List 如何在Prolog中将split/3转换为split/4？

以下代码适用于

拆分列表/3

：

split_list([], _, [[],[]]).
split_list(T, 0, [[],T]).
split_list([H|T], N, [[H|Y],Z]) :-
   N1 is N-1,
   split_list(T, N1, [Y,Z]).

例如：

?- split_list([a,s,d,f,g,h,j], 3, R).
R = [[a, s, d], [f, g, h, j]] .              % observed answer

但是，我想将

split_list/3

转换为

split/4

：

?- split([a,s,d,f,g,h,j], 2, R1, R2).
R1 = [a,s], R2 = [d,f,g,h,j].                % expected answer

split(AsBs, N, As, Bs) :- append(As, Bs, AsBs), length(As, N).

---

我怎样才能得到我想要的答案？有什么建议吗？谢谢：）

这里是对

split/4

的直接定义：

?- split([a,s,d,f,g,h,j], 2, R1, R2).
R1 = [a,s], R2 = [d,f,g,h,j].                % expected answer

split(AsBs, N, As, Bs) :- append(As, Bs, AsBs), length(As, N). 把上述问题概括一下如何

?- split([a,s,d,f,g,h,j], I, R1, R2).
   I = 0, R1 = [],              R2 = [a,s,d,f,g,h,j]
;  I = 1, R1 = [a],             R2 =   [s,d,f,g,h,j]
;  I = 2, R1 = [a,s],           R2 =     [d,f,g,h,j]
;  I = 3, R1 = [a,s,d],         R2 =       [f,g,h,j]
;  I = 4, R1 = [a,s,d,f],       R2 =         [g,h,j]
;  I = 5, R1 = [a,s,d,f,g],     R2 =           [h,j]
;  I = 6, R1 = [a,s,d,f,g,h],   R2 =             [j]
;  I = 7, R1 = [a,s,d,f,g,h,j], R2 =              [].

---

如果您对保留现有定义的语义感兴趣，可以通过这种方式重用它

split_list(L,N,R1,R2) :- split_list(L,N,[R1,R2]).

在这个答案中，我们使用约束来表示声明性整数算法

:- use_module(library(clpfd)). 现在，如果再看一下

split\u5的代码，我们可以看到其中的代码：

%%拆分（[]，N，N0，Zs，Zs）：-%%追加（[]，Zs，Zs）。
%%N#=N0，%
%%拆分（[X | Xs]，N，N0，Ys，[X | Zs]）：-%追加（[X | Xs]，Ys，[X | Zs]）：-
%%N1#=N0+1，%
%%N#>=N1，%
%%拆分（Xs、N、N1、Ys、Zs）。%附加（Xs、Ys、Zs）。
示例查询：

?- N = 2, split(XsYs, N, Xs, Ys).
   XsYs = [_A,_B|Ys], N = 2, Xs = [_A,_B]
;  false.

?- N = 2, XsYs = [a,b,c,d,e], split(XsYs, N, Xs, Ys).
   XsYs = [a,b,c,d,e], N = 2, Xs = [a,b], Ys = [c,d,e]
;  false.

?- XsYs = [a,s,d,f,g,h,j], split(XsYs, N, Xs, Ys).
   XsYs = [a,s,d,f,g,h,j], N = 0, Xs = []             , Ys = [a,s,d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 1, Xs = [a]            , Ys =   [s,d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 2, Xs = [a,s]          , Ys =     [d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 3, Xs = [a,s,d]        , Ys =       [f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 4, Xs = [a,s,d,f]      , Ys =         [g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 5, Xs = [a,s,d,f,g]    , Ys =           [h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 6, Xs = [a,s,d,f,g,h]  , Ys =             [j]
;  XsYs = [a,s,d,f,g,h,j], N = 7, Xs = [a,s,d,f,g,h,j], Ys =              []
;  false.

?- Xs = [a,b,c], Ys = [d,e,f], split(XsYs, N, Xs, Ys).
XsYs = [a,b,c,d,e,f], N = 3, Xs = [a,b,c], Ys = [d,e,f].
？-N=2，拆分（XsYs，N，Xs，Ys）。
XsYs=[[uA，[uB]Ys]，N=2，Xs=[[uA，[uB]
;  错。
？-N=2，XsYs=[a，b，c，d，e]，拆分（XsYs，N，Xs，Ys）。
XsYs=[a，b，c，d，e]，N=2，Xs=[a，b]，Ys=[c，d，e]
;  错。
？-XsYs=[a，s，d，f，g，h，j]，拆分（XsYs，N，Xs，Ys）。
XsYs=[a，s，d，f，g，h，j]，N=0，Xs=[]，Ys=[a，s，d，f，g，h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=1，Xs=[a]，Ys=[s，d，f，g，h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=2，Xs=[a，s]，Ys=[d，f，g，h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=3，Xs=[a，s，d]，Ys=[f，g，h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=4，Xs=[a，s，d，f]，Ys=[g，h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=5，Xs=[a，s，d，f，g]，Ys=[h，j]
;  XsYs=[a，s，d，f，g，h，j]，N=6，Xs=[a，s，d，f，g，h]，Ys=[j]
;  XsYs=[a，s，d，f，g，h，j]，N=7，Xs=[a，s，d，f，g，h，j]，Ys=[]
;  错。
？-Xs=[a，b，c]，Ys=[d，e，f]，拆分（XsYs，N，Xs，Ys）。
XsYs=[a，b，c，d，e，f]，N=3，Xs=[a，b，c]，Ys=[d，e，f]。

脚注1：为了强调这些相似之处，对
fd_length/2
（优化变量）的程序文本进行了轻微修改：

L
被
Xs
取代，
参数对
N，N0
由
N，N0
和
（is）/2通过
（#=）/2
。
是的，对不起，你是对的-@repeat没什么大不了的。第一个列表的长度总是正好等于N吗？是的，我想@repeat我建议把
length
放在前面，因为
N
很可能是给定的。@false。这样做，但为了保留有利的终止属性而使用延迟实现如何？对于简单的预期用例，您的版本会留下一个CP，并且不会因为：
split（XsYs，2，Xs，Ys）
?- N = 2, split(XsYs, N, Xs, Ys).
   XsYs = [_A,_B|Ys], N = 2, Xs = [_A,_B]
;  false.

?- N = 2, XsYs = [a,b,c,d,e], split(XsYs, N, Xs, Ys).
   XsYs = [a,b,c,d,e], N = 2, Xs = [a,b], Ys = [c,d,e]
;  false.

?- XsYs = [a,s,d,f,g,h,j], split(XsYs, N, Xs, Ys).
   XsYs = [a,s,d,f,g,h,j], N = 0, Xs = []             , Ys = [a,s,d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 1, Xs = [a]            , Ys =   [s,d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 2, Xs = [a,s]          , Ys =     [d,f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 3, Xs = [a,s,d]        , Ys =       [f,g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 4, Xs = [a,s,d,f]      , Ys =         [g,h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 5, Xs = [a,s,d,f,g]    , Ys =           [h,j]
;  XsYs = [a,s,d,f,g,h,j], N = 6, Xs = [a,s,d,f,g,h]  , Ys =             [j]
;  XsYs = [a,s,d,f,g,h,j], N = 7, Xs = [a,s,d,f,g,h,j], Ys =              []
;  false.

?- Xs = [a,b,c], Ys = [d,e,f], split(XsYs, N, Xs, Ys).
XsYs = [a,b,c,d,e,f], N = 3, Xs = [a,b,c], Ys = [d,e,f].