Question

我在CLIPS中有两个规则，如果它们都是真的我想要合并......但不确定如何做到这一点。我有一个名为grant-eligible的属性....我在想是否将其设置为TRUE然后我可以阅读下一条规则，然后将'grant-eligible'设置为{{1 ....但是，当我这样做时，似乎我的代码进入了无限循环...

所以这是我的规则：

FALSE

这是我现在尝试实施的那些

    (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))

  )

如果这两条规则都属实，则授予的奖励应为9500，或者可能为5000或者可能为4500 ......任何想法？

解决方案:(其中(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) ) )和ff-grant-eligible是我的控制事实......他们代表ff =女性finaid，es =优秀学生）

es-grant-eligible

Answer 1

您可以通过多种方式控制程序的执行（例如，控制事实，突出性，模块）。这个答案将使用控制事实（有突出性）来处理应用程序的各个阶段。我还假设您有一个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 deffact定义应用程序的阶段序列，go-to-next-stage规则推进应用程序阶段。由于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会执行该规则。因此，如果您的计划中没有其他规则，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相关的任何规则，它们将首先执行。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)))

您同样会添加great-student-finaid规则。然后，app-apply-grants阶段有一条规则：

(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))))

以这种方式对其进行建模的一个好处是，您不必在应用程序的数据中包含控制事实（例如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），您可以获得与我在此处所做的相同的效果，这通常是可取的，但它需要更长的答案（并且这个已经相当长）。 / p>

如何将这两个规则与CLIPS结合起来？

1 个答案: