Question

测试用例

?- decompose([[1,2,8],[3,4],[5,6]], L1, L2).
L1 = [1,3,5], L2 = [[2,8],[4],[6]] ? ;
no

我尝试了另一种实现，但给出的反馈是效率低下。

效率低下的实施

listFirst([], []).
listFirst([H1|T1], [H2|Z]):-
    H1 = [H2|_],
    listFirst(T1, Z).

listFollowers([], []).
listFollowers([H1|T1], [T2|Z]):-
    H1 = [H2|T2],
    listFollowers(T1, Z).

decompose(A,L1,L2) :-
    listFollowers(A, L2),
    listFirst(A, L1).

Answer 1

跟进@findall's previous answer ...如何使用meta-predicate maplist/4？

list_head_tail([X|Xs], X, Xs).

decompose(Mss, Hs, Ts) :-
   maplist(list_head_tail, Mss, Hs, Ts).

示例查询：

?- decompose([[a,b,c],[d,e,f]], Heads, Tails).
Heads = [a,d], Tails = [[b,c],[e,f]].

?- decompose([[1,2,8],[3,4],[5,6]], L1, L2).
L1 = [1,3,5], L2 = [[2,8],[4],[6]].

Answer 2

正如潜伏者所说，listFirst和listFollowers的功能可以合并为谓词，以便同时执行这些功能。像这样;

decompose([[H|T]|T0], [H|L1], [T|L2]) :- decompose(T0, L1, L2).
decompose([], [], []).

将矩阵分解为prolog中第一个元素列表和剩余元素的子列表

2 个答案: