"UBIVERSM OF BOM'H CAROLINA
Department ot Statistics
-
Chapel Bill, B. C.
1
Ma:tbematical SCieDCeS Directorate
Air Force Ott1ce ~ SCientu1c Researoh
Wash ng1;oD
'
25,
D. C.
AJ'OSR Report Bo •
..,-:1)1- 0 <)/;14/
B. C. Bose, S.
s.
Sbri.kbame aDd B. !I. Parker
this is u .2lP08i1ioZ'y JlllP8r descr1b1D& the recent
work on • OrthosoDal LaUD Squares and Bta1er' 8 CoDJectuJ:e t by the au1lJlore. Scae ~ tbe praota haw
baen a1mplU':1e4 8D4 eDItNIb back SI"OUD4 _ten&1 baa
been 1Dclwtecl 1;0 _ _ it possible tR pad.te level
atw1eDts ot _~t1C8, or ptl'8ODII nth equivalent
training to tollow tbe work.
, QualJ1'1e4 l'8q\ll!llston _1' obta1D copies ot tbla NlDI't baa the AS'lIA
~t 8en1ce center, ArJ.1DStOD Jra1l station, ArJ.1DStOD 12 V1z'&1D1&.
~Dt or DBteDae OOD~ aauat be established tor AS'nA service.,
or have their -Dee4-t.o-Jaaov" ceJ'tUiecl by the cop1zant II1Utary _ncy
at their pro-1eCt or CQ1ltnct.
Institute
or Statla'tiea
M1~' 8eftn Bo. ~
(
,
,"
ORTHOGONAL LATIN SQUARES AND EULER'S CONJECTURE*
b:'t
R~
10
c.
BOSE, S. S. SHRIKHANDE and E. T., PARKER
Introdu crt,ion
Latin squares were first studied by Euler [16J towards the end of
the eighteenth century in a lengthy memoir entitled "Reoherches sur une
nouvelle sspsce de quarres magiques."
4 x 4 squares
Consider the two
exhibited in Figure 1.
(II)
(I)
b
c
d
"
~
y
(5
a
d
c
y
6
a.
~
C
d
a
b
(5
y
~
a
d
c
b
a
~
a.
6
y
a
b
/
'-
The square (I) has the rour Latin letters a, b,
0,
d ocoupy-lng its
16 cells in such a way that each letter occurs once in every rOv1 and once
in every column.
The rragical property (now called orthogonality) consists
in the fact that when the two squares are superposed as in Figure 2, each
Latin letter occurs with each Greek
sa
b~
cy
d6
there were four regiments a, b, c, d
by
a5
da.
c~
and in each regiment there were
c6
dy
a~
ba.
d~
ca-
bo
ay
letter once and only once.,
Thus i f
Figo 2
*This research was supported in part by the United States Air Force
through the Air Foroe Office or Soientific Research of the Air Research and
Development Connnand, under Contract Noo AF49(638)-2l3" Reproduction in whole
or in part is permitted for any purpose of the United States Government.
2.
four officers of ranks as
~,
y, 6, then Figo 2 may be regarded as a solution
of the problem of arranging these 16 officers in a 4x4 square, such that
each regiment and each rank is represented in every row and every column"
A similar problem for 25 officers can easily be solvedo
The problem of
56 officers is, however, another story.
In general, a Latin square of order n is defined as an nxn square,
the n
2
cells of which are occupied by n distinct symbols (which may be Latin
or Greek letters, or just plain integers) such that each symbol occurs
exactly once in each row and once in each column.
and j=th column is called the cell (i,j).
The cell in the i-th row
Two Latin squares are said to be
orthogonal if on superposition each symbol of the first square occurs
exactly once with each symbol of the second square.
The word Latin square
derives from the fact that Euler used Latin letters as the
first of his pair of orthogonal Latin squares.
s~nbo1s
for the
There may exist a set of
more than two Latin squares such that any pair is orthogonaL
W(;;l
exhibit
in Figure 5 three mutually orthogonal Latin squares of order 4, the first
tvro
being the same as squares (I) and (II) of Fig. 1, with the symbols
ch8.L"1ged from Latin and Greek letters to integers.
l:-L:J
5
4
1
2
5
4
1
2
5
4
:3
5
4
1
2
4
5
2
1
1
I4
5
1,:,
1
:2
2
1
4
5
4
321
3
4
1
2
2
I
Euler showed that the problem of n
2
officers, requiring the construc-
tion of a pair of orthogonal Latin squares of order n, can always be solved
ahen n is odd 9 or n is an evenly even number (Leo,
On the basis of extensive trials Euler stated
If
11
is divisible by 4),.
I do not hesitate to conclude
that it is impossible to produce any complete square of 56 entries, and the
Sa1"l6
impossibility extends to the cases of n ,., lOJ)
all unevenly even numbers. It
may be stated as follows:
squares of order n a 4t
~
11 ="
14 and in general to
This became famous as Euler t 3 conjecture and
There does not exist a pair of orthogonal Latin
2 for any positive integer to
In 1901 the 1'rench mathematician G" Tarry
[3:{1
proved by exhaustive
I
enumeration that Euler's conjecture was indeed true for the case n
The labor involved in the method of exhaustive
'farry to settle the case 11
Itt
VIe quote from -c.11e art,icle on
enume:cation.~
~
60
which enabled
6, goes up extremely rapidly as n increaseso
~A
Hall [17]: "Extensive searches
survey of combinatorial analysis 9 by Marshall
011
S'...rAG~ the computer at UCLA, h8.ve failed
to produce an orthogonal lOxIO pairo
But even with more than 100 hours of
hlgh speed search, the part of t.he possible ca.ses tried is so microscopic
that no conclusion may be drawno ll
Several attempts? necessarily erroneous, have been made in tbe past
to prove Euler's conjecture~ c"g, Peterson
i"lacNetsh [21]
0
[oOJ,
T;Jernicke [3SJ, and
Ievi [18 J p. JA] points out the inadmissibility of the
argument used by Peterson and r..rncN'eisho
be fallacious by 11acNeish [2~1
Wernicke f; S 8.l"gl.l.111ent was shmm
Jc,o
v
:fe shall try 'GO tra.cH :In t.hiiJ int.roduction the suc;cessive steps by
which a complete solution of the problem of constFQcting a pair of orthogonal
Latin squares of any order 4t + 2, t > 1, was achieved.
It is hoped that
with the help of the references given here the interested reader will be
able to trace the development of :f.deas leading to this result"
However!]
40
in the exposition which follows (sections 2=7) we shall confine ourselves
to giving some counterexamples to Euler t s conjecture j with enough background
material to make the chapter self contained"
An excellent account of early rese.moches on l.atin squares is found
in a paper by Norton
[g4]
on the 7x7 squares"
However, the first general
results on the construction of mutually orthogonal Latin squares (moo.Ls)
n
of a given order were due to NacNeish [21J
where PI' P2'
U0.'
&
n
2
l
If v ." PI P
'
2
&
v.,
n
Pu
u
Pu are distinct primes.9 we define the arithmetic funct.ion
n(y) by
n{v)
:::=
min
)
=
1 "
MacNeish showed that we can always conotruct at least n(v) moool"so of
order v, and conjectu:t'ed that n{v) is the maximum possible number.
denote by
N(v) the maximum number of
m~ool.s. of order
to construct, then MacNeish's theorem may be stated as
MacNeish's conjecture
to two reasons.
as N(v)
= n(v).
