Quasigroups and Related Systems
15 (2007, xx − yy )
Transversals in Latin Squares
Ian M. Wanless
Abstra t A latin square of order
n is an n × n array of n symbols in whi h ea h symbol o
urs n entries
exa tly on e in ea h row and olumn. A transversal of su h a square is a set of
su h that no two entries share the same row, olumn or symbol. Transversals are losely related to the notions of omplete mappings and orthomorphisms in (quasi)groups, and are fundamental to the on ept of mutually orthogonal latin squares. Here we provide a brief survey of the literature on transversals. We over (1) existen e and enumeration results, (2) generalisations of transversals in luding partial transversals and plexes, (3) the spe ial ase when the latin square is a group table, (4) a onne tion with overing radii of sets of permutations. The survey in ludes a number of onje tures and open problems.
1. Introdu tion A
latin square of order n is an n×n array of n symbols in whi h ea h symbol diagonal of su h
o
urs exa tly on e in ea h row and in ea h olumn. By a
a square we mean a set of entries whi h ontains exa tly one representative of ea h row and olumn. A
transversal
is a diagonal in whi h no symbol is
repeated. Histori ally, interest in transversals arose from the study of orthogonal latin squares. A pair of latin squares said to be
orthogonal mates
if the
n2
is simple to see that if we look at all the orresponding positions in
A
A = [aij ]
B = [bij ] of order n are (aij , bij ) are distin t. It n o
urren es of a given symbol in B , and
ordered pairs
must form a transversal. Indeed,
Theorem 1. A latin square has an orthogonal mate i it has a de omposition into disjoint transversals.
2000 Mathemati s Subje t Classi ations: 05B15 20N05 Keywords: transversal, partial transversal, Latin square, plex, nqueens, turnsquare, Cayley table, quasigroup, omplete mapping, orthomorphism, overing radius
I. M. Wanless
2
For example, below there are two orthogonal latin squares of order 8. Subs ripted letters are used to mark the transversals of the left hand square whi h orrespond to the positions of ea h symbol in its orthogonal mate (the right hand square).
1a 7b 2c 8d 4f 6e 3h 5g
2b 8a 1d 7c 3e 5f 6g 4h
3c 5d 6a 4b 1g 7h 2e 8f
4d 6c 3b 5a 2h 8g 1f 7e
5e 2f 4g 6h 7a 1b 8c 3d
6f 4e 5h 2g 8b 3a 7d 1c
7g 1h 8e 3f 5c 2d 4a 6b
8h 3g 7f 1e 6d 4c 5b 2a
a b c d f e h g
b a d c e f g h
c d a b g h e f
d c b a h g f e
More generally, there is interest in sets of
squares
e f g h a b c d
f e h g b a d c
g h e f c d a b
h g f e d c b a
(1)
mutually orthogonal latin
(MOLS), that is, sets of latin squares in whi h ea h pair is or
thogonal in the above sense. The literature on MOLS is vast (start with [15, 16, 37℄) and provides ample justi ation for an interest in transversals. Subsequent investigations have ranged far beyond the initial justi ation of Theorem
1 and have proved that transversals are interesting obje ts in their
own right. Despite this, a number of basi questions about their properties remain unresolved, as will be ome obvious in the subsequent pages. Orthogonal latin squares exist for all orders
n 6∈ {2, 6}.
For
n=6
there
is no pair of orthogonal squares, but we an get lose. Finney [25℄ gives the following example whi h ontains 4 disjoint transversals indi ated by the subs ripts
a, b, c
and
d. 1a 2c 3 4 5d 6b
2 1d 4b 6a 3c 5
3b 6 1 5c 2a 4d
Table 1 shows the squares of order to their maximum number
m
4c 5b 2d 1 6 3a n,
5 4a 6c 3d 1b 2 for
6d 3 5a 2b 4 1c 4 ≤ n ≤ 8,
of disjoint transversals.
table are ounts of main lasses (A
ounted a
ording The entries in the
main lass, or spe ies is an equivalen e
lass of latin squares ea h of whi h has essentially the same stru ture. See [15, 37℄ for the denition.) Eviden e su h as that in Table 1 led van Rees [54℄ to onje ture that, as
n → ∞,
a vanishingly small proportion of latin squares have orthogonal
Transversals in Latin Squares
Table 1: Number
3
m
n=4
5
6
7
8
0
1
0
6
0
33
1
0
1
0
1
0
2
0
0
2
5
7
3

0
0
24
46
4
1

4
68
712
5

1

43
71330
6


0

209505
7



6

8




2024
Total
2
2
12
147
283657
m
of disjoint transversals in latin squares of order
n ≤ 8.
mates. However, the trend seems to be quite the reverse (see [57℄), although no rigorous way of establishing this has yet been found. A point that Table 1 raises is that some latin squares have no transversals at all. We now look at some results in this regard.
mq q×q
A latin square of order represented by a matrix of
is said to be of blo ks
A11 A21
A12 A22
. . .
. . .
Aij
··· ··· ..
where ea h blo k
A
i′ j ′
Aij
i+j ≡
if it an be
A1m A2m . . .
Amm q and two + j ′ mod m.
is a latin subsquare of order
ontain the same symbols i
type
as follows
.
Am1 Am2 · · ·
q step
i′
blo ks
Aij
and
The following
lassi al theorem is due to Maillet [39℄.
Theorem 2. Suppose that q is odd and m is even. No q step type latin square of order mq possesses a transversal. As we will see in 4, this rules out many group tables having transversals.
In parti ular, no y li group of even order has a transversal.
By
ontrast, there is no known example of a latin square of odd order without transversals.
Conje ture 1.
Ea h latin square of odd order has at least one transversal.
I. M. Wanless
4
This onje ture is known to be true for
n≤9
(see 3). It is attributed
to Ryser [46℄ and has been open for forty years. In fa t, Ryser's original
onje ture was somewhat stronger: for every latin square of order number of transversals is ongruent to
n mod 2.
n,
the
In [2℄, Balasubramanian
proved the even ase.
Theorem 3.
is even.
In any latin square of even order the number of transversals
Despite this, it has been noted in [8℄ (and other pla es) that there are many ounterexamples of odd order to Ryser's original onje ture. Hen e the onje ture has now been weakened to Conje ture
1
as stated.
One
obsta le to proving this onje ture was re ently revealed in [57℄.
For every n > 3 there exists a latin square of order n whi h
ontains an entry that is not in luded in any transversal.
Theorem 4.
Given Theorem
1,
this latest theorem showed existen e for all
n>3
of
a latin square without an orthogonal mate. The same result was obtained in [24℄ without showing Theorem
4.
2. Partial transversals We have seen in 1 that not all latin squares have transversals, whi h prompts the question of how lose we an get to nding a transversal in su h ases. We dene a
partial transversal of length k
to be a set of
k
en
tries, ea h sele ted from dierent rows and olumns of a latin square su h that no two entries ontain the same symbol. (e.g. [50℄) a partial transversal of length be a diagonal on whi h
k
k
Note that in some papers
is dened slightly dierently to
dierent symbols appear.
Sin e not all squares of order
n
n n − 1.
have a partial transversal of length
(i.e. a transversal), the best we an hope for is to nd one of length
The following onje ture has been attributed by Brualdi (see [15, p.103℄).
Conje ture 2.
of length n − 1.
Every latin square of order n possesses a partial transversal
A laimed proof of this onje ture by Derienko [18℄ ontains a fatal error [8℄. Re ently, a paper [32℄ has appeared in the maths arXiv laiming a proof of Conje ture
2.
However, given the history of the problem su h a
laim should be treated autiously, at least until the paper has been refereed.
Transversals in Latin Squares
5
The best reliable result to date states that there must be a partial transversal of length at least
n − O(log2 n).
and the impli it onstant in the `big
O'
This was shown by Shor [50℄,
was very marginally improved by
Fu et al. [26℄. Subsequently Hatami and Shor [29℄ dis overed an error in [50℄ (dupli ated in [26℄) and orre ted the onstant to a higher one. Nonetheless,
n−O(log2 n). This improved 2 3 3 n+O(1) (Koksma [35℄), 4 n+O(1)
the important thing remains that the bound is on a number of earlier bounds in luding
n−
(Drake [19℄) and
√
n
(Brouwer et al. [4℄ and Woolbright [59℄).
Erd®s and Spen er [21℄ showed that any o
urs more than diagonal with
(n − 1)/16
n×n
array in whi h no entry
times has a transversal (in the sense of a
n dierent symbols on it).
It has also been shown by Cameron
and Wanless [8℄ that every latin square possesses a diagonal in whi h no symbol appears more than twi e. Conje ture
2
has been well known and open for de ades. A mu h sim
pler problem is to onsider the shortest possible length of a maximal partial transversal (maximal in the sense that it is ontained in no partial transversal of greater length). It is easy to see that no partial transversal of length stri tly less than
1 2 n an be maximal, sin e there are not enough `used'
symbols to ll the submatrix formed by the `unused' rows and olumns. However, for all
n > 4,
maximal partial transversals of length
easily be onstru ted using a square of order
S
of order
1 2n
n
1 2n
an
S
but
whi h ontains a subsquare
and a partial transversal ontaining the symbols of
not using any of the same rows or olumns as
S.
3. Number of transversals In this se tion we onsider the question of how many transversals a latin square an have. We dene
t(n)
and
T (n)
to be respe tively the minimum
and maximum number of transversals among the latin squares of order
n.
We have seen in 1 that some latin squares have no transversals but it is not settled for whi h orders su h latin squares exist. Thus for lower
t(n) we annot do any better than to observe that t(n) ≥ 0, with o
urring at least when n is even. A related question, for whi h seems to have been published, is to nd an upper bound on t(n)
bounds on equality no work when
n
is odd.
Turning to the maximum number of transversals, it should be lear that
T (n) ≤ n!
sin e there are only
n!
dierent diagonals. An exponential
improvement on this trivial bound was obtained by M Kay et al. [42℄:
I. M. Wanless
6
Theorem 5.
where c =
q
For n ≥ 5,
√ 3− 3 6
√
e
3/6
√ 15n/5 ≤ T (n) ≤ cn n n!
≈ 0.61354.
The lower bound in Theorem
5
is very simple and would not be too dif
ult to improve. The upper bound took onsiderably more work, although it too is probably far from the truth. In the same paper the authors reported the results of an exhaustive
omputation of the transversals in latin squares of orders up to and in luding 9. Table 2 lists the minimum and maximum number of transversals over all latin squares of order
n
n ≤ 9,
for
2 de imal pla es.
and the mean and standard deviation to
n
t(n)
Mean
Std Dev
T (n)
2
0
0
0
0
3
3
3
0
3
4
0
2
3.46
8
5
3
4.29
3.71
15
6
0
6.86
5.19
32
7
3
20.41
6.00
133
8
0
61.05
8.66
384
9
68
214.11
15.79
2241
Table 2: Transversals in latin squares of order Table 2 onrms Conje ture
1
for
n ≤ 9.
n ≤ 9.
The following semisymmetri
squares (see [15℄ for a denition of semisymmetri ) are representatives of the unique main lass with
t(n)
transversals for
the largest subsquares are shown in
1 2 3 4 5
2 1 5 3 4
3 4 1 5 2
4 5 2 1 3
5 3 4 2 1
3 2 1 5 4 7 6
2 1 3 6 7 4 5
1 3 2 7 6 5 4
5 6 7 4 1 2 3
4 7 6 1 5 3 2
bold.
7 4 5 2 3 6 1
6 5 4 3 2 1 7
2 1 3 9 8 4 5 6 7
n ∈ {5, 7, 9}. 1 3 2 5 4 7 8 9 6
3 2 1 4 6 9 7 8 5
6 5 4 3 2 8 9 7 1
7 4 9 2 5 3 6 1 8
In ea h ase
8 9 5 1 7 6 2 4 3
9 6 7 8 1 5 3 2 4
5 7 8 6 9 1 4 3 2
4 8 6 7 3 2 1 5 9
Transversals in Latin Squares
n
Lower Bound
7
Upper Bound
10
5504
75000
11
37851
528647
12
198144
3965268
13
1030367
32837805
14
3477504
300019037
15
36362925
2762962210
16
244744192
28218998328
17
1606008513
300502249052
18
6434611200
3410036886841
19
87656896891
41327486367018
20
697292390400
512073756609248
21
5778121715415
6803898881738477
Table 3: Bounds on
T (n)
for
10 ≤ n ≤ 21.
In Table 3 we reprodu e from [42℄ bounds on
T (n) for 10 ≤ n ≤ 21. The 5, though
upper bound is somewhat sharper than that given by Theorem
proved by the same methods. The lower bound in ea h ase is onstru tive and likely to be very lose to the true value.
When
n 6≡ 2 mod 4
the
lower bound omes from the group with the highest number of transversals (see Table 4).
When
n ≡ 2 mod 4
the lower bound omes from a so
alled turnsquare, many of whi h were analysed in [42℄. A
turnsquare
is
obtained by starting with the Cayley table of a group (typi ally a group of the form
Z2 ⊕ Zm
for some
m)
and turning some of the inter alates (that
is, repla ing a subsquare of order 2 by the other possible subsquare on the same symbols). For example,
5 6 2 3 4 0 1 7 8 9 a hieves
5504
6 2 3 4 0 1 7 8 9 5
2 3 4 0 1 7 8 9 5 6
3 4 0 1 2 8 9 5 6 7
4 0 1 2 3 9 5 6 7 8
0 1 7 8 9 5 6 2 3 4
1 7 8 9 5 6 2 3 4 0
7 8 9 5 6 2 3 4 0 1
8 9 5 6 7 3 4 0 1 2
9 5 6 7 8 4 0 1 2 3
transversals. The `turned' entries have been marked in
(2)
bold.
