structured/0710.0805.structured.json

{"paper_meta":{"paper_id":"arxiv:0710.0805","title":"0710.0805","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0710.0805v3 [cs.CC] 31 Aug 2008\nOn the satisfiability threshold and clustering of solutions\nof random 3-Sat formulas\nElitza Maneva∗\nAlistair Sinclair†\nAbstract\nWe study the structure of satisfying assignments of a random 3-Sat formula. In particular,\nwe show that a random formula of density α ≥4.453 almost surely has no non-trivial “core”\nassignments. Core assignments are certain partial assignments that can be extended to satisfying\nassignments, and have been studied recently in connection with the Survey Propagation heuristic\nfor random Sat. Their existence implies the presence of clusters of solutions, and they have\nbeen shown to exist with high probability below the satisfiability threshold for k-Sat with\nk ≥9 [1]. Our result implies that either this does not hold for 3-Sat or the threshold density\nfor satisfiability in 3-Sat lies below 4.453. The main technical tool that we use is a novel simple\napplication of the first moment method.\n1\nIntroduction\nThe study of random instances of 3-Sat has been a major research focus in recent years, both\nbecause of its inherent interest and because it is a natural test case for the wider understanding of\nthe complexity of computational tasks on random inputs. In random 3-Sat the input is a formula\ndrawn uniformly at random from all formulas of fixed density α, i.e., formulas with αn clauses on n\nvariables. Friedgut [7] proved that there exists a function αc(n), known as the satisfiability threshold,\nsuch that for any positive ǫ, random formulas of density αc(n) −ǫ have satisfying assignments with\nhigh probability, and random formulas of density αc(n) + ǫ have no satisfying assignment with\nhigh probability. It is conjectured that αc(n) is a constant (and that its value is about 4.27), but\ncurrently all that is known is that, for large n, 3.520 ≤αc(n) ≤4.506 [11, 8, 6].\nIn the range of densities for which the formula is satisfiable with high probability, the inter-\nesting algorithmic question is whether we can find even one of the many satisfying assignments in\npolynomial time. The lower bound on αc(n) is a result of the analysis of such a polynomial time\nalgorithm [11, 8]. This algorithm belongs to a family of algorithms known as “myopic” because\nthey assign variables greedily one by one in an order that is based only on the number of positive\nand negative occurrences of each variable.\nAn apparently much more powerful algorithm is Survey Propagation [17, 18]. In experiments on\nvery large instances (say, with n = 106 variables) it finds solutions for formulas of densities only just\nbelow the conjectured threshold value α = 4.27; however, a rigorous analysis of its performance is\nstill far from our reach. Like the myopic algorithms it also assigns variables one by one in a greedy\nmanner, but its choices are based on more global information about the role of each variable in\n∗IBM Almaden Research Center, San Jose, CA, enmaneva@us.ibm.com. Some of this research was done while\nthis author was a PhD student at UC Berkeley, supported in part by NSF grant DMS-0528488.\n†Computer Science Division, University of California at Berkeley, sinclair@cs.berkeley.edu. Supported in part by\nNSF grants DMS-0528488 and CCR-0635153.\n1\n\nthe formula. That information is provided by the fixed point of a sophisticated message passing\ndynamics between variables and clauses.\nThe message passing procedure is based on an intriguing (informal) picture of the properties\nof the solution space of a random formula, which is derived from the 1-step Replica Symmetry\nBreaking ansatz of statistical physics [16]. A key postulate of this physical picture is that, for\nformulas of density higher than a certain value (estimated to be 3.92), the space of solutions is split\ninto “clusters.” Within the same cluster it is possible to reach any satisfying assignment from any\nother by flipping one variable at a time, while always keeping the formula satisfied. On the other\nhand, in order to get from a solution in one cluster to a solution in another, the values of a linear\nnumber of variables have to be flipped at the same time. Loosely speaking, the message passing\nprocedure of Survey Propagation was proposed as a way to collect information about the clusters\nof assignments, rather than individual assignments.\nThe remarkable performance of Survey Propagation is a compelling reason to explore properties\nof the solution space of typical formulas as a way to further our understanding of random 3-Sat,\nits hardness, and the satisfiability threshold, and also (looking much further ahead) in order to\nsystematically design algorithms for more general problems with distributional inputs.\nA detailed study of the Survey Propagation algorithm undertaken in [14] and [3] led to an inter-\npretation of the message passing procedure as the more familiar Belief Propagation algorithm [19]\napplied to a particular probability distribution on partial assignments, i.e., assignments of values\nfrom the set {0, 1, ∗}. Here ∗is to be interpreted as “unassigned,” and a variable is allowed to be\nunassigned only if it is not forced to be assigned 0 or 1 in order to satisfy a clause.\nAmong these partial assignments, the set of “core” assignments plays a central role. These are\npartial assignments that are obtained from a satisfying assignment by successively replacing each\nunconstrained variable by ∗. (A variable is unconstrained if changing its value does not make any\nclause unsatisfied.) Any satisfying assignment has a unique corresponding core. Moreover, since\nall assignments in a cluster have the same core, cores can be viewed (informally) as “summaries”\nof clusters. Of course, this view is useful only if different clusters tend to have distinct non-trivial\ncores. (The trivial core assignment is the one without any assigned variables.) Recently, Achlioptas\nand Ricci-Tersenghi [1] showed that, in random k-Sat for k ≥9, for some range of densities up to\nthe satisfiability threshold, with high probability every satisfying assignment has a non-trivial core\nassignment associated with it. This implies that clusters have a large number of frozen variables;\nindeed, for large k the fraction of frozen variables comes arbitrarily close to 1. (The clustering\npicture has also been confirmed for 8-Sat by [15] and [1] using a different method that does not\nsay anything about cores.)\nThe above results hold only for random k-Sat with k ≥8 or 9. In this paper we investigate\nsimilar questions for the apparently harder case of random 3-Sat. Our main result is the following:\nTheorem 1 For random instances of 3-Sat with density greater than 4.453, with high probability\nthere exist no non-trivial core assignments.\nThis theorem requires some interpretation.\nNote first that the density 4.453 lies above the\nconjectured threshold value of 4.27 but below the current best known upper bound of 4.506. Thus\nwe may deduce:\nCorollary 2 One of the following statements holds for random 3-Sat:\n• αc(n) ≤4.453; or\n2\n\n• there is a range of densities immediately below the satisfiability threshold for which with high\nprobability there are no non-trivial core assignments.\nOne interpretation of Theorem 1 is as evidence for an improved upper bound on the threshold of\nthe form αc(n) ≤4.453. On the other hand, if in fact αc(n) > 4.453 then the theorem establishes\na range of densities immediately below the threshold for which with high probability the formula\nis satisfiable but has no non-trivial core assignments. This would represent a surprising difference\nbetween the properties of random 9-Sat and random 3-Sat. Interestingly, experiments with 3-Sat\nand solutions of large formulas found by Survey Propagation do not find cores (see [14]). It is quite\nconceivable that both of the statements in the above corollary hold.\nWe now briefly discuss our proof technique, which involves a novel application of the first mo-\nment method and is, we believe, perhaps more noteworthy than the result itself. A straightforward\napplication of the first moment method to the set of core assignments only allows us to show that\nwith high probability there are no cores of small size or of large size. To handle core assignments\nof intermediate size, it is further necessary to bound the probability that a partial assignment can\nbe extended to a satisfying assignment. This probability is equivalent to the probability that a\nrandom formula with a given density of 2-clauses and 3-clauses is satisfiable. For this sub-problem,\nit is natural to use one of the methods that have been previously introduced for bounding the prob-\nability of satisfiability of random 3-Sat formulas [10, 5, 13, 9, 12, 6]. However, the most powerful\nmethod, from [6], is very heavy numerically, and it does not seem possible to carry it out for the\nwhole range of required densities; on the other hand, simpler methods such as those in [13] are\napparently not powerful enough. Instead we introduce a new method, which we now briefly outline\n(for a detailed description, see Section 3.2).\nTraditionally, bounds on the probability of satisfiability such as those mentioned above are\nobtained by applying the first moment method to a random variable which counts the number of\nsatisfying assignments of a particular kind. Indeed, much of the work on bounding the satisfiability\nthreshold has been directed towards identifying a set of satisfying assignments that is a strict subset\nof the set of all satisfying assignments (so that its expected size is much smaller), but is always\nnon-empty if the formula is satisfiable. The novelty in our approach lies in identifying a new random\nvariable which depends not only on satisfying assignments but also on partial assignments. This\nrandom variable is at least 1 for every satisfiable formula, it exploits the clustering structure of the\nsolution space, and most importantly, it is significantly easier to compute than many alternative\napproaches.\nFinally we note that the proof of our theorem depends on a claim that a particular analytic\nfunction takes only negative values in a given range. We do not provide a complete proof of this\nnumerical claim, but we outline the steps needed for completing the proof, and give very strong\nnumerical evidence that all the corresponding statements hold. The difficulty with making the\nproof completely rigorous is that this would require too much computational effort, which we feel is\nnot justified at this stage given that the bound we obtain is still some distance from the satisfiability\nthreshold. The results presented here should rather be considered as a proof of concept.\nThe remainder of this paper is structured as follows. In Section 2 we give necessary background\nand precise definitions of the various concepts used in the paper. Section 3 is devoted to the proof\nof our main result, Theorem 1. We conclude in Section 4 with some final remarks and suggestions\nfor future work.\n3\n\n2\nTechnical definitions\nLet x1, . . . , xn be a set of n Boolean variables. A literal is either a variable or its negation. A 3-Sat\nformula is a formula in CNF, where each clause is a disjunction of 3 distinct literals (on different\nvariables). For every clause c we will denote the set of variables that appear in the clause by V (c).\nThe distribution that we consider is the following: for a given density α we choose uniformly at\nrandom and with replacement m = ⌊αn⌋clauses out of all possible clauses on 3 distinct variables.\nFor a clause c and a variable xi ∈V (c) we denote by sc,i and uc,i the value for variable xi that is\nrespectively satisfying and unsatisfying for clause c.\nSuppose that the variables x = (x1, . . . , xn) are allowed to take values in {0, 1, ∗}, which we refer\nto as a partial assignment. A variable taking value ∗(star) should be thought of as unassigned.\nDefinition 1 A partial assignment to x is invalid for a clause c if either (a) all variables are\nunsatisfying; or (b) all variables are unsatisfying except for one index j ∈V (c), for which xj = ∗.\nOtherwise, the partial assignment is valid for clause c. We say that a partial assignment is valid\nfor a formula (or just “valid”) if it is valid for all of its clauses.\nFor a valid partial assignment, the subset of variables that are assigned either 0 or 1 values can\nbe divided into constrained and unconstrained variables in the following way:\nDefinition 2 We say that a variable xi is the unique satisfying variable for a clause c if it is\nassigned sc,i whereas all other variables in the clause (i.e., the variables {xj : j ∈V (c)\\{i}}) are\nassigned uc,j. A variable xi is constrained by clause c if it is the unique satisfying variable for c.\nA variable is unconstrained if it has 0 or 1 value, and is not constrained by any clause. Thus for any\npartial assignment σ ∈{0, 1, ∗}n the variables are divided into stars, constrained and unconstrained\nvariables. Let S∗(σ) be the set of unassigned variables, and n∗(σ) and no(σ) denote respectively\nthe number of stars and the number of unconstrained variables. We define the weight of a valid\npartial assignment to be\nW(σ) := ρn∗(σ)(1 −ρ)no(σ),\n(1)\nwhere ρ is a parameter in the interval [0, 1]. The weight of an invalid partial assignment is 0.\nIn [14] the Survey Propagation algorithm is interpreted as a special case of a larger family of\nBelief Propagation algorithms applied to a family of distributions on valid partial assignments. This\nfamily of distributions, parameterized by ρ, is defined as Pr[σ] ∝W(σ). At one extreme, ρ = 0,\nthis becomes just the uniform distribution over (full) satisfying assignments. The other extreme,\nρ = 1, corresponds to Survey Propagation. The pure version of Survey Propagation corresponds to\nsetting ρ = 1. Intermediate values of ρ interpolate between these extremes.\nNext we define a natural partial order (represented by an acyclic directed graph) on valid partial\nassignments. The vertex set of the directed graph G consists of all valid partial assignments. The\nedge set is defined in the following way: for a given pair of valid partial assignments σ and τ, the\ngraph includes a directed edge from σ to τ if there exists an index i ∈{1, . . . , n} such that (i)\nσj = τj for all j ̸= i; and (ii) τi = ∗and σi ̸= ∗.\nValid partial assignments can be separated into levels based on their number of star variables,\ni.e., the assignment σ is in level n∗(σ). Thus every edge goes from an assignment in level l −1 to\none in level l, where 1 ≤l ≤n. G is acyclic and we write τ < σ if there is a directed path in G\nfrom σ to τ. In this case we will also say that the assignment τ is consistent with the assignment\nσ. The outgoing edges of any valid partial assignment σ correspond to its unconstrained variables,\nand therefore its outdegree is equal to no(σ). The minimal assignments in this ordering are the\nassignments without unconstrained variables, i.e., the positive weight assignments for ρ = 1.\n4\n\nDefinition 3 A core of a satisfying assignment σ is a minimal assignment τ such that τ < σ.\nThe following proposition about cores is proved in [14].\nProposition 3 [14] Any satisfying assignment σ has a unique core. Furthermore, if satisfying\nassignments σ1, σ2 ∈{0, 1}n belong to the same cluster of solutions then they have the same core.\nIn the above proposition a cluster is simply a connected component of the graph on solutions,\nin which two solutions are connected by an edge if and only if they are at Hamming distance 1.\nDefinition 4 A cover is a valid partial assignment that contains no unconstrained variables.\nIn particular, the core of any satisfying assignment is a cover. On the other hand not all cover\nassignments are cores (because they may not be extendable to satisfying assignments). We say that\na cover assignment τ is non-trivial if n∗(τ) < n, so that it has at least one assigned variable.\nThe proof of Theorem 1 uses a surprising property of the weights (1), which was observed\nin [14] but was not utilized there. (A more general statement and a connection to a combinatorial\nobject known as “convex geometry” was developed in [2].) Specifically, the total weight of partial\nassignments consistent with a given satisfying assignment is exactly 1. This fact implies that the\nprobability of satisfiability is at most the expected total weight of partial assignments.\nTheorem 4 [14] For every ρ ∈[0, 1], P\nτ≤σ W(τ) = ρn∗(σ) for any valid partial assignment σ ∈\n{0, 1, ∗}n. In particular, if σ is a satisfying assignment then P\nτ≤σ W(τ) = 1.\nIn our proof, we will apply Theorem 4 with various values for ρ ∈(0.8, 1). In each application\nwe will chose the value of ρ to get the best bound possible for the probability that a formula chosen\nfrom some distribution has a satisfying assignment. In particular, if a satisfying assignment exists,\nthe total weight of valid partial assignments is at least 1. Therefore, the probability of satisfiability\nis at most as large as the expected value of the total weight of partial assignments.\nApplying\nthis idea directly to the original distribution on 3-Sat formulas leads to an upper bound on the\nthreshold αc that is weaker than the currently best known bound of 4.506. However, the derivation\nis simpler than other approaches, which makes it possible to apply the same method to bound\nthe probability that a fixed valid partial assignment can be extended to a satisfying assignment.\n(This amounts to applying the method to random formulas coming from a variety of distributions\non formulas with both 2-clauses and 3-clauses.) This allows us to estimate the probability of the\nexistence of a non-trivial core, and thus to prove the theorem.\n3\nProof of the main theorem\nThis section contains a proof of Theorem 1. We begin with an overview of the entire proof, in the\ncourse of which we will state various technical lemmas; these lemmas will be proved in the three\nsubsections that follow.\nNote first that for α ≥4.506 the statement of the main theorem follows from the fact that\nrandom 3-Sat formulas of density at least 4.506 are known to be unsatisfiable with high proba-\nbility [6]. Hence from now on we focus on the case that α ∈[4.453, 4.506]. Our goal is to prove\nthat, for densities above 4.453, with high probability there are no non-trivial covers that can be\nextended to satisfying assignments, i.e., there are no non-trivial cores.\nTo this end, define the size of a cover (or of a core) to be the number of variables assigned value 0\nor 1. The following lemma, proved in [14], establishes that with high probability all non-trivial\ncovers (and consequently cores) are of linear size.\n5\n\nLemma 5 [14] For a random 3-Sat formula of density α, with high probability there are no non-\ntrivial covers of size strictly less than\n1\nαe2 n.\nThis lemma implies that it is sufficient to consider core assignments of size an for a ∈[1/(αe2), 1].\nLet Xa denote the number of cover assignments of size ⌊an⌋, and Ya denote the number of core\nassignments of size ⌊an⌋.\nIn addition to the density α, the size measure a, and the weight parameter ρ ∈[0, 1] from\nTheorem 4, we will use two further parameters, d and b, whose precise definitions will be given\nin the proofs in the following subsections.\nRoughly speaking, d is the fraction of clauses that\nare constraining with respect to a given partial assignment, and b is the fraction of constrained\nvariables with respect to a given partial assignment. The ranges of these parameters are d ∈(a/α, 1]\nand b ∈[0, 1 −a]. We will also make use of a derived parameter r which is defined by d, a and\nα. It appears in connection to the event that the constraining clauses succeed in constraining all\nconstrained variables. Specifically, r is the value satisfying the equation d = ar\nα ln\nr\nr−1. (Note that\nsuch a value r always exists and is unique as the right-hand side is monotonic in r.) In fact, because\nof the form of this expression, it will be more convenient to think of d as being determined by r,\na, and α. Whenever we take the supremum over one of these parameters, we always mean the\nsupremum over its allowed range.\nWe now define two functions that play a central role in our analysis:\nf(α, a, r)\n:=\na ln(2) + H(a) + αH(d) + αd ln\n 3a3\n8\n \n+ α(1 −d) ln\n \n1 −a2(3 −a)\n4\n \n+ αd ln(r/e) −a ln(r −1);\nh(α, a, r, ρ, b)\n:=\nb ln(2) + (1 −a −b) ln(ρ) + (1 −a)H(b/(1 −a))\n−α(1 −d)b b(6 −5b −15a) + 12(1 −a)a\n2(4 −a2(3 −a))\n+ b ln\n \n1 −ρe\n−3α(1−d)\nb(b+2a)\n2(4−a2(3−a))\n \n.\nHere H denotes the entropy function H(x) = −x ln(x) −(1 −x) ln(1 −x).\nThe first ingredient in the proof of Theorem 1 is the following lemma, whose proof is presented\nin Section 3.1.\nLemma 6 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1],\nlim\nn→∞\n1\nn ln (E[Xa]) ≤sup\nr f(α, a, r).\nA simple application of Markov’s inequality immediately yields:\nCorollary 7 If α ≥1 and a ∈[0, 1] are such that, for every r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds\nthat f(α, a, r) < 0, then with high probability random 3-Sat formulas of density α do not have\ncovers of size an.\nThe second ingredient in the proof of Theorem 1 is the following lemma, which is proved in\nSection 3.2.\nLemma 8 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1] and ρ ∈[0, 1),\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr (f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ, b)}).\n6\n\nRemark: This lemma can be strengthened by allowing ρ to depend on r; however, we will not\nneed to use this stronger version.\nSince every core is a cover we know that Ya ≤Xa. Hence (in light of Lemma 6) Lemma 8 is\ninteresting only when supb h(α, a, r, ρ, b) is negative. In fact, we shall prove Lemma 8 by showing\nthat this supremum bounds the logarithm of 1/n times the probability that a particular cover of\nsize a can be extended to a satisfying assignment. (For a precise statement, see Lemma 11 in\nSection 3.2.) An immediate corollary of Lemma 8 is the following.\nCorollary 9 If α ≥1, a ∈[0, 1] and there exists ρ ∈[0, 1) such that for every r > 1 with\nd = ar\nα ln\nr\nr−1 ≤1, and for every b ∈[0, 1 −a], it holds that f(α, a, r) + h(α, a, r, ρ, b) < 0, then with\nhigh probability random 3-Sat formulas of density α do not have cores of size an.\nThe final ingredient in the proof of Theorem 1 is the following numerical claim, whose proof we\ndiscuss in Section 3.3.\nClaim 10 For every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1], and r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it\nholds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nRemark: In Claim 10 we have used the weight ρ ≡ρ(a) = 0.4a+0.7. We arrived at this particular\nchoice of ρ by first optimizing numerically for ρ at 100 values for a ∈[1/4.506, 1], and then fitting\na simple function to the values for ρ that were found. Since Lemma 8 holds for every value of ρ,\nwe may choose any convenient function. Using a simple analytic function guarantees that h is also\nanalytic, which makes the numerical analysis of h easier.\nFinally, combining Claim 10 with Corollary 9 completes the proof of Theorem 1.\n3.1\nThe expected number of covers: Proof of Lemma 6\nLet s = ⌊an⌋. Then we have\nE[Xa]\n=\nE\n \n \nX\nσ∈{0,1,∗}n\nInd [σ is valid ∩(n∗(σ) = n −s) ∩(no(σ) = 0)]\n \n \n=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is valid ∩(no(σ) = 0)]\n=\n n\ns\n \n2s Pr[σ = (0s ∗n−s) is valid and x1, . . . , xs are constrained].\nWe denote by P the probability that σ = (0s ∗n−s) is valid and all of its assigned variables\nare constrained. P is equivalent to the probability of the following event in an experiment with\nm = αn balls thrown uniformly and independently at random into 23 n\n3\n \nbins. There are 3 kinds\nof bins:\n1. Bins of type 1 should be empty. These correspond to clauses that are not allowed:\n• (xi1 ∨xi2 ∨xi3), with i1, i2, i3 ∈{1, 2, . . . , s};\n• (xi1 ∨xi2 ∨ ̄xj), with i1, i2 ∈{1, 2, . . . , s} and j > s;\n• (xi1 ∨xi2 ∨xj), with i1, i2 ∈{1, 2, . . . , s} and j > s.\n7\n\nThe total number of these is\n s\n3\n \n+ 2(n −s)\n s\n2\n \n= n3\n a3\n6 + a2(1 −a)\n \n+ o(n3).\n2. Bins of type 2 correspond to constraining clauses: (xi1 ∨xi2 ∨ ̄xt), with i1, i2, t ∈{1, 2, . . . , s}.\nFor each variable xt there are\n s−1\n2\n \n= n2 a2\n2 +o(n2) clauses that could constrain it and at least\none has to be included. Equivalently, there has to be at least one ball in one of those bins for\nevery xt with t ∈{1, 2, . . . , s}. The total number of these clauses is: s\n s−1\n2\n \n= n3 a3\n2 + o(n3).\n3. There are no constraints for the remaining bins, of type 3. Their total number is\n23\n n\n3\n \n−n3\n a3\n6 + a2(1 −a)\n \n−n3 a3\n2 + o(n3) = n3\n3 (4 −a2(3 −a)) + o(n3).\nSuppose m′ = dm of the clauses we choose are of type 2, and the remaining m −m′ are of\ntype 3. The probability of this event is\npm′ =\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\n.\nThe probability that the m′ clauses of type 2 are such that there is at least one of each kind is\nthe same as the coupon collectors probability of success, with s = ⌊an⌋different coupons, and\nm′ = (dα)n trials.\nWe will use the following general fact from [4], which was previously used\nin a very similar context in [12]: Let q(cN, N) denote the probability of collecting N coupons\nwithin cN trials.\nIf c < 1, q(cN, N) = 0.\nOtherwise, as N goes to infinity q(cN, N) grows\nlike g(c)N, where g(c) =\n r0\ne\n c\n1\nr0−1, and r0 is the solution of r ln\n \nr\nr−1\n \n= c.\nMore precisely,\nlimN→∞1\nN ln (q(cN, N)) = ln(g(c)). We have\nP\n=\nm\nX\nm′=0\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (6 −2a)\n8\n+ o(1)\n m−m′\nq(m′, s)\n≤\nm max\nm′\n( m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\nq(m′, s).\n)\nFinally,\nE[Xa] ≤\n n\ns\n \n2sm max\nm′\n m\nm′\n \n3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3−a)\n4\n+ o(1)\n m−m′\nq(m′, s)\n \n,\nand hence\nlim\nn→∞\n1\nn ln (E[Xa])\n≤\nln\n \n2a\naa(1 −a)1−a\n \n+ sup\nd\n \n \n \n \n \nα ln\n \n \n \n \n3 a3\n8\n d \n1 −a2 (3−a)\n4\n 1−d\ndd (1 −d)1−d\n \n \n + a ln\n \ng\n dα\na\n \n \n \n \n \n \n=\nsup\nd\nf(α, a, r) = sup\nr\nf(α, a, r).\nThis completes the proof of Lemma 6.\n8\n\n3.2\nThe probability that a cover assignment is a core: Proof of Lemma 8\nFor a partial assignment σ, it will be convenient to denote by φσ(xS∗(σ)) the formula that is obtained\nby substituting the variables that have 0/1 assignments in σ, i.e., removing from φ clauses that\nare satisfied by at least one of the assigned variables and removing all remaining appearances of\nassigned variables. This is a formula on n∗(σ) variables. Notice that if σ is a valid assignment\nfor φ then the formula φσ contains no empty clauses and no unit clauses. Furthermore all clauses\nof type 2 (from the previous section) are removed, because they are satisfied by their constrained\nvariables. Among the clauses of type 3, there are clauses that are removed, there are clauses that\nbecome two-variable clauses, and there are clauses that remain untouched. Since there is no simple\nway to describe the resulting distribution on formulas, we will keep referring to the set of all clauses\nof type 3, even the ones that are removed in φσ. When we condition on the fact that σ is a cover\nand m′ of the clauses are of type 2, as in the previous section, we know that the set of clauses we\nare interested in are distributed exactly as a uniform set of (m −m′) clauses of type 3. Thus we\ncan express the expected number of cores as\nE[Ya]\n=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is a cover] × Pr[σ is a core | σ is a cover ]\n=\n n\ns\n \n2s Pr[σ = (0s ∗n−s) is a cover] × Pr[σ = (0s ∗n−s) is a core | σ is a cover]\n=\n n\ns\n \n2s ×\nm\nX\nm′=s\npm′ q(m′, s)\n× Pr[φσ(xs+1, . . . , xn) is satisfiable | σ = (0s ∗n−s), m −m′ clauses are of type 3].\nWe will bound this probability using the Poisson approximation, which is a standard technique\nin this area. There are two related random models. In the first model, which we call the exact\nmodel, (m −m′) clauses are chosen uniformly at random with replacement from all M = n3(4 −\na2(3 −a))/3 + o(n3) clauses of type 3. In the second model, which we call the Poisson model, each\nof the M clauses is included in the formula with probability p = (m −m′)/M −1/(n2√log n) =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2\n \n. The expected number of clauses in both models is the same up to a term\nδ = M/(n2√log n) = Θ(n/√log n) which is sub-linear in n.\nThe Poisson model has been studied before in the context of random 3-Sat. It can be shown,\nas the example below demonstrates, that whenever a property holds with high probability in the\nexact model, it also holds with high probability in the corresponding Poisson model. Applying first\nmoment techniques to the Poisson model is usually easier, because the clauses are independently\nchosen; however the bounds obtained are usually weaker.\nNext, we relate the probability that φσ is satisfiable under the two models. Let Prp denote\nprobability in the Poisson model, and Pre denote probability in the exact model. Let the random\nvariable J denote the number of distinct clauses included in the formula. Then\nPrp[φσ is satisfiable]\n=\nM\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i]\n≥\nm−m′−δ\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i].\n9\n\nSince this conditional probability decreases as i increases, and Prp[J ≤Ep[J]] ≥1/2, where Ep[J] =\nm −m′ −δ, we have\nPrp[φσ is satisfiable]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] ×\nm−m′−δ\nX\ni=0\nPr[J = i]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] × 1\n2.\nOn the other hand, for the exact model\nPre[φσ is satisfiable]\n=\nm−m′\nX\ni=0\nPre[J = i] × Pr[φσ is satisfiable | J = i]\n≤\n m−m′−δ\nX\ni=0\nPre[J = i]\n!\n+ Pr[φσ is satisfiable | J = m −m′ −δ]\n≤\nPre[at least δ clauses repeated] + 2 Prp[φσ is satisfiable]\n≤\n m −m′\nδ\n m −m′\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n≤\n (m −m′)2\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n=\nθ(2−n√log n) + 2 Prp[φσ is satisfiable].\nThus if the probability of satisfiability in the Poisson model is bounded above by c−n for some\nconstant c, then limn→∞1\nn ln (Pre[ φσ is satisfiable]) ≤c. Therefore, it suffices to get a bound for\nthe Poisson model. Any of the first moment techniques for bounding the satisfiability threshold of 3-\nSat can be adapted to bound the probability that φσ is satisfiable. Of the ones that are technically\napplicable (i.e., result in an explicit analytic expression for every setting of the parameters s and\nm′), we obtain the strongest result using the novel approach of applying the first moment method\nto the distribution on partial assignments defined by the weights in equation (1).\nLemma 11 In the Poisson model with parameters n, m = αn, m′ = dm, s = ⌊an⌋and for\nσ = (0s ∗n−s), and r such that d = ar\nα ln\nr\nr−1,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤\ninf\nρ∈[0,1)\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b).\nProof: We will apply Theorem 4 to the formula φσ(xs+1, . . . , xn). The theorem implies that if φσ\nhas a satisfying assignment then P\nτ∈V W(τ) ≥1 where W(τ) = ρn∗(τ)(1 −ρ)no(τ) and the sum is\nover the set V of all partial assignments τ ∈{0, 1, ∗}n−s that are valid for φσ. Thus the probability\nof satisfiability is bounded from above by the expected value of P\nτ∈V W(τ). This holds for any\nvalue of ρ ∈[0, 1]. We bound this expectation.\nFor any t ∈{0, 1, . . . , n −s} let Zt denote the sum of the weights of valid assignments τ for\nφσ(xs+1, . . . , xn) such that n∗(τ) = n −s −t. Then\nE[Z] =\nn−s\nX\nt=0\nE[Zt] ≤n\nmax\nt∈{0,1,...,n−s} E[Zt].\n10\n\nE[Zt]\n=\nt\nX\nu=0\nρn−s−t(1 −ρ)u\nX\nτ∈{0,1,∗}n−s\nPrp [τ is valid ∩(n∗(τ) = n −s −t) ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\nX\nτ∈{0,1,∗}n−s : n∗(τ)=n−s−t\nPrp[τ is valid ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\n n −s\nt\n t\nu\n \n2t\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\nNext we derive the probability that the assignment (xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid\nand the first u variables are unconstrained. Recall that φσ is obtained from φ by setting its first s\nvariables according to σ. Only clauses of type 3 influence φσ and according to the Poisson model,\neach of them is included independently with probability p. The probability that the assignment\n(xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid and the first u variables are unconstrained is equivalent\nto the probability that among the included clauses of type 3, there are no clauses of the following\nkinds:\n• (xi1 ∨xi2 ∨xi3), with i1 ∈{1, 2, . . . , s + t}, i2, i3 ∈{s + 1, s + 2, . . . , s + t},\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s+t}, i2 ∈{s+1, s+2, . . . , s+t} and j ∈{s+1, . . . , s+u},\nand for every j ∈{s + u + 1, . . . , s + t}, the formula contains a clause (xi1 ∨xi2 ∨ ̄xj), with\ni1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, . . . , s + t}.\nIn the Poisson model all clauses are independent, so it is easy to put these events together to\nobtain:\nQ\n=\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\n=\n(1 −p)((t\n3)+s(t\n2)) + 2(n−s−t)((t\n2) + st) + u((t\n2)+st) ×\n \n1 −(1 −p)(t\n2)+st t−u\nThe expression for the expectation can be simplified:\nE[Zt]\n=\nρn−s−t\n n −s\nt\n \n2t\nt\nX\nu=0\n(1 −ρ)u\n t\nu\n \nQ\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×\nt\nX\nu=0\n t\nu\n \n(1 −ρ)u (1 −p)u((t\n2)+st) ×\n \n1 −(1 −p)(t\n2)+st t−u\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×\n \n(1 −ρ)(1 −p)(t\n2)+st +\n \n1 −(1 −p)(t\n2)+st t\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n2)(6n−5t−3s−2)/3+2(n−s−t)st \n1 −ρ(1 −p)(t\n2)+st t\n11\n\nLet t = bn, and recall that s = ⌊an⌋, p =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2\n \n. Then\nlim\nn→∞\n1\nn ln (E[Zt])\n=\nln\n (1 −a)1−a 2b ρ1−a−b\nbb (1 −a −b)1−a−b\n \n−α (1 −d) b (b(6 −5b −3a) + 12(1 −a −b)a)\n2(4 −a2(3 −a))\n+ b ln\n \n1 −ρe\n−3α(1−d)b(b+2a)\n2(4−a2(3−a))\n \n=\nh(α, a, r, ρ, b).\nFinally,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤lim\nn→∞\n1\nn ln (E[Z]) ≤\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b),\nwhich is the statement of Lemma 11.\nSubstituting the bound of Lemma 11 into the expression for the expectation yields\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr\n(f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ(a, r), b)}),\ncompleting the proof of Lemma 8.\n3.3\nNumerical analysis: Steps to the proof of Claim 10\nWhat remains is to verify the numerical claim that for every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1],\nand r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a],\nf(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0. We outline the steps towards a proof.\nThere are two stages. First, we identify the (a, r) pairs such that for every α ∈[4.453, 4.506] it\nholds that f(α, a, r) < 0. Second, for the remaining range of values of (a, r) we show that for every\nb ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nThe derivative of f with respect to r is\n∂f\n∂r = a\n \nln\n \n1 +\n1\nr −1\n \n−\n1\nr −1\n \n× ln\n \n \n3\n \nα −ar ln\n \nr\nr−1\n \n2\n 4\na2 −3 + a\n \nln\n \nr\nr−1\n \n \n .\nSince the first factor is always negative the derivative is 0 when\n3α = ln\n \nr\nr −1\n 8\na2 −6 + 2a + 3ar\n \n.\n(2)\nThis equation has a unique root for every a ∈[1/(4.506e2), 1], α ∈[4.453, 4.506] (the derivative of\nthe right-hand side is negative). Therefore we can conclude that for every a and α in the above\nrange, f is at first increasing with r then decreasing.\nIts maximum is achieved at the root of\nequation (2).\nFigure 1 shows the values for r where f is maximized, with respect to a, when\nα = 4.453.\nSince the root of (2) is monotone decreasing with respect to a and α, we can find values r1\nand r2 such that for every a and α in the range, if r < r1 then ∂f/∂r > 0, and if r > r2 then\n∂f/∂r < 0. Values satisfying this condition are r1 = 1.2 and r2 = 670. Thus if we show that\nf(α, a, r) is negative for r = r1 and r = r2 then it is negative for any r outside the range (1.2, 670).\n12\n\n5\n 10\n 15\n 20\n 25\n 0\n 0.2\n 0.4\n 0.6\n 0.8\n 1\nr\na\narg maxr f (4.453, a, r)\nFigure 1: The pairs (a, r) such that f(4.453, a, r) > −0.0001, and the value of r that maximizes\nf(4.453, a, r) for every a.\nWe will use the shorthand notation q := r ln\nr\nr−1. For r ∈(1.2, 670) we have q ∈(1.0007, 2.16).\nThe whole range satisfies the condition that d = ar\nα r ln\nr\nr−1 ≤1.