Question

在底部，我提供了整个程序（ciao）的链接，以帮助更轻松。我尝试在Prolog函数中创建像

这样的问题列表

 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最好（具有最大的关键字点数）。我的问题是所有看起来都很好（做了一些write（）来看看发生了什么），直到它去了最后一个函数：

count_pnt_keys()

Prolog检查所有单词，如果它们相等但什么时候离开关键字应该回到函数调用它，但它只是'不'。也许我应该检查列表是否为空，然后我再次调用与Tail相同的功能？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).

链接到代码（ciao） http://wklej.org/id/754478/

Answer 1

使用三个空闲参数调用count_pnt，这意味着count_pnt将首先用空列表统一其所有参数。在回溯时，会调用递归count_pnt子句，这会导致count_pnt_keys再次出现三个自由参数，这将导致Ah与[]等统一而不是失败。< / p>

您是否真的按照count_pnt的代码建议致电test？

Answer 2

count_pnts(_,_,[],_).
count_pnt_word(_,[],_).
count_pnt_keys([],_,_).

应该看起来像是一个问题

Prolog递归停止

2 个答案: