Prolog - 通过推导减少知识库 [英] Prolog - Reduce the knowledge base by deduction

问题描述

我需要创建一个规则来搜索与 my_rule 匹配的事实.这些事实将用于改变知识库.(my_rule (Conclusion, Premise)).

I need to create a rule that will search for the facts that matches my_rule. These facts will be use to change the knowledge base. (my_rule (Conclusion, Premise)).

我有这个知识基础:

:- dynamic( is/2 ).

is( m1, house ).
is( m1, thing ).
is( m2, house ).
is( m2, thing ).
is( m3, niche ).
is( m3, house ).
is( m3, thing ).
is( m4, car ).
is( m4, mobile ).
is( m4, house ).
is( m4, thing ).

my_rule( is( X, thing ), is( X, house ) ).
my_rule( is( X, house ), is( X, niche ) ).

找到规则后，代码将搜索数据库中是否存在结论及其前提.

When a rule is found, the code will search if a Conclusion and its Premise exist in the database.

我不知道如何实现这一点，是的，这是一个家庭作业.我只想有人指出从哪里开始.

I have no idea how to achieve this, and yes this is for a homework. I just want someone to point out where to start.

谢谢.

推荐答案

首先，您需要选择一个不同的谓词名称，因为 is/2 是计算算术表达式的内置函数，例如

Firstly, you need to pick a different predicate name, since is/2 is a built-in to evaluate arithmetic expressions, e.g.

?- X is 3+2.
X = 5.

如果您尝试查阅源文件，您的代码会导致以下错误:

Your code leads to the following error if you try to consult the sourcefile:

?- [myfile].
ERROR: /home/someuser/code/myfile.pl:1:
        dynamic/1: No permission to modify static procedure `(is)/2'

让我们将其重命名为 is_a/2.然后你的代码看起来像:

Let's rename it to is_a/2. Then your code looks like:

:- dynamic( is_a/2 ).

is_a(m1, house).
is_a(m1, thing).
is_a(m2, house).
is_a(m2, thing).
is_a(m3, niche).
is_a(m3, house).
is_a(m3, thing).
is_a(m4, car).
is_a(m4, mobile).
is_a(m4, house).
is_a(m4, thing).

然后您可以定义一个谓词来描述成对的结论和前提，如下所示:

Then you could define a predicate to describe pairs of conclusions and premises like so:

conclusion_premise(is_a(X, thing), is_a(X, house)).
conclusion_premise(is_a(X, house), is_a(X, niche)).

在此基础上，您可以定义 my_rule/2 来描述 C 和 P 必须是一对相应的结论和前提，并且随后将两者都称为目标:

Building on this you could define my_rule/2 to describe that C and P must be a corresponding pair of conclusion and premise and to subsequently call both as goals:

my_rule(C,P) :-
   conclusion_premise(C,P),
   call(C),
   call(P).

现在您可以查询 my_rule/2 以搜索相应的结论-前提对:

Now you can query my_rule/2 to search for corresponding conclusion-premise pairs:

?- my_rule(Conclusion,Premise).
Conclusion = is_a(m1, thing),
Premise = is_a(m1, house) ;
Conclusion = is_a(m2, thing),
Premise = is_a(m2, house) ;
Conclusion = is_a(m3, thing),
Premise = is_a(m3, house) ;
Conclusion = is_a(m4, thing),
Premise = is_a(m4, house) ;
Conclusion = is_a(m3, house),
Premise = is_a(m3, niche) ;
false.

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