Question

在我的程序中，我遇到以下情况：

模板：

(deftemplate MAIN::travel-banchmark
    (slot name)
    (slot value)
)

事实：

(travel-banchmark (name location) (value torino))
(travel-banchmark (name location) (value roma))
(travel-banchmark (name location) (value milano))
(travel-banchmark (name location) (value venezia))

我必须（按规则）创建k=3（nb，k是变量）位置（k组合）的所有可能子集，并为每个子集I必须置换其中的元素（以及每个置换assert。）

示例组合：

torino roma milano
torino roma venezia
roma milano venezia
venezia milano torino

排列示例：

torino roma milano -> roma torino milano -> torino milano roma ...

我想知道是否可以在规则的LHS中执行某些操作，以避免在RHS中编写所有逻辑？

有没有建议我可以使用的有用语法？

Answer 1

这是使用规则进行操作的一种方法：

         CLIPS (6.31 6/12/19)
CLIPS> 
(deftemplate travel-banchmark
   (slot name)
   (slot value))
CLIPS>    
(deftemplate permutation
   (multislot values))
CLIPS>    
(deffacts initial
   (k-combination 3)
   (travel-banchmark (name location) (value torino))
   (travel-banchmark (name location) (value roma))
   (travel-banchmark (name location) (value milano))
   (travel-banchmark (name location) (value venezia)))
CLIPS>    
(defrule first-in-permutation
   (k-combination ~0)
   (travel-banchmark (name location) (value ?city))
   =>
   (assert (permutation (values ?city))))
CLIPS>    
(defrule next-in-permutation
   (k-combination ?k)
   ?p <- (permutation (values $?cities))
   (test (< (length$ ?cities) ?k))
   (travel-banchmark (name location) (value ?city))
   (test (not (member$ ?city ?cities)))
   =>
   (assert (permutation (values ?cities ?city))))
CLIPS>    
(defrule cleanup
   (declare (salience -5))
   (k-combination ?k)
   ?p <- (permutation (values $?cities))
   (test (< (length$ ?cities) ?k))
   =>
   (retract ?p))   
CLIPS> (reset)
CLIPS> (run)
CLIPS> (facts)
f-0     (initial-fact)
f-1     (k-combination 3)
f-2     (travel-banchmark (name location) (value torino))
f-3     (travel-banchmark (name location) (value roma))
f-4     (travel-banchmark (name location) (value milano))
f-5     (travel-banchmark (name location) (value venezia))
f-8     (permutation (values venezia milano roma))
f-9     (permutation (values venezia milano torino))
f-11    (permutation (values venezia roma milano))
f-12    (permutation (values venezia roma torino))
f-14    (permutation (values venezia torino milano))
f-15    (permutation (values venezia torino roma))
f-18    (permutation (values milano venezia roma))
f-19    (permutation (values milano venezia torino))
f-21    (permutation (values milano roma venezia))
f-22    (permutation (values milano roma torino))
f-24    (permutation (values milano torino venezia))
f-25    (permutation (values milano torino roma))
f-28    (permutation (values roma venezia milano))
f-29    (permutation (values roma venezia torino))
f-31    (permutation (values roma milano venezia))
f-32    (permutation (values roma milano torino))
f-34    (permutation (values roma torino venezia))
f-35    (permutation (values roma torino milano))
f-38    (permutation (values torino venezia milano))
f-39    (permutation (values torino venezia roma))
f-41    (permutation (values torino milano venezia))
f-42    (permutation (values torino milano roma))
f-44    (permutation (values torino roma venezia))
f-45    (permutation (values torino roma milano))
For a total of 30 facts.
CLIPS>

为进行比较，请使用函数生成排列：

CLIPS> (clear)
CLIPS> 
(deftemplate travel-banchmark
   (slot name)
   (slot value))
CLIPS>    
(deftemplate permutation
   (multislot values))
CLIPS>    
(deffacts initial
   (k-combination 3)
   (travel-banchmark (name location) (value torino))
   (travel-banchmark (name location) (value roma))
   (travel-banchmark (name location) (value milano))
   (travel-banchmark (name location) (value venezia)))
CLIPS> 
(deffunction gen-permutation (?k ?cities $?result)
   (if (= ?k 0)
      then
      (assert (permutation (values ?result)))
      (return))
   (foreach ?c ?cities
      (gen-permutation (- ?k 1) (delete-member$ ?cities ?c) ?result ?c)))
CLIPS>       
(defrule generate
   (k-combination ?k)
   =>
   (bind ?cities (create$))
   (do-for-all-facts ((?tb travel-banchmark)) (eq ?tb:name location)
      (bind ?cities (create$ ?cities ?tb:value)))
   (gen-permutation ?k ?cities))
CLIPS> (reset)
CLIPS> (run)
CLIPS> (facts)
f-0     (initial-fact)
f-1     (k-combination 3)
f-2     (travel-banchmark (name location) (value torino))
f-3     (travel-banchmark (name location) (value roma))
f-4     (travel-banchmark (name location) (value milano))
f-5     (travel-banchmark (name location) (value venezia))
f-6     (permutation (values torino roma milano))
f-7     (permutation (values torino roma venezia))
f-8     (permutation (values torino milano roma))
f-9     (permutation (values torino milano venezia))
f-10    (permutation (values torino venezia roma))
f-11    (permutation (values torino venezia milano))
f-12    (permutation (values roma torino milano))
f-13    (permutation (values roma torino venezia))
f-14    (permutation (values roma milano torino))
f-15    (permutation (values roma milano venezia))
f-16    (permutation (values roma venezia torino))
f-17    (permutation (values roma venezia milano))
f-18    (permutation (values milano torino roma))
f-19    (permutation (values milano torino venezia))
f-20    (permutation (values milano roma torino))
f-21    (permutation (values milano roma venezia))
f-22    (permutation (values milano venezia torino))
f-23    (permutation (values milano venezia roma))
f-24    (permutation (values venezia torino roma))
f-25    (permutation (values venezia torino milano))
f-26    (permutation (values venezia roma torino))
f-27    (permutation (values venezia roma milano))
f-28    (permutation (values venezia milano torino))
f-29    (permutation (values venezia milano roma))
For a total of 30 facts.
CLIPS>

如何在CLIPS中对事实进行组合和排列

1 个答案: