John wiley sons alon n spencer j h the probabilistic method in combinatorics (2ed2000)

The Probabilistic Method has recently been developed intensively and became one of the most powerful and widely used tools applied in Combinatorics.. The interplay between Discrete Mathe

THE PROBABILISTIC

METHOD

Tel Aviv University,

Tel Aviv, Israel.

Joel H Spencer,

Courant Institute of Mathematical Sciences,

New York University,

New York, USA

A Wiley-Interscience Publication

JOHN WILEY & SONS, INC.

New York / Chichester / Weinheim / Brisbane / Singapore / Toronto

To Nurit and Mary Ann

The Probabilistic Method has recently been developed intensively and became one

of the most powerful and widely used tools applied in Combinatorics One ofthe major reasons for this rapid development is the important role of randomness inTheoretical Computer Science, a field which is recently the source of many intriguingcombinatorial problems

The interplay between Discrete Mathematics and Computer Science suggests analgorithmic point of view in the study of the Probabilistic Method in Combinatoricsand this is the approach we tried to adopt in this book The manuscript thus includes adiscussion of algorithmic techniques together with a study of the classical method aswell as the modern tools applied in it The first part of the book contains a description

of the tools applied in probabilistic arguments, including the basic techniques thatuse expectation and variance, as well as the more recent applications of Martingalesand Correlation Inequalities The second part includes a study of various topics inwhich probabilistic techniques have been successful This part contains chapters ondiscrepancy and random graphs, as well as on several areas in Theoretical ComputerScience; Circuit Complexity , Computational Geometry, and Derandomization ofrandomized algorithms Scattered between the chapters are gems described underthe heading "The Probabilistic Lens" These are elegant proofs that are not necessarilyrelated to the chapters after which they appear and can be usually read separately.The basic Probabilistic Method can be described as follows: in order to provethe existence of a combinatorial structure with certain properties, we construct anappropriate probability space and show that a randomly chosen element in this spacehas the desired properties with positive probability This method has been initiated

vii

viii PREFACE

by Paul Erd}os, who contributed so much to its development over the last fifty years,that it seems appropriate to call it "The Erd}os Method" His contribution cannot bemeasured only by his numerous deep results in the subject, but also by his manyintriguing problems and conjectures that stimulated a big portion of the research inthe area

It seems impossible to write an encyclopedic book on the Probabilistic Method;too many recent interesting results apply probabilistic arguments, and we do not eventry to mention all of them Our emphasis is on methodology, and we thus try todescribe the ideas, and not always to give the best possible results if these are tootechnical to allow a clear presentation Many of the results are asymptotic, and weuse the standard asymptotic notation: for two functionsf andg, we writef = O (g )

iff  cgfor all sufficiently large values of the variables of the two functions, where

cis an absolute positive constant We writef = ifg = O (f )andf = (g )if

f = O (g )andf = If the limit of the ratiof =gtends to zero as the variables

of the functions tend to infinity we writef = o(g ) Finally,f  gdenotes that

f = (1 + o(1))g, i.e., thatf =gtends to1when the variables tend to infinity Eachchapter ends with a list of exercises The more difficult ones are marked by a().The exercises, which have been added to this new edition of the book, enable thereader to check his/her understanding of the material, and also provide the possibility

of using the manuscript as a textbook

Besides these exercises, the second edition contains several improved resultsand covers various topics that have not been discussed in the first edition Theadditions include a continuous approach to discrete probabilistic problems described

in Chapter 3, various novel concentration inequalities introduced in Chapter 7, adiscussion of the relation between discrepancy and VC-dimension in Chapter 13 andseveral combinatorial applications of the entropy function and its properties described

in Chapter 14 Further additions are the final two probabilistic lenses and the newextensive appendix on Paul Erd}os, his papers, conjectures and personality

It is a special pleasure to thank our wives, Nurit and Mary Ann Their patience,understanding and encouragment have been a key-ingredient in the success of thisenterprise

NOGAALON, JOELH SPENCER

We are very grateful to all our students and colleagues who contributed to the creation

of this second edition by joint research, helpful discussions and useful comments.These include Greg Bachelis, Amir Dembo, Ehud Friedgut, Marc Fossorier, Dong

Fu, Svante Janson, Guy Kortzers, Michael Krivelevich, Albert Li, Bojan Mohar,Janos Pach, Yuval Peres, Aravind Srinivasan, Benny Sudakov, Tibor Sz´abo, GregSorkin, John Tromp, David Wilson, Nick Wormald and Uri Zwick, who pointed outvarious inaccuracies and misprints, and suggested improvements in the presentation

as well in the results Needless to say, the responsibility for the remaining mistakes,

as well as the responsibility for the (hopefully very few) new ones, is solely ours

It is a pleasure to thank Oren Nechushtan, for his great technical help in thepreparation of the final manuscript

ix

The Probabilistic Lens: High Girth and

5.2 PropertyBand Multicolored Sets of Real Numbers 65

The Probabilistic Lens: Weierstrass Approximation Theorem 113

xiv CONTENTS

11.2 Random Restrictions and Bounded Depth Circuits 185

The Probabilistic Lens: Crossing Numbers,

Incidences, Sums and

xvi CONTENTS

The Probabilistic Lens: Triangle-free graphs have

Part I

METHODS

The Basic Method

What you need is that your brain is open

– Paul Erd}os

1.1 THE PROBABILISTIC METHOD

The probabilistic method is a powerful tool in tackling many problems in discretemathematics Roughly speaking, the method works as follows: Trying to prove that astructure with certain desired properties exists, one defines an appropriate probabilityspace of structures and then shows that the desired properties hold in this space withpositive probability The method is best illustrated by examples Here is a simple one

The Ramsey-numberR(k ; `)is the smallest integernsuch that in any two-coloring

of the edges of a complete graph onnverticesK

nby red and blue, either there is aredK

k(i.e., a complete subgraph onkvertices all of whose edges are colored red) orthere is a blueK

` Ramsey (1929) showed thatR(k ; `)is finite for any two integers

kand` Let us obtain a lower bound for the diagonal Ramsey numbersR(k ; k )

Proposition 1.1.1 If n

k

2 1 ( 2 )

< 1thenR(k ; k ) > n ThusR(k ; k ) > b2

k =2

cfor allk  3.

Proof Consider a random two-coloring of the edges ofK

nobtained by coloringeach edge independently either red or blue, where each color is equally likely Forany fixed setRofkvertices, letA

Rbe the event that the induced subgraph ofK

non

Ris monochromatic (i.e., that either all its edges are red or they are all blue) Clearly,

1

2 THE BASIC METHOD

Since there are n

< 1 Thus, withpositive probability, no eventA

Roccurs and there is a two-coloring ofK

nwithout amonochromaticK

k, i.e.,R(k ; k ) > n Note that ifk  3and we taken = b2

k =2 c

k 2

k !

n k 2 2

of P Erd}os from 1947 Although Szele applied the probabilistic method to anothercombinatorial problem, mentioned in Chapter 2, already in 1943, Erd}os was certainlythe first one who understood the full power of this method and has been applying itsuccessfully over the years to numerous problems One can, of course, claim that theprobability is not essential in the proof given above An equally simple proof can bedescribed by counting; we just check that the total number of two-colorings ofK

nisbigger than the number of those containing a monochromaticK

k.Moreover, since the vast majority of the probability spaces considered in thestudy of combinatorial problems are finite spaces, this claim applies to most ofthe applications of the probabilistic method in discrete mathematics Theoretically,this is, indeed, the case However, in practice, the probability is essential Itwould be hopeless to replace the applications of many of the tools appearing in thisbook, including, e.g., the second moment method, the Lov´asz Local Lemma and theconcentration via martingales by counting arguments, even when these are applied

to finite probability spaces

The probabilistic method has an interesting algorithmic aspect Consider, forexample, the proof of Proposition 1.1.1 that shows that there is an edge two-coloring

ofK

nwithout a monochromatic K

2 log 2

n Can we actually find such a coloring?This question, as asked, may sound ridiculous; the total number of possible colorings

is finite, so we can try them all until we find the desired one However, such aprocedure may require2

( 2 )

steps; an amount of time which is exponential in the size(=

n

2

) of the problem Algorithms whose running time is more than polynomial

in the size of the problem are usually considered unpractical The class of problemsthat can be solved in polynomial time, usually denoted byP(see, e.g., Aho, Hopcroftand Ullman (1974) ), is, in a sense, the class of all solvable problems In thissense, the exhaustive search approach suggested above for finding a good coloring

of K

n is not acceptable, and this is the reason for our remark that the proof ofProposition 1.1.1 is non-constructive; it does not suply a constructive, efficient anddeterministic way of producing a coloring with the desired properties However, acloser look at the proof shows that, in fact, it can be used to produce, effectively, acoloring which is very likely to be good This is because for largek, ifn = b2

k =2 c

k 2

k ! n 2 k=2

k

 2 1+

k 2

k !

 1 Hence, a random coloring of

K

n is very likely not to contain a monochromatic K

2 logn This means that if,

for some reason, we must present a two coloring of the edges ofK

1024 without amonochromatic we can simply produce a random two-coloring by flipping a

; probably much smaller thanour chances of making a mistake in any rigorous proof that a certain coloring is good!Therefore, in some cases the probabilistic, non-constructive method, does supplyeffective probabilistic algorithms Moreover, these algorithms can sometimes beconverted into deterministic ones This topic is discussed in some detail in Chapter15.

The probabilistic method is a powerful tool in Combinatorics and in Graph Theory

It is also extremely useful in Number Theory and in Combinatorial Geometry Morerecently it has been applied in the development of efficient algorithmic techniques and

in the study of various computational problems In the rest of this chapter we presentseveral simple examples that demonstrate some of the broad spectrum of topics inwhich this method is helpful More complicated examples, involving various moredelicate probabilistic arguments, appear in the rest of the book

players is very likely to have propertyS

k By a random tournament we mean here atournamentT onV obtained by choosing, for each1  i < j  n, independently,either the edge(i; j or the edge(j; i), where each of these two choices is equallylikely Observe that in this manner, all the2

( 2 )

possible tournaments onV are equallylikely, i.e., the probability space considered is symmetric It is worth noting that weoften use in applications symmetric probability spaces In these cases, we shall

sometimes refer to an element of the space as a random element, without describing

explicitly the probability distribution Thus, for example, in the proof of Proposition1.1.1 random 2-edge-colorings ofK

nwere considered, i.e., all possible coloringswere equally likely Similarly, in the proof of the next simple result we study randomtournaments on

4 THE BASIC METHOD

Theorem 1.2.1 If n

k

(1 2

, and all thesen kevents corresponding to the various possible choices of

vare independent It follows that



 X

K V

jK j=k

Pr (A K ) =

 n k

 (1 2

k n k

< 1 :

Therefore, with positive probability no eventA

K occurs, i.e., there is a tournament

onnvertices that has the propertyS

k

and(1 2

k n k

< e (n k )=2 k

, Theorem1.2.1 implies thatf (k )  k

2

2 k

(ln 2) 1+ o(1)

It is not too difficult to check that

f (1) = 3andf (2) = 7 As proved by Szekeres (cf Moon (1968) ),f(k )  c
k
2

k

.Can one find an explicit construction of tournaments with at most c verticeshaving propertyS

k? Such a construction is known, but is not trivial; it is described

in Chapter 9

A dominating set of an undirected graphG = (V; E)is a setU  V such thatevery vertexv 2 V U has at least one neighbor inU

Theorem 1.2.2 Let G = (V; E) be a graph onn vertices, with minimum degree

Æ > 1 ThenGhas a dominating set of at mostn

Xbe the random set of all vertices inV X that donot have any neighbor inX The expected value ofjXjis clearlynp For each fixedvertexv 2 V,Pr (v 2 Y ) = Pr (v and its neighbors are not inX)  (1 p)

Æ +1

.Since the expected value of a sum of random variables is the sum of their expectations(even if they are not independent) and since the random variablejY jcan be written

as a sum ofn indicator random variables

Xis clearly a dominating set

ofGwhose cardinality is at most this size

The above argument works for anyp 2 [0; 1] To optimize the result we useelementary calculus For convenience we bound1 p  e

p

(this holds for allnonnegativepand is a fairly close bound whenpis small) to give the simpler bound

GRAPH THEORY 5

Take the derivitive of the right hand side with respect topand set it equal to zero.The right hand side is minimized at

p = ln(Æ + 1)

Æ + 1 :

Formally, we setpequal to this value in the first line of the proof We now have

“alteration" principle which is discussed in Chapter 3 The random choice did notsupply the required dominating setU immediately; it only supplied the setX, whichhas to be altered a little (by adding to it the setY

X) to provide the required dominatingset The third involves the optimal choice ofp One often wants to make a randomchoice but is not certain what probabilitypshould be used The idea is to carry outthe proof withpas a parameter giving a result which is a function ofp At the end that

pis selected which gives the optimal result There is here yet a fourth idea that might

be called asymptotic calculus We wanted the asymptotics ofmin np + n(1 p)

a clean bound A good part of the art of the probabilistic method lies in finding

suboptimal but clean bounds Did we give away too much in this case? The answerdepends on the emphasis for the original question ForÆ = 3our rough bound gives

jU j  0:596nwhile the more precise calculation givesjU j  0:496n, perhaps asubstantial difference ForÆlarge both methods give asymptoticallyn

ln Æ Æ

It can be easily deduced from the results in Alon (1990b) that the bound inTheorem 1.2.2 is nearly optimal A non-probabilistic, algorithmic, proof of thistheorem can be obtained by choosing the vertices for the dominating set one byone, when in each step a vertex that covers the maximum number of yet uncoveredvertices is picked Indeed, for each vertexvdenote byC(v )the set consisting ofv

together with all its neighbours Suppose that during the process of picking verticesthe number of verticesuthat do not lie in the union of the setsC(v )of the verticeschosen so far isr By the assumption, the sum of the cardinalities of the setsC(u)

over all such uncovered vertices u is at leastr (Æ + 1), and hence, by averaging,there is a vertexv that belongs to at leastr (Æ + 1)=nsuch setsC(u) Adding this

vto the set of chosen vertices we observe that the number of uncovered vertices isnow at mostr (1

Æ +1 n ) It follows that in each iteration of the above procedure thenumber of uncovered vertices decreases by a factor of1 (Æ + 1)=nand hence after

n

Æ +1

ln(Æ + 1)steps there will be at mostn=(Æ + 1)yet uncovered vertices which cannow be added to the set of chosen vertices to form a dominating set of size at mostthe one in the conclusion of Theorem 1.2.2

Combining this with some ideas of Podderyugin and Matula, we can obtain a veryefficient algorithm to decide if a given undirected graph onnvertices is, say, n

2

-edge

connected A cut in a graph is a partition of the set of vertices into

6 THE BASIC METHOD

two nonempty disjoint setsV = V

1 [ V

2 Ifv 1

2 V

1andv 2

1and another end inV

2 In fact, we sometimes identify the cut with the set

of these edges The edge-connectivity ofGis the minimum size of a cut ofG Thefollowing lemma is due to Podderyugin and Matula (independently)

Lemma 1.2.3 LetG = (V; E)be a graph with minimum degreeÆand letV = V

1 [ V 2

be a cut of size smaller thanÆinG Then every dominating setU ofGhas vertices

inV

1and inV

2.

Proof Suppose this is false andU  V

1 Choose, arbitrarily, a vertexv 2 V

2 andletv

i

2 V

1thene i

= fv ; v i

i

= fu; v i

LetG = (V; E)be a graph onnvertices, and suppose we wish to decide ifGis

n=2edge-connected, i.e., if its edge connectivity is at leastn=2 Matula showed, byapplying Lemma 1.2.3, that this can be done in timeO (n

3

By the remark followingthe proof of Theorem 1.2.2, we can slightly improve it and get anO (n

8=3 log n)

algorithm as follows We first check if the minimum degreeÆofGis at leastn=2 Ifnot,Gis notn=2-edge connected, and the algorithm ends Otherwise, by Theorem1.2.2 there is a dominating setU = fu

1

; : ; u k

gofG, wherek = O (log n), and itcan in fact be found inO (n

By Lemma 1.2.3 the edge connectivity ofGis simply the minimum betweenÆ and

1.3 COMBINATORICS

A hypergraph is a pairH = (V; E), whereV is a finite set whose elements are called

vertices andE is a family of subsets ofV, called edges It isn-uniform if each of

its edges contains preciselynvertices We say thatHhas propertyB, or that it is

2-colorable if there is a 2-coloring ofV such that no edge is monochromatic Let

m(n) denote the minimum possible number of edges of ann-uniform hypergraphthat does not have propertyB

Proposition 1.3.1 [Erd}os (1963a) ] Everyn-uniform hypergraph with less than2

 X e2E

Pr (A e

< 1

and there is a 2-coloring without monochromatic edges 

In Chapter 3, Section 3.5 we present a more delicate argument, due to hakrishnan and Srinivasan, and based on an idea of Beck, that shows thatm(n)  n

argu-withapoints in one color,b = v apoints in the other LetS  V be a uniformlyselectedn-set Then

P r (S is monochr omatic under ) =

a n

+ b n

v n

v =2 n

v n

 )  2 v (1 p) m

When this quantity is less than1there exist S

mis not2-colorable - and hencem(n)  m

The asymptotics provide a fairly typical example of those encountered whenemploying the probabilistic method We first use the inequality1 p  e

 1som(n)  m Now we need findvto minimize We may interpret as twice the probability of picking white balls from

8 THE BASIC METHOD

an urn withv =2 white and v =2black balls, sampling without replacement It istempting to estimate pby2

n+1

, the probability for sampling with replacement.This approximation would yieldm  v 2

n 1 (ln 2) Asvgets smaller, however, theapproximation becomes less accurate and, as we wish to minimizem, the tradeoffbecomes essential We use a second order approximation

p = 2

v =2 n

v n

= 2

1 n

n 1 Y i=0

v 2i

v i

 2

1 n e n 2

Elementarycalculus givesv = n

2

=2for the optimal value The evenness ofv may require achange of at most2which turns out to be asymptotically negligible This yields thefollowing result of Erd}os (1964) :

Theorem 1.3.2

m(n) < (1 + o(1))

e ln 2 4 n 2

be a family of pairs of subsets of an arbitrary set WecallFa(k ; `)-system ifjA

1  i  k, letX

ibe the event that all the elements ofA

iprecede all those ofB

iinthis order ClearlyPr (X

i ) = 1 Æ

iprecede those ofB

iand all the elements ofA

jprecede those ofB

j.Without loss of generality we may assume that the last element ofA

idoes not appearafter the last element ofA

j But in this case, all elements ofA

iprecede all those of

B

j, contradicting the fact thatA

i

\ B j 6= ; Therefore, all the eventsX

iare pairwisedisjoint, as claimed It follows that1  Pr

h W i=1 X i

= h P i=1 Pr(X i ) = h
1

Æ

k +` k

,completing the proof 

Theorem 1.3.3 is sharp, as shown by the familyF =

 (A; X n A) : A  X ; jAj = k

, whereX = f1; 2; : : k + `g

1.4 COMBINATORIAL NUMBER THEORY

A subsetAof an abelian groupGis called sum-free if(A + A) \ A = ;, i.e., if there

1 3

n.

Proof Letp = 3k + 2be a prime, which satisfiesp > 2 max

1in jb

>

1 3

Let us choose at random an integerx,1  x < p,according to a uniform distribution onf1; 2; : : p 1g, and defined

Therefore, the expected number of elementsb

i

such thatd

i

2 C is more thann=3 Consequently, there is anx,1  x < pand

a subsequenceA of B of cardinality jAj >

n 3

, such thatxa(mo d p) 2 C for all

a 2 A ThisAis clearly sum-free, since ifa

1 + a 2

= a

3for somea

1

; a 2

; a 3

Z

p This completes the proof.

In Alon and Kleitman (1990) it is shown that every set ofnnonzero elements of

an arbitrary abelian group contains a sum-free subset of more than2n=7elements,and that the constant2=7is best possible The best possible constant in Theorem1.4.1 is not known

1.5 DISJOINT PAIRS

The probabilistic method is most striking when it is applied to prove theorems whosestatement does not seem to suggest at all the need for probability Most of theexamples given in the previous sections are simple instances of such statements Inthis section we describe a (slightly) more complicated result, due to Alon and Frankl(1985) , which solves a conjecture of Daykin and Erd}os

LetF be a family ofmdistinct subsets ofX = f1; 2; : : ng Letd(F )denotethe number of disjoint pairs inF, i.e.,

d(F ) =

 (F ; F 0 ) : F ; F 0

2 F ; F \ F

0

= ; :

Daykin and Erd}os conjectured that ifm = 2

( 1 +Æ)n

, then, for every fixed Æ > 0,

subsets ofX = f1; 2; : : ng, whereÆ > 0 Then

d(F ) < m

2 Æ

Proof Suppose (1.1) is false and pick independentlytmembersA

1

; A 2

; : ; A

tofwith repetitions at random, where is a large positive integer, to be chosen later

10 THE BASIC METHOD

We will show that with positive probabilityjA

1 [ A 2 [ : : [ A

t

j > n=2and stillthis union is disjoint to more than2

Pr (A i

 S; i = 1; : : t)

 2 (2 n=2

=2 ((1=2)+Æ )n

) t

= 2 n(1 Æ t) : (1.2)Define

v (B ) =

fA 2 F : B \ A = ;g

:

LetY be a random variable whose value is the number of membersB 2 F whichare disjoint to all theA

i (1  i  t) By the convexity ofz

= 1 m t

m

 P

v (B) t m



 1 m t

m

 2d(F) m

 t

 2m

1 tÆ 2

=2 )  m tÆ 2

=2

> 2 n=2

and the right-hand side

of (1.4) is greater than the right-hand side of (1.2) Thus, with positive probability,

 p ( 2 ) +

 n t

 (1 p) ( t 2 )

2 Suppose n  4and letH be an n-uniform hypergraph with at most 4

1 3

edges Prove that there is a coloring of the vertices ofHby4colors so that inevery edge all colors are represented

5 (*) LetG = (V; E)be a graph onn  10vertices and suppose that if we add

toGany edge not inGthen the number of copies of a complete graph on10

vertices in it increases Show that the number of edges ofGis at least8n 36:

6 (*) Theorem 1.2.1 asserts that for every integerk > 0there is a tournament

(k +l) k+l

k l :

8 (Prefix-free codes; Kraft Inequality) LetF be a finite collection of binarystrings of finite lengths and assume no member ofFis a prefix of another one.LetN

idenote the number of strings of lengthiinF Prove that

X i

N i 2 i

 1:

9 (*) (Uniquely decipherable codes; Kraft-McMillan Inequality) LetF be afinite collection of binary strings of finite lengths and assume that no twodistinct concatenations of two finite sequences of codewords result in the samebinary sequence LetN

idenote the number of strings of lengthiinF Provethat

X i

N i 2 i

THE PROBABILISTIC LENS:

Theorem

A familyFof sets is called intersecting ifA; B 2 FimpliesA \ B 6= ; Suppose

n  2k and letF be an intersecting family ofk-element subsets of an n-set, fordefinitenessf0; : : n 1g The Erd}os-Ko-Rado Theorem is thatjF j 

g,(1  i  k 1), and the members of each such pair aredisjoint The result follows, sinceFcan contain at most one member of each pair.

Now we prove the Erd}os-Ko-Rado Theorem Let a permutationoff0; : : n 1g

andi 2 f0; : : n 1gbe chosen randomly, uniformly and independently and set

A = f (i);  (i + 1); : :  (i + k 1)g, addition again modulon Conditioning onany choice ofthe Lemma givesPr [A 2 F]  k =n HencePr [A 2 F ]  k =n But

Ais uniformly chosen from allk-sets so

k n

 Pr[A 2 F] =

jFj n k

and

jFj  k n

 n k



12

n X

n Linearity ofExpectation states that

E[X] = c E[X

1 ] + : : + c

n E[X n ]

The power of this principle comes from there being no restrictions on the dependence

or independence of theX

i In many instancesE[X]can be easily calculated by ajudicious decomposition into simple (often indicator) random variablesX

i.Letbe a random permutation onf1; : : ng, uniformly chosen LetX ( )bethe number of fixed points of To findE[X]we decomposeX = X

1 + : : + X

n

whereX

iis the indicator random variable of the event (i) = i Then

E[X i

= Pr[ (i) = i] =

1 n

so that

E[X] =

1 n + : : +

1 n

= 1:

In applications we often use that there is a point in the probability space for which

X  E[X] and a point for whichX  E[X] We have selected results with a

13

14 LINEARITY OF EXPECTATION

purpose of describing this basic methodology The following result of Szele (1943) ,

is oftimes considered the first use of the probabilistic method

Theorem 2.1.1 There is a tournament T with n players and at least n!2

and

E[X] =

X E[X

 ] = n!2

(n 1)

Thus some tournament has at leastE[X]Hamiltonian paths.

Szele conjectured that the maximum possible number of Hamiltonian paths in atournament onnplayers is at most n!

(2 o(1))

n This was proved in Alon (1990a) and

is presented in the Probabilistic Lens: Hamiltonian Paths, (following Chapter 4)

2.2 SPLITTING GRAPHS

Theorem 2.2.1 LetG = (V; E)be a graph withnvertices andeedges ThenG

contains a bipartite subgraph with at leaste=2edges.

Proof LetT  V be a random subset given byPr [x 2 T ] = 1=2, these choicesmutually independent SetB = V T Call an edgefx; y gcrossing if exactly one

ofx; yare inT LetXbe the number of crossing edges We decompose

X = X fx;y g2E X xy

whereX

xyis the indicator random variable forfx; y gbeing crossing Then

E[X xy ] = 1=2

as two fair coin flips have probability1=2of being different Then

E[X] =

X fx;y g2E E[X xy ] = e 2

ThusX  e=2 for some choice ofT and the set of those crossing edges form abipartite graph.

