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Unformatted text preview: 43(4 1:) 45m hataflom W M )zéwvs? [aH‘EV [P MA Amte afiawba 62% W5» ear» We 6W6: («fifkmwmf ' hm (cow’s (WWWMJ 0“) ‘éwm g‘WsV ‘0‘ col/«>ke' XAC‘J [6% me #30? x3 61v? ‘ “ok- a” (/7, (ALAQ 5} Hot,» {mad he >(CWONVXQASQ 4 he“ Qad’x 37va W («41 a‘ggug‘e') Mvg (5 0‘ V1] 'C’N‘WW: 6% a 5.. “Mfg j; l \‘WYbOMMfiQjY \‘S f 4 indeponwa (91$ evm*§. DEE. 1:; & (wk Canaan» AHAQHIALQZ‘ wz La»; $1M, MM. we! mam? MMW “1 I W(‘A¢3\A41\w aAgg): lg: P(A~) fmwmfi “3:31;: '6? “We (5 HAMMW, a}: 3M6} {Ly/Lire wig! MMMM‘I ¥¥’”AgmNé ' PfLflfi;fifliflflflfL (“5&2 ‘ WW “N EYWMQWQ Tame, jaw ECL/dclwzm Q/aCV‘ 9—f_ WWC‘A c9 QMW AKth 5m be “53% or R fiuc‘ \miePMaU/wh’) as, lbw. o M ‘ v Clfll‘AVQ/v‘ .4312 MSW”? 7;; ,SL «M a? 2L 7: Q 3 EL MM [9 Ma$r\" We, C F/V‘g’ m gawk.“ Chc‘wxw 4’00»! 3&4; L15 skow Aflg \WOQZP', M (g) Cam/“14%) I 1. Afij’flin-‘r (“00.2013 /p A .0 p ‘L A n“. C’Wb‘aé) I'd W ’PEV‘H S’=I4M,Mb,ccc/ «EC/Each; / ‘ 56“be («5,645 , H Ai:§,zh‘1’ex‘f.evig aj)§(,\ow SA ; “WW Ejfifbgl® W A] arm} : IN Amgk) M n W Wk) \gA—Drr L\ A A7/ / A“ £39. emu/J1 in 10M «6’ l } 62v A' WA W2 W A-C CQWWKM 6°? M V C 5 &\CA ‘Aok CDCQ‘N(>' QC‘AC‘Z AL :@I if ‘21 (WAN 4\ / W (PM R370. ‘TarV \P((\K\->_ K" K‘?(K£JAL) 5/ \eitP(Ac> 70,b~gk:j:f' I fl) [ avfileaa/cfiv“ W‘bn’“ ? I4()’47-)A?>):/P{Al:). Lem AWN/3‘“ new/W43 WI}L M6“ “‘2 jvéw G M “J (A A 4/ c M #6 t 4, f L / ghG am“ MUN—I rélPM‘L)>o PM - gag “Wow? UM WWW) ”” r /.—~——-—/' Mum) 4-??? WM. U, M EX?“ “u 69 / M all ice the zed l is are [so ige )er lflS H Ml- .r tic we en :nt II'S S a 161' .nd whs LECTURE 8 The Lovész Local Lemma The Lemma. Let A1, - - - , An be events in a probability space. In combinatorial applications the ‘A,- are “bad” events. We wish to show Pr [All] > 0 so that there is a point (coloring, tournament, configuration) x which is good. The basic probabilistic method of Lecture 1 may be written: Counting sieve. If X Pr [A‘-] < 1, then Pr [AA] > O- _ There are other simple conditions that ensure Pr [AAi] > 0. Independence sieve. If A1 , ‘ - v , A" are mutually independent and all Pr [Ai] < 1, then Pr [/\A,-] > O. The Lovész Local Lemma is a sieve method which allows for some dependence among the A;. A graph G on vertices [n] (the indices for the Ag) is called a dependency graph for A1, - - ‘ , A,1 if for all i A is mutually independent of all Aj with {i,j} E G. (That is, A. is independent of any Boolean function ofthese Aj.) LovAsz LOCAL LEMMA (Symmetric case). LetA; , - ‘ ~ , A" be events with depen- dency graph G such that Pr [A]: p for all i, de'g (i): d for all i and 4dp < 1. Then Pr [AA] > 0. Proof We show by induction on s that if [SI 5 s, then for any i Pr [Ai For S =Q this is immediate. Renumber for convenience so that i = n, S '= {1, - - ~ , s} and {i,x}£ G for x>d. Now /\ S 2p. jeS Pr [An/i1 ‘ ' ’ A—di/A—d-kl ‘ ' ‘ A-s] PrLAl ' ’ ' AdlAd+l ‘ ' ' As] ' 57 “[AniA-l . ‘ I 14—5] 58 LECTURE 8 We bound the numerator Fri/Mil ‘ - - All“, ' ‘ fiJSPrMnl/LH - - A] = Pr [An] Sp as A" is mutually independent of Ad“, ~ - - , AS. We bound the denominator d PITA) ' ' ‘ AdlAd+1 ' ‘ ‘ A3121” 2 Pr[AilAd+l ' ' ' As] i=1 (1 _>.1—- 2 2p (Induction) Hence we have the quotient Pr [AA/Tl ~ - - KslSp/%=2p, completing the induction. Finally Pr[/§1“'5n]=ll Pr[5il51"‘/§s~i]ZH (1—2P)>0~ D i=1 i=1 The proof is so elementary that it could, and I think» should, be taught in a first course in probability. It has had and continues to have a profound efiect on probabilistic methods. The diagonal Ramsey function. A lower bound for R(k, k), our first use of the probabilistic method in Lecture 1, provides a Simple application of the Lovész Loeal Lemma. Consider a random two—coloring of K,1 with AS the event “S is monochromatic,” S ranging over the k—setsof vertices. Define G by placing {5, T}E G if and only if [SH T122. Then AS is mutually independent of all AT with T not adjacent to G, since the AT give information only about edges outside of S. Hence G is a dependency graph. (When Isn TI =2 the events A5, A7- are independent; note however that mutual independence from a family of AT is far stronger than pairwise independence with each AT. Recall the old chestnut: I have two children and at least one is a girl. What is the probability they are both girls. Conditional on the younger being a girl it is one half. Conditional on the older being a girl it is one half. Conditional on the disjunction it is one third. But I digress.) We apply the Lovasz Local Lemma with p = Pr [AS] = 21%;) and d=|{T: |Sfl Tl22}|s(k>( n 2 k~2 k n k 4 2‘“(2)< <2)(k—2) 1’ then R(k, k) > n. The asymptotics are somewhat disappointing. COROLLARY. If ttor t in a :ct on )f the ovasz “S is acing [11 AT itside T are is far tut: I both n the third. and “w... W “WA... a- . THE LOVASZ LOCAL LEMMA COROLLARY. R(k, k)>—e@k2k/2(1+o(1)). This improves the lower bound given in Lecture 1 by a factor of 2 and the improvement via the deletion method in Lecture 2 (which, oddly, was only published after the better bound) by a factor of x/Z The gap between the upper and lower bounds has not really been effectively decreased. The lower bound of Erdos was found in April 1946 (published in 1947) and progress on this diflicult problem has been slow. The van der Waerden function. Here the improvement is more impressive Color [n] randomly. For each arithmetic progression S of size k let As he the event that S is monochromatic. Let S, T be adjacent in G if they intersect. (In all our applications the probability space will be a random Coloring of some set (2. For Ramsey’s Theorem (I was the edge set of K". Events will be not adjacent if they deal with the coloring on disjoint sets.) Now 172—2‘4‘ and d5 nk as one pro- gression intersects (exercise) at most nk others. Hence we have the following theorem. THEOREM. If 4nk21rk< 1, then W(k)> n. That is, W(k)>2k/8lc This greatly improves the bound W( k) > 2”2 of Lecture 1. Still we must in all honesty mention that W(p)2p2”, for p prime, has been shown by completely constructive means! Algorithm? The Lovasz Local Lemma proves the existence of an x satisfying /\ 15,» even when Pr [/\ Kijrnay be exponentially small- We have seen in Lecture 4 that when X Pr [Ai]<1 there often is an algorithm to find a specific “good” x. Open Problem. Can the Lovasz Local Lemma be implemented by a Good Algorithm? ' Let us be more specific. Suppose 51, ~ ‘ - , SnC[n] with all 15;]: 10 and all deg (j) :- 10. Two color [21] randomly and let A be the event that S,- is monochro- matic. Let i, i’ be adjacent in the dependency graph if and only if their correspond— ing sets intersect. Each S,- is intersected by at most 90 other Sir. We apply the Lovasz Local Lemma with p = Pr [A] = 2~9 and. d r: 90. As 4dp <1, Pr [A [if] > O and so there is a two~coloring X for which no S,» is monochromatic. Is there a polynomial (in n) time algorithm for finding such a coloring? Notice that the Lovasz Local Lemma here guarantees the existence of a “needle in a haystack." If, say, S], ‘ - - , Sn/w are disjoint a random X is good with probability at most (1 ~2”9)”/1°. Can we actually find this exponentially small needle in polynomial time? Addendum: Joke. Here we give an ironic demonstration of the power of the Lovasz Local Lemma. THEOREM. Let S, T be finite sets with |T128IS[. Then there exists a function f: S —> T which is injective. 60 LECTURE 8 Proof. Let f be a random function. For each {x, y}C S let Axy be the event f(x) =f(y). Then Pr [AW]=1/|T|=p. Let {x, y} and {u, v} be adjacent in the dependency graph if and only if they intersect. Then the maximal degree in the dependency graph is d = 2([Sl — 1). As 4dp < 1, Fr [A/lxy]> 0 and so there exists an f for which A” for all x, y. That is, f is injective. When IT! 2 365 and I WI = 23 the “birthday problem” says that f has probability less than % of being injective. When lTI = 8|S| the probability of a random f being injective is exponentially small. The Counting Sieve proves the existence of an injective f only when (I?!) < Cl Anti van der Waerden- Here is the original use of the Lovasz Local Lemma. THEOREM. Let k, m satisfy 4m(m —1)(1 —1/k)"'< 1. Let Sc R with Isl 2 m, Then there exists a kvcoloring x: R —> [k] so that every translate S+t is k—colored. That is, for all te T and 15 is k there exists s e S with x(s+ t) = 1'. Without use of the Lovasz Local Lemma no proof is known that gives the existence of an m = m(k) with this property. Notice a fundamental difference between the translation and homothety groups. Gallai’s Theorem, a consequence of van der Waerden’s Theorem, states that for all finite S and all finite colorings of R there is a monochromatic S’ = aS+ t. Proof First we let BC R be an arbitrary finite set and k—color B so that all S+ t C B have all k colors. Color B randomly. For each I such that S+ t C B let A, be the event that S+t does not have all k colors. Then i Pr [AJS k(1« l/k)m =p. Let t, t’ be adjacent in the dependency graph if and only if S+t and S+t’ intersect. With given t this occurs only if t': +s—s" for distinct s, s'e S, so that the dependency graph has degree at most d : m(m — 1). The conditions on k, m are precisely that 4dp < 1. The Lovasz Local Lemma applies and /\ A, ¢ Q; there is a k-coloring of B for which all translates of S lying in B have all k colors. Compactness. To color all of R we need the Compactness Principle. We state this in a form convenient for us. COMPACTNESS PRINCIPLE. Let Q be an infinite set, k a positive integer, and let U be afamily ofpairs (B, x) where B C Q is finite, x: B——> [k] such that (i) U is closed under restriction. That is, if (B, x) e U and B’C B then (BI,XIB')€ U; (ii) For all B some (B,X) E U. Then there exists A40“) [k] such that (B,X[B)e U for allfinite BC U. Proof Let X be the topological space of all XtQ—9[k]. Here we consider [k] discrete and X has the usual product topology. That is, a basis for the open sets. is given by the sets {X2 x(bi)= ai, IS is s} over all s, b1, ~ ~ - , b5, a1, - - -, as. For every finite B let X3 be the set ofX e X with (B, 143) e U. By (ii) ,XB ¢ @. Splitting XEX according to xlB gives a finite (IBlk) partition of X into sets both open am prc Sin prc all on me inv be i all We as He ...
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