Question

我想问一下prolog编程。我有：

byCar(auckland,hamilton).
byCar(valmont,metz).

byTrain(metz,frankfurt).

byPlane(frankfurt,bangkok).
byPlane(bangkok,auckland).

travell(From,To,go(From,To,car)) :- byCar(From,To).
travell(From,To,go(From,To,train)) :- byTrain(From,To).
travell(From,To,go(From,To,plane)) :- byPlane(From,To).
travell(From,To,go(From,Step,Via,Go)) :- travell(From,Step,go(From,Step,Via)),travell(Step,To,Go).

然后我在winprolog中问：

?- travell(valmont,hamilton,Go).

回答

Go = go(valmont,metz,car,go(metz,frankfurt,train,go(frankfurt,bangkok,plane,go(bangkok,auckland,plane,go(auckland,hamilton,car)))))

我的问题是，是否可以计算它有多少'去'？就像我的例子之前它有5'去'

Answer 1

如果你可以统一go / 3和go / 4 arity，那么

会更容易。无论如何，试试

count_last_arg(Struct,Func,Count) :-
     count_last_arg(Struct,Func,0,Count).

count_last_arg(Struct,Func,N,Count) :-
     Struct =.. L, last(L, LL), LL =.. [Func|_]
  -> N1 is N+1,
     count_last_arg(LL,Func,N1,Count)
  ;  Struct =.. [Func|_]
  -> Count is N+1
  ;  Count is N.

计算prolog中的重复术语

1 个答案: