Prolog 中 bagof/3 的问题
据我了解，谓词 setof/3 和 bagof/3 可以用来生成问题的解决方案列表。 （链接到 gprolog 手册）。
正如预期的那样，以下查询的解决方案是 a、b 和 c。
?- nth(_, [a,b,c], X).
X = a ? ;
X = b ? ;
X = c ? ;
yes
现在我尝试这个：
?- setof(X, nth(_, [a,b,c], X), ListOfSolutions).
ListOfSolutions = [a] ? ;
ListOfSolutions = [b] ? ;
ListOfSolutions = [c]
yes
在我看来，解决方案应该是[a，b，c]。我做错了什么？
我正在 Mac OS 上使用 gprolog 1.4.0。
编辑：解决方案
我真正需要的是 (^)/2 运算符，但这里给出的答案是完全正确的，非常感谢您的帮助。如果有人有类似的问题，这里是我当前从 3 维网格中选择单元格的代码。
% selectFLR(?Grid, ?ClassId, ?TDayIdD, ?HourId, -ListOfFLR)
% ---------------------------------------------------------
selectFLR(Grid, ClassId, DayId, HourId, ListOfFLR) :-
bagof(FLR, ClassId^DayId^HourId^selectSingleFLR(Grid, ClassId, DayId, HourId, FLR), ListOfFLR).
selectSingleFLR(Grid, ClassId, DayId, HourId, FLR) :-
nth(ClassId, Grid, Class),
nth(DayId, Class, Day),
nth(HourId, Day, FLR).
As far as I understand the predicates setof/3 and bagof/3 can be used generate a list of solutions to a problem. (Link to gprolog manual).
As expected the solutions to the following query are a, b and c.
?- nth(_, [a,b,c], X).
X = a ? ;
X = b ? ;
X = c ? ;
yes
And now I try this:
?- setof(X, nth(_, [a,b,c], X), ListOfSolutions).
ListOfSolutions = [a] ? ;
ListOfSolutions = [b] ? ;
ListOfSolutions = [c]
yes
The solution should have been [a,b,c] in my opinion. What am I doing wrong?
I am using gprolog 1.4.0 for Mac OS.
Edit: Solution
What I really needed was the (^)/2 operator, but the answer given here was completely correct, thank you very much for your help. If anyone has a similar problem here is my current code to select cells from a 3-dimensional grid.
% selectFLR(?Grid, ?ClassId, ?TDayIdD, ?HourId, -ListOfFLR)
% ---------------------------------------------------------
selectFLR(Grid, ClassId, DayId, HourId, ListOfFLR) :-
bagof(FLR, ClassId^DayId^HourId^selectSingleFLR(Grid, ClassId, DayId, HourId, FLR), ListOfFLR).
selectSingleFLR(Grid, ClassId, DayId, HourId, FLR) :-
nth(ClassId, Grid, Class),
nth(DayId, Class, Day),
nth(HourId, Day, FLR).
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不，不应该。 nth(_,[a,b,c],X) 每次给出 X 的 1 个解。 setof（和 bagof）的工作原理如下：
如果您将 Things 指定为 X，并且 nth/3 中的 X（如上例所示）每次都只是一个变量，setof 将只创建该单个变量的列表。其他可能的统一也是可能的，但每次只会制作一个包含 1 件物品的袋子。
形式：谓词 bagof 和 setof 产生目标中自由变量的单独绑定的集合。 setof 产生没有重复项的集合的排序版本... findall 的作用类似于 bagof，所有自由变量都自动存在量化。另外findall返回一个空列表[]没有目标满足，而bagof失败。
长话短说：使用 findall :P
No, it should not. nth(_,[a,b,c],X) gives 1 solution for X every time. setof (and bagof) work like:
If you specify Things as X, and the X from nth/3 is (as you show in the above example) just a single variable each time, setof will just create a list of that single variable. Other possible unifications will be possible, but it will just make a Bag of 1 item that is in Things each time.
Formal: The predicates bagof and setof yield collections for individual bindings of the free variables in the goal. setof yields a sorted version of the collection without duplicates ... findall acts like bagof with all free variables automatically existentially quantified. In addition findall returns an empty list [] there is no goal satisfaction, whereas bagof fails.
To make a long story short: use findall :P
尝试这样写：
Try writing like this: