人工智能不確定性推理部分參考答案.doc

不確定性推理部分參考答案1 設(shè)有如下一組推理規(guī)則: 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 max0,CF(E1) =0.6 max0,0.5=0.3(2) 再由r2求CF(E4) CF(E4)=0.7 max0, minCF(E2 ), CF(E3 ) =0.7 max0, min0.3, 0.6=0.21(3) 再由r3求CF1(H)CF1(H)= 0.8 max0,CF(E4) =0.8 max0, 0.21)=0.168(4) 再由r4求CF2(H)CF2(H)= 0.9 max0,CF(E5) =0.9 max0, 0.7)=0.63(5) 最后對(duì)CF1(H )和CF2(H)進(jìn)行合成，求出CF(H) CF(H)= CF1(H)+CF2(H)+ CF1(H) CF2(H) =0.6922 設(shè)有如下推理規(guī)則 r1: IF E1 THEN (2, 0.00001) H1 r2: IF E2 THEN (100, 0.0001) H1 r3: IF E3 THEN (200, 0.001) H2 r4: IF H1 THEN (50, 0.1) H2且已知P(E1)= P(E2)= P(H3)=0.6, P(H1)=0.091, P(H2)=0.01, 又由用戶告知： P(E1| S1)=0.84, P(E2|S2)=0.68, P(E3|S3)=0.36請(qǐng)用主觀Bayes方法求P(H2|S1, S2, S3)=? 解：(1) 由r1計(jì)算O(H1| S1) 先把H1的先驗(yàn)概率更新為在E1下的后驗(yàn)概率P(H1| E1) P(H1| E1)=(LS1 P(H1) / (LS1-1) P(H1)+1) =(2 0.091) / (2 -1) 0.091 +1) =0.16682 由于P(E1|S1)=0.84 P(E1)，使用P(H | S)公式的后半部分，得到在當(dāng)前觀察S1下的后驗(yàn)概率P(H1| S1)和后驗(yàn)幾率O(H1| S1) P(H1| S1) = P(H1) + (P(H1| E1) P(H1) / (1 - P(E1) (P(E1| S1) P(E1) = 0.091 + (0.16682 0.091) / (1 0.6) (0.84 0.6) =0.091 + 0.18955 0.24 = 0.136492 O(H1| S1) = P(H1| S1) / (1 - P(H1| S1) = 0.15807 (2) 由r2計(jì)算O(H1| S2) 先把H1的先驗(yàn)概率更新為在E2下的后驗(yàn)概率P(H1| E2) P(H1| E2)=(LS2 P(H1) / (LS2-1) P(H1)+1) =(100 0.091) / (100 -1) 0.091 +1) =0.90918 由于P(E2|S2)=0.68 P(E2)，使用P(H | S)公式的后半部分，得到在當(dāng)前觀察S2下的后驗(yàn)概率P(H1| S2)和后驗(yàn)幾率O(H1| S2) P(H1| S2) = P(H1) + (P(H1| E2) P(H1) / (1 - P(E2) (P(E2| S2) P(E2) = 0.091 + (0.90918 0.091) / (1 0.6) (0.68 0.6) =0.25464 O(H1| S2) = P(H1| S2) / (1 - P(H1| S2) =0.34163 (3) 計(jì)算O(H1| S1,S2)和P(H1| S1,S2) 先將H1的先驗(yàn)概率轉(zhuǎn)換為先驗(yàn)幾率O(H1) = P(H1) / (1 - P(H1) = 0.091/(1-0.091)=0.10011 再根據(jù)合成公式計(jì)算H1的后驗(yàn)幾率 O(H1| S1,S2)= (O(H1| S1) / O(H1) (O(H1| S2) / O(H1) O(H1) = (0.15807 / 0.10011) (0.34163) / 0.10011) 0.10011 = 0.53942 再將該后驗(yàn)幾率轉(zhuǎn)換為后驗(yàn)概率P(H1| S1,S2) = O(H1| S1,S2) / (1+ O(H1| S1,S2) = 0.35040(4) 由r3計(jì)算O(H2| S3) 先把H2的先驗(yàn)概率更新為在E3下的后驗(yàn)概率P(H2| E3) P(H2| E3)=(LS3 P(H2) / (LS3-1) P(H2)+1) =(200 0.01) / (200 -1) 0.01 +1) =0.09569 由于P(E3|S3)=0.36 P(H1)，使用P(H | S)公式的后半部分，得到在當(dāng)前觀察S1,S2下H2的后驗(yàn)概率P(H2| S1,S2)和后驗(yàn)幾率O(H2| S1,S2) P(H2| S1,S2) = P(H2) + (P(H2| H1) P(H2) / (1 - P(H1) (P(H1| S1,S2) P(H1) = 0.01 + (0.33557 0.01) / (1 0.091) (0.35040 0.091) =0.10291 O(H2| S1,S2) = P(H2| S1, S2) / (1 - P(H2| S1, S2) =0.10291/ (1 - 0.10291) = 0.11472 (6) 計(jì)算O(H2| S1,S2,S3)和P(H2| S1,S2,S3) 先將H2的先驗(yàn)概率轉(zhuǎn)換為先驗(yàn)幾率O(H2) = P(H2) / (1 - P(H2) )= 0.01 / (1-0.01)=0.01010 再根據(jù)合成公式計(jì)算H1的后驗(yàn)幾率 O(H2| S1,S2,S3)= (O(H2| S1,S2) / O(H2) (O(H2| S3) / O(H2) O(H2) = (0.11472 / 0.01010) (0.00604) / 0.01010) 0.01010 =0.06832 再將該后驗(yàn)幾率轉(zhuǎn)換為后驗(yàn)概率P(H2| S1,S2,S3) = O(H1| S1,S2,S3) / (1+ O(H1| S1,S2,S3) = 0.06832 / (1+ 0.06832) = 0.06395 可見(jiàn)，H2原來(lái)的概率是0.01，經(jīng)過(guò)上述推理后得到的后驗(yàn)概率是0.06395，它相當(dāng)于先驗(yàn)概率的6倍多。3 設(shè)有如下推理規(guī)則 r1: IF E1 THEN (100, 0.1) H1 r2: IF E2 THEN (50, 0.5) H2 r3: IF E3 THEN (5, 0.05) H3且已知P(H1)=0.02, P(H2)=0.2, P(H3)=0.4，請(qǐng)計(jì)算當(dāng)證據(jù)E1，E2，E3存在或不存在時(shí)P(Hi | Ei)或P(Hi |Ei)的值各是多少(i=1, 2, 3)？ 解：(1) 當(dāng)E1、E2、E3肯定存在時(shí)，根據(jù)r1、r2、r3有P(H1 | E1) = (LS1 P(H1) / (LS1-1) P(H1)+1) = (100 0.02) / (100 -1) 0.02 +1) =0.671P(H2 | E2) = (LS2 P(H2) / (LS2-1) P(H2)+1) = (50 0.2) / (50 -1) 0.2 +1) =0.9921P(H3 | E3) = (LS3 P(H3) / (LS3-1) P(H3)+1) = (5 0.4) / (5 -1) 0.4 +1) =0.769 (2) 當(dāng)E1、E2、E3肯定存在時(shí)，根據(jù)r1、r2、r3有P(H1 | E1) = (LN1 P(H1) / (LN1-1) P(H1)+1) = (0.1 0.02) / (0.1 -1) 0.02 +