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Table of contents :
Contents
1 Chvétal’sConjecture................L ....................... 1
1.1 Chvétal’s Conjecture .................................... 2
1.2 Latent Subsets of Intersecting Families .......................... 8
2 Intersecting Families with Restricted Intersection Values ............... 13
2.1 Introduction ......................................... 14
2.2 The Case n 5 2k + 3 .................................... 16
2.3 The Case When 1: Is Sufficiently Large .......................... 17
2.4 On Possible Counterexamples ............................... 22
2.5 Conclusions and Further Problems ............................ 25
e
3 Cyclically Invariant Matchings of the Middle Level of the Boolean Lattice . . . 27
3.1 Introduction ......................................... 28
3.2 Lexical and Modular Matchings ........... . ................. 28
3.3 Distribution Vectors .................................... 40
3.4 Orbits and Uniqueness of iModular Matchings ..................... 42
3.5 Tweaking.......................... ................ 50
3.6 Labeling Zeros in Cyclic 01 Arrangements . . .. ..... i ............... 52
3.7 Odd Graph Matchings ................................... 54
3.8 aInvariant Hamiltonian Cycles .............................. 55
3.9 A Strong Form of the Cycle Lemma ........................... 58
4 Graph Decompositions and alabelings ............................ 63
4.1 Introduction ......................................... 64
4.2 On Prisms D,’; and Cm1 x Cm, x    x Cm, ....................... 67
4.3 Decompositions into Nonisomorphic Trees ........................ 73
4.4 alabelings ......................................... 77
Bibliography ............................................... 88
Vita ..................................................... 93
Combinatorics of finite sets Snevily, Hunter Saint Clair ProQuest Dissertations and Theses; 1991; ProQuest
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Order Number 9210996
Combinatorics of ﬁnite sets
Snevily, Hunter Saint Clair, Ph.D. University of Illinois at UrbmaChampaign, 1991
UMI
300 N. Zeeb Rd. Ann Arbor, MI 48106
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COMBINATORICS OF FINITE SETS
BY HUNTER SAINT CLAIR SNEVILY
B.S., Emory University, 1981 M.S., University of Illinois, 1988
THESIS Submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate College of the University of Illinois at UrbanaChampaign, 1991
Urbana, Illinois
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UNIVERSITY OF ILLINOIS AT URBANACHAMPAIGN
THE GRADUATE COLLEGE
JULY, 1991
WE HEREBY RECOMMEND THAT THE THESIS BY
HUNTER SAINT CLAIR SNEVILY COMBINATORICS OF FINITE SETS ENTITLED
BE ACCEPTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
M
g
é a
Director of Thesis Research
Head of Department
Committee on Final Examinationi‘
ﬁTXAC¢IZLebt.OA
/ 4.5k3 + 7.5%:2 + 316 + 1. We prove necessary conditions for possible counterexamples to Conjecture 2 when n is sufficiently large.
In Chapter 3, we consider structural aspects of 2["] for n = 2k + 1 that are motivated by the
“revolving door conjecture” of Erdtis. Let B(k) denote the bipartite graph whose vertices are the k and k+ 1 sets of [2k + 1], with edges speciﬁed by the inclusion relationship. Erdiis conjectured that B(k) contains a Hamitonian cycle. Any such cycle must be composed of two matchings between
the middle levels of the Boolean lattice. In Chapter 3, we study such matchings that are invariant under cyclic permutations of the ground set. We show how to construct the lexical matchings
of Kierstead and Trotter in a simple manner. We then construct a new class of matchings called modular matchings and show that these are. nonisomorphic to the lexical matchings. We describe the orbits of the modular matchings under automorphisms of B(k), and we also construct an example of a matching that is neither lexical nor modular. We then extend the concept of lexical and modular
matchings to other levels (not all levels) of the Boolean lattice. In sections 3.8 and 3.9, we discuss cyclically invariant Hamiltonian cycles, offer a conjecture concerning their structure, and give a iv
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generalization of the Cycle Lemma. In Chapter 4, we consider the problem of isomorphic graph decomposition, in which a graph G is to be decomposed into edgedisjoint copies of some smaller graph H. More precisely, consider the
following graphs. An (n + 2)dimensional prism D3, is the Cartesian product K2 x K2 x    x Cm, where K2 is taken n times and Cm is a cycle with m vertices. We prove that if T is any tree having
12 + 2 edges (11 Z 0), then the graph D3", can be decomposed into edgedisjoint copies of T. In the last section of Chapter 4, we generalize some results about special vertex labelings that, by a theorem of Rosa, yield decompositions of the complete graph Kn into isomorphic copies of certain speciﬁed graphs.
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Dedication To Harriet and Maddie
vi
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Acknowledgment
I would like to thank André Kézdy, Daniel Lichtblau, Matthew Markert, Steve Penrice, and
Bruce Reznick for their helpful comments and encouragement.
Special thanks go to Douglas West my advisor. I constantly hounded him to listen to me expound the virtues of my newest proof; sometimes it was correct and somtimes wasn’t, but he was
always available and managed to offer me sensible advice whether I was right or wrong. I would also like to thank the Department of Mathematics, the Ofﬁce of Naval Research
(N0001485k0570), and the National Security Agency (MDA90490H4011) for their continued ﬁnancial support and the faculty for their willingness to answer any questions I had. Finally I would like to thank my wife for her constant encouragement, support, and love. Being
the spouse of a graduate student is never an easy task.
vii
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Contents
1
Chvétal’sConjecture................L .......................
1
1.1
Chvétal’s Conjecture ....................................
2
1.2
Latent Subsets of Intersecting Families ..........................
8
2 Intersecting Families with Restricted Intersection Values ............... 13 2.1
Introduction .........................................
14
2.2
The Case n 5 2k + 3 ....................................
16
2.3 The Case When 1: Is Sufﬁciently Large .......................... 17 2.4
On Possible Counterexamples ............................... 22
2.5
Conclusions and Further Problems ............................
25
e
3
Cyclically Invariant Matchings of the Middle Level of the Boolean Lattice . . . 27 3.1
Introduction ......................................... viii
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28
3.2
28 Lexical and Modular Matchings ........... . .................
3.3
40 Distribution Vectors ....................................
3.4
. 42 Orbits and Uniqueness of iModular Matchings ....................
3.5
...... Tweaking.......................... ..........
3.6 Labeling Zeros in Cyclic 01 Arrangements
4
50
. . .. ..... i............... 52
3.7
Odd Graph Matchings ...................................
54
3.8
aInvariant Hamiltonian Cycles ..............................
55
3.9
A Strong Form of the Cycle Lemma ...........................
58
................ 63 Graph Decompositions and alabelings ............
4.1
Introduction ......................................... 64
4.2
On Prisms D,’; and Cm1 x Cm, x    x Cm, .......................
67
4.3
Decompositions into Nonisomorphic Trees ........................
73
4.4
alabelings
.........................................
77
Bibliography ............................................... 88 93 Vita .....................................................
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Chapter 1
Chvétal’s Conjecture
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1.1
Chvétal’s Conjecture
t the subset Throughout this chapter, X will denote a ﬁnite set with n elements. We let 2x represen an ideal, or lattice of X, with sets ordered by inclusion. A collection I of subsets of X is called
is called'an hereditary family, if A e I and B C A imply B e I. A collection of subsets G Q I
I is called intersecting subfamily of I if no two‘sets in G are disjoint. An intersecting subfamily S of that a star if there exists an element 1' 6 X such that j e A for all A e S. Chvatal [8] conjectured size if I is an ideal, then the maximum size of an intersecting subfamily of I equals the maximum
of a star in I. Chvatal [8] proved that if the elements of X are given a linear order and I is an ideal
in 2x with the property that {a1,a2, . . .,ak} e I and b, _ 2 bases, and q — l bases form a strong star, then I satisﬁes Chvdtal’s conjecture.
A family of sets F’ is subordinate to a family of sets F if for every A’ e F’ there exists A e F such that A’ g .4.
Theorem 4. (Kleitman and Magnanti [29]) Let F be an intersecting family of subsets of X with the property that each set in F contains at least one of the elements when E X. Then there is an
intersecting family F' subordinate to F that satisﬁes:
(1') IF’I .>. IFI (ii) F’ is a star with 1:1 or $2 in its center.
We are now in the position to prove the following theorem.
Theorem 5. Let I be an ideal in 2M, and let B = {31, . . .,B.,} be the basis elements of I. If B can be partitioned into two 1stars, then I satisﬁes Chvdtal’s conjecture.
Proof: Let 01 and 02 be any partition of B into two 1stars. By Corollary 1, we may assume that their centers are distinct. Without loss of generality, assume that {1} is the center of C1 and
{2} is the center of 02 (we may also assume that n > 2). If either 01 = 1 or Cg = 1, we are done by Theorem 3. Thus we may assume that I05] 2 2. Let I. be the idea] whose bases are the
elements of 0;. Let F be any maximum intersecting subfamily of I. By Theorem 3 we may assume
that IF n Cgl 2 2. Now observe that if D,E e F n I., with D, E distinct elements of 0;, then {i} 6
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is contained in D n E. Thus every element of F contains either 1 or 2, and Theorem 4 completes the proof.El
Example 2. Let I be the ideal (I g 2(201) with the following basis elements: {1,2,10,14,15}, {1,2,6,11,14,19}, {4,10,11,19}, {3,5,6}, {3, 5, 10}, {6,7,8,9,10,12,13,16,17,18,20}. Then the basis elements of I can be partitioned into the following 1stars, with {10} as the center of C; and . {6} as the center of 02:
01 = {{1,2,10,14,15},{4,10,11,19},{3,5,10},{6,7,8,9,10,12,13,16,17,18,20}} 02 = {{1,2,6,11,14,19},{3,5,6}} By Theorem 5 we see that I satisﬁes Chvatal’s conjecture. None of the other theorems we have mentioned apply to this ideal.
Theorem 5 can be used to give a new short proof of the following result of Stein [47]. Theorem 6. (Stein [47]) Let I be an ideal in 2H with basis elements {Bl,...,Bq} such that
IE; n l e {0, 1} when i 96 3'. Then I satisﬁes Chvdtal’s conjecture. Proof: Suppose our claim is false. Let I be a. counterexample of minimal size. If F is a maximumsized intersecting family in I, then B, E F for all i (by the minimality of I), and hence
IBaOBjI=1fori#j.
Deﬁne C. = Uj¢i(Bi n Bj). Choose C = C, for some 2' such that C,~ is minimal. By Theorem 5 we may assume that C 2 3. Assume that C = {a1,a2,. ..,a,,}, and for a. e 0 deﬁne D“, =
{j : a. 6 .35}. Choose D = D“, for some a, so that 2151;“ 2'Bil'lcil is maximal. Note that F 5 21156062q 2Bil‘cil), since in Bj there are 2B,>—0, sets that contain C5, and any 7
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subset of Bj in F must contain all of Cj. Let I’ be the ideal generated by the 35’s such that j E D. Now the star S in I’ with center
{ag}, where D“ = D, satisﬁes
l3I = Z 2'3"“  (IDI  1) :‘en
2 2l0l—1(Z 2IlIGJ‘I) _ (IDI _. 1), .‘iED
But 2c—1 2 C + 1, since C 2 3. Hence
ISI 2 (ICI + 1)(2 213,140,.) — (IDI  1) 1'60
2 2x 2 2ll0jl) + (z 2ll‘ll _ IDI) + 1 “SEC .1.q
1'60
2 F + 1. And this gives us our contradiction. D
1.2
Latent Subsets of Intersecting Families
In this section we will give a new proof of a. fundamental identity concerning latent subsets of an
intersecting family and show how it can be used to generalize an old result of Kleitman. Let F be an intersecting family of subsets of X. A set A C X is called a latent subset of F if there exists
B E F such that A Q B and A ¢ F. Let Fi‘ denote the collection of all latent subsets of F and let
f = IF] and fL = IFLI. Using a result of Marica and Schtinheim [34], we will show that fL 2 f for any intersecting family F. We will prove their theorem using a result of Daykin [9] since this method will be useful for another application. Throughout this section the symbol “ I ” will be
8
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used to mean complementation within X.
Deﬁnition: For any collections of subsets AB of X, deﬁne:
AVB:{E§X:E=AUBforsomeAe.Aa.ndBGB}
AAB={E§X:E=AaorsomeAEAand 3613}
.A’ = {A’ : A e A}
Theorem 7. (Daykin [9]) If .A, B are any two collections of subsets of X, then IAIIBI 5
IA V BIIA A BI.
The following proof appears in [1](p.94). Theorem 8. (Marica and Scht'mheim [34]) Let A — B = {A — B : A 6 A,B E B}. IfA is a
collection of subsets of X, then IA  AI 2 IAI.
Proof: For any two collections A,B of subsets of X we have, by Daykin’s result, IAIIBI =
IAIIB’I S IAVB’IIAAB’I = I(.AVB’)’I.AAB’ = IA’ABIIAVB’I = IBAIIA—BI. Now takeB = .A
to obtain IA — AI 2 IA]. El
Corollary 2. If F is an intersecting family of subsets of X, then fL 2 f.
Proof: F — F Q FL. El
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It is important to note that the latent subset inequality f” 2 f also follows immediately from
Berge’s Theorem. We include the approach above because the technique in the proof of Theorem 8 is also used in our next proof. We say that two collections F1, F2 of subsets of X are mutually intersecting if for any A 6 F1 and any B 6 F2 we have A n B aé 0. We are now in the position to give an easy proof of the following result of Kleitman and Magnanti.
Theorem 9. (Kleitman and Magnanti [29]) Let F1 and F2 be mutually intersecting collections of subsets of X. Then 1314f; 2 f1 f2. Proof: In the proof of Theorem 8, it was shown that F1F2 S F1  F2F2 — Fll. Since
F1 — F2 9 F15 and that F2 — F1 g F214, the inequality follows. E!
In [27], Kleitman showed that if A1, . . ”Ah are disjoint intersecting families of subsets of an
nset X, then UA. S 2”  2""‘. We will generalize Kleitman’s result by proving the following theorem.
Theorem 10. If A1,. . .,Ak are disjoint intersecting families of an ideal I, then UA. 5 I —
I/2".
Proof: Let A = I and for i > 1 let I. = I — U§;11Aj. Since we are trying to maximize A. = U321 Aj for each i, we may assume that each A. is a dual ideal in I, and hence I. is an ideal
for all i. Since each A. is an intersecting family, Corollary 2 implies A. S I.~/2. At each stage, the least that can be omitted from Ag.“ is half of what remains in I;, so greedily, at least II /2" must be omitted overall. D
10
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We can also use the identity fL 2 f to give another proof of Theorem A from the previous section.
Proof of Theorem A. Since .A = LA’ S IALI = (AL)’, it sufﬁces to show that A, A’, AL, and (AL)’ are pairwise disjoint. By deﬁnition .A and .A’ are disjoint, and also A and .4"; hence .A’ and (AL)’ are disjoint. First we will show that A’ and AL are disjoint. Suppose B is in AL and in
A'. Then there exists 0 E .A such that B C 0; hence B n C’ = 0 and C’ e A’. This is impossible since .A’ also satisﬁes the hypothesis of our theorem. Therefore .AL and A’ are disjoint, whence .A
and (AL)’ are also disjoint. It remains to show that AL and (AL)’ are disjoint. Suppose B is in
.AL and in (AL)’. Then there exists C e .A such that B C C and also D E .A such that B’ (_Z D. But then C U D = X, which is impossible. CI
The identity fL 2 f can also be used to give a new proof of a more general result concerning
two intersecting families. Theorem A follows from this when we set 8 = A.
