多个列表的组合 - Prolog [英] Combinations of multiple lists - Prolog

问题描述

我需要在列表列表中找到组合.例如，给出以下列表，

列表 = [[1, 2], [1, 2, 3]]

这些应该是输出，

梳 = [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3]]

另一个例子:

列表 = [[1,2],[1,2],[1,2,3]]梳 = [[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3]....等等]

我知道如何为具有两个子列表的列表执行此操作，但它需要适用于任意数量的子列表.

我是 Prolog 新手，请帮忙.

解决方案

这个答案寻找提供的赏金一个纯粹的解决方案，也考虑到 Ess".这里我们概括这个前面的像这样回答:

list_crossproduct(Xs, []) :-成员([]，Xs).list_crossproduct(Xs, Ess) :-Ess = [E0|_],相同长度(E0，Xs)，地图列表(maybelonger_than(Ess)，Xs)，list_comb(Xs, Ess).可能比(Xs，Ys)更长:-可能比(Ys，Xs)更短.也许比([]，_)更短.可能比([_|Xs]，[_|Ys])更短:-可能比(Xs，Ys)更短.

list_crossproduct/2 通过提前关联 Xs 和 Ess 获得双向.

<上一页>?- list_comb(Xs, [[1,2,3],[1,2,4],[1,2,5]]).不终止 % 糟糕！?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5]]).Xs = [[1],[2],[3,4,5]] % 现在也可以了;错误的.

有多个答案的示例查询:

?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5],X,Y,Z]).X = [1,2,_A]，Y = [1,2,_B]，Z = [1,2,_C], Xs = [[1],[2],[3,4,5,_A,_B,_C]];X = [1,_A,3],Y = [1,_A,4],Z = [1,_A,5], Xs = [[1],[2,_A],[3,4,5]];X = [_A,2,3]，Y = [_A,2,4]，Z = [_A,2,5], Xs = [[1,_A],[2],[3,4,5]];错误的.

I need to find the combinations in a list of lists. For example, give the following list,

List = [[1, 2], [1, 2, 3]]

These should be the output,

Comb = [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3]]

Another example:

List = [[1,2],[1,2],[1,2,3]]

Comb = [[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3]....etc]

I know how to do it for a list with two sublists but it needs to work for any number of sublists.

I'm new to Prolog, please help.

解决方案

This answer hunts the bounty offered "for a pure solution that also takes into account for Ess". Here we generalize this previous answer like so:

list_crossproduct(Xs, []) :-
   member([], Xs).
list_crossproduct(Xs, Ess) :-
   Ess = [E0|_],
   same_length(E0, Xs),
   maplist(maybelonger_than(Ess), Xs),
   list_comb(Xs, Ess).

maybelonger_than(Xs, Ys) :-
   maybeshorter_than(Ys, Xs).

maybeshorter_than([], _).
maybeshorter_than([_|Xs], [_|Ys]) :-
   maybeshorter_than(Xs, Ys).

list_crossproduct/2 gets bidirectional by relating Xs and Ess early.

?- list_comb(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
nontermination                                % BAD!

?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
   Xs = [[1],[2],[3,4,5]]      % this now works, too
;  false.

Sample query having multiple answers:

?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5],X,Y,Z]).
   X = [1,2,_A],
   Y = [1,2,_B],
   Z = [1,2,_C], Xs = [[1],[2],[3,4,5,_A,_B,_C]]
;  X = [1,_A,3],
   Y = [1,_A,4],
   Z = [1,_A,5], Xs = [[1],[2,_A],[3,4,5]]
;  X = [_A,2,3],
   Y = [_A,2,4],
   Z = [_A,2,5], Xs = [[1,_A],[2],[3,4,5]]
;  false.

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