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Correção dos exercícios de engenharia do conhecimento em Prolog Jacques Robin, DI-UFPE www.di.ufpe.br/~jr.

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Apresentação em tema: "Correção dos exercícios de engenharia do conhecimento em Prolog Jacques Robin, DI-UFPE www.di.ufpe.br/~jr."— Transcrição da apresentação:

1 Correção dos exercícios de engenharia do conhecimento em Prolog Jacques Robin, DI-UFPE

2 Estudo de caso: a terrível novela Requisitos em Inglês 1. A soap opera is a TV show whose characters include a husband, a wife and a mailman such that: 2. the wife and the mailman blackmail each other 3. everybody is either alcoholic, drug addict or gay 4. Dick is gay, Jane is alcoholic and Harry is a drug addict 5. the wife is always an alcoholic and the long-lost sister of her husband 6. the husband is always called Dick and the lover of the mailman 7. the long-lost sister of any gay is called either Jane or Cleopatra 8. Harry is the lover of every gay 9. Jane blackmails every drug addicted lover of Dick 10. soap operas are invariably terrible! 0. Who are the characters of a terrible TV show?

3 Correção do exercício 1: A terrível novela em L1

4 Correção do exercício 2: A terrível novela em Prolog tvShow(Cast,Qual) :- soapOpera(Cast,Qual). soapOpera([dick,W,M],terrible) :- soapCast([dick,W,M]), blackmail(W,M), alcoholic(W), longLostSister(W,dick), lover(M,dick). soapCast([]). soapCast([H|T]) :- soapChar(H), soapCast(T). soapChar(C) :- alcoholic(C). soapChar(C) :- drugAddict(C). soapChar(C) :- gay(C). gay(dick). alcoholic(jane). drugAddict(harry). lover(harry,G) :- gay(G). longLostSister(jane,G) :- gay(G). longLostSister(cleopatra,G) :- gay(G). blackmail(jane,M) :- lover(M,dick), drugAddict(M). ?-tvShow(Cast,terrible). * Cast = [dick,jane,harry].

5 Estudo de caso: o BD acadêmico Requisitos em Inglês 1. Bob is 40 and the manager of the CS department. 2. His assistants are John and Sally. 3. Marys highest degree is an MS and she works at the CS department. 4. She co-authored with her boss and her friends, John and Sally, a paper published in the Journal of the ACM. 5. Phil is a faculty, who published a paper on F- Logic at a Conference of the ACM, jointly with Mary and Bob. 6. Every faculty is a midaged person who writes article, makes in the average $50,000 a year and owns a degree of some kind, typically a PhD. 7. One is considered midage if one is between 30 and 50 years old. 8. A facultys boss is both a faculty and a manager. 9. Friends and children of a person are also persons. 10. Every department has a manager who is an employee and assistants who are both employees and students 11. A boss is an employee who is the manager of another employee of the same department. 12. A joint work is a paper that is written by two faculties 13. There are three types of papers: technical reports, journal papers and conference papers 0a: Who are the midaged employees of the CS department and who are their boss? 0b: Who published jointly with Mary in the Journal of the ACM? 0c: Where did Mary published joint work with Phil?

6 Correção do exercício 3: O banco de dados acadêmico em Prolog 1/ fatos ground person(bob). age(bob,40). works(bob,cs,faculty). manager(cs,bob). dept(cs). works(john,cs,assistant). study(john,cs). works(sally,cs,assistant). study(sally,cs). hiDeg(mary,ms). works(mary,cs,faculty). friends(mary,bob). friends(mary,sally). works(phil,cs,faculty). degree(phd). degree(ms). journal(jacm). conf(cacm). article(flogic,[john,sally,mary,bob],jacm). article(florid,[phil,mary,bob],cacm).

7 Correção do exercício 3: O banco de dados acadêmico em Prolog 2/ regras de dedução hiDeg(F,phd) :- works(F,_,faculty), not hiDeg(F,ms). salary(P,5000) :- works(F,_,faculty), not salary(F,_). midaged(F) :- age(F,A), !, integer(A), A >= 30, A =< 50. midaged(F) :- works(F,_,faculty). works(B,D,faculty) :- manager(D,B), works(E,D,faculty), !. activity(F,paperWriting) :- works(F,_,faculty). person(P2) :- friends(P1,P2), person(P1). person(C) :- parent(A,C), person(A). person(P) :- study(P,D), dept(D). person(P) :- works(P,_,D), dept(D). works(S,D,assistant) :- study(S,D), dept(D), works(S,D,_), !. works(M,D,_) :- manager(M,D). boss(B,E) :- manager(D,B), works(E,D,_). jointWork(W,F1,F2,P) :- works(F1,_,faculty), works(F2,_,faculty), F1 \= F2, report(W,Fl,P), member(F1,Fl), member(F2,Fl). member(H,[H|T]). member(X,[H|T]) :- member(X,T). report(T,Al,J) :- article(T,Al,J), journal(J). report(T,Al,C) :- article(T,Al,C), conf(C). report(T,Al,D) :- techrep(T,Al,D), dept(D).

8 Correção do exercício 3: O banco de dados acadêmico em Prolog 3/ consultas midagedWorkerOf(E,D) :- works(E,D,_), midaged(E). bossOfMidagedWorkerOf(B,D) :- midagedWorkerOf(E,D), boss(B,E). ? setof(E,midagedWorkerOf(E,cs),Le), setof(B,bossOfMidagedWorkerOf(B,cs),Lb). * E = _20, Le = [bob,mary], B = _51, Lb = [bob]; * no. ? setof(F,jointWork(_,F,mary,jacm),Lf). * F = _20, Lf = [bob]; * no. ? setof(P,jointWork(_,phil,mary,P),Lp). * P = _20, Lp = [cacm]; * no

9 Correção do exercício 4: O banco de dados acadêmico em L1 1/ formulas ground

10 Correção do exercício 4: O banco de dados acadêmico em L1 2/ formulas quantificadas

11 Correção do exercício 4: O banco de dados acadêmico em L1 3/ consultas

12 Estudo de caso: A curiosidade matou o gato? * Requisitos em inglês 1. Jack owns a dog. 2. Every dog owner is an animal lover. 3. No animal lover kills an animal. 4. Either Jack or curiosity killed Tuna 5. Tuna is a cat A. Did curiosity kill the cat? B. Quem matou o gato? * Em L1 1. x Dog(x) Owns(Jack,x) 2. x ( y Dog(y) Owns(x,y)) AnimalLover(x) 3. x,y (AnimalLover(x) Animal(y)) Kills(x,y) 4.(Kills(Jack, Tuna) Kills(Curiosity, Tuna)) Kills(Jack, Tuna) Kills(Curiosity, Tuna)) 5.Cat(Tuna) 6. x Cat(x) Animal(x ) 0. Kills(Curiosity,Tuna) B. X, Kills(X,Tuna)

13 Exercício 5: A curiosidade matou o gatou? em Prolog owns(jack,dog1). /* 1 */ dog(dog1). /* 1 */ animalLover(H) :- owns(H,A), animal(A). /* 2 */ notKills(X,A) :- animalLover(X), animal(A), !. /* 3 */ notKills(X,A) :- not kills(X,A). /* 3 */ kills(curiosity,tuna) :- notKills(jack,tuna). /* 4 */ kills(jack,tuna) :- notKills(curiosity,tuna). /* 4 */ cat(tuna). /* 5 */ animal(A) :- cat(A). /* 6 */ animal(A) :- dog(A). /* 7 */ ?- kills(curiosity,tuna). yes ?- kills(curiosity,X). X = tuna More (y/n)? y no ?

14 Exercício 6: A curiosidade matou o gato? em LIFE * Arquivo da curiosidade e tuna load("glOnto.lf")? animal := {cat; dog}. % 6-7 dog1 <| dog. % 1 tuna <| cat. % 5 feeling <| abstObj. curiosity <| feeling. owns(jack,dog1). % 1 love(H,animal) :- owns(H,dog). % 2 notKills(X,Y) :- love(X,Y), !. % 3 notKills(X,Y) :- not kills(X,Y). % 3 kills(curiosity,tuna) :- notKills(jack,tuna). % 4 kills(jack,tuna) :- notKills(curiosity,tuna). % 4 * Arquivo da ontologia geral entities := {situation; object; quality; quantity; place; time}. situation := {event; relation}. event := {action; happening}. object := {physObj; abstObj}. physObj := {liveBeing; artefact}. person <| liveBeing. checkList([],Sort) -> true. checkList([Sort|Tail],Sort) -> checkList(Tail). Consultas: > kills(curiosity,tuna)? *** Yes --1> *** No > kills(X,Y)? *** Yes, X = curiosity, Y = tuna. --1> *** No

15 Estudo de caso: Coloração de mapa * Colorir mapa tal que: países adjacentes de cores diferentes A B C F D E * Instância de problema de resolução de restrições A B C F D E

16 Exercício 7: Coloração de mapa em Prolog

17 Exercício 8: Coloração de mapa em LIFE


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