人工智能第6章课后习题参考答案

第6章 不确定性推理部分参考答案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 × 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) =0.6926.10  设有如下推理规则 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, 又由用户告知： P(E1| S1)=0.84, P(E2|S2)=0.68, P(E3|S3)=0.36请用主观Bayes方法求P(H2|S1, S2, S3)=? 解：(1) 由r1计算O(H1| S1) 先把H1的先验概率更新为在E1下的后验概率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，得到在当前观察S1下的后验概率P(H1| S1)和后验几率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计算O(H1| S2) 先把H1的先验概率更新为在E2下的后验概率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，得到在当前观察S2下的后验概率P(H1| S2)和后验几率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) 计算O(H1| S1,S2)和P(H1| S1,S2) 先将H1的先验概率转换为先验几率O(H1) = P(H1) / (1 - P(H1) = 0.091/(1-0.091)=0.10011 再根据合成公式计算H1的后验几率 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 再将该后验几率转换为后验概率P(H1| S1,S2) = O(H1| S1,S2) / (1+ O(H1| S1,S2) = 0.35040(4) 由r3计算O(H2| S3) 先把H2的先验概率更新为在E3下的后验概率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(E3)，使用P(H | S)公式的前半部分，得到在当前观察S3下的后验概率P(H2| S3)和后验几率O(H2| S3) P(H2| S3) = P(H2 | ¬ E3) + (P(H2) P(H2| ¬E3) / P(E3) × P(E3| S3) 由当E3肯定不存在时有 P(H2 | ¬ E3) = LN3 × P(H2) / (LN3-1) × P(H2) +1) = 0.001 × 0.01 / (0.001 - 1) × 0.01 + 1) = 0.00001因此有P(H2| S3) = P(H2 | ¬ E3) + (P(H2) P(H2| ¬E3) / P(E3) × P(E3| S3) =0.00001+(0.01-0.00001) / 0.6) × 0.36 =0.00600O(H2| S3) = P(H2| S3) / (1 - P(H2| S3) =0.00604(5) 由r4计算O(H2| H1) 先把H2的先验概率更新为在H1下的后验概率P(H2| H1) P(H2| H1)=(LS4 × P(H2) / (LS4-1) × P(H2)+1) =(50 × 0.01) / (50 -1) × 0.01 +1) =0.33557 由于P(H1| S1,S2)=0.35040 > P(H1)，使用P(H | S)公式的后半部分，得到在当前观察S1,S2下H2的后验概率P(H2| S1,S2)和后验几率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) 计算O(H2| S1,S2,S3)和P(H2| S1,S2,S3) 先将H2的先验概率转换为先验几率O(H2) = P(H2) / (1 - P(H2) )= 0.01 / (1-0.01)=0.01010 再根据合成公式计算H1的后验几率 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 再将该后验几率转换为后验概率P(H2| S1,S2,S3) = O(H1| S1,S2,S3) / (1+ O(H1| S1,S2,S3) = 0.06832 / (1+ 0.06832) = 0.06395 可见，H2原来的概率是0.01，经过上述推理后得到的后验概率是0.06395，它相当于先验概率的6倍多。6.11设有如下推理规则 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，请计算当证据E1，E2，E3存在或不存在时P(Hi | Ei)或P(Hi |Ei)的值各是多少(i=1, 2, 3)？ 解：(1) 当E1、E2、E3肯定存在时，根据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) 当E1、E2、E3肯定存在时，根据r1、r2、r3有P(H1 | ¬E1) = (LN1 × P(H1) / (LN1-1) × P(H1)+1) = (0.1 × 0.02) / (0.1 -1) × 0.02 +1) =0.002P(H2 | ¬E2) = (LN2 × P(H2) / (LN2-1) × P(H2)+1) = (0.5 × 0.2) / (0.5 -1) × 0.2 +1) =0.111P(H3 | ¬E3) = (LN3 × P(H3) / (LN3-1) × P(H3)+1) = (0.05 × 0.4) / (0.05 -1) × 0.4 +1) =0.0326.13 设有如下一组推理规则: r1: IF E1 AND E2 THEN A=a (CF=0.9) r2: IF E2 AND (E3 OR E4) THEN B=b1, b2 (CF=0.8, 0.7) r3: IF A THEN H=h1, h2, h3 (CF=0.6, 0.5, 0.4) r4: IF B THEN H=h1, h2, h3 (CF=0.3, 0.2, 0.1)且已知初始证据的确定性分别为：CER(E1)=0.6, CER(E2)=0.7, CER(E3)=0.8, CER(E4)=0.9。假设|=10，求CER(H)。 解：其推理过程参考例6.9 具体过程略6.15 设U=V=1，2，3，4且有如下推理规则： IF x is 少 THEN y is 多其中，“少”与“多”分别是U与V上的模糊集，设 少=0.9/1+0.7/2+0.4/3 多=0.3/2+0.7/3+0.9/4已知事实为 x is 较少“较少”的模糊集为 较少=0.8/1+0.5/2+0.2/3请用模糊关系Rm求出模糊结论。 解：先用模糊关系Rm求出规则 IF x is 少 THEN y is 多所包含的模糊关系Rm Rm (1,1)=(0.90)(1-0.9)=0.1 Rm (1,2)=(0.90.3)(1-0.9)=0.3 Rm (1,3)=(0.90.7)(1-0.9)=0.7 Rm (1,4)=(0.90.9)(1-0.9)=0.7 Rm (2,1)=(0.70)(1-0.7)=0.3 Rm (2,2)=(0.70.3)(1-0.7)=0.3 Rm (2,3)=(0.70.7)(1-0.7)=0.7 Rm (2,4)=(0.70.9)(1-0.7)=0.7 Rm (3,1)=(0.40)(1-0.4)=0.6 Rm (3,2)=(0.40.3)(1-0.4)=0.6 Rm (3,3)=(0.40.7)(1-0.4)=0.6 Rm (3,4)=(0.40.9)(1-0.4)=0.6 Rm (4,1)=(00)(1-0)=1 Rm (4,2)=(00.3)(1-0)=1 Rm (4,3)=(00.7)(1-0)=1 Rm (3,4)=(00.9)(1-0)=1即：因此有即，模糊结论为 Y=0.3, 0.3, 0.7, 0.86.16 设 U=V=W=1,2,3,4且设有如下规则： r1：IF x is F THEN y is G r2：IF y is G THEN z is H r3：IF x is F THEN z is H其中，F、G、H的模糊集分别为： F=1/1+0.8/2+0.5/3+0.4/4 G=0.1/2+0.2/3+0.4/4 H=0.2/2+0.5/3+0.8/4请分别对各种模糊关系验证满足模糊三段论的情况。解：本题的解题思路是：由模糊集F和G求出r1所表示的模糊关系R1m, R1c, R1g再由模糊集G和H求出r2所表示的模糊关系R2m, R2c, R2g再由模糊集F和H求出r3所表示的模糊关系R3m, R3c, R3g 然后再将R1m, R1c, R1g分别与R2m, R2c, R2g合成得R12 m, R12c, R12g 最后将R12 m, R12c, R12g分别与R3m, R3c, R3g比较6