\nNext we take care of the boundary region with respect to a. Notice that the derivative of f is\nnegative for a large enough (close to 1), because of the entropy term in f. The derivative of f with\nrespect to a is\n∂f\n∂a\n=\nln(2) −ln\na\n1 −a −q ln\n3a2(α −aq)\n2q(4 −a2(3 −a)) + 3q −3a(α −aq)(2 −a)\n4 −a2(3 −a)\n+ q ln(r) −ln(r −1).\nThe following observations are helpful for bounding this derivative:\n• 3q + q ln(r) −ln(r −1) is maximized in the interval r ∈[1.2, 670] at r = 1.2.\n• a2(α −aq)/(4 −a2(3 −a)) is an increasing function of a.\n• a(α −aq)(2 −a)/(4 −a2(3 −a)) is an increasing function of a.\nFor a > 0.999, using the above facts, the derivative is negative. Thus if we show that f is\nnegative for a = 0.999, r ∈(1.2, 670), α ∈[4.453, 4.506] then f is negative also for every a > 0.999.\nWe are left with the region a ∈[1/(4.506e2), 0.999], r ∈(1.2, 670).\nIn this region all the\nderivatives of f can be bounded, and a sufficiently fine grid can be chosen over which to evaluate\nf in order to identify the grid sections where the function can take positive values. In particular,\nthe derivative with respect to a is at most 28.2, and with respect to α it is at most 1. Furthermore,\nsince we know that for every a and α, f is maximized as a function of r at the root of equation (2),\none can find this maximum and the range of r where f(α, a, r) is positive using binary search. The\npoints with f(4.453, a, r) > −0.0001 are depicted in Figure 1.\nIn the second stage we need to analyze f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) for the remaining\nregion of values for (a, r).\n13\n\nFirst, notice that we can take care of the boundary regions with respect to b by taking advantage\nof the entropy term, which has very large slope for b close to 0, and very steep negative slope for b\nclose to 1 −a. Specifically, the derivative with respect to b is\n∂h\n∂b\n=\nln(2) −ln(ρ) −ln\n \nb\n1 −a −b\n \n−α(1 −d)12b −15b2 −30ab + 12a −12a2\n2(4 −a2(3 −a))\n+ ln(1 −ρe−A) + 3α(1 −d)b\nρe−A(a + b)\n(1 −ρe−A)(4 −a2(3 −a))\nwhere A = 3α(1−d)b(b+2a)\n2(4−a2(3−a)) .\nUsing the bounds on all parameters: α ∈[4.453, 4.506], a ∈(0.28, 0.7), r ∈(1.5, 14.3), and the\nones that follow from those: d ∈(0.06, 0.26), A ∈[0, 1.59], ρ ∈(0.02, 0.21), we can conclude that\n−3.04 −ln\n \nb\n1 −a −b\n \n< ∂h\n∂b < 5.38 −ln\n \nb\n1 −a −b\n \n.\nThe lower bound can be made positive by setting b/(1−a −b) ≤0.04 and the upper bound can\nbe made negative by setting b/(1 −a −b) ≥220. Therefore it suffices to show that h is negative\nfor b in the range [0.01, 0.996(1 −a)].\nAgain, using the bounds for the parameters, all the derivatives of f + h can be bounded, and a\nsufficiently fine grid can be chosen over which to evaluate f +h. We did not perform the evaluation\non a grid that is as fine as is required for the rigorous proof because, based on the current bounds on\nthe derivatives, we would need to evaluate the function at more than 1010 points. (As we discussed\nin the introduction, we feel that the computational effort is not yet justified.)\nFor illustration, Figure 2 shows the estimated values of supr f(α, a, r) and of supr,b(f(α, a, r) +\nh(α, a, r, ρ(a), b)) for α = 4.453, 4.470, 4.490, and 4.506. The maximum is taken over evaluations\nat a grid of step size 0.001 for the parameters r and b. We estimated with better precision the\nlocation of the maximum of f + h by evaluating the function at a fine grid in the region where\nthe evaluations on the coarse grid give the largest values. The maximum found in this way is at\nα = 4.453, a = 0.62566, b = 0.03568 and r = 2.00134 (d = 0.19473). The value of f + h at this\npoint is −0.000058.\n4\nConcluding remarks\nAs mentioned in the introduction, all of the known rigorous upper bounds for the satisfiability\nthreshold of 3-Sat are based on the first moment method [10, 5, 13, 9, 12, 6]; the corresponding\nupper bounds in this sequence of results are: 4.758, 4.64, 4.601, 4.596, 4.571, 4.506. The general\nmethod is to consider a random variable Z that is equal to the number of satisfying assignments\nof a particular kind. These satisfying assignments are such that at least one exists if the formula is\nsatisfiable. Showing that above a certain density E[Z] →0 thus implies (by Markov’s inequality)\nthat Pr[Z = 0] →1, and consequently the probability that the formula is unsatisfiable also goes\nto 1. For example, [5] takes Z to be the number of negatively prime solutions, i.e., solutions for\nwhich every variable assigned 1 is constrained.\nHere, we have used the same idea but with Z taken to be the total weight of partial assignments\nunder the weight function (1) inspired by Survey Propagation. Given the dramatic success of Survey\nPropagation for random 3-Sat, it seems plausible that this approach can potentially yield rather\ntight upper bounds on the threshold. We were able to achieve only partial progress in this direction,\nbut we are quite hopeful that extensions of our approach could lead to further progress.\n14\n\n-0.08\n-0.06\n-0.04\n-0.02\n 0\n 0.02\n 0.1\n 0.2\n 0.3\n 0.4\n 0.5\n 0.6\n 0.7\n 0.8\n 0.9\n 1\nsupr f, supr,b (f+h)\na\nα = 4.453\nα = 4.470\nα = 4.490\nα = 4.506\nFigure 2: The values of supr f(α, a, r) and supr,b(f(α, a, r) + h(α, a, r, 0.4a + 0.7, b)) for α = 4.453,\n4.470, 4.490, and 4.506.\nOne natural extension would be to use a different weight ρ for each variable, depending for\nexample on the number of positive and negative occurrences of the variable. The corresponding\ngeneralization of Theorem 4 is proved in [2]. It is quite possible that the value of E[Z] in this case\nis significantly smaller.\n5\nAcknowledgments\nWe would like to thank the anonymous referees for their detailed and very useful suggestions.\nReferences\n[1] D. Achlioptas and F. Ricci-Tersenghi. 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Kirousis, E. Kranakis, D. Krizanc, and Y. C. Stamatiou. Approximating the unsatisfi-\nability threshold of random formulas. Random Struct. Algorithms, 12(3):253–269, 1998.\n[14] E. Maneva, E. Mossel, and M. J. Wainwright. A new perspective on survey propagation. J.\nACM, 54(4):2–41, 2007.\n[15] M. M ́ezard, T. Mora, and R. Zecchina. Clustering of solutions in the random satisfiability\nproblem. Phys. Rev. Lett., 94(197205), 2005.\n[16] M. M ́ezard, G. Parisi, and M. A. Virasoro. Spin Glass Theory and Beyond. World Scientific,\nSingapore, 1987.\n[17] M. M ́ezard, G. Parisi, and R. Zecchina. Analytic and algorithmic solution of random satis-\nfiability problems. Science, 297, 812, 2002. (Scienceexpress published online June 27, 2002;\n10.1126/science.1073287).\n[18] M. M ́ezard and R. Zecchina. Random k-satisfiability: from an analytic solution to an efficient\nalgorithm. Phys. Rev. E, 66, 2002.\n[19] J. Pearl. Probabilistic reasoning in intelligent systems: networks of plausible inference. Morgan\nKaufmann, Palo Alto, CA, 1988.\n16","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0710.0805v3 [cs.CC] 31 Aug 2008\nOn the satisfiability threshold and clustering of solutions\nof random 3-Sat formulas\nElitza Maneva∗\nAlistair Sinclair†\nAbstract\nWe study the structure of satisfying assignments of a random 3-Sat formula. In particular,\nwe show that a random formula of density α ≥4.453 almost surely has no non-trivial “core”\nassignments. Core assignments are certain partial assignments that can be extended to satisfying\nassignments, and have been studied recently in connection with the Survey Propagation heuristic\nfor random Sat. Their existence implies the presence of clusters of solutions, and they have\nbeen shown to exist with high probability below the satisfiability threshold for k-Sat with\nk ≥9 [1]. Our result implies that either this does not hold for 3-Sat or the threshold density\nfor satisfiability in 3-Sat lies below 4.453. The main technical tool that we use is a novel simple\napplication of the first moment method.\n1\nIntroduction\nThe study of random instances of 3-Sat has been a major research focus in recent years, both\nbecause of its inherent interest and because it is a natural test case for the wider understanding of\nthe complexity of computational tasks on random inputs. In random 3-Sat the input is a formula\ndrawn uniformly at random from all formulas of fixed density α, i.e., formulas with αn clauses on n\nvariables. Friedgut [7] proved that there exists a function αc(n), known as the satisfiability threshold,\nsuch that for any positive ǫ, random formulas of density αc(n) −ǫ have satisfying assignments with\nhigh probability, and random formulas of density αc(n) + ǫ have no satisfying assignment with\nhigh probability. It is conjectured that αc(n) is a constant (and that its value is about 4.27), but\ncurrently all that is known is that, for large n, 3.520 ≤αc(n) ≤4.506 [11, 8, 6].\nIn the range of densities for which the formula is satisfiable with high probability, the inter-\nesting algorithmic question is whether we can find even one of the many satisfying assignments in\npolynomial time. The lower bound on αc(n) is a result of the analysis of such a polynomial time\nalgorithm [11, 8]. This algorithm belongs to a family of algorithms known as “myopic” because\nthey assign variables greedily one by one in an order that is based only on the number of positive\nand negative occurrences of each variable.\nAn apparently much more powerful algorithm is Survey Propagation [17, 18]. In experiments on\nvery large instances (say, with n = 106 variables) it finds solutions for formulas of densities only just\nbelow the conjectured threshold value α = 4.27; however, a rigorous analysis of its performance is\nstill far from our reach. Like the myopic algorithms it also assigns variables one by one in a greedy\nmanner, but its choices are based on more global information about the role of each variable in\n∗IBM Almaden Research Center, San Jose, CA, enmaneva@us.ibm.com. Some of this research was done while\nthis author was a PhD student at UC Berkeley, supported in part by NSF grant DMS-0528488.\n†Computer Science Division, University of California at Berkeley, sinclair@cs.berkeley.edu. Supported in part by\nNSF grants DMS-0528488 and CCR-0635153.\n1"},{"paragraph_id":"p2","order":2,"text":"the formula. That information is provided by the fixed point of a sophisticated message passing\ndynamics between variables and clauses.\nThe message passing procedure is based on an intriguing (informal) picture of the properties\nof the solution space of a random formula, which is derived from the 1-step Replica Symmetry\nBreaking ansatz of statistical physics [16]. A key postulate of this physical picture is that, for\nformulas of density higher than a certain value (estimated to be 3.92), the space of solutions is split\ninto “clusters.” Within the same cluster it is possible to reach any satisfying assignment from any\nother by flipping one variable at a time, while always keeping the formula satisfied. On the other\nhand, in order to get from a solution in one cluster to a solution in another, the values of a linear\nnumber of variables have to be flipped at the same time. Loosely speaking, the message passing\nprocedure of Survey Propagation was proposed as a way to collect information about the clusters\nof assignments, rather than individual assignments.\nThe remarkable performance of Survey Propagation is a compelling reason to explore properties\nof the solution space of typical formulas as a way to further our understanding of random 3-Sat,\nits hardness, and the satisfiability threshold, and also (looking much further ahead) in order to\nsystematically design algorithms for more general problems with distributional inputs.\nA detailed study of the Survey Propagation algorithm undertaken in [14] and [3] led to an inter-\npretation of the message passing procedure as the more familiar Belief Propagation algorithm [19]\napplied to a particular probability distribution on partial assignments, i.e., assignments of values\nfrom the set {0, 1, ∗}. Here ∗is to be interpreted as “unassigned,” and a variable is allowed to be\nunassigned only if it is not forced to be assigned 0 or 1 in order to satisfy a clause.\nAmong these partial assignments, the set of “core” assignments plays a central role. These are\npartial assignments that are obtained from a satisfying assignment by successively replacing each\nunconstrained variable by ∗. (A variable is unconstrained if changing its value does not make any\nclause unsatisfied.) Any satisfying assignment has a unique corresponding core. Moreover, since\nall assignments in a cluster have the same core, cores can be viewed (informally) as “summaries”\nof clusters. Of course, this view is useful only if different clusters tend to have distinct non-trivial\ncores. (The trivial core assignment is the one without any assigned variables.) Recently, Achlioptas\nand Ricci-Tersenghi [1] showed that, in random k-Sat for k ≥9, for some range of densities up to\nthe satisfiability threshold, with high probability every satisfying assignment has a non-trivial core\nassignment associated with it. This implies that clusters have a large number of frozen variables;\nindeed, for large k the fraction of frozen variables comes arbitrarily close to 1. (The clustering\npicture has also been confirmed for 8-Sat by [15] and [1] using a different method that does not\nsay anything about cores.)\nThe above results hold only for random k-Sat with k ≥8 or 9. In this paper we investigate\nsimilar questions for the apparently harder case of random 3-Sat. Our main result is the following:\nTheorem 1 For random instances of 3-Sat with density greater than 4.453, with high probability\nthere exist no non-trivial core assignments.\nThis theorem requires some interpretation.\nNote first that the density 4.453 lies above the\nconjectured threshold value of 4.27 but below the current best known upper bound of 4.506. Thus\nwe may deduce:\nCorollary 2 One of the following statements holds for random 3-Sat:\n• αc(n) ≤4.453; or\n2"},{"paragraph_id":"p3","order":3,"text":"• there is a range of densities immediately below the satisfiability threshold for which with high\nprobability there are no non-trivial core assignments.\nOne interpretation of Theorem 1 is as evidence for an improved upper bound on the threshold of\nthe form αc(n) ≤4.453. On the other hand, if in fact αc(n) > 4.453 then the theorem establishes\na range of densities immediately below the threshold for which with high probability the formula\nis satisfiable but has no non-trivial core assignments. This would represent a surprising difference\nbetween the properties of random 9-Sat and random 3-Sat. Interestingly, experiments with 3-Sat\nand solutions of large formulas found by Survey Propagation do not find cores (see [14]). It is quite\nconceivable that both of the statements in the above corollary hold.\nWe now briefly discuss our proof technique, which involves a novel application of the first mo-\nment method and is, we believe, perhaps more noteworthy than the result itself. A straightforward\napplication of the first moment method to the set of core assignments only allows us to show that\nwith high probability there are no cores of small size or of large size. To handle core assignments\nof intermediate size, it is further necessary to bound the probability that a partial assignment can\nbe extended to a satisfying assignment. This probability is equivalent to the probability that a\nrandom formula with a given density of 2-clauses and 3-clauses is satisfiable. For this sub-problem,\nit is natural to use one of the methods that have been previously introduced for bounding the prob-\nability of satisfiability of random 3-Sat formulas [10, 5, 13, 9, 12, 6]. However, the most powerful\nmethod, from [6], is very heavy numerically, and it does not seem possible to carry it out for the\nwhole range of required densities; on the other hand, simpler methods such as those in [13] are\napparently not powerful enough. Instead we introduce a new method, which we now briefly outline\n(for a detailed description, see Section 3.2).\nTraditionally, bounds on the probability of satisfiability such as those mentioned above are\nobtained by applying the first moment method to a random variable which counts the number of\nsatisfying assignments of a particular kind. Indeed, much of the work on bounding the satisfiability\nthreshold has been directed towards identifying a set of satisfying assignments that is a strict subset\nof the set of all satisfying assignments (so that its expected size is much smaller), but is always\nnon-empty if the formula is satisfiable. The novelty in our approach lies in identifying a new random\nvariable which depends not only on satisfying assignments but also on partial assignments. This\nrandom variable is at least 1 for every satisfiable formula, it exploits the clustering structure of the\nsolution space, and most importantly, it is significantly easier to compute than many alternative\napproaches.\nFinally we note that the proof of our theorem depends on a claim that a particular analytic\nfunction takes only negative values in a given range. We do not provide a complete proof of this\nnumerical claim, but we outline the steps needed for completing the proof, and give very strong\nnumerical evidence that all the corresponding statements hold. The difficulty with making the\nproof completely rigorous is that this would require too much computational effort, which we feel is\nnot justified at this stage given that the bound we obtain is still some distance from the satisfiability\nthreshold. The results presented here should rather be considered as a proof of concept.\nThe remainder of this paper is structured as follows. In Section 2 we give necessary background\nand precise definitions of the various concepts used in the paper. Section 3 is devoted to the proof\nof our main result, Theorem 1. We conclude in Section 4 with some final remarks and suggestions\nfor future work.\n3"},{"paragraph_id":"p4","order":4,"text":"2\nTechnical definitions\nLet x1, . . . , xn be a set of n Boolean variables. A literal is either a variable or its negation. A 3-Sat\nformula is a formula in CNF, where each clause is a disjunction of 3 distinct literals (on different\nvariables). For every clause c we will denote the set of variables that appear in the clause by V (c).\nThe distribution that we consider is the following: for a given density α we choose uniformly at\nrandom and with replacement m = ⌊αn⌋clauses out of all possible clauses on 3 distinct variables.\nFor a clause c and a variable xi ∈V (c) we denote by sc,i and uc,i the value for variable xi that is\nrespectively satisfying and unsatisfying for clause c.\nSuppose that the variables x = (x1, . . . , xn) are allowed to take values in {0, 1, ∗}, which we refer\nto as a partial assignment. A variable taking value ∗(star) should be thought of as unassigned.\nDefinition 1 A partial assignment to x is invalid for a clause c if either (a) all variables are\nunsatisfying; or (b) all variables are unsatisfying except for one index j ∈V (c), for which xj = ∗.\nOtherwise, the partial assignment is valid for clause c. We say that a partial assignment is valid\nfor a formula (or just “valid”) if it is valid for all of its clauses.\nFor a valid partial assignment, the subset of variables that are assigned either 0 or 1 values can\nbe divided into constrained and unconstrained variables in the following way:\nDefinition 2 We say that a variable xi is the unique satisfying variable for a clause c if it is\nassigned sc,i whereas all other variables in the clause (i.e., the variables {xj : j ∈V (c)\\{i}}) are\nassigned uc,j. A variable xi is constrained by clause c if it is the unique satisfying variable for c.\nA variable is unconstrained if it has 0 or 1 value, and is not constrained by any clause. Thus for any\npartial assignment σ ∈{0, 1, ∗}n the variables are divided into stars, constrained and unconstrained\nvariables. Let S∗(σ) be the set of unassigned variables, and n∗(σ) and no(σ) denote respectively\nthe number of stars and the number of unconstrained variables. We define the weight of a valid\npartial assignment to be\nW(σ) := ρn∗(σ)(1 −ρ)no(σ),\n(1)\nwhere ρ is a parameter in the interval [0, 1]. The weight of an invalid partial assignment is 0.\nIn [14] the Survey Propagation algorithm is interpreted as a special case of a larger family of\nBelief Propagation algorithms applied to a family of distributions on valid partial assignments. This\nfamily of distributions, parameterized by ρ, is defined as Pr[σ] ∝W(σ). At one extreme, ρ = 0,\nthis becomes just the uniform distribution over (full) satisfying assignments. The other extreme,\nρ = 1, corresponds to Survey Propagation. The pure version of Survey Propagation corresponds to\nsetting ρ = 1. Intermediate values of ρ interpolate between these extremes.\nNext we define a natural partial order (represented by an acyclic directed graph) on valid partial\nassignments. The vertex set of the directed graph G consists of all valid partial assignments. The\nedge set is defined in the following way: for a given pair of valid partial assignments σ and τ, the\ngraph includes a directed edge from σ to τ if there exists an index i ∈{1, . . . , n} such that (i)\nσj = τj for all j ̸= i; and (ii) τi = ∗and σi ̸= ∗.\nValid partial assignments can be separated into levels based on their number of star variables,\ni.e., the assignment σ is in level n∗(σ). Thus every edge goes from an assignment in level l −1 to\none in level l, where 1 ≤l ≤n. G is acyclic and we write τ < σ if there is a directed path in G\nfrom σ to τ. In this case we will also say that the assignment τ is consistent with the assignment\nσ. The outgoing edges of any valid partial assignment σ correspond to its unconstrained variables,\nand therefore its outdegree is equal to no(σ). The minimal assignments in this ordering are the\nassignments without unconstrained variables, i.e., the positive weight assignments for ρ = 1.\n4"},{"paragraph_id":"p5","order":5,"text":"Definition 3 A core of a satisfying assignment σ is a minimal assignment τ such that τ < σ.\nThe following proposition about cores is proved in [14].\nProposition 3 [14] Any satisfying assignment σ has a unique core. Furthermore, if satisfying\nassignments σ1, σ2 ∈{0, 1}n belong to the same cluster of solutions then they have the same core.\nIn the above proposition a cluster is simply a connected component of the graph on solutions,\nin which two solutions are connected by an edge if and only if they are at Hamming distance 1.\nDefinition 4 A cover is a valid partial assignment that contains no unconstrained variables.\nIn particular, the core of any satisfying assignment is a cover. On the other hand not all cover\nassignments are cores (because they may not be extendable to satisfying assignments). We say that\na cover assignment τ is non-trivial if n∗(τ) < n, so that it has at least one assigned variable.\nThe proof of Theorem 1 uses a surprising property of the weights (1), which was observed\nin [14] but was not utilized there. (A more general statement and a connection to a combinatorial\nobject known as “convex geometry” was developed in [2].) Specifically, the total weight of partial\nassignments consistent with a given satisfying assignment is exactly 1. This fact implies that the\nprobability of satisfiability is at most the expected total weight of partial assignments.\nTheorem 4 [14] For every ρ ∈[0, 1], P\nτ≤σ W(τ) = ρn∗(σ) for any valid partial assignment σ ∈\n{0, 1, ∗}n. In particular, if σ is a satisfying assignment then P\nτ≤σ W(τ) = 1.\nIn our proof, we will apply Theorem 4 with various values for ρ ∈(0.8, 1). In each application\nwe will chose the value of ρ to get the best bound possible for the probability that a formula chosen\nfrom some distribution has a satisfying assignment. In particular, if a satisfying assignment exists,\nthe total weight of valid partial assignments is at least 1. Therefore, the probability of satisfiability\nis at most as large as the expected value of the total weight of partial assignments.\nApplying\nthis idea directly to the original distribution on 3-Sat formulas leads to an upper bound on the\nthreshold αc that is weaker than the currently best known bound of 4.506. However, the derivation\nis simpler than other approaches, which makes it possible to apply the same method to bound\nthe probability that a fixed valid partial assignment can be extended to a satisfying assignment.\n(This amounts to applying the method to random formulas coming from a variety of distributions\non formulas with both 2-clauses and 3-clauses.) This allows us to estimate the probability of the\nexistence of a non-trivial core, and thus to prove the theorem.\n3\nProof of the main theorem\nThis section contains a proof of Theorem 1. We begin with an overview of the entire proof, in the\ncourse of which we will state various technical lemmas; these lemmas will be proved in the three\nsubsections that follow.\nNote first that for α ≥4.506 the statement of the main theorem follows from the fact that\nrandom 3-Sat formulas of density at least 4.506 are known to be unsatisfiable with high proba-\nbility [6]. Hence from now on we focus on the case that α ∈[4.453, 4.506]. Our goal is to prove\nthat, for densities above 4.453, with high probability there are no non-trivial covers that can be\nextended to satisfying assignments, i.e., there are no non-trivial cores.\nTo this end, define the size of a cover (or of a core) to be the number of variables assigned value 0\nor 1. The following lemma, proved in [14], establishes that with high probability all non-trivial\ncovers (and consequently cores) are of linear size.\n5"},{"paragraph_id":"p6","order":6,"text":"Lemma 5 [14] For a random 3-Sat formula of density α, with high probability there are no non-\ntrivial covers of size strictly less than\n1\nαe2 n.\nThis lemma implies that it is sufficient to consider core assignments of size an for a ∈[1/(αe2), 1].\nLet Xa denote the number of cover assignments of size ⌊an⌋, and Ya denote the number of core\nassignments of size ⌊an⌋.\nIn addition to the density α, the size measure a, and the weight parameter ρ ∈[0, 1] from\nTheorem 4, we will use two further parameters, d and b, whose precise definitions will be given\nin the proofs in the following subsections.\nRoughly speaking, d is the fraction of clauses that\nare constraining with respect to a given partial assignment, and b is the fraction of constrained\nvariables with respect to a given partial assignment. The ranges of these parameters are d ∈(a/α, 1]\nand b ∈[0, 1 −a]. We will also make use of a derived parameter r which is defined by d, a and\nα. It appears in connection to the event that the constraining clauses succeed in constraining all\nconstrained variables. Specifically, r is the value satisfying the equation d = ar\nα ln\nr\nr−1. (Note that\nsuch a value r always exists and is unique as the right-hand side is monotonic in r.) In fact, because\nof the form of this expression, it will be more convenient to think of d as being determined by r,\na, and α. Whenever we take the supremum over one of these parameters, we always mean the\nsupremum over its allowed range.\nWe now define two functions that play a central role in our analysis:\nf(α, a, r)\n:=\na ln(2) + H(a) + αH(d) + αd ln\n 3a3\n8"},{"paragraph_id":"p7","order":7,"text":"+ α(1 −d) ln"},{"paragraph_id":"p8","order":8,"text":"1 −a2(3 −a)\n4"},{"paragraph_id":"p9","order":9,"text":"+ αd ln(r/e) −a ln(r −1);\nh(α, a, r, ρ, b)\n:=\nb ln(2) + (1 −a −b) ln(ρ) + (1 −a)H(b/(1 −a))\n−α(1 −d)b b(6 −5b −15a) + 12(1 −a)a\n2(4 −a2(3 −a))\n+ b ln"},{"paragraph_id":"p10","order":10,"text":"1 −ρe\n−3α(1−d)\nb(b+2a)\n2(4−a2(3−a))"},{"paragraph_id":"p11","order":11,"text":".\nHere H denotes the entropy function H(x) = −x ln(x) −(1 −x) ln(1 −x).\nThe first ingredient in the proof of Theorem 1 is the following lemma, whose proof is presented\nin Section 3.1.\nLemma 6 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1],\nlim\nn→∞\n1\nn ln (E[Xa]) ≤sup\nr f(α, a, r).\nA simple application of Markov’s inequality immediately yields:\nCorollary 7 If α ≥1 and a ∈[0, 1] are such that, for every r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds\nthat f(α, a, r) < 0, then with high probability random 3-Sat formulas of density α do not have\ncovers of size an.\nThe second ingredient in the proof of Theorem 1 is the following lemma, which is proved in\nSection 3.2.\nLemma 8 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1] and ρ ∈[0, 1),\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr (f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ, b)}).\n6"},{"paragraph_id":"p12","order":12,"text":"Remark: This lemma can be strengthened by allowing ρ to depend on r; however, we will not\nneed to use this stronger version.\nSince every core is a cover we know that Ya ≤Xa. Hence (in light of Lemma 6) Lemma 8 is\ninteresting only when supb h(α, a, r, ρ, b) is negative. In fact, we shall prove Lemma 8 by showing\nthat this supremum bounds the logarithm of 1/n times the probability that a particular cover of\nsize a can be extended to a satisfying assignment. (For a precise statement, see Lemma 11 in\nSection 3.2.) An immediate corollary of Lemma 8 is the following.\nCorollary 9 If α ≥1, a ∈[0, 1] and there exists ρ ∈[0, 1) such that for every r > 1 with\nd = ar\nα ln\nr\nr−1 ≤1, and for every b ∈[0, 1 −a], it holds that f(α, a, r) + h(α, a, r, ρ, b) < 0, then with\nhigh probability random 3-Sat formulas of density α do not have cores of size an.\nThe final ingredient in the proof of Theorem 1 is the following numerical claim, whose proof we\ndiscuss in Section 3.3.\nClaim 10 For every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1], and r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it\nholds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nRemark: In Claim 10 we have used the weight ρ ≡ρ(a) = 0.4a+0.7. We arrived at this particular\nchoice of ρ by first optimizing numerically for ρ at 100 values for a ∈[1/4.506, 1], and then fitting\na simple function to the values for ρ that were found. Since Lemma 8 holds for every value of ρ,\nwe may choose any convenient function. Using a simple analytic function guarantees that h is also\nanalytic, which makes the numerical analysis of h easier.\nFinally, combining Claim 10 with Corollary 9 completes the proof of Theorem 1.\n3.1\nThe expected number of covers: Proof of Lemma 6\nLet s = ⌊an⌋. Then we have\nE[Xa]\n=\nE"},{"paragraph_id":"p13","order":13,"text":"X\nσ∈{0,1,∗}n\nInd [σ is valid ∩(n∗(σ) = n −s) ∩(no(σ) = 0)]"},{"paragraph_id":"p14","order":14,"text":"=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is valid ∩(no(σ) = 0)]\n=\n n\ns"},{"paragraph_id":"p15","order":15,"text":"2s Pr[σ = (0s ∗n−s) is valid and x1, . . . , xs are constrained].\nWe denote by P the probability that σ = (0s ∗n−s) is valid and all of its assigned variables\nare constrained. P is equivalent to the probability of the following event in an experiment with\nm = αn balls thrown uniformly and independently at random into 23 n\n3"},{"paragraph_id":"p16","order":16,"text":"bins. There are 3 kinds\nof bins:\n1. Bins of type 1 should be empty. These correspond to clauses that are not allowed:\n• (xi1 ∨xi2 ∨xi3), with i1, i2, i3 ∈{1, 2, . . . , s};\n• (xi1 ∨xi2 ∨ ̄xj), with i1, i2 ∈{1, 2, . . . , s} and j > s;\n• (xi1 ∨xi2 ∨xj), with i1, i2 ∈{1, 2, . . . , s} and j > s.\n7"},{"paragraph_id":"p17","order":17,"text":"The total number of these is\n s\n3"},{"paragraph_id":"p18","order":18,"text":"+ 2(n −s)\n s\n2"},{"paragraph_id":"p19","order":19,"text":"= n3\n a3\n6 + a2(1 −a)"},{"paragraph_id":"p20","order":20,"text":"+ o(n3).\n2. Bins of type 2 correspond to constraining clauses: (xi1 ∨xi2 ∨ ̄xt), with i1, i2, t ∈{1, 2, . . . , s}.\nFor each variable xt there are\n s−1\n2"},{"paragraph_id":"p21","order":21,"text":"= n2 a2\n2 +o(n2) clauses that could constrain it and at least\none has to be included. Equivalently, there has to be at least one ball in one of those bins for\nevery xt with t ∈{1, 2, . . . , s}. The total number of these clauses is: s\n s−1\n2"},{"paragraph_id":"p22","order":22,"text":"= n3 a3\n2 + o(n3).\n3. There are no constraints for the remaining bins, of type 3. Their total number is\n23\n n\n3"},{"paragraph_id":"p23","order":23,"text":"−n3\n a3\n6 + a2(1 −a)"},{"paragraph_id":"p24","order":24,"text":"−n3 a3\n2 + o(n3) = n3\n3 (4 −a2(3 −a)) + o(n3).\nSuppose m′ = dm of the clauses we choose are of type 2, and the remaining m −m′ are of\ntype 3. The probability of this event is\npm′ =\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\n.\nThe probability that the m′ clauses of type 2 are such that there is at least one of each kind is\nthe same as the coupon collectors probability of success, with s = ⌊an⌋different coupons, and\nm′ = (dα)n trials.\nWe will use the following general fact from [4], which was previously used\nin a very similar context in [12]: Let q(cN, N) denote the probability of collecting N coupons\nwithin cN trials.\nIf c < 1, q(cN, N) = 0.\nOtherwise, as N goes to infinity q(cN, N) grows\nlike g(c)N, where g(c) =\n r0\ne\n c\n1\nr0−1, and r0 is the solution of r ln"},{"paragraph_id":"p25","order":25,"text":"r\nr−1"},{"paragraph_id":"p26","order":26,"text":"= c.\nMore precisely,\nlimN→∞1\nN ln (q(cN, N)) = ln(g(c)). We have\nP\n=\nm\nX\nm′=0\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (6 −2a)\n8\n+ o(1)\n m−m′\nq(m′, s)\n≤\nm max\nm′\n( m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\nq(m′, s).\n)\nFinally,\nE[Xa] ≤\n n\ns"},{"paragraph_id":"p27","order":27,"text":"2sm max\nm′\n m\nm′"},{"paragraph_id":"p28","order":28,"text":"3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3−a)\n4\n+ o(1)\n m−m′\nq(m′, s)"},{"paragraph_id":"p29","order":29,"text":",\nand hence\nlim\nn→∞\n1\nn ln (E[Xa])\n≤\nln"},{"paragraph_id":"p30","order":30,"text":"2a\naa(1 −a)1−a"},{"paragraph_id":"p31","order":31,"text":"+ sup\nd"},{"paragraph_id":"p32","order":32,"text":"α ln"},{"paragraph_id":"p33","order":33,"text":"3 a3\n8\n d \n1 −a2 (3−a)\n4\n 1−d\ndd (1 −d)1−d"},{"paragraph_id":"p34","order":34,"text":"+ a ln"},{"paragraph_id":"p35","order":35,"text":"g\n dα\na"},{"paragraph_id":"p36","order":36,"text":"=\nsup\nd\nf(α, a, r) = sup\nr\nf(α, a, r).\nThis completes the proof of Lemma 6.\n8"},{"paragraph_id":"p37","order":37,"text":"3.2\nThe probability that a cover assignment is a core: Proof of Lemma 8\nFor a partial assignment σ, it will be convenient to denote by φσ(xS∗(σ)) the formula that is obtained\nby substituting the variables that have 0/1 assignments in σ, i.e., removing from φ clauses that\nare satisfied by at least one of the assigned variables and removing all remaining appearances of\nassigned variables. This is a formula on n∗(σ) variables. Notice that if σ is a valid assignment\nfor φ then the formula φσ contains no empty clauses and no unit clauses. Furthermore all clauses\nof type 2 (from the previous section) are removed, because they are satisfied by their constrained\nvariables. Among the clauses of type 3, there are clauses that are removed, there are clauses that\nbecome two-variable clauses, and there are clauses that remain untouched. Since there is no simple\nway to describe the resulting distribution on formulas, we will keep referring to the set of all clauses\nof type 3, even the ones that are removed in φσ. When we condition on the fact that σ is a cover\nand m′ of the clauses are of type 2, as in the previous section, we know that the set of clauses we\nare interested in are distributed exactly as a uniform set of (m −m′) clauses of type 3. Thus we\ncan express the expected number of cores as\nE[Ya]\n=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is a cover] × Pr[σ is a core | σ is a cover ]\n=\n n\ns"},{"paragraph_id":"p38","order":38,"text":"2s Pr[σ = (0s ∗n−s) is a cover] × Pr[σ = (0s ∗n−s) is a core | σ is a cover]\n=\n n\ns"},{"paragraph_id":"p39","order":39,"text":"2s ×\nm\nX\nm′=s\npm′ q(m′, s)\n× Pr[φσ(xs+1, . . . , xn) is satisfiable | σ = (0s ∗n−s), m −m′ clauses are of type 3].\nWe will bound this probability using the Poisson approximation, which is a standard technique\nin this area. There are two related random models. In the first model, which we call the exact\nmodel, (m −m′) clauses are chosen uniformly at random with replacement from all M = n3(4 −\na2(3 −a))/3 + o(n3) clauses of type 3. In the second model, which we call the Poisson model, each\nof the M clauses is included in the formula with probability p = (m −m′)/M −1/(n2√log n) =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2"},{"paragraph_id":"p40","order":40,"text":". The expected number of clauses in both models is the same up to a term\nδ = M/(n2√log n) = Θ(n/√log n) which is sub-linear in n.\nThe Poisson model has been studied before in the context of random 3-Sat. It can be shown,\nas the example below demonstrates, that whenever a property holds with high probability in the\nexact model, it also holds with high probability in the corresponding Poisson model. Applying first\nmoment techniques to the Poisson model is usually easier, because the clauses are independently\nchosen; however the bounds obtained are usually weaker.\nNext, we relate the probability that φσ is satisfiable under the two models. Let Prp denote\nprobability in the Poisson model, and Pre denote probability in the exact model. Let the random\nvariable J denote the number of distinct clauses included in the formula. Then\nPrp[φσ is satisfiable]\n=\nM\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i]\n≥\nm−m′−δ\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i].\n9"},{"paragraph_id":"p41","order":41,"text":"Since this conditional probability decreases as i increases, and Prp[J ≤Ep[J]] ≥1/2, where Ep[J] =\nm −m′ −δ, we have\nPrp[φσ is satisfiable]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] ×\nm−m′−δ\nX\ni=0\nPr[J = i]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] × 1\n2.\nOn the other hand, for the exact model\nPre[φσ is satisfiable]\n=\nm−m′\nX\ni=0\nPre[J = i] × Pr[φσ is satisfiable | J = i]\n≤\n m−m′−δ\nX\ni=0\nPre[J = i]\n!\n+ Pr[φσ is satisfiable | J = m −m′ −δ]\n≤\nPre[at least δ clauses repeated] + 2 Prp[φσ is satisfiable]\n≤\n m −m′\nδ\n m −m′\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n≤\n (m −m′)2\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n=\nθ(2−n√log n) + 2 Prp[φσ is satisfiable].\nThus if the probability of satisfiability in the Poisson model is bounded above by c−n for some\nconstant c, then limn→∞1\nn ln (Pre[ φσ is satisfiable]) ≤c. Therefore, it suffices to get a bound for\nthe Poisson model. Any of the first moment techniques for bounding the satisfiability threshold of 3-\nSat can be adapted to bound the probability that φσ is satisfiable. Of the ones that are technically\napplicable (i.e., result in an explicit analytic expression for every setting of the parameters s and\nm′), we obtain the strongest result using the novel approach of applying the first moment method\nto the distribution on partial assignments defined by the weights in equation (1).\nLemma 11 In the Poisson model with parameters n, m = αn, m′ = dm, s = ⌊an⌋and for\nσ = (0s ∗n−s), and r such that d = ar\nα ln\nr\nr−1,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤\ninf\nρ∈[0,1)\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b).\nProof: We will apply Theorem 4 to the formula φσ(xs+1, . . . , xn). The theorem implies that if φσ\nhas a satisfying assignment then P\nτ∈V W(τ) ≥1 where W(τ) = ρn∗(τ)(1 −ρ)no(τ) and the sum is\nover the set V of all partial assignments τ ∈{0, 1, ∗}n−s that are valid for φσ. Thus the probability\nof satisfiability is bounded from above by the expected value of P\nτ∈V W(τ). This holds for any\nvalue of ρ ∈[0, 1]. We bound this expectation.\nFor any t ∈{0, 1, . . . , n −s} let Zt denote the sum of the weights of valid assignments τ for\nφσ(xs+1, . . . , xn) such that n∗(τ) = n −s −t. Then\nE[Z] =\nn−s\nX\nt=0\nE[Zt] ≤n\nmax\nt∈{0,1,...,n−s} E[Zt].\n10"},{"paragraph_id":"p42","order":42,"text":"E[Zt]\n=\nt\nX\nu=0\nρn−s−t(1 −ρ)u\nX\nτ∈{0,1,∗}n−s\nPrp [τ is valid ∩(n∗(τ) = n −s −t) ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\nX\nτ∈{0,1,∗}n−s : n∗(τ)=n−s−t\nPrp[τ is valid ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\n n −s\nt\n t\nu"},{"paragraph_id":"p43","order":43,"text":"2t\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\nNext we derive the probability that the assignment (xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid\nand the first u variables are unconstrained. Recall that φσ is obtained from φ by setting its first s\nvariables according to σ. Only clauses of type 3 influence φσ and according to the Poisson model,\neach of them is included independently with probability p. The probability that the assignment\n(xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid and the first u variables are unconstrained is equivalent\nto the probability that among the included clauses of type 3, there are no clauses of the following\nkinds:\n• (xi1 ∨xi2 ∨xi3), with i1 ∈{1, 2, . . . , s + t}, i2, i3 ∈{s + 1, s + 2, . . . , s + t},\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s+t}, i2 ∈{s+1, s+2, . . . , s+t} and j ∈{s+1, . . . , s+u},\nand for every j ∈{s + u + 1, . . . , s + t}, the formula contains a clause (xi1 ∨xi2 ∨ ̄xj), with\ni1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, . . . , s + t}.\nIn the Poisson model all clauses are independent, so it is easy to put these events together to\nobtain:\nQ\n=\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\n=\n(1 −p)((t\n3)+s(t\n2)) + 2(n−s−t)((t\n2) + st) + u((t\n2)+st) ×"},{"paragraph_id":"p44","order":44,"text":"1 −(1 −p)(t\n2)+st t−u\nThe expression for the expectation can be simplified:\nE[Zt]\n=\nρn−s−t\n n −s\nt"},{"paragraph_id":"p45","order":45,"text":"2t\nt\nX\nu=0\n(1 −ρ)u\n t\nu"},{"paragraph_id":"p46","order":46,"text":"Q\n=\nρn−s−t\n n −s\nt"},{"paragraph_id":"p47","order":47,"text":"2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×\nt\nX\nu=0\n t\nu"},{"paragraph_id":"p48","order":48,"text":"(1 −ρ)u (1 −p)u((t\n2)+st) ×"},{"paragraph_id":"p49","order":49,"text":"1 −(1 −p)(t\n2)+st t−u\n=\nρn−s−t\n n −s\nt"},{"paragraph_id":"p50","order":50,"text":"2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×"},{"paragraph_id":"p51","order":51,"text":"(1 −ρ)(1 −p)(t\n2)+st +"},{"paragraph_id":"p52","order":52,"text":"1 −(1 −p)(t\n2)+st t\n=\nρn−s−t\n n −s\nt"},{"paragraph_id":"p53","order":53,"text":"2t (1 −p)(t\n2)(6n−5t−3s−2)/3+2(n−s−t)st \n1 −ρ(1 −p)(t\n2)+st t\n11"},{"paragraph_id":"p54","order":54,"text":"Let t = bn, and recall that s = ⌊an⌋, p =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2"},{"paragraph_id":"p55","order":55,"text":". Then\nlim\nn→∞\n1\nn ln (E[Zt])\n=\nln\n (1 −a)1−a 2b ρ1−a−b\nbb (1 −a −b)1−a−b"},{"paragraph_id":"p56","order":56,"text":"−α (1 −d) b (b(6 −5b −3a) + 12(1 −a −b)a)\n2(4 −a2(3 −a))\n+ b ln"},{"paragraph_id":"p57","order":57,"text":"1 −ρe\n−3α(1−d)b(b+2a)\n2(4−a2(3−a))"},{"paragraph_id":"p58","order":58,"text":"=\nh(α, a, r, ρ, b).\nFinally,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤lim\nn→∞\n1\nn ln (E[Z]) ≤\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b),\nwhich is the statement of Lemma 11.\nSubstituting the bound of Lemma 11 into the expression for the expectation yields\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr\n(f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ(a, r), b)}),\ncompleting the proof of Lemma 8.\n3.3\nNumerical analysis: Steps to the proof of Claim 10\nWhat remains is to verify the numerical claim that for every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1],\nand r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a],\nf(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0. We outline the steps towards a proof.\nThere are two stages. First, we identify the (a, r) pairs such that for every α ∈[4.453, 4.506] it\nholds that f(α, a, r) < 0. Second, for the remaining range of values of (a, r) we show that for every\nb ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nThe derivative of f with respect to r is\n∂f\n∂r = a"},{"paragraph_id":"p59","order":59,"text":"ln"},{"paragraph_id":"p60","order":60,"text":"1 +\n1\nr −1"},{"paragraph_id":"p61","order":61,"text":"−\n1\nr −1"},{"paragraph_id":"p62","order":62,"text":"× ln"},{"paragraph_id":"p63","order":63,"text":"3"},{"paragraph_id":"p64","order":64,"text":"α −ar ln"},{"paragraph_id":"p65","order":65,"text":"r\nr−1"},{"paragraph_id":"p66","order":66,"text":"2\n 4\na2 −3 + a"},{"paragraph_id":"p67","order":67,"text":"ln"},{"paragraph_id":"p68","order":68,"text":"r\nr−1"},{"paragraph_id":"p69","order":69,"text":".\nSince the first factor is always negative the derivative is 0 when\n3α = ln"},{"paragraph_id":"p70","order":70,"text":"r\nr −1\n 8\na2 −6 + 2a + 3ar"},{"paragraph_id":"p71","order":71,"text":".\n(2)\nThis equation has a unique root for every a ∈[1/(4.506e2), 1], α ∈[4.453, 4.506] (the derivative of\nthe right-hand side is negative). Therefore we can conclude that for every a and α in the above\nrange, f is at first increasing with r then decreasing.\nIts maximum is achieved at the root of\nequation (2).\nFigure 1 shows the values for r where f is maximized, with respect to a, when\nα = 4.453.\nSince the root of (2) is monotone decreasing with respect to a and α, we can find values r1\nand r2 such that for every a and α in the range, if r < r1 then ∂f/∂r > 0, and if r > r2 then\n∂f/∂r < 0. Values satisfying this condition are r1 = 1.2 and r2 = 670. Thus if we show that\nf(α, a, r) is negative for r = r1 and r = r2 then it is negative for any r outside the range (1.2, 670).\n12"},{"paragraph_id":"p72","order":72,"text":"5\n 10\n 15\n 20\n 25\n 0\n 0.2\n 0.4\n 0.6\n 0.8\n 1\nr\na\narg maxr f (4.453, a, r)\nFigure 1: The pairs (a, r) such that f(4.453, a, r) > −0.0001, and the value of r that maximizes\nf(4.453, a, r) for every a.\nWe will use the shorthand notation q := r ln\nr\nr−1. For r ∈(1.2, 670) we have q ∈(1.0007, 2.16).\nThe whole range satisfies the condition that d = ar\nα r ln\nr\nr−1 ≤1.\nNext we take care of the boundary region with respect to a. Notice that the derivative of f is\nnegative for a large enough (close to 1), because of the entropy term in f. The derivative of f with\nrespect to a is\n∂f\n∂a\n=\nln(2) −ln\na\n1 −a −q ln\n3a2(α −aq)\n2q(4 −a2(3 −a)) + 3q −3a(α −aq)(2 −a)\n4 −a2(3 −a)\n+ q ln(r) −ln(r −1).\nThe following observations are helpful for bounding this derivative:\n• 3q + q ln(r) −ln(r −1) is maximized in the interval r ∈[1.2, 670] at r = 1.2.\n• a2(α −aq)/(4 −a2(3 −a)) is an increasing function of a.\n• a(α −aq)(2 −a)/(4 −a2(3 −a)) is an increasing function of a.\nFor a > 0.999, using the above facts, the derivative is negative. Thus if we show that f is\nnegative for a = 0.999, r ∈(1.2, 670), α ∈[4.453, 4.506] then f is negative also for every a > 0.999.\nWe are left with the region a ∈[1/(4.506e2), 0.999], r ∈(1.2, 670).\nIn this region all the\nderivatives of f can be bounded, and a sufficiently fine grid can be chosen over which to evaluate\nf in order to identify the grid sections where the function can take positive values. In particular,\nthe derivative with respect to a is at most 28.2, and with respect to α it is at most 1. Furthermore,\nsince we know that for every a and α, f is maximized as a function of r at the root of equation (2),\none can find this maximum and the range of r where f(α, a, r) is positive using binary search. The\npoints with f(4.453, a, r) > −0.0001 are depicted in Figure 1.\nIn the second stage we need to analyze f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) for the remaining\nregion of values for (a, r).\n13"},{"paragraph_id":"p73","order":73,"text":"First, notice that we can take care of the boundary regions with respect to b by taking advantage\nof the entropy term, which has very large slope for b close to 0, and very steep negative slope for b\nclose to 1 −a. Specifically, the derivative with respect to b is\n∂h\n∂b\n=\nln(2) −ln(ρ) −ln"},{"paragraph_id":"p74","order":74,"text":"b\n1 −a −b"},{"paragraph_id":"p75","order":75,"text":"−α(1 −d)12b −15b2 −30ab + 12a −12a2\n2(4 −a2(3 −a))\n+ ln(1 −ρe−A) + 3α(1 −d)b\nρe−A(a + b)\n(1 −ρe−A)(4 −a2(3 −a))\nwhere A = 3α(1−d)b(b+2a)\n2(4−a2(3−a)) .\nUsing the bounds on all parameters: α ∈[4.453, 4.506], a ∈(0.28, 0.7), r ∈(1.5, 14.3), and the\nones that follow from those: d ∈(0.06, 0.26), A ∈[0, 1.59], ρ ∈(0.02, 0.21), we can conclude that\n−3.04 −ln"},{"paragraph_id":"p76","order":76,"text":"b\n1 −a −b"},{"paragraph_id":"p77","order":77,"text":"< ∂h\n∂b < 5.38 −ln"},{"paragraph_id":"p78","order":78,"text":"b\n1 −a −b"},{"paragraph_id":"p79","order":79,"text":".\nThe lower bound can be made positive by setting b/(1−a −b) ≤0.04 and the upper bound can\nbe made negative by setting b/(1 −a −b) ≥220. Therefore it suffices to show that h is negative\nfor b in the range [0.01, 0.996(1 −a)].\nAgain, using the bounds for the parameters, all the derivatives of f + h can be bounded, and a\nsufficiently fine grid can be chosen over which to evaluate f +h. We did not perform the evaluation\non a grid that is as fine as is required for the rigorous proof because, based on the current bounds on\nthe derivatives, we would need to evaluate the function at more than 1010 points. (As we discussed\nin the introduction, we feel that the computational effort is not yet justified.)\nFor illustration, Figure 2 shows the estimated values of supr f(α, a, r) and of supr,b(f(α, a, r) +\nh(α, a, r, ρ(a), b)) for α = 4.453, 4.470, 4.490, and 4.506. The maximum is taken over evaluations\nat a grid of step size 0.001 for the parameters r and b. We estimated with better precision the\nlocation of the maximum of f + h by evaluating the function at a fine grid in the region where\nthe evaluations on the coarse grid give the largest values. The maximum found in this way is at\nα = 4.453, a = 0.62566, b = 0.03568 and r = 2.00134 (d = 0.19473). The value of f + h at this\npoint is −0.000058.\n4\nConcluding remarks\nAs mentioned in the introduction, all of the known rigorous upper bounds for the satisfiability\nthreshold of 3-Sat are based on the first moment method [10, 5, 13, 9, 12, 6]; the corresponding\nupper bounds in this sequence of results are: 4.758, 4.64, 4.601, 4.596, 4.571, 4.506. The general\nmethod is to consider a random variable Z that is equal to the number of satisfying assignments\nof a particular kind. These satisfying assignments are such that at least one exists if the formula is\nsatisfiable. Showing that above a certain density E[Z] →0 thus implies (by Markov’s inequality)\nthat Pr[Z = 0] →1, and consequently the probability that the formula is unsatisfiable also goes\nto 1. For example, [5] takes Z to be the number of negatively prime solutions, i.e., solutions for\nwhich every variable assigned 1 is constrained.\nHere, we have used the same idea but with Z taken to be the total weight of partial assignments\nunder the weight function (1) inspired by Survey Propagation. Given the dramatic success of Survey\nPropagation for random 3-Sat, it seems plausible that this approach can potentially yield rather\ntight upper bounds on the threshold. We were able to achieve only partial progress in this direction,\nbut we are quite hopeful that extensions of our approach could lead to further progress.\n14"},{"paragraph_id":"p80","order":80,"text":"-0.08\n-0.06\n-0.04\n-0.02\n 0\n 0.02\n 0.1\n 0.2\n 0.3\n 0.4\n 0.5\n 0.6\n 0.7\n 0.8\n 0.9\n 1\nsupr f, supr,b (f+h)\na\nα = 4.453\nα = 4.470\nα = 4.490\nα = 4.506\nFigure 2: The values of supr f(α, a, r) and supr,b(f(α, a, r) + h(α, a, r, 0.4a + 0.7, b)) for α = 4.453,\n4.470, 4.490, and 4.506.\nOne natural extension would be to use a different weight ρ for each variable, depending for\nexample on the number of positive and negative occurrences of the variable. The corresponding\ngeneralization of Theorem 4 is proved in [2]. It is quite possible that the value of E[Z] in this case\nis significantly smaller.\n5\nAcknowledgments\nWe would like to thank the anonymous referees for their detailed and very useful suggestions.\nReferences\n[1] D. Achlioptas and F. Ricci-Tersenghi. On the solution-space geometry of random constraint\nsatisfaction problems. In Proc. 38th ACM Symp. Theory of Computing (STOC), pages 130–139,\n2006.\n[2] F. Ardila and E. Maneva. Pruning processes and a new characterization of convex geometries.\nDiscrete Mathematics, 2008. To appear.\n[3] A. Braunstein and R. Zecchina. Survey propagation as local equilibrium equations. J. Stat.\nMech. : Theory and Experiments (JSTAT), page 06007, 2004.\n[4] V. Chvatal. Almost all graphs with 1.44 edges are 3-colourable. Random Struct, Algorithms,\n2:11–28, 1991.\n[5] O. Dubois and Y. Boufkhad. A general upper bound for the satisfiability threshold of random\nr-SAT formulae. J. Algorithms, 24:395–420, 1997.\n15"},{"paragraph_id":"p81","order":81,"text":"[6] O. Dubois, Y. Boufkhad, and J. Mandler. Typical random 3-SAT formulae and the satisfiability\nthreshold. In Proc. 11th ACM-SIAM Symp. on Discrete Algorithms (SODA), pages 126–127,\n2000. Extended version at arXiv:cs/0211036.\n[7] E. Friedgut. Neccesary and sufficient conditions for sharp threhsolds of graph properties and\nthe k-problem. J. Amer. Math. Soc., 12:1017–1054, 1999.\n[8] M. Hajiaghayi and G. Sorkin. The satisfiability threshold of random 3-SAT is at least 3.52.\nTechnical report, 2003. Preprint at arXiv:math/0310193.\n[9] S. Janson, Y. C. Stamatiou, and M. Vamvakari. Bounding the unsatisfiability threshold of\nrandom 3-SAT. Random Struct. Algorithms, 17(2):103–116, 2000.\n[10] A. Kamath, R. Motwani, K. V. Palem, and P. G. Spirakis. Tail bounds for occupancy and the\nsatisfiability threshold conjecture. Random Struct. Algorithms, 7(1):59–80, 1995.\n[11] A. C. Kaporis, L. M. Kirousis, and E. G. Lalas. The probabilistic analysis of a greedy satisfi-\nability algorithm. Random Struct. Algorithms, 28(4):444–480, 2006.\n[12] A. C. Kaporis, L. M. Kirousis, Y. C. Stamatiou, M. Vamvakari, and M. Zito. The unsatisfia-\nbility threshold revisited. Discrete Applied Mathematics, 155(12):1525–1538, 2007.\n[13] L. M. Kirousis, E. Kranakis, D. Krizanc, and Y. C. Stamatiou. Approximating the unsatisfi-\nability threshold of random formulas. Random Struct. Algorithms, 12(3):253–269, 1998.\n[14] E. Maneva, E. Mossel, and M. J. Wainwright. A new perspective on survey propagation. J.\nACM, 54(4):2–41, 2007.\n[15] M. M ́ezard, T. Mora, and R. Zecchina. Clustering of solutions in the random satisfiability\nproblem. Phys. Rev. Lett., 94(197205), 2005.\n[16] M. M ́ezard, G. Parisi, and M. A. Virasoro. Spin Glass Theory and Beyond. World Scientific,\nSingapore, 1987.\n[17] M. M ́ezard, G. Parisi, and R. Zecchina. Analytic and algorithmic solution of random satis-\nfiability problems. Science, 297, 812, 2002. (Scienceexpress published online June 27, 2002;\n10.1126/science.1073287).\n[18] M. M ́ezard and R. Zecchina. Random k-satisfiability: from an analytic solution to an efficient\nalgorithm. Phys. Rev. E, 66, 2002.