不确定性推理复习

6.8  设有如下一组推理规则:

    r₁:  IF  E₁  THEN  E₂ (0.6)

    r₂:  IF  E₂  AND  E₃  THEN  E₄ (0.7)

    r₃:  IF  E₄  THEN  H (0.8)

    r₄:  IF  E₅  THEN  H (0.9)

且已知CF(E₁)=0.5,  CF(E₃)=0.6,  CF(E₅)=0.7。求CF(H)=?

    解：(1) 先由r₁求CF(E₂)

        CF(E₂)=0.6 × max{0,CF(E₁)}

             =0.6 × max{0,0.5}=0.3

(2) 再由r₂求CF(E₄)

        CF(E₄)=0.7 × max{0, min{CF(E₂ ), CF(E₃ )}}

             =0.7 × max{0, min{0.3, 0.6}}=0.21

(3) 再由r₃求CF₁(H)

CF₁(H)= 0.8 × max{0,CF(E₄)}

      =0.8 × max{0, 0.21)}=0.168

(4) 再由r₄求CF₂(H)

CF₂(H)= 0.9 ×max{0,CF(E₅)}

      =0.9 ×max{0, 0.7)}=0.63

(5) 最后对CF₁(H )和CF₂(H)进行合成，求出CF(H)

        CF(H)= CF₁(H)+CF₂(H) — CF₁(H) × CF₂(H)

             =0.028