Question

我有一棵家谱。我需要使用CLIPS找到一个人的姐夫。

家谱

寻找兄弟会的规则

Rule-1（寻找配偶的兄弟）

(defrule MAIN::fnd_BrthrsNLaw1 ;spouse's brother
   (findRelative (person ?pn) (relationship b_i_lw)) ;Brothers-in-Law
   (and
        (and (and (or (marriage-between (personA ?pn) (personB ?sp))
                      (marriage-between (personA ?sp) (personB ?pn))
                  )
                  (child-of (relationship ?x) (person ?sp) (mother ?m) (father ?f))
             )
             (child-of (relationship son-of) (person ?pn2) (mother ?m) (father ?f))
        )
        (not (marriage-between (personA ?pn) (personB ?pn2)))
   )
   =>
   (printout t ?pn "'s brothers-in-Law: (spouse's brother) " ?pn2 crlf) 
)

规则-2（寻找配偶的姐姐的丈夫）

(defrule MAIN::fnd_BrthrsNLaw2 ;spouse's sister's husband
   (findRelative (person ?pn) (relationship b_i_lw)) ;Brothers-in-Law
   (and
        (and 
              (and 
                    (or 
                        (marriage-between (personA ?pn) (personB ?sp))
                        (marriage-between (personA ?sp) (personB ?pn))
                    )
                    (and 
                         (child-of (relationship ?x) (person ?sp) (mother ?a) (father ?b))
                         (child-of (relationship daughter-of) (person ?p2) (mother ?a) (father ?b))
                    )
               )
               (and
                    (not (eq ?sp ?p2))
                    (marriage-between (personA ?p2) (personB ?px2))
               )
        )
    =>
    (printout t ?pn "'s brothers-in-Law: (spouse's sister's husband) " ?px2 " <sister = " ?p2 " ; spouse = " ?sp " ; spouse's parents: " ?a " , " ?b ">" crlf)
)

Rule-3（寻找姐姐的丈夫）

(defrule MAIN::fnd_BrthrsNLaw3 ;sister's husband
   (findRelative (person ?p) (relationship b_i_lw)) ;Brothers-in-Law
   (and 
          (and
                (child-of (relationship ?y) (person ?p) (mother ?c) (father ?d))
                (child-of (relationship daughter-of) (person ?sist) (mother ?c) (father ?d))
          )
          (and 
                (not (eq ?p ?sist))
                (marriage-between (personA ?sist) (personB ?p2))
          )
    )
    =>
   (printout t ?p "'s brothers-in-Law: (sister's husband) " ?p2 <sister = " ?sist " ; " ?p "'s parent = " ?sp " ; spouse's parents: " ?c " , " ?d ">"crlf) 
   (reset)
)

模板

(deftemplate MAIN::marriage-between
   (slot personA)
   (slot personB))

(deftemplate MAIN::child-of
   (slot relationship (type SYMBOL) (allowed-symbols son-of daughter-of))
   (slot person)
   (slot mother)
   (slot father))

(deftemplate MAIN::findRelative
   (slot relationship (type SYMBOL) 
         (allowed-symbols g_g_p ;Great-GrandParents 
                    g_ch ;GrandChildren 
                    b_i_lw)) ;Brothers-in-Law 
   (slot person))

现在 - 1）戴安娜，姐夫（或b_i_lw）必须是安德鲁，马克和爱德华 2）马克，b_i_lw是安德鲁，爱德华和查尔斯。 3）Charles，b_i_lw'必须是Mark。

但是我的输出并不完全正确，因为一些错误的信息也被推断为配偶的兄弟姐妹（规则名为“fnd_BrthrsNLaw2”）和姐姐的丈夫（规则名为“fnd_BrthrsNLaw3”）。

Mark，Diana和Charles的输出如下：

搜索戴安娜的输出：

CLIPS> (assert (findRelative (person Diana) (relationship b_i_lw)))
<Fact-20>
CLIPS> (run)
Diana's brothers-in-Law: (spouse's brother) Edward
Diana's brothers-in-Law: (spouse's brother) Andrew
Diana's brothers-in-Law: (spouse's sister's husband) Mark <sister = Anne ; spouse = Charles ; Charles's parents: Elizabeth , Phillip>
Diana's brothers-in-Law: (sister's husband) Charles <sister = Diana ; Diana's parent: Kydd , Spencer>

搜索标记的输出：

CLIPS> (assert (findRelative (person Mark) (relationship b_i_lw)))
<Fact-20>
CLIPS> (run)
Mark's brothers-in-Law: (spouse's brother) Edward
Mark's brothers-in-Law: (spouse's brother) Andrew
Mark's brothers-in-Law: (spouse's brother) Charles
Mark's brothers-in-Law: (spouse's sister's husband) Mark <sister = Anne ; spouse = Anne ; Anne's parents: Elizabeth , Phillip>

搜索Charles的输出：

CLIPS> (assert (findRelative (person Charles) (relationship b_i_lw)))
<Fact-20>
CLIPS> (run)
Charles's brothers-in-Law: (spouse's sister's husband) Charles <sister = Diana ; spouse = Diana ; Diana's parents: Kydd , Spencer>
Charles's brothers-in-Law: (sister's husband) Mark <sister = Anne ; Anne's parent: Elizabeth , Phillip>

我的假设是在规则-2中使用（不是（eq））而规则-3不会影响它们所在的'和'块，尽管我已经测试过（不是（eq））在当它们是平等的时候是不平等和错误的，我打算得到它。

Answer 1

您需要使用测试CE来评估规则条件下的表达式。例如，而不是

(not (eq ?sp ?p2))

使用

(test (not (eq ?sp ?p2)))

或

(test (neq ?sp ?p2))

您现有的规则试图将事实与关系名称eq匹配，而不是调用eq函数来比较值。

此外，在fnd_BrthrsNLaw3规则操作中调用reset命令是您可能不想做的事情。

不能让'不'和'eq'一起工作

1 个答案: