깡통 'not'와 'eq'가 함께 작동합니다.

Question

나는 가계도를 가지고 있습니다. CLIPS를 사용하여 사람의 며느리를 찾아야합니다. 처남이

규칙 1

(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) 
) 

(배우자의 형제를 찾기 위해) 찾을 수

규칙

가계도

규칙 2 (찾을 수 배우자의 여동생의 남편)

(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) 
) 

규칙 3 (여동생의 남편을 찾을 수)

(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) 
) 

템플릿

(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)) 

,451,515,

지금 - 1) 다이애나, 처남 (또는 b_i_lw) 앤드류 마크 에드워드 2) 마크해야, b_i_lw의 앤드류, 에드워드와 찰스이다. 3) Charles, b_i_lw '는 Mark 여야합니다. 일부 잘못된 정보도 배우자의 형제 자매 (규칙 "fnd_BrthrsNLaw2"라는 이름의)와 자매의 남편 ("fnd_BrthrsNLaw3"라는 규칙)의 규칙에 대한 올바른 사람과 추론 될 때

하지만 내 출력은 완전히 정확하지 않습니다. 다이애나를 검색하는

출력 : 마크, 다이애나와 찰스에 대한

출력 아래에서 언급

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> 

출력 마크를 검색하는 :

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> 

Charles 검색을위한 출력 : 나는 (이 없습니다 (당량))입니다 테스트 한하지만

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> 

내 가정은 그들 안에있는 '및'블록에 영향을 미치는되지 않는 규칙 2, 규칙 3 (하지 (EQ))의 사용이다 그들이 동등 할 때 언제나 진실하고 동등하지 않고 거짓이되는 것을 얻습니다.

Answer 1

테스트 조건을 사용하여 규칙 조건에서 표현식을 평가해야합니다.대신

(not (eq ?sp ?p2)) 

사용

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

또는

(test (neq ?sp ?p2)) 

기존 규칙

는 관계 이름 EQ를 가진 사실을 일치하도록 시도보다는 값을 비교하는 EQ 기능을 호출의 예를 들면, .

또한 fnd_BrthrsNLaw3 규칙 동작에서 reset 명령을 호출하면 수행하고 싶지 않을 수도 있습니다.