形式语言与自动机课后习题答案
完整版形式语言与自动机课后习题答案部分.ppt
• pp.84：习题 7(1)
用自然语言描述下列文法定义的语言
G: AaaA|aaB
BBcc|D#cc
DbbbD|#
• 解题思路
– 观察每个产生式及其组合产生的子语言的特点； – 根据开始符的产生式将它们并起来就是整个文法产生的语言；
• 解答
(1) D产生式：DbbbD|# – 使用DbbbD可产生句型：(bbb)mD (m1); – 进一步使用D#可得：L(D)={(bbb)m#| m0}
• A|0A|1A；
– 产生语言{0x|x{0, 1}*}的文法
• S0A；
– G: S0A
A|0A|1A
精心整理
11
G FH
课后作业二 (cont.)
• 习题8(3)的解答
– 分析：语言的特点
• {11x11|x*}{111, 11}；
– 产生语言{x|x{0, 1}*}的文法
• A|0A|1A；
– 习题 22 --- 前/后缀
– 习题 23 --- 前/后缀
– 习题 28(1)(2)(10) --- L的描述
精心整理
3
G FH
课后作业一 (cont.)
• pp.40：习题 21
– 判断集合是否字母表的依据
• 非空性
• 有穷性
• 可区分性：字母表中的字符两两互不相同
• 整体性或不可分性
– 解答：(1)、(2) 和(6) 是字母表，其它不是
– 产生子语言{11x11|x*}的文法
• S11A11 ；
– 产生子语言{111, 11}的文法
• S111|11；
– G: S11A11|111|11
A|0A|1A
其它答案 (1) G: S11A|111|11
形式语言与自动机Chapter7练习参考解答
Chapter 7 练习参考解答Exercise 7.1.3 从以下文法出发：S → 0A0 | 1B1 | BBA → CB → S | AC → S | εa) 有没有无用符号？如果有的话去除它们。
b) 去除ε-产生式。
c) 去除单位产生式。
d) 把该文发转化为乔姆斯基范式。
参考解答：a)没有无用符.b) 所有符号S,A,B,C都是可致空的，消去ε-产生式后得到新的一组产生式:S → 0A0 | 1B1 | BB | B | 00 | 11A → CB → S | AC → Sc) 单元偶对包括：（A,A）,（B,B）,（C,C）,（S,S）,（A,C）,（A,S）,（A,B）,（B,A）,（B,C）,（B,S）,（C,A）,（C,B）,（C,S）,（S,A）,（S,B）,（S,C）,消去单元产生式后得到新的一组产生式S → 0A0 | 1B1 | BB | B | 00 | 11A → CB → S | AC → SS → 0A0 | 1B1 | BB | 00 | 11A → 0A0 | 1B1 | BB | 00 | 11B → 0A0 | 1B1 | BB | 00 | 11C → 0A0 | 1B1 | BB | 00 | 11d)先消去无用符号C，得到新的一组产生式:S → 0A0 | 1B1 | BB | 00 | 11A → 0A0 | 1B1 | BB | 00 | 11B → 0A0 | 1B1 | BB | 00 | 11引入非终结符C，D，增加产生式C → 0和D → 1，得到新的一组产生式:S → CAC | DBD | BB | CC | DDA → CAC | DBD | BB | CC | DDB → CAC | DBD | BB | CC | DDC → 0D → 1引入非终结符E，F，增加产生式E → CA和F → DB，得到满足Chomsky范式的一组产生式:S → EC | FD | BB | CC | DDA → EC | FD | BB | CC | DDB → EC | FD | BB | CC | DDE → CAF → DBC → 0D → 1Exercise 7.2.1(b)用CFL泵引理来证明下面的语言都不是上下文无关的：b) {a n b n c i | i ≤n}。
形式语言与自动机理论-蒋宗礼-第四章参考答案
1.写出表示下列语言的正则表达式。
（吴贤珺02282047）⑴{0, 1}*。
解：所求正则表达式为：(0+1)*。
⑵{0, 1}+。
解：所求正则表达式为：(0+1)+。
⑶{ x│x∈{0,1}+ 且x中不含形如00的子串 }。
解：根据第三章构造的FA，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷{ x│x∈{0,1}*且x中不含形如00的子串 }。
解：根据上题的结果，可得所求正则表达式为：ε+1*(01+)*(01+0+1)。
⑸{ x│x∈{0,1}+ 且x中含形如10110的子串 }。
解：所求正则表达式为：(0+1)*10110(0+1)*。
⑹ { x│x∈{0,1}+ 且x中不含形如10110的子串 }。
解：根据第三章的习题，接受x的FA为：要求该FA对应的正则表达式，分别以q0、q1、q2、q3、q4为终结状态考虑：q为终态时的正则表达式：(0*(11*0(10)*(ε+111*11*0(10)*)0)*)*q为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)*1q为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*2q为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)*3q为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*4将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}+ 且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
形式语言与自动机理论-蒋宗礼-第四章参考答案
1.写出表示下列语言的正则表达式。
（吴贤珺02282047）⑴{0, 1}*。
解：所求正则表达式为：(0+1)*。
⑵{0, 1}+。
解：所求正则表达式为：(0+1)+。
⑶{ x│x∈{0,1}+ 且x中不含形如00的子串 }。
解：根据第三章构造的FA，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷{ x│x∈{0,1}*且x中不含形如00的子串 }。
解：根据上题的结果，可得所求正则表达式为：ε+1*(01+)*(01+0+1)。
⑸{ x│x∈{0,1}+ 且x中含形如10110的子串 }。
解：所求正则表达式为：(0+1)*10110(0+1)*。
⑹ { x│x∈{0,1}+ 且x中不含形如10110的子串 }。
解：根据第三章的习题，接受x的FA为：要求该FA对应的正则表达式，分别以q0、q1、q2、q3、q4为终结状态考虑：q为终态时的正则表达式：(0*(11*0(10)*(ε+111*11*0(10)*)0)*)*q为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)*1q为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*2q为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)*3q为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*4将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}+ 且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
形式语言与自动机Chapter5练习参考解答
Chapter 5 练习参考解答Exercise 5.1.2 (c) 下面的文法产生了正则表达式0*1(0+1)*的语言：εε|1|0|01B B B A A BA S →→→试给出下列串的最左推导和最右推导：c) 00011。
参考解答：一个最左推导：S ⇒lm A1B ⇒lm 0A1B ⇒lm 00A1B ⇒lm 000A1B ⇒lm 0001B ⇒lm 00011B ⇒lm 00011一个最右推导：S ⇒rm A1B ⇒rm A11B ⇒rm A11 ⇒rm 0A11⇒rm 00A11⇒rm 000A11 ⇒rm 00011! Exercise 5.1.3 证明任何正则语言都是上下文无关语言。
提示：通过对正则表达式中的运算符的数目进行归纳的方法来构造CFG 。
参考解答：对于任何正规表达式R ，归纳于R 中算符的数目n 构造如下产生式集合P(R)，相应的开始符号为S(R)：基础：n=0.（1）R 为ε，则任选非终结符A ，令P(R)只包含A →ε，以及S(R)为A ；（2）R 为φ，令P(R) 为空集；（3）R 为a ，则任选非终结符A ，令P(R)只包含A →a ，以及S(R)为A ； 基础：n>0.（1）R为R1+R2，则适修改非终结符的名字，使得P(R1)与P(R2)中的所有非终结符没有重名，任选不出现在P(R1)⋃P(R2)中的非终结符A，令P(R)= P(R1) ⋃P(R2)⋃{ A→ S(R1), A→ S(R2) }，并且，令S(R)为A；（2）R为R1R2，则适修改非终结符的名字，使得P(R1)与P(R2)中的所有非终结符没有重名，任选不出现在P(R1)⋃P(R2)中的非终结符A，令P(R)= P(R1) ⋃P(R2)⋃{ A→ S(R1)S(R2) }；并且，令S(R)为A；（3）R为R1*，任选不出现在P(R1) 中的非终结符A，令P(R)= P(R1) ⋃{ A→ AS(R1) , A→ε }；并且，令S(R)为A.设L为正规语言，R为正规表达式，且有L=L(R). 令上下文无关文法G 的产生式集合为上述归纳过程所得到的P(R)，以及G的开始符号为S(R). 可以归纳证明L(G)=L(R)=L.! Exercise 5.1.4 (选做)如果一个CFG的每个产生式的体都最多只有一个变元，并且该变元总在最右端，那么该CFG称做右线性的。
形式语言与自动机Chapter3练习参考解答
Chapter 3 练习参考解答Exercise 3.1.1 写出下列语言的正规表达式:c） The set of strings of 0' s and1' s with at most one pair of consecutive 1' s.c）最多包含两个相继的 1 的所有0， 1 字符串的集合.参考解答：该题的翻译有二义性（Sorry ）.1）按原题意的解法对不包含相继的 1 的所有0， 1 字符串的集合，正规表达式可以为：1*（0+01）* 或（0+10 ）*1*；包含最多一对相继的1，正规表达式可以为：（0+10）*11（0+01）*；所以，结果正规表达式可以为：1*（0+01）* + （0+10 ）*11（0+01）2）若理解为11 可以出现多次的解法正规表达式可以为：1*（0+01+011 ）* 或（0+10+110）*1*；等等Exercise 3.1.2 写出下列语言的正规表达式:b） 0 的个数能够被 5 整除的所有0， 1 字符串的集合. 参考解答：该题问题不多.正规表达式可以为：（1*01*01*01*01*01* ）*Exercise 3.2.1（2）c）给出所有正规表达式R（i2j）, 并尽可能简化之.d）给出该自动机的语言的一个正规表达式.参考解答：该题问题反映较多的是关于正规表达式的简化. 本科程较侧重原理，技术方面不够细致，因此对于“最简“的正规表达式没有作明确的规定，也没有类似于命题演算中有关”范式“的讨论.同学们在做题时除了应用有关定律外， 还可以自己总结出一些规律，比如一个表达式的语言 R 是另一个表达式S 所代 表语言的一个子集，则对于 R+S 就可以消去R ，例如 +1*可以简化为1*.再 如，在已做过的习题中出现的公式，例如Exercise3.4.1(g)我们可以验证(汁R)*= R*，因此 (+1)*可以简化为1*.综合已阅过的作业，推荐以下结果：C) R (121)= 1*+1*0 ( +11*0)*11* = 1*+1*0(11*0)*11*R (1 2 = 1*0+1*0 ( ；+11*0)* ( ；+11*0) = 1*0(11*0)*R (123)= * +1*0 2+11*0)*0 = 1*0(11*0)*0R (221)= 11* + ( +11*0) ( +11*0)*11* = (11*0)*11* R (222)= +11*0 + ( +11*0) ( ；+11*0)* ( ；+11*0) = (11*0)*R (223)= 0 + ( +11*0) ( +11*0)*0 = (11*0)*0R (321) = * + 1 (<+11*0)*11* =1(11*0)*11*R (322)= 1 + 1 ( +11*0)* ( ；+11*0) =1(11*0)*R (323)= ； + 0 + 1 ( +11*0)* 0 = ； + 0 + 1(11*0)*0d)该自动机的语言的一个正规表达式为R (133) =1*0(11*0)*0 +1*0(11*0)*0 ( ； + 0 + 1(11*0)*0)* ( ； + 0 + 1(11*0)*0)=1*0(11*0)*0 (0 + 1(11*0)*0)*Exercise 3.2.3 使用状态消去技术，将如下 DFA 转化为一个正规表达式.参考解答：该题问题不多，状态消去的次序不同，结果形式上可能有所不同，但 相互之间是等价的.以下是一个解法:/1+0(01+10*11)*(00+10*10) Start 5结果正规表达式可以为：(1+0(01+10*11)*(00+10*10))* 原状态图:消去状态r :消去状态s :1Start 0Exercise 324 将下列正规表达式转化为带 &转移的NFA.b) (0+1)01 c) 00(0+1)*参考解答：若严格按照所介绍的算法构造，则结 果如下:b)Exercise 3.4.1验证下列包含正规表达式的等式 c) (RS) T = R (ST) .g)(+R)* = R* . 参考解答：c)将两个表达式具体化，将R 替换为a ，将S 替换为b.(RS)T 具体化为(ab)a ，R(ST)具体化为 a(ba)，而 L((ab)a)=L(a(ba))={abc}， 所以原等式成立；g)将两个表达式具体化，将R 替换为a.(+R)* 具体化为(；+a)*，R*具体化为 a*，而 L(( +a)*)=L(a*)={ ,a ,aa,aaa,…， (注：严格证明L(( ；+a)*)=L(a*)，可以在归纳证明：对任意k>=0，{；，a}k ={a}k 的基础上进行)，所以原等式成立;Exercise 342证明或否证下列关于正规表达式的命题 b) (RS+R)*R = R (SR+R)*.d) (R+S)*S = (R*S)* .Start参考解答：b)将两个表达式具体化，将R替换为a,将S替换为b.(RS+R)*R 具体化为(ab+a)*a，R (SR+R)*具体化为a(ba+a)*，可以证明L((ab+a)*a)=L(a(ba+a)*) (注：同上，可以先归纳证明：对任意k>=0，{ab，a}k{a}={a}{ba ，a}k，而由连接运算对-运算的分配律，—k —k 可知L((ab+a)*a)= *0,1,2, -({ab，a} {a}), L(a(ba+a)*)= k=o,1,2, ・({a}{ba，a})，由此证得L((ab+a)*a)=L(a(ba+a)*))，所以原等式成立；g)将两个表达式具体化，将R替换为a，将S替换为b.(R+S)*S 具体化为(a+b)*b，(R*S)* 具体化为(a*b)*，由于〉L((a*b)*)，而>■' L((a*b)*)，所以原等式不成立。
形式语言与自动机理论_哈尔滨工业大学中国大学mooc课后章节答案期末考试题库2023年
形式语言与自动机理论_哈尔滨工业大学中国大学mooc课后章节答案期末考试题库2023年1.令字母表【图片】, 则克林闭包【图片】中元素的长度为？参考答案:只能是有限的2.由字符0和1构成且含有奇数个1的DFA，至少需要几个状态？参考答案:23.双栈PDA可以接受任意图灵机接受的语言。
参考答案:正确4.由某字母表【图片】中的字符构成的全部正则表达式的集合，也可以看做是一个语言，则该语言为：参考答案:上下文无关语言5.由字符0和1构成且含有奇数个1和偶数个0的DFA，至少需要几个状态？参考答案:46.字符串的长度可以是任意的，那么也可以是无穷长的。
参考答案:错误7.设【图片】和【图片】是字母表【图片】上的任意语言且【图片】是无穷的，则两个语言的连接【图片】一定是无穷的。
参考答案:错误8.每一个有穷的语言都是正则语言。
参考答案:正确9.任何正则语言都是上下文无关语言。
参考答案:正确10.任意有穷集合的克林闭包一定是无穷集合。
参考答案:错误11.递归可枚举语言是可判定的语言。
参考答案:错误12.任何有限的语言都是上下文无关语言。
参考答案:正确13.NFA处于某个状态q且输入某字符a时，如果状态转移函数未定义，则NFA会：参考答案:停止自动机的运行，并拒绝该串。
14.有穷自动机有了空转移（不消耗输入串的状态跳转）, 改变了它识别语言的能力。
参考答案:错误15.对同一个语言，可能存在两个不同的有穷自动机识别。
参考答案:正确16.带有空转移的非确定有穷自动机中，对于某一个状态，是否可以同时存在“对某字符a的非确定性”和“空转移”？参考答案:可以。
17.图灵机是算法的好模型。
参考答案:错误18.确定的图灵机与非确定的图灵机等价。
参考答案:正确19.由字符0和1构成且含有偶数个1的DFA，至少需要几个状态？参考答案:220.如果一个语言是不可判定的，那么它的补也一定是不可判定的参考答案:错误21.确定的有穷自动机中，“确定的”含义是：参考答案:状态转移是确定的22.由字符0和1构成且长度为偶数的全部字符串的DFA，至少需要几个状态？参考答案:223.集合的克林闭包与正比包一定不相等参考答案:错误24.设【图片】是字母表【图片】上的任意语言，则语言【图片】的闭包【图片】一定是无穷的。
形式语言与自动机蒋宗礼答案
形式语言与自动机蒋宗礼答案形式语言与自动机蒋宗礼答案【篇一：形式语言第四章参考答案(蒋宗礼)】p> 解：所求正则表达式为：(0+1)*。
+⑵ {0, 1}。
解：所求正则表达式为：(0+1)+。
⑶ { x│x∈{0,1}且x中不含形如00的子串 }。
解：根据第三章构造的fa，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷ { x│x∈{0,1}*且x中不含形如00的子串 }。
++ +q1为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)* q2为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*q3为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)* q4为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)* 将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
由题设，x=0时，│x│=1，模5是1，不符合条件，所以不必增加关于它的状态。
下面对每一个状态考虑输入0和1时的状态转移。
q: 输入1，模5是1，进入q1。
+q0: 设x=5n。
输入0，x=5n*2=10n，模5是0，故进入q0输入1，x=5n*2+1=10n+1，模5是1，故进入q1q1：设x=5n+1。
输入0，x=(5n+1)*2=10n+2，模5是2，故进入q2输入1，x=(5n+1)*2+1=10n+3，模5是3，故进入q3 q2：设x=5n+2。
《形式语言与自动机》第二版英文版课后题答案
Exercise 2.3.1Here are the sets of NFA states represented by each of the DFA states A through H: A = {p}; B = {p,q}; C = {p,r}; D = {p,q,r}; E = {p,q,s}; F = {p,q,r,s}; G = {p,r,s}; H = {p,s}.Exercise 2.3.4(a)The idea is to use a state qi, for i= 0,1,...,9 to represent the idea that we have seen an input i and guessed that this is the repeated digit at the end. We also have state qs, the initial state, and qf, the final state. We stay in state qs all the time; itSolutions for Section 2.4Exercise 2.4.1(a)We'll use q0 as the start state. q1, q2, and q3 will recognize abc; q4, q5, and q6 will recognize abd, and q7 through q10 will recognize aacd. The transition table is:Exercise 2.4.2(a)The subset construction gives us the following states, each representing the subset of the NFA states indicated: A = {q0}; B = {q0,q1,q4,q7}; C = {q0,q1,q4,q7,q8}; D = {q0,q2,q5}; E = {q0,q9}; F = {q0,q3}; G = {q0,q6}; H = {q0,q10}. Note that F, G and H can be combined into one accepting state, or we can use these three state to signal the recognition of abc, abd, and aacd, respectively.Solutions for Section 2.5Exercise 2.5.1For part (a): the closure of p is just {p}; for q it is {p,q}, and for r it is {p,q,r}.For (b), begin by noticing that a always leaves the state unchanged. Thus, we can think of the effect of strings of b's and c's only. To begin, notice that the only ways to get from p to r for the first time, using only b, c, and ε-transitions are bb, bc, and c. After getting to r, we can return to r reading either b or c. Thus, every string of length 3 or less, consisting of b's and c's only, is accepted, with the exception of the string b. However, we have to allow a's as well. When we try to insert a's in these strings, yet keeping the length to 3 or less, we find that every string of a's b's, and c's with at most one a is accepted. Also, the strings consisting of one c and up to 2 a's are accepted; other strings are rejected.There are three DFA states accessible from the initial state, which is the ε closure of p, or {p}. Let A = {p}, B = {p,q}, and C = {p,q,r}. Then the transition table is:Solutions for Section 3.1Exercise 3.1.1(a)The simplest approach is to consider those strings in which the first a precedes the first b separately from those where the opposite occurs. The expression:c*a(a+c)*b(a+b+c)* + c*b(b+c)*a(a+b+c)*Exercise 3.1.2(a)(Revised 9/5/05) The trick is to start by writing an expression for the set of strings that have no two adjacent 1's. Here is one such expression: (10+0)*(ε+1)To see why this expression works, the first part consists of all strings in which every 1 is followed by a 0. To that, we have only to add the possibility that there is a 1 at the end, which will not be followed by a 0. That is the job of (ε+1).Now, we can rethink the question as asking for strings that have a prefix with no adjacent 1's followed by a suffix with no adjacent 0's. The former is the expression we developed, and the latter is the same expression, with 0 and 1 interchanged. Thus, a solution to this problem is (10+0)*(ε+1)(01+1)*(ε+0). Note that the ε+1 term in the middle is actually unnecessary, as a 1 matching that factor can be obtained from the (01+1)* factor instead.Exercise 3.1.4(a)This expression is another way to write ``no adjacent 1's.'' You should compare it with the different-looking expression we developed in the solution to Exercise 3.1.2(a). The argument for why it works is similar. (00*1)* says every 1 is preceded by at least one 0. 0*at the end allows 0's after the final 1, and (ε+1) at the beginning allows an initial 1, which must be either the only symbol of the string or followed by a 0.Exercise 3.1.5The language of the regular expression ε. Note that ε* denotes the language of strings consisting of any number of empty strings, concatenated, but that is just the set containing the empty string.Solutions for Section 3.2Exercise 3.2.1Part (a): The following are all R0expressions; we list only the subscripts. R11 = ε+1; R12 = 0; R13 = phi; R21 = 1; R22 = ε; R23 = 0; R31 = phi; R32 = 1; R33 = ε+0.Part (b): Here all expression names are R(1); we again list only the subscripts. R11 = 1*; R12 = 1*0; R13 = phi; R21 = 11*; R22 = ε+11*0; R23 = 0; R31 = phi; R32 = 1; R33 = ε+0.Part (e): Here is the transition diagram:If we eliminate state q2 we get:Applying the formula in the text, the expression for the ways to get from q1 to q3 is: [1 + 01 +00(0+10)*11]*00(0+10)*Exercise 3.2.4(a)Exercise 3.2.6(a)(Revised 1/16/02) LL* or L+.Exercise 3.2.6(b)The set of suffixes of strings in L.Exercise 3.2.8Let R(k)ijm be the number of paths from state i to state j of length m that go through no state numbered higher than k. We can compute these numbers, for all states i and j, and for m no greater than n, by induction on k.Basis: R0ij1 is the number of arcs (or more precisely, arc labels) from state i to state j. R0ii0 = 1, and all other R0ijm's are 0.Induction: R(k)ijm is the sum of R(k-1)ijm and the sum over all lists (p1,p2,...,pr) of positive integers that sum to m, of R(k-1)ikp1 * R(k-1)kkp2 *R(k-1)kkp3 *...* R(k-1)kkp(r-1) * R(k-1)kjpr. Note r must be at least 2.The answer is the sum of R(k)1jn, where k is the number of states, 1 is the start state, and j is any accepting state.Solutions for Section 3.4Exercise 3.4.1(a)Replace R by {a} and S by {b}. Then the left and right sides become {a} union {b} = {b} union {a}. That is, {a,b} = {b,a}. Since order is irrelevant in sets, both languages are the same: the language consisting of the strings a and b.Exercise 3.4.1(f)Replace R by {a}. The right side becomes {a}*, that is, all strings of a's, including the empty string. The left side is ({a}*)*, that is, all strings consisting of the concatenation of strings of a's. But that is just the set of strings of a's, and is therefore equal to the right side.Exercise 3.4.2(a)Not the same. Replace R by {a} and S by {b}. The left side becomes all strings of a's and b's (mixed), while the right side consists only of strings of a's (alone) and strings of b's (alone). A string like ab is in the language of the left side but not the right.Exercise 3.4.2(c)Also not the same. Replace R by {a} and S by {b}. The right side consists of all strings composed of zero or more occurrences of strings of the form a...ab, that is, one or more a's ended by one b. However, every string in the language of the left side has to end in ab. Thus, for instance, ε is in the language on the right, but not on the left.Solutions for Section 4.1Exercise 4.1.1(c)Let n be the pumping-lemma constant (note this n is unrelated to the n that is a local variable in the definition of the language L). Pick w = 0n10n. Then when we write w = xyz, we know that |xy| <= n, and therefore y consists of only 0's. Thus, xz, which must be in L if L is regular, consists of fewer than n0's, followed by a 1 and exactly n0's. That string is not in L, so we contradict the assumption that L is regular.Exercise 4.1.2(a)Let n be the pumping-lemma constant and pick w = 0n2, that is, n2 0's. When we write w = xyz, we know that y consists of between 1 and n 0's. Thus, xyyz has length between n2 + 1 and n2 + n. Since the next perfect square after n2 is (n+1)2 = n2 + 2n + 1, we know that the length of xyyz lies strictly between the consecutive perfect squares n2 and (n+1)2. Thus, the length of xyyz cannot be a perfect square. But if the language were regular, then xyyz would be in the language, which contradicts the assumption that the language of strings of 0's whose length is a perfect square is a regular language.Exercise 4.1.4(a)We cannot pick w from the empty language.Exercise 4.1.4(b)If the adversary picks n = 3, then we cannot pick a w of length at least n.Exercise 4.1.4(c)The adversary can pick an n > 0, so we have to pick a nonempty w. Since w must consist of pairs 00 and 11, the adversary can pick y to be one of those pairs. Then whatever i we pick, xy i z will consist of pairs 00 and 11, and so belongs in the language.Solutions for Section 4.2Exercise 4.2.1(a)aabbaa.Exercise 4.2.1(c)The language of regular expression a(ab)*ba.Exercise 4.2.1(e)Each b must come from either 1 or 2. However, if the first b comes from 2 and the second comes from 1, then they will both need the a between them as part of h(2) and h(1), respectively. Thus, the inverse homomorphism consists of the strings {110, 102, 022}.Exercise 4.2.2Start with a DFA A for L. Construct a new DFA B, that is exactly the same as A, except that state q is an accepting state of B if and only if δ(q,a) is an accepting state of A. Then B accepts input string w if and only if A accepts wa; that is, L(B) = L/a.Exercise 4.2.5(b)We shall use D a for ``the derivative with respect to a.'' The key observation is that if epsilon is not in L(R), then the derivative of RS will always remove an a from the portion of a string that comes from R. However, if epsilon is in L(R), then the string might have nothing from R and will remove a from the beginning of a string in L(S) (which is also a string in L(RS). Thus, the rule we want is:If epsilon is not in L(R), then D a(RS) = (D a(R))S. Otherwise, D a(RS) = D a(R)S + D a(S).Exercise 4.2.5(e)L may have no string that begins with 0.Exercise 4.2.5(f)This condition says that whenever 0w is in L, then w is in L, and vice-versa. Thus, L must be of the form L(0*)M for some language M (not necessarily a regular language) that has no string beginning with 0.In proof, notice first that D0(L(0*)M = D0(L(0*))M union D0(M) = L(0*)M. There are two reasons for the last step. First, observe that D0 applied to the language of all strings of 0's gives all strings of 0's, that is, L(0*). Second, observe that because M has no string that begins with 0, D0(M) is the empty set [that's part (e)].We also need to show that every language N that is unchanged by D0is of this form. Let M be the set of strings in N that do not begin with 0. If N is unchanged by D0, it follows that for every string w in M, 00...0w is in N; thus, N includes all the strings of L(0*)M. However, N cannot include a string that is not in L(0*)M. If x were such a string, then we can remove all the 0's at the beginning of x and get some string y that is also in N. But y must also be in M.Exercise 4.2.8Let A be a DFA for L. We construct DFA B for half(L). The state of B is of the form [q,S], where:∙q is the state A would be in after reading whatever input B has read so far.