I. M. Wanless
8
The study of turnsquares was pioneered by Parker (see [5℄ and the referen es therein) in his unsu
essful quest for a triple of MOLS of order 10. He noti ed that turnsquares often have many more transversals than is typi al for squares of their order, and used this as a heuristi in the sear h for MOLS. It is has long been suspe ted that
T (10)
is a hieved by (2). This sus
pi ion was strengthened by M Kay et al. [41℄ who examined several billion squares of order 10, in luding every square with a nontrivial symmetry, and found none had more than
5504
transversals. Parker was indeed right that
the square (2) is ri h in orthogonal mates (it has 12265168 of them [38℄, whi h is an order of magnitude greater than he estimated). However, using the number of transversals as a heuristi in sear hing for MOLS is not failsafe. For example, the turnsquare of order 14 with the most transversals (namely, 3477504) does not have any orthogonal mates [42℄. there are squares of order the bare minimum of
n
Meanwhile
with orthogonal mates but whi h possess only
n transversals (the left hand square in (1) is one su h).
Nevertheless, the number of transversals does provide a useful invariant for squares of small orders where this number an be omputed in reasonable time (see, for example, [34℄ and [55℄). It is straightforward to write a ba ktra king algorithm to ount transversals in latin squares of small order, though this method urrently be omes impra ti al if the order is mu h over 20. See [30℄, [31℄ for some algorithms and omplexity theory results on the problem of ounting transversals. It seems very di ult to nd theoreti al estimates for the number of transversals (unless, of ourse, that number is zero). a ute that there are not even good estimates for sals of the y li group of order
Conje ture 3.
n.
This di ulty is so
zn , the number of transver
Vardi [52℄ makes the following predi tion:
There exist real onstants 0 < c1 < c2 < 1 su h that cn1 n! ≤ zn ≤ cn2 n!
for all odd n ≥ 3. Vardi makes this onje ture while onsidering a variation on the toroidal
nqueens problem.
The toroidal
how many dierent ways
n×n
nqueens problem
is that of determining in
n nonatta king queens an be pla ed on a toroidal
hessboard. Vardi onsidered the same problem using semiqueens in
pla e of queens, where a semiqueen is a pie e whi h moves like a toroidal queen ex ept that it annot travel on righttoleft diagonals. The solution to
Transversals in Latin Squares
9
Vardi's problem provides an upper bound on the toroidal
nqueens problem.
The problem an be translated into one on erning latin squares by noting that every onguration of
n
nonatta king semiqueens on a toroidal
n×n
hessboard orresponds to a transversal in a y li latin square
L
n,
problem is
where
Lij ≡ i − j mod n.
Note that the toroidal
nqueens
of order
equivalent to ounting diagonals whi h simultaneously yield transversals in
L′ij = i + j mod n. As a orollary of Theorem 5 we an infer that the je ture 3 is true (asymptoti ally) with c2 = 0.614.
L
and
L′ ,
where
upper bound in ConThis also yields an
nqueens probConje ture 3 has also
upper bound for the number of solutions to the toroidal lem.
Theorem
5
is valid for all latin squares, but
been atta ked by methods whi h are spe i to the y li square. Cooper and Kovalenko [12℄ rst showed that Vardi's upper bound is asymptoti ally true with
c2 = 0.9153,
and this was then improved to
√ c = 1/ 2 ≈ 0.7071
in [36℄. Finding a lower bound of the form given in Conje ture 3 is still an open problem. However, [10℄ and [45℄ do give some lower bounds, ea h of whi h applies only for some
n. Cooper et al. [11℄ estimated zn is around 0.39n n!.
that perhaps
the orre t rate of growth for
4. Finite Groups By using the symbols of a latin square to index its rows and olumns, ea h latin square an be interpreted as the Cayley table of a quasigroup.
In
this se tion we onsider the important spe ial ase when that quasigroup is asso iative; in other words, it is a group. Mu h of the study of transversals in groups has been phrased in terms of the equivalent on epts of omplete mapping and orthomorphisms. Mann [40℄ introdu ed omplete mappings for groups, but their denition works just as well for quasigroups. of a quasigroup
η(x) = x ⊕ θ(x)
orthomorphism
(Q, ⊕)
is a
is also a permutation. The
of
(Q, ⊕),
θ of the elements η : Q → 7 Q dened by permutation η is known as an
It is this: a permutation
omplete mapping
if
following terminology introdu ed in [33℄. All of
the results of this paper ould be rephrased in terms of omplete mappings and/or orthomorphisms be ause of our next observation.
Let (Q, ⊕) be a quasigroup and LQ its Cayley table. Then θ : Q 7→ Q is a omplete mapping i we an lo ate a transversal of LQ by sele ting, in ea h row x, the entry in olumn θ(x). Similarly, η : Q 7→ Q Theorem 6.
I. M. Wanless
10
is an orthomorphism i we an lo ate a transversal of LQ by sele ting, in ea h row x, the entry ontaining symbol η(x). Having noted that transversals, omplete mappings and orthomorphisms are essentially the same thing, we will adopt the pra ti e of expressing our results in terms of transversals even when the original authors used one of the other notions. As mentioned, this se tion is devoted to the ase when our latin square
LG ,
is
the Cayley table of a nite group
G.
The extra stru ture in this
ase allows for mu h stronger results. For example, suppose we know of a transversal of
gi .
Let
g
LG
that omprises a hoi e from ea h row
be any xed element of
the element
gi g
G.
i
of an element
Then if we sele t from ea h row
this will give a new transversal and as
g
ranges over
G
i
the
transversals so produ ed will be mutually disjoint. Hen e
If LG has a single transversal then it has a de omposition into disjoint transversals.
Theorem 7.
We saw in 1 that the question of whi h latin squares have transversals has not been settled. The same is true for group tables, but we are getting mu h loser to answering the question, building on the pioneering work of Hall and Paige. Consider the following ve propositions: (i)
LG
has a transversal.
(ii)
LG
an be de omposed into disjoint transversals.
(iii) There exists a latin square orthogonal to
LG .
(iv) There is some ordering of the elements of that
a1 a2 · · · an = ε,
where
(v) The Sylow 2subgroups of
ε G
G,
say
a1 , a2 , . . . , an , G.
su h
denotes the identity element of are trivial or non y li .
The fa t that (i), (ii) and (iii) are equivalent omes dire tly from Theorem
1
and Theorem
7.
Paige [43℄ showed that (i) implies (iv). Hall and
Paige [28℄ then showed that (iv) implies (v). They also showed that (v) implies (i) if
G is a soluble,
symmetri or alternating group. They onje tured
that (v) is equivalent to (i) for all groups.
Transversals in Latin Squares
11
It was subsequently noted in [17℄ that both (iv) and (v) hold for all nonsoluble groups, whi h proved that (iv) and (v) are equivalent. A mu h more dire t and elementary proof of this fa t was given in [53℄. To summarise:
(i)⇔(ii)⇔(iii)⇒(iv)⇔(v)
Theorem 8.
Conje ture 4.
(i)⇔(ii)⇔(iii)⇔(iv)⇔(v)
As mentioned above, Conje ture
4
ble, symmetri and alternating groups. other groups in luding the linear groups
P SL(2, q)
is known to be true for all soluIt has also been shown for many
GL(2, q), SL(2, q), P GL(2, q)
and
(see [23℄ and the referen es therein).
After de ades of in remental progress on Conje ture
4 there has re ently
been what would appear to be a very signi ant breakthrough. In a preprint Wil ox [58℄ has laimed to redu e the problem to showing it for the sporadi simple groups (of whi h the Mathieu groups have already been handled in [13℄). See [15℄, [22℄ or [58℄ for further reading and referen es on the HallPaige onje ture.
7 is that for any G LG is independent of
An immediate orollary of the proof of Theorem number of transversals through a given entry of
the the
entry hosen. Hen e (see Theorem 3.5 of [16℄) we get:
Theorem 9.
of G.
The number of transversals in LG is divisible by G, the order
M Kay et al. [42℄ also showed the following simple results, in the spirit of Theorem 3:
The number of transversals in any symmetri latin square of order n is ongruent to n modulo 2.
Theorem 10.
Corollary 1. Let G be a group of order n. If G is abelian or n is even then the number of transversals in G is ongruent to n modulo 2. Corollary 1 annot be generalised to nonabelian groups of odd order, given that the nonabelian group of order 21 has 826814671200 transversals.
If G is a group of order n 6≡ 1 mod 3 then the number of transversals in G is divisible by 3.
Theorem 11.
I. M. Wanless
12
We will see below that the y li groups of small orders
n ≡ 1 mod 3
have a number of transversals whi h is not a multiple of three.
zn , the number zn′ = zn /n denote the
The semiqueens problem in 3 led to an investigation of of transversals in the y li group of order
n.
Let
number of transversals through any given entry of the y li square of order
n.
Sin e
zn = zn′ = 0
for all even
the following dis ussion that
n
n
by Theorem 8 we shall assume for
is odd.
zn′ are known from [47℄ and [48℄. They are z1′ = z3′ = ′ = 3441, z ′ = 79259, z ′ = 2424195, z ′ = 1, z5′ = 3, z7′ = 19, z9′ = 225, z11 13 15 17 ′ ′ = 275148653115, z ′ = 19686730313955 94471089, z19 = 4613520889, z21 23 ′ and z25 = 1664382756757625. Interestingly, if we take these numbers modThe initial values of
ulo 8 we nd that this sequen e begins 1,1,3,3,1,1,3,3,1,1,3,3,1. from Theorem 10 that
zn′
is always odd for odd
n,
We know
but it is an open ques
tion whether there is any deeper pattern modulo 4 or 8. We also know from Theorem 11 that
′ of {zn
mod 3}
zn′
is divisible by 3 when
are 1,1,0,1,0,0,2,0,0,1,0,0,2.
An interesting fa t about
zn
n ≡ 2 mod 3.
is that it is the number of
The initial terms
diagonally y li
latin squares of order n (in other words, the number of quasigroups on the set
{1, 2, . . . , n}
whi h have the transitive automorphism
[56℄ for a survey on su h obje ts.
(123 · · · n)).
See
We now dis uss the number of transversals in general groups of small order.
For groups of order
by Theorem
8.
n ≡ 2 mod 4 there an be no transversals, n ≤ 23 the number of transversals
For ea h other order
in ea h group is given in Table 4.
The groups are ordered a
ording to
the atalogue of Thomas and Wood [51℄. The numbers of transversals in abelian groups of order at most
16
were obtained by Shieh et al [49℄.
and y li groups of order at most
21
The remaining values in Table 4 were
omputed by Shieh [47℄. M Kay et al. [42℄ then independently onrmed all
ounts ex ept those for y li groups of order in Shieh [47℄.
≥ 21,
orre ting one misprint
Bedford and Whitaker [3℄ oer an explanation for why all the non y li groups of order 8 have 384 transversals. The groups of order 4, 9 and 16 with the most transversals are the elementary abelian groups of those orders. Similarly, for orders 12, 20 and 21 the group with the most transversals is the dire t sum of y li groups of prime order. It is an open question whether su h a statement generalises to all
n.
By Corollary 1 we know that in ea h ase overed by Table 4 (ex ept the nonabelian group of order 21), the number of transversals must have
Transversals in Latin Squares
n
Number of transversals in groups of order
3
3
4
0, 8
5
15
7
133
8
0, 384, 384, 384, 384
9
2025, 2241
13
n
11
37851
12
0, 198144, 76032, 46080, 0
13
1030367
15
36362925
16
0, 235765760, 237010944, 238190592, 244744192, 125599744, 121143296, 123371520, 123895808, 122191872, 121733120, 62881792, 62619648, 62357504
17
1606008513
19
87656896891
20
0, 697292390400, 140866560000, 0, 0
21
5778121715415, 826814671200
23
452794797220965 Table 4: Transversals in groups of order
n ≤ 23.
the same parity as the order of the square. It is remarkable though, that the groups of even order have a number of transversals whi h is divisible by a high power of 2. Indeed, any 2group of order transversals whi h is divisible by this is true for general
2n−1 .
n ≤ 16
has a number of
It would be interesting to know if
n.
5. Generalised transversals There are several ways to generalise the notion of a transversal. We have already seen one of them, namely the partial transversals in 2.
In this
se tion we olle t results on another generalisation, namely plexes. A
k
kplex
in a latin square of order
n is a set of kn entries whi h
in ludes
representatives from ea h row and ea h olumn and of ea h symbol. A
transversal is a
1plex.
The marked entries form a
3plex
in the following
I. M. Wanless
14
square:
1∗ 2∗ 3 4 5∗ 6 The name
kplex
2 1 5∗ 6 4∗ 3∗
3 4 1 2∗ 6∗ 5∗
4∗ 3∗ 6 5 2 1∗
5 6∗ 2∗ 3∗ 1 4
6∗ 5 4∗ 1∗ 3 2
(3)
was oined in [55℄ only re ently. It is a natural extension
of the names duplex, triplex, and quadruplex whi h have been in use for many years (prin ipally in the statisti al literature, su h as [25℄) for 2, 3 and 4plexes. The entries not in luded in a an
(n − k)plex
of
L.
kplex of kplex
a latin square
Together the
plex are an example of what is alled an
L
n form (n − k)of L. For
of order
and its omplementary
orthogonal partition
dis ussion of orthogonal partitions in a general setting see Gilliland [27℄
L is de omposed into disjoint parts K1 , K2 , . . . , Kd where Ki is a ki plex then we all this a (k1 , k2 , . . . , kd )partition of L. A ase of parti ular interest is when all parts are the same size, k . We
all su h a partition a k partition . For example, the marked 3plex and its
omplement form a 3partition of the square in (3). By Theorem 1, nding and Bailey [1℄. For our purposes, if
a 1partition of a square is equivalent to nding an orthogonal mate. Some results about transversals generalise dire tly to other plexes, while others seem to have no analogue. Theorem
3
and Theorem
7
seem to be in
the latter lass, as observed in [42℄ and [55℄ respe tively. However, Theorems 2 and 8 showed that not every square has a transversal, and exa tly the same arguments work for any
kplex
where
k
is odd [55℄.
Suppose that q and k are odd integers and m is even. No q step type latin square of order mq possesses a kplex. Theorem 12.
Theorem 13. Let G be a group of nite order n with a nontrivial y li Sylow 2subgroup. The Cayley table of G ontains no kplex for any odd k but has a 2partition and hen e ontains a kplex for every even k in the range 0 ≤ k ≤ n. The situation for even
k
is quite dierent to the odd ase. Rodney [9,
p.105℄ onje tures that every latin square has a duplex. This onje ture was strengthened in [55℄ to the following:
Transversals in Latin Squares
15
Conje ture 5. Every latin square has the maximum possible number of disjoint duplexes. In parti ular, every latin square of even order has a 2partition and every latin square of odd order has a (2, 2, 2, . . . , 2, 1)partition.
1. It also implies kplexes for every even value of k up to the order
Note that this onje ture also strengthens Conje ture that every latin square has of the square.
5 is true for all latin squares of orders ≤ 8 and for all soluble
Conje ture
groups (see [53, 55℄).
Depending on whether a soluble group has a non
2subgroup, possible even k
trivial y li Sylow
it either has a
has them for all
but no odd
k.
kplex
for all possible
k,
or
If the HallPaige onje ture
ould be proved it would ompletely resolve the existen e question of plexes in groups, and these would remain the only two possibilities. It is worth noting that other s enarios o
ur for latin squares whi h are not based on groups. For example, the square in (3) has no transversal but learly does have a
3plex.
It is onje tured in [55℄ that there exist arbitrarily large latin
squares of this type.
For all even n > 4 there exists a latin square of order n whi h has no transversal but does ontain a 3plex. Conje ture 6.
Another possibility was shown by a family of squares onstru ted in [20℄.
For all even n there exists a latin square of order n whi h has kplexes for every odd value of k between 14 n − 12 and 34 n + 12 , but not for any odd value of k outside this range.
Theorem 14.
a
Interestingly, there is no known example of odd integers a latin square whi h has an
aplex
and a
cplex
but no
and
The union of an aplex and a disjoint bplex of a latin square L is an (a + b)plex of L. However, it is not always possible to split an (a + b)plex into an aplex and a disjoint bplex. Consider a duplex whi h onsists of 1 2 n disjoint inter alates (latin subsquares of order 2). Su h a duplex does 1 not ontain a partial transversal of length more than n, so it is a long way 2 from ontaining a 1plex. We say that a
kplex
is
indivisible
if it ontains no
The duplex just des ribed is indivisible. indivisible
kplex
cplex
for
0 < c < k.
k
there is a
Indeed, for every
in some su iently large latin square.
This was rst
shown in [55℄, but su iently large in that ase meant quadrati in This was improved to linear in [6℄ as a orollary of the following result.
k.
I. M. Wanless
16
For every k ≥ 2 there exists a latin square of order 2k whi h
ontains two disjoint indivisible kplexes.
Theorem 15.
Theorem
15 means that some squares an be split in half
in a way that
makes no further division possible. Experien e with latin squares suggests that they generally have a vast multitude of partitions into various plexes, whi h in one sense means that latin squares tend to be a long way from being indivisible. This makes Theorem
15
slightly surprising.
It is a wide open question for what values of
n ontaining an indivisible kplex. k is small relative to n.
square of order
k
and
n
there is a latin
However, Bryant et al. [6℄
found the answer when
Theorem 16. Let n and k be positive integers satisfying 5k ≤ n. Then there exists a latin square of order n ontaining an indivisible kplex. So far we have essentially looked at questions where we start with a latin square and ask what sort of plexes it might have.
To omplete the
se tion we onsider the reverse question. We want to start with a plex and ask what latin squares it might be ontained in. Stri tly speaking this is a silly question, sin e we dened a plex in terms of its host latin square, whi h therefore is the only possible answer. However, suppose we dene a
khomogeneous
partial latin square
of order
n to be an n × n array
in whi h
ea h ell is either blank or lled (the latter meaning that it ontains one of
{1, 2, . . . , n}),
and whi h has the properties that (i) no symbol o
urs twi e
within any row or olumn, (ii) ea h symbol o
urs
(iii) ea h row and olumn ontains exa tly
k
k
times in the array,
lled ells.
(The standard
denition of a homogeneous partial latin square is slightly more general. However, on e empty rows and olumns have been deleted, it agrees with ours.) We an then sensibly ask whether this square is a
kplex.
khomogeneous
If it is then we say the partial latin square is
partial latin
ompletable
be ause the blank entries an be lled in to produ e a latin square.
If 1 < k < n and k > 41 n then there exists a khomogeneous partial latin square of order n whi h is not ompletable.
Theorem 17.
Burton [7℄, and Daykin and Häggkvist [14℄ independently onje ture that if
k ≤ 41 n
kplex is ompletable. It seems ertain that for k n, every kplex is ompletable. This has already n ≡ 0 mod 16 in [14℄. The following partial extension
then every
su iently small relative to been proved when
result due to Burton [7℄ also seems relevant.
Transversals in Latin Squares
17
For k ≤ 14 n every khomogeneous partial latin square of order n is ontained in a (k + 1)homogeneous partial latin square of order n. Theorem 18.
6. Covering radii for sets of permutations A novel approa h to Conje ture
1 and Conje ture 2 has re ently been opened
up by Andre Kézdy and Hunter Snevily.
To explain this interesting new
approa h, we need to introdu e some terminology. Consider the
ming distan e.
symmetri group Sn as a metri spa e equipped with Ham
That is, the distan e between two permutations
g, h ∈ Sn
is
the number of points at whi h they disagree (n minus the number of xed
gh−1 ). Let P be a subset of Sn . The overing radius cr(P ) of P is the smallest r su h that the balls of radius r with entres at the elements of P over the whole of Sn . In other words every permutation is within distan e r of some member of P , and r is hosen to be minimal with this points of
property.
Let P ⊆ Sn be a set of permutations. If P  ≤ n/2, then However, there exists P with P  = ⌊n/2⌋ + 1 and cr(P ) < n.
Theorem 19.
cr(P ) = n.
s, what is the smallest m su h that there is a set S of permutations with S = m and cr(S) ≤ n − s? We let f (n, s) denote this minimum value m. This problem
an also be interpreted in graphtheoreti language. Dene the graph Gn,s on the vertex set Sn , with two permutations being adja ent if they agree in at least s pla es. Now the size of the smallest dominating set in Gn,s is f (n, s). Theorem 19 shows that f (n, 1) = ⌊n/2⌋ + 1. Sin e any two distin t permutations have distan e at least 2, we see that f (n, n−1) = n! for n ≥ 2. Moreover, f (n, s) is a monotoni in reasing fun tion of s (by denition). The next ase to onsider is f (n, 2). Kézdy and Snevily made the folThis result raises an obvious question.
Given
n
and
lowing onje ture in unpublished notes.
Conje ture 7.
If n is even, then f (n, 2) = n; if n is odd, then f (n, 2) > n.
The KézdySnevily onje ture has several onne tions with transversals. The rows of a latin square of order
n
form a
sharply transitive set
permutations (that is, exa tly one permutation arries
j );
i
to
j,
for any
i
and every sharply transitive set is the set of rows of a latin square.
of
and
I. M. Wanless
18
Let S be a sharply transitive subset of Sn . Then S has
overing radius at most n − 1, with equality if and only if the orresponding latin square has a transversal. Corollary 2. If there exists a latin square of order n with no transversal, then f (n, 2) ≤ n. In parti ular, this holds for n even. Theorem 20.
Hen e Conje ture
7
implies Conje ture
1,
as Kézdy and Snevily ob
served. In fa t a stronger result holds:
Theorem 21. If S is the set of rows of a latin square L of order n with no transversal, then S has overing radius n − 2. The following result is due to Kézdy and Snevily. See [8℄ for a proof.
Theorem 22.
Conje ture 7 implies Conje ture 2.
In other words, to solve the longstanding Ryser and Brualdi onje tures it may su e to answer this: How small an we make a subset has the property that every permutation in of
S
Sn
S ⊂ Sn whi h
agrees with some member
in at least two pla es?
In Corollary 2 we used latin squares to nd an upper bound for when
n
is even. For odd
n
f (n, 2)
we an also nd upper bounds based on latin
squares. The idea is to hoose a latin square with few transversals, or whose transversals have a parti ular stru ture, and add a small set of permutations meeting ea h transversal twi e. For
n = 5, 7, 9,
we now give a latin square
for whi h a single extra permutation su es, showing that in these ases.
1 2 3 4 5 1
2 1 5 3 4 3
3 4 1 5 2 4
4 5 2 1 3 2
5 3 4 2 1 5
1 2 3 4 5 6 7 3
2 3 1 5 4 7 6 2
3 1 2 6 7 4 5 1
4 5 6 7 1 2 3 7
5 4 7 1 6 3 2 6
6 7 4 2 3 5 1 5
7 6 5 3 2 1 4 4
1 2 3 4 5 6 7 8 9 5
3 1 2 6 4 5 9 7 8 4
2 3 1 5 6 4 8 9 7 6
4 5 7 9 8 2 1 3 6 1
f (n, 2) ≤ n + 1 6 4 9 8 7 1 3 2 5 3
5 6 8 7 9 3 2 1 4 2
7 8 4 1 3 9 5 6 2 9
9 7 6 3 2 8 4 5 1 8
8 9 5 2 1 7 6 4 3 7
In general, we have the following:
Theorem 23. f (n, 2) ≤ 43 n + O(1)
for all n.
The reader is en ouraged to seek out [8℄ and the survey by Quistor [44℄ for more information on overing radii for sets of permutations.
Transversals in Latin Squares
19
7. Con luding Remarks We have only been able to give the briefest of overviews of the fas inating subje t of transversals in this survey.
Spa e onstraints have for ed the
omission of mu h worthy material, in luding proofs of the theorems quoted. However, even this brief skim a ross the surfa e has shown that many basi questions remain unanswered and mu h work remains to be done.
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S hool of Mathemati al S ien es Monash University Vi 3800, Australia email: ian.wanlesss i.monash.edu.au