使用剪輯的條件或選擇性規則

Question

我希望你做得很好？我是CLIPS方面的初學者。我有幾個節點（開始節點（輸入）和結束節點（輸出）的圖形）我想創建一個規則，當輸入的數量等於輸出時，規則必須幫助我選擇幾個組合（開始和結束節點）爲每個組合的最短路徑

如果我的輸入數大於輸出數或相反的數，可以在之前添加另一條規則。還需要相結合，考慮到輸入或輸出，並推導其朝向最近點？

(deftemplate path 
(slot start) 
(slot end) 
(multislot path) 
(slot cost(type NUMBER))) 

(deftemplate info 
(slot start) 
(slot end) 
(multislot path) 
(slot cost)) 




(deffacts variable_table 

(path (start A)(end B)(path A,a,b,B)   (cost 7)) 
(path (start A)(end C)(path A,a,b,c,C)  (cost 10)) 
(path (start A)(end D)(path A,a,m,n,d,D)  (cost 8.5)) 
(path (start A)(end E)(path A,a,m,n,E)  (cost 5.5)) 
(path (start A)(end F)(path A,a,m,n,k,g,f,F) (cost 9.4)) 
(path (start A)(end CS) (path A,a,m,n,k,CS) (cost 6.7)) 
(path (start A)(end G)(path A,a,m,n,k,g,G) (cost 8)) 
(path (start A)(end H)(path A,a,m,n,d,e,H) (cost 10.3)) 
(path (start A)(end I)(path A,a,m,I)   (cost 3.5)) 

(path (start B)(end A) (path B,b,a,A)  (cost 7 )) 
(path (start B)(end C) (path B,b,c,C)  (cost 4)) 
(path (start B)(end D) (path B,b,c,d,D)  (cost 7)) 
(path (start B)(end E) (path B,b,g,k,n,E) (cost 5.8 )) 
(path (start B)(end F) (path B,b,g,f,F)  (cost 4.7)) 
(path (start B)(end CS)(path B,b,g,k,CS)  (cost 4.6)) 
(path (start B)(end G) (path NG2,b,g,G)  (cost 3.3)) 
(path (start B)(end H) (path NG2,b,g,f,e,H) (cost 6.5)) 
(path (start B)(end I) (path NG2,b,g,k,n,m,I)(cost 7.8)) 

(path (start C)(end A) (path C,c,b,a,A)  (cost 10)) 
(path (start C)(end B) (path C,c,b,B)  (cost 5)) 
(path (start C)(end D) (path C,c,d,D)  (cost 4)) 
(path (start C)(end E) (path C,c,d,n,E)  (cost 5.3)) 
(path (start C)(end F) (path C,c,d,e,f,F) (cost 7.6)) 
(path (start C)(end CS)(path C,c,b,g,k,CS) (cost 7.6)) 
(path (start C)(end G) (path C,c,b,g,G)  (cost 6.3)) 
(path (start C)(end H) (path C,c,d,e,H)  (cost 5.8)) 
(path (start C)(end I) (path C,c,d,n,m,I) (cost 9)) 

(path (start D)(end A) (path D,d,n,m,a,A) (cost 8.5)) 
(path (start D)(end B) (path D,d,c,b,B)  (cost 7)) 
(path (start D)(end C) (path D,d,c,C)  (cost 4)) 
(path (start D)(end E) (path D,d,n,E)  (cost 5)) 
(path (start D)(end F) (path D,d,e,f,F)  (cost 5.6)) 
(path (start D)(end CS)(path D,d,n,k,CS)  (cost 6.2)) 
(path (start D)(end G) (path D,d,e,f,g,G) (cost 6)) 
(path (start D)(end H) (path D,d,e,H)  (cost 3.8)) 
(path (start D)(end I) (path D,d,n,m,I)  (cost 7)) 

(path (start E)(end A) (path E,n,m,a,A)  (cost 5.5)) 
(path (start E)(end B) (path E,n,k,g,b,B) (cost 5.8)) 
(path (start E)(end C) (path E,n,d,c,C)  (cost 7)) 
(path (start E)(end D) (path E,n,d,D)  (cost 5)) 
(path (start E)(end F) (path E,n,k,g,f,F) (cost 5.9)) 
(path (start E)(end CS)(path E,n,k,CS)  (cost 3.2)) 
(path (start E)(end G) (path E,n,k,g,G)  (cost 4.5)) 
(path (start E)(end H) (path E,n,d,e,H)  (cost 6.8)) 
(path (start E)(end I) (path E,n,m,I)  (cost 4)) 

(path (start F)(end A) (path F,f,g,k,n,m,a,A)(cost 9.4)) 
(path (start F)(end B) (path F,f,g,b,B)  (cost 4.7)) 
(path (start F)(end C) (path F,f,e,d,c,C) (cost 7.6)) 
(path (start F)(end D) (path F,f,e,d,D)  (cost 5.6)) 
(path (start F)(end E) (path F,f,g,k,n,E) (cost 5.9)) 
(path (start F)(end CS)(path F,f,g,k,CS)  (cost 4.7)) 
(path (start F)(end G) (path F,f,g,G)  (cost 3.4)) 
(path (start F)(end H) (path F,f,e,H)  (cost 3.8)) 
(path (start F)(end I) (path F,f,g,k,n,m,I) (cost 7.9)) 


(path (start CS)(end A) (path CS,k,n,m,a,A) (cost 6.7)) 
(path (start CS)(end B) (path CS,k,g,b,B) (cost 4.6)) 
(path (start CS)(end C) (path CS,K,g,b,c,C) (cost 7.6)) 
(path (start CS)(end D) (path CS,k,n,d,D) (cost 6.2)) 
(path (start CS)(end E) (path CS,K,N,E)  (cost 3.2)) 
(path (start CS)(end F) (path CS,k,g,f,F) (cost 4.7)) 
(path (start CS)(end G) (path CS,k,g,G)  (cost 3.3)) 
(path (start CS)(end H) (path CS,k,g,f,e,H) (cost 6.5)) 
(path (start CS)(end I)(path CS,k,n,m,I)  (cost 5.2)) 

(path (start G)(end A) (path G,g,k,n,m,a,A) (cost  8)) 
(path (start G)(end B) (path G,g,b,B)  (cost 3.3)) 
(path (start G)(end C) (path G,g,b,c,C)  (cost 6.3)) 
(path (start G)(end D) (path G,g,f,e,d,D) (cost 6)) 
(path (start G)(end E) (path G,g,k,n,E)  (cost 4.5)) 
(path (start G)(end F) (path G,g,f,F)  (cost 3.4)) 
(path (start G)(end CS)(path G,g,k,CS)  (cost 3.3)) 
(path (start G)(end H) (path G,g,f,e,H)  (cost 5.2)) 
(path (start G)(end I) (path G,g,k,n,m,I) (cost 6.5)) 

(path (start H)(end A) (path H,e,d,n,m,a,A) (cost 10.3)) 
(path (start H)(end B) (path H,e,f,g,B)  (cost 6.5)) 
(path (start H)(end C) (path H,e,d,C)  (cost 3.8)) 
(path (start H)(end D) (path H,e,d,c,D)  (cost 5.8)) 
(path (start H)(end E) (path H,e,d,n,E)  (cost 6.8)) 
(path (start H)(end F) (path H,e,f,F)  (cost 3.8)) 
(path (start H)(end CS)(path H,e,f,g,k,CS) (cost 6.5)) 
(path (start H)(end H) (path H,e,f,g,H)  (cost 5.2)) 
(path (start H)(end I) (path H,e,d,n,m,I) (cost 8.8)) 

(path (start I)(end A) (path I,m,a,A)  (cost 3.5)) 
(path (start I)(end B) (path I,m,n,k,g,b,B) (cost 7.8)) 
(path (start I)(end C) (path I,m,n,d,c,C) (cost 9)) 
(path (start I)(end D) (path I,m,n,d,D)  (cost 7)) 
(path (start I)(end E) (path I,m,n,E)  (cost 4)) 
(path (start I)(end F) (path I,m,n,k,g,f,F) (cost 7.9)) 
(path (start I)(end CS)(path I,m,m,k,CS)  (cost 5.2)) 
(path (start I)(end G) (path I,m,n,k,g,G) (cost 6.5)) 
(path (start I)(end H) (path I,m,n,d,e,I) (cost 8.8))) 

(defrule lancesaisiepoint 
?f1 <- (debut node) 
=> 
    (retract ?f1) 
    (printout t "how many node start") 
    (bind ?x (read)) 
    (assert (startnode ?x)) 
) 

(defrule saisiepoint 
?f <- (startnode ?a) 
    (test (> ?a 0)) 
=> 
(printout t "Enter nbr node start " crlf) 
(printout t "start node name ") 
(bind ?start (read)) 
(printout t "node start value ") 
(bind ?y1 (read)) 
(assert (startnode ?start ?y1)) 
(retract ?f) 
(assert (startnode (- ?a 1))) 
) 

(defrule lancesaisiemoins 
?f1 <- (startnode 0) 
=> 
(retract ?f1) 
(printout t "how many node end ") 
(bind ?x (read)) 
(assert (endnode ?x)) 
) 

(defrule saisiemoins 
?f <- (endnode ?a) 
(test (> ?a 0)) 
=> 
(printout t "Enter nbr node end" crlf) 
(printout t "end node name ") 
(bind ?end (read)) 
(printout t "node end value ") 
(bind ?y2 (read)) 
(assert (endnode ?end ?y2)) 
(retract ?f) 
(assert (endnode (- ?a 1)))) 


(defrule supprnbrpoint ;compter 
?x<- (startnode 0) 
=> 
    (retract ?x) 
    ) 

(defrule supprnbrmoins 
?x<- (endnode 0) 
=> 
    (retract ?x) 
    ) 

(defrule info 

?M <- (startnode ?start ?y1) 
?K <- (endnode ?end ?y2) 
    => 
(assert (start ?start end ?end)) 
) 

(defrule varvar 

?L<- (start ?start end ?end) 

(path (start ?start)(end ?end)(path $?path)(cost ?cost)) 
    => 
(retract ?L) 
(printout t " PATH " ?start " to " ?end " by " ?path " with a cost of " ?cost crlf) 
(assert (info (start ?start) (end ?end) (path $?path) 
(cost ?cost))) 
) 

組合的結果，如果我有3個的StartNode和2終端節點

how many node start 3 
Enter nbr node start 
start node name A 
node start value 3 
Enter nbr node start 
start node name C 
node start value 5 
Enter nbr node start 
start node name E 
node start value 5 
how many node end 2 
Enter nbr node end 
end node name B 
node end value -5 
Enter nbr node end 
end node name I 
node end value -5 
PATH A to I by (A,a,m,I) with a cost of 3.5 
PATH C to I by (C,c,d,n,m,I) with a cost of 9 
PATH E to I by (E,n,m,I) with a cost of 4 
PATH A to B by (A,a,b,B) with a cost of 7 
PATH C to B by (C,c,b,B) with a cost of 5 
PATH E to B by (E,n,k,g,b,B) with a cost of 5.8 

，然後作爲最終的結果，我希望得到

PATH A to I by (A,a,m,I) with a cost of 3.5 */ because it is the low cost between the 3 first combinaison */ 
PATH E to B by (E,n,k,g,b,B) with a cost of 5.8 */ because it is the low cost between the 3 second combinaison */ 
PATH C to B by (C,c,b,B) with a cost of 5 */ because it is the low cost between (C to B) and (C to I) */ 

我希望它是明確的？ 感謝你的幫助

Answer 1

嘗試規則信息和varvar結合以下幾點：

(defrule varvar 
    (startnode ?start ?) 
    (endnode ?end ?) 
    (path (start ?start) (end ?end) (path $?path) (cost ?cost)) 
    (not (and (endnode ?end2 ?) 
      (path (start ?start) (end ?end2) (cost ?cost2&:(< ?cost2 ?cost))))) 
    => 
    (printout t " PATH " ?start " to " ?end " by " ?path " with a cost of " ?cost crlf) 
    (assert (info (start ?start) (end ?end) (path $?path) (cost ?cost)))) 

，你會得到的結果是：

PATH A to I by (A,a,m,I) with a cost of 3.5 
PATH E to I by (E,n,m,I) with a cost of 4 
PATH C to B by (C,c,b,B) with a cost of 5 

它爲什麼你希望目前尚不清楚由於路徑E到I具有較低的成本，因此路徑E到B被選擇。