Question

我有像这样的Prolog声明

verb('part_of-8').
noun('doctor_investigation_system-2').
noun('dis-4').
berelation('be-6').
verb('be-6').
noun('hospital_information_system-11').
noun('his-13').
rel('part_of-8', 'doctor_investigation_system-2').
rel('doctor_investigation_system-2', 'dis-4').
rel('part_of-8', 'be-6').
rel('part_of-8', 'hospital_information_system-11').
rel('hospital_information_system-11', 'his-13').

associatedWith(X,Y,Z) :-
   verb(Y),
   noun(X),
   noun(Z),
   X\=Y, Y\=Z, Z\=X,
   rel(X,Y), rel(Y,Z),
   not(beralation(X)), not(beralation(Z)), not(beralation(Y)).

我的目标是获得associationWith（X，Y，Z），其中X，Y，Z不是“be”项（berelation），但我写的上述规则不起作用，怎么办它的工作

Answer 1

我相信你正在寻找\+ "is not provable" operator

因此：

associatedWith(X,Y,Z) :-
  verb(Y),
  noun(X),
  noun(Z),
  X\=Y,
  Y\=Z,
  Z\=X,
  rel(X,Y),
  rel(Y,Z),
  \+ beralation(X),
  \+ beralation(Z),
  \+ beralation(Y).

还有另一种方式（没有\+，!“cut”运算符）：

associatedWith(X,_,_) :-
  berelation(X), !, fail.
associatedWith(_,Y,_) :-
  berelation(Y), !, fail.
associatedWith(_,_,Z) :-
  berelation(Z), !, fail.
associatedWith(X,Y,Z) :-
  verb(Y),
  noun(X),
  noun(Z),
  X\=Y,
  Y\=Z,
  Z\=X,
  rel(X,Y),
  rel(Y,Z).

用非运算符写prolog语句

1 个答案: