Question

我目前正在尝试学习一点Prolog。作为练习，我试图解决以下问题：

这些规则是：

*Every person that has neither a car nor a plane, has a bike.
*Every person that doesn't have a plane but has a bike, has a car
*Every person that doesn't have a plane but has a car, has a truck
*Every person that doesn't have a truck but has a boat, doesn't have a plane
*Every person that doesn't have a boat but has a plane, doesn't have a car

现在有四个人：

*Person1 doesn't have a car but has a boat
*Person2 doesn't have a boat but has a plane
*Person3 doesn't have a plane but has a bike
*Person4 doesn't have a bike but has a car

哪个人没有卡车？

到目前为止，我想出的是：

doesnthave(car,pa).
has(boat,pa).
doesnthave(boat,pb).
has(plane,pb).
doesnthave(plane,pc).
has(bike,pc).
doesnthave(bike,pd).
has(car,pd).

has(bike,X) :- doesnthave(car,X),doesnthave(plane,X).
has(car,X) :- doesnthave(plane,X),has(bike,X).
has(truck,X) :- doesnthave(plane,X),has(car,X).
doesnthave(plane,X) :- doesnthave(truck,X),has(boat,X).
doesnthave(car,X) :- doesnthave(boat,X),has(plane,X).

现在这似乎还不够。或者这不是在prolog中解决这样的谜题的方法吗？

编辑：看来前两个陈述是矛盾的。他们共同产生：每个既没有汽车也没有飞机的人都有车。我不确定是否有合理的解决方案。

Answer 1

不确定这个解决方案，但更简单的知识表示肯定会有所帮助：

has(car,   pa, n). has(boat,  pa, y).
has(boat,  pb, n). has(plane, pb, y).
has(plane, pc, n). has(bike,  pc, y).
has(bike,  pd, n). has(car,   pd, y).

has(bike,  X,  y) :- has(car,   X, n), has(plane, X, n).
has(car,   X,  y) :- has(plane, X, n), has(bike,  X, y).
has(truck, X,  y) :- has(plane, X, n), has(car,   X, y).
has(plane, X,  n) :- has(truck, X, n), has(boat,  X, y).
has(car,   X,  n) :- has(boat,  X, n), has(plane, X, y).

现在我们可以查询所有权（指出它是一对多的关系）

?- setof((P,T,R), has(T,P,R), L), maplist(writeln, L).
pa,boat,y
pa,car,n
pb,boat,n
pb,car,n
pb,plane,y
pc,bike,y
pc,car,y
pc,plane,n
pc,truck,y
pd,bike,n
pd,car,y
L = [ (pa, boat, y), (pa, car, n), (pb, boat, n), (pb, car, n), (pb, plane, y), (pc, bike, y), (pc, car, y), (pc, ..., ...), (..., ...)|...].

注意我在P ransport之前放置了T erson，然后我们可以直观地检查结果......

似乎解决方案是三重...

Prolog - 试图解决文本难题

1 个答案: