在底部，我给出了整个程序（ciao）的链接，以使帮助更容易。我尝试在 Prolog 函数中创建问题列表，例如
questions([[[What, are,you,doing,?],[Where,am,I,Incorrect,?]]]).
answers([[Im,doing,exercise],[I,do,nothing]],[[You,are,incorrect,at,'..'],[i,dont,know]]]).
wordkeys([[[Incorrect,50],[doing,20]]]).
我知道它看起来很乱，但我真的需要帮助，我将不胜感激。主要功能是检查哪个答案是最好的（具有最大的关键字点总和）。我的问题是一切看起来都很好（做了一些 write() 来看看发生了什么），直到它进入最后一个函数：
count_pnt_keys()
Prolog 检查所有单词是否相等，但何时从关键字中退出应该返回到调用它的函数，但它只是 'no' 。也许我应该在只用 Tail 再次调用相同的函数之前检查列表是否为空？这个怎么做？
规则：
count_pnt([],[],[]).
count_pnt([Ah|At],Keys,RList) :- %choose one answer from answer list and go further
count_pnt_word(Ah,Keys,Pnts), %return sum of points for one answer
count_ADD_POINT(RList,Pnts), %not important here
count_pnt(At,Keys,RList). %call again with next question
/*vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv*/
count_pnt_word([],[],0)
count_pnt_word([Ah|At],Keys,Pnts) :- %choose one WORD from answer and go further
count_pnt_keys(Ah,Keys,Pnts),
count_pnt_word(At,Keys,Pnts). %call recursion with next word from answer
/*vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv*/
count_pnt_keys([],[],0)
count_pnt_keys(AWord,[Kh|Kt],Pnts) :- %check word with all keys for first question
get_tail_h(Kh,KWORD,KPOINTS), %just return head and tail from first key
AWord==KWORD, %check if they are equal
/*counting points*/ !, %counting not important when end counting points go out to next
count_pnt_keys(AWord,Kt,Pnts). %call again if not equal and try with next key
我称之为：
test :-
write(AnswerFirst),
count_pnt(FirstPackOfAnswer,FirstPackofKeys,ReturnedListwithpoints),
write(ReturnedListwithpoints).
代码链接 (ciao) http://wklej.org/id/754478/