Non-commutativity and combinatorial problems [lecture notes]

45 Pages·2002·0.269 MB·English

by  David M. R. Jackson

Non
ommutativity
and
Combinatorial Problems
Professor D.M. Ja
kson
Department of Combinatori
s and Optimization
Fa
ulty of Mathemati
s
University of Waterloo
June 12, 2002
2
Contents
1 Introdu
tion 5
1.1 An example of a sequen
e problem . . . . . . . . . . . . . . . . . 5
1.2 Sequen
es as en
oding devi
es for
ombinatorialstru
tures . . . . 6
1.3 Non
ommutativity,rings, homomorphisms . . . . . . . . . . . . . 7
1.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5 Review of some familiarresults . . . . . . . . . . . . . . . . . . . 8
2 Non
ommutation 9
2.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3 Rings 15
3.1 The formalpower series ring Q[[x℄℄ . . . . . . . . . . . . . . . . . 16
3.2 The ring Qhha;bii. . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3 The general approa
h . . . . . . . . . . . . . . . . . . . . . . . . 19
4 Using the ring Qhha;bii 21
4.1 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4.2 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.3 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5 Conditions on adja
ent symbols 33
5.1 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.2 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.3 Summaryand
losing
omments . . . . . . . . . . . . . . . . . . 41
6 Problems to work on 43
3
4 CONTENTS
Chapter 1
Introdu
tion
This talk is about methods of solution of a wide
lass of problems in Combina-
tori
s(orDis
rete Mathemati
s)
alledsequen
e problems. Thegeneralquestion
in this
lass is to determine the number of sequen
es in a given set of symbols
(
alled the alphabet) that satisfy pres
ribed
onditions. It is also a talk about
an approa
h that uses a
ombinationof
ombinatorialideas and ideas fromab-
stra
t algebra, but in a
on
rete setting. Finally,itis a talk about the
on
rete
bene(cid:12)ts that o

ur when good abstra
tions are made.
I have prepared this set of Notes so that you will have a full and a

urate
re
ord of the talk,and so that youwillbe ableto study the materialfurther on
your own, if the ideas attra
t you. There are problems for you to work on in
the last
hapter.
1.1 An example of a sequen
e problem
Itisreasonabletoaskwhyisthe
ountingofsequen
es interesting,soIwillgive
you a very famous problemthat is easy to state and that no one has solved.
The Ex
luded Volume Problem
On a re
tangular grid with origin O, you are allowed to take a walk starting
at O and (cid:12)nishing at O. A walk is a sequen
e of steps, ea
h of unit length, in
the dire
tions east, north, west or south. The
ondition is that the walk is not
allowed to
ross itself, and the problem is to (cid:12)nd a formula for the number of
walks with n steps.
Its origin
Theproblem
omesfromthephysi
als
ien
esand
on
erns
hemi
al
ompounds

alled polymers (plasti
is an example). The
hemi
al diagrams for polymers
are longandstringy. It is alawofphysi
s thatno site inspa
e
an be o

upied
twi
e: on
e an in(cid:12)nitesimal volume in spa
e is o

upied, it is ex
luded from
5
6 CHAPTER 1. INTRODUCTION
further o

upan
y. The number is needed for the determination of the energy
fun
tion. This perplexing problem is
alled the Ex
luded Volume Problem, and
there is a Noble Prize awaitingthe person who
an solve it.
The
orrespondingsequen
e problem
This problem
an be en
oded as a sequen
e as follows. Let E;N;W and S
denote the four types of step. Then a sequen
e in the alphabet fE;N;W;Sg
de(cid:12)nes a walk, starting by
onvention from O. For example, SWNE is a walk
around the boundary of a unit square in a
lo
kwise dire
tion, with O at the
northeast
orner. Sin
e a walk starts and (cid:12)nishes at the same point, #N's =
#S's, and#W's=#E's. Sin
eawalkisnotpermittedto
ross itself,SWNE;
ENWS; NEESWSW; amongothers, are forbiddenas subsequen
es. Thus the
Ex
luded Volume Problem
an be viewed as a question of
ounting sequen
es
that possess none of the forbidden subsequen
es.
Best known results
The best known result so far is an approximate one that takes a

ount of all
forbidden subsequen
es up to a parti
ular length. The idea is that the longer
forbidden subsequen
es rule out, relatively, fewer and fewer sequen
es. The
result was obtained quite re
ently by a
olleague, Doron Zeilberger, and his
do
toral student, who used a theorem that Ian Goulden and I obtained many
years ago in an entirely di(cid:11)erent
onnexion.
1.2 Sequen
es as en
oding devi
es for
ombina-
torial stru
tures
It has been seen how sequen
es
an be used to en
ode the Ex
luded Volume
Problem. This is not an ex
eptional
ase. There are many instan
es where
sequen
es play this role.
An applied problem
For example, I have in
luded a problem that happened to be sent to me from
oneofthe
omputerresear
h organisations. Itisaquestionthat
omesfromthe
optimization of a
omputer
ode for a (cid:12)ling system, and involves the determi-
nation of the probability that a parti
ular bran
h of the
ode will be exe
uted.
It appears as Problem 8.
General theory
Partof
ombinatori
s
on
ernsitselfwith
ountingthenumberofsequen
esthat
satisfy various
onditions. Even quite mild
onditions
an produ
e questions
that are very hard indeed. In this talk I amgoing to explore what
an be done
1.3. NONCOMMUTATIVITY, RINGS, HOMOMORPHISMS 7
when the
onditions use only the relationship between adja
ent symbols. The
Ex
luded Volume Problem is in this
lass, so this
lass is a signi(cid:12)
ant one.
1.3 Non
ommutativity, rings, homomorphisms
Firstofall,donotbe intimidatedbythese terms. Youarealreadyfamiliarwith
operations that do not
ommute. For example, \add 1 and multiply by 2" is
not the sameas \multiplyby 2 and add 1." \Putting on hikingboots and then
walking twenty miles" is very di(cid:11)erent from \walking twenty miles and then
putting on hikingboots" (pa
hyderms ex
luded).
You already have some experien
e with rings without realising it, be
ause
the rules (we
all them axioms) for a ring list the permissible operations, su
h
as multiplyingout bra
kets and
olle
ting terms, in High S
hool algebra.
Homomorphisms will be new to you, but they are fun
tions, and they al-
low you to do legally what you have always be tempted to do with fun
tions,
whi
h is to assume that f(x+y) = f(x)+f(y) and f(xy) = f(x)f(y): Not
all fun
tions are homomorphisms. For example, sin(x+ y) 6= sin(x)+sin(y)
p p p p p p
and sin(xy) 6= sin(x)sin(y); x+y 6= x+ y; although xy = x y: Far
from being diÆ
ult, homomorphisms put one into that blissful world where
everything begins to simplify.
Ideas
Iamgoingtousenooperationsmore
omplexthan+(addition)and(cid:1)(multipli-

ation),but themultipli
ationistobe non
ommutative (so ab6=ba):Moreover,
these operations will be used in a fairly sophisti
ated way. The variables are
to be a and b (we
all them indeterminates) and all of our expressions will be
formed fromthese. They willbe polynomialsand in(cid:12)nite series.
There will be familiar rules that spe
ify the operations that are legal on
expressions. These youhaveseen before, but Iamgoingtobe
arefulaboutex-
pressingthemindetail. Theobje
tprodu
edbytheserulesisanon
ommutative
ring.
In addition, I am going to use three very
arefully designed fun
tions, one
(Æ) fordeletingunwantedinformation,another((cid:30)) forre
ording
ountinginfor-
mation, and a third ((cid:18)) for pulling out a parti
ular
lass of sequen
es (
alled
permutations) that are of spe
ial interest. Ea
h of these fun
tions turn out to
be a homomorphism. These fun
tions a
t on rings.
These, then, are the names of the tools to be used.
What I do
Perhaps I should let you know a little of what I do. My resear
h deals with
dis
rete stru
ture and its analysis through
ombinatorial means and through
algebrai
and analyti
te
hniques, with appli
ationsto a variety ofquestions in
topology,the theory of fun
tions and to mathemati
alphysi
s. Su
h things are
only seen at the university level.
8 CHAPTER 1. INTRODUCTION
1.4 Problems
It is useful to have a
olle
tion of problems to work on and to assist in the
explanation ofthe ideas. These are Problems1 to 9, and they are solved in the
Notes. AttheendoftheNotesaresomequestionsforyoutoworkonyourselves.
They
an allbe solved by the theory explained here.
1.5 Review of some familiar results
I am going to assume that you are familiar with the binomial theorem, the
solution of linear equations by Cr(cid:19)amer's Rule, multipli
ationofsmallmatri
es,
x (cid:0)1
the expansions of series su
h as e and (1(cid:0)x) ; and elementary properties of
determinants.
Chapter 2
Non
ommutation
2.1 Problem 1
Find the number
2 of f0;1g-sequen
es of length 2.
The usual solution
The set of f0;1g-sequen
es of length 2 is f00;01;10;11g;so
n =4:
En
oding the sequen
es
Iamgoingtolookatthisproblemingreaterdetailtosqueezesomemoregeneral
ideas out of it. The (cid:12)rst idea is to represent this set in a more
onvenient way
by using the translation
0$a; 1$b:
Thenthisset be
omesfaa;ab;ba;bbgand,en
oded aspolynomial,thisbe
omes
2 2
a +ab+ba+b :
Clearly, we require that a and b do not
ommute, for otherwise ab = ba so
the sequen
es 01 and 10 would be indistinguishable, and this would defeat our
purpose. Inotherwords,thealgebrai
statementthataandbtonot
ommuteis
adire
t
onsequen
eofthe
ombinatorialstatementthatyou
annotinter
hange
di(cid:11)erent symbols in a sequen
e without
hanging the sequen
e into a di(cid:11)erent
one.
Deleting unwanted information
The se
ond idea is to use the substitution fun
tion Æ su
h that
a7!1; b7!1:
9
10 CHAPTER 2. NONCOMMUTATION
Then
(cid:0) (cid:1)
2 2 2 2
Æ a +ab+ba+b = Æ(a )+Æ(ab)+Æ(ba)+Æ(b )
= Æ(a)Æ(a)+Æ(a)Æ(b)+Æ(b)Æ(a)+Æ(b)Æ(b)
2 2 2 2
= 1 +1 +1 +1
= 4:
This explains why the fun
tion Æ has been introdu
ed in the (cid:12)rst pla
e. It
preserves the relevant information (the number of sequen
es), and deletes the
rest (what the sequen
es a
tually were).
Æ as a homomorphism for deletingsequen
es
Letusformallyre
ordthethreevery
onvenientpropertiesthatthesubstitution
fun
tion Æ has. For any polynomialsf and g in the a's and b's
9
Æ(f+g) = Æ(f)+Æ(g) (additivityproperty); >
=
Æ(fg) = Æ(f)Æ(g) (multipli
ativeproperty);
>
;
Æ(1) = 1:
The(cid:12)rststatesthatÆ
anbeappliedtoea
hsequen
einaset. These
ondstates
thatit
anbedonesymbolbysymbol. ThisiswhatImeantintheIntrodu
tion
by saying that we will use fun
tions that preserve
ounting information and
that also behave ni
ely with respe
t to additionand multipli
ation. In general,
a fun
tion that satis(cid:12)es the above three
onditions is
alled a homomorphism,
and Æ is a
on
rete example of su
h a fun
tion. We have just made use of the
(cid:12)rst two of these properties in the above example.
The solution
Observe that
(cid:0) (cid:1) (cid:0) (cid:1)
2 2 2 2
a +ab+ba+b = a +ab + ba+b =a(a+b)+b(a+b)
2
= (a+b) :
In this simpli(cid:12)
ation,only standard properties of expanding bra
kets has been
used, although
are has been taken with non
ommutativity.
The solution to the problem is therefore
(cid:0) (cid:1)
2

2 = Æ (a+b) (en
oding the problem)
2
= (Æ(a)+Æ(b)) (Æ is a homomorphism)
2
= (1+1) (de(cid:12)nition of Æ)
= 4: