自動機理論、語言和計算導(dǎo)論課后習(xí)題答案(中文版)

1、.Solutions for Section 2.2Exercise 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 th

2、e 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

3、 out that only 13 are accessible from the initial state, 000r. Here is the transition table:杠桿可能出現(xiàn) 8 種情況，影響著最終狀態(tài)。 并且也要說明， 前面一個大理石球是否從 D 滾出，也就是說，前一個輸入是否被接受。令0 代表向左方的狀態(tài)（如圖表）， 1 代表向右方。這三個杠桿的每一個狀態(tài)都可以用三個數(shù)（ 0 或 1）組成的序列表示。這個序列后面跟著字母 a 或者 r。a 代表接受狀態(tài)， r 代表拒絕狀態(tài)。16 種可能的狀態(tài)中， 只有 13 種是從初始狀態(tài) 000r 可達的。下面它的有窮自動機的轉(zhuǎn)移表

4、。AB->000r 100r 011r*000a 100r 011r*001a 101r 000a010r 110r 001a*010a 110r 001a011r 111r 010a100r 010r 111r*100a 010r 111r101r 011r 100a*101a 011r 100a110r 000a 101a*110a 000a 101a111r 001a 110aExercise 2.2.2The statement to be proved is -hat(q,xy) = -hat( -hat(q,x),y), and we proceed by induction

5、 on the length of y.證明：通過對 |y|進行歸納，來證明?(q , xy)= ?( ?(q , x) , y) ，具體過程如下：Basis: If y = ,then the statement is-hat(q,x) = -hat( -hat(q,x), . This) statement follows from the basis in the definition of -hat. Note that in applying this definition, we must treat -hat(q,x) as if it were just a state, sa

6、y p. Then, the statement to be proved is p = -hat(p, ,)which is easy to recognize as the basis in the definition of -hat.基礎(chǔ) :y =0，則 y=。那么需證?(q,x)= ?( ?(q ,x), )，記 p= ?(q,x),命題變?yōu)閜= ?(p , ),由 ?的定義知這顯然成立。Induction: Assume the statement for strings shorter than y, and break y = za, where a is the last s

7、ymbol of y. The steps converting-hat( -hat(q,x),y) to -hat(q,xy) are summarized in the following table:歸納 : 假設(shè)命題對于比y短的串成立 , 且 y = za, 其中 a 是 y 的結(jié)尾符號。?( ?(q,x),y) 到 ?(q,xy) 的變換總結(jié)在下表中 :Expression 表達式Reason 原因?( ?(q,x),y)Start開始?( ?(q,x),za)y=za by assumption由假設(shè) y=zaDefinition of -hat, treating -hat(q,

8、x) as a state?(q,x),z),a)(?的定義 , 把 ?(q,x) 看作是一個狀態(tài)(?Inductive hypothesis歸納假設(shè)(q,xz),a)?(q,xza)Definition of -hat?的定義?(q,xy)y=zaExercise 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.狀態(tài) A, B，C 分別表示以 ,和 00 結(jié)尾的串的狀態(tài)。.0 1 ->ABA B C

9、 A *C CAExercise 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

10、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

11、 want to consider what happens to 5a+b if we multiply by 2 and add 1.對于一個二進制整數(shù)，如果讀入一個比特，其值等于原數(shù)乘以；否則等于原數(shù)乘以再加以 1。而任意一個數(shù)均可寫成形如 5a+b，其中 a 任意， 0<= b <=4，那么輸入 0，原數(shù)變?yōu)?2(5a+b) = 10a+2b，由于 10a 是 5 的倍數(shù) ,，因此 10a+2b 除以的余數(shù)與 2b 相同。輸入 1，則得 2(5a+b)+1 類似。因此對于所有的數(shù)只要記住它被除的余數(shù)就可以。由于 b 是 0, 1, 2, 3 或者 4, 我們可以容易得到該

12、DPA 的轉(zhuǎn)移表，具體如下：The table below shows this automaton. State qi means that the input seen so far has remainder i when divided by 5.其中狀態(tài) qi 代表輸入串被除的余數(shù)i 的狀態(tài)。01->*q0 q0 q1q1 q2 q3q2 q4 q0q3 q1 q2q4 q3 q4There is a small matter, however, that this automaton accepts strings with leading 0's. Since th

13、e 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

14、to state d, which we never leave. The complete automaton is:.但是上述自動機仍接受以開頭的字符串。因為題目要求只接受以開頭的串，可增加一個初始狀態(tài) s 和“死亡狀態(tài)” d。在狀態(tài)初始狀態(tài) s, 若看到，則轉(zhuǎn)到狀態(tài) q1；若看到 , 則直接轉(zhuǎn)到狀態(tài) d，識別終止。所求自動機如下 :01->s dq1*q0 q0 q1q1 q2 q3q2 q4 q0q3 q1 q2q4 q3 q4d ddExercise 2.2.9Part (a) is an easy induction on the length of w, starting

15、at length 1.Basis: |w| = 1. Then -hat(q0,w) = -hat(qf,w), because w is a single symbol, and-hat agrees with on single symbols.Induction: Let w = za, so the inductive hypothesis applies to z. Then-hat(q0,w) = -hat(q0,za) = -hat(q0,z),a) = -hat(qf,z),a) by the inductive hypothesis = -hat(qf,za) = -hat

16、(qf ,w).證明： a) 通過對 w 長度的歸納證明?；A(chǔ) : 若|w| = 1，則 w 是一個符號，此時需證?0?f,w),而對于單個符(q ,w) =(q號擴展轉(zhuǎn)移函數(shù)?與轉(zhuǎn)移函數(shù) 的作用是一樣的，得證。歸納 :令 w = za, 假設(shè)對于 z 命題 ?(q0,z) =?(qf ,z)成立。那么 ?(q0,w) = ?(q0,za)?(qf,za) =?(qf,w).= (q0,z),a) =(qf,z),a) 由歸納假設(shè) =For part (b), we know that -hat(q0,x) = qf. Since x , we know by part (a) that -h

17、at(qf,x) = qf. It is then a simple induction on k to show that -hat(q0,xk ) = qf. Basis: For k=1 the statement is given.Induction: Assume the statement for k1;- i.e., -hat(q0,xSUP>k-1) = qf. Using Exercise 2.2.2, -hat(q0,xk) = -hat( -hat(q0,xk-1),x) = -hat(qf ,x) by the inductive hypothesis = qf

18、by (a).b) x 是屬于 L(A) 的非空串，也即串 x 被接收，因此 ?(q0,x) = qf ，則由 a)知 ?(qf,x) = ?(q0,x)= qf 。現(xiàn)在通過對 k 的歸納來證明 ?(q0 ,xk) = qf ?；A(chǔ) : k=1 時，需證 ?(q0,x) = qf ，由已知可得。歸納：假設(shè)對于 k-1 命題成立，也就是說， ?(q0,xk-1) = qf 。由練習(xí) 2.2.2,?(q0,xk)= ?( ?(q0,xk-1 ),x) =?(qf ,x) 由歸納假設(shè) = qf 由(a)。Exercise 2.2.10The automaton tells whether the n

19、umber 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, namelyzero 1's, and ?(A,w)

20、 = 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 oddnumber of 1's, then so does

21、z. By the inductive hypothesis,-hat(A,z) = B, and the transitions of the DFA tell us -hat(A,w) =B. Thus, in this case, -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, -hat

22、(A,z) = B. The transitions of the DFA tell us -hat(A,w) = A. If w has an odd number of 1's, then z has an even number of 1's. By the inductivehypothesis, -hat(A,z) = A, and the transitions of the DFA tell us -hat(A,w) = B. Thus, in this case as well, -hat(A,w)= A if and only if w has an even

23、 number of 1's.這個自動機表示，狀態(tài)A 表示偶數(shù)個 1，狀態(tài) B 表示奇數(shù)個 1，不管串有偶數(shù)個還是奇數(shù)個 1，都會被接受。當(dāng)且僅當(dāng)串w 中有偶數(shù)個 1 時， ? (A,w) = A. 。用.歸納法證明如下基礎(chǔ) : |w| = 0?？沾?dāng)然有偶數(shù)個1 ，即 0 個 1，且 ? (A,w) = A.歸納：假設(shè)對于比w 短的串命題成立。令w = za, 其中 a 為 0 或 1。情形 1： a = 0. 如果 w 有偶數(shù)個 1, 則 z 有偶數(shù)個 1。由歸納假設(shè)， ? (A,z) = A 。由轉(zhuǎn)移表的 DFA 知 ?(A,w) = A. 如果 w 有奇數(shù)個 1, 則 z 有奇

24、數(shù)個 1. 由歸納假設(shè)，? (A,z) = B, 由轉(zhuǎn)移表的 DFA 知 ? (A,w) = B. 因此這種情況下 ?(A,w) = A當(dāng)且僅當(dāng)w 有偶數(shù)個1。情形 2： a = 1. 如果 w 有偶數(shù)個1, 則 z 有奇數(shù)個 1。由歸納假設(shè)， ? (A,z) = B.由轉(zhuǎn)移表的 DFA 知 ?(A,w) = A.如果w有奇數(shù)個1,則z有偶數(shù)個 。由歸納1假設(shè) , ?(A,z) = A, 由轉(zhuǎn)移表的 DFA 知 ?(A,w) = B.因此這種情況下 ?(A,w) =A 當(dāng)且僅當(dāng) w 有偶數(shù)個 1.綜合上述情形，命題得證。Solutions for Section 2.3Exercise 2.3

25、.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.下表就是利用子集構(gòu)造法將NFA 轉(zhuǎn)化成的 DFA。其中構(gòu)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->ABAB D CC E AD

26、F C*EFG*FFG*GEH*HEH.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 repre

27、sents no guess having been made. The transition table:記狀態(tài) qi 為已經(jīng)看到 i 并猜測 i 就是結(jié)尾將要重復(fù)的數(shù)字， i = 0,1,.,9 。初始狀態(tài)為 qs，終止?fàn)顟B(tài)為 qf。我們可以一直停留在狀態(tài) qs，表示尚未猜測。轉(zhuǎn)移表如下：01.9->qs qs,q0 qs,q1. qs,q9q0 qfq0. q0q1 q1qf. q1. . .q9 q9q9. qf*qf . Solutions for Section 2.4Exercise 2.4.1(a)We'll use q0 as the start state.

28、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:記 q0 為初始狀態(tài)。 q1, q2 和 q3 識別 abc; q4, q5 和 q6 識別 abd, q7 到 q10 識別 aacd. 轉(zhuǎn)移表如下：abcd->q0 q0,q1,q4,q7q0q0q0q1q2q2q3*q3q4q5q5q6*q6q7q8q8q9.q9 q10*q10 Exercise 2.4.2(a)Th

29、e 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, Gand H can be combined into one accepting state, or we can use these thr

30、ee state to signal the recognition of abc, abd, and aacd, respectively.由子集構(gòu)造法可得以下 DFA 的狀態(tài)，其中每一個狀態(tài)都是 NFA 狀態(tài)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 可以整合到一個接受狀態(tài)中，或者我們可以用這三個狀態(tài)來分別標(biāo)記已識別abc, abd 和 aacd。a b c d->ABAAABCDAAC

31、CDEADBAFGEBAAH*FBAAA*GBAAA*HBAAASolutions 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.(a): 根據(jù)狀態(tài)的 閉包的的性質(zhì)。求得，p 的 閉包： p ； q 的 閉包： p,q ； r 的 閉包： p,q,r 。For (b), begin by noticing that a always leaves the state unchanged. Thus, we can thin

