discrete mathematics - yale lecture notes - l lovasz (1999) ww

2.3 The number of subsets Nowthatwehave introducedthe notionof subsets,we can formulate our rst general combi-natorialproblem: what isthe numberof all subsetsof aset withn elements?. No

Lecture Notes, YaleUniversity, Spring1999

L Lovasz and K Vesztergombi

Parts of these lecture notes are based on

(Tankonyvkiado,Budapest,1972);

Chapter 13 is based on a section in

L Lov



asz { M.D Plummer: Matching theory

(Elsevier, Amsterdam,1979)

1 Introduction 5

2.1 A party 6

2.2 Sets and thelike 8

2.3 The numberof subsets 10

2.4 Sequences 14

2.5 Permutations 16

3 Induction 19 3.1 The sum ofoddnumbers 19

3.2 Subsetcounting revisited 21

3.3 A fewmore inductionproofs 22

3.4 Counting regions 24

4 Counting subsets 27 4.1 The numberof orderedsubsets 27

4.2 The numberof subsets ofa given size 27

4.3 The BinomialTheorem 28

4.4 Distributingpresents 30

4.5 Anagrams 31

4.6 Distributingmoney 32

5 Pascal's Triangle 34 5.1 IdentitiesinthePascalTriangle 34

5.2 A bird'seye viewat the PascalTriangle 37

6 Combinatorial probability 44 6.1 Events andprobabilities 44

6.2 Independentrepetitionof an experiment 45

6.3 The Lawof LargeNumbers 46

7 Fibonacci numbers 48 7.1 Fibonacci's exercise 48

7.2 Lots of identities 49

7.3 A formulafortheFibonacci numbers 51

8 Integers, divisors, and primes 53 8.1 Divisibilityof integers 53

8.2 Primesand theirhistory 53

8.3 Factorization into primes 56

8.4 Onthe setof primes 57

8.5 Fermat's \Little"Theorem 61

8.6 The EuclideanAlgorithm 62

8.7 Testingforprimality 66

9.1 Even and odddegrees 69

9.2 Paths,cycles,and connectivity 73

10Trees 77 10.1 How to growa tree? 78

10.2 Rootedtrees 80

10.3 How manytreesare there? 80

10.4 How to storeatree? 81

11Finding the optimum 88 11.1 Findingthebesttree 88

11.2 TravelingSalesman 91

12Matchings in graphs 93 12.1 A dancingproblem 93

12.2 Anothermatching problem 94

12.3 The maintheorem 96

12.4 How to nda perfectmatching? 98

12.5 Hamiltoniancycles 101

For moststudents, the rst and often onlyareaof mathematics incollege is calculus Andit

is true that calculus is the single most important eld of mathematics, whose emergence in

the17thcenturysignalledthebirthofmodernmathematicsandwasthekeyto thesuccessful

applicationsof mathematicsinthesciences

Butcalculus(oranalysis)isalso verytechnical Ittakesalotofworkeven tointroduceits

fundamental notionslike continuityor derivatives (after all, ittook 2 centuries just to de ne

these notions properly) To get a feeling forthe power of its methods, say b describing one

of itsimportant applicationsindetail,takes years ofstudy

Ifyouwantto become amathematician,computerscientist, orengineer,thisinvestment is

necessary But ifyourgoal isto develop afeeling forwhat mathematicsisall about,where is

itthat mathematical methodscan behelpful,and what kindofquestions do mathematicians

work on,you maywant to lookfortheanswerinsome other eldsof mathematics

Therearemanysuccessstoriesofappliedmathematicsoutsidecalculus Arecenthottopic

ismathematicalcryptography,whichisbasedonnumbertheory(thestudyofpositiveintegers

1;2;3;: :),andis widelyapplied,among others, incomputersecurityand electronicbanking

Other important areas in applied mathematics include linear programming, coding theory,

theory of computing The mathematics in these applications is collectively called discrete

mathematics (\Discrete" here isusedastheoppositeof \continuous";it isalsooften usedin

themore restrictive senseof\ nite".)

