1 / 32

# For Wednesday - PowerPoint PPT Presentation

For Wednesday. Read chapter 10 Prolog Handout 4. Exam 1. Monday Take home due at the exam. Try It. reverse(List,ReversedList) evenlength(List) oddlength(List). The Anonymous Variable. Some variables only appear once in a rule Have no relationship with anything else

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'For Wednesday ' - archie

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Prolog Handout 4

• Monday

• Take home due at the exam

• reverse(List,ReversedList)

• evenlength(List)

• oddlength(List)

• Some variables only appear once in a rule

• Have no relationship with anything else

• Can use _ for each such variable

• Basic arithmetic operators are provided for by built-in procedures:+, -, *, /, mod, //

• Note carefully:?- X = 1 + 2.X = 1 + 2?- X is 1 + 2.X = 3

• Comparison operators:><>==< (note the order: NOT <=)=:=(equal values)=\=(not equal values)

• Retrieving people born 1950-1960:?- born(Name, Year), Year >= 1950, Year =< 1960.

• Difference between = and =:=?- 1 + 2 =:= 2 + 1.yes?- 1 + 2 = 2 + 1.no?- 1 + A = B + 2.A = 2B = 1

• Definition of length/2length([], 0).length([_ | Tail], N) :-length(Tail, N1),N is 1 + N1.

• Note: all loops must be implemented via recursion

• 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).

• A way to prevent backtracking.

• Used to simplify and to improve efficiency.

• Can’t say something is NOT true

• Use a closed world assumption

• Not simply means “I can’t prove that it is true”

• A way to write self-modifying code, in essence.

• Typically just storing data using Prolog’s built-in predicate database.

• Dynamic predicates must be declared as such.

• assert and variants

• retract

• Fails if there is no clause to retract

• retractall

• Doesn’t fail if no clauses

• 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)

• 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

• 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).

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)

• 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)

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)

• 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

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

• 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.

• 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

• 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))

• 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.

• 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

"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)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))))

• 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?

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)

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