∙S is the set of states of A such that A can get from exactly these states to an accepting state by reading any input string whose length is the same as the length of the string B has read so far.It is important to realize that it is not necessary for B to know how many inputs it has read so far; it keeps this information up-to-date each time it reads a new symbol. The rule that keeps things up to date is: δB([q,S],a) = [δA(q,a),T], where T is the set of states p of A such that there is a transition from p to any state of S on any input symbol. In this manner, the first component continues to simulate A, while the second component now represents states that can reach an accepting state following a path that is one longer than the paths represented by S.To complete the construction of B, we have only to specify:∙The initial state is [q0,F], that is, the initial state of A and the accepting states of A. This choice reflects the situation when A has read 0 inputs: it is still in its initial state, and the accepting states are exactly the ones that can reach anaccepting state on a path of length 0.∙The accepting states of B are those states [q,S] such that q is in S. The justification is that it is exactly these states that are reached by some string of length n, and there is some other string of length n that will take state q to an accepting state.Exercise 4.2.13(a)Start out by complementing this language. The result is the language consisting of all strings of 0's and 1's that are not in 0*1*, plus the strings in L0n1n. If we intersect with 0*1*, the result is exactly L0n1n. Since complementation and intersection with a regular set preserve regularity, if the given language were regular then so would be L0n1n. Since we know the latter is false, we conclude the given language is not regular.Exercise 4.2.14(c)Change the accepting states to be those for which the first component is an accepting state of A L and the second is a nonaccepting state of A M. Then the resulting DFA accepts if and only if the input is in L - M.Solutions for Section 4.3Exercise 4.3.1Let n be the pumping-lemma constant. Test all strings of length between n and 2n-1 for membership in L. If we find even one such string, then L is infinite. The reason is that the pumping lemma applies to such a string, and it can be ``pumped'' to show an infinite sequence of strings are in L.Suppose, however, that there are no strings in L whose length is in the range n to 2n-1. We claim there are no strings in L of length 2n or more, and thus there are only a finite number of strings in L. In proof, suppose w is a string in L of length at least 2n, and w is as short as any string in L that has length at least 2n. Then the pumping lemma applies to w, and we can write w = xyz, where xz is also in L. How long could xz be? It can't be as long as 2n, because it is shorter than w, and w is as short as any string in L of length 2n or more. n, because xz is at most n shorter than w. Thus, xz is of length between n and 2n-1, which is a contradiction, since we assumed there were no strings in L with a length in that range.Solutions for Section 4.4Exercise 4.4.1Revised 10/23/01.B|xC|x xD|x x xE|x x xF|x x x xG| x x x x xH|x x x x x x x---------------A B C D E F GNote, however, that state H is inaccessible, so it should be removed, leaving the first four states as the minimum-state DFASolutions for Section 5.1Exercise 5.1.1(a)S -> 0S1 | 01Exercise 5.1.1(b)S -> AB | CDA -> aA | εB -> bBc | E | cDC -> aCb | E | aAD -> cD | εE -> bE | bTo understand how this grammar works, observe the following:∙A generates zero or more a's.∙D generates zero or more c's.∙E generates one or more b's.∙B first generates an equal number of b's and c's, then produces either one or more b's (via E) or one or more c's (via cD).That is, B generates strings in b*c* with an unequal number of b's and c's.∙Similarly, C generates unequal numbers of a's then b's.∙Thus, AB generates strings in a*b*c* with an unequal numbers of b's and c's, while CD generates strings in a*b*c* with an unequal number of a's and b's.Exercise 5.1.2(a)Leftmost: S => A1B => 0A1B => 00A1B => 001B => 0010B => 00101B => 00101Rightmost: S => A1B => A10B => A101B => A101 => 0A101 => 00A101 => 00101Exercise 5.1.5S -> S+S | SS | S* | (S) | 0 | 1 | phi | eThe idea is that these productions for S allow any expression to be, respectively, the sum (union) of two expressions, the concatenation of two expressions, the star of an expression, a parenthesized expression, or one of the four basis cases of expressions: 0, 1, phi, and ε.Solutions for Section 5.2Exercise 5.2.1(a)S/ | \A 1 B/ | / |0 A 0 B/ | / |0 A 1 B| |e eIn the above tree, e stands for ε.Solutions for Section 5.3Exercise 5.3.2B -> BB | (B) | [B] | εExercise 5.3.4(a)Change production (5) to:ListItem -> <LI> Doc </LI>Solutions for Section 5.4Exercise 5.4.1Here are the parse trees:S S/ | / / | \a S a Sb S/ | \ \ | \ |a Sb S a S e| | |e e eThe two leftmost derivations are: S => aS => aaSbS => aabS => aab and S => aSbS => aaSbS => aabS => aab. The two rightmost derivations are: S => aS => aaSbS => aaSb => aab and S => aSbS => aSb => aaSb => aab. Exercise 5.4.3The idea is to introduce another nonterminal T that cannot generate an unbalanced a. That strategy corresponds to the usual rule in programming languages that an ``else'' is associated with the closest previous, unmatched ``then.'' Here, we force a b to match the previous unmatched a. The grammar:S -> aS | aTbS | εT -> aTbT | εExercise 5.4.6Alas, it is not. We need to have three nonterminals, corresponding to the three possible ``strengths'' of expressions:1. A factor cannot be broken by any operator. These are the basis expressions, parenthesized expressions, and theseexpressions followed by one or more *'s.2. A term can be broken only by a *. For example, consider 01, where the 0 and 1 are concatenated, but if we follow it bya *, it becomes 0(1*), and the concatenation has been ``broken'' by the *.3.An expression can be broken by concatenation or *, but not by +. An example is the expression 0+1. Note that if weconcatenate (say) 1 or follow by a *, we parse the expression 0+(11) or 0+(1*), and in either case the union has been broken.The grammar:E -> E+T | TT -> TF | FF -> F* | (E) | 0 | 1 | phi | eSolutions for Section 6.1Exercise 6.1.1(a)(q,01,Z0) |- (q,1,XZ0) |- (q,ε,XZ0) |- (p,ε,Z0)|- (p,1,Z0) |- (p,ε,ε)Solutions for Section 6.2Exercise 6.2.1(a)We shall accept by empty stack. Symbol X will be used to count the 0's on the input. In state q, the start state, where we have seen no 1's, we add an X to the stack for each 0 seen. The first X replaces Z0, the start symbol. When we see a 1, we go to state p, and then only pop the stack, one X for each input 1. Formally, the PDA is ({q,p},{0,1},{X,Z0},δ,q,Z0). The rules:1.δ(q,0,Z0) = {(q,X)}2.δ(q,0,X) = {(q,XX)}3.δ(q,1,X) = {(p,ε)}4.δ(p,1,X) = {(p,ε)}Exercise 6.2.2(a)Revised 6/20/02.Begin in start state q0, with start symbol Z0, and immediately guess whether to check for:1.i=j=0 (state q1).2.i=j>0 (state q2).3.j=k (state q3).We shall accept by final state; as seen below, the accepting states are q1 and q3. The rules, and their explanations:∙δ(q0,ε,Z0) = {(q1,Z0), (q2,Z0), (q3,Z0)}, the initial guess.∙δ(q1,c,Z0) = {(q1,Z0)}. In case (1), we assume there are no a's or b's, and we consume all c's. State q1 will be one of our accepting states.∙δ(q2,a,Z0) = {(q2,XZ0)}, and δ(q2,a,X) = {(q2,XX)}. These rules begin case (2). We use X to count the number of a's read from the input, staying in state q2.∙δ(q2,b,X) = δ(q4,b,X) = {(q4,ε)}. When b's are seen, we go to state q4 and pop X's against the b's.∙δ(q4,ε,Z0) = {(q1,Z0)}. If we reach the bottom-of-stack marker in state q4, we have seen an equal number of a's and b's.We go spontaneously to state q1, which will accept and consume all c's, while continuing to accept.∙δ(q3,a,Z0) = {(q3,Z0)}. This rule begins case (3). We consume all a's from the input. Since j=k=0 is possible, state q3 must be an accepting state.∙δ(q3,b,Z0) = {(q5,XZ0)}. When b's arrive, we start counting them and go to state q5, which is not an accepting state.∙δ(q5,b,X) = {(q5,XX)}. We continue counting b's.∙δ(q5,c,X) = δ(q6,c,X) = {(q6,ε)}. When c's arrive, we go to state q6 and match the c's against the b's.∙δ(q6,ε,Z0) = {(q3,ε)}. When the bottom-of-stack marker is exposed in state q6, we have seen an equal number of b's and c's. We spontaneously accept in state q3, but we pop the stack so we cannot accept after reading more a's.Exercise 6.2.4Introduce a new state q, which be comes the initial state. On input ε and the start symbol of P, the new PDA has a choice of popping the stack (thus accepting ε), or going to the start state of P.Exercise 6.2.5(a)Revised 6/6/06.(q0,bab,Z0) |- (q2,ab,BZ0) |- (q3,b,Z0) |- (q1,b,AZ0) |- (q1,ε,Z0) |- (q0,ε,Z0) |- (f,ε,ε)Exercise 6.2.8Suppose that there is a rule that (p,X1X2...X k) is a choice in δ(q,a,Z). We create k-2 new states r1,r2,...,r k-2 that simulate this rule but do so by adding one symbol at a time to the stack. That is, replace (p,X1X2...X k) in the rule by (r k-2,X k-1X k. Then create new rules δ(r k-2,ε,X k-1) = {(r k-3,X k-2X k-1)}, and so on, down to δ(r2,ε,X3) = {(r1,X2X3)} and δ(r1,X2) = {(p,X1X2)}.Solutions for Section 6.3Exercise 6.3.1({q},{0,1),{0,1,A,S},δ ,q ,S) where δ is defined by:1.δ(q,ε,S) = {(q,0S1), (q,A)}2.δ(q,ε,A) = {(q,1A0), (q,S), (q,ε)}3.δ(q,0,0) = {(q,ε)}4.δ(q,1,1) = {(q,ε)}Exercise 6.3.3In the following, S is the start symbol, e stands for the empty string, and Z is used in place of Z0.1.S -> [qZq] | [qZp]The following four productions come from rule (1).2.[qZq] -> 1[qXq][qZq]3.[qZq] -> 1[qXp][pZq]4.[qZp] -> 1[qXq][qZp]5.[qZp] -> 1[qXp][pZp]The following four productions come from rule (2).6.[qXq] -> 1[qXq][qXq]7.[qXq] -> 1[qXp][pXq]8.[qXp] -> 1[qXq][qXp]9.[qXp] -> 1[qXp][pXp]The following two productions come from rule (3).10.[qXq] -> 0[pXq]11.[qXp] -> 0[pXp]The following production comes from rule (4).12.[qXq] -> eThe following production comes from rule (5).13.[pXp] -> 1The following two productions come from rule (6).14.[pZq] -> 0[qZq]15.[pZp] -> 0[qZp]Exercise 6.3.6Convert P to a CFG, and then convert the CFG to a PDA, using the two constructions given in Section 6.3. The result is a one-state PDA equivalent to P.Solutions for Section 6.4Exercise 6.4.1(b)Not a DPDA. For example, rules (3) and (4) give a choice, when in state q, with 1 as the next input symbol, and with X on top of the stack, of either using the 1 (making no other change) or making a move on ε input that pops the stack and going to sta te p.Exercise 6.4.3(a)Suppose a DPDA P accepts both w and wx by empty stack, where x is not ε (i.e., N(P) does not have the prefix property). Then (q0,wxZ0) |-* (q,x,ε) for some state q, where q0 and Z0 are the start state and symbol of P. It is not possible that (q,x,ε) |-* (p,ε,ε) for some state p, because we know x is not ε, and a PDA cannot have a move with an empty sta ck. This observation contradicts the assumption that wx is in N(P).Exercise 6.4.3(c)Modify P' in the following ways to create DPDA P:1.Add a new start state and a new start symbol. P, with this state and symbol, pushes the start symbol of P' on top of thestack and goes to the start state of P'. The purpose of the new start symbol is to make sure P doesn't accidentally accept by empty stack.2.Add a new ``popping state'' to P. In this state, P pops every symbol it sees on the stack, using ε input.3.If P' enters an accepting state, P enters the popping state instead.As long as L(P') has the prefix property, then any string that P' accepts by final state, P will accept by empty stack.Solutions for Section 7.1Exercise 7.1.1A and C are clearly generating, since they have productions with terminal bodies. Then we can discover S is generating because of the production S->CA, whose body consists of only symbols that are generating. However,B is not generating. Eliminating B, leaves the grammarS -> CAA -> aC -> bSince S, A, and C are each reachable from S, all the remaining symbols are useful, and the above grammar is the answer to the question.Exercise 7.1.2Revised 6/27/02.a)Only S is nullable, so we must choose, at each point where S occurs in a body, to eliminate it or not. Since there is no body that consists only of S's, we do not have to invoke the rule about not eliminating an entire body. The resulting grammar:S -> ASB | ABA -> aAS | aA | aB -> SbS | bS | Sb | b | A | bbb)The only unit production is B -> A. Thus, it suffices to replace this body A by the bodies of all the A-productions. The result:S -> ASB | ABA -> aAS | aA | aB -> SbS | bS | Sb | b | aAS | aA | a | bbc)Observe that A and B each derive terminal strings, and therefore so does S. Thus, there are no useless symbols.d)Introduce variables and productions C -> a and D -> b, and use the new variables in all bodies that are not a single terminal:S -> ASB | ABA -> CAS | CA | aB -> SDS | DS | SD | b | CAS | CA | a | DDC -> aD -> bFinally, there are bodies of length 3; one, CAS, appears twice. Introduce new variables E, F, andG to split these bodies, yielding the CNF grammar:S -> AE | ABA -> CF | CA | aB -> SG | DS | SD | b | CF | CA | a | DDC -> aD -> bE -> SBF -> ASG -> DSExercise 7.1.10It's not possible. The reason is that an easy induction on the number of steps in a derivation shows that every sentential form has odd length. Thus, it is not possible to find such a grammar for a language as simple as {00}.To see why, suppose we begin with start symbol S and try to pick a first production. If we pick a production with a single terminal as body, we derive a string of length 1 and are done. If we pick a body with three variables, then, since there is no way for a variable to derive epsilon, we are forced to derive a string of length 3 or more.Exercise 7.1.11(b)The statement of the entire construction may be a bit tricky, since you need to use the construction of part (c) in (b), although we are not publishing the solution to (c). The construction for (b) is by induction on i, but it needs to be of the stronger statement that if an A i-production has a body beginning with A j, then j > i (i.e., we use part (c) to eliminate the possibility that i=j).Basis: For i = 1 we simply apply the construction of (c) for i = 1.Induction: If there is any production of the form A i -> A1..., use the construction of (a) to replace A1. That gives us a situation where all A i production bodies begin with at least A2 or a terminal. Similarly, replace initial A2's using (a), to make A3 the lowest possible variable beginning an A i-production. In this manner, we eventually guarantee that the body of each A i-production either begins with a terminal or with A j, for some j >= i. A use of the construction from (c) eliminates the possibility that i = j.Exercise 7.1.11(d)As per the hint, we do a backwards induction on i, that the bodies of A i productions can be made to begin with terminals.Basis: For i = k, there is nothing to do, since there are no variables with index higher than k to begin the body.Induction: Assume the statement for indexes greater than i. If an A i-production begins with a variable, it must be A j for some j > i. By the induction hypothesis, the A j-productions all have bodies beginning with terminals now. Thus, we may use the construction (a) to replace the initial A j, yielding only A i-productions whose bodies begin with terminals.After fixing all the A i-productions for all i, it is time to work on the B i-productions. Since these have bodies that begin with either terminals or A j for some j, and the latter variables have only bodies that begin with terminals, application of construction (a) fixes the B j's.Solutions for Section 7.2Exercise 7.2.1(a)Let n be the pumping-lemma constant and consider string z = a n b n+1c n+2. We may write z = uvwxy, where v and x, may be ``pumped,'' and |vwx| <= n. If vwx does not have c's, then uv3wx3y has at least n+2a's or b's, and thus could not be in the language.If vwx has a c, then it could not have an a, because its length is limited to n. Thus, uwy has n a's, but no more than 2n+2b's and c's in total. Thus, it is not possible that uwy has more b's than a's and also has more c's than b's. We conclude that uwy is not in the language, and now have a contradiction no matter how z is broken into uvwxy.Exercise 7.2.1(d)Let n be the pumping-lemma constant and consider z = 0n1n2. We break Z = uvwxy according to the pumping lemma. If vwx consists only of 0's, then uwy has n2 1's and fewer than n 0's; it is not in the language. If vwx has only 1's, then we derive a contradiction similarly. If either v or x has both 0's and 1's, then uv2wx2y is not in 0*1*, and thus could not be in the language.Finally, consider the case where v consists of 0's only, say k 0's, and x consists of m 1's only, where k and m are both positive. Then for all i, uv i+1wx i+1y consists of n + ik 0's and n2 + im 1's. If the number of 1's is always to be the square of the number of 0's, we must have, for some positive k and m: (n+ik)2 = n2+ im, or 2ink + i2k2= im. But the left side grows quadratically in i, while the right side grows linearly, and so this equality for all i is impossible. We conclude that for at least some i, uv i+1wx i+1y is not in the language and have thus derived a contradiction in all cases.Exercise 7.2.2(b)It could be that, when the adversary breaks z = uvwxy, v = 0k and x = 1k. Then, for all i, uv i wx i y is in the language.Exercise 7.2.2(c)The adversary could choose z = uvwxy so that v and x are single symbols, on either side of the center. That is, |u| = |y|, and w is either epsilon (if z is of even length) or the single, middle symbol (if z is of odd length). Since z is a palindrome, v and x will be the same symbol. Then uv i wx i y is always a palindrome.。
形式语言与自动机答案蒋宗礼
形式语言与自动机答案蒋宗礼【篇一：形式语言第四章参考答案(蒋宗礼)】p> 解：所求正则表达式为：(0+1)*。
+⑵ {0, 1}。
解：所求正则表达式为：(0+1)+。
⑶ { x│x∈{0,1}且x中不含形如00的子串 }。
解：根据第三章构造的fa，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷ { x│x∈{0,1}*且x中不含形如00的子串 }。
++ +q1为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)* q2为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*q3为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)* q4为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
由题设，x=0时，│x│=1，模5是1，不符合条件，所以不必增加关于它的状态。
下面对每一个状态考虑输入0和1时的状态转移。
q: 输入1，模5是1，进入q1。
+q0: 设x=5n。
输入0，x=5n*2=10n，模5是0，故进入q0输入1，x=5n*2+1=10n+1，模5是1，故进入q1q1：设x=5n+1。
输入0，x=(5n+1)*2=10n+2，模5是2，故进入q2输入1，x=(5n+1)*2+1=10n+3，模5是3，故进入q3 q2：设x=5n+2。
形式语言与自动机理论-蒋宗礼-第四章参考答案
1.写出表示下列语言的正则表达式。
（吴贤珺02282047）⑴{0, 1}*。
解：所求正则表达式为：(0+1)*。
⑵{0, 1}+。
解：所求正则表达式为：(0+1)+。
⑶{ x│x∈{0,1}+ 且x中不含形如00的子串 }。
解：根据第三章构造的FA，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷{ x│x∈{0,1}*且x中不含形如00的子串 }。
解：根据上题的结果，可得所求正则表达式为：ε+1*(01+)*(01+0+1)。
⑸{ x│x∈{0,1}+ 且x中含形如10110的子串 }。
解：所求正则表达式为：(0+1)*10110(0+1)*。
⑹ { x│x∈{0,1}+ 且x中不含形如10110的子串 }。
解：根据第三章的习题，接受x的FA为：要求该FA对应的正则表达式，分别以q0、q1、q2、q3、q4为终结状态考虑：q为终态时的正则表达式：(0*(11*0(10)*(ε+111*11*0(10)*)0)*)*q为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)*1q为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*2q为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)*3q为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*4将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}+ 且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
形式语言与自动机课后习题答案
形式语言与自动机课后习题答案第二章4•找出右线性文法，能构成长度为1至5个字符且以字母为首的字符串。
答：G={N,T,P,S}其中N={S,A,B,C,D} T={x,y} 其中x€ {所有字母} y € {所有的字符} P如下:4 x S f xA A f y A f yB4 y B f yC C f y C f yD D f y 6 •构造上下文无关文法能够产生L={ 3 / {a,b}*且3中a的个数是b的两倍}答: G={N,T,P,S}其中N={S} T={a,b} P 如下：S f aab S f aba S f baaS f aabS S f aaSbS f aSab S f SaabS f abaS S f abSaS f aSba S f SabaS f baaS S f baSaS f bSaa S f Sbaa7 •找出由下列各组生成式产生的语言(起始符为S)⑴ S f SaS S f b⑵ S f aSb S f c(3) S f a S f aE E f aS答：(1) b(ab) n /n > 0}或者L={(ba) n b /n > 0}(2)L={a n cb n /n > 0}(3)L={a2n+1 /n > 0}第三章1.下列集合是否为正则集，若是正则集写出其正则式。
(1)含有偶数个a和奇数个b的{a,b}*上的字符串集合(2)含有相同个数a和b的字符串集合(3)不含子串aba的{a,b}*上的字符串集合答：(1)是正则集，自动机如下b b b b⑵ 不是正则集，用泵浦引理可以证明，具体见17题(2)(3) 是正则集先看L '为包含子串aba的{a,b}*上的字符串集合显然这是正则集，可以写出表达式和画出自动机。
(略) 则不包含子串aba的{a,b}*上的字符串集合L是L'的非。
根据正则集的性质，L也是正则集。
4．对下列文法的生成式，找出其正则式(1)G=({SAB,C,D},{a,b,c,d},P,S), 生成式P如下：S f aA S —BA f abS A —bBB f b B f cCC fD D f bBD f d(2)G=({SAB,C,D},{a,b,c,d},P,S), 生成式P如下：S f aA S f BA f cC A f bBB f bB B f aC fD C f abBD f d答：(1) 由生成式得：S=aA+B ①A=abS+bB ②B=b+cC ③C=D ④D=d+bB ⑤③④⑤式化简消去CD得到B=b+c(d+bB)即B=cbB+cd+b =>B=(cb)*(cd+b) ⑥将②⑥代入①S=aabS+ab(cb)*(cd+b)+(cb)*(cd+b) =>S=(aab)*(ab+ & )(cb)*(cd+b)(2) 由生成式得：S=aA+B ①A=bB+cC②B=a+bB ③C=D+abB④D=dB ⑤由③得B=b*a ⑥将⑤⑥代入④ C=d+abb*a=d+ab+a ⑦将⑥⑦代入② A=b+a+c(d+b+a) ⑧将⑥⑧代入① S=a(b +a+c(d+ab +a))+b*a++=ab a+acd+acaba+b*a5. 为下列正则集，构造右线性文法：(1){a,b}*(2)以abb 结尾的由a 和b 组成的所有字符串的集合(3)以b为首后跟若干个a的字符串的集合(4)含有两个相继a和两个相继b的由a和b组成的所有字符串集合答：(1)右线性文法G=({S},{a,b},P,S)P: S f aS S f bS S(2)右线性文法G=({S},{a,b},P,S)P: S f aS S f bS S f abb(3)此正则集为{ba*}右线性文法G=({S,A},{a,b},P,S)P: S f bA A f aA A f£(4)此正则集为{{a,b}*aa{a,b}*bb{a,b}*, {a,b}*bb{a,b}*aa{a,b}*} 右线性文法G=({SAB,C},{a,b},P,S)P: S f aS/bS/aaA/bbBA f aA/bA/bbCB f aB/bB/aaCC f aC/bC/ &7.设正则集为a(ba)*(1)构造右线性文法⑵找出(1)中文法的有限自b动机答：(1)右线性文法G=({S,A},{a,b},P,S)P: S f aA A f bS A f&(2)自动机如下：(p2是终结状态)9.对应图(a) (b)的状态转换图写出正则式。
形式语言与自动机 习题答案(部分)
a）语言{ww:w∈{a,b}*}的文法G＝（V，∑，R，S）如下： V＝{S，a,b,A,B,C,Ta,Tb,Tc} ∑＝{a,b} R={ S -> CC, C -> AC | BC | Tc BTc -> Tcb ATc -> Tca Tc -> e } 或 文法G[S]： S→CD Ab→bA C→aCA Ba→aB C→bCB Bb→bB AD→aD C→ε BD→bD D→ε Aa→bD 或 S→aAS|bBS|aAE|bBE Aa→aA Ab→bA Ba→aB AE→Ea|a BE→Eb|b L(G)={ww|w∈{a,b}*}
4331把两个带头分别移动第一个带头向左移动到带头第二个带头向右移动直到发现空格为2向右移动第一个带头记录下字符的值
2.2.3
2.2.9
2.3.4
3.5.14
• (a):是上下文相关的，因为m,n,p之间必存在二者 相等的情况，如果a的个数确定了，则b或者c的个 数也是确定的。 • (b):是上下文相关的，因为当a的个数确定时，b和 c的个数是不能和a相同的，即n和p的取值有了限 制。 • (c):是上下文相关的，a的个数限定了b和c的个数。 • (d):是上下文相关的，当a和b的个数相等的情况下， 则限定了c的个数不能和a，b相同。 • (e):是上下文相关的，若要满足 |w1|=|w2|…..=|wn|>=2,则|w|的值限定了必须是可 被整除的数（素数）。
4.1.10
• 功能是： • 首先找到第二个和第三个字符，要求不能 是空格，然后记录下，再分别填入之后的 第一个空格和第二个空格所位置。
4.3.3
• （1）把两个带头分别移动，第一个带头向左移动 到带头，第二个带头向右移动，直到发现空格为 止。 • （2）向右移动第一个带头，记录下字符的值。判 断有没有遇到空格，若遇到空格，则跳至（）， 否则，跳至（3）。 • （3）向右移动第二个带头，并复制下已记录的字 符，再跳至（2）。 • （4）第一个带头遇到空格，说明w已经复制完， 故停机。
《形式语言与自动机》(王柏、杨娟编著)课后习题答案
形式语言与自动机课后习题答案第二章4．找出右线性文法，能构成长度为1至5个字符且以字母为首得字符串。
答：G={N,T,P,S}其中N={S,A,B,C,D} T={x,y} 其中x∈{所有字母} y∈{所有得字符} P如下: S→x S→xA A→y A→yBB→y B→yC C→y C→yD D→y6．构造上下文无关文法能够产生L={ω/ω∈{a,b}*且ω中a得个数就是b得两倍}答：G={N,T,P,S}其中N={S} T={a,b} P如下:S→aab S→aba S→baaS→aabS S→aaSb S→aSab S→SaabS→abaS S→abSa S→aSba S→SabaS→baaS S→baSa S→bSaa S→Sbaa7．找出由下列各组生成式产生得语言（起始符为S）(1)S→SaS S→b(2)S→aSb S→c(3)S→a S→aE E→aS答：（1）b(ab)n /n≥0}或者L={(ba)n b/n≥0}(2) L={a n cb n /n≥0}(3)L={a2n+1 /n≥0}第三章1．下列集合就是否为正则集，若就是正则集写出其正则式。
（1）含有偶数个a与奇数个b得{a,b}*上得字符串集合（2）含有相同个数a与b得字符串集合（3）不含子串aba得{a,b}*上得字符串集合答：（1）就是正则集，自动机如下题（2）。
(3) 就是正则集先瞧L’为包含子串aba得{a,b}*上得字符串集合显然这就是正则集，可以写出表达式与画出自动机。
（略）则不包含子串aba得{a,b}*上得字符串集合L就是L’得非。
根据正则集得性质，L也就是正则集。
4．对下列文法得生成式，找出其正则式（1）G=({S,A,B,C,D},{a,b,c,d},P,S),生成式P如下：S→aA S→BA→abS A→bBB→b B→cCC→D D→bBD→d（2）G=({S,A,B,C,D},{a,b,c,d},P,S),生成式P如下：S→aA S→BA→cC A→bBB→bB B→aC→D C→abBD→d答：(1) 由生成式得：S=aA+B ①A=abS+bB ②B=b+cC ③C=D ④D=d+bB ⑤③④⑤式化简消去CD，得到B=b+c(d+bB)即B=cbB+cd+b =>B=(cb)*(cd+b) ⑥将②⑥代入①S=aabS+ab(cb)*(cd+b)+(cb)*(cd+b) =>S=(aab)*(ab+ε)(cb)*(cd+b) (2) 由生成式得：S=aA+B ①A=bB+cC ②B=a+bB ③C=D+abB ④D=dB ⑤由③得 B=b*a ⑥将⑤⑥代入④ C=d+abb*a=d+ab+a ⑦将⑥⑦代入② A=b+a+c(d+b+a) ⑧将⑥⑧代入① S=a(b+a+c(d+ab+a))+b*a=ab+a+acd+acab+a+b*a5、为下列正则集，构造右线性文法：(1){a,b}*(2)以abb结尾得由a与b组成得所有字符串得集合(3)以b为首后跟若干个a得字符串得集合(4)含有两个相继a与两个相继b得由a与b组成得所有字符串集合答：（1）右线性文法G=({S},{a,b},P,S)P: S→aS S→bS S→ε(2) 右线性文法G=({S},{a,b},P,S)P: S→aS S→bS S→abb(3) 此正则集为{ba*}右线性文法G=({S,A},{a,b},P,S)P: S→bA A→aA A→ε(4) 此正则集为{{a,b}*aa{a,b}*bb{a,b}*, {a,b}*bb{a,b}*aa{a,b}*}右线性文法G=({S,A,B,C},{a,b},P,S)P: S→aS/bS/aaA/bbBA→aA/bA/bbCB→aB/bB/aaCC→aC/bC/ε7、设正则集为a(b a)*(1)构造右线性文法(2)找出（1）中文法得有限自b动机答：（1）右线性文法G=({S,A},{a,b},P,S)P: S→aA A→bS A→ε（2）自动机如下：)9、对应图（a）(b)得状态转换图写出正则式。
形式语言与自动机理论-蒋宗礼-第四章参考答案
1.写出表示下列语言的正则表达式。
（吴贤珺02282047）⑴{0, 1}*。
解：所求正则表达式为：(0+1)*。
⑵{0, 1}+。
解：所求正则表达式为：(0+1)+。
⑶{ x│x∈{0,1}+ 且x中不含形如00的子串 }。
解：根据第三章构造的FA，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷{ x│x∈{0,1}*且x中不含形如00的子串 }。
解：根据上题的结果，可得所求正则表达式为：ε+1*(01+)*(01+0+1)。
⑸{ x│x∈{0,1}+ 且x中含形如10110的子串 }。
解：所求正则表达式为：(0+1)*10110(0+1)*。
⑹ { x│x∈{0,1}+ 且x中不含形如10110的子串 }。
解：根据第三章的习题，接受x的FA为：要求该FA对应的正则表达式，分别以q0、q1、q2、q3、q4为终结状态考虑：q为终态时的正则表达式：(0*(11*0(10)*(ε+111*11*0(10)*)0)*)*q为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)*1q为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*2q为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)*3q为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*4将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}+ 且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
形式语言与自动机理论-蒋宗礼-第四章参考答案
1.写出表示下列语言的正则表达式。
（吴贤珺02282047）⑴{0, 1}*。
解：所求正则表达式为：(0+1)*。
⑵{0, 1}+。
解：所求正则表达式为：(0+1)+。
⑶{ x│x∈{0,1}+ 且x中不含形如00的子串 }。
解：根据第三章构造的FA，可得所求正则表达式为：1*(01+)*(01+0+1)。
⑷{ x│x∈{0,1}*且x中不含形如00的子串 }。
解：根据上题的结果，可得所求正则表达式为：ε+1*(01+)*(01+0+1)。
⑸{ x│x∈{0,1}+ 且x中含形如10110的子串 }。
解：所求正则表达式为：(0+1)*10110(0+1)*。
⑹ { x│x∈{0,1}+ 且x中不含形如10110的子串 }。
解：根据第三章的习题，接受x的FA为：要求该FA对应的正则表达式，分别以q0、q1、q2、q3、q4为终结状态考虑：q为终态时的正则表达式：(0*(11*0(10)*(ε+111*11*0(10)*)0)*)*q为终态时的正则表达式：0*1(1*(0(10)*111*1)*(0(10)*00*1)*)*1q为终态时的正则表达式：0*11*0((10)*(111*11*0)*(00*11*0)*)*2q为终态时的正则表达式：0*11*0(10)*1(11*11*0((10)*(00*11*0)*)*1)*3q为终态时的正则表达式：0*11*0(10)*11(1*(11*0((00*11*0)*(10)*)*11)*)*4将以上5个正则表达式用“+”号相连，就得到所要求的正则表达式。
⑺ { x│x∈{0,1}+ 且当把x看成二进制数时，x模5与3同余和x为0时，│x│=1且x≠0时，x的首字符为1}。
解：先画出状态转移图，设置5个状态q0、q1、q2、q3、q4，分别表示除5的余数是0、1、2、3、4的情形。
另外，设置一个开始状态q.由于要求x模5和3同余，而3模5余3,故只有q3可以作为终态。
形式语言与自动机理论--第三章参考答案
第三章作业答案1．已知DFA M1与M2如图3－18所示。
(xxxx 02282068) (1) 请分别给出它们在处理字符串1011001的过程中经过的状态序列。
(2) 请给出它们的形式描述。
Sq q1q q图3－18 两个不同的DFA解答：(1)M1在处理1011001的过程中经过的状态序列为q0q3q1q3q2q3q1q3;M2在处理1011001的过程中经过的状态序列为q0q2q3q1q3q2q3q1;(2)考虑到用形式语言表示,用自然语言似乎不是那么容易,所以用图上作业法把它们用正则表达式来描述:M1: [01+(00+1)(11+0)][11+(10+0)(11+0)]* M2: (01+1+000){(01)*+[(001+11)(01+1+000)]*} *******************************************************************************2．构造下列语言的DFA( xx02282085 ) （1）{0，1}*，1（2）{0，1}+，1（3）{x|x{0，1}+且x 中不含00的串}（设置一个陷阱状态，一旦发现有00的子串，就进入陷阱状态）（4）{ x|x{0，1}*且x中不含00的串}（可接受空字符串，所以初始状态也是接受状态）（5）{x|x{0，1}+且x中含形如10110的子串}（6）{x|x{0，1}+且x中不含形如10110的子串}（设置一个陷阱状态，一旦发现有00的子串，就进入陷阱状态）（7）{x|x{0，1}+且当把x看成二进制时，x模5和3同余，要求当x为0时，|x|=1,且x0时，x的首字符为1 }1.以0开头的串不被接受，故设置陷阱状态，当DFA在启动状态读入的符号为0，则进入陷阱状态2.设置7个状态：开始状态qs,q0:除以5余0的等价类，q1：除以5余1的等价类,q2:除以5余2的等价类，q3:除以5余3的等价类，q4:除以5余4的等价类，接受状态qt3.状态转移表为（8）{x|x{0，1}+且x的第十个字符为1}（设置一个陷阱状态，一旦发现x的第十个字符为0，进入陷阱状态）（9）{x|x{0，1}+且x以0开头以1结尾}（设置陷阱状态，当第一个字符为1时，进入陷阱状态）（10）{x|x{0，1}+且xxx至少含有两个1}（11）{x|x{0，1}+且如果x以1结尾，则它的xx为偶数；如果x以0结尾，则它的xx为奇数}可将{0，1}+的字符串分为4个等价类。
形式语言与自动机理论-蒋宗礼-参考答案
ij求索-百度文库2.1回答下面的问题：（周期律02282067）（1）在文法中，终极符号和非终极符号各起什么作用？/终结符号是一个文法所产生的语言中句子的中出现的字符,他决上了一个文法的产生语言中字符的范围。
/ 非终结符号又叫做一个语法变量，它表示一个语法范畴，文法中每一个产生式的左部至少要还有一个非终结符号，（二，三型文法要求更严，只允许左部为一个非终结符号）他是推导或归约的核心。
（2）文法的语法范畴有什么意义？开始符号所对应的语法范畴有什么特殊意义？/ 文法的非终结符号A所对应的语法范畴代表着一个集合L （A）,此集合由文法产生式中关于A的产生式推导实现的/ 开始符号所对应的语法范畴则为文法G = （V, T, P, S}所产生的语言L （G）*={ vvl w e 厂且S =► w }（3）在文法中，除了的变量可以对应一个终极符号行的集合外，按照类似的对应方法，一个字符串也可以对应一个终极符号行集合，这个集合表达什么意义？/字符串对应的终极符号行集合表示这个字符串所能推导到的终极字符串集合，为某个句型的语言。
（4）文法中的归约和推导有什么不同？/ 推导：文法G = {V, T, P, S},如果则称gd在G中推导岀了汐5。
/ 归约：文法G={V, T, P, S},如果则称汐5在G中归约到*7》。
/ 这他们的左义，我个人理解两个槪念从不同角度看待文法中的产生式，推导是自上而下（从产生式的左边到右边），而归约是自下而上（从产生式的右边到左边），体现到具体实际中，如编译中语法分析时语法树的建立，递归下降，LL （1）等分析法采用自开始符号向下推导识別输入代码生成语法树，对应的LR （1）, LALR等分析法则是采用自输入代码（相当于文法中语言的句子）自底向上归约到开始符号建立语法树，各有优劣。
（5）为什么要求左义语言的字母表上的语言为一个非空有穷集合？/ 菲空：根据字母表幕的立义：工°={£},£为字母表中0个字符组成的。
形式语言与自动机理论-蒋宗礼-第三章参考答案
第三章作业答案1．已知DFA M1与M2如图3－18所示。
(敖雪峰 02282068)(1) 请分别给出它们在处理字符串1011001的过程中经过的状态序列。
(2) 请给出它们的形式描述。
Sq q1图3－18 两个不同的DFA解答：(1)M1在处理1011001的过程中经过的状态序列为q 0q 3q 1q 3q 2q 3q 1q 3; M2在处理1011001的过程中经过的状态序列为q 0q 2q 3q 1q 3q 2q 3q 1;(2)考虑到用形式语言表示,用自然语言似乎不是那么容易,所以用图上作业法把它们用正则表达式来描述:M1: [01+(00+1)(11+0)][11+(10+0)(11+0)]*M2: (01+1+000){(01)*+[(001+11)(01+1+000)]*}******************************************************************************* 2．构造下列语言的DFA( 陶文婧 02282085 ) （1）{0，1}*，1（2）{0，1}+，1（3）{x|x {0，1}+且x 中不含00的串}（设置一个陷阱状态，一旦发现有00的子串，就进入陷阱状态）（4）{ x|x∈{0，1}*且x中不含00的串}（可接受空字符串，所以初始状态也是接受状态）（5）{x|x∈{0，1}+且x中含形如10110的子串}（6）{x|x∈{0，1}+且x中不含形如10110的子串}（设置一个陷阱状态，一旦发现有00的子串，就进入陷阱状态）（7）{x|x∈{0，1}+且当把x看成二进制时，x模5和3同余，要求当x为0时，|x|=1,且x≠0时，x的首字符为1 }1.以0开头的串不被接受，故设置陷阱状态，当DFA在启动状态读入的符号为0，则进入陷阱状态2.设置7个状态：开始状态q s,q0:除以5余0的等价类，q1：除以5余1的等价类,q2:除以5余2的等价类，q3:除以5余3的等价类，q4:除以5余4的等价类，接受状态q t（8）{x|x∈{0，1}+且x的第十个字符为1}（设置一个陷阱状态，一旦发现x的第十个字符为0，进入陷阱状态）（9）{x|x∈{0，1}+且x以0开头以1结尾}（设置陷阱状态，当第一个字符为1时，进入陷阱状态）（10）{x|x∈{0，1}+且x中至少含有两个1}（11）{x|x∈{0，1}+且如果x以1结尾，则它的长度为偶数；如果x以0结尾，则它的长度为奇数}可将{0，1}+的字符串分为4个等价类。
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形式语言与自动机课后习题答案第二章4．找出右线性文法，能构成长度为1至5个字符且以字母为首的字符串。
答：G={N,T,P,S}其中N={S,A,B,C,D} T={x,y} 其中x∈{所有字母} y∈{所有的字符} P如下: S→x S→xA A→y A→yBB→y B→yC C→y C→yD D→y6．构造上下文无关文法能够产生L={ω/ω∈{a,b}*且ω中a的个数是b的两倍}答：G={N,T,P,S}其中N={S} T={a,b} P如下:S→aab S→aba S→baaS→aabS S→aaSb S→aSab S→SaabS→abaS S→abSa S→aSba S→SabaS→baaS S→baSa S→bSaa S→Sbaa7．找出由下列各组生成式产生的语言（起始符为S）(1)S→SaS S→b(2)S→aSb S→c(3)S→a S→aE E→aS答：（1）b(ab)n /n≥0}或者L={(ba)n b/n≥0}(2) L={a n cb n /n≥0}(3)L={a2n+1 /n≥0}第三章1．下列集合是否为正则集，若是正则集写出其正则式。
（1）含有偶数个a和奇数个b的{a,b}*上的字符串集合（2）含有相同个数a和b的字符串集合（3）不含子串aba的{a,b}*上的字符串集合答：（1）是正则集，自动机如下b b b b(2) 不是正则集，用泵浦引理可以证明，具体见17题（2）。
(3) 是正则集先看L’为包含子串aba的{a,b}*上的字符串集合显然这是正则集，可以写出表达式和画出自动机。
（略）则不包含子串aba的{a,b}*上的字符串集合L是L’的非。
根据正则集的性质，L也是正则集。
4．对下列文法的生成式，找出其正则式（1）G=({S,A,B,C,D},{a,b,c,d},P,S),生成式P如下：S→aA S→BA→abS A→bBB→b B→cCC→D D→bBD→d（2）G=({S,A,B,C,D},{a,b,c,d},P,S),生成式P如下：S→aA S→BA→cC A→bBB→bB B→aC→D C→abBD→d答：(1) 由生成式得：S=aA+B ①A=abS+bB ②B=b+cC ③C=D ④D=d+bB ⑤③④⑤式化简消去CD，得到B=b+c(d+bB)即B=cbB+cd+b =>B=(cb)*(cd+b) ⑥将②⑥代入①S=aabS+ab(cb)*(cd+b)+(cb)*(cd+b) =>S=(aab)*(ab+ε)(cb)*(cd+b) (2) 由生成式得：S=aA+B ①A=bB+cC ②B=a+bB ③C=D+abB ④D=dB ⑤由③得 B=b*a ⑥将⑤⑥代入④ C=d+abb*a=d+ab+a ⑦将⑥⑦代入② A=b+a+c(d+b+a) ⑧将⑥⑧代入① S=a(b+a+c(d+ab+a))+b*a=ab+a+acd+acab+a+b*a5.为下列正则集，构造右线性文法：(1){a,b}*(2)以abb结尾的由a和b组成的所有字符串的集合(3)以b为首后跟若干个a的字符串的集合(4)含有两个相继a和两个相继b的由a和b组成的所有字符串集合答：（1）右线性文法G=({S},{a,b},P,S)P: S→aS S→bS S→ε(2) 右线性文法G=({S},{a,b},P,S)P: S→aS S→bS S→abb(3) 此正则集为{ba*}右线性文法G=({S,A},{a,b},P,S)P: S→bA A→aA A→ε(4) 此正则集为{{a,b}*aa{a,b}*bb{a,b}*, {a,b}*bb{a,b}*aa{a,b}*}右线性文法G=({S,A,B,C},{a,b},P,S)P: S→aS/bS/aaA/bbBA→aA/bA/bbCB→aB/bB/aaCC→aC/bC/ε7.设正则集为a(ba)*(1)构造右线性文法(2)找出（1）中文法的有限自 b动机答：（1）右线性文法G=({S,A},{a,b},P,S)P: S→aA A→bS A→ε（2）自动机如下：(p2是终结状态)9.对应图（a）(b)的状态转换图写出正则式。
（图略）（1）由图可知q0=aq0+bq1+a+εq1=aq2+bq1q0=aq0+bq1+a=>q1=abq1+bq1+aaq0+aa=(b+ab) q1+aaq0+aa=(b+ab) *( aaq0+aa)=>q0=aq0+b(b+ab) *( aaq0+aa ) +a+ε= q0(a+b (b+ab) *aa)+ b(b+ab) *aa+a+ε=(a+b (b+ab) *aa) *((b+ab) *aa+a+ε)=(a+b (b+ab) *aa) *(3)q0=aq1+bq2+a+bq1=aq0+bq2+bq0=aq1+bq0+a=>q1=aq0+baq1+bbq0+ba+b=(ba)*(aq0 +bbq0+ba+b)=>q2=aaq0+abq2+bq0+ab+a=(ab)*(aaq0 +bq0+ ab+a)=>q0=a(ba)*(a+bb) q0 + a(ba)*(ba+b)+b(ab)*(aa+b)q0+ b(ab)*(ab+a)+a+b =[a(ba)*(a+bb) +b(ab)*(aa+b)]* (a(ba)*(ba+b)+ b(ab)*(ab+a)+a+b)10.设字母表T={a,b},找出接受下列语言的DFA：（1）含有3个连续b的所有字符串集合（2）以aa为首的所有字符串集合（3）以aa结尾的所有字符串集合答：（1）M=({q0,q1 q2,q3},{a,b},σ,q0,{q3}),其中σ如下：（2）M=({q0,q1 q2 },{a,b},σ,q0,{q2}),其中σ如下：（3）M=({q0,q1 q2 },{a,b},σ,q0,{q2}),其中σ如下：14构造DFA M1等价于NFA M，NFA M如下：（1）M=({q0,q1 q2,q3},{a,b},σ,q0,{q3}),其中σ如下：σ(q0,a)={q0,q1} σ(q0,b)={q0}σ(q1,a)={q2} σ(q1,b)= {q2 }σ(q2,a)={q3} σ(q2,b)= Φσ(q3,a)={q3} σ(q3,b)= {q3 }（2）M=({q0,q1 q2,q3},{a,b},σ,q0,{ q1,q2}),其中σ如下：σ(q0,a)={q1,q2} σ(q0,b)={q1}σ(q1,a)={q2} σ(q1,b)= {q1,q2 }σ(q2,a)={q3} σ(q2,b)= {q0}σ(q3,a)= Φσ(q3,b)= {q0}答：（1）DFA M1={Q1, {a,b},σ1, [q0],{ [q0,q1,q3]，[q0,q2,q3]，[q0, q1,q2,q3]}其中Q1={[q0],[q0,q1], [q0,q1,q2],[ q0,q2],[ q0,q1, q2,q3],[ q0,q1, q3],[ q0,q2, q3],[ q0,q3]}σ1满足（2）DFA M1={Q1, {a,b},σ1, [q0],{ [q1],[q3], [q1,q3],[q0,q1,q2],[q1,q2] ,[q1,q2,q3],[q2,q3]}其中Q1 ={[q0],[q1,q3], [q1],[q2],[ q0,q1,q2],[q1,q2],[q3], [q1,q2,q3],[q2,q3]}σ1满足15. 15.对下面矩阵表示的ε-NFA(1)给出该自动机接收的所有长度为3的串(2)将此ε-NFA转换为没有ε的NFA答：（1）可被接受的的串共 23个，分别为aac, abc, acc, bac, bbc, bcc, cac, cbc, ccc, caa, cab, cba, cbb, cca, ccb, bba, aca, acb, bca, bcb, bab, bbb, abb （2）ε-NFA：M=({p,q,r},{a,b,c},σ,p,r) 其中σ如表格所示。
因为ε-closure(p)= Φ则设不含ε的NFA M1=({p,q,r},{a,b,c},σ1,p,r)σ1(p,a)=σ’(p,a)=ε-closure(σ(σ’(p,ε),a))={p}σ1(p,b)=σ’(p,b)=ε-closure(σ(σ’(p,ε),b))={p,q}σ1(p,c)=σ’(p,c)=ε-closure(σ(σ’(p,ε),c))={p,q,r}σ1(q,a)=σ’(q,a)=ε-closure(σ(σ’(q,ε),a))={p,q}σ1(q,b)=σ’(q,b)=ε-closure(σ(σ’(q,ε),b))={p,q,r}σ1(q,c)=σ’(q,c)=ε-closure(σ(σ’(q,ε),c))={p,q,r}σ1(r,a)=σ’(r,a)=ε-closure(σ(σ’(r,ε),a))={p,q,r}σ1(r,b)=σ’(r,b)=ε-closure(σ(σ’(r,ε),b))={p,q,r}σ1(r,c)=σ’(r,c)=ε-closure(σ(σ’(r,ε),c))={p,q,r}图示如下：(r为终止状态)a,b,c16．设NFA M=({q0,q1},{a,b},σ,q0,{q1}),其中σ如下：σ(q0,a)={q0,q1} σ(q0,b)={q1}σ(q1,a)= Φσ(q1,b)= {q0, q1}构造相应的DFA M1，并进行化简答：构造一个相应的DFA M1={Q1, {a,b},σ1, [q0],{ [q1]，[q0,q1]}其中Q1 ={[q0]，[q1]，[q0,q1]}σ1满足由于该DFA已是最简，故不用化简17.使用泵浦引理，证明下列集合不是正则集：（1）由文法G的生成式S→aSbS/c产生的语言L(G)（2）{ω/ω∈{a,b}*且ω有相同个数的a和b}（3）{a k ca k/k≥1}（4）{ωω/ω∈{a,b}*}证明：（1）在L(G)中，a的个数与b的个数相等假设L(G)是正则集，对于足够大的k取ω= a k (cb)k c令ω=ω1ω0ω2因为|ω0|>0 |ω1ω0|≤k 存在ω0使ω1ω0iω2∈L所以对于任意ω0只能取ω0=a n n∈(0,k)则ω1ω0iω2= a k–n(a n)i(cb)k c 在i不等于0时不属于L与假设矛盾。
则L(G)不是正则集（2）假设该集合是正则集，对于足够大的k取ω= a k b k令ω=ω1ω0ω2因为|ω0|>0 |ω1ω0|≤k 存在ω0使ω1ω0iω2∈L所以对于任意ω0只能取ω0=a n n∈(0,k)则ω1ω0iω2= a k–n(a n)i b k在i不等于0时a与b的个数不同，不属于该集合与假设矛盾。