Introduction to Higher Mathematics: Combinatorics and Graph Theory [lecture notes]

48 Pages·2017·0.383 MB·English

by  Melody Chan

Introduction to Higher Mathematics:
Combinatorics and Graph Theory
MelodyChan
(cid:13)c 2017byMelodyChan
VersionDate:November15,2017
Contents
1 Combinatorics 1
1.1 Thepigeonholeprinciple . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Puttingthingsinorder . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Bijections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Theprincipleofinclusion-exclusion . . . . . . . . . . . . . . . . . 11
1.6 TheErdo˝s-Ko-Radotheorem . . . . . . . . . . . . . . . . . . . . . 14
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2 Graphtheory 21
2.1 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Trees. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.3 Graphcoloring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.4 Ramseytheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.5 Theprobabilisticmethod . . . . . . . . . . . . . . . . . . . . . . . 35
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
A Appendix:Classexercises 41
A.1 Randomizedplaylists . . . . . . . . . . . . . . . . . . . . . . . . . 41
A.2 Catalanbijections . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
A.3 Howmanytrees? . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Draft:November15,2017 2 (cid:13)c2017,M.Chan
Chapter 1
Combinatorics
Combinatoricsisthestudyoffinitestructuresinmathematics.Sometimespeoplere-
fertoitas“theartofcounting,”andindeed,countingisatthecoreofcombinatorics,
thoughthere’smoretoitaswell.
1.1 The pigeonhole principle
Letusstartwithoneofthesimplestcountingprinciples.Thissaysthatifweput41
ballsinto40boxesanywaywewant,thenthereissomeboxcontainingatleasttwo
balls.Moregenerally:
Theorem1.1. (Pigeonholeprinciple)Letbandnbepositiveintegerswithb>n.If
weplacebballsintonboxes,thensomeboxmustcontainatleasttwoballs.
If that seems obvious, good. It is worth pausing for a moment and asking our-
selveshowwewouldprovesuchastatement:i.e.convinceadoubtfulpersonbeyond
anydoubtwhatsoeverusingareasonedargument.Howwouldyoudothat?
Proof. We prove this statement by contradiction. Suppose we place b balls into n
boxes,butthateachboxcontainsatmostoneball.Sinceeachboxcontainsatmost
oneballandthereareatmostnboxes,wehavemostnballsintotal.Butthenumber
ofballsbwasassumedstrictlygreaterthann,sowehavearrivedatacontradiction.
Thereforesomeboxcontainsatleasttwoballs.
Theorem 1.2. (Pigeonhole principle, general version) Suppose b,k,n are positive
integerswithb > nk.Ifweplacebballsintonboxes,thensomeboxmustcontain
atleastk+1balls.
Proof. Thisproofisanexercise.TrytoimitatetheproofofTheorem1.1.
This silly-sounding principle is actually quite useful. We just need to free our
mindsfromthinkingliterallyaboutboxesandballs(orpigeonsandpigeonholes,asI
Draft:November15,2017 1 (cid:13)c2017,M.Chan
2 1. Combinatorics
prefer)andmoregenerallyaboutplacingsomefinitesetofthingsintoafixednumber
ofcategories.Herearesomeexamples.
Byanintegerwemeanawholenumber...,−2,1,0,1,2,....
Example1.3. Nomatterhowyouchoose6positiveintegers,twoofthemwilldiffer
byamultipleof5.
Proof. Weobservethattwonumbersdifferbyamultipleof5preciselywhentheir
remaindersupondivisionby5arethesame.Thereareonly5possibleremainders
0,1,2,3,4
so the Pigeonhole Principle implies that given any six numbers, some pair of them
havethesameremainder.
Example1.4. Givenanyfivepointsinasquareofsidelength1,sometwoofthem
areatdistance<0.75ofeachother.
Youmightenjoyplayingwiththisexample.Forexample,ifyouareallowedto
chooseonlyfourpointsinsteadoffive,youcanputthematthefourcornersofthe
square.Butifyou havetochoosefivepoints,we’reclaimingthattwoofthemwill
bewithin<0.75ofeachothernomatterwhat.Tryit!
How can we prove this using the Pigeonhole Principle? What are the pigeons?
Whatarethepigeonholes?Whatwewouldreallylikeistobeabletoidentifyfour
regionsofthesquarethatcoverit,suchthateachregionisrelativelysmall:anytwo
pointsinagivenregionhavedistance<0.75.
Proof. Dividethesquareintofoursquaresofsidelength1/2.Givenfivepoints,the
PigeonholePrincipleimpliesthattwoofthepointsliewithinthesmallsquare,and
√
hence are at distance at most 2/2 ≈ 0.707 from each other (this number is the
lengthofthediagonalofthesmallsquare).
1.2 Putting things in order
Supposenisapositiveinteger.
Howmanywaysaretheretoputndistinctsymbolsinanyorder?
Let’strywritingdownallthewaystoorder1,...,nforsmallvaluesofn.Writ-
ingdownsmallexamplesisoftenagoodstrategytogetstarted.
Draft:November15,2017 (cid:13)c2017,M.Chan
1.2. Puttingthingsinorder 3
n=1: 1
n=2: 12, 21
n=3: 123, 132, 213, 231, 312, 321
n=4: 1234, 1243, 1324, 1342, 1423, 1432,
2134, 2143, 2314, 2341, 2413, 2431,
3124, 3142, 3214, 3241, 3412, 3421,
4123, 4132, 4213, 4231, 4312, 4321
Weget1,2,6,24,...
Definition. Letnbeapositiveinteger.Thefactorialofnis
n·(n−1)·····3·2·1
andiswrittenn!.
Let us notice that n · (n − 1)! = n!. (Incidentally, for this reason, we define
0!=1.)Ourevidencesuggests:
Proposition 1.5. Let n be a positive integer. There are n! ways to put n distinct
symbolsinorder.
Informally,wemightargue:Therearendifferentwaystopickthefirstsymbol.
Havingdonethat,n−1unusedsymbolsremain,sotherearen−1waystopickthe
secondsymbol.Proceedinginthisway,weconcludethattherearen(n−1)···1=n!
waystoputallnsymbolsinorder.
That’sprettygood,althoughformytaste,it’snotquiteaproof.Whatdoes“pro-
ceedinginthisway”reallymean?Howdidweknowtomultiplyallthenumbers1
throughn,insteadof,say,addingthem,ordoingsomethingevenweirderwiththem?
Proof. (ProofofProposition1.5)Byinductiononn.Whenn=1,thereisonlyone
orderingpossible,sothepropositionholds.
Next,assumethatn>1andthatwehavealreadyshownthatthereare(n−1)!
waystoputn−1distinctsymbolsinorder.Nowwewishtocountthewaystoputn
distinctsymbolsinorder.Let’sgroupthesewaysaccordingtowhatthefirstsymbol
is. There are n such groups. Within each group, we may chop off the first symbol
andnoticethattheremainingstringsconstituteallwaystoordertheremainingn−1
symbols.Thuseachgrouphassize(n−1)!.So,thereare
(n−1)!+···+(n−1)!=n·(n−1)!=n!
(cid:124) (cid:123)(cid:122) (cid:125)
n
ways.
Draft:November15,2017 (cid:13)c2017,M.Chan
4 1. Combinatorics
Hereisanillustrationoftheinductivestepoftheproof.Supposewealreadyknow
thatthereare6waystoput3symbolsinorder,andwewishtodeducethatthereare
24waystoputsymbols1,2,3,4inorder.Wedividetheorderingsaccordingtowhat
thefirstsymbolis.Thereare4groups,andeachgrouphassize6.
1234, 1243, 1324, 1342, 1423, 1432
2134, 2143, 2314, 2341, 2413, 2431
3124, 3142, 3214, 3241, 3412, 3421
4123, 4132, 4213, 4231, 4312, 4321
Let’sgeneralize:
Proposition 1.6. (Multiplication rule) Suppose m ,...,m are positive integers,
1 n
forsomen ≥ 1.Supposewehaveacollection C ofstringsofsymbolsoflengthn,
suchthat
• therearem symbolsappearingasthefirstsymbolofastringinC,and,
1
• foreachi=2,...,n,anyinitialsubstringofi−1symbolsappearinginCmay
beextendedinexactlym waystoaninitialsubstringofisymbolsappearing
i
inC.
Then
|C|=m ·····m .
1 n
Proof. Exercise.Proveitbyinductiononn,imitatingtheproofofProposition1.5.
That sounds more complicated than it should. It’s basically saying that if you
needtomakeasequenceofnchoices,andthenumberofchoicesyouhaveateach
juncturedoesn’tdependonpastactions,thenallinallyoumaymultiplythenumber
of choices you have at each juncture. (Think of the strings of symbols as all of the
possiblewrittenrecordsofthechoicesyoumade.)
Example1.7. Abinarystringisasequenceof0sand1s.Howmanybinarystrings
oflengthkarethere?
Solution. Bythemultiplicationrule,it’s
2·2···2=2k.
(cid:124) (cid:123)(cid:122) (cid:125)
k
Example1.8. Supposethereare100studentsinaclass.Howmanywaysarethere
tochooseapresident,avice-president,andatreasurer?Nostudentmayholdmore
thanoneposition.
Draft:November15,2017 (cid:13)c2017,M.Chan
1.3. Bijections 5
Solution. There are 100 choices of president. Having chosen a president, there are
99 choices of vice-president. Having made those choices, there are 98 choices of
treasurer.Thereare100·99·98choicesinall.Notice,bytheway,that
100!
100·99·98=
97!
usingawholelotofcancellation.
Generalizing:
Proposition 1.9. Suppose we have n distinct symbols and suppose k is a positive
integerwithk ≤n.Thereare
n·(n−1)···(n−k+1)=n!/(n−k)!
(cid:124) (cid:123)(cid:122) (cid:125)
k
waystochoosekofthesesymbolsandplacetheminorder.
Proof. ThisisaspecialcaseofProposition1.6.Thisnumberissometimesdenoted
P(n,k), and referred to as the number of permutations of n objects taken k at a
time.
1.3 Bijections
This is as good a time as any to mention that bijections between finite sets are a
combinatorialist’sfavoritetoolforcounting.
Definition.
1. Afunctionf: X →Y iscalledaninjectionifforallx,x(cid:48) ∈X,f(x)=f(x(cid:48))
implies x = x(cid:48). (“Different elements of X are sent to different elements of
Y.”)Suchafunctionisalsocalledone-to-one.
2. Afunctionf: X →Y iscalledasurjectionifforeveryy ∈Y,thereissome
x∈X withf(x)=y.Suchafunctionisalsocalledonto.
3. Afunctionf: X →Y iscalledabijectionifitisbothinjectiveandsurjective.
Drawing some cartoons of injective/surjective/bijective functions convinces us
thatbijectionsareone-to-onecorrespondences.Alternatively,supposef isbothin-
jective and surjective. Surjectivity implies that every element y ∈ Y is the image
underf ofatleastonex ∈ X;injectivityimpliesthateveryelementy ∈ Y isthe
imageunderf ofatmostonex∈X.
Thus,wenoticethatcompositionsofbijectionsarebijections.Itisagoodexer-
cisetoconvinceyourself(andthenprove)thefollowingbasicfactaboutbijections:
Youcancheck:
Draft:November15,2017 (cid:13)c2017,M.Chan
6 1. Combinatorics
Proposition1.10. Afunctionf: X → Y isabijectionifandonlyiff hasatwo-
sidedinverse,i.e.afunctiong: Y →X suchthatf ◦g =1 andg◦f =1 .
Y X
Now:
IfX andY arefinitesetsinbijectivecorrespondencethentheyhavethe
samenumberofelements.
(“Bijective correspondence” simply means that there exists a bijection from one to
theother.)Theclaimisobvious,right?Thenweshouldbeabletoproveit...
Proof. Let|X| = n.Well,whatdoesthatactuallymean?Presumably,itmeansthat
thereisabijectionbetweenX andtheset{1,...,n}.Ifso,thenbycomposingthat
bijection with a bijection between X and Y, we obtain a bijection between Y and
{1,...,n},so|Y|=nalso.
Thinkoftheset{1,...,n}asthen-elementsetontheshelfattheNationalBureau
ofStandards.
ThepointisthatonewaytocounttheelementsinasetY issimplytoestablish
a bijection between Y and some set X whose cardinality you already know. This
comesupconstantly,andisextremelysatisfying!Infact,onecouldarguethatthisis
thebest,mostexplicitwaystocountthings—thevariousotherwayswediscussare
buttricksweusewhenunabletoproduceexplicitbijections.
Wewillseetonsofexamples,startingstraightawayinthenextsection.
1.4 Subsets
Definition. Forn≥0aninteger,wedenotetheset{1,...,n}by[n].
Somewhatstrangely,thisconventionincludes[0]=∅.Weask:
Howmanysubsetsofann-elementsetarethere?
Solution. We claim the answer is 2n. It’s good enough to count, by the bijective
correspondence principle, subsets of your favorite n-element set. Mine is [n] =
{1,...,n}.
Notethatthereisanaturalbijectionbetweenthesubsetsof[n]andthelengthn
binarystrings.Namely,givenA ⊆ [n],lete bethebinarystringwhoseith digitis
A
1ifi∈Aand0ifi(cid:54)∈A.(Ithinkofe asa“membership-recordingstring.”)Argue
A
on your own that this is a bijection. Finally, we are done by Example 1.7, because
wealreadycountedthebinarystringsoflengthn:thereare2nofthem.
Definition. LetX beanyset.ThepowersetofX isthesetofallsubsetsofX;itis
denotedP(X).
Soif|X|=nthen|P(X)|=2n.
Nowweconsidersubsetsofafixedsize.Fixk ≥0.
Draft:November15,2017 (cid:13)c2017,M.Chan
1.4. Subsets 7
Howmanysubsetsofann-elementsethavesizek?
Howmany3-elementsubsetsofa4-elementsetarethere,forexample?Let’slist
them:
{1,2,3},{1,2,4},{1,3,4},{2,3,4}.
(Eventually, laziness will overtake us and we will simply write 123,124,134,234
instead.)
So there are four of them. Now one way to compute that number, four, is as
follows.Wealreadyknowthatthereare4·3·2=24waystochoose3elementsof
the4andplacetheminorder.Now,considerthefunctionfrom
{orderedtuples(a,b,c)ofdistinctelementsof{1,2,3,4}}
to
{3-elementsubsetsof{1,2,3,4}}
that forgets the ordering. (This kind of function really is referred to as a forgetful
function:afunctionwhosejobitistoforgetsomestuff.)Hereitis:
(1,2,3) (1,2,4) (1,3,4) (2,3,4)
(1,3,2) (1,4,2) (1,4,3) (2,4,3)
(2,1,3) (2,1,4) (3,1,4) (3,2,4)
(2,3,1) (2,4,1) (3,4,1) (3,4,2)
(3,1,2) (4,1,2) (4,1,3) (4,2,3)
(3,2,1) (4,2,1) (4,3,1) (4,3,2)
↓ ↓ ↓ ↓
{1,2,3} {1,2,4} {1,3,4} {2,3,4}
Ingeneral,eachsetdownstairshas6=3·2·1elementssittingaboveit:that6is
thenumberofwaystoorderthethreegivensymbols.Sowecompute4as?·6=24.
Moregenerally:
Definition. Let k,n ≥ 0 be integers with k ≤ n. Let (cid:0)n(cid:1) denote the number of
k
k-elementsubsetsofann-elementset.Suchexpressionsarecalledbinomialcoeffi-
cients;wewillseewhyinabit.
Proposition1.11. Wehave
(cid:18) (cid:19)
n
·k!=P(n,k).
k
(Recall that P(n,k) denotes the number of ways to choose k elements from an n-
elementsetandorderthem.Therefore,
(cid:18) (cid:19)
n n!
= . (1.1)
k k!(n−k)!
Draft:November15,2017 (cid:13)c2017,M.Chan
8 1. Combinatorics
Proof. Thisproofisanexerciseforyou,imitatingtheargumentabove.
YoucanalsowriteC(n,k)insteadof(cid:0)n(cid:1),forthenumberofcombinationsofnthings
k
takenkatatime,butthesedaysthenotation(cid:0)n(cid:1)ismorestylish.
k
Corollary1.12. Foreachn≥0,wehavethefollowingidentity:
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
n n n n
2n = + +···+ + .
0 1 n−1 n
Proof. Bothsidescountthenumberofsubsetsof[n].
Thereissomuchtosayaboutthebinomialcoefficients!Let’sstartbyarranging
themvisuallyonthepageinabigtriangle,whereineachrownstartingatn = 0,
likethis:
(cid:18) (cid:19)
0
0
(cid:18) (cid:19) (cid:18) (cid:19)
1 1
0 1
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
2 2 2
2 1 0
ThisiscalledPascal’striangle;itwasknowntotheancients.Aportionofitisshown
inFigure1.1.
LookingatFigure1.1,letmakesomeobservations.Thesearejustguessesforthe
moment:
Conjecture1.13.
1. Eachnumber(besidesallthe1s)isthesumofthetwonumbersdirectlyabove
it.
2. Thetriangleisleft-rightsymmetric.
3. Everynumberdownthemiddleiseven.
4. Everyrowisunimodal,meaningitfirstincreasesandthendecreases.
5. Ifnisprime,theneach(cid:0)n(cid:1),otherthan(cid:0)n(cid:1)and(cid:0)n(cid:1),isamultipleofn.
k 0 n
Itturnsoutthatthesestatementsaretrueingeneral.
Proof. Let us prove the first statement, leaving the rest as homework. The claim
amountstoprovingtheidentity
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
n n−1 n−1
= + .
k k k−1
Draft:November15,2017 (cid:13)c2017,M.Chan