北京邮电大学-形式语言与自动机-课后习题答案

形语言与动第章41至：其中x{所有母{所有的字}P下S→SxAA→A→yBByB→yCyDy6/ω∈且ω中a是b倍：P下S→S→S→S→SS→SS→SS→SS→S→S→bSaaS7S）S→SbS→aSbS→SE→aS（1）n/n≥nn/n≥0}

Sc

/n0}11个个{上的字符2数b3串aba的a,b}*上（1）是正则偶a偶b

aa

奇a偶bb偶a奇b

aa

奇a奇见2）。

L为包含子串aba的{上的字符串的a,b}*上合LL4（1）生成式S→BAA→BbB→CDDd（2）生成式S→BAABBaCDCDd：(1)由生成式①②③④⑤去CD，得到*⑥(cd+b)+(cb)*=>S=(aab)ε)(cb)*①②③④B=ba⑥C=d+abb

a⑦A=ba+c(d+ba)S=a(ba+c(d+ab*a

a+acd+acab

a5.(2)由和组成的所

001(3)b为首后跟aa继b的由和（1）右线性法SS→ε法SS为法S→aAA→ε为法SA→BCε7.为a(ba)1（1）右线性法S→A→2a

b

是终结状态9.（）的（1）由图知+a+εq121q+a0011110q10*(0*()+a+ε000=qε0)*q012q+b102q+a010101002020+0

012303012020123030120201202012030120121100302301231010120201230132303q++000表的：（1）含有续b（2）以aa为首（3）以aa结尾（1）q,q}),其abqqqq

0123

qqqq

0003

qqqq

1233（2）,q,{q}),如下：abqq

01

qq

12

ΦΦq

2

q

2

q

2（3）,q,{q}),如下：abqqq

012

qqq

122

qqq

000M等价于MM如1（1）,q,q}),如下：,q}(q}0100}}11}Φ22}}33（2）,q,q,{q,q}),如下：,q}(q}0201}}112}{q}220Φ(q{q}330（1）M,,,q,q]，[q,q,q]}Q],,q,q],[qq,q,qq,q],[q,q,q],[q,q1ab][q,q][q]010],q,q][,q]010102,q][q,q,q][,q]0121202

1111111131301212123231131201212312323[q]02[q,q,q]0123[q,q]013[q,q]023[q]03

[q,q]03[q,q]0123[q,q]0123[q,q]03[q,q]03

]0[q,q,q]023[q,q,q]023[q]03[q]03（2）DFAM,,[q],,q],[q,q]}Q],,q,q],,q1ab][q,q]]031]][q,q]1320]]]1212]]]230[q,q][q,q][q,q,q]01223012]][q,q,q]1223012]3

Φ

]0,q],q][,q,q]12323012]]]2330的εε起始态φ

abcq

φr(态

{r}

φ

为的串ε-NFA转ε可被接23为aac,（2）-NFA：其为ε含ε的NFAM

’(p,a)=(’(p,ε

’(p,b)=ε(’(p,ε

’(p,c)=ε(’(p,ε

’ε(’(q,ε

’ε(’(q,ε

’ε(’(q,ε

’ε(’(r,ε

’(r,b)=ε(’(r,ε

’(r,c)=ε(’(r,ε：态pq

1111010111110101111cr．设NFA}),如下：0101,q}(q}0101Φ(q,q}110的的DFAM,,[q]，]}Q]，[q][q]}1ab][q,q]]011]1

Φ

]01]],q]010101已是最简，故不用化简（1）由文式→言L(G)（2）{ω/ω{a,b}*有a（3）ca/k≥1}（4）{ω/ω（中，与bL(G)kωa

(cb)cω=ωω||ω|≤存在ωωL意取

n∈ωω=a)(cb)c不于0不属于L则2的k取=abω=ωω||ω|≤存在ωωL意取n∈(0,k)ωω=a)b在于0时a的个3的k取=aω=ωω||ω|≤存在ωωL

意取n∈(0,k)ωω=a

)

ca

i不于0时后a的4的k取=ababωω=ωωω||ω|≤存在ωωL意取n∈(0,k)ωω=a)bab于0时足ω的．构{a,b}*的如果以结尾，则输1结尾，2态表示

aabaabab

abbbε生成式、S→|A,A|,A→A|,AS|b|ε,A→S|a，S|1214245345|ε

11111111S1SAB111111111S1SAB11解:⑴由算无N’={S,A}12345G=({S,S,A,A},{,S)式111345111Sε1S→|A,12A→|A,134A→|A,245AS|b,3A4AS|,5由算4，消单生N={S,A,A},S1112345N=N=N=N==N={S,A,},SA1A2A3A4A512345法4，则PS1a|ε,S→|,Aa|,1Aa|,2Aa|,3Aa|,4Aa|5由算1和算2消除G=({S},{},P,S)式P：a|ε.111111设2型文G=({},{},P,),其中P：S→ASBε;A→aAS|a;B|A|bbG变换为无生成式,无单,没有无,Chomsky范式解:⑴由算无N’={S}SS|,出A→|,得B→||BS|B,S∈’出S→ε的法G=({S},{},S),如下：Sε1S→ASB|AB,A→||a,B|SB||B|A|,由算4，消单生N={S,S},N=S},={A},N={}S→|AB∈且式故中有→||,S→|,A||a,→||BS||aA|a|

G=({S,S,},{},S)P如下：11111Sε|ASB|,1S→ASB|AB,A→||a,B|SB||||a|由算1和算2消除号此号;转化的范式法:→,,S变,|A→ED|D→,Ea,||aA|a|,变换为BCS||Fb,由此G=({S,S,},{},P,S)其式如下：1111Sε|AC|AB,1S→|,A→ED||a,B||BS|||a|,C,D,Ea,Fb.的范式:S|a,SS|b解:将非为为位,D高位,对于SS用S|a→|aS|,4.2.4,为D→|b||bD',→DS|DSD’,将D生成入S→||’D|a,将D生成入’D’→|bS|||D'|D'||D',由此的范G=({S,D,D’},{},P,S),式P11S→||||a,D|b|aSD'|,D’→|bS|||D'|D'||D'.A→b|Aa,A→b|Aa|b,AAa|Ab|a132212313解:⑴转化的范式文法:A→|A,13425A→|A|b,21426A→|A|a,31537Ab,4Aa,5

1234552141234552141345234425431534233452576252673437562473A→,625A→,734转化的范法为,,A位A为,A→AA,AAA|AA代入得→A|AAA|A|b,264.2.4,A→A|b|AA’|’,23443442A’→A’|A’|AA|A,25426546AAA,AAA|AAAAAA|A|AA|a,A生符生成A323A→|AAAAA|bA|AAA’A|A|A334534455553442552537|a,4.2.4,AbAA|A|a|bAA’|’A’|’,35525555325533A’→A|AA|A’A|A|AA’|AAA’|34544554425545344553AAA’|A’,44255373于→入生成式得A→AA||AA’A|A,6344534252A生符生成A636AAAAA|bAAAA|aAAA|’AAA|654452545445553445AAA|aAA|AAAA|’AA’25534453445554422554425|A|AAAAA|AAAA|425534252534425A’A|bA’A|bA,34425255于→,将入A生成AbAA|AA|aA|bAAA’A|AA|7554254455342534,34入’A’→aAA’|bAAAAA’|AA’|A’|225544522554452445AA’|A’AA’|aAAAA’|534452255344523452AA’|AAA’|’A’|5442525442524252AAA’|A’A’A’|aAAAA’|5344252255344252344252A’|bAA’|aA|bAAAA|’AAA|aAA|2525245544525544544AA|AAA|AAA|AA’A|53445255344344554425AAA|aA|bAAAAA|255442544255534425AAA’|aAA’A|bAA|bA,255344253442255入AA’→|aAA|’A|A’|aAAAA’|AAAA’34552555345534253|bA|AA||AA’A|bA’AA’A|A|542545534253434A’|bAAA’|aAA’|bAAA’|AAA’543255434355332554A’A’,334

由此的范法G=({S,D,D’},{},P,S),式P11A→|A,13425A→A|b|AA’|’,23443442AbAA|A|a|A’|’AAA’|’,3552555532553Ab,4Aa,5AAAAA|bAAAA|aAAA|’AAA|654452545445553445AAA|aAA|AAAA|’AA’25534453445554422554425|A|AAAAA|AAAA|425534252534425A’A|bA’A|bA,34425255AbAA|AA|aA|bAAA’A|AA|7554254455342534,34A’→aAA’|bAAAAA’|AA’|A’|225544522554452445AA’|A’AA’|aAAAA’|534452255344523452AA’|AAA’|’A’|5442525442524252AAA’|A’A’A’|aAAAA’|5344252255344252344252A’|bAA’|aA|bAAAA|’AA|aAA|252524554425444AA|AAA|AAA|AA’A|53445255344344554425AAA|aA|bAAAAA|255442544255534425AAA’|aAA’A|bAA|bA,255344253442255A’→|aAA|’A|A’|aAAAA’|AAAA’34552555345534253|bA|AA||AA’A|bA’AA’A|A|542545534253434A’|bAAAA’|aAA’|AAA’|bA’AAA543255343553432534A’A’.334:S,aS|bS|a,机解:根据P机使按文法的.=(Q,T,Г,δ,q,F00Q={q,q},0fT={,Г={},Z=S,0F={q},fδ定义如下δ(qδ(q

00

ε,S)={(q,)},0ε)={(q,),(qbS),(,a)},00δ(q

0

)={(q,ε)},0δ(q

0

ε,ε)={(q,ε)}.f={ijck|i,j,k0且=j=k}的.性?为什么:G=({S,A,B,C,D,E},{a,b,c}PS)PS→A→aAbε,B→|ε,D→ε,Eε

00000000110000000001101100000子中个数相同时，对ω最ADabDabcC及SEBabBc=(,X},δ,,Zφ中如下01000δ(q,b,Z)=,)}δ(,ε,Z)={(q,ε)}δ(q,b,=,,δ(q={(q,ε)},δ(q,b,=,,δ(,a,Z)=,Z)},法产=解:在G中,N={

,q],[q],],],[q,q],[q,Z],],0000100011001010]}.11S生成式有S→,q],000S→,q],001δ(q)=,)},则有000,q]→b[q][q,q],00000000,q]→b[q][q,q],00001100,q]→b[q][q,q],00100001,q]→b[q][q,q],00101101δ(q={(q,XX)},则有00]→][q],000000,]][q,],00110,]][q,],00001,]][q,],00111δ(q={(q,X)},则01]→],001]→],011δ(q,,Z)=,Z则有1000,q]a[q,q],100000,q]a[q,q],101001δ(q,ε,)=,ε)},,q]→,000δ(q={(q,ε)},11]→ε11利用12,,G={}={,Z,q],[q]={生成式下001001101S→,q],000,q]→b[q][q,q],00001100,]][q,],00111]→],011,q]a[q,q],100000,q]→,000,q]→.000

000言:{anbncm|};明:假设言,由理,数p,当ω∈L且|≥ω=ap

b

p

c

p

,将ω写为ωωωω,同时满|ω|ωω含a和c,因下,|ω如果ωωωω=aj

(b

)≤则当i≠,ωωω现a的数与b的个数不;ωω含cωω=cj(j≤p),当于1时ωωω现c的于a(b的)如果ωa和(b和c)当i=0时ωωb于或盾L言{ak

|k是质数明:假L言,由理数p,当ω∈L且ω|时,可取ω=k(k≥k1)