自动机答案—翻译.doc

Solutions for Section 2.2
Exercise 2.2.1(a)
States correspond to the eight combinations of switch positions, and also must indicate whether the previous roll came out at D, i.e., whether the previous input was accepted. Let 0 represent a position to the left (as in the diagram) and 1 a position to the right. Each state can be represented by a sequence of three 0's or 1's, representing the directions of the three switches, in order from left to right. We follow these three bits by either a indicating it is an accepting state or r, indicating rejection. Of the 16 possible states, it turns out that only 13 are accessible from the initial state, 000r. Here is the transition table:

杠杆可能出现8种情况，影响着最终状态。并且也要说明，前面一个大理石球是否从D滚出，也就是说，前一个输入是否被接受。令 0 代表向左方的状态（如图表）， 1 代表向右方。这三个杠杆的每一个状态都可以用三个数（0或1）组成的序列表示。这个序列后面跟着字母a或者r。a代表接受状态，r代表拒绝状态。16种可能的状态中，只有13种是从初始状态000r可达的。下面它的有穷自动机的转移表。

A
B
->000r
100r
011r
*000a
100r
011r
*001a
101r
000a
010r
110r
001a
*010a
110r
001a
011r
111r
010a
100r
010r
111r
*100a
010r
111r
101r
011r
100a
*101a
011r
100a
110r
000a
101a
*110a
000a
101a
111r
001a
110a
Exercise 2.2.2
The statement to be proved is δ-hat(q,xy) = δ-hat(δ-hat(q,x),y), and we proceed by induction on the length of y.
证明：通过对|y|进行归纳，来证明(q , xy)=((q , x) , y) ，具体过程如下：
Basis: If y = ε, then the statement is δ-hat(q,x) = δ-hat(δ-hat(q,x),ε). This statement follows from the basis in the definition of δ-hat. Note that in applying this definition, we must treat δ-hat(q,x) as if it were just a state, say p. Then, the statement to be proved is p = δ-hat(p,ε), which is easy to recognize as the basis in the definition of δ-hat.
基础: =0，则y=ε。那么需证(q,x)=((q ,x),ε)，记p=(q,x),命题变为 p=(p ,ε), 由的定义知这显然成立。
Induction: Assume the statement for strings shorter than y, and break y = za, where a is the last symbol of y. The steps converting δ-hat(δ-hat(q,x),y) to δ-hat(q,xy) are summarized in the following table:
归纳: 假设命题对于比 y　短的串成立, 且y = za, 其中 a 是y的结尾符号。((q,x),y) 到(q,xy) 的变换总结在下表中:

Expression 表达式
Reason 原因
((q,x),y)

Start 开始
((q,x),za)
y=za by assumption 由假设y=za
δ(((q,x),z),a)
Definition of δ-hat, treating δ-hat(q,x) as a state
的定义, 把(q,x) 看作是一个状态
δ((q,xz),a)
Inductive hypothesis 归纳假设
(q,xza)
Definition of δ-hat 的定义
(q,xy)
y=za
Exercise 2.2.4(a)
The intuitive meanings of states A, B, and C are that the string seen so far ends in 0, 1, or at least 2 zeros.
状态 A, B，C分别表示以１,０和00结尾的串的状态。

0
1
->A
B
A
B
C
A
*C
C
A
Exercise 2.2.6(a)
The trick is to realize that reading another bit either multiplies the number seen so far by 2 (if it is a 0), or multiplies by 2 and then adds 1 (if it is a 1). We don't need to remember the entire number seen --- just its remainder when divided by 5. That is, if we have any number of the form 5a+b, where b is the remainder, between 0 and 4, then 2(5a+b) = 10a+2b. Since 10a is surely divisible by 5, the remainder of 10a+2b is the same as the remainder of 2b when divided by 5. Since b, is 0, 1, 2, 3, or 4, we can tabulate the answers easily. The same idea holds if we want to consider what happens to 5a+b if we multiply by 2 and add 1.
对于一个二进制整数，如果读入一个比特０，其值等于原数乘以２；否则等于原数乘以２再加以1。而任意一个数均可写成形如5a+b，其中a任意，0<= b <=4，那么输入0，原数变为2(5a+b) = 10a+2b，由于10a 是5的倍数,，因此10a+2b 除以５的余数与2b相同。输入1，则得2(5a+b)+1类似。因此对于所有的数只要记住它被５除的余数就可以。由于b是 0, 1, 2, 3或者 4, 我们可以容易得到该DPA的转移表，具体如下：
The table below shows this automaton. State qi means that the input seen so far has remainder i when divided by 5.
其中状态qi 代表输入串被５除的余数i 的状态。

0
1
->*q0
q0
q1
q1
q2
q3
q2
q4
q0
q3
q1
q2
q4
q3
q4
There is a small matter, however, that this automaton accepts strings with leading 0's. Since the problem calls for accepting only those strings that begin with 1, we need an additional state s, the start state, and an additional ``dead state'' d. If, in state s, we see a 1 first, we act like q0; i.e., we go to state q1. However, if the first input is 0, we should never accept, so we go to state d, which we never leave. The complete automaton is:
但是上述自动机仍接受以０开头的字符串。因为题目要求只接受以１开头的串，可增加一个初始状态s和“死亡状态”d。在状态初始状态s, 若看到１，则转到状态q1；若看到０, 则直接转到状态d，识别终止。所求自动机如下:

0
1
->s
d
q1
*q0
q0
q1
q1
q2
q3
q2
q4
q0
q3
q1
q2
q4
q3
q4
d
d
d
Exercise 2.2.9
Part (a) is an easy induction on the length of w, starting at length 1.
Basis: |w| = 1. Then δ-hat(q0,w) = δ-hat(qf,w), because w is a single symbol, and δ-hat agrees with δ on single symbols.
Induction: Let w = za, so the inductive hypothesis applies to z. Then δ-hat(q0,w) = δ-hat(q0,za) = δ(δ-hat(q0,z),a) = δ(δ-hat(qf,z),a) [by the inductive hypothesis] = δ-hat(qf,za) = δ-hat(qf,w).
证明：a) 通过对w长度的归纳证明。
基础: 若|w| = 1，则w 是一个符号此时需证(q0,w) = (qf,w), 而对于单个符号扩展转移函数与转移函数δ的作用是一样的，得证。
归纳: 令w = za, 假设对于z命题(q0,z) = (qf,z)成立。那么(q0,w) = (q0,za) = δ((q0,z),a) = δ( (qf,z),a) [由归纳假设] = (qf,za) = (qf,w).
For part (b), we know that δ-hat(q0,x) = qf. Since xε, we know by part (a) that δ-hat(qf,x) = qf. It is then a simple induction on k to show that δ-hat(q0,xk) = qf.
Basis: For k=1 the statement is given.
Induction: Assume the statement for k-1; i.e., δ-hat(q0,xSUP>k-1) = qf. Using Exercise 2.2.2, δ-hat(q0,xk) = δ-hat(δ-hat(q0,xk-1),x) = δ-hat(qf,x) [by the inductive hypothesis] = qf [by (a)].
b) x是属于L(A)的非空串，也即串x被接收，因此(q0,x) = qf ，则由 a)知(qf,x) =(q0,x)= qf 。现在通过对k 的归纳来证明(q0,xk) = qf 。
基础: k=1 时，需证(q0,x) = qf ，由已知可得。
归纳：假设对于k-1命题成立，也就是说，(q0,xk-1) = qf 。由练习 2.2.2, (q0,xk) =((q0,xk-1),x) = (qf,x) [由归纳假设] = qf [由(a)]。
Exercise 2.2.10
The automaton tells whether the number of 1's seen is even (state A) or odd (state B), accepting in the latter case. It is an easy induction on |w| to show that dh(A,w) = A if and only if w has an even number of 1's.
Basis: |w| = 0. Then w, the empty string surely has an even number of 1's, namely zero 1's, and (A,w) = A.
Induction: Assume the statement for strings shorter than w. Then w = za, where a is either 0 or 1.
Case 1: a = 0. If w has an even number of 1's, so does z. By the inductive hypothesis, (A,z) = A. The transitions of the DFA tell us (A,w) = A. If w has an odd number of 1's, then so does z. By the inductive hypothesis, δ-hat(A,z) = B, and the transitions of the DFA tell us δ-hat(A,w) = B. Thus, in this case, δ-hat(A,w) = A if and only if w has an even number of 1's.
Case 2: a = 1. If w has an even number of 1's, then z has an odd number of 1's. By the inductive hypothesis, δ-hat(A,z) = B. The transitions of the DFA tell us δ-hat(A,w) = A. If w has an odd number of 1's, then z has an even number of 1's. By the inductive hypothesis, δ-hat(A,z) = A, and the transitions of the DFA tell us δ-hat(A,w) = B. Thus, in this case as well, δ-hat(A,w) = A if and only if w has an even number of 1's.
这个自动机表示，状态A表示偶数个1，状态B表示奇数个1，不管串有偶数个还是奇数个1，都会被接受。当且仅当串w中有偶数个1时， (A,w) = A.。用归纳法证明如下
基础: |w| = 0。空串当然有偶数个 1 ，即0个 1，且 (A,w) = A.
归纳：假设对于比w 短的串命题成立。令 w = za, 其中 a 为 0 或1。
情形1： a = 0. 如果w有偶数个 1, 则z有偶数个1。由归纳假设， (A,z) = A。由转移表的DFA知(A,w) = A.如果w有奇数个1, 则z有奇数个1. 由归纳假设， (A,z) = B, 由转移表的 DFA 知 (A,w) = B. 因此这种情况下(A,w) = A 当且仅当 w 有偶数个 1。
情形2： a = 1. 如果w有偶数个 1, 则z有奇数个1。由归纳假设， (A,z) = B. 由转移表的DFA知 (A,w) = A. 如果w有奇数个 1, 则z有偶数个1。由归纳假设, (A,z) = A, 由转移表的DFA知 (A,w) = B. 因此这种情况下(A,w) = A 当且仅当 w 有偶数个 1.
综合上述情形，命题得证。
Solutions for Section 2.3
Exercise 2.3.1
Here 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}.
下表就是利用子集构造法将NFA转化成的DFA。其中构造的子集有：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}.

0
1
->A
B
A
B
D
C
C
E
A
D
F
C
*E
F
G
*F
F
G
*G
E
H
*H
E
H
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; it represents no guess having been made. The transition table:
记状态qi为已经看到i并猜测i就是结尾将要重复的数字，i = 0,1,...,9 。初始状态为qs，终止状态为qf。我们可以一直停留在状态qs，表示尚未猜测。转移表如下：

0
1
...
9
->qs
{qs,q0}
{qs,q1}
...
{qs,q9}
q0
{qf}
{q0}
...
{q0}
q1
{q1}
{qf}
...
{q1}
...
...
...
...
.
q9
{q9}
{q9}
...
{qf}
*qf
{}
{}
...
{}
Solutions for Section 2.4
Exercise 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:

记q0为初始状态。q1, q2和q3识别 abc; q4, q5和q6 识别abd, q7 到q10 识别aacd. 转移表如下：

a
b
c
d
->q0
{q0,q1,q4,q7}
{q0}
{q0}
{q0}
q1
{}
{q2}
{}
{}
q2
{}
{}
{q3}
{}
*q3
{}
{}
{}
{}
q4
{}
{q5}
{}
{}
q5
{}
{}
{}
{q6}
*q6
{}
{}
{}
{}
q7
{q8}
{}
{}
{}
q8
{}
{}
{q9}
{}
q9
{}
{}
{}
{q10}
*q10
{}
{}
{}
{}
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.

由子集构造法可得以下DFA的状态，其中每一个状态都是NFA状态的子集： 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}.注意到 F, G 和H 可以整合到一个接受状态中，或者我们可以用这三个状态来分别标记已识别abc, abd 和aacd。

a
b
c
d
->A
B
A
A
A
B
C
D
A
A
C
C
D
E
A
D
B
A
F
G
E
B
A
A
H
*F
B
A
A
A
*G
B
A
A
A
*H
B
A
A
A
Solutions for Section 2.5
Exercise 2.5.1
For part (a): the closure of p is just {p}; for q it is {p,q}, and for r it is {p,q,r}.
(a): 根据状态的ε闭包的的性质。求得，
p的ε闭包：{p}； q的ε闭包：{p,q}； r的ε闭包：{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.
b) 由于输入a状态总是保持不变，因此只需考虑输入b和c的情况。可以看出，从状态p第一次到r且只经过b，c和ε转移的路径为bb, bεc和 c ；到r之后，读入b仍可回到r，读入c回到p ，则可通过继续读入串bb, bc和 c回到r。
因此，每一个由b和c组成的长度小于等于3的串可以被接受，除了串b不能接受。向这些串中插入a，并保持长度小于等于3，就会得到所有由a，b，c组成的，至多含有一个a的可被接受的串。由一个c和两个a组成的任意串也是可以被接受的。其它的串均被拒绝。
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:
由初始状态，即p的ε闭包或者{p}，有3个状态可以达到。令A = {p}, B = {p,q}, C = {p,q,r}。转移表如下：

a
b
c
->A
A
B
C
B
B
C
C
*C
C
C
C
Solutions for Section 3.1
Exercise 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)*

首先考虑第一个a在第一个b的前面，然后再考虑相反的情况。表达式为： 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).
首先写出没有两个1相邻的串的集合，如下：(10+0)*(ε+1) 。表达式的第一部分表示每个1之后都紧跟一个0的这样的串组成。为了表示结尾可能是1的情况，则可在串尾处加上 (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.
题目要求的串可由两部分组成，也就是，前缀没有相邻的1，后缀没有相邻的0。前半部分也就是已经给出的(10+0)*(ε+1)，根据对称性后半部分可将上式的1和0交换得到。所求即为(10+0)*(ε+1)(01+1)*(ε+0)。注意中间的ε+1 项没有作用，因为1可以由后面的(01+1)* 项得到。因此最后得到的正则表达式为(10+0)*(01+1)*(ε+0)
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. 你可以与练习3.1.2(a)中我们给出的不同样子的表达式作比较。为什么起作用的原因是类似的。

这个表达式是 “没有相邻的1”的另一种描述方式。(00*1)* 表示每个 1 的前面都至少有一个0做前缀。最后的0* 允许在最后一个1后面有0。开头的(ε+1) 允许初始为1，要么串就只有这一个符号，要么后面跟着的就是0。
Exercise 3.1.5
The 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.

正则表达式 ε。ε*表示由任意多个空串组成的串，也是只包含空串的集合。
Solutions for Section 3.2
Exercise 3.2.1
Part (a): The following are all R0 expressions; 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.

a) 下面就是所有 R0 的表达式；我们只写出下标： 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.
b) 下面就是所有 R(1) 的表达式；我们只写出下标：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转移图:

If we eliminate state q2 we get:
如果消除状态q2，有:

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)*
由课本中的公式，q1到q3的正则表达式：[1 + 01 + 00(0+10)*11]*00(0+10)*
Exercise 3.2.4(a)
利用定理3。7每个用正则表达式来定义的语言也可用穷自动机来定义

Exercise 3.2.6(a)
(Revised修改 1/16/02) LL* or L+.
Exercise 3.2.6(b)
The set of suffixes of strings in L. (以)L中串(作为)后缀/下标的集合。
Exercise 3.2.8
Let 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