Chasing the k-sat Threshold

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1 Chasing the k-sat Threshold Amin Coja-Oghlan Goethe University Frankfurt

2 Random Discrete Structures In physics, phase transitions are studied by non-rigorous methods. New mathematical ideas are necessary for a rigorous theory. Impact on discrete maths, computer science, physics. Amin Coja-Oghlan (Frankfurt) Random k-sat 2 / 28

3 The k-sat problem The P NP problem [Cook, Karp 1970s] Many fundamental problems are NP-hard: Boolean Satisfiability [model checking, VLSI,... ], Graph Colouring [timetabling/allocation,... ], Travelling Salesman [scheduling problems,... ],... P NP-conjecture: it is impossible to solve any of these efficiently. Amin Coja-Oghlan (Frankfurt) Random k-sat 3 / 28

4 The k-sat problem Fix k 3 and let x 1, x 2,..., x n be Boolean variables. Given an expression of the form (x 1 x 17 x 29 ) ( x }{{} 11 x 2 x 1 ) }{{} k-clause k-clause... find a Boolean assignment that makes the entire formula true. Amin Coja-Oghlan (Frankfurt) Random k-sat 4 / 28

5 The k-sat problem Fix k 3 and let x 1, x 2,..., x n be Boolean variables. Given an expression of the form (x 1 x 17 x 29 ) ( x }{{} 11 x 2 x 1 ) }{{} k-clause k-clause... find a Boolean assignment that makes the entire formula true. NP-complete in the worst case. Exhaustive search takes time 2 n. Amin Coja-Oghlan (Frankfurt) Random k-sat 4 / 28

6 Random k-sat The random k-sat model Generate k-clauses Φ 1,..., Φ m uniformly and independently. Let Φ = Φ 1 Φ m. Amin Coja-Oghlan (Frankfurt) Random k-sat 5 / 28

7 Random k-sat Kirkpatrick, Selman (experimental) Science 1994 There occurs a satisfiability phase transition. Industrial SAT solvers require exponential time. Amin Coja-Oghlan (Frankfurt) Random k-sat 5 / 28

8 First analytical results Theorem If m/n > 2 k log 2 log 2 2, then Φ is unsatisfiable w.h.p. [folklore] Amin Coja-Oghlan (Frankfurt) Random k-sat 6 / 28

9 First analytical results Theorem [KKKS 1998] If m/n > 2 k log 2 1+log o k (1), then Φ is unsatisfiable w.h.p. Amin Coja-Oghlan (Frankfurt) Random k-sat 6 / 28

10 First analytical results Theorem [KKKS 1998] If m/n > 2 k log 2 1+log o k (1), then Φ is unsatisfiable w.h.p. Theorem [Friedgut 1999] There is a non-uniform threshold for satisfiability. Amin Coja-Oghlan (Frankfurt) Random k-sat 6 / 28

11 Algorithmic lower bounds Algorithm Density Culprit LLL Θ(2 k /k 2 ) Moser, Tardos 2010 Walksat Ω(2 k /k) ACO, Frieze 2012 e Unit Clause 2 2k /k Chao, Franco 1990 SC+backtracking k /k Frieze, Suen 1996 Fix 2 k log(k)/k ACO 2009 Amin Coja-Oghlan (Frankfurt) Random k-sat 7 / 28

12 Non-constructive bounds Theorem [Achlioptas, Moore 2002] Φ is satisfiable w.h.p. if m/n 2 k 1 log 2 O(1). Based on the second moment method. Off by a factor of 2. Amin Coja-Oghlan (Frankfurt) Random k-sat 8 / 28

13 Non-constructive bounds Theorem [Achlioptas, Peres 2003] Φ is satisfiable w.h.p. if Weighted second moment. m/n 2 k log 2 k log 2 2 Leaves a gap of k log 2/2. O(1). Amin Coja-Oghlan (Frankfurt) Random k-sat 9 / 28

14 Enter the physicists The replica and the cavity method [Mézard, Parisi, Zecchina 2001] Exact predictions as to the k-sat threshold. New algorithms: Belief/Survey Propagation guided decimation. Mathematically highly non-rigorous. Amin Coja-Oghlan (Frankfurt) Random k-sat 10 / 28

15 Enter the physicists The replica trick Goal: to compute E log Z. Generally, it is not true that E log Z log E [Z]. However, 1 E log Z = lim j 0 j log E [ Z j]. Now, compute the integer moments E [ Z j] and guess the correct extension to non-integer j. Amin Coja-Oghlan (Frankfurt) Random k-sat 10 / 28

16 Enter the physicists Replica symmetry breaking [Mézard, Parisi, Zecchina 2001] As m/n increases, the solution space clusters. uniquenessα d,+ symmetry αdynamic d RSB α c one-step RSBα s Fig. 2. Pictorial representation of the different phase transitions Goal solutions of a rcsp. At α d,+ some clusters appear, but for α d,+ < To comprise turn the only physics anresults exponentially into rigorous small fraction of solutions. Forα d < solutions are split among about e NΣ theorems. clusters of size e Ns.Ifα the set of solutions is dominated by a few large clusters (with strong Amin Coja-Oghlan (Frankfurt) Random k-sat 11 / 28

17 New (rigorous) result Theorem [ACO, Panagiotou 2012] Φ is satisfiable w.h.p. if m/n < 2 k log 2 3 log 2 2 o k (1). Guided by the physicists Belief Propagation technique. Matches the replica symmetric bound. Remaining gap of Amin Coja-Oghlan (Frankfurt) Random k-sat 12 / 28

18 The vanilla second moment Suppose Z 0, and Z > 0 Φ is satisfiable. By the Paley-Zygmund inequality, P [Φ satisfiable] P [Z > 0] E [Z]2 E [Z 2 ]. If E [ Z 2] C E [Z] 2, then Φ is satisfiable w.h.p. [Friedgut 99] Vanilla random variable: Z = #satisfying assignments. Amin Coja-Oghlan (Frankfurt) Random k-sat 13 / 28

19 The vanilla second moment Goal: to compute E [ Z 2]. We have E [ Z 2] = P [both σ, τ satisfy Φ] σ,τ {0,1} n Amin Coja-Oghlan (Frankfurt) Random k-sat 14 / 28

20 The vanilla second moment Goal: to compute E [ Z 2]. We have E [ Z 2] = P [both σ, τ satisfy Φ] σ,τ {0,1} n n = P [both σ, τ satisfy Φ] σ {0,1} n d=0 τ:dist(σ,τ)=d Amin Coja-Oghlan (Frankfurt) Random k-sat 14 / 28

21 The vanilla second moment Goal: to compute E [ Z 2]. We have E [ Z 2] = P [both σ, τ satisfy Φ] σ,τ {0,1} n n = P [both σ, τ satisfy Φ] σ {0,1} n d=0 τ:dist(σ,τ)=d n = 2 n P [both 1, τ satisfy Φ 1 ] m d=0 τ:dist(1,τ)=d Amin Coja-Oghlan (Frankfurt) Random k-sat 14 / 28

22 The vanilla second moment Goal: to compute E [ Z 2]. We have E [ Z 2] = = σ,τ {0,1} n P [both σ, τ satisfy Φ] n σ {0,1} n d=0 τ:dist(σ,τ)=d = 2 n n = 2 n n d=0 τ:dist(1,τ)=d d=0 ( n d P [both σ, τ satisfy Φ] P [both 1, τ satisfy Φ 1 ] m ) (1 2 1 k + 2 k (1 d/n) k ) m Amin Coja-Oghlan (Frankfurt) Random k-sat 14 / 28

23 The vanilla second moment We have E [ Z 2] = 2 n With α = d/n let n d=0 ( ) n (1 2 1 k + 2 k (1 d/n) k ) m. d f (α) = ln 2 + h(α) + m n ln [1 2 1 k + 2 k (1 α) k], where h(α) = α ln α (1 α) ln(1 α). Then 1 n ln E [ Z 2] max 0<α<1 f (α). Amin Coja-Oghlan (Frankfurt) Random k-sat 15 / 28

24 panneal := a/ln 1 K 2 1 K k C 2 Kk 1 K a k > r d 2 The vanilla second k $ln 2 K 2; > k d 7; moment r := 128 ln 2 K 2 k := 7 > f d a/ln 2 C h a C r$panneal a ; f := a/ln 2 C h a C r panneal a 1 > plot f a, f, a = ; 2 (2) (3) (4) (5) a K0.01 K0.02 K0.03 > plot f a, f 1 2, a = ; We have 1 n ln E [ Z 2] max 0<α<1 f (α). Moreover, f (1/2) 2 n ln E [Z]. Thus, E [ Z 2] exp(ω(n)) E [Z] 2. How come? Amin Coja-Oghlan (Frankfurt) Random k-sat 16 / 28

25 Asymmetry and the majority assignment Satisfying assignments are correlated. For each literal let d l =degree of l in Φ. Let d = (d l ) l be the degree sequence. Amin Coja-Oghlan (Frankfurt) Random k-sat 17 / 28

26 Asymmetry and the majority assignment Satisfying assignments are correlated. For each literal let d l =degree of l in Φ. Let d = (d l ) l be the degree sequence. The majority assignment is σ maj (x) = { 1 if dx d x, 0 otherwise. Amin Coja-Oghlan (Frankfurt) Random k-sat 17 / 28

27 Asymmetry and the majority assignment Satisfying assignments are correlated. For each literal let d l =degree of l in Φ. Let d = (d l ) l be the degree sequence. The majority assignment is The majority weight is σ maj (x) = { 1 if dx d x, 0 otherwise. w maj = 1 max {d x, d x }. km x V Amin Coja-Oghlan (Frankfurt) Random k-sat 17 / 28

28 Asymmetry and the majority assignment a K0.2 K a K0.6 K0.8 K0.5 K1 K1.2 K1 K1.4 K1.6 K1.5 K1.8 K2 K2 > > ptnae d a/r$ln 1K 1Ktheta a k K theta a k ; k k ptnae := a/r ln 1 K 1 K q a K q a > plot ptnae a C h a, a = 0..1 ; σ maj has maximum probability of being SAT. Conjecture: a typical SAT assignment leans towards σ maj. (11) [AP 03] Amin Coja-Oghlan (Frankfurt) Random k-sat 18 / 28

29 Asymmetry and the majority assignment Corollary [ACO, Panagiotou 2012] 1 There is δ = δ(k) > 0 such that w.h.p. 1 n 2 E [ln(z + 1)] ln E [Z] δn. dist(σ, σ maj ) 1 2 δ. σ sat. Even the first moment is off! Amin Coja-Oghlan (Frankfurt) Random k-sat 19 / 28

30 Asymmetry and the majority assignment Corollary [ACO, Panagiotou 2012] W.h.p. E [ln(z + 1)] ln E [Z] δn. Proof We have But P [w maj > E [w maj ] + ε] = exp [ Θ(ε 2 n) ]. E [Z w maj > E [w maj ] + ε] = exp (Ω(εn)) E [Z w maj E [w maj ]]. Amin Coja-Oghlan (Frankfurt) Random k-sat 20 / 28

31 Asymmetry and the majority assignment Proof We have But > > > xi evalf 5 (15) x plot ptsat a, xi, a = 0..1 ; P [w maj > E [w maj ] + ε] = exp [ Θ(ε 2 n) ]. (15) (16) E [Z w maj > E [w maj ] + ε] = exp (Ω(εn)) E [Z w maj E [w maj ]]. Amin Coja-Oghlan (Frankfurt) Random k-sat 21 / 28

32 Symmetrizing the problem Idea To work with a special type of symmetric satisfying assignments. Not-All-Equal assignments Both σ and σ are satisfying. Succeeds up to m/n 2 k 1 ln 2. [AM 02] a K0.2 K0.4 K0.6 K0.8 K1 > > r d 2 k $ln 2 K 2 r := 128 ln 2 K 2 Amin Coja-Oghlan (Frankfurt) Random k-sat 22 / 28 ptmax d x/evalf RootOf diff ptsat a, x C h a, a, a, 0..1 ; (12)

33 Symmetrizing the problem Idea To work with a special type of symmetric satisfying assignments. Balanced assignments The total number of true literal occurrences is km/2 ± 1. Succeeds up to m/n = 2 k ln 2 k ln 2 2. [AP 03] Amin Coja-Oghlan (Frankfurt) Random k-sat 22 / 28

34 Tackling asymmetry ACO, Panagiotou [2012] We need to accommodate asymmetric assignments. Key idea: to quantify asymmetry via Belief Propagation. Amin Coja-Oghlan (Frankfurt) Random k-sat 23 / 28

35 Tackling asymmetry ACO, Panagiotou [2012] We need to accommodate asymmetric assignments. Key idea: to quantify asymmetry via Belief Propagation. Belief Propagation [MPZ 2001] The marginal of x is p(x) = #sat assignments where x is true. #sat assignments Amin Coja-Oghlan (Frankfurt) Random k-sat 23 / 28

36 Tackling asymmetry ACO, Panagiotou [2012] We need to accommodate asymmetric assignments. Key idea: to quantify asymmetry via Belief Propagation. Belief Propagation [MPZ 2001] The marginal of x is p(x) = According to Belief Propagation, #sat assignments where x is true. #sat assignments p(x) p BP (x) = d x d x 2 k+1 + O Call p BP (x) the type of x. ( ) dx 2 d x. 2 k Amin Coja-Oghlan (Frankfurt) Random k-sat 23 / 28

37 Tackling asymmetry Fixing the marginals We fix the degree sequence d. Let T be the set of all types. An assignment σ has p BP -marginals if d l (σ(l) t) = O(1) for all t T. l:p BP (l)=t Amin Coja-Oghlan (Frankfurt) Random k-sat 24 / 28

38 Tackling asymmetry Clause types The type of a clause l 1 l k is (p BP (l 1 ),..., p BP (l k )). Judicious assignments Suppose σ has p BP -marginals. σ is judicious if for all l = (l 1,..., l k ) T k, j [k] (σ(φ ij ) l j ) = O(1). i:φ i has type l Let Z J = #judicious satisfying assignments. Amin Coja-Oghlan (Frankfurt) Random k-sat 25 / 28

39 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

40 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

41 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Create d l clones of each literal l and pile all clones of each type t. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

42 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Create d l clones of each literal l and pile all clones of each type t. A t-fraction of all clones are true under σ, τ. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

43 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Create d l clones of each literal l and pile all clones of each type t. A t-fraction of all clones are true under σ, τ. Match the clones to the clauses judiciously. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

44 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Create d l clones of each literal l and pile all clones of each type t. A t-fraction of all clones are true under σ, τ. Match the clones to the clauses judiciously. The problem factorizes over clause types. Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

45 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. Intuition: clones and clauses Let σ, τ be assignments with p BP -marginals. Create d l clones of each literal l and pile all clones of each type t. A t-fraction of all clones are true under σ, τ. Match the clones to the clauses judiciously. The problem factorizes over clause types. Satisfiability just requires a particular shuffling for each clause type! Amin Coja-Oghlan (Frankfurt) Random k-sat 26 / 28

46 Tackling asymmetry Lemma [ACO, Panagiotou 2012] If r < 2 k ln ln 2, then E [ Z 2 J ] C E [ZJ ] 2. To obtain the main theorem, we need to control the cluster size. Technical problems: large deviations phenomena fixed point problem, enumeration of assignments with a given overlap, analytical hazards. Amin Coja-Oghlan (Frankfurt) Random k-sat 27 / 28

47 Conclusion First successful analysis of an asymmetric problem. These also arise within message passing algorithms such as Belief Propagation, Survey Propagation. We obtain the replica symmetric bound. To improve the result, one needs to cope with condensation. Amin Coja-Oghlan (Frankfurt) Random k-sat 28 / 28

Phase transitions in discrete structures

Phase transitions in discrete structures Amin Coja-Oghlan Goethe University Frankfurt Overview 1 The physics approach. [following Mézard, Montanari 09] Basics. Replica symmetry ( Belief Propagation ).