1 Agents that Think Rationally. 2 Propositional Logic. 3 Logical Implication. 4 Resolution. 5 The Davis-Putnam Procedure.

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1 Agenda Artificial Intelligence 4. Propositional Reasoning On Computers that Think about What is True or False 1 Agents that Think Rationally 2 Propositional Logic Jörg Hoffmann Wolfgang Wahlster 3 Logical Implication 4 Resolution 5 The Davis-Putnam Procedure 6 Clause Learning Summer Term 2012 Thanks to Profs. Burgard, Nebel, Riedmiller (Univ. Freiburg) for slide sources 7 Phase Transitions 8 Local Search Methods Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 1/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 2/102 Agents that Think Rationally Often, rational action requires rational (logical) thought on the agent s part. Here, thought = reasoning about a knowledge base (KB). A KB is composed of sentences in a formal language (syntax). A truth theory (logic) interprets the sentences as precise statements about the world (semantics). Interaction with the KB through Ask and Tell (simplified): Ask(KB,α) = yes Tell(KB,α) = KB Forget(KB,α) = KB exactly when α follows from the KB so that α follows from KB (will not be discussed) Levels of Knowledge Knowledge level: Most abstract level. For example, the automated DB information system knows that a trip from Saarbrücken to Nancy costs 10 EUR. Logical level: Encoding of knowledge in a formal language. Price(SB, Nancy, 10.00) Implementation level: The internal representation of the sentences, for example: As a string Price(SB, Nancy, 10.00) As a value in a matrix When Ask and Tell work as desired, it is possible to remain on the knowledge level. Effective user interface! The user has his/her own mental model of the world and uses it for communication with the agent. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 4/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 5/102

2 7 LOGICAL AGENTS A Knowledge-Based Agent The Wumpus World: Example A knowledge-based agent uses its knowledge base to represent its background knowledge store its observations store its executed actions... derive actions 4 3 Stench Breez e Stench Breez e PIT PIT Breez e function KB-AGENT(percept ) returns an action persistent: KB, a knowledge base t, a counter, initially 0, indicating time 2 Stench Gold Breez e TELL(KB, MAKE-PERCEPT-SENTENCE( percept, t)) action ASK(KB, MAKE-ACTION-QUERY(t)) TELL(KB, MAKE-ACTION-SENTENCE(action, t)) t t + 1 return action 1 Breez e PIT Breez e START Figure 7.1 A generic knowledge-based agent. Given a percept, the agent adds the percept to its knowledge base, asks the knowledge base for the best action, and tells the knowledge base that it has in fact taken that action. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 6/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 7/102 The Wumpus World (1) The Wumpus World (2) A 4 4 grid In the square containing the wumpus and in the directly adjacent squares, the agent perceives a stench. In the squares adjacent to a pit, the agent perceives a breeze. In the square where the gold is, the agent perceives a glitter. When the agent walks into a wall, it perceives a bump. When the wumpus is killed, its scream is heard everywhere. Percepts are represented as a 5-tuple, e.g., [Stench, Breeze, Glitter, None, None] Actions: Go forward, turn right by 90, turn left by 90, pick up an object in the same square (grab), shoot (there is only one arrow), leave the cave (only works in square [1,1]). The agent dies if it falls down a pit or meets a live wumpus. Initial situation: The agent is in square [1,1] facing east. Somewhere exists a wumpus, a pile of gold and 3 pits. Goal: Find the gold and leave the cave. means that it stinks, there is a breeze and a glitter, but no bump and no scream. The agent cannot perceive its own location. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 8/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 9/102

3 Reasoning in the Wumpus World (1) Reasoning in the Wumpus World (2) [1,2] and [2,1] are safe: The wumpus is in [1,3]! How do we know? 1,4 2,4 3,4 4,4 1,3 2,3 3,3 4,3 1,2 2,2 3,2 4,2 OK 1,1 2,1 3,1 4,1 A OK OK A = Agent B = Breeze G = Glitter, Gold OK = Safe square P = Pit S = Stench V = Visited W = Wumpus 1,4 2,4 3,4 4,4 1,3 2,3 3,3 4,3 1,2 2,2 3,2 4,2 P? OK 1,1 2,1 A 3,1 P? 4,1 V B OK OK 1,4 2,4 3,4 4,4 1,3 W! 2,3 3,3 4,3 1,2 A 2,2 3,2 4,2 S OK OK 1,1 2,1 B 3,1 P! 4,1 V V OK OK A = Agent B = Breeze G = Glitter, Gold OK = Safe square P = Pit S = Stench V = Visited W = Wumpus 1,4 2,4 3,4 4,4 P? 1,3 W! 2,3 A 3,3 P? 4,3 S G B 1,2 S 2,2 3,2 4,2 V V OK OK 1,1 2,1 B 3,1 P! 4,1 V V OK OK (a) (b) (a) (b) Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 10/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 11/102 Questionnaire Questionnaire, ctd. Why do we want the agent to think? (A): Sounds intelligent. (C): Thinking helps to take better actions. (B): Killing a wumpus. (D): Convenient interface (for humans) to KB. The Wumpus... (A): Is one of those AI illustrative examples... (C): Eats gold. (B): Can shoot arrows at the agent. (D): Emits a stench. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 12/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 13/102