Theorem 11. (Hilton [25]) Let A and B be intersecting families of X = {1,2,...,n} such that AEA and 368 implyAUByéX. ThenA+B52"‘1.
Proof: Let C = A08, A0 = A—6, and 30 = 86. Let A“ = .AonCL, and let 3" = Bo 06’".
Without loss of generality assume that lA‘I _>_ B*. Since C U .A" is an intersecting family, we have
ICL — A‘l Z C + IA‘I, which implies CL 2 C + IA‘I + B". Clearly, A0, A5, Ba, 85, C, C' are pairwise disjoint. From our previous proof, we know that C, 6’, CL, and (CL)’ are pairwise disjoint. It is also easy to see that A0, 30, and (CL)’ are pairwise disjoint. Now (CL)’ = CL 2
11
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ICI + I(.A")’ + (B")’I. Let ’D’ = (CL)’ — ((A‘)’U (B‘)’), so that I'D’I 2 ICI. Since C,C’,CL,(CL)’ are pairwise disjoint, D and D’ are disjoint from C and C’. By construction, ’D’ is disjoint from A0 and
Bo, and 'D is disjoint from A5 and 36. Suppose A 6 A3 n 'D’. Then A’ 6 A0 n 6" = A“ , since
’D’ C (CL)’, and this gives us a contradiction. Therefore, 'D’ is disjoint from A5, and by symmetry, also from 83. Thus 'D, 'D’, A0, A3, Bo, 86,6, and C’ are all pairwise disjoint. We are done since
A, B and 'D are disjoint from 8 = D’ U (Ao)’ U (Bo)’ U6’ and IS] 2 [Al + IBI. CI
12
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Chapter 2
Intersecting Families with Restricted Intersection Values
13
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2.1
Introduction
In this chapter, we focus on the following conjecture, due originally to Frankl and Fiiredi [18].
Throughout this chapter, we will always assume that X is a ﬁnite set with IX  = n and that tc is deﬁned as below.
Conjecture 1. Let .A be a collection of subsets of X such that 1 S A n BI S k for all 11,3 6 A. Then lAI S tn'k, where
tn=C3ﬁ+c;j+m+C;ﬁ_ We prove this conjecture for n S 21: + 3 and n > 4.51:3 + 7.51s:2 + 31: + 1. The case k = 1 is a
wellknown theorem of deBruijn and Erdtis [10]. It can be shown that if .7: is an intersecting family of X such that maxFe; F S k + 1, then [.77I S tmk. For example, see Theorem 1 of Chapter 1 here or Theorem 2 of [24]. To form an intersecting family of size M: with intersection sizes bounded by k, just consider all sets of size
k + 1 or less in X containing a. ﬁxed element 1‘. Thus Conjecture 1 asserts that if we restrict the size of intersections in an intersecting family to k or less, than we need only look at sets of size k + 1 or less to ﬁnd a maximumsized intersecting family. However, it is interesting to note that when arbitrary set sizes are allowed, there is another family satisfying the same conditions; this is
obtained by replacing the set {at} with the set X — {z}. A stronger version of Conjecture 1, due to the present author, is the following.
14
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Conjecture 2. Let S = {11, . . .,lk} be a collection of positive integers such that l1 < l2 <    < lk.
IfA be a collection of subsets ofX such that [A n BI E S for all A,B E A, then MI 5 tmk.
We show that any counterexample to Conjecture 2 (subject to the restrictions Ink 2 2) has no set of size _ r)_< (17:23“) If 1' = k + 3 in (2) and Mel 2 k + 3, then the coefﬁcient of ("L“) is less than 1. Hence by (1) we may assume that A(k + 1) U A(k + 2)] > 0. Therefore, Ao may be chosen from .A(S k + 2). By
substituting into (2), for arbitrary r we have
(3) A(>r)l< (—_)(“; 1)
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In particular, for r = k + 3 we have
(4) IA(2k+3)I s EEC?) Now let 9 = A(k + 1) U .A(Ic + 2). We ﬁrst consider the case where ng = (0 (Le. no element of X is contained in every element of g) and no two sets in .A(< k + 3) intersect in exactly one element. By the FW Theorem, we have
kl
(5) Mk 1: +3): 5 :3 (n). Hence IAI 5 tn+1,k_1 + 154330?) It sufﬁces to show that the inequality below is satisﬁed when
n > 2192. (6)
2
—1
tn+1,k—1 + m (n k ) S tc
Subtracting tn_k_1 + #9?) from both sides in (6), we obtain (7)
k—l
n—l
tn.k2Sm( k )
The left side of (7) is less than (k — 1) 22;). Thus (7) is implied by (k — 1) :3 S 715%v which is true when n > 2,672, as is easily veriﬁed (actually, n need only exceed k3” + k + Isl/2 for this).
Thus for n > 214:2 and my = 0, we may assume that there are at least two sets in .A(< k + 3), say C; and Cg, such that 01 n 02 = {p}. Furthermore, ng = 0 guarantees some 03 E g such that
p ¢ 03. Let Q = 01 U 02 U 03; note that Q 5 316+ 3. If A G A(2 16+ 1), then IQ n AI 2 2. Since
each set in A(2 1: + 1) must contain a (k + 1)subset having at least two elements of Q, and these
(k + 1)sets are all distinct, we conclude (even with overcounting) that A(2 k + 1)] S (3";3) (2:? .
An easy computation shows that this is less than (mil) when n > 4.51:3 + 7.51:2 + 3k + 1. 19
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For the remainder of the proof, we consider the case n0 ;6 0 and argue as Frankl and Furedi
did in [18]. For the sake of completeness we present the details, most of which are omitted in [18].
Choose 1) 6 ng. Let A1 be a minimal set in .A(2 1:: + 3) not containing p, and let 1‘ = A1 (note
that r 2 k + 3). Then
{AeA(2k+1)=peA}s(";1)—(n';_1). because each such set must have a distinct (k+ 1)subset containing 11, but the (k + 1)sets avoiding
A1 are forbidden. Using (1) for the lower bound, and for the upper bound using (3) and the fact
that the choice of A1 implies HA 6 A(Z k + 1) : p ¢ A} = HA 6 A(2 7‘) :1) ¢ A}, we have k 2 n—1 ,( k ) (s)(n — kr — 1 ) n — k — 3 until later. For this range of r, we clear fractions in (8) so the left side is a
function of r and the right side is ﬁxed, which we express as
(9) f(r)= ('gl) (11—2—1) tc requires A(S k + 1) = tn,k_1. However, the
ErdiisKoRado Theorem (see [1] for various proofs) implies that this maximal size can be achieved only by the intersecting family consisting of all sets of size at most k containing a ﬁxed element 3:.
Since the singleton set {2:} must also intersect each set in B, we have a: = p. But now {2:} n A1 = 0), and the contradiction completes the proof. D
Note that if in the last paragraph of the proof we omit the set {11} and set A1 = X — {p}, then
we obtain the second extremal family mentioned earlier. 21
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On Possible Counterexamples
2.4
In this section we show that if 11,]: 2 2 and n is sufﬁciently large, then a counterexample to
Conjecture 2 cannot contain sets of size 5 lg + 2. Daniel Lichtblau [31] proved the special case
of Theorem 14 below that is obtained by setting 5' = {2,...,k}. The following straightforward numerical lemma is required for our results.
Lemma 1. Let S = {11,. . .,lk} be a collection of k (k 2 2) positive integers such that 11 < lg
(4(lk+21)
5H
1)
2.
(lk'l'l)
5+1!
“—11‘2
n
k. Inequalities (4) and (5) follow from using (6) to bound the left side by a geometric sum. El
Theorem 14. Given lc 2 2, let S = {l1,.. .,lk} be a set of positive integers such that 2 S (1
tmk. We claim that .A contains fewer than 4(lk + 1)5+'1 subsets of size at least W. Otherwise, we
may choose Q Q {A E A i Mi 2 W} such that IQ] = 4(lk+ 1)5+'1. Each set in Q intersects any other set in Q in at most I}, elements. Thus the sets in Q contain at least IQIW —(3)1,,
distinct elements, which by formula (1) of Lemma 1 is larger than n. Deﬁne A" = .A  {A 6 .A : IA] 2 W}. If two or more elements were common to every
set in .A“, then the FW Theorem would imply
um(";2)+(;;§)+...+(n32), which with the remark above and formula (1) of Lemma would yield
MI s M‘l + 40:. + 05+“ < t..“ +’} < w . Recall that AD is a minimal set sharing at least ll elements with every member of .A. If all sets in .A“ intersecting A0 in exactly 11 elements have the same intersection S with A0, then we
23
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can discard an element of A0 outside S to obtain a smaller set having at least (1 elements in
common with every set in .A", because the preceding paragraph guarantees that the sets in A“ do
not all contain S. Continuing, we obtain a minimal set A1 (_2 A0 intersecting every set in A“ in at
least 11 elements. The minimality guarantees 3,0 6 A“ such that IE n All = 11 = IC n AI and
B 0.41 95 CnAl. Let B n A1: T= {21,22,” men}, and choose 3/ e C n A1 such that y ¢T.
Suppose D E A“ and 2:1 ¢ D. Then we can select 11 elements from D n A; and one element from D n (B — T). Together these elements form an (11 + 1)set; call it R. Let ‘R. = {A : A 6 .A*,R Q A,x1 ¢ A}. Since the pairwise intersection sizes of sets in ‘R, belong to {12,. . .,lk}, the
FW Theorem implies that
M >+< )++< o >11—112
n11—2
n—11—2
There are at most (“‘11") [BI sets of size (1 + 1 that can play the role of 72. above. Hence there are
(“vw 2>+ +)
at most
sets in .A" that fail to contain $1. By our choice of n, A1 and B, this quantity is bounded above by
n—I1—nn—l2
n
1
(1"+1)l4(zk+1)4+h(2( k—12))—2(—l—__k+1)4(nk—11 ) The same bound holds for (82, $3, . . ”an, and y. Thus at most n—ll—2
n
n—11—2
("+1)2(1+1)4( k—l )52(zk+1)3( k—l ) sets in .A‘ fail to contain the set T U {3/}, which implies that at least ___n__
5+,
_
lAI
4(lk+1)
‘
n—l1_l
2(lk+1)3(
k— 1 2)>AI'%
lk(
n—ll—l
k
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l
>2tmk
sets in .A contain T U {y}, where we have used formulas (1) and (3) of Lemma 1 and the deﬁnition of tc
By the FW Theorem, we know that at most
n11—1 n—11—1 0 ) (n—11—1 (k_1)+(k_2)+...+
sets in A contain T U {y}. This is bounded by 2("2‘11'1) < if”; '1) < %t,.,k, by our choice of n. Hence there can be no such family. CI
2.5
Conclusions and Further Problems
If we allow k = f(n), then we can obtain results such as the following, which gives us a better
estimate of [AI than the FW Theorem.
Proposition 1. Fix It 2 n35, and let .A be a collection of subsets of X such that l 5 IA n BI S k
for all 11,3 6 A. Then .A _ W: + (2:1). Let Q = {A e A : [AI 2 k + 2}, and choose A,B 6 Q. Then all the subsets of size k + 1 in A are distinct from all the subsets of size 1:: + 1 in B. By
Theorem 1 in chapter 1 (or by the ErdiisKoRado Theorem) we see that IQI > (2:). Hence Q generates at least (k + 2)(2::) distinct subsets of size k + 1, and this is bounded by S (1:11)‘ For k 2 11%, we obtain a contradiction. El Frankl and Wilson in [19] presented an elegant argument using linear algebra in proving their result. It is our hope that linear algebra can also be used to resolve Conjectures 1 and 2.
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Given the preceding theorem, an afﬁrmative answer to the following problem would prove
Conjecture 2 for sufﬁciently large n. The question is also interesting in its own right, since it would
also generalize a theorem of RayChaudhuri and Wilson.
Problem 1. Let S = {l1, . . .,lk} be a collection of k 2 2 nonnegative integers (again we assume I; < l2 <    < lk). Let A be a collection of subsets of X, all having more than 1,, elements, such
that M n BI E S for all A,B e .A. Then is it true that MI 5 (2)?
We already know that a minimal counterexample to Conjecture 2 contains only sets larger . than lk when n is sufﬁciently large. Problem 1 asks whether any such collection has size bounded
by (2) = (";1) + (2:1) 5 in,» The theorem of RayChaudhuri and Wilson given in [40] concerns the special case where the sets all have the same size.
Theorem 15. Let .A be a collection of psubsets of X such that A n Bl 6 {ll,...,lk} for all
A,B E .A, wherep) lk >
>11. Then MI 5 (2).
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Chapter 3
Cyclically Invariant Matchings of the Middle Level of the Boolean Lattice
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Introduction
3.1
Let [n] = {1,2, . . . , n}. Following the notation of Kierstead and Trotter in [26], we let B(lc) denote the bipartite graph whose vertices consist of the kelement subsets and k + lelement subsets of
[n], with adjacency deﬁned by inclusion. The set of all kelement subsets of [n] is denoted by ([2]);
similarly (£111) denotes the (k + 1)subsets. We will think of matchings in B(k) as bijections from
(I?) to (£111) or as collections of edges in B(k), whichever is more convenient. Thus if M is a matching in B(k) and S E ([2]), then (S, M(5)) is an edge in M. Let a be the cyclic permutation
(1,2, .. .,n) of [n], acting on the sets of [n] in the natural way. A lfactor M of B(k) is called a
invariant if every edge (5, M(5)) is mapped to another edge in M by 0' (Le. (0(3), a(M(S))) 6 M). In this chapter we are primarily concerned with ainvariant 1factors of B(k) and their relationship to the following wellknown open question, attributed to Erdiis: Is 8( k) hdmiltonian ? Clearly, any
hamiltonian cycle in B(k) will consist of two complete matchings. In the next section, we describe two explicit classes of ainvariant matchings in B(lc); the
“lexical matchings” introduced by Kierstead and Trotter in [26], and a new class we call the “modular matchings”.
3.2
Lexical and Modular Matchings
Since lexical and modular matchings are ainvariant 1factors, we will digress for a moment and
discuss general ainvariant lfactors in more detail. Throughout this chapter, .7‘1, denotes the set of all cyclic arrangements of k 1’s and k+ 1 0’8. Similarly, 6;, denotes the set of all cyclic arrangements
of k 0’s and k + 1 1’3. Given 1) in 1"], (or in 91,) we will let 5 denote a ﬁxed linear representative of 28
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v, with the understanding that if we allow the nth position (last) of 5 to precede the lat we get 1).
Note that 5 is just a 01 sequence, and that 0 acts on 5 in a natural way by permuting positions of 2). Throughout this chapter, we will see that by working with linear repesentatives of v e .7‘1, we
will come to a better understanding of a invariant lfactors of 81,.
The notation '17, will mean the (n — 1)subsequence of 5 that begins at position 1 and ends at position n  1. Normally we will restrict our discussion to elements of 1:1,, with the understanding
that a dual remark applies to elements of 9],. The following is an example of this duality. For n e .7],
(w e 9],), we say that 5 (17)) is in standard position if 5 (17)) has a 0 (1) in the nth position. Given 1: in $1,, any corresponding 5 can be interpreted as the kelement subset of [n] whose elements are
the positions in '13 containing 1’s. We will refer to the linear representatives '1? as 01 sequences or as sets in ([2]), whichever is convenient. As an example of this ﬂexibility, consider the following. Given I
v e .‘Fk, let '13 be any linear representation of 1). Let A; be the set of all sequences 05(5), where i = 1, . . . , n, and let 1;], be a set of 01 sequences that contains exactly one linear representative 5
roreach um ' f.Th A_= ["1. I: enUBefp. u (1,) Given 0 6 .77, and w e 6],, we say that 1: ﬁts into m if there exists linear representatives 17
and 17) (of u and w, repectively) such that 5:17)], meaning that, after a rotation, 11 and 10 agree in all but one position. We are interested in functions I from .77, to Q'k such that 0 ﬁts into f(v) for all u 6 77,. We call such a function f a special function or sfunction. In particular, we are
interested in bijections from f], to Gk that are special (sbijections). We show next that there is a
11 correspondence between (Jrinvariant lfactors of B(k) and sbijections from f], to 91,. Let M be a matching in B(k). A set S 6 ([2]) is an m — vertex of M if M(S) — S = {z}. (This
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term was introduced by Kierstead and Trotter in [26]).
Lemma 2. There is a 11 correspondence between oinvariant Ifactors of B(k) and sbijections from .‘Fk to 0],.
Proof: Suppose M is a ainvariant lfactor of 80:). The ainvariance implies that M is completely determined by its nvertices. Let Q = {51,. . .,Sq} be the collection of all nvertices
of M, and let R = {T. : T. = S. U {n},S.' 6 Q}. Hence, ([2]) = U;e[n]a‘(Q), where 0"(Q) = {ad(SI)! ' ' ' 1 0'15”}.
Note that Ifkl = nl = Q = Iii(2:) is C1,, the kth Catalan number. (This formula is reviewed in more detail in Section 3.9). Applying a' repeatedly to the edges (S;,T.) generates all of
M. Counting arguments thus imply that Q has one linear representative of each u e .75}. Similarly, R has one lnear representative of each w e 91,. Therefore M yields an sbijection fM between .77, and 9],. Conversely, an sbijection f between f], and (J,‘ generates a ainvariant lfactor Mj of
B(k) by repeated application of a. This establishes our 11 correspondence. EJ Notation. Given '0 6 .77," the labels 010,011, . . ”on. will be reserved for labeling distinct 0’s of
1). Similarly, given in E 9)., the labels [30,ﬁ1, . . .,p,, will be used for the 1’s of m. We will use the
notation 5a.. to denote the linear representative of v in which the 0 labeled a. is in the nth position
(last). A sequence (2ktuple) of k 0’s and k 1’s is called a ballot sequence if for any initial subsequence, there are always at least as many 0’s as 1’s. A well known and frequently rediscovered fact is the
following lemma. 30
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I
Lemma 3. CYCLE LEMMA Given v 6 17,, there exists a unique 0, labeled ag, in v such that 50“.
is a ballot sequence.
Before generalizing the Cycle Lemma, we must introduce some additional notation. From now
on we will only consider linear representatives 5 of v e 1} that are in standard position (having a 0 in the last position). Let jl, . . . ,jk denote the k positions of 5: a1, . . . , a" that contain a 1. We will call the subsequence a1 mail. a 1interval (of a). The subsequence aj, . . .agk is called a reverse 1interval. We deﬁne 0intervals and reverse 0intervals analogously. Note that reverse intervals do not contain the last position. It is not difﬁcult to see that the following is a direct consequence of Lemma 1 in [26]. Indeed, a further generalization of this appears in Section 3.9, so we postpone
proving this until then.
' Lemma 4. GENERALIZED CYCLE LEMMA (Kierstead, Trotter [26]) Given v 6 17,, for each I
0 _ 1, we determine 1; by scanning positions leftward from «1,3 until we ﬁnd a 1 that has not previously been designated one of 11,. . ”111; call this position It, paired with jg. Such a position must exist, because the 0interva1 a1, . . . ,aj, has at least as many 1’s as 0’s, so these 1’s cannot all have been paired already. By construction, every I encountered during the scan is already
paired with a 0 between a, and a5,. Also, if the scan encounters a 0 in {(151 , . . .,a,,}, then it also encounters the mate of this 0, or else position It would have been selected earlier. Finally, if the
scan encounters any 0 not in {a_,,, . . .,a,,}, then the corresponding Ointerval has more 0’s than
1’s, which means we have already seen more 1’s than 0’s in the scan and found 1;. These three statements imply that am. . .,a,, has a equal number of 1’s and 0’s, which in turn guarantees that al . . .a;, is a 1interval with more 1’s than 0’8.
Since the reverse Ointervals for the 0’s in positions 91,. . ., gk, have at least as many 1’s as 0’s, we can apply the same argument to the reverse of a1, . . . , a;" to obtain k — 2' 1’6 whose reverse
intervals in 5mm. have more 1’s than 0’s. Since these 1’s end 1intervals in Bax—a itself that have at least as many 0’s as 1’s, they are distinct from the 1’s found earlier, and we have proved that
17”“. has exactly 1' 1’s ending 1intervals with more 1’s than 0’s. E!
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It is worth noting that the 1’s constructed in the proof above are in fact the 1’s paired with d to a1, . . . , aj, in the usual “bracketing” where 1’s correspond to left parentheses and 0’s correspon right parentheses.
The next result is the restatement of Lemma 4 with 0’s and 1’s interchanged.
I
Lemma 6. Given w E 91,, for each 0 S i S k there exists a unique 1, labeled ﬁg, such that 1715..
contains exactly 1' 1intervals having more 1’s than 0’8.
By combining Lemmas 5 and 6 (i.e., by replacing the 0 labeled a. in u with a 1 that becomes the 1 labeled ﬂ. in the resulting element of 9],), we obtain a natural sbijection Lg. Example 1
below is an example of such an sbijection for i = 2 and k = 4. Note that this generates two 0;. x n rectangular arrays consisting of 0’s and 1’s. The Ck x n
array with its rows corresponding to elements of f], will be called the nmatrix of L. or simply the matrix of L5. Any row from this matrix is called an ilexical sequence. By summing up the number of 1’s that appear each column (for columns 1 through 21:) of either array, we obtain a vector (with
2]: entries) that we will call the distribution vector of Lg. For Example 1, the distribution vector is (7,7,7,7,7,7,7,7). The deﬁnitions of nmatrix and distribution vector extend to all sbijections. Given the correspondence between sbijections and ainvariant matchings, with the choice between
these terms made only for convenience, we will normally use the same notation for an sbijection and the corresponding ainvariant lfactor. With this convention, L; is the ilexical matching
constructed by Kierstead and Trotter in [26]. In order to discuss modular matchings we will need the following deﬁnition.
33
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ft:
91:
110001010
110001011
110000110
110000111
001110100
001110101
011100010
011100011
100110010
100110011
100101100
100101101
100011100
100011101
011001100
011001101
010111000
010111001
001111000
001111001
101001010
101001011
101000110
101000111
011010010
011010011
010110100
—»
010110101
L2 (’6 = 4)
Example 1.
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Deﬁnition. Let Z; represent the residue classes of the integers modulo 1. Deﬁne a function f;
from 2[”] to Z; by the following rule: for S 6 2M let f1(S) = (EGGS a) mod I.
We claim that for any j e Zk+1 there are exactly Xvi—1.6:) kelement subsets S of [2k] such that
fk+1(5) =jActually, we will prove the following stronger result.
Theorem 16. Suppose n,lc, and l are chosen so that n > 2, k+l=n, and gcd(k,l)=1. Then for any
1' e Z; there are exactly “'71) kelement subsets S of [n — 1] such that f1(S) = j. Proof: All arithmetic will be done modulo 1. Let .7: be the set of all cyclic arrangements
of I 0’s and k 1’s. Then [fl = Hm?) = E; 2). Given 1: E 1", we will write 5 to indicate a linear representative of v, with the understanding that if we allow the nth position (last) of 5 to
precede the ﬁrst we get 1). Note that 5 corresponds to a ksubset of [n]. Let a = (12 . . .n) be the cyclic permutation that acts on '13 by permuting the positions of 5. Let A; be the set of all cyclic
permutations of 5 (Le. .45 = {a(5),az(5), . . . ,a"(5) =5}) and let B; = {ajl(5),aj2(5), . . .,a5l(5 ) : 1 5 1'1 < jz <    < j; S n} be the I linear representatives of v in A; that contain a 0 in the
nth position.
Since Uuef B; = (“I“), it sufﬁces to show that f; :8; A Z; is a bijection. Compute f1(5). Note that f1(a(5)) =
f1(5) + h if there is a 0 in the nth position of 5; otherwise
f((a(75)) = f1(5), since k  1 — (n — 1) _=_ 0 mod I. In general, we have f1(a‘+1(5)) = f1(a"(5)) + k if there is a 0 in the nth position of 0‘5), otherwise fl(0‘+l(5)) = f;(a‘(5)). This implies that
f;(aji+1(1__))) = f1(aj‘(5)) + k = ﬂ(aj1(5))+ ik. Since gcd(lc,l)=1, we see that f; : B; > Z; is a bijection. El
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a1(5)=0110100 f4(0110100) = f4({2,3, 5}) = 2 02(5)=0011010 f4(0011010) = f4({3,4, 6}) = 1 03(5)=0001101 f4(0001101) = f4({4,5, 7}) = o
a4(5)=100011o f4(1000110)= f4({1, 5, 6}) = o a5(6)=0100011 f4(0100011) = f4({2,6, 7}) = 3
a6(5)=1010001 [4(1010001) = f4({1,3, 7}) = 3 07(6)=1101000 f4(1101000) = f4({1,2,4}) = 3 k=3,t=k+1;n=7
Example 2.
An instance of the bijection described in Theorem 16 is given in the example above, where
k=3,l=k+1_,andn=7. A close inspection of the proof of Theorem 16 shows us that we have in fact proved the following
corollary. (Here, when we apply fk+1 to a 01 sequence, we weight the 1’s according to what position
they occur in.)
Corollary 3. Given 1) 6 17,, for each 0 g i 5 Is there exists a unique zero, labeled 01;, such that fk+1(5aa) = 5
Using techniques similar to those given in Theorem 16 we can also prove the following result.
Lemma 7. Given w E 91:; for each 0 5 i 5 I: there exists a unique 1, labeled [35, such that
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fk+1(i”ﬁa) = i'
Let M. be the sbijection that arises naturally by combining Corollary 3 and Lemma 7 (Le. by
replacing the 0 labeled a. with a 1, which becomes the 1 labeled ﬂ. in the resulting element of 91:)
Example 3 below is an example of such an sbijection when i = 2 and k = 4. Note that the distribution vector for this example is (7,7,7,6,8,7,7,7). The modular matchings
 are the ainvariant matchings that arise from the sbijections Mg. We now give a more explicit description of modular matchings. This was done ﬁrst by Ihrig and
Snevily, and later by Duﬁ'us and Kierstead. The following deﬁnitions and results (except for Claim
3) appear in [13]. Given a jsubset S of [n], we let S = {411,412, . . .,a_,}, with the convention that al < (12 <    < a5, while for E: [n] — S = {31,52,. ..,En_,~} we assume that 51>52>   >5,,_,. Also, 25' denotes the sum of the elements in S. For n = 2k + 1 and 0 S i S 1:, deﬁne a map
G, : ([21) u+ (i111) by aim) = Su 6,, where p a (i+ :3 3) mod(k + 1). We will show that G. = Mj, where i+j E 1 mod(k +1).
Claim 1. For i = 0,1, . . .,k, G; is a matching in B(k), and 00,01, . . .,G’k is a lfactorization
of B(k). Proof: The second assertion is an immediate consequence of the ﬁrst and the deﬁnition of G;. To see that G. is a matching, it sufﬁces to exhibit a map F, : (£11) —+ ([2]) such that F. o G.
is the identity. Deﬁne F; by F.(T) = T — {1),}, where T = {b1,...,bk+1} and p E (i + 2T) mod(k + 1). Suppose that for some S 6 ([2]), G;(S) = S U {63}. There are k + 1 — j elements of 5' less than 55, so there are Ej —2 — k +j elements of S less than 71,. By the deﬁnition of Ga,
37
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TI:
91:
111001000
111001001
110110000
110110001
101010010
101010011
110001010
110001011
011100010
011100011
101001100
101001101
100110100
100110101
011010100
011010101
010111000
010111001
010010110
010010111
001100110
001100111
100001110
100001111
001011010
001011011
000111100
r
000111101
M2 (16:4)
Example 3.
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1'5 i+ )3 5 mod(k +1). Computing modulo k+ 1, we have: (35 +3)  (k + 2) 55, +(i +2 S) — 1 E (i + 2(5 U {Ej})) — 1. Hence, F;(G.'(5)) = F,(5U {ll5}) = 5, as 13, is the pth smallest element
of 5 U {Tej}, where p E (2' + 2(5 U {aj})) mod(k + 1). El
Claim 2. For i = 0,1,. .., k, G. is a 0invariant matching, i.e. Gﬁ’ = G's.
Proof: Let 5 e ([2]), write 5 = {01,02,” .,ak}, and suppose 0(5) = {a’1,. . Hat}. Suppose
045) = s u {5,} and G',(0(5)) = 0(5) u {2;}, meaning that j a (i + :3) mod(k + 1) and l E (i + 20(5)) mod(k + 1). We must show that 0(55) = {iii} in every case.
If 2k + 1 e S, then 2 0(5) 5 E 5 mod(k + 1), sol = j mod (I: + 1). Therefore, 0(35) =5} +1
is the jth largest element of 0(3) and 0(Ej) =a_, +1 =df. If 2k + 1 ¢ 5, then 20(5) E (25 — 1) mod(k + 1), so I E (j — 1) mod(k + 1). Ifj > 1, then 0(3j) is the j — lst largest element of 0(5);
thus, 0(Hj) =di. Otherwise j = 1, which means 35: 2k + 1 and 0(5_,) = 1 = {tif}, since I: + 1 E 0 mod (k + 1). El
Claim 3. G. = M}, where i+j = 1 mod(k + 1). Proof: Since G; and M, are 0invariant matchings and have the same number of nvertices, it sufﬁces to show that every nvertex of G. is an nvertex of M5. Let 5 be an nvertex of G., meaning
that n =51 is added to 5 by 0;. From the deﬁnition of 0,, we see that 1 = (i + 25) mod(k + 1), which implies 2: 5 = j mod(k + 1). This is precisely the condition for 5 to be an nvertex of M5. El
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3.3
Distribution Vectors
From now on, we will always think of sbijections as being displayed in tabular form as in Examples
1 and 3, specifying the nvertices. Keeping this in mind, we can now discuss distribution vectors in a meaningful way. The following surprising result was ﬁrst conjectured by the author and later
proven by Mathew Markert [33] for general 2' and independently by the author for the case i = 1. Markert’s original proof ran several pages. Below we give a simpliﬁed proof valid when gcd(i, n) = 1, based on the proof for i = 1.
Theorem 17. (Markert [33]) Let (dhdg, . . .,d2k) be the distribution vector of an sbijectian. Then
d. + d(2,,+1)_. = F176;) for an 1 g is 1:.
Proof: (For the case gcd(i, n) = 1.) Let M be a ainvariant matching. As proved earlier, M is completely determined by its nvertices. First, we will prove that (11 + dn_1 = F116;“). For any 11 in
$1,, the number of blocks of contiguous 1’s equals the number of blocks of contiguous 0’s. Clearly, the same property holds for any 11) e 91,. Let b(v) equal the number of blocks of contiguous 1’s in 12.
Then by complementation we have 205,“ b(v) = Ewen; b(w). In order to preserve this property,
we see that for every nvertex A that contains both 1 and n — 1, there must be an nvertex A’ that contains neither 1 nor n — 1. Using such a correspondence, we have d1 + d,,_1 = ﬁl(zkk). The proof can be extended to general 2' by using a simple observation about powers of a. It
is well known that 0' can be written as the disjoint union of products of cycles. If a' is written in this manner, for 1 5 I S n — 1, then the sequence (n — I, n, 1) appears in one of the cycles. This
observation, together with a detailed look at what 0' does to nvertices, can be used to complete
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the proof for i = I. If gcd(l, n) = 1, the proof can be made precise using an interesting observation about powers
of a. When gcd(l, n) = 1, 0’ is a cycle, and we can write a’ in cyclic form with the further property that n is placed on the right (when written in this way we will say that o" is properly ordered).
For example, if n = 7, then we want to write a3 as (3625147) and not as (1473625). Now take the nmatrix M of M and permute the columns according to the proper ordering given by a’; for the example above, column 3 is moved to column 1, column 1 is moved to column 5, column 7 remains ﬁxed, etc. Because k and n are relatively prime, 0" acts naturally on the elements of .77," meaning
that each element of 7:1: is represented exactly once as a row of the new matrix M’. Since the dual remark applies to the analogous relabeling of the matrix obtained by changing the last column of
M to 1’3, we obtain another ainvariant matching M’ by viewing the columns of M’ under the usual ordering 1,. ..,n. Since columns 1 and n — 1 of M’ are columns I and n — l of M, applying
the ﬁrst paragraph of the proof to M’ yield the desired equality for i = 1. Cl
It is possible that the transformation of the nmatrix used in this proof will yield further conclusions about the possible distribution vectors or about the sets that can appear together as
nvertices of the same ainvariant matching. Such conclusions would use the fact that two I:subsets of [n — 1] that map into the same element of g when the element n is added cannot both be nvertices in an sbijection. We leave this as a subject for future investigation. Meanwhile, we observe Theorem 17 imposes an upper bound on the number of possible distribution vectors for
the sbijections when k is ﬁxed.
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Corollary 4. Forﬁxed I: there are at most (Ck+1)" distinct distribution vectors for the sbijections, where Ck is the kth Catalan number.
We close this section with a structural conjecture that generalizes Theorem 17.
Conjecture 3. Let M be a ainuariant matching, and let M be its matrix. Let R = a1a2 . . .41s
bearowofM. Then R’=a’2k a’2k _ 1 ...a’2a’10isalsoarowofM. (Herea§=1—a. .
This conjecture holds for both lexical and modular matchings. For modular matchings, it
holds because S is an nvertex of G. if and only if 25' has a ﬁxed value (1 — i) modulo k + 1, and indeed 2(R’) = n(n  1)/2 — Ezeﬂm — :c) = 23(R). For modular matchings, it depends on the arguments in the proof of Lemma 5. We know that R is an nvertex of L. if and only if R has exactly i lintervals having more 1’s than 0’s. In the proof of Lemma 5, we pair these 1’s with
the i 0’s that end 0intervals having at least as many 1’s as 0’s. These are the i 0’s whose reverse Ointervals have more 0’s than 1’s. Hence these are the i 1’s in R’ that end 1intervals having more 1’s than 0’8, and R’ is also an nvertex of Lg.
3.4
Orbits and Uniqueness of iModular Matchings
In this section we will show that for k 2 4 no modular matching is equivalent to a lexical matching,
and we will describe the orbits of the modular matchings under the automorphisms of 30:). First we explain the notion of “equivalence” between matchings. Following [26], we generalize the deﬁnitions
of ilexical and imodular matchings. In Section 3.2, we deﬁned ilexical and imodular matchings
with respect to the standard order on [n]. We now extend those deﬁnitions. Let Aut be the group 42
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of automorphisms of B(k) that maps ksets to ksets. Viewing a matching M as a set of edges, we deﬁne M to be ilezical (or imodular) if there exists an automorphism a EAut such that for
every edge (S,T)EM, (0(3), a(T)) E L. (or (a(S),a(T)) 6 M5). If M is viewed as a function from
ksets to k + lsets, then M is ilexical (respectively, imodular) iﬁ‘ M = a“ o L. o a (respectively,
or1 o Mgoa). If M is such a matching, we will use the notation L5” or Mf” , respectively, to represent M. Let E. (M;) denote the set of all ilexical (imodular) matchings formed in this manner. (Note:
in this section, 0: denotes a permutation of [u], not a label on the 0’s of a cyclic arrangement.) In [14] it is shown that Ant is isomorphic to S", the symmetric group on [it], via the map that
sends 1r to a,” where a,(S) = {7r(s) : s E S}, for any subset S of [n]. Thus given L? 6 £5, we can view a as a permutation on [n]. This tells us that L? is completely determined by its nvertices.
In fact, most remarks made about the nvertices of L. also apply to the nvertices of any L?. For
example, Kierstead and Trotter [26] introduced the notion of an zﬁlter of a ainvariant matching, meaning a subset S of [n] that intersects every :cvertex of the matching. If L, has an nﬁlter of size 1', then so must Lf‘, since ainvariance implies that L, has an lﬁlter of size j for any I E [n]. Thus
we will conﬁne ourselves to making remarks about the nvertices of L5, with the understanding that similar remarks apply to the nvertices of any L? 6 [3,. A similar discussion applies to any M? E M. and M5.
In this section, our main task is to show that if Mf’ = Mg, then a is a power of a, or possibly a
power of o‘ composed with p, where p = (1 2]: + 1)(2 2k) . . . (k k + 2). We begin with the following technical result.
Theorem 18. If M,“ = M, and a(n) = t, then one of the following three possibilities holds, except
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when i = k = 2: a) {a(k),a(k+ 1)} = {t — (12+ 1),t— k}, or b) oz(k) = t— (k + 1), or c) a(k+1)=t—k.
Proof: For k = 1, this holds by inspection, so we assume k > 1. In this discussion, unless otherwise stated, all congruences are given modulo 1: + 1. Partition [n  1] into k special pairs,
deﬁned to be (k,k+ 1) and {(p,p+lc+ 1) : 1 S p S k}. If an n—vertex of M; contains exactly one element of the special pair of the form (p, p+ Ic+ 1), then by the congruence deﬁnition of Mg, it has another n—vertex having the other element of the special pair instead. We call this the substitution principle.
Let S" = {1,2, . . . , k + 1}. Because 5“ contains one element from each congruence class mod 16 + 1, there is a unique 1' 6 S“ such that fk+1(S*  {j}) = i. Let S = S“ — {j}; this choice is made so that S is an nvertex of M;. The case 1' e {k,k + 1} is easier, so we deal with that ﬁrst;
it yields possibilities (b) and (c) in the statement of the theorem. Let 3" = n — j. We say that an
nvertex T of M, has Property A if there are k — 1 other nvertices (its “mates”) that each have I: — 1 elements in common with T. Note that there must be a single remaining element, either k
or k + 1, belonging to T and all its mates. By the substitution principle, S has Property A. By the deﬁnition of Mg, any nvertex satisfying Property A must contain 3" and one element of each special pair of the form (p,p + k + 1). Since M, is oinvariant, we obtain the tvertices of Mf‘ by applying a or by applying a‘ to the nvertices of Mg. This maps position 3" to some ﬁxed position.
Hence there is only one position whose pattern of intersections with other tvertices of M; behaves like 3", meaning that it is the ﬁxed special position in satisfying Property A for tvertices. We must
have a""‘(a(j’)) = j’, or a(j’) = j’ +t E t — j mod (2k + 1). 44
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With S deﬁned as above, we are left with the case j ¢ {k,k+ 1}; note that {k,k + 1} g S. In this case, we say that an n—vertex T of M. has Property B if there are k — 2 other nvertices (its
“mates”) that each share k — 1 elements with T. There must be two remaining elements belonging to T and all its mates. By the substitution principle, S has Property B, there being exactly k  2
special pairs other than {k, k + 1} of which S contains one element. Property B requires that T contain a single element from each of at least k — 2 special pairs other than (k, Ic+ 1). If T contains one element from each such special pair, then j < It implies fk+1 (T) 75 fk+1(S) ? i. Hence any
nvertex T other than S that has Property B is disjoint from {k, k + 1} and consists of one special pair and one element from each of k — 2 other special pairs.
As above, applying a‘ or a to nvertices of M. preserves their intersection pattern and turns
them into tvertices. Hence a""(a({k, k + 1})) is a. special pair for some nvertex of M. satisfying Property B (the one arising from S). If we can show that this special pair must be {k, k + 1}, then
we obtain a({k,lc + 1}) = {t + k,t + k + 1} E {t — k — 1,t  k} mod 2k + 1, as desired. Since application of 0"“ o a preserves the intersection sizes for pairs of nvertices (rows of the matrix of
M5), it sufﬁces to show that {k, k + 1} has a different intersection pattern with the nvertices from that of any other special pair.
If any other special pair {1), p’} with p’ = p+ It + 1 intersects an nvertex A exactly once, then the substitution principle yields another nvertex A’ that intersects A n — 1 times and {p, 1”} once. However, this is not true for the pair {k, k+ 1}, so it sufﬁces to show that there exists an nvertex A0 of M, that intersects {k, k+1} exactly once. Ifj 75 k— 1, then let A0 = S—{j+1, k+1}U{j,k+2},
otherwise let A0 = S — {1, k} U {k — 1, k + 2}. This .40 has the desired property, since 2 A0 5 2 S mod(k + 1), except that when i = k = 2 we havej = 1 and S = {2,3} = {k,k + 1}, and there is
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no such nvertex A0. CI The claim of the theorem need not hold when i = h = 2, as when a is the permutation taking 1, 2, 3,4, 5 to 3, 1, 4, 2, 5, respectively. Hence our results about equivalence of matchings will usually
omit this case. By considering the effects of a, we can strengthen Theorem 18, and this stronger
result is what we will need to characterize permutations a that leave M, unchanged.
Theorem 19. If M,“ = Mg, then one of the following three possibilities holds (except when k =
i = 2), with the some choice holding for each s E [n]: a) {a(s — (k + 1)), a(s — k)} = {0(8) — (k + 1),a(s) — k}, or b) a(s — (k + 1)) = (1(3) — (k + 1), or c) a(s — k) = 0(3) — k.
Proof: Theorem 18 is the special case of this statement for s = n. We use a" to obtain the
general case, with the added property that the same alternative holds for each s. The svertices of
M, are obtained from the nvertices of M, by relabeling (or reading) the columns of the nmatrix as s + 1,3 + 2, . . ., n, 1, . . . , s in order; this is equivalent to rotating each row by 3 positions. Since we have not changed the rows, we have not changed their intersections. Hence the svertices have the same patterns of intersection described by Properties A and B. Since these properties for nvertices led us to the partition of columns 1, . . . , n— 1 into special pairs as described in the proof of Theorem
18, the same intersection patterns in the relabeled columns give the same partition, but the special
pairs of columns have been relabeled to (s + k,s + k+ 1) and {(s +1), 3 + 11+ 16+ 1) : 1 5 p S k  1} (addition here and throughout this proof is modulo n).
There is exactly one nvertex whose 1’s appear in the ﬁrst k + 1 columns of the nmatrix. The cases in Theorem 18 are speciﬁed by which column is not included in this set; the missing column
is (a) in the ﬁrst k — 1 columns, or (b) is the kth column, or (c) is the k + 1th column. Given the 46
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choice of ((a), (b), or (c)) that holds for the nvertices, the same choice holds for all 5, because we are using the same (columnrelabeled) matrix. It remains only to translate the conclusions into
the language of general 3. If the missing element is in the “special” special pair (.9 + k,s + k + 1),
let 1" be its mate (this was case (b) or case (c)). For clarity, let t = 0(3). As before, the svertices become the tvertices under a or under a”"+‘, and the uniqueness of j’ in the intersection patterns
implies that a"‘(a(j’)) = 3", or a(j’) = j’ +t — .9. Since 1" = s — k for all s or j’ = s — (k + 1) for all s, we have the desired conclusion for case (b) or (c). Finally, we consider case (a) where the missing element j in Theorem 18 lies in the ﬁrst It — 1 columns.
Using Property B, we ﬁnd that the “special” special pair must be preserved
when a or 0‘" is applied to the svertices, where again we are using t = 0(3). In this case
a"°(‘)(a({s + k,s + k + 1})) = {s + k,s + k + 1} for all s, or {a(s — (k + 1)),a(s — 16)} = {11(3) — (k + 1),a(s)  k}. [:1
Our aim now is to use this to characterize those a such that M,°‘ = Mg. The result was proved by the author and independently by Duffus and Kierstead [12]. We will present their proof, since it is simpler. It sufﬁces to prove the result for G., since we know by Claim 3 that Mj = G. when j E 1 —2'
mod k + 1. Let p be the permutation that reverses a sequence: p = (1 2k + 1)(2 2k). . .(k k + 2).
Lemma 8. G? = Gj where i +j E 1 (mod I: + 1).
Proof: First observe that 211(5) 5 (E S) (mod Ic + 1) for all .5' e ([2]). It sufﬁces to
prove that p(G.(S)) = Gj(p(S)) for S e ([2]). Take p(S) = {s’1,.. .,s;,}, ordered as in the earlier discussion of G.. Let G;(S) = Suﬁ1,} and Gj(p(S)) = p(S)U{.;’,}, where y 5 6+2 S) mod(k+1) 47
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and z E (j + 2p(S)) mod(k + 1). We must show that pa”) =;’,. Computing modulo k + 1, we
get: zEj+Zp(S)Ej—:S E (1—i)ZS E 1—(i+ZS) : 1—y. Since 3,, is the yth largest element of 5', pay) is the yth smallest element of p(§) =p(—S). Since 2 + y E l mod(lc + 1),
we conclude that p('§y) is the zth largest element of p(—S). Hence p(§y) =92. El Theorem 20. If M,“ = M; and k = i = 2 does not hold, then a is a power of a', or a power of a
composed with p.
Proof: All arithmetic is done modulo 2k + 1. Deﬁne t by a(t) = n. We will use Theorem 19 above to prove by induction on 3' that a(t + jk) = n + jlc for all 3', or a(t — jk) = n — jk for all 1', or a(t — jk) = n + jk for all 1'. Since k and n are relatively prime, this speciﬁes a completely.
In the ﬁrst two possibilities mentioned, the resulting permutation is a = 0"“; in the third, it is
a = 0"‘1 op. We ﬁrst consider alternatives (b) and (c) in Theorem 19, where the proofs are quite easy. For alternative (b), we prove a(t + j k) = n + jk, which holds for j = 0 by the deﬁnition of t. For j > 0,
apply Theorem 19 for s = t + (j — 1)k. Since s — (k + 1) = s + k, Theorem 19 and the induction hypothesis yield a(t + jk) = a(t + (j — 1)k) — (k + 1) = n + jk. For alternative (c), the induction step applies Theorem 19 with s = t — (j — 1)]: to obtain a(t  jk) = a(t — (j — 1)k) — k = n — jk.
Finally, alternative (a) in Theorem 19 is {a(s — (k + 1)), a(s — k)} = {0(3) — k, (1(8) — (k + 1)} forall .9. When .9: t, we havetwocases: a(t—k) = n—(k+ 1) = n+k and a(t—k) = n—k. In the ﬁrst case, we prove by induction on 3' that a(t — jk) = n + jk. In the second case, we
prove by induction on j that a(t — jk) = n — jk. We can write these formulas simultaneously as
a(t — jk) = n:i:jk. 48
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For j = 0, the claim holds by the deﬁnition of t. For j = 1, we have a(t — k) = n :i: k in either
case by choice. For j > 1, we use the knowledge of a on t — (j — 1)]: and t — (j — 2)k. Note that
t — (j — 1)k — (k + 1) = t — (j — 2)k. Hence applying Theorem 19 with s = t — (j — 1)]: yields
{0‘0  (J'  2)k),0t(t  116)} = {Ot(t  (J'  1)k)  k,a(t  (1'  1)k)  (k + 1)} = {n i (J'  1)k k,n :i: (j — 1)]: + k} = {n d: (j — 2)k,n :l: jk}. Since we know by the induction hypothesis that a(t _ (j — 2)k) = n :l: (j  2)k, we conclude that a(t :i:jk) = n i jk. E3
Consider the orbit M. = {M3 : a 6 Sn} of Mg. Combining Lemma 8 and Theorem 20, we see
that if 22' aé k+2, then the stabilizer of M. is the cyclic group Zn, and thus IMg = (2k)!; otherwise,
the stabilizer of M. is the dihedral group Dm and thus M. = (2k)!/2, except when k = 2 and i=2.
Our next aim is to show that modular matchings are distinct from lexical matchings. Kierstead and Trotter [26] showed that ,6. = ﬂk—i Together, Lemma 8 and Theorem 20 imply that M. = Mj
when i+j E 1 mod(k+1). Duﬂ'us and Kierstead proved that if i+j i 1 mod(k+1), then M. 76 M5;
the proof appears in [13].
Lemma 9. For k > 4 and any i, or for k = 4 and 2' 96 2, the matching L; contains at least three
nuertices 5051,52 such that I T20 n 51 n 52  = k — 1.
Proof: Let i and k be given. Without loss of generality we may assume that i S k/2.
Now consider the following three 01 sequences: i) '50: 0"“1"0‘+1; ii) 751: 0"".1 101k‘10‘“; iii) 52: 0"";2 1001*‘10‘H. Each of 50, 51, and 172 appears as a row in the nmatrix of Li. D Deﬁnition We say that a modular matching M. is equivalent to the lexical matching Lj if M. E [.j.
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Theorem 21. For k 2 4, no modular matching is equivalent to a lexical matching.
Proof: Let i and k > 4 be given. By the previous lemma, it sufﬁces to show that the matrix of M. fails to contain three rows that have k — 1 1’s in exactly the same columns. This follows from
the fact that the set of integers [2k] contains each congruence class of Zk+1 at most twice. The remaining case I: = 4 must be handled differently. Note that the distribution vector of a a
invariant matching can be viewed as a multiset (change the parentheses to brackets); when viewed
in this manner we call the distribution vector of an sbijection a distribution set. If Mj 6 Ba,
then the distribution sets of Mj and L; must be the same. When k = 4, the distribution sets of Lo,L1 and L2 are, respectively, {0,5, 5, 7, 7,9,9, 14},{5,5,5,7, 7,9,9,9} and {7, 7, 7,7,7,7, 7,7}. Furthermore, the distribution sets for L3 and L4 equal those for L1 and Lo, since £2; = £11,4.
0n the other hand, for k = 4 the distribution sets of M1,M2 and M3 are, respectively, {5,7, 7,7,7,7,7,9}, {6,7,7,7,7,7, 7,8} and {7,7,7,7,7,7,7,7}, and those of Mo and M4 equal
those for M1 and M2, since M. = M, when i+ j E 1 mod(k + 1). This tells us that these lexical
and modular matchings are not equivalent, with the possible exception of L2 and M3. By examining all 56 of the possible 3e1ement subsets of [8], we can show that the L2 has ten 9iilters of size three, but that the M3 has eleven 9ﬁlters of size three. Thus M3 ¢ £2 when k = 4. D
3.5
Tweaking
In this section we will give an example of a ainvariant matching in B(k) for k _>_ 5 that is neither lexical nor modular. The method used to generate this matching is quite simple; we slightly alter
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(tweak) the sbijection Lo. EXAMPLE A. Let 1) E fk be the cyclic 01 arrangement that contains 1—): 0"1"0 as a linear
representative. Let u 6 IF), be the cyclic 01 arrangement that contains 17: 0k'1101k‘10. Let
61: 01*0" and let 5;: 1k‘10k10. Note that 351 is a linear representative of v and 171 is a linear representative of u. Note that both 5 and 17 are Olexical sequences. Let L3 be the sbijection
defined by letting L5(v) = Lo(u), L3(u) = Lo(v), and Lf,(w) = Lo(w) for all 10 76 mu in .7}. There exist cyclic permutations of 17 and "1! so that 5 ﬁts inside Lo(u) and 27 ﬁts inside Lo('v),
namely 31 and 17.1, so L3 is in fact an sbijection. Let in: 010k’11k‘10, 1722: 0210k‘21k'10, and 17,3: 0310k‘31k‘10. These are 0lexical sequences and appear as rows in the nmatrix of L0 and also also in that of L6, by construction. Since these rows have common 1’s in k — 1 locations, the
argument given in Theorem 21 shows that L6 is not equivalent to a modular matching.
The set F = {2, 2k} is an nﬁlter of L3, since every nvector of L0 has a 1 in position 2k and the linear representatives of v and u (51 and 511) that are nvectors of L6 have a 1 in position 2. Lemma 9 in [26] tells us that for i 5 k/2 the size of the smallest nﬁlter of L. is i+ 1. Thus the only possible lexical matchings equivalent to L5 are Lo and L1. Now Lo contains a 0 in its distribution set (position 1) and L3 does not, because L3 contains 51, 171, and 1711 as rows of its
matrix. Therefore LE, 9! ﬁg. 011 the other hand, the value 1 is in the distribution set of L5, arising in the ﬁrst position from El. We will show that 1 is not in the distribution set of L1 when k 2 6. Let 0 S j < 16—2, and con
sider the following 01 sequences. Let 51: 0515+10k'5 1"5'10, and let 5,: 0j1j+10k'“5‘1101"‘5‘10. We see that 5, and 5:3 are 1lexical sequences and together they show us that for 1 5 i 5 21:, d. _>_ 2, where d. is the ith entry in the distribution vector D = (d1,d2, . . . ,dn) of L1. These results imply
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that L5 9! [.1 (when k 2 6). D
Finally, We remark that the tweaked sbijection L5 does in fact satisfy Conjecture 3. Its n
matrix is the same as that of L0, except that the rows 0"1"0 and ok11o1klo are replaced by
01*0" and 1*‘10k10. In Lo, each of these two rows turns into itself under complementation and l'\i
i reversal of the ﬁrst 2k positions. In L6, the two corresponding rows turn into each other under
these operations. Hence L6 satisﬁes Conjecture 3 if and only if Lo satisﬁes Conjecture 3, which it does.
3.6
Labeling Zeros in Cyclic 01 Arrangements
Throughout this section we will always assume that gcd(lc, l)=1, and we will let J7}: denote the set of all cyclic 01 arrangements consisting of 1: 1’5 and l 0’s. In section 2 we saw that the ainvariant
1factorizations of B(k) given by the lexical and modular matchings arise from distinguishing the 0’s of each 1) 6 77. by a particular rule. The method used in Theorem 16 to label the zeros for the modular matchings also applies to the zeros of any 1: 6 11. In this section we will give a method equivalent to that of Kierstead and Trotter for labeling zeros of any vector v e .7], and show how
this method also applies to the zeros of any 1) E 5". This allows us to extend the concept of a lexical matching in a natural way to other consecutive levels of the Boolean lattice, but does not
always guarantee to yield a matching.
We deﬁne v“ to be the 01 sequence consisting of 1: 1’3 followed by 1 0’6. Given 0 e 1)", we label the zeros of v with the labels 010,021,. ..,ou_1. As before, we let 3a,. to denote the linear
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two representative of v in which the zero labeled a. is in the nth position, where n = k + 1. Given two 01 sequences a: and y we say that y is obtained from :1: via a switch if a: and 31 agree in all but consecutive positions, say j and j + 1, and in :1: these positions contain 0 and 1, respectively, but
in y these values have been switched to 1 and 0, respectively. Given 1) e 1'," and 0 _ i this 0interval contains more 0’s than 1’5. We will use this fact, which also follows from the Strong Cycle Lemma proved in section 3.9, to prove our next theorem.
Theorem 23. Using the labels ‘10. . . .,'y;_1 and Theorem 22 above, for each a e .771, label the zeros
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of u so that 9(510) > 9(511) > . . . > 951,“). For ﬁxed 1; e .77)" we claim that 7. = 01,, where a, is the zero guaranteed by Lemma 5 which generates the nvertices for the ilexical matching.
Proof: It sufﬁces to show that g(1_)a..) > gag,+1)' Throughout this proof, the label a. will be used to represent the zero given in Lemma. 5. Let '5“; a1a2 . . .an, and let 1;, . . .,lk_, be the
k — i positions of 5:,“ that end 0intervals containing more 0’s than 1’8. From our remarks above,
we see that the zeros labeled on“, . . . , a], occupy the positions 11, . . . , lk_.. For each 1 S p 5 k — i, the reverse 0interval 01,, mp“, . . . ,agk contains more 1’s than 0’s. Thus for some j > i we have
a1p+1,...,a2k,a,.,a1,...,arp =50”. On the other hand, for j > i, we see that 960,1.) < 9(7),"), because each 0 in position it moved to position 1 adds I: to the value of g and each 1 in position it moved to position 1 subtracts k + 1 from the value of g, and shifting more 1’s than 0’s guarantees that we have subtracted more than we added. D
Using the above labelings and our discussion in Section 2, we may extend the concept of lexical
and modular matchings to other consecutive levels of the Boolean lattice. First, extend the notion
of an sbijection between 1:), and G], to that of an sinjection between .77,“ and fif, and given 1) e f)", use a labeling scheme guaranteed by Theorem 16 or Theorem 22 to replace a speciﬁc zero in v with
a one. Note that this does not always guarantee a matching (i.e. an sinjection) if I at k + 1.
3.7
Odd Graph Matchings
The odd graph 0(k) = 0 is the graph whose vertices are the letsubsets of [2k + 1] and whose edges are pairs of disjoint subsets. This section demonstrates the utility of the modular matchings by
using them to give explicit matchings in 0(k). Let m be a matching in 0. In [14] it is observed
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that m can be lifted to a matching M in B(lc) by setting M(S) = [n] — m(S), for each 5' e ([2]). In this section we go in the other direction: we show that when k is even, say I: = 21', the modular
matching Mg.“ can be lowered to a matching m in 0 by setting m(S) = [n] — Ma+1(S), for each
S e ([2]). This technique was ﬁrst used by Kierstead and Trotter in [26] to show that the lexical matchings L., can be lowered if lc = 21‘; no other lexical or modular matching can be lowered in this manner.
Theorem 24; If k = 21' then m is a matching in the odd graph 0 (k).
Proof: Clearly m(S) n S = 0. We must show that (m o m)(S) = S, for all 5' e ([2]). First, observe that 2?:12' E 1 (mod k + 1) and that 1 — (i+ 1) E i+ 1 (mod k + 1). This tells us that if
S is an nvertex of M544, then so is m(S); i.e. S U m(S) = [2k]. Since Me.” is determined by its nvertices, because it is cyclically invariant, we are done. El
3.8
aInvariant Hamiltonian Cycles
A subgraph H C B(k) is said to be ainvariant if my 6 E(H) implies that a(x)a(y) E E(H). In this section we brieﬂy consider, properties of ainvariant Hamiltonian cycles of 30:). Several of the
observations made here are known to some researchers (most notably Duﬂ'us, Hanlon, and Roth
[14]) but not to others; for the sake of completeness we present them here. Given a ainvariant Hamiltonian cycle 0 of B(Ic), we deﬁne the cycle matrix of C to be the
2(2kz'1) by (2k + 1) 01 matrix determined by listing the binary representation of successive vertices of C' as rows. We may determine C in several ways. The ﬁrst way is to specify the ﬁrst column of the
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cycle matrix; this column is called the “master” column. Our next lemma shows that every column of the cycle matrix is a cyclic shift of the master column. The master column can be described by
listing a sequence of PET (2:) integers that represent the lengths of alternate blocks of ones and zeros. To see this, ﬁrst note that because 30:) is bipartite, the integers describing the master column must
be odd and greater than 1. There must be Ck (Catalan number) blocks of ones, because each block contributes 1 to the difference between the number of (k + 1)sets with a ﬁxed element and the
number of ksets with the same ﬁxed element. The total is, (2:) — (131) = #6:) = 0],. The other way to describe the ainvariant Hamiltonian cycle C is to write the ﬁrst ﬁﬁf) rows of the
cycle matrix; we call this the master block of C. The master block of C will contain ﬁfe?) linear representatives, one for each 1) in J‘k and one for each 11) in 9],.
Lemma 10. Every column of the cycle matrix: of a ainvariant Hamiltonian 0 is a cyclic shift of the master column.
Proof: For '0 E fk, let 5 be a linear representative of a, and let A; = {01(5), . . .,a"(5) =5} be the set of all cyclic permutations of ‘5. Consider all pairs of vertices {2, y} in A; with dist(z, 3/)
minimal. Without loss of generality, we may assume that {5,05(5)} is such a pair. Using the fact that C is isomorphic to Cm, where m = 2(2 , number the vertices of C in a clockwise fashion
from 1 to m, with 5 in position 1, and (75(5) in position I (l < m/2). We will show that the elements of A; are evenly distributed throughout C; i.e. they occur in positions 1, l + F256;)!
1 + 51—10:)“ .., 1 + 3,54,26,13. This is equivalent to showing that each v E .7: and each m E G are represented exactly once in positions 1 through 1311(35) in 0. Since C is ainvariant, repeated application of 01' shows that there is a member of A; in position 1+ i(l — 1), for each 1 S i S n — 1.
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These are equallyspaced members of A; in 0. Since 0"” is the identity and and {17, 01(5)} were a
closest pair of elements from A3, these must be all the members of A; in C. Hence the n elements of A; are equally spaced in C. CI Note that the argument given above tells us that each column of the cycle matrix of C can be
obtained by rotating the master column of C by a distinct integer multiple of 7%, (2f). Therefore, we can relabel the elements to obtain another ainvariant cycle such that the master column is the concatenation of the columns of the ﬁrst master block, simply by placing in the second column the
column that was the ﬁrst shifted up by #6:), and so on. In example 4 below, we present a Hamiltonian cycle for 8(2), its corresponding cycle matrix,
master column (in two ways), and its master block. Note that we can relabel the elements (permute the columns of the cycle matrix) in Example 4 so that the master block appears as below.
11010 11000 11100
10100 If we concatenate the columns of the above master block we get the master column of Example 4. Furthermore, the reverse complement of row 4 gives row 1, and the reverse complement of row
3 gives row 2. In general, if the master column is a palindrome, as in Example 4, and the columns have been permuted so the master column is the concatenation of the columns of the master block,
then the rows of the master block pair up from opposite ends of the block as reverse complements.
We list here examples of master columns for ainvariant Hamiltonian cycles that are palindromes
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fork=3andk=4.
h=3
331577771533
k=4
11537571311355335131353355311137537511
Master columns that are palindromes have also been found for k = 5 and k = 6. Based on this
(admittedly limited) data we make the following conjecture.
Conjecture 4. For each Is there exists a cyclically invariant Hamiltonian cycle C in B(k) such that the master column of C' is a palindrome.
3.9
A Strong Form of the Cycle Lemma
In this section we generalize the Cycle Lemma and apply this to problems involving generalizations
of the Catalan numbers. For the sake of readability some of the deﬁnitions of the earlier sections will be repeated here.
The most wellknown special case of the Cycle Lemma of A. Dvoretsky and T. Motzkin [11] states that in any cyclic arrangement of k 1’s and k + 1 0’5, there is exactly one position such
that every clockwise segment beginning at that position has more 0’s than 1’3. Since this position
must contain a 0, we immediately obtain a formula for the Catalan numbers Ck counting the ballot sequences (binary sequences of length 2k in which every initial sequence has as many 0’s as 1’s). There are (2:) sequences of length 2k + 1 starting with 0. Viewing these cyclically, we see that the
cycle lemma guarantees that exactly one of the rotations that begin with a 0 will have more 0’s
than 1’s in all initial segments, so Ck = FlTeb' 58
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Hamiltonian Cycle
Cycle Matrix
Master Block
Master Column
Master Column
{1,4, 5}
10011
10011
1
7
{1, 4}
10010
10010
1
{1,2,4}
11010
11010
1
{1, 2}
11000
11000
1
{1, 2, 3}
11100
1
{1, 3}
10100
1
{1,3,4}
10110
1
{3, 4}
00110
0
{3, 4, 5}
00111
0
{3, 5}
00101
0
{1, 3, 5}
10101
1
{1, 5}
10001
1
{1, 2, 5}
11001
1
{2, 5}
01001
0
{2, 3, 5}
01101
0
{2,3}
01100
0
{2, 3, 4}
01110
0
{2,4}
01010
0
{2, 4, 5}
01011
0
{4, 5}
00011
0 Example 4.
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3
3
7
Hence the simple form of the Cycle Lemma can be viewed as giving special meaning to one of
the 0’s in a cyclic arrangement of 0’s and 1’s. Kierstead and Trotter [26] generalized this by giving
meaning to each of the 0’s in a cyclic arrangement of k 1’s and k + 1 0’s. They observe that the
result seems to be implicit in the work of Feller [15] and Narayana [36] . The full statement of the original Cycle Lemma asserts that any circular arrangement of k 1’s
and qk + p 0’s has exactly 1) positions such that every clockwise segment starting at that position has more than q times as many 0’s as 1’s (1) 2 0). In this section we extend the lemma of Kierstead and Trotter distinguishing the 0’s of a cyclic arrangement to the more general setting of k 1’s and qk + 1 0’3. We also collect several results on Catalan numbers and their generalizations that follow
readily from this.
Given a cyclic arrangement ao,...,a,._1 of 0’s and 1’s, we denote the interval a544,. ..,a,
(indices modulo n) by any of [i + 1,j], (i,j], [i + 1,j + 1), (i,j + 1). Hence (i,z’] denotes the linearization of a ending at position 2'. Given a ﬁxed linearization, we use 0interval to mean an initial segment ending at a 0 and Iinterval to mean an initial segment ending at a 1. If a; = 0,
then (i, i] is a. 0linearization, and the full sequence is the trivial 0interval. The number of 0’s and 1’s in an interval are the 0count too and Icount 101, respectively. An interval [r,s] is qgood if
wo([r,s]) > qw1(['r,s]), i.e. its Ocount is more than q times its 1count. For q = 1 and p = 1, Kierstead and Trotter [26] proved that for any cyclic arrangement and any choice of i E {0, . . . ,k}, there is a. unique 0linearization of a such that exactly 1' of the nontrivial
Ointervals are lgood, thus distinguishing the 0’s of the cyclic arrangement. In fact, they proved a stronger statement, and we extend this to arbitrary q. The proof is a straightforward extension of
the KiersteadTrotter proof. 60
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Lemma 11. (Strong Cycle Lemma, special case) Suppose a is a cyclic arrangement of k 1’s and
qk + 1 0’8, and choose 2' E {0, . . .,qk}. Then there is a unique 0linearization (jg, jg] of a in which exactly i+ 1 0interuals are qgood. Furthermore, fori < qk, the 0’s that end qgood 0intervals in
(jg, jg] also end qgood 0intervals in (j.+1,j.+1] (we call this the nesting property). Proof: Let a be a cyclic arrangement ao,. . .,a,._1 of qk + 1 0’8 and 1: 1’8. For an arbitrary
interval [r, s] in a, deﬁne the deﬁciency of [r, s] to be 6[r, s] = qw1([r, 8]) — wo([r,s]). For a given 0linearization (r, r] of a, let D(r) = {j : 6(r,j] < 0 and a5 = 0}; this is the set of positions ending qgood 0intervals in (r, 7‘]. Note that r e D(1'). We claim that the sets D(r) are strictly ordered by inclusion.
Let r,s be the positions of two 0’s. Since 60, s] + 6(s, r] = —1, exactly one of (r, s] and (s,r] is qgood. If (r,s] is qgood, then we claim D(s) C D(r). First, note that s e D(r), but 1 ¢ D(s).
Now consider 3' E D(s); we have two cases. Ifj e (r,s], then 6(r,j] = 6(s,j] — 6(s,r] < 0. If
3' E (s,r], then 6(r,j] = 6(s,j] + 6(r,s] < 0. D
As an immediate corollary, we obtain the following probabilistic result of Chung and Feller (this was originally proved using analytic methods).
Corollary 5. (Chung and Feller [7]) In a random sequence of n A’s and n B’s, the probability of
the event that there are exactly k values of i’s such that the ith B precedes the ith A is 1/(n + 1).
The following result is also immediate from the lemma. This is a generalization of a wellknown
result due to G. Raney [39] (he proved the case k = n).
Corollary 6. (Montagh [35]) Let (11,. ..,a,. be a sequence of integers with sum 1. For each 1 5
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has exactly the cyclic shift a;,a.+1, . . . ,am a1, . . . ,a...1 k S n, there is exactly one value i such that 16 positive partial sums.
g West. If a. < 0, then replace a; with 1“il'l'10
This follows from a transformation suggested by Dou
replace a; with 10““. Finally, if a. = 0, then (a.~ + 1 ones followed by a zero). If a. > 0, then 1’s and k + 1 . . ,an is replaced by a 01 sequence with k replace a. with 10. Thus the sequence a1, .
ing 01 representations of the integers and the nest 0’s for some 1:. Now use the n 0’s that end the erty concerning cyclic shifts of a1, . . ., an. property of these 0’s to prove Montagh’s prop
y lt of A. D. Sands. Let A be a set with a kar We will now give a very simple proof of a resu keted. To which a product of n + 1 terms can be brac operation. We want to count the ways in previous where the occurrence of it indicates that the avoid brackets we use reverse Polish notation, and 5 terms we have the following possibilities: 1: terms are to be multiplied. Thus for k = 3
= (abc)de. abcdepp = ab(cde); abcdpeu = a(bcd)e; abcudep h do ents of A represents a product. Those whic Clearly not every sequence of p’s and elem and ences. We see that a sequence involving ru’s correspond to products will be called legal sequ eding (k — 1)r and the number of members of A prec n + 1 members of A is legal if and only if n = can be s the number of It’s preceding this point (this any point in the sequence exceeds k — 1 time . ng Cycle Lemma immediately yields the following proved easily by induction on r). Hence the Stro
involving ru’s and (k — 1)r + 1 members of A, Corollary 7. (Sands [43]) Given any sequence it by cylic permutation is legal. exactly one of the kr + 1 sequences obtained from
pute the product of n + 1 members of A. This tells us that there are 335“?) ways to com
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Chapter 4
Graph Decompositions and
a—labelings
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4.1
Introduction
If G is a graph and L = {G1,G2,. . .,G,,} is a list of graphs, not necessarily distinct, then an [,decomposition of G is a family of edgedisjoint subgraphs H; of G, at least one of which is isomorphic
to G, for each 1', such that G = UH.', where the union of graphs as usual means V(G) = UV(H.) and E(G) = UE(H.). If G has a [.decomposition, then we say that G is .Cdecomposable and write [.G. The most interesting and most frequently studied special case is that where I. is a single graph
H, in which case we write LIG as H IG and call the LIdecomposition an Hdecomposition. We also
say that G is Hdecomposable or that H decomposes G. Given graphs G, H in general, it is very difﬁcult to determine whether G has an Hdecomposition. For example, the special case “Can the complete graph Kn2+n+1 be expressed as the edgedisjoint union of copies of Kn+1?” is equivalent to the wellknown difﬁcult question of whether a projective plane of order n exists. (A projective plane of order n is deﬁned to be a collection of subsets from a
set of n2 +n+ 1 points, each subset having size n+ 1, such that each pair of points appears together in exactly one of the subsets (Deﬁnition 1.1.2 of [2]). This is precisely the property required for the vertex sets of the (n + 1)cliques in a Kn+1decomposition of Kn2+n+1.)
In Section 2 of this chapter, we will consider Hdecompositions of special classes of graphs, where H is an arbitrary tree T of a. given size. In Section 3, we consider Ldecompositions, where
I. is a list of trees of the same size that have a special common subgraph. In Section 4, we consider special labelings of the vertices of a graph II, where the existence of such a labeling guarantees
that H decomposes complete graphs of special sizes. Before continuing, we summarize some of the terminology of graph theory that we will use.
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Additional elementary deﬁnitions can be found in [4] or other texts. All our graphs are simple graphs, meaning there are no loops or multiple edges and the edges are unordered pairs of vertices. The degree of a vertex in a graph is the number of edges incident to it, and a kregular graph is
a graph in which every vertex has degree 1:. A vertexlabeled or simply labeled graph is a graph in
which the vertices are given distinct labels; the label of vertex 'v is usually denoted by f(v). An
edgelabeled graph is a graph in which the edges are given distinct labels. (The use of “labeled” is distinguishable from “weighted”, which, in the literature of graph theory, is usually used when labels need not be distinct.) The girth of a graph is the length of its shortest cycle, with the girth of a graph with no cycles being inﬁnite. The diameter of a graph is the maximum over all vertex
pairs of the length of the shortest path between the two vertices, being inﬁnite if the graph is not connected. A graph G is vertex transitive if given any two vertices any in V(G), there exits an
automorphism 1/) of G such that Ms) = 31. Since a graph is speciﬁed by its vertices and edges, we may write G = (V, E), where V denotes
the vertex set V(G) and E denotes the edge set E(G). A bipartite graph is a graph G whose vertices can be partitioned into two sets X and Y, called the partite sets of G, such that every edge of G consists of an element of X and an element of Y. In order to refer conveniently to the partite sets of a bipartite graph by name, we may write G = (X, Y, E) for such a graph,
following the practice of numerous authors. The cartesian product of graphs G and H, denoted
G x H, is the graph whose vertex set is the cartesian product V(G) x V(H), with edges given
by E(G x H) = {((u,v),(u’,v)) : (u,u’) E E(G)} U {((u,v),(u,v’)) : (v,v’) 6 E(H)}. We will
implicitly use the fact that E(G x H) consists of V(G) copies of E(H) and V(H) copies of E(G).
Cartesian product is associative and commutative for unlabeled graphs. In the next two sections,
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we will study cartesian products of cycles and iterated cartesian products of cycles with edges.
More precisely, we use the following deﬁnition and notation. Deﬁnition 1. Let Cm be a cycle, and let x represent the cartesian product. We deﬁne the
(n + 2)dimensionai prism (over 0...) to be the graph D; = K2 x K2 x  .  x K2 x 0",, where K2 is taken n times.
For n = 0 we have D9,, = 0,... The indexing of the dimensions is chosen so that the ndimensional prism DI,"2 is the graph commonly called the ncube. It is important to note that
Dr" is vertextransitive. Each vertex can be viewed as a binary nvector together with an element of the integers modulo m. The cycle Cm can be rotated and the other coordinates complemented
as needed to produce an isomorphism taking any vertex to any other. We also note that D3, is
(n + 2)regular, has m2" vertices and m2" + m2""’1 edges (n > 0), and is bipartite if m is even.
Our investigation of graph decompositions is motivated by the following two conjectures stated
in [23], and by possible generalizations of them to £decompositions.
Conjecture 5. (Graham and Haggkvist) If G is a 2mregular graph, then G is decomposable by any tree T with m edges.
Conjecture 6. (Haggkvist) If G is an mregular bipartite graph, then G is decomposable by any
tree T with m edges.
In the next section, we prove these conjectures for several special classes of graphs, as indicated below.
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4.2
Graphs
Conjecture(s) Satisﬁed
lxC'mxump
5
All prisms D?" with m even
5,6
All 2m—regular graphs with girth > m
5
On Prisms D3, and Gm1 x s x   x Gmp
Our proofs will be facilitated by constructing a more detailed object than a T—decomposition, which we now introduce.
Deﬁnition 2. Let G be a graph, and let T be a vertexlabeled tree that decomposes G. Let S be the set of (distinct) labels used on the vertices of T, with f(v) being the label on ’0. Let T1, . . . , Tn
be the edgedisjoint subgraphs in some Tdecomposition of G. For each T5, choose an isomorphism
(13. from T; to T; this associates with each vertex v E V(T.) the unique label f;('v) = f(¢.~(v)). The resulting collection of labeled trees T1,. . .,T,,, with labelings f1, ..., f", is called a labeled
Tdecomposition of G. Given a labeled Tdecomposition, we are interested in the list 5., of labels
associated with each vertex v of G. Given 1) E V(G), let 3., = {f.(v) : v e T.}; we call this the multilabel of 1) induced by the labeled Tdecomposition.
We seek labeled decompositions for which the induced multilabels are wellbehaved, as described next.
Property Po. Let T be a labeled tree with label set S. A labeled Tdecomposition of G
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is full if every induced multilabel S” is equal to the label set S of T. A 2mregular graph G is said to satisfy property Po if, for every labeled tree T with m edges, there exists a full labeled
Tdecomposition of G. Property P1. Let T be a labeled tree with label set S, where 5*, S' are the disjoint subsets
of the labels used on the two partite sets of T (i.e., S = 5"" U .5"). A labeled Tdecomposition of a bipartite graph G with partite sets X and Y is bifull if 3., = 5+ for every 1) E X and 5., = S"
for every '0 6 Y. An mregular bipartite graph G = (X, Y, E) is said to satisfy property P1 if, for every labeled tree T with m edges, there exists a bifull labeled Tdecomposition of G.
Figure 1(a) demonstrates that every cycle satisﬁes property Po; we simply arrange the labeled
edges around the cycle in a consistent orientation, so that both labels appear at each vertex. As
Figure 1(b) demonstrates, every even cycle satisﬁes property P1; we arrange isomorphically labeled twoedge paths around an even cycle in a consistent orientation, so that the labels of both endpoints appear at the vertices of one partite set on the cycle, and the label of the midpoint appears at the
vertices of the other partite set. Note that if T is a tree with m edges and G is a 2mregular graph of order n, then a full
labeled Tdecomposition of G must assign a total of (m + 1)n labels to the various vertices, if T has m edges Since there are mn edges in G and m in T, there will be exactly n copies of T, and
hence each label appears at each vertex exactly once. That is, no vertex of G can serve as the same labeled vertex of T for two different copies T. of T in the decomposition. By similar counting
arguments, a bifull labeled decomposition also cannot assign a label more than once at a vertex.
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{1.2}
2 1
.
1
2
(1,2)
12
\2
/2 ‘
.
'
[1,2]
1 {192]
l 2
Figure 1(a). Labeled decomposition (any cycle).
{1,3}
3 1
[2)
/\
31 .
\\ /
131
r21
.
1 3
{1,3}
Figure 1(b). Labeled decomposition (even cycle).
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We verify Conjecture 5 on special classes of graphs by proving that they satisfy property Po;
we verify Conjecture 6 by proving that the graphs satisfy property P1. Our simple technique is induction on m, using for the basis the fact that cycles satisﬁes these properties; all cycles in the
case of P0, even cycles in the case of P1. Our ﬁrst result strengthens and generalizes that of Fink
in [16]; he proved that the ncube satisﬁes Conjecture 5.
Theorem 25. If m is even, then the prism Dr" satisﬁes property P1. In particular, every tree with
n + 2 edges decomposes Dg.
Proof: We proceed by induction on 11. As noted earlier the claim is immediate for n = 0. Now suppose our claim is true for n = k 2 0. Let T be any labeled tree with k +3 edges, with labeling f. Let T’ be the labeled tree with k + 2 edges obtained by deleting from T a vertex :2: of degree 1; let
3/ be the neighbor of a: in T. Let 5+, 5" denote the label sets of the two partite sets in T’; we may
assume f(y) 6 5+. By our inductive hypothesis, Dl‘n satisﬁes property P1. Write D1; = (X, Y, E) to indicate the partite sets of Dl‘n. By Property P1, D5; has a. labeled T’decomposition A such that Su = 5"” for v E X and S” = S“ for v E Y. By the vertex transitivity of Din, it also has a
labeled T’decomposition B such that S” = S" for v 6 X and S” = 5+ for v 6 Y.
The graph Dig“ consists of two copies G1,G2 of D5,, plus an edge joining the two copies of each vertex. To form the desired labeled decomposition for D13“, begin by using A on G1 and B on 02. For each edge uw between 01 and 02, the multilabels contributed to the endpoints by
these decompositions are {34", 5"}; suppose a is the endpoint with label 5+ and w is the endpoint with label 5‘. To the copy T; of T’ such that Mu) = f(y), we add the edge aw. Now we extend
f. so that f.(w) = f(2:); this augmentation of T; is isomorphic to T, because 10 ¢ V(T.). This adds
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f(z) to each 1) such that S” = .5". The vertex transitivity of D5,“ implies that it doesn’t matter which partite set is labeld 5+. Thus We have constructed a bifull labeled Tdecomposition of G,
which establishes property P1 since T was arbitrary. D In the next theorem, we show that the cartesian product of cycles satisﬁes Conjecture 5; note
that prisms D; with n even belong to this class.
Theorem 26. Every cartesian product of p cycles has property P0. In particular, every tree with
p edges decomposes Hp = Cm, X C'm2 X    x Cmp.
Proof: We will proceed by induction on p. As noted earlier the claim is immediate for p = 1.
Now suppose our claim is true for p = k _>_ 1. Let T be any labeled tree with k + 1 edges, with
labeling f. Let T’ be the labeled tree with k edges obtained by deleting from T a vertex :1: of degree 1; let y be the neighbor of a: in T. Let S denote the set of labels used on T’ by f. By
our inductive hypothesis, Hk satisﬁes property Po; we will construct an appropriate labeling for Hk+1 = H], x k+1. By Property Po, Hk has a labeled T’decomposition A such that S" = S for each vertex V.
The graph H1,“ is isomorphic to my,“ copies of Hi, arranged in a cycle, with corresponding vertices of neighboring copies of H], ’15 joined by an edge. To build the desired labeled decomposition,
begin by taking a copy of the labeled decomposition A for each copy of Hk. Let no, . . ”k+11 be the copies of a particular vertex v e V(Hk); these induce a cycle. For each such 1), we use these
edges to expand copies of T’. Since 3., = S, we can select for each vj the copy T. of T’ such that
f.('v,) = f(y). We add the edge 12,i to this subtree (where addition is modulo my“) to obtain a
copy of T, and we extend f. so that f.(v,+1) = f(z) The new vertex was not part of T5, because T.
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is entirely conﬁned to the jth copy of H1, in the construction. When this is done for all j and all 1), we have added f(z) to the multilabel on each vertex and completed a full labeled Tdecornposition of H"+1. This establishes property P1, since T was arbitrary. D
Finally, we show that 2mregular graphs with large girth satisfy Conjecture 5. The next
theorem establishes full labeled decompositions for trees with small diameter, but if the girth of a 2mregular graph G exceeds m, then it will apply to all trees with m edges and establish property Pa for G.
Theorem 27. Let G be a 2m—regular graph, and let T be a tree with m edges and diameter less than the girth of G. Then G has a full labeled Tdecompasition.
Proof: We will proceed by induction on m. As noted earlier and illustrated in Figure 1(a), the claim is immediate for m = 1. Now suppose our claim is true for m = k 2 1. Let T be a labeled
tree with k + 1 edges and labeling f. Let T’ be the labeled tree with k edges obtained by deleting from T a vertex 2: of degree 1; let y be the neighbor of a: in T. Let S denote the set of labels used
on T’ by f.
By Petersen’s Theorem (see [4]), every 2mregular graph has a 2factor (a spanning 2regular subgraph). Let H be a 2factor of G, and let G’ be the graph obtained from G by deleting the edges of II. By the induction hypothesis, there is a full labeled T’decomposition A of G’, since the girth of G’ is at least the girth of G, each copy T. of T’ in A has its copy of g on exactly one
cycle in H. If so, . . . , vp_1 is one of the cycles in H, we use the edges of this cycle to expand copies of T’ that contribute f(y) to the multilabels on vertices on this cycle. We can select for each vi
the copy T. of T’ such that fi(1)j) = f(y). We add the edge 0515+; to this subtree (where addition 72
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‘is modulo 1)) to obtain a copy of T, and we extend f. so that f.(v,+1) = f(z). The new vertex was not part of Ti, because the girth of G exceeds the diameter of T. When this is done for all vertices
of all cycles of H, we have added f(2:) to the multilabel on each vertex and completed a. full labeled Tdecomposition of G. U
Haggkvist has proven (unpublished) the following slightly stronger result.
Theorem 28. (Haggkvist [23]) Every tree T of order mII with diameter at most I: decomposes every 2mregular graph of girth at least k, for k 2 3.
4.3
Decompositions into Nonisomorphic Trees
Our motivation in considering £~decompositions is the possibility that a 2mregular graph has an Ldecomposition where E is a list of trees, each having m edges. We introduce new notation to
simplify the discussion.
Suppose T is the union of two trees R and H, which have exactly one vertex b in common. We
describe the formation of T in this manner by writing T = (R, b,H). Given a collection of trees L3 = {T1,. ..,T,,}, we say that a ﬁxed tree F is a common rootstock of LI, if for each tree T. e L we can write T. = (R;,b., Hg) such that each R; is isomorphic to F under an isomorphism 1/); such
that 1b;(b.') is the ﬁxed vertex b of F. This notion of rootstock (and the term) is due to Fink [16]. Note that every decomposition of D?" into trees with n + 2 edges consists of m2""1 trees. The motivation for considering common rootstock is the intuition that the more trees we specify in B,
the more alike they must be to guarantee an Ldecomposition. Fink [16] proved that a family I. of 73
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1; trees with n + 2 edges decomposes the cube D2 if it has a common rootstock with k = [1092p] edges. We extend this to arbitrary prisms by proving a stronger result.
Our technique is an extension of our earlier technique for labeled decomposition. Indeed, our theorem that D3, has property P; can be taken as a basis case. Note that if the trees in .C are isomorphic, then I, has a common rootstock that is all of each tree, as considered earlier. Now
we extend the notions of labeled decomposition to Edecomposition. Although these deﬁnitions are straightforward generalizations of those for Tdecomposition, their technical complexity is such
that it was helpful to work with the special case of isomorphic decomposition ﬁrst. Deﬁnition 3. Let .6 = {T1,. ..,T,,} be a list of trees with a common rootstock F, where
T. = (R;,b., Hg) and each R, is isomorphic to F (via the isomorphism «#5). Suppose f is a labeling of F by distinct labels, with label set S. We obtain a natural labeling f. of the rootstock of each
T; e E by 12(1)) = f(¢;(v)) for 1) e V(R.'). We leave the remaining vertices of each T. unlabeled, obtaining what we call a rootlabeled family LI. Let G be a graph, and let .6 = {T1, . . .,T,,} be a list of trees that decomposes G, with common
rootstock F. Suppose E has a rootlabeling as described above. Let U1, . . . , U., be the edgedisjoint subtrees in some £decomposition of G. For each U5, choose an isomorphism ¢j from U; to one of
the trees in L1. If this tree is T;, the mapping associates with each vertex v 6 V(Uj) the unique label 91(1)) = f,(¢,(v)) if «155(0) 6 Rg; otherwise it associates no label with 'v. The resulting collection of rootlabeled trees U1, . . . , Uq, with labelings 91,. . .,g,,, is called a rootlabeled Ldecomposition of
G.
Given a rootlabeled Bdecompusition of G and a vertex v e V(G), let 5., be the list of labels
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contributed to v by trees U, containing 12; such a contribution occurs only when 95, takes 1) to a vertex in some 12,. We call S,, the multilabel of 1) induced by the rootlabeled ﬂdecomposition. A
rootlabeled Ldecomposition of G is full if every induced multilabel S” is equal to the label set S
of the rootstock of L‘. and each label in S" is contributed by only one subtree Uj. A rootlabeled
£decomposition of a bipartite graph G = (X, Y, E) is bifull if S” = 5+ for every I: E X and S” = S" for every 1) E Y, where 3"”,5" is the partition of S induced by the partite sets of the common rootstock F under the labeling f, and again each label in S” is contributed by only one subtree U, .
Theorem 29. Suppose ﬂ is a rootlabeled family of p trees having a common rootstock with k 2 0 edges. If p 5 2", each tree in 13 has n + 2 edges, and m is even, then D9,, has a bifull rootlabeled £decomposition.
Proof: The proof is by induction on k. If k = 0, then there is only one tree. If we label it
arbitrarily, then Theorem 25 provides the proof. Also, if k 2 n + 1, then all trees in I. are the
same labeled tree (we can give the same label to the one unlabeled vertex in each if k = n + 1), and Theorem 25 again provides the proof. For k 5 n, we may assume p = 2"; otherwise we have the freedom to list trees in L: repeatedly until p = 2", because the total number of trees in an
Ldecomposition of D3, is 2” + 2"'1 > 2". Suppose that k > 0 and the claim holds for smaller values of the parameter 1:. Let a: be a
vertex of degree 1 in F other than I), and let 1: be the neighbor of z in F. Partition L‘. arbitrarily
into two sets [,1 and £12, each of size 2"“. From each T. in E1 or £2, delete the vertex 2. such that
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1/2(a:.) = a: to obtain T;’. The resulting families (2’1 and .02 satisfy the conditions of the theorem. Write Dru1 = (X, Y, E) to indicate the partite sets of Dru—1 , and let 5'", S" be the partition of S — f(c) induced by the partite sets of F—z under the root labeling f. By our inductive hypothesis,
D3,"1 has a bifull rootlabeled Liadecomposition A such that S” = 5+ for v 6 X and 5,, = S" for v 6 Y. By the vertex transitivity of D3,“, it also has a bifull rootlabeled B’zdecomposition B such that S”: S" forv 6 X and 5'": 5+ forv EY.
The graph D3,, consists of two copies G1,02 of D3,“, plus an edge joining the two copies of each vertex. To form the desired labeled decomposition for D3,, begin by using A on Gl and B on G2. For each edge uw between 01 and 02, the multilabels contributed to the endpoints by these
decompositions are {S+, 5‘}; suppose u is the endpoint with label 5+ and w is the endpoint with label .5". By symmetry, we may assume f(y) 6 5*. Suppose U, is the subtree in the decomposition
(A or B) that contributes f(y) to 5'“. To this subtree, which is isomorphic to some T,’, we add the edge MD. This turns it into a copy of T,, because Uj is entirely contained in one of {Gg}, and w
belongs to the other. We set qSJ(w) = z, and add f(a:) to Sm. When we have done this with each new edge, we have augmented the rootstock of each subtree, because each subtree U5 is a copy of
some T. and has a vertex 1) such that ¢g(¢,(v)) = y. We have added f(:c) to each 5,, such that .5'" = .S". The vertex transitivity of D3, implies that it doesn’t matter which partite set was which.
Thus we have constructed a bifull rootlabeled Ldecomposition of D3, as desired. El
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4.4
alabelings
Given a graph G, an injective function f :V(G) > N has been called a vertex labeling, a valuation,
or a vertex numbering of G, by various authors. We use the terminology of vertex labeling, consistent
with the earlier part of this chapter. Rosa [41] called such a function f on a graph G with q edges a ﬁlabeling if f is an injection from the vertices of G to the set {0, 1, . . . ,q} such that the values
I f(z)— f(g) for the (1 pairs of adjacent vertices z, y are distinct. A ﬂlabeling is now more commonly called a graceful labeling. An alabeling [41] is a graceful labeling having the additional property
that there exists an integer A so that for each edge :cy either f(:r) S A < f(y) or f(y) S A < f(a:)
(i.e., the larger label on the edge is bigger than A and the smaller label is at most A. We will call this value A the critical value of the alabeling. Note that a graph with an alabeling is necessarily
bipartite. We ﬁnd alabelings particularly attractive because of the following theorem of Rosa.
Theorem 30. (Rosa [41])Let G be a graph with e edges, and suppose that G has an alabeling.
Then the complete graph Kgpe+1 can be decomposed into isomorphic copies of G, where p is any positive integer.
In this section, we construct alabelings for the graphs D2", and for the graphs Of}; (a cycle
C4", with a pendant path Pn attached to each of its vertices). This generalizes work of Kotzig [30],
Bodendiek et al.[6], Frucht and Gallian [21], and Frucht [20]. Finally, we deﬁne a binary operation é (called the weak tensor product) and prove that if G,H have alabelings, then G é H also has an alabeling. This considerably enlarges the class of graphs known to have alabelings.
First, we will construct alabelings for the graphs D2,. 77
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Remark. Let G be a graph with an alabeling. Since G may have several alabelings (e.g. a
with path P,” n > 2), we avoid confusion and write G = (H,L,E) to indicate the sets of vertices high and low labels (H and L may have sub or superscripts, E = E(G)), and assume that we have
chosen some ﬁxed ozlabeling fa (with critical value )9) such that for all h E H, fG(h) > AG and that fa(l) 5 A9 for all I 6 L. When G is understood, we will ignore the subscript, i.e. fa = f
(AG = A). Furthermore, we will usually abuse notation and let h E H (I E L) not only designate a. vertex in H (L) but also its label, i.e. f(h) = h (f(1) = 1).
Theorem 31. For every nonnegative integer n there exists an alabeling of DEF.
