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Propositional Satisfiability (SAT) Toby Walsh Cork Constraint Computation Centre University College Cork Ireland 4c.ucc.ie/~tw/sat/

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Presentation on theme: "Propositional Satisfiability (SAT) Toby Walsh Cork Constraint Computation Centre University College Cork Ireland 4c.ucc.ie/~tw/sat/"— Presentation transcript:

1 Propositional Satisfiability (SAT) Toby Walsh Cork Constraint Computation Centre University College Cork Ireland 4c.ucc.ie/~tw/sat/

2 Outline What is SAT? How do we solve SAT? Why is SAT important?

3 Propositional satisfiability (SAT) Given a propositional formula, does it have a “model” (satisfying assignment)? 1st decision problem shown to be NP- complete Usually focus on formulae in clausal normal form (CNF)

4 Clausal Normal Form Formula is a conjunction of clauses C1 & C2 & … Each clause is a disjunction of literals L1 v L2 v L3, … Empty clause contains no literals (=False) Unit clauses contains single literal Each literal is variable or its negation P, -P, Q, -Q, …

5 Clausal Normal Form k-CNF Each clause has k literals 3-CNF NP-complete Best current complete methods are exponential 2-CNF Polynomial (indeed, linear time)

6 How do we solve SAT? Systematic methods Truth tables Davis Putnam procedure Local search methods GSAT WalkSAT Tabu search, SA, … Exotic methods DNA, quantum computing,

7 Procedure DPLL(C) (SAT) if C={} then SAT (Empty) if empty clause in C then UNSAT (Unit) if unit clause, {l} then DPLL(C[l/True]) (Split) if DPLL(C[l/True]) then SAT else DPLL(C[l/False])

8 GSAT [Selman, Levesque, Mitchell AAAI 92] Repeat MAX-TRIES times or until clauses satisfied T:= random truth assignment Repeat MAX-FLIPS times or until clauses satisfied v := variable which flipping maximizes number of SAT clauses T := T with v’s value flipped

9 WalkSAT [Selman, Kautz, Cohen AAAI 94] Repeat MAX-TRIES times or until clauses satisfied T:= random truth assignment Repeat MAX-FLIPS times or until clauses satisfied c := unsat clause chosen at random v:= var in c chosen either greedily or at random T := T with v’s value flipped Focuses on UNSAT clauses

10 Why is SAT important? Computational complexity 1st problem shown NP-complete Can therefore be used in theory to solve any NP-complete problem Many direct applications

11 Some applications of SAT Hardware design Signals Hi = True Lo = False Gates AND gate = and connective INVERTOR gate = not connective..

12 Some applications of SAT Hardware design State of the art HP verified 1/7th of the DEC Alpha chip using a DP solver 100,000s of variables 1,000,000s of clauses Modelling environment is one of the biggest problems

13 Some applications of SAT Planning But planning is undecidable in general Even propositional STRIPS planning is PSPACE complete! How can a SAT solver, which only solves NP- hard problems be used then?

14 Some applications of SAT Planning as SAT Put bound on plan length If bound too small, UNSAT Introduce new propositional variables for each time step

15 Some applications of SAT Diagnosis as SAT Otherwise know as “SAT in space” Deep Space One spacecraft Propositional theory to monitor, diagnose and repair faults Runs in LISP!

16 Computational complexity Study of “problem hardness” Typically worst case Big O analysis Sorting is easy, O(n logn) Chess and GO are hard, EXP-time “Can I be sure to win?” Need to generalize problem to n by n board Where do things start getting hard?

17 Computational complexity Hierarchy of complexity classes Polynomial (P), NP, PSpace, …. NP-complete problems mark boundary of tractability No known polynomial time algorithm Though open if P=/=NP

18 NP-complete problems Non-deterministic Polynomial time If I guess a solution, I can check it in polynomial time But no known easy way to guess solution correctly! Complete Representative of all problems in this class If this problem can be solved in polynomial time, all problems in the class can be solved Any NP-complete problem can be mapped into any other

19 NP-complete problems Many examples Propositional satisfiability (SAT) Graph colouring Travelling salesperson problem Exam timetabling …

20 SAT is NP-complete Cook (1971) showed that all non-deterministic Turing machines can be reduced to SAT => There is a polynomial reduction of any problem in NP to SAT But not all SAT problems are equally hard!

21 Random k-SAT sample uniformly from space of all possible k-clauses n variables, l clauses Rapid transition in satisfiability 2-SAT occurs at l/n=1 [Chavatal & Reed 92, Goerdt 92] 3-SAT occurs at 3.26 < l/n < 4.598 SAT phase transition [Mitchell, Selman, Levesque AAAI-92]

22 Random 3-SAT Which are the hard instances? around l/n = 4.3 What happens with larger problems? Why are some dots red and others blue?

23 Random 3-SAT Complexity peak coincides with solubility transition l/n < 4.3 problems under-constrained and SAT l/n > 4.3 problems over- constrained and UNSAT l/n=4.3, problems on “knife-edge” between SAT and UNSAT

24 Random 3-SAT Varying problem size, n Complexity peak appears to be largely invariant of algorithm backtracking algorithms like Davis-Putnam local search procedures like GSAT

25 3SAT phase transition Lower bounds (hard) Analyse algorithm that almost always solves problem Backtracking hard to reason about so typically without backtracking Complex branching heuristics needed to ensure success But these are complex to reason about

26 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions

27 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X]

28 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] No assumptions about the distribution of X except non-negative!

29 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem

30 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem The expected value of X can be easily calculated

31 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem E[X] = 2^n * (7/8)^l

32 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem E[X] = 2^n * (7/8)^l If E[X] =1) = prob(SAT) < 1

33 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem E[X] = 2^n * (7/8)^l If E[X] < 1, then 2^n * (7/8)^l < 1

34 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem E[X] = 2^n * (7/8)^l If E[X] < 1, then 2^n * (7/8)^l < 1 n + l log2(7/8) < 0

35 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions E.g. Markov (or 1st moment) method For any statistic X prob(X>=1) <= E[X] Let X be the number of satisfying assignments for a 3SAT problem E[X] = 2^n * (7/8)^l If E[X] < 1, then 2^n * (7/8)^l < 1 n + l log2(7/8) < 0 l/n > 1/log2(8/7) = 5.19…

36 3SAT phase transition Upper bounds (easier) Typically by estimating count of solutions To get tighter bounds than 5.19, can refine the counting argument E.g. not count all solutions but just those maximal under some ordering

37 SAT phase transition Shape of transition “sharp” both for 2-SAT and 3-SAT [Friedut 99] Backbone (dis)continuity 2-SAT transition is "2nd order", continuous 3-SAT transition is "1st order", discontinuous backbone = truth assignments that are fixed when we satisfy as many clauses as possible [Monasson et al. 1998],…

38 2+p-SAT Morph between 2-SAT and 3-SAT fraction p of 3-clauses fraction (1-p) of 2-clauses [Monasson et al 1999]

39 2+p-SAT Maps from P to NP NP-complete for any p>0 Insight into change from P to NP, continuous to discontinuous, …? [Monasson et al 1999]

40 2+p-SAT

41 Observed search cost linear for p<0.4 exponential for p>0.4 But NP-hard for all p>0!

42 2+p-SAT Continuous 2SAT like Discontinuous 3SAT like

43 Simple bound Are the 2-clauses UNSAT? 2-clauses are more constraining than 3- clauses For p<0.4, transition occurs at lower bound! 3-clauses are not contributing

44 2+p-SAT trajectories

45 The real world isn’t random? Very true! Can we identify structural features common in real world problems? Consider graphs met in real world situations social networks electricity grids neural networks...

46 Real versus Random Real graphs tend to be sparse dense random graphs contains lots of (rare?) structure Real graphs tend to have short path lengths as do random graphs Real graphs tend to be clustered unlike sparse random graphs L, average path length C, clustering coefficient (fraction of neighbours connected to each other, cliqueness measure) mu, proximity ratio is C/L normalized by that of random graph of same size and density

47 Small world graphs Sparse, clustered, short path lengths Six degrees of separation Stanley Milgram’s famous 1967 postal experiment recently revived by Watts & Strogatz shown applies to: actors database US electricity grid neural net of a worm...

48 An example 1994 exam timetable at Edinburgh University 59 nodes, 594 edges so relatively sparse but contains 10-clique less than 10^-10 chance in a random graph assuming same size and density clique totally dominated cost to solve problem

49 Small world graphs To construct an ensemble of small world graphs morph between regular graph (like ring lattice) and random graph prob p include edge from ring lattice, 1-p from random graph real problems often contain similar structure and stochastic components?

50 Small world graphs ring lattice is clustered but has long paths random edges provide shortcuts without destroying clustering

51 Small world graphs

52

53

54 Other bad news disease spreads more rapidly in a small world Good news cooperation breaks out quicker in iterated Prisoner’s dilemma

55 Other structural features It’s not just small world graphs that have been studied Large degree graphs Barbasi et al’s power-law model Ultrametric graphs Hogg’s tree based model Numbers following Benford’s Law 1 is much more common than 9 as a leading digit! prob(leading digit=i) = log(1+1/i) such clustering, makes number partitioning much easier

56 Open questions Prove random 3-SAT occurs at l/n = 4.3 random 2-SAT proved to be at l/n = 1 random 3-SAT transition proved to be in range 3.26 < l/n < 4.598 random 3-SAT phase transition proved to be “sharp” 2+p-SAT heuristic argument based on replica symmetry predicts discontinuity at p=0.4 prove it exactly!

57 Open questions Impact of structure on phase transition behaviour some initial work on quasigroups (alias Latin squares/sports tournaments) morphing useful tool (e.g. small worlds, 2-d to 3-d TSP, …) Optimization v decision some initial work by Slaney & Thiebaux

58 Open questions Does phase transition behaviour give insights to help answer P=NP? it certainly identifies hard problems! problems like 2+p-SAT and ideas like backbone also show promise But problems away from phase boundary can be hard to solve over-constrained 3-SAT region has exponential resolution proofs under-constrained 3-SAT region can throw up occasional hard problems (early mistakes?)

59 Conclusions SAT is fundamental problem in logic, AI, CS, … There exist both complete and incomplete methods for solving SAT We can often solve larger problems than (worst- case) complexity would suggest is possible!


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