If we
v that it is possible
N(v)
~
nev), and
This conjecture seemed plausible due
Since n(4t + 2) ~ l~ the correctness of MacNeish's conjec=
tUX'e implied the correctness or Euler' b conject.ure.
existence of a finite projecti·;.re plane oi.'o:,:'aer
existence of a set of v=lm.o.Ls, a.s sho~!n
by
"IT
Again .. since the
is equivalent ·to the
Bose
[2J,
no projective planes
of non=prllile=po'Y;"€l' order could Gxist if the }\'iacNei.sh cCln.iectu.:('B "t-Jere true
This
talli,~d,·dth
the fact t.h,:"f, Hll know;; 1 fi:nhi6
pro;j~~cthe
planes are of
Toward the end of 1958 Parker [25J shotied that i f there exists a
balanced incomplete block (BIB) design (cfo seCD 5) with v treatments,
). .. 1 and block size
k
a prime=power , then there exists a set of k-2
moooLSa of order Va
This result was improved to k=l v.lhen the BIB design
0
was a cyclic projective plane [26J
0
Taken in conjunction with the known
2
existence of cyclic projective planes [52] with v "" m
"A. "'"
1 when m is a prime pmler, this shows that N(m
ill
1 are bo"i:.h
<}
prime~powers
2
<}
.... m
J9
<}
'If
1; k
1)
...;> m,
i f m and
and disproves MacNeish's conjecture eugc if
4, N(21) ~ 4 whereas n(~.l) '" 20
1Jl ""
Since it is not known whether there are an infinite number of values
of m for vrhich both nl and m+1 are both prLile=powers, this left undec:tded the
question whether MacNeishPs conjecture is false for an infinity of values"
Bose
[5J
and Shr:tkhande
[31]by
directions were able to obtain
generalizing Parker's result in certain
all
infinUe number of counter examples to
MacNeish's conjec"c,ure, but the methods so far used by them were not pOwer=
e
They realized that a more general
£ul enough to disprove Euler's conjecture.
type of design than they were using was necessary.
This led them to give up
the c0ndition of constancy' of block size and to introduce the concept of
pairwise balanced designs of lndex unity and type (v; k p
(of
0
sec
0
4:).,
2
S'
00 q
•
k ),
1.1
They shmred that if there exist-s such a design toget,her uUh
of order v where q "" min (ql~ Q2'
obtain a pair of
TIl. 0
0
1. S
0
CL ))
m
and
used this theorem
the first to be d:i.scovered (ef"
Euler~ s
to
sel:':"
(Actually the result N(50) > 5 was
""
5) ).
Bose and Shrilthande [8, ~ u.sed
the theoreUi just mentioned together with a certain
ing to "the improV'ement
[7]
of order 22, thus providing t.he first published
counter example "1:.0 Euler's co:n;)ec-c.ure"
to show that
k
obt.ai:o(~d
b;)r
Park~n~
lll~rovement correspond~
for symmetric cycl:Lc BIB designs 9
conjeet1m2: t<Jas false for an :Lnfinity' of values of v
(the lowest of which was 22) and obtained a large number of general theorems
giving lower bounds for the flmotion N(v)o
Meanwhile, Parker, influenced by [7]jl used the method of differences
to show [27, 2S]-that if q is a prime~ower ~ 3 mod (4)., then there exists
a pair of 1Uoooloso of order (5q=1)/20
This includes the important case of
order 100
The next step was taken by Bose and Shrikhande who refined their
main theorem by introducing the concept of a set of clear equiblock components
(efo sec" 4).
Using this theorem, wh:i,ch is given in section 4,» and the re=
suIts of [8, 9~ 27, 28J they showed [10,
Moo.l.s. of order v for v
= 18
1]J
that there exists a pair of
and all v > 260
This left undecided only the
cases v "" 14 and 26 which were settled by Parker [29]
We thus reach the
Q
surprising result that Euler's conjecture is false for all numbers of the
e
form 4t ·r 2 if t > 1.
In other til"ords, the problem of n 2 officers can be
solved for all n :.> 2, except for the case n
=:
60
The results of Ill] and
[29J, together with a simple const:ruct1on by' Parker (cf. seco 7) for obtaining
a pair of moooLso of order 12t
<?
10 for all non-negative integral values'-9! t,
y,Tll1 appear :l:n a joint paper by Bose, Shrikhande and Parker in the Canadian
Journal'6fMathematics ~~.
Finally}' using an ineqnality first proved by Bose and Shrikhandein
[11],
114]
Chowla$ Erdos and Straus
"tdt.h v"
In fact ll they show that there exists an int,egar
"!<T{
) > ~J
1
i~ v
i-;
20
have shown that l>!(v) tends to infinity
Orthogonal Latin
'V
o
such that
1/91
~'f<
squaI'e~'l. .or-,:thogonal
arrays and transitive arrayso
t
Let us consider a matrix A • (a ) with m • >..v columns and crows,
ij
where each a
ij
represents one of the symbols 1, 2, 000" Vo
t=roued submatrices formed from A by choosing trows t
~ Co
Consider all
Each column of
'7
• 0
the submatrix oan be regarded as an ordered t=pleto
The matrix A 1s called
an orthogonal array (m, v, c, t) of size mS' c constraints,9 v levels,9
strength t and index >..; i f each t-rowed submatrix which can be formed from
A contains every one of the v
t
'
possible t=plets each repeated>" times"
Orthogonal arrays of strength 2 and index lll1ity playa special part in the
theory of orthogonal Latin squares, and in the seque1 we shall only consider
orthogonal ar-rays of this type
0
Such arrays depend only on two parameters ~
vizo the number of levels (symbols) v J and the number of constraints (rows)
Co
The size
01"
the number of columns is then v
2
Every ordered pair
0
formed from the v symbols occurs exactly once in each two=rowed 5ubmatrix
of the arrayo
it OA Lv ~ cJ
t <:
C
0
We shall use a special notation for such an array
e~d
call
It is clear frOID the dafmi tion that if we drop a.ny t rOtiS jl
r
J we
from OA v, c
Thaoremlo
get an orthogonal array OA [v <q c=tJ of c=t constraints.
The existence of a set of q=l mvooloso of order k
implies the existence of a.n orthogonal array OA Dc;> q+l] and cOilverselyo
Suppose there exist.s a set of q=l moo~loso (1 ]
1
of order k i1 with k symbols 1 9 2,
i~th 1'0-\11
Ut1 be
00,'
jl
ko
Let
[I'Q 1 be
0
[L
J,
2
o.
0
[j..q-~lJ.
a kxk square, the
'"
Similarly ~ let
of l':rhich conta.ins the symbol i in every positiono
a lock square, the j=th column of which contains t,ne symbol j is."l
e'lJ9ry position"
fL. 2J,
For example: :If k "" 4'} t,hen (i'lJ,
as the squares exhibited in Figo 5 f) and
[J.n]
el
and [1
[L ] :may be taken
5
are the squares
shovrn in Fig 0 40
1
1
1
.
J.
