人工智能課后習(xí)題答案

1、第1章1.1解圖如下： (1 1->2(2 1->3(3 2->3(6 3->2(5 3->1(4 2->18數(shù)碼問題啟發(fā)函數(shù)為不在位的將牌數(shù)啟發(fā)函數(shù)為不在位的將牌數(shù)距離和S(4S(5 第2章2.1解圖： 第3章3.18(1證明：待歸結(jié)的命題公式為( P Q P ，合取范式為：P Q P ，求取子句集為, , S P Q P =，對子句集中的子句進行歸結(jié)可得：1P QP 歸結(jié)由上可得原公式成立。(2證明：待歸結(jié)的命題公式為( ( ( P Q R P Q P R （，合取范式為：( ( P Q R P Q P R ，求取子句集為, , , S P Q R P

2、Q P R =，對子句集中的子句進行歸結(jié)可得：1P Q R P QP4R Q歸結(jié)P R歸結(jié)7R 歸結(jié)歸結(jié)由上可得原公式成立。(3證明：待歸結(jié)的命題公式為( ( Q P Q P Q ，合取范式為：( ( Q P Q P Q ，求取子句集為, , S Q P Q P Q =，對子句集中的子句進行歸結(jié)可得：Q PQQ P P 歸結(jié)P 歸結(jié)歸結(jié)由上可得原公式成立。3.19答案(1/, /, /mgu a x b y b z =(2( /, ( /mgu g f v x f v u =(3不可合一(4/, /, /mgu b x b y b z =3.23證明R1：所有不貧窮且聰明的人都快樂：( ( (

3、 x Poor x Smart x Happy x R2：那些看書的人是聰明的：( ( x read x Smart x R3：李明能看書且不貧窮：( (read Li Poor Li R4：快樂的人過著激動人心的生活：( ( x Happy x Exciting x 結(jié)論李明過著激動人心的生活的否定：(Exciting Li 將上述謂詞公式轉(zhuǎn)化為子句集并進行歸結(jié)如下：由R1可得子句：( ( ( Poor x Smart x Happy x 由R2可得子句：( ( read y Smart y 由R3可得子句：( read Li (Poor Li 由R4可得子句：( ( Happy z Exc

4、iting z 有結(jié)論的否定可得子句：( Exciting Li 根據(jù)以上6條子句，歸結(jié)如下：( Happy Li Li /z 8( ( Poor Li Smart Li Li /x ( Smart Li (read Li Li /y由上可得原命題成立。第4章4.9答案 4.11答案 第5章5.9答案解：把該網(wǎng)絡(luò)看成兩個部分，首先求取(1|12 P T S S 。1首先求取(1|1 P T S ，因為(1|1 0.7(1 0.2P S F P F =>=，所以(1|1 (1 (1|1 (1(1|1 (11(1P T S P T P T F P T P F S P F P F =+

5、5;假設(shè)(1|1 1P S F =，(1|1 (1 20.10.1818(1 (1 1(21 0.11P T F LS P T LS P T =××=×+×+(1|1 0.18180.10.1(0.70.2 0.151110.2P T S =+×=2然后求取(1|2 P T S ，因為(2|2 0.6(2 0.4P S F P F =>=，所以(1|2 (1 (1|2 (1(2|2 (21(2P T S P T P T F P T P F S P F P F =+×假設(shè)(2|2 1P S F =，(1|2 (1 1000.10.

6、9174(1 (1 1(1001 0.11P T F LS P T LS P T =××=×+×+(1|2 0.10.91740.1(0.60.4 0.372510.4P T S =+×=3求取(1|1 O T S 和(1|2O T S (1|1(1|1 (1|10.15110.1780110.1511P T S O T S P T S =(1|2(1|2 (1|20.37250.5936110.3725P T S O T S P T S =4求取(1|12P T S S (1|12 (10.11110.9510(1|12 (1 0.1(1 0

7、.11111(1 10.1(1|1 (1|2(1 (10.17800.59360.11110.1111(1|12 0.95100.48741(1|12 10.9510O T S S O T P T S S P T O T P T O T S O T S O T O T O T S S O T S S =××××=+5求取(|12 P H S S ，因為(1|12 0.4874(1 0.1P T S S P T =>=, 所以(|12 ( (1 (|1 (1|121(1P H S S P H P T P H T P H P T S S P T =+&

8、#215;假設(shè)(1|12 1P T S S =，(650.01(|1 0.3963(1 ( 1(651 0.011LS P H P H T LS P H ××=×+×+(|12 0.010.39630.01(0.48740.1 0.176210.1P H S S =+×=6求取(|3 P H S ，因為(|2 0.0001( 0.01P H T P H =<=，所以(|3 (|2 ( (|2(3|2(2P H S P H T P H P H T P S T P T =+×假設(shè)(2|3 0P T S =，則(|2 ( 0.0001

9、0.01(1 ( 1(0.00011 0.0011P H T LN P H LN P H =××=×+×+(|3 00.0100.020.00670.03P H S =+×=7求(|12 O H S S 和(|3O H S (|12 (|3 (|12 0.39630.21381(|12 10.3963(|3 0.00670.00671(|3 10.0067O H S S O H S P H S S P H S S P H S P H S =8求(|123P H S S S ( (|123 (0.01010.1418(|123 ( 0.010.0

10、1011( 10.01(|12 (|3( (0.21380.00670.01010.0101(|123 0.14180.12411(|123 10.1418O H O H S S S O H P H S S S P H P H O H S S O H S O H O H O H S S S O H S S S =××××=+5.10答案解：12(56 max(5, (60.8(4(56 min(4, (560.5(1 max0,(4(56(1, 4(56 0.50.80.4( max0,(1(, 1 0.40.90.36( max0,(2(, 2CF E E CF E CF E CF E E E CF E CF E E CF E CF E E E CF E E E E CF H CF E CF H E CF H CF E CF H E =×=×=×=×=×3121212123 0.80.60.48(