完整版人工智能参考答案

1、第6章不确定性推理局部参考答案6.8设有如下一组推理规那么r：IFE1THENE2 (0.6)2：IFE2ANDE3 THEN E4 (0.7)r3：IFE4THENH (0.8)r4：IFE5THENH (0.9)且 CF(Ei)=0.5, CF(E3)=0.6, CF(E5)=0.7.求 CF(H)=？解：(1)先由ri求CF(E2)CF(E2)=0.6 x max(0,CF(Ei)=0.6 x max(0,0.5=0.3(2) 再由 r2 求 CF(E4)CF(E4)=0.7 x max(0, minCF(E 2 ), CF(E3 )=0.7 x max0, min0.3, 0.6=0.

2、21(3) 再由 r3求 CF1(H)CF1(H)= 0.8 x max0,CF(E4)=0.8 x max0, 0.21)=0.168(4) 再由 r4 求 CF2(H)CF2(H)= 0.9 x max0,CF(E5)=0.9 x 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) X CF2(H)=0.6926.10 设有如下推理规那么r1：IF E1 THEN(2, 0.00001)H1r2：IF E2 THEN(100, 0.0001)H1r3：IF E3 THEN(200, 0.00

3、1)H2r4：IF H1 THEN(50, 0.1) H2且 P(E1)= P(E2)= P(H 3)=0.6, P(H1)=0.091, P(H2)=0.01,又由用户告知:P(E| Si)=0.84, P(E2|&)=0.68, P(E3|S3)=0.36请用主观Bayes方法求P(H2|Si, S2, &)=？解：(1)由 r1 计算 O(H| Si)先把H1的先验概率更新为在 £下的后验概率P(H1| E1)P(H1| E1)=(LS 1 X P(H1) / (LS1-1) X P(H1)+1)=(2 X 0.091) / (2 -1) X 0.091 +1)

4、=0.16682S1下的后验概由于P(E1|S1)=0.84 > P(E),使用P(H | S)公式的后半局部,得到在当前观察 率P(H1| S1)和后验几率O(H1| S1)P(Hi| Si) = P(Hi) + (P(Hi| Ei) - P(Hi) / (1 - P(Ei) X (P(Ei| Si) - P(Ei)=0.09I + (0.I6682 -0.09I) / (i -0.6) X (0.84 -0.6)=0.09I + 0.I8955 乂 0.24 = 0.I36492O(Hi| Si) = P(Hi| Si) / (i - P(Hi| Si) =0.I5807(2) 由

5、r2 计算 O(H 11 S2)先把Hi的先验概率更新为在 E2下的后验概率P(Hi| E2)P(Hi| E2)=(LS2 X P(Hi) / (LS2-i) X P(Hi)+i) =(i00 x 0.09i) / (i00 -i) x 0.09i +i) =0.909i8由于P(E2|S2)=0.68 > P(E2),使用P(H | S)公式的后半局部,得到在当前观察S2下的后验概率P(Hi| S2)和后验几率O(Hi| S2)P(Hi| S2) = P(Hi) + (P(Hi| E2) - P(Hi) / (i - P(E2) X (P(E2| S2) - P(E2) =0.09i

6、+ (0.909i8 -0.09i) / (i -0.6) X (0.68 -0.6) =0.25464O(Hi| S2) = P(Hi| S2) / (i - P(Hi| S2) =0.34i63(3) 计算 O(Hi| Si,S2)和 P(Hi| Si,S2)先将Hi的先验概率转换为先验几率O(Hi) = P(Hi) / (i - P(Hi) = 0.09i/(i-0.09i)=0.i00ii 再根据合成公式计算Hi的后验几率O(Hi| Si,S2)= (O(Hi| Si) / O(H i) X (O(Hi| S2) / O(Hi) X O(Hi) =(0.i5807/0.i00ii) X

7、 (0.34i63) / 0.i00ii) X 0.i00ii =0.53942再将该后验几率转换为后验概率P(Hi| Si,S2) = O(Hi| Si,S2) / (i+ O(Hi| Si,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)=(LS 3 X P(H2) / (LS 3-i) x P(H2)+i) =(200 X 0.0i) / (200 -i) X 0.0i +i) =0.09569由于P(E3|S3)=0.36 < P(E3),使用P(H | S)公式的前半局部,得到在当

8、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) X P(E3| S3) 由当E3肯定不存在时有P(H2 | ？ E3) = LN 3 X P(H2) / (LN3-i) X P(H2) +i) =0.00i X 0.0i / (0.00i - i) X 0.0i + i) =0.0000i因此有P(H2| S3) = P(H2 | ？ E3) + (P(H2)- P(H2| ？E3) / P(E3) X P(E3| S3) =0.00001+(0.01-0.0000

9、1) / 0.6) X 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 H下的后验概率P(H2| H1)P(H2| H1)=(LS4 X P(H2) / (LS4-1) X P(H2)+1)=(50 X 0.01) / (50 -1) X 0.01 +1)=0.33557由于P(Hi| Si,S2)=0.35040 > P(Hi),使用P(H | S)公式的后半局部,得到在当前观察Si,S2下H2的后验概率P(H2| Si,S2)和后验几率O(H2|

10、 Si,S2)P(H2| Si,S2)= P(H2) + (P(H 2| Hi) - P(H2) / (1 - P(H 1) X (P(Hi| Si,S2)- P(Hi) =0.01 + (0.33557 -0.01) / (1 -0.091) X (0.35040 -0.091) =0.10291O(H2| Si,S2)= P(H2| Si, S2) / (1 - P(H2| Si, 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)

11、 / (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| Si,S2,S3) = (O(H 2| Si,S2)/ O(H2) X (O(H2| S3) / O(H2) X O(H2) =(0.11472 / 0.01010) X (0.00604) / 0.01010) X 0.01010 =0.06832再将该后验几率转换为后验概率P(H2| Si,S2,S3)= O(Hi| Si,S2,S3)/ (1+ O(H i| Si,S2,S3) =0.06832 / (1+ 0.06832) = 0.06395可见,H2原来的概率是

12、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设有如下推理规那么1:IFE1THEN(100, 0.1)H12：IFE2THEN(50, 0.5)H23：IFE3THEN(5, 0.05)H3且P(Hi)=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1、3有P(Hi | E1) = (LS1 X P(H1) / (LS1-1) X P(H1)+1)=(100 X 0.02) /

13、 (100 -1) X 0.02 +1)=0.671P(H2 | E2)= (LS2 X P(H2) / (LS2-1) X P(H2)+1) =(50 X 0.2) / (50 -1) X 0.2 +1) =0.9921P(H3 | E3) = (LS3 X P(H3) / (LS3-1) X P(H3)+1) =(5 X 0.4) / (5 -1) X 0.4 +1) =0.769 当Ei、E2、E3肯正存在时,根据 卅、2、3有P(Hi | ？E1) = (LN 1 X P(H1) / (LN 1-1) X P(H)+1) =(0.1 X 0.02) / (0.1 -1) X 0.02

14、+1) =0.002P(H2 | ？E2) = (LN 2 X P(H2) / (LN 2-1) X P(H2)+1) =(0.5 X 0.2) / (0.5 -1) X 0.2 +1) =0.111P(H3 | ？E3) = (LN 3 X P(H3) / (LN 3-1) X P(H3)+1) =(0.05 X 0.4) / (0.05 -1) X 0.4 +1) =0.0326.13设有如下一组推理规那么：1:IFE1AND E2 THEN A=a(CF=(0.9)2:IFE2AND (E3 OR E4) THEN B=b1, b2(CF=0.8, 0.7)3:IFATHEN H=h1,

15、 h2, h3(CF=0.6, 0.5, 0.4)4:IFBTHEN H=h1, h2, h3(CF=0.3, 0.2, 0.1)且初始证据确实定性分别为：CER(E1)=0.6, CER(E2)=0.7, CER(E3)=0.8, CER(E4)=0.9.假设 | Q |=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且有如下推理规那么：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事实为x is 较少“较少的模糊集为较

16、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求出规那么IF x is 少 THEN y is 多 所包含的模糊关系RmRm (1,1)=(0.9 A 0) V (1-0.9)=0.1 Rm (1,2)=(0.9 A 0.3) V (1-0.9)=0.3 Rm (1,3)=(0.9 A 0.7) V (1-0.9)=0.7 Rm (1,4)=(0.9 A 0.9) V (1-0.9)=0.7 Rm (2,1)=(0.7 A 0) V (1-0.7)=0.3 Rm (2,2)=(0.7 A 0.3) V (1-0.7)=0.3 Rm (2,3)=(0.7

17、 A 0.7) V (1-0.7)=0.7 Rm (2,4)=(0.7 A 0.9) V (1-0.7)=0.7 Rm (3,1)=(0.4 A 0) V (1-0.4)=0.6 Rm (3,2)=(0.4 A 0.3) V (1-0.4)=0.6 Rm (3,3)=(0.4 A 0.7) V (1-0.4)=0.6 Rm (3,4)=(0.4 A 0.9) V (1-0.4)=0.6 Rm (4,1)=(0 A 0) V (1-0)=1 Rm (4,2)=(0 A 0.3) V (1-0)=1 Rm (4,3)=(0 A 0.7) V (1-0)=1 Rm (3,4)=(0 A 0.9) V (1-0)=1即：0.1 0.3 0.7 0.90.3 0.3 0.7 0.7 Rm =0.6 0.6 0.6 0.61111-因此有='：0.8,0.5,0.2,00.1 0.30.3 0.30.6 0.6110.70.70.610.90.70.61T.0.3,0.3.0.7,0.8 /即,模糊结论为Y '=0.3, 0.3, 0.7, 0.86.16 设U=V=W=1,2,3,4且设有如下规那么：r： IF x is F THEN y is Gr2：IFyisGTHENzisHr3：IFxisFTHENzisH其中,F、G、H的模糊集分别为：F=1/1+0.