Undergraduate Texts in Mathematics
Apostol: Introduction to Analytic Number Theory.
Halmos:
Armstrong: Basic Topolog...

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Undergraduate Texts in Mathematics

Apostol: Introduction to Analytic Number Theory.

Halmos:

Armstrong: Basic Topology.

Halmos: Naive Set Theory.

Bak/Newman: Complex Analysis.

boss/Joseph: Elementary Stability and Bifurcation Theory.

Banchoff/Wermer: Linear Algebra Through Geometry.

Childs: A Concrete Introduction to Higher Algebra.

Vector

Spaces. Second edition.

Janich: Topology.

Kemeny/Snell: Finite Markov Chains. Klarnbauer: Aspects of Calculus.

Chung: Elementary Probability Theory with Stochastic Processes.

Lang: Undergraduate Analysis.

Croom: Basic Concepts of Algebraic Topology.

Lang: A First Course in Calculus. Fifth Edition.

Curtis: Linear Algebra: An Introductory Approach.

Lang: Calculus of One Variable. Fifth Edition.

Dixmier: General Topology.

Lang: Introduction to Linear Algebra. Second Edition.

Driver: Why Math? Ebbinghaus/FlumjThomas Mathematical Logic.

Lax/Burstein/Lax: Calculus with Applications and Computing, Volume I. Corrected Second Printing. LeCuyer: College Mathematics with APL.

Fischer: Intermediate Real Analysis. Lidl/Pilz: Applied Abstract Algebra.

Fleming: Functions of Several Variables. Second edition. Foulds: Optimization Techniques: An Introduction. Foulds: Combination Optimization for Undergraduates.

Macki/Strauss: Introduction to Optimal Control Theory. Malitz: Introduction to Mathematical Logic.

MarsdenlWeinstein: Calculus I, II, Ill. Second edition.

Franklin: Methods of Mathematical Economics.

continued after Index

Dennis Stanton Dennis White School of Mathematics University of Minnesota Minneapolis, MN 55455 U.S.A.

Editorial Board

F. W. Gehring Department of Mathematics University of Michigan Ann Arbor, Ml 48109 U.S.A.

P. R. Halnios Department of Mathematics University of Santa Clara Santa Clara, CA 95053 U.S.A.

AMS Classifications: 05—01, 05—A05, 05—A IS

Library of Congress Cataloging in Publication Data Stanton, Dennis. Constructive combinatorics. (Undergraduate texts in mathematics) Bibliography: p. Includes index. 1. Combinatorial analysis. 1. White, Dennis, 1945— II. Title. 111. Series. QA164.S79 1986 511.6 86-6585 © 1986 by Springer-Verlag New York Inc. All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag, 175 Fifth Avenue. New York. New York 10010, U.S.A.

The use of general descriptive names, trade names, trademarks, etc. in this publication, even if the former are not especially indentilied, is not to be taken as a sign that such names, as by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Printed and bound by R.R. Donnelley & Sons, Harrisonburg, Virginia. Printed in the United States of America.

987654321 ISBN 0-387-96347-2 Springer-Verlag New York Berlin Heidelberg Tokyo ISBN 3-540-96347-2 Springer-Verlag Berlin Heidelberg New York Tokyo

vi

complexity, lattice theory, group theory, representation theory, special functions or mathematical physics. In this book we use combinatorial algorithms for two purposes. First, a

constructive proof of a theorem can be an algorithm. These algorithms often describe a bijection between two finite sets. So we concentrate on interesting mathematical

theorems which are proved by bijections. The other purpose is interactive: use the algorithms to investigate interesting mathematical examples. Here the examples are our main focus. An algorithm can be used to generate data related to a problem. It is then up to the students to study these data, formulate as many conjectures as they can, and then prove them. They are not told what the theorems are in advance.

Unfortunately, this kind of 'research" is usually impossible in most undergraduate mathematics courses. The material here is more than what can be covered in a 10 week course. Two sections of peripheral interest are 1.4 and 2.4. Moreover, some of the material in

could be considered graduate material. and Chapters 3 and 4 Strictly speaking, each chapter can be presented independently, although we frequently tie together material from different chapters. There are many other topics which would have been suitable for inclusion. One such topic we regretted omitting was the Lagrange inversion formula (see [La] and IRa]). The notes are organized in the following way. In Chapter 1 algorithms which list fundamental combinatorial objects are given. They are written in a shorthand version of Pascal (no declaration or i/o statements are given). It is assumed that the students are familiar with a programming language, though not necessarily Pascal. In Chapter 2, a partially ordered set is defined for each object. We concentrate on the Boolean algebra. A number of interesting bijections are given in Chapter 3 for these objects. Finally, we generalize bijections to involutions in Chapter 4. There is some

emphasis on tableaux in these last two chapters. Thus they can serve as a combinatorial forerunner to the theory of representations of the symmetric group. We have included more complete Pascal programs in the Appendix. Furthermore, we would be happy to provide disks (Apple Macintosh Pascal© or Turbo Pascal©) with source code for these programs to interested readers. The exercises vary from true exercises to very difficult problems. We have assigned each exercise a number from one to four, which we believe is some indication of its difficulty (one is easy, four is hard). Exercises involving a computer

are marked with a 'C". Exercises labeled 3C or 4C might be suitable for a term project. We feel strongly that anyone using this book as a text should assign one or more of these

Notes

102

Exercises

102

4 Involutions

110

The Euler Pentagonal Number Theorem Vandermonde's Determinant

ill

4.2 4.3

The Cayley-Haxriilton Theorem

120

4.4

The Matrix-Tree Theorem

125

4.5

Lattice Paths

130

4.6

The Involution Principle

141

Notes Exercises

147

4.1

114

148

Bibliography

156

Appendix

159 159

A.3

Permutations Subsets Set Partitions

A.4

Integer Partitions

166

A.5

Product Spaces

167

A.6

Match to First Available

169

A.7

The Schensted Correspondence

171

A.8

The Prüfer Correspondence The Involution Principle

176

A.1

A.2

A.9

Index

162

164

178

180

2

he used for virtually any combinatorial object. We shall see in Chapter 2 that it also has many remarkable and surprising theoretical properties.

§1.1 Permutations

A permutation of n distinct objects of length k is an ordered anangement of any k of the objects. For instance, the permutations of {a, b, c, d} of length two

are ab, ac, ad, ba, bc, bd, Ca, cb, cd, da, db and dc. The next proposition is clear.

PRoPosmoN 1.1 The number of permutations of n objects of length k is n(n—l)"(n—k+1).

Sometimes we shall write (n)k (called the falling factorial) for

n(n—l)"(n—k+1). A permutation of n objects of length n is frequently called a permutation of n objects (or simply a permutation of n). It is clear that we can take the set [n] = { 1,2, ... , n} for the n objects. We shall frequently use this notation. Proposition 1.1 shows that the number of permutations of n is = n!. Perhaps the most natural ordering of the permutations of n is lexicographic (lex) order. We say that it precedes a in lex order, if, for some i, the first i

entries of it and a are the same, and the (i+l)th entry of It is less than the (i+l)th entry of a. The lex list of the permutations of 3 is 123, 132, 213, 231, 312 and 321. This ordering is quite simple. You are asked to consider it in Exercises 2 and 3. We shall return to lex order in § 1.2. We consider instead an algorithm to list all permutations of n that is due to Johnson [Joh] and Trotter [1']. It is based on a 'combinatorial proof' of n! = n (n—i)!: for each of the (n—i)! permutations of [n—i], there are n "positions" into

which n may be inserted. The algorithm has the property that each permutation differs from its predecessor by only a transposition of adjacent symbols. The lex list does not have this property. How does the algorithm work? Suppose we have the list for permutations of

Then we construct the list for permutations of [nJ by The insertions go inserting n into each of the n possible positions of each from left to right if i is odd and right to left if i is even. The lists for n = 1, 2, 3 [n—li:

and 4 are given below with the recursive structure indicated.

4

1

Done 4— false

white not Done do PrintØt)

if A 0 then m4—max{i: iEAJ ,r[j]

+ d[m]] xli + d[m]] m 4—

+ d[m]

il:-1[7t[j]] 4—j

if m

Let

be the collection of complements of the subsets of Certainly consists (n—k)-element subsets of [nJ. Furthermore, no member of can be a subset of any member of (Suppose A B and A B. Then A n = 0,

of

and A and

are two members of which would be disjoint.) The picture below describes this situation, since k n — k,

We now obtain a lower bound on the number of k-element subsets which lie below If we apply the map a n —2k times to (call this iterated map an-2k), we obtain all of these subsets. Let (4.2)

1) + ... + (ai —

= I

n—k

1

1

Since

(4.3)

= VFI

=

>

afl_k must be n. So by the Kruskal-Katona Theorem (Theorem 4.3),

(4.4)

>

(n-ii).

Repeating this n —2k —

1

more times gives

48

whichmeans A—{1}isin

and A—{j}, So

Therefore, all of

is in

(A—{j})u{1}

Let A—{k}€

E

A, 1

Finally we assume that j

Since (A—{k,j})u{1} SJ(A — {k})

the switching map S3 fixes A — {k}, so A — {k} = case is A —

{j} = SJ(A — {j})

A and The last

c

then JE A, A and subsets (A—{k,j])u{1} and A—{j} in If

The

can be checked as in the

previous paragraph.

Suppose we iterate the switching maps S3 for various j, until

toacollection

suchthat

is converted

forall j. Thisispossiblebecause S3 either

or gives

fixes

more members which contain 1. Since the switching maps do not change the size of the collection, Lemma 4.5 implies that mis completes Step (1). Step (2) Let First we show that

u

as indicated. Clearly

=

(4.7)

where a1 is the operation of deleting 1 from a set. Let B = A — (j} E j 1, A F(T). Since 1' is fixed under B u { i} so B E This establishes (4.7), which clearly implies (4.8)

We now separate the collection deleted,

into two subcollections: those subsets with 1 and those subsets with some element 1 deleted,

'u { 1 }. (This notation means that we first delete 1 from each member of

next apply a, and then reinsert I into each member.) Because these two are disjoint, (4.9)

=

This completes Step (2).

Step (3)

Suppose that

+

50

(4.16) Taken together, (4.15) and (4.16) contradict (4.13). So (4.10) must hold, and the

proof is complete.

Notes Three good general references for posets are [Ai], [Be] and [Gre-K 1]. Spemer theorems are included in §3 of Chapter Vifi of [Ai]. They are also a central topic of

[Gre-Ki]. Exercise 7 below, and Exercises 18 of Chapter 3 and Exercise 20 of and qj are Chapter 4 establish the entries of the chart for The entries for

given in Exercises 8 and 10. For 11 and 12, and for

they are Exercises 9 and 22, for

Exercises

Exercises 13 and 29. A matching between two adjacent

levels of the Boolean algebra can easily be shown to exist from Halls Theorem. The decomposition into symmetric chains is not guaranteed from this technique. The fact that matching to first available in the Boolean algebra works is due to Aigner [Au. The can be found in [Wh-Wi]. relationship between the various matching schemes in

Kleitman's solution of the Littlewood-Offord problem appears in [Gre-K 1]. The

proof of the Kruskal-Katona Theorem is due to Frankl, [Fr].

Exercises Given a finite poset (P, ), show that there is at least one way to totally so that a1 order P, that is, label all of the elements of P with [n]: a1, ... , 1.[2]

implies i j. 2.[2C]

Such a labeling is called a linear extension of (P,

Write a program which takes as input a poset P and gives as output a

linear extension of (P, ). For each poset (P, ) pictured below, write a program which finds the total number of linear extensions of (F, ). Can you prove any theorems here? 3.[3C]

52 12.[3]

1

and ii.

is a lattice by finding the meet of two partitions

Prove that

(Hint: Take the smaller of the two partial sums to define the partial sum of the meet.

Reason similarly for the join.) 13.[2]

Prove that

is rank unimodal, rank symmetric and order symmetric.

(The Spemer property follows from Exercise 29.)

14.[l]

What is the rank of (n, n—i, ... , 1) in

A chain C is called a maximal chain if C is not contained in any other (n, n—i,..., In chain. What is the length of a maximal chain in 15.[4C]

1)? Write a program which gives the number of maximal chains each poset has. What are your conclusions? [Sta2]

have?

and

16.[2]

How many maximal chains do

17.[2]

How many edges do the Hasse diagrams of

1 8.[3C]

Write a program to count the number of edges in the Hasse diagram of

and

have?

Formulate a conjecture, based upon an appropriate combination of Bell numbers.

(Hint: Try

a

—

a combinatorial proof of this result?) = nm.

Write a program to compute the Whitney numbers of

19.[4C]

Formulate and prove as many conjectures as you can. §3.3 may be useful.

Suppose v0,

20.[2]

...

is a finite sequence of positive real numbers. We

,

is log-concave if Vk2 Vk_1Vk+1 for 1 k n —

say v0, v1, ...

,

v0, v1, ...

is log-concave, prove that it is also unimodal.

21.[3}

,

Suppose the polynomial v(x) =

roots. Show that v0, v1, ...

,

x1 + ... + v0

If

has negative real

is log-concave. Possible hint: This result has a

combinatorial proof. Let {—r1, ...

w(A) of a subset A C [n] to be

+

1.

,

be the roots of v(x). Define the weight

54

kind are unimodal. 23.[2]

Let S be a subset of [999] u {0}. If

76, show that S must

IS!

contain at least two numbers n and m, so that the difference n — m can be computed with no "borrowing'. (Hint: consider the poset C310.) 24.[3]

{a1,... ,

Let A =

{a1,

...

a1+ 1, a1÷1, ...

c [n], where a1 < ... <ar. Define g(A)

,

where t is the largest i for which a1 — 2i =is

,

minimal (assume a0 = 0). Prove that g(A) = f(A), where f is defined in Algorithm 14. [Au

Suppose A c [n] is represented as a sequence of n parentheses, where parenthesis i is right if i E A and left otherwise. Thus, 25.[3]

A = {1, 3,4,6,7, 10, 12, 13, 16, 19} c [21]

corresponds to ) 1

(

)

)

(

2

3

4

5 6

)

) 7

(

( 8

)

)

)

(

(

(

and if we pair off the parentheses in the usual way,

.. —

I

( 1

2

) 3

)

(

)

)

)

(

(

)

(

(

9 10 11 12 13 14 15 16 17 18 19 20 21

(

(

)

(

..

I

)

)

(

(

)

(

(

)

(

(

9 10 11 12 13 14 15 16 17 18 19 20 21

4 5 6 7 8

we are left with a string of unpaired right parentheses followed by a string of unpaired left parentheses. ) 1

))7 14( 17( 20 21 (

(

4

Now take the first unpaired left parenthesis and turn it around. ) 1

4

)

(

(

(

2021

The resulting string of parentheses

123456789101112131415161718192021

56 32.[4]

Generalize the Erdös-Ko-Rado Theorem to collections such that A, B E implies Al k and A B

of subsets of [n]

58 n

(0.3)

f(x) =

X

k= 0

Let B be the set of n-tuples of 0's and l's, and let the weight of any element = B be where has k l's. Define the bijection ip : A B by 1

=

if i a

{ 0 if

a.

The bijection p is weight-preserving, because w(a)

generating function for B is (1 + (0.4)

of

w(p(a)). It is clear that the

so (0.2) becomes

f(x) = (1+x)".

Another kind of combinatorial proof of the binomial theorem can be given when x is a non-negative integer. In this case, the right-hand side of (0.4) counts functions from [n} to a set S of size 1 + x. The right-hand side of (0.3) counts the same

functions by classifying by the number of members of [n] which get sent to the first x members of S. The theorem can easily be extended to all real x. In this chapter we give examples of all these phenomena. The Catalan numbers provide bijections between apparently unrelated sets. The Prilfer correspondence associates a simple B with a complicated A. Partitions and permutations illustrate these ideas. The most difficult bijection that we consider is the Schensted correspondence between permutations and tableaux. Several of the constructions in this chapter and the next involve graphs. While

none of the graph theory concepts used are difficult, we will state here some of the key definitions and results involving graphs.

A graph is a set of vertices and a set of edges. The edges are usually a collection of 2-element subsets of the vertex set (such graphs are called simple), but sometimes one-element subsets (loops) or repetitions (multiple edges) are allowed. If the edges are ordered pairs of vertices, the graph is called directed (or a digraph) and the edges are directed edges. The degree of a vertex is the number of vertices incident to it. A vertex in a directed graph has an in-degree and an out-degree.

A path in a graph is a sequence of adjacent vertices. A path is simple if no vertex is repeated (except that the first may equal the last). A cycle is a simple path which starts and ends at the same vertex. We sometimes will refer to the unordered set of vertices in a cycle as a cycle. A graph is connected if there is a path from

60

(3) Full binary tree Afull binary tree is a binary tree where every vertex has

either 0 or 2 sons. (4) Well-formed parentheses A sequence of parentheses is called well-formed

if, at any point in the sequence, the number of right parentheses up to this point does not exceed the number of left parentheses up to the same point. Moreover, the total number of left parentheses equals the total number of right parentheses.

(5) Ballot problem Suppose Alice and Barbara are candidates for office. The result is a tie. In how many ways can the ballots be counted so that Alice is always ahead of or tied with Barbara?

(6) Standard tableaux Given a partition of n, a standard tableau T is an arrangement of [n] in the n cells of the Ferrers diagram of which increase across rows and down columns. These objects will be discussed in greater detail in §3.5. THEOREM 1.1 The following sets of objects all have the same number of elements,

and this number is (1) (2) (3) (4) (5) (6)

binary trees on n vertices; ordered trees on n + 1 vertices; full binary trees on 2n + 1 vertices; well-formed sequences of 2n parentheses; solutions to the ballot problem when 2n votes are cast; and standard tableaux in a 2 x n rectangular Ferrers diagram.

Proof We use bijections to show (1 )-(6) are equinuinerous. Then we show that (4) yields the Catalan number. (1) = (2): We give a bijection

from binary trees to ordered trees. Let B be a

binary tree. Here is how we construct T = (p(B). (a)

The vertices of B are the vertices of T with the root deleted.

(b)

Therootof B is the first son of the root ofT. Vertex v is a left son of vertex w in B if and only if v is the first son of w

(c)

in T. (d)

Vertex v is a right son of vertex w in B if and only if v is the brother to the

right of w in T. In the examples below, we have labeled the vertices to help the reader trace what happens under (p.

62 0

2

2

)(nn)n II

5

I

3

1

P

T To define

6

4

[3

5

1

let qr1(Ø) = the tree consisting of only a root, Again, by = T has been defined for all sequences

induction, assume

and ordered trees T with k +

1

length 2n. Write P = P1 P2...

vertices, k

investigate this statistic in Exercise 29.

82

while

2, (1, 2, 1, 1)) has

11121 1214!

34

1112141 12131

Itis not an accidentthat the numberoftableaux in each set is the same. Itis ournext theorem.

THEOREM 5.1

There is bijection between

p'), where p is

p) and

obtainedfrom p by applying an adjacent transposition.

Proof Let p' agree with p, except for interchanged. Let Tn

and Pk+1' which have been

p) so that T has Pk k's and Pk+1 k+l's. We need

to produce a tableau T' n

p') which has Pk k+l's and Pk+1 k's. We will do

this by switching some k's in T to k-i-i's, and also switching some k-i-i's in T to k's. The k and k+1 entries in row i of T have the following structure.

row

'

'

1

row i+1

k

'

k k+1

... k k k-i-1'--k-i-l k

i

c

b

a

row i -

k+1 k-i-i

-.. k k+1 d

For

with k+l's below them. Immediately to the right of these k's, there will be some number b 0 of k's with no k-i-i 's below, then some number c 0 of k+ l's with no k above, and finally some number d 0 of k+l's with k's above them. We change T to 1' by changing each row i to this form, this row, we let a

0 be the number of k's

a row i -

row

1

i

row i + 1

c

b

'

k k k+ 1 ... k+ 1 k

' ...

k k+1

-'

k÷1

k

k

k+1 ...

k+1

d

Note that

the k's and k+l's which are paired with k+i's below and k's are left unchanged. However, the b k's have become c k's, and the c k+l's have become b k+l's. So the total number of unpaired k's in T is equal to the total number of unpaired k-i-I's in T'. This implies that T' has Pk+1 k's and

above

k-i-i's.

It is easy to see that T' is column strict,

84

THEOREM 5.3 The number of standard tableaux of shape X,

equals the number

of maximal chains in Young's lattice

It is simple to state and difficult to prove.

There is an amazing formula for

We will give a proof of an equivalent formula in §4.5. To state this formula, we need The We write c E to define a hook. Let c be a cell of the Ferrers diagram of

hook of c, HC, consists of the cells to the right of c, below c, and c itself. In the bijection proving Theorem 3.3 of this chapter, we used the hooks of the major In the example below the hook of is diagonal. The length of the hook of c,

c is shaded and its length

= 7.

=1

I

I

I

I

C

is called the hook formula.

The formula for

be a partition of n. Then

THEOREM 5.4 Let

n!

—

CE

4 22. We insert the hook lengths into the cells of ?..

As an example, take

6 5 2111

32 21 Then d422

=

8!

=

56.

86

entries insert the number 4 into P" and Q" (in some way) and add one to all of the of P" and Q" which are 4. This will give us P and Q. Of course, the difficult part of this is to determine exactly how to insert the number 4. This algorithm is called the Schensted column insertion algorithm. First, we should remark that it is not necessary to do the addition and subtraction of one to the entries of P" and Q". The Schensted correspondence will and the produce tableaux (P', Q') for it', where the entries of F' are [8] —

entries of Q

are

[7]. For it = 35186724,

it = 3518672,

P'=

The Schensted column insertion algorithm inserts 4 into P' and 8 into Q'. The natural position of 4 in the first column of F is between the 2 and the 6. The 4 takes the place of the 6, or bumps the 6 out of the first column of F'. The 6 now is inserted by the same method into the second column of P'. This time the 6 is placed after the 5 and does not bump any entry of the second column. The column insertion algorithm has been completed. The resulting tableau P is 81

LiV that the shape of P differs from that of P' by the addition of exactly one cell (the cell 6 of P). We place an 8 in that cell for the definition of Q'. N1ote

—

111371 1215

48 6

A given insertion could cause several bumps. An example in the general case is given later.

To find P and Q from it, just apply the column insertion algorithm successively to the entries of it. If we use two line notation for it,

88

Removing c from Q gives Q. Next, we start with the largest entry of Q',

and do

this inverse bumping procedure until another entry is bumped from the first column of

P., and continue. We will not give a formal proof by induction of this case. Instead, we will consider the generalized Schensted correspondence for multiset permutations, Let m be a multiset with content (p1, ...

that m contains p1 is. A permutation

so

of m is any sequence of the elements of 111. For (p1,..., such

is

2322132. Clearly if each p1 =

1,

= (1,

4, 2), one

IlL has no repetitions so that multiset

permutations are just usual permutations. The generalized Schensted correspondence will associate to 7t the pair of tableaux (P, Q), where P is column strict of content p, and Q is a standard tableau of the same shape as P. For = 2322132 we will

seethat

(P,Q)=

(

13451 267

12221

233

'

To produce P and Q, we use the same bumping procedure with a minor modification. Suppose we are inserting k into a column. The number which is bumped out of the column is the smallest number k. (In the previous case, it was >k because repeats were not allowed.) For example, suppose we are inserting 4 into the column strict tableau P. 1

P— —

1

1

2 4 4 6 16

35567

I

7

Then the 4 will bump the 4 in the first column, so that the new first column is the

old first column.

Lii

The bumped 4 now bumps the 5 in the second column, for the following first two

90

The 7 from column 6 is larger than everything in column 7. It is therefore placed at the end of column 7, and the insertion of 4 into P has been completed. 1

1

1

244

23 3357 34 556 47 77

6 61 7

6

The new cell has been marked in bold face. We would place the next entry of Q (here

a 25) in that cell. The reader should verify that = 2322132 gives the (P, Q) that was previously claimed. The entries of Q are 1234567, the first row of the two line notation for it. The inverse Schensted correspondence (P, Q) —* it is as before. We use the largest entry of Q to find the entry e of P which bumps to the left. This time the largest number of the column which is e is bumped to the left (Note that the entry directly to the left of e is e, so that this set is non-empty.) The number eventually bumped out of the first column is again the last entry of it. We call this inverse procedure column deletion. Because column deletion is the inverse to column insertion, it is easy to see by induction that the generalized Schensted correspondence is a bijection. THEOREM 6.2 The generalized Schensted correspondence is a bijection between all

multiset permutations it of content p, and pairs of tableaux (P, Q), where P is column strict of content p, and Q is standard with the same shape as P.

Suppose that p =

(p1,

Pm)' so that the multinomial coefficient gives the

number of multiset permutations it. The number of ordered pairs (P, Q) is a sum of Kostka numbers times so that we have this corollary. and

COROLLARY 6.3

Pl+...+Pm=n, then

=

where the summation is over all partitions

of n.

92

Cell(P, k) 4— (i,

end.

The column deletion algorithm begins at row r and column c of the tableau P. The tableau after deletion is Del((r, c), F) and the value bumped out is Val((r, c), P). ALGORITHM 18: Schensted Column Deletion

begin

F

P

x 4-- P(r, c)

for j

*— c

— 1

downto 1 do

i+-1 repeat i4.— i+ 1

until P'(i, j) > x or i >

if-i-i y

P'(i,j)

x

x*—y

F Val((r,c),P)—y

Del((r, c), P) end.

Now we give the Schensted encode algorithm. We use Ins(k, P) and Cell(k, P) from Algorithm 17. The permutation is it, whose two-line notation has top row j1, j2, ... , and bottom row It1, It2, ... , Ire. The resulting pair of tableaux is (Schp(it), SchQOt)). ALGORiTHM 19: Schensted Encode

begin

t P

4— Jns(1c1,

P)

Q(CeIlØt1, P)) 4—

Schp(it)

4—

IlJ

P

I

94

from the generalized Schensted correspondence. The answer to the first question is given by this theorem. THEOREM 7.1

The number of rows of P (or Q)

is

the length of the longest

increasing subsequence of it. In the example it = 35186724 of §3.6, P had 4 rows, so it has an increasing

subsequence of length 4, 3567, and none longer. To prove Theorem 7,1, we need to describe what happens to P and Q first column only at each stage. Suppose, as in the previous section, it = 35186724. Entry into 0

Column 1 of P

Action Taken

1

Insert3;Q114—1

3

1

2

InsertS;

3

1

5

2

1

1

5

2

1

1

5

2

8

4

1

1

5

2

6

4

1

1

5

2

6

4

7

6

3

4

5

6

7

8

Insert 1; bump 3

Insert 8; Q31

4

Insert 6; bump 8

Insert 7; Q41

+— 6

Insert 2; bump 5

Insert 4; bump 6

1

1

2

2

6

4

7

6

1

1

2

2

r 96

inserted below row t. This property characterizes the class of a pair.

LEMMA 7,2 The pair (i, itt) belongs to class t

and only

the length of the

largest increasing subsequence of it ending at it1 is t. Proof It remains to show if the length of the largest increasing subsequence of it ending at is t, then (i, it1) belongs to class t. We do this by induction on t. = 1, then iç is smaller than all of the preceding

row of column and (i, it1) belongs to class

If t

so it1 is inserted into the first

1.

Now suppose t> 1, and choose an increasing subsequence S of it ending at it1 of length t. Let be the predecessor to it1 in S. Then the subsequence S = S

—

{it1} which ends at

to class t — — 1

was

1.

is also of maximal length. By induction, (j,

belongs

So when it1 was inserted into the first column, the entry v in row < it1. Thus,

be the entry in row t when

it1 was inserted either below row t or in row t. Let w was inserted. The entry w (which precedes it1) is a

member of a pair in class t. By the first part of the theorem, there is an increasing

subsequence of it of length t ending at w. If w < itt, we could attach

to this

subsequence and have an increasing subsequence of length t + 1. This contradicts our hypothesis, so w > and was inserted into row t.

Proof of Theorem 7.1 Suppose S is one of the longest increasing subsequences of it and has length t and suppose that P has r rows. By Lemma 7.2, the last member

it1 of S is inserted into row t of P, so t r. Conversely, the (r, 1) entry of P is in class r, so Lemma 7.2 implies that r t.

The answer to our second question is provided by Theorem 7.3.

THEOREM 7.3 Suppose that it corresponds to (P, Q) in the Schensted correspondence. Then ir1 (the inverse of it) corresponds to (Q, P). Proof In fact, Theorem 7.3 holds for all two line arrays with distinct entries. In this case ic' is the two line array obtained by interchanging the two lines, then sorting the columns according to the first line. Our proof will be by induction on the number of columns of P. First we show that the first columns agree, and then apply induction. We initially consider the permutation case, but the reader should have no difficulty

98

\ 1k

For example, the class 3 pairs of it = 35186724 give

(458\ 6[4J1 '..

while the class 3 pairs of

158\

give

(rn6 8 \

(68

k85

This proves that the inverse of itb is ic1b, and completes the proof of Theorem 7.3.

I COROLLARY 7.4 The number of standard tableaux with n entries is equal to the number of involutions it of [n].

Proof An involution of [n] is a permutation it of [n] such that it = ir'. By Theorem 7.3, the Schensted correspondence is a bijection between all pairs of standard

tableaux (P, F), and all permutations it of [n] such that it = ic'.

The third application involves the matching f in the Boolean algebra of §2.2.

We shall use the generalized Schensted correspondence to obtain the matching f.

Let Ac[n], IAI=p, andwrite A asan n-tupleof n—p 0's andp l's. Written this way, we may consider A as a multiset permutation of p l's and n — p Os. Now apply Algorithm 19 to A to obtain (P, Q). Since P is column strict with

0's and p l's, P must have either one or two rows. If there is no 1 below the last 0 in row 1, change that 0 to a 1 to obtain a new tableau P. (For 34.[4C]

> 0 if and

and

then show that Kx

if P

ri.)

Write a program which finds the number of column strict tableaux of shape

(

A. whose entries are N. State and prove your conjectures. 35.[1]

Suppose that it 443511242. Find the P and Q tableaux of it in the

generalized Schensted correspondence. 36.[1]

Suppose

121214141 p

13141

,

14151

Q=

111415191 12161 13171

1

Find the multiset permutation it which cormsponds to (P, Q). From the Schensted correspondence prove the ErdOs-Szekeres Theorem: 37.[2] any sequence of n2 + 1 distinct real numbers has either an increasing subsequence or

I

decreasing subsequence of length n + 1. Use Algorithm 19 to list all involutions of [n] and the corresponding tableaux. Is there any relationship between the shape of the tableau and the cycle structure of the permutation? 38.[3C]

39.[2]

A lattice permutation is a multiset permutation it of

l's, A.2 2's,...,

n's, such that for any initial segment of it, the number of l's the number of

2's ...

the number of n's. Thus, 1121233213 is a lattice permutation while

I

4

Involutions

Many combinatorial formulas include positive and negative values. It might first seem that a bijection is not the proper tool for dealing with these formulas. This is not the case, however. Such formulas can sometimes be proved by using an involution on

a signed set. In fact, involutions may be used to prove theorems seemingly unrelated to combinatorics. This will be done in Section 3 for the Cayley-Harnilton Theorem. A signed set A is a set which has been partitioned into two subsets, A+ and A

with A+uA=A and

Theelementsof A+ and A

are

call ed positive and negative, respectively. We are interested in the value hAil = — IA1.

if some of the elements of A can be paired with some of the elements of

A+,

then the total size of the sets that we have to count to compute hAil is reduced. In fact, A and if we pair up all of A, then hAil is just the number of if

elements of A+ which are unpaired. More formally, such a matching is an involution on A, that is, a pennutation (p on A such that p2 = id. This involution has the if and only if (p(x) e A. Note that x, then x E property that whenever

this means that if x e A

then ç(x)

A+.

minus the precisely the number of fixed points of p in number of fixed points of (p in A. if we write F((p) for the fixed point set of (p, a new signed set and F(p) = F((p) and F(p) is empty. Such an = hAil. Typically, one or both of

Notice that

hAil

is

involution ip is called sign-reversing, for if x is not fixed by ç, then (p(x) has the sign opposite from x. An important formula from elementary enumeration theory is the principle of

inclusion-e.xclarion. This principle can easily be proved with a sign-reversing

involution. Suppose X is some finite set of objects and each of these objects is endowed with certain properties. A property may be thought of as a subset of X. Suppose P denotes the collection of properties. So associated with x E X there is a subset

P of properties with which x is endowed. For any subset T c P of

propertieslet

and ND(T)=.(xE

Thus,

112

(1.1)

[1 (1—x1) = 1=1

1

+

(_l)kx(3k2±k)/2

k=1

which is a special case of the Jacobi triple product identity [An]. It is called the arise when constructing

pentagonal number theorem because the numbers larger regular pentagons from smaller ones.

Proof Let PD(n) be the set of all partitions of n into distinct parts. Let PD(n) = PDO(n). This makes PD(n) into a signed set with sgn(A) = )# (—1 The idea of this proof will be to construct a sign-reversing involution p on PD(n) with no fixed points, unless n is of the form n = (3k2±k)/2, in which case p will have exactly one fixed point. The sign of this fixed point will be Clearly, if p has these properties, Theorem 1.1 follows. Suppose A PD(n). Recall that we write A = (A1, ... ) with > A2>....

(The inequalities are strict here because the parts of A are distinct.)

Let a(A) = max{j

:

+ 1 —j} and b(A) =

=

Thus b(A) is the smallest

part of A and a(A) is the length of the 'staircase" on the border of the Ferrers diagram of A. For A = (7, 6, 5, 3, 2), a(A) = 3 and b(A) = 2. If b(A) a(A), create a new partition p(A) of n by moving the b(A) part adjacent to a(A). Thus p('l, 6, 5, 3, 2) = (8, 7, 5, 3):

a(A)

• • • • • •

9

. .

•

•./>Y

• • • • • •

p(X)= : : : : : • • •

b(A)

Note that since A has distinct parts, b(p(A)) > a(p(A)) = b(A). The reader should carefully check this. If b(A)> a(A), then p(A) is obtained by creating a "part" consisting of the a(A) cells at the end of the first a(A) rows. This part is placed under the b(A) part, i. e., this new part is now the smallest part of p(A). For example, p(9, 8, 6, 3) =

(8,7,6,3,2):

114

As an application of Theorem 1.1, recall that the generating function for all partitions is given by

(1.2)

H (1— x'Y' =

p(n)

1

Clearing the fraction and substituting (1.1) gives

(1.3)

(_l)kx(31c2+ k)/2

p(n)

1.

Equating coefficients in (1.3) gives, for n > 0, (_1)k p(n—(3k2+k)/2) = 0.

(1.4)

For any given value of n, this sum is actually finite and gives a recurrence for p(n).

§4.2 Vandermonde's Determinant Sign-reversing involutions are a natural tool for handling identities involving determinants because the terms in the expansion automatically have signs attached. In this section we give a proof due to Gessel [Ge] which establishes Vandermonde's determinant using a sign-reversing involution. Vandermonde's determinant is (2.1)

=

H

J

It is clear that the product side of (2.1) has (II) 2

2

terms, while the determinant has only n! terms. We need a sign-reversing involution that cancels

Ifl\ 2

—n!

r 116

2

T=

F

3

1

4

5

on non-transitive tournaments To prove (2.1), we need a sign-reversing involution which preserves the weight. If such a q exists, (2.4) becomes

II

(2.5)

w(T)sgn(T).

=

T

transitive

Any transitive T corresponds to a ranking permutation 7t =

such that

; wins n — i games. For such T, w(T) =

n-i

Also, sgn(T) = (_l)m, where m is the number of inversions of so (2.5) becomes the definition of the sign of the permutation

II

(2.6)

=

sgn(it)

4'... 1

This is precisely

fl

But the right-hand side of (2.6) is the definition of the determinant in (2.1).

The proof then hinges on the involution ç. We need first a characterization of non-transitive tournaments that you are asked to show in Exercise 17. PROPOSITION 2.1 T is a non-transitive tournament and only with equal out-degree.

T has two vertices

Proof Exereise. The sequence of out-degrees, or wins, (a1, a2, ... , a.) is called the score vector

of T. Choose the lexicographically first pair (i, i),

such that

=

For example, the tournament below has score vector (2, 3, 0, 3, 2, 6, 5); choose i = 1

andj=5.

118

1

7 33

5

The involution (p reverses the edges of the 1-5-4 triangle:

7

5

Thus (p produces this tournament:

120

§4.3 The Cayley-Hamilton Theorem In this section we give a combinatorial proof of the Cayley-Harnilton Theorem. This proof once again uses a sign-reversing involution. It is due to Straubing [Str}. THEOREM 3.1 Let A be any n x n matrix over any field. Let

= det(?J

—

A)

be the characteristic polynomial of A. Then PA(A) =0. It might seem surprising that a theorem from linear algebra has a combinatorial

proof. However, sign-reversing involutions are a perfectly suitable tool for handling determinants, because determinants are signed sums of products of the entries of a matrix. In fact, it can be argued that the combinatorial proof we give here is the most

"natural" proof because it does not depend upon the field of scalars. Proofs of this theorem from algebra usually first prove a weak version for diagonal or triangular

matrices and then "extend" to all matrices. However, this extension requires that the scalars be the complex numbers, and some major theorem, such as the Fundamental Theorem of Algebra or Taylor's Theorem, must be used to eliminate the dependence on the complex numbers.

Proof We begin by writing the characteristic polynomial as a signed sum of products n

(3.1)

pA(X) =

sgn(it) H (AJ i—i

Each fixed-point i of it (i. e., lt(i) = i) will contribute either product, while each non-fixed-point i will contribute

or

to the

Equation (3.1) can now

be written (3.2)

=

S

(3.3)

[n] — S

c

sgn(it)

subset which satisfies

c fixed points of it = F(it).

Note that since it fixes everything in [n] outside of S, we may regard It as a permutation of the elements of S. Let P(S) denote the set of permutations of S. The sign of it will be the same when it is regarded as a permutation in P(S). So we may

122

Next, replace X in (3.5) with A. We wish to describe the ij-th entry of the resulting matrix.

(3,6)

(p (A)).. = A

'J

k0

IrEP(S)

(_1)d(7t) H tES

This is clearly

This involves describing (3.7)

alklj.

.

We visualize a term in this sum combinatorially as a directed path P of length n — k

from i to j on the vertices [n]. Moreover, let the weight of P, w(P), be the product of the weights of the edges of P where the weight of an edge is as above: w(e) aij if e = i —* j. Then w(P) is exactly a term in this sum. We may now give a complete combinatorial description of the right hand side of

(3.6). Let (S, it, P) be a triple such that (a) (b) (c)

S is a subset of En]; it is a permutation on S; and P is a directed path from i to j of length n —

with vertices in [n].

The weight of the triple, w(S, it, P), is w(P) w(it) and the sign of the triple, sgn(S, it, P), is sgn*(it). Then the ij-th entry of pA(A) is the generating function (3.8) =

(S

it, P) sgn(S, it, P).

Clearly, the set of such triples,

is a signed set. To prove the

Cayley-Hamilton Theorem, we need to show that (3.8) is zero. Thus, we require a

weight-preserving, sign-reversing involution q' on We can visualize a triple (S, it, P) E

as a directed multi-graph on the

vertices [n] with two kinds of edges: the edges from it and the edges used in P. Note that the path P may use an edge more than once and may also use edges in it; hence the graph is a multi-graph. As an example, let n = 9, i = 2, k = 4, and j = 5. Choose S ={l, 4, 6, 9}, it =(146) (9) and P = 211515.

124

the cycle of it containing v. Let P be P with C inserted at the position of v. Let be S with the vertices in C removed. Let be it with C removed. Note that the second occurrence of v in P now completes a cycle in P. No earlier vertex in P satisfies (2): those up to and including the first occurrence of v are the same as in satisfy (2) in P; those between the two occurrences of v were on P and they C and therefore could not have been encountered before v in P. Therefore

(S,,P) satisfies (2). Inourfirstexainple,

=

{9} and P = 214611515:

0 Now suppose v satisfies (2); let C be the cycle in P just completed. Then C

will be a cycle from v to v in P. Let P be P with C removed (including one be it with the be S with the vertices in C added. Let occurrence of v). Let cycle C added. This construction is legal because no vertex before the second occurrence of v could be in it or could be a repetition in P. Note that the first occurrence of v in P now satisfies (1) because v is now in it. In our second {1, 2, 3,4,5, 6, 9} and P=215:

example,

/\__

S

7

2=v

3

5

8

6

126

T =

1

"s. 6

2

Thus, w(T)

then w(T) =

1 i * j n. Let I denote the variables trees in

is

a monomial in the variables

1 i, j n. The generating function for the

is given by

Z

=

(4.2)

w(T).

T

=

To count the number of spanning trees of G, merely put

=

1

for each pair

foreachpafr (i,j) thatisnotanedgein

(i,j) thatisanedgein G and 0.

as an n x n determinant.

Let us now identify PROPOSITION 4.2 Let 1

i,j

Note: For n =2, Proposition 4.2 is —a12 1

(4.3)

f3(j) =

= a12a23+ a13 a21+ a13 a23,

det

a21+a23j

—a21

which is the generating function for the thee rooted labeled trees on [3]. 3

2

1

1

I

I

21

1

2

128

The weight of a triple, w(S, it, f), is w(f) w(lt); the sign of a triple, sgn(S, it, f), is sgn*(it) = as in §4.3. To prove Proposition 4.2, we must find a weight-preserving, sign-reversing involution p on these triples. Such a triple can be represented as a directed graph on [n+1] with two kinds of edges: those which represent it, as in §4.3; and those which represent f, that is, an edge i j if f(i) = j, or the functional digraph of f (see Exercise 13 of Chapter 3). For example, if n = 10, S = {3, 5, 7, 8, 9], it = (398) (5) (7) and —

(1

246

—

71'

then this triple can be represented:

-.

[ni-S

As another example, let n

10, S

1111

4

{3, 5, 7, 8, 9], it = (398) (5) (7) and

( 1 246 l0\ 41118 iol'

130

Clearly, (p is sign-reversing and weight-preserving; also (p2 = id because C will move back to its original position. What are the fixed-points of (p? They will be

cycles. Thus, it is empty, S = 0, and the graph of f: [ni —* [n-I-i] has no cycles. This graph must then be a tree, with every edge directed along the path toward n + 1, that is, a in Conversely, any tree in those triples (S, it, f) with

naturally defines such a function f. The example at the beginning of this

section corresponds to the function

(123456

f= So

47734

the right-hand side of (4.8) becomes

(4.9)

w(T)

=

which is exactly what Proposition 4.2 claimed.

§4.5 Lattice Paths Franklin's proof of the pentagonal number theorem appeared in 1881. Another early sign-reversing involution, called the reflection principle, was given by Andre in 1887. This time the signed set S = S consists of certain lattice paths in the plane. The involution reflects a lattice path through a line to obtain a new lattice path. In this section we use the reflection principle for the generalized ballot problem. We also use a related idea (due to Gessel and Viennot) to relate determinants of binomial

coefficients to non-intersecting lattice paths and column strict tableaux. We prove a

formula of Frobenius, which is a precursor of the hook formula for standard tableaux (Theorem 5.4 of Chapter 3). In Chapter 3, we saw that the Catalan number

was the solution to the ballot

problem: if candidates A and B both receive n votes, how many ways are there to count the votes so that A is never behind B? We give an alternative proof here, which uses the reflection principle. Represent any sequence of 2n votes as a lattice path (up for A, down for B) with unit steps, beginning at the origin. For example, ABAABABBAAABBABB is represented by

132

(P(P)

(2n, 1)

(0, 1) (0,—i) PE S

It is clear that this principle works for a more general ballot problem. Suppose

A and B receive n and m votes respectively, n m. How many ways are there to count the votes so that A is never behind B? This time we consider lattice paths from (0, 1) to (n+m, n—m+l). The reflection principle gives us the answer immediately:

f n+m\ n

(n+m\

n+ =

1—

n+l

m (n+m '..

The reflection principle was generalized by Gessel and Viennot [Ge-V] to allow

k-tuples of lattice paths. They showed that there are many relationships between lattice

paths, determinants and tableaux. We will present a few of these. It is more convenient if we "tilt" our pictures A lattice path P will no

longer consist of steps (1, 1) (up) and (1,—i) (down), but horizontal (1, 0) and vertical (0, 1) steps. For example, a lattice path P from (1, 1) to (4, 3) could be:

(4, 3)

(0, 0)

It is easy to see that such lattice paths are equivalent to the up-down lattice paths of the ballot problem. In fact, a solution to the ballot problem corresponds to a lattice path P

from (0, 0) to (n, n) which always lies at or above the line y =

x.

The lattice paths we will consider always begin on the line y = 1. Let us write P: (a, 1) —* (b, N) to mean a lattice path from (a, 1) to (b, N). How many P are there such that P : (a, 1) —+ (b, N)? Clearly, this number is

134

general. There are certainly

(4\(5 2

1

pairs of paths (P1, P2) such that P1 (3, 1) -4(4,4) and P2: (4, 1) -4 (6,4). From these pairs we must exclude all pairs (there will be

(3 \( 6 '..O)\ 3

of them) (P1, P2) which intersect. If P1 and P2 intersect, let m be the last point of intersection. Construct (P1, P2) = ç(P1, P2) by interchanging the paths from m to the endpoints. Then

(3, 1) —* (6, 4) and P2: (4, 1) -4 (4, 4).

(4,4)

(6,4)

(4,4)

(4,1)

(3,1)

Any such pair (P1,

(3,1)

(6,4)

(4,1)

must intersect so sp is a bijection to all such

There are

clearly

(3 \( 6 3

pairs (P1, P2).

Recall that S is the set of k-tuples of paths (P1, P2, ...

,

together with

the permutation a. The sign of a makes S into a signed set. Clearly,

(5.2)

IIS1I

=

IS—I

=

fN+b0(1)—a1 —1 ba(j)—aj

sgn(a) H 's

which is det M in Theorem 5.1. All that is necessary is a sign-reversing involution on S whose fixed point set F((p) is given by (i) and (ii) of Theorem 5.1. The 'bad' elements of S are those with paths which intersect. These are the ones on which we must define p. Let (a, (P1. ... , Pt)) E S be such that at least

two paths of (P1, ...

,

intersect. Choose the first pair i <j in lex order such that

136

w(P) =

(5.3)

x1. H Ic HoyP)

In this example

(7, 7)

33 1

(3, 1)

3,3, 6} and w(P) =

= {1,

x11 x32 x61.

Now extend w to S by w(a, (P1, ... , Pt)) =

w(P1) w(P2)

w(Pk). By the

preceding remarks, w is preserved by p. We can now replace the binomial coefficients that appear in Theorem 5.1 by generating functions. Given (a, 1) and (b, N), we saw that there are

(N+b—a—1

\

b—a

lattice paths P : (a, 1) —* (b, N). But we can also describe the generating function of all such paths: (5.4)

E

w(P).

P

Any path P is uniquely determined by its horizontal steps. Thus the terms in (5.4) will be monomials m

n,

m

xNN

such that m1 + rn2 + ... + mN =

xpx2, ...

b

and b—a. Write

—

a.

Note in particular that (5.4) depends only on

138

6

6 5

4

P1

P2

4

}'3

P4

Each horizontal edge is labeled with its y-coordinate. Place these entries of path

into row i of the tableau T,

1

T=

2

4

335

Why is T column strict? Certainly it has shape X and is wealdy increasing across rows. Let be the entry in row i, column j of T. Then the paths and

(which correspond to rows i + 1

and

i respectively in T) begin their

jth horizontal edge at x=k—i+j—1 and x=k—i+j respectively. Since to the left of

the jth horizontal edge of

must be strictly above the jth

horizontal edge of 1'k—i+I' Thus The

inverse of p is also easy. Given a tableau T, the entries in row i of T

determine the horizontal steps of

and thus the entire path

again do not intersect because I is column strict. The weight of a column strict tableau T, w(T), is defined by (5.7)

w(T) =

11

entries

lot T

In our example,

The paths

is

140 k

det

(5.8)

xN)] =

1

a

S k

fi i=

h

,..., XN).

1

Such a monomial can be created by selecting a square-free monomial from each

hk, k =

with the product of these monomials equal to x1 x2'" XN. The

i+

number of ways of doing this is easily seen to be the multinomial coefficient

N

(

+ 1

— 1

'

)

'

So (5,8) implies

1 i,j

d,,=N! det

(5.9)

By factoring out entries in the last column, we get

(5.10)

= N!

...

i+ k)],

where 1 i, j k. Let PkJ(x) be the polynomial m x of degree k — j

(x+k)

(5.11)

so that k

(5.12)

)!]4det

= N!

—

i )],

1

i,j k.

By column operations which do not change the value of the determinant, k

(5.13)

= N!

—i

)kJ]

1

J k.

This determinant is Vandermonde's; the evaluation given in (2.1) gives Proposition 5.5.

Thereadercanirytoevaluate detM for x1=x2= ... 5.2. The result is the number of column strict tableaux of shape [N].

1 inTheorem with entries in

142

else else

(x is not a fixed point of p or w} end. Proof We construct a graph 0 whose vertices are elements of A. The edges of G with the p-edges given by (x, p(x)) for x E A — F(p) and the are labeled p or

by (x, NJ(x)) for x

A — F(W). For example, if the involution (p is given

by this picture:

and the involution

by this picture:

'V

then

the graph G looks like this:

144

these two sets. Thus, they had found a bijection between the two sets. As an example of Theorem 6.2, if n = 12, there are 9 partitions in each set:

111, 9 i3, 62, 64 12, 6 16, 43, 42

4 18 and 112 with parts congruent to

1

or 4 mod 5; and 12, 111, 102, 93, 84, 831, 75, 741 and 642 whose parts differ by at least two. We will now give three examples of the involution principle. First we prove Eulers theorem. THEOREM 6.3 The number of

partitions of n into odd parts equals the nwnber of partitiotrs of n into distinct parts.

Proof Let

be

the set of all partitions of k and let

be the set of partitions

of n — k into even, distinct parts. Let n

A= U

PkxEDk

and define the sign of an element x =

(pa,

p2) e A by

(_1)number of parts of

sgn(x)

For the involution (p, take the smallest even part e of p1 or p2 and move e from p1 to if e is not in p2. Otherwise move e from p2 to p1. Clearly changes

the number of parts of p2 by one, so q) is sign-reversing. Also, F(p)

{(p1,ø):p1 has only

P

As an example, let n=26 and k= 12:

• • • • • • • . •

. P

• . . • • • • •

.

• •

.

. . . . . . . •

• • •

.

(P) • • •

. . . . .

• •

146

Remniel [Re] has used this version of the involution principle to give several

bijections. Here are two. The first is a theorem due to Schur. The number of partitions of n into parts congruent to 1 or 5 mod 6 equals the number of partitions of n into distinct parts congruent to 1 or 2 mod

THEOREM 6.3

3.

Proof The idea of the proof is to defme two signed sets, A and B: A related to the mod 6 condition and B to the mod 3 condition. Let us begin with A. Since F(cp) c A+, the involution should move any partition in A with parts 2,3,4, 6, 8 For any partition of n, let be the set of these "illegal' parts of For example,if then S={9,8,6,2}. Now define A = S) : X partitions n, S C SA(?,,)} and put S) = (_i)ISI, The involution

p on A merely changes S. Suppose

S) E A. If

is non-empty, insert (delete) its largest element into (from) S to obtain S; cp(A,, S) = (X, S). For example, 12, {9, 8, 2}) = (9 82736 2 12, {8, 2}).

p( 9 82

The fixed points of cp are those

S) such that

is

empty, that is,

has no

"illegal" parts.

The set B is defined in a similar way. The "illegal' parts are somewhat more

complicated. There are two kinds of illegal parts: multiples of three, i. e., 3, 6, 9, ..; and of equal parts congruent to 1 or 2 mod 3, i. e., 12, 22, 42, For any partition of n, let =

Let

=

:

k is a part of

0 mod 3] and isarepeatedpartof X, or 2 mod3}.

'2(k)- Put B =

and k

partitions n, S C

S) :

and put

sgn(A., S) = (_i)ISI. Again, the involution N' either inserts or deletes an element from

S. To see which element, we need a weight w on SB(?.). If k k; if j2 e 12(X), let w(j2) =j

+j. Given

delete or insert the element of

let w(k) =

S) E B, if SB(X) is non-empty, either

with largest weight from S. For example, if

31 andS= {6,42}, then {6, 42, 3] and w(42) =

j6}), since

8 is the largest weight in SB(X). The fixed points of those (X, 5) such that S8(?.) is empty.

It remains to construct the signed bijection

are

between A and B. But all that is

148

Remmel [Re] has given several other applications of the involution principle.

New applications of the principle are being found with increasing frequency. Many of these are too involved to be given here.

Exercises 1.[2]

with sgn(a) = (—1)Ial. Find

Let A be the Boolean algebra

a

on A such that al and 1p(a)I differ by exactly one for

sign-reversing involution

all a E A. Use the involution given in the introduction to this chapter which proved the principle of inclusion-exclusion. What famous identity involving binomial coefficients have you proved?

Let A be the k-element subsets of [n], with n even. For each a 2.[2] define val(a) to be

val(a)=

A,

i.

lEa

Let = A: val(a) is even} and A = {a A: val(a) is odd}. Construct a sign-reversing involution on A which will prove

=

{

ifkisodd

0

n/2 \

if k is even.

In each of the next five exercises, give a generating function proof of the identity

and then prove it combinatorially with a sign-reversing involution. In each case all the parameters may be considered positive integers. Recall from Chapters 1 and 3 the combinatorial interpretations of binomial coefficients and Stirling numbers. 3.[2]

v

/n\ 'k k)

p)

j

( m—ji

(_i)P.

=

o

=

{(

ifmisodd, if m = 2k.

150

sgn(p)

= (1— x)(1—x2) ...

p e PD

Prove that (1— x)(1— x2) ... =

(1_x)(1_x2)...

1,

by finding an appropriate involution on the set of ordered pairs (p1, p2),

p and

P2E PD. l2.[2]

It is also clear that (1— x2 )(1— x4) ...

1

(1—x)(1—x2)...

=

Define a signed set and an involution which proves this identity. You should consider ordered pairs of partitions, as in Exercise 11. 13.[2]

Define a signed set and an involution p which proves this identity: (1 —x)(l —x2).•. (1—

1

(1—x)(1—x2)...

=

...

As in Exercises 11 and 12, consider ordered pairs of partitions. 14.[3]

Write down eight different versions of the identity in Exercise 13. For example, here is another:

...

(1 (1— x)(1— x2)

1

(1—x11)

=

Interpret each of these as an identity involving partitions and give a combinatorial proof of each. Some will require sign-reversing involutions. a signed set and an involution p which proves: 1

(1—x)(l—x2)...

1

1

152

(a) Define a weight-preserving, sign-reversing mvolution on

proves det A = 0 if a1 =

which

for 1 i n.

I

(b) Give a combinatorial proof of det (AB) = (det A) (det B).

(c) Define A' =

with y1.=

det

where

is the ji-th

minor. Give a combinatorial proof that A' A = (det A) I. (d) Give a combinatorial proof of the expansion formula for det A along the

jth row of A.

22.[2]

Prove Cayleys Theorem (Theorem 2.1 of Chapter 3) using Theorem 4.1.

23.[1]

be the complete homogeneous symmetric function in Let hk(xj, ... , of degree k. From Section 5 we see that hk(l, 1, ... , 1) =

( What is

24.[2]

q2,

Ni-k k

...

,

).

qN)?

Use Exercise 23 and (5.8) to find a determinantal expression for q0111

where the sum is over all column strict tableaux P of shape ?.. and IIPII entries in P. Can you evaluate your determinant? 25.[1]

Let p.t be a partition of n,

is

the sum of

Define the homogeneous

symmetric function =

Interpret 26.[2]

...

as a generating function for a class of multiset permutations.

,

Use the Schensted correspondence (Chapter 3) to conclude h1.L

=

154

29.[3]

Prove the hook formula (Theorem 5.4 of Chapter 3) using the Frobenius formula (Proposition 5.5). 30.[1]

Show that the existence of sign-reversing involutions p and immediately implies F((p)j = IF(NI)l, without the explicit bijection given by the involution principle, 31.[2]

Suppose A, B and C are three signed sets and suppose p and signed bijections between A and B and between B and C, respectively,

are

(a) Prove that there is a signed bijection between A and C, which might be considered the "composition" of p and

(b) The signed bijections p and N' might be degenerate in some sense. For example, p might be a "pure" bijection, i. e., p a sign-reversing involution on A or B: p(B+) c and p(B) c A. Determine under what conditions the composition described in (a)

requires the involution principle. 32.[4CJ

Program the involution principle and apply your program to Euler's theorem. Run your program for various values of n. How does the bijection compare with the bijection given in Chapter 3? Guess and prove the theorem. 33.[4C1

Suppose A is a signed set and X, Y c A, with lxi = IYI = hAil and Write a program to construct two random involutions p and 'qi such that F(p) = X and F(Nc) = Y. Investigate

X Y = 0.

(a) the average length of a path from an element in X to an element in Y; and

(b) the average number of cycles in the graph of p and iv. What conjectures can you make and what theorems can you prove?

34.[4]

This problem is due to Blas and Sagan [Bi-Sa] and Zeilberger [Z4]. Let G be a simple graph with vertex set V(G) and edge set E(G). A

proper coloring of G with k colors is a function from V(G) to the colors such that no two adjacent vertices are given the same color. The chromatic polynom ial of G is the function p0(x), the number of ways of properly coloring V(G) with x colors. For

example,

Bibliography

Undergraduate Texts in Combinatorics [Bo] K. Bogart, Introductory Combinatorics, Pitman, Pitman, Massachusetts, 1983. [Br) R. Brualdi,Introductory Combinatorics, North Holland, New York, 1978. [Li] C. Liu, Introduction to Combinatorial Mathematics, McGraw-Hill, New York, 1968. [Ro] F. Roberts, Applied Combinarorics, Prentice-Hall, Englewood Cliffs, New Jersey, 1984. [Tu] A. Tucker, Applied Combinatorics, Second Edition, Wiley, New York, 1984.

Graduate Texts in Combinatorics [Al] M. Aigner, Combinatorial Theory, Springer, New York, 1979. [Be] C. Berge, Principles of Combinarorics, Academic Press, New York, 1971. [Co] L. Comtet, Advanced Combinatorics, Reidel, Dordrect, Boston, 1974. [G-I] I. Goulden and D. Jackson, Combinatorial Enumeration, Wiley-Interscience, New York, 1983. [Wi] S. 0. Williamson, Combinatorics for Computer Science, Computer Science Press, Rockville, Maiyland, 1985.

Texts on Combinatorial Algorithms [B] S. Even, Combinatorial Algorithms, Macmillan, New York, 1973. [Kn] D. Knuth, The Art of Computer Programming, Vol. 3, Addison-Wesley, Reading, Massachusetts, 1973. [N-W] A. Nijenhuis and H. Wilf, Combinatorial Algorithms, Academic Press, New York, 1978. [R-N-D] E. Reingold, J. Nievergelt, and N. Deo, Combinatorial Algorithms, Prentice-Hall, Englewood Cliffs, New Jersey, 1977.

Other References [Au M. Aigner, Lexicographic matching in Boolean algebras, J. Comb. Th. B 14 (1973), 187-194. [An] G. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Massachusetts, 1976. [B-Z] E. Bender and D. Zeilberger, Some asymptotic bijections, J. Comb. Th. A 38 (1985), 96-98. [BI-Sa] A. Blas and B. Sagan, Bijective proofs of two broken circuit theorems, J. Graph Th., to appear.

158 [P0] M. Pouzet, Application dune proprieté combinatoire des parties dun ensemble aux groupes et

aux relations, Math Z. 150 (1976), 117-134. [Ra] G. Raney, Functional composition patterns and power series reversion, Trans. Amer. Math. Soc. 94(1960), 441-451. [Rel J. Remmel, Bijective proofs of classical partition identities, J. Comb. Th. A 33 (1982), 273-286. [Ri] J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1980. [Sch] C. Schensted, Longest increasing and decreasing subsequences, Canad. J. Math. 13 (1961), 179- 19 1.

[Sc] M. P. SchUtzenberger, La correspondance de Robinson, Combinatoire etRepresentation du Groupe Symdtrique, Strasbourg, 1976 (D. Foata, Ed.), 59-113, Lecture Notes in Mathematics, No. 579, Springer-Verlag, Berlin, 1977. [Stal] R. Stanley, Weyl groups, the hard Lefsheftz theorem, and the Sperner property, SIAM J. Aig. Disc. Meth. 1 (1980), 168-184. [Sta2] R. Stanley, On the number of reduced decompositions of elements of Coxeter groups, Europ. J. Combinawrics 5 (1984), 359-372. [Sta3] R. Stanley, Theory and applications of plane partitions: part 1, Stud. AppI. Math 1(1971), 167- 188.

[Sta4] R. Stanley, Theory and applications of plane partitions: part 2, Stud. Appi. Math 1(1971), 259-279. [StrJ H. Straubing, A combinatorial proof of the Cayley-Hamilton theorem, Disc. Math. 43 (1983), 273-279. [T] H. Trotter, Algorithm 115: Perm, Comm. ACM 5 (1962), 434-435. [Wh-Wi] D. White and S. G. Williamson, Recursive matching algorithms and linear orders on the subset lattice, J. Comb. Th. A 23(1977), 117-127. [Whn] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc. 38 (1932), 572-579. [ZI] D. Zeilberger, A combinatorial approach to matrix algebra, Disc. Math. 56 (1985), 61-72. [Z2] D. Zeilberger, Garsia's and Milne's bijective proof of the inclusion-exclusion principle, Disc. Math. 51(1984), 109-110. [Z3] D. Zeilberger, A truly refmed bijection among trees, to appear. [Z4] D. Zeilberger, personal communication.

160

procedure UnrankPerm (Rnk : integer; var P1: Permutation);

var j, Dir, PrevRank, Remainder, Count, PrevN : integer; begin with N do begin

forj —1 toNdo

(initialize permutation]

ValueIj} -0; PrevRank :- Rnk; for PrevN —N downto I do begin PrevRank mod PrevN; PrevRank := PrevRank dlv PrevN; if (PrevRank mod 2— 1) then begin

j:=0;

{amount moved up or down] {rank of FtevN-1] {even means PrevN moving left; odd means right}

(initialize at left] {moving right]

Dir:- 1; end {if then] else

begin j :—N+

(initialize at right} (moving left]

1;

Dir:--l; end; {else} Count :— 0; repeat

j —j + Dir;

{advance left or right one position]

if (Value[j] —0) then Count:— Count + 1; until (Count — Remainder + 1); Value[j] :— PrevN;

(advance count for each index not assigned} (quit when count reaches amount to be moved]

end; (for} end; (with] end; {UnrankPerm}

procedure RankPerm (Pi: Permutation; var Rnk: integer); var i, Moves, Remainder: integer; function MoveCount (p: integer) : integer;

{returns number of numbers

Apostol: Introduction to Analytic Number Theory.

Halmos:

Armstrong: Basic Topology.

Halmos: Naive Set Theory.

Bak/Newman: Complex Analysis.

boss/Joseph: Elementary Stability and Bifurcation Theory.

Banchoff/Wermer: Linear Algebra Through Geometry.

Childs: A Concrete Introduction to Higher Algebra.

Vector

Spaces. Second edition.

Janich: Topology.

Kemeny/Snell: Finite Markov Chains. Klarnbauer: Aspects of Calculus.

Chung: Elementary Probability Theory with Stochastic Processes.

Lang: Undergraduate Analysis.

Croom: Basic Concepts of Algebraic Topology.

Lang: A First Course in Calculus. Fifth Edition.

Curtis: Linear Algebra: An Introductory Approach.

Lang: Calculus of One Variable. Fifth Edition.

Dixmier: General Topology.

Lang: Introduction to Linear Algebra. Second Edition.

Driver: Why Math? Ebbinghaus/FlumjThomas Mathematical Logic.

Lax/Burstein/Lax: Calculus with Applications and Computing, Volume I. Corrected Second Printing. LeCuyer: College Mathematics with APL.

Fischer: Intermediate Real Analysis. Lidl/Pilz: Applied Abstract Algebra.

Fleming: Functions of Several Variables. Second edition. Foulds: Optimization Techniques: An Introduction. Foulds: Combination Optimization for Undergraduates.

Macki/Strauss: Introduction to Optimal Control Theory. Malitz: Introduction to Mathematical Logic.

MarsdenlWeinstein: Calculus I, II, Ill. Second edition.

Franklin: Methods of Mathematical Economics.

continued after Index

Dennis Stanton Dennis White School of Mathematics University of Minnesota Minneapolis, MN 55455 U.S.A.

Editorial Board

F. W. Gehring Department of Mathematics University of Michigan Ann Arbor, Ml 48109 U.S.A.

P. R. Halnios Department of Mathematics University of Santa Clara Santa Clara, CA 95053 U.S.A.

AMS Classifications: 05—01, 05—A05, 05—A IS

Library of Congress Cataloging in Publication Data Stanton, Dennis. Constructive combinatorics. (Undergraduate texts in mathematics) Bibliography: p. Includes index. 1. Combinatorial analysis. 1. White, Dennis, 1945— II. Title. 111. Series. QA164.S79 1986 511.6 86-6585 © 1986 by Springer-Verlag New York Inc. All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag, 175 Fifth Avenue. New York. New York 10010, U.S.A.

The use of general descriptive names, trade names, trademarks, etc. in this publication, even if the former are not especially indentilied, is not to be taken as a sign that such names, as by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Printed and bound by R.R. Donnelley & Sons, Harrisonburg, Virginia. Printed in the United States of America.

987654321 ISBN 0-387-96347-2 Springer-Verlag New York Berlin Heidelberg Tokyo ISBN 3-540-96347-2 Springer-Verlag Berlin Heidelberg New York Tokyo

vi

complexity, lattice theory, group theory, representation theory, special functions or mathematical physics. In this book we use combinatorial algorithms for two purposes. First, a

constructive proof of a theorem can be an algorithm. These algorithms often describe a bijection between two finite sets. So we concentrate on interesting mathematical

theorems which are proved by bijections. The other purpose is interactive: use the algorithms to investigate interesting mathematical examples. Here the examples are our main focus. An algorithm can be used to generate data related to a problem. It is then up to the students to study these data, formulate as many conjectures as they can, and then prove them. They are not told what the theorems are in advance.

Unfortunately, this kind of 'research" is usually impossible in most undergraduate mathematics courses. The material here is more than what can be covered in a 10 week course. Two sections of peripheral interest are 1.4 and 2.4. Moreover, some of the material in

could be considered graduate material. and Chapters 3 and 4 Strictly speaking, each chapter can be presented independently, although we frequently tie together material from different chapters. There are many other topics which would have been suitable for inclusion. One such topic we regretted omitting was the Lagrange inversion formula (see [La] and IRa]). The notes are organized in the following way. In Chapter 1 algorithms which list fundamental combinatorial objects are given. They are written in a shorthand version of Pascal (no declaration or i/o statements are given). It is assumed that the students are familiar with a programming language, though not necessarily Pascal. In Chapter 2, a partially ordered set is defined for each object. We concentrate on the Boolean algebra. A number of interesting bijections are given in Chapter 3 for these objects. Finally, we generalize bijections to involutions in Chapter 4. There is some

emphasis on tableaux in these last two chapters. Thus they can serve as a combinatorial forerunner to the theory of representations of the symmetric group. We have included more complete Pascal programs in the Appendix. Furthermore, we would be happy to provide disks (Apple Macintosh Pascal© or Turbo Pascal©) with source code for these programs to interested readers. The exercises vary from true exercises to very difficult problems. We have assigned each exercise a number from one to four, which we believe is some indication of its difficulty (one is easy, four is hard). Exercises involving a computer

are marked with a 'C". Exercises labeled 3C or 4C might be suitable for a term project. We feel strongly that anyone using this book as a text should assign one or more of these

Notes

102

Exercises

102

4 Involutions

110

The Euler Pentagonal Number Theorem Vandermonde's Determinant

ill

4.2 4.3

The Cayley-Haxriilton Theorem

120

4.4

The Matrix-Tree Theorem

125

4.5

Lattice Paths

130

4.6

The Involution Principle

141

Notes Exercises

147

4.1

114

148

Bibliography

156

Appendix

159 159

A.3

Permutations Subsets Set Partitions

A.4

Integer Partitions

166

A.5

Product Spaces

167

A.6

Match to First Available

169

A.7

The Schensted Correspondence

171

A.8

The Prüfer Correspondence The Involution Principle

176

A.1

A.2

A.9

Index

162

164

178

180

2

he used for virtually any combinatorial object. We shall see in Chapter 2 that it also has many remarkable and surprising theoretical properties.

§1.1 Permutations

A permutation of n distinct objects of length k is an ordered anangement of any k of the objects. For instance, the permutations of {a, b, c, d} of length two

are ab, ac, ad, ba, bc, bd, Ca, cb, cd, da, db and dc. The next proposition is clear.

PRoPosmoN 1.1 The number of permutations of n objects of length k is n(n—l)"(n—k+1).

Sometimes we shall write (n)k (called the falling factorial) for

n(n—l)"(n—k+1). A permutation of n objects of length n is frequently called a permutation of n objects (or simply a permutation of n). It is clear that we can take the set [n] = { 1,2, ... , n} for the n objects. We shall frequently use this notation. Proposition 1.1 shows that the number of permutations of n is = n!. Perhaps the most natural ordering of the permutations of n is lexicographic (lex) order. We say that it precedes a in lex order, if, for some i, the first i

entries of it and a are the same, and the (i+l)th entry of It is less than the (i+l)th entry of a. The lex list of the permutations of 3 is 123, 132, 213, 231, 312 and 321. This ordering is quite simple. You are asked to consider it in Exercises 2 and 3. We shall return to lex order in § 1.2. We consider instead an algorithm to list all permutations of n that is due to Johnson [Joh] and Trotter [1']. It is based on a 'combinatorial proof' of n! = n (n—i)!: for each of the (n—i)! permutations of [n—i], there are n "positions" into

which n may be inserted. The algorithm has the property that each permutation differs from its predecessor by only a transposition of adjacent symbols. The lex list does not have this property. How does the algorithm work? Suppose we have the list for permutations of

Then we construct the list for permutations of [nJ by The insertions go inserting n into each of the n possible positions of each from left to right if i is odd and right to left if i is even. The lists for n = 1, 2, 3 [n—li:

and 4 are given below with the recursive structure indicated.

4

1

Done 4— false

white not Done do PrintØt)

if A 0 then m4—max{i: iEAJ ,r[j]

+ d[m]] xli + d[m]] m 4—

+ d[m]

il:-1[7t[j]] 4—j

if m

Let

be the collection of complements of the subsets of Certainly consists (n—k)-element subsets of [nJ. Furthermore, no member of can be a subset of any member of (Suppose A B and A B. Then A n = 0,

of

and A and

are two members of which would be disjoint.) The picture below describes this situation, since k n — k,

We now obtain a lower bound on the number of k-element subsets which lie below If we apply the map a n —2k times to (call this iterated map an-2k), we obtain all of these subsets. Let (4.2)

1) + ... + (ai —

= I

n—k

1

1

Since

(4.3)

= VFI

=

>

afl_k must be n. So by the Kruskal-Katona Theorem (Theorem 4.3),

(4.4)

>

(n-ii).

Repeating this n —2k —

1

more times gives

48

whichmeans A—{1}isin

and A—{j}, So

Therefore, all of

is in

(A—{j})u{1}

Let A—{k}€

E

A, 1

Finally we assume that j

Since (A—{k,j})u{1} SJ(A — {k})

the switching map S3 fixes A — {k}, so A — {k} = case is A —

{j} = SJ(A — {j})

A and The last

c

then JE A, A and subsets (A—{k,j])u{1} and A—{j} in If

The

can be checked as in the

previous paragraph.

Suppose we iterate the switching maps S3 for various j, until

toacollection

suchthat

is converted

forall j. Thisispossiblebecause S3 either

or gives

fixes

more members which contain 1. Since the switching maps do not change the size of the collection, Lemma 4.5 implies that mis completes Step (1). Step (2) Let First we show that

u

as indicated. Clearly

=

(4.7)

where a1 is the operation of deleting 1 from a set. Let B = A — (j} E j 1, A F(T). Since 1' is fixed under B u { i} so B E This establishes (4.7), which clearly implies (4.8)

We now separate the collection deleted,

into two subcollections: those subsets with 1 and those subsets with some element 1 deleted,

'u { 1 }. (This notation means that we first delete 1 from each member of

next apply a, and then reinsert I into each member.) Because these two are disjoint, (4.9)

=

This completes Step (2).

Step (3)

Suppose that

+

50

(4.16) Taken together, (4.15) and (4.16) contradict (4.13). So (4.10) must hold, and the

proof is complete.

Notes Three good general references for posets are [Ai], [Be] and [Gre-K 1]. Spemer theorems are included in §3 of Chapter Vifi of [Ai]. They are also a central topic of

[Gre-Ki]. Exercise 7 below, and Exercises 18 of Chapter 3 and Exercise 20 of and qj are Chapter 4 establish the entries of the chart for The entries for

given in Exercises 8 and 10. For 11 and 12, and for

they are Exercises 9 and 22, for

Exercises

Exercises 13 and 29. A matching between two adjacent

levels of the Boolean algebra can easily be shown to exist from Halls Theorem. The decomposition into symmetric chains is not guaranteed from this technique. The fact that matching to first available in the Boolean algebra works is due to Aigner [Au. The can be found in [Wh-Wi]. relationship between the various matching schemes in

Kleitman's solution of the Littlewood-Offord problem appears in [Gre-K 1]. The

proof of the Kruskal-Katona Theorem is due to Frankl, [Fr].

Exercises Given a finite poset (P, ), show that there is at least one way to totally so that a1 order P, that is, label all of the elements of P with [n]: a1, ... , 1.[2]

implies i j. 2.[2C]

Such a labeling is called a linear extension of (P,

Write a program which takes as input a poset P and gives as output a

linear extension of (P, ). For each poset (P, ) pictured below, write a program which finds the total number of linear extensions of (F, ). Can you prove any theorems here? 3.[3C]

52 12.[3]

1

and ii.

is a lattice by finding the meet of two partitions

Prove that

(Hint: Take the smaller of the two partial sums to define the partial sum of the meet.

Reason similarly for the join.) 13.[2]

Prove that

is rank unimodal, rank symmetric and order symmetric.

(The Spemer property follows from Exercise 29.)

14.[l]

What is the rank of (n, n—i, ... , 1) in

A chain C is called a maximal chain if C is not contained in any other (n, n—i,..., In chain. What is the length of a maximal chain in 15.[4C]

1)? Write a program which gives the number of maximal chains each poset has. What are your conclusions? [Sta2]

have?

and

16.[2]

How many maximal chains do

17.[2]

How many edges do the Hasse diagrams of

1 8.[3C]

Write a program to count the number of edges in the Hasse diagram of

and

have?

Formulate a conjecture, based upon an appropriate combination of Bell numbers.

(Hint: Try

a

—

a combinatorial proof of this result?) = nm.

Write a program to compute the Whitney numbers of

19.[4C]

Formulate and prove as many conjectures as you can. §3.3 may be useful.

Suppose v0,

20.[2]

...

is a finite sequence of positive real numbers. We

,

is log-concave if Vk2 Vk_1Vk+1 for 1 k n —

say v0, v1, ...

,

v0, v1, ...

is log-concave, prove that it is also unimodal.

21.[3}

,

Suppose the polynomial v(x) =

roots. Show that v0, v1, ...

,

x1 + ... + v0

If

has negative real

is log-concave. Possible hint: This result has a

combinatorial proof. Let {—r1, ...

w(A) of a subset A C [n] to be

+

1.

,

be the roots of v(x). Define the weight

54

kind are unimodal. 23.[2]

Let S be a subset of [999] u {0}. If

76, show that S must

IS!

contain at least two numbers n and m, so that the difference n — m can be computed with no "borrowing'. (Hint: consider the poset C310.) 24.[3]

{a1,... ,

Let A =

{a1,

...

a1+ 1, a1÷1, ...

c [n], where a1 < ... <ar. Define g(A)

,

where t is the largest i for which a1 — 2i =is

,

minimal (assume a0 = 0). Prove that g(A) = f(A), where f is defined in Algorithm 14. [Au

Suppose A c [n] is represented as a sequence of n parentheses, where parenthesis i is right if i E A and left otherwise. Thus, 25.[3]

A = {1, 3,4,6,7, 10, 12, 13, 16, 19} c [21]

corresponds to ) 1

(

)

)

(

2

3

4

5 6

)

) 7

(

( 8

)

)

)

(

(

(

and if we pair off the parentheses in the usual way,

.. —

I

( 1

2

) 3

)

(

)

)

)

(

(

)

(

(

9 10 11 12 13 14 15 16 17 18 19 20 21

(

(

)

(

..

I

)

)

(

(

)

(

(

)

(

(

9 10 11 12 13 14 15 16 17 18 19 20 21

4 5 6 7 8

we are left with a string of unpaired right parentheses followed by a string of unpaired left parentheses. ) 1

))7 14( 17( 20 21 (

(

4

Now take the first unpaired left parenthesis and turn it around. ) 1

4

)

(

(

(

2021

The resulting string of parentheses

123456789101112131415161718192021

56 32.[4]

Generalize the Erdös-Ko-Rado Theorem to collections such that A, B E implies Al k and A B

of subsets of [n]

58 n

(0.3)

f(x) =

X

k= 0

Let B be the set of n-tuples of 0's and l's, and let the weight of any element = B be where has k l's. Define the bijection ip : A B by 1

=

if i a

{ 0 if

a.

The bijection p is weight-preserving, because w(a)

generating function for B is (1 + (0.4)

of

w(p(a)). It is clear that the

so (0.2) becomes

f(x) = (1+x)".

Another kind of combinatorial proof of the binomial theorem can be given when x is a non-negative integer. In this case, the right-hand side of (0.4) counts functions from [n} to a set S of size 1 + x. The right-hand side of (0.3) counts the same

functions by classifying by the number of members of [n] which get sent to the first x members of S. The theorem can easily be extended to all real x. In this chapter we give examples of all these phenomena. The Catalan numbers provide bijections between apparently unrelated sets. The Prilfer correspondence associates a simple B with a complicated A. Partitions and permutations illustrate these ideas. The most difficult bijection that we consider is the Schensted correspondence between permutations and tableaux. Several of the constructions in this chapter and the next involve graphs. While

none of the graph theory concepts used are difficult, we will state here some of the key definitions and results involving graphs.

A graph is a set of vertices and a set of edges. The edges are usually a collection of 2-element subsets of the vertex set (such graphs are called simple), but sometimes one-element subsets (loops) or repetitions (multiple edges) are allowed. If the edges are ordered pairs of vertices, the graph is called directed (or a digraph) and the edges are directed edges. The degree of a vertex is the number of vertices incident to it. A vertex in a directed graph has an in-degree and an out-degree.

A path in a graph is a sequence of adjacent vertices. A path is simple if no vertex is repeated (except that the first may equal the last). A cycle is a simple path which starts and ends at the same vertex. We sometimes will refer to the unordered set of vertices in a cycle as a cycle. A graph is connected if there is a path from

60

(3) Full binary tree Afull binary tree is a binary tree where every vertex has

either 0 or 2 sons. (4) Well-formed parentheses A sequence of parentheses is called well-formed

if, at any point in the sequence, the number of right parentheses up to this point does not exceed the number of left parentheses up to the same point. Moreover, the total number of left parentheses equals the total number of right parentheses.

(5) Ballot problem Suppose Alice and Barbara are candidates for office. The result is a tie. In how many ways can the ballots be counted so that Alice is always ahead of or tied with Barbara?

(6) Standard tableaux Given a partition of n, a standard tableau T is an arrangement of [n] in the n cells of the Ferrers diagram of which increase across rows and down columns. These objects will be discussed in greater detail in §3.5. THEOREM 1.1 The following sets of objects all have the same number of elements,

and this number is (1) (2) (3) (4) (5) (6)

binary trees on n vertices; ordered trees on n + 1 vertices; full binary trees on 2n + 1 vertices; well-formed sequences of 2n parentheses; solutions to the ballot problem when 2n votes are cast; and standard tableaux in a 2 x n rectangular Ferrers diagram.

Proof We use bijections to show (1 )-(6) are equinuinerous. Then we show that (4) yields the Catalan number. (1) = (2): We give a bijection

from binary trees to ordered trees. Let B be a

binary tree. Here is how we construct T = (p(B). (a)

The vertices of B are the vertices of T with the root deleted.

(b)

Therootof B is the first son of the root ofT. Vertex v is a left son of vertex w in B if and only if v is the first son of w

(c)

in T. (d)

Vertex v is a right son of vertex w in B if and only if v is the brother to the

right of w in T. In the examples below, we have labeled the vertices to help the reader trace what happens under (p.

62 0

2

2

)(nn)n II

5

I

3

1

P

T To define

6

4

[3

5

1

let qr1(Ø) = the tree consisting of only a root, Again, by = T has been defined for all sequences

induction, assume

and ordered trees T with k +

1

length 2n. Write P = P1 P2...

vertices, k

investigate this statistic in Exercise 29.

82

while

2, (1, 2, 1, 1)) has

11121 1214!

34

1112141 12131

Itis not an accidentthat the numberoftableaux in each set is the same. Itis ournext theorem.

THEOREM 5.1

There is bijection between

p'), where p is

p) and

obtainedfrom p by applying an adjacent transposition.

Proof Let p' agree with p, except for interchanged. Let Tn

and Pk+1' which have been

p) so that T has Pk k's and Pk+1 k+l's. We need

to produce a tableau T' n

p') which has Pk k+l's and Pk+1 k's. We will do

this by switching some k's in T to k-i-i's, and also switching some k-i-i's in T to k's. The k and k+1 entries in row i of T have the following structure.

row

'

'

1

row i+1

k

'

k k+1

... k k k-i-1'--k-i-l k

i

c

b

a

row i -

k+1 k-i-i

-.. k k+1 d

For

with k+l's below them. Immediately to the right of these k's, there will be some number b 0 of k's with no k-i-i 's below, then some number c 0 of k+ l's with no k above, and finally some number d 0 of k+l's with k's above them. We change T to 1' by changing each row i to this form, this row, we let a

0 be the number of k's

a row i -

row

1

i

row i + 1

c

b

'

k k k+ 1 ... k+ 1 k

' ...

k k+1

-'

k÷1

k

k

k+1 ...

k+1

d

Note that

the k's and k+l's which are paired with k+i's below and k's are left unchanged. However, the b k's have become c k's, and the c k+l's have become b k+l's. So the total number of unpaired k's in T is equal to the total number of unpaired k-i-I's in T'. This implies that T' has Pk+1 k's and

above

k-i-i's.

It is easy to see that T' is column strict,

84

THEOREM 5.3 The number of standard tableaux of shape X,

equals the number

of maximal chains in Young's lattice

It is simple to state and difficult to prove.

There is an amazing formula for

We will give a proof of an equivalent formula in §4.5. To state this formula, we need The We write c E to define a hook. Let c be a cell of the Ferrers diagram of

hook of c, HC, consists of the cells to the right of c, below c, and c itself. In the bijection proving Theorem 3.3 of this chapter, we used the hooks of the major In the example below the hook of is diagonal. The length of the hook of c,

c is shaded and its length

= 7.

=1

I

I

I

I

C

is called the hook formula.

The formula for

be a partition of n. Then

THEOREM 5.4 Let

n!

—

CE

4 22. We insert the hook lengths into the cells of ?..

As an example, take

6 5 2111

32 21 Then d422

=

8!

=

56.

86

entries insert the number 4 into P" and Q" (in some way) and add one to all of the of P" and Q" which are 4. This will give us P and Q. Of course, the difficult part of this is to determine exactly how to insert the number 4. This algorithm is called the Schensted column insertion algorithm. First, we should remark that it is not necessary to do the addition and subtraction of one to the entries of P" and Q". The Schensted correspondence will and the produce tableaux (P', Q') for it', where the entries of F' are [8] —

entries of Q

are

[7]. For it = 35186724,

it = 3518672,

P'=

The Schensted column insertion algorithm inserts 4 into P' and 8 into Q'. The natural position of 4 in the first column of F is between the 2 and the 6. The 4 takes the place of the 6, or bumps the 6 out of the first column of F'. The 6 now is inserted by the same method into the second column of P'. This time the 6 is placed after the 5 and does not bump any entry of the second column. The column insertion algorithm has been completed. The resulting tableau P is 81

LiV that the shape of P differs from that of P' by the addition of exactly one cell (the cell 6 of P). We place an 8 in that cell for the definition of Q'. N1ote

—

111371 1215

48 6

A given insertion could cause several bumps. An example in the general case is given later.

To find P and Q from it, just apply the column insertion algorithm successively to the entries of it. If we use two line notation for it,

88

Removing c from Q gives Q. Next, we start with the largest entry of Q',

and do

this inverse bumping procedure until another entry is bumped from the first column of

P., and continue. We will not give a formal proof by induction of this case. Instead, we will consider the generalized Schensted correspondence for multiset permutations, Let m be a multiset with content (p1, ...

that m contains p1 is. A permutation

so

of m is any sequence of the elements of 111. For (p1,..., such

is

2322132. Clearly if each p1 =

1,

= (1,

4, 2), one

IlL has no repetitions so that multiset

permutations are just usual permutations. The generalized Schensted correspondence will associate to 7t the pair of tableaux (P, Q), where P is column strict of content p, and Q is a standard tableau of the same shape as P. For = 2322132 we will

seethat

(P,Q)=

(

13451 267

12221

233

'

To produce P and Q, we use the same bumping procedure with a minor modification. Suppose we are inserting k into a column. The number which is bumped out of the column is the smallest number k. (In the previous case, it was >k because repeats were not allowed.) For example, suppose we are inserting 4 into the column strict tableau P. 1

P— —

1

1

2 4 4 6 16

35567

I

7

Then the 4 will bump the 4 in the first column, so that the new first column is the

old first column.

Lii

The bumped 4 now bumps the 5 in the second column, for the following first two

90

The 7 from column 6 is larger than everything in column 7. It is therefore placed at the end of column 7, and the insertion of 4 into P has been completed. 1

1

1

244

23 3357 34 556 47 77

6 61 7

6

The new cell has been marked in bold face. We would place the next entry of Q (here

a 25) in that cell. The reader should verify that = 2322132 gives the (P, Q) that was previously claimed. The entries of Q are 1234567, the first row of the two line notation for it. The inverse Schensted correspondence (P, Q) —* it is as before. We use the largest entry of Q to find the entry e of P which bumps to the left. This time the largest number of the column which is e is bumped to the left (Note that the entry directly to the left of e is e, so that this set is non-empty.) The number eventually bumped out of the first column is again the last entry of it. We call this inverse procedure column deletion. Because column deletion is the inverse to column insertion, it is easy to see by induction that the generalized Schensted correspondence is a bijection. THEOREM 6.2 The generalized Schensted correspondence is a bijection between all

multiset permutations it of content p, and pairs of tableaux (P, Q), where P is column strict of content p, and Q is standard with the same shape as P.

Suppose that p =

(p1,

Pm)' so that the multinomial coefficient gives the

number of multiset permutations it. The number of ordered pairs (P, Q) is a sum of Kostka numbers times so that we have this corollary. and

COROLLARY 6.3

Pl+...+Pm=n, then

=

where the summation is over all partitions

of n.

92

Cell(P, k) 4— (i,

end.

The column deletion algorithm begins at row r and column c of the tableau P. The tableau after deletion is Del((r, c), F) and the value bumped out is Val((r, c), P). ALGORITHM 18: Schensted Column Deletion

begin

F

P

x 4-- P(r, c)

for j

*— c

— 1

downto 1 do

i+-1 repeat i4.— i+ 1

until P'(i, j) > x or i >

if-i-i y

P'(i,j)

x

x*—y

F Val((r,c),P)—y

Del((r, c), P) end.

Now we give the Schensted encode algorithm. We use Ins(k, P) and Cell(k, P) from Algorithm 17. The permutation is it, whose two-line notation has top row j1, j2, ... , and bottom row It1, It2, ... , Ire. The resulting pair of tableaux is (Schp(it), SchQOt)). ALGORiTHM 19: Schensted Encode

begin

t P

4— Jns(1c1,

P)

Q(CeIlØt1, P)) 4—

Schp(it)

4—

IlJ

P

I

94

from the generalized Schensted correspondence. The answer to the first question is given by this theorem. THEOREM 7.1

The number of rows of P (or Q)

is

the length of the longest

increasing subsequence of it. In the example it = 35186724 of §3.6, P had 4 rows, so it has an increasing

subsequence of length 4, 3567, and none longer. To prove Theorem 7,1, we need to describe what happens to P and Q first column only at each stage. Suppose, as in the previous section, it = 35186724. Entry into 0

Column 1 of P

Action Taken

1

Insert3;Q114—1

3

1

2

InsertS;

3

1

5

2

1

1

5

2

1

1

5

2

8

4

1

1

5

2

6

4

1

1

5

2

6

4

7

6

3

4

5

6

7

8

Insert 1; bump 3

Insert 8; Q31

4

Insert 6; bump 8

Insert 7; Q41

+— 6

Insert 2; bump 5

Insert 4; bump 6

1

1

2

2

6

4

7

6

1

1

2

2

r 96

inserted below row t. This property characterizes the class of a pair.

LEMMA 7,2 The pair (i, itt) belongs to class t

and only

the length of the

largest increasing subsequence of it ending at it1 is t. Proof It remains to show if the length of the largest increasing subsequence of it ending at is t, then (i, it1) belongs to class t. We do this by induction on t. = 1, then iç is smaller than all of the preceding

row of column and (i, it1) belongs to class

If t

so it1 is inserted into the first

1.

Now suppose t> 1, and choose an increasing subsequence S of it ending at it1 of length t. Let be the predecessor to it1 in S. Then the subsequence S = S

—

{it1} which ends at

to class t — — 1

was

1.

is also of maximal length. By induction, (j,

belongs

So when it1 was inserted into the first column, the entry v in row < it1. Thus,

be the entry in row t when

it1 was inserted either below row t or in row t. Let w was inserted. The entry w (which precedes it1) is a

member of a pair in class t. By the first part of the theorem, there is an increasing

subsequence of it of length t ending at w. If w < itt, we could attach

to this

subsequence and have an increasing subsequence of length t + 1. This contradicts our hypothesis, so w > and was inserted into row t.

Proof of Theorem 7.1 Suppose S is one of the longest increasing subsequences of it and has length t and suppose that P has r rows. By Lemma 7.2, the last member

it1 of S is inserted into row t of P, so t r. Conversely, the (r, 1) entry of P is in class r, so Lemma 7.2 implies that r t.

The answer to our second question is provided by Theorem 7.3.

THEOREM 7.3 Suppose that it corresponds to (P, Q) in the Schensted correspondence. Then ir1 (the inverse of it) corresponds to (Q, P). Proof In fact, Theorem 7.3 holds for all two line arrays with distinct entries. In this case ic' is the two line array obtained by interchanging the two lines, then sorting the columns according to the first line. Our proof will be by induction on the number of columns of P. First we show that the first columns agree, and then apply induction. We initially consider the permutation case, but the reader should have no difficulty

98

\ 1k

For example, the class 3 pairs of it = 35186724 give

(458\ 6[4J1 '..

while the class 3 pairs of

158\

give

(rn6 8 \

(68

k85

This proves that the inverse of itb is ic1b, and completes the proof of Theorem 7.3.

I COROLLARY 7.4 The number of standard tableaux with n entries is equal to the number of involutions it of [n].

Proof An involution of [n] is a permutation it of [n] such that it = ir'. By Theorem 7.3, the Schensted correspondence is a bijection between all pairs of standard

tableaux (P, F), and all permutations it of [n] such that it = ic'.

The third application involves the matching f in the Boolean algebra of §2.2.

We shall use the generalized Schensted correspondence to obtain the matching f.

Let Ac[n], IAI=p, andwrite A asan n-tupleof n—p 0's andp l's. Written this way, we may consider A as a multiset permutation of p l's and n — p Os. Now apply Algorithm 19 to A to obtain (P, Q). Since P is column strict with

0's and p l's, P must have either one or two rows. If there is no 1 below the last 0 in row 1, change that 0 to a 1 to obtain a new tableau P. (For 34.[4C]

> 0 if and

and

then show that Kx

if P

ri.)

Write a program which finds the number of column strict tableaux of shape

(

A. whose entries are N. State and prove your conjectures. 35.[1]

Suppose that it 443511242. Find the P and Q tableaux of it in the

generalized Schensted correspondence. 36.[1]

Suppose

121214141 p

13141

,

14151

Q=

111415191 12161 13171

1

Find the multiset permutation it which cormsponds to (P, Q). From the Schensted correspondence prove the ErdOs-Szekeres Theorem: 37.[2] any sequence of n2 + 1 distinct real numbers has either an increasing subsequence or

I

decreasing subsequence of length n + 1. Use Algorithm 19 to list all involutions of [n] and the corresponding tableaux. Is there any relationship between the shape of the tableau and the cycle structure of the permutation? 38.[3C]

39.[2]

A lattice permutation is a multiset permutation it of

l's, A.2 2's,...,

n's, such that for any initial segment of it, the number of l's the number of

2's ...

the number of n's. Thus, 1121233213 is a lattice permutation while

I

4

Involutions

Many combinatorial formulas include positive and negative values. It might first seem that a bijection is not the proper tool for dealing with these formulas. This is not the case, however. Such formulas can sometimes be proved by using an involution on

a signed set. In fact, involutions may be used to prove theorems seemingly unrelated to combinatorics. This will be done in Section 3 for the Cayley-Harnilton Theorem. A signed set A is a set which has been partitioned into two subsets, A+ and A

with A+uA=A and

Theelementsof A+ and A

are

call ed positive and negative, respectively. We are interested in the value hAil = — IA1.

if some of the elements of A can be paired with some of the elements of

A+,

then the total size of the sets that we have to count to compute hAil is reduced. In fact, A and if we pair up all of A, then hAil is just the number of if

elements of A+ which are unpaired. More formally, such a matching is an involution on A, that is, a pennutation (p on A such that p2 = id. This involution has the if and only if (p(x) e A. Note that x, then x E property that whenever

this means that if x e A

then ç(x)

A+.

minus the precisely the number of fixed points of p in number of fixed points of (p in A. if we write F((p) for the fixed point set of (p, a new signed set and F(p) = F((p) and F(p) is empty. Such an = hAil. Typically, one or both of

Notice that

hAil

is

involution ip is called sign-reversing, for if x is not fixed by ç, then (p(x) has the sign opposite from x. An important formula from elementary enumeration theory is the principle of

inclusion-e.xclarion. This principle can easily be proved with a sign-reversing

involution. Suppose X is some finite set of objects and each of these objects is endowed with certain properties. A property may be thought of as a subset of X. Suppose P denotes the collection of properties. So associated with x E X there is a subset

P of properties with which x is endowed. For any subset T c P of

propertieslet

and ND(T)=.(xE

Thus,

112

(1.1)

[1 (1—x1) = 1=1

1

+

(_l)kx(3k2±k)/2

k=1

which is a special case of the Jacobi triple product identity [An]. It is called the arise when constructing

pentagonal number theorem because the numbers larger regular pentagons from smaller ones.

Proof Let PD(n) be the set of all partitions of n into distinct parts. Let PD(n) = PDO(n). This makes PD(n) into a signed set with sgn(A) = )# (—1 The idea of this proof will be to construct a sign-reversing involution p on PD(n) with no fixed points, unless n is of the form n = (3k2±k)/2, in which case p will have exactly one fixed point. The sign of this fixed point will be Clearly, if p has these properties, Theorem 1.1 follows. Suppose A PD(n). Recall that we write A = (A1, ... ) with > A2>....

(The inequalities are strict here because the parts of A are distinct.)

Let a(A) = max{j

:

+ 1 —j} and b(A) =

=

Thus b(A) is the smallest

part of A and a(A) is the length of the 'staircase" on the border of the Ferrers diagram of A. For A = (7, 6, 5, 3, 2), a(A) = 3 and b(A) = 2. If b(A) a(A), create a new partition p(A) of n by moving the b(A) part adjacent to a(A). Thus p('l, 6, 5, 3, 2) = (8, 7, 5, 3):

a(A)

• • • • • •

9

. .

•

•./>Y

• • • • • •

p(X)= : : : : : • • •

b(A)

Note that since A has distinct parts, b(p(A)) > a(p(A)) = b(A). The reader should carefully check this. If b(A)> a(A), then p(A) is obtained by creating a "part" consisting of the a(A) cells at the end of the first a(A) rows. This part is placed under the b(A) part, i. e., this new part is now the smallest part of p(A). For example, p(9, 8, 6, 3) =

(8,7,6,3,2):

114

As an application of Theorem 1.1, recall that the generating function for all partitions is given by

(1.2)

H (1— x'Y' =

p(n)

1

Clearing the fraction and substituting (1.1) gives

(1.3)

(_l)kx(31c2+ k)/2

p(n)

1.

Equating coefficients in (1.3) gives, for n > 0, (_1)k p(n—(3k2+k)/2) = 0.

(1.4)

For any given value of n, this sum is actually finite and gives a recurrence for p(n).

§4.2 Vandermonde's Determinant Sign-reversing involutions are a natural tool for handling identities involving determinants because the terms in the expansion automatically have signs attached. In this section we give a proof due to Gessel [Ge] which establishes Vandermonde's determinant using a sign-reversing involution. Vandermonde's determinant is (2.1)

=

H

J

It is clear that the product side of (2.1) has (II) 2

2

terms, while the determinant has only n! terms. We need a sign-reversing involution that cancels

Ifl\ 2

—n!

r 116

2

T=

F

3

1

4

5

on non-transitive tournaments To prove (2.1), we need a sign-reversing involution which preserves the weight. If such a q exists, (2.4) becomes

II

(2.5)

w(T)sgn(T).

=

T

transitive

Any transitive T corresponds to a ranking permutation 7t =

such that

; wins n — i games. For such T, w(T) =

n-i

Also, sgn(T) = (_l)m, where m is the number of inversions of so (2.5) becomes the definition of the sign of the permutation

II

(2.6)

=

sgn(it)

4'... 1

This is precisely

fl

But the right-hand side of (2.6) is the definition of the determinant in (2.1).

The proof then hinges on the involution ç. We need first a characterization of non-transitive tournaments that you are asked to show in Exercise 17. PROPOSITION 2.1 T is a non-transitive tournament and only with equal out-degree.

T has two vertices

Proof Exereise. The sequence of out-degrees, or wins, (a1, a2, ... , a.) is called the score vector

of T. Choose the lexicographically first pair (i, i),

such that

=

For example, the tournament below has score vector (2, 3, 0, 3, 2, 6, 5); choose i = 1

andj=5.

118

1

7 33

5

The involution (p reverses the edges of the 1-5-4 triangle:

7

5

Thus (p produces this tournament:

120

§4.3 The Cayley-Hamilton Theorem In this section we give a combinatorial proof of the Cayley-Harnilton Theorem. This proof once again uses a sign-reversing involution. It is due to Straubing [Str}. THEOREM 3.1 Let A be any n x n matrix over any field. Let

= det(?J

—

A)

be the characteristic polynomial of A. Then PA(A) =0. It might seem surprising that a theorem from linear algebra has a combinatorial

proof. However, sign-reversing involutions are a perfectly suitable tool for handling determinants, because determinants are signed sums of products of the entries of a matrix. In fact, it can be argued that the combinatorial proof we give here is the most

"natural" proof because it does not depend upon the field of scalars. Proofs of this theorem from algebra usually first prove a weak version for diagonal or triangular

matrices and then "extend" to all matrices. However, this extension requires that the scalars be the complex numbers, and some major theorem, such as the Fundamental Theorem of Algebra or Taylor's Theorem, must be used to eliminate the dependence on the complex numbers.

Proof We begin by writing the characteristic polynomial as a signed sum of products n

(3.1)

pA(X) =

sgn(it) H (AJ i—i

Each fixed-point i of it (i. e., lt(i) = i) will contribute either product, while each non-fixed-point i will contribute

or

to the

Equation (3.1) can now

be written (3.2)

=

S

(3.3)

[n] — S

c

sgn(it)

subset which satisfies

c fixed points of it = F(it).

Note that since it fixes everything in [n] outside of S, we may regard It as a permutation of the elements of S. Let P(S) denote the set of permutations of S. The sign of it will be the same when it is regarded as a permutation in P(S). So we may

122

Next, replace X in (3.5) with A. We wish to describe the ij-th entry of the resulting matrix.

(3,6)

(p (A)).. = A

'J

k0

IrEP(S)

(_1)d(7t) H tES

This is clearly

This involves describing (3.7)

alklj.

.

We visualize a term in this sum combinatorially as a directed path P of length n — k

from i to j on the vertices [n]. Moreover, let the weight of P, w(P), be the product of the weights of the edges of P where the weight of an edge is as above: w(e) aij if e = i —* j. Then w(P) is exactly a term in this sum. We may now give a complete combinatorial description of the right hand side of

(3.6). Let (S, it, P) be a triple such that (a) (b) (c)

S is a subset of En]; it is a permutation on S; and P is a directed path from i to j of length n —

with vertices in [n].

The weight of the triple, w(S, it, P), is w(P) w(it) and the sign of the triple, sgn(S, it, P), is sgn*(it). Then the ij-th entry of pA(A) is the generating function (3.8) =

(S

it, P) sgn(S, it, P).

Clearly, the set of such triples,

is a signed set. To prove the

Cayley-Hamilton Theorem, we need to show that (3.8) is zero. Thus, we require a

weight-preserving, sign-reversing involution q' on We can visualize a triple (S, it, P) E

as a directed multi-graph on the

vertices [n] with two kinds of edges: the edges from it and the edges used in P. Note that the path P may use an edge more than once and may also use edges in it; hence the graph is a multi-graph. As an example, let n = 9, i = 2, k = 4, and j = 5. Choose S ={l, 4, 6, 9}, it =(146) (9) and P = 211515.

124

the cycle of it containing v. Let P be P with C inserted at the position of v. Let be S with the vertices in C removed. Let be it with C removed. Note that the second occurrence of v in P now completes a cycle in P. No earlier vertex in P satisfies (2): those up to and including the first occurrence of v are the same as in satisfy (2) in P; those between the two occurrences of v were on P and they C and therefore could not have been encountered before v in P. Therefore

(S,,P) satisfies (2). Inourfirstexainple,

=

{9} and P = 214611515:

0 Now suppose v satisfies (2); let C be the cycle in P just completed. Then C

will be a cycle from v to v in P. Let P be P with C removed (including one be it with the be S with the vertices in C added. Let occurrence of v). Let cycle C added. This construction is legal because no vertex before the second occurrence of v could be in it or could be a repetition in P. Note that the first occurrence of v in P now satisfies (1) because v is now in it. In our second {1, 2, 3,4,5, 6, 9} and P=215:

example,

/\__

S

7

2=v

3

5

8

6

126

T =

1

"s. 6

2

Thus, w(T)

then w(T) =

1 i * j n. Let I denote the variables trees in

is

a monomial in the variables

1 i, j n. The generating function for the

is given by

Z

=

(4.2)

w(T).

T

=

To count the number of spanning trees of G, merely put

=

1

for each pair

foreachpafr (i,j) thatisnotanedgein

(i,j) thatisanedgein G and 0.

as an n x n determinant.

Let us now identify PROPOSITION 4.2 Let 1

i,j

Note: For n =2, Proposition 4.2 is —a12 1

(4.3)

f3(j) =

= a12a23+ a13 a21+ a13 a23,

det

a21+a23j

—a21

which is the generating function for the thee rooted labeled trees on [3]. 3

2

1

1

I

I

21

1

2

128

The weight of a triple, w(S, it, f), is w(f) w(lt); the sign of a triple, sgn(S, it, f), is sgn*(it) = as in §4.3. To prove Proposition 4.2, we must find a weight-preserving, sign-reversing involution p on these triples. Such a triple can be represented as a directed graph on [n+1] with two kinds of edges: those which represent it, as in §4.3; and those which represent f, that is, an edge i j if f(i) = j, or the functional digraph of f (see Exercise 13 of Chapter 3). For example, if n = 10, S = {3, 5, 7, 8, 9], it = (398) (5) (7) and —

(1

246

—

71'

then this triple can be represented:

-.

[ni-S

As another example, let n

10, S

1111

4

{3, 5, 7, 8, 9], it = (398) (5) (7) and

( 1 246 l0\ 41118 iol'

130

Clearly, (p is sign-reversing and weight-preserving; also (p2 = id because C will move back to its original position. What are the fixed-points of (p? They will be

cycles. Thus, it is empty, S = 0, and the graph of f: [ni —* [n-I-i] has no cycles. This graph must then be a tree, with every edge directed along the path toward n + 1, that is, a in Conversely, any tree in those triples (S, it, f) with

naturally defines such a function f. The example at the beginning of this

section corresponds to the function

(123456

f= So

47734

the right-hand side of (4.8) becomes

(4.9)

w(T)

=

which is exactly what Proposition 4.2 claimed.

§4.5 Lattice Paths Franklin's proof of the pentagonal number theorem appeared in 1881. Another early sign-reversing involution, called the reflection principle, was given by Andre in 1887. This time the signed set S = S consists of certain lattice paths in the plane. The involution reflects a lattice path through a line to obtain a new lattice path. In this section we use the reflection principle for the generalized ballot problem. We also use a related idea (due to Gessel and Viennot) to relate determinants of binomial

coefficients to non-intersecting lattice paths and column strict tableaux. We prove a

formula of Frobenius, which is a precursor of the hook formula for standard tableaux (Theorem 5.4 of Chapter 3). In Chapter 3, we saw that the Catalan number

was the solution to the ballot

problem: if candidates A and B both receive n votes, how many ways are there to count the votes so that A is never behind B? We give an alternative proof here, which uses the reflection principle. Represent any sequence of 2n votes as a lattice path (up for A, down for B) with unit steps, beginning at the origin. For example, ABAABABBAAABBABB is represented by

132

(P(P)

(2n, 1)

(0, 1) (0,—i) PE S

It is clear that this principle works for a more general ballot problem. Suppose

A and B receive n and m votes respectively, n m. How many ways are there to count the votes so that A is never behind B? This time we consider lattice paths from (0, 1) to (n+m, n—m+l). The reflection principle gives us the answer immediately:

f n+m\ n

(n+m\

n+ =

1—

n+l

m (n+m '..

The reflection principle was generalized by Gessel and Viennot [Ge-V] to allow

k-tuples of lattice paths. They showed that there are many relationships between lattice

paths, determinants and tableaux. We will present a few of these. It is more convenient if we "tilt" our pictures A lattice path P will no

longer consist of steps (1, 1) (up) and (1,—i) (down), but horizontal (1, 0) and vertical (0, 1) steps. For example, a lattice path P from (1, 1) to (4, 3) could be:

(4, 3)

(0, 0)

It is easy to see that such lattice paths are equivalent to the up-down lattice paths of the ballot problem. In fact, a solution to the ballot problem corresponds to a lattice path P

from (0, 0) to (n, n) which always lies at or above the line y =

x.

The lattice paths we will consider always begin on the line y = 1. Let us write P: (a, 1) —* (b, N) to mean a lattice path from (a, 1) to (b, N). How many P are there such that P : (a, 1) —+ (b, N)? Clearly, this number is

134

general. There are certainly

(4\(5 2

1

pairs of paths (P1, P2) such that P1 (3, 1) -4(4,4) and P2: (4, 1) -4 (6,4). From these pairs we must exclude all pairs (there will be

(3 \( 6 '..O)\ 3

of them) (P1, P2) which intersect. If P1 and P2 intersect, let m be the last point of intersection. Construct (P1, P2) = ç(P1, P2) by interchanging the paths from m to the endpoints. Then

(3, 1) —* (6, 4) and P2: (4, 1) -4 (4, 4).

(4,4)

(6,4)

(4,4)

(4,1)

(3,1)

Any such pair (P1,

(3,1)

(6,4)

(4,1)

must intersect so sp is a bijection to all such

There are

clearly

(3 \( 6 3

pairs (P1, P2).

Recall that S is the set of k-tuples of paths (P1, P2, ...

,

together with

the permutation a. The sign of a makes S into a signed set. Clearly,

(5.2)

IIS1I

=

IS—I

=

fN+b0(1)—a1 —1 ba(j)—aj

sgn(a) H 's

which is det M in Theorem 5.1. All that is necessary is a sign-reversing involution on S whose fixed point set F((p) is given by (i) and (ii) of Theorem 5.1. The 'bad' elements of S are those with paths which intersect. These are the ones on which we must define p. Let (a, (P1. ... , Pt)) E S be such that at least

two paths of (P1, ...

,

intersect. Choose the first pair i <j in lex order such that

136

w(P) =

(5.3)

x1. H Ic HoyP)

In this example

(7, 7)

33 1

(3, 1)

3,3, 6} and w(P) =

= {1,

x11 x32 x61.

Now extend w to S by w(a, (P1, ... , Pt)) =

w(P1) w(P2)

w(Pk). By the

preceding remarks, w is preserved by p. We can now replace the binomial coefficients that appear in Theorem 5.1 by generating functions. Given (a, 1) and (b, N), we saw that there are

(N+b—a—1

\

b—a

lattice paths P : (a, 1) —* (b, N). But we can also describe the generating function of all such paths: (5.4)

E

w(P).

P

Any path P is uniquely determined by its horizontal steps. Thus the terms in (5.4) will be monomials m

n,

m

xNN

such that m1 + rn2 + ... + mN =

xpx2, ...

b

and b—a. Write

—

a.

Note in particular that (5.4) depends only on

138

6

6 5

4

P1

P2

4

}'3

P4

Each horizontal edge is labeled with its y-coordinate. Place these entries of path

into row i of the tableau T,

1

T=

2

4

335

Why is T column strict? Certainly it has shape X and is wealdy increasing across rows. Let be the entry in row i, column j of T. Then the paths and

(which correspond to rows i + 1

and

i respectively in T) begin their

jth horizontal edge at x=k—i+j—1 and x=k—i+j respectively. Since to the left of

the jth horizontal edge of

must be strictly above the jth

horizontal edge of 1'k—i+I' Thus The

inverse of p is also easy. Given a tableau T, the entries in row i of T

determine the horizontal steps of

and thus the entire path

again do not intersect because I is column strict. The weight of a column strict tableau T, w(T), is defined by (5.7)

w(T) =

11

entries

lot T

In our example,

The paths

is

140 k

det

(5.8)

xN)] =

1

a

S k

fi i=

h

,..., XN).

1

Such a monomial can be created by selecting a square-free monomial from each

hk, k =

with the product of these monomials equal to x1 x2'" XN. The

i+

number of ways of doing this is easily seen to be the multinomial coefficient

N

(

+ 1

— 1

'

)

'

So (5,8) implies

1 i,j

d,,=N! det

(5.9)

By factoring out entries in the last column, we get

(5.10)

= N!

...

i+ k)],

where 1 i, j k. Let PkJ(x) be the polynomial m x of degree k — j

(x+k)

(5.11)

so that k

(5.12)

)!]4det

= N!

—

i )],

1

i,j k.

By column operations which do not change the value of the determinant, k

(5.13)

= N!

—i

)kJ]

1

J k.

This determinant is Vandermonde's; the evaluation given in (2.1) gives Proposition 5.5.

Thereadercanirytoevaluate detM for x1=x2= ... 5.2. The result is the number of column strict tableaux of shape [N].

1 inTheorem with entries in

142

else else

(x is not a fixed point of p or w} end. Proof We construct a graph 0 whose vertices are elements of A. The edges of G with the p-edges given by (x, p(x)) for x E A — F(p) and the are labeled p or

by (x, NJ(x)) for x

A — F(W). For example, if the involution (p is given

by this picture:

and the involution

by this picture:

'V

then

the graph G looks like this:

144

these two sets. Thus, they had found a bijection between the two sets. As an example of Theorem 6.2, if n = 12, there are 9 partitions in each set:

111, 9 i3, 62, 64 12, 6 16, 43, 42

4 18 and 112 with parts congruent to

1

or 4 mod 5; and 12, 111, 102, 93, 84, 831, 75, 741 and 642 whose parts differ by at least two. We will now give three examples of the involution principle. First we prove Eulers theorem. THEOREM 6.3 The number of

partitions of n into odd parts equals the nwnber of partitiotrs of n into distinct parts.

Proof Let

be

the set of all partitions of k and let

be the set of partitions

of n — k into even, distinct parts. Let n

A= U

PkxEDk

and define the sign of an element x =

(pa,

p2) e A by

(_1)number of parts of

sgn(x)

For the involution (p, take the smallest even part e of p1 or p2 and move e from p1 to if e is not in p2. Otherwise move e from p2 to p1. Clearly changes

the number of parts of p2 by one, so q) is sign-reversing. Also, F(p)

{(p1,ø):p1 has only

P

As an example, let n=26 and k= 12:

• • • • • • • . •

. P

• . . • • • • •

.

• •

.

. . . . . . . •

• • •

.

(P) • • •

. . . . .

• •

146

Remniel [Re] has used this version of the involution principle to give several

bijections. Here are two. The first is a theorem due to Schur. The number of partitions of n into parts congruent to 1 or 5 mod 6 equals the number of partitions of n into distinct parts congruent to 1 or 2 mod

THEOREM 6.3

3.

Proof The idea of the proof is to defme two signed sets, A and B: A related to the mod 6 condition and B to the mod 3 condition. Let us begin with A. Since F(cp) c A+, the involution should move any partition in A with parts 2,3,4, 6, 8 For any partition of n, let be the set of these "illegal' parts of For example,if then S={9,8,6,2}. Now define A = S) : X partitions n, S C SA(?,,)} and put S) = (_i)ISI, The involution

p on A merely changes S. Suppose

S) E A. If

is non-empty, insert (delete) its largest element into (from) S to obtain S; cp(A,, S) = (X, S). For example, 12, {9, 8, 2}) = (9 82736 2 12, {8, 2}).

p( 9 82

The fixed points of cp are those

S) such that

is

empty, that is,

has no

"illegal" parts.

The set B is defined in a similar way. The "illegal' parts are somewhat more

complicated. There are two kinds of illegal parts: multiples of three, i. e., 3, 6, 9, ..; and of equal parts congruent to 1 or 2 mod 3, i. e., 12, 22, 42, For any partition of n, let =

Let

=

:

k is a part of

0 mod 3] and isarepeatedpartof X, or 2 mod3}.

'2(k)- Put B =

and k

partitions n, S C

S) :

and put

sgn(A., S) = (_i)ISI. Again, the involution N' either inserts or deletes an element from

S. To see which element, we need a weight w on SB(?.). If k k; if j2 e 12(X), let w(j2) =j

+j. Given

delete or insert the element of

let w(k) =

S) E B, if SB(X) is non-empty, either

with largest weight from S. For example, if

31 andS= {6,42}, then {6, 42, 3] and w(42) =

j6}), since

8 is the largest weight in SB(X). The fixed points of those (X, 5) such that S8(?.) is empty.

It remains to construct the signed bijection

are

between A and B. But all that is

148

Remmel [Re] has given several other applications of the involution principle.

New applications of the principle are being found with increasing frequency. Many of these are too involved to be given here.

Exercises 1.[2]

with sgn(a) = (—1)Ial. Find

Let A be the Boolean algebra

a

on A such that al and 1p(a)I differ by exactly one for

sign-reversing involution

all a E A. Use the involution given in the introduction to this chapter which proved the principle of inclusion-exclusion. What famous identity involving binomial coefficients have you proved?

Let A be the k-element subsets of [n], with n even. For each a 2.[2] define val(a) to be

val(a)=

A,

i.

lEa

Let = A: val(a) is even} and A = {a A: val(a) is odd}. Construct a sign-reversing involution on A which will prove

=

{

ifkisodd

0

n/2 \

if k is even.

In each of the next five exercises, give a generating function proof of the identity

and then prove it combinatorially with a sign-reversing involution. In each case all the parameters may be considered positive integers. Recall from Chapters 1 and 3 the combinatorial interpretations of binomial coefficients and Stirling numbers. 3.[2]

v

/n\ 'k k)

p)

j

( m—ji

(_i)P.

=

o

=

{(

ifmisodd, if m = 2k.

150

sgn(p)

= (1— x)(1—x2) ...

p e PD

Prove that (1— x)(1— x2) ... =

(1_x)(1_x2)...

1,

by finding an appropriate involution on the set of ordered pairs (p1, p2),

p and

P2E PD. l2.[2]

It is also clear that (1— x2 )(1— x4) ...

1

(1—x)(1—x2)...

=

Define a signed set and an involution which proves this identity. You should consider ordered pairs of partitions, as in Exercise 11. 13.[2]

Define a signed set and an involution p which proves this identity: (1 —x)(l —x2).•. (1—

1

(1—x)(1—x2)...

=

...

As in Exercises 11 and 12, consider ordered pairs of partitions. 14.[3]

Write down eight different versions of the identity in Exercise 13. For example, here is another:

...

(1 (1— x)(1— x2)

1

(1—x11)

=

Interpret each of these as an identity involving partitions and give a combinatorial proof of each. Some will require sign-reversing involutions. a signed set and an involution p which proves: 1

(1—x)(l—x2)...

1

1

152

(a) Define a weight-preserving, sign-reversing mvolution on

proves det A = 0 if a1 =

which

for 1 i n.

I

(b) Give a combinatorial proof of det (AB) = (det A) (det B).

(c) Define A' =

with y1.=

det

where

is the ji-th

minor. Give a combinatorial proof that A' A = (det A) I. (d) Give a combinatorial proof of the expansion formula for det A along the

jth row of A.

22.[2]

Prove Cayleys Theorem (Theorem 2.1 of Chapter 3) using Theorem 4.1.

23.[1]

be the complete homogeneous symmetric function in Let hk(xj, ... , of degree k. From Section 5 we see that hk(l, 1, ... , 1) =

( What is

24.[2]

q2,

Ni-k k

...

,

).

qN)?

Use Exercise 23 and (5.8) to find a determinantal expression for q0111

where the sum is over all column strict tableaux P of shape ?.. and IIPII entries in P. Can you evaluate your determinant? 25.[1]

Let p.t be a partition of n,

is

the sum of

Define the homogeneous

symmetric function =

Interpret 26.[2]

...

as a generating function for a class of multiset permutations.

,

Use the Schensted correspondence (Chapter 3) to conclude h1.L

=

154

29.[3]

Prove the hook formula (Theorem 5.4 of Chapter 3) using the Frobenius formula (Proposition 5.5). 30.[1]

Show that the existence of sign-reversing involutions p and immediately implies F((p)j = IF(NI)l, without the explicit bijection given by the involution principle, 31.[2]

Suppose A, B and C are three signed sets and suppose p and signed bijections between A and B and between B and C, respectively,

are

(a) Prove that there is a signed bijection between A and C, which might be considered the "composition" of p and

(b) The signed bijections p and N' might be degenerate in some sense. For example, p might be a "pure" bijection, i. e., p a sign-reversing involution on A or B: p(B+) c and p(B) c A. Determine under what conditions the composition described in (a)

requires the involution principle. 32.[4CJ

Program the involution principle and apply your program to Euler's theorem. Run your program for various values of n. How does the bijection compare with the bijection given in Chapter 3? Guess and prove the theorem. 33.[4C1

Suppose A is a signed set and X, Y c A, with lxi = IYI = hAil and Write a program to construct two random involutions p and 'qi such that F(p) = X and F(Nc) = Y. Investigate

X Y = 0.

(a) the average length of a path from an element in X to an element in Y; and

(b) the average number of cycles in the graph of p and iv. What conjectures can you make and what theorems can you prove?

34.[4]

This problem is due to Blas and Sagan [Bi-Sa] and Zeilberger [Z4]. Let G be a simple graph with vertex set V(G) and edge set E(G). A

proper coloring of G with k colors is a function from V(G) to the colors such that no two adjacent vertices are given the same color. The chromatic polynom ial of G is the function p0(x), the number of ways of properly coloring V(G) with x colors. For

example,

Bibliography

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Other References [Au M. Aigner, Lexicographic matching in Boolean algebras, J. Comb. Th. B 14 (1973), 187-194. [An] G. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Massachusetts, 1976. [B-Z] E. Bender and D. Zeilberger, Some asymptotic bijections, J. Comb. Th. A 38 (1985), 96-98. [BI-Sa] A. Blas and B. Sagan, Bijective proofs of two broken circuit theorems, J. Graph Th., to appear.

158 [P0] M. Pouzet, Application dune proprieté combinatoire des parties dun ensemble aux groupes et

aux relations, Math Z. 150 (1976), 117-134. [Ra] G. Raney, Functional composition patterns and power series reversion, Trans. Amer. Math. Soc. 94(1960), 441-451. [Rel J. Remmel, Bijective proofs of classical partition identities, J. Comb. Th. A 33 (1982), 273-286. [Ri] J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1980. [Sch] C. Schensted, Longest increasing and decreasing subsequences, Canad. J. Math. 13 (1961), 179- 19 1.

[Sc] M. P. SchUtzenberger, La correspondance de Robinson, Combinatoire etRepresentation du Groupe Symdtrique, Strasbourg, 1976 (D. Foata, Ed.), 59-113, Lecture Notes in Mathematics, No. 579, Springer-Verlag, Berlin, 1977. [Stal] R. Stanley, Weyl groups, the hard Lefsheftz theorem, and the Sperner property, SIAM J. Aig. Disc. Meth. 1 (1980), 168-184. [Sta2] R. Stanley, On the number of reduced decompositions of elements of Coxeter groups, Europ. J. Combinawrics 5 (1984), 359-372. [Sta3] R. Stanley, Theory and applications of plane partitions: part 1, Stud. AppI. Math 1(1971), 167- 188.

[Sta4] R. Stanley, Theory and applications of plane partitions: part 2, Stud. Appi. Math 1(1971), 259-279. [StrJ H. Straubing, A combinatorial proof of the Cayley-Hamilton theorem, Disc. Math. 43 (1983), 273-279. [T] H. Trotter, Algorithm 115: Perm, Comm. ACM 5 (1962), 434-435. [Wh-Wi] D. White and S. G. Williamson, Recursive matching algorithms and linear orders on the subset lattice, J. Comb. Th. A 23(1977), 117-127. [Whn] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc. 38 (1932), 572-579. [ZI] D. Zeilberger, A combinatorial approach to matrix algebra, Disc. Math. 56 (1985), 61-72. [Z2] D. Zeilberger, Garsia's and Milne's bijective proof of the inclusion-exclusion principle, Disc. Math. 51(1984), 109-110. [Z3] D. Zeilberger, A truly refmed bijection among trees, to appear. [Z4] D. Zeilberger, personal communication.

160

procedure UnrankPerm (Rnk : integer; var P1: Permutation);

var j, Dir, PrevRank, Remainder, Count, PrevN : integer; begin with N do begin

forj —1 toNdo

(initialize permutation]

ValueIj} -0; PrevRank :- Rnk; for PrevN —N downto I do begin PrevRank mod PrevN; PrevRank := PrevRank dlv PrevN; if (PrevRank mod 2— 1) then begin

j:=0;

{amount moved up or down] {rank of FtevN-1] {even means PrevN moving left; odd means right}

(initialize at left] {moving right]

Dir:- 1; end {if then] else

begin j :—N+

(initialize at right} (moving left]

1;

Dir:--l; end; {else} Count :— 0; repeat

j —j + Dir;

{advance left or right one position]

if (Value[j] —0) then Count:— Count + 1; until (Count — Remainder + 1); Value[j] :— PrevN;

(advance count for each index not assigned} (quit when count reaches amount to be moved]

end; (for} end; (with] end; {UnrankPerm}

procedure RankPerm (Pi: Permutation; var Rnk: integer); var i, Moves, Remainder: integer; function MoveCount (p: integer) : integer;

{returns number of numbers

0) then

1

in S)) or Ci — 0);

{j-0 means no more subsets]

begin Done

:—

false;

Count:— 0;

for

to N do If (i in S) then Count:—Count+ 1;

S :=(S - [j..N])+ Ii +

(count number in subset past this element}

1..j + Count];

(remove these andj and add Count contiguous elements)

end (if then) else

Done

:—

true

end; {OetNextSubset} function Matched (A: Subset;

var B : Subset) : boolean;

(returns true and subset B if match}

var ListDone, StopLoop : boolean; p

: integer;

begin GetFirstSubset(K + 1, B, StopLoop); p —1; while not StopLoop do (search for first match or end of list) if (A x) or (i > Conjugate.Part[j]);

{dont go past end of column}

i:=i-1;

{valuetobumpinrowabove}

y := CellEntry[i, ii;

f bump it]

CellEntry[i,j] := x; x := y;

(repeat for next column down]

end; (for} end; {with] end; {SchenstedDelete}

(convert permutation pi in two-line form to pair of tableaux] procedure SchenstedEncocje (p1: TwoLinePerm; var Z: TableauPair);

var i, Row, Col : integer;

procedure Empty (var T: Tableau); var

(make T the empty tableau]

integer; begin with T do begin Shape.NumberOfParts := 0; Conjugate.NumberOfParts 0; for i := Ito MaxiniumNumparts do begin CellEntry[i, 0] := 0; CellEntry[0, ii := 0; Shape.Part[i] := 0; Conjugate.Part[i] := 0; end; {for] CellEntry[0, 01 := 0;

end; (with] end; (Empty] begin Empty('Z.Bumping); Empty(Z,Template);

{start with empty tableaux}

fori:— itoNdo begin Schenstedlnsert(Z.Bumping, pi.BottomRow[i], Row, Col); (Schensted insert permutation value] Z.Template.CellEntry[Row, Col] := pi.TopRow[i]; {insert top row value into new cell}

end; (for] SameShape(Z.Template, Z.Bulnping); end; {SchenstedEncode}

{Bumping has right shapes; make Template same}

(convert pair of tableaux to a permutation in two-line form] procedure SchensteciDecode (Y: TableauPair; var pi : TwoLinePerm);

176

A.8 The Prufer Correspondence The data type for a tree is a vector of subset The vth subset is the set of vertices adjacent to vertex v. This is called the adjacency list, The number of vertices, N, is a global variable,

const MaxTreeSize - 20; type Vertex - 0..MaxTreeSize; VertexSet — set of Vertex; Tree —

array[l..MaxTreeSize] of VertexSet; (tree Is adjacency list)

Vector — array[l.,MaxTreeSize] of Vertex;

var N: integer; procedure DecodeVector (a: Vector; var T: Tree); var i, v, w: Vertex; Degree: procedure AddAnEdge (var T: Tree; v, w : Vertex); begin

(add edge (v,w) toT)

T[v] :— T(v] + Lw];

T[w] :— T[w] + [v];

end; (AddAnEdge}

function Largest (Degree: Vector) Vertex;

(returns largest terminal vertex of tree with given degrees}

var Vertex; begin i :—N;

while(i >- l)and (Degree[i] o l)do

i:—i-l; i;

end; (Largest) function Smallest (Degree Vector) : Vertex;

var i:Vertex; begin

i—I; while (i

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