4 Declarative Languages General Problem Solving using Propositional Logic Before a system that is capable of thinking can be built, one must find a way to express knowledge. We need a precise, declarative language. Declarative: Rather than telling the system what to do ( procedural ) the language just describes properties of the world (and leaves the system to figure it out on its own). Precise: We must know, - which symbols represent sentences, - what it means for a sentence to be true, and - when a sentence follows from other sentences. One possibility: Propositional Logic. (some new problem) model problem in propositional logic use off-the-shelf reasoning tool (its solution) Any problem that can be formulated as drawing conclusions from a propositional formula Very successful using modern SAT solvers! ( later in this chapter) Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 15/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 16/102 Basics of Propositional Logic (1) Basics of Propositional Logic (2) We are interested in knowing the following: Propositions: The building blocks of propositional logic are indivisible, atomic statements (atomic propositions), e.g., The block is red The wumpus is in [1,3] and the logical connectives and, or, and not, which we can use to build formulas. When is a proposition true? When does a proposition follow from a knowledge base (KB)? (symbolically: KB = ϕ) Can we syntactically define a concept of derivation, (symbolically: KB ϕ) such that it is equivalent to the concept of logical implication? Meaning and implementation of Ask. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 17/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 18/102

5 Syntax of Propositional Logic Countable alphabet Σ of atomic propositions: P, Q, R,... Logical formulas: P Σ ϕ ϕ ψ ϕ ψ ϕ ψ ϕ ψ atomic formula falseness truth negation conjunction disjunction implication equivalence Atom: Atomic formula. Literal: Possibly negated atomic formula. Clause: Disjunction of literals. Operator precedence: > > > > (we ll be using brackets except for negation). Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 19/102 Semantics: Intuition Atomic propositions can be true (T ) or false (F ). The truth of a formula follows from the truth of its atomic propositions (truth assignment or interpretation) and the connectives. Example: (P Q) R If P and Q are false and R is true, the formula is? If P and R are true, the formula is? Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 20/102 Semantics: Formally Example A truth assignment of the atoms in Σ, or an interpretation over Σ, is a function I : Σ {T, F } Interpretation I satisfies (or models) a formula ϕ, written I = ϕ: I = I = I = P iff P I = T I = ϕ iff I = ϕ I = ϕ ψ iff I = ϕ and I = ψ I = ϕ ψ iff I = ϕ or I = ψ I = ϕ ψ iff if I = ϕ, then I = ψ I = ϕ ψ iff if I = ϕ if and only if I = ψ P T Q T I : R F S F ϕ = ((P Q) (R S)) ( (P Q) (R S)) Question: I = ϕ? Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 21/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 22/102

6 Terminology The Truth Table Method An interpretation I is called a model of ϕ if I = ϕ. An interpretation is a model of a set of formulas if it fulfils all formulas of the set. A formula ϕ is satisfiable if there exists I that satisfies ϕ, unsatisfiable if ϕ is not satisfiable, falsifiable if there exists I that doesn t satisfy ϕ, and valid (a tautology) if I = ϕ holds for all I. Two formulas are logically equivalent (ϕ ψ) if, for all I: I = ϕ iff I = ψ. How can we decide if a formula is satisfiable, valid, etc.? Generate a truth table Example: Is ϕ = ((P H) H) P valid? P H P H (P H) H (P H) H P F F F F T F T T F T T F T T T T T T F T Satisfiability, falsifiability, unsatisfiability likewise. Is this a good method for answering these questions? Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 23/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 24/102 Normal Forms Producing an Equivalent CNF A formula is in conjunctive normal form (CNF) if it consists of a conjunction of disjunctions of literals l i,j, i.e., if it has the following ( form: n mi ) i=1 j=1 l i,j A formula is in disjunctive normal form (DNF) if it consists of a disjunction of conjunctions of literals: ( n mi ) i=1 j=1 l i,j For every formula, there exists at least one equivalent formula in CNF and one in DNF. A formula in DNF is satisfiable iff A formula in CNF is valid iff 1 Eliminate and : α β ( α β) etc. 2 Move inwards: (α β) ( α β) etc. 3 Distribute over : ((α β) γ) (α γ) (β γ) 4 Simplify: α α α etc. The result is a conjunction of disjunctions of literals. Analogous conversion of any formula to an equivalent formula in DNF. During conversion, formulas can expand exponentially. Note: Conversion to CNF formula can be done polynomially if only satisfiability should be preserved. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 25/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 26/102

7 Questionnaire Logical Implication: Intuition P Q is true under interpretation I if? (A): P I = F, Q I = F. (C): P I = T, Q I = F. (B): P I = F, Q I = T. (D): P I = T, Q I = T. A set of formulas (a KB) usually provides an incomplete description of the world, i.e., leaves the truth values of a proposition open. Example: KB = {P Q, R P, S} is definitive with respect to S, but leaves P, Q, R open (although they cannot take on arbitrary values). A CNF formula is... (A): An AND over OR s. (C): Satisfiable if every disjunction is satisfiable. (B): An OR over AND s. (D): Satisfiable if at least one disjunction is satisfiable. Models of the KB: P Q R S F T F T F T T T T F T T T T T T In all models of the KB, Q R is true Q R follows logically from KB. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 27/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 29/102 Logical Implication: Formal Proof of the Deduction Theorem The formula ϕ follows logically from the KB if ϕ is true in all models of the KB (symbolically KB = ϕ): KB = ϕ iff for all I, if I models KB then I models ϕ Note: The = symbol is a meta-symbol (used between formulas, not inside them). Some properties of = relationships: Deduction theorem: KB {ϕ} = ψ iff KB = ϕ ψ Contraposition theorem: KB {ϕ} = ψ iff KB {ψ} = ϕ Contradiction theorem: KB {ϕ} is unsatisfiable iff KB = ϕ Assumption: KB {ϕ} = ψ, i.e., every model of KB {ϕ} is also a model of ψ. Let I be any model of KB. If I is also a model of ϕ, then it follows that I is also a model of ψ. This means that I is also a model of ϕ ψ, i.e., KB = ϕ ψ. Assumption: KB = ϕ ψ. Let I be any model of KB that is also a model of ϕ, i.e., I = KB {ϕ}. From the assumption, I is also a model of ϕ ψ and thereby also of ψ, i.e., KB {ϕ} = ψ. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 30/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 31/102

8 Questionnaire KB = (ϕ ψ) is the case if and only if? (A): For all I, if I = KB and I = ϕ, then I = ψ. (C): Either for all I, I = KB; or for all I, I = ϕ ψ. (B): For all I, if I = KB, then I = (ϕ ψ). (D): For all I, either I = KB or I = ϕ or I = ψ. Questionnaire, ctd. The contraposition to ϕ ψ is ψ ϕ. What is the contraposition to Pech im Spiel, Glück in der Liebe? (A): Glück in der Liebe, Pech im Spiel. (C): Glück in der Liebe, Glück im Spiel. (B): Pech in der Liebe, Glück im Spiel. (D): Pech in der Liebe, Pech im Spiel. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 32/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 33/102 Inference Rules, Calculi, and Proofs We can often derive new formulas from formulas in the KB. These new formulas should follow logically from the KB. Example: If the KB is {..., (ϕ ψ),..., ψ,...} then ψ is a logical consequence of KB. Inference rules, e.g., ϕ, ϕ ψ ψ Calculus: Set of inference rules. Proof step: Application of an inference rule on a set of formulas. Proof: Sequence of proof steps where every newly-derived formula is added, and in the last step, the goal formula is produced. Soundness and Completeness In the case where in the calculus C there is a proof for a formula ϕ, we write KB C ϕ (optionally without subscript C). A calculus C is sound (or correct) if all formulas that are derivable from a KB do follow logically: KB C ϕ implies KB = ϕ This usually follows from the soundness of the inference rules and the logical axioms. A calculus is complete if every formula that follows logically from the KB is also derivable with C from the KB: KB = ϕ implies KB C ϕ Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 35/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 36/102

9 Resolution: Basic Ideas Resolution is a calculus, i.e., a way to derive new formulas that follow logically from the KB. Goal: We attempt to show that the KB (a conjunction/set of formulas) is unsatisfiable. Prerequisite: All formulas must be in CNF (but recall that there exists a fast satisfiability-preserving transformation to CNF). Computational Complexity: In the worst case, the derivation process requires an exponential amount of time (probably unavoidable because it is co-np-complete to decide whether a CNF is unsatisfiable). Our hope is to be better in typical cases, and thus beat the truth-table method. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 37/102 Reminder: Resolution Notation Assumption: All formulas in the KB are in CNF (equivalently: the KB is one CNF formula). KB is handled as a set of clauses. Each clause is handled as a set of literals. Notational conventions: Set of clauses: Set of literals: C, D Empty set of literals: Literal: l; Inverse (negation) of a literal: l An interpretation I satisfies a clause C iff there exists l C such that I = l. I satisfies iff, for all C, we have I = C. For any I, I =? For any I, with = { } do we have I =? For any I, with = do we have I =? Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 38/102 The Resolution Rule Derivations and Soundness With exclusive union meaning that the two sets are disjoint: C 1 {l}, C 2 {l} C 1 C 2 C 1 C 2 is called the resolvent of the parent clauses C 1 {l} and C 2 {l}. l and l are the resolution literals. Example: {P, Q, R} resolves with {P, S, R} to Is the resolvent equivalent to the parent clauses? Notation: R( ) = {C C is a resolvent of two clauses from } We say D can be derived from using resolution, i.e., D, if there exist C 1, C 2, C 3,..., C n = D such that C i R( {C 1,..., C i 1 }), for 1 i n. We say that n is the length of the derivation. Lemma. If D, then = D. Proof. Since all D R( ) follow logically from, the lemma follows from induction over the length of the derivation. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 39/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 40/102

10 Completeness Using Resolution: Example (A) Is resolution also complete? I.e., does = ϕ imply ϕ? Resolution is refutation-complete! is unsatisfiable implies. Theorem. is unsatisfiable iff. (Proof omitted.) So, can we use resolution to test whether = ϕ? Run resolution on ϕ. By the contradiction theorem, ϕ is unsatisfiable iff = ϕ. Input: KB = {Q P, P ( Q R S), Q S, R S}, φ = S. Question: Do we have KB = φ? Step 1: Transform KB and φ into CNF. Step 2: Write as set of clauses. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 41/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 42/102 Using Resolution: Example (A), ctd. Resolution: Conclusions So Far Resolution is a refutation-complete proof method. There are others (Davis-Putnam Procedure, Tableaux Procedure,... ). In order to implement the method, a strategy must be developed to determine which resolution steps will be executed and when. In the worst case, a resolution proof can take exponential time. This, however, very probably holds for all other proof procedures (the problem is co-np-hard). For CNF formulas in propositional logic, the Davis-Putnam Procedure (later in this chapter) is currently the most effective complete reasoning tool in practice; we will see that this procedure can be understood as a variant of resolution. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 43/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 44/102

11 Where is the Wumpus? The Situation Where is the Wumpus? Knowledge of the Situation B = Breeze, S = Stench, B i,j = there is a breeze in (i, j) S 1,1 B 1,1 S 2,1 B 2,1 S 1,2 B 1,2 Knowledge about the wumpus and smell: R 1 : S 1,1 W 1,1 W 1,2 W 2,1 R 2 : S 2,1 W 1,1 W 2,1 W 2,2 W 3,1 R 3 : S 1,2 W 1,1 W 1,2 W 2,2 W 1,3 R 4 : S 1,2 W 1,3 W 1,2 W 2,2 W 1,1 To show: KB = W 1,3 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 45/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 46/102 And Now in Clausal Form Resolution Proof for the Wumpus World Resolution: Situational knowledge: S 1,1, S 2,1, S 1,2 Knowledge about the wumpus and smell: R 1 : S 1,1 W 1,1, S 1,1 W 1,2, S 1,1 W 2,1 R 2 :..., S 2,1 W 2,2,... R 3 :... R 4 : S 1,2 W 1,3 W 1,2 W 2,2 W 1,1 Negated goal formula: W 1,3 W 1,3, S 1,2 W 1,3 W 1,2 W 2,2 W 1,1 S 1,2 W 1,2 W 2,2 W 1,1 S 1,2, S 1,2 W 1,2 W 2,2 W 1,1 W 1,2 W 2,2 W 1,1 S 2,1, S 2,1 W 2,2 W 2,2 W 2,2, W 1,2 W 2,2 W 1,1 W 1,2 W 1,1 S 1,1, S 1,1 W 1,1... S 1,1, S 1,1 W 1,2... W 1,1, W 1,1 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 47/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 48/102

12 Problems with Propositional Logic General Problem Solving using Propositional Logic (some new problem) Although propositional logic suffices to represent the wumpus world, it is rather involved Rules must be set up for each square. R 1 : S 1,1 W 1,1 W 1,2 W 2,1 R 2 : S 2,1 W 1,1 W 2,1 W 2,2 W 3,1 R 3 : S 1,2 W 1,1 W 1,2 W 2,2 W 1,3... More powerful logics: quantification over object variables. First-Order Predicate Logic (next chapter of lecture). model problem in propositional logic use off-the-shelf reasoning tool (its solution) Criticism here: model problem in propositional logic an activity carried out by a human is often infeasible. Does not mean that the reasoning becomes infeasible, too! (SAT solvers tackle formulas with millions of clauses... ) Still, millions may not be enough (in particular, quantifying over an infinite set of objects). Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 49/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 50/102 Questionnaire Questionnaire, ctd. What are resolvents of {P, Q, R} and { P, Q, R}? (A): {Q, R}. (C): {P, P, R, S}. (B): {Q, Q, P, R}. (D): {Q, Q, R}. Which statements are true? (A): {Q, Q, R} R({{P, Q, R}, {P, Q, R}}). (C): {Q, R} R({{ Q, R}, {P, Q, R}, { P, Q, R}}). (B): {Q, Q, P, R} R({{P, Q, R}, { P, Q, R}}). (D): {R} R(R({{ Q, R}, {P, Q, R}, { P, Q, R}})). Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 51/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 52/102

13 Summary Motivation Rational agents require knowledge of their world in order to make rational decisions. With the help of a declarative (knowledge-representation) language, this knowledge is represented and stored in a knowledge base. We use propositional logic for this (for the time being). Logical implication can be mechanized by using an inference calculus resolution. Propositional logic is impractical when the world becomes too large (or infinite). We next look at the most effective practical methods for reasoning in propositional logic. Actually, Davis-Putnam-Logemann-Loveland hence DPLL. SAT solving: Given set of clauses, return interpretation I so that I =, if such I exists; complete SAT solver: if such I does not exist, return unsatisfiable. Use like a calculus: Question: Does a formula ϕ logically follow from? Reduction to unsatisfiability of ϕ. Complexity: co-np. Use like a search for solutions: Wanted: Model of, i.e., I =. Complexity: NP; can be easier in practice because we can stop once we found the first I that works. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 53/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 55/102 General Problem Solving using SAT and DPLL The DPLL Procedure (some new problem) model problem as SAT use off-the-shelf DPLL solver Very successful! (its solution) In particular: checking the correctness of software ( bounded model checking, encodes program execution into SAT). DPLL Function Given a set of clauses defined over a set of variables Σ, return satisfiable if is satisfiable. Otherwise return unsatisfiable. 1. If = return satisfiable 2. If return unsatisfiable 3. Unit-propagation (UP) Rule: If contains a unit-clause C, assign a truth-value to the proposition in C that satisfies C, simplify to and return DPLL( ). 4. Splitting Rule: Select from Σ a proposition P which has not been assigned a truth-value. Assign one truth value to it, simplify to and call DPLL( ) a. If the call returns satisfiable, then return satisfiable. b. Otherwise assign the other truth-value to P in, simplify to and return DPLL( ). Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 56/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 57/102

14 Example (B) Example (A) = {{P, Q, R}, { P, Q}, {R}, {P, Q}} = {{ Q, P }, {P, Q, R, S}, {Q, S}, {R, S}, {S}} 2a. P F {{Q}, { Q}} 3a. UP rule: Q T { } 1. UP rule: R T {{P, Q}, { P, Q}, {P, Q}} 2. Splitting rule: 2b. P T {{ Q}} 3b. UP rule: Q F {} 1. UP rule: S T {{ Q, P }, {P, Q, R}, {Q}, {R}} 2. UP rule: Q T {{ P }, {P, R}, {R}} 3. UP rule: R T {{ P }, {P }} 4. UP rule: P T { } Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 58/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 59/102 Properties of DPLL Example (A): DPLL Tree DPLL is complete, correct, and guaranteed to terminate. DPLL constructs a model, if one exists. = {{ Q, P }, {P, Q, R, S}, {Q, S}, {R, S}, {S}} In general, DPLL requires exponential time (splitting rule!) Heuristics are needed to determine which variable should be instantiated next and which value should be used. DPLL is polynomial on Horn clauses, i.e., clauses with at most one positive literal P 1,... P n Q i P i Q Any DPLL tree for UNSAT is a resolution refutation of! S T F Q T F R T F P T F { Q, P } {S} {Q, S} {R, S} {P, Q, R, S} Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 60/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 61/102

15 Example (A): Resolution Proof from DPLL Tree DPLL vs. Resolution = {{ Q, P }, {P, Q, R, S}, {Q, S}, {R, S}, {S}} T F {S} { S} T F {Q, S} { Q, S} T F {R, S} { Q, R, S} Theorem. If DPLL returns unsatisfiable on after applying the unit-propagation and splitting rules n times, then with a derivation of length n. Proof Sketch. Consider first DPLL without the unit-propagation rule. Consider any leaf node N, for variable v, both of whose truth values x {v, v} directly result in an empty clause C(N, x). Then v C(N, v) and v C(N, v), so we can resolve these clauses to a clause C(N) that does not contain v. C(N) can contain only the negations of the decision literals l 1,..., l k above N. { Q, P } {P, Q, R, S} Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 62/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 63/102 DPLL vs. Resolution, So What? Questionnaire Why would be interested in this theorem? DPLL is an effective practical method for conducting resolution proofs. DPLL constructs tree resolution proofs: each derived clause C (not in ) is used in only a single application of the resolution rule. If we need C several times, we must derive it anew. This is a fundamental weakness! There exist whose shortest tree-resolution proof is exponentially longer than their shortest (general) resolution proof! To the rescue: Clause Learning. Which are true? (A): If can be derived from by UP, then =. (C): If can be derived from by resolution, then =. (B): If =, then can be derived from by UP. (D): If =, then DPLL returns unsatisfiable on. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 64/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 65/102

16 Questionnaire Which clauses can be derived by UP from {{P, Q, R}, { P, Q, R}, { S, S }, {S, R}, { R}}? (A): { S, P }. (C): {Q, R}. (B): {S }. (D): {Q}. Example (C) = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, {X 1,..., X 20 }, { X 1,..., X 20 }} Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 66/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 68/102 Implication Graphs Example (C) State of search during DPLL is set of time-stamped literals: what literal was set to TRUE at what time (level in the search tree)? Choice and implied literals are distinguished. = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, {X 1,..., X 20 }, { X 1,..., X 20 }} X X Q@12, R@12, R@12 Implication graphs capture this, along with the reasons for UP: Vertices are all time-stamped literals l@t, and if an empty clause was obtained. Arcs (l@t, ) indicate that l participated in a unit propagation deriving l. Each has k arcs (l i, ) where {l 1,..., l k, l } is a clause. This is a DAG (directed acyclic graph). Why? What are the nodes with indegree 0? Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 69/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 70/102

17 Example (D) Conflict Graphs = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, { Q, S}, { Q, S} {X 1,..., X 20 }, { X 1,..., X 20 }} X X Q@12, R@12, R@12, S@12, S@12 A proposition P where both P and P are in the implication graph is called conflict variable; P and P are called conflict literals. A conflict graph G C is a sub-graph of the implication graph that captures a conflict and its reasons: is a vertex in G C, and there is exactly one conflict variable whose literals are nodes in G C. All vertices in G C have a path to. Any vertex l j in G C either is a choice literal l j ch(g C ), or has the predecessors l 1,..., l k where {l i1,..., l ik, l j }. We start at a conflict variable then backchain over the clauses until we reach choice literals. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 71/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 72/102 Example (C) Example (D) = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, {X 1,..., X 20 }, { X 1,..., X 20 }} X X Q@12, R@12, R@12 = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, { Q, S}, { Q, S} {X 1,..., X 20 }, { X 1,..., X 20 }} X X Q@12, R@12, R@12, S@12, S@12 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 73/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 74/102

