Bounds on the Satisfiability Threshold for Power Law Distributed Random SAT
Tobias Friedrich, Anton Krohmer, Ralf Rothenberger, Thomas Sauerwald, Andrew M. Sutton
aa r X i v : . [ c s . D M ] J un Bounds on the Satisfiability Threshold for Power LawDistributed Random SAT
Tobias Friedrich , Anton Krohmer , Ralf Rothenberger , Thomas Sauerwald ,and Andrew M. Sutton Hasso Plattner Institute, Potsdam, Germany, [email protected] University of Cambridge, Cambridge, United Kingdom, [email protected]
Abstract
Propositional satisfiability (SAT) is one of the most fundamental problems in computerscience. The worst-case hardness of SAT lies at the core of computational complexity theory.The average-case analysis of SAT has triggered the development of sophisticated rigorousand non-rigorous techniques for analyzing random structures.Despite a long line of research and substantial progress, nearly all theoretical workon random SAT assumes a uniform distribution on the variables. In contrast, real-worldinstances often exhibit large fluctuations in variable occurrence. This can be modeled bya scale-free distribution of the variables, which results in distributions closer to industrialSAT instances.We study random k -SAT on n variables, m = Θ( n ) clauses, and a power law distributionon the variable occurrences with exponent β . We observe a satisfiability threshold at β =(2 k − / ( k − β (2 k − / ( k − − ε for any constant ε > unsatisfiable with high probability (w. h. p.).For β > (2 k − / ( k −
1) + ε , the picture is reminiscent of the uniform case: instancesare satisfiable w. h. p. for sufficiently small constant clause-variable ratios m/n ; they are unsatisfiable above a ratio m/n that depends on β . Introduction
Satisfiability of propositional formulas (SAT) is one of the most researched problems in the-oretical computer science. SAT is widely used to model practical problems such as boundedmodel checking, hardware and software verification, automated planning and scheduling, andcircuit design. Even large industrial instances with millions of variables can often be solved veryefficiently by modern SAT solvers. The structure of these industrial SAT instances appears toallow a much faster processing than the theoretical worst-case of this NP-complete problem. Itis an open and widely discussed question which structural properties make a SAT instance easyto solve for modern SAT solvers.
Random SAT:
For modeling typical inputs, we study random propositional formulas. Inrandom satisfiability, we have a distribution over Boolean formulas in conjunctive normal form(CNF). The degree of a variable in a CNF formula is the number of disjunctive clauses in whichthat variable appears either positively or negatively. Two interesting properties of randommodels are its degree distribution and its satisfiability threshold . The degree distribution F ( x ) ofa formula Φ is the fraction of variables that occur more than x times (negated or unnegated). Asatisfiability threshold is a critical value around which the probability that a formula is satisfiablechanges from 0 to 1. Uniform random SAT:
In the classical uniform random model, the degree distribution isbinomial. On uniform random k -SAT, the satisfiability threshold conjecture [1] asserts if Φ is aformula drawn uniformly at random from the set of all k -CNF formulas with n variables and m clauses, there exists a real number r k such thatlim n →∞ Pr { Φ is satisfiable } = ( m/n < r k ;0 m/n > r k . A well-known result of Friedgut [22] establishes that the transition is sharp, even though itslocation is not known exactly for all values of k (and may also depend on n ). For k = 2, thecritical threshold is r = 1 [14, 18, 23]. Recently, Coja-Oghlan and Panagiotou [16, 17] gave asharp bound (up to lower order terms) with r k = 2 k log 2 − (1 + log 2) ± o k (1). Ding, Sly, andSun [20] derive an exact representation of the threshold for all k > k , where k is a large enoughconstant. Explicit bounds also exist for low values of k , e.g., 3 . r . r ≈ . Other random SAT models:
In the regular random model [11], formulas are constructedat random, but the degree distribution is fixed: each literal appears exactly ⌊ km n ⌋ or ⌊ km n ⌋ + 1times in the formula. Similarly, Bradonjic and Perkins [12] considered a random geometric k -SAT model in which 2 n points are placed at random in [0 , d . Each point corresponds to aunique literal, and clauses are formed by all k -sets of literals that lie together within a ball ofdiameter Θ( n − /d ). Again, this model has a binomial variable distribution. Power law random SAT:
Recently, there has been a paradigm shift when modeling real-worlddata. In many applications, it has been found that certain quantities do not cluster around aspecific scale as suggested by a uniform distribution, but are rather inhomogeneous [15, 30]. Inparticular, the degree distribution in complex networks often follows a power law [31]. Thismeans that the fraction of vertices of degree k is proportional to k − β , where the constant β depends on the network. To mathematically study the behavior of such networks, randomgraph models that generate a power law degree distribution have been proposed [2, 10, 27, 32].While there has been a large amount of research on power law random graphs in the past fewyears [33], there is little previous work on power law SAT formulas. Nevertheless, the observationthat quantities follow a power law in real-world data has also emerged in the context of SAT [11].As all aforementioned random SAT models assume strongly concentrated degree distributions,it was conjectured that this property might be modeled well by random formulas with a powerlaw degree distribution.To address this conjecture, and to help close the gap between the structure of uniform randomand industrial instances, Anstegui, Bonet, and Levy [4] recently proposed a power-law randomSAT model. This model has been studied experimentally [4–7], and empirical investigationsfound that (1) indeed the constraint graphs of many families of industrial instances obey a power-law and (2) SAT solvers that are constructed to specialize on industrial instances perform better2 k − k − Theorem 5.1Theorem 3.1Theorem 4.2 satisfiableunsatisfiable unknown β m / n . . . . satisfiableunsatisfiable unknown (solver timeout) β m / n Figure 1:
Illustration of our asymptotic results for the power law satisfiability threshold loca-tion when n → ∞ (left) compared with empirical results for randomly generated power law3-SAT formulas on n = 10 variables checked with the SAT solver MiniSAT (right). Thetimeout was set to one hour.on power-law formulas than on uniform random formulas. To complement these experimentalfindings, we contribute with this paper the first theoretical results on this model. Our results:
We study random k -SAT on n variables and m = Θ( n ) clauses. Each clausecontains k = Θ(1) different, independently sampled variables. Each variable x i is chosen withnon-uniform probability p i and negated with probability / . A formal definition can be foundin Section 2. We first study sufficient conditions under which the resulting k -SAT instances are unsatisfiable . Assume a probability distribution ~p on the variables where p i is non-decreasing in i ∈ { , . . . , n } . If the k most frequent variables are sufficiently common, we prove in Section 3the following statement: Theorem 3.1.
Let Φ be a random k -SAT formula with probability distribution ~p on the variables(c.f. Definition 2.1), with k > and mn = Ω(1) . If p n − k +1 = Ω(( log nn ) /k ) , then Φ is w. h. p.unsatisfiable. Our focus are power law distributions with some exponent β . Theorem 3.1 implies thatpower law random k -SAT formulas with β = k − k − − ε for an arbitrary constant ε > , cf. Corollary 3.1.In Section 4 we show that something similar holds for the clause-variable ratio mn , i.e. powerlaw random k -SAT formulas with mn bigger than some constant are unsatisfiable with highprobability. Although this already follows from basic observations, we derive a better bound onthe value of the constant. Theorem 4.1.
Let Φ be a random k -SAT formula with probability distribution ~p on the variables(c.f. Definition 2.1), with k > and r = mn . With high probability, Φ is unsatisfiable if (cid:0) − k (cid:1) r " n Y i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m n < . In Section 5 we prove the following positive result, which complements our picture of thesatisfiability landscape:
Theorem 5.1.
Let Φ be a random k -SAT formula whose variable probabilities follow a powerlaw distribution (c.f. Definition 2.2). If the power law exponent is β > k − k − + ε for an arbitrary ε > , Φ is satisfiable with high probability if mn is a small enough constant. Together our main theorems prove that random k -SAT instances whose variables followpower law distributions do not only exhibit a phase transition for some clause-variable ratio We say that an event E holds w. h. p. , if there exists a δ > E ] > − O ( n − δ ). = mn , but also around the power law exponent β = k − k − . Figure 1 contains an overviewof our results. To prove these statements, we borrow tools developed for the uniform randomSAT model. Note, however, that many of their common techniques like the differential equationmethod seem difficult to apply to non-uniform distributions; as removing a variable results in amore complex rescaling of the rest of the distribution. It is therefore crucial to perform carefuloperations on the formulas that leave the distribution of variables intact. To this end, we usetechniques known from the analysis of power law random graphs. Clause length:
We focus on power law variable distributions but fix the length of every clauseto k >
2. Power law models have also been proposed in which clause length is distributed bya power law as well [4, 5]. As long as there is a constant minimum clause length k min >
2, ourresults can be extended to this case in the following way.If the clause lengths are distributed as a power law, there will appear Θ( n ) clauses of length k min , and all other clauses are of larger size. In that case, Theorems 3.1 and 5.1 are directlyapplicable to the linear number of clauses with size k min (obtaining different hidden constants);and we have that the formula is satisfiable with high probability if β > k min − k min − + ε and m/n isa small enough constant. On the other hand, the formula is unsatisfiable with high probability,if β k min − k min − − ε . Consequently, the satisfiability of the formula does (asymptotically) notdepend on the second power law. We analyze random k -SAT on n variables and m = Θ( n ) clauses, where k >
2. The constant r := mn is called clause-variable ratio or constraint density . We denote by x , . . . , x n the Booleanvariables. A clause is a disjunction of k literals ℓ ∨ . . . ∨ ℓ k , where each literal assumes a (possiblynegated) variable. Finally, a formula Φ in conjunctive normal form is a conjunction of clauses c ∧ . . . ∧ c m . We conveniently interpret a clause c both as a Boolean formula and as a setof literals. Following standard notation, we write | ℓ | to refer to the indicator of the variablecorresponding to literal ℓ . We say that Φ is satisfiable if there exists an assignment of variables x , . . . , x n such that the formula evaluates to 1. Definition 2.1 (Random k -SAT) . Let m, n be given, and consider any probability distribution ~p on n variables with P ni =1 p i = 1 . To construct a random SAT formula Φ , we sample m clausesindependently at random. Each clause is sampled as follows:1. Select k variables independently at random from the distribution ~p . Repeat until no vari-ables coincide.2. Negate each of the k variables independently at random with probability / . Observe that by setting p i = n for all i , we obtain again the uniform random SAT model.The probability to draw a specific clause c is Q ℓ ∈ c p | ℓ | k P J ∈P k ( { , ,...,n } ) Q j ∈ J p j , (2.1)where P k ( · ) denotes the set of cardinality- k elements of the power set. The factor 2 k in the denominator comes from the different possibilities to negate variables. Note that k ! P J ∈P k ( { , ,...,n } ) Q j ∈ J p j is the probability of choosing a k -clause that contains no variablemore than once. To see that this probability is almost 1 for most distributions, we apply thefollowing result from [3]. Lemma 2.1 (Non-Uniform Birthday Paradox) . Let ~p = ( p , . . . , p n ) be any probability distri-bution on n items. Assume you sample t items from ~p . Let E ( t ) be the event that there is acollision, i.e. that at least 2 of t items are equal. Then, Pr[ E ( t )] t k ~p k = t n X i =1 p i . k -clause thereby has collisions is at most k k ~p k ; so for k ~p k = o (1) and constant k we obtain that the probability to draw a specific clause c is(1 + o (1)) k !2 k Y ℓ ∈ c p | ℓ | . (2.2) Power law Distributions.
In this paper, we are mostly concerned with distributions p i thatfollow a power law. To this end, we define two models: A general model to capture most powerlaw distributions (which is harder to analyze), and a concrete model that gives us one instanceof ~p depending only on n that can be used to compute precise leading constants. We use thegeneral model to derive some asymptotic results; and the concrete model to compare with theuniform random SAT model and for the experiments.Before we define these two models, let us establish the concept of a weight w i of a variable x i .The weight gives us (roughly) the expected number of times the variable appears in the formula.That is, p i := w i P j w j . Thus, fixing the weights ~w = ( w , . . . , w n ) also fixes the probability distribution ~p . It is importantto distinguish between the initial distribution of variables ~p and modified distributions that mayarise as a result of stochastic considerations. For instance, the smallest-weight variable in a clauseis clearly not distributed according to ~p (except in 1-SAT). To avoid confusion, we identify avariable with its weight, as the weights stay fixed throughout the analysis. For convenience, wefurther assume w. l. o. g. that the variables are ordered increasingly by weight, i. e. for i j we have w i w j . Note that our definition of power law ensures that for β >
2, we have P j w j = Θ( n ).We are now ready to define the two models. Definition 2.2 (General Power Law) . Let the weights ~w := w , . . . , w n be given, and let W bea weight selected uniformly at random. We say that ~w follows a power law with exponent β , if w = Θ(1) , w n = Θ( n β − ) , and for all w ∈ [ w , w n ] it holds F ( w ) := Pr[ W > w ] = Θ( w − β ) (2.3)Whenever we need the explicit constants bounding the distribution function, we refer tothem by α , α as in α w − β F ( w ) α w − β . (2.4)We point out that Definition 2.2 assumes a deterministic weight sequence; but it can be easilygeneralized to also support randomly generated weights.For the concrete model, we define the weights as follows. Definition 2.3 (Concrete Power Law) . Given a power law exponent β , we call ~w the concretepower law sequence, if w n − i +1 := ( ni ) β − . (2.5)One can check that for these concrete weights, it holds n · F ( w ) = ⌊ nw − β ⌋ , so in a sense,they are a canonical choice for producing a power law weight distribution.To analyze power law distributions, we often make use of the following result of Bringmann,Keusch, and Lengler [13, Lemma B.1], which allows replacing sums by integrals. Theorem 2.1 ([13]) . Let f : R → R be a continuously differentiable function, and let F > ( w ) :=Pr[ W > w ] . Then, for any w ¯ w , X i ∈ [ n ] ,w w i ¯ w n f ( w i ) = f ( w ) · F ( w ) − f ( ¯ w ) · F > ( ¯ w ) + Z ¯ ww f ′ ( w ) · F ( w ) d w. Using this theorem, the following corollary can be shown:
Corollary 2.1.
Let the variables w i be power law distributed with exponent β > , and define W > w := P i ∈ [ n ]: w i > w w i . Then, W > w = Θ( nw − β ) . roof. Observe that P w ′ > w w ′ = P i ∈ [ n ] ,w i > w w i . We apply Theorem 2.1 to obtain n X i ∈ [ n ] ,w i > w w i = w · F ( w ) + Z w n w F ( v ) d v α w − β + [ α − β v − β ] w n w α β − β − w − β . In a similar fashion, one may show that n P i ∈ [ n ] ,w i > w w i > α β − β − (1 − o (1)) w − β .Hence, P j w j = W > w = Θ( n ) and therefore p i = Θ( w i n ). Finally, we denote by V therandom variable describing the weight of a SAT variable chosen according to a power law distri-bution p i , that is, Pr[ V = w ] = P i p i · [ w i = w ], where denotes the indicator variable of theevent. Note that this is not equivalent to W , since there is a subtle difference in the two randomprocesses: W is a random variable drawn uniformly at random from w , . . . , w n , whereas V isa random variable drawn from the same set, but with the non-uniform distribution p , . . . , p n .Hence, by Corollary 2.1, Pr[ V > w ] = Θ( w − β ) . (2.6)Using Theorem 2.1, we can show that the probability to draw a certain clause c is as givenby Equation (2.2) for Definition 2.2 with exponent β >
2, since k ~p k = n X i =1 p i = Θ( n − ) n X i =1 w i = Θ( n − ) · n − ββ − = o (1) . It remains to show that using a power law distribution in Definition 2.1 indeed results in apower law distribution of variable occurrences. Anstegui et al. [5] provide a proof sketch for thisfact, we prove it rigorously.
Theorem 2.2.
