/i9/

Yo, 6oZ3

ALGORITHMS OF SCHENSTED AND HILLMAN-GRASSL

AND OPERATIONS ON STANDARD BITABLEAUX

THESIS

Presented to the Graduate Council of the

North Texas State University in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF ARTS

by

David C. Sutherland, B.A.

Denton, Texas

August, 1983

Sutherland, David C., Algorithms of Schensted and

Hillman-Grassi and Operations on Standard Bitableaux. Master

of Arts (Mathematics), August, 1983, 101 pp., bibliography,

7 titles.

In this thesis, we describe Schensted's algorithm for

finding the length of a longest increasing subsequence of a

finite sequence. Schensted's algorithm also constructs a

bijection between permutations of the first N natural numbers

and standard bitableaux of size N. We also describe the

Hillman-Grassl algorithm which constructs a bijection between

reverse plane partitions and the solutions in natural numbers

of a linear equation involving hook lengths. Pascal programs

and sample output for both algorithms appear in the appendix.

In addition, we describe the operations on standard

bitableaux corresponding to the operations of inverting and

reversing permutations. Finally, we show that these operations

generate the dihedral group D4 .

TABLE OF CONTENTS

PageLIST OF TABLES... . . . ........... . . . . . ... iP

Chapter

I. INTRODUCTION ...... ........ 1

II. THE ALGORITHMS OF SCHENSTED AND HILLMAN-GRASSL 3

Schensted's AlgorithmThe Hillman-Grassi Algorithm

III. OPERATIONS ON STANDARD BITABLEAUX .. . . . . . 24

Schensted's Algorithm: A Graphical ApproachThe Inverse PermutationThe P-tableau for the Reverse PermutationThe Q-tableau for the Reverse PermutationThe Dihedral Group D4

APPENDIX.-...-...... . . . . . . . . . . . . . ..... 67

BIBLIOGRAPHY.-....... ....................... .. 101

iii

LIST OF TABLES

Table Page

I. Schensted's Algorithm.. .... . . .... .. .67

II. Reverse Schensted's Algorithm . .. . . . . . 68

III. Schensted's Algorithm . . . . . . . . . . . 69

IV. Data for Schensted's Algorithm . . . . . . . . 71

V. Reverse Schensted's Algorithm . . . . . 84

VI. Data for Reverse Schensted's Algorithm . 86

VII. Hillman-Grassl Algorithm . . . . . . . . 87

VIII. Data for the Hillman-Grassl Algorithm . 90

IX. Reverse Hillman-Grassl Algorithm . . . . . .. 94

X. Data for Reverse Hillman-Grassl Algorithm 97

iv

CHAPTER I

INTRODUCTION

In 1959, Craige Schensted developed an algorithm for

finding the length of a longest increasing subsequence of

a finite sequence. In the second chapter, we describe this

algorithm as it is applied to permutations of natural numbers.

It is shown that Schensted's algorithm constructs a bijection

between the set of permutations of the first N natural numbers

and the set of standard bitableaux of size N. We then gener-

alize Schensted's algorithm from permutations to finite

sequences of natural numbers.

The second chapter also contains a description of an

algorithm of Abraham P. Hillman and Richard M. Grassl which

constructs a bijection between the set of reverse plane

partitions of a natural number N on a frame F and the solutions

in the natural numbers of a linear equation involving hook

lengths.

Pascal programs and sample output for both these algorithms

appear in the appendix. In particular, Table IV contains all

the permutations on the first N natural numbers for N=1,2,3,4,5

and the standard bitableaux associated with them by Schensted's

algorithm.

In the third chapter, we examine several operations on

permutations and the operations on standard bitableaux associated

1

2

with them by Schensted's algorithm. The chapter begins with

a graphical approach to Schensted's algorithm which is due

to Donald E. Knuth. This approach leads to a proof of the

statement that if (P,Q) is the standard bitableau for 7,then (Q,P) is the standard bitableau for Tr1.

The second section of this chapter describes the char-

acterization of the P-tableau for the reverse permutation in

terms of the P-tableau for the original permutation. In

particular, a detailed proof is given of Schensted's lemma

that row and column insertion commute. This lemma is used to

show that the P-tableau for the reverse permutation is the

conjugate of the P-tableau for the original permutation.

As Schensted noted in his paper in 1959, conjugation

does not suffice to describe the Q-tableau for the reverse

permutation. This problem was later solved by Marcel P.

Schitzenberger through the development of a procedure for

constructing the Q-tableau for the reverse permutation by

conjugation and evacuation, which involves filling a missing

square in a frame to obtain a standard tableau.

Finally, in the last section of the third chapter, we

show that the operations of inverting and reversing permutations

generate the dihedral group D4 . Then, the results of the

preceding sections are used to obtain a representation of D4

as operations acting on the set of standard bitableaux of a

given size.

CHAPTER II

THE ALGORITHMS OF SCHENSTED

AND HILLMAN-GRASSL

In this chapter, we will define and describe two algorithms.

The first is Schensted's algorithm which defines a bijection

between permutations of the first N natural numbers and standard

bitableaux of size N. The second is the Hillman-Grassl algo-

rithm which defines a bijection between the set of reverse plane

partitions of a natural number N on a frame F and the solutions

in the natural numbers of a linear equation involving hook lengths.

Schensted's Algorithm

We begin with several basic definitions. A permutation 7of the first N natural numbers is a bijection whose domain and

range is the set {1,2,...,N} consisting of the first N natural

numbers, The size of the permutation is the natural number N.

We denote a permutation in column notation by

1 2 3 ... N

7 (1)T(2) 7(3) ... 7(N)

where 7(i) is the image of i under 7. A generalized permutation7 (of the same size) is a bijection whose range and domain aresubsets of the natural numbers. We denote a generalized permu-

tation by

a1 a2 a3 ... aN

7(a1) 7(a2) 7(a3) ... T(aN)

3

4

where {a ,a2 '...'aN} is the domain of 71 arranged in increasing

order and rr(a.) is the image of a.. A multiset is a set where

each element occurs with a certain multiplicity; i.e., it is

a set {a1 ,a 2 ... ,an} with a function m from {a,a2'''an} to

the natural numbers such that the value m(a.) of m on a. is

the multiplicity of ai. For example, S = {1,1,3,4,3,5} is a

multiset on the set {1,3,4,5} and the function m is given by

m(1)=2, m(3)=2, m(4)=1 and m(5)=1.

A frame F is a finite subset of the Cartesian product of

the natural numbers. A numbering a is a function from a frame

F onto some subset of the natural numbers. Frames are usually

depicted as a pattern of squares where the frame element (i,j)

appears as a square in row i and column j of a grid which is

ordered in the same way as a matrix. For example, if

F = {(1,1), (2,2), (2,3), (4,2), (3,1)1,

then we depict F by

Numberings are usually depicted by placing a((i,j)) in the (i,j)

square of F. For example, if a: F+N' (N' a subset of the natural

numbers) is defined by a((i,j)) = i+j, then we depict the

numbering of F by

2

4|574

6

The multiset of all entries in F is called the content of a.

5

Now let us examine one natural source of frames. Given

a natural number N, we define a partition of N to be a

sequence (X.)= of natural numbers A. such that X.>A. for-+1each i andn=1= N. For example, {5,3,1,1} and {4,3,2,1}

are partitions of 10, A partition of N defines a frame with

N squares in the following manner: Row i of the frame consists

of the squares (i,j), for j=1,2,...,A.. Thus, the earlier

examples of partitions define the frames

and

Observe that a frame defined by a partition satisfies the

following properties:

i) If (I,J) is in F, then (I,j) is in F for j=1,2,...,J-1.

ii) If (I,J) is in F, then (i,J) is in F for i=1, 2,..,I-1.

We now define a tableau to be an ordered pair (F,a) where

F is a frame determined by a partition and a is a numbering.

If T is a tableau, then the frame of T is denoted by ITI andthe content of T is denoted by {T} . For example, if T is a

tableau with F the frame defined by the partition {5,3,1,1}

and a the numbering sending (i,j) to i+j, we depict T by

2 3 4 5 6

4

5

The size of a tableau T is the cardinality of ITJ. A tableauis said to be standard if its numbering a satisfies the

following:

6

i) For every i,j, a((i,j)) < a((i+1,j)).

ii) For every i,j, a((i,j)) < a((i,j+1)).

In other words, the elements of the tableau are non-decreasing

along rows and down columns.

Finally, we define a bitableau to be an ordered pair of

tableaux having the same frame and a standard bitableau to be

a bitableau where both tableaux are standard.

In 1959, Craige Schensted developed an algorithm which

showed that there exists a one-to-one correspondence between

the set of permutations of the first N natural numbers and the

set of standard bitableaux (P,Q) where both tableaux are of

size N.

We begin the description of Schensted's algorithm by de-

fining the insertion of a number s into row i of the tableau T

which has I rows. Consider i

7

Algorithm 2,1 (Schensted). Let

1 2 3 ... N

IT = ( I1) 71(2) 71(3) ... I (N)be a permutation of the first N natural numbers. Create a

tableau P, which will be called the P-tableau for ' by the

following: Define P1 to be the tableau of size one with 71(1)

in square (1,1). Define Pi recursively to be the tableau

obtained by inserting 71(i) into the rows of tableau P.-. We

refer to the tableau PN as P.

The frame of squares that are involved in the insertion of

7(i) is called the bumping path of 71(i) in Pi,

Table I in the appendix is a pidgin Pascal version of

Schensted's algorithm.

Example 2.1. Let

1 2 3 4 5

2 4 3 5 1.

Then, we have

P1 = 2 P2 = 2 4

P3 = 2 3 P4 = 2 3 5

4 ,4 4,1 3 5

P = P = 24

The second tableau Q, which is called the Q-tableau, is

obtained by numbering the squares of IP I from 1 to N in the

order they were added in the construction of the P-tableau.

Example 2.2. Let

1 2 3 4 5

IT2 4 3 5 1 .

8

In Example 2.1, observe the order in which squares were added

to form P. Then we see that the Q-tableau for Tr is

1 2 4

Q = 35

Thus, the bitableau associated with 7r is

1 3 5 1 2 4

(P,Q) = 2 34 , 5.

Table IV in the appendix contains the bitableaux for all

permutations of the first N natural numbers for N = 1,2,...,5.

The bitableaux were generated by the Pascal program which

appears in Table III in the appendix.

To see that Schensted's algorithm has the required

properties, we need the following lemmas.

Lemma 2.1. Let (P.) be the sequence of P-tableaux

formed by applying Schensted's algorithm to a permutation fr.

Then each Pi is standard.

Proof. Pi is standard since it has only one square

containing (1). Suppose Pi_1 is standard. If r(i) is addedto the end of row one, then Pi is standard. Suppose 7(i)

bumps 7(i1) from row one and is inserted in square (1,2) .We now induct on the rows to show that the bumping path

of 7T(i) moves down and left. Suppose T(i 1) is inserted into

square (2,k 2 ) where 22 1. Denote by 7r(k') the number insquare (2,21) . By the induction hypothesis on P._ , Tr 1)

9

Therefore, 7r(i1) is inserted into square (2,2 ) where , j_1 and 7T(i._ ) was bumped from square(j-1, _). By the induction hypotheses, Tr(i. )

10

Lemma 2.3. If 7 is a permutation of 1,2,...,N, then the

bitableau (P,Q) resulting from Schensted's algorithm is

standard. Also, IPI=IQI and {P} = {Q}.

Proof. Since P=PN and Q=QN, by Lemma 2.1 and Lemma 2.2,

(P,Q) is an ordered pair of standard tableaux. By definition,

IQI=IPI and we have (P,Q) a standard bitableau. Since Tr is apermutation of the first N natural numbers, we immediately

have {P}={Q}.

We have now shown that the tableaux for P and Q obtained bySchensted's algorithm are standard. Next, we show that

Schensted's algorithm can be reversed, so that given a standard

bitableau we can produce a permutation.

Algorithm 2.2 (Reverse Schensted). Let (P,Q) be a

bitableau of size N. Construct the permutation 7 by the

following: First locate the square in Q containing N. Findthe number k in the same square of P. Then reverse the bumping

procedure up the rows of P beginning with k. More specifically,

if k is in the first row, then Tr (N)=k. If k is not in the

first row, then k bumps the last number in the row above the

one containing k that is smaller than k. Continuing this process

recursively until we reach the first row, Tr(N) is the number

that was bumped out of the first row.

Repeat the above process from N-i down to one on the

tableau resulting from the previous step.

Table II in the appendix is a pidgin Pascal program of

Schensted's reverse algorithm.

11

Example 2.3. Let P=(a,F), Q=(,G) and

1(3 5 1 2 4(P,Q) = 2 3

4 , 5.

i) Observe that 5=x((3,1)). Thus reverse the bumping

on a((3,1))=4. The four bumps the two which in turn bumps the

one. Thus, 7(5)=1. Note that we have reconstructedP4 = 2 3 5

4

ii) Observe that 4=9((1,3)). Because 5=a((1,3)) and (1,3)

is in row one, we have 7(4)=5. We now have

P 3 = 2 34

iii) Observe that 3=x((2,1)). Reverse the bumping procedure

on a((2,1)) in P3 . The four bumps the three; thus, 7(3)=3 andwe have

P2 = 2 4

iv) Observe that 2=8 ((1,2)). Therefore, from P2 , 7(2)=4.

v) Finally, 7(1)=2.

Therefore,

r= 1 2 3 4 52 4 3 5 1 .

Table VI in the appendix contains examples of the reverse

Schensted algorithm applied to standard bitableaux. The examples

were generated by the Pascal program in Table V.

Theorem 2.1 (Schensted). There exists a one-to-one

correspondence between the set of permutations of the first N

natural numbers and the set of standard bitableaux (P,Q) where

P and Q are of size N.

12

Proof. Given a permutation of 1,2,...,N, the previous

lemmas show that Schensted's algorithm uniquely determines a

standard bitableau with the desired properties. Given a

standard bitableau (P,Q) constructed for the permutation 7where the size of its tableaux is N, we can reverse Schensted's

algorithm to obtain the permutation Tr. Therefore, Schensted'salgorithm provides a one-to-one correspondence between the

two sets.

Schensted's algorithm also provides a proof of the formula

2N! = f

where f is the number of standard tableaux of shape A and

where the sum is over all partitions A of N. The above formula

2follows immediately from Theorem 2.1 noting that f is the

number of standard bitableaux of size N with frame A and that

there are N! permutations of size N.

We now consider a simple generalization of Schensted's

algorithm. A permutation of the first N natural numbers can be

considered as a finite sequence of natural numbers. For

example, the permutation

(1 2 3 4 5)2 3 5 4 1)

is identified with the sequence 23541. Thus, Schensted's

algorithm may be applied to finite number sequences arising

from permutations. To extend it to any finite sequence, we

modify Schensted's algorithm as follows:

If 'r(i) is larger than, or equal to, all entries in row

j of P and row j of P has J entries, then a((j,J+1))=r(i) .

13

Now Schensted's algorithm is defined

sequences s of real numbers,

Example 2.4. Let s= 12553213. Then

is found as follows:

P = 1 , p = 1 2

P4 = 1 2 5 5, P = 1 2 3 55 5

on all finite

the bitableau (P,Q)

P3 =p3

p6

1 2 5

1 2 2 5

35

I

1 1 2 5P = 2

35

11

P8 = 2 535

1 2 3 4Q = 5 8

67

Therefore,

1 1 2 3 1 2 3 4(P,Q) = 2 5 5 8

3 65 ,I 7

This modification causes only one change in the above

discussion: We no longer have {P}={Q}.

Another way of describing this extension of Schens ted's

algorithm, due to Schensted, is by replacing the a1 occurrences

of the smallest number b1 in the sequence with 1, 2,,. a , .

Then replace the acccurrences of the second smallest number

b2 with a 1 +l,a 1 +2,...,a1+a2 Continue this process for each

b. in the sequence. The result is a sequence with no repeated

numbers whose increasing subsequences are in a one-to-one

correspondence with the non-decreasing subsequences

and

2 3

14

of the original sequence. The decreasing subsequences of the

two sequences are in a one-to-one correspondence.

Example 2.5. Let s= 12553213. Then the new nonrepeating

sequence is 13785426. The bitableau for the new sequence is

1 2 4 6 1 2 3 4

3 8 5 8

5 6

7 , 7

Thus, the bitableau for the original sequence is

1 1 2 3 1 2 3 42 5 5 8

3 6

5 , 7

which is the same as the bitableau obtained in Example 2.4.

The Hillman-Grassl Algorithm

We now consider a special class of tableaux called reverse

plane partitions (rpp). A reverse plane partition R of the natural

number N is a tableau with frame F and a number N where N is the

sum of all entries in {R} and where zeros are allowed as entries.

We also define the hook length of a square of a frame F. The

hook length h of (IJ) in F is the number of squares of F

in the set

{ (ij) : i>I, j>J }.

The hook lengths of F define a new tableau H, called the hook

tableau, with frame F and numbering a,defined by a((i,j))=h...

Example 2.6. Let

F =

15

Then

6 4 1H = 4 2

3 11

It is immediate from the definition that the rows and

columns of H are non-decreasing.

In 1975, A. P. Hillman and R. M. Grassl described a bi-

jection between the set of rpp's of N on F and the set of allsolutions in the natural numbers of

h..x. . = NF 1J 1J

where h 1 is the hook length of (i,j) in F. This bijection

is described below as an algorithm. For clarity, an example

is given parallel to the algorithm.

Algorithm2.3 (Hillman-Grassl). Example2.7. Let

Given R, an rpp of N on F, we construct 1 1 2R = 2 4

a solution in the natural numbers of 3 54

G h..x.. = NF Ji

as follows:

i) Compute the hook tableau H

for R.

ii) Locate the last non-zero en-

try of the first non-zero column of R.

Then locate the zigzag path z for this

entry. The zigza path, like the

bumping path in Schensted's algorithm,

is a frame of squares found by beginning

6 4 l

i) H = 4 23 1

1

ii) 1 1 2R = 2 4

3 5

4

with the given square of R and moving

up to the above square if the above

entry is the same as the given entry.

Otherwise, move to the square to the

right. Repeat this process with the

new square until we reach a square

not in the tableau. Next, subtract

one from each entry lying in the zig-

zag path. This new tableau is called

R2 ,

The pivot square in R2 is the

square with first coordinate dame as

the first square added to z and second

coordinate same as the last square

added to z3

After determining the piyot

square, add one to the same square of

the multiplicity tableau M which is a

tableau of frame F with initial entries

all zero.

iii) Continue ii) recursiyely until

R has been reduced to a tableau with all

entries zero,

iv) M contains the x.

iii) R =

16

1R 2 = 2

33

0M = 0

0

1

1222

1 245

0 0

00

1 244

0 0

00

1 234

0 00

0

0M = 0

11

1R4 = 1

11

0M = 1

11

1 0 0

M = 1 01 0

1

1 0 0

M = 1 1111

1 0 0M = 1 3

1 1

0 0

R6 = 0 3

0 30

0 0

R8 = 0 1

0 10

0 0

R1 0 = 0 0

0 00

1

1

MlooM = 1 o1 1

1

1 0 0

M = 1 2

1 1

1

111

1

1

1

N = 6.1 + 4#0

= 22,

+ 1.1 + 4-1 + 2-3 + 3;1 + 1-1 + 1.1

we see that the tableau M does yield a solution of

J h..x.. = N.1J 1J

Table VIII in the appendix contains examples of the

Hillman-Grassl algorithm which were generated by the Pascal

program in Table VII.

Next, we examine the reverse algorithm; i.e., given a

solution in the natural numbers of

G h. .x. . = N,F

produce an rpp of N on F. Again, an example is developed

parallel to the algorithm.

Algorithm 2.4 (Reverse Hillman-Grassi). Example 2.8.

Given a solution in the natural numbers the exnns io

Let

nof

22 in the above

example be the

given solution.,

of

h..x. = N,F 1s r p

we construct an rpp of N on F as follows:

0R5 = 0

00

0

R9 = 000

0

R9 = 000

17

034

022

000

0

As,

i) Use the given h. elements to

define a hook tableau H.

ii) Use the given x.. elements to

define a multiplicity tableau M.

iii) Rn+i is the zero tableau of

frame F where n = a. where a. is1 1

in {M}.

iv) Construct Rn, Rn-',,R =R by

using the information in M to reconstruct

the zigzag paths. (The order that the

pivots are used is crucial.) Begin

with the entries in the last column of

M from top to bottom, then use the entries

in the next to last column from top to

bottom. Continue back through column

one.

Next reconstruct the zigzag paths

by exactly reversing the rules used to

construct them in the previous algorithm.

Begin with the rightmost position of the

column containing the particular pivot

element from M under consideration.

Next move down a square if the lower entry

is the same. Otherwise, move left. Conti

until reaching the column containing the

pivot element.

i)

ii)

iii)

iv)

nue

18

6H = 4

31

1M = 1

11

4 121

0 131

0

1

1

1

1

1

0= 0RI = o0

0

0R9 = 0

00

0

R8 = 000

0= 0

00

0= 0

0

0

0

R = o

000

000

000

011

022

033

034

19

After each path is reconstructed, 1 1 2R = 1 3add one to the entries along the path 1 4

1to obtain the next Rn-i. Continue until

all the pivots from the multiplicity 1 1 2R3 = 2 4

matrix have been utilized. 2 42

112 1 1 2R2 = 2 4 R= R1 = 2 4

3 5 3 5

3 . 4

Table X in the appendix contains examples of the reverse

Hillman-Grassl algorithm which were generated by the Pascal

program in Table IX.

Lemma 2.4. If p.. is the pivot square for a zigzag path

z of the rpp R, then the hook length h.. is also the path13

length of z.

Proof. The lemma follows immediately from the fact that,

by definition, the zigzag path moves up and right, begins

at the end of a column, and ends at the end of a row. D

Lemma 2.5. The sequence of tableaux (R.)! that arei =1

produced by applying the Hillman-Grassl algorithm on the rpp

R is a sequence of rpp's.(However, the tableaux R. are never

rpp's of the same number.)

Proof. Let (R.)= be the sequence of tableaux produced

by applying the Hillman-Grassl algorithm on the rpp R. If R2

is anrpp, then by induction so will R3 , R4 , ... , R2,. In the

algorithm, the zigzag path moves up a column only when equal

entries are encountered. Therefore, in the tableau obtained

by subtracting one from each entry in the zigzag path, the

columns are still non-decreasing.

20

The entries below the zigzag path entries in a specific

column are already larger entries and the entries above are

already smaller. A similar argument follows for the rows of

R2 . Therefore, R2 is an rpp. 0

Recall that in step (iv) of the reverse Hillman-Grassl

algorithm, we use what seems to be an arbitrary ordering of

the pivots. In fact, this arbitrary ordering is the reverse

of the order in which the pivots were constructed in the

forward algorithm.

Lemma 2.6. Let (I.,Ji) be the pivot square constructed

at step i of the Hillman-Grassl algorithm. If the multiset

of pivot squares is ordered according to the order in which

the zigzag paths were constructed, then J.

21

paths and (Ii,J.)=(Ii ,Ji+). Next consider the case where

zi and zi+1 intersect at some square (k,m) and zi+ 1 begins

strictly between (Ii,J ) and the first square of z1. This also

produces a contradiction since from square (k,m) onward, the

two paths must coincide.

Finally, note that the paths must intersect. If not, then

all entries along zi must have become zero. In particular,

the first square of zi must have become zero. Since the -new

tableaux arerpp's(Lemma 2.6), then the entire column containing

the first square of zi must have zero entries. This contradicts

zi+1 beginning in the same column, Therefore, Ii+1

22

Hillman-Grassi algorithm reconstructs the rpp, given the

solution of

h. .x.. = N.F L J

Therefore, there is a one-to-one correspondence between

the two sets, D

CHAPTER BIBLIOGRAPHY

1. Hillman, Abraham P. and Grassi, Richard M., "Reverse planepartitions and tableau hook numbers," Journal ofCombinatorial Theory, Series A, 21 (1976),216~221.

2. Schensted, Craige, "Longest increasing and decreasingsubsequences," Canadian Journal of Mathematics,13 (1961), 179-191--

23

CHAPTER III

OPERATIONS ON STANDARD BITABLEAUX

In this chapter, we will examine several operations on

permutations and describe these operations in terms of the

standard bitableaux associated with the permutations by

Schensted's algorithm.

To begin, let

7T = 1 2 3 ... N

Tr(1) Tr(2) ir(3) ... n(N)

be a permutation of the first N natural numbers. We define

the inverse 7-r of 7 to be the permutation

1 2 3 ... N

where r(i)=j if and only if Tr1(j)=i. We define the reverse

of n by

7Tr _ 1 2 3 ... N

Tr(N) Tr(N-1) Tr(N-2) ... r(l)

As unary operations on the set of permutations, inverse is

denoted by (in) and reverse is denoted by (r): thus, (in)('r)='rK

and (r)(Tr)=err.

The main results of this chapter are the explicit

descriptions of the standard bitableaux for 7-r and err interms of the standard bitableaux for rr. We also examine a

representation of the dihedral group of order 8, D4, in terms

of operations on standard bitableaux.

24

25

Schensted's Algorithm:

A Graphical Approach

In 1970, Donald E. Knuth described a graphical approach

to Schensted's algorithm which can be utilized as a proof

technique. Let 7 be a permutation of the first N naturalnumbers. Write r as a sequence of ordered pairs {(i, i(i))}N

Considering each pair as a vertex, we define a graph Grr in

the following way: Connect (i,lr(i)) to (j,Tr(j)) if and only

if i

26

Proof. Let (j,lr(j)) and (k,'r(k)) be in C. and let k7r(k), then by definition, there is a directed edge

from (k,Tr(k)) to (j,7r(j)) contradicting the supposition that

(j,Tr(j)) is in Ci. Therefore, 7r(j)1 and the theorem is true for all natural numbers

less than i.

Let (i ,7r (i)) be in Cv,. By the induction hypothesis, the

first row in the tableau Pi_1 constructed so far is as follows:

Square (1,m) contains 7(iM) where (im , )(i)) is the pair in Cmsuch that im is the largest number among all the first coordinates

in Cm less than X. Now let m

27

7 (i)

28

Then the graph G,. is depicted by

C 1 : (1, 5) o o (2,1)

C2 : (37 o (6,2)

C3: (5,4) o

C4-0

(7,6)

Also, the P-tableau for 7 is1 2 4 6

P = 3 7

5

Notice that 3 bumps 5 in row two, but (4,3) is in C2 while

(1,5) is in C1 .

We next generalize the graph G,. so that the remaining

rows of the P- and Q-tableaux for 7 can be determined. Remove

the first coordinate from the first pair in each C. in G andthe second coordinate from the last pair in each C-. From

Corollaries 3.1 and 3.3, we see that we have removed the numbers

that appear in the first row of the P- and Q-tableaux for Tr.

Next create new subsets C2, C 2, Ck byrealigning

the remaining elements into new pairs without changing the

order in which they appear. More precisely, if

C. = { (m 1 ,71 (m)), (m2 ,'(m2)), ... , (mQ,7r(m )) },

then

C. = { (m2'(m1)), (m3 '7r(m9), ... , (Mk1 )) }.

Repeat this process recursively to form a finite sequence

of sequences of subsets

( C , C , ... , C )1k.J=1

J

29

Each term of the sequence defines a generalized permutation

. in column notation formed by arranging the pairs in increasing

order of first coordinates. The first coordinates form the

domain of 11. and the second coordinates form the image of yr..J J

Finally, construct new graphs G from TT. as G71 was

constructed from -rr.

Example 3.3. Let

1 2 3 4 5 6)

1 5 6 4 2 3

Then we depict G7r by

C1: (1,1) o

C2 : (5) ,) o o (5,2)

C3 : (3,6) o o (6,3)

Then

__ 4 5 67r2 5 4 6

and we depict G by

2

C2 : (4,5) o o (5,4)

C2: (6,6)

Then

73 = 5and G is depicted by

3

C3 : (5,5) o1

Now we use Schensted's algorithm to obtain the P- and Q-tableaux for 7, 72, and 7T 3 . The standard bitableau for 7 is

30

1 2 3 1 2 34 6 4 6

5 , 5.

The standard bitableau for 2 is

4 6 4 6)5 ,5

and the standard bitableau for 7V3 is

(5 , 5 .

Observe in the above example that the standard bitableau

for v2 is the same as the standard bitableau for 7 withoutthe first rows of the P- and Q-tableaux. The same is true

for 7r 3 with respect to Tr 2 . We next show that this is truein general; i.e., the standard bitableau for Tv. can be obtained

from the standard bitableau for 7r by removing the first j-1rows of each tableau.

Lemma 3.2. If 7(i) is inserted into row j of theP-tableau for 7 at step k1 of Schensted's algorithm and Tr(i 2 )is inserted into row j of the P-tableau for 7 at step k 2 wherek2> 9, then T(i1 ) precedes 7(i 2) in the image of T. (i.e.,

Jif 9(h1)=r1) and Ty(h 2 )=7(2), then h1

J-31Ck where k1 is not necessarily distinct from k . Then2 2

(2t (2 )) immediately follows (1 , Tr(i 1)) in Ck-1 by Theorem3.1 on G . Similarly, ( 22r(' 2)) immediately follows

J-1

(i2 2T(i2)) in Ck2 (2 could equal k ). Therefore, by definitionof , (2i, T (i)) and ( 2 , 2))are in G . From the definition

Jof r and because k

It suffices to consider only the second row of the Q-tableau

as the result for any row will follow by induction.

If the first time a number is inserted into square (2,k)

of the P-tableau for r is at step Zk, then 2 is entered into

square (2,k) of the Q-tableau for 7r. By Corollary 3.1 and

Lemma 3.2, Zk is in the domain of 7 2. Since k is in thedomain of 72 and step kk is the first step when a number is

inserted into square (2,k) of the P-tableau for rr, it follows

that Ek is the number in the domain of 7r2 that maps to the

number inserted into square (2,k) of the P-tableau for and

thus is inserted into square (1,k) of the Q-tableau for r2aThus, the second row of the Q-tableau for 7 equals the firstrow of the Q-tableau for 7r2'

Therefore, row j of the Q-tableau for 7r equals the firstrow of the Q-tableau for rr2'

We now state the following corollary which restates

Theorem 3.2 in a manner similar to Theorem 3.1 and Corollary 3.1.

Corollary 3.4 (Knuth). Row j of the P- and Q-tableauxfor i can be constructed from G. . More precisely, the entry

in square (j,Z) of the P-tableau is the largest number occurringas a second coordinate of a pair in C. Also, the entry in

square (j,9) of the Q-tableau for ir is the smallest numberoccurring as the first coordinate of a pair in C.

32

The Inverse Permutation

We now use the methods of the previous section to obtain

the characterization of the standard bitableau for 7r- interms of the standard bitableau for 7T.

Theorem 3,3. Let (P,Q) be the standard bitableau associated

with the permutation 7r by Schensted's algorithm. Then thestandard bitableau associated with the inverse permutation n-1

by Schensted's algorithm is (Q,P),

Proof. Consider the graphs GTV and G,-1. Observe that

by definition of the inverse permutation, the two graphs are

the same if the vertex (i,7r(i)) in G, is identified with thevertex (7(i),i) in G-1. Therefore, on interchanging firstand second coordinates, the subsets Ci are the same for the

two graphs. However, since the pairs are reversed in thegraphs, each Ci in G,-1 is ordered in reverse to its counter-

part in GT

Therefore, if Tr(iJ) is the second coordinate from C. inJ JG that is in square (1,j) of the P-tableau for 7r, then T(i.)

is the first coordinate from C in G,-1 that is in the (1,j)

square of the Q-tableau for n- . Similarly, if i. is the firstcoordinate from C.inj C that is in square (1,j) of the Q-tableaufor Tr, then i. is the second coordinate from C. in G -1 that

3 3ris in the (1,j) square of the P-tableau for 7r-

Thus, using Theorem 3.1 and Corollary 3.1, the standard

bitableau associated with 7V1 is (Q,P).

33

34

Example 3.4. If

7r= 1 2 3 4 5

3 1 4 2 5 ,

then

- 1 2 3 4 52 4 1 3 5).

After applying Schensted's algorithm, we obtain the P- and Q-tableaux for 7r to be

p = 1 2 5 and Q= 1 3 53 4 2 4

Similarly, the P- and Q-tableaux for rV are

P = 2 4 and Q =

We see that P=Q and P=Q as predicted by Theorem 3.3.

An immediate consequence of Theorem 3.3 is the following

corollary:

Corollary 3.5. Let (P,Q) be the standard bitableau

associated with the permutation 7 by Schensted's algorithm.Then P=Q if and only if r=7r-1.

Theorem 3.4 (Schensted). The number of squares in row

one of the P-tableau for T is the length, k, of a longest

increasing subsequence of T.

Proof, By Theorem 3.1, the number of squares in row one

of the P-tableau equals the number, k, of non-empty subsets

C. in G.

Let (ik,V(ik)) be a pair from Ck. Since (ik' r(ik)) is in

Ck, there is a pair (ik-1,(ik-1)) from Ck_1 such that there

is a directed edge from (ik-1r(k-1)) to (ik',r(ik)).

35

Continuing backwards inductively, we obtain a sequence of

pairs f(i.,7r (i.))}k where (i.,7r (i.)) is in C. for each

j=1,2,...,k and where there is a directed edge from (i. ,7r(i. ))7-1 -1

to (i.,7r(i )) for j=2,3,...,k. Observe that I(i1),2r(i),...,7r( iis an increasing subsequence of 7. Hence k

36

when k is even. Similarly, Ck is of the form

1 2 1m-1 1m m+1 12-1 1

iQiQ1 m+1 m m-1 i2 i1

when k is odd. Thus, if Ck contains an even number of vertices,then Ck contains no fixed points of T and if Ck contains an oddnumber of vertices, then Ck contains one and only one fixed

point of Tf.

Now, Tr2 is the permutation obtained by removing the firstcoordinate from the first vertex in each Ck and the second

coordinate from the last vertex in each Ck. Thus, if Ck contains

an even number of vertices, then we obtain the pairs

12 13 1m+1/1-1 1Q

1Q 1-1 ' ' m+1 i '3 ' i2

in G while if Ck contains an odd number of vertices, we obtain

the pairs

12 i i2m 1m+1 Q

it 'm+1 ' lm ' 2

in G . Thus, we see that i is a fixed point of 7 if and onlyif i is not a fixed point of 7 2 . Then for each Ck containing

an even number of vertices, we obtain a fixed point of 7T2 'Similarly, for each Ck containing an odd number of vertices,

we do not obtain a fixed point of n2. We also see that thedomain and range of T2 are the same. Thus, after relabelling,72 can be regarded as a permutation and not just a generalizedpermutation. Also, 7T 2 =7T 2 1

37

We now proceed by induction on the size of the permutation.

If the size of ar is one, then the theorem follows trivially.

Suppose the theorem is true for permutations of size less

than N and let w be a permutation of size N such that r=ir-1We have that

number of columns of odd length in P

= number of columns in P- number of columns of even length in P

where P is the P-tableau for IT. Denote by P, the tableau Pwithout the first row. By Theorem 3.2, the P-tableau for72

2

is P; Then,

number of columns of even length in P= number of columns of odd length in P.

Also,

number of columns in P = number of subsets Ck for w.Therefore,

number of columns of odd length in P

= number of subsets Ck- number of columns of odd length in P.

Since 72 is a permutation of size less than N,number of columns of odd length in P

= number of fixed points in r2.

But,

number of fixed points in 72= number of subsets Ck of even length.

38

Thus ,

number of columns of odd length in P= number of subsets Ck

- number of subsets Ck of even length= number of subsets Ck of odd length= number of fixed points of n.

Therefore, the theorem follows by induction. L

The P-tableau for the Reverse Permutation

In this section, we describe the P-tableau for 7rr interms of the P-tableau for 7. The main result follows froma technical lemma first proven by Schensted. To state this

lemma, we begin with the description of a variation of

Schensted's algorithm.

As described in Chapter II, in Schensted's algorithm,

elements from 7 are inserted into the rows of the P-tableau.By replacing row with column, we obtain a variation of Schensted's

algorithm in which elements are inserted into the columns ofthe P-tableau. We denote by Pi, the P-tableau after the

insertion of 7(i) into the rows of the P-tableau constructedfrom Tr(1) , Tr(2), ... , r(i-1) and we denote by iP the P-tableauafter the insertion of 7r(i) into the columns of the P-tableauconstructed from Tr (1) , Tr(2), ... , m i-i).

Lemma 3.3. Let (p)N be the sequence of P-tableaux1i1iformed by applying Schensted's algorithm with column insertion

to a permutation ff. Then each iP is standard.

39

Proof. The proof is similar to that of Lemma 2.1 where

(Pi) is shown to be a sequence of standard tableaux. C

Define the conjugate tableau P for the standard tableau

P to be the tableau obtained by interchanging the rows and

columns of P. The next two lemmas follow immediately by the

interchanging of the rows and columns of a tableau.

Lemma 3.4. If P is a standard tableau, then P is a

standard tableau.

Lemma 3.5. If NP is the P-tableau for 7 constructed bySchensted's algorithm with column insertion, then NP is the

conjugate of the P-tableau for Tr constructed by Schensted'salgorithm with row insertion; i.e., NP = PN'

In order to state the next two lemmas, we need notation

for the insertion of an arbitrary number into a standard

tableau. Let a be a natural number and P a standard tableau.

Then a-+P denotes the tableau obtained by inserting a into the

columns of P and P-a denotes the tableau obtained by inserting

a into the rows of P.

Example 3,5. Let a=6 and

p = 1 2 3

4 5

Then

a+P =1.2 34 56

and

Pia = 1 2 3 64 5

40

Recall that the row bumping path of 7r(i) in the tableau

P is the finite sequence of squares of P' (i) consisting of

all squares into which a number is bumped, or inserted, during

the insertion of 'r(i). The last member of the row bumping

path (by order of insertion) is the square of P{-ir(i) not in P.

The number in this square is denoted by ii(i) and the square by

17r(i) . The column bumping path is defined similarly and thenumber in the last square is denoted by T(i) and the square

by 171(i)|.

In Lemma 2.1 we showed that the row bumping path moves

down and left. In a similar manner, the column bumping path

moves up and right.

Lemma 3.6, Suppose (Tr(i)+P)+-7r(j) = (i)+(p+,r )).

Then the shapes of (Tr(i)+P)+ r(j) and Tr(i)+(P(7r (j)) can be

reconstructed from the following data: the shapes Hr(i)+PI

and IPA-r(j)i, and whether i(i)(j).

Proof. We consider two cases.

i) First, suppose the squares 1IF(i) j and Ir(j) [ are notthe same. Then the shapes j'ir(i)+}PJ and I P I differ by thesquare 17(i)JI while the shapes JP+r (j)J Iand IP! differ bythe square Il (j)I. Also, |(7T(i)+P)- r(j)I has one more squarethan ITr(i)-PI and 17(i) +(P÷.1(j))j has one more square thanIP+r (j)fJ . Since fI(i)jI is different from |T(j) I and

1T(i)+(P+7 r(j) = ( i)P+r(j),

both their shapes equal

IT (i)+(P Tr(j))I U |(x i)+}P)+g (j).

41

In pictures, if

and

J'i(i)+)PJ = 7~

then

Ir(i)+(P+1r(j))I = 1r(i) = I(T(i)+P)+T(j) I.7T(j)

ii) Now suppose 1f(i) = 17(j) 1 and 7r(i)

S7T(i)+(P+ (j))[( and (7 (i)+P)+-( j)(. Thus, we have

7()+(P- 7T~)

If Tr(i)>i(j), then we have a similar result with the

final picture

I7ri+P-

construction of r(i)+(P-ir(j)). Now consider 'r(i)+P. M is

the number in the last square of column one of 7r(i)-+P. When

7n(j) is inserted into the rows of Tr(i)+P, M will not beaffected unless f(j) is bumped from the next to last row of

7r(i)-P. However, when 7(j) is inserted into the last row ofTr(i)+P, M is bumped and is inserted into the first square of

a new last row of Tr(i)+P. In either case, the construction

of (Tr(i)+P)+1(j) is the same as Tr(i)+(Pf1n(j)) . Thus,

(r(iy )*P)-i(j) = Tr(i)+(Pw (j)) .

The case when ' (j)=M is similar to the case when 7(i)=M.Now consider the remaining case where M is in {P} (and is

neither rr(i) or Tr(j)) . Before proceeding, we make the followingobservation: Let P* denote P without the square containing M.Observe that since M is the maximum of the elements in {P},

M is at the end of a row and at the end of a column. Thus, P*is also a standard tableau. Also, M can bump no other number.Even if the insertion of another number causes M to be bumpedfrom its square, M will simply be inserted into a new square

at the end of the next row or column. The number bumping Mwould be inserted into the square containing M even if M werenot already in that square. Therefore, Tr(i)-*(P7r (j)) andTr (i)+(P*+Tr(j)) differ only by the square containing M. Asimilar observation holds for (Tr(i)+P)+-Trn(j) and ( (i )p*)f7 (j) .

We now use induction on M. If M=1, then the lemma followseasily. Suppose the lemma is true for M

44

(7r(i)+P*)+ r(j) = 7 (i)+(P*+- ( j))

and by the above observation ' (i)t-P+w-(j)) and T(i)+(P*+].(j))

differ only by the square containing M. Similarly, (r(i)+P)4-wr(j)

and (7r(i)+-P*)+w7r (j) differ only by the square containing M.

Therefore, the proof of the lemma would be completed if we

could prove that M occupies the same square in both (r(i)+-P)+rr (j)

and Tr(i)+(P+'r(j)).

We prove this by using the induction hypothesis to apply

Lemma 3.6 to P*. This says that r(7r (i)+P*)+E r(j) equals

IT (i)+(P*+w r(j))l which can be constructed from knowing J7r(i)+P*[

and IP*+r(j) J, and which is the larger of 7(i) and 7(j). Wethen proceed by considering the cases where M in P is, or is not,

in the same square as 'r(i) in 7r(i)+P* or in the same square as

r(j) in P*+Tr(j).

First consider the case when M in P is not in the same

square as '(i) in 7r(i)+P* or the same square as 7r(j) in P*+r(j).Since M is the largest number involved, if the insertion of Tr(i)

or 7i(j) into P caused M to be bumped, then M would be ii(i) or7r(j) . Since this is not the case, M is not bumped and remains

in the same position in both. ('r(i)+P)+fir(j) and 7r(i)+(P+7(j)).

Thus, the lemma is true for this case.

Next, we consider all other possible cases where the

square containing M in P coincides with either the square

containing Tr(i) in 7r(i)+P* or the square containing 7r(j) inP*+7 1(j). However, because inserting into rows and inserting

into columns are symmetric, we need only consider all cases

45

where the square containing M in P coincides with the square

containing r(i) in 7r(i)+}P*.

We distinguish the following cases:

i) 7r(j)| in P*+w7(j) lies in the same column as Tr(i)in 7r (i)+-P*.

ii) |F(j)I in P*+Tr(j) lies in a column to the left of jr(i)in 'r(i)+P*.

iii) I(j)( in P*+Tr(j) lies one column to the right of V,(i)in T (i)+P*.

iv) Jl(j)J in P*+7r (j) lies more than one column to theright of lr(i)j in Tr (i)+}P*.

We now proceed to verify each case.

i) First suppose 7F(j)f in P*+

46

while if Sr(i)>r(j),

I )+T P)+7(j) r(j) M

7ti)

Now consider how 7(j) is inserted into P. Referring againto case (ii) in the proof of Lemma 3.6, we proceed as follows:

By supposition, F(j) will bump M into a new square at the end

of the next row. Now consider '(i)+(P 'n(j)). Thus, we concludethat if W(i)

47

and i(i) will be in the square from which M was bumped; i.e.,

I7(i)I is the same in Tr(i)+P as in 7(i)+P*. Now insert 71(j)into the rows of 7r(i)+P. Observe that f7r(i)+-PI and IP*I

differ by the squares containing M and W(i). Finally, observe

that 7(j) cannot cause the bumping of M. If it did, then Mwould be bumped into a new square at the end of the next row

and this would contradict the conclusion of Lemma 3.6 which

says 1(T(i)+P*)+I(jj) = iy+(p* )

Now consider inserting 7(j) into the rows of P. Bysupposition, 7r(j) is in a square to the left of the square

containing M. Now insert 7(i) into the columns of P+f7(j).Again, by Lemma 3.6, this causes M to be bumped to the end of

the next column. In pictures, we have

17 (i) ->P *I =I71(i)

and

fP*+71(j)| =

while

48

I ((i)+P)+wruj) = MI)+(P = (j))

0

iii) Now suppose that Jr(j) ( in P*+7f(j) lies one columnto the right of Vii(i) ( in Tr(i)+P*. Then when 7r(i) is insertedinto the columns of P, r(i) will bump M from its square. Mwill be bumped into a new square located at the end of the

next column and 7i(i) will be in the square from which M wasbumped. Also, M is in the same square of Tr(i)+P as 7r(j) isin P*+7r (j) . This is because of the supposition and because Tr(j)is in the square at the end of its column and that square is

not in Tr(i)+P*. Thus, when Tr(j) is inserted into the rows ofn (i)+P, Tr(j) will bump M into the square at the end of thenext row. Again by Lemma 3.6, the square containing F(i) is

not affected.

Now consider inserting 7(j) into the rows of P. Since Mis in a square left of Tr(j) in P* and below 7r(j) in P*, theinsertion of 7(j) into P is the same as the insertion of 7(j)into P*. Now insert 7(i) into the columns of P--7(j). Again,since the square containing M is left and below the square

containing 7r(j), the insertion of Tr(i) into P+-7(j) is the sameas in P*. Thus, Tr(i) bumps M into the next column in the squarebelow the one containing 71(j). In pictures, we have

49

=7 '(j) = 1Tr(i)-+(P-Tr(j)[.

M

7T(i)

iv) Lastly, suppose 17r(j)i in P*+ .r(j) lies more than onecolumn to the right of HI(i)[ in 7(i)+P*. Then when T(i) isinserted into the columns of P, i(i) will bump M from its

square. M will be bumped into a new square located at the end

of the next column and Tr(i) will be in the square from which

M was bumped. However, when 7(j) is inserted into the rowsof 7r (i)+P, 7n(j) is inserted into the same square as it was

inserted in P*-+-7(j) since M cannot affect it.

Now consider inserting Tr(j) into the rows of P. Since

M is in a square left of r(j) in P* and below r(j) in P*, theinsertion of 7r(j) into P is the same as the insertion of 7r(j)

into P*. Now insert 7r(i) into the columns of P*-Tr(j). Again,

since the square containing M is left and below the square

containing r (j), the insertion of 7(i) into P*+-T(j) is thesame as in P*. Thus, F(i) bumps M into the next column inthe last square. In pictures, we have

( i +P)+ (j) = = I (i)+(P (j)) .

71(j)

7T(i) M

50

We have verified that, in all cases, M is in the same

square in I (r(i)+P)+'r(j) I and I (i+(P r(j))I. Therefore,

the lemma is true. C

Theorem 3.6 (Schensted). Let 7 be a permutation of thefirst N natural numbers. If P is the P-tableau for 7r, then

P is the P-tableau for ,rrr

Proof. Let rr be a permutation of the first N natural

numbers and let P be the P-tableau for 7. We proceed byinduction on N. If N=1, then P=1=F. If N=2, then

7r = (1 2)or 7r =( 1~In the first case,

P= 12, r = 1 2 ' _-i1 ' =2

and the theorem is true. Similarly, the theorem is true for

the second case.

Let k be a natural number and assume the theorem is

true for N

51

Thus, by the induction hypothesis, R*=P* and N-iR = PN-1'

Hence,

P =P = (P )N N-1iN

= (N-1R)N by the induction hypothesis,

= (7(1)+R*)+ r(N)

= 7r(1)+(R*+ r(N)) by Lemma 3.7,

= Tr(1)+(P*+7r (N)) by the induction hypothesis,

,n()+(P*)

= Tr(1)+(NR*) by the induction hypothesis,

= NR = R. Q

This theorem can be restated as follows:

Corollary 3.6. The tableau (('i()(2))+wr(3))...+11(N))

is the transpose of the tableau (r (1)+. ..(7T(N-2)+(1T(N-1)+w (N)) ).

Also, since reversing a permutation changes increasing

subsequences to decreasing subsequences, we have another

corollary to Theorems 3.4 and 3.6.

Corollary 3.7. The number of squares in column one of

the P-tableau for 7r is the length of a longest decreasing

subsequence of 7.Example 3.6. Let

1 2 3 4 5

4 3 1 5 2 .

Then

r _ 1 2 3 4 5)2 5 1 3 4 / .

Applying Schensted's algorithm to 7r, we obtain

52

1 2 1 4

(P,Q) = 3 5 2 54 , 3 .

Applying Schensted's algorithm to Tr, we obtain

(R,S) = 1 3 4 1 2 52 5 , 3 4 .

Observe that as Theorem 3.6 states, P=R. However, notice

that q S.

The Q-tableau for the Reverse Permutation

The previous section describes the relationship between

the P-tableaux for Tr and 7r. However, Example 3.6 shows

that conjugation is not sufficient to describe the relationship

between the Q-tableaux for Tr and rr. To begin this description,

we describe Marcel P. Schutzenberger's A-operation on tableaux.

Let Q be a standard tableau of size N, a natural number.

Define a new tableau AQ by the following rules: Remove the

number from square (1,1) of Q. Then move the smaller of the

numbers in squares (1,2) and (2,1) of Q into square (1,1) of

Q. This leaves an empty square which is filled by repeating

the first rule in the following process until the second rule

can be applied:

i) If the (i,j) square of Q is empty and one of (i,j+1)

or (i+1,j) is in Q, then fill square (i,j) with the smaller

of the numbers in squares (i,j+1) or (i+1,j) of Q. If either

(i,j+1) or (i+1,j) is not in Q, then the other is chosen for

square (i,j).

53

ii) If the (i,j) square of Q is empty and both (i, j+1)

and (i+1,j) do not belong to Q, then remove the square (i,j)

from Q. The process terminates and the tableau obtained is

AQ.

Observe that, because the tableau Q contains a finite

number of squares and the empty square moves down or right,

the above process terminates in a finite number of steps.

Example 3.7. Let

1 2 3 8Q = 4 6 7

5 9

Construct AQ as follows:

1 2 3 8 Q 2 3 8 2C 3 8Q= 4 6 7 -++ 467 -+-+ 4 6 7

5 9 5 9 5 9

2 3E 8 2 3 7 8 2 3 7 8-+ 4 6 7 +-+ 4 6L -- 46 = AQ.

5 9 5 9 5 9

Lemma 3.8. If Q is a standard tableau of size N, then

AQ is a standard tableau of size N-1.

Proof. Let Q be a standard tableau of size N. During

the intermediate steps of the construction of AQ, the frames

obtained are not tableaux because there is a missing square.

We will prove that the numberings associated with these

frames are standard in the sense that entries increase along

rows and down columns. From this we conclude that AQ is

standard.

Remove square (1,1) from Q. As Q is standard, the

resulting numbering is also standard. We fill the missing

54

square in Q at (1,1) by moving either square (1,2) or (2,1)

into the (1,1) position. If square (1,2) is moved, then

by choice its entry is smaller than the entry in square (2,1)

and the first column is still increasing. If square (2,1) is

moved, then we have a similar result.

Suppose that after each of the first k-1 steps of removing

and filling squares, we obtain a standard numbering of a frame.

Now consider the missing square created by filling the position

of the (k-l)th missing square. Call that square (ik'jk)'

We fill (ikjk) by moving either square (ik,jk+1) or (ik+ljk)

into the (ikjk) position. If square (ikjk+1) is moved,

then by choice its entry is smaller than the entry in square

(ik+l,jk). Thus, column jk is increasing below and including

square (ikjk). Also, column jk+1 and row ik are still

increasing. Finally, recall that all entries in column jkabove the square (ikjk) were smaller than the previous entry

in square (ikjk) by the induction hypothesis. Also, that

entry was smaller than the entry now in square (ik'jk)'Therefore, column jk is still increasing. If square (ik+l,jk)is moved, we have a similar result. Therefore, by induction,

AQ is standard.

Finally, observe that since only one square is actually

removed and never replaced, AQ has size N-1. C

Let Q and R be standard tableaux where the frame of Q

is a subset of the frame of R. Then we define the difference

of Q and R, IRI\IQI, to be the frame containing all ordered

pairs in JRI but not in IQI. If IRI=IQJ, then IR!\Q!=p.

55

Applying the A-operation repeatedly to Q, we obtain a

sequence A'Q of standard tableaux where AQ=A0Q and A Q+1=A(AQ).

If the size of AQ is N, then the size of A'Q is N-i. The

sequence ends when N=i at ANQ. The tableau ANQ which has

size zero is called the empty tableau.

For the next lemma, let Q be the Q-tableau for 7, apermutation of the first N natural numbers. We write QN toshow the dependency of Q on the size of 7w. In a similarmanner, define QN* to be the Q-tableau for 7r* where 7f* is

the permutation

2 3 4 N

(7(2) 7(3) 7(4) ' '.7(N))

Define QN** to be the Q-tableau for the permutation 7**

where 7** is the permutation

1 2 3 N-1

7(1) 71(2) 7(3) '.. 7(N-1))

Lemma 3.9 (Schutzenberger). AQN - QN**

Proof. The proof is by induction on N. If N=1, then

7(1)=1 and AQ and Q * are both the empty tableau. Suppose

that for k1.

By definition, QN*=Q (N-1)* U S J(N) where sJ(N) is thesquare of QN* containing N. Similarly, AQN = AQN-1 U s.(N)

1where si(N) is the square of AQN containing N. Combining

the above equalities, we obtain

AQN = Q(N-1)* U s.(N)

= (QN* U Si(N))\(s.(N))

56

It remains to be shown that Is.(N)I= 1s.(N)I. From Schensted'slemma ,

PN = (7r1(1)+P(N- 1) *)+ T(N)

= Tr( )+(P (N-1) *+ T(N) ) .

Since the P and Q tableaux have the same shapes for a given

permutation, we have

IQN-11\Q(N-1)*!=

for some square sk. Since it is not known what relationship

exists among sk, ISi (N)I, and Is.(N)I, we can only concludefor now from the previous statement that

I(N-1)*( U S s (N) I, Is "(N) I } 1-:. QN1

However, either sk=Isi(N)I or sk Isi(N)[. In the former

case,

QN1 = Q(N-1)* 1 U U !si(N)I, s }

where s =fQNI\ Q(N)* 1. Thus, s, is directly under or on the

right side of 1si(N) . However, this contradicts the standardness

of QN. Hence, Isi(N)I=Is.(N)I and AQNQN*

In the latter case, we obtain

QNJ = Q(N-1)*J U f 15i(N)I, s

Again, by Lemma 3.7, we have QN and QN-1 differing by si(N)and QN* and Q (N-l) * differing by (sI(N)). Since Q (N-1) N-1

(by induction hypothesis), we have |si(N)I=si(N)I in the two

tableaux. Therefore, AQN=QN*.

Let Q be the Q-tableau for a permutation fr of size N.

Let Is (k) I be the square (i ,j )=IAN-+lQI\IAN-QI. Thecollection of all such squares forms a frame F. Define a

57

numbering a on F by a(Is () I)=Q. Then we define a new tableau

Q=(F,a). Schiitzenberger refers to the process of creating this

new tableau as evacuation.

Lemma 3.10. IPI=iPI, {P}={p}, and P=P.

Proof. These identities follow immediately from theA

definitions of Q and A Q and by the symmetry of inserting into

rows and inserting into columns.

Lemma 3.11. Define $: {Q}\1 -+ {Q}\N by ii(i)=i-1 for

i=1,2,3,...,N. Define the extension of p to tableaux by

e : AQ + R where R is a tableau with IAQ I=IR I and *p(i) appearsin the square containing i in AQ. Then, we have

AA

Q = (AQ) U sN(N)

Proof. It is clear that ' is well-defined and is a

bijection. Thus, 'p is also a well-defined bijection. The

equalityA A

Q = ie(AQ) sN(N)

follows from the following observations:

i) By construction, AnQ=An-1(AQ) for n=1,2, ...,N.

ii) Because of (i), the insertion of squares in the

construction of Q is the same as the insertion of squares

in the cons truction of AQ (except that Q has one more squareA

than AQ) .

Aiii) The numbers inserted into the squares of AQ are one

more than those in Q since' {AQ}={Q}\1. LI

58

Example 3.8. Consider

Q=1243 5.

Then

AQ = 2 43 5 .

Continuing, we obtain

2 3 4 3 4AQ = 5 , A Q = 5 ,

4 5A Q = 5, and A Q is the empty tableau. Thus,

Q =13 5 and AQ =24

Also,

( Q) =1 3

and the identity from Lemma 3.11 holds.

The next theorem characterizes the Q-tableau for 71r given

the permutation 7r.

Theorem 3.7 (Schiitzenberger). If 7 is a permutation onN elements with Q-tableau Q, then the Q-tableau for 'rr is

Proof. If N=1, then Q=1=Q and the theorem is true. Now

suppose the theorem is true for all permutations of size less

than N. Let R denote the Q-tableau for 7r and define R**

analogously to Q**.

By the induction hypothesis, R** = Q** = e(Q*) and by

definition R = R** U s(7(N)) for some square (ij By

the previous lemma,

Q = e (AQ) U sN (Tr (N))

59

and it follows thatA A

Q = ip (AQ) U N(7T(N))A N

By Lemma 3.9, AQ=Q* and OQ=Q* . Thus,

R = e (Q *) U s(T(N))

A= eC (Q) U sP(r(N))

= (Q\(SN('r(N)))) U s (7(N))

However, since I Q= Qi=jPI=IRI(where P is the P-tableau for 7r),

we must have R=Q.

Therefore, by induction, the theorem is true. Q

Example 3.9. Let

_1 2 3 4 5S4 3 1 5 2 .

Then

r 1 2 3 4 52 5 1 3 4.

Applying Schensted's algorithm to 7, we obtain

1 2 1 4

(P,Q) = 3 5 2 54 , 3 .

Applying Schensted's algorithm to 7r, we obtain1 3 4 1 2 5

(R,S) = 2 5 , 3 4

Observe that

S= 1 2 5 = S.3 4

The Dihedral Group D4

In this section, we show that the operations of inverting

and reversing, acting on the set of permutations of the first

N natural numbers, generate the dihedral group D4 . We then

60

apply the earlier results in this chapter to obtain a

representation of D4 as operations on standard bitableaux.

Recall that D4 is the group of symmetries of a square

and its eight elements are generated by

61

AA

Lemma 3.12. T = T, T = T, T = T, and T = T.

Proof. The first equality follows immediately from the

definition of conjugate tableaux. The second equality follows

from the observation that (r) (r) = (id) and the third is

stated in Lemma 3.10. Finally, the fourth equality follows

from the first three equalities sinceA

A - A

T = T = T = T.

The eight bitableaux listed earlier can now be written

as (P,Q), (P,Q), (Q,P), (P,Q), (Q,P), (Q,P), (Q,P) and (P,Q).

In general, these eight bitableaux are distinct, Of course,

if N (P,Q) = 2 2

3 ,4,

62

-1 = 27T 4 2

T r 1 2 3 4

(,Tr)1 -

3(Q,P) = 2

4

(QP) = 1

(P) =

(1 2 3413 4 2/

(( ) r -1 =1 2

(1 2(( 71 r)-1)r = C2 4

( (7,r ) 1 ) 1~ (4= 2

3 41

3 4

3 41

3 1

3 43}2

Even though the

distinct, we observe

either Q=Q or P=P in

2 3

2 4

2 4

(P,) = 3 4

AA '13(Q,P) = 2

4 ,

AA 12(P,Q) =3

12

,3

1 2 4)

1 2 3)

1

, ~ J

1 234.

1 3

4.

3 4)

2 4)

1'

eight bitableaux in Example 3.10 are

that Q=Q. It is easy to verify that

all the permutations of size four with

exactly one fixed point. However, when N>4, examples can be

found where even the tableaux P, P, P, P, Q, Q, Q, and Qare distinct.

Example 3.11. Let

'12 3 4 57 23 4 1 5 .

Then we obtain the following eight distinct permutations and

standard bitableaux when we act on 7 with the eight operations

listed earlier:

T UU=(pQ)= 1 3 4 5 1 2 3 5

3 4

1 3

63

r

-1

-1 r(7T 1)

(Tr -1

((rr -1 r

((7T ) )) -

5

_

4

1

_ (1

"4

1 2

4 f(P, Q) =4

5

41

5

2 3 4 51 4 3 2)

2 3 4 51

1 2 3 5)

2 3 4 513 21 4)

2 3 4 55 4 3 1)

2 3 4 5

3 4 5 2

23453251i

(Q,P) = 3

1

~3

4

2

5

3

1245

5

12

3,4

134

'5

4 5

1235

3

1 3 4 5

, 2

5

,

2

1 2 3 4

4

,r)_1)r)-1 1 2 3 4 5 _ 1 2 3 4l 5 2 3 45 2 5, 3.

We denote a function f:x+f(x) by f(x). Thus, for example,

the operation (in): (P,Q) (Q,P) on standard bitableaux will

be written simply as (Q,P). The order, OR(P,Q), of an operation

(P,Q) is defined to be the smallest natural number n for which

the operation composed with itself n times is the identity

operation.

Lemma 3.13. OR(P,Q)=1, OR(P,Q)=0R(Q,P)=OR(P,)=0R(QP)

=OR(P,Q)=2, and OR(Q,P)=OR(Q,P)=4.

Proof. By definition, OR(P,Q)=1. Since

(P,Q) (r) (P,) (r) (PQ) (PQ)

and

1 3

(QP) = 24

5

,I

1

64

(P,Q) (in)> (Q,P) (in)> pQ(PQ)(PQ)

we have OR(P,Q)=OR(Q,P)=2. We have OR(Q,P)=2 since

(P,Q) (r) (in) (r)> (QP) Cr) (in)>(r) ^(PQ) = (Q

Similarly, OR(P,Q)=2=OR(P,Q).

Now, since

A

(P,Q) (r) (in)> (r) (in)> A _-

A

Cr) (in)> (Q,P) = APA

(r) (in) (P,Q) = (P,Q),A A

we have OR(Q,P)=4. Similarly, OR(Q,P)=4. 0

Theorem 3.8. The eight bitableaux listed earlier form

a representation of D4 .

Proof. Recall D4 is generated by

65

where x4=1, =1 and xy ylx. Therefore, we have the followingrepresentation of D4

{ (P,Q), (QP), (P,Q), (Q,P)A A A A

(P,Q), (Q,P), (P,Q), (Q,P) }. E

CHAPTER BIBLIOGRAPHY

1. Greene, Curtis, "An extension of Schensted's theorem,"Advances in Mathematics, 14 (1974), 254-265.

2. Knuth, Donald E., "Permutations, matrices, and generalizedYoung tableaux, " Pacific Journal of Mathematics,34 (1970), 709-727T~~

3. Schensted, Craige, "Longest increasing and decreasingsubsequences," Canadian Journal of Mathematics,13 (1961), 179-1

4. Schutzenberger, Marcel P., "La correspondence de Robinson,"in Combinatoire et Repr6sentation du Group ym6trique(D. Foata, ed.), Lecture Notes in Mathematics 579,Springer-Verlag, New York/Heideierg/Berlin, (T977),61-116.

5. , "Quelque remarques sur uneconstruction de Schensted, " Mathematica Scandinavica,12 (1963), 117-128,

6. Thomas, Glanffrwd P., "On a construction of Schutzenberger, "Discrete Mathematics, 17 (1977), 107-118.

66

APPENDIX

TABLE I

SCHENSTED'S ALGORITHM

(Tableaux are represented as matrices.)

BEGINREAD PIP(1,1):=PI(1);Q(1,1):=1;FOR I:=2 TO N DO

BEGINFOR J:=1 TO N DO

BEGINLOCATE K WHERE (P(J,K)O AND (P(J,K+1)0);IF PI(I) > P(J,K) THEN

BEGINP(J,K+1):=PI(I);Q(J,K+1):=I;J:=NEND

ELSEBEGINLOCATE FIRST M WHERE P(J,M) > PI(I);SWITCH:=P(J,M);P(J,M):=PI(I);PI(I):=SWITCHEND

ENDEND;

WRITE P;WRITE QEND.

67

TABLE II

REVERSE SCHENSTED'S ALGORITHM

(Tableaux are represented as matrices.)

BEGINREAD P;READ Q;FOR K:=N DOWNTO 1 DO

BEGINLOCATE I,J WHERE Q(I,J)=K;IF I=1 THEN

BEGINPI(K):=P(I,J);P(I,J) :=0END

ELSEBEGINSWITCH1:=P(I,J);P(I,J):=0;FOR L:=I-1 DOWNTO 1 DO

BEGINLOCATE LAST M WHERE (P(L,M) < SWITCH)

AND (P(L,M) (> 0);SWITCH2:=P(L,M);P(L,M):=SWITCH1;SWITCH 1:=SWITCH2;PI(K):=SWITCH1END

ENDEND

WRITE PIEND.

68

TABLE III

SCHENSTED'S ALGORITHM

PROGRAM SCH (DCS,SEQ);VAR DCS,SEQ:

PI:P,Q:I,J,K,M,N:SW:

PROCEDURE LOCATEK;BEGIN

REPEAT K:=K+1UNTIL P(/J,K/)K:=K-1END;

PROCEDURE LOCATEM;BEGIN

REPEAT M:=M+1;UNTIL P(/J,M/)END;

TEXT;ARRAY (/1..20/) OF INTEGER;ARRAY (/1..20,1..20/) OF INTEGER;INTEGER;INTEGER;

= 0;

> PI(/I/)

BEGINREWRITE (DCS); RESET (SEQ);WRITELN (DCS); WRITELN (DCS);WRITELN (DCS,'THE SEQUENCE IS');FOR I:=1 TO N DO

BEGINREAD (SEQ,PI(/I/)); WRITE (DFOR J:=1 TO N DO

BEGINP(/I,J/) :=O;Q(/I,J/) :=OEND

END;P(/1,1/):=PI(/1/);

READLN (SEQ,N);

WRITELN (DCS);

CSPI (/I/):3);

69

Q(/1,1/):=1;

70

FOR I:=2 TO N DOBEGINFOR J:=1 TO N DO

BEGINLOCATE K;IF PI(/I/) > P(/J,K/) THEN

BEGINP(/J, K+1/) :=PI(/I/) ;Q(/J,K+1/) :=I;J:=NEND

ELSEBEGINLOCATEM;SW:=P(/J,M/);

P(/J,M/):=PI(/I/);PI(/I/):=SWEND

ENDEND;

WRITELN (DCS); WRITELN (DCS);WRITELN (DCS,'THE P-TABLEAU IS');FOR I:=1 TO N DO

BEGIN

WRITELN (DCS);

FOR J:=1 TO N DOBEGINIF P(/I,J/) 0 THEN WRITE (DCS,P(/I,J/):3)END;

WRITELN (DCS)END;

WRITELN (DCS,'THE Q-TABLEAU IS');WRITELN (DCS);FOR I:=1 TO N DO

BEGINFOR J:=1 TO N DO

BEGINIF Q(/I,J/) 0 THEN WRITE (DCS,Q(/I,J/):3)END;

WRITELN (DCS)END;

END.

TABLE IV

DATA FOR SCHENSTED'S ALGORITHM

----------------------------------------------SEQUENCE P-TABLEAU Q-TABLEAU

-----------------------------------------------------111

--------------------------------------------1 2 1 2 1 2

------------------------------------------------1 1

2 1 2 2---------------------------------------------

1 2 3 1 2 3 1 2 3-----------------------------------------------------

1 3 1 32 1 3 2 2

----------------------------------------1 2 1 2

1 3 2 3 3-------------------------------------

1 3 1 22 3 1 2 3

-----------------------------------------1 2 1 3

3 1 2 3 2------------------------------------------------------

1 13 2 1 2 2

3 3----------------------------------------

1 2 3 1 3 44 1 2 3 4 2

1 3 1 44 2 1 3 2 2

4 3-------------------------------------------------

4 1 3 2 3 24 4

---------------------------------------1 3 1 3

4 2 3 1 2 24 4

------------------------------------------------

71

72

SEQUENCE P-TABLEAU Q-TABLEAU-----------------------------------------------------

1 2 1 44 3 1 2 3 2

4 3-----------------------------------------------------

1 14 3 2 1 2 2

3 34 4

-----------------------------------------------------1 2 3 1 2 4

1 4 2 3 4 3-----------------------------------------------------

1 3 1 22 4 1 3 214 314

-----------------------------------------------------1 2 1 2

1 4 3 2 3 34 4

-----------------------------------------------------1 3 1 2

2 4 3 1 2 34 4

-----------------------------------------------------1 2 1 2

3 4 1 2 3 4 314-----------------------------------------------------

114 1 23 4 2 1 2 3

3 4-----------------------------------------------------

1 2 3 1 2 31 2 4 3 4 4

-----------------------------------------------------1 3 1 3

2 1 4 3 214 214-----------------------------------------------------

1 2 4 1 2 31 3 4 2 3 4

-----------------------------------------------------1 3 4 1 2 3

2 3 4 1 2 4-----------------------------------------------------

1 2 1 33 1 4 2 314 214

----------------------------------------------------

73

SEQUENCE P-TABLEAU Q-TABLEAU

3 2 4 11 1423

1 324

1 2 314 1 2 3 4 1 2 314

2 1 3 14

1 3 2 4

1 3 42

1 2143

1 3 42

1 2 43

2 3 1 LI

j 1 2 4

3 2 1 4

1 3 4 1 21 4

2

1 2 4 1 3 43

3

2

1 4L

2

3

1 4

2

3

1 2 3 1 4 5

2

35

1 3

2

4

5

1 2

34

5

1 3

2

4

5

1 2

34

5

1 5

2

34I

1 4

2

3

5

1 42

3

5

1 5

2

34

5 4 1 2 3

5 4 2 1 3

5 4 1 3 2

5 4 2 3 1

5 4 3 1 2

SEQUENCE P-TABLEAU Q-TABLEAU

1 15 4 3 2 1 2 2

3 34 45 5

1 2 3 1 3 55 1 14 2 3 4 2

5

5 2 4 1 3

5 1 4 3 2

5 2 4 3 1

4

1 32 45

1 2345

1 3245

1 25 3 4 1 2 3,4

5

5 3 4 2 11 423

1 32 54

1 3245

1 3245

13254

1324

5 5------------------------------------------------

1 23 13 45 1 2 4 3 4

525

1 3 1 45 2 1 4 3 24 2 5

5 3------------------------------------------------

1 2 4 1 3 45 1 3 14 2 3

525

74

75

SEQUENCE P-TABLEAU Q-TABLEAU--------------------------------------

1 3 4 1-3-45 2 3 4 1 2 2

5 5--------------------------------------------------

5 3 1 4 2 3 4 2 55 3

-----------------------------------------------------14 1 4

5 3 2 4 1 2 23 35 5

----------------------------------------1 2 3 4 1 3 4 5

5 1 2 3 4 5 2

-----------------------------------------1 3 4 1 4 5

5 2 1 3 4 2 25 3

-----------------------------------------1 2 4 1 3 55 1 3 2 4 3 25 4

-------------------------------------1 3 4 1-3-5

5 2 3 1 4 2 25 4

-------------------------------------1 2 4 1-4- 5

5 3 1 2 4 3 25 3

-----------------------------------------------------1 4 1 5

5 3 2 1 4 2 23 35 4

-------------------------------------1 2 3 1-2-5

4 5 1 2 3 4 5 3 4

----------------------------------------1 3 1 24 5 2 1 3 2 5 3 54 4

76

SEQUENCE P-TABLEAU Q-TABLEAU-------------------------------------------------

1 2 3 1 2 4

1 5 2 4 3 4 3

5 5

1 3 1 2

2 5 1 4 3 214 314

5 5

--------------------------------------

1 2 4 1 2 4

1 5 3 4 2 3 3

5 5

--------------------------------------

1 3 4 1 2 4

2 5 3 4 1 2 3

5 5

1 2 1 2

3 5 1 4 2 314 314

5 5

--------------------------------------

114 1 2

3 5 2 4 1 2 5 314

3 5

-----------------------------------------

1 2 3 4 1 2 4 5

1 5 2 3 4 5 3

----------------------------------------

1 3 4 1 2 5

2 5 1 3 4 2 5 314

--------------------------------------

1 2 4 1 2 5

1 5 3 2 4 3 3

5 4

-----------------------------

1 3 4 1 2 5

2 53114 2 3

5 4

---------------------------

1 2 4 1 2 5

3 5 1 2 4 3 5 314

------------------------------------

114 1 2

3 5 2 1 4 2 5 3 5

3 3

77

SEQUENCE P-TABLEAU Q-TABLEAU

4 5 1 3 21 23 5

1 23 45

- - - - - - - - - - v qwlo lb - - - - - - - - - - --m- - - - - m a

145 2 3 11 32 54

1 24 5 3 1 2 3 5

14

1 5

1 23 45

1 23 54

1 2145 3 2 1 2 3

3 1414 5

---------------------------------------------------1 2 3 1 2 5

1 514 2 3 14

5

1 32 5 1 3 2 4

5

1 5 4 3 21 23

5

2 5 14 3 1

3 5 14 1 2

3 5 4 2 1

1 3245

1 23 145

1 14235

1 23 514

1 2345

1 23145

123514

12345

4

78

SEQUENCE P-TABLEAU Q-TABLEAU------------------------------------------

1 3 1 44 2 1 5 3 2 5 2 5

4 3------------------------------------------

1 2 5 1 3 44 1 3 5 2 3 24 5

------------------------------------------1 3 5 1 3 4

4 2 3 5 1 2 24 5

---------------------------------------------------

4 3 1 5 2 3 5 2 54 3

------------------------------------------------1 5 124

4 3 2 5 1 2 23 34 5

- --------------------------- -------1 2 3 1 2 423-

1 4 2 5 3 4 5 3 5-------------------------------------

1 3 5 1 2242 4 1 5 3 2 4 3 5

1 2 5 14-------

1 4 3 5 223 34 5

-------------------------------------

1 3 5 1 2242 4 3 5 1 2 3

4 5--------------------------------------

1 2 5 1 2 243 4 1 5 2 3 4 3 5

--------------------------------------

1 4 5 1 2 43 4 2 5 1 2 3

3 5------------------------------------

1 2 3 5 1 2-3241 2 4 5 3 4 5

-----------------------------------------------------

79

SEQUENCE P-TABLEAU Q-TABLEAU--------------------------------------

1 2 3 1 3-5

4 1 5 2 3 4 5 2 4

1 3 1 3

4 2 5 1 3 2 5 2 5

4 4

--------------------------

1 2 1 3

4 1 5 3 2 3 5 2 4

4 5

1 3 1 3

4 2 5 3 1 2 5 2 4

4 5

---- --- ---- --- --- - ---------

1 2 1 3

4 3 5 1 2 3 5 2 5

4 4

---- ---- ---- --- -----------

1 5 1 3

4 3 5 2 1 2 2

3 4

4 5

--------------------------

1 2 3 1-2-31 4 5 2 3 4 5 4 5

---- --- --- --- --- - ---------

13 5 1 2 3

2 4 5 1 3 2 4 4 5

--------------------------

1 2 5 1 2-3

1 4 5 3 2 3 4

4 5

----------------- - - - - - - - - -

1 3 5 1 2-3

2 4 5 3 1 2 4

4 5

--------------------------------

1 2 5 1 2 3

3 4 5 1 2 3 4 4 5

---- --- ---- --- --- -------

1 4 5 1 2 33 4 5 2 1 2

3 5

80

SEQUENCE P-TABLEAU Q-TABLEAU---------------------------------------------------

1 2 3 1 2 31 2 5 4 3 4 4

5 5--------------------------------------

1 3 1 32 1 5 4 3 214 214

5 5-------------------------------------------------

1 2 4 1 2 31 3 5 4 2 3 4

5 5-------------------------------------------------

1 3 4 1 2 32 3 5 4 1 2 4

5 5--------------------------------------------------

1 2 1 33 1 5 4 2 314 214

5 5----------------------------------------------------

114 1 33 2 5 4 1 2 5 214

3 5-------------------------------------------------

1 2 3 4 1 2 3 51 2 5 3 4 5 4

----------------------------------------------------

1 3 4 1 3 52 1 5 3 4 2 5 214

------------------------------------------------1 2 4 1 2 3

1 3 5 2 4 3 5 4 5------------------------------------------------

1 3 4 1 2 32 3 5 1 4 2 5 4 5

------------------------------------------------1 2 4 1 3 5

3 1 5 2 4 3 5 214-------------------------------------

114 1 33 2 5 1 4 2 5 2 5

3 4------------------------------------------------

1 2 3 1 3 44 1 2 5 3 4 5 2 5

-----------------------------------------------------

81

SEQUENCE P-TABLEAU Q-TABLEAU------------------------------------------

1 3 5 1 3 42 1 4 5 3 2 4 2 5------------------------------------1 2 4 5 1-2-3-4

1 3 4 5 2 3 5-------------------------------------------------------

2345121345 1 2 3 42 5

----------------------------------------23 125 13251 2 5 1 3 5

3 1 4 5 2 3 4 2 5-------------------------------------------

1 4 5 1 3 43 2 4 5 1 2 2

3 5------------------------------------

1 2 3 4 -12-3-41 2 3 5 4 5 5

-----------------------------------------1 3 4 1 3 4

2 1 3 5 4 2 5 2 5-------------------------------------1 2 4 1-2-4

1 3 2 5 4 3 5 3 5-------------------------------------

1 3 4 1--2---4--

2 3 1 5 4 2 5 3 5

1 2 4 1 3 43 1 2 5 4 3 5 2 5---------------------------------------

1 4 1 43 2 15 4 2 5 2 5

3 3--------------------------------------

1 2 3 5 -13-4- 54 1 2 3 5 4 2

--------------------------------------------------------1 3 5 1 4 5

4 2 1 3 5 2 24 3

--------------------------------------1 2 5 1-3-5

4 1 3 2 5 3 2

4 4-----------------------------------------------------

82

SEQUENCE P-TABLEAU Q-TABLEAU-----------------------------------------------------

1 3 5 1 3 54 2 3 1 5 2 2

4 4-------------------------------------------------------------

1 2 5 1 4 54 3 1 2 5 3 2

4 3-----------------------------------------------------

1 5 1 54 3 2 1 5 2 2

3 34 4

-----------------------------------------------------1 2 3 5 1 2 4 5

1 4 2 3 5 4 3-----------------------------------------------------

1 3 5 1 2 52 4 1 3 5 2-4 314

-----------------------------------------------------1 2 5 1 2 5

1 4 3 2 5 3 34 4

-----------------------------------------------------1 3 5 1 2 5

2 4 3 1 5 2 34 4

----------------------------------------------------1 2 5 1 2 5

3 4 1 2 5 314 314-----------------------------------------------------

1 4 5 1 2 53 4 2 1 5 2 3

3 4-----------------------------------------------------

1 2 3 5 1 2 3 51 2 4 3 5 4 4

-----------------------------------------------------1 3 5 1 3 5

2 1 4 3 5 214 214-----------------------------------------------------

1 2 4 5 1 2 3 51 3 4 2 5 3 4

-----------------------------------------------------1 3 4 5 1 2 3 5

2 3 4 1 5 2 4----------------------------------------------------

83

SEQUENCE P-TABLEAU Q-TABLEAU

1 2 5 1 3 53 1 4 2 5 314 214

----------------------------1 4 5 1 3 5

3 2 4 1 5 2 23 4

---- -------------------------------------------------1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

1 3 4 5 1 3 4 521 314 5 2 2

13 451 2 4 5 1 2 4 51 3 2 4 5 3 3

1 3 4 5 1 2 4 52 3 1 4 5 2 3

-----------------------------------------------------1 2 4 5 1 3 4 5

3 1 2 4 5 3 2

1 4 5 1 4 53 2 1 4 5 2 2

3 3-----------------------------------------

TABLE V

REVERSE SCHENSTED'S ALGORITHM

PROGRAM RSA (TAB,DCS);VAR TAB,DCS: TEXT;

I,II,J,JJ,K,L,LL,N: INTEGER;SW1,SW2,SW3: INTEGER;A: ARRAY (/1..20/) OF INTEGER;P,Q: ARRAY (/1..20,1..20/) OF INTEGER;

(* THE TABLEAUX ARE INPUT AS N BY N MATRICES(* WHERE N IS THE SIZE OF THE PERMUTATION.PROCEDURE QLOCATE;

BEGINFOR I:=1 TO N DO

BEGINFOR J:=1 TO N DO

BEGINIF Q(/I,J/)=K THEN

BEGINII:=I; I:=N;

JJ:=J;J:=NEND

ENDEND

END;PROCEDURE PLOCATE;

BEGINFOR L:=1 TO N DO

BEGINIF P(/I,L/) 0 THEN

BEGINIF P(/I,L/)

85

FOR I:=1 TO N DOBEGINFOR J:=1 TO N DO

BEGINREAD (TAB,P(/I,J/));IF P(/I,J/)0 THEN WRITE (DCS,P(/I,J/):3)END;

READLN (TAB);WRITELN (DCS)END;

WRITELN (DCS);WRITELN (DCS); WRITELN (DCS,' THE Q TABLEAU IS:');FOR I:=1 TO N DO

BEGINFOR J:=1 TO N DO

BEGINREAD (TAB,Q(/I,J/));IF Q(/I,J/)O THEN WRITE (DCS,Q(/I,J/):3)END;

READLN (TAB);WRITELN (DCS)END;

FOR K:=N DOWNTO 1 DOBEGINQLOCATE;IF II=1 THEN

BEGINA(/K/):=P(/II,JJ/)P(/II,JJ/):=0

ENDELSE

BEGINSW1:=P(/II,JJ/); P(/II,JJ/):=0;FOR I:=II-1 DOWNTO 1 DO

BEGINPLOCATE;N:=N+1-1

END;A(/K/):=SW1END

END;WRITELN (DCS);WRITELN (DCS,' THE ORIGINAL PERMUTATION IS:');FOR I:=1 TO N DO WRITE (DCS,A(/I/):3);WRITELN (DCS);WRITELN (DCS)END.

TABLE VI

DATA FOR REVERSE SCHENSTED'S ALGORITHM

THE P TABLEAU IS:1 3 52 467

THE Q TABLEAU IS:1 2 63 457

THE ORIGINAL PERMUTATION IS:2 7 1 6 4 5 3

THE P TABLEAU IS:1 2 3 9 15 204 6 8 16 175 7 14

10 1112 181319

THE Q TABLEAU IS:1 2 3 10 11 124 8 9 13 145 15 176 167 18

1920

THE ORIGINAL PERMUTATION IS:10 13 14 12 5 4 1 7 8 16 19 20 11 18 17 2 15 9 6 3

86

TABLE VII

HILLMAN-GRASSL ALGORITHM

PROGRAM HGA (DCS,INPUT);LABEL 2;VAR DCS: TEXT;

R,H,M,Z: ARRAY (/1..10,1..10/) OF INTEGER;E, I,J,K,L,N,S,T,X,Y: INTEGER;CK,MAX, SS, SUM, SZ , TT: INTEGER;

PROCEDURE LOCATE;LABEL 1;BEGINJ:=O; E:=0;

1: J:=J+1; L:=O;REPEAT L:=L+1UNTIL (R(/L,J/) 0) OR (L=N);IF (L=N) AND (J=N) THEN E:=1;IF (L=N) AND (E=O) THEN GOTO 1END;

PROCEDURE LOCATEK;BEGINK:=1;FOR Y:=1 TO N DO IF R(/Y,J/) 0 THEN CK:=Y;K:=CKEND;

PROCEDURE MULTIPLICITYBEGINSS:=N; TT:=N;FOR S:=1 TO N DO

BEGINFOR T:=1 TO N DO

BEGINIF Z(/T,S/) 0 THEN

BEGINIF T < TT THEN TT:=T;IF S < SS THEN SS:=SEND

ENDEND;

M(/TT,SS/):=M(/TT,SS/)+1END;

BEGINREWRITE (DCS); READ (N); READLN; WRITELN (DCS);WRITELN (DCS,'THE REVERSE PLANE PARTITION R IS'); WRITELN (DCS);

87

88

FOR I:=1 TO N DOBEGINFOR J:=1 TO N DO

BEGINREAD (R(/I,J/));IF R(/I,J/) 0 THEN WRITE (DCS,R(/I,J/):3)END;

READLN;WRITELN (DCS)END;

WRITELN (DCS,'THE HOOK TABLEAU H IS');WRITELN (DCS);FOR I:=1 TO N DO

BEGINFOR J:=1 TO N DO

BEGINREAD (H(/I,J/)); M(/I,J/):=0; Z(/I,J/):=0;IF H(/I,J/) 0 THEN WRITE (DCS,H(/I,J/):3)END;

READLN;WRITELN (DCS)END;

WRITELN (DCS);I:=0;REPEAT

I:=I+1; LOCATEJ; LOCATEK; IF E=1 THEN GOTO 2;Z(/K,J/):=R(/K,J/);REPEAT

X:=_0;IF K1 THEN

BEGINIF R(/K,J/)=R(/K-1,J/) THEN

BEGINZ(/K-1,J/):=R(/K-1,J/); K:=K-1END

ELSEBEGINIF R(/K,J+1/) 0 THEN

BEGINZ(/K,J+1/) :=R(/K,J+1/); J:=J+1; X:=1END

ENDEND

ELSEBEGINIF R(/K,J+1/) 0 THEN

BEGINZ(/K,J+1=R(/KJ+1END

END

UNTIL (R(/K,J+1/)=O) AND (X=O);MULTIPLICITY;FOR S:=1 TO N DO

BEGINFOR T:=1 TO N DO

BEGINIF Z(/S,T/)O THEN

BEGINZ(/S,T/):=Z(/S,T/)-1R(/S,T/):=Z(/S,T/)END

J:=J+1; X::1

END

END;FOR 3:=1 TO N DO

BEGINFOR T:=1 TO N DO Z(/S,T/):=OEND;

WRITELN (DCS,'R',I:2,' IS'); WRITELN (DCS);FOR S:=1 TO N-1 DO

BEGINFOR T:=1 TO N-1 DO WRITE (DCS,R(/S,T/):3);WRITELN (DCS)END;WRITELN (DCS)

UNTIL E=1;WRITELN (DCS);WRITELN (DCS,'THE MULTIPLICITY TABLEAU IS');FOR S:=1 TO N-1 DO

BEGINFOR T:=1 TO N-1 DO

BEGINWRITE (DCS,M(/S,T/):3)END;

WRITELN (DCS)END;

SUM: =0;FOR S:=1 TO N DO

BEGINFOR T:=1 TO N DO

BEGINSUM:=SUM+H(/S,T/)*M(/S,T/)END

END;WRITELN (DCS);WRITELN (DCS,'THE SUMMATION IS',SUM:5)END.

89

2:

TABLE VIII

DATA FOR THE HILLMAN-GRASSL ALGORITHM

THE REVERSE PLANE PARTITION R IS

1 2 43 5 54

THE HOOK TABLEAU H IS

5411

32

21

R 1 IS

133

250

4

50

R 2 IS

122

240

440

R 3 IS

111

230

330

R 4 IS

000

1

30

230

R 5 IS

000

1 22 20 0

90

91

R 6 IS

0 1 10 1 10 0 0

R 7 IS

0 0 00 0 10 0 0

R 8 IS

0 0 00 0 00 0 0

THE MULTIPLICITY TABLEAU IS

2 2 01 1 11 0 0

THE SUMMATION IS 24

THE REVERSE PLANE PARTITION R IS

1 1 22 43 54

THE HOOK TABLEAU H IS

6)41423 11

92

R 1 IS

1233

1450

2000

0000

R 2 IS

1222

1440

2000

0000

R 3 IS

1340

2000

0000

R 4 IS

0000

0340

1000

0000

R 5 Is

0000

0330

1000

0000

R 6 IS

0000

0220

1000

0000

R 7 IS

0000

0110

1000

0000

1111

93

R 8 IS

0000

0 10 00 00 0

0000

R 9 IS

0000

0 00 00 00 0

0000

THE MULTIPLICITY TABLEAU IS

1 0 1 01 3 0 01 1 0 01 0 0 0

THE SUMMATION IS 22

TABLE IX

REVERSE HILLMAN-GRASSL ALGORITHM

PROGRAM HGR (DCS,INPUT);VAR DCS: TEXT;

H,M,R,Z: ARRAY (/1..5,1..5/) OF INTEGER;COLVECT,ROWVECT: ARRAY (/1..5/) OF INTEGER;III, J, JJ,K,L,S,T,W: INTEGER;SUM,NROW, NCOL: INTEGER;

PROCEDURE WRITER;BEGINWRITELN (DCS);WRITELN (DCS,'R',SUM:2,' IS');SUM:=SUM-1;WRITELN (DCS);FOR II:=1 TO NROW DO

BEGINFOR JJ:=1 TO NCOL DO

BEGINWRITE (DCS,R(/II,JJ/))END;

WRITELN (DCS)END

END;BEGINREWRITE (DCS); READ (NROW,NCOL); READLN;FOR I:=1 TO NCOL DO READ (COLVECT(/I/)); READLN;FOR I:=1 TO NROW DO READ (ROWVECT(/I/)); READLN;WRITELN (DCS,'THE HOOK TABLEAU IS');WRITELN (DCS);FOR I:=1 TO NROW DO

BEGINFOR J:=1 TO NCOL DO

BEGINREAD (H(/I,J/));IF H(/I,J/) 0 THEN WRITE (DCS,H(/I,J/))END;

READLN; WRITELN (DCS)END;

WRITELN (DCS); WRITELN (DCS,'THE MULTIPLICITY TABLEAU IS');WRITELN (DCS);

94

95

FOR I:=1 TO NROW DOBEGINFOR J:=1 TO NCOL DO

BEGINREAD (M(/I,J/));WRITE (DCS,M(/I,J/));R(/I,J/):=0; Z(/I,J/) :=-1END;

READLN; WRITELN (DCS)END;

SUM:=1;FOR I:=1 TO NROW DO

BEGIN