18 Clause Learning Example (C) Conflict graphs encode logical implications: = ( l = ) l@t ch(g C ) = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, {X 1,..., X 20 }, { X 1,..., X 20 }} X X Q@12, R@12, R@12 Given, setting all choice literals in a conflict graph results in a conflict. We can re-write this to: = l l@t ch(g C ) The negation of the choice literals is a valid clause! Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 75/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 76/102 Clause Learning and Non-Chronological Backtracking Example (C) Say we obtain conflict G C upon choosing l@t, and learn a clause C. We can backtrack to the maximal time stamp in C! Why? Since l@t ch(g C ), C = {l 1,..., l k, l@t}... :-( = {{ P, Q, R}, { P, Q, R}, { P, Q, R}, { P, Q, R}, {X 1,..., X 20 }, { X 1,..., X 20 }},{ P, Q} Q@1, R@1, R@1 BUT: Given the choices l 1,..., l k and the new clause C, l@t is now enforced by unit propagation! Set l@t as implied literal! Re-run unit propagation and conflict analysis; if we obtain a conflict, then we learn a clause whose maximal time stamp is < t. Actually, common clause learning implementations (CDCL, see below) always erase the choices down to level max{t 1,..., t k }, and set l@t as implied literal at that level. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 77/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 78/102

19 Clause Learning vs. Resolution The learned clauses follow from, correct? But doesn t that mean they re useless? Remember? DPLL constructs tree resolution proofs, in which each derived clause C (not in ) is used in only a single application of the resolution rule. If we need C several times, we must derive it anew. Is this still the case with clause learning? We add the derived clauses to, so we can use them as often as we like. Clause learning renders DPLL equivalent to full resolution! [Beame et al. (2004); Pipatsrisawat and Darwiche (2009)] Clause learning avoids making the same mistakes over again. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 79/102 Clause Learning, Remarks DPLL+Clause Learning called conflict directed clause learning (CDCL) in literature. (Non-chronological backtracking corresponds to the Remove N from the tree, as well as l i+1,..., l k where i is maximal with l i C(N). from our proof DPLL vs. Resolution above.) While we just select ch(g C ), much more can be done: For any cut through G C, where ch(g C ) lies on one side and on the other, the literals on the left cut border yield a valid clause. Must take care to not learn too many clauses... Clause learning originates from dependency-directed backtracking and explanation-based (no-good) learning developed in the CSP community. The distinguishing feature here is that the no-good is a clause the exact same type of constraint as the rest of. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 80/102 Questionnaire Which clauses can we derive by UP from {{P, Q, R}, { P, Q, R}, {S}, { S, S }, {S, R}, { R}}? (A): { S, R}. (C): {R}. (B): {Q, R}. (D):. Questionnaire, ctd. Which clauses can we learn after choosing R@1 and running UP on {{P, Q, R}, { P, Q, R}, {S}, { S, S }, {S, R}}? (A): { S, R}. (C): {R}. (B): {Q, R}. (D):. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 81/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 82/102

20 Questionnaire, ctd. Which clauses can we learn after choosing and running UP on {{P, Q, R}, { P, Q, R}, {S}, { S, S }, {S, R}}? (A): { S, R}. (C): {R}. (B): {Q, R}. (D):. How Good is DPLL in the Typical Case? We know that SAT is NP-complete, i.e., in the worst case, it takes exponential time. This is clearly also true for DPLL and for CDCL. But what about typical cases? (whatever they are) For CNF-formulas in which the probability for a positive appearance, negative appearance and non-appearance in a clause is 1/3, DPLL needs on average quadratic time [Goldberg (1979)] The probability that these formulas are satisfiable is very high. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 83/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 85/102 Phase Transitions (Mostly) empirical approach to identifying typical problem hardness. Cheeseman et al. (1991) came up with the following plausible conjecture: Phase Transitions with 3-SAT Constant clause length model [Mitchell et al. (1992)]: Clause length k is given. Uniformly choose variables for every clause k, and use the complement with probability 0.5 for each variable. Phase transition for 3-SAT with a clause/variable ratio of approx. 4.3: Phase Transition Conjecture All NP-complete problems have at least one order parameter and the hard to solve problems are around a critical value of this order parameter. This critical value (a phase transition) separates one region from another, such as over-constrained and under-constrained regions of the problem space. Confirmation for Graph Coloring and Hamiltonian path... later also for numerous other NP-complete problems. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 86/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 87/102

21 And DPLL? The Davis-Putnam (DPLL) Procedure shows extreme runtime peaks at the phase transition: Interpreting the Phase Transition Solution probability close to 1 (under-constrained), many solutions, first DPLL search path usually successful. Solution probability close to 0 (over-constrained), most DPLL search paths short (empty clause after branching on few variables). At the phase transition, there are many near successes! ( close, but no cigar ) Note: Long runs can occur also in the regions of the more easily satisfiable/unsatisfiable instances! Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 88/102 These are properties of this particular distribution of formulas! Predicting solver performance (within this distribution). Benchmarks are located in the phase transition region. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 89/102 Questionnaire Say I encode a Wumpus problem into that turns out to have clause/variable ratio 10. Which is true? (A): is very likely to be unsatisfiable. (B): is very likely to be satisfiable. Questionnaire, ctd. Say I sit down tonight and write a random with clause/variable ratio 1.1. Which are true? (A): I m bored. (B): is satisfiable. (C): All slides for next week are (D): is very likely to be prepared already. satisfiable. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 90/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 91/102

22 Local Search Methods for Solving SAT General Problem Solving using SAT and Local Search Say we are interested in finding a satisfying assignment, and we can sacrifice completeness if we can solve much large instances this way. Local Search! (as in earlier chapter in this lecture) Start with a random interpretation. Through local modifications, we hope to produce better interpretations. Main difficulties: local maxima, plateaus. Compared to DPLL: If unsatisfiable? Potentially quicker in finding solutions to satisfiable formulas! (some new problem) model problem as SAT use off-the-shelf local search (its solution, for solvable instances) Note the for solvable instances. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 93/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 94/102 Good interpretations, and Local Maxima GSAT As a measure of the quality of an interpretation I, we use the number of satisfied clauses. However, this is not an optimization problem: if I does not satisfy all clauses, then it is completely useless. Randomized restart and/or inject noise to escape local maxima. Procedure GSAT INPUT: a set of clauses, Max-Flips, and Max-Tries OUTPUT: a satisfying truth assignment of, if found begin for i := 1 to Max-Tries I := a randomly-generated truth assignment for j := 1 to Max-Flips if I satisfies then return I X := a variable such that a change in its truth assignment gives the largest increase in the number of clauses of that are satisfied by T I := I with the truth assignment of X reversed end for end for return no satisfying assignment found end Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 95/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 96/102

23 Example (B) The Search Behavior of GSAT = {{P, Q, R}, { P, Q}, {R}, {P, Q}} We must allow sideways movements! Most time is spent searching on plateaus. 1. I := {(P, F ), (Q, T ), (R, T )} number satisfied clauses: 2. Swap X := R? Number satisfied clauses: 3. Swap X := P? Number satisfied clauses: 4. Swap X := Q? Number satisfied clauses: 5. Swap X := P. I := {(P, T ), (Q, T ), (R, T )} number satisfied clauses: 6. Swap X := Q. number satisfied clauses: Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 97/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 98/102 Questionnaire Say we run GSAT on an unsatisfiable formula with n clauses. What is the best quality we may ever attain? (A): 0. (C): n 1. (B): 1. (D): n. Questionnaire, ctd. Say = {{P, Q, R}, { P, Q, R}, { S, S }, {S, R}, { R}} and GSAT starts with I : P T, Q F, R T, S T, S T. Which statements are true? (A): Quality now: 0. (C): Quality after swap R: 3. (B): Quality now: 3. (D): Quality after swap R, S: 4. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 99/102 Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 100/102

24 State of the Art SAT competitions: Since beginning of the 90s Distinguish random vs. industrial vs. handcrafted benchmarks. Largest industrial instances: > variables. State of the art DPLL: Vastly superior on handcrafted and industrial benchmarks. Key techniques: Clause Learning! Also, efficient implementation (unit propagation!), good branching heuristics. State of the art local search: Better on random instances. No dramatic progress in last decade. Parameters are difficult to adjust. Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 101/102 References I Paul Beame, Henry A. Kautz, and Ashish Sabharwal. Towards understanding and harnessing the potential of clause learning. Journal of Artificial Intelligence Research, 22: , Peter Cheeseman, Bob Kanefsky, and William M. Taylor. Where the really hard problems are. In J. Mylopoulos and R. Reiter, editors, Proceedings of the 12th International Joint Conference on Artificial Intelligence (IJCAI-91), pages , Sydney, Australia, August Morgan Kaufmann. A. Goldberg. On the complexity of the satisfiability problem. In 4th Workshop on Automated Deduction, pages 1 6, David Mitchell, Bart Selman, and Hector J. Levesque. Hard and easy distributions of SAT problems. In Proceedings of the 10th National Conference of the American Association for Artificial Intelligence (AAAI-92), pages , San Jose, CA, July MIT Press. Knot Pipatsrisawat and Adnan Darwiche. On the power of clause-learning SAT solvers with restarts. In Ian P. Gent, editor, Proceedings of the 15th International Conference on Principles and Practice of Constraint Programming (CP 09), volume 5732 of Lecture Notes in Computer Science, pages Springer, Jörg Hoffmann and Wolfgang Wahlster Artificial Intelligence 102/102