A more subtle probability space gives a small improvement

Theorem 2.2.2 IfGhas2nvertices andeedges then it contains a bipartite subgraph with at least en

2n 1

edges IfGhas2n + 1vertices andeedges then it contains a bipartite subgraph with at least e(n+1)

edges.

Here is a more complicated example in which the choice of distribution requires

a preliminary lemma LetV = V

1 [ : : [ V

Theorem 2.2.3 Supposeh(E) = +1for all crossing k-setsE Then there is an

of degree kwith all coefficients having absolute value at most one andp

1 p 2

p k

having coefficient one Then for allf 2 P

kthere existp

1

; : ; p k

2 [0; 1]with

jf(p 1

; : ; p k )j  c

Herec is positive and independent off.

Proof Set

M (f) = max

p1;:::;pk2[0;1]

jf(p 1

; : ; p k )j:

! Ris continuous,M must assume its minimumc 

Proof.[Theorem 2.2.3] Define a randomS  V by setting

Pr[x 2 S] = p

i

x 2 V i

these choices mutually independent,p

ito be determined SetX = h(S) For each

k-setEset

X E

=

 h(E) ifE  S

0 otherwiseSayEhas type(a

a 1 1

p a k

Combining terms by type

E[X] =

X p a 1 1

p a k

X

of type

h(E):

For any other type there are fewer thann

 n k :

Thus

E[X] = n

k f(p 1

; : ; p k

2 [0; 1]withjf(p

1

; : ; p k )j  c Then

E[jXj]  E[X]  c n

k :

Some particular value ofjXjmust exceed or equal its expectation Hence there is aparticular setS  V with

1=(k 1) ), so that all crossingk-sets are thesame color From Theorem 2.2.3 there then exists a set of size((ln n)

1=(k 1) ), atleast 1

2

+  of whosek-sets are the same color This is somewhat surprising since it

is known that there are colorings in which the largest monochromatic set has size atmost thek 2-fold logarithm ofn

2.3 TWO QUICKIES

Linearity of Expectation sometimes gives very quick results

Theorem 2.3.1 There is a two-coloring ofK

nwith at most

 n a

 2 1 ( 2 )

monochromaticK

a.

Proof.[outline] Take a random coloring LetX be the number of monochromatic

K

aand findE[X] For some coloring the value ofXis at most this expectation 

In Chapter 15 it is shown how such a coloring can be found deterministically andefficiently

BALANCING VECTORS 17

Theorem 2.3.2 There is a two-coloring ofK

m;nwith at most

 m a



n b

 2

The next result has an elegant nonprobabilistic proof, which we defer to the end of

this chapter Herejv jis the usual Euclidean norm

Theorem 2.4.1 Letv

1

; : ; v n

2 R n

, alljv i

= 1 Then there exist ; : ; 

n v n

j  p n;

and also there exist ; : ; 

n

= 1so that

j

1 v 1 + : : + 

n v n

j  p n:

Proof Let ; : ; 

nbe selected uniformly and independently fromf 1; +1g Set

X = j

1 v 1 + : : + 

n v n j 2

Then

X = n X i=1

n X j=1

 i

 j v i v j

Thus

E[X] =

n X i=1

n X j=1 v i v j E[

i

 j ]

Wheni 6= j,E[

i

 j ] = E[

i ]E[

j ] = 0 Wheni = j,

i

= 1soE[

2 i

= 1 Thus

E[X] =

n X i=1 v i v i

18 LINEARITY OF EXPECTATION

Theorem 2.4.2 Let v

1

; : ; v n

2 R n

, all jv i

 1 Let p

1

; : ; p 2 [0; 1]be arbitrary and setw = p

1 v 1 + : : + p v

n Then there exist ; : ; 

n

2 f0; 1gso that, settingv =  v

1 + : : + 

n v

n,

jw v j 

p n 2 :

The random choice of

igives a randomvand a random variable

n X i=1 (p i

 i )v i 2

= n X i=1

n X j=1 v i v j (p i

 i )(p j

 j )

so that

E[X] =

n X i=1

n X j=1 v i v j E[(p i

 i )(p j

 j )]

Fori 6= j

E[(p

i

 i )(p j

 j )] = E[p

i

 i ]E[p j

 j ] = 0

Fori = j

E[(p

i

 i ) ] = p i (p i 1) 2 + (1 p

i )p 2 i

= p i (1 p i )  1 4

 1 4

n X i=1 jv i 2

 n 4

and the proof concludes as in that of Theorem 2.4.1.

= 1,

1  i; j  nso that

n X n X a ij x i y j

 ( r 2 + o(1))n

3=2 :

q 2

 + o(1))n

= 1be selected dently and uniformly and set

indepen-R i

= n X j=1 a ij y j

R = n X i=1 jR i

Fixi Regardless ofa

ij,a ij y

jis+1or 1with probability1=2and their values (over

j) are independent (I.e., whatever thei-th row is initially after random switching

it becomes a uniformly distributed row, all2 possibilities equally likely.) ThusR

i

has distributionS

n- the distribution of the sum ofnindependent uniformf 1; 1g

random variables - and so

E[jR i j] = E[jS

n j] = ( r 2

 + o(1)) p n:

These asymptotics may be found by estimatingS

nby

p

nN whereN is standardnormal and using elementary calculus Alternatively, a closed form

E[jS n j] = n2

1 n



n 1 b(n 1)=2c

 + o(1))n

3=2 :

= n X i=1 x i R i

= n X i=1 jR i

= R  (

r 2

 + o(1))n

3=2 :



Another result on unbalancing lights appears in the Probabilistic Lens: ing Lights, (following Chapter 12.)

Unbalanc-20 LINEARITY OF EXPECTATION

2.6 WITHOUT COIN FLIPS

A nonprobabilistic proof of Theorem 2.2.1 may be given by placing each vertex ineitherTorBsequentially At each stage placexin eitherTorBso that at least half

of the edges fromxto previous vertices are crossing With this effective algorithm

at least half the edges will be crossing

There is also a simple sequential algorithm for choosing signs in Theorem 2.4.1When the sign forv

i is to be chosen a partial sumw =  v

1 + : : + 

i 1 v

i 1hasbeen calculated Now if it is desired that the sum be small select

p

n, otherwise it is either less than

p

norgreater than

i, the partial sum Select so that

w s

= w

s 1 + (p s

 )v s

= s X i=1 (p i

 i )v i

has minimal norm A random 2 f0; 1gchosen withPr [

s 1

v s E[p s

 ] + jv s j 2 E(p s

 )

= jw

s 1 j 2 + p s (1 p s )jv s j 2

(2.1)

so for some choice of 2 f0; 1g,

jw s j 2

 jw

s 1 j 2 + p s (1 p s )jv s j 2 :

As this holds for all1  s  n(takingw

0

= 0), the final

jw n j 2

 n X i=1 p i (1 p i )jv i 2 :

While the proofs appear similar, a direct implementation of the proof of Theorem 2.4.2

to find ; : ; 

nmight take an exhaustive search with exponential time In applyingthe greedy algorithm at thes-th stage one makes two calculations ofjw

s j 2

, depending

on whether = 0or1, and picks that giving the smaller value Hence there areonly a linear number of calculations of norms to be made and the entire algorithmtakes only quadratic time In Chapter 15 we discuss several similar examples in amore general setting

= 2b 3 + 2b 4 :

3 Prove that every set ofnnon-zero real numbers contains a subsetAof strictly

more than n=3 numbers such that there are no a

1

; a 2

; a 3

are pairwise distinct

5 LetH be a graph, and letn > jV (H)jbe an integer Suppose there is agraph onnvertices andtedges containing no copy ofH, and suppose that

n c n!

2 :

7 LetFbe a family of subsets ofN = f1; 2; : : ng, and suppose there are no

:

8 (*) LetXbe a collection of pairwise orthogonal unit vectors inR

n

and supposethe projection of each of these vectors on the firstkcoordinates is of Euclideannorm at least Show thatjXj  k =

THE PROBABILISTIC LENS:

ijbethe number of ones in thei-th row LetS be the set of permutations 2 S

Theorem 1 [Br´egman’s Theorem]

per (A) 

Y 1in (r i

!) 1=ri

denoteA with rows 1; : :  (i 1) and columns

 1; : :   (i 1)deleted and letR

 i denote the number of ones of row iinA

i

.(This is nonzero as the  i-th column has a one.) Set

L = L(;  =

Y 1in R

22

n E[X n ]

The power of this principle comes from there being no restrictions on the dependence

or independence of the< small>X

i In many instancesE[X]can...  of whosek-sets are the same color This is somewhat surprising since it

is known that there are colorings in which the largest monochromatic set has size atmost the< small>k...

n< /small> This was proved in Alon (1990a) and

is presented in the Probabilistic Lens: Hamiltonian Paths, (following Chapter 4)

2.2 SPLITTING GRAPHS

Theorem