prolog - 如何在 Prolog 中制作一个程序来识别用户输入的名词的性别、复数和大小写？

Question

我有以下词典规则

gr_noun_suffix(masculine,singular,nominative,10,'ος').
gr_noun_suffix(masculine,singular,nominative,10,'ης').
gr_noun_suffix(masculine,singular,possessive,10,'η').
gr_noun_suffix(masculine,singular,accusative,10,'η').
gr_noun_suffix(masculine,plural,nominative,10,'ες').
gr_noun_suffix(masculine,plural,possessive,10,'ων').
gr_noun_suffix(masculine,plural,accusative,10,'ες').

例如，我需要提供输入

gr_noun(X,F,Gender,Plurality,Singularity,Case,Code,[σκυλος],[]).

并得到

F = σκυλ(X),
Gender= masculine,
Singularity= singular,
Case= possessive,
Code= 10 .

换句话说，我需要打破单词的最后一个音节，并与规则进行比较，以便喜欢适用的内容。

我似乎被困在如何打破这个词的最后一个音节。

Answer 1

不要将原子拆分为字符列表。用于sub_atom/5查找后缀或将其断开。

我无法理解所有论点应该是什么，但这对于一个起点来说应该足够了：

gr_noun(X, F, Gender, Singularity, Case, Code, Noun) :-
    gr_noun_suffix(Gender, Singularity, Case, Code, X),
    sub_atom(Noun, Before, Len, 0, X),
    sub_atom(Noun, 0, Before, Len, F0),
    F =.. [F0, X].

有了这个，我得到：

?- gr_noun(X, F, Gender, Singularity, Case, Code, σκυλος).
X = ος,
F = σκυλ(ος),
Gender = masculine,
Singularity = singular,
Case = nominative,
Code = 10 ;