by Lemma 4.2, there is a ∗ SSSODLS(8(2k + 1)) by Construction 3.1. The proof is now complete. To give the proof of Theorem 1.4 we first prove the following. Lemma 4.3. For n ≡ 2(mod 4), there is no YMS (n, t ) whenever t ≥ 3. Proof. Let n = 34k + 2. We need only to prove that there is no YMS(n, 3). Suppose on the contrary that C is a YMS(n, 3). Let T = S11 ∪S12 (ci,j − ci,j ). Let St (n) =
(
1 n
i∈I 2 n
it . We have S1 (n) =
1 n 2
(n2 − 1), S3 (n) =
1 3 n 4
(n2 − 1)2 . Noting that
S11 ∪S12
(ci3,j − ci,j ) =
− ci,j ) since C is a YMS(n, 3), we have n n 1 3 2 1 T = (S3 (n) − S1 (n)) = n (n − 1)2 − n(n2 − 1)
3 S21 ∪S22 ci,j
2
2
4
2
= (2k + 1)(2k + 1)(4k + 1)(4k + 3)(16k + 16k + 5)(8k2 + 8k + 1), 2
which is an odd integer. On the other hand, ci3,j − ci,j = ci,j (ci,j − 1)(ci,j + 1), which is even. It implies that T is also even, a contradiction. So there is no YMS(4k + 2, 3). Proof of Theorem 1.4. For n ≡ 2(mod 4), there does not exist a YMS(n, 4) by Lemma 4.3. For n ≡ 0(mod 4), let n = 4m. If m = 1, by using computer search, we know that there is no YMS(4, 3), hence there is no YMS(4, 4). For m ≥ 2, there is a ∗ SSSODLS(4m) by Theorem 1.2 and hence there is a YMS(4m, 4) by Construction 2.1. Remark. An MS(n) M is symmetric if mi,j + mn−1−i,n−1−j = n2 − 1, i, j ∈ In . Let M = nA + B, M is elementary if A, B are both Latin squares. Otherwise, it is nonelementary. If A, B are strongly symmetric, then M is symmetric. The YMS(n, 4)s obtained in this paper are all symmetric and elementary. So we have the following. Corollary 4.4. There exists a symmetric elementary YMS (n, 4) if and only if n ≡ 0(mod 4) and n ̸= 4.
Y. Zhang et al. / Discrete Mathematics 328 (2014) 79–87
Acknowledgment The authors would like to thank Professor L. Zhu of Suzhou University for helpful discussions and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
G. Abe, Unsolved problems on magic squares, Discrete Math. 127 (1994) 3–13. M. Ahmed, Algebric combinatorics of magic squares. Ph.D. Dissertation, University of California, Davis, 2004. W.S. Andrews, Magic Squares and Cubes, second ed., Dover, New York, 1960. N. Cao, K. Chen, Y. Zhang, Existence of Yang Hui type magic squares, preprint. H. Cao, W. Li, Existence of strongly symmetric self-orthogonal diagonal Latin squares, Discrete Math. 311 (2011) 841–843. S.v.R. Cammann, Magic squares, in: Encyclopæ dia Britannica, fourteenth ed., Chicago, 1973. K. Chen, W. Li, Existence of normal bimagic squares, Discrete Math. 312 (2012) 3077–3086. S. Chikaraishi, M. Kobayashi, N. Mutoh, G. Nakamura, Magic squares with powered sum, http://usr.u-shizuoka-ken.ac.jp/kn/AN10118525201111001030.pdf. C.J. Colbourn, J.H. Dinitz (Eds.), Handbook of Combinatorial Designs, second ed., Chapman and Hall/CRC, Boca Raton FL, 2007. H. Derksen, C. Eggermont, A.v.D. Essen, Multimagic squares, Amer. Math. Monthly 114 (2007) 703–713. B. Du, H. Cao, Existence of strongly symmetric self-orthogonal diagonal Latin squares, Acta Math. Appl. Sin. (Chin. Ser.) 25 (2002) 187–189. T. Harmuth, Über magische Quadrate und ähniche Zahlenfiguren, Arch. Math. Phys. 66 (1881) 286–313. T. Harmuth, Über magische Rechtecke mit ungeraden Seitenzahlen, Arch. Math. Phys. 66 (1881) 413–447. H. Yang, Yang Hun suan fa, 1274, http://www.tulips.tsuknba.acjp/limedio/dlam/B14B1407158/1.pdf, reprint, 1433. Y. Zhang, K. Chen, J. Lei, Large sets of orthogonal arrays and multimagic squares, J. Combin. Des. 21 (2013) 390–403.
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