Question

我使用Constraint Handling Rules在SWI-Prolog中编写了一组简单的约束。它使用两个相对简单的推理规则：

%If A means B, then B means A.
means(A,B) ==> means(B,A).    
%If A means B and A means C, then B means C.
means(A,B),means(A,C) ==> means(B,C).

我希望means([3,is,equal,to,4],[3,equals,4])为true，但它似乎会导致无限递归：

:- use_module(library(chr)).
:- chr_constraint means/2.
:- initialization(main).


means([A,equals,B],[A,'=',B]).
means([A,is,equal,to,B],[A,'=',B]).
means([A,equals,B],[A,and,B,are,equal]).


%These are the rules of inference for this program.
    %If A means B, then B means A.
    means(A,B) ==> means(B,A).    
    %If A means B and A means C, then B means C.
    means(A,B),means(A,C) ==> means(B,C).

main :-
    %This part works as expected. X = [3,'=',4].
    means([3,is,equal,to,4],X),writeln(X),

    %This statement should be true, so why does it produce an infinite recursion?
    means([3,is,equal,to,4],[3,and,4,are,equal]).

我在此程序中添加了一个简单规则，但它仍然导致Out of local stack错误：

:- use_module(library(chr)).
:- chr_constraint means/2.
:- initialization(main).


%These are the rules of inference for this program.
    %If A means B, then B means A.
    means(A,B) ==> means(B,A).    
    %If A means B and A means C, then B means C.
    means(A,B),means(A,C) ==> means(B,C).
    means(A,B) \ means(A,B) <=> true.
    means(A,A) <=> true.

means([A,equals,B],[A,'=',B]).
means([A,is,equal,to,B],[A,'=',B]).
means([A,equals,B],[A,and,B,are,equal]).

main :-
    %This part works as expected. X = [3,'=',4].
    means([3,is,equal,to,4],X),writeln(X),

    %This statement should be true, so why does it produce an infinite recursion?
    means([3,is,equal,to,4],[3,and,4,are,equal]).

是否可以重写推理规则，以便它们不会产生无限递归？

Answer 1

请阅读可用的CHR文献，了解有关CHR的这些方面的更多详细信息。

例如，Tips for CHR programming包含在编程提示：

中

设置语义

CHR系统允许存在相同的约束，即具有多个约束   同样的算符，arity和论点。对于大多数约束求解器，这是不可取的：它会影响   效率和可能终止。因此应该添加适当的简化规则   表格：constraint \ constraint <=> true

因此，如果添加CHR simpagation规则，您的示例将按预期工作：

<强> means(A,B) \ means(A,B) <=> true.

示例查询和结果：

?- means([3,is,equal,to,4],[3,and,4,are,equal]).
means([3, and, 4, are, equal], [3, is, equal, to, 4]),
means([3, is, equal, to, 4], [3, and, 4, are, equal]).

使用约束处理规则避免无限递归

1 个答案: