具有更多确定性的 Prolog append/3 实现? [英] Prolog append/3 realization with more determinism?

问题描述

append(X,[Y],Z) 找到最后一个元素是民间常识列表Z和剩余列表X的Y.

It is folk knowledge that append(X,[Y],Z) finds the last element Y of the list Z and the remaining list X.

但是拥有自定义谓词last/3有一些优势，即它可以在不离开选择点的情况下做出反应:

But there is some advantage of having a customized predicate last/3, namely it can react without leaving a choice point:

?- last([1,2,3],X,Y).
X = 3,
Y = [1,2]

?- append(Y,[X],[1,2,3]).
Y = [1,2],
X = 3 ;
No

有没有办法实现不同的实现append/3 这也不会在上面的例子?

Is there a way to realize a different implementation of append/3 which would also not leave a choice point in the above example?

P.S.:我在比较:

/**
 * append(L1, L2, L3):
 * The predicate succeeds whenever L3 unifies with the concatenation of L1 and L2.
 */
% append(+List, +List, -List)
:- public append/3.
append([], X, X).
append([X|Y], Z, [X|T]) :- append(Y, Z, T).

和(如格特扬·范·诺德):

And (à la Gertjan van Noord):

/**
 * last(L, E, R):
 * The predicate succeeds with E being the last element of the list L
 * and R being the remainder of the list.
 */
% last(+List, -Elem, -List)
:- public last/3.
last([X|Y], Z, T) :- last2(Y, X, Z, T).

% last2(+List, +Elem, -Elem, -List)
:- private last2/4.
last2([], X, X, []).
last2([X|Y], U, Z, [U|T]) :- last2(Y, X, Z, T).

推荐答案

一种方法是使用 foldl/4 和适当的帮助谓词:

One way to do it is to use foldl/4 with the appropriate help predicate:

swap(A, B, B, A).

list_front_last([X|Xs], F, L) :-
    is_list(Xs),
    foldl(swap, Xs, F, X, L).

应该是这样:

?- list_front_last([a,b,c,d], F, L).
F = [a, b, c],
L = d.

?- list_front_last([], F, L).
false.

?- list_front_last([c], F, L).
F = [],
L = c.

?- Ys = [y|Ys], list_front_last(Ys, F, L).
false.

尝试看看是否可以从定义中省略 is_list/1.

Try to see if you can leave out the is_list/1 from the definition.

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