The Satisfiability problem Quest throughout history to establish an - PowerPoint PPT Presentation

Heuristic Optimization The Satisfiability problem Quest throughout history to establish an effective process (e.g., a mechanical process ) for human reasoning. Heuristic Optimization Lecture 8 4th century BC 1600s 1800s 1910s 1930s A L B R T r e o u a s i i d n r s c b o P s i s a e s e m t n l r k o l d e e c i f c i n i p t z u , l o e u l l l c r s – e c F Algorithm Engineering Group e l – i a s u e p “ r n a d p – s e i F r c a a d e r g y r h fi Hasso Plattner Institute, University of Potsdam o d l a l e M a W i a n p e s t r m o d i – n e a a o h t d s u c a t c B r a i c t h i t t t u i h s i t n e o e e e s o i u r m h e o a o ” ffi n i f t n t s r l e e a s o t o a c o r i a t 9 June 2015 b c m d i f n i e u a c a – n i a l u a t t x l n g i c f i o e o r v o m b e n m r d r s a s i , t a o i l o i f s n l o s g i c 9 June 2015 1 / 22 Heuristic Optimization Heuristic Optimization The Satisfiability problem Satisfiability : A formal definition In the 20th century, the advent of computers A propositional logic formula is built from inspired mathematicians to • variables that can take on one of two values (true/false) x, y, z, . . . • try to understand what people do when they • operators {∧ , ∨ , ¬} create proofs − conjunction (logical AND), e.g., x ∧ y • reduce logical reasoning to some canonical − disjunction (logical OR), e.g., x ∨ y form that can be implemented by an − negation (logical NOT), e.g., ¬ x algorithm • parentheses that can group expressions, e.g., ( x ) ∧ ( ¬ x ∨ y ) A formula F is said to be satisfiable if it can be made true by assigning UNIVAC ( www.computerhistory.org ) appropriate logical values (true or false) to its variables. Problem: given a formula, F , decide whether F is satisfiable. Given a statement S in some well-defined logical syntax • is there an algorithm to prove S is true (or false)? Many applications: theoretical computer science, complexity theory, • what is the complexity of such an algorithm? algorithmics, cryptography and artificial intelligence. DEC VT100 ( commons.wikimedia.org ) 9 June 2015 2 / 22 9 June 2015 3 / 22

Heuristic Optimization Heuristic Optimization Satisfiability : Basics Definitions A well-formed Boolean expression can be described by the grammar: The assignment of n Boolean variables can be represented as x ∈ { 0 , 1 } n . � expr � ::= � variable � | � expr � ∧ � expr � Let F be a formula on n variables. We write F [ x ] ∈ { 0 , 1 } as the evaluation of F | � expr � ∨ � expr � under the assignment x ∈ { 0 , 1 } n . | ( � expr � ) | ¬� expr � Given a Boolean expression F on n Boolean variables, we say an assignment x ∈ { 0 , 1 } n satsifies F if F [ x ] = 1 . The assignment of a Boolean variable v is a binding to a value in { 0 , 1 } . Example If all variables in an expression are bound, the evaluation can be done recursively: F = ( ¬ x 1 ∨ x 2 ) ∧ ¬ x 1 ∧ ( ¬ x 3 ∨ ¬ x 1 ) E F E ∧ F E ∨ F ( E ) ¬ E x = (0 , 0 , 0) , F [ x ] = ? 1 0 0 0 0 0 1 x = (1 , 0 , 1) , F [ x ] = ? 0 0 1 0 1 0 1 1 0 0 1 1 0 1 1 1 1 1 0 9 June 2015 4 / 22 9 June 2015 5 / 22 Heuristic Optimization Heuristic Optimization Definitions Is Satisfiability easy or hard? Horn formulas Let F be the set of all admissible formulas. We consider some subsets of F : Two Boolean formulas E and F on n Boolean variables are said to be equivalent F 1 formulas satisfied when all variables are set true (false). if ∀ x ∈ { 0 , 1 } n , F [ x ] = E [ x ] . In this case we write F ≡ E F 2 formulas F ≡ E , where E is in CNF and each clause contains at most one positive (resp., negative) literal. A literal: a variable v or its negation ¬ v . A clause: a disjunction of literals, e.g., F 3 formulas F ≡ E , where E is in CNF and each clause contains ≤ 2 literals. ( x 1 ∨ ¬ x 2 ∨ ¬ x 3 ∨ · · · ∨ x i ) F 4 formulas F ≡ E , where E is conjunction of exclusive-or clauses. A formula F is said to be in conjunctive normal form (CNF) when F is written as 2-CNF formulas Affine formulas a conjunction of clauses. Schaefer’s Dichotomy Theorem (1978) Lemma 1. Every formula F ∈ ( F 1 ∪ F 2 ∪ F 3 ∪ F 4 ) can be decided in time polynomial For every well-formed formula F , there is a formula E such that (1) E is in CNF, in the length of F . and (2) F ≡ E . 2. The class F \ ( F 1 ∪ F 2 ∪ F 3 ∪ F 4 ) is NP-complete. a a Technical note: Schaefer’s approach is constrained to classes that can be CNF form is much easier to work with! recognized in log space. 9 June 2015 6 / 22 9 June 2015 7 / 22

Heuristic Optimization Heuristic Optimization Resolution for first-order logics Davis-Putnam procedure 1958 Martin Davis & Hilary Putnam developed a resolution procedure for Rule 1: unit rule first-order logic (quantifiers allowed) ( x 1 ∨ ¬ x 2 ∨ x 3 ) ∧ ( x 2 ∨ ¬ x 3 ) ∧ ( x 1 ) ∧ ( ¬ x 1 ∨ x 4 ∨ x 3 ) ∧ ( ¬ x 2 ) Herbrand’s theorem : if a first-order formula is unsatisfiable then it has some ( x 2 ∨ ¬ x 3 ) ∧ ( x 4 ∨ x 3 ) ∧ ( ¬ x 2 ) ground formula in propositional logic (quantifier-free) that is unsatsifiable. ( ¬ x 3 ) ∧ ( x 4 ∨ x 3 ) Reduced formula: Davis-Putnam procedure 1. Generate all propositional ground instances set x 1 = 1 set x 2 = 0 2. Check if each instance F is satisfiable For each unit clause ( ℓ ) The main innovation is in (2), where we must solve Satisfiability • set ℓ to 1 , ¬ ℓ to 0 • remove clauses containing ℓ , delete occurrences of ¬ ℓ Given a propositional logic formula F in CNF, assign variables using three • repeat until no unit clauses exist reduction rules . 9 June 2015 8 / 22 9 June 2015 9 / 22 Heuristic Optimization Heuristic Optimization Davis-Putnam procedure Davis-Putnam procedure Rule 2: pure literal rule pure literal Rule 3: rule for eliminating atomic formulas (ground resolution) ( x 1 ∨ ¬ x 3 ∨ x 4 ) ∧ ( ¬ x 3 ∨ ¬ x 2 ) ∧ ( x 1 ∨ x 2 ) ∧ ( ¬ x 2 ∨ ¬ x 4 ) ∧ ( x 3 ∨ ¬ x 4 ) ( x ∨ ℓ 1 , 1 ∨ · · · ∨ ℓ 1 ,k 1 ) ∧ ( ¬ x ∨ ℓ 2 , 1 ∨ · · · ∨ ℓ 2 ,k 2 ) ∧ C pure literal ( ℓ 1 , 1 ∨ · · · ∨ ℓ 1 ,k 1 ∨ ℓ 2 , 1 ∨ · · · ∨ ℓ 2 ,k 2 ) ∧ C ( x 3 ∨ ¬ x 4 ) Reduced formula: When x = 0 , ( ℓ 1 , 1 ∨ · · · ∨ ℓ 1 ,k 1 ) When ( x = 1 , ℓ 2 , 1 ∨ · · · ∨ ℓ 2 ,k 2 ) must be true must be true set x 1 = 1 set x 2 = 0 For each pure literal ℓ Replace ( x ∨ A ) ∧ ( ¬ x ∨ B ) ∧ C with ( A ∨ B ) ∧ C as long as there are no • set ℓ to 1 , ¬ ℓ to 0 ℓ 1 ,i ∈ A , ℓ 2 ,j ∈ B that are complementary • remove clauses containing ℓ • repeat until no pure literals exist 9 June 2015 10 / 22 9 June 2015 11 / 22

Heuristic Optimization Heuristic Optimization Using memory wisely A closer look at the splitting rule: In 1962, Loveland and Logemann tried to split implement DP procedure on an IBM 704, ( x ∨ A ) ∧ ( ¬ x ∨ B ) ∧ C = = ⇒ ( A ∧ C ) , ( B ∧ C ) but found that it used too much RAM. split ⇒ F ′ , F ′′ F = = L&L insight: keep a stack for formulas in external storage (tape A formula containing at drive) so the formulas in RAM F least one occurrence of a variable x don’t get too large. set x ← 1 set x ← 0 remove clauses containing x remove clauses containing ¬ x IBM 704 at NASA in 1957 ( commons.wikimedia.org ) remove occurrences of literal ¬ x remove occurrences of literal x Rule 3a: splitting rule F ′ := F � F ′′ := F � � � From ( x ∨ A ) ∧ ( ¬ x ∨ B ) ∧ C , create a pair of separate formulas a � � x =1 x =0 ( A ∧ C ) , ( B ∧ C ) . Observation: Recursively check ( A ∧ C ) and ( B ∧ C ) for satisfiability. • If F ′ or F ′′ contain an empty clause: then unsatisfied • If F ′ or F ′′ contain no clauses: then satisfied a where A , B and C don’t contain any occurrences of the variable x 9 June 2015 12 / 22 9 June 2015 13 / 22 Heuristic Optimization Heuristic Optimization Davis-Putnam-Logemann-Loveland (DPLL) DPLL search space Davis-Putnam procedure with Logemann-Loveland enhancement (splitting rule) F DPLL ( F ) F ′ F ′′ Input : A set of clauses F Output : A truth value if F is a consistent set of literals then return true; . . . . . . . . if F contains an empty clause then return false ; . . . . . . . . . . . . . . . . for each unit clause ( ℓ ) in F do F ← unit-propagate ( ℓ, F ) ; end for each pure literal ℓ in F do n 2 i = 2 n +1 − 1 F ← pure-literal-assign ( ℓ, F ) ; � Total size of search tree? end i =0 ℓ ← choose-literal ( F ) ; How can we reduce the total number of nodes expanded? return DPLL ( F ∧ ℓ ) ∨ DPLL ( F ∧ ¬ ℓ ) ; 9 June 2015 14 / 22 9 June 2015 15 / 22