Question

我希望你做得很好？我是CLIPS的初学者。我有几个节点的图形（起始节点（输入）和结束节点（输出）。我想在我输入的数量等于输出的情况下创建规则，规则必须帮助我在几个节点之间进行选择组合（开始和结束节点）每个组合的最短路径。

如果输入数量大于输出数量或相反，则可以使用相同的规则添加另一个规则。在组合之间进行选择之后，还需要考虑输入或输出并将其导出到最近的点？

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

组合结果如果我有3个startnode和2个endnode

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) */

然后作为最终结果我希望得到

case class Person(batch: Long, age: Long, name: String)
val df = spark.read.json("/home/white/tmp/a.json").as[Person]

我希望很清楚？谢谢你的帮助

Answer 1

尝试将规则信息和varvar组合到以下内容中：

(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))))

您将获得的结果是：

 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

目前尚不清楚为什么你期望选择路径E到B，因为路径E到I的成本较低。

使用剪辑的条件或选择性规则

1 个答案: