Prolog:从谓词列表中生成一组所有基本术语 [英] Prolog: generate a set of all ground terms from a list of predicates

问题描述

鉴于以下事实:

female(alice).
male(bob).
male(charlie).
lives(alice,oxford).
lives(bob,oxford).
lives(charlie,cambridge).

我要生成集合:

[alice,bob,charlie,oxford,cambridge].

我知道如何处理单个谓词:

I know how to this for a single predicate:

?- setof(X, male(X), S).
S = [bob, charlie].

但我不知道如何概括这一点.目的是使用谓词列表查询程序:

But I do not know how to generalise this. The aim is to query the program with a list of predicates:

f([male/1,female/1,lives/2]).

我试图概括上面的setof"语句:

I tried to generalise the above 'setof' statements:

g(P,1) :-
  Atom = [P,X],
  Goal =..Atom,
  setof(X, Goal(X, Y), S),
  write(S).

g(P,2).

f([]).

f([H|T]) :-
  H = P/A,
  g(P,A),
  f(T).

但这没有用.

有人可以帮忙吗?

推荐答案

大功告成.Prolog 中的高阶编程可以通过call 或 apply 谓词.这是一个带有 call/1 的版本，由 ISO 标准化:

You're almost there. Higher-order programming in Prolog can be achieved with the call or apply predicates. Here's a version with call/1, which is standardized by ISO:

get_ground_args(Goal, Args) :-
    findall(X, (call(Goal), Goal =.. [_|GoalArgs], member(X, GoalArgs)), Args).

示例:

?- get_ground_args(lives(_,_), Args).
Args = [alice, oxford, bob, oxford, charlie, cambridge].

现在要将其与您的原始设置联系起来，使用 length/2 在嵌套的 setof/findall 中生成初始参数:

Now to tie this in to your original setup, use length/2 to generate the initial arguments inside a nested setof/findall:

all_ground_args(Preds, Args) :-
    setof(X, (findall(Y, (member(Pred/Arity, Preds),
                          length(GoalArgs, Arity),
                          Goal =.. [Pred|GoalArgs],
                          get_ground_args(Goal, GroundArgs),
                          member(Y, GroundArgs)),
                      AllArgs),
              member(X, AllArgs)),
          Args).

演示:

?- all_ground_args([male/1,female/1,lives/2], Args).
Args = [alice, bob, cambridge, charlie, oxford].

(当谓词具有非基础参数时，所有这些都可能会失败.)

(All of this will probably fail when a predicate has non-ground args.)

这篇关于Prolog:从谓词列表中生成一组所有基本术语的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！