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Gábor Kusper University of Linz RISC Austria

Proving by Assignment Trees that SAT Solvers are Non-Polynomial in Unit Propagation Framework with 1 Selection and Cache. Gábor Kusper University of Linz RISC Austria. 200 4. Outline. Representations of SAT Unit Propagation and Resolution DPLL and GUS Unit Propagation Framework.

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Gábor Kusper University of Linz RISC Austria

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  1. Proving by Assignment Trees that SAT Solvers are Non-Polynomialin UnitPropagation Frameworkwith 1 Selection and Cache GáborKusper University of Linz RISC Austria 2004

  2. Outline • Representations of SAT • Unit Propagation and Resolution • DPLL and GUS • Unit Propagation Framework

  3. Motivation • Theorema, http://www.theorema.org/ • A1, ..., An G. • A1, ..., An, G is unsatisfiable.

  4. Boolean Satisfiability (SAT) • Identify truth assignment that satisfies boolean formula or prove it does not exist. • SAT: Boolean Satisfiability if the boolean formula is in CNF form. • Well-known NP-complete problem.

  5. Pozitive Literal Negative Literal Clause Conjunctive Normal Form (CNF) F = ( a c )  ( b c )  (¬a ¬b ¬c )

  6. Representations F = ( a c )  ( b c )  (¬a ¬b ¬c ) S = {{ a,c }, { b,c }, {¬a,¬b, ¬c }} + x + M = x + + - - - True True True b c a False False False

  7. ClauseClassification x - + x x x x-x + - + 2-Clause EmptyClause Unit Clause ClearClause

  8. Clause Set Classification x - + x x x x-x + - + Unsatisfiable Clause Set Empty Clause Set, Satisfiable

  9. Simplifying SAT Resolution Unit Propagation

  10. A - + + x + + x + x + + + B res(A, B, a) a Resolution • if A, B are clauses and a is a literal, then • res(A, B, a) := A B \ {a, ¬a}.

  11. x - + x- x x - + -- x +xx + ++ UP by +xx Unit Clause Unit Propagation (UP) • If S is a clause set and a is a literal, then • UP(S, {a}) := {C \{a} | C  S  a C}.

  12. SAT Solvers DPLL GUS

  13. x - + x- x x - + -- x +xx + ++ UP by +xx UP by x-x BCP • Boolean Constraint Propagation (BCP): Iterated application of unit propagation.

  14. DPLL Procedure • Davis & Putnam & Logemann & Loveland Procedure • Function DPLL(S : Clause Set) : Boolean • Begin • S := BCP(S); • if S =  then return True; • if   S then return False; • Letabe a variablein S; • return DPLL( S  {{a}} )  DPLL( S  {{ ¬a}} ); • End

  15. + x x x + x - x x Example - - x x + + - x - - - x x + + - x - + x x - - x x + + - x - - x x x + + x + x

  16. Assignment • We say for short • clause disjunctive set of literals. • assignment conjunctive set of literals. • The intended meaning of assignment { a, b } is that we assign true to a and false to b. • If C is a clause, then C is an assignment. • If A is an assignment, then A is a clause. • If S is a clause set, then S is an assignment set. • If T is an assignment set, then T is a clause set.

  17. Hyper-Unit Propagation (HUP) • If S is a clause set and A is an assignment, then • HUP(S, A) := {C \A | C  S  C  A =  }. x x + xx x x - + -- x + ++ HUP by ++x x x + xx x x - + -- x + ++ x - + x- x UP by x+x UP by +xx

  18. + + + x -- + x C sm(C, c) c Sub-Model (sm) • If C is a clause and c is a literal in C, then • sm(C, c) := C \ { c }  { c }. • sm(C, c) is an assignment. • (a  b  c)  (a  b)  c.

  19. Expansion Rule • If HUP(S, A) is unsat, then {A}  S  S. • If C  S, c  C and HUP(S, sm(C, c)) is unsat, then {sm(C, c)}  S  S, i.e., • {C \ {c}}  S  S, • because res(C, sm(C, c)) = C \ {c}.

  20. GUS • General Unicorn-SAT(GUS) • FunctionGUS(S : Clause Set) : Boolean • Begin • if S =  then return True; • Let C be a minimal clause in S; • For each c  C do • If ( GUS( HUP(S, sm(C, c) ) ) ) then return True; • od • return False; • End

  21. + - x x x+ x x + - x x x- x Example - - x x + + - x - x x + x x - x x + - x - -x x x x x

  22. Unit Propagation Framework Unit Propagation Framework with 1 Selection and Cache Limitation Lemma

  23. Strategy Set & Function • If C is a clause and T is an assignment set, then • T is a strategy set generated by C, • if { C }  T is unsatisfiable. • If C is a clause and F is a function, that has the type • “function F( C : clause) : assignment set”, and • F(C) is a strategy set generated by C, then • F is a strategy function.

  24. + + + + + x x x - x x x + + + + - - - + - - + x - + x x + x x x C C T T + x x x + x x x C T Example x x x x C T

  25. Strategy Function of DPLL • Function DPLL_STF( C : Clause ) : Assignment Set • Begin • if ( C is empty ) then return ; • if ( C is unit ) then return { C }; • Let c be a literal in C; • return { { c }, { c } }; • End

  26. + + + + + x x x - x x x C T + x x x +x x x C T Example x x x x C T

  27. Strategy Function of GUS • Function GUS_STF( C : Clause ) : Assignment Set • Begin • T := ; • While (C is not empty) do • Let c be a literal in C; • T := T { sm(C, c) }; • C := C \ { c }; • od • return T; • End

  28. + + + + - - - + - - + x - + x x + x x x C T + x x x + x x x C T Example x x x x C T

  29. UPFw1S • Function UPFw1S(S : Clause Set, STF : Strategy Function) : Boolean • Begin • if ( S =  ) then return True; • if (  S ) then return False; • Let C be a clause in S; // 1 Selection • T := STF ( C ); // T is a strategy set • For each AT do • if ( UPFw1S( HUP(S, A), STF ) ) then return True; • od • return False; • End

  30. HUP by A1 HUP by A2 HUP by Ak UPFw1S is a SAT solver • We know that if T is a strategy set generated by C, • then { C }  T is unsatisfiable. • Assume T = { A1, A2, ..., Ak } is strategy set generated by C S. S S1 C S2 A1 A2 Ak Sk

  31. Simulated DPLL & GUS • Function Simulated_DPLL( S : Clause Set) : Assignment • Begin • return UPFw1S(S, DPLL_STF); • End • Function Simulated_GUS( S : Clause Set) : Assignment • Begin • return UPFw1S(S, GUS_STF); • End

  32. + + x [4]= + - x - + x - - x + x + [4]= + x - - x + - x - x + + [4]= x + - x - + x - - or or k-Clause Set • [2^k] := the set of all possible k-clauses on the same variables. • For all k : [2^k] is unsatisfiable. • Assume we have 3 variables. [1]= x x x

  33. HUP by ++x HUP on k-Clause Set • HUP([2^k], A) = [2^(k-m)], where m = |A|. [2] [8] + + + + + - + - + [8]= + - - - + + - + - - - + - - - HUP by ++x x x + =[2] x x -

  34. HUP by + HUP by + DPLL and GUS on [2] [2] [1] DPLL: UP steps: 1 + - [2] [1] GUS: UP steps: 1 + -

  35. HUP by +x HUP by -x HUP by -+ HUP by +x DPLL and GUS on [4] 1 [2] [4] DPLL: ++ UP steps: 4 -x 1 [2] +x [1] [4] GUS: UP steps: 4 ++ +- 1 [2] -x

  36. HUP by +xx HUP by -xx HUP by --+ HUP by -+x HUP by +xx DPLL and GUS on [8] 4 [4] [8] DPLL: +++ UPs: 10 -xx 4 [4] +xx [1] [8] GUS: +++ -1 UPs: 11 1 [2] ++- +-x 4 [4] -xx

  37. Limit for Strategy Sets • If T is strategy set generated by C and C is a k-clause, then T subsumes at least 2^k – 1 k-clauses on the variables of C. • This holds, because { C }  T is unsatisfiable. - - + - - + T = - + x = - + + - x x - + - + + + + + - + - + + - - + + + - - + - + x + x x C T

  38. HUP by --+ HUP by -+x HUP by +xx Limit for Children • If we break down [2^k] by a strategy set to • [2^k1], [2^k2], ..., [2^km] then • 2^k1 + 2^k2 + ... + 2^km 2^k – 1. • This holds, because of Limit for Strategy Sets. [1] [8] +++ [2] ++- +-x [4] -xx

  39. HUP by --+ HUP by -+x HUP by +xx Assignment Tree (A-tree) • A, Cs is an A-tree if A is an assignment and Cs is a set of A-trees. • We can represent a run of UPFw1S by an A-tree. xxx [1] [8] -+x --+ +xx +++ [2] ++- x+x x-+ xx+ +-x [4] -xx xx+

  40. Unit Assignment Tree (a), Cs is a unit A-tree if Cs is a set of unit A-trees. (b)A, Cs is a unit A-tree if A is a unit and Cs is a set of unit A-trees. convert to unit A-tree xxx xxx -+x -xx --+ -xx +xx +xx x+x x+x x-+ x-x x-x xx+ x+x xx+ xx+ xx+ xx+ xx+

  41. Cached Unit Assignment Tree xxx convert to cached unit A-tree -xx -xx +xx xxx x+x x-x x-x x+x -xx +xx xx+ xx+ xx+ xx+ x+x x-x x-x x+x xx+ xx+ xx+ xx+

  42. Auxiliary Lemmas • The number of edges of a cached unit A-tree equals to the number of used UP steps in the run of UPFw1SC, if this cached unit A-tree corresponds to this run of the framework. • A well-know graph theory states that if a tree has N nodes, then it has N-1 edges.

  43. Number of Nodes of a CUA-Tree • Let AT be a A-tree, which represent a run of UPFw1SC on [2^n], n > 0. • We convert AT to a cached unit A-tree. Let CUAT this cached unit A-tree. • Then CUAT has 2^n + 2^(n-1) – 1 nodes. • Proof by Induction: • For k = 1 is true, because [2] has 2 nodes. • Let us assume that it holds for k = n-1. • We show that it holds for k = n. We assume that n > 1. • The child nodes of root correspond to a run of UPFw1SC on [2^(n-1)]. • Since n > 1, the root has at least two child nodes. • Hence, it has at least 2 * (2^(n-1)+2^(n-2) – 1) + 1 nodes.

  44. Limitation Lemma • There is no SAT solver algorithm in the Unit Propagation Framework with 1 Selection and Cache, which can show that [2^n] is unsatisfiable using fewer UP steps than 2^n + 2^(n-1) - 2. • It holds, because each cache unit A-tree, which correspond to a run of this framework on [2^n] has at least 2^n + 2^(n-1) – 1 nodes.

  45. Hence • Therefore, each SAT solver in this framework uses at least 2^n + 2^(n-1) – 2 UP steps to show that [2^n] is unsatisfiable. • Hence, SAT solvers are non-polynomial in Unit Propagation Framework with 1 Selection and Cache.

  46. Thank you for your attention! Download:http://www.risc.uni-linz.ac.at/people/gkusper/UPF.ppt

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