编译原理第三版课后习题解答

第二章习题解答

P36-6

(1)

L G ()1是0~9组成的数字串

(2)

最左推导:

568

56534

30127012010⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒D DD DDD NDD ND N D DD ND N D DD DDD DDDD NDDD NDD ND N

最右推导:

N ND N ND N ND N D N ND N D N ND N ND N D ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒77272712712701274434

886868568

P36-7

G(S)

O N O D N S O AO A AD N

→→→→→1357924680|||||||||||

P36-8

文法：

E T E T E T T

F T F T F F E i

→+-→→|||*|/()| 最左推导:

E E T T T

F T i T i T F i F F i i F i i i E T T F F F i F i E i E T i T T i F T i i T i i F i i i ⇒+⇒+⇒+⇒+⇒+⇒+⇒+⇒+⇒⇒⇒⇒⇒⇒+⇒+⇒+⇒+⇒+⇒+********()*()*()*()*()*()*()

最右推导:

E E T E T

F E T i E F i E i i T i i F i i i i i E T F T F F F E F E T F E F F E i F T i F F i F i i i i i ⇒+⇒+⇒+⇒+⇒+⇒+⇒+⇒+⇒⇒⇒⇒⇒+⇒+⇒+⇒+⇒+⇒+⇒+**********()*()*()*()*()*()*()*()

语法树：/********************************

E

E F

T

E +

T F F T +i

i

i

E

E

F

T

E

-T F F T -i

i

i

E E

F

T

+T F F

T

i

i

i

*i+i+i

i-i-i

i+i*i

*****************/

P36-9

句子iiiei 有两个语法树：

S iSeS iSei iiSei iiiei S iS iiSeS iiSei iiiei ⇒⇒⇒⇒⇒⇒⇒⇒

P36-10

/**************

)

(|)(|S T T

TS S →→

***************/

P36-11

/*************** L1:

ε

||cC C ab aAb A AC S →→→ L2:

bc

bBc B aA A AB S ||→→→ε

L3:

ε

ε||aBb B aAb A AB S →→→ L4:

A

B B A A B A S |01|10|→→→ε ***************/

第三章习题参考答案

P64–7

(1)

101101(|)*

1 ε ε 1 0 1 1

确定化：

0 1 {X} φ {1,2,3} φ φ φ {1,2,3} {2,3} {2,3,4} {2,3} {2,3} {2,3,4} {2,3,4} {2,3,5} {2,3,4}

{2,3,5} {2,3} {2,3,4,Y} {2,3,4,Y}

{2,3,5}

{2,3,4,}

1 0

0 0 1 1 0

0 1 0 1 1 1 最小化：

X 1 2 3 4 Y

5 X

Y

6

0 1

2 3

5 4

{,,,,,},{}

{,,,,,}{,,}{,,,,,}{,,,}

{,,,,},{},{}{,,,,}{,,}

{,,,},{},{},{}

{,,,}{,0123456012345135012345124601234560123413501234560123101

0==== 301231240123456011011223323401234561

01

01}{,,,}{,,}

{,},{,}{},{},{}

{,}{}{,}{,}

{,}{}{,}{}{},{},{,},{},{},{}

===== 0 1

0 0 1 0

0 1 0 1 1 1

P64–8

(1)

01)0|1(*

(2)

)5|0(|)5|0()9|8|7|6|5|4|3|2|1|0)(9|8|7|6|5|4|3|2|1(*

(3)

******)110|0(01|)110|0(10

P64–12

(a)

a,b a a b {0} {0,1} {1} {0,1} {0,1} {1} {1} {0} φ φ

φ

φ

给状态编号：

a b 0

1

2

5 0

1 2 4 3 0

1