1
2
5
4
2
2
2
2
1
2
5
4
3
3
3
5
1
2
5
4
4
4
4-eJ
1
2
5
4
• -rm=
We write the elements of each square in a single row such that
the symbol in the i=,th row and j=th column occupies the n=th position :in the
row where n
[La],
~cll
= k(l=l)+
jo
[11],
[tq=l}'
He,
It is convenient to take the squares in the orQ.ar
a (q+l) x n
vIe then get
an orthogonal array OA (k,9 q....l) c
2
matrix A which is
This f'ollows directly from the definition
of a set of Dl"ooLsa as the ordered pair occ'l.l:l'Jing in the n=t.h colu,'1l1'l in the
submatrix formed by the SFGh and t=th rows (3, t "" R, C, 1, 2, "" ~, q=l) is
the pair consisting of the symbols occurring in the cell (i, j) of the
squares
[l~a
[LtJ·
and
For example, the orthogonal array OA (4, 5) corre~>
sponding to the set of moo.1.s. in Figo 3, is
Row
1
1
1
1
1
2
5
4
1
2
5
1
2
1
2
4
4
4,
4
R
1
:5
4
C
4:
2
:;
2
1
1
1
2
1
4:
5
:; 4:
1
:J
2
Conversely i f an OA [k, q<;j>l] is given$ we can take the t,,10 inUial
r01fS
e
(201)
A ...
2
2
2
2
1
2
:5
4
4
2
1
4:
:5
:;
4
:;
4
1
2
5
4:
4:
3
2
1
.
"
(3
:5
5
5
·•
1
2
5
4
:;
4
1
2
·
-4
5
2
2
1
4
·"
0
the rows Rand C, and number the other rcr!ils,l,
r~,
"0 0
.
~
q=lo
The pair
'i<
v
(~ )
occurs in just one colurnn of the sUbmatrix formed by the rows Rand Co
rr.a.y be called the column
rL]
L S
S ....
1, 2, ,,<>., q=l
(1).
mutuall~'
This
1'19 now fill the cells of the lock square
by Pl1~liting in the cell (i,~ j) the symbol occurring
in the column (}) of OA(k, q.tr-l).
q=l
as
It is eas:r to
S8eJ
that
Tile
orthogonal Latin squares
get
ft
set of
This proves
the theorem"
It is clear that i f the symbols of a Latin square are renamed it
still remains a Latin squareo
If there is a set of
MeOol-So
then this
renaming can be done independently for each square without doatroy:L""1g their
mutual
orthogonali;~yo
Henoe it is possible to transform the set into a
form in which the j=th symbol occupies the cell (l~ j) for each squareo
Such a set of moo.Lso is said to be in the standard form.
For example,
the set of Latin squares exhibited in Fig" 5 is in the standard form"
A q x k(k=l) matrix T j whose eleruents are k symbols say
k, is ca.lled a doubly t,ransitive array with k levels and q
constraints, if each
two~rowed
all the ordered pairs (
~)
, i
'!
Elubmatrix formed from T contains as columns
j formed from 'Ghe symbols 1:$ 2:;
Such an array may be denoted by T [k,
T[k)
qJ
q]"
oo~,
l'
k~
The doubly transitive array
Tk =l each of order q x k such that each row of T consists of'
i
all the syrabols 1, 2,
uo~» k in some order or other
and Illay be denoted by RT [k, qJ
con6traini~s
called an E=il'.atrix with q
(i
0
1 9 2 s "o.~ k~l),
e
A q x k matrix whose elements are
qJ.
•
is said to be resolvable i f i'!~ can be subdivided into the 8ubmatrices
T1 , T2 ,
E DC$
<; 0
11:
uymbols say 1$ 2,
ou"p
k, is
and k levels, and is denoted b:r
1$:' each element of the j~th colunm is the symbol j"
Ik!s q1
The existenoe of a doubly transitive arra.y T
of ordm' k
0
matrix obtained by prefixing t;e T the E-rr.atx'ix B [k 9 q] Tiith the sarrie
sytflbols "
'rhus
1~
.,:>
[E,
'I'J
It follows from definition that A is an orthogonal array 01\
from theorem 1 is equivalent to a set of q=2 meool.so
l!<J
qJ, which
10"
Theorem 5.
The existence of a set of q=l m.oel.s. of order k
implies the existence of resolvable doubly transitive array RT [k, ql and
conversely.
Take the q=l
[L1], (12]'
ffi"O.l.S.
.0.,
fq=l] in the standard
form and obtain the corresponding orthogonal arra3r A ... OA [k, q+l] after
prefixing the squares
~le
['La]
and
[l'c1as in the proof of theorem 1
Then
0
can write
...
A ""
9
where a. j is a lxk roW'=vsctor each of whose elements is j? and E, Tl' T2' ."",
Tk=l are Jil.at,rices of order qxk.
E is the E=matrix E [k, q].
Since the squares are in the standard form
If we drop the initial
rO"lfJ
of A the restdual
matrix
If we t.ake anJr
pair formed from 1, 2, "
t~1TO
1fo
TOl<led submatri.1C of T
..
every ordered
.',' k appears as a column exactly once.
pa.irs (~) , j ... 1, 2 9 ~. O!J k arise from '"he columns in Eo
J
Hence
is a doubly transitive array of k levels and q constraintso
consider the
two~'rowed
of A~ (i "" C~ 1» 2 j
submatrh: formed by the
0009
~!.nitia.l rON'
But the
Again, if we
and the i=th roo
q=l).~ it, fol1o~rs from the fa.ct that no ordered
pair can occur as a column vectol' more than once $ that every row of Tj
consists of the symbols 1, 2$
resolvable
0
000
The submatrices 'fp
JI
k in some order or othero
T2 ,
OO>!)
Hence T is
Tk=l may be called the components
of T"
For example!J when k ... 4)) q
4 the structure of A is shown by the
OIl
partitions of the matrix A in (201)0
Conversely suppose a doubly transitive resolvable aITay
is giveno
By prefixing an E=matrixl/ E [k~ ql with the same symbols, and
I
,
~
then adding a new initial row 0.19 <l2 J1 " •• , ~ we obtain the orthogonal
array A "'" OA ~, q+1] given by (202), which from Theorem 1 is equivalent
to the existence of a set of q=l moool.so
Cor
There cannot exist more than k-l mooolos. of order ko
0
Suppose there exist q=l m"o"l.se of order ko
e
We then obtain a
corresponding resolvable doubly transitive array RT [k,
qJ.
Since the
~
elements in any column of this array must be all different q
i-le
shall conclude this section with the following theorem due
If v ~
Theorem 4.
Let
11 11
1 2
, N(v) ~ min [N(n ), N(n )
l
2
min U~(nl) <7 N(n ) ] ...
2
0
We then
(s ~ t), s ." 1, 2,j
rem and
0 •• ,
Let the symboll!\ in these squares be 1, 2,
respectively.
j='~h
"0 0 '
column
co~sider
D
1
;
Of~-'uJ
t
m
the set of v
1 3 2,
<>"
obtained by replacing each symbol bkl
If now each
at;)
a
0
II
n
2
m in any definite
n
000;;
= D1n
n
"
l
and 1, 2,
0
<>
0'
il
2
composite symbols
Let the symbol in the 1 o~th
0
the u=th square of order 11
.
.
be "'Che u=t,h
square
of order n • Let
2•
~t u_-r""I'
(u)
(a~;), ~t»o
square I a.t~):J
J
Retain only m squares of order ~
rllo
and of order n , and number these squares 1, 2,
2
order
k.
