如何将这两个规则与CLIPS结合？

Question

我在CLIPS中有两个规则，如果它们都是真的，我想将它们组合在一起……虽然不确定如何去做。 我有一个称为grant-eligible的属性....我在想如果将其设置为TRUE那么我可以阅读下一条规则，然后将'grant-eligible'为FALSE ....但是似乎当我这样做时，我的代码正陷入无限循环...

所以这是我的规则：

    (defrule complete "rule for app completeness"
  ?f <- (application (transcript-received Yes) (app-complete FALSE)
    (gpa
                ?v_gpa&:(
                    > ?v_gpa 0)))

  =>
  (modify ?f (app-complete TRUE)))


    (defrule denied "rule for admission - DENIED"
  ?f <- (application (app-complete TRUE) (app-decision FALSE)
    (gpa
                ?v_gpa&:(
                    < ?v_gpa 3.0))

    (ssat
                ?v_ssat&:(
                    >= ?v_ssat 0.0))



        )


  =>
  (modify ?f (app-decision DENIED))

  )

(defrule accepted "rule for admission - ACCEPTED"
  ?f <- (application (app-complete TRUE) (app-decision FALSE)
    (gpa
                ?v_gpa&:(
                    >= ?v_gpa 3.5))

    (ssat
                ?v_ssat&:(
                    >= ?v_ssat 1500))


        )

  =>
  (modify ?f (app-decision ACCEPTED))

  )

这就是我现在要实施的

(defrule female-finaid "rule for finaid applications for female students"
  ?f <- (application (app-decision ACCEPTED) 
    (gender F) (grade-entry Freshman) (country USA)
    (grant-eligible TRUE)
    (grant ?v_grant)
    )

  =>
  (modify ?f
            (grant (+ ?v_grant 5000))
            (grant-eligible TRUE)
        )
    )

    (defrule great-students-finaid "rule for finaid applications for female students"
  ?f <- (application (app-decision ACCEPTED) 
    (country USA)
    (grant-eligible TRUE)
    (grant ?v_grant)
    (gpa
                ?v_gpa&:(
                    >= ?v_gpa 4.0))
    )

  =>
  (modify ?f
            (grant (+ ?v_grant 4500))
            (grant-eligible FALSE)
        )
    )

如果这两个规则都成立，那么授予的补助金应该是9500，或者是5000，或者是4500...。

解决方案：（其中ff-grant-eligible grant- ff-grant-eligible和es-grant-eligible grant- es-grant-eligible是我的控制权...它们代表ff =女最终成绩，es =优秀学生）

    (defrule female-finaid "rule for finaid applications for female students"
  ?f <- (application (app-decision ACCEPTED) (ff-grant-eligible TRUE)
    (gender F) (grade-entry Freshman) (country USA)

    (grant ?v_grant)
    )

  =>
  (modify ?f
            (grant (+ ?v_grant 5000))
            (ff-grant-eligible FALSE)
        )
    )

    (defrule great-students-finaid "rule for finaid applications for female students"
  ?f <- (application (app-decision ACCEPTED) (es-grant-eligible TRUE)
    (country USA)

    (grant ?v_grant)
    (gpa
                ?v_gpa&:(
                    >= ?v_gpa 4.0))
    )

  =>
  (modify ?f
            (grant (+ ?v_grant 4500))
            (es-grant-eligible FALSE)
        )
    )

Answer 1

您可以通过多种方式来控制程序的执行（例如，控制事实，显着性，模块）。 该答案将在应用程序处理阶段使用控制事实（具有显着性）。 我还将假设您具有与每个应用程序关联的唯一id插槽。

考虑以下事实和规则：

(deffacts application-stages "ordered sequence of stages for an application"
  (stages app-received app-accept-reject app-evaluate-grants
          app-apply-grants app-complete))

(defrule go-to-next-stage "Advances to the next application stage"
  ?stage <- (app-stage ?id ?current-stage)
  (stages $? ?current-stage ?next-stage $?)
  =>
  (retract ?stage)
  (assert (app-stage ?id ?next-stage))
  (printout t "Application " ?id " moved from stage " ?current-stage
              " to " ?next-stage "." crlf))

application-stages定义定义了application-stages顺序，而go-to-next-stage规则则推进了应用程序阶段。 由于该规则的显着性低于默认值（0），因此只有在没有与当前阶段对应的其他规则时才执行该规则。 因此，在程序中没有其他规则的情况下，您将获得以下内容：

CLIPS> (reset)
CLIPS> (assert (app-stage app-001 app-received))
<Fact-2>
CLIPS> (run)
Application app-001 moved from stage app-received to app-accept-reject.
Application app-001 moved from stage app-accept-reject to app-evaluate-grants.
Application app-001 moved from stage app-evaluate-grants to app-apply-grants.
Application app-001 moved from stage app-apply-grants to app-complete.
CLIPS> 

但是，如果您有与特定阶段关联的任何规则，则将首先执行它们。 因此，您可以将规则添加到app-evaluate-grants阶段，如下所示：

(defrule female-finaid "rule for finaid applications for female students"
  (app-stage ?id app-evaluate-grants)
  (application (app-decision ACCEPTED) (id ?id)
    (gender F) (grade-entry Freshman) (country USA)
  =>
  (assert (grant ?id female 5000)))

同样，您将添加一个great-student-finaid规则。 然后，对于app-apply-grants阶段只有一条规则：

(defrule apply-grant "Adds the amount of a grant to an application"
  (app-stage ?id app-apply-grants)
  ?grant <- (grant ?id ? ?amount)
  ?app <- (application (id ?id) (grant ?v_grant))
  =>
  (retract ?grant)
  (modify (?app (grant (+ ?v_grant ?amount))))

用这种方法建模的好处之一是，您不必在应用程序的数据中包括控制事实（例如，可grant-eligible ）。 也就是说，您的控制逻辑与数据模型是分开的。 请注意，通过使用CLIPS模块（通过defmodule ），您可以达到与我在这里所做的相同的效果，通常这是更可取的，但它需要更长的答案（并且这个答案已经相当长了）。