Proof: First, we will describe alabelings of D2,, and D3,. Let {111, v2, . . . , 04?} be the vertices of C4,, and assign the value a. (z' = 1,...,4p) to the vertex 12,, where a. = (i — 1)/2 if 1' odd,
a, = 4p+ 1 — i/2 if i is even and S 4p/2, and a; = 412— i/2 ifz' is even and > 4p/2. This gives us a. labeled bipartite graph W =(H, L, E) where the vertices in H = {v. : i is even} have been assigned
values 2 2p, and the vertices in L = {1), : i is odd} have been assigned values 5 2p — 1. Note that W is an alabeling of 04p. Now consider D3,, = K2 x 049. Take two copies of W. In one copy increase the value of all vertices in H by 81); call this new labeled graph W'. In the second copy of W, increase the value of all vertices by 4p; call this new labeled graph W”. Note that the edges in W’ are assigned the values
81) + 1, . . ., 12p, and the edges in W’ are assigned the values 1,. . .,4p. In order to avoid confusion between the vertices of W’ and W”, we will let an, 112, . . . ,mp (u. = 1);) denote the vertices of W', and we will let 31,33, . . .,:r4,, (2:, = 11,) denote the vertices in W”. Connect W’ and W” according
to the following rule. For i = 1,. . .,4p, place an edge between 2:, and 14.4.1, where all arithmetic
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is done modulo 41). This gives us an alabeling of D3,, = (H1,L1,E) (see Figure 2(a)), because the edges running between the two copies of D2,, (= C4,) in our alabeling of D}, are assigned the values 41) + l, . . .,8p. Furthermore, note that the vertices of T = {11; : i even} in W’ are adjacent
only to vertices of S = {$. : 1' odd} in W” and that the vertices of R = {us : i odd} in W’ are adjacent only to vertices of Q = {c. : i even} in W”. To simplify our proof we will make use of the following notation and observations. For n 2 0
let q” = E(Df,',,) = 2"+1p(n + 2) and deﬁne q_1 = 0. Let rn = q,, — and. Thus 70 = 4p, and
1'" = 2"p(n + 3) when n 2. 1. Let mo = 21) and let mn = 2p+ 2?;011'; = 2p+ (1,.1 when n 2 1. Note that rn > q,._1 and that r" > mn — 1 when n 2 1. Let If, 5‘, and I; represent the integers in
the intervals [1, 2"p(n + 1)], [2"p(n + 1) + 1, 2"p(n + 3)], [2"p(n + 3) + 1, 2”+1p(n + 2)], respectively. Now suppose we have an alabeling of D2,, = (11", L", E) with the following properties: i) all vertices in Hn have been assigned values 2 ma, ii) all vertices in L" have been assigned values
< m", iii) D2,, consists of two isomorphic copies of D3,:1 (call them W1 and W2) such that the edges connecting W1 and W2 have been assigned the values in I3. To obtain an alabeling of
D2,?1 = (H"+1,L"+1,E) with the same properties, we proceed as follows. Take two copies of D2,, = (H", L", E) with the above alabeling. In one copy, add r,,+1 to all the vertices in H", and
call this new labeled graph W"; the edges in W" have been assigned the values in IQ“. In the second copy, add 1'” to all the vertices, and call this new labeled graph W"; the edges in W"
have been assigned the values in If“. We will now show how to connect W“ with W” and get
an alabeling of D2,?1. We will make use of the plane of symmetry between W1 and W2 in D2,. Let Wi“ be the labeled graph in W"' obtained from W1. Deﬁne Wﬁ', W1", W5" similarly (see Figure 2(b)). We will place the four graphs Wf,W;*,Wf‘,W; in a cycle and join them together to get
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3
8
3
24
ll
16
5
1
21
1
13
9
WI
D°=C3
15
10
23
Wu
.
8
12
. 9
13 “
15
10 Wn
Figure 2(a). alabeling of D} = K; x Ga.
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D2,";H = C, x D31. Place edges between vertices of W; and W5“ in exactly the same manner as they appear between W1 and W2 in D2,. Similarly, let Wf“ and W; also be connected in the
same manner as W1 and W2 are in D2,,. Thus the labeled subgraph induced by Wf‘ and W5” is
isomorphic to the graph obtained from our alabeling of D2,, by adding r...” to the vertices of H" in W; and by adding rn to the vertices of W2. Similarly, the labeled subgraph induced by Wf" and
W; is isomorphic to the graph obtained from our alabeling of D2,, = (H", L", E) by adding rn+1 to the vertices of H" in W2 and by adding 1‘" to the vertices of W1 . With the above additional edges
between W" and W“ we now have a graph isomorphic to D2:1' Indeed, D3,;H can be written as D2,",H = (H"+1,L"+1,E) where the vertices in H"+1 have been assigned values 2 mu“, and the
vertices in L"+1 have been assigned values < mm“. Thus, properties i) and ii) are satisﬁed (see
Figure 2(c)). The edges running between Wf‘ and W2" (with the vertex in H"+1 contained in Wf) and the edges running between W1" and W; (with the vertex in H"+1 contained in W5) are assigned the labels 2"+1p(n + 3) + 1,. ..,2"+‘p(n + 4). Similarly, the edges between Wf and W5“ (with the
vertex in H""'1 contained in W5“) and the edges between W1“ and W; (with the vertex in 11"“
contained in Wf“) are assigned the labels 2"+1p(n + 2) + 1, . . . , 2"+1p(n + 3). Together these sets of labels give us L3“, thus property iii) is satisﬁed.
We have shown that all edges in D3,,"l have been assigned a unique value between 1 and 2"+2p(n + 3). Thus all that remains is to show that each vertex is assigned a unique value. This follows from the fact that 7'" is greater than gn_1 and mn  1. In particular, the intervals in which
the new labels are placed are disjoint. El
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W}
53 2
W;
W2
:
29
3
:
.
l
1
0
1
24 +r,(:o all)
16 9
g 1
1 2 .
10
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:
2
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3
i

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+
:
‘ :
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19
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1
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10
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Woo
Figure 2(c). alabeling of D3 = K2 x K2 x Ca.
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37
Let 6'5" designate the graph formed by adding a pendant path Pn to each vertex of the cycle
gs. This Cm (see Figure 3(a)). We will show next that all graphs of the form Cf}; have alabelin
he called generalizes Frucht’s work; Frucht [20] showed that all graphs of the form 05?, which crowns, are graceful.
Theorem 32. All graphs of the form Cf; have alabeh'ngs.
Proof: Note that if we remove a lfactor from the cycle C4", in 05;; we are left with 2m copies new of P2,. (see Figure 3(a)). Consider the following a—labeling given to P2,, listed below. Call this labeled graph P. 0
2n—1
l
2n2
v
'
v
v
nl to.
n
—_.————.
This labeling partitions the vertices of P = (H, L, E) into two sets H and L, where the vertices in H have been assigned the labels 1:, n+ 1, n+2, . . . , 2n— 1 and the vertices in L have been assigned the labels 0, 1, .. .,n — 1. We will now describe a procedure for labeling 2m copies of P2“, using each of the numbers 0, 1,2, . . .,3mn  1,3mn+ 1,. . .,4mn exactly once. For 1 S i S m, let P(i)
be the labeled path obtained from P by adding (2m — i)n to the label of each vertex in L and
by adding (2m + (i  2))n to the label of each vertex in H. For m < z' 5 2m, let P(i) be the labeled graph obtained from P by adding (2m — i)n to the label of each vertex in L and by adding
(2m+ (2' — 2))n + 1 to the label of each vertex in H . We see that, for 1 S i _ A. for all h. in H. (i = 1,2). Furthermore, we will abuse notation and
designate vertices by their labels. Keeping this in mind, we claim that the following function f’
is an alabeling of G1 é G2. For (h1,h2) 6 H1 x H2 let f’((h1,h2)) = elhg — (e1 — h) and for (11,12) 6 L1 x L2 let f’((11,lz)) = all; +11. Clearly, f’ is a function from V(G1 é G2) into the set {0,1,...,e1e2}. If (h1,h2) 6 H1 x H2, then f’((h1,h2)) > elAg + A1. Similarly, for any
(11,12) 6 L1 x L2, we have f’((11,12)) 5 ed; + A1. We will show that f’ is injective. Suppose that
f’((h1,h2)) = f’((h’1,h'2)). Then elhg + h1 = elh’z + M, so e1(h2 — h’,) E h.’1 — h] mod e1 which implies h’l = h]. Hence also h; = h'z. The case f’((11,12)) = f’((l§, 5)) is handled similarly, proving that f’ is injective.
It remains to show that the edges of G1 é G2 are assigned distinct values. Suppose that two
edges (h1,h2)(l1,12) and (hi, 901,15) are assigned the same value, i.e. suppose that e1(h2 —12)— el + h — I; = e1(h’2  1’2)  e1 + h’l — 11. Then as before, we have h: — l; E H, — 11 mod el and this implies that hl  I; = h’,  (’1. Thus h; = ’1’, and 11 = 15, since I; is an alabeling of GI, whence
h2 = h’2 and 12 = 15. This tells us that the edges of GI ® G2 are assigned distinct values under the
labeling f’. El
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‘5’
l
i \‘4 O
~ \
M
V 69
50
Figure 4. alabeling of G é H for various G, H.
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Bibliography [1]
Ian Anderson, Combinatorics of Finite Sets, Oxford University Press, 1987.
[2]
Ian Anderson, Combinarorial Designs, Halsted Press, 1990.
[3]
torica. 8 L. Babai, A short proof of the nonuniform RayChauduriWilson inequality, Combina
(1988), 133135.
[4]
M. Behzad, G. Chartrand, and L. LesniakFoster, Graphs and Diagraphs, Wadsworth International, Belmont CA 1981.
[5]
C. Berge, A theorem related to the Chvdtal conjecture , Proc. 5th British Combinatorial Conference.
[6]
R. Bodendiek, H. Schumacher, and H. Wegner, Uber grazio'se Graphen, MathPhys. Semester
berichte 24 (1977), 103126.
[7]
K.L. Chung and W. Feller, Fluctuations in coin tossing, Proc. Nat. Acad. Sci. USA 35 (1949), 605608.
[8]
V. Chvétal, Intersecting families of edges in hypergraphs having the hereditary property, Hy
pergraph Seminar, Lecture Notes in Math, SpringerVerlag Berlin 411 (1974), 6166.
88
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
Math. 10 (1977), [9] DE. Daykin, A lattice is distibutiue if AB 5 AVBAAB., Nanta. 5860.
Math. 10 (1948), 421[10] N.G. deBruijn and P. Erdiis, On a combinatorial problem, Indagationes 423.
J. 14 (1947), 305313. [11] A. Dvoretzky and Th. Motzkin, A problem of arrangements, Duke Math.
[12] D.A. Duﬁ'us, personal communication. Boolean lattice, J. [13] DA. Duﬁ'us, ILA. Kierstead, H.S. Snevily, Matchings in the middle of the
Combinatorial Theory (A), submitted. cycles in some [14] Dwight Duffus, Phil Hanlon, and Robert Roth, Matchings and Hamiltonian families of symmetric graphs, Nokanoma Math. J ., to appear.
I, 3rd Ed., Wiley, [15] W. Feller, An Introduction to Probability Theory and its Applications Vol. 1968.
Theory 14 [16] John F. Fink, 0n the decomposition of ncubes into isomorphic trees, J. Graph
(1990) 405411. 30 [17] P. C. Fishburn, Combinatorial optimization problems for systems of subsets, SIAM Review
(1988), 578588. Colloq. [18] P. Frank] and Z. Fiiredi, Families of ﬁnite sets with a missing intersection, Proc.,
Math. Soc. J. Bolyai, Eger, Hungary, 37 (1981), 305318. [19] P. Frank] and RM. Wilson, Intersection theorems with geometric consequences, Combinatorica.
1(4) (1981), 357368. 89
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
, Ann. NY Acad. of Sci. [20] Roberto W. Frucht, Graceful numbering of wheels and related graphs
319 (1979), 219—229. Combinatoria 26 (1988), 6982. [21] Roberto W. Frucht and Joseph A. Gallian, Labeling prisms, Ars problems in labeling graphs, [22] Joseph A. GaJJian, .4 Survey: Recent results, conjectures, and open
J. of Graph Theory 13 (1989), 491504.
ys in Combinatorics, [23] Roland Haggkvist, Decompositions of complete bipartite graphs, in “Surve sity Press (1989), 1989” (J. Siemons ed.), LMS Lecture Note Series Vol. 141, Cambridge Univer 115147.
on a Family of Finite Sets, [24] A.J.W. Hilton, Analogues of a theorem of Erdos, K0, and Rado
Quart. J. Math. Oxford (2), 25 (1974), 1928. of ﬁnite sets, Math[25] A.J.W. Hilton, Some intersection and union theorems for several families
ematika 25 (1978), 125128.
Boolean [26] H. A. Kierstead and W. T. Trotter, Ezplicit matchings in the middle levels of the lattice, Order 5 (1988), 163171.
. [27] DJ. Kleitman, .Families of nondisjoint subsets, J. Combinatorial Theory 1 (1966), 153155 , (B. Bollobas [28] DJ. Kleitman, Extremal hypergraph problems, in “Surveys in Combinatorics” ed.), LMS Lecture Note Series Vol. 38, Cambridge University Press (1979), 4465.
[29] D. J. Kleitman and T. L. Magnanti, 0n the latent subsets of intersecting collections, J. Com
binatorial Theory (A) 16 (1974), 215220. 90
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
rphic cubes, J. Combinatorial [30] Anton Kotzig, Decompositions of complete graphs into isomo
Theory (B) 31 (1981), 292296. [31] Daniel Lichtblau, personal communication. restricted intersection values, [32] Daniel Lichtblau and Hunter Snevily, Intersecting families with
Combinatorica, submitted.
[33] Mathew Markert, personal communication. , Can. Math. Bull. 12 [34] J. Marica and J. Schonheim, Differences of sets and problem of Graham
(1969), 63537. result of Chung and Feller, [35] Balazs Montagh, A simple proof and a generalization of an old
Discrete Math. 87 (1991) , 105108. Expositions 23, [36] T. Narayana, Lattice Path Combinatorics with Statictical Applications, Math. University of Toronto Press, 1979.
s Thesis, [37] Steve Penrice, Lexical Matchings in the Middle Levels of the Boolean Lattice, Master’ University of South Carolina, 1988.
253268. [38] L. Pyber, An extension of a alrlFiiredi theorem, Discrete Mathematics 52 (1984) Math. Soc. 94 [39] G.N. Raney, Functional composition and power series reversion, Trans. Amer.
(1960), 441451. 737744. [40] D.K. RayChaudhuri and R.M. Wilson, 0n tdesigns, Osaka J. Math 12 (1975), , Journées [41] A. Rosa, On certain valuations of the vertices of a graph, Théorie des graphes
internationales d étude, Rome, (1966), 349355. 91
Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.
designs, Journal [42] HJ. Ryser, An extension of a theorem of deBruijn and Erdos on combinatorial
of Algebra 10 (1968), 246261. 219221. [43] AD. Sands, 0n generalized catalan numbers, Discrete Math. 21 (1978), Combinatorial [44] J. Scht'mheim, Hereditary systems and Chmital’s conjecture, Proc. 5th British Conference, 1974.
to appear. [45] HS. Snevily, A new result on Chudtal’s conjecture, J. Combinatorial Theory (A), Math. 43 (1983), [46] P. Stein, 0n Chvdtal’s conjecture related to a hereditary system, Discrete 97105.
[47] P. Stein, Chudtal’s conjecture and point intersections, Discrete Math. 43 (1983), 321323.
[48] D. L. Wang and P. Wang, Some results about the Chvdtal conjecture, Discrete Math. 24 (1978), 94101. Reidel, [49] D. B. West, Extremal problems in partially ordered sets, Ordered Sets, I. Rival, ed.,
Dordrecht, The Netherlands, 1982, pp. 473521.
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Vita Hunter Saint Clair Snevily was born in Summit, New Jersey. He recieved his Bachelor’s of
. Science in mathematics from Emory University in 1981. After receiving his degree from Emory he
was awarded a scholarship by the Air Force Institute of Technology to study electrical engineering at the University of New Mexico. After two years of study he had decided that he was better suited to be a mathematician.
He entered The University of Illinois at UrbanaChampaign in August, 1985 and worked as a teaching/research assistant until January, 1991. He is married to Harriet McQuarie and has a daughter named Madison.
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