Theaimofthisbookisnottocover\discretemathematics"indepth(itshouldbeclearfrom

the descriptionabove that such a taskwouldbe ill-de nedand impossibleanyway) Rather,

we discussa numberof selected resultsand methods,mostly fromtheareas ofcombinatorics,

graphtheory,and combinatorial geometry, witha littleelementarynumbertheory

At the same time, it is important to realize that mathematics cannot be done without

proofs Merely stating the facts, without saying something about why these facts are valid,

wouldbeterriblyfarfromthespiritofmathematicsandwouldmakeitimpossibletogiveany

idea about how it works Thus, wherever possible, we'll give the proofs of the theorems we

state Sometimesthisisnotpossible;quitesimple,elementaryfactscanbeextremelydiÆcult

toprove,andsomesuch proofsmaytake advancedcoursesto gothrough Inthesecases,we'll

state at leastthat theproof ishighly technicaland goesbeyond thescopeof thisbook

Another important ingredient of mathematics is problem solving You won't be able to

learnanymathematicswithoutdirtyingyourhands and tryingouttheideasyoulearnabout

in the solution of problems To some, this may sound frightening, but in fact most people

pursuethistypeofactivityalmosteveryday: everybodywho playsa gameof chess, orsolves

a puzzle,issolvingdiscretemathematicalproblems Thereaderisstrongly advisedto answer

thequestions posed inthetext and to go throughtheproblemsat theendof each chapter of

thisbook Treat it aspuzzlesolving, and ifyou ndsome ideathat you come upwith inthe

solutionto playsome role later,be satis edthatyouare beginning to getthe essenceof how

mathematicsdevelops

We hopethat we can illustrate that mathematicsis abuilding, whereresults arebuilton

earlierresults,oftengoingbackto thegreatGreek mathematicians;thatmathematicsisalive,

withmorenewideasandmorepressingunsolvedproblemsthanever;andthatmathematicsis

anart,wherethebeautyofideasandmethodsisasimportantastheirdiÆcultyorapplicability

2.1 A party

Alice invites six guests to her birthday party: Bob, Carl, Diane, Eve, Frank and George

When they arrive,they shake handswith each other (strange european costum) This group

isstrange anyway, because one of them asks: \How manyhandshakesdoesthismean?"

\I shook6 handsaltogether" says Bob, \andIguess,sodid everybodyelse."

\Sincethere areseven of us,thisshouldmean 7
6=42 handshakes"ventures Carl

\This seems too many" says Diane \The same logic gives 2 handshakes if two persons

meet,whichis clearlywrong."

\This is exactly the point: every handshake was counted twice We have to divide42 b

2,to get therightnumber: 21." settlesEve theissue

When they go to thetable,Alice suggests:

\Let's changetheseating every halfan hour, untilweget every seeting."

\But you stay at theheadof thetable"says George, \sinceyou have yourbirthday."

How long is thisparty going to last? How many dierent seatingsare there (with Alice's

place rst10places?

(c)Showthattheresultabo eforthenumberofpossibleoutcomesforthe rst10places

canbealsoobtainedusing(a)and(b)

There is nothing special about the numbers 100 and 10 in the problem above; we could

carryoutthe same fornathletes withthe rst k placesrecorded

To give a more mathematicalform to theresult,we can replacetheathletes b anysetof

sizen The listof the rst k placesis given b a sequence of k elements of n, whichall have

to be dierent We mayalso view this as selecting a subset of the athletes with k elements,

and thenordering them Thuswe have thefollowing theorem

Theorem 4.1 The number of ordere k-subsets of an n-set is n(n 1): :(n k+1)

(Note thatifwestart withnand count downk numbers,the lastone willbe n k+1.)

4.3Ifyougeneralizethesolutionofexercise4.1,yougettheanswerin theform

n!

(n k)!

Checkthatthisisthesamenumberasgivenin theorem4.1

4.4Explainthesimilarityandthedierencebetweenthecountingquestionsansweredb

theorem4.1andtheorem2.2

4.2 The number of subsets of a given size

From here,we can easilyderiveone of themostimportant counting results

Theorem 4.2 The number of k-subsets of an n-set is

n(n 1): :(n k+1)

k!

=n!

k!(n k)!

