在prolog中拆分列表

Question

我对Prolog很新,我需要一些小问题的帮助:

我试图在两个列表中拆分一对夫妇.第一个列表包含具有给定键的所有对,第二个列表包含所有其他对象.

这是我到目前为止的代码:

splitList([],_,[],[]).
splitList([(A,X)|Rest], B, [Elem1|List1], [Elem2|List2]):-
    (
        A == B
    ->
        Elem1 = (A,X),
        splitList(Rest, B, List1, [Elem2|List2])
    ;
        Elem2 = (A,X),
        splitList(Rest, B, [Elem1|List1], List2)
    ).

当我尝试测试时,这就是我得到的:

[trace] [3] 143 ?- splitList([(1,yellow),(1,blue),(2,yellow),(2,blue)],1,X,Y).
   Call: (37) splitList([ (1, yellow), (1, blue), (2, yellow), (2, blue)], 1, _G4821, _G4822) ? creep
   Call: (38) 1==1 ? creep
   Exit: (38) 1==1 ? creep
   Call: (38) _G4928= (1, yellow) ? creep
   Exit: (38) (1, yellow)= (1, yellow) ? creep
   Call: (38) splitList([ (1, blue), (2, yellow), (2, blue)], 1, _G4929, [_G4931|_G4932]) ? creep
   Call: (39) 1==1 ? creep
   Exit: (39) 1==1 ? creep
   Call: (39) _G4940= (1, blue) ? creep
   Exit: (39) (1, blue)= (1, blue) ? creep
   Call: (39) splitList([ (2, yellow), (2, blue)], 1, _G4941, [_G4931|_G4932]) ? creep
   Call: (40) 2==1 ? creep
   Fail: (40) 2==1 ? creep
   Call: (40) _G4931= (2, yellow) ? creep
   Exit: (40) (2, yellow)= (2, yellow) ? creep
   Call: (40) splitList([ (2, blue)], 1, [_G4949|_G4950], _G4932) ? creep
   Call: (41) 2==1 ? creep
   Fail: (41) 2==1 ? creep
   Call: (41) _G4958= (2, blue) ? creep
   Exit: (41) (2, blue)= (2, blue) ? creep
   Call: (41) splitList([], 1, [_G4949|_G4950], _G4959) ? creep
   Fail: (41) splitList([], 1, [_G4949|_G4950], _G4959) ? creep
   Fail: (40) splitList([ (2, blue)], 1, [_G4949|_G4950], _G4932) ? creep
   Fail: (39) splitList([ (2, yellow), (2, blue)], 1, _G4941, [_G4931|_G4932]) ? creep
   Fail: (38) splitList([ (1, blue), (2, yellow), (2, blue)], 1, _G4929, [_G4931|_G4932]) ? creep
   Fail: (37) splitList([ (1, yellow), (1, blue), (2, yellow), (2, blue)], 1, _G4821, _G4822) ? creep
false.

显而易见的解决方案应该是X = [(1,黄色),(1,蓝色)]和Y = [(2,黄色),(2,蓝色)],但我得到假.有人能告诉我我做错了什么吗？

提前致谢,

WALLE

Answer 1

让我们看一下倒数第二个递归调用:

splitList([(2,blue)], 1, [Elem1|List1], [Elem2|List2])
                          ^^^^^^^^^^^

在我标记的地方,您希望第一个列表仍然至少有一个元素.但是,最后一次递归调用无法满足此条件.这就是它失败的原因.跟踪的相关部分是:

# penultimate call:
Call: (40) splitList([ (2, blue)], 1, [_G4949|_G4950], _G4932) ? creep
[…]
# last call:
Call: (41) splitList([], 1, [_G4949|_G4950], _G4959) ? creep
Fail: (41) splitList([], 1, [_G4949|_G4950], _G4959) ? creep

没有替代允许_G4949变量在最后一次调用中存在.

我这样写:

splitList([], _, [], []).
splitList([(A, X)|Rest], A, [(A, X)|Rest1], Rest2) :-
        splitList(Rest, A, Rest1, Rest2).
splitList([(A, X)|Rest], B, Rest1, [(A, X)|Rest2]) :-
        A =\= B,
        splitList(Rest, B, Rest1, Rest2).

BTW,问题很好!