编译原理课后答案 - 图文

第二章

2.3叙述由下列正规式描述的语言 (a) 0(0|1)*0

在字母表{0, 1}上,以0开头和结尾的长度至少是2的01串 (b) ((ε|0)1*)*

(c) (0|1)*0(0|1)(0|1)

在字母表{0, 1}上,倒数第三位是0的01串 (d) 0*10*10*10*

在字母表{0, 1}上,所有的01串,包括空串 在字母表{0, 1}上,含有3个1的01串 (e) (00|11)*((01|10)(00|11)*(01|10)(00|11)*)*

在字母表{0, 1}上,含有偶数个0和偶数个1的01串

2.4为下列语言写正规定义

C语言的注释,即以 /* 开始和以 */ 结束的任意字符串,但它的任何前缀(本身除外)不以 */ 结尾。 [解答] other → a | b | … other指除了*以外C语言中的其它字符 other1 → a | b | …

other1指除了*和/以外C语言中的其它字符 comment → /* other* (* ** other1 other*)* ** */ (f) 由偶数个0和偶数个1构成的所有0和1的串。

[解答] 由题目分析可知,一个符号串由0和1组成,则0和1的个数只能有四种情况: x 偶数个0和偶数个1(用状态0表示); x 偶数个0和奇数个1(用状态1表示); x 奇数个0和偶数个1(用状态2表示); x 奇数个0和奇数个1(用状态3表示); 所以,

x 状态0(偶数个0和偶数个1)读入1,则0和1的数目变为:偶数个0和奇数个1(状态1) x 状态0(偶数个0和偶数个1)读入0,则0和1的数目变为:奇数个0和偶数个1(状态2) x 状态1(偶数个0和奇数个1)读入1,则0和1的数目变为:偶数个0和偶数个1(状态0) x 状态1(偶数个0和奇数个1)读入0,则0和1的数目变为:奇数个0和奇数个1(状态3) x 状态2(奇数个0和偶数个1)读入1,则0和1的数目变为:奇数个0和奇数个1(状态3) x 状态2(奇数个0和偶数个1)读入0,则0和1的数目变为:偶数个0和偶数个1(状态0) x 状态3(奇数个0和奇数个1)读入1,则0和1的数目变为:奇数个0和偶数个1(状态2) x 状态3(奇数个0和奇数个1)读入0,则0和1的数目变为:偶数个0和奇数个1(状态1)

因为,所求为由偶数个0和偶数个1构成的所有0和1的串,故状态0既为初始状态又为终结状态,其状态转换图:

由此可以写出其正规文法为:

S0 → 1S1 | 0S2 | ε S1 → 1S0 | 0S3 | 1 S2 → 1S3 | 0S0 | 0 S3 → 1S2 | 0S1

在不考虑S0 → ε产生式的情况下,可以将文法变形为: S0 = 1S1 + 0S2 S1 = 1S0 + 0S3 + 1 S2 = 1S3 + 0S0 + 0 S3 = 1S2 + 0S1 所以: S0 = (00|11) S0 + (01|10) S3 + 11 + 00 (1) S3 = (00|11) S3 + (01|10) S0 + 01 + 10

(2) 解(2)式得: S3 = (00|11)* ((01|10) S0 + (01|10)) 代入(1)式得:

S0 = (00|11) S0 + (01|10) (00|11)*((01|10) S0 + (01|10)) + (00|11) => S0 = ((00|11) + (01|10) (00|11)*(01|10))S0 + (01|10) (00|11)*(01|10) + (00|11) => S0 = ((00|11)|(01|10) (00|11)*(01|10))*((00|11) + (01|10) (00|11)* (01|10)) => S0 = ((00|11)|(01|10) (00|11)* (01|10))+

因为S0→ε所以由偶数个0和偶数个1构成的所有0和1的串的正规定义为: S0 → ((00|11)|(01|10) (00|11)* (01|10))* (g) 由偶数个0和奇数个1构成的所有0和1的串。

[解答] 此题目我们可以借鉴上题的结论来进行处理。

对于由偶数个0和奇数个1构成的所有0和1的串,我们分情况讨论:

1

(1) 若符号串首字符为0,则剩余字符串必然是奇数个0和奇数个1,因此我们必须在上题偶数个0和偶数个1的符号串基础上再读入10(红色轨迹)或01(蓝色轨迹),又因为在0→1和1→3的过程中可以进行多次循环(红色虚线轨迹),同理0→2和2→3(蓝色虚线轨迹),所以还必须增加符号串(00|11)*,我们用S0表示偶数个0和偶数个1, 用S表示偶数个0和奇数个1则其正规定义为: S → 0(00|11)*(01|10) S0 S0 → ((00|11)|(01|10) (00|11)* (01|10))*

