我想要建立一个专家系统,在有几层楼的大楼发生紧急情况时(它需要为任何楼层工作),电梯应该把人们带到地面。问题是,把电梯送到任何楼层的缺陷从来没有出现在议事日程上,所以系统什么也不做。正确的行动应该是发射规则,然后是另一个规则,把人们从地板上带走。
破坏规则的代码如下:
(defrule move_to_floor "elevator moves to any floor "
?i <- (elevator is_at floor ?x has ?y adults and ?z minors)
(floor ?fl&~?x has ?n adult and ?m minor people)
(test (> (+ ?n ?m) 0))
=>
(retract ?i)
(assert (elevator is_at floor ?fl has ?y adults and ?z minors))
)在上述另一个缺陷中从用户初始化的事实如下:
f-0 (initial-fact)
f-1 (elevator is_at 0 has 0 adults and 0 minors)
f-3 (capacity 4)
f-4 (floors 3)
f-5 (initCanEnter 0) ;At 0 this prevents from entering the init_defrule again
f-6 (floor 3 has 2 adult and 1 minor people)
f-7 (floor 2 has 4 adult and 5 minor people)
f-8 (floor 1 has 1 adult and 2 minor people)我似乎找不到解决办法。另外,我使用的是deffacts,而不是脱节模板,因为我已经看到很多人在互联网上使用。
发布于 2019-01-17 18:16:04
您可以使用patterns命令查看规则中的哪些模式是匹配的。
CLIPS (6.31 2/3/18)
CLIPS>
(defrule move_to_floor "elevator moves to any floor "
?i <- (elevator is_at floor ?x has ?y adults and ?z minors)
(floor ?fl&~?x has ?n adult and ?m minor people)
(test (> (+ ?n ?m) 0))
=>
(retract ?i)
(assert (elevator is_at floor ?fl has ?y adults and ?z minors)))
CLIPS>
(deffacts initial
(elevator is_at 0 has 0 adults and 0 minors)
(capacity 4)
(floors 3)
(initCanEnter 0) ;At 0 this prevents from entering the init_defrule again
(floor 3 has 2 adult and 1 minor people)
(floor 2 has 4 adult and 5 minor people)
(floor 1 has 1 adult and 2 minor people))
CLIPS> (reset)
CLIPS> (matches move_to_floor)
Matches for Pattern 1
None
Matches for Pattern 2
f-5
f-6
f-7
Partial matches for CEs 1 - 2
None
Activations
None
(3 0 0)
CLIPS> 在这种情况下,第一个模式不匹配。这是因为您的模式需要is_at ?x,但是您的事实包含is_at 0 (在事实中缺少符号)。如果你纠正这个问题,规则将被列入议程。
CLIPS>
(deffacts initial
(elevator is_at floor 0 has 0 adults and 0 minors)
(capacity 4)
(floors 3)
(initCanEnter 0) ;At 0 this prevents from entering the init_defrule again
(floor 3 has 2 adult and 1 minor people)
(floor 2 has 4 adult and 5 minor people)
(floor 1 has 1 adult and 2 minor people))
CLIPS> (reset)
CLIPS> (agenda)
0 move_to_floor: f-1,f-7
0 move_to_floor: f-1,f-6
0 move_to_floor: f-1,f-5
For a total of 3 activations.
CLIPS>如果此时发出一个(运行)命令,规则将在从一层到另一层的循环中不断地触发,所以接下来需要解决的问题。
如果您使用的是缺陷模板事实而不是有序事实,那么如果您拼写错了槽名,那么如果您有多个属性的事实,最好使用这些事实。
CLIPS> (clear)
CLIPS>
(deftemplate elevator
(slot at_floor (type INTEGER))
(slot adults (type INTEGER))
(slot minors (type INTEGER)))
CLIPS>
(deftemplate floor
(slot # (type INTEGER))
(slot adults (type INTEGER))
(slot minors (type INTEGER)))
CLIPS>
(deffacts initial
(elevator (at_floor 0))
(capacity 4)
(floors 3)
(initCanEnter 0)
(floor (# 3) (adults 2) (minors 1))
(floor (# 2) (adults 4) (minors 5))
(floor (# 1) (adults 1) (minors 2)))
CLIPS>
(defrule move_to_floor
?i <- (elevator (at_floor ?x))
(floor (# ?fl&~?x) (adults ?n) (minors ?m))
(test (> (+ ?n ?m) 0))
=>
(modify ?i (at_floor ?fl)))
CLIPS> (reset)
CLIPS> (facts)
f-0 (initial-fact)
f-1 (elevator (at_floor 0) (adults 0) (minors 0))
f-2 (capacity 4)
f-3 (floors 3)
f-4 (initCanEnter 0)
f-5 (floor (# 3) (adults 2) (minors 1))
f-6 (floor (# 2) (adults 4) (minors 5))
f-7 (floor (# 1) (adults 1) (minors 2))
For a total of 8 facts.
CLIPS> (agenda)
0 move_to_floor: f-1,f-7
0 move_to_floor: f-1,f-6
0 move_to_floor: f-1,f-5
For a total of 3 activations.
CLIPS>https://stackoverflow.com/questions/54241466
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