prolog 中的存在限定符，使用 setof/bagof [英] existential qualifier in prolog, using setof / bagof

问题描述

我有一个简短的问题.在序言中使用 setof 的存在限定符(即 ^).

I had a quick question re. existential qualifier using setof in prolog (i.e. ^).

使用 SICStus 似乎(尽管许多网站声称)，S 确实似乎在下面的代码中被量化(使用沼泽标准，事实之母/事实之子，我没有包括在这里):

using SICStus it seems that (despite what a number of websites claim), S does indeed appear to be quantified in the code below (using the bog standard, mother of / child of facts, which i havent included here):

child(M,F,C) :- setof(X,(mother(S,X)),C).

我使用以下方法检查统一:

i check the unification using:

child(M,F,C) :- setof(X-S,(mother(S,X)),C).

所以下面的代码，与存在运算符似乎没有区别:

so the following code, with the existential operator seem to make no difference:

child(M,F,C) :- setof(X,S^(mother(S,X)),C).

知道这是为什么吗?那么在什么情况下您需要统一器?

Any ideas why this is? What would be a situation where you would need the unifier then?

谢谢！

推荐答案

好吧，我不确定我能不能完美地解释它，但让我试试.

Ok, I'm not sure I can explain it perfectly, but let me try.

这与您正在查询二元关系 mother/2 的事实有关.在那种情况下，使用 XS 作为模板对结果集 C 的影响与使用 S^ 在球门前.在 XS 中，您在模板中使用了两个变量，因此 X 和 S 的每个可能的绑定都包含在 C 中.使用 S^ 在前面获得相同的效果目标，因为这是说在构造结果时忽略 S 的绑定".

It has to do with the fact that you are querying over a 2-ary relation, mother/2. In that case using X-S as the template has a similar effect on the result set C as using S^ in front of the goal. In X-S you are using both variables in the template, and therefore each possible binding of X and S is included in C. You get the same effect using S^ in front of the goal, as this is saying "ignore bindings of S when constructing the result".

但是当您查询三元关系时，两者之间的区别变得更加清晰.SWI手册有这个例子:

But the difference between the two becomes clearer when you query over a 3-ary relation. The SWI manual has this example:

foo(a, b, c).
foo(a, b, d).
foo(b, c, e).
foo(b, c, f).
foo(c, c, g).

现在执行与示例中类似的查询

Now do similar queries as in your example

setof(X-Z, foo(X,Y,Z), C).

和

setof(Z, X^foo(X,Y,Z), C).

你会得到不同的结果.

这不仅仅是检查统一性，X-Z 有效地改变了你的结果集.

It's not just checking unification, X-Z effectively changes your result set.

希望有所帮助.

编辑:当我包含上面两个查询的结果时，也许它会澄清一些事情.第一个是这样的:

Edit: Maybe it clarifies things when I include the results of the two queries above. The first one goes like this:

?- setof(X-Z, foo(X,Y,Z), C).   
Y = b
C = [a-c, a-d] ;
Y = c
C = [b-e, b-f, c-g] ;
No

第二个产生:

?- setof(Z, X^foo(X,Y,Z), C).
Y = b
C = [c, d] ;
Y = c
C = [e, f, g] ;
No

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