Theorem 4.1 Of course, if we want to know thenumber of unordere subsets, then we have

overcounted; every subset was counted exactly k! times (with every possible ordering of its

elements) Sowehavetodividethisnumberb k!togetthenumberofsubsetswithkelements

(withoutordering)

The numberofk-subsetsofann-set issuchanimportantquantitythatone hasaseparate

notation forit:

4.5Whichproblemsdiscussedduring thepartywerespecialcasesoftheorem4.2?

4.6Tabulate thevaluesof

n

k

forn;k5

In thefollowingexercises, try to prove theidentitiesb usingtheformulaintheorem 4.2,

and also without computation, b explaining both sides of the equation as the result of a

also have aname,binomial c eÆcients, which comesfroma very important

formulainalgebrainvolvingthem We arenowgoing to discussthistheorem

The issue is to compute powers of thesimple algebraicexpression(x+y) We start with

+3x2

y+3xy

2

+y3

y+6x2

y2

+4xy3

+y4

:

n

k Letusmakethisobservationprecise Weillustratetheargument

forthenext valueof n, namely n=5,butit worksingeneral

Thinkof expanding

(x+y) =(x+y)(x+y)(x+y)(x+y)(x+y)

sothatwegetridofallparanthesis Wegeteachtermintheexpansionb selectingoneofthe

two termsineach factor,and multiplyingthem If wechoosex,say,2times thenwe choosey

3 times,andwegetx

2

y3

Howmanytimesdo we getthissame term? Clearlyasmanytimes

as the number of ways to select the two factors that supply x (the remaining factors supply

y) Thuswehave to choosetwo factorsoutof 5,which can bedonein

+5

1

!

xy4

+5

2

!

x2

y3

+5

3

!

x3

y2

+5

4

!

x4

y+5

5

!

x5

:

We can applythisargument ingeneralto obtain

Theorem 4.3 (The Binomial Theorem) The c eÆcient of x

k

yk

n

!

xn

:

This importanttheorem iscalled theBinomialTheorem;thename comesfrom theGreek

wordbinome foranexpressionconsistingof twoterms,inthiscase, x+y The appearenceof

thenumbers

n

k

inthistheorem isthesourceof theirname: binomial c eÆcients

The BinomialTheoremcanbeappliedinmanyways togetidentitiesconcerningbinomial

coeÆcients For example,let ussubstitutex=y =1,thenweget

2 =n

0

!

+n

1

!

+n

n

!

Later on we are going to see trickier applications of this idea For the time being, another

twist onit iscontainedinthenext exercise

4.12GiveaproofoftheBinomialTheoremb inducion,basedonexercise4.2

4.13(a)Provetheidentity

=1,withthelastdependingontheparityofn.)

(b)Thisidentityisobviousifnisodd Why?

4.14Proveidentity4,usingacombinatorialinterpretationofthetwosides(recallexercise

4.2)

Suppose we have n dierent presents, which we want to distributeto k children For some

reason, we are told how many presents should each child get; so Adam should get n

Adam

presents, Barbara, n

Barbara

presents etc In a mathematically convenient (though not very

friendly) way, we call the children 1;2;: : k; thus we are given the numbers (non-negative

1+n

2+: :+n

k

= n, else there is no way to

distributethepresents

The questionis,of course, how manyways can these presentsbedistributed?

Wecanorganize thedistributionofpresentsasfollows Welayoutthepresentsinasingle

row of length n The rst child comes and picks up the rst n

1presents, starting from the

left Then thesecond comes, and picks up the next n

Itisclearthatwecandeterminewhogetswhatb choosingtheorderinwhichthepresents

arelaidout Therearen!waystoorderthepresents But,ofcourse,thenumbern!overcounts

thenumberof waysto distributethepresents, sincemanyof theseorderingsleadto thesame

results(that is,every childgetsthe same setof presents) The questionis,how many?

Soletusstartwithagivendistributionofpresents,andlet'saskthechildrentolayoutthe

presentsforus, nicelyinarow, startingwiththe rstchild,thencontinuingwith thesecond,

third, etc Thisway we getback one possibleordering that leadsto thecurrent distribution

The rstchildcanlayouthispresentsinn

1

!possibleorders; nomatterwhichorder hechoses,

thesecondchildcanlayoutherpresentsinn

2

!possibleways,etc Sothenumberof ways the

presentscan belaidout(giventhedistributionofthepresentstothechildren)isaproductof

4.15Wecandescribetheprocedureofdistributingthepresentsasfollows First,weselect

Completethisargumentandshowthat itleadstothesameresultasthepreviousone

4.16Thefollowingspecialcasesshouldbefamiliarfrompreviousproblemsandtheorems

4.17 (a) How manyways can you place n rooks on achessboard so that no twoattack

each other(Figure9)? Weassumethat therooks areidentical, soe.g interchangingtwo

rooksdoesnotcountasaseparateplacement

(b)Howmanywayscanyoudothisifyouhave4blackand4whiterooks?

(c)Howmanywayscanyoudothis ifallthe8rooksaredierent?

4.5 Anagrams

Quiteprobablyyouhaveplayedwithanagrams Oneselectsaword(say,COMBINATORICS)

and triesto composefrom its lettersmeaningful,oftenfunnywords orexpressions

Howmanyanagramscanyoubuildfromagivenword? Ifyoutrytoanswerthisquestionb

playingaroundwiththeletters,you willrealizethatthequestionisbadlyposed;itisdiÆcult

to draw the line between meaningful and non-meaningful anagrams For example, it could

easilyhappen that A CROCBIT SIMON.Andit may be true thatNapoleon always wanted

aTOMBINCORSICA Itisquestionable,butcertainlygrammaticallycorrect, toassert that

COB ISROMANTIC.Some universitiesmayhave a courseon MAC INROBOTICS

But one would have to write a book to introducean exciting character, ROBIN

COSMI-CAT,who enforcesa COSMICRIOTBAN,whileappealingTOCOSMICBRAIN

Andit wouldbe terriblydiÆcultto explainan anagramlikeMTBIRASCIONOC

To avoid this contraversy, let's accept everything, i.e., we don't require the anagram to

be meaningful (or even pronouncible) Of course, the productionof anagrams becomes then

uninteresting;butat least we can tellhowmanyof them arethere!

4.18HowmanyanagramscanyoumakefromthewordCOMBINATORICS?

4.19Which word givesriseto moreanagrams: COMBINATORICSorCOMBINA

TOR-ICA?(ThelatteristheLatin nameofthesubject.)

4.20Whichwordwith13lettersgivesriseto themostanagrams? Whichwordgivesrise

totheleast?

So let's see the general answer to the question of counting anagrams If you have solved

theproblemsabove,itshouldbeclearthatthenumberof anagramsn-letter worddependson

howmanytimeslettersofthewordarerepeated SosupposethatthewordcontainsletterNo

1 n

1

times,letter No 2n

2times,etc.,letter No kn

ktimes Clearly,n

1+n

2+: :+n

k

=n

Now to form an anagram, we have to selectn

1positionsforletter No 1,n

2positionsfor

letterNo 2,etc.,n

kpositionsfroletterNo 3 Havingformulateditthisway,wecan seethat

thisisnothingbutthequestionof distributingnpresents to k children,whenit isprescribed

howmanypresentseach childgets Thuswe know from theprevioussection that theanswer

6!=(2!2!) =180) We say that these wordsare \essentiallythe same" (at least as faras

counting anagramsgoes): theyhave twoletters repeated twice and two lettersoccuring

onlyonce

(a)Howmany6-letterwordsarethere? (Asbefore,thewordsdon'thavetobemeaningful

Thealphabet has26letters.)

(b)Howmanywordswith6lettersare\essentiallythesame"asthewordLETTER?

(c)Howmany\essentiallydierent"6-letterwordsarethere?

(d)Tryto ndageneralanswertoquestion(c)(thatis,howmany\essentiallydierent"

wordsare there onnletters?) If youcan't nd it,read thefollowingsectionand return

tothisexerciseafterit

4.6 Distributing money

Instead of distributing presents, let's distribute money Let us formulate the question in

general: we have n pennies that we want to distribute among k kids Each child must get

at least one penny (and, of course, an integer number of pennies) How many ways can we

distributethemoney?

Beforeansweringthisquestion,wemustclarifythedierencebetweendistributingmoney

and distributing presents If you are distributing presents, you have to decidenot only how

manypresentseachchildgets,butalsowhicharethesepresents Ifyouaredistributingmoney,

onlythe quantity matters In other words, thepresents aredistinguishable whilethepennies

arenot (Aquestionlikeinsection 4.4, wherewe specifyinadvence howmanypresentsdoes

a given childget, would be trivialformoney: thereisonlyone wayto distributenpenniesso

thatthe rstchildgets n

1,thesecond childgets n

2,etc.)