\n[19] J. Pearl. Probabilistic reasoning in intelligent systems: networks of plausible inference. Morgan\nKaufmann, Palo Alto, CA, 1988.\n16"}],"pages":[{"page":1,"text":"arXiv:0710.0805v3 [cs.CC] 31 Aug 2008\nOn the satisfiability threshold and clustering of solutions\nof random 3-Sat formulas\nElitza Maneva∗\nAlistair Sinclair†\nAbstract\nWe study the structure of satisfying assignments of a random 3-Sat formula. In particular,\nwe show that a random formula of density α ≥4.453 almost surely has no non-trivial “core”\nassignments. Core assignments are certain partial assignments that can be extended to satisfying\nassignments, and have been studied recently in connection with the Survey Propagation heuristic\nfor random Sat. Their existence implies the presence of clusters of solutions, and they have\nbeen shown to exist with high probability below the satisfiability threshold for k-Sat with\nk ≥9 [1]. Our result implies that either this does not hold for 3-Sat or the threshold density\nfor satisfiability in 3-Sat lies below 4.453. The main technical tool that we use is a novel simple\napplication of the first moment method.\n1\nIntroduction\nThe study of random instances of 3-Sat has been a major research focus in recent years, both\nbecause of its inherent interest and because it is a natural test case for the wider understanding of\nthe complexity of computational tasks on random inputs. In random 3-Sat the input is a formula\ndrawn uniformly at random from all formulas of fixed density α, i.e., formulas with αn clauses on n\nvariables. Friedgut [7] proved that there exists a function αc(n), known as the satisfiability threshold,\nsuch that for any positive ǫ, random formulas of density αc(n) −ǫ have satisfying assignments with\nhigh probability, and random formulas of density αc(n) + ǫ have no satisfying assignment with\nhigh probability. It is conjectured that αc(n) is a constant (and that its value is about 4.27), but\ncurrently all that is known is that, for large n, 3.520 ≤αc(n) ≤4.506 [11, 8, 6].\nIn the range of densities for which the formula is satisfiable with high probability, the inter-\nesting algorithmic question is whether we can find even one of the many satisfying assignments in\npolynomial time. The lower bound on αc(n) is a result of the analysis of such a polynomial time\nalgorithm [11, 8]. This algorithm belongs to a family of algorithms known as “myopic” because\nthey assign variables greedily one by one in an order that is based only on the number of positive\nand negative occurrences of each variable.\nAn apparently much more powerful algorithm is Survey Propagation [17, 18]. In experiments on\nvery large instances (say, with n = 106 variables) it finds solutions for formulas of densities only just\nbelow the conjectured threshold value α = 4.27; however, a rigorous analysis of its performance is\nstill far from our reach. Like the myopic algorithms it also assigns variables one by one in a greedy\nmanner, but its choices are based on more global information about the role of each variable in\n∗IBM Almaden Research Center, San Jose, CA, enmaneva@us.ibm.com. Some of this research was done while\nthis author was a PhD student at UC Berkeley, supported in part by NSF grant DMS-0528488.\n†Computer Science Division, University of California at Berkeley, sinclair@cs.berkeley.edu. Supported in part by\nNSF grants DMS-0528488 and CCR-0635153.\n1"},{"page":2,"text":"the formula. That information is provided by the fixed point of a sophisticated message passing\ndynamics between variables and clauses.\nThe message passing procedure is based on an intriguing (informal) picture of the properties\nof the solution space of a random formula, which is derived from the 1-step Replica Symmetry\nBreaking ansatz of statistical physics [16]. A key postulate of this physical picture is that, for\nformulas of density higher than a certain value (estimated to be 3.92), the space of solutions is split\ninto “clusters.” Within the same cluster it is possible to reach any satisfying assignment from any\nother by flipping one variable at a time, while always keeping the formula satisfied. On the other\nhand, in order to get from a solution in one cluster to a solution in another, the values of a linear\nnumber of variables have to be flipped at the same time. Loosely speaking, the message passing\nprocedure of Survey Propagation was proposed as a way to collect information about the clusters\nof assignments, rather than individual assignments.\nThe remarkable performance of Survey Propagation is a compelling reason to explore properties\nof the solution space of typical formulas as a way to further our understanding of random 3-Sat,\nits hardness, and the satisfiability threshold, and also (looking much further ahead) in order to\nsystematically design algorithms for more general problems with distributional inputs.\nA detailed study of the Survey Propagation algorithm undertaken in [14] and [3] led to an inter-\npretation of the message passing procedure as the more familiar Belief Propagation algorithm [19]\napplied to a particular probability distribution on partial assignments, i.e., assignments of values\nfrom the set {0, 1, ∗}. Here ∗is to be interpreted as “unassigned,” and a variable is allowed to be\nunassigned only if it is not forced to be assigned 0 or 1 in order to satisfy a clause.\nAmong these partial assignments, the set of “core” assignments plays a central role. These are\npartial assignments that are obtained from a satisfying assignment by successively replacing each\nunconstrained variable by ∗. (A variable is unconstrained if changing its value does not make any\nclause unsatisfied.) Any satisfying assignment has a unique corresponding core. Moreover, since\nall assignments in a cluster have the same core, cores can be viewed (informally) as “summaries”\nof clusters. Of course, this view is useful only if different clusters tend to have distinct non-trivial\ncores. (The trivial core assignment is the one without any assigned variables.) Recently, Achlioptas\nand Ricci-Tersenghi [1] showed that, in random k-Sat for k ≥9, for some range of densities up to\nthe satisfiability threshold, with high probability every satisfying assignment has a non-trivial core\nassignment associated with it. This implies that clusters have a large number of frozen variables;\nindeed, for large k the fraction of frozen variables comes arbitrarily close to 1. (The clustering\npicture has also been confirmed for 8-Sat by [15] and [1] using a different method that does not\nsay anything about cores.)\nThe above results hold only for random k-Sat with k ≥8 or 9. In this paper we investigate\nsimilar questions for the apparently harder case of random 3-Sat. Our main result is the following:\nTheorem 1 For random instances of 3-Sat with density greater than 4.453, with high probability\nthere exist no non-trivial core assignments.\nThis theorem requires some interpretation.\nNote first that the density 4.453 lies above the\nconjectured threshold value of 4.27 but below the current best known upper bound of 4.506. Thus\nwe may deduce:\nCorollary 2 One of the following statements holds for random 3-Sat:\n• αc(n) ≤4.453; or\n2"},{"page":3,"text":"• there is a range of densities immediately below the satisfiability threshold for which with high\nprobability there are no non-trivial core assignments.\nOne interpretation of Theorem 1 is as evidence for an improved upper bound on the threshold of\nthe form αc(n) ≤4.453. On the other hand, if in fact αc(n) > 4.453 then the theorem establishes\na range of densities immediately below the threshold for which with high probability the formula\nis satisfiable but has no non-trivial core assignments. This would represent a surprising difference\nbetween the properties of random 9-Sat and random 3-Sat. Interestingly, experiments with 3-Sat\nand solutions of large formulas found by Survey Propagation do not find cores (see [14]). It is quite\nconceivable that both of the statements in the above corollary hold.\nWe now briefly discuss our proof technique, which involves a novel application of the first mo-\nment method and is, we believe, perhaps more noteworthy than the result itself. A straightforward\napplication of the first moment method to the set of core assignments only allows us to show that\nwith high probability there are no cores of small size or of large size. To handle core assignments\nof intermediate size, it is further necessary to bound the probability that a partial assignment can\nbe extended to a satisfying assignment. This probability is equivalent to the probability that a\nrandom formula with a given density of 2-clauses and 3-clauses is satisfiable. For this sub-problem,\nit is natural to use one of the methods that have been previously introduced for bounding the prob-\nability of satisfiability of random 3-Sat formulas [10, 5, 13, 9, 12, 6]. However, the most powerful\nmethod, from [6], is very heavy numerically, and it does not seem possible to carry it out for the\nwhole range of required densities; on the other hand, simpler methods such as those in [13] are\napparently not powerful enough. Instead we introduce a new method, which we now briefly outline\n(for a detailed description, see Section 3.2).\nTraditionally, bounds on the probability of satisfiability such as those mentioned above are\nobtained by applying the first moment method to a random variable which counts the number of\nsatisfying assignments of a particular kind. Indeed, much of the work on bounding the satisfiability\nthreshold has been directed towards identifying a set of satisfying assignments that is a strict subset\nof the set of all satisfying assignments (so that its expected size is much smaller), but is always\nnon-empty if the formula is satisfiable. The novelty in our approach lies in identifying a new random\nvariable which depends not only on satisfying assignments but also on partial assignments. This\nrandom variable is at least 1 for every satisfiable formula, it exploits the clustering structure of the\nsolution space, and most importantly, it is significantly easier to compute than many alternative\napproaches.\nFinally we note that the proof of our theorem depends on a claim that a particular analytic\nfunction takes only negative values in a given range. We do not provide a complete proof of this\nnumerical claim, but we outline the steps needed for completing the proof, and give very strong\nnumerical evidence that all the corresponding statements hold. The difficulty with making the\nproof completely rigorous is that this would require too much computational effort, which we feel is\nnot justified at this stage given that the bound we obtain is still some distance from the satisfiability\nthreshold. The results presented here should rather be considered as a proof of concept.\nThe remainder of this paper is structured as follows. In Section 2 we give necessary background\nand precise definitions of the various concepts used in the paper. Section 3 is devoted to the proof\nof our main result, Theorem 1. We conclude in Section 4 with some final remarks and suggestions\nfor future work.\n3"},{"page":4,"text":"2\nTechnical definitions\nLet x1, . . . , xn be a set of n Boolean variables. A literal is either a variable or its negation. A 3-Sat\nformula is a formula in CNF, where each clause is a disjunction of 3 distinct literals (on different\nvariables). For every clause c we will denote the set of variables that appear in the clause by V (c).\nThe distribution that we consider is the following: for a given density α we choose uniformly at\nrandom and with replacement m = ⌊αn⌋clauses out of all possible clauses on 3 distinct variables.\nFor a clause c and a variable xi ∈V (c) we denote by sc,i and uc,i the value for variable xi that is\nrespectively satisfying and unsatisfying for clause c.\nSuppose that the variables x = (x1, . . . , xn) are allowed to take values in {0, 1, ∗}, which we refer\nto as a partial assignment. A variable taking value ∗(star) should be thought of as unassigned.\nDefinition 1 A partial assignment to x is invalid for a clause c if either (a) all variables are\nunsatisfying; or (b) all variables are unsatisfying except for one index j ∈V (c), for which xj = ∗.\nOtherwise, the partial assignment is valid for clause c. We say that a partial assignment is valid\nfor a formula (or just “valid”) if it is valid for all of its clauses.\nFor a valid partial assignment, the subset of variables that are assigned either 0 or 1 values can\nbe divided into constrained and unconstrained variables in the following way:\nDefinition 2 We say that a variable xi is the unique satisfying variable for a clause c if it is\nassigned sc,i whereas all other variables in the clause (i.e., the variables {xj : j ∈V (c)\\{i}}) are\nassigned uc,j. A variable xi is constrained by clause c if it is the unique satisfying variable for c.\nA variable is unconstrained if it has 0 or 1 value, and is not constrained by any clause. Thus for any\npartial assignment σ ∈{0, 1, ∗}n the variables are divided into stars, constrained and unconstrained\nvariables. Let S∗(σ) be the set of unassigned variables, and n∗(σ) and no(σ) denote respectively\nthe number of stars and the number of unconstrained variables. We define the weight of a valid\npartial assignment to be\nW(σ) := ρn∗(σ)(1 −ρ)no(σ),\n(1)\nwhere ρ is a parameter in the interval [0, 1]. The weight of an invalid partial assignment is 0.\nIn [14] the Survey Propagation algorithm is interpreted as a special case of a larger family of\nBelief Propagation algorithms applied to a family of distributions on valid partial assignments. This\nfamily of distributions, parameterized by ρ, is defined as Pr[σ] ∝W(σ). At one extreme, ρ = 0,\nthis becomes just the uniform distribution over (full) satisfying assignments. The other extreme,\nρ = 1, corresponds to Survey Propagation. The pure version of Survey Propagation corresponds to\nsetting ρ = 1. Intermediate values of ρ interpolate between these extremes.\nNext we define a natural partial order (represented by an acyclic directed graph) on valid partial\nassignments. The vertex set of the directed graph G consists of all valid partial assignments. The\nedge set is defined in the following way: for a given pair of valid partial assignments σ and τ, the\ngraph includes a directed edge from σ to τ if there exists an index i ∈{1, . . . , n} such that (i)\nσj = τj for all j ̸= i; and (ii) τi = ∗and σi ̸= ∗.\nValid partial assignments can be separated into levels based on their number of star variables,\ni.e., the assignment σ is in level n∗(σ). Thus every edge goes from an assignment in level l −1 to\none in level l, where 1 ≤l ≤n. G is acyclic and we write τ < σ if there is a directed path in G\nfrom σ to τ. In this case we will also say that the assignment τ is consistent with the assignment\nσ. The outgoing edges of any valid partial assignment σ correspond to its unconstrained variables,\nand therefore its outdegree is equal to no(σ). The minimal assignments in this ordering are the\nassignments without unconstrained variables, i.e., the positive weight assignments for ρ = 1.\n4"},{"page":5,"text":"Definition 3 A core of a satisfying assignment σ is a minimal assignment τ such that τ < σ.\nThe following proposition about cores is proved in [14].\nProposition 3 [14] Any satisfying assignment σ has a unique core. Furthermore, if satisfying\nassignments σ1, σ2 ∈{0, 1}n belong to the same cluster of solutions then they have the same core.\nIn the above proposition a cluster is simply a connected component of the graph on solutions,\nin which two solutions are connected by an edge if and only if they are at Hamming distance 1.\nDefinition 4 A cover is a valid partial assignment that contains no unconstrained variables.\nIn particular, the core of any satisfying assignment is a cover. On the other hand not all cover\nassignments are cores (because they may not be extendable to satisfying assignments). We say that\na cover assignment τ is non-trivial if n∗(τ) < n, so that it has at least one assigned variable.\nThe proof of Theorem 1 uses a surprising property of the weights (1), which was observed\nin [14] but was not utilized there. (A more general statement and a connection to a combinatorial\nobject known as “convex geometry” was developed in [2].) Specifically, the total weight of partial\nassignments consistent with a given satisfying assignment is exactly 1. This fact implies that the\nprobability of satisfiability is at most the expected total weight of partial assignments.\nTheorem 4 [14] For every ρ ∈[0, 1], P\nτ≤σ W(τ) = ρn∗(σ) for any valid partial assignment σ ∈\n{0, 1, ∗}n. In particular, if σ is a satisfying assignment then P\nτ≤σ W(τ) = 1.\nIn our proof, we will apply Theorem 4 with various values for ρ ∈(0.8, 1). In each application\nwe will chose the value of ρ to get the best bound possible for the probability that a formula chosen\nfrom some distribution has a satisfying assignment. In particular, if a satisfying assignment exists,\nthe total weight of valid partial assignments is at least 1. Therefore, the probability of satisfiability\nis at most as large as the expected value of the total weight of partial assignments.\nApplying\nthis idea directly to the original distribution on 3-Sat formulas leads to an upper bound on the\nthreshold αc that is weaker than the currently best known bound of 4.506. However, the derivation\nis simpler than other approaches, which makes it possible to apply the same method to bound\nthe probability that a fixed valid partial assignment can be extended to a satisfying assignment.\n(This amounts to applying the method to random formulas coming from a variety of distributions\non formulas with both 2-clauses and 3-clauses.) This allows us to estimate the probability of the\nexistence of a non-trivial core, and thus to prove the theorem.\n3\nProof of the main theorem\nThis section contains a proof of Theorem 1. We begin with an overview of the entire proof, in the\ncourse of which we will state various technical lemmas; these lemmas will be proved in the three\nsubsections that follow.\nNote first that for α ≥4.506 the statement of the main theorem follows from the fact that\nrandom 3-Sat formulas of density at least 4.506 are known to be unsatisfiable with high proba-\nbility [6]. Hence from now on we focus on the case that α ∈[4.453, 4.506]. Our goal is to prove\nthat, for densities above 4.453, with high probability there are no non-trivial covers that can be\nextended to satisfying assignments, i.e., there are no non-trivial cores.\nTo this end, define the size of a cover (or of a core) to be the number of variables assigned value 0\nor 1. The following lemma, proved in [14], establishes that with high probability all non-trivial\ncovers (and consequently cores) are of linear size.\n5"},{"page":6,"text":"Lemma 5 [14] For a random 3-Sat formula of density α, with high probability there are no non-\ntrivial covers of size strictly less than\n1\nαe2 n.\nThis lemma implies that it is sufficient to consider core assignments of size an for a ∈[1/(αe2), 1].\nLet Xa denote the number of cover assignments of size ⌊an⌋, and Ya denote the number of core\nassignments of size ⌊an⌋.\nIn addition to the density α, the size measure a, and the weight parameter ρ ∈[0, 1] from\nTheorem 4, we will use two further parameters, d and b, whose precise definitions will be given\nin the proofs in the following subsections.\nRoughly speaking, d is the fraction of clauses that\nare constraining with respect to a given partial assignment, and b is the fraction of constrained\nvariables with respect to a given partial assignment. The ranges of these parameters are d ∈(a/α, 1]\nand b ∈[0, 1 −a]. We will also make use of a derived parameter r which is defined by d, a and\nα. It appears in connection to the event that the constraining clauses succeed in constraining all\nconstrained variables. Specifically, r is the value satisfying the equation d = ar\nα ln\nr\nr−1. (Note that\nsuch a value r always exists and is unique as the right-hand side is monotonic in r.) In fact, because\nof the form of this expression, it will be more convenient to think of d as being determined by r,\na, and α. Whenever we take the supremum over one of these parameters, we always mean the\nsupremum over its allowed range.\nWe now define two functions that play a central role in our analysis:\nf(α, a, r)\n:=\na ln(2) + H(a) + αH(d) + αd ln\n 3a3\n8\n \n+ α(1 −d) ln\n \n1 −a2(3 −a)\n4\n \n+ αd ln(r/e) −a ln(r −1);\nh(α, a, r, ρ, b)\n:=\nb ln(2) + (1 −a −b) ln(ρ) + (1 −a)H(b/(1 −a))\n−α(1 −d)b b(6 −5b −15a) + 12(1 −a)a\n2(4 −a2(3 −a))\n+ b ln\n \n1 −ρe\n−3α(1−d)\nb(b+2a)\n2(4−a2(3−a))\n \n.\nHere H denotes the entropy function H(x) = −x ln(x) −(1 −x) ln(1 −x).\nThe first ingredient in the proof of Theorem 1 is the following lemma, whose proof is presented\nin Section 3.1.\nLemma 6 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1],\nlim\nn→∞\n1\nn ln (E[Xa]) ≤sup\nr f(α, a, r).\nA simple application of Markov’s inequality immediately yields:\nCorollary 7 If α ≥1 and a ∈[0, 1] are such that, for every r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds\nthat f(α, a, r) < 0, then with high probability random 3-Sat formulas of density α do not have\ncovers of size an.\nThe second ingredient in the proof of Theorem 1 is the following lemma, which is proved in\nSection 3.2.\nLemma 8 For a random 3-Sat formula of density α ≥1, and for every a ∈[0, 1] and ρ ∈[0, 1),\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr (f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ, b)}).\n6"},{"page":7,"text":"Remark: This lemma can be strengthened by allowing ρ to depend on r; however, we will not\nneed to use this stronger version.\nSince every core is a cover we know that Ya ≤Xa. Hence (in light of Lemma 6) Lemma 8 is\ninteresting only when supb h(α, a, r, ρ, b) is negative. In fact, we shall prove Lemma 8 by showing\nthat this supremum bounds the logarithm of 1/n times the probability that a particular cover of\nsize a can be extended to a satisfying assignment. (For a precise statement, see Lemma 11 in\nSection 3.2.) An immediate corollary of Lemma 8 is the following.\nCorollary 9 If α ≥1, a ∈[0, 1] and there exists ρ ∈[0, 1) such that for every r > 1 with\nd = ar\nα ln\nr\nr−1 ≤1, and for every b ∈[0, 1 −a], it holds that f(α, a, r) + h(α, a, r, ρ, b) < 0, then with\nhigh probability random 3-Sat formulas of density α do not have cores of size an.\nThe final ingredient in the proof of Theorem 1 is the following numerical claim, whose proof we\ndiscuss in Section 3.3.\nClaim 10 For every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1], and r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it\nholds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nRemark: In Claim 10 we have used the weight ρ ≡ρ(a) = 0.4a+0.7. We arrived at this particular\nchoice of ρ by first optimizing numerically for ρ at 100 values for a ∈[1/4.506, 1], and then fitting\na simple function to the values for ρ that were found. Since Lemma 8 holds for every value of ρ,\nwe may choose any convenient function. Using a simple analytic function guarantees that h is also\nanalytic, which makes the numerical analysis of h easier.\nFinally, combining Claim 10 with Corollary 9 completes the proof of Theorem 1.\n3.1\nThe expected number of covers: Proof of Lemma 6\nLet s = ⌊an⌋. Then we have\nE[Xa]\n=\nE\n \n \nX\nσ∈{0,1,∗}n\nInd [σ is valid ∩(n∗(σ) = n −s) ∩(no(σ) = 0)]\n \n \n=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is valid ∩(no(σ) = 0)]\n=\n n\ns\n \n2s Pr[σ = (0s ∗n−s) is valid and x1, . . . , xs are constrained].\nWe denote by P the probability that σ = (0s ∗n−s) is valid and all of its assigned variables\nare constrained. P is equivalent to the probability of the following event in an experiment with\nm = αn balls thrown uniformly and independently at random into 23 n\n3\n \nbins. There are 3 kinds\nof bins:\n1. Bins of type 1 should be empty. These correspond to clauses that are not allowed:\n• (xi1 ∨xi2 ∨xi3), with i1, i2, i3 ∈{1, 2, . . . , s};\n• (xi1 ∨xi2 ∨ ̄xj), with i1, i2 ∈{1, 2, . . . , s} and j > s;\n• (xi1 ∨xi2 ∨xj), with i1, i2 ∈{1, 2, . . . , s} and j > s.\n7"},{"page":8,"text":"The total number of these is\n s\n3\n \n+ 2(n −s)\n s\n2\n \n= n3\n a3\n6 + a2(1 −a)\n \n+ o(n3).\n2. Bins of type 2 correspond to constraining clauses: (xi1 ∨xi2 ∨ ̄xt), with i1, i2, t ∈{1, 2, . . . , s}.\nFor each variable xt there are\n s−1\n2\n \n= n2 a2\n2 +o(n2) clauses that could constrain it and at least\none has to be included. Equivalently, there has to be at least one ball in one of those bins for\nevery xt with t ∈{1, 2, . . . , s}. The total number of these clauses is: s\n s−1\n2\n \n= n3 a3\n2 + o(n3).\n3. There are no constraints for the remaining bins, of type 3. Their total number is\n23\n n\n3\n \n−n3\n a3\n6 + a2(1 −a)\n \n−n3 a3\n2 + o(n3) = n3\n3 (4 −a2(3 −a)) + o(n3).\nSuppose m′ = dm of the clauses we choose are of type 2, and the remaining m −m′ are of\ntype 3. The probability of this event is\npm′ =\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\n.\nThe probability that the m′ clauses of type 2 are such that there is at least one of each kind is\nthe same as the coupon collectors probability of success, with s = ⌊an⌋different coupons, and\nm′ = (dα)n trials.\nWe will use the following general fact from [4], which was previously used\nin a very similar context in [12]: Let q(cN, N) denote the probability of collecting N coupons\nwithin cN trials.\nIf c < 1, q(cN, N) = 0.\nOtherwise, as N goes to infinity q(cN, N) grows\nlike g(c)N, where g(c) =\n r0\ne\n c\n1\nr0−1, and r0 is the solution of r ln\n \nr\nr−1\n \n= c.\nMore precisely,\nlimN→∞1\nN ln (q(cN, N)) = ln(g(c)). We have\nP\n=\nm\nX\nm′=0\n m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (6 −2a)\n8\n+ o(1)\n m−m′\nq(m′, s)\n≤\nm max\nm′\n( m\nm′\n 3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3 −a)\n4\n+ o(1)\n m−m′\nq(m′, s).\n)\nFinally,\nE[Xa] ≤\n n\ns\n \n2sm max\nm′\n m\nm′\n \n3 a3\n8\n+ o(1)\n m′ \n1 −a2 (3−a)\n4\n+ o(1)\n m−m′\nq(m′, s)\n \n,\nand hence\nlim\nn→∞\n1\nn ln (E[Xa])\n≤\nln\n \n2a\naa(1 −a)1−a\n \n+ sup\nd\n \n \n \n \n \nα ln\n \n \n \n \n3 a3\n8\n d \n1 −a2 (3−a)\n4\n 1−d\ndd (1 −d)1−d\n \n \n + a ln\n \ng\n dα\na\n \n \n \n \n \n \n=\nsup\nd\nf(α, a, r) = sup\nr\nf(α, a, r).\nThis completes the proof of Lemma 6.\n8"},{"page":9,"text":"3.2\nThe probability that a cover assignment is a core: Proof of Lemma 8\nFor a partial assignment σ, it will be convenient to denote by φσ(xS∗(σ)) the formula that is obtained\nby substituting the variables that have 0/1 assignments in σ, i.e., removing from φ clauses that\nare satisfied by at least one of the assigned variables and removing all remaining appearances of\nassigned variables. This is a formula on n∗(σ) variables. Notice that if σ is a valid assignment\nfor φ then the formula φσ contains no empty clauses and no unit clauses. Furthermore all clauses\nof type 2 (from the previous section) are removed, because they are satisfied by their constrained\nvariables. Among the clauses of type 3, there are clauses that are removed, there are clauses that\nbecome two-variable clauses, and there are clauses that remain untouched. Since there is no simple\nway to describe the resulting distribution on formulas, we will keep referring to the set of all clauses\nof type 3, even the ones that are removed in φσ. When we condition on the fact that σ is a cover\nand m′ of the clauses are of type 2, as in the previous section, we know that the set of clauses we\nare interested in are distributed exactly as a uniform set of (m −m′) clauses of type 3. Thus we\ncan express the expected number of cores as\nE[Ya]\n=\nX\nσ∈{0,1,∗}n:n∗(σ)=n−s\nPr[σ is a cover] × Pr[σ is a core | σ is a cover ]\n=\n n\ns\n \n2s Pr[σ = (0s ∗n−s) is a cover] × Pr[σ = (0s ∗n−s) is a core | σ is a cover]\n=\n n\ns\n \n2s ×\nm\nX\nm′=s\npm′ q(m′, s)\n× Pr[φσ(xs+1, . . . , xn) is satisfiable | σ = (0s ∗n−s), m −m′ clauses are of type 3].\nWe will bound this probability using the Poisson approximation, which is a standard technique\nin this area. There are two related random models. In the first model, which we call the exact\nmodel, (m −m′) clauses are chosen uniformly at random with replacement from all M = n3(4 −\na2(3 −a))/3 + o(n3) clauses of type 3. In the second model, which we call the Poisson model, each\nof the M clauses is included in the formula with probability p = (m −m′)/M −1/(n2√log n) =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2\n \n. The expected number of clauses in both models is the same up to a term\nδ = M/(n2√log n) = Θ(n/√log n) which is sub-linear in n.\nThe Poisson model has been studied before in the context of random 3-Sat. It can be shown,\nas the example below demonstrates, that whenever a property holds with high probability in the\nexact model, it also holds with high probability in the corresponding Poisson model. Applying first\nmoment techniques to the Poisson model is usually easier, because the clauses are independently\nchosen; however the bounds obtained are usually weaker.\nNext, we relate the probability that φσ is satisfiable under the two models. Let Prp denote\nprobability in the Poisson model, and Pre denote probability in the exact model. Let the random\nvariable J denote the number of distinct clauses included in the formula. Then\nPrp[φσ is satisfiable]\n=\nM\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i]\n≥\nm−m′−δ\nX\ni=0\nPrp[J = i] × Pr[φσ is satisfiable | J = i].\n9"},{"page":10,"text":"Since this conditional probability decreases as i increases, and Prp[J ≤Ep[J]] ≥1/2, where Ep[J] =\nm −m′ −δ, we have\nPrp[φσ is satisfiable]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] ×\nm−m′−δ\nX\ni=0\nPr[J = i]\n≥\nPr[φσ is satisfiable | J = m −m′ −δ] × 1\n2.\nOn the other hand, for the exact model\nPre[φσ is satisfiable]\n=\nm−m′\nX\ni=0\nPre[J = i] × Pr[φσ is satisfiable | J = i]\n≤\n m−m′−δ\nX\ni=0\nPre[J = i]\n!\n+ Pr[φσ is satisfiable | J = m −m′ −δ]\n≤\nPre[at least δ clauses repeated] + 2 Prp[φσ is satisfiable]\n≤\n m −m′\nδ\n m −m′\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n≤\n (m −m′)2\nM\n δ\n+ 2 Prp[φσ is satisfiable]\n=\nθ(2−n√log n) + 2 Prp[φσ is satisfiable].\nThus if the probability of satisfiability in the Poisson model is bounded above by c−n for some\nconstant c, then limn→∞1\nn ln (Pre[ φσ is satisfiable]) ≤c. Therefore, it suffices to get a bound for\nthe Poisson model. Any of the first moment techniques for bounding the satisfiability threshold of 3-\nSat can be adapted to bound the probability that φσ is satisfiable. Of the ones that are technically\napplicable (i.e., result in an explicit analytic expression for every setting of the parameters s and\nm′), we obtain the strongest result using the novel approach of applying the first moment method\nto the distribution on partial assignments defined by the weights in equation (1).\nLemma 11 In the Poisson model with parameters n, m = αn, m′ = dm, s = ⌊an⌋and for\nσ = (0s ∗n−s), and r such that d = ar\nα ln\nr\nr−1,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤\ninf\nρ∈[0,1)\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b).\nProof: We will apply Theorem 4 to the formula φσ(xs+1, . . . , xn). The theorem implies that if φσ\nhas a satisfying assignment then P\nτ∈V W(τ) ≥1 where W(τ) = ρn∗(τ)(1 −ρ)no(τ) and the sum is\nover the set V of all partial assignments τ ∈{0, 1, ∗}n−s that are valid for φσ. Thus the probability\nof satisfiability is bounded from above by the expected value of P\nτ∈V W(τ). This holds for any\nvalue of ρ ∈[0, 1]. We bound this expectation.\nFor any t ∈{0, 1, . . . , n −s} let Zt denote the sum of the weights of valid assignments τ for\nφσ(xs+1, . . . , xn) such that n∗(τ) = n −s −t. Then\nE[Z] =\nn−s\nX\nt=0\nE[Zt] ≤n\nmax\nt∈{0,1,...,n−s} E[Zt].\n10"},{"page":11,"text":"E[Zt]\n=\nt\nX\nu=0\nρn−s−t(1 −ρ)u\nX\nτ∈{0,1,∗}n−s\nPrp [τ is valid ∩(n∗(τ) = n −s −t) ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\nX\nτ∈{0,1,∗}n−s : n∗(τ)=n−s−t\nPrp[τ is valid ∩(no(τ) = u)]\n=\nρn−s−t\nt\nX\nu=0\n(1 −ρ)u\n n −s\nt\n t\nu\n \n2t\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\nNext we derive the probability that the assignment (xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid\nand the first u variables are unconstrained. Recall that φσ is obtained from φ by setting its first s\nvariables according to σ. Only clauses of type 3 influence φσ and according to the Poisson model,\neach of them is included independently with probability p. The probability that the assignment\n(xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid and the first u variables are unconstrained is equivalent\nto the probability that among the included clauses of type 3, there are no clauses of the following\nkinds:\n• (xi1 ∨xi2 ∨xi3), with i1 ∈{1, 2, . . . , s + t}, i2, i3 ∈{s + 1, s + 2, . . . , s + t},\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨xj), with i1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, s + 2, . . . , s + t} and j > s + t,\n• (xi1 ∨xi2 ∨ ̄xj), with i1 ∈{1, 2, . . . , s+t}, i2 ∈{s+1, s+2, . . . , s+t} and j ∈{s+1, . . . , s+u},\nand for every j ∈{s + u + 1, . . . , s + t}, the formula contains a clause (xi1 ∨xi2 ∨ ̄xj), with\ni1 ∈{1, 2, . . . , s + t}, i2 ∈{s + 1, . . . , s + t}.\nIn the Poisson model all clauses are independent, so it is easy to put these events together to\nobtain:\nQ\n=\nPrp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]\n=\n(1 −p)((t\n3)+s(t\n2)) + 2(n−s−t)((t\n2) + st) + u((t\n2)+st) ×\n \n1 −(1 −p)(t\n2)+st t−u\nThe expression for the expectation can be simplified:\nE[Zt]\n=\nρn−s−t\n n −s\nt\n \n2t\nt\nX\nu=0\n(1 −ρ)u\n t\nu\n \nQ\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×\nt\nX\nu=0\n t\nu\n \n(1 −ρ)u (1 −p)u((t\n2)+st) ×\n \n1 −(1 −p)(t\n2)+st t−u\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n3) + s(t\n2) + 2(n−s−t)((t\n2) + st)\n×\n \n(1 −ρ)(1 −p)(t\n2)+st +\n \n1 −(1 −p)(t\n2)+st t\n=\nρn−s−t\n n −s\nt\n \n2t (1 −p)(t\n2)(6n−5t−3s−2)/3+2(n−s−t)st \n1 −ρ(1 −p)(t\n2)+st t\n11"},{"page":12,"text":"Let t = bn, and recall that s = ⌊an⌋, p =\n3α(1−d)\nn2(4−a2(3−a)) + o\n 1\nn2\n \n. Then\nlim\nn→∞\n1\nn ln (E[Zt])\n=\nln\n (1 −a)1−a 2b ρ1−a−b\nbb (1 −a −b)1−a−b\n \n−α (1 −d) b (b(6 −5b −3a) + 12(1 −a −b)a)\n2(4 −a2(3 −a))\n+ b ln\n \n1 −ρe\n−3α(1−d)b(b+2a)\n2(4−a2(3−a))\n \n=\nh(α, a, r, ρ, b).\nFinally,\nlim\nn→∞\n1\nn ln (Prp[ φσ(xs+1, . . . , xn) is satisfiable]) ≤lim\nn→∞\n1\nn ln (E[Z]) ≤\nsup\nb∈[0,1−a]\nh(α, a, r, ρ, b),\nwhich is the statement of Lemma 11.\nSubstituting the bound of Lemma 11 into the expression for the expectation yields\nlim\nn→∞\n1\nn ln (E[Ya]) ≤sup\nr\n(f(α, a, r) + min{0, sup\nb\nh(α, a, r, ρ(a, r), b)}),\ncompleting the proof of Lemma 8.\n3.3\nNumerical analysis: Steps to the proof of Claim 10\nWhat remains is to verify the numerical claim that for every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1],\nand r > 1 with d = ar\nα ln\nr\nr−1 ≤1, it holds that either f(α, a, r) < 0 or for every b ∈[0, 1 −a],\nf(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0. We outline the steps towards a proof.\nThere are two stages. First, we identify the (a, r) pairs such that for every α ∈[4.453, 4.506] it\nholds that f(α, a, r) < 0. Second, for the remaining range of values of (a, r) we show that for every\nb ∈[0, 1 −a], f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) < 0.\nThe derivative of f with respect to r is\n∂f\n∂r = a\n \nln\n \n1 +\n1\nr −1\n \n−\n1\nr −1\n \n× ln\n \n \n3\n \nα −ar ln\n \nr\nr−1\n \n2\n 4\na2 −3 + a\n \nln\n \nr\nr−1\n \n \n .\nSince the first factor is always negative the derivative is 0 when\n3α = ln\n \nr\nr −1\n 8\na2 −6 + 2a + 3ar\n \n.\n(2)\nThis equation has a unique root for every a ∈[1/(4.506e2), 1], α ∈[4.453, 4.506] (the derivative of\nthe right-hand side is negative). Therefore we can conclude that for every a and α in the above\nrange, f is at first increasing with r then decreasing.\nIts maximum is achieved at the root of\nequation (2).\nFigure 1 shows the values for r where f is maximized, with respect to a, when\nα = 4.453.\nSince the root of (2) is monotone decreasing with respect to a and α, we can find values r1\nand r2 such that for every a and α in the range, if r < r1 then ∂f/∂r > 0, and if r > r2 then\n∂f/∂r < 0. Values satisfying this condition are r1 = 1.2 and r2 = 670. Thus if we show that\nf(α, a, r) is negative for r = r1 and r = r2 then it is negative for any r outside the range (1.2, 670).\n12"},{"page":13,"text":"5\n 10\n 15\n 20\n 25\n 0\n 0.2\n 0.4\n 0.6\n 0.8\n 1\nr\na\narg maxr f (4.453, a, r)\nFigure 1: The pairs (a, r) such that f(4.453, a, r) > −0.0001, and the value of r that maximizes\nf(4.453, a, r) for every a.\nWe will use the shorthand notation q := r ln\nr\nr−1. For r ∈(1.2, 670) we have q ∈(1.0007, 2.16).\nThe whole range satisfies the condition that d = ar\nα r ln\nr\nr−1 ≤1.\nNext we take care of the boundary region with respect to a. Notice that the derivative of f is\nnegative for a large enough (close to 1), because of the entropy term in f. The derivative of f with\nrespect to a is\n∂f\n∂a\n=\nln(2) −ln\na\n1 −a −q ln\n3a2(α −aq)\n2q(4 −a2(3 −a)) + 3q −3a(α −aq)(2 −a)\n4 −a2(3 −a)\n+ q ln(r) −ln(r −1).\nThe following observations are helpful for bounding this derivative:\n• 3q + q ln(r) −ln(r −1) is maximized in the interval r ∈[1.2, 670] at r = 1.2.\n• a2(α −aq)/(4 −a2(3 −a)) is an increasing function of a.\n• a(α −aq)(2 −a)/(4 −a2(3 −a)) is an increasing function of a.\nFor a > 0.999, using the above facts, the derivative is negative. Thus if we show that f is\nnegative for a = 0.999, r ∈(1.2, 670), α ∈[4.453, 4.506] then f is negative also for every a > 0.999.\nWe are left with the region a ∈[1/(4.506e2), 0.999], r ∈(1.2, 670).\nIn this region all the\nderivatives of f can be bounded, and a sufficiently fine grid can be chosen over which to evaluate\nf in order to identify the grid sections where the function can take positive values. In particular,\nthe derivative with respect to a is at most 28.2, and with respect to α it is at most 1. Furthermore,\nsince we know that for every a and α, f is maximized as a function of r at the root of equation (2),\none can find this maximum and the range of r where f(α, a, r) is positive using binary search. The\npoints with f(4.453, a, r) > −0.0001 are depicted in Figure 1.\nIn the second stage we need to analyze f(α, a, r) + h(α, a, r, 0.4a + 0.7, b) for the remaining\nregion of values for (a, r).\n13"},{"page":14,"text":"First, notice that we can take care of the boundary regions with respect to b by taking advantage\nof the entropy term, which has very large slope for b close to 0, and very steep negative slope for b\nclose to 1 −a. Specifically, the derivative with respect to b is\n∂h\n∂b\n=\nln(2) −ln(ρ) −ln\n \nb\n1 −a −b\n \n−α(1 −d)12b −15b2 −30ab + 12a −12a2\n2(4 −a2(3 −a))\n+ ln(1 −ρe−A) + 3α(1 −d)b\nρe−A(a + b)\n(1 −ρe−A)(4 −a2(3 −a))\nwhere A = 3α(1−d)b(b+2a)\n2(4−a2(3−a)) .\nUsing the bounds on all parameters: α ∈[4.453, 4.506], a ∈(0.28, 0.7), r ∈(1.5, 14.3), and the\nones that follow from those: d ∈(0.06, 0.26), A ∈[0, 1.59], ρ ∈(0.02, 0.21), we can conclude that\n−3.04 −ln\n \nb\n1 −a −b\n \n< ∂h\n∂b < 5.38 −ln\n \nb\n1 −a −b\n \n.\nThe lower bound can be made positive by setting b/(1−a −b) ≤0.04 and the upper bound can\nbe made negative by setting b/(1 −a −b) ≥220. Therefore it suffices to show that h is negative\nfor b in the range [0.01, 0.996(1 −a)].\nAgain, using the bounds for the parameters, all the derivatives of f + h can be bounded, and a\nsufficiently fine grid can be chosen over which to evaluate f +h. We did not perform the evaluation\non a grid that is as fine as is required for the rigorous proof because, based on the current bounds on\nthe derivatives, we would need to evaluate the function at more than 1010 points. (As we discussed\nin the introduction, we feel that the computational effort is not yet justified.)\nFor illustration, Figure 2 shows the estimated values of supr f(α, a, r) and of supr,b(f(α, a, r) +\nh(α, a, r, ρ(a), b)) for α = 4.453, 4.470, 4.490, and 4.506. The maximum is taken over evaluations\nat a grid of step size 0.001 for the parameters r and b. We estimated with better precision the\nlocation of the maximum of f + h by evaluating the function at a fine grid in the region where\nthe evaluations on the coarse grid give the largest values. The maximum found in this way is at\nα = 4.453, a = 0.62566, b = 0.03568 and r = 2.00134 (d = 0.19473). The value of f + h at this\npoint is −0.000058.\n4\nConcluding remarks\nAs mentioned in the introduction, all of the known rigorous upper bounds for the satisfiability\nthreshold of 3-Sat are based on the first moment method [10, 5, 13, 9, 12, 6]; the corresponding\nupper bounds in this sequence of results are: 4.758, 4.64, 4.601, 4.596, 4.571, 4.506. The general\nmethod is to consider a random variable Z that is equal to the number of satisfying assignments\nof a particular kind. These satisfying assignments are such that at least one exists if the formula is\nsatisfiable. Showing that above a certain density E[Z] →0 thus implies (by Markov’s inequality)\nthat Pr[Z = 0] →1, and consequently the probability that the formula is unsatisfiable also goes\nto 1. For example, [5] takes Z to be the number of negatively prime solutions, i.e., solutions for\nwhich every variable assigned 1 is constrained.\nHere, we have used the same idea but with Z taken to be the total weight of partial assignments\nunder the weight function (1) inspired by Survey Propagation. Given the dramatic success of Survey\nPropagation for random 3-Sat, it seems plausible that this approach can potentially yield rather\ntight upper bounds on the threshold. We were able to achieve only partial progress in this direction,\nbut we are quite hopeful that extensions of our approach could lead to further progress.\n14"},{"page":15,"text":"-0.08\n-0.06\n-0.04\n-0.02\n 0\n 0.02\n 0.1\n 0.2\n 0.3\n 0.4\n 0.5\n 0.6\n 0.7\n 0.8\n 0.9\n 1\nsupr f, supr,b (f+h)\na\nα = 4.453\nα = 4.470\nα = 4.490\nα = 4.506\nFigure 2: The values of supr f(α, a, r) and supr,b(f(α, a, r) + h(α, a, r, 0.4a + 0.7, b)) for α = 4.453,\n4.470, 4.490, and 4.506.\nOne natural extension would be to use a different weight ρ for each variable, depending for\nexample on the number of positive and negative occurrences of the variable. The corresponding\ngeneralization of Theorem 4 is proved in [2]. It is quite possible that the value of E[Z] in this case\nis significantly smaller.\n5\nAcknowledgments\nWe would like to thank the anonymous referees for their detailed and very useful suggestions.\nReferences\n[1] D. Achlioptas and F. Ricci-Tersenghi. On the solution-space geometry of random constraint\nsatisfaction problems. In Proc. 38th ACM Symp. Theory of Computing (STOC), pages 130–139,\n2006.\n[2] F. Ardila and E. Maneva. Pruning processes and a new characterization of convex geometries.\nDiscrete Mathematics, 2008. To appear.\n[3] A. Braunstein and R. Zecchina. Survey propagation as local equilibrium equations. J. Stat.\nMech. : Theory and Experiments (JSTAT), page 06007, 2004.\n[4] V. Chvatal. Almost all graphs with 1.44 edges are 3-colourable. Random Struct, Algorithms,\n2:11–28, 1991.\n[5] O. Dubois and Y. Boufkhad. A general upper bound for the satisfiability threshold of random\nr-SAT formulae. J. Algorithms, 24:395–420, 1997.\n15"},{"page":16,"text":"[6] O. Dubois, Y. Boufkhad, and J. Mandler. Typical random 3-SAT formulae and the satisfiability\nthreshold. In Proc. 11th ACM-SIAM Symp. on Discrete Algorithms (SODA), pages 126–127,\n2000. Extended version at arXiv:cs/0211036.\n[7] E. Friedgut. Neccesary and sufficient conditions for sharp threhsolds of graph properties and\nthe k-problem. J. Amer. Math. Soc., 12:1017–1054, 1999.\n[8] M. Hajiaghayi and G. Sorkin. The satisfiability threshold of random 3-SAT is at least 3.52.\nTechnical report, 2003. Preprint at arXiv:math/0310193.\n[9] S. Janson, Y. C. Stamatiou, and M. Vamvakari. Bounding the unsatisfiability threshold of\nrandom 3-SAT. Random Struct. Algorithms, 17(2):103–116, 2000.\n[10] A. Kamath, R. Motwani, K. V. Palem, and P. G. Spirakis. Tail bounds for occupancy and the\nsatisfiability threshold conjecture. Random Struct. Algorithms, 7(1):59–80, 1995.\n[11] A. C. Kaporis, L. M. Kirousis, and E. G. Lalas. The probabilistic analysis of a greedy satisfi-\nability algorithm. Random Struct. Algorithms, 28(4):444–480, 2006.\n[12] A. C. Kaporis, L. M. Kirousis, Y. C. Stamatiou, M. Vamvakari, and M. Zito. The unsatisfia-\nbility threshold revisited. Discrete Applied Mathematics, 155(12):1525–1538, 2007.\n[13] L. M. Kirousis, E. Kranakis, D. Krizanc, and Y. C. Stamatiou. Approximating the unsatisfi-\nability threshold of random formulas. Random Struct. Algorithms, 12(3):253–269, 1998.\n[14] E. Maneva, E. Mossel, and M. J. Wainwright. A new perspective on survey propagation. J.\nACM, 54(4):2–41, 2007.\n[15] M. M ́ezard, T. Mora, and R. Zecchina. Clustering of solutions in the random satisfiability\nproblem. Phys. Rev. Lett., 94(197205), 2005.\n[16] M. M ́ezard, G. Parisi, and M. A. Virasoro. Spin Glass Theory and Beyond. World Scientific,\nSingapore, 1987.\n[17] M. M ́ezard, G. Parisi, and R. Zecchina. Analytic and algorithmic solution of random satis-\nfiability problems. Science, 297, 812, 2002. (Scienceexpress published online June 27, 2002;\n10.1126/science.1073287).\n[18] M. M ́ezard and R. Zecchina. Random k-satisfiability: from an analytic solution to an efficient\nalgorithm. Phys. Rev. E, 66, 2002.\n[19] J. Pearl. Probabilistic reasoning in intelligent systems: networks of plausible inference. Morgan\nKaufmann, Palo Alto, CA, 1988.\n16"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"very large instances (say, with n = 106 variables) it finds solutions for formulas of densities only just","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"below the conjectured threshold value α = 4.27; however, a rigorous analysis of its performance is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"random and with replacement m = ⌊αn⌋clauses out of all possible clauses on 3 distinct variables.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"Suppose that the variables x = (x1, . . . , xn) are allowed to take values in {0, 1, ∗}, which we refer","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"unsatisfying; or (b) all variables are unsatisfying except for one index j ∈V (c), for which xj = ∗.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"W(σ) := ρn∗(σ)(1 −ρ)no(σ),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"family of distributions, parameterized by ρ, is defined as Pr[σ] ∝W(σ). At one extreme, ρ = 0,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"ρ = 1, corresponds to Survey Propagation. The pure version of Survey Propagation corresponds to","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"setting ρ = 1. Intermediate values of ρ interpolate between these extremes.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"σj = τj for all j ̸= i; and (ii) τi = ∗and σi ̸= ∗.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"assignments without unconstrained variables, i.e., the positive weight assignments for ρ = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"τ≤σ W(τ) = ρn∗(σ) for any valid partial assignment σ ∈","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"τ≤σ W(τ) = 1.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"constrained variables. Specifically, r is the value satisfying the equation d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"Here H denotes the entropy function H(x) = −x ln(x) −(1 −x) ln(1 −x).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"Corollary 7 If α ≥1 and a ∈[0, 1] are such that, for every r > 1 with d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"Claim 10 For every α ∈[4.453, 4.506], a ∈[1/(4.506e2), 1], and r > 1 with d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"Remark: In Claim 10 we have used the weight ρ ≡ρ(a) = 0.4a+0.7. We arrived at this particular","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"Let s = ⌊an⌋. Then we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"Ind [σ is valid ∩(n∗(σ) = n −s) ∩(no(σ) = 0)]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"σ∈{0,1,∗}n:n∗(σ)=n−s","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"Pr[σ is valid ∩(no(σ) = 0)]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"2s Pr[σ = (0s ∗n−s) is valid and x1, . . . , xs are constrained].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"We denote by P the probability that σ = (0s ∗n−s) is valid and all of its assigned variables","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"m = αn balls thrown uniformly and independently at random into 23 n","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"= n3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"= n2 a2","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"= n3 a3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"2 + o(n3) = n3","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"Suppose m′ = dm of the clauses we choose are of type 2, and the remaining m −m′ are of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"pm′ =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"the same as the coupon collectors probability of success, with s = ⌊an⌋different coupons, and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"m′ = (dα)n trials.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"If c < 1, q(cN, N) = 0.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"like g(c)N, where g(c) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"= c.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"N ln (q(cN, N)) = ln(g(c)). We have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"m′=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"f(α, a, r) = sup","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"σ∈{0,1,∗}n:n∗(σ)=n−s","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"2s Pr[σ = (0s ∗n−s) is a cover] × Pr[σ = (0s ∗n−s) is a core | σ is a cover]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"m′=s","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"× Pr[φσ(xs+1, . . . , xn) is satisfiable | σ = (0s ∗n−s), m −m′ clauses are of type 3].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"model, (m −m′) clauses are chosen uniformly at random with replacement from all M = n3(4 −","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"of the M clauses is included in the formula with probability p = (m −m′)/M −1/(n2√log n) =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"δ = M/(n2√log n) = Θ(n/√log n) which is sub-linear in n.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"Prp[J = i] × Pr[φσ is satisfiable | J = i]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"Prp[J = i] × Pr[φσ is satisfiable | J = i].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"Since this conditional probability decreases as i increases, and Prp[J ≤Ep[J]] ≥1/2, where Ep[J] =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"Pr[φσ is satisfiable | J = m −m′ −δ] ×","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"Pr[J = i]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"Pr[φσ is satisfiable | J = m −m′ −δ] × 1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"Pre[J = i] × Pr[φσ is satisfiable | J = i]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"i=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"Pre[J = i]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"+ Pr[φσ is satisfiable | J = m −m′ −δ]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"Lemma 11 In the Poisson model with parameters n, m = αn, m′ = dm, s = ⌊an⌋and for","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"σ = (0s ∗n−s), and r such that d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"τ∈V W(τ) ≥1 where W(τ) = ρn∗(τ)(1 −ρ)no(τ) and the sum is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"φσ(xs+1, . . . , xn) such that n∗(τ) = n −s −t. Then","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"E[Z] =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"t=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"u=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"Prp [τ is valid ∩(n∗(τ) = n −s −t) ∩(no(τ) = u)]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"u=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"τ∈{0,1,∗}n−s : n∗(τ)=n−s−t","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"Prp[τ is valid ∩(no(τ) = u)]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"u=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"Prp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"Next we derive the probability that the assignment (xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"(xs+1, xs+2, . . . , xn) = (0t ∗n−s−t) is valid and the first u variables are unconstrained is equivalent","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"Prp[τ = (0t ∗n−s−t) is valid and xs+1, . . . , xs+u are unconstrained]","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"u=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"u=0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"Let t = bn, and recall that s = ⌊an⌋, p =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"and r > 1 with d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"∂r = a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"3α = ln","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"∂f/∂r < 0. Values satisfying this condition are r1 = 1.2 and r2 = 670. Thus if we show that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"f(α, a, r) is negative for r = r1 and r = r2 then it is negative for any r outside the range (1.2, 670).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"We will use the shorthand notation q := r ln","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"The whole range satisfies the condition that d = ar","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"• 3q + q ln(r) −ln(r −1) is maximized in the interval r ∈[1.2, 670] at r = 1.2.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"negative for a = 0.999, r ∈(1.2, 670), α ∈[4.453, 4.506] then f is negative also for every a > 0.999.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"where A = 3α(1−d)b(b+2a)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"h(α, a, r, ρ(a), b)) for α = 4.453, 4.470, 4.490, and 4.506. The maximum is taken over evaluations","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"α = 4.453, a = 0.62566, b = 0.03568 and r = 2.00134 (d = 0.19473). The value of f + h at this","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"that Pr[Z = 0] →1, and consequently the probability that the formula is unsatisfiable also goes","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"Figure 2: The values of supr f(α, a, r) and supr,b(f(α, a, r) + h(α, a, r, 0.4a + 0.7, b)) for α = 4.453,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":44427,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}