为什么我删除了想要的号码列表后，又为什么要重新获得原件? [英] Why do i get the original again as a result after i get the list with the desired number removed?

问题描述

我正在尝试实施一个序言规则，该规则将从某个列表中删除数字X并返回不包含X的列表.但是，在获得正确的结果之后，我也可以得到原始列表的替代结果，并且可以弄清楚代码中的问题出在哪里.

i was trying to implement a prolog rule which removes the number X from a certain list and returns the list without the X . However after getting the correct result i get as an alternative the original list as a result as well and i get figure out where is the problem in my code.

remove1(X,[],[]).
remove1(X,[X|T],T).
remove1(X,[H|T],[H|R]):-
    X \= H,
    remove1(X,T,R).
remove1(X,[H|T],[H|T]):-
    \+member(X,T).

推荐答案

基本上，第一个和最后一个子句都是不必要的.最后一个子句是错误结果的出处.

Basically, the first and the last clause are unnecessary. The last clause is where the wrong results come from.

这似乎是另一个教科书"谓词，通常称为select/3. [Sterling& Shapiro]，第67页中的教科书定义与

This seems to be another "textbook" predicate, and it is usually called select/3. The textbook definition in [Sterling&Shapiro], page 67, is identical to the SWI-Prolog library implementation and goes something like this:

%!  select(?Elem, ?List1, ?List2)
%
%   Is true when List1, with Elem removed, results in List2.

select(X, [X|Tail], Tail).
select(Elem, [Head|Tail], [Head|Rest]) :-
    select(Elem, Tail, Rest).

您会注意到它不需要任何辅助谓词，并且是真实的关系.尝试例如:

You will notice that it does not need any helper predicates and is a true relation. Try for example:

?- select(X, [1,2,3], R).

甚至:

?- select(X, List, Rest), numbervars(List).

(numbervars/1只是使解决方案更具可读性)

(numbervars/1 just makes the solutions a bit more readable)

此定义之所以是如此笼统，是因为……嗯，这是一个笼统的定义.第一个子句执行选择"，第二个子句定义递归步骤.

This definition is as general as it is because... well, it is a general definition. The first clause does the "selecting", the second clause defines the recursive step.

[Sterling& Shapiro]:Sterling L，Shapiro EY. Prolog的艺术:先进的编程技术.麻省理工学院出版社； 1994.

[Sterling&Shapiro]: Sterling L, Shapiro EY. The art of Prolog: advanced programming techniques. MIT press; 1994.

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