1
be
ai~).
Let
[(u)
a:., 1 *1
j' denote the square
J.J
u'
of t u* by the composite SJDllbol
in [:-ou] is replaced by the corresponding
L: J, we get a square
obtained in this way by taking u ." 1:1
[tu'
2~
L:J of order n1n2 "
00.'
!n
The m squares
form a set of ffioooLso of
e
order v '" nl n
2
0
Details of the verification are left to the reader as an
exercise"
Finite fields and comolete sets of Latin sQuares.
+
A set of s
-
elements is said to form a finite field if there
exist two laws of composition, vizQ addition and multiplication, for which
the following laws are valid:
(1)
a
and
(i1)
a+b"~
h-!>a,
(iv)
If a and b belong to the set there exists an element x
If
such that
(v)
If a
and
b
a~x
b
belong to the set, the sum a+b belongs to the set 9
Q
b,
belong to the set the product ab
belongs to
the set,
(vi)
(vii)
("~'i';~)
'\ v ...... J- .....
'. ,)
( ~i.X
a(bc)
::>
(ab)o,
(a+o)o '" ab+ac, c{a+b)
= ca+cb~
ab '" ba j
If a
and
b belong to the set and a
f
0 there exists an
element. 'dT of the set, such that ay "'" b.
From (i)~(:l.v) it can be shown -iihat 'c,he:ce exist.s a unique element
called the zero element such that
ai-O'" a
fol" every a
belonging to the
set,o
A set of
8
elements for wh:tch the
la1>tS
(i )=( viii) are valid is
For a finite field it is poss1.ble to carry out the usual operations
involving addition, subtraction, multiplication and divisionD
x
III particular
is uniquely determined from a+x • b and -a can be defined by a+( =8) ... 0"
Than
ay
OIl
b-1p(=a)
b
can be written as
uniquely determines
the unit element such that
field
0
b=a.
Oa. '" aO = O.
Also
Again if a ., 0,
y, and there exists a unique element 1
a.l '" lea" a
for all
a
called
belonging to the
Of special importance is the fact that the theorems governing the
solution of linear equations remain the same as for the field of real numbers.
An excellent account of the theory of finite fields is found in [15.\1 Chapter IX]
and [25, Chapter VIII] to which the reader is referred for further detaE.s
A simple exanlple of
be a positive integero
a finite field is obtained as follows:
Two integers
to be congruent (mod p), i f a-b
writing a · b(mod p)o
e
a
and
Let
0
p
b (+ve, =ve or 0) are said
is divisible by po
This 1s denoted by
The set of integers is nov subdivided into p
disjoint subsets such that the integers in the same subset are all congruent
to one another (mod p).
CaL
Clearly
The subset to which
(a)'" (a+rp) where
called residue classes (mod p).
i
such that 0
~
i < po
r
a
belongs can be denoted by
is any integer.
These subsets are
In each residue class there is one integer
This integer may be called the standard representa-
tive of the residue classo
Thel'e are then exactly p
residue classes
(0), (1), (2), ••• , (p-l) with standard representatives 6, 1, 2, vo., p=lo
~1Te
now define the addition and multiplication of these residue classes by
(a)(b) ... (ab)
WE> then get a finite cOlmTmt.ative ringe
the residue classes form a field.
denoted by GF (p )
0
p
is pr:ime then
The field of' residue classes (mod p), is
For example, if
classes (0), (1), (2), (5), (4).
If in particular
p "" 5,
We have
GF ( 5) consists of the f1 va
If it is understood that the modulus is a given prime it is
convenient to drop the parenthesiso
We may then write (302) as
(mod 5)
It is well known that if v '" p n where p
exists a finite field wit.h
pn
elements
is a
prL~~
then there
It is denoted by GF(pn).
0
finite field is also called a. Ga.lois fieldG
A
Conversely the number of
elements in a finite field1 must be of the form
pn ~
~Je will not discuss
here the general finite field· bu'!; shall give the rules of addition and
2
GF(2 ) wi-ilh
multiulication for the field
elements 0, IJl
e
spectively~
(I;)
and ro
*
4,
elementso
The field has four
where 0 and 1 a.re zero and unit elements re=
The addition and multiplication tables of the field are shown
below (Fig" 5)
Multiplication table of GF(2 2 )
Addition table of GF(Z2)
o
1
m
w*
11
0
«)
ro
ro
w
10
I *-.
:~
*
* ro*
ro
ro
(1)
* *
I 1
ro
L.-
I !
1
(l)
*
0)
1
1
(l),
0
Figo 5
the element 1.."1 the rom headed tYy a
addition table,.
2
&2
and colunm headed by / &2
Thus
with zero is zero"
a
1
in the
The product of any element
To find the product of two non=zero elements
we take the element in the row headed by
in the multiplication table"
Thus {J,),,~*
~
and
81. and the column headed by
s
1,
,l' . . 00*
The reader
150
e
should verify that the laws (i)c,(:tx) are in fa.ct satisfiedo
We have already seen that there cannot exist more than
of order
Vo
A set of v-J. IDoo"los" of order v
a complete set
Vfe shall now show that if
0
set of M"ooLs" of order
v
v
(if it exists) is called
is a prime power a complete
of the function
In terms
exists"
v=l rnoool.so
N(v)
defined in the introduction this may be stated as
Theorem
50
,
N( V)
c:
v~l
n
:if v· P
Consider the Galois .field GF(pn) vrlth elements <:to • 0,
0,1
7.>1 jJ 1J..
2
jl
a:v~'l"
,,') o!)
~rake a
"'lXV
square ,!lnd in the ce11 (i, j) put the
,.
element ai_ determined by
where
1 ~ u -< v=l
is fixed and iJlj "" I, 2 9 "<J",
oO'jjained will be a Lp,tin square"t<::!.th s;ymbols
follc)'~"1S
of
t.h('~
fielclo
rOi~'
Hence each
Taking u "" l~ 2~
Hill 'have all
[1JJ:J f2]!) ,
c.:>
fI,_
I:- \7=
v
v=l
vo 'J;J
are mutually orthogonaL
of ':-'he
o;o.? O'.p
0
u.
J)
fr01TI the properties of the field since for a fi>:ed
Sii'l1ilarly each column has all the
squares
The square [LuJ thus
Vo
_j"1
-,
...1
-
"
'~he
:.l
as
ranges
j
symbols exactly onceo
symbolfJ eX13.ctly once c
in (5,..'1) we can obtai!l.
\oJ0
she,ll
).10,';1 61'10('3'
Consider the squnl'0
L..'X' 1>1:1.11 contain the BJi1llbol
u
This
av=lo
a.""X'
.'\..
v=l
Latin
that these Latin squares
. '*
[tu?~l ' U rUG
giv'en b"l)"
~
(5. 5)
If'
<tt
and <tt
*
are "'ny- two elements of the field ,the S1Jnbol ~ of
,.nIl occur trlth the symbol
ct
t
*
of
[tu *]
Ii J
tu
in t.he cell (i.llj) when the
160
squares are superposed, if and only if the equations (504) and (3,,5) are
simultaneously satisfiedo
.?\
u yi
Given
*
u, u ,
a.t~
a.
t
*; u,u* •
l~
2t
OOOj
v=l,
t,t*. 0, 1, 2, ooo~ v-I, the equations (5,,4) and (50S) uniquely
In fact
(1.0;
u t*
,
Thus the integers i,j, (1 ~ i ~ v,
1 ~ j ~ v)
This shows that when superposed any symbol of
~
of
u
*]
just enoe
e
1)) 2 11
~u.."
v
a.
u* t
are uniquely determined"
~~
occurs with a s;ymbol
This proves the theorem.
0
The symbols
=(1
(1.0'
Ctp
"0",
av=l may be replaced by the integers
in the squares constructed above i f desiredo
The reader may
verify that the set of 5 m..,ool.so exhibited in Figure 5 can be obtained in
2·
GF ( 2 ) by taking
this way from the Galois field
0:
5
0
C'lo'" 0, a
1
llIl
1.11 (12
81
Cl)
ro*
S:llnilarly by considering the field of residue classes (mod 5) and
(1), (1;2
(2)$ (f,5 ... (5), 0. = (4) we can construct
o := (O)p ~
4
the complete set of 4 m.o"Ls.o of order 50 The actual construction is
taking a
10"1
:&
lef.t to the reader as un exercise"
For further details on the construction
of complete sets of oTthogonal Latin squares using finite fields reference
should be made
-[;0
[2J, [6] and [25, Chapter VIII
J.
Combining 'J:heorem ,1 and Theorem 5" we get the following result due·
to r-lacNeish [21J and Harm [22]
0
n~'
Theorem 6.
N(v)
p"
If
u
is the maximum nmnber of mo 0 10 s
0
0
of order
then
N(v);;- n(v)1I where
v ~ and
n (v)
is defined by
170
4"
Pairwise balanced designs of index unity and the main theorem.
An arrangement of
v
objects (called treatments) in b
sets
(called blocks) will be called a pairwise balanced design of index unity and
km) i f each block contains either k l , k , """ or k
m
2
treatments which are all distinct (k :: v, k .; k j ), and every pair of
i
i
type (v; kl' k: 2 , ""'"
distinct treatments occurs in exactly one block of the design"
number of blocks containing k
If the
treatments is b , then clearly
i
i
m
v(v-l) • E biki(k i
i-l
1)
=
Consider a pairwise balanceddesign (D) of index unity and type
e
(v; kl » k , "Q'"
2
k ,"
m
The subdesign (D i ) formed by the blocks of size k
i
"0.' m.
(D I)' I < m,
will be called the i=th equiblock component of (D), i - I , 2,
The set of equiblock components (D l ), (D 2 ), Q"" Jl
be said to form a clear set if the
(D
t)
t
:z
will
b.
blocks comprising (D ) JI (D 2 ),
I
are disjoint~ Le., no two blocks contain a common treatment" Clearly
1=1
l.
':5
v
000'
a necessary condition for this is
I
Z biki
1'"'1
Let 8 be a k x 1 column vector, then following the notation used
in (7)s we shall denote by
'.r(o), the q x k(k=l) matrix obtained from a
doubly transitive array T Clr,qj
on replacing the symbol
occurring in the i=th position in 5"
'1'1' 'l'211 ,,.,,~ '1'k =l
where
cj
11
If
i
by the element
T is resolva.ble into components
a similar meaning will be a.ssigned to '1' (6) and 71 (6)
i
cj
denotes the j=th column of T.
If D is a k x b matrix defined by
c
~\J'here 6j
is a k x 1 colmnn vector, then we define T(D )
T (D)
[p(61 )
B
Ti (n)
Theorem 7.
000'
(D
P ( 6 ) JI
2
[Pi( ~ ), Pi (62 )>>
D
t ), t
QUO»
0
7
0 0
P ( l\,)
0
J
]
Pi( ~) ]
0
k ), such that the set of equib10ck components
m
< m, form a clear set.
mutually orthogonal Latin squares of order
then there exist at least
Proof.
".
Ti (D) by
Let there exist a pairwise balancei design (D) of index
unity and tJ'Pe (v; kl' k ,
2
(D ), (D ),
l
2
$
and
q *=2
k
i
If there exist
qi=l
and if
mutually orthogonal I,atin squares of order v
0
Let us define
\
and
Then
Let Oil' 512 ,
vOOy
0io,
:1
orthogonal La tin squares of order
array
0;\ [k p
qi+1
J
be the blocks of the equiblock component (D i )
k.
~
0
menea rle can construct an
or~(ihoCfona.l
whose symbols are the treatments occurring in 0ij
Denote this array by A
ij
0
Let
c
0
~i
Let
first q
be the q
* rows,
~(1)
Then
that i f t
c
*x
bik2 matrix obtained from Ai by retaining only the
i
and let
has q* rows and
and t
bik~
2':
columns 0
Clearly
fI(l)
has the property
are any two treatments identical or distinct contained in
d
t
any block of (D1)~
(D 2 ),
00",
(D ), then the ordered pair (t c ) occurs as
or
t
a column exactly once in any two=rowed submatrix of ti(l).
~u
Let
first
be the matrix obtained from
q* rows 1 u •
K~
1,
has the property that if
tamed in any block of
000'
ta
rna
and
(D,f+l)'
A(2)
tbare any two distinct treatments con=
t
000'
(Dm)
the: the ordered pair (~) occurs
2':
d'l'
u .,,""
Again let
t
8(''')
•
The number of columns
is
m
where
T (D ) by retaining only the
u u
Then
exactly once in any two=rowec1 submatrix of
in
d
n (n
13
1, 2,
[}(5)
000'
b k (k ~1)
"t1 u
U
be the E=mB.trix .E(q*, v ) with. symbols tl"t,2'
2
v )
2
v
is anyone of the
t
"'v= Ebk
2
isl i i
..."reat ment s not contsme
· d in (D)
l ' (D)
2»
0 ,. 0 ,
or (D)
f..
Then
[Oo
1\(1), L
A(2)
.. LIA(S) ]
1..
Y
is an orthogonal array OA [v, q'h~ _.
1" Hence from Theorem 1 we get a set of
.-
*
q -2 m.o01.s. of order
Va
20.
50
Use of balanced incomplete block designs.
A balanced incomplete block (BIB) design with parameters
v, b, r, k, '" is an arrangement of v objeots or 'treatments into b sets
or blocks such that (i) eaoh block contains k < v different treatments,
(ii) eaoh treatment occurs in
r
different blocks, (1ii) eaoh pair of treat-
ments occurs together in exactly '" blocks.
The parameters satisfy the
relations
",(v-I)
r(k-l),
1:1
b>v.
bk • vr,
These conditions are neoessary but not sufficient for the existence of a BIB
e
design"
BIB designs were first introduoed into statistical studies by
Yates' [36] but occur in earlier lit.erature in conneotion with various combinatorial problems.
problem of
Subsequent to Yates many authors have dealt with the
constr~cting these
designs.
bibliography we shall only refer to
A BIB design with
'"
1:1
1
Wi~hout
attempting a complete
[5].
is clearly a pairwise balanced design of
index unity and type (vjk) and will be denoted by BIB(v;k)o
is said to be symmetric i f b = v o
A BIB design
A BIB design is said to be resolvable [4J
if the blocks oan be divided into subsets such that each treatment occurs
exactly once in each set"
replication.
~le
Each such subset is said to give a complete
shall establish below a connection
bet~leen
set of m.o.Ls. and an important class of BIB designs.
a complete
For the details of
the proofs of some of the other results that will be used, we shall
(without attempting a complete bibliography) refer the reader to
[31
and
[23]
0
Given a complete set of moo.Lso of order
s
we can derive a BIB
design with parmfJeters
A ... 1
in the fol1ovJing mannsY't
Adjoin to the
[:R]
[tc]
and
moo"Lsofl]'
}1
[L 2
F'oI' example, when
s
6
2
treatment,s in
----
2
5
5
6
7
8
9
10
11
12
114 JI5
16
From each of the squares ~uJ'
the two sque.res
8
4, this arrangement w.ay be taken as
S'"
1
15
1
[L '=lJ
Arrange the
as :I.n the proof of 'l'!1eore:m 10
an s x s square a
set of
U 00'
u
4:
Iill
R ,l
es
1, 2 11 "00' s=1 we get a
blocks by superposing the square of the treatments on
~uJ
and
taking those treatments in the same block which occur together with the same
symbol of
,!!,u]"
The blocks thus obtained for the case
S'"
4.
by using
the squares in Figures 5 a.nd 4 are shown in (502) within the boxo
Each block
is given by a column, and blocks arising from the same square are partitioned
orf from the others by a dotted lineo
presentQ)
(The
€JiS
should be neglected for the
220
1
2
5
5 9 15:
6 10 14:
7 11 15 ~
.
1
2
5 4 ~
~
2
1
5 4 ~
1
2
5 4 ~
5 6 7 8
6 5 8 7: 7 8 5 6
9 10 11 12 : 11 12 9 10 : 12 11 10 9
1
0
2
5 4
8 7 6 5
10 9 12 11
:
. 16 15 14 15 .• 14 15 16 15 • 15 16 15 14
~
i
.~!
~
4 8 12 16 . 15 14 15 16
&
'------~-----~:__------"li_·-----__;8:-------~
e5 e5 8 5 e5 :
0
blocks
It is clear that the number of treatments v· 52, the number of
2
b c 5 +S, and that each block has size k
So
Also each set of blocks
III
obtained from one square gives a complete replication.
two
Hence
r-
blocks of the same replication have no treatment in common.
Ill
8+1.
Any
Any two
blocks of different replications must have exactly one treatment in
common~
since any symbol belonging to one Latin square occurs with a given symbol of
the second Latin square exactly once.
This shows that no pair of treatments
can occur in a block more than once The total number of pairs formed by the
2
2 2
8
treatments is 6 (8 =1)/2. SL~ce each block gives 5(5=1)/2 pairs, the
2
8(5+1) blocks give rise to exactly s2(8 =1)/2 pairs. This shows that every
0
pair of treatments occm'"s in a block exactly once.
From the BIB design wit,h pa.rameters
~iven
Thus},
III
1.
by (5.1), we can obtain
another BIB design with parameters
A.'" 1
(505)
in the follcr.nng manner:
treatments 91" 82 , ••• , 9s +1 : Add 9 to each
i
block of the i=th replication, and take a new block consisting of the added
Take
treatments~
(5.5).
8+1
new
It is easy to see "Gha.t we get the
BIB design with parmlleters
The actual design is shown in (5.5) for the case
s · 4.
25.
Since we have shown that a complete set of moool.so exists whenever
s
is a prime power, the
(6. S)
mat
vel"
designs with parameters given by (Sol) and
18. priM POINI'.
ftDite.t S 01
01
subsets of
•
BIB
ob~0t8
oa11ed
'po1Du'
aDd CtR'Uin
S called lines, this system is called a finite projective plane
when the following three conditions are satisfied.
(i)
Any two points belong to one and only one line.
(ii)
There is just one point belonging to each of two distinct lines.
(iii)
There exist four points, no three of which belong to the same
line.,
It can be shown thatYone line contains
contain
s+l pointso
s+l points then all lines
The finite projective plane is then said to be of order so
It is easy to see that if a
Bm desi(m with parameters (5.S)
exists, and we take the treatments to be our
9
points' and the blocks as our
'lines' then the above conditions are satisfied.
s
complete set of muo.l.s. of order
Thus the existence of a
implies the existence of
BIB designs
(5.1) and (503) and the existence of the finite projective plane.
shown by Bose
[2J
It was
that conversely from the fini-;:'e projective plamof order
s, we can arriva at a complete set of' moo.1. s c of' order
s.
Thus the
existence of any of the foUl" mathematical constructs implies the existence
of the others c
However? ill
c~J.J.
the moon examples
s
is a prime power,
though the existence of fin its projective plenes and complete sets of ffioool.sa
not based on finite fields is kncr.m
plane of any
non-priw~=power order
fl'i,)
34J c
\'Jhether a. finite projective
exists is an important unsolved problem.
As already noted, a
BIB
design with
balanced design, it follows from Theorem 7
Theorem 8.
The existence of a
Corollary.
If further
k
A.
1
iIlI
is a pairwise
0
BIB(v,k)
implies
N(v) ~ N(k) = 1
is a prime=power N{v) ~ k
2.
=
The corollary was proved by Parker [2SJ and gave the first counter
N(s2+s +l ) ~ s
example to !·YacNeishis conjecture, since from (5oB) we get
if
s
n(21)
and
2D
s -+ 1
min{7,5)
are both prime=powerso
=
1 ... 2.,
N(s2+s +l ) ~ s
powers
N(2l) ~ 40
In particular
Theorem 9.
:I
4
when
s
and 8+1 are both prime=
If there exists a resolvable
BIB(v,k)
with
r
replications then
N(vo}l) ~ min [N{k) ~ N(k+l) ]
N(v+r) ~ min [N(k<;Jol), N(r)
(a)
J
=
1
=
1
To prove part (1) we take a new treatment x, and add it to
each block of the first replication of the
BIB
design.
We then get a
pair-Illise balanced design of :tndex unity and type (v+l; k, k+l).
follows from Theorem 7.
there exists a resolvable
If in particnl..'U'
BIB
"Jith parameters given by (5.1).
that N(s2 v1 ) ;; [ 6=1, N(s+l) ] ~- L
Taking
The result
is a prime or a prime power
8
8
m
7 we get
It follows
N(50) ~
50
Historically this .fas the first counter exal11ple to Euler's conjecture to be
discovered
G
However, the actual squares of order 50 would have been too
unwieldy to be pUblishedo
1
N(21) ~ BJI
we have
As mentioned in the introduction, Parker [26]
improved his result to
0
Taking s
=
(b)
To prove part (ii) we take
1"
new treatments
and construct a new design by adding the treatment
000'
xl"
to each block of the
G
i
We also add a new block consisting of all the new treat-
i-th replicationo
mentso
x p x 211
This gives rise to a pairwise balanceidesign of index unity and
type (v+r; k+l,
The result follows from Theorem 7 0
1")0
As early as 1847 Kirkman [t8] proposed his famous school girl problem,
which in its most general terms can be stated as follows:
Given v .. 6t
school girls, it is proposed to take them out for a walk for
secuti.ve days, the girls marching three abreast.
2t + 1
~
3
con=
Find an arrangement so that
every pair of girls marches abreast exactly one day.
There is an extensive
liteTature on this problem and a complete bi.bliography of all work up to
1911 is given in [lSJ, and an excellent summary of known results is contained
in
1).,
Chapter X
J
0
The girls can be considered as treatments, each triplet
can be considered as a block, and each day' 5 walk can be considered as a
oOL~lete
replication of the resolvable
v
Let
:=
6t.,'FS,
t == 4m-v2, then
4, since the smallest number
must exceed 40
b
:=
BIB design with parameters
(2t<,,1)(5t+l),
r
g:
5t";\-1,
k'" 5,
H(r) "" N(12nr>7) > n(12m+7) > 4
""
iJ.l
:=
A'" 1
from Theorem
the pd.me=poT>Jer decomposition of 12m9 7
It follows th&.t frOTH a solution (504) for any
t
tIO
4m-+2
we can obtain a pair of mutually Ot'thogonal Latin squares of order
36m+220
As stated in the introduction the pair of orthogonal squares of order 22 was
the first counter exronple of Euler 1 s conjecture to be published [7]
0
The
construction depends on the 15 school girl problem, a solution of which is
shown in Figure 7
0
The. plocks or triplets of girls are given by the columns
and each replication i or a
day~
s walk.;; is included in a boxo
e
260
Rapo 5
Rep 0 2
Rap" 1
1
5
6
5
7
2
4:
7
6
1
5
2
8
9
11
14
5
9
10
12
8
4
4
12
10
15
15
5
15
11
14
15
6
2
8
4
11
6
10
15
2
Rap
0
4
Rep
4
6
2
1
5
5
11
14
10
7
8
12
15
9
15
0
575
6 12 15
1
9 14
1
7
2
11
12
15
9
8
15
4
14
5
5
9
8
11
12
15
5
1
15
10
7
RepQ 7
,...
e
5
10
14
----,-
7
2
5
4.:
6
1
14
8
10
15
5
11
9
12
15
Suppose ve ofl1it 5 treatments
~,
a ;
2
0:
5
not occurring in the same
block of a BIB(v;k), then we get a pairwise balanced design of index unity
Since in t.he original BIB(v;k), no two blocks
and type (v; k J k=l, k=2L
can have m,ore than one treatment in common '"he -c.brae blocks of (D) of size
obviously no treatmen:t, :!.n
Theorem 10
0
COlTuilO:t1
and form a clear
If there exist,s a
N(v-S)
~
Hence
t.J8
get
BIB {v,;ld j
min(N(k), N(k=l), 1
Hence we can use the BIB design with v
seto
ft
~
b • 21,
N(k=2) )
=
r · k • 5,
1
0
A· 1
whose blocks are given by (5 2) to obtain a pair of moooLs .. of order 18.
0
270
Similarly, from. knmm Bm(v;k)
designs for
k .... 5
for
v
liB
25, 41, 45, 61, 65
85 and 125 given in [3], we can obtain two Moo o1 oB. of orders 22, 38, 42)il 58,
62, 82 and 1220
70
of the ring R
of residue classes (mod n)
whose elements either belong to R
x 19 x '
2
pair
0
0
•
J
x
(~), where
have
c
i and j belong to
c
(~ : ~)
are reduced (mod Xl).
is
R
of R
where
of
X
m indefinites
c where i=j
'=
there correspond
c(mod n), 0
n
~ c
< no
ordered pairs which
If (~) is one of these pairs then
e ... O~
1, 2,
OH,
n-l, and i
"if
e
j -+ S
and
i
The ordered pair (j) both members of which belong to
R will be called an R=pairo
to be
or to the set
as their associated difference.
the other pairs are
X
We shall consider matrices
0
We shall say that the difference associated with the ordered
0
m
Conversely to each element
to
Let 03 1, 2, 000' n=l be the elements
Use of the method of differences.
A pair (i )
where
xj
belongs to
j.
R
and
x
j
is called an RX-pail'" and the difference associated with it is defined
xj
c
If
is any element of R
€I
we shall forma.lly define x
With this definition, corresponding to any illdefinite
x
pairs the difference associated with each of 'PJhioh is
x
of these pairs then the other pairs are (
;j: ~)
(tx ),
These pairs are of course all the pairs
j
some order or other.
associated with the
~..re
XR
may' similarly' define
lX-l i )
pair \
is
where
Xi
:m
1
a
"*'
there are
j
j
j
If
(>
(;)
G· 0, 1, : •• ,
0, 1,
pairs
0
000'
e ...
x.
.J
n ax
is one
n~l.
n=l in
The difference
0
<)
We shall now prove t.he following thE:..orem:
Theorem 110
squares of order
If
3m"*' 10
m is odd there exist at least two orthogonal Latin
Taking m
9
4t
</0
3
this implies the existence or
a pair of orthogonal Latin squares for all orders
12t "" 10
0
(>
Consider the 4 x 4m matrix Ao
of
ito
it is divided into 4 parts), whose elements belong to R the ring
of residue classes mod (2m
o o
1
A() "'"
given below (tio exhibit the structure
2
...
0:
..
m.
() 0- It
m+l:
"
2m 2111=1
1)
4
or
X
the set of indefinites xl" x 2 ' .,,,,,, xm•
o
1
m:
2
o
0
2m 2m=1 .• om+1 : xl
.,
0
.
0
9
0
2m 2111=1
o•
•
•o
X
xl x 2
m
o
•
"
(j
0
0
1
• " c
2
o
~
1
2
m
m.
0
0
o
"c
"o
\ole note that. of the
l·owed
subma~Gr:ix
of A,
o
4nl pa.irs occurring as columns in any two=
2m are R=pai-i's~ the differences associated with
which are all the non=null elements of R,
Til
associated with which are all the elements of
XR pairso
a.s
X.o
l.
are
o
RXc~pairs
the differences
X and the same is true of
Let A,..,~ be the matrix derived from A0
to every element ofA
x2
and reducing mod(2m + 1);
by adding
xi
*e
G, 0
~
G ~ 2m;;
being considered
Let
Then it is evident that in any two=rowed submatrix of A, any
R<~pair
formed by two di8t1..11ct elements of R, or any RX
occurs
exactly onceo
Let
or
XR~pair
A* be an orthogonal array· OA~, 4J corresponding to
E '" E(4,2m+l) be an E~trix ldth symbols 0" 1~ 2,
is an OA(4, 3m+1), which proves the resu1to
V"
0'
2m"
Then
29c
Taking t
that
N(v)
~
fi!I
0, 1, 2, 3, 4, 8, 9 and 12 respectivelYll it follows
2 for v • 109 22, 54, 46,
Two superposed 10
106, 118 and 1540
x 10 orthogonal squares obtained by this method
The symbols xl'
are exhibited below.
58~
~,
%3 have been replaced by 7, 8, 9.
A Pair of 10 x 10 Orthogonal Squares
8"
00
67
58
49
91
85
75
12
24
56
76
11
07
68
59
92
84
25
55
40
85
70
22
17
08
69
95
54
46
51
94
86
71
55
27
18
09
45
50
62
19
95
80
72
44
57
28
56
61
05
58
29
96
81
75
55
47
60
02
14
57
48
59
90
82
74
66
01
15
25
21
32
45
54
65
05
10
77
88
99
42
55
64
05
16
20
51
89
97
78
65
04
15
26
50
41
52
98
79
87
Concluding remarks
We have now demonstrated that
nUlllbers
e
v
of the form
4t
-i"
conjecture is false for all
alternatively
[11J
and [29]
20
conjecture is false for many
Act,ually, it ha.s been shown tha.t the
t:> 1"
0
Euler~s
For details of the proof see [12J or
Finally, as mentioned in the introduction, Chowla 9 Erdos and
Straus
[14] have shown that N(v)
there exists an integer
v0
N(v)
tends to infinity with v o
such that
>13 v
1/91
if v> v
All the methods so far lmown f'ail to give
non=prime=power order
Vo
In fact,
0
l 2
N(v) > v /
for any
It is interesting to speculate whether this
reflects an actual upper bound on
N(v)
when
v
is not a prime-power,
or is due to JAck of sufficient power in the methods now availableo
bound, if
t~e,
This
i-lould imply the non-e:dstence of any finite projective plane
of non=prime=povrer order
0
A.lso of great interest is the best result of the Chawla, Erdo's,
Straus
type"
This question may be formulated as fo110,('18:
upper bound of positive numbers
to find a positive integer
v
N(v) >
o
is the
Q corresponding to which it is possible
and a positive number
Q
C
~Vhat
l'
if
v>v
o
?
c, such that
References
10
w" Wo Ro
Bal1~
Mathematical recreations and essays, revised by
Ho So f-L Coxeter" Macmillan and Coo Ltd", London (1942)0
2
0
Ro Co Bose,ll
!tOn the application of the properties of Galois fields
to the problem of construction of hyper=Graeco=Latin squares" 9
Sankh1!, Vol o 5 (1958), ppo 525=5580
,
:iOn the construction of balanced incomplete block designs",
Anno of Eugeno, London, VoL 9 (1959)3 ppo 555=-0990
,
!IA note on the resolvability of balanced incomplete block
designsll,
Sankhya, VoL 6 (1942)J ppo 105=110
0
\lNata on Parkeris method of constructing pairwise orthogonal
sets of Latin squares",
Notices~" ~o ~o, VOl o
6 (1959)9
po 179 (abstract)"
60
Ro Co Bose and Ko Ro Nair, liOn complete sets of Latin squaresit »
7"
Roe" Bose and S" So ShrikhEmde y,
about the non=existenc8 of
'btO
nOn the falsity or Euler i s conjecture
ortlwgonal Latin squares of order
4t + 2li , .!]'oco ~" AC_8£0 92." U"S.f;."1 VoL 45 (1959)$ PPo 754=7570
--=- ,
If
A theorem on the construction of orthogonal Latin squares
and the fals:i.ty of
Enler~s
conjectu:.."s!l,
Notic~
Amero Math •.Soc.,
VoL 6 (1959)., pp" 5?8·=579 o
ffOn the c:cnst1.''Uction of sErts of pairwise Latin squares
Mimeoc Series No o 222 (1959L
~Q
Soco
To be pUblished in Trans"
Amero
"On the falsity of Eulerie conjecture for all orders
exceeding 26", Notices ~o Matho Soc cp Vol" 6 (1959), p" 579,
(abstract)
,
110
0
"On the construction of pairwise orthogonal sets of
Latin squares and the falsity of a conjecture of Euler, II",
~o of Statistic~$ ,!!oNoCo
ll
M1Jne~o ~eries No o 225 (1959)
120 Ro Co Bosel' So So Shrikhande and Eo To Parker,
"Further results
on the construction of mutually orthogonal Latin squares and the
falsity of Euler's conjecture."
To be published in the Canadian
Journal of Mathematicso
150
R o Do Carmichael, Introduction
1£ ~
theory of &,oups of finite
- - - - - - - -order, -Ginn and Co 0 , Boston, U0 S oA 0 and London, England (1957 L
140
So Chowla, Po Erdos and Eo Go Straus,
"On the maximal number of
pairwise orthogonal Latin squares of a given order.\t"
To be
published 0 .
150
0 0 Eckenstein lJ
"Bibliogxoaphy of Kirkman's school girl problem,11
Messenger ofIJIatho]l VoL 41 (1911=12), PPo 55=560
160
L~
"Recherches sur u.11e nouvelle espece des quarres magiquea" ~
Euler,
Verho zeeuwsch Genooto ~o Vli~;., VoL 9 (1782)$ PPo 85=2590
170
Mo Ha.ll,
Surveys in App1iei Mathematic::'!, IV,
analysis and probabilitYl
Some a.spects of
A survey of combinatorial analysisjI
John Wiley and Sons) ~lCus New York (1958)0
"On a. problem in comb1Ylations fl ,
Volo 2 (1847), ppo 191~204o
CaIilb
0
and Dub
-
190
Fo Wo Levi,
200
Ho F MacNeish, "Das problem del" 36 ofriziere" ~ Jber. Deutso
v
Y.tath
0
Jc
Finite geometrical systems, Uo of Calcutta, (1942)0
0
Volo 50 (1921)11 PPo 151=1550
~.
!!!>,
, "EulerUs squares", ~ of Matho.\1 Vola 25 (1922)>> ppo 221 ,2270
21"
"The construction of orthogonal Latin squares",
220
~o ~o,
,
Anno of
- -
Vola 15 (1942).\1 ppo 418=423.
Analysis
~d
Desigg of
Experiment8~ Dover
Publications,
Inc 0' New York (1949).
24c W" Ho Norton,
"The 7 x 7 squares",
Anno of Eugeno, London, Vol. 9,
(1959l ppo 269=5070
250
Eo To Parker,
"Construct-ion of some sets of pairwise orthogonal
Latin squares",
Notices
~o
Matho
~o,
Vol. 5 (1958)
po 815 (abstract)o
,
"Construction of some sets of mutually orthogonal Latin
squares".
,
To be published in
~roco ~o ~.
"Orthogonal Latin squares",
Vol 6 (1959), po 276
0
,
Soco
Notices Amero l.fath. Soco
(abstract) 0
"Orthogonal Latin squares",
Froco
~o ~o
SCo
U~o,
Volo 45 (1959).\1 PPo 859-8620
- ,
"Completion of disproof of Evlerts conjecture", Notices
~o ~o ~O)) Vole 6 (1959).\1
Peterson,
ilLes 56 officers ll ,
500
J
51
So S. Sl't..rikhande,
0
o
po
591.
Anno !:!atho (1901=2), PPo 415=4270
"Group divisible designs and the construction of
pairwise orthogonal sets of Latin squares",
~~ho ~~o,
520
J
0
Singer,
Notices Amero
Volo 6 (1959), PPo 181=1820
"A theorem in finite projective geometry and some
applications to number theorylt,
Vol 45 (1958), ppo 577-585.
0
Transo Amero Matho Soc",
"Le probleme des 56 officiers",
!!!o
Franco !!o
~o ~
Vola 29 (1900)9 ppo 170=2050
Me
Ou
Vebl~n
and F
0
Ho
\oledderburn~
If
Non-Desarguesian and
non~Pascalian
Jeometries", Trans" Amero Matho ~co, Vole 8 (1907), ppo 379=3880
55e
Pu
~rernickes
"Das problem dar 36 of'fiziere", Jbero D3utso Mat o
~Q'
VOl o 19 (1910)~ ppa 284=2870
360
Fa Yates}'
If
Incomplete randomized blocks ll ,
Anno of Eugeno, London 9
Vol u 7 (1956)9 ppu 121=1400
University of North Carolina and Case Institute of Technology
University of North Carolina
e.
Remington Rand
Univac~
Division of Sperry RB.nd Corporationl"Sto Paul, Minno