Combinatorics of Finite Sets (Dover Books on Mathematics)

Ii

Combinatorics of Finite Sets Ian Anderson University of Glasgow

DOUER PUBLICATIONS, INC. Mineola, new York

Copyright Copyright © 1987 by Ian Anderson

All rights reserved under Pan American and International Copyright Conventions.

Published in Canada by General Publishing Company, Ltd, 895 Don Mills Road, 400-2 Park Centre, Toronto, Ontario M3C 1 W3. Published in the United Kingdom by David & Charles, Brunei House, Forde Close, Newton Abbot, Devon TQ12 4PU.

Bibliographical Note This Dover edition, first published in 2002, is a corrected republication of the work as published by Oxford University Press, Oxford, England, and New York, in 1989 (first publication: 1987).

Library of Congress Cataloging-in-Publication Data Anderson, Ian, Ph, D. Combinatorics of finite sets / Ian Anderson. p. cm

a corrected republication of the work as published by Oxford University Press, Oxford, England, and New York, in 1989 (first publication: 1987)"- -T.p. verso. Includes index. ISBN 0-486-42257-7 (pbk ) 1. Set theory. 2. Combinatorial analysis. I. Title.

QA248 .A657 2002 511.3'22 - dc2l 2002019228

Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

To Margaret Grace

Preface

The past quarter century has seen the remarkable rise of combinatorics as a distinctive and important area of mathematics. Combinatorial topics have found a place in many university degree courses, and, since the founding of the Journal of Combinatorial

Theory in 1966, there has been a flood of publications in the combinatorial area. Some branches of combinatorics are already well established, with their own unified body of theory and applications: examples are graph theory, with its topological flavour, coding theory and design theory, with their algebraic connections, and enumeration theory, concerned with the techniques of counting. These areas are

already supplied with many textbooks at both undergraduate and postgraduate levels. The purpose of the present work, however, is to advertise another area of combinatorics, where a body of theory, at one time very scattered and ad hoc, is gradually being moulded into an elegant unity. Without attempting to be exhaustive, the book is intended to be a readable introduction to what is, for the author, a fascinating subject. The origins of the theory can perhaps be traced back to 1928, when Sperner published a simple theorem which has had repercussions far beyond his wildest dreams. Sperner's theorem simply asserts that if

you want to find as many subsets of an n-element set as possible, subject to the condition that no subset is contained in another, then you cannot do better than to choose all the subsets of size [in]. This theorem has been reproved and generalized to such an extent that it has given rise to a whole branch of the theory of partially ordered sets

(or posets) called Sperner theory. Although we shall go some way down this path, we shall not restrict ourselves to this area. Instead, we shall use Sperner's theorem as a springboard and a signpost, discovering where the ideas involved in its various proofs lead to. We shall be led to consider the structure of the set of subsets of a finite set viewed as a poset, its chain decompositions and its antichains, its rank levels, and the inclusion relations between these levels. This will take

us, for example, to the famous Kruskal-Katona theorem which answers the following question: given r subsets of S, all of size k, vii

viii

I

Preface

what is the least possible number of sets of size k - 1 contained in them? The answer to this question involves a nice interplay between two different orderings, namely the partial ordering of the subsets of S by inclusion, and the total ordering of the subsets of S of a given size by what we shall call the squashed ordering, which is a variation on the more familiar lexicographic or alphabetical ordering. Anyone writing a book in this area is faced right at the start by a fundamental problem. Many results for the poset of subsets of a set

can be extended to more general posets. Sometimes one of the available proofs for sets easily extends to more general posets; sometimes, however, the nicest proof for sets does not generalize. So should the results be presented in their most general form (thereby

sometimes losing out on clarity) or should they be presented for subsets of a set (thereby losing out on generality but perhaps gaining in clarity)? An example of this problem arises in connection with the Kruskal-Katona theorem. Several proofs are available, but a more difficult proof due to Clements and Lindstrom establishes the result in

the more general context of the poset of divisors of a number (or subsets of a multiset). Since we include the Clements-Lindstrom theorem in Chapter 9, there is strictly speaking no need to include a

separate proof of the Kruskal-Katona theorem. However, in the simpler context of sets, the Kruskal-Katona theorem has such an elegant theory surrounding it that it would be almost criminal to omit the simpler case. So we present both proofs. On many other occasions we prove results in more than one way because the different proofs

illustrate different ideas and different techniques. On the whole, I have taken the view that I should present results in their simplest forms, concentrating mainly on sets and multisets. Accordingly, the reader will not find a discussion of, say, geometric lattices, although their `prototype', the poset of partitions of a set, is discussed. In a few

places we look at posets more generally; the final chapter, for example, discusses extensions of the theorem of Dilworth concerning chain decompositions of a general poset. In searching out the material for this book I was greatly helped by several survey articles, two of which deserve special mention: the first

is that by Greene and Kleitman (1978), and the second is the more recent one by D. B. West (1982). As will be seen from a glance at the extensive list of references at the end of the book, a number of more

recent results have been included. Inevitably some of the results presented here will have been improved upon by the time this book appears in print, but in a sense this does not matter for the aim of the book is not to provide an exhaustive survey but to present some of

Preface

I

ix

the ideas and techniques which go to make up the subject. Inevitably, also, my choice of material will not meet with the approval of all, but a number of interesting results not in the text have been included in the exercises at the end of each chapter with hints or outlines of their solutions at the end of the book. A number of people have helped and encouraged me in the writing

of this book. In particular, I gladly acknowledge the helpful comments of Professor George Clements and Dr Hazel Perfect. I should also like to thank the Oxford University Press for encouraging me to write, and the University of Glasgow for granting me a period of study leave during which the final compilation of the book was accomplished. Glasgow December 1985

I.A.

Contents Notation

xv

1. Introduction and Sperner's theorem 1.1 A simple intersection result 1.2 Sperner's theorem

1.3 A theorem of Bollobas

4

Exercises 1 2. Normalized matchings and rank numbers 2.1

1

2

Sperner's proof

7

2.2 Systems of distinct representatives

10 12

2.3 LYM inequalities and the normalized matching property 2.4 Rank numbers: some examples

13 17

Exercises 2 3. Symmetric chains 3.1 Symmetric chain decompositions 3.2 Dilworth's theorem 3.3 Symmetric chains for sets 3.4 Applications 3.5 Nested chains 3.6 Posets with symmetric chain decompositions Exercises 3 4. Rank numbers for multisets 4.1 Unimodality and log concavity 4.2 The normalized matching property 4.3 The largest size of a rank number Exercises 4

23 27 30 32 36 43 45 53

56 60 63

68

5. Inter secting systems and the ErdOs-Ko-Rado

theo rem 5.1

The EKR theorem

70

5.2 Generalizations of EKR 5.3 Intersecting antichains with large members

73 77

5.4 A probability application of EKR

79

5.5

Theorems of -Milner and Katona xi

81

xii I Contents

5.6 Some results related to the EKR theorem Exercises 5 6. Ideals and a lemma of Kleitman 6.1 Kleitman's lemma 6.2 The Ahlswede-Daykin inequality 6.3 Applications of the FKG inequality to probability theory 6.4 Chvatal's conjecture Exercises 6

83 85 87 90 97 103 108

7. The Kruskal-Katona theorem 7.1

Order relations on subsets

7.2 The 1-binomial representation of a number 7.3 The Kruskal-Katona theorem

7.4 Some easy consequences of Kruskal-Katona Compression Exercises 7 8. Antichains 8.1 Squashed antichains 8.2 Using squashed antichains 8.3 Parameters of intersecting antichains Exercises 8 9. The generalized Macaulay theorem for multisets 9.1 The theorem of Clements and Lindstrom 7.5

112 115 119 124 126 127 131 135 139 143

9.2 Some corollaries 9.3 A minimization problem in coding theory

145 151 154

9.4 Uniqueness of maximum-sized antichains in multisets Exercises 9

157 160

10. Theorems for multisets 10.1 Intersecting families 10.2 Antichains in multisets 10.3

Intersecting antichains

Exercises 10

11. The Littlewood-Offord problem 11.1 Early results 11.2 M-part Sperner theorems 11.3 Littlewood-Offord results Exercises 11

162 168 172 174

176 178 183 185

Contents I xiii

12. Miscellaneous methods 12.1 The duality theorem of linear programming 12.2 Graph-theoretic methods 12.3 Using network flow Exercises 12 13. Lattices of antichains and saturated chain partitions 13.1 Antichains 13.2 Maximum-sized antichains 13.3 Saturated chain partitions 13.4 The lattice of k-unions Exercises 13 Hints and solutions

187 192 194 198

201 203 205 212

212 214

References

241

Index

249

Some notation used in the text

(n)

ad e(k)d Vsi Cs/I

(k)S Ni <S

Fk(n) 2C(P)

0 i(m) dk(P)

binomial coefficient (n-choose-r) the shadow of s/4 the shadow of s/I at level k

the shade of si the compression of s/I the set of complements of members of s/I the set of k-subsets of S the number of members of rank i the squashed ordering the first n k-sets (or k-vectors) the set of antichains of the poset P the empty set the number of divisors of m the size of the largest k-union in P

xv

1 Introduction and Sperner's theorem

1.1

A simple intersection result

The main theme of this book is the study of collections s4 of subsets of a finite set S, where s4 is described in terms of intersection, union, or inclusion conditions. An amazing richness and variety of results will be discovered, developed, and extended in various directions.

Although our main initial theme will be a study of a theorem of Sperner which could be said to be the inspiration of all that follows, we get into training by first of all asking what must surely be one of the simplest questions possible.

Problem Let 4 be a collection of subsets of an n-element set S (or an n-set S) such that A, fl A; 0 0 for each pair i, j. How big can IsiI be? The answer, and more besides, is given by the following theorem.

Theorem 1.1.1 If s4 is a collection of distinct subsets of the n-set S

such that A; fl A; # 0 for all A,, A; E 4, then I4 2n '. Further, if I si I < 2"-', s4 can be extended to a collection of 2"-' subsets also satisfying the given intersection property.

Proof If A E 4 then the complement A' = S - A is certainly not in A since A fl A' = 0. So we immediately obtain IdI -- 22" = 2"-' This bound cannot be improved upon since the collection of all subsets of {1, . . . , n} containing 1 satisfies the intersection condition and has 2"-' members. Now suppose IsiI 0. Then associate with each sum Y,(2e1-1)x1 the corresponding set of indices i for which (2s,-1)x1>0.) (Erdos 1945) 1.8

Let A1, ... , Am, B1, ... , Bm be subsets of an n-set S such that

8 1 Combinatorics of finite sets

A, c B; for each i and A; = B1 whenever i *j. Show that m

1

;_, (n - IB; -A;I

S(n, r + 1) if r > k(n). Further, k(n) = k(n - 1) or k(n) = k(n - 1) + 1. Proof We proceed by induction on n, assuming that the theorem is true for n and considering the case n + 1. First of all, suppose that k -- k(n). Then, by Theorem 2.4.1(ii),

S(n + 1, k) -S(n + 1, k - 1)

=S(n, k-1)+kS(n, k)-S(n, k - 2) - (k - 1)S(n, k-1) = S(n, k - 1) - S(n, k-2)+k(S(n, k) - S(n, k - 1)) + S(n, k - 1) > 0 by the induction hypothesis. Next suppose that k -- k(n) + 2. Then, by Theorem 2.4.1(iii),

S(n+1,k)-S(n+1,k-1)=> (')(so,k-1)-S(j,k-2)) r.

3 Symmetric chains

3.1

Symmetric chain decompositions

The problem of generalizing Sperner's theorem on subsets of a set (or

divisors of a square-free number) to multisets (or non-square-free

numbers) was posed in 1949 as a prize problem by the Dutch Wiskundig Genootschap. The solution, due to de Bruijn, Tengbergen, and Kruyswijk (1951), is not only very simple but exhibits a structural aspect of the poset of divisors of a number which turns out to be of great usefulness. The aim of this chapter is to describe this structural property and to give a number of varied applications. The basic idea in their solution was to decompose the poset into chains with certain symmetry properties. If P is a ranked poset with

rank function r, we say that the elements x1, ... , x,, of P form a symmetric chain if (i) x;+1 covers x, for each i < h (ii) r(x1) + r(x,,) = r(P)

where r(P) is the largest rank in P. Note that (i) requires that the chains are saturated and (ii) requires that each chain finishes as far above the middle rank as it started below, i.e. it is `symmetrically positioned' about the middle rank. In the case of subsets of an n-set S, A,,. . . , A,, form a symmetric chain if (i) A; c Ai+1 and JAt+1I = 1 + IA,l for each i < h (ii) IAII + IAhI = n.

In the case of divisors of a number m, the divisors d1, ... , d,, form a symmetric chain if (i) d; divides d;+1 and d;+1/d; is prime for each i < h (ii) r(di) + r(d,,) = r(m), where r(d;) is the number of prime factors of d; counted according to multiplicity. de Bruijn et al. showed that the set of divisors (subsets) of a number m(set S) can be partitioned into symmetric chains. Here are two examples to illustrate the idea. Example 3.1.1 Subsets of {1, 2, 3, 4). 27

28 1 Combinatorics of finite sets

(3, 4) {2, 4}

{4} c {1, 4} {1, 2, 4} {3} c {1, 3} c {1, 3, 4} f2)1-- {2, 3} c {2, 3, 4}

0c{1}c{1,2}c{1,2,3}c{1,2,3,4}. Example 3.1.2 Divisors of 180 = 22325.

2.5 22.5 32. 5

32

1

5

3.5 2.3.5 22.3.5

3

2.3 2.32

2

22

22.3

2.32.5 22.32

22.32.5

In terms of multisets, Example 3.1.2 partitions the set of subsets (or more strictly submultisets) of the multiset {2, 2, 3, 3, 5} as follows: {2, 5} c {2, 2, 5} {3, 3} c {3, 3, 5) (5) c {3, 5} c {2, 3, 5} c {2, 2, 3, 5}

{3} c {2,3}c{2,3,3}c{2,3,3,5} 0c{2}c {2, 2)c{2,2,3}c{2,2,3,3}c{2,2,3,3,5}. Theorem 3.1.1 (de Bruijn et al. 1951) The set of divisors of a number (or subsets of a set) can be expressed as a disjoint union of symmetric chains.

