Analyzing Walksat on random formulas

Analyzing W alksat on Random F ormulas ∗ Amin Coja-Oghlan † Alan Frieze ‡ October 22, 2018 Abstract Let Φ be a uniformly distributed random k -SA T formula with n v ariables and m clauses. W e prov e that the Walksat algo rithm from Papadimitriou (FOCS 1991)/Sch ¨ oning (FOCS 19 99) ﬁnds a satisfying assignment of Φ in polynomial time w .h.p. if m/n ≤ ρ · 2 k /k for a certain constant ρ > 0 . T his is an improv ement by a factor o f Θ( k ) o ver the best pre vious analysis o f Walksat from Coja-Ogh lan, Feige, Frieze, Kriv elevich , V ilenchik (SODA 2 009). K ey wor ds: random structures, phase transitions, k -SA T , loc al search algorithms. 1 Introd uction Let Φ = Φ k ( n, m ) be a k -CNF on n Boolean variables x 1 , . . . , x n with m clauses chosen uniform ly at random ( k ≥ 3 ). T he inter est in random k -SA T stems largely fro m the e xperimental observation th at for certain de nsities r the random f ormula Φ is a ch allenging alg orithmic benchmark [7, 15]. Howe ver , analyzing algorithm s on rand om fo rmulas is notoriou sly difﬁcult. Inde ed, the current r igorous results for random k -SA T mostly deal with alg orithms that are extrem ely si mp le both to state and to analyze, or with algorithm s that wer e sp eciﬁcally designed so as to allow f or a rigoro us analysis. More precisely , the present analysis tech niques are essentially conﬁned to simple algo rithms that aim to construc t a satisfying assign- ment by dete rmining the value of one variable at a time for goo d , without any backtrackin g or reassignin g variables at a later time. By contrast, most ‘real-life’ satisﬁability algorithm s actually rely substantially on reassigning variables. Maybe the simplest example of a natu ral algor ithm that eludes the standard analysis tec hniques is Walksat [1 7, 1 8]. Similar lo cal search algorith ms are q uite successfu l in practical SA T -solving [ 19]. Starting from the all-true assignmen t, W alksat trie s to ﬁnd a satisfying assi gn ment of its inp ut k -CNF formu la Φ = Φ 1 ∧ · · · ∧ Φ m as follows. If the current assi gn ment σ is satisfyin g, then clearly there is nothing to do and the algor ithm terminates. Otherwise, the algo rithm picks an index i such that c lause Φ i is un satisﬁed unifo rmly at r andom am ong all such indices. Clause Φ i is a disjun ction of k literals Φ i 1 ∨ · · · ∨ Φ ik . Walks at picks an index j ∈ { 1 , . . . , k } u niformly at random and ﬂips the v alue assigned to the variable underlying the literal Φ ij . Of c ourse, this ensu res that under the new assignment clause Φ i is satisﬁed, but ﬂippin g Φ ij may create new unsatisﬁed clau ses. I f after a certain num ber T max of iterations n o satisfying assignment is found , Walksat g i ves up and c oncedes failure. The pseudocod e is shown in Figure 1. In the worst case, it c an be shown that (2 − 2 /k ) (1+ o (1)) n executions o f Walksat with indepen dent coins tosses will ﬁnd a satisfying assignment of a satis ﬁable input formula Φ on n variables with probability 1 − o (1) , for a suitable T max = T max ( k ) = O ( n ) [18]. Although Walksat is con ceptually very simple, analyz ing this algor ithm on rand om formulas is a challenge. Ind eed, W alksat does not follow the nai ve tem plate of the previously analysed algorithms that assign on e variable at a time for g ood, bec ause its random choices may (an d will) lead Walksat to ﬂipping quite a few variables several times over . This causes stoc hastic dep endencies that seem to ren der the ∗ An e xtended abstract version of this w ork appeare d in the procee dings of ANALCO 2012. † Supported by E PSRC grant E P/G039070 /1. Goethe Univ ersity , Mathematics Instit ute, 10 Robert Mayer St, Frankfurt 60325, Germany , a coghlan@math.uni-f rankfurt.de ‡ Carne gie Mellon Uni ve rsity , Departmen t of Mathe matical Sciences, Pitt sburgh, P A 1521 3, USA, alan@random.ma th.cmu.edu 1 Algorithm 1.1 Walk sat (Φ , T max ) Input: A k -CNF Φ = Φ 1 ∧ · · · ∧ Φ m over the variables x 1 , . . . , x n and a number T max ≥ 0 . Output: An assignm ent σ : V → { 0 , 1 } . 0. Initially , let σ ( x i ) = 1 f or i = 1 , . . . , n . 1. Repeat the following T max times (with independe nt random choices) 2. If σ is a satisfying assignment, then halt and output σ . 3. Otherwise, choose an index i such that clause Φ i is unsatisﬁed under σ uniform ly at random . 4. Suppose that Φ i = Φ i 1 ∨ · · · ∨ Φ ik . Choose an index j ∈ { 1 , . . . , k } un iformly at random. Flip the value of the variable underly ing the literal Φ ij in the assignment σ . 5. Return ‘failure’. Figure 1: T he Walksat algo rithm. differential equation metho d, the main stay of th e previous analyses of random k -SA T algo rithms, useless. The goal of the present paper is to present an analysis of Walksat via a different approach that allo ws us to deal with the stochastic depende ncies. Our main result is as follows. Theorem 1.2 Ther e is a constant k 0 > 3 such that for any k ≥ k 0 and 0 < m/n ≤ 1 25 · 2 k /k , Walksat ( Φ , ⌈ n/k ⌉ ) o utputs a satisfying assignment w .h.p. 1.0.1 Related work. T o put Theor em 1.2 in per spectiv e, let u s comp are it with o ther results on random k -SA T algorith ms. The simplest co nceiv able one is presumably UnitCl ause . Considerin g all variables unassigned initially , UnitClause sets one v ariable at a time as follows. I f there is a clause in which k − 1 variables have been assigned already without satis fyin g that clause (a ‘unit clause’), the algorithm has to assign the k th v ariable so as to satisfy the unit c lause. If there is no unit clau se, a currently un assigned variable is chosen randomly and is assigned a random truth v alue. As UnitCla use is extremely simple and does not backtrack, i t can be analyzed via the method of differential equations [1]. Th e result is that UnitClause ﬁnds a satisfying assignment with a non-vanishing prob ability so long as m/n < (1 − o k (1)) e 2 · 2 k /k , wh ere o k (1) hide s a term that tends to 0 as k g ets large [6]. Furtherm ore, Sh ortestClause , a na tural genera lization of UnitClause , succeeds for m/ n < (1 − o k (1))e 2 / 8 · 2 k /k with high proba bility [8] . I ndeed, th e algor ithm can be m odiﬁed s o as to s ucce ed with high probability even for m/n < (1 . 81 7 − o k (1)) · 2 k /k by allowing a very limited amount of back tracking [11]. Finally , the algorithm Fix from [9], wh ich was speciﬁcally designed f or solving random k -SA T instances, succeeds u p to m/n < (1 − o k (1))2 k ln( k ) /k . By com- parison, non-co nstructive arguments sho w that the thresho ld for the e xistence of a satisfyin g assignment is (1 + o k (1)) · 2 k ln 2 [2, 3]. In summar y , Theo rem 1.2 shows that W alksat is br oadly competitive with the other k nown algo rithms for ran dom k -SA T . That said, the m ain point of th is paper is not to p roduce a better alg orithmic boun d for random k -SA T , but to a ddress the method ological challenge of analyzing algorithms such as Walksat that may re assign variables. This dif ﬁcult aspect did n ot occur or was sidestep ped in the afor ementioned previous analyses [1, 8, 9, 11]. Indeed , th e l ack of techniques for such analyses is arguably one of the most importan t shortcomin gs of the current theory of random discrete structures. Theorem 1 .2 improves substan tially on the previous analyses of Walksat , at least for general k . The b est p revious result for this case showed that w .h.p. Walksat will ﬁnd a satisfying a ssignment with T max = n if m/n < ρ ′ · 2 k /k 2 , for a certain constant ρ ′ > 0 [10]. The proo f of this result is based o n a rather simple observation that allo ws to sidestep the analysis of th e stochastic d ependenc ies that arise in the 2 execution of Walksat . Ho wever , it is not d if ﬁcult to see that this argum ent is conﬁn ed to clause/variable densities m/n < 2 k /k 2 . Theorem 1.2 improves this r esult by a f actor o f Θ( k ) . Furthermo re, the techniques of Alekhnovich and Ben- Sasson [4] show that f or any k Walksat will w .h.p. ﬁnd a satisfying assignmen t within O ( n ) iterations if m/n < r k − pur e , where r k − pur e is the ‘pure literal threshold’. The an alysis in [4] depends hea vily on the fact that th e combinatorial structur e o f the hypergrap h underlyin g the rand om k -CNF Φ is extremely simple for m/n < r k − pur e . Furthermor e, because r k − pur e → 0 in the lim it of l arge k [16], this result is quite weak for general k . Y et [4] remains the best known result for ‘small’ k . For instance, in th e case k = 3 the pure literal bo und is r 3 − pure ≈ 1 . 6 3 [5 ]. Monasson and Semerjian [20] ap plied non-rig orous techniques from statistical m echanics to study the Walksat algorith m on random formulas. Th eir work suggests that Walksat ( Φ , O ( n )) will ﬁnd a satis- fying a ssignment w .h.p. if m/n < (1 − o k (1))2 k /k . Theorem 1.2 conﬁrms this claim, up to the con stant factor 1 / 25 . In contr ast to the pr e vio us ‘ indirect’ attemp ts at analyzin g Walksat on r andom form ulas [4, 10], in the present p aper we develop a tech nique for trac ing the execution of the algorithm directly . This allows us to keep track of the arising stoch astic dependen cies explicitly . Befor e we ou tline our analysis, we need some notation and prelimina ries. 2 Pr eliminaries W e let Ω k ( n, m ) be the set of all k -SA T formulas with variables from V = { x 1 , . . . , x n } th at contain exactly m clauses. T o be precise, we consider each formula an ordered m -tuple of clauses and each clause an o rdered k -tup le of literals, allo wing both literals to o ccur repeatedly in one clause and c lauses to occur repeatedly in the fo rmula. Th us, | Ω k ( n, m ) | = (2 n ) km . Let Σ k ( n, m ) be the power set o f Ω k ( n, m ) , and let P = P k ( n, m ) be the uniform probab ility measure. Through out, we assume that m = ⌈ rn ⌉ f or a ﬁxed number r > 0 , the density . As in dicated above, we denote a unifo rmly rand om elem ent of Ω k ( n, m ) by Φ . In addition, we use the symbol Φ to den ote speciﬁc (i.e ., non- random ) ele ments of Ω k ( n, m ) . If Φ ∈ Ω k ( n, m ) , then Φ i denotes the i th clause of Φ , an d Φ ij denotes the j th literal of Φ i . If Z ⊂ [ m ] is a set of indices, th en we let Φ Z = V i ∈ Z Φ i . If l ∈ { x 1 , ¯ x 1 , . . . , x n , ¯ x n } is a literal, th en we deno te its u nderlying v ariab le by | l | . Furthermo re, we deﬁne sign( l ) = − 1 if l is a negative liter al, and sign( l ) = 1 if l is positive. Recall th at a ﬁltration is a sequence ( F t ) 0 ≤ t ≤ τ of σ - algebras F t ⊂ Σ k ( n, m ) such that F t ⊂ F t +1 for all 0 ≤ t < τ . For a random variable X : Ω k ( n, m ) → R we let E [ X |F t ] denote th e conditional expectation . Thus, E [ X |F t ] : Ω k ( n, m ) → R is a F t -measurab le random variable such that fo r any A ∈ F t we have X Φ ∈ A E [ X |F t ] (Φ) = X Φ ∈ A X (Φ) . Also rememb er that P [ ·|F t ] assigns a probability measure P [ ·|F t ] (Φ) to any Φ ∈ Ω k ( n, m ) , nam ely P [ ·|F t ] (Φ) : A ∈ Σ k ( n, m ) 7→ E [ 1 A |F t ] (Φ) , where 1 A is the indicator of the e vent A . W e n eed the following well-known bound. Lemma 2.1 Let ( F t ) 0 ≤ t ≤ τ be a ﬁ ltration and let ( X t ) 1 ≤ t ≤ τ be a sequ ence of non- ne gative random vari- ables such th at each X t is F t -measurable. Assume that ther e ar e numbers ξ t ≥ 0 such that E [ X t |F t − 1 ] ≤ ξ t for all 1 ≤ t ≤ τ . Then E[ Q 1 ≤ t ≤ τ X t |F 0 ] ≤ Q 1 ≤ t ≤ τ ξ t . Pr oof. For 1 ≤ s ≤ τ we let Y s = Q s t =1 X t . Let s > 1 . Since Y s − 1 is F s − 1 -measurab le, we obtain E [ Y s |F 0 ] = E [ Y s − 1 X s |F 0 ] = E [E [ Y s − 1 X s |F s − 1 ] |F 0 ] = E [ Y s − 1 E [ X s |F s − 1 ] |F 0 ] ≤ ξ s E [ Y s − 1 |F 0 ] , whence the assertion follows by inductio n. ✷ W e also need the following tail bound (“ Az uma-Hoeffding”, e.g. [13, p. 37]). 3 Lemma 2.2 Let ( M t ) 0 ≤ t ≤ τ be a supe r -marting ale with respect to a ﬁltr ation ( F t ) 0 ≤ t ≤ τ such tha t M 0 = 0 . Suppo se that ther e e xist numbers c t such that | M t − M t − 1 | ≤ c t for all 1 ≤ t ≤ τ . Then for any λ > 0 we have P [ M τ > λ ] ≤ exp  − λ 2 / (2 P τ t =1 c 2 t )  . A k - CNF Φ = Φ 1 ∧ · · · ∧ Φ m giv es rise to a bipartite graph whose vertices are the v ariables V an d the clauses { Φ i : i ∈ [ m ] } , and in which ea ch clause is adjacent to all the variables that occur in it. This is the factor graph o f Φ . For a vertex v of the factor graph we denote by N ( v ) = N Φ ( v ) th e neighbo rhood of v in the factor graph. For a set Z ⊂ [ m ] we let N (Φ Z ) = S i ∈ Z N (Φ i ) be th e set o f all variables that occur in the sub-fo rmula Φ Z . Let A, B be two d isjoint sets of vertices of the factor g raph. Recall that a l -fold ma tching fr om A to B is a set M o f A - B -edges such that each a ∈ A is incident with precisely l edg es fr om M , while each b ∈ B is incident with at most one e dge f rom M . W e will make u se of the following simple expansion prop erty of the factor graph of random formulas. Lemma 2.3 There is a constan t k 0 > 0 such that for a ll k ≥ k 0 and fo r m/n ≤ 2 k ln 2 the random formula Φ h as the following pr operty w .h.p. F or a ny set Z ⊂ [ m ] of size | Z | ≤ n/k 2 ther e is a 0 . 9 k -fold matching fr om Φ Z to N ( Φ Z ) . (1) Pr oof. W e start by proving that w .h.p. the random formu la Φ has the following property . For any set U of ≤ n/k variables we ha ve |{ i ∈ [ m ] : N ( Φ i ) ⊂ U }| ≤ 1 . 1 | U | /k . (2) T o prove (2) we u se a ‘ﬁrst m oment’ argument. For set U ⊂ V we let X U = 1 if |{ i ∈ [ m ] : N ( Φ i ) ⊂ U }| > 1 . 1 | U | /k , and we set X U = 0 oth erwise. Th en E [ X U ] = P [ X U = 1] ≤  m 1 . 1 | U | /k  ( | U | /n ) 1 . 1 | U | . Furthermo re, for any 1 ≤ u ≤ n/k we let X u = P U ⊂ V : | U | = u X U . Assum ing th at k ≥ k 0 is sufﬁciently large, we obtain E [ X u ] ≤ X U ⊂ V : | U | = u E [ X U ] ≤  n u  m 1 . 1 u/k   u n  1 . 1 u ≤   e n u · "  e m 1 . 1 u/k  1 /k · u n # 1 . 1   u ≤   e n u "  e2 k k ln 2 1 . 1 · n u  1 /k · u n # 1 . 1   u ≤ " e  u n  0 . 1 − 1 /k  e2 k k ln 2 1 . 1  1 . 1 /k # u ≤  e 2  u n  0 . 09  u . Summing the last expression over 1 ≤ u ≤ n / k an d assuming that k ≥ k 0 is large enough, we see that E X 1 ≤ u ≤ n/k X u ≤ X 1 ≤ u ≤ ln 2 n  e 2  u n  0 . 09  u + X ln 2 n 0 such tha t for all k ≥ k 0 and for any ε > 0 , λ > 4 satisfying ε ≤ k − 3 and ε λ ≤ 1 e (2e) − 4 k the random formula Φ with m/n ≤ 2 k ln 2 ha s the following pr operty w .h.p. Let Z ⊂ [ m ] be any set o f size | Z | ≤ εn . I f i 1 , . . . , i l ∈ [ m ] \ Z is a sequenc e of pairwise distinct indices such that | N ( Φ i s ) ∩ N ( Φ Z ∪{ i j :1 ≤ j 3 such that for all k ≥ k 0 and all r ≤ 1 6 · 2 k /k 2 w .h.p . Walksat ( Φ , n ) will ﬁn d a satisfying assignment. 5 3 Outline of the analysis Thr oughou t this section we assume that k ≥ k 0 for some lar ge enough constant k 0 > 0 , an d that r = m/n ∼ ρ · 2 k /k with k − 2 ≤ ρ < ρ 0 = 1 / 25 . W e can m ake these assumptio ns as o therwise the assertion of Theor em 1.2 alr ead y follows fr om Theor em 2.5. Furth ermor e, let λ = √ k and ε = exp( − k 2 / 3 ) . (5) The standard app roach to analyzing an algorithm on r andom k -SA T form ulas is the method of deferred decisions , whic h often r educes the analysis to the study of a sy stem of o rdinary differential equatio ns that capture the dynamics of the algorithm [1]. Rou ghly speaking, the method of deferr ed dec isions applies where the state of the algorithm after a giv en number of steps can be described by a simple probability distribution, depending only o n a very fe w parameter s d etermined b y the past decisions of the algorithm. This is typically so in the case of simple backtrac k-free algorithms such as UnitClause . Howe ver , in the case of Walksat , this approach does not apply because the algorithm is bound to ﬂi p many variables more than once. This entails that the a lgorithms’ futur e steps depe nd on pa st events in a more complicated way than the method of deferred decisions can accommodate. Hence, our approach will be to use the method of deferred decisions to trace the effect of ﬂipping a variable for the ﬁr st time . But we will n eed addition al arguments to de al with the depen dencies that arise out of ﬂipping the same variable se veral times. T o get started , let us in vestigate the effect of the ﬁrst ﬂip that Walksat per forms. Let σ = 1 be the assignment that sets every variable to true. Clearly , a clause Φ i is unsatisﬁed u nder σ iff it consists of n egati ve literals o nly . As Φ con sists of m unifor mly random an d indepen dent clauses, the numb er of unsatisﬁed cla uses has a binomial distribution Bin( m, 2 − k ) , and thus ther e will b e (1 + o (1))2 − k m ∼ ρn/ k all-negativ e clau ses w .h.p . T o perfo rm its ﬁrst ﬂip, Walksat chooses an index i ∈ [ m ] such that Φ i is all-negativ e unifor mly at random, then chooses a literal index j ∈ [ k ] uniform ly , and sets σ ( | Φ ij | ) to false, thereby satisfying clause Φ i . But, of course, ﬂipping | Φ ij | may well generate n e w unsatisﬁed clauses. W e need to stud y their num ber . As Φ i is just a uniformly rando m a ll-negati ve clause, the ra ndom variable | Φ ij | is uniform ly distributed over the set of all n variables, and thus we may assume witho ut loss th at | Φ ij | = x 1 . Furth ermore, if a clause Φ l becomes unsatisﬁed be cause v ariab le x 1 got ﬂipped , th en x 1 must have been the only variable that appears positively in Φ l . Now , the number of clauses whose only positive literal is x 1 has distribution Bin( m, k / ( n 2 k ) + O (1 /n 2 )) . Ind eed, the probability that a rand om clause has prec isely one positi ve literal is k / 2 k , and the probab ility that this positi ve literal happens to be x 1 is 1 /n ; the O (1 / n 2 ) accounts for the number of clauses in which variable x 1 occurs mo re than o nce. Hence, the e xpec ted number of newly created unsatisﬁed clauses equals (1 + o (1)) km 2 k n ∼ ρ. In summa ry , as we ar e assuming tha t ρ ≤ ρ 0 = 1 / 25 < 1 , th e expected change in the nu mber of unsatisﬁed clauses as a result of the ﬁrst ﬂip is bound ed from above by ρ − 1 + o (1) < 0 . (The pr ecise value is even smaller because x 1 may occu r in fur ther all-negativ e clauses.) Th us, we expect that the ﬁrst ﬂip will indeed reduce the n umber of un satisﬁed clauses. Of course, th is simple calculation does not extend to the further steps of Walksat b ecause knowing the outcome of the ﬁrst ﬂip render s the various above s tatemen ts about clauses/literals being uniformly distrib uted in valid. T o analyze the further ﬂips, we will describe Walksat as a stoch astic proc ess. Our time parameter will b e the number of iterations of the main loop (Step s 2–4 in Figure 1), i.e ., th e n umber of ﬂips per formed. T o represen t the condition ing of the random input formula imposed up to time t , we will deﬁne a sequence of random maps ( π t ) t ≥ 0 . The se maps reﬂect for each pair ( i, j ) ∈ [ m ] × [ k ] the cond itional distribution of the literals Φ ij , given the in formation that Walksat has rev ealed after performing th e ﬁrst t ﬂips. More precisely , the value of π t ( i, j ) will either be just the sign of the literal Φ ij , or the a ctual literal Φ ij itself. In the initial map π 0 , we have π 0 ( i, j ) = sign( Φ ij ) for all ( i, j ) ∈ [ m ] × [ k ] . At times t ≥ 1 the m ap π t will f eature the occurren ces of all variables that have been ﬂipped thus far . That is, f or any pair ( i , j ) such that Walksat has ﬂipped th e variable | Φ ij | at least once b y tim e t , we 6 PI0. If the assignment σ t − 1 satisﬁes Φ , then the process terminates. PI1. Otherwise, choose an index i t such that Φ i t is unsatisﬁed under σ t − 1 unifor mly at random from the set of all such indices. In addition, choose j t ∈ [ k ] unif ormly at random. Deﬁne σ t : V → { 0 , 1 } by letting σ t ( | Φ i t j t | ) = 1 − σ t − 1 ( | Φ i t j t | ) and σ t ( x ) = σ t − 1 ( x ) for all x 6 = | Φ i t j t | . PI2. Initially , let Z t = Z t − 1 and N t = N t − 1 . While there is an index i ∈ [ m ] \ Z t such that Φ i is ( A t − 1 ∪ N t ∪ {| Φ i t j t |} ) -negative and either • ther e are at least k 1 indices j ∈ [ k ] with | Φ ij | ∈ A t − 1 ∪ {| Φ i t j t |} , or • ther e are more than λ indices j ∈ [ k ] with | Φ ij | ∈ N t , add the least such index i min to Z t and add the variables {| Φ i min j | : j ∈ [ k ] } to N t . PI3. Let A t = ( A t − 1 ∪ {| Φ i t j t |} ) \ N t . Deﬁne the map π t : [ m ] × [ k ] → {− 1 , 1 } ∪ L by letting π t ( i, j ) =  Φ ij if | Φ ij | ∈ A t ∪ N t , sign( Φ ij ) otherwise . Figure 2: the co nstruction of the maps π t let π t ( i, j ) = Φ ij . This inf ormation will be necessary f or u s to inves tigate th e effect o f ﬂipp ing th e same variable more than once. In additio n, we n eed to pay p articular attention to clauses that con tain many variables that hav e be en ﬂipped at least onc e. The reason is that these clau ses ha ve ‘too little random ness’ left for a direc t analysis, and thu s we will need to stu dy them sep arately . More precisely , in o ur map π t we will fu lly reveal all clauses Φ i in which at least k 1 = 0 . 5 7 k (6) literals Φ ij have been ﬂipp ed at least on ce. Furtherm ore, we will also recursively reveal all clauses that contain at lea st λ variables from clauses that were fully re vealed b efore. This recursi ve p rocess ensures that we can separate th e analy sis o f clauses that are ‘ heavily co nditioned’ by the past steps of Walksat from the bulk of the formula. Throu ghout this process that mirrors t he execution of Walksat , all variables whose occurre nces hav e been revealed will be labeled eith er with an asterisk or with a zero . Those variables th at g ot rev ealed because they oc cur either in a ‘heavily conditioned’ clause o r in an other clause that got r e vealed by the recursive process described in the p revious paragraph will be lab eled 0 . All o ther variables that have been ﬂipped by Walk sat at least on ce are labeled ∗ . W e will let A t denote the set of all variables lab eled ∗ , and N t the set of all variables labeled 0 . Let us now deﬁne the ma ps π t and the sets A t , N t formally . E ach π t is a map [ m ] × [ k ] → {− 1 , 1 } ∪ L , with L = { x 1 , ¯ x 1 , . . . , x n , ¯ x n } the set of literals. As mentioned above, we let π 0 ( i, j ) = sign( Φ ij ) for all ( i, j ) ∈ [ m ] × [ k ] . Additionally , let A 0 = N 0 = Z 0 = ∅ , and let σ 0 : V → { 0 , 1 } , x 7→ 1 be the all-true assignment. F or a set S ⊂ V we call a clause Φ i S -ne gative if for all j ∈ [ k ] with sign( Φ ij ) = 1 we h a ve Φ ij ∈ S . (I n other word s, Φ i is S -negative if all of its positive literals lie in S .) For t ≥ 1 , we deﬁne th e maps π t along with the sets A t , N t , Z t inductively via the p rocess shown in Figur e 2. Intuiti vely , the set Z t contains the clauses that are ‘ heavily conditioned’ at time t , and N t is the set of variables that occur in such c lauses. Moreover, A t is th e set of all variables that have been ﬂipped at least once by time t except the ones that belong to N t . Let T be the stopping time of this process, i.e., the minimum t such that σ t satisﬁes Φ (or ∞ if there is no such t ). For t > T , we deﬁne π t = π T , σ t = σ T , A t = A T , N t = N T , and Z t = Z T . Steps PI0 – PI1 mirror the main loop of the Wal ksat algorith m; in particular, the stoppin g time T equals the to tal number of iterations of the main loop of Walksat before a satisfying assignment is found. The purpo se o f th e remaining steps is to ‘u pdate’ the sets A t and Z t and the map π t as describ ed above. Before we continue, it may be useful to illustrate the construction of the maps π t with an example. Example 3.1 Let u s go thr ough the example of a 5 -SAT formula with 6 cla uses on 10 variab les. F or the 7 sake of th is e xamp le, we will work with k 1 = 2 an d λ = 2 . (Recall that in our pr oof we actually assume that k ≥ k 0 is lar ge eno ugh, k 1 is as in (6) a nd λ = √ k .) W e will repr esent the maps π t by ta bles whose columns corr espond to the clauses Φ i . Thus, the j th entry in column i r epresents the value π t ( i, j ) . T o impr ove r eadability , we ju st write + and − in stead of ± 1 . Suppo se that the initial ma p π 0 , co ntaining the signs of all literals, r eads π 0 = − − − + + + − + − + − + − − − − − + − − − − + + − − − − + + The in itial assignmen t σ 0 is th e all-true assignment, a nd A 0 = N 0 = Z 0 = ∅ . Thr ou ghout, we will mark the variables in A t by an asterisk ∗ a nd the variables in N t by a 0 . Being all-ne ga tive, clau ses Φ 1 and Φ 3 ar e unsatisﬁed under σ 0 . Ther efor e, at time t = 1 step PI1 chooses i 1 ∈ { 1 , 3 } randomly ; say , the outco me is i 1 = 1 . In addition , PI1 chooses j 1 ∈ [ k ] = { 1 , 2 , 3 , 4 , 5 } u niformly at rando m. Su ppose the result is j 1 = 5 . T o c arry o n, we need to re veal the variable | Φ 15 | . Thus far , th e p r ocess ha s n ot imp osed a ny conditionin g on | Φ 15 | , and ther efor e this vari- able is u niformly distributed over the set o f a ll ou r n = 10 variables. Assume tha t ind eed | Φ 15 | = x 1 . Then PI1 sets σ 1 ( x 1 ) = 0 and σ 1 ( x ) = 1 for a ll x 6 = x 1 . T o implement PI2 we need to r eveal all occu rr ences of x 1 in our rando m formula. As ther e is n o pr evious cond itioning on any of variables | Φ ij | with ( i, j ) 6 = (1 , 5) , these variables r emain independe ntly uniformly distributed over th e set of all va riables, an d thus th e events {| Φ ij | = x 1 } occur indep endently with pr obability 1 /n . Su ppose that x 1 occurs at the following positions: π 0 = − − − x 1 + + − x 1 − + − + − − − − ¯ x 1 + − − − − + + ¯ x 1 − − − + x 1 Then ther e is no clause with at least k 1 occurr ences of a variable fr om A 0 ∪ N 0 ∪ { x 1 } = { x 1 } , and thus step PI2 is void. Hen ce, at the end of the ﬁrst iteration w e ha ve A 1 = { x 1 } , N 1 = Z 1 = ∅ , and π 1 = − − − x ∗ 1 + + − x ∗ 1 − + − + − − − − ¯ x ∗ 1 + − − − − + + ¯ x ∗ 1 − − − + x ∗ 1 At time t = 2 ther e ar e two unsatisﬁed clauses: Φ 2 , whose only positive literal got ﬂipped to false, and Φ 3 , which was unsatisﬁed initially . Step PI1 choo ses one of th em randomly , say i 2 = 2 , and also chooses a rando m position j 2 ∈ [ k ] , say j 2 = 2 . As we a lr eady kno w fr om the ﬁrst step, the literal in this po sition is Φ 22 = π 1 (2 , 2) = x 1 . In effect, the second iteration re verses the ﬂip mad e in the ﬁrst o ne and th us σ 2 is th e a ll-true assign ment. Sin ce we have r evealed all the o ccurr ences of x 1 alr eady , step PI2 is void a nd π 2 = π 1 , A 2 = { x 1 } , and N 2 = Z 2 = ∅ . At the start o f th e third iteration the u nsatisﬁed clau ses ar e Φ 1 , Φ 3 . Sup pose PI1 chooses i 3 = 1 and j 3 = 1 . Th en we n eed to r eveal the v ariable | Φ 11 | . At this point, the only conditioning imposed o n this variable is that it is differ ent fr om x 1 , becau se all occurr ences of x 1 have been re vealed alr ead y . Thu s, | Φ 11 | is uniformly distrib uted over x 2 , . . . , x 10 . Supp ose th at | Φ 11 | = x 2 . Then σ 3 ( x 2 ) = 0 and σ 3 ( x ) = 1 for all x 6 = x 2 . T o r eveal the occurrences of x 2 all over the formula, note that b y the same argument we applied to | Φ 11 | all spots marked ± in π 2 hide variables that ar e uniformly distributed over x 2 , . . . , x 10 . 8 Let us assume that x 2 occurs in the following positions. π 1 = ¯ x 2 − − x ∗ 1 + + − x ∗ 1 − + − + − − ¯ x 2 − ¯ x ∗ 1 + − − − − + x 2 ¯ x ∗ 1 − − − + x ∗ 1 As c lause Φ 1 is A 2 ∪ N 2 ∪ { x 2 } = { x 1 , x 2 } -negative and contains k 1 = 2 occurrences of variables fr om A 2 ∪ { x 2 } = { x 1 , x 2 } , PI2 sets Z 3 = { 1 } , r e vea ls the r emaining thr ee variables in Φ 1 , and adds all variables that occur in Φ 1 to N 3 . Su ppose th at the r emaining v ariables in Φ 1 ar e | Φ 12 | = x 3 , | Φ 13 | = x 4 , | Φ 13 | = x 5 . Then N 3 = { x 1 , x 2 , x 3 , x 4 , x 5 } ; in particular , x 1 , x 2 ar e now la beled 0 . The new 0 label ‘overwrites’ the ∗ becau se PI3 ensur es that A 3 = ( A 2 ∪ { x 2 } ) \ N 3 = ∅ . In order to carry out PI2 , we need to r eveal all occurr ences of variables fr om N 3 . Su ppose this yields π 1 = ¯ x 0 2 − − x 0 1 + + ¯ x 0 3 x 0 1 − x 0 5 − + ¯ x 0 4 − − − ¯ x 0 1 + ¯ x 0 5 − − ¯ x 0 4 x 0 3 x 0 2 ¯ x 0 1 − ¯ x 0 5 ¯ x 0 3 x 0 4 x 0 1 Then clause Φ 4 has become A 2 ∪ N 3 ∪ { x 2 } = { x 1 , . . . , x 5 } -ne gative (as ther e is no + -sign left in column four), and thus PI2 sets Z 3 = { 1 , 4 } . T o pr oceed, we need to re veal the remaining − -sign o f Φ 4 , add th e underlying variable to N 3 , and r eveal all of its occurr ences. Supp ose that this yields π 1 = ¯ x 0 2 − − x 0 1 + + ¯ x 0 3 x 0 1 − x 0 5 − + ¯ x 0 4 − − ¯ x 0 6 ¯ x 0 1 x 0 6 ¯ x 0 5 − − ¯ x 0 4 x 0 3 x 0 2 ¯ x 0 1 − ¯ x 0 5 ¯ x 0 3 x 0 4 x 0 1 At this po int PI2 stops, becau se clauses Φ 5 , Φ 6 have + -signs left and clauses Φ 2 , Φ 3 contain only on e variable lab eled 0 . Thus, a t the end o f the third iteration we have A 3 = ∅ , N 3 = { x 1 , . . . , x 6 } , Z 3 = { 1 , 4 } , and π 3 = ¯ x 0 2 − − x 0 1 + + ¯ x 0 3 x 0 1 − x 0 5 − + ¯ x 0 4 − − ¯ x 0 6 ¯ x 0 1 x 0 6 ¯ x 0 5 − − ¯ x 0 4 x 0 3 x 0 2 ¯ x 0 1 − ¯ x 0 5 ¯ x 0 3 x 0 4 x 0 1 As th e fourth iter atio n commen ces, the only u nsatisﬁed clau se left is Φ 3 , whence i 4 = 3 . Mor eover , assume that j 4 = 1 . A s we have r evealed all occurrences of x 1 , . . . , x 6 , at this point we know that | Φ 31 | is unifo rmly distrib uted over { x 7 , x 8 , x 9 , x 10 } . Suppo se that inde ed | Φ 31 | = x 7 . Thus, PI1 sets σ 4 ( x 2 ) = σ 4 ( x 7 ) = 0 and σ 4 ( x ) = 1 for a ll x 6 = x 2 , x 7 . Suppose t ha t r evealing all occurr ences of x 7 yields π 3 = ¯ x 0 2 − ¯ x 7 x 0 1 + x 7 ¯ x 0 3 x 0 1 − x 0 5 ¯ x 7 + ¯ x 0 4 ¯ x 7 − ¯ x 0 6 ¯ x 0 1 x 0 6 ¯ x 0 5 − − ¯ x 0 4 x 0 3 x 0 2 ¯ x 0 1 − ¯ x 0 5 ¯ x 0 3 x 0 4 x 0 1 Then there ar e no A 3 ∪ N 3 ∪ { x 7 } -negative clauses Φ i with i 6∈ Z 3 that hav e at least two occu rr ences of a variable fr om A 3 ∪ { x 7 } . Th er efor e, PI2 is void, and at the end of the fourth iter ation we ha ve π 4 = ¯ x 0 2 − ¯ x ∗ 7 x 0 1 + x ∗ 7 ¯ x 0 3 x 0 1 − x 0 5 ¯ x ∗ 7 + ¯ x 0 4 ¯ x ∗ 7 − ¯ x 0 6 ¯ x 0 1 x 0 6 ¯ x 0 5 − − ¯ x 0 4 x 0 3 x 0 2 ¯ x 0 1 − ¯ x 0 5 ¯ x 0 3 x 0 4 x 0 1 , 9 A 4 = { x 7 } , N 4 = { x 1 , . . . , x 6 } , and Z 4 = { 1 , 4 } . As σ 4 is satisfying the pr ocess stops and T = 4 . ✷ T o trace the process PI0 – P I3 over time we deﬁn e a ﬁltratio n ( F t ) t ≥ 0 by le tting F t be th e σ -alg ebra generated by th e random variables i s , j s and π s ( i, j ) with s ≤ t and ( i, j ) ∈ [ m ] × [ k ] . Then intuitively , a random variable X is F t -measurab le if its value is deter mined by the ﬁrst t steps of the p rocess PI0 –PI3 . In particular, we ha ve the following. Fact 3.2 F or any t ≥ 1 , any x ∈ V , and a ny i ∈ [ m ] the events { σ t ( x ) = 1 } , { Φ i is satisﬁed under σ t } , { x ∈ A t } , { i ∈ Z t } , { x ∈ N t } , and { T = t } ar e F t -measurable. Pr oof. The co nstruction in steps PI2 and PI3 ensures that for any t ≥ 1 we hav e Φ i t j t ∈ A t ∪ N t and thus π t ( i t , j t ) = Φ i t j t This imp lies that for any variable x ∈ V the event { σ t ( x ) = 1 } is F t -measurab le. In fact, we ha ve σ t ( x ) = 1 iff the nu mber |{ 1 ≤ s ≤ t : | π t ( i s , j s ) | = x }| of times x has b een ﬂipp ed is ev en (because σ 0 is the all-true assignment). This implies that for any i ∈ [ m ] th e e vent { Φ i is satisﬁed under σ t } is F t -measurab le. I n f act, if there is an ind ex j ∈ [ k ] such that π t ( i, j ) = 1 , then Φ ij is a positive literal whose u nderlying variable ha s no t been ﬂippe d before, when ce σ t satisﬁes Φ i . Moreover , if there is an index j ∈ [ k ] su ch that Φ ij 6 = ± 1 , then by the previous par agraph the event that the literal Φ ij = π t ( i, j ) is true un der σ t is F t -measurab le. If there is such a satisﬁ ed liter al Φ ij , then Φ i is satisﬁed. Co n versely , if there is no j ∈ [ k ] su ch that either π t ( i, j ) = 1 or π t ( i, j ) is a literal that is satisﬁed under σ t , then clau se Φ i is unsatisﬁed. Hence , the e vent { σ t is satisfying } is F t -measurab le as well, and therefor e so is the e vent { T = t } . Furthermo re, observe that i ∈ Z t iff for all j ∈ [ k ] we hav e π t ( i, j ) 6∈ {− 1 , 1 } . F or if i ∈ Z t , then for all j ∈ [ k ] we have | Φ ij | ∈ N t and thus π t ( i, j ) = Φ ij 6 = ± 1 due to PI3 . Conv ersely , if k ≥ k 0 is large en ough, any i ∈ [ k ] such that π t ( i, j ) 6∈ {− 1 , 1 } fo r all j ∈ [ k ] must satisfy one of the two condition s that lead PI2 to add i to Z t . Hence, fo r any i ∈ [ m ] the event { i ∈ Z t } is F t -measurab le. As by co nstruction N t = { π t ( i, j ) : i ∈ Z t , j ∈ [ k ] } , we conclud e that for any v ariable x ∈ V the event { x ∈ N t } is F t -measurab le. Finally , the con struction in PI3 ensures that A t = {| π t ( i s , j s ) | : 1 ≤ s ≤ t } \ N t . As for any x the ev ents { x ∈ {| π t ( i s , j s ) | : 1 ≤ s ≤ t }} and { x ∈ N t } a re F t -measurab le, so is the e vent { x ∈ A t } . ✷ If π t ( i, j ) = ± 1 , then up to time t the proc ess PI0 – PI3 has only taken the sign of the liter al Φ ij into account, but has b een oblivious to the un derlying variable. The only con ditioning is that | Φ ij | 6∈ A t ∪ N t (because otherwise PI3 would ha ve rep laced the ± 1 by the actual literal). Since the inpu t fo rmula Φ is random , this implies that | Φ ij | is unif ormly distributed ov er V \ ( A t ∪ N t ) . In fact, f or all ( i, j ) suc h that π t ( i, j ) = ± 1 the und erlying variables a re indepen dently unifor mly distributed over V \ ( A t ∪ N t ) . Formally , we can state this key observation as follows. Fact 3.3 Let t ≥ 0 . Let E t be th e set o f all pairs ( i, j ) such th at π t ( i, j ) ∈ {− 1 , 1 } . The cond itional joint distribution o f the variables ( | Φ ij | ) ( i,j ) ∈E t given F t is uniform over ( V \ ( A t ∪ N t )) E t . Tha t is, for any map f : E t → V \ ( A t ∪ N t ) we have P [ ∀ ( i, j ) ∈ E t : | Φ ij | = f ( i, j ) |F t ] = | V \ ( A t ∪ N t ) | −|E t | . Let T ∗ = θ n with θ = 0 . 3 8 /k . Our overall go al is to prove that the stopping time of the pro cess PI0–PI3 satisﬁes T ≤ T ∗ w .h.p. (The number θ is chosen somewhat arbitrarily; for the analysis to work it seems to be essential that θ = c/k fo r some c > 0 that is neither “too small” nor “to o large”. The c oncrete constant ab ove happens to work.) T o prove this, we will deﬁne non-negative random v ariab les S t , H t such that S t + H t = 0 imp lies that σ t is a satisfying assignment. W e will then trace S t , H t for 1 ≤ t ≤ T ∗ . For an y t ≥ 1 le t D t = { i ∈ [ m ] : Φ i is A t ∪ N t -negative } . As PI3 en sures that Φ i is A t ∪ N t -negative iff π t ( i, j ) 6 = 1 for all j ∈ [ k ] , the event { i ∈ D t } is F t - measurable for any i ∈ [ m ] . W e deﬁne S 0 = |D 0 | and S t = |D t | − |A t | for t ≥ 1 . (7) 10 Any clause Φ i with i 6∈ D t is satisﬁed u nder σ t . F or if j ∈ [ k ] is such that π t ( i, j ) = 1 , then Φ ij is a positive literal and σ t ( Φ ij ) = 1 , bec ause Walksat starts with the all-true assignment σ 0 and the variable Φ ij has not been ﬂipped u p to time t . Clearly , in ord er to study the random variable S t it is crucial to estimate |D t | . This is th e purpose of the following proposition, whose proof we defer to Section 4. Proposition 3.4 W .h .p. we have |D t | ≤ 2 2 − k m for a ll t ≤ T ∗ . T o deﬁne the random variables H t , let us call an assignment τ : N t → { 0 , 1 } rich f or Z t if in each clause Φ i with i ∈ Z t at least 0 . 8 k literals Φ ij are satisﬁed under τ . Proposition 3.5 W .h .p. ther e is a sequence ( τ t ) 1 ≤ t ≤ T ∗ with the following pr operties. 1. F or any 1 ≤ t ≤ T ∗ , τ t is a rich assi gn ment for Z t . 2. F or any 1 < t ≤ T ∗ and any x ∈ N t − 1 we have τ t ( x ) = τ t − 1 ( x ) . Mor eover , τ t is F t -measurable for all t . Assuming that there is a sequence ( τ t ) 1 ≤ t ≤ T ∗ as in Proposition 3.5, we deﬁne H 0 = 0 and H t = |{ x ∈ N t : σ t ( x ) 6 = τ t ( x ) }| for 1 ≤ t ≤ T ∗ , and H t = |N t | for t > T ∗ . For th e sake of c ompleteness, we also let H t = |N t | if there is no such sequence ( τ t ) 1 ≤ t ≤ T ∗ . The proof of Proposition 3.5 hinges upo n the following fact. Proposition 3.6 W .h .p. we have |Z t | ≤ εn for all t ≤ T ∗ . W e defer the pro of of Prop osition 3.6 to Section 5. Assuming Proposition 3.6, we can d eriv e Propo sition 3 .5 rather easily . Pr oof of Pr oposition 3.5 ( assuming Pr oposition 3 .6). By Lem ma 2.3, we may assume tha t Φ h as the expansion prop erty (1). Furth ermore, by Proposition 3.6 we may assume that |Z t | ≤ ε n for all t ≤ T ∗ . Under these assumptions we will construct the sequence ( τ t ) 1 ≤ t ≤ T ∗ by induction on t ≥ 1 . Thus, suppose that 1 ≤ t ≤ T ∗ and that we have already got assignments τ s with 1 ≤ s < t th at satisfy 1.–2. The set Z = Z t \ Z t − 1 of indices that Z t gained at time t has size | Z | ≤ |Z t | ≤ εn . There fore, (1) ensures that there is a 0 . 9 k -fold matching M from Z to the set N = N ( Φ Z ) = {| Φ ij | : ( i, j ) ∈ Z × [ k ] } ⊂ N t of variables that o ccur in the clauses Φ i with i ∈ Z . The constructio n in PI2 ensures that none of these clauses Φ i has mo re than λ occurren ces of a v ariable from N t − 1 (as otherwise i ∈ Z t − 1 ). There fore, in the matching M ′ obtained from M b y omitting all edges e = { i, x } with i ∈ Z and x ∈ N t − 1 each clause Φ i with i ∈ Z is in cident w ith a t least 0 . 9 k − λ ≥ 0 . 8 k ed ges. Now , f or each ed ge e = { i, x } ∈ M ′ let τ t ( x ) be the truth value that makes the correspon ding literal in Φ i ev aluate to true. Furthermore , for a ll y ∈ N t − 1 let τ t ( y ) = τ t − 1 ( y ) , and for all other variables x ′ ∈ N t let τ t ( x ′ ) = 1 . T his en sures that τ t satisﬁes the condition s in Proposition 3.5. ✷ Having deﬁned the random variables S t , H t , we are now go ing to v erif y that they suit their intend ed purpo se, i.e., that S t + H t = 0 implies th at σ t is satisfying. Proposition 3.7 Let 1 ≤ t ≤ T ∗ . If S t + H t = 0 , then σ t is a satisfying assignment. Pr oof. Le t U t be the n umber of clause indices i ∈ [ m ] \ Z t such that Φ i is unsatisﬁed u nder σ t . W e claim that U t ≤ S t = |D t | − |A t | . (8) T o see this, r ecall that any in dex i ∈ [ m ] such th at Φ i is u nsatisﬁed under σ t belongs to D t . Therefo re, to prove (8) i t sufﬁces to construct injecti ve maps s t : A t → D t such that for any x ∈ A t the clause Φ s t ( x ) is 11 satisﬁed und er σ t . In fact, the map s t will have the property that for each x ∈ A t there is an in dex j ∈ [ k ] such that x = | Φ s t ( x ) j | and such that the literal Φ s t ( x ) j is true under σ t . The construction of the maps s t is induc ti ve. For t = 0 we have A 0 = ∅ and thus there is noth ing to do. Thu s, suppo se that 1 ≤ t ≤ T an d that we have deﬁned s t − 1 already . Let y = | Φ i t j t | be the variable ﬂipped at tim e t . If i t 6∈ Z t , then y ∈ A t and we d eﬁne s t ( y ) = i t . Moreover , we let s t ( x ) = s t − 1 ( x ) for all x ∈ A t \ { y } ⊂ A t − 1 . (No te that it is p ossible that y ∈ A t − 1 as y may hav e been ﬂip ped before. ) For t > T we set s t = s t − 1 . T o verify that s t has the desired pr operties, assume that T ≥ t an d observe that P I1 ensures that Φ i t was unsatisﬁed under σ t − 1 . Thus, i t ∈ D t − 1 ⊂ D t . But as P I1 sets σ t ( y ) = 1 − σ t − 1 ( y ) , Φ i t is satisﬁed under σ t . Fur thermore, for all x ∈ A t \ { y } we have σ t ( x ) = σ t − 1 ( x ) , an d th us e ach of these variables contributes a tru e literal to its c lause Φ s t ( x ) = Φ s t − 1 ( x ) by ind uction. Since s t − 1 is in jecti ve but Φ i t was unsatisﬁed under σ t − 1 , we have i t 6∈ Im ( s t − 1 ) , whence s t is injectiv e. Th is establishes (8). As ( 8) shows, S t = 0 imp lies U t = 0 , i.e., σ t satisﬁes all clauses Φ i with i 6∈ Z t . T o complete the proof , we need to show that if H t = 0 , then σ t also satisﬁes all clau ses Φ i with i ∈ Z t . But if H t = 0 , then σ t ( x ) = τ t ( x ) fo r all x ∈ N t , and τ t is a satisfying assignmen t of Φ Z t . ✷ Finally , we hav e all the pieces in place to prove Theorem 1.2. Pr oof of Theor em 1.2 (assuming Pr opositions 3.4 and 3.6). Pro position 3.7 sho ws that P [ T ≥ T ∗ ] = P [ T ≥ T ∗ ∧ ∀ 1 ≤ t ≤ T ∗ : S t + H t > 0] . W e are going to bound the probability on the r .h.s. T o this end, we work with tw o rand om variables S ′ t , H ′ t that are easier to analyze than the original S t , H t . Namely , we let S ′ 0 = H ′ 0 = 0 , an d S ′ t = S ′ t − 1 −  1 if π t − 1 ( i t , j t ) = − 1 , 0 other wise ( t ≥ 1) . In other words, we let S ′ t = S ′ t − 1 − 1 if the variable ﬂipped at time t h ad not been ﬂipped befo re and does not occur in any of the ‘exceptional’ clauses Φ Z t − 1 . Other wise, S ′ t = S ′ t − 1 . W e claim that S t ≤ |D t | + k | Z t | + S ′ t for any t ≥ 0 . (9) T o see this, recall fr om (7) that S t = |D t | − |A t | . By PI3 , the set A t contains all v ariables | Φ i s j s | such that π s − 1 ( i s , j s ) = − 1 with s ≤ t , except the ones that belong to N t . Since |N t | ≤ k |Z t | , we obtain (9). Furthermo re, we let H ′ 0 = 0 and H ′ t = H ′ t − 1 +    − 1 if | Φ i t j t | ∈ N t − 1 and σ t ( | Φ i t j t | ) = τ t ( | Φ i t j t | ) , 1 if | Φ i t j t | ∈ N t − 1 and σ t ( | Φ i t j t | ) 6 = τ t ( | Φ i t j t | ) , 0 otherwise ( t ≥ 1) . Thus, starting at 0 , we decrease the value of H ′ t by one if the variable ﬂipped at time t lies in N t − 1 and its new v alue coincides with the ‘ideal’ assignment τ t , while we increase by one if these values differ . W e claim that H t ≤ k |Z t | + H ′ t for any t ≥ 0 . (10) For H 0 = H ′ 0 and H t − H t − 1 = |{ x ∈ N t : σ t ( x ) 6 = τ t ( x ) }| − |{ x ∈ N t − 1 : σ t − 1 ( x ) 6 = τ t − 1 ( x ) }| ≤ |N t \ N t − 1 | + H ′ t − H ′ t − 1 ≤ k |Z t \ Z t − 1 | + H ′ t − H ′ t − 1 for any t ≥ 1 . Combining (9) and (10) with Propositions 3.4 and 3.6, we see that w .h.p. S t + H t ≤ |D t | + 2 k |Z t | + S ′ t + H ′ t ≤ 2 2 − k m + 2 k |Z t | + S ′ t + H ′ t ≤ 4 ρn k + 2 k εn + S ′ t + H ′ t for any t ≤ T ∗ . (11 ) 12 Hence, we are left to analyze S ′ t + H ′ t . The s equ ence ( S ′ t + H ′ t ) t is a super -mar tingale. More precisely , we claim that with γ = 0 . 429 we hav e E [ S ′ t + H ′ t |F t − 1 ] < S ′ t − 1 + H ′ t − 1 − γ for all t ≤ min { T , T ∗ } . (12) There are two cases to consider . Case 1: i t 6∈ Z t − 1 . The con struction in step PI2 ensures that there are fe wer than λ indices j such that | Φ i t j | ∈ N t − 1 . Further more, PI2 e nsures that there are less than k 1 indices j such that | Φ i t j | ∈ A t − 1 . Moreover, ther e is no in dex j such th at π t − 1 ( i t , j ) = 1 , because o therwise clause Φ i t would have been satisﬁed under σ t − 1 . Th is means that for at least k − k 1 − λ indices j ∈ [ k ] we have π t − 1 ( i t , j ) = − 1 . Ther efore, as j t ∈ [ k ] is c hosen u niformly at random, with prob ability at least 1 − ( k 1 + λ ) /k ≥ 0 . 4 3 − λ/k we hav e S ′ t = S ′ t − 1 − 1 . In addition , as Φ i t contains at most λ variables fro m N t − 1 , the probability that H ′ t = H ′ t − 1 + 1 is bounded from above by λ /k < 0 . 0 0001 . Thus, (12) holds. Case 2: i t ∈ Z t − 1 . As the assignm ent τ t − 1 is rich, there are at least 0 . 8 k indices j such that τ t ( Φ i t j ) = τ t − 1 ( Φ i t j ) = 1 . Howev er, for all o f these in dices j we have σ t − 1 ( Φ i t j ) = 0 , b ecause Φ i t is unsatisﬁed und er σ t − 1 . Hence, the pro bability that τ t ( Φ i t j t ) = 1 and σ t − 1 ( Φ i t j t ) = 0 is at lea st 0 . 8 , and if this event indeed occur s then σ t ( Φ i t j t ) = τ t ( Φ i t j t ) = 1 . Th erefore, H ′ t − H ′ t − 1 has expectation ≤ − 0 . 8 + 0 . 2 ≤ − 0 . 6 . Moreover, S ′ t ≤ S ′ t − 1 with certainty . This implies (12). T o complete the p roof, we are going to ap ply Azuma’ s inequality (Lemma 2 .2 in Section 2) to the random variable S ′ T ∗ + H ′ T ∗ . The inequa lity applies because (12) shows that ( S ′ t + H ′ t ) t ≥ 0 is a super- martingale. Howe ver , there is a minor technical intric acy: to use the inequality , we need an upper bound on the expectation E [ S ′ T ∗ + H ′ T ∗ ] . But as (12) only holds for t ≤ min { T , T ∗ } , this w ou ld r equire knowledge of the probab ility that T ≥ T ∗ , the very quantity that we want to estimate. T o cir cumvent this p roblem, we deﬁne furth er rand om variables R t by letting R t = S ′ t + H ′ t for t ≤ min { T ∗ , T } and R t = R t − 1 − γ fo r t > min { T ∗ , T } . Then R 0 = 0 and E [ R t |F t − 1 ] ≤ R t − 1 − γ for all t ≥ 0 . Thus, E [ R T ∗ ] ≤ − γ T ∗ . Recalling the d eﬁnition (5) of ε , we obtain for k ≥ k 0 sufﬁciently large and ρ ≤ ρ 0 = 1 / 25 the bo und E [ R T ∗ ] ≤ − γ · T ∗ ≤ − 4 ρn/k − 1 0 k εn. (13) Furthermo re, | R t − R t − 1 | ≤ 2 f or all t ≥ 0 by the d eﬁnitions of S ′ t , H ′ t . Th erefore, Azuma’ s inequality and (13) yield P [ R T ∗ > − 4 ρn/k − 2 k εn ] ≤ P h R T ∗ > E [ R T ∗ ] + n 2 / 3 i ≤ exp  − n 4 / 3 8 T ∗  = o (1) . (14) Finally , we obtain from (9), (10), and Proposition 3.7 P [ T > T ∗ ] ≤ P [ ∀ t ≤ T ∗ : |D t | + 2 k |Z t | + R t > 0] ≤ P [ |D T ∗ | + 2 k |Z T ∗ | + R T ∗ > 0] ≤ P [ |D T ∗ | + 2 k |Z T ∗ | > 4 ρn/k + 2 k εn ] + P [ R T ∗ > − 4 ρn/k − 2 k εn ] (11), (14) = o (1 ) , thereby completin g the proof. ✷ Our r emaining task is to establish Pro positions 3.4 and 3 .6. From a formal point of vie w , we should start with Proposition 3.6 becau se the proof of Proposition 3.4 depends on it. Howe ver , th e a rgument that is used in the p roof of Proposition 3.4 is conceptually similar to b ut techn ically far simpler than the one that we use to prove Prop osition 3.6. Hence, for d idactical reasons we will start with the pro of of Proposition 3 .4 in Section 4 and postpone the proof of Proposition 3.6 to Section 5. 4 Pr oof of Proposition 3.4 In this section we keep the notation and the assumptions fr o m Pr oposition 3.4. 13 Our goal is to bound the number |D T ∗ | of A T ∗ ∪ N T ∗ -negative cla uses Φ i , i.e., clauses whose positiv e literals all belong to A T ∗ ∪ N T ∗ . Thus, we need to study h ow the process PI0–PI3 ‘h its’ the po sitions ( i, j ) ∈ [ m ] × [ k ] that represent positi ve literals by adding their underlying v ariable to A T ∗ ∪ N T ∗ . T o this end, we consider the two random v ariab les K ∗ t ( i, j ) =  1 if π t − 1 ( i, j ) = 1 and Φ ij ∈ A t , 0 other wise, (15) K 0 t ( i, j ) =  1 if π t − 1 ( i, j ) = 1 and Φ ij ∈ N t , 0 other wise, (16) for any ( i, j ) ∈ [ m ] × [ k ] and t ≥ 1 . Recall that π t − 1 ( i, j ) = sign( Φ ij ) iff Φ ij is a literal such that | Φ ij | 6∈ A t − 1 ∪ N t − 1 (cf. PI3 ). T o simplify the notation, we deﬁne for a s et I ⊂ [ m ] × [ k ] K ∗ t ( I ) = Y ( i,j ) ∈I K ∗ t ( i, j ) , K 0 t ( I ) = Y ( i,j ) ∈I K 0 t ( i, j ) . If I ∗ , I 0 ⊂ [ m ] × [ k ] are both non-empty , then K ∗ t ( I ∗ ) · K 0 t ( I 0 ) = 0 . (17) Indeed , suppose that K 0 t ( I 0 ) 6 = 0 . The n PI2 mu st hav e add ed at least one clause to Z t . But the construction in PI2 ensures th at the ﬁrst clause that gets added to Z t contains the v ariable | Φ i t j t | ﬂipped at time t . Thus, A t ⊂ A t − 1 by PI3 , and thus there canno t be a pair ( i, j ) with K ∗ t ( i, j ) = 1 . In effect, K ∗ t ( I ∗ ) = 0 . Lemma 4.1 Let t ≥ 1 a nd ∅ 6 = I ∗ ⊂ [ m ] × [ k ] . Let E ∗ t ( I ∗ ) be the event that | Φ ij | = | Φ i t j t | 6∈ A t − 1 ∪ N t − 1 for all ( i, j ) ∈ I ∗ , and that ( i t , j t ) 6∈ I ∗ . Then P [ E ∗ t ( I ∗ ) |F t − 1 ] ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) |} −|I ∗ | . (18) Pr oof. Since clause Φ i t is unsatisﬁed under σ t − 1 , Φ i t is A t − 1 ∪ N t − 1 -negative and thus π t − 1 ( i t , j t ) 6 = 1 . Hence, PI3 ensures th at either | Φ i t j t | ∈ A t − 1 ∪ N t − 1 or π t − 1 ( i t , j t ) = − 1 . If E ∗ t ( I ∗ ) occurs, then | Φ i t j t | 6∈ A t − 1 ∪ N t − 1 and thus π t − 1 ( i t , j t ) = − 1 . Furth ermore, if I ∗ occurs, then | Φ ij | 6∈ A t − 1 ∪ N t − 1 for all ( i, j ) ∈ I ∗ , and thus π t − 1 ( i, j ) ∈ {− 1 , 1 } by PI3 . Thus, by Fact 3.3 | Φ i t j t | an d | Φ ij | with ( i, j ) ∈ I ∗ are independ ently uniformly distributed over V \ ( A t − 1 ∪ N t − 1 ) . Ther efore, P [ E ∗ t ( I ∗ ) |F t − 1 ] ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) |} −|I ∗ | , as claimed. ✷ Corollary 4.2 F or an y t ≥ 1 , I ∗ ⊂ [ m ] × [ k ] we have E [ K ∗ t ( I ∗ ) |F t − 1 ] ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) |} −|I ∗ | . Pr oof. If Q ( i,j ) ∈I ∗ K ∗ t ( i, j ) = 1 , then the e vent E ∗ t ( I ∗ ) occurs. Hen ce, Lemma 4.1 implies that E   Y ( i,j ) ∈I ∗ K ∗ t ( i, j ) |F t − 1   ≤ P [ E ∗ t ( I ∗ ) |F t − 1 ] ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) |} −|I ∗ | , (19) as claimed. ✷ Lemma 4.3 F or any t ≥ 1 , δ t ≥ 0 an d I 0 ⊂ [ m ] × [ k ] we have E  K 0 t ( I 0 ) · 1 {|Z t \ Z t − 1 | ≤ δ t } |F t − 1  ≤  k δ t max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) | − k δ t }  | I 0 | . 14 Pr oof. W e may assume that I 0 6 = ∅ . W e may also assume th at π t − 1 ( i, j ) = 1 fo r all ( i, j ) ∈ I 0 as otherwise K 0 t ( I 0 ) = 0 . W e are going to work with the conditional distrib utio n p [ · ] = P [ ·|F t − 1 ] . Let E 0 be the e vent that K 0 t ( I 0 ) = 1 and |Z t \ Z t − 1 | ≤ δ t . Then our goal is to estimate p  E 0  . If th e event E 0 occurs, then π t − 1 ( i t , j t ) = − 1 and | Φ i t j t | ∈ N t . In deed, being un satisﬁed unde r the assign ment σ t − 1 , clause Φ i t is A t − 1 ∪ N t − 1 -negative, and thus π t − 1 ( i t , j t ) 6 = 1 . Fu rthermore , if π t − 1 ( i t , j t ) = Φ i t j t , then | Φ i t j t | ∈ A t − 1 ∪ N t − 1 by PI3 , and thus Z t = Z t − 1 and N t = N t − 1 by th e construction in step PI2 . But if N t = N t − 1 , then K 0 t ( I 0 ) = 0 b y deﬁnition. Thus, assume that π t − 1 ( i t , j t ) = − 1 an d | Φ i t j t | ∈ N t . W e need to trace the process descr ibed in PI2 that enhances the sets N t and Z t . Th is process m ay add a sequen ce o f clause in dices to th e set Z t and the variables that occur in th ese c lauses to N t . As these variables get added to the set N t one by on e, we will study the p robability that they occu r in one of the positions ( i, j ) ∈ I 0 . The ﬁrst clause that PI2 ad ds to Z t necessarily contains the newly ﬂipped variable | Φ i t j t | , and thu s we may assume that th is is the ﬁrst variable tha t gets added to N t . In addition, if |Z t \ Z t − 1 | ≤ δ t , PI2 m ay add up to k δ t − 1 furth er variables to N t . T o track this process, we need a bit of notation. Let s 1 , . . . , s y be th e clau se indices that PI2 adds to Z t , in th e ord er in wh ich they get added b y the process. Let y ∗ = min { y , δ t } . F or each 1 ≤ i ≤ y ∗ let 1 ≤ j i, 1 < · · · < j i,l i ≤ k be the unique sequence of indices such that π t − 1 ( s i , j i,q ) = − 1 an d | Φ s i j i,q | 6∈ {| Φ i t j t |} ∪ N t − 1 ∪ i − 1 [ h =1 N ( Φ s h ) ∪  | Φ s i j i,u | : u < q  for all q ≤ l i . This me ans that  | Φ s i j i,q | : 1 ≤ q ≤ l i  are th e new variables that Φ s i contributes to N t and that did not belong to A t − 1 already . Let ξ 0 = | Φ i t j t | and let ξ 1 , . . . , ξ L be the sequence of variables | Φ s i j i,q | with q = 1 , . . . , l i and i = 1 , . . . , y ∗ . Hence, ξ 0 , . . . , ξ L is th e sequence o f variables not in A t − 1 that PI2 adds to N t , in the order in which the process adds these variables to N t . By our choice o f y ∗ , the total n umber of these variables satis ﬁes L + 1 ≤ k y ∗ ≤ k δ t . Of course, L and ξ 0 , . . . , ξ L are random variables. If E 0 occurs, then each of the variables Φ ij with ( i, j ) ∈ I 0 occurs in the sequence ξ 0 , . . . , ξ L . Henc e, there e xists a map f : I 0 → { 0 , 1 , . . . , k δ t − 1 } such that f ( i, j ) ≤ L and Φ ij = ξ f ( i,j ) for all ( i, j ) ∈ I 0 . For a gi ven f let E 0 ( f ) de note this e vent. Then by the union bou nd, p  E 0  ≤ X f : I 0 →{ 0 , 1 ,...,kδ t − 1 } p  E 0 ( f )  ≤ ( k δ t ) |I 0 | max f : I 0 →{ 0 , 1 ,...,kδ t − 1 } p  E 0 ( f )  . (20) W e claim that p  E 0 ( f )  ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) | − k δ t } −|I 0 | (21) for any f . T o p rove (21), let I 0 l = f − 1 ( l ) be the set o f po sitions ( i, j ) ∈ I 0 where the variable ξ l occurs ( 0 ≤ l ≤ L ) . More over , let E 0 l ( f ) be the event that a. Φ ij = ξ l for all ( i, j ) ∈ I 0 l , and b . Φ ij 6 = ξ l for all ( i, j ) ∈ I 0 \ I 0 l . As π t − 1 ( i, j ) = 1 fo r a ll ( i, j ) ∈ I 0 l , given F t − 1 the v ariables Φ ij with ( i, j ) ∈ I 0 l are independ ently unifor mly distributed over V \ ( A t − 1 ∪ N t − 1 ) by Fact 3.3. Henc e, given the e vent T ν εn or |D T ∗ | ≤ 2 2 − k m . Pr oof. Let E be the event that |Z T ∗ | ≤ εn but |D T ∗ | > 2 2 − k m . Our goal is to show that P [ E ] = o (1) . T o this end , we will decom pose E into various ‘sub-events’ that are sufﬁciently de tailed for u s to b ound their probab ilities via Corollary 4.4. In ord er to bound the probability of E we will then use the union bound. As a ﬁrst step, we need to decomp ose E accor ding to the sequence ( |Z t \ Z t − 1 | ) t ≥ 1 of increments of the sets Z t . Mor e precisely , let ∆ be the set of all s equ ences δ = ( δ t ) 1 ≤ t ≤ T ∗ of non-negative integers with P T ∗ t =1 δ t ≤ εn . Let E ( δ ) be the ev ent that |Z t \ Z t − 1 | ≤ δ t for all 1 ≤ t ≤ T ∗ and |D T ∗ | > 2 2 − k m . If the ev ent E occurs, then there is a sequen ce δ such that the event E ( δ ) o ccurs. Hence, by the union bou nd P [ E ] ≤ X δ ∈ ∆ P [ E ( δ )] ≤ | ∆ | · max δ ∈ ∆ P [ E ( δ )] . As it is well known that | ∆ | =  εn + T ∗ − 1 T ∗ − 1  ≤  εn + T ∗ εn  , we obtain P [ E ] ≤  εn + T ∗ εn  max δ ∈ ∆ P [ E ( δ )] . (22) Fixing any sequence δ ∈ ∆ , we now decompose the ev ent E ( δ ) f urther accord ing to the precise set M of cla uses tha t end up in D T ∗ , and accord ing to the precise ‘reason’ why each clause i ∈ M b elongs to D T ∗ . More precisely , let M ⊂ [ m ] be a set of size µ = 2 2 − k m . Moreover , f or disjoint Q ∗ , Q 0 ⊂ M × [ k ] let E 0 ( Q ∗ , Q 0 ) be the e vent that π 0 ( i, j ) = 1 for ( i, j ) ∈ Q ∗ ∪ Q 0 , while π 0 ( i, j ) = − 1 fo r ( i, j ) ∈ M × [ k ] \ ( Q ∗ ∪ Q 0 ) . Furthermo re, for maps τ ∗ : Q ∗ → [ T ∗ ] , τ 0 : Q 0 → [ T ∗ ] let E ( δ, τ ∗ , τ 0 ) be the e vent that |Z t \ Z t − 1 | ≤ δ t for all 1 ≤ t ≤ T ∗ and π τ ∗ ( i,j ) − 1 ( i, j ) = 1 wh ile Φ ij ∈ A τ ∗ ( i,j ) for all ( i, j ) ∈ Q ∗ , π τ 0 ( i,j ) − 1 ( i, j ) = 1 wh ile Φ ij ∈ N τ 0 ( i,j ) for all ( i, j ) ∈ Q 0 . If the ev ent E ( δ ) o ccurs, then ther e exist Q ∗ , Q 0 and τ ∗ , τ 0 such that the events E 0 ( Q ∗ , Q 0 ) and E ( δ, τ ∗ , τ 0 ) o ccur . In fact, if E ( δ ) occu rs, then |D T ∗ | ≥ µ . Th us, select a sub set M ⊂ D T ∗ of size µ . By the deﬁnition of D T ∗ , each i ∈ M is A T ∗ ∪ N T ∗ -negative. Th us, for any j ∈ [ k ] such that Φ ij is 16 a positi ve literal there is a time 1 ≤ t = t ( i, j ) ≤ T ∗ such th at π t − 1 ( i, j ) = 1 but π t ( i, j ) ∈ A t ∪ N t . If π t ( i, j ) ∈ A t ( i,j ) , then includ e ( i, j ) in Q ∗ and set τ ∗ ( i, j ) = t . Otherwise, add ( i, j ) to Q 0 and let τ 0 ( i, j ) = t . Then indeed both E 0 ( Q ∗ , Q 0 ) and E ( δ, τ ∗ , τ 0 ) occur . Thus, by the unio n bound, P [ E ( δ )] ≤ X Q ∗ ,Q 0 ,τ ∗ ,τ 0 P  E 0 ( Q ∗ , Q 0 ) ∩ E ( δ, τ ∗ , τ 0 )  . (23) The event E 0 ( Q ∗ , Q 0 ) depe nds only o n the sign s of the liter als an d is the refore F 0 -measurab le. Fur- thermor e, as signs of the literals Φ ij are mutually independ ent, we get P  E 0 ( Q ∗ , Q 0 )  = 2 − kµ . Therefo re, (23) yields P [ E ( δ )] ≤ 2 − kµ X Q ∗ ,Q 0 ,τ ∗ ,τ 0 P  E ( δ, τ ∗ , τ 0 ) |F 0  . (24) Thus, we are left to estimate P  E ( δ, τ ∗ , τ 0 ) |F 0  . W e deﬁned the random variables K ∗ t ( · , · ) , K 0 t ( · , · ) so that if the e vent E ( δ, τ ∗ , τ 0 ) occurs, then Y ( i,j ) ∈ Q ∗ K τ ∗ ( i,j ) ( i, j ) · Y ( i,j ) ∈ Q 0 K τ 0 ( i,j ) ( i, j ) · T ∗ Y t =1 1 {|Z t \ Z t − 1 | ≤ δ t } = 1 . In order to apply Corollar y 4.4 to th e a bove expression, we are going to reor der th e p roduct according to the time parame ter . More pr ecisely , let Q ∗ t = τ ∗ − 1 ( t ) and Q 0 t = τ 0 − 1 ( t ) . T hen P  E ( δ, τ ∗ , τ 0 ) |F 0  ≤ E   Y ( i,j ) ∈ Q ∗ K τ ∗ ( i,j ) ( i, j ) Y ( i,j ) ∈ Q 0 K τ 0 ( i,j ) ( i, j ) T ∗ Y t =1 1 {|Z t − Z t − 1 | ≤ δ t } = 1 |F 0   = E " T ∗ Y t =1 K t ( Q ∗ t ) K t ( Q 0 t ) · 1 {|Z t \ Z t − 1 | ≤ δ t } = 1 |F 0 # . If |Z t \ Z t − 1 | ≤ δ t for all t ≤ T ∗ , then |N t − 1 | + k δ t ≤ k P s ≤ t δ t ≤ k εn for all t ≤ T ∗ . Further more, |A t | ≤ t ≤ T ∗ = n k for all t ≥ 0 . Hence, | V \ ( A t − 1 ∪ N t − 1 ) | − k δ t ≥ n (1 − kε − 1 / k ) ≥ n/ 1 . 0 1 for all t ≤ T ∗ , provided that k ≥ k 0 is large enough. Thus, Corollary 4.4 entails in combination with Lemma 2.1 P  E ( δ, τ ∗ , τ 0 ) |F 0  =  1 . 01 n  | Q ∗ | · Y ( i,j ) ∈ Q 0 1 . 01 k δ τ 0 ( i,j ) n . (25) For an y M ⊂ [ m ] o f size µ and any two disjoint Q ∗ , Q 0 ⊂ M × [ k ] let S ( M , Q ∗ , Q 0 ) = X τ ∗ ,τ 0  1 . 01 n  | Q ∗ | · Y ( i,j ) ∈ Q 0 1 . 01 k δ τ 0 ( i,j ) n , with the sum ranging over all maps τ ∗ : Q ∗ → [ T ∗ ] , τ 0 : Q 0 → [ T ∗ ] . Recall th at θ = T ∗ /n . As P t ≤ T ∗ δ t ≤ εn , we o btain S ( M , Q ∗ , Q 0 ) ≤  1 . 01 T ∗ n  | Q ∗ |  1 . 01 k n  | Q 0 | X τ 0 Y ( i,j ) ∈ Q 0 δ τ 0 ( i,j ) =  1 . 01 T ∗ n  | Q ∗ |  1 . 01 k n  | Q 0 | T ∗ X t =1 δ t ! | Q 0 | ≤ (1 . 01 θ ) | Q ∗ | (1 . 01 εk ) | Q 0 | . (26) 17 Combining (24), (25), and (26), we thus get for any δ ∈ ∆ P [ E ( δ )] ≤ 2 − kµ X M ⊂ [ m ]: | M | = µ X Q ∗ ,Q 0 ⊂ M × [ k ]: Q ∗ ∩ Q 0 = ∅ S ( M , Q ∗ , Q 0 ) ≤ 2 − kµ  m µ  X q ∗ ,q 0 : q ∗ + q 0 ≤ kµ X Q ∗ ,Q 0 : | Q ∗ | = q ∗ , | Q 0 | = q 0 (1 . 01 θ ) q ∗ (1 . 01 εk ) q 0 ≤ 2 − kµ  m µ  X q ∗ ,q 0 : q ∗ + q 0 ≤ kµ  k µ q ∗ , q 0 , k µ − q ∗ − q 0  (1 . 01 θ ) q ∗ (1 . 01 k ε ) q 0 ≤  m µ   1 + 1 . 01( θ + k ε ) 2  kµ ≤ " e m µ ·  1 + 1 . 01( θ + k ε ) 2  k # µ ≤ " e2 k − 2 ·  1 + 1 . 01( θ + k ε ) 2  k # µ ≤ 0 . 999 µ , (27) provided that k ≥ k 0 is sufﬁciently big. Fin ally , combining (22) and (27), we obtain P [ E ] ≤  εn + T ∗ εn  0 . 999 µ ≤  e( εn + θn ) εn  εn 0 . 999 µ ≤ (e(1 + θ/ε )) εn 0 . 999 µ . (28) By ou r assumption th at ρ ≥ k − 3 (cf. the ﬁrst parag raph in Section 3), we ha ve µ = 2 2 − k m ≥ ρn/k ≥ k − 4 n . Hen ce, recalling that θ ≤ 1 /k and ε = exp( − k 2 / 3 ) (cf. (5)), we obtain from (28) P [ E ] ≤ exp  n  ε ln(2e /ε ) − k − 4  ≤ exp h n  k ex p( − k 2 / 3 ) + k − 4 ln 0 . 999 i = exp( − Ω( n )) = o (1) , provided that k ≥ k 0 is sufﬁciently large. ✷ Finally , Proposition 3.4 is immedia te from Propositions 3.6 and 4.5. 5 Pr oof of Proposition 3.6 Thr oughou t this section we keep the notation and the assumptions of Pr oposition 3.6. 5.1 O utline The g oal in this section is to bound th e size of the set Z T ∗ . Ther e are two reaso ns why step PI2 may a dd a clause index i ∈ [ m ] to the set Z t for some 1 ≤ t ≤ T ∗ . First, the clause Φ i may feature at lea st k 1 variables fro m the set A t − 1 ∪ {| Φ i t j t |} , i.e., v ariab les that ha ve been ﬂipp ed at least once. Secon d, Φ i may contain at lea st λ variables that also o ccur in clauses that were added to Z t previously . The key issue is to deal with the ﬁr st case. Once that is do ne, we can bound the number o f clauses that g et includ ed for the second reason via Lemma 2.4, i.e., via the expansion properties of the random formula. Thus, we need to in vestigate how a clause Φ i comes to co ntain a lot of variables from A t − 1 ∪ {| Φ i t j t |} for some t ≤ T ∗ . There are two ways in which this ma y occur . First, Walksat may have tried to satisfy Φ i ‘actively’ s everal times, i.e., i s = i for several s ≤ t . Second, Φ i may con tain sev eral o f the variables | Φ i s j s | ﬂipped at times s < t ‘accid entally’, i.e., witho ut Walksat try ing to activ ely satisfy i . More precisely , for any t ≥ 0 we call a pair ( i, j ) ∈ [ m ] × [ k ] • t -a ctive if there is 1 ≤ s ≤ t such that ( i, j ) = ( i s , j s ) and π s − 1 ( i, j ) = − 1 . • t -p assive if ther e is 1 ≤ s ≤ t such that ( i, j ) 6 = ( i s , j s ) but | Φ ij | = | Φ i s j s | and π s − 1 ( i, j ) ∈ {− 1 , 1 } . Furthermo re, we say that i ∈ [ m ] is t -active if there are k 2 = k 1 − 10 − 6 k ind ices j such that ( i, j ) is t -active. Similarly , we say that i is t - passive if there are k 3 = 1 0 − 6 k ind ices j such that ( i, j ) is t - passiv e. These d eﬁnitions ensur e that any i ∈ [ m ] for w hich there are at least k 1 indices j ∈ [ k ] such that | Φ ij | ∈ A t − 1 ∪ {| Φ i t j t |} is eith er t -activ e or t -passi ve. 18 T o prove Prop osition 3 .6, we will deal separately with t -acti ve and t -p assi ve clauses. Let A t be the number of t -activ e clauses, and let P t be the number of t -passi ve clauses. Lemma 5.1 F or any 1 ≤ t ≤ T ∗ we have P [ A t < εn/ 4 ∨ |Z t | > εn ] ≥ 1 − 1 /n 2 . W e defer the pro of of Lemma 5.1 to Section 5.2. Lemma 5.2 F or any 1 ≤ t ≤ T ∗ we have P [ P t < εn/ 4 ∨ |Z t − 1 | > εn ] ≥ 1 − 1 /n 2 . Pr oof. As in the proof of Propo sition 4.5, we are going to break the ev ent of interest, i.e., E = { P t ≥ εn/ 4 ∧ |Z t − 1 | ≤ εn } , down into sub-events whose probabilities can be estimated via Lemm a 4.1. Then we will u se the un ion bound to estimate the probab ility of E . For a set M ⊂ [ m ] of µ = εn/ 4 clause ind ices let E ( M ) be the ev ent th at |Z t − 1 | ≤ εn and all i ∈ M are t -passi ve. If E occurs, then there is a set M such that the e vent E ( M ) occur s. Hence, by the unio n bound P [ E ] ≤ X M ⊂ [ m ]: | M | = µ P [ E ( M )] ≤  m µ  max M P [ E ( M )] . (29) Thus, ﬁx a set M ⊂ [ m ] of size µ . Let Q ⊂ M × [ k ] be a set such th at for e ach i ∈ M there are precisely k 3 indices j ∈ [ k ] such that ( i, j ) ∈ Q . Let E ( M , Q ) be th e ev ent that |Z t − 1 | ≤ εn and all p airs ( i, j ) ∈ Q are t -passive. I f the e vent E ( M ) occur s, th en ther e exists a set Q such that E ( M , Q ) occurs. Therefo re, again by the union bound P [ E ( M )] ≤ X Q P [ E ( M , Q )] ≤  k k 3  µ max Q P [ E ( M , Q )] . (30 ) For a map τ : Q → [ t ] let E ( M , Q , τ ) be the event that |Z t − 1 | ≤ εn and τ ( i, j ) = min { s ∈ [ t ] : ( i, j ) is s - passi ve } for all ( i, j ) ∈ Q. If the e vent E ( M , Q ) occu rs, then ther e is a map τ such that th e event E ( M , Q , τ ) occu rs. Con sequently , for any M , Q we h av e P [ E ( M , Q )] ≤ X τ P [ E ( M , Q, τ )] ≤ t | Q | max τ P [ E ( M , Q, τ )] . (31) Combining (29), (30), and (31), we see that P [ E ] ≤  m µ  k k 3  µ t k 3 µ max M ,Q,τ P [ E ( M , Q, τ )] . (32) Hence, ﬁx any M , Q, τ . Let Q s = τ − 1 ( s ) fo r any 1 ≤ s ≤ t , and let E ∗ s ( Q s ) be the event that | Φ ij | = | Φ i t j t | 6∈ A t − 1 ∪ N t − 1 for all ( i, j ) ∈ Q s , an d ( i t , j t ) 6∈ Q s . If E ( M , Q, τ ) occurs, then the ev ents E ∗ s ( Q s ) occu r f or all 1 ≤ s ≤ t . Moreover , the constructio n P I0–PI3 ensure s that |A s | ≤ s , and that |N s − 1 | ≤ k |Z s − 1 | ≤ k ε n fo r all 1 ≤ s ≤ t . Therefor e, Lemma 4.1 implies P [ E ( M , Q, τ )] ≤ P " t \ s =1 E ∗ s ( Q s ) ∩ {|N s − 1 | ≤ k εn } # ≤ t Y s =1 max { 1 , n − s + 1 − k εn } −| Q s | . (33) As s ≤ t ≤ T ∗ ≤ n/k , ε = exp( − k 2 / 3 ) , and because we a re a ssuming th at k ≥ k 0 is sufﬁciently large, we have n − s + 1 − k εn ≥ n/ 1 . 0 01 . Hence, (33) yields P [ E ( M , Q, τ )] ≤ t Y s =1 max { 1 , n − s + 1 − k εn } −| Q s | ≤ (1 . 001 /n ) µk 3 . (34) 19 Finally , combinin g (32) and (34) and recalling that θ = T ∗ /n , we get P [ E ] ≤  m µ  k k 3  µ t k 3 µ (1 . 001 / n ) µk 3 ≤ " e m µ ·  1 . 001e k θ k 3  k 3 # µ ≤ " 4e2 k ρ εk  1 . 001e k θ k 3  k 3 # µ . By our choice of θ we h av e 1 . 001 e k θ ≤ 10 . Hence, we obtain for k ≥ k 0 large enough P [ E ] ≤  4e2 k ρ εk k − k 3 / 2 3  µ ≤ exp( − µ ) = o (1) , thereby completin g the proof. ✷ Pr oof of Pr opo sition 3 .6. In o rder to bound |Z t | for 0 ≤ t ≤ T ∗ , we are go ing to consider a superset Y t ⊃ Z t whose size is easier to estimate. T o d eﬁne Y t , we let Y ∗ t be the set of all i that are either t -activ e or t -passiv e. Now , Y t is the outcome of the following process. Initially , let Y t = Y ∗ t . While there is a clause i ∈ [ m ] \ Y t such that |{ j ∈ [ k ] : | Φ ij | ∈ N ( Φ Y t ) }| ≥ λ, ad d i to Y t . Comparing the above process with the construction in PI2 , we see that indeed Y t ⊃ Z t . (35) Also note that Y t ⊃ Y t − 1 for all t ≥ 1 . T o bound |Y t | , we proceed by inductio n on t . Let Y t be the event that either the ran dom for mula Φ violates the property (3), or |Y t | > εn . W e claim that P [ Y 0 ] = o (1) an d that P [ Y t ] ≤ P [ Y t − 1 ] + 2 n − 2 for all 1 ≤ t ≤ T ∗ . (36) Since trivially Y 0 = ∅ , Y 0 is simply th e e vent that Φ v iolates (3). Hence, L emma 2.4 shows directly that P [ Y 0 ] = o (1) . (37) Now , consider some 1 ≤ t ≤ T ∗ . Lem mas 5.1 (applied to t − 1 ) and Lemma 5.2 (applied to t ) sho w that P [ A t + P t ≤ εn/ 2 ∨ |Z t − 1 | > εn ] ≥ 1 − 2 /n 2 . Furthermo re, if Y t − 1 does n ot occur, then we know that |Z t − 1 | ≤ |Y t − 1 | ≤ εn and that (3 ) is satisﬁed. If in addition A t + P t ≤ εn/ 2 , then (3) ensures that |Y t | ≤ εn , and thus Y t does not occur . Theref ore, P [ Y t ] = P [ Y t − 1 ] + P [ Y t \ Y t − 1 ] ≤ P [ Y t − 1 ] + P h A t + P t > εn 2 ∧ |Z t − 1 | ≤ εn i ≤ P [ Y t − 1 ] + 2 /n 2 . Finally , (36) and (37) yield P [ |Y T ∗ | > εn ] ≤ P [ Y T ∗ ] ≤ P [ Y 0 ] + T ∗ X t =1 2 /n 2 = o (1) + 2 T ∗ /n 2 = o (1) . In combina tion with (35), this implies the assertion. ✷ 5.2 P r oof of Lemma 5.1 How can a clause Φ i become t -active? If this occurs, then Walksat m ust have tried ‘acti vely’ to satisfy Φ i at least k 2 times by ﬂipping one of its variables. But each tim e, the variable that Walksat ﬂipped to satisfy Φ i got ﬂipped again because ﬂipping it rendered another clause unsatisﬁed. More precisely , if Φ i is t -active, th en there exist distinct ‘slots’ j 1 , . . . , j k 2 ∈ [ k ] and times s 1 , . . . , s k 2 ∈ [ t ] such that ( i, j l ) is s l -active for l = 1 , . . . , k 2 . T his means that at the times s l , Walksat actively tried 20 to satisfy Φ i by ﬂipping | Φ ij l | ( l = 1 , . . . , k 2 ). H owe ver , as Walksat had to m ake k 2 attempts, each of the variables | Φ ij l | with l < k 2 must hav e been ﬂipped once more b y time s l +1 . Hence, | Φ ij l | occurs positively in a clause Φ h l that is unsatisﬁed at some time s l < q l < s l +1 . In particular , h l ∈ D q l ⊂ D t . Thus, in o rder to prove Lemma 5.1 we ar e g oing to bou nd th e probab ility that there are at least εn / 4 clauses Φ i that admit j 1 , . . . , j k 2 ∈ [ k ] such that f or each 1 ≤ l < k 2 there is a nother clause Φ h l with the following properties. A1. W e have sign( Φ ij l ) = − 1 , and there is an index j ∈ [ k ] such that sign( Φ h l j ) = 1 and Φ h l j = | Φ ij l | . A2. h l ∈ D t , i.e., Φ h l is A t ∪ N t -negative. In order to d eal with A1 we will need to reﬁne ou r ﬁltration. Gi ven a subset Q ⊂ [ m ] × [ k ] and a map g : Q → [ m ] × [ k ] , we let Ω g be the event that sign( Φ ij ) = − 1 , sign( Φ g ( i,j ) ) = 1 and | Φ ij | =   Φ g ( i,j )   for all ( i, j ) ∈ Q. Since the literals of the random formu la Φ a re independen tly uniformly distrib uted , we see that P [Ω g ] ≤ 2 −| Q ∪ g ( Q ) | n −| Q | . (38) W e consider Ω g as a prob ability space equipped with the uniform distribution (in other words, we are g oing to co ndition on Ω g ). Further, we deﬁne a ﬁltration ( F g,t ) t ≥ 0 on Ω g by letting F g,t = {E ∩ Ω g : E ∈ F t } . In other words, F g,t is the projection of F t onto Ω g . Henc e, Fact 3.2 directly implies the following. Fact 5.3 F or any t ≥ 0 , any x ∈ V , and a ny i ∈ [ m ] the events { σ t ( x ) = 1 } , { Φ i is satisﬁed under σ t } , { x ∈ A t } , { i ∈ Z t } , { x ∈ N t } , and { T = t } ar e F g,t -measurable. Moreover , since the only cond itioning we impose in Ω g concern s the literals Φ ij with ( i, j ) ∈ Q ∪ g ( Q ) , Fact 3.3 yields the follo wing. Fact 5.4 Let t ≥ 0 . Let E t be the set of a ll pairs ( i, j ) ∈ [ m ] × [ k ] \ ( Q ∪ g ( Q )) such that π t ( i, j ) ∈ {− 1 , 1 } . The condition al joint distrib ution of the variables ( | Φ ij | ) ( i,j ) ∈E t given F t,g is uniform over ( V \ ( A t ∪ N t )) E t . That is, for any map f : E t → V \ ( A t ∪ N t ) we have P [ ∀ ( i, j ) ∈ E t : | Φ ij | = f ( i, j ) |F t,g ] = | V \ ( A t ∪ N t ) | −|E t | . Similarly , with respect to the rando m v ariab les K ∗ t ( · , · ) and K 0 t ( · , · ) deﬁned in (15) an d (16), Corol- lary 4.4 implies the following. Corollary 5.5 F or any t ≥ 1 , δ t ≥ 0 an d I ∗ , I 0 ⊂ [ m ] × [ k ] \ ( Q ∪ g ( Q )) we have E  K ∗ t ( I ∗ ) K 0 t ( I 0 ) 1 { |Z t \ Z t − 1 | ≤ δ t } |F g,t − 1  ≤ max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) |} −|I ∗ | ·  k δ t max { 1 , | V \ ( A t − 1 ∪ N t − 1 ) | − k δ t }  | I 0 | . As a further preparatio n, we need the following lemma. Lemma 5.6 Let 1 ≤ t ≤ T ∗ . Set µ = εn/ 4 and let M ⊂ [ m ] b e a set o f size | M | = µ . Fu rthermor e, let Q ⊂ M × [ k ] , let I ⊂ [ m ] be a set of size | I | ≤ | Q | , and let g : Q → I × [ k ] . Let E ( M , Q, I , g ) denote the event that |Z t | ≤ εn an d the following thr ee statements hold. a. F or all ( i, j ) ∈ Q we ha ve sign( Φ ij ) = − 1 , sign( Φ g ( i,j ) ) = 1 , an d Φ g ( i,j ) = | Φ ij | . b . I ⊂ D t . c. F or each i ∈ I ther e is j ∈ [ k ] su ch that ( i, j ) ∈ g ( Q ) . Then P [ E ( M , Q, I , g )] ≤ 2  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | exp(1 . 011 k θ | I | ) . 21 Pr oof. T o estimate P [ E ( M , Q, I , g )] , we ne ed to decompo se the event E ( M , Q , I , g ) into ‘m ore d etailed’ sub-events wh ose pro babilities can be bo unded directly via Coro llary 5.5. T o this end, let I ∗ , I 0 be two dis- joint sub sets of I × [ k ] \ g ( Q ) , and let t ∗ : I ∗ → [ T ∗ ] , t 0 : I 0 → [ T ∗ ] be two maps. L et E ( M , Q, I , g , t ∗ , t 0 ) be the e vent that |Z t | ≤ εn and tha t the following statements are true. a. F or all ( i, j ) ∈ Q we have sign( Φ ij ) = − 1 , sign( Φ g ( i,j ) ) = 1 , and Φ g ( i,j ) = | Φ ij | . b . i. If ( i, j ) ∈ I × [ k ] \ ( g ( Q ) ∪ I ∗ ∪ I 0 ) , then sign( Φ ij ) = − 1 . ii. If ( i, j ) ∈ I ∗ , then sign( Φ ij ) = π t ∗ ( i,j ) − 1 ( i, j ) = 1 an d Φ ij ∈ A t ∗ ( i,j ) . iii. If ( i, j ) ∈ I 0 , then sign( Φ ij ) = π t 0 ( i,j ) − 1 ( i, j ) = 1 an d Φ ij ∈ N t 0 ( i,j ) . c. F or each i ∈ I ther e is j ∈ [ k ] such that ( i, j ) ∈ g ( Q ) . If the event E ( M , Q, I , g ) occu rs, then there exist I ∗ , I 0 , t ∗ , t 0 such that the event E ( M , Q, I , g , t ∗ , t 0 ) occurs. In deed, the deﬁnition of the set D T ∗ is such that if i ∈ D T ∗ , then for any ( i, j ) ∈ I × [ k ] such that sign( Φ ij ) = 1 we have Φ ij ∈ A T ∗ ∪ N T ∗ . Thus, by the union boun d, P [ E ( M , Q, I , g )] ≤ X I ∗ , I 0 X t ∗ ,t 0 P  E ( M , Q, I , g , t ∗ , t 0 )  . (39) Furthermo re, let δ = ( δ 1 , . . . , δ t ) be a sequen ce such that P t s =1 δ s ≤ εn . Let E ( δ , M , Q, I , g , t ∗ , t 0 ) be th e event that |Z s \ Z s − 1 | ≤ δ s for all 1 ≤ s < t and th at E ( M , Q, I , g , t ∗ , t 0 ) occurs. Then b y the union bound , P  E ( M , Q, I , g , t ∗ , t 0 )  ≤ X δ P  E ( δ, M , Q , I , g , t ∗ , t 0 )  ≤  T ∗ + εn εn  max δ P  E ( δ, M , Q , I , g , t ∗ , t 0 )  . (40) The event E ( δ, M , Q, I , g , t ∗ , t 0 ) is sufﬁciently speciﬁc so th at we c an estimate its proba bility easily . Namely , if E ( δ, M , Q , I , g , t ∗ , t 0 ) occurs, then Ω g occurs and Y ( i,j ) ∈I ∗ K ∗ t ∗ ( i,j ) ( i, j ) Y ( i,j ) ∈I 0 K 0 t ∗ ( i,j ) ( i, j ) t Y s =1 1 {|Z s \ Z s − 1 | ≤ δ s } = 1 . (41) T o bound the probability that ( 41) occu rs, we reorder the product by th e time parameter . That is, letting I ∗ s = t ∗ − 1 ( s ) , I 0 s = t 0 − 1 ( s ) , we get P  E ( δ, M , Q , I , g , t ∗ , t 0 ) |F g, 0  ≤ E   Y ( i,j ) ∈I ∗ K t ∗ ( i,j ) ( i, j ) Y ( i,j ) ∈I 0 K t ∗ ( i,j ) ( i, j ) t Y s =1 1 {|Z s \ Z s − 1 | ≤ δ s } = 1 |F g, 0   ≤ E " t Y s =1 K ∗ s ( I ∗ s ) K 0 s ( I 0 s ) 1 {|Z s \ Z s − 1 | ≤ δ s } |F g, 0 # . (42) Since for any s ≤ t ≤ T ∗ we have |A s | ≤ s ≤ T ∗ ≤ n k , and as |N s | ≤ k P s q =1 δ s ≤ k ε n , we see that |A s ∪ N s | + k δ s ≤ 0 . 0 01 n fo r all s ≤ t . Hen ce, (42) and Corollary 5.5 yield P  E ( δ, M , Q , I , g , t ∗ , t 0 ) |F g, 0  ≤ t Y s =1  1 . 01 n  |I ∗ s |  1 . 01 k δ s n  |I 0 s | ≤  1 . 01 n  |I ∗ | + |I 0 | Y ( i,j ) ∈I 0 k δ t 0 ( i,j ) . (43) 22 Furthermo re, if the event E ( δ, M , Q , I , g , t ∗ , t 0 ) o ccurs, then f or all ( i, j ) ∈ I × [ k ] \ ( g ( Q ) ∪ I ∗ ∪ I 0 ) we have sig n( Φ ij ) = − 1 , while sign( Φ ij ) = 1 for all ( i, j ) ∈ I ∗ ∪ I 0 . This event is F 0 ,g -measurab le. Hence, as the signs of the literals Φ ij are independ ently uniformly distributed, we obtain from (43) P  E ( δ, M , Q , I , g , t ∗ , t 0 ) | Ω g  ≤ 2 −| I × [ k ] \ g ( Q ) |  1 . 01 n  |I ∗ ∪I 0 | Y ( i,j ) ∈I 0 k δ t 0 ( i,j ) . (44) Combining (38) and (44), we get P  E ( δ, M , Q , I , g , t ∗ , t 0 )  = P [Ω g ] P  E ( δ, M , Q , I , g , t ∗ , t 0 ) | Ω g  ≤ (2 n ) −| Q | 2 −| I × [ k ] |  1 . 01 n  |I ∗ ∪I 0 | Y ( i,j ) ∈I 0 k δ t 0 ( i,j ) . (45) As ( 39) and (40) show , in ord er to obtain P [ E ( M , Q, I , g )] , we n eed to sum (45) over all possible choices of δ, I ∗ , I 0 , t ∗ , t 0 : P [ E ( M , Q, I , g )] ≤  T ∗ + εn εn  (2 n ) −| Q | X I ∗ , I 0 X t ∗ : I ∗ → [ t ] X t 0 : I 0 → [ t ] P  E ( δ, M , Q , I , g , t ∗ , t 0 )  ≤  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | X I ∗ , I 0  1 . 01 n  |I ∗ ∪I 0 | t |I ∗ | X t 0 : I 0 → [ t ] Y ( i,j ) ∈I 0 k δ t 0 ( i,j ) ≤  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | X I ∗ , I 0  1 . 01 n  |I ∗ ∪I 0 | t |I ∗ | t X s =1 k δ s ! |I 0 | ≤  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | X I ∗ , I 0 (1 . 01 t/n ) |I ∗ | (1 . 01 k ε ) |I 0 | [ as P t s =1 δ s ≤ εn ] ≤  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | (1 + 1 . 01( θ + k ε )) k | I | Hence, P [ E ( M , Q, I , g )] ≤ 2  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | exp(1 . 011 k | I | θ ) . as desired. ✷ Pr oof of Lemma 5.1. Let µ = ε n/ 4 an d ﬁx some 1 ≤ t ≤ T ∗ . Let E be the event that |Z t | ≤ εn and A t ≥ µ . For a set M ⊂ [ m ] of size | M | = µ we let E ( M ) signify the e vent that all clauses i ∈ M are t -active. If E o ccurs, then there is a set M o f size µ such that E ( M ) o ccurs. Hence, by the union boun d P [ E ] ≤ X M ⊂ [ m ]: | M | = µ P [ E ( M )] ≤  m µ  max M P [ E ( M )] . (46) T o bound the e xp ression o n th e r .h .s., ﬁx some set M ⊂ [ m ] of size µ . Let Q ( M ) be the set of all Q ⊂ M × [ k ] such that for each i ∈ M we have | { j ∈ [ k ] : ( i, j ) ∈ Q } | = k 2 − 1 . For a set Q ∈ Q ( M ) let E ( M , Q ) be the event that |Z t | ≤ εn and a. all pairs ( i, j ) ∈ Q are s ( i, j ) -active for some s ( i, j ) ≤ t , and b . for each i ∈ M ther e is j ′ ∈ [ k ] su ch that ( i, j ′ ) is s -activ e at some time s satisfying max j :( i,j ) ∈ Q s ( i, j ) < s ≤ t. 23 If the event E ( M ) o ccurs, then ther e exists Q ∈ Q ( M ) such that E ( M , Q ) occurs. (In f act, if E ( M ) o ccurs, then by the deﬁnition of t -active, fo r any i ∈ M ther e are at least k 2 indices j su ch that ( i, j ) is s -activ e for some s ≤ t . W e can thu s let Q contain th e pairs ( i, j ) for the ‘earliest’ k 2 − 1 such ind ices j .) Hence, by the union bound P [ E ( M )] ≤ X Q ∈Q P [ E ( M , Q )] ≤  k k 2 − 1  µ max Q ∈Q P [ E ( M , Q )] . (47) Now , ﬁx a set M ⊂ [ m ] , | M | = µ , and a set Q ∈ Q ( M ) . I f the e vent E ( M , Q ) occurs, then there exist I , g such that the ev ent E ( M , Q , I , g ) as in Lemma 5.6 occurs. Inde ed, this is precisely what we po inted out in A1 , A2 above. Thus, by the union bound P [ E ( M , Q )] ≤ X I , g P [ E ( M , Q, I , g )] ≤ ( k 2 − 1) µ X ν =1 X I ⊂ [ m ]: | I | = ν X g : Q → I × [ k ] P [ E ( M , Q, I , g )] ≤ ( k 2 − 1) µ X ν =1  m ν  ( k ν ) ( k 2 − 1) µ max I , g : | I | = ν,g : Q → I × [ k ] P [ E ( M , Q, I , g )] . (48) According to Lemma 5.6, P [ E ( M , Q, I , g )] ≤ 2  T ∗ + εn εn  (2 n ) −| Q | 2 −| I × [ k ] | exp(1 . 011 k θν ) . (49) Combining (48) and (49), we obtain P [ E ( M , Q )] ≤ 2  T ∗ + εn εn  (2 n ) −| Q | ( k 2 − 1) µ X ν =1  m ν  ( k ν ) ( k 2 − 1) µ 2 − kν exp(1 . 011 k θν ) ≤ 2  T ∗ + εn εn  (2 n ) −| Q | ( k 2 − 1) µ X ν =1  e m ν 2 k  ν ( k ν ) ( k 2 − 1) µ exp(1 . 011 k θν ) ≤ 2  T ∗ + εn εn  (2 n ) −| Q | ( k 2 − 1) µ X ν =1  e ρn k ν  ν ( k ν ) ( k 2 − 1) µ exp(1 . 011 k θν ) . Since the largest summand is the one with ν = ( k 2 − 1) µ and as | Q | = ( k 2 − 1) µ , we obtain P [ E ( M , Q )] ≤ 2 k µ  T ∗ + εn εn   exp(1 + 1 . 011 k θ ) ρ 2  ( k 2 − 1) µ (50) Let ξ > 0 be such tha t  k k 2 − 1  = (2 ξ ) k 2 − 1 and let ζ = exp(1 + 1 . 0 11 k θ ) . Plugging (50) into (47), we get P [ E ( M )] ≤ 2 k µ  T ∗ + εn εn  k k 2 − 1  µ  ζ ρ 2  ( k 2 − 1) µ ≤ 2 k µ  T ∗ + εn εn  ( ξ ζ ρ ) ( k 2 − 1) µ . (51) Finally , (46) and (51) yield P [ E ] ≤ 2 k µ  T ∗ + εn εn  m µ  ( ζ ξ ρ ) ( k 2 − 1) µ ≤ 2 k µ  T ∗ + εn εn   e m µ ( ζ ξ ρ ) ( k 2 − 1)  µ ≤ 2 k µ  T ∗ + εn εn   4e2 k ρ k ε ( ζ ξ ρ ) ( k 2 − 1)  µ . (52) If ρ ≤ ρ 0 = 1 / 25 , then 4e2 k ρ k ε ( ζ ξ ρ ) ( k 2 − 1) < exp( − k 2 / 100) (53) 24 for k ≥ k 0 large enough. Hence, (52) and (53) yield for k ≥ k 0 large enough P [ E ] ≤ 2 k µ  T ∗ + εn εn  exp( − k 2 µ/ 100) ≤ 2 k µ  e( T ∗ + εn ) εn  εn +1 exp( − k 2 µ/ 100) ≤ 2 k µ  e ( 1 /k + ε ) ε  εn +1 exp( − k 2 µ/ 100) ≤ exp [2 εn − εn ln ε − k 2 µ/ 100 + o ( n )] ≤ exp [ n (2 ε − ε ln ε − k 2 ε/ 400 + o (1))] [by our choice of µ ] ≤ exp [ − nk 2 ε/ 401] = o (1) , [by our choice of ε , cf. (5)] , as desired. ✷ Refer ences [1] D. Achlioptas: Lower bo unds for random 3-SA T via differential eq uations. The oretical Compu ter Science 265 (2001) 159–1 85. [2] D. Achliop tas, C. Moor e: Rand om k -SA T : two moments sufﬁce to cross a sharp threshold. SIAM Journal on Computing 36 (2006 ) 740–76 2. [3] D. Achlioptas, Y . Peres: T he threshold for rand om k -SA T is 2 k ln 2 − O ( k ) . Jou rnal of th e AMS 17 (2004 ) 947–97 3. [4] M. Alek hnovich and E. Ben-Sasson: Lin ear upper bou nds for random walk on small density rando m 3-CNFs. SIAM J. Comput. 36 (2007) 1248–1 263. [5] A. Broder, A. Fr ieze, E. Upfal: On the satisﬁability and maximum satisﬁability of random 3-CNF formu las. Proc. 4th SOD A (1993) 322–3 30. [6] M.-T . Ch ao, J. Franco: Probabilistic analysis of a generalization of the unit-clause literal selection heuristic for the k -satisﬁability problem. Infor m. Sci. 51 (1990) 289–31 4. [7] P . Cheeseman, B. Kanefsky , W . T aylor: Where the really hard pro blems ar e. Proc. IJCAI (1991) 331–3 37. [8] V . Chv ´ atal, B. Reed: Mick g ets some (the odds are on his side). Proc. 33th FOCS (1992) 620–62 7. [9] A. Coja-Ogh lan: A better algorithm for rando m k -SA T . SIAM J . Com puting 39 (2010) 2823–28 64. [10] A. Coja -Oghlan, U. Feige, A. Frieze, M. Krivele vich, D. V ilenchik : On smoothed k -CNF f ormulas and the W alksat algor ithm. Proc. 20th SOD A (200 9) 451–460. [11] A. Frieze, S. Suen : Analysis of two simp le heur istics on a random instance of k -SA T . Journ al of Algorithms 20 (1996 ) 312–3 55. [12] M. Hajiaghayi, G. Sorkin: The satisﬁability threshold of random 3-SA T is at least 3 . 52 . IBM Research Report RC22942 (2003 ). [13] S. Janson, T . Łuczak, A. Ruci ´ nski: Rand om Graphs, W iley 2000. [14] A. Kaporis, L. Kirousis, E. Lalas: T he pro babilistic analysis o f a greedy satisﬁability algorithm. Random Structures and Algorithms 28 (2006 ) 444–48 0. [15] D. Mitchell, B. Selman, H. Le vesque: Hard and easy distrib utio n of SA T problem s. Proc. 10th AAAI (1992 ) 459–46 5. [16] M. Molloy: Cores in r andom hypergra phs and Boolean for mulas. Rand om Str uct. Algor ithms 27 (2005 ) 124–13 5. 25 [17] C. H. Papadimitrio u: On selecting a satisfying truth assignment. Proc. 32nd FOCS (1991) 163–169 . [18] U. Sch ¨ oning: A prob abilistic algorith m for k -SA T and constraint satisfaction pr oblems. Proc. 40 th FOCS (1999) 410–41 4. [19] B. Selman, H. Kau tz, B. Cohen: Loca l search strate gies for satisﬁability testing. In Da vid S. John- son, Michael A. T rick (eds.): Cliques, co loring, and satisﬁability: second DIMA CS implementation challenge, October 11-13, 1993. DIMA CS Series in Dis crete Mathe matics and Theoretical Computer Science 26 (199 6). [20] G. Semerjian , R. Monasson: A study of p ure random w alk on ran dom satisﬁability pro blems with “physical” methods. Proc. 6th SA T (2003 ) 120–13 4. 26

Analyzing Walksat on random formulas

Original Paper

Comments & Academic Discussion

Leave a Comment

Original Paper

Related Papers

Comments & Academic Discussion

Leave a Comment