32、k of the effect of strings of b's and c's only. To begin, notice that the only ways toget from p to r for the first time, using only b, c, and -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

33、'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, th

34、e strings consisting of one c and up to 2 a's are accepted; other strings are rejected.b) 由于輸入 a 狀態(tài)總是保持不變，因此只需考慮輸入b 和 c 的情況?？梢钥闯?，從狀態(tài) p 第一次到 r 且只經(jīng)過 b，c 和 轉(zhuǎn)移的路徑為 bb, b c和 c ；到 r 之后，讀入 b 仍可回到 r，讀入 c 回到 p ，則可通過繼續(xù)讀入串 bb, bc 和 c 回到 r。因此，每一個由 b 和 c 組成的長度小于等于3 的串可以被接受， 除了串 b 不能接受。向這些串中插入 a，并保持長度小于等于3，就會

35、得到所有由 a， b， c 組成的，至多含有一個 a 的可被接受的串。由一個 c 和兩個 a 組成的任意串也是可以被接受的。其它的串均被拒絕。There are three DFA states accessible from the initial state, which is the closureor p. Let A = p, B = p,q, and C = p,q,r. Then the transition table is:由初始狀態(tài)，即 p 的 閉包或者 p ，有 3 個狀態(tài)可以達到。令 A = p, B = p,q,C = p,q,r 。轉(zhuǎn)移表如下：a bc->AA

36、BCBBCC*C CCCSolutions 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)*首先考慮第一個a 在第一個b 的前面，然后再考慮相反的情況。表達式為：c*a(a+c)*b(a+b

37、+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)*( +1)To see why this expression works, the first part consists of all strings in which every 1 is fol

38、lowed by a 0. To that, we have only to add the possibility that there is a 1 at theend, which will not be followed by a 0. That is the job of ( +1).首先寫出沒有兩個 1 相鄰的串的集合，如下： (10+0)*( +1) 。表達式的第一部分表示每個 1 之后都緊跟一個 0 的這樣的串組成。為了表示結(jié)尾可能是 1 的情況，則可在串尾處加上 ( +1)。Now, we can rethink the question as asking for stri

39、ngs 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, asolution to this problem is (10+0)*( +1)(01+1)*( +0). Note that the +1 term in the middle

40、 is actually unnecessary, as a 1 matching that factor can be obtained from the (01+1)* factor instead.題目要求的串可由兩部分組成， 也就是，前綴沒有相鄰的 1，后綴沒有相鄰的 0。前半部分也就是已經(jīng)給出的 (10+0)*( +1)，根據(jù)對稱性后半部分可將上式的 1 和 0 交換得到。所求即為 (10+0)*( +1)(01+1)*( +0)。注意中間的 +1項沒有作用，因為 1 可以由后面的 (01+1)* 項得到。因此最后得到的正則表達式為 (10+0)*(01+1)*( +0)Exerci

41、se 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

42、 allows 0's after the final 1, and ( 1) at the+beginning allows an initial 1, which must be either the only symbol of the string or followed by a 0.你可以與練習(xí) 3.1.2(a)中我們給出的不同樣子的表達式作比較。為什么起作用的原因是類似的。這個表達式是 “沒有相鄰的 1”的另一種描述方式。 (00*1)* 表示每個 1 的前面都至少有一個 0 做前綴。最后的 0* 允許在最后一個 1 后面有 0。開頭的 ( +1) 允許初始為 1，要么串

43、就只有這一個符號，要么后面跟著的就是 0。Exercise 3.1.5The language of the regular expression . Note that * denotes the language of string consisting of any number of empty strings, concatenated, but that is just the set containing the empty string.正則表達式。*表示由任意多個空串組成的串，也是只包含空串的集合。Solutions for Section 3.2Exercise 3.2.1

44、Part (a): The following are all R 0 expressions; we list only the subscripts. R11 =1;+R12 = 0; R13 = phi; R21 = 1; R22 = ; R230; =R31 = phi; R32 = 1; R33 =+0.a) 下面就是所有R0 的表達式；我們只寫出下標(biāo)：R11 = +1；R12 = 0; R13 =(phi); R21 = 1; R22 = ; R230; =R31 =(phi); R32 = 1; R33 =+0.(1)Part (b): Here all expression n

45、ames are R; we again list only the subscripts. R11 = 1*; R12 = 1*0; R13 = phi; R21 = 11*; R22 = +11*0; R23 = 0; R31 = phi; R32 = 1; R33 = +0.b) 下面就是所有 R(1) 的表達式；我們只寫出下標(biāo)： R11 = 1*; R12 = 1*0; R13 = phi; R21 = 11*; R22 = +11*0; R23 = 0; R31 = phi; R32 = 1; R33 = +0.Part (e): Here is the transition dia

46、gram轉(zhuǎn)移圖 :If we eliminate state q2 we get:如果消除狀態(tài) 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)*由課本中的公式， q1 到 q3 的正則表達式： 1 + 01 + 00(0+10)*11*00(0+10)*Exercise 3.2.4(a)利用定理 3。7 每個用正則表達式來定義的語言也可用窮自動機來定義Exercise 3.2.6(a)(Revised

47、 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(作為)后綴/下標(biāo)的集合。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 gr

48、eater than n, by induction on k.令 R(k)ijm 為從狀態(tài) i 到狀態(tài) j ，長度為 m，且沒有經(jīng)過編號大于 k 的路徑的個數(shù)。對于所有狀態(tài) I 和 j，以及 m（mn），通過對 k 歸納來計算這個個數(shù)。0Basis: R ij1 is the number of arcs (or more precisely, arc labels) from state i to state j. 0 00基礎(chǔ) : k=0，R ij1 是由狀態(tài) i 到狀態(tài) j 的箭?。ǜ鼫?zhǔn)確的說， 是箭弧標(biāo)號）的個數(shù)。00R ii0 = 1,其他的 R ijm 's 都為 0。I

49、nduction: R(k)ijm is the sum of R(k-1)ijmand the sum over all lists (p1,p2,.,pr) of(k-1)(k-1)(k-1)(k-1)*positive integers that sum to m, of Rikp1 * Rkkp2 *Rkkp3 *.* Rkkp(r-1)R(k-1)kjpr. Note r must be at least 2.歸納: R(k)ijm 是 R(k-1) ijm 的和， R(k-1)ikp1* R (k-1) kkp2*R(k-1) kkp3 *.* R (k-1)kkp(r-1) *R

50、(k-1)kjpr。 (p1,p2,.,pr)是所有和為 m 的正整數(shù)序列， r 大于等于 2。(k)The answer is the sum of R 1jn, where k is the number of states, 1 is the start state, and j is any accepting state.答案就是 R(k)1jn 的總和，其中 k 是狀態(tài)個數(shù)， 1 為開始狀態(tài)， j 是任意接受狀態(tài)。Solutions for Section 3.4Exercise 3.4.1(a)Replace R by a and S by b. Then the left an

51、d 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.將 R 替換為 a，S 替換為 b 。 等式變?yōu)?a + b = b + a.也就是a,b= b,a. 因為集合中元素的順序是無關(guān)緊要的，所以，等式兩邊是一樣的：由串 a 和 b 構(gòu)成的語言。Exercise 3.4.1(f)Replace R by

52、 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.將 R 替換為 a 。右邊變?yōu)?a*,

53、 代表 a 組成的所有串，包含空串。左邊是 (a*)*, 代表由 a 組成的串構(gòu)成的串，也就是由 a 構(gòu)成的串。當(dāng)然相等。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 th