Proof We shall use the notation of divisors of m, and proceed by induction on the number n of distinct primes dividing m. If m = p" then 1, p, p2, ... , p" is a symmetric chain containing all the divisors

of m; thus the theorem is true for n = 1. Suppose now that the theorem is true for n = k, and consider a number with k + 1 distinct prime factors. Such a number can be written as m = m,p" where m,

has k distinct prime factors and where p 4. ml. We show how to construct symmetric chains for m from symmetric chains for m,. Let d,, ... , d,, be one of the symmetric chains in the chain decomposition

of mand consider all the divisors of m of the form dps, 1 -- i -- h, 0 -- 0 = a. We put these divisors of m into symmetric chains by the following procedure. Write them in a rectangular array as shown, and

Symmetric chains 1 29

then peel off chains as indicated:

d1 d2

... dh_2

dh-1

dh

dip dzp

... dh-zP

dh-lp

dhP

d1p 2 d2p2

...

dh-1P

dh_zpz

dipa-1 d2pa-1

dh-2P«-i

dlpadzpa

dh-2Pa The outer layer gives the first chain tl

d1, dz,

z

dhpz

CI-1

dhpa-1

dh-iP dh-1pa

... , dh-I, dh, dhP, dhP 2, .

dhP a

. .

dhPa

,

which clearly satisfies condition (i) and is symmetric since

r(d1) + r(dhp a) = r(d1) + r(dh) + a = r(m1) + r(p a) = r(m). Similarly the second layer gives a symmetric chain, and so on. Finally, every divisor of m will be obtained in this way from some symmetric chain for m 1.

Example 3.1.3 We can use Example 3.1.1 and the construction given in the proof of the theorem to obtain a symmetric chain decomposition of the set of subsets of {1, 2, 3, 4, 5}. For example, starting with the chain {3} c {1, 3} c (1, 3, 4) we consider the rectangular array {3} {1, 3} {3, 5} {1, 3, 5)

{1, 3, 4} {1, 3, 4, 5}

and obtain the chains

{3}c{1,3}c{1,3,4}c{1,3,4,5} and {3,5}c(1,3,5). Having introduced the idea of symmetric chains, we can now give our third proof of Sperner's theorem. Proof of Theorem 1.2.1 Partition the poset of subsets of the n-set S

into symmetric chains. Now the number of chains must be

n ,n/2]) since each chain must contain exactly one subset of size

30 1

Combinatorics of finite sets

[n/2]. However, if si is an antichain then s4 can contain at most n one member of each chain; so Isif - ( \[n/2]

3.1.1 Rank numbers for multisets

Consider a symmetric chain decomposition of m = pi' ... p;' and put

K = k, + ... + k,. Since each chain starting at rank h ends at rank K - h, we must have N,, = NK_,,. Also, if h < JK then all chains containing a divisor of rank h must contain a divisor of rank h + 1, and so we must have N,,+, , N,,. Thus we have the following theorem.

Theorem 3.1.2 If No, ... , NK are the rank numbers for the poset of divisors of m = pi' ... where K = E; k;, then (i) No -- N,

...

N[Knj;

.1....

N[K,7

NK

and

(ii) N,, = NK_,,

for each h.

In the next chapter we shall find out where we have strict inequalities in (i). Meanwhile we note that the N, form a unimodal sequence and that the largest value taken by any Ni is N[K/21. The proof of Theorem 1.2.1 given above therefore generalizes directly to give the following theorem.

Theorem 3.1.3 If si is an antichain in the poset of divisors of m = pi' ... p;,, then 141 --N(Kr2, where K = E, k,. Thus the poset of divisors of m possesses the Sperner property. 3.2

Diiworth's theorem

Before looking further at symmetric chains it is of interest to compare Theorem 3.1.1 with the theorem of Dilworth concerning chains in a general poset. It is clear that if a poset can be expressed as a union of w chains then any antichain in P can have at most w members, one from each chain. Thus maximum size of an antichain in P -- c(P)

(3.1)

where c(P) denotes the minimum number of chains into which P can

be decomposed. Dilworth's theorem asserts that there is in fact

Symmetric chains

1

31

equality; c(P) is the largest size of an antichain in P. If there is a symmetric chain decomposition, then we acquire some extra information, namely that c(P) is the largest rank number, but we emphasise

that Dilworth's theorem holds in any poset, even one without any rank function. There are now many different proofs of Dilworth's theorem in the literature. The proof we now present is one of the simplest, and is due to Perles (1963). Theorem 3.2.1 (Dilworth 1950)

In any poset P, the maximum size of

an antichain is equal to the minimum number of chains in a chain decomposition of P.

Proof (Perles 1963) In view of (3.1), we have to show that the minimum number of chains in a chain decomposition of P is less than or equal to the size of a maximum antichain. We use induction, and assume that the required result is true for all posets Q with IQI < IPI. Let m be the maximum size of an antichain in P. We now consider two cases.

a(i) Suppose first that there is an antichain of size m which is neither the set of all maximal elements nor the set of all minimal elements of P. Let W be such an antichain, and define

P+ = {x E P: x- y for some YEW} P- = {x EP:x -- y for some y c- W).

Since there is at least one minimal element not in W, P+ * P. Similarly P- * P. Thus IP+I < IPI and IP I < IPI, so we shall be able to apply the induction hypothesis to them. Observe that P+ U P- = P, for if there were an element of P not in P+ U P- then we could add it

to W to obtain a larger antichain. Also, P+ fl P-= W. Now, as already mentioned, P+ and P- are, by induction, the union of m chains, say P+ = U" , C, and P- = U?=1 D;. The elements of W, being the minimal elements of P+ and the maximal elements of P-, are the minimal elements of the chains C, and the maximal elements of the chains D.. We can therefore glue the chains C, and D; together in pairs to form m chains which form a chain decomposition of P.

(ii) Next suppose that there are at most two maximum-sized antichains, one or both of the set of all maximal elements and the set of all minimal elements. Choose a maximal element a and a minimal element b with b _ a. Then P - {a, b } contains an antichain of size m - 1 but none of size m, so that by the induction hypothesis it can be

32 1

Combinatorics of finite sets

decomposed into m - 1 chains. These chains, together with {a, b }, give a decomposition of P into m chains.

3.2.1 The marriage problem

Dilworth's theorem will be considered further in Chapter

13.

Meanwhile we close this section by showing how it can be used to give an alternative proof of the marriage theorem (Theorem 2.2.1).

Proof Let A...... A. be m , n subsets of S = {x,, .

. . , x,, } such that, for each k -- m, any k sets A; contain at least k elements in their union. Define a poset as follows: take objects u,, ... , u and

v,, ... , vand define an order relation by

u;;v; u;

u;

if and only if x;EA;

ifi0j

v,4v; ifi*j. It can be seen that u; is really representing x; and v; is representing A;. Now {u1, ... , U.1 is an antichain in P, and we now show that it is a

maximum-sized antichain. Suppose, after a suitable relabelling if necessary, that {U1.... , Ur, v,, ... , v5} is an antichain. Then none of x,, ... , Xr is in any of A,,

.

. .

, As, so the only possible elements of

Thus thessets A...... A.have --n - r elements in their union. However, they have ;s elements in their union, so we must have s _ n - r, i.e. r + s , n. Therefore n is indeed the maximum size of an antichain. It now follows from Dilworth's theorem that P can be decomposed into n chains. Each chain must contain exactly one u;. Since the v; are all in different chains, there corresponds to each v; a u; in the same chain with u; -- v,; thus to each A; there corresponds an xi with x, E A;. 3.3 Symmetric chains for sets

Having looked at symmetric chains in a more general setting, we are

now going to return to the particular case of subsets of a set and examine the structure of the chains more closely. Suppose we wish to find a symmetric chain decomposition of the set of subsets of a given

set. The construction given in the proof of Theorem 3.1.1 is a step-by-step construction, altering the chains at each step. This makes

for a tedious construction if the given set is large. Is there a neater way of writing down the chains without having to find the chains for

Symmetric chains 133

all the smaller sets in the process? A positive answer to this question has been given by Leeb (pers. comm.) and, independently, by Greene and Kleitman (1976b). we associate a Corresponding to each subset A of S = {x1, ... , sequence of n parentheses. If x; E A we put a right parenthesis in the

ith place; if x, I A we put a left parenthesis in the ith place. For example, if S = (1, ... , 9) and A = {2, 3, 4, 7, 81 we represent A by Having done this we can now `close' the brackets as far as possible. In the example given, we obtain

()))(())(

Elements of S which correspond to right parentheses which are paired

in the closing process are called the basic elements of A. In the example 2, 7 and 8 are the basic elements of A. The unpaired right parentheses represent non-basic elements of A. Note that all unpaired

left parentheses must occur to the right of all the unpaired right parentheses, for otherwise further closing of brackets would be possible. These unpaired left parentheses represent unpaired elements of the complement A'. Now define A- to be the set of basic elements of A, and A+ to be the union of A with the set of unpaired elements of A'. Then IA+I = IAI + IA'I - number of paired elements of A' = ISI - number of paired elements of A

=ISI - IAI so that Al I+ IA+I = ISI Now form a chain containing A by starting with A-, the set of basic elements of A, and adding one by one the non-basic elements of A in order, and then adding one by one the unpaired elements of A' in order. The chain therefore starts with A-, contains A, and ends with A+. Since Al I+ IA+I = ISI, the chain is a symmetric chain containing A. In the example above, the basic elements of A are 2, 7, 8, and the elements 3, 4, 9 are added in turn to give the chain

{2,7,8}c{2,3,7,8}c{2,3,4,7,8}c(2,3,4,7,8,9).

(3.2)

The reader could check that this is one of the chains which the proof of Theorem 3.1.1 would construct! In terms of parentheses, the basic elements determine all the closed parentheses uniquely and leave all the unpaired parentheses to be

34 1 Combinatorics of finite sets

fitted in. The chains are constructed by first taking all the unpaired parentheses as left ones (giving A-), then changing the first of them to a right parenthesis, then the second, and so on. The chain (3.2) is, in terms of parentheses,

Observe how the closed parts remain unchanged while the remaining parts change step by step from left to right parentheses. If two subsets of S have the same basic elements, they will have the same closed parts and hence will give rise to (and be contained in) the

same chain. The set of all chains obtained in this way forms a symmetric chain decomposition of the set of subsets of S; every set

lies in a chain, and the chains are disjoint since they are each determined by basic elements and each subset has a unique set of basic elements.

We can now see that the chains formed in this way are precisely those obtained in Theorem 3.1.1. For small values of n we could find the chains explicitly and verify this; therefore we proceed with an induction argument. Suppose that the chains of Theorem 3.1.1 for an

n-set are the same as those of the parenthesis construction, and consider how the chains for an (n + 1)-set are formed. The first new

chain in the de Bruijn construction is obtained by extending an to its final set:

existing chain by adjoining

A, A, U

A2 U

...

A,, A,, U

This corresponds to appending `(' to each string of parentheses representing the sets A,, . , Ah and then changing it to `)' to represent A,, U The resulting chain is precisely a chain of subsets of {x,, ... , with the same basic elements as those of the original chain. The second new chain formed from A, c ... c A,, is obtained by appending to each of A,,. . . , A,_,, i.e. appending `)' to the strings of parentheses representing A,, ... , Ah_,. The fact . .

that A,, is omitted corresponds to making the last non-basic element, say x;, of Ah no longer available. This is achieved by forming a closed

pair of brackets, `(' at the jth place and `)' at the (n + 1)th place;

Symmetric chains 1 35

therefore the resulting chain consists of all subsets of {x1, . . . , whose basic elements are precisely those of the original chain together with Thus again the new chain consists of subsets with the same basic elements. The chains possess some elementary properties which are worth recording. Theorem 3.3.1 Let A, c ... c A, be one of the symmetric chains of subsets of the n-set S constructed as above. (i) If A;+1= A; U {x,,,;}, then m; < m;+1 for each i, and the m; are alternatively odd and even. (ii) The number of basic elements in each of the A. is 2(n + 1 - r). (iii) For each i, there is a set B. such that Ae_1 c B; cA;+1 and with B; in a shorter chain.

Proof (i) That m; < m;+1 is immediate from the construction. Note

also that any block of closed parentheses must contain an even number of elements, so that m;+1 = mi + (even number) + 1. (ii) We have

n = (number of paired elements) + (number of unpaired elements) = 2(number of basic elements) + (number of unpaired elements).

Now the chain is constructed by adding one unpaired element at a

time, so that the number of unpaired elements must be r - 1. Therefore

n = 2(number of basic elements) + r - 1. (iii) The sequences of parentheses representing Ai_1 and A;+, differ in two places. We have

Ai-1:... (XX(...

A1:...)XX(...

A;+1: ... )XX)... where XX denotes a (possibly empty) closed part. Consider the subset B; represented by

B,:... (XX).... It clearly satisfies A;_, c Bi c A;+1. Also, since a further bracketing is now possible in the sequence for Bi it follows that B; has more basic

36 1 Combinatorics of finite sets

elements than A;, and so, by part (ii), the chain in which it lies is shorter.

There are yet more equivalent ways of constructing the same symmetric chain decompositions as those described above, and we briefly mention two of these. If we look at, for example, the 2-subsets of (1, . . . , 4) in lexicographic order (alphabetical, with 1 in place of a, 2 in place of b, etc.), and assign to each the first available 3-set containing it we obtain {1, 2)-{1, 2, 3} {1, 3)-{1, 3, 4} {1, 4}-{1, 2, 4) (2, 3)-{2, 3, 4) {2, 4) -none left {3, 4)-

Note that this is precisely the pairing between 2-subsets and 3-subsets in the symmetric chain decomposition of Example 3.1. Aigner (1973) has shown that this always holds, so that the chains can be built up rank by rank. Another approach is due to White and Williamson (1977). To find out which set covers the set {a,, . . . , a,,) in its chain, work out a; - 2i

for each i (taking ao = 0) and define t to be the largest interger i, 0 < i , h, for which a; - 2i is minimal; then the set covering {a,, ... , a,,} in its chain is {a...... a,,, I +ar}. For example, to find

the set covering {2, 3, 7, 8}, work out 0 - 2.0 = 0, 2 - 2. 1= 0,

3-2.2=-1, 7-2.3=1, and 8-2.4=0; take t=2 to obtain the set {2, 3, 4, 7, 81 since 1 + a2 =1 + 3 = 4. 3.4

Applications

We now give some applications of the symmetric chain structure.

3.4.1 The number of antichains Given an n-set S, how many antichains of subsets of S are there? We shall accept as antichains two rather dubious candidates, namely the empty antichain which contains no subsets and the antichain consist-

ing of only the empty set. This is to enable us to formulate an equivalent version of the question posed above. We show that antichains of subsets of S are in one-one correspondence with increasing functions from 9'(S), the set of subsets of S, to

Symmetric chains 1 37

{O, 1). Here we say that a function f : 91(S)--- {O, 1} is increasing if

f (X) --f (Y) whenever X c Y. Given an antichain d, we can define f : P(S) - {O, 1} by f (B) = 0 if B is a subset of a member of d and f(B) = 1 otherwise. To check that this f is increasing we need only check that if B c D and f (D) = 0 then f (B) = 0. However, if f (D) = 0 then D c A for some A E .saf ; then B c A also, so f (B) = 0. It is clear that under this correspondence the sets in the antichain

are precisely the maximal sets on which f takes the value zero. Conversely, given an increasing function f : 6P(S) -3 {0, 1), the maximal sets on which f takes the value zero will form an antichain. Thus the required correspondence has been exhibited. Note that the empty antichain corresponds to the function f which takes the value unity everywhere.

The number of antichains is therefore the number of ways of defining an increasing function f : 9'(S)-+ {0, 1}. We therefore consider the problem of constructing as many increasing functions as we

can. Suppose that we decompose the set of subsets of S into symmetric chains, and that we attempt to construct f by defining it on the sets in one chain at a time. More precisely, suppose that we have

succeeded in defining an increasing function f consistently on all chains of size less than k and that we now attempt to extend the definition off to a given chain of size k. If k = 2, any chain of size k is of the form X c Y. Here there are at most three possible choices of values for f(X) and f(Y): we could have f (X) =f(Y) = 0, or f (X) =f(Y) = 1, or f (X) = 0 and f (Y) = 1. Certainly the choice f (X) =1, f (Y) = 0 is not allowed. Thus there are at most three possibilities. If k > 2, consider any two sets X, Y in the chain under consideration with IYI = IXJ + 2. We then have

..XcZaY... for some set Z. Now by Theorem 3.3.1(iii) there is a set U, X c U c Y, with U in a shorter chain, and hence with f (U) already defined. It follows that if the chain X,CXZC.. . CXk is being considered, then, for each i = 2, . . . , k - 1, there is a set U; such that X_, c U c X+, and with f(U;) already defined. If f(U) = 0 for all i, 2 -_ i , k - 1, then we must take f (X _,) = 0 for each i , k -1 and we are possibly free to define f on Xk_, and Xk. As before, there are at most three possible choices here, f(Xk_,) =1, f(Xk) = 0 not

being allowed. So again there are at most three possible ways of

38 1 Combinatorics of finite sets

extending f. If f (U,) = 1 for all i, 2 -- i , k - 1, then similarly there is only possible freedom of choice at f (XI) and f(X2), and again there are at most three possible ways of extending f. Finally, if f is not constant on the U,, let j be the first i at which f (U,) = 1. Then f (X,) = 1 for all i > j and f (X,) = 0 for all i < j -1, so we are only free to define f on X _1 and X j. As before there are at most three possible choices.

Thus as each chain is considered in turn, there are at most three ways of extending the definition of f consistently to that chain. The total number of possible increasing functions f is therefore at most 3'

where m is the number of chains. Since m =

we therefore

have the following theorem.

Theorem 3.4.1 (Hansel 1966) The number of antichains of subsets of an n-set is at most

0

3([.2J).

By using a much more complicated argument based on the same ideas, Kleitman and Markowsky (1974) later reduced this upper bound to 2(1+0(log n/n))((:OJ)

More recently, Korshunov (1981) has published a complicated asymptotic formula for the number of antichains. In the case of even n, the estimate is

p (n/2). )(2-' + n22 n5 _ n X 2 -n-4)

2(-"12) ex p

Even in the case of n = 6 this gives remarkably good results: 7 996 118

instead of the exact value 7 828 354. The first few exact values are tabulated below. n

Number of antichains

5

3 6 20 168 7581

6 7

7 828 354 2 414 682 040 998

1

2 3

4

Symmetric chains 1 39

3.4.2 A storage problem The following problem arises in the mathematical theory of information storage and retrieval. Given a collection 99 of subsets of an n-set S, find the shortest sequence T of elements of S such that every set in 90 appears at least once as a subsequence of consecutive terms of T. In the case where A is the set of all subsets of S, the problem is to

find as short a sequence as possible containing every subset as a subsequence of consecutive terms. If n = 5, for example, and S = {1, 2, 3, 4, 5), the sequence

1234512413524 satisfies the requirements.

A trivial upper bound for the length of the shortest sequence is obtained by considering the sequence obtained simply by writing down each subset one after the other. Since there are 2" subsets of average size 2n, the length of the resulting sequence is 2In2". However, an improvement on this can be achieved by exploiting the existence of a symmetric chain decomposition of the poset of subsets.

Theorem 3.4.2 (Lipski 1978) If S is an n-set, there exists a sequence of length (asymptotically) E(2/r)2" which contains every subset of S as a subsequence of consecutive terms.