Let Φ be a random k -SAT formula that follows an arbitrary power law distribu-tion with exponent β (c.f. Definition 2.2) and m = Θ( n ) . Then, there are d min = Θ ( w min ) and d max = Θ ( w max ) , such that for all d min d d max w. h. p. it holds that N > d = Θ( n · d − β ) , where N > d is the number of variables that appear at least d times in Φ .Proof. Let f x be the number of appearances of x or ¯ x in Φ. Observe that E [ f x ] k · m · p x , sinceeach variable can appear at most once in a clause. On the other hand, it holds E [ f x ] > m · p x ,since this is the expected number of appearances of x in a 1-SAT formula. Thus, since m = Θ( n )and k = Θ(1) by assumption, it holds that E [ f x ] = Θ( w x ) . (2.7)We first prove the statement for d > c ln n , where c > x with E [ f x ] < d appearfewer than d times; and all variables x with E [ f x ] > d appear at least d times. The requirement d min d d max is needed so that the Chernoff bounds work, which might not be the case if d is too close to w min or w max . Due to Definition 2.2 and Equation (2.7) this implies N > d = Θ (cid:0) n · d − β (cid:1) . Now let us consider the case d < c ln n and let Y i be random variables indicating if f x i > d for i = 1 , , . . . , n . To show a lower bound on N > d , we again look at variables x with E [ f x ] > d .For those it holds thatPr ( Y i = 0) Pr (cid:18) f x i < E [ f x i ] (cid:19) e − E [ fxi ] e − d , (2.8)again due to Chernoff Bounds. Also, by Equation (2.7), it holds for variables x with E [ f x ] > d that w x = Ω(2 d ). By the requirements on the weight distribution from Definition 2.2 there are6( n · d − β ) such variables. Therefore, it holds that E [ N > d ] > E X i ∈ [ n ]: E [ f xi ] > d Y i > c ′ (cid:16) − e − d (cid:17) · n · d − β for a suitable constant c ′ >
0. Observe that if we condition on Y i = 1, i. e. that x i appears atleast d times, this lowers the probability of all other variables to appear d times, and vice versa.Thus, the random variables Y , . . . , Y n are negatively correlated and we may apply a Chernoffbound [9, Theorem 1.16]. Since 1 − e − d/ = Ω(1) and d = O (log n ), we obtain that w. h. p. N > d > c ′ · (cid:16) − e − d (cid:17) · n · d − β = Ω( n · d − β ) . To show an upper bound on N > d we consider variables x with E [ f x ] d e of which thereare n · (1 − Θ( d − β )) due to Definition 2.2. For these variables, by a Chernoff bound [21] it holdsthat Pr ( Y i = 1) Pr ( f x i > e · E [ f x i ]) − d . Now let N ′ > d be the number of variables with E [ f x ] d e and f x > d . Thus, there exists aconstant c ′′ > E (cid:2) N ′ > d (cid:3) n · − d (cid:0) − c ′′ · d − β (cid:1) . Due to negative association of the Y i ’s we can again use a Chernoff bound, yielding that w. h. p., N ′ > d n · − d (cid:0) − c ′′ · d − β (cid:1) . (2.9)If E h N ′ > d i is very small, for example E h N ′ > d i = O (log n ), then we can use negative associ-ation to apply the Chernoff bound with t > e · E h N ′ > d i to achieve Equation (2.9) with highprobability, since t = n · d − β = Ω (cid:16) n polylog( n ) (cid:17) . Observe that 1 − c ′′ d − β = O (1). Furthermore,for variables x with E [ f x ] > d e , we pessimistically assume f x > d . This gives us N > d = O (cid:0) n · ( d − β + 2 − d ) (cid:1) = O (cid:0) n · d − β (cid:1) , since 2 − d = O (cid:0) d − β (cid:1) for constant β . For small power law exponents, one can show that they result in formulas that are unsatisfiable(for large n ) for all constant clause-variable ratios. The rationale behind this is that largevariables with weight Θ( w n ) appear polynomially often together in a clause. For constant k ,they thus appear in all 2 k configurations (negated and non-negated), making the formula triviallyunsatisfiable. Theorem 3.1, already stated in the introduction, gives a sufficient condition onthe variable distribution to make a random k -SAT formula unsatisfiable. Theorem 3.1.
Let Φ be a random k -SAT formula with probability distribution ~p on the variables(c.f. Definition 2.1), with k > and mn = Ω(1) . If p n − k +1 = Ω(( log nn ) /k ) , then Φ is w. h. p.unsatisfiable.Proof. Recall that p i is without loss of generality increasing in i . Consider the k largest variables n − k + 1 , . . . , n . We call E i the event that clause i consists of these variables. Then,Pr[ E i ] = Ω( p kn − k +1 ) = Ω( log nn ) . Since each clause is drawn independently at random, we obtain by a Chernoff bound (see forexample Theorem 1.1 in [21]) that with high probability, the total number of clauses consistingof these variables is 7 E| := m X i =1 [ E i ] = Ω(log n ) . In other words, the number of clauses in which the k largest variables appear together increasesas a logarithm in n . Since in each of these clauses, the literals appear negated or non-negatedwith constant probability / , we have that all 2 k possible combinations of negated and non-negated literals appear in the formula with probability at least1 − k · ( k − k ) |E| = 1 − n − Ω(1) by the union bound. Since all 2 k combinations cannot be satisfied at once, the resulting formulais unsatisfiable.By applying Theorem 3.1 to a power law distribution on the variables, we obtain the followingpower law threshold for unsatisfiability. Corollary 3.1.
Let Φ be a random k -SAT formula that follows an arbitrary power law distri-bution fulfilling Definition 2.2. If the power law exponent is β k − k − − ε for an arbitrary ε > , Φ is unsatisfiable with high probability.Proof. Observe that from β = k − k − − ε it follows k = β − β − − ε ′ for some constant ε ′ . Bysetting nF ( w ) k we obtain that the largest k variables all have weight Θ( w n ) = Θ( n β − ) . Consequently, when β > p n − k ) k = Θ( n − k β − β − ) = Θ( n − ε ′ β − β − ) = ω ( log nn ) , and the statement follows from Theorem 3.1. For the case where β
2, one can show usingTheorem 2.1 that P i w i = Θ( n β − ), and therefore p n − k = Ω(1). Again, the statement followsfrom Theorem 3.1. It is a well-known result that random SAT on any probability distribution will result in unsat-isfiable formulas if the clause-variable ratio is high. This follows from the probabilistic method:The expected number of assignments that satisfy a formula is 2 n (1 − − k ) m . This is independentfrom the variable distribution as long as each variable is negated with probability / . Hence,if the clause-variable ratio exceeds ln(2) / ln( k k − ), the resulting formula will be unsatisfiablewith high probability. This constant is rather large, however: In the case of k = 3 this yieldsan upper bound on the clause-variable ratio of ≈ . Theorem 4.1.
Let Φ be a random k -SAT formula with probability distribution ~p on the variables(c.f. Definition 2.1), with k > and r = mn . With high probability, Φ is unsatisfiable if (cid:0) − k (cid:1) r " n Y i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m n < . The following is a corollary from this theorem:
Corollary 4.1.
Let Φ be a random k -SAT formula that follows Definition 2.1 with k > , r = mn and k ~p k = o (1) . With high probability, Φ is unsatisfiable if (cid:0) − k (cid:1) r (cid:18) − exp (cid:18) − (cid:18) k k − r (cid:19) (1 + o (1)) (cid:19)(cid:19) < . roof. We can upper-bound the left-hand side of the inequality as follows (cid:0) − k (cid:1) r " n Y i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m n (cid:0) − k (cid:1) r " n n X i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m = (cid:0) − k (cid:1) r " − n n X i =1 − k · p i k − (cid:0) − k k ~p k (cid:1) ! m by applying the inequality of arithmetic and geometric means. Since k · p i k − (cid:16) − k k ~p k (cid:17) is upper-bounding a probability, we can assume it to be at most 1. It now holds that − k · p i k − (cid:0) − k k ~p k (cid:1) ! m > exp − m (2 k − (cid:16) − k k ~p k (cid:17) k · p i − = exp − k · p i k − m (cid:0) − k k ~p k (cid:1) (1 + o (1)) ! , since k ~p k = o (1) implies max i ( p i ) = o (1). By plugging this into the inequality from before andapplying the inequality of arithmetic and geometric means again, we get (cid:0) − k (cid:1) r " n Y i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m n (cid:0) − k (cid:1) r " − n n X i =1 exp − k · p i k − m (cid:0) − k k ~p k (cid:1) ! (1 + o (1)) ! (cid:0) − k (cid:1) r − n Y i =1 exp − k · p i k − m (cid:0) − k k ~p k (cid:1) ! (1 + o (1)) !! n = (cid:0) − k (cid:1) r " − exp − k k − r (cid:0) − k k ~p k (cid:1) ! (1 + o (1)) ! . For k ~p k = o (1) this is roughly (cid:0) − k (cid:1) r (cid:18) − exp (cid:18) − (cid:18) k k − r (cid:19) (1 + o (1)) (cid:19)(cid:19) . Interestingly, the above corollary gives the same inequality as the Single-Flip Method foruniform random SAT [26]. This shows that the uniform distribution resembles a worst-case forthis method; and all other distributions can only improve this bound.If ~p follows a power law distribution as in Definition 2.3, we can derive the following theorem,which gives an upper bound independent of n . Theorem 4.2.
Let Φ be a random k -SAT formula with k > and r = mn that follows a powerlaw distribution fulfilling Definition 2.3. Let further N ∈ N + be any constant. If the power lawexponent is β > , then Φ is w. h. p. unsatisfiable if (cid:0) − k (cid:1) r N N − Y l =1 " − exp − (1 + o (1)) r k k − β − β − (cid:18) Nl (cid:19) β − ! N < . Proof.
We apply Theorem 4.1. If ~p follows a power law distribution as in Definition 2.3, we canfurther simplify E [ N SF ] to E [ N SF ] (cid:0) − k (cid:1) m n Y i =1 " − exp − k · p i k − m (cid:0) − k k ~p k (cid:1) − k · p i k − ! − x > e − x − x which holds for all x <
1. We upper bound the probabilities p i by choosing an integer N > N buckets of equal size.For i ∈ (cid:2)(cid:6) l − N n (cid:7) + 1 , (cid:6) lN n (cid:7)(cid:3) and 1 l N − p i (cid:18) nn − ⌈ lN n ⌉ +1 (cid:19) β − P ni =1 (cid:0) ni (cid:1) β − = (cid:16) NN − l (cid:17) β − P ni =1 (cid:0) ni (cid:1) β − . The last bucket i ∈ (cid:2)(cid:6) N − N n (cid:7) + 1 , n (cid:3) is simply bounded by 2 nN in the overall product. W.l.o.g.we assume that there are exactly nN variables in each bucket, as we could split the factor for (cid:6) lN n (cid:7) into appropriate parts which both obey the upper bound on p i for bucket l and l + 1respectively. We can now upper-bound E [ N SF ] by (cid:0) − k (cid:1) m nN N − Y l =1 − exp − k k − · m · (cid:0) Nl (cid:1) β − (cid:16)P ni =1 (cid:0) ni (cid:1) β − (cid:17) (cid:0) − k k ~p k (cid:1) − k k − (cid:0) Nl (cid:1) β − nN We are now interested in what happens to the expression in the exponent when n tends toinfinity. First, P ni =1 (cid:0) ni (cid:1) β − → β − β − n for β >
2. Second, by Theorem 2.1 we have that k ~p k jj =Θ (cid:16) n − j β − β − (cid:17) → j > β − k ~p k jj = Θ (cid:0) n − j +1 (cid:1) → j β −
1. Finally,for every constant N we have that k k − (cid:0) Nl (cid:1) β − is also constant. Using m = r · n we can thussimplify k k − · m · (cid:0) Nl (cid:1) β − (cid:16)P ni =1 (cid:0) ni (cid:1) β − (cid:17) (cid:0) − k k ~p k (cid:1) − k k − (cid:0) Nl (cid:1) β − = (1 + o (1)) r k k − β − β − (cid:18) Nl (cid:19) β − . Plugging this into our inequality we get E [ N SF ] (cid:0) − k (cid:1) m nN N − Y l =1 " − exp − (1 + o (1)) r k k − β − β − (cid:18) Nl (cid:19) β − ! nN = (cid:0) − k (cid:1) r N N − Y l =1 " − exp − (1 + o (1)) r k k − β − β − (cid:18) Nl (cid:19) β − ! N n This establishes Theorem 4.2.The bound from this Theorem improves as N → ∞ . As this expression is rather terse, we alsonumerically determine in Table 1 the smallest constant r such that the formula is unsatisfiable.We compare these values to the upper bounds for uniform random SAT obtained from theSingle-Flip Method. In the remainder of this section, we show Theorem 4.1 and Theorem 4.2. Definition 4.1 (Single-Flip Property) . For a random formula Φ a truth assignment A has the single-flip property iff A satisfies Φ and every assignment A ′ obtained from A by flipping exactlyone zero to one does not satisfy Φ . Let N SF be the number of truth assignments with the single-flip property for Φ. As arguedin [26], such an assignment exists if Φ is satisfiable. From Markov’s Inequality, we thus knowPr[Φ satisfiable] E [ N SF ] . In the following, we derive a bound on E [ N SF ]. By Lemma 2.1, the probability of choosinga clause c is at most k !2 k · Q ℓ ∈ c p | ℓ | − k k ~p k . To bound the number of assignments with the single-flip property, we use the following result.
Lemma 4.1 ([26]) . The expected number of assignments with the single-flip property is E [ N SF ] = (cid:0) − k (cid:1) m X assignment A Pr[ A single-flip | A satisfying ] . ower law distribution with exponent β uniformdist. k Table 1:
Numerical upper bounds on the density threshold obtained from the Single-FlipMethod (cf. Theorems 4.1 and 4.2). Empty fields indicate unsatisfiability for all constantdensities by Theorem 3.1. To the best of our knowledge, the bounds for uniform randomSAT with k > k = 3 the bestknown unconditional numerical upper bound is 4 . Proof.
Note that for a certain truth assignment A , the probability of choosing a clause which isnot satisfied by A is / k . Therefore, the probability that A is a satisfying assignment for Φ isexactly (cid:0) − k (cid:1) m .We next bound the probability that a satisfying assignment A has the single-flip property. Lemma 4.2.
For a satisfying assignment A = ( a , a , . . . , a n ) ∈ { , } n it holds that Pr[ A single-flip | A satisfying ] Y i : a i =0 − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m . Proof.
For a satisfying assignment A to have the single-flip property, all assignments A i obtainedby flipping a bit a i = 0 of A must not satisfy Φ. To fulfill this property for A i , we have to chooseat least one clause which contains ¯ X i and k − A i does not satisfy the clause. Let S i ( c ) denote the event that a clause c is satisfied by A , butnot by A i . Then,Pr[ S i ( c )] = k ! · p i P J ∈P k − ([ n ] \{ i } ) Q j ∈ J p j k (cid:0) − k k ~p k (cid:1) k · p i k (cid:0) − k k ~p k (cid:1) since P J ∈P k − ([ n ] \{ i } ) Q j ∈ J p j k ~p k k − ( k − . The probability of choosing a clause not satisfied by A i under the condition that A is satisfying is thenPr[ S i ( c ) | A sat] = Pr[ S i ( c ) | A satisfies c ] k · p i k − (cid:0) − k k ~p k (cid:1) as the probability of choosing a clause which is satisfied by any assignment is exactly k − k . Fora fixed assignment A i we concludePr[ A i unsat | A sat] = 1 − (cid:0) − Pr[ S i ( c ) | A sat] (cid:1) m − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m . (4.1)It remains to find the joint probability that all single-flipped assignments A i for 1 i n with a i = 0 are not satisfying. We show this using a correlation inequality by Farr [28]. The sets ofclauses which are not satisfied by the A i ’s are pairwise disjoint as each clause in the set for A i hasto contain ¯ X i , whereas each clause in the set for A j ( j = i ) can not contain ¯ X i . In the contextof the correlation inequality from [28] we set V = { , , . . . , m } , I = { i ∈ { , , . . . , n } | a i = 0 } , X v = i iff the v -th clause is satisfied by A , but not by A i , and F i the “increasing” collection ofnon-empty subsets of V . The application of the Theorem then directly yieldsPr[ A single-flip | A sat] = Pr[ \ i : a i =0 A i unsat | A sat] Y i : a i =0 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m . lgorithm 1 Clause Shrinking Algorithm
Input: k -SAT formula Φ; weight distribution ~w for all c ∈ Φ do ℓ ← argmin ℓ ∈ c { w | ℓ | } ℓ ← argmin ℓ ∈ c \{ ℓ } { w | ℓ | } c ← ( ℓ ∨ ℓ ) Solve Φ using any polynomial time 2-SAT algorithmCombining Lemmas 4.1 and 4.2 we get that the expected number of assignments with single-flip property is at most E [ N SF ] (cid:0) − k (cid:1) m X I ⊆{ , ,...,n } Y i ∈ I " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m = (cid:0) − k (cid:1) m n Y i =1 " − − k · p i k − (cid:0) − k k ~p k (cid:1) ! m . This establishes Theorem 4.1.
In this section, we provide a complementary result to Theorems 3.1 and 4.2 proving that if β > k − k − + ε and the clause-variable ratio r = mn does not exceed some small constant, then arandom k -SAT formula with exponent β is satisfiable with high probability. Let us first restatethe main result: Theorem 5.1.
Let Φ be a random k -SAT formula whose variable probabilities follow a powerlaw distribution (c.f. Definition 2.2). If the power law exponent is β > k − k − + ε for an arbitrary ε > , Φ is satisfiable with high probability if mn is a small enough constant. We show this statement by constructing an algorithm that satisfies Φ w. h. p. if the clause-variable ratio is small. Algorithm 1 contains a formal description. The main idea is to shrink allclauses to size 2 by selecting the literals with smallest weight in each clause; and then runningany well-known (polynomial time) 2-SAT algorithm (e. g. [8]).In the following, we seek to establish that Algorithm 1 will find a satisfying assignment (forsmall constraint densities) with high probability. To this end, we first analyze the probabilitydistribution of a clause c after it has been shrunk. Lemma 5.1.
Let ℓ , ℓ be the selected literals of an arbitrary clause c ∈ Φ in Algorithm 1. Then, Pr[ | ℓ | = i, | ℓ | = j ] + Pr[ | ℓ | = j, | ℓ | = i ] O ( n ( w i w j ) − ( k − β − ) . Proof.
W. l. o. g., we assume that w i w j . Then, Pr[ | ℓ | = j, | ℓ | = i ] = 0 by the definition ofAlgorithm 1. For the event | ℓ | = i, | ℓ | = j to happen, all other k − | ℓ | = i, | ℓ | = j ] = 12 · (cid:18) k (cid:19) · (1 + o (1)) · p i · p j · Pr[ V > w j ] k − = Θ( n ) · w i w − ( k − β − j O ( n ) · ( w i w j ) − ( k − β − . The last statement holds since w i w j .Having derived a bound on the probability distribution of a shrunk clause, it is possible tocompute the probability that the resulting 2-SAT formula is satisfiable. We use that the clausesare sampled independently. To avoid confusion, we write Φ ′ and c ′ , whenever we talk about theshrunk formula and clauses. To upper bound the probability of Φ not being satisfiable, we lookat so-called bi-cycles in Φ ′ . 12 efinition 5.1. A bi-cycle of length l is a sequence of l + 1 clauses of the form ( u, ℓ ) , (cid:0) ¯ ℓ , ℓ (cid:1) , . . . , (cid:0) ¯ ℓ l − , ℓ l (cid:1) , (cid:0) ¯ ℓ l , v (cid:1) , where ℓ , . . . , ℓ l are literals of distinct variables and u, v ∈ (cid:8) ℓ , . . . , ℓ l , ¯ ℓ , . . . , ¯ ℓ l (cid:9) . Chvatal and Reed [14, Theorem 3] show that if the formula Φ ′ is unsatisfiable, it mustcontain a bi-cycle. Consequently, by upper bounding the probability that a bi-cycle appears, weimmediately obtain an upper bound on the probability that Φ ′ and henceforth Φ is unsatisfiable. Theorem 5.2 ([14]) . Let Φ ′ be any -SAT formula. If Φ ′ contains no bi-cycle, it is satisfiable. Before we are able to prove the main Theorem, we need the following auxiliary Lemma.
Lemma 5.2.
Let β = δ + 1 + ε for some ε > . For all l n , there is a constant c with X S ⊆ [ n ]: | S | = l Y i ∈ S w δi n l · c l l ! . Proof.
We begin by observing that the term on the left side of the equation is obviously monotonein w i : If δ > δ < w i increases the sum. Thus, instead ofconsidering the true distribution function F ( w ), we may consider the upper (lower) bound on F ( w ), see Equation (2.4). For the sake of brevity, we consider the distribution b F ( w ) = αw − β ,where α is chosen to be either α if δ >
0, or α otherwise.To estimate this sum, we arrange the elements of S increasingly by weight, such that w s
1. For i we get n − i X s l − i = s l − i − +1 w δs l − i n − i +1 X s l − i +1 = s l − i +1 w δs l − i +1 . . . n X s l = s l − +1 w δs l n i · d i i ! n − i X s l − i = s l − i − +1 w i · ( δ +1 − β )+ δs l − i (5.1)13o bound the sum, we distinguish two cases. If i ( δ + 1 − β ) + δ >
0, then n n − i X s l − i = s l − i − +1 w i · ( δ +1 − β )+ δs l − i αw ( i +1) · ( δ +1 − β ) s l − i − + Z w n w sl − i − α ( i · ( δ + 1 − β ) + δ ) w i · ( δ +1 − β )+ δ − β d w = αw ( i +1) · ( δ +1 − β ) s l − i − + h α ( i · ( δ +1 − β )+ δ )( i +1) · ( δ +1 − β ) w ( i +1) · ( δ +1 − β ) i w n w sl − i − = (cid:16) α + α ( i · ( δ +1 − β )+ δ )( i +1)( β − δ − (cid:17) w ( i +1) · ( δ +1 − β ) s l − i − − α ( i · ( δ +1 − β )+ δ )( i +1)( β − δ − w ( i +1) · ( δ +1 − β ) n = α ( β − i +1)( β − δ − w ( i +1) · ( δ +1 − β ) s l − i − · (cid:16) − i · ( δ +1 − β )+ δβ − ( w n w sl − i − ) ( i +1) · ( δ +1 − β ) (cid:17) . For the integration, we need to make sure that the special case − i · ( δ + 1 − β ) + δ − β doesnot occur. By rearranging, one can see that this is only true for i = −
1, however, our weightsbegin at i = 1.We now bound the error term that appears from the integration limit w n . Note that we onlyneed to consider the case where i ( δ + 1 − β ) < − δ , otherwise the error term is smaller than 1and may simply be omitted.Observe from Equation (5.1) that w s l − i − w n − i . Further, by Equation (2.4) we have that w n = Θ( n β − ). Similarly, in = b F ( w n − i ) = αw − βn − i , (5.2)therefore w n − i = Θ(1) · ( ni ) β − . Therefore, we have w n w s l − i − w n w n − i = Θ( i β − ) . (5.3)Recall that we are in the case where the error term is positive; and that the exponent ( δ + 1 − β )is negative. Substituting the above inequality, we obtain1 − i · ( δ +1 − β )+ δβ − ( w n w sl − i − ) ( i +1) · ( δ +1 − β ) − i · ( δ +1 − β ) β − i i +1 β − · ( δ +1 − β ) = 1 − δ +1 − ββ − i i +1 β − · ( δ +1 − β ) By inspecting the exponent 1 + i +1 β − · ( δ + 1 − β ), we observe that it is of order O (1). In particular,once i is a large enough constant, the exponent becomes negative. Therefore, we may concludethat 1 − δ +1 − ββ − i i +1 β − · ( δ +1 − β ) = O (1) , (5.4)where the constant is not dependent on the iteration i . Thus, as d was chosen large enough, n n − i X s l − i = s l − i − +1 w i · ( δ +1 − β )+ δs l − i di +1 w ( i +1) · ( δ +1 − β ) s l − i − , (5.5)Plugging this into inequality (5.1) proves the induction step.Choosing i = l − s = 0 yields X S ⊆ [ n ]: | S | = l Y i ∈ S w δi n l · w l · ( δ +1 − β )1 · d l l ! . Since w δ +1 − β = Θ(1), we can choose an appropriate constant c such that the statement holds.We are now able to show Theorem 5.1. As discussed above, we do this by upper boundingthe probability that a bi-cycle appears in Φ ′ . To this end, we calculate the expected numberof bi-cycles in Φ ′ , observe that it is poly( n ) − , and apply Markov’s inequality. This yields thatw. h. p., Φ ′ and thus Φ are satisfiable. 14 roof of Theorem 5.1. We calculate the expected number of bi-cycles in Φ ′ . First, we fix a set S ⊆ [ n ] of l > X B denote the random variable countinghow many times a specific bi-cycle B with the variables from S appears in F . Then E [ X B ] (cid:18) ml + 1 (cid:19) ( l + 1)! · Pr[ u ∨ x ] Pr[ ¯ x l ∨ v ] · l − Y i =1 Pr[ ¯ x i ∨ x i +1 ] . The factor (cid:0) ml +1 (cid:1) ( l + 1)! counts the possible positions of B in F . By Lemma 5.1, E [ X B ] m l +1 · (cid:0) c n (cid:1) l +1 · w | u | w | v | Y i ∈ S w i ! − ( k − β − for some suitable constant c . Now let X S denote the random variable counting how many times any bi-cycle with the variables from S appears in F . There are l ! permutations of the l variables;and 2 l combinations of literals on l variables. Similarly, literals u and v have 4 possible signcombinations. Thus, E [ X S ] m l +1 · l ! · l · (cid:0) c n (cid:1) l +1 · X i ∈ S w − ( k − β − i ! Y i ∈ S w − ( k − β − i . To estimate the sum, we upper bound w i w n for all sets up to a certain size l , which we willdetermine later. We set δ := 2 − ( k − β −
2) and define α ( l ) as X i ∈ S w δ/ i ! α ( l ) := O ( l ) , if δ l · w δn , if δ > l l , O ( n ) , otherwise.Now let X denote the random variable counting the number of bi-cycles that appear in F . E [ X ] n X l =2 l +2 · m l +1 · l ! · ( c n ) l +1 · α ( l ) X S ⊆ [ n ] | S | = l Y i ∈ S w δi . Since δ +1 = 2 − ( k − β − < β by our assumption β > k − k − + ε , we can apply Lemma 5.2.Using r := m/n , we obtain that the right-hand side is at most E [ X ] n X l =2 l +2 · m l +1 · l ! · ( c n ) l +1 · α ( l ) · n l · c l l ! n n X l =2 c l · r l · α ( l ) , (5.6)for some suitable constant c . Since r is a small enough constant we thus have c · r <
1. If δ
0, we are finished, since then n n X l =2 c l · r l · α ( l ) n n X l =2 ( c · r ) l · l O ( n ) . Otherwise, if δ >
0, we choose l := − · ln − ( c r ) ln( n ) , which ensures ( r · c ) l = O ( n − ) for all l > l . For l = 2 , . . . , l , equation (5.6) sums up to at most n l X l =2 ( c r ) l · l · w δn = O (log ( n ) · n − k β − β − ) , where we substituted w n = Θ( n β − ) and δ = 2 − ( k − β − β > k − k − + ε , theexponent 1 − k β − β − < − ε ′ is negative, and we thus have E [ X ] n n X l =2 c l r l α ( l ) O (log ( n ) · n − ε ′ ) + O ( n ) , which proves the Theorem by Markov’s inequality.15 Discussion of the Results
In this work, we have shown that with high probability, a power law random k -SAT formulais satisfiable, if β > k − k − + ε and the clause-variable ratio is not too large; and that it isunsatisfiable if β k − k − − ε , or if the clause-variable ratio is too large. Here, we give a fewobservations following these results.First, as explained in Section 1 our results translate directly to the model where clauselengths are power law distributed. This observation might help to explain a phenomenon thatarose in [5]: The authors experimentally observed that a random-sat formula with double powerlaw distribution (both variables and clause lengths are drawn from a power law) can be solvedextremely fast by MiniSAT. Although the formula was of length 5 · , MiniSAT already gavean answer after 4 seconds! Using our results, we are now able to provide a potential explanationfor this phenomenon: Disregarding the double power law distribution, the smallest clause length k min occurring in their generated formulas is one. Thus, there will be Θ( n ) clauses of length oneand by Theorem 3.1 the formula is likely unsatisfiable.Second, we observe a sharp threshold in the sense of Friedgut [22] (for small constraintdensities r ) for β at the point k − k − . In contrast, it is unclear whether such a sharp thresholdexists (and can be analytically derived) for fixed β but variable r . Considering however, thatdecades of research were dedicated to the same question in the uniform case—an arguablysimpler model—it is unlikely that we obtain a satisfying answer any time soon; at least for all k . As in the uniform model, however, it might be more tractable to get sharp thresholds for k → ∞ . References [1] D. Achlioptas, A. Coja-Oghlan, and F. Ricci-Tersenghi. On the solution-space geometry ofrandom constraint satisfaction problems.
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