Random Graphs and Erdős Magic Joel Spencer Asperen, October 30- November 1, 2018

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1 Radom Graphs ad Erdős Magic Joel Specer Aspere, October 30- November 1, 2018 A Special Note: These otes are oly a approximatio to the actual lectures. There is some extra material here, ad there will almost certaily be some material i the lectures ot covered i the otes. I particular, the split ito specific days should be regarded oly as a estimate. JS The Probabilistic Method. Lecture I: Tuesday, October 30 1 History The probabilistic method prove the existece of some object with a give property, by showig that it occurs with o-zero probability whe we select a radom object from a appropriate probability space. Usig the probabilistic method, we ca prove theorems that themselves ivolve o probability. We ca ofte avoid complicated costructios. The probabilistic method was iveted by Paul Erdős. It is hard to say that it was his chief discovery, as he has cotributed so may thigs to may differet braches of mathematics. But it is bears most deeply the mar of his character, ad he urtured it as its sole practitioer for 25 years. The first theorem we will prove has a story associated with it that goes bac to well before Erdős developmet of the probabilistic method. Three frieds were walig i the hills outside of Budapest, as was their habit, talig about mathematics: G. Szeeres, Esther Klei, ad the 17 year old Paul Erdős, who was already ow o the uiversity campus as a mathematical sorcerer. 1 Esther had brought a mathematical problem bac with her. Allthough they did t realize it, Ramsey had already prove it, ad it is ow ow as a (versio) of Ramsey s theorem: 2 Ramsey s Theorem Let K be the complete graph with odes. 1 Soo after Erdős was dubbed Die Zauberer vo Budapest 1

2 Theorem 2.1 Ramsey s Theorem (limited versio): For all, l, there exists so that the edges of K are colored red ad blue, either there exists a red complete sub-graph of poits or there exists a blue complete sub-graph of l poits. It turs out, Szeeres was able to prove this, ad get a simple boud o the growth rate of, though, at the time, he was majorig i chemical egieerig. It may seem amazig that a chemical egieer could tacle this problem. I fact, as he was to recall i later years, he had a powerful icetive: he was married with Esther shortly afterwards. For a example of Ramseys theorem with = l = 3, the graph = 6 is satisfactory. Cosider, for istace six people sittig listeig to this lecture. We ll call a perso gregarious if they ow at least 3 other people. (Let us assume that owig is reflexive, as we are cocered with udirected graphs.) Suppose there is a gregarious perso. If ay two of people ow also ow each other, the we have a gregarious triagle. If all three of them are mutually uacquaited, the we have a complete subgraph of stragers. Obviously, if there is o gregarious perso, the there must be a retirig perso who does t ow at least three people, ad the argumet wors symmetrically. Defie the Ramsey fuctio R(,l) is the least with this property. It ca be show by a simple iductio that: + l 2 R(,l) 1 (The method is to show R(,l) R( 1,l) + R(,l 1).) I , Erdős proved both a upper ad a lower boud for R(,). The lower boud ivolved essetially the first use of the probabilistic method. 2 Theorem 2.2 (Erdős 47) 3 R(,) 2 2 c4 1/2 1 Theorem 2.3 (Erdős 47) If ( ) 2 1 ( 2), the R(,) >. We will prove this later theorem with the probabilistic method. I order to prove that Ramsey s fuctio is greater tha, we must prove there exists a two-colorig χ of K with o moo χ of K (where moo is short for moochromatic, ad moo χ meas moochromatic uder χ ). To prove this, cosider a radom colorig o poits. (So the probability space cotais 2 ( 2) poits ). For ay S {1,...,}, S =, let AS be the evet that S is moo χ. The: Pr[A S ] = 2 1 ( 2) 2 There was aother paper a bit earlier by Schütte that used somethig lie the probabilistic method, but its sigificace was ot appreciated. 3 I this course, we will be iterested i the asymptotic behavior. My recet boo Asymptopia deals with this viewpoit. 2

3 Let BAD = S = A S. BAD is the evet that there is a set of poits that is moo. We are looig for a set that is ot moo, so set GOOD = BAD. By the uio boud, we have: Pr[BAD] Pr[A S ] = 2 1 ( 2) < 1 Therefore, S = Pr[GOOD] > 0 Thus, by Erdős magic we ca coclude there is a good χ. 4 Ed of Proof. Aalysis Now let us aalyze this result, as Erdős did i his paper, to see what explicit asymptotic boud this gives for. This is a typical example of what might be called asymptotic calculus. Defie: f() := max such that 2 1 ( 2) < 1 As a rough estimate, lets try: 5! The, as a first cut, we get:! 22 /2 2 (igorig!(!)) Now, a little more formally, this ca be coverted ito a upper boud: <! < 2) 1 2( <! 1/ where asymptotically, ad ca be igored. Now we ca apply Stirlig s formula to get a approximatio for!:! e 2π (!) 1 e Pluggig bac i: < e This o-costructive proof leads to algorithmic questios: how log does it tae to fid a solutio? 5 Note that meas roughly lie, with o formal iterpretatio. f() g(), o the other had, meas lim f() g() = 1. 3

4 We saw the 2 term whe we dropped!, so we ca coclude that is at least expoetial i. I fact, we have a lower boud for that turs out to be just the same: > ( )! (1! if ) 1 This coditio is fulfilled wheever >> 2 (if ot, we have a additioal e 2 term). So certaily, with > c2, 1 ( )!. Therefore f() e 2 2 Ed of Proof. Remar: what is the ituitio behid usig radomess? We wat a set that does ot have structure, because structure ca be used agaist us. Radomess is a way to get such a set. 3 Touramets (Begiig) A touramet T is a roud-robi competitio betwee players (vertices). For all i,j, either i beats j or j beats i (directed edges). Defiitio (Schütte) T has property S if for all distict players x 1,...,x there exists y who beats all of them. Some examples: S 1 - every player is beate by all others. T 3 provides a example 1 beatig 2 beatig 3 beatig 1. T 7 provides a example of a touramet with property S 2 The rule for the edges is: i beats j if i j is a square mod 7. What about S 10? This problem seems to get pretty complex as it scales up. Nevertheless, Erdős came up with a simple proof for the existece of touramets usig the probabilistic method. Theorem 3.1 (Erdős, 1963) For all there exists a (fiite) touramet T with property S. I fact, well show the followig result, where we ll fill i the...later! Theorem 3.2 (Erdős 1963) If... with vertices that has the property S. the there exists a touramet T Proof: Cosider a radom touramet T. Let K = {x 1,...,x } be a set of players. Call y is a witess for K if y beats {x 1,...,x }. The: Pr[y is a witess for K] = 2 The let A K be the evet that K has o witess. Pr[A K ] = (1 2 ) Why? Each y K has probability 2 of beig a witess ad therefore probability 1 2 of ot beig a witess. As differet y K ivolve separate coi flips, the evets y is ot a witess for K are mutually idepedet. 4

5 Now, let BAD = K = A K ad GOOD = BAD. Clearly, GOOD is the evet that the touramet has the property S. Computig the probability of the evet BAD is ot easy. However, we ca boud this probability usig the fact that the probability of a disjuctio of evets is less tha or equal to the sum of the probabilities of idividual evets. Therefore, Pr[BAD] K = Pr[A K ] = (1 2 ). Therefore, if we replace..... i the statemet of the theorem by the coditio ( ) (1 2 ) < 1, we will obtai Pr[GOOD] > 0. Now, Erdős Magic: If the probability of a evet is positive, there exists some poit of the probability space for which the evet occurs. For example, sice Pr[GOOD] > 0, there exists some touramet T with the property S. Ed of Proof. The first corollary of the above theorem is that for every, there are touramets with property S. The reaso is that for every fixed, ( ) is a polyomial i, while (1 2 ) is expoetially small. Therefore, the coditio ( ) (1 2 ) < 1 is satisfied if is large eough. Now, let f() be the smallest such that ( ) (1 2 ) < 1. What are the asymptotics for f()? I order to fid the asymptotics, we estimate ( ) usig the iequality ( ). This is ot a good estimate i geeral (For example,! is a better estimate), but it turs out that i this problem usig more accurate estimates wo t give us better results. Also, we ow that for small ǫ, we have 1 ǫ e ǫ, ad for all positive ǫ, 1 ǫ < e ǫ. Therefore, it suffices to have e 2 ( ) < 1. Also, it turs out that we wo t lose much by replacig ( ) by. Therefore, we eed < e 2, or equivaletly 2 log <. Now, the problem is to fid the smallest such that α log <, where α = 2. If = α log α(1 + ǫ) for some positive ǫ, we have α log = α log (αlog α(1 + ǫ)) = α log α+α log((1+ǫ)log α) < α log α(1+ǫ) =. Thus, for = αlog α(1 + o(1)), we have αlog <. Therefore, f() = (2 )log(2 )(1 + o(1)) = 2 2 l 2(1 + o(1)). Therefore, if ANS() deotes the miimum for which there is a touramet T with the property S, we have ANS() = O( 2 2 ). The followig theorem provides a lower boud for ANS(). Theorem 3.3 If 2 1, the every touramet T with vertices does ot have the property S. 5

6 Proof: It is easy to see that there must be a player i the touramet that wis at least half its games. Let x 1 be such a player, ad A 1 be the set of players that have beate x 1. Therefore, A Now cosider a player x 2 i A 1 that wis at least half its games with other players i A 1, ad let A 2 be the set of players i A 1 that have beate x 2 (Thus, A ). Similarly, we ca defie x 3,x 4,... Now, it is ot difficult to see that the set x 1,x 2,... is a set of at most players with o witess. Ed of Proof. Usig a more complicated o-probabilistic argumet it ca be show that ANS() = Ω(2 ). The exact asymptotic for ANS() is ot ow. Asymmetric Ramsey Numbers Let s assume we wat to fid R(1000, 2000). Therefore, we eed to costruct a 2-colorig of the edges of a complete graph with o red K 1000 or blue K Sice the situatio is ot symmetric betwee red ad blue, we should ot use a fair coi for colorig. Istead, we should bias the coi to color more edges with blue. Sice we do t ow how to bias the coi, we use a techique called parameterizatio. To boud R(,l) (from below), try a colorig of K with Pr[χ({i,j}) = Red] = p, where p is umber that will be fixed later. There are two ids of bad evets: For every set S with S =, A S is the evet that all the edges betwee the vertices i S are colored red, ad for every set T with T = l, B T is the evet that all the edges betwee the vertices i T are colored blue. Let BAD = S = A S T =l B T. Clearly, Pr[A S ] = p ( 2) ad Pr[BT ] = (1 p) (l 2). Therefore, Pr[BAD] p ( 2) + (1 p) (l 2). l Thus, we have proved the followig theorem. Theorem 3.4 If there exists a p [0,1] with ( ) p ( 2) + ( l) (1 p) ( l 2) < 1, the R(,l) >. 4 Alteratio Theorem 4.1 For all iteger m ad p [0,1] m m R(,l) m p ( 2) (1 p) (l 2) l Proof: Tae a radom colorig of K m with edges Red with probability p. For each red K ad each blue K l remove oe of their vertices. The remaiig graph will have either red K or blue K l. As the expected umber of poits removed (we may remove the same poit more tha oce but this just goes i our favor) is m m p ( 2) + (1 p) (l 2) l 6

7 so the expected umber of poits remaiig is at least m m m p ( 2) (1 p) (l 2) l ad by Erdős Magic this ca be achieved. Asymptotic Calculus: How good does this theorem do? That is, give,l, what is the maximum over m,p of m m m p ( 2) (1 p) (l 2) l Lets tae = 4 ad l large. We have m (1 p) (l 2) [me pl/2 ] l l so this will be small if we tae, say, l = (3/p)l m. We wat m ( m 4) p 6 to be big. If we tae p = m 1/2 the we are subtractig less tha m/2 so we still have at least m/2. That is, we would wat l = 3m 1/2 l m or m = cl 2 /l 2 l. So the Theorem gives R(4,l) = Ω(l 2 l 2 l). This is t the best ow but its ot bad. 2-Colorable Families Let {A 1,A 2,...,A m } be a family of subsets of a set Ω. We say that this family is 2-colorable (or has the property B), if there is a colorig χ : Ω {Red,Blue} such that o A is moochromatic. For example, the family {{1,2}, {2,3}, {3,1}} is ot 2-colorable Also, the family {{1,2,3}, {1,5,6}, {3,4,5}, {1,4,7}, {3,6,7},{2,5,7}, {2,4,6}} is ot 2-colorable. This family is costructed from the Fao projective plae Theorem 4.2 (Erdős 1963) Let A i = for 1 i m. If m < 2 1, the {A 1,...,A m } is 2-colorable. Proof: Color radomly. Let BAD i be the evet that A i is moochromatic. Sice A i =, Pr[BAD i ] = 2 1. Let BAD = i BAD i. Therefore, Pr[BAD] m2 1 < 1. Therefore, there is a colorig which is ot BAD, i.e., o A i is moochromatic uder this colorig. Ed of Proof. Let m() be the least such that there is a family {A 1,...,A m } with A i = which is ot 2-colorable. Usig this otatio, the above theorem ca be stated as follows. Theorem 4.3 (Erdős 1963) m() 2 1. I order to fid a upper boud for m(), we eed to costruct families that are ot 2-colorable. The followig o-probabilistic costructio gives such a family. Set Ω = 2 1, ad cosider the family of all -elemet 7

8 subsets of Ω. This family is ot 2-colorable, because for every 2-colorig of Ω, there is a set of elemets that are colored the same. This shows that 2 1 m() c4 1/2. I most problems, the probabilistic method ca be applied to fid lower bouds or upper bouds, but ot both. However, i this problem Erdős also used the probabilistic method to improve the above upper boud. The idea is to pic the sets A 1,A 2,...,A m idepedetly at radom from the set of all -elemet subsets of a uiverse Ω of size v, where v is a parameter. Let χ : Ω {Red, Blue} be a 2-colorig of the uiverse with a red ad b = v a blue poits. It is easy to see that ( a ) ( + b Pr[A i is moochromatic uder χ] = ( ) v ) 2( v/2) ( v. ) The last iequality follows from the covexity of the fuctio ( x ) = x(x 1) (x +1)! (this fuctio is defied for all real x) for x. Let BAD χ be the evet that o A is moochromatic uder χ. Sice A s are chose idepedetly, we have ( ( a ) ( + b ) m ( Pr[BAD χ ] = 1 ( ) v 1 ) 2( v/2) ) m ( v. ) Let BAD = χ BAD χ ad GOOD = BAD. By the uio boud, sice there are 2 v differet colorigs, ( Pr[BAD] 2 v 1 2( v/2) ) m ( v. ) Therefore, we have proved the followig theorem. Theorem 4.4 (Erdős 1964) If there exists v with ( 2 v 1 2( v/2) ) m ( v < 1, ) the m() > m, i.e., there is a family A 1,A 2,...,A m {1,...,v} with o 2-colorig. 8