Proof. We consider first n even, say n = 2k, and we write S = T U R

where ITI = I R I = k. Then both T and R have symmetric chain decompositions of their posets of subsets into m = ([k2]) symmetric

chains: P(T) = 01 U ... U Pm, P(R) = yi, U ... U RV,,,. Corresponding to the chain 4,, Oi : {x,, ... , xi} c {x1i .. . , xj, xi+1} c ... c {x...... x,,}

(h !j)

we associate the sequence U = x1x2 ... x,,. Then every subset of T occurs as an initial part of one of the sequences U1, . . . , U,". Similarly let V1, ... , V. be sequences corresponding to the chains `Y1, ... , V. such that every subset of R occurs as an initial part of some V. If we

let V denote the sequence obtained by writing V in reverse order, then every subset of R occurs as a final part of one of the V . Now consider the sequence V1 UI V, U2 .

.

. V1 U. V2 U1 V2 U2 ... V2 Um ... V. U1 ... V. Um.

If A is any subset of S we can write A = E U F where E c T and F c R. Now E occurs as the initial part of some Ue and F occurs as the

40 1 Combinatorics of finite sets

final part of some 11f: thus A occurs in the above sequence as part of YfU,. Thus the sequence contains every subset of S as required. The 2 length of the sequence is F. 2m2(k + 1) = 2(k + 1)( \[k/2]1 However, (k2ir)"22k

(tk12k

by Stirling's approximation, so that

2m2(k+1) =k

. 22'`.

k=42".

This estimate for the length of the sequence can be halved by using Exercise 3.10, thus proving the theorem. The proof for n odd is similar.

3.4.3 A probability result related to Sperner's theorem Another application of the symmetric chain structure leads to an elegant probabilistic version of Sperner's theorem. We investigate the

following problem. Suppose that two subsets A, B are chosen independently and at random from the set of subsets of an n-set S, and suppose that the underlying probabilities of each being chosen are known. Then what can be said about the probability that A C- B?

We shall show that, no matter what the underlying probability distribution is, the probability that A c B is at least 1 Lemma 3.4.3 Let p1, ... , p,, be a probability distribution on an m-element set T. Then the probability that two elements of T chosen at random are identical is ,1/m, irrespective of the p;. Proof The required probability is E7=1 p; . Now we clearly have M

1

(P' so

P+-2 r=1

0

l

- M/

m '=1

P'+m.1 2,0 m

whence 1 m ,2 YIPI -m m,_1Pi-m =1

2

1

m

m=m

1

0

Symmetric chains

1

41

Lemma 3.4.4 If {q(x)} is a probability distribution on a poset P and P is partitioned into m chains, then the probability that two elements of P chosen at random lie in the same chain is --1/m, no matter what the values of q are.

Proof In Lemma 3.4.3 take T to be the set of chains. The required probability is the probability of choosing two chains at random and obtaining the same chain twice. The probability p, of choosing the ith chain is given by pi = E q(x) where the summation is over all x in the ith chain. The result is immediate from Lemma 3.4.3.

Note that the probability that two elements chosen at random lie in the same chain can be written as

Y,

xeP

q(x)( Ey

x,ym

q(y)) _ Z q2(x) + 2 x

same chain

E q(x)q(y)

x Nk since otherwise (3.4) is trivial. Now, by the normalized matching property, U AJ

iEE

Nk+1 IEI

Nk

Nk+'IFI %U BJ

Nk

so that

(uA1)n(uB1)I=IUAiI+ UB' JEF !EE jEF !EE

l

l

UA,)U(UB') JEP

iEE

Nk'(IEI+IFI)-Nk+j

-_Nk+1 (IEI +IFI -Nk) Nk

IEI + IFI - Nk.

Thus the Ford-Fulkerson condition is satisfied, and the chains can be extended to include all elements of ranks k and n - k. Repeating the process at each rank level gives a construction of the chains for P. If n is even the whole process begins with the elements of the middle rank in place of the pairs (u, v), and if n is odd the first step is to pair off

the elements of rank 2(n - 1) with those of rank 2(n + 1), which is possible since the normalized matching condition reduces to the Hall condition between the two ranks in this case.

52 1 Combinatorics of finite sets

Quantitative results in this area are also possible: here we ask how many symmetric chain decompositions does P have? Recall from

Exercise 2.20 that if the sets A,, ... , A. possess an s.d.r. and if I A; I , t for each i, then there are at least f (m, t) different ways of choosing an s.d.r. for the A;, where

fm,t

ifm - t, -{ t!t(t-1)... (t-m+1) _

if mPk(1_P)n-k(Mk-\k k=1 _

pk(1 _ p)n-k(Mk-\n

k

k

+lE Pn-i(1_pyM"-;=

pk(1

k

1

-j - J) 1

_ p)n-k(Mk-(-111 n-1

+ I pn-i(1- py ((n) i