ON THE CONSTRUCTION OF PAIRWISE ORTHOGONAL SETS OF LATIN SQUARES
AND THE FALSITY OF A CONJECTURE OF EULER. II
by
R. C. Bose and S. S. Shrikbande
University of North Carolina
This research was supported by the United States Air
Force through,the Air Force Office of Scientific Research of the Air Research and Development Command"
under Contract No. AF 49(6,38)-213. Reproduction in
whole or in part is permitted for any purpose of the
United States Government.
Institute of Statistics
Mimeograph series No. 225
May 1959
•
ON THE CONSTRUCTION OF PAIRWISE ORTHOGONAL SETS OF LATIN SQUARES
AND THE FALSITY OF A CONJECTURE OF EULER. III
by
R. C. Bose and S. S. Shrikhande
University of North Carolina
1.
Introduction.
under the same title
be adhered to.
This report is a sequel to our earlier report
Ll_7,
the notation and terminology of which will
Since the earlier report was written Parker ~3_7 has
proved that if 4t+3 is a prime power there exist at least two orthogonal
Latin squares of order 6t+4.
The particular case t=l leads to orthogo-
nal squares of order 10 from which follows the existence of two orthogonal Latin squares of all orders which are multiples of 10.
here prove that the main theorem of
canditions are satisfied.
Ll_7 can be
We shall
improved when certain
With the help of this we can in many cases
improve lower bounds for N(v) obtained in
Ll_7.
We can also demon-
strate (using Parker's result) that there exist at least two orthogonal Latin squares of order v for v > 6 except for v
= 14
and 26 which
do not follow from our theorems.
1
This research was supported by the United States Air Force through
the Air Force Office of Scientific Research of the Air Research and
Development Command, under Contract No. AF 49(638)-213. Reproduction in whole or in part is permitted for any purpose of the United
States GovertuILent.
- 2 -
2.
£1_7.
Improvement of the main theorem of
Consider a pairwise balanced design (D) of index unity and type
k ,k2 , ••• ,km). The set of equiblock components (Dl ),
l
••• , (Do), £< m, will be said to form a clear set if the
(Vj
A
blocks comprising
(Dl ), (D2 ), ••• , (D£)
blocks contain a common treatment.
(12 )'
~ b
i=l i
are disjoint, i.e
no two
Q ,
Clearly a necessary condition for
this is
£
~
i=l
bik. < v
1.
=
Q
We shall now prove:
Theorem 1.
Let there exist a pairwise balanced design (D) of index
unity and type (v; k ,k , u,km), such that the set of equiblock coml 2
ponents (Dl ), (D2 ), '.0, (D£), £ < m, form a clear set. If there exist
0
qi-l p.o.los. of order k and i f
i
then there exist at least q*-2 p.o.l.s. of order v.
Proof.
Let us define
and
Then
Let 5il, 5i2 , ••• , 5ib be the blocks of the equiblock component
i
(Di ) written out as columns (i ~ l). By hypothesis there exist qi-l
p.o.l.s. of order kit
Hence we can construct an orthogonal array A
ij
- 3 with ~+l rows and k~ columns, whose symbols are the treatments occurring in 8ij • Let
Ai
Let 6
i
be the q (1) x
Ai2 , ••• , Aib _7 ·
i
matrix obtained from Ai by retaining only the
= LAil,
bik~
first q* rows, and let
6(1) = 1:61, 6 , ... ,
2
Then 6(1) has q* rows and Z
b.k~
J. J.
columns.
6t_7 .
Clearly 6(1) has the property
that if t c and t are any two treatments identical or distinct cond
tained in any block of (Dl ), (D2 ), ... , or (Dt), then the ordered pair
t
(1)
(t c ) occurs as a column exactly once in any two rowed submatrix of 6
•
d
Also let Pu (Du ), u = t+l,.. o,m be defined as in LrlJ and let 6 u
be the matrix obtained from P (D ) by retaining only the first q*
u u
rows, then
has the property that if t
and ~ are any two distinct treatments cona
t
tained in any block of (D + ), ••• , (Dm) then the ordered pair (t~)
t l
occurs exactly once in any two rowed submatrix of 6(2). The number
of columns in 6(2) is
m
Z
b k (k -1)
u=t+l u u u
Again let 6(3) be the q*xv2 matrix whose n-th column
every position the treatment t
v
2
n
=, where t
=v
-
t
Z
i=l
n
conta~ns
is anyone of the
b.ki
l.
treatments not contained in (Dl ), (D ), ••• , or (Dt)~
2
Then
in
- 4 -
any two rows for coordinatization we get q*-2 p.o.l.s. of order v.
