D-S方法应用举例

例 在专家系统的推理过程中,原始证据的确定性是由用户回答的。例如,有以下的推理规则:

   rule1 IF E1 AND E2 THEN A={a1,a2},cF={0.3,0.5} 

   rule2 IF E3 AND (E4 OR E5) THEN N={n1},cF={0.7} 

   rule3 IF A THEN H={h1,h2,h3},cF={0.1,0.5,0.3} 

   rule4 IF N THEN H={h1,h2,h3},cF={0.4,0.2,0.1} 

这些规则形成的推理网络如下图

      

                            推理网络图

设用户给出的原始证据的确定性是

     cER(E1)=0.8,cER(E2)=0.6,cER(E3)=0.9,cER(E4)=0.5,cER(E5)=0.7 并假定:|U|=20

解:下面依次求出A的确定性cER(A),N的确定性cER(N)和H的确定性cER(H)。

    1.求cER(A)

    先求出rule1条件部分的确定性,利用公式有

       cER(E)=cER(E1 AND E2)

             =min{cER(E1),cER(E2)}

             =min{0.8,0.6}=0.6

再求假设A的基本概率赋值函数m为

        m({a1},{a2})=(0.6*c1,0.6*c2)

                    =(0.6*0.3,0.6*0.5)

                    =(0.18,0.3)

于是有

       bel(A)=m({a1})+m({a2})=0.18+0.3=0.48

               Pl(A)=1-bel(ØA)=1-0=1

       f(A)=bel(A)+(|A|/|U|)[Pl(A)-bel(A)]

           =0.48+(2/20)*(1-0.48)=0.53

      cER(A)=MD(A,E')*f(A)=1*0.53=0.53

其中MD(A,E')是因为E'为已证实的命题(此例中为A的前提E1和E2均已证实)。 

  2.求cER(N)

  利用公式,有

      cER(E)=cER(E3 AND (E4 OR E5))

            =min{cER(E3),max{cER(E4),cER(E5)}}

            =min{0.9,max{0.5,0.7}}=0.7   

      m({n1})=(0.7*c1)=(0.7*0.7)=0.49

          bel(N)=m({n1})=0.49

          Pl(N)=1-bel(ØN)=1

       f(N)=bel(N)+(|N|/|U|)[Pl(N)-bel(N)]

           =0.49+(1/20)*(1-0.49)=0.52

      cER(N)=MD(N,E')*f(N)=1*0.52=0.52

  3.求cER(H)

  利用公式,有

      m1({h1},{h2},{h3})=(cER(A)*c1,cER(A)*c2,cER(A)*c3)

                        =(0.53*0.1,0.53*0.5,0.53*0.3)

                        =(0.053,0.265,0.159)

      m1(U)=1-(0.053+0.265+0.159)=0.524

      m2({h1},{h2},{h3})=(cER(N)*c1,cER(N)*c2,cER(N)*c3)

                        =(0.52*0.4,0.52*0.2,0.52*0.1)

                        =(0.208,0.104,0.052)

      m2(U)=1-(0.208+0.104+0.052)=0.636

然后求出m1与m2的正交和m1m2。为此,先求出 K-1

   K-1=m1({h1})*m2({h1})+m1({h1})*m2({U})+m1({h2})*m2({h3})

       +m1({h2})*m2({U})+m1({h3})*m2({h3})+m1({h3})*m2({U})

       +m1({U})*m2({h1})+m1({U})*m2({h2})+m1({U})*m2({h3})+m1({U})*m2({U})

      =0.053*0.208+0.053*0.636+0.263*0.104+0.265*0.636+0.159*0.052

       +0.159*0.636+0.524*0.208+0.524*0.104+0.524*0.052+0.524*0.636

      ≈0.874

于是

    m({h1})=(1/0.874)*[m({h1})*m2({h2})+m1({h1})*m2({U})+m1({U})*m2({h1})]

           =(1/0.874)*[0.053*0.208+0.053*0.636+0.524*0.208]

           ≈0.176

同理,我们可以计算得到:

    m({h2})≈0.299

    m({h3})≈0.157

       m(U)=1-(0.176+0.299+0.157)=0.368

   bel(H)=m({h1})+m({h2})+m({h3})

         =0.176+0.299+0.157=0.632

   Pl(H)=1     

       f(H)=bel(H)+(|H|/|U|)[Pl(H)-bel(H)]

           =0.632+(3/20)*(1-0.632)=0.687

      cER(H)=MD(H,E')*f(H)=0.687

 至此,我们求出了H的确定性为0.687。

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