Even thoughtheproblemisquite dierentfrom thedistributionof presents, we can solve

it b imagininga similardistribution method We lineup thepennies(it does notmatter in

which order, they are all alike), and then let child No 1 begin to pick them up from left to

right After a whilewe stop him and let the second child pick up pennies, etc (Figure 10)

Thedistribution of the money isdetermined by specifying where to startwith a new child

Nowthere aren 1points(betweenconsecutive pennies)wherewecan letanewchildin,

and we have to select k 1 of them (since the rst child alsways starts at the beginning,we

havenochoicethere) Thuswehavetoselecta(k 1)-elementsubsetfroman(n 1)-element

set The numberof possibilitiesto do sois

n 1

k 1

To sumup,weget

Theorem 4.4 Thenumberofwaystodistributenidenticalpennies tok children,sothat each

child gets at least one, is

n 1

k 1

It is quite surprising that the binomial coeÆcients give the answer here, in a quite

non-trivialandunexpected way

Let's alsodiscussthenatural (though unfair)modi cationofthisquestion, wherewe also

allow distributions in which some children get no money at all; we consider even giving all

themoneytoone child Withthefollowingtrick, wecan reducetheproblemofcountingsuch

distribnutions to the problem we just solved: we borrow 1 penny from each child, and the

distribute the whole amount (i.e., n+k pennies) to the children so that each child gets at

leastone penny Thiswayevery childgetsbackthemoneyweborrowedfrom himorher,and

P P P P P P P P P P

Alice

Figure 10: Howto distributenpenniesto k children?

thelucky onesgetsome more The \more"isexactlynpenniesdistributedto k children We

already know that thenumberof ways to distributen+k pennies to k childrensothat each

childgetsat least onepennyis

4.23Wedistributenpenniestokb ysand`girls,sothat(tobereallyunfair)werequire

thateachofthegirlsgetsatleastonepenny Inhowmanywayscanwedothis?

4.24kearlsplaycards Originally,theyallhaveppennies Attheendofthegame,they

count how much money they have They do not borrow from each other, sothat they

cannotloosemorethantheirppennies Howmanypossibleresultsarethere?

To study various properties of binomial coeÆcients, the followingpicture is very useful We

arrange all binomialcoeÆcients into a triangularscheme: inthe \zeroeth" row we put

1

!

and2

1

!

; : ;n

n

!

We shift these rows so that their

midpointsmatch; this way we get a pyramid-like scheme, called the Pascal Triangle (named

after theFrench mathematicianand philosopherBlaisePascal, 1623-1662) TheFigure below

shows onlya nitepiece of thePascalTriangle

5.2ProvethateachrowinthePascalTrianglestartsandendswith1

5.1 Identities in the Pascal Triangle

Looking at the Pascal Triangle, it is not hard to notice its most important property: every

numberinit(otherthanthe1'sontheboundary)isthesumofthetwonumbersimmediately

above it This in fact is a property of the binomial coeÆcients you have already met: it

translates into therelation

ThispropertyofthePascalTriangle enablesusto generatethetrianglevery fast,building

it up row b row, using (5) It also gives usa tool to prove many properties of the binomial

coeÆcients, aswe shallsee

As a rst application, let us give a new solution of exercise 4.3 There the task was to

prove theidentity

whichisclearly0,sincethesecondtermineachbracketcancelswiththe ... subsets,we can formulate ourrst general

combi-natorialproblem: what isthe numberof all subsetsof aset withn elements?

Westartwithtrying outsmallnumbers It playsno role whattheelementsofthesetare;... (Besides,they would winmuch less Andto ll

outsomanyticketswouldspoil theparty:: )

So they decideto playcards instead Alice,Bob, Carl and Dianeplaybridge Lookingat

his cards,Carlsays:...

doistostartwith;,thenlistallsubsetswith1elements,thenlistallsubsetswith2elements,

etc Thisis thewaythelist(1) isputtogether

We could order the subsets asin aphone book This methodwill bemore