Note that
t
t
m
2
2
v = E b i ki + E b k (k -1) + v - .E biki
i=l
u=t+l u u u
~=l
from (3.1) of
3.
L1J.
Applications to BIB designs.
By omitting a single treatment from the design BIB (v;k) we get a
pairwise balanced design of index unity and type (v; k, k-l) where the
r blocks of size k-l form a clear set.
Again if from a BIB (v;k) we
delete x treatments belonging to the same block 2
S
x
Sk
we get a
pairwise balanced design of index unity and type (v; k, k-l, k-x)
where the equiblock component consisting of the single blocks of size
k-x is clear.
Hence we have
Theorem 2.
(i)
and if 2
Existence of a BIB (V;k) implies
N(v-l) > min(N(k), 1 + N(k-l»
S x S k,
(ii)
=k
then
N(v-x) ~ min(N(k), N(k-l), 1 + N(k-x»
Example (1).
r
- 1,
=
= s+l, A.
- 1.
Consider the BIB design with parameters v = b = s2 + s + 1,
= 1,
With s
= 16.
Taking x
= 6, 8, 9 respectively we
get N(267) ~ 10, N(265) ~ 8, N(264) ~ 7 where as n(267)
= 2,
n(265)
=4
and n(264) = 2.
Suppose we omit 3 treatments 01' 02'
~
not occurring in the same
block of a BIB (v;k), then we get a pairwise balanced design (n) of
- 5 index unity and type (v; k, k-l, k-2).
Since in the original BIB (v;k)
no two blocks can have more than one treatment in common the three
blocks of (D) of size k-2 which have been obtained by deleting (0:1,0: ),
2
(0:2'C)' (0:1'C)
have obviously no treatment in common and form a clear
equiblock component.
Theorem 3.
Hence we get
The existence of a BIB (v;k) implies that
N(v-3) ~ min(N(k), N(k-l), 1 + N(k-2»
Example (2).
v
= 21,
Consider the BIB (v;k) designs
25, 41, 45, 61, 65, 85, 125.
- 1
1:2_7 with
•
k
=5
and
It follows that there exist at
least two p.o.l.s. of the following orders:
18, 22, 38, 42, 58, 62,
82 and 122.
Example (3).
From the designs BIB (81;9) and BIB (73;9) we get
N(78) ~ 6, N(70) ~ 6.
From the design BIB (273i17) we get N(270) ~ 2 since
Example (4).
N(15) ~ n(15)
= 2c
SUppose there exists a resolvable BIB design with parameters v, b,
r, k, A
= 1.
Let x
new treatment Qi' i
< r.
=
To each block of the i-th replication add a
= 1,2, ••• ,x,
and add a new block Ql' Q2' ••• , Qx •
We then get a pairwise balanced design of index unity and type
(v+x; k+l, k, x) if x < r, and type (v+x; k+l, r) if x
block component formed by the new block is clear.
= r.
When x
The equi-
= r-l,
the
set of equiblock components consisting of the new block, and the blocks
of the r-th replication form a clear set.
Theorem 4.
Hence we have
The existence of a resolvable BIB (Vik) implies
(i) N(v+x) ~ min(N(k), N(k+l), 1 + N(x»
- 1
if x ~ r-2 ,
'.
- 6 N(v+r-1) ~ min(l + N(k), N(k+1), 1 + N(r-1»
(ii)
N(v+r) ~ min(N(k+1)j 1 + N(r»
(iii)
Corollary.
v
= 6t+3,
b
As noted in
= (2t+1)(3t+1),
L1_7 a
- 1,
•
resolvable solution of the design
r = 3t+1, k
= 3,
A.
=1
always exists.
It
follows from the second part of the theorem that N(9t+3) ~ 2 if
~
N(3t)
2, which is certainly true for all odd values of t.
Example (5).
r
= 8,
k
= 7,
'».
Taking x
=1
=5
in the BIB design With v
4.
b
= 56,
we have
N(54) ~ (N(7), N(8), 1 + N(5»
Example (6).
= 49,
Taking t
=3
- 1
=4 •
in the corollary we get N(30) ~ 2.
Use of group divisible designs.
Theorem 5.
If' there exists a GD(v; k,m; 0,1) then
N(v) ~ min(N(k), 1 + N(m»
- 1
•
This follows from the fact that in the pairwise balanced design of index
unity and type (v; k, m) obtained from the GD designj the blocks of
size m form a clear equib10ck component.
Corollary.
N(s2_ l ) ~ N(s-l), if s is a prime or a prime power.
This follows from the existence of a resolvable GD(s2_ l ; s, s-l; 0,1).
Theorem 6.
If' there exists a GD(v; k,m; 0,1) then
N(v-l) ~ min(N(k), N(k-l), 1 + N(m), 1 + N(m-l»
- 1
and if the design is resolvable then
N(v-l) ~ min(N(k), N(k-l), N(m), N(m-1»
•
The first part follows from the fact that if we omit any particular
treatment from the corresponding pairwise balanced design of index unity
and type (v; k,m; 0,1) we get a design of the type (V-I; k,k-l; m,m-l),
in which the equib10ck components with blocks of size m and m-l form a
clear set.
The second part has already been proved in Ll_7 and is
given here for completeness.
- 7 Theorem 7.
Suppose there exists a resolvable Gn(v; k,m; 0,1) with
r replications, then
(i)
N(v+l) ~ min(N(k), N(k+l), 1 + N(m»
- 1
,
N(v+x) ~ min(N(k), N(k+l), 1+ N(m), 1+ N(x» - 1 if 1 < x < r,
(11)
(a) N(v+r) ~ min(N(k+l), 1 + N(m), 1 + N(r»
(iii)
- 1
,
(b) N(v+r) ~ m1n(N(k+l), N(m+l), 1+ N(k), 1+ N(r» - 1
(c) N(v+r+l) ~ min(N(k+l), N(m+l), 1 + N(r+l»
- 1
,
,
where in part (iii) we choose whichever lower bound is better for N(v+r).
To prove parts (i) we add a new treatment 9
replication.
1
to each block of one
To prove part (ii) we add a new treatment 9
block of the i-th replication, i
= 1,2, ••• ,x,
i
to each
and take a new block
(91 , 9 , ... , 9). For the first part we note that the equiblock
2
component given by the groups form a clear set. For the second part
we note that the set of equiblock components glvcn by the groups and
the new block forms a clear set.
treatment 9
i
To prove part (11i)(a) we add a new
to each block of the i-th replication, i
new block (9 , 9 , ••• , 9 ).
r
1
2
= 1,2, ••• ,r,
and
To prove part (iii)(b) we add a new
treatment 9. to each block of the i-th replication for the first r-l
J.
replications, and a new treatment 9
0
new block
(Go'
9 , ••• , 9 _ ).
1
r l
to each of the groups, and add a
We note in this case that the set of
equiblock components given by the r-th replication, and the neWly
added block forms a clear set.
To prove (iii)(c) we add one new treat-
ment to the blocks of each replication, one new treatment to each of
the blocks corresponding to the groups, and one block containing the new
treatments.
- 8 The group designs most useful to us are the semi-regular group divisible designs with A.
l
= 0,
A.
2
= 1.
For such a design the number of
replications r is equal to the group size m, and v = kIn.
In the nota-
tion used in ["1_7 such a design is denoted by SRGD(km; k,m; 0,1).
Also from the corollary to Lemma 7 of ["1_7 there exists a resolvable
SRGD(km; k,m; 0,1) if k
:s
N(m) + 1.
Combining this with theorem 7 we
have
Theorem 8.
(i)
(1i)
(i11)
If k
< N(m) + 1, then
N(km+1) ~ m1n(N(k), N(k+l), 1 + N(m»
- 1
,
N(km+x) ~ m1n(N(k), N(k+l), 1+ N(m), 1 +N(x»
N{kmT.m) ~ min(N(k+l), 1 + N(m»
Example (7).
- 1
-1
if 1 < x < m ,
•
Taking k, m and x as shown we derive the lower bound
for N(km+x), noting that N(24) ~ :5 from Table 1 of
["lJ and N(10)
from Parker's result.
(1)
k
(11)
k
(iii)
k
(iv)
k
= 7,
= 8,
m
= 7,
= 7,
m
= 5;
N(82) ~ 4 ,
m = 11, x = 7;
N(95) ~ 6 ,
= 11,
= 19,
m=
x
= 5;
N(1:58)~4 ,
8, x = 4;
N(60) ~ 3 ,
= 24,
x
x =10;
N(106) ~2 ,
(Vi)
k = 8, m = 13, x = 7;
N(lll)~ 6 ,
(vii)
k = 4, m = 27, x =10;
N(118)~ 2 ,
(v)
(Viii)
(ix)
(x)
k = 4" m
k
= 7,
m = 16" x =10;
N(122)~ 2
,
= 5;
N(124) ~4 ,
k = 7, m = 19, x = 5J
N(138) ~4 •
k = 7, m = 17, x
~ 2
- 9 -
5.
Improved lower bounds for N(v), v
~
154.
We give here those values of v ~ 154 for which the lower bound of
N(v) can be improved over the bound given in Table
I
of ["1_7.
Table I
v
n(v)
leb. for
N(v)
Remarks
10
1
2
18
1
2
30
1
2
34
38
42
1
2
1
2
1
2
46
1
2
Parker
Ex. (2),
Parker.
Parker
Ex.. (2),
Ex. (2),
Parker
54
1
4
Ex.~
60
2
3
62
1
2
70
78
82
1
6
1
6
1
4
90
1
2
95
4
6
106
1
2
111
114
118
2
6
1
2
1
2
122
1
2
124
138
154
3
4
1
4
1
2
Ex. (7)(iv), Th. 8
Ex. (2), Th. 3
Ex.• (3), Th. 3
Ex. (3), Th. 3
Ex. (7)(1), Th. 8
Parker
Ex. (7)(ii), Th. 8
Parker. Also Ex~7)(v), Th. 8
Ex, (7)(vi), Th. 8
114 = 38.3, Lemma 1 of L1J
Parker. Also Ex. (7)(vii), Th.8
EXo(2), Th.3. Also Ex. (7)(vii1),Th.8
Ex. (7)(ix), Tll. 8
Ex. (7)(x), Th. 8
Parker. Also 154=22.7, Lemma 1 of £i1
Th. 3
Also Ex. (6), Th. 4
Th. 3
Th. 3
(5), Th. 4
- 10 6.
The existence of at least two orthogonal Latin
s~uares
of
order v for all v > 26.
N(v) ~ 2 if2 < v ~ 270; v ~ 6, 14, 26.
Lemma 1.
Proof.
L1J
This result has already been checked in Table I of
supplemented by the improvements noted in the last section, up to
v
= 154.
Since the
r~su1t
is true for all odd v and all v divisible
by 4, we need only check for values of v of the form 4t+2.
We note that 162
= 9.18,
170
= 10.17,
174
= 3.58,
186
= 3.62,
190
= 10.19, 198 = 9.22, 210 = 10.21, 222 = 3.74, 230 = 10.23,
234
= 13.18,
238 = 7.34, 242
258 = 3.86, 266 = 7.38, 270
= 11022, 246 = 3.82, 250 = 10.25,
= 10.27.
Since from Lemma 1 of
Ll-7,
»
N(v1v2 ) ~ min(N(v1 ), N(V
we have N(v) f 2 for all values of v
2
the factors of which have been noted above.
We have thus only to check the cases v = 158, 166, 178, 182, 194,
202, 206, 214, 218,
226~
254 and 262.
If v can be expressed in the
fonn
v=lml+x
where k ~ N(m) + 1 and 1< x<m, then we can apply part (i1) of Theorem 8.
-
In particular if k = 4, N(m) > 3, N(x) = 2 and 1 < x
N(v) ~ 2.
218
= 4.44
= 4.49
m, then
The proof of the Lemma is completed by noting that
158 = 4.37 + 10, 166 = 4.37 + 18, 178
194
~
+ 18, 202
= 4.45
+ 22, 226
= 4.49 + 30, 254
Theorem 9.
+ 22, 206
= 4.40
= 4.49
+ 18, 182
= 4.40 + 22,
+ 10, 214
= 4.49 + 18,
+ 4.61 + 10, 262
= 4.60
+ 22.
There exist at least two orthogonal Latin squares for
all orders v > 2, v ~ 6, 14, 26.
- 11 The theorem has already been proved for all values of v
We shall now prove the theorem by induction for all v > 270.
~
270.
For this
we shall assume that the theorem is true for all v* < v, where v > 270
and show that this implies N(v) ~ 2.
can take v
wise.
E
Without loss of generality we
2 (mod 4), since the required result would be true other-
Let
n > 60
v = 4n + 30,
•
Put
where
a
~
2, Pl
= 3,
P2
= 5,
P3 = 7, and in general Pi is the i-th odd
prime in the ascending order of magnitude.
Also P is the largest
k
prime occurring in the prime power decomposition of n.
are integers positive or zero, and c
k
Then c ,c , ••• ,c
k
l 2
is a non-zero positive integer.
We shall consider 6 distinct mutually exclusive and collectively exhaustive cases:
Case I.
Then v
N(3)
=2
c
~ O.
l
= 3(nl +10),
where n l
= 2a
Cl-l c 2
PI
P2".
and N(nl+lO) ~ 2 by assumption N(v) ~ 2.
Case II.
c
l
= 0,
c
2
~ O.
and N(n +6)
2
?
Case III.
=2
2 by assumption N(v)
cl
= 0,
c
2
c 2-1
a
Then v = 5(n2 +6), where n2
?
P2
•••
Since N(5) = 4,
2.
= 0, a = 2.
c
c
·
k
3
Then v = 4ml +30, ml = P3
Pk •
c
.
c
k
3
Now min(P3 ' 0'" Pk ) ~ 7<> Hence N(ml ) ~ n(ml ) ~ 6.
0 ••
v > 270, ml > 60.
It follows from Theorem 8 that N(v) ~ 2.
Also since
- 12 -
= 0, c2 = 0, a = 3, c3 = o.
c4
Ck
Then v = 8m2+30 where ~ = P4 • •• Pk •
Case IV.
c1
c4
.
ck
Now min(P4 ' ••• , Pk ) f 11. Hence N(m1 ) ~ n(m1 ) ~ 10.
m > 30, since v > 270. Hence N(v) ~ 2 from Theorem 8.
2
Case V.
Then v
c1
= 0,
= 7~ +30
C
Now min(23 , P3 3
c2
= 0,
where
a
= 3,
. ck
' ••• , Pk )
N(v) ~ 2 from Theorem 8.
Case VI.
Then v
c1
= 0,
= 4m4+30
c2
c ~ 1.
3
c -1
P33
•••
~ =~
-1
= 0,
where m4
a
Also
f
~
7.
•
Hence N(m ) ~ 6, ~ > 30.
3
4.
c
= 2 P33
t3
c
ck
P 3 ' • eo, Pk ) ~ 4.
3
N(v) > 2 from Theorem 8.
t3
Now min(2
Hence
1
Hence
This completes the proof of the theorem.
N.B.
We know that for v
order v do not exist.
=2
or 6 two orthogonal Latin squares of
The cases 14 and 26 are still undecided.
Bibliography
£:1_7 Bose, Ho C. and Shrikhande, S. S., "On the construction of
pairwise orthogonal sets of Latin squares and the falsity
of a conjecture of Euler," University of North Carolina,
Institute of Statistics Mlmeograph Series No. 222, 1959.
1:2_7
Bose, H. C., "On the construction of balanced incomplete block
designs," Ann. of Eugen. London, Vol. 9 (1939), pp. 353-399.
,
.
- 13 -
L3_7
Parker, E. T." "Orthogona.l Latin squares," appearing in the
June issue of Proceedings of National
Acad~y
of Science,
U.S.A., Vol. 45 (1959).
Addendum
As this report was being readied for mimeographing we received
from Parker a communication giving a construction for obtaining two
orthogonal Latin squares of orders 14 and 26.
Our '!heorem 9 should
now read:
'!here exist at least two orthogonal Latin squares for all order
v > 2, v
~
6.
The question raised inthe concluding remarks of ~1_7 is completely
answered.
If a positive integer v
> 2 is called Eulerian if two
orthogonal Latin squares of order v do not exist, then 6 is the only
Eulerian number.