(2) 若符号串首字符为1,则剩余字符串必然是偶数个0和偶数个1,其正规定义为: S → 1S0

S0 → ((00|11)|(01|10) (00|11)* (01|10))* 综合(1)和(2)可得,偶数个0和奇数个1构成的所有0和1串其正规定义为: S → 0(00|11)*(01|10) S0|1S0 S0 → ((00|11)|(01|10) (00|11)* (01|10))* 2.7(c) ((ε|a)b*)* start ε ε a ε ε εε ε εbε ε ε

ababbab:s->4->0->1->5->6->7->8->4->0->1->5->6->7->6->7->8->4->0->1->5->6->7->8->f

2.12 为下列正规式构造最简的DFA (b) (a|b)* a (a|b) (a|b)

(1) 根据算法2.4构造该正规式所对应的NFA,如图所示。

(2) 根据算法2.2(子集法)将NFA转换成与之等价的DFA(确定化过程) 初始状态 S0 = ε-closure(0) = {0, 1, 2, 4, 7} 标记状态S0

2

S1 = ε-closure(move(S0, a)) = ε-closure({5, 8}) = {1, 2, 4, 5, 6, 7, 8, 9, 11} S2 = ε-closure(move(S0, b)) = ε-closure({3}) = {1, 2, 3, 4, 6, 7} 标记状态S1

S3 = ε-closure(move(S1, a)) = ε-closure({5, 8, 12}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 16} S4 = ε-closure(move(S1, b)) = ε-closure({3, 10}) = {1, 2, 4, 5, 6, 7, 10, 13, 14, 16} 标记状态S2

S1 = ε-closure(move(S2, a)) = ε-closure({5, 8}) = {1, 2, 4, 5, 6, 7, 8, 9, 11} S2 = ε-closure(move(S2, b)) = ε-closure({3}) = {1, 2, 3, 4, 6, 7} 标记状态S3

S5 = ε-closure(move(S3, a)) = ε-closure({5, 8, 12, 17}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18}

S6 = ε-closure(move(S3, b)) = ε-closure({3, 10, 15}) = {1, 2, 4, 5, 6, 7, 10, 13, 14, 15, 16, 18} 标记状态S4 S7 = ε-closure(move(S4, a)) = ε-closure({5, 8, 17}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 17, 18}

S8 = ε-closure(move(S4, b)) = ε-closure({3, 15}) = {1, 2, 3, 4, 6, 7, 15, 18} 标记状态S5

S5 = ε-closure(move(S5, a)) = ε-closure({5, 8, 12, 17}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18} S6 = ε-closure(move(S5, b)) = ε-closure({3, 10, 15}) = {1, 2, 4, 5, 6, 7, 10, 13, 14, 15, 16, 18} 标记状态S6 S7 = ε-closure(move(S6, a)) = ε-closure({5, 8, 17}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 17, 18}

S8 = ε-closure(move(S6, b)) = ε-closure({3, 15}) = {1, 2, 3, 4, 6, 7, 15, 18} 标记状态S7 S3 = ε-closure(move(S7, a)) = ε-closure({5, 8, 12}) = {1, 2, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 16}

S4 = ε-closure(move(S7, b)) = ε-closure({3, 10}) = {1, 2, 4, 5, 6, 7, 10, 13, 14, 16} 标记状态S8

S1 = ε-closure(move(S8, a)) = ε-closure({5, 8}) = {1, 2, 4, 5, 6, 7, 8, 9, 11} S2 = ε-closure(move(S8, b)) = ε-closure({3}) = {1, 2, 3, 4, 6, 7}

由以上可知,确定化后的DFA的状态集合S = {S0, S1, S2, S3, S4, S5, S6, S7, S8},输入符号集合Σ = {a, b},状态转换函数move如上,S0为开始状态,接收状态集合F = {S5, S6, S7, S8},其状态转换图如下所示: (3) 根据算法2.3过将DFA最小化

第一次划分:{S0, S1, S2, S3, S4} {S5, S6, S7, S8} {S0, S1, S2, S3, S4}a = {S1, S3, S1, S5, S7} 第二次划分:{S0, S1, S2} {S3, S4} {S5, S6, S7, S8} {S0, S1, S2}a = {S1, S3, S1} 第三次划分:{S0, S2} {S1} {S3, S4} {S5, S6, S7, S8}

{S0, S2}a = {S1} {S0, S2}b = {S2} S0, S2不可区分,即等价。 {S5, S6, S7, S8}a = {S5, S7, S3, S1} 第四次划分:{S0, S2} {S1} {S3, S4} {S5, S6} {S7, S8} {S3, S4}a = {S5, S7} 第五次划分:{S0, S2} {S1} {S3} {S4} {S5, S6} {S7, S8} {S5, S6}a = {S5, S7} 第六次划分:{S0, S2} {S1} {S3} {S4} {S5} {S6} {S7, S8} {S7, S8}a = {S3, S1} 第七次划分:{S0, S2} {S1} {S3} {S4} {S5} {S6} {S7} {S8} 集合不可再划分,

所以S0, S2等价,选取S0表示{S0, S2},其状态转换图,即题目所要求的最简DFA如下所示:

第三章

3.1

3

3.2

3.10

3.11

4

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