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clips - CLIPS 条件元素不匹配但规则已激活

问题描述

我有以下 CLIPS 脚本。我正在尝试获取如果未知且已知，则应激活p1, p2, p3 最后一条规则的值。get-p2-2p2p3

(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)))

而是p2在规则中广为人知get-p2。因此，该规则get-p2-2不应该被激活。但它确实被激活了，我得到了输出

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

为什么被get-p2-2激活？

标签： clips

解决方案

您不会撤回任何未知事实，因此 p1、p2 和 p3 都是已知和未知的，这允许激活 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>

在 get-p1、get-p2 和 get-p3 中提取未知的事实，你会得到你想要的结果。

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

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