240 likes | 443 Views
For Monday. Read 10.1-10.2 Prolog handout 4 Homework: Chapter 9, exercises 9-11. Exam 1. Wednesday Covers chapters 1-9 Focus on search and logic No Lisp or Prolog Take home handed out that day, due the following Monday. Note: no class on March 3. Homework. Prolog homework
E N D
For Monday • Read 10.1-10.2 • Prolog handout 4 • Homework: • Chapter 9, exercises 9-11
Exam 1 • Wednesday • Covers chapters 1-9 • Focus on search and logic • No Lisp or Prolog • Take home handed out that day, due the following Monday. • Note: no class on March 3
Homework • Prolog homework • Logic homework
Counting Loops • Definition of sum/3sum(Begin, End, Sum) :- sum(Begin, End, Begin, Sum).sum(X, X, Y, Y).sum(Begin, End, Sum1, Sum) :- Begin < End, Next is Begin + 1, Sum2 is Sum1 + Next, sum(Next, End, Sum2, Sum).
Negation • Can’t say something is NOT true • Use a closed world assumption • Not simply means “I can’t prove that it is true”
Resolution • Propositional version. {a Ú b, ¬b Ú c} |- a Ú c OR {¬aÞ b, b Þ c} |- ¬a Þ c Reasoning by cases OR transitivity of implication • Firstorder form • For two literals pj and qk in two clauses • p1Ú ... pj ... Ú pm • q1Ú ... qk ... Ú qn such that q=UNIFY(pj , ¬qk), derive SUBST(q, p1Ú...pj1Úpj+1...ÚpmÚq1Ú...qk1 qk+1...Úqn)
Implication form • Can also be viewed in implicational form where all negated literals are in a conjunctive antecedent and all positive literals in a disjunctive conclusion. ¬p1Ú...Ú¬pmÚq1Ú...ÚqnÛ p1Ù... Ù pmÞ q1Ú ...Ú qn
Conjunctive Normal Form (CNF) • For resolution to apply, all sentences must be in conjunctive normal form, a conjunction of disjunctions of literals (a1Ú ...Ú am) Ù (b1Ú ... Ú bn) Ù ..... Ù (x1Ú ... Ú xv) • Representable by a set of clauses (disjunctions of literals) • Also representable as a set of implications (INF).
Example Initial CNF INF P(x) Þ Q(x) ¬P(x) Ú Q(x) P(x) Þ Q(x) ¬P(x) Þ R(x) P(x) Ú R(x) True Þ P(x) Ú R(x) Q(x) Þ S(x) ¬Q(x) Ú S(x) Q(x) Þ S(x) R(x) Þ S(x) ¬R(x) Ú S(x) R(x) Þ S(x)
Resolution Proofs • INF (CNF) is more expressive than Horn clauses. • Resolution is simply a generalization of modus ponens. • As with modus ponens, chains of resolution steps can be used to construct proofs. • Factoring removes redundant literals from clauses • S(A) Ú S(A) -> S(A)
Sample Proof P(w) Q(w) Q(y) S(y) {y/w} P(w) S(w) True P(x) R(x) {w/x} True S(x) R(x) R(z) S(z) {x/A, z/A} True S(A)
Refutation Proofs • Unfortunately, resolution proofs in this form are still incomplete. • For example, it cannot prove any tautology (e.g. PÚ¬P) from the empty KB since there are no clauses to resolve. • Therefore, use proof by contradiction (refutation, reductio ad absurdum). Assume the negation of the theorem P and try to derive a contradiction (False, the empty clause). • (KB Ù ¬P Þ False) Û KB Þ P
Sample Proof P(w) Q(w) Q(y) S(y) {y/w} P(w) S(w) True P(x) R(x) {w/x} True S(x) R(x) R(z) S(z) {z/x} S(A) False True S(x) {x/A} False
Resolution Theorem Proving • Convert sentences in the KB to CNF (clausal form) • Take the negation of the proposed theorem (query), convert it to CNF, and add it to the KB. • Repeatedly apply the resolution rule to derive new clauses. • If the empty clause (False) is eventually derived, stop and conclude that the proposed theorem is true.
Conversion to Clausal Form • Eliminate implications and biconditionals by rewriting them. p Þ q -> ¬p Ú q p Û q > (¬p Ú q) Ù (p Ú ¬q) • Move ¬ inward to only be a part of literals by using deMorgan's laws and quantifier rules. ¬(p Ú q) -> ¬p Ù ¬q ¬(p Ù q) -> ¬p Ú¬q ¬"x p -> $x ¬p ¬$x p -> "x ¬p ¬¬p -> p
Conversion continued • Standardize variables to avoid use of the same variable name by two different quantifiers. "x P(x) Ú$x P(x) -> "x1 P(x1) Ú $x2 P(x2) • Move quantifiers left while maintaining order. Renaming above guarantees this is a truthpreserving transformation. "x1 P(x1) Ú $x2 P(x2) -> "x1$x2 (P(x1) Ú P(x2))
Conversion continued • Skolemize: Remove existential quantifiers by replacing each existentially quantified variable with a Skolem constant or Skolem function as appropriate. • If an existential variable is not within the scope of any universally quantified variable, then replace every instance of the variable with the same unique constant that does not appear anywhere else. $x (P(x) Ù Q(x)) -> P(C1) Ù Q(C1) • If it is within the scope of n universally quantified variables, then replace it with a unique nary function over these universally quantified variables. "x1$x2(P(x1) Ú P(x2)) -> "x1 (P(x1) Ú P(f1(x1))) "x(Person(x) Þ$y(Heart(y) Ù Has(x,y))) -> "x(Person(x) Þ Heart(HeartOf(x)) Ù Has(x,HeartOf(x))) • Afterwards, all variables can be assumed to be universally quantified, so remove all quantifiers.
Conversion continued • Distribute Ù over Ú to convert to conjunctions of clauses (aÙb) Ú c -> (aÚc) Ù (bÚc) (aÙb) Ú (cÙd) -> (aÚc) Ù (bÚc) Ù (aÚd) Ù (bÚd) • Can exponentially expand size of sentence. • Flatten nested conjunctions and disjunctions to get final CNF (a Ú b) Ú c -> (a Ú b Ú c) (a Ù b) Ù c -> (a Ù b Ù c) • Convert clauses to implications if desired for readability (¬a Ú ¬b Ú c Ú d) -> a Ù b Þ c Ú d
Sample Clause Conversion "x((Prof(x) Ú Student(x)) Þ($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x(¬(Prof(x) Ú Student(x)) Ú($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù$y(Book(y) Ù Has(x,y)))) "x((¬Prof(x) Ù ¬Student(x)) Ú ($y(Class(y) Ù Has(x,y)) Ù$z(Book(z) Ù Has(x,z)))) "x$y$z((¬Prof(x)Ù¬Student(x))Ú ((Class(y) Ù Has(x,y)) Ù (Book(z) Ù Has(x,z)))) (¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù Book(g(x)) Ù Has(x,g(x)))) (¬Prof(x) Ú Class(f(x))) Ù (¬Prof(x) Ú Has(x,f(x))) Ù (¬Prof(x) Ú Book(g(x))) Ù (¬Prof(x) Ú Has(x,g(x))) Ù (¬Student(x) Ú Class(f(x))) Ù (¬Student(x) Ú Has(x,f(x))) Ù (¬Student(x) Ú Book(g(x))) Ù (¬Student(x) Ú Has(x,g(x))))
Clause Conversion (¬Prof(x)Ù¬Student(x))Ú (Class(f(x)) Ù Has(x,f(x)) Ù Book(g(x)) Ù Has(x,g(x)))) (¬Prof(x) Ú Class(f(x))) Ù (¬Prof(x) Ú Has(x,f(x))) Ù (¬Prof(x) Ú Book(g(x))) Ù (¬Prof(x) Ú Has(x,g(x))) Ù (¬Student(x) Ú Class(f(x))) Ù (¬Student(x) Ú Has(x,f(x))) Ù (¬Student(x) Ú Book(g(x))) Ù (¬Student(x) Ú Has(x,g(x))))
Sample Resolution Problem • Jack owns a dog. • Every dog owner is an animal lover. • No animal lover kills an animal. • Either Jack or Curiosity killed Tuna the cat. • Did Curiosity kill the cat?
In Logic Form A) $x Dog(x) Ù Owns(Jack,x) B) "x ($y Dog(y) Ù Owns(x,y)) Þ AnimalLover(x)) C) "x AnimalLover(x) Þ ("y Animal(y) Þ ¬Kills(x,y)) D) Kills(Jack,Tuna) Ú Kills(Cursiosity,Tuna) E) Cat(Tuna) F) "x(Cat(x) Þ Animal(x)) Query: Kills(Curiosity,Tuna)
In Normal Form A1) Dog(D) A2) Owns(Jack,D) B) Dog(y) Ù Owns(x,y) Þ AnimalLover(x) C) AnimalLover(x) Ù Animal(y) Ù Kills(x,y) Þ False D) Kills(Jack,Tuna) Ú Kills(Curiosity,Tuna) E) Cat(Tuna) F) Cat(x) Þ Animal(x) Query: Kills(Curiosity,Tuna) Þ False