Question

我有以下CLIPS脚本。我正在尝试获取p1, p2, p3的值如果未知get-p2-2并且已知p2，则应激活最后一个规则p3。

(defrule main
(initial-fact) 
=>
(assert(fact (read))) ; user enters 1
(assert(p1 unknown))
(assert(p2 unknown))
(assert(p3 unknown))
)

;;;=====================================================
(defrule get-p1
(fact 1)
(p1 unknown)
=>
(printout t"p1 known"crlf)
(assert (p1 known)))

;;;======================================================
(defrule get-p2
(fact 1)
(p1 known)
(p2 unknown)
=>
(printout t "p2 known"crlf)
(assert (p2 known))
(assert (fact 2)))



;;;======================================================
(defrule get-p3
(fact 2)
(p3 unknown)
=>
(printout t"p3 known"crlf)
(assert (p3 known)))

;;;======================================================
(defrule get-p2-2
(fact 2)
(p2 unknown)
(p3 known)
=>
(printout t "p2 known"crlf)
(assert (p2 known)))

但是p2在规则get-p2中是已知的。因此，规则get-p2-2应该永远不会被激活。但是它确实被激活了，我得到了输出

 p1 known
 p2 known
 p3 known
 p2 known ; this should not be here

为什么激活get-p2-2？

Answer 1

您不会撤回任何未知的事实，因此p1，p2和p3都是已知的，也未知，因此可以激活get-p2-2。

CLIPS> (reset)
CLIPS> (run)
1
p1 known
p2 known
p3 known
p2 known
CLIPS> (facts)
f-0     (initial-fact)
f-1     (fact 1)
f-2     (p1 unknown)
f-3     (p2 unknown)
f-4     (p3 unknown)
f-5     (p1 known)
f-6     (p2 known)
f-7     (fact 2)
f-8     (p3 known)
For a total of 9 facts.
CLIPS>

撤消get-p1，get-p2和get-p3中的未知事实，您将获得所需的结果。

(defrule get-p1
  (fact 1)
  ?f <- (p1 unknown)
   =>
   (retract ?f)
   (printout t "p1 known" crlf)
   (assert (p1 known)))

CLIPS条件元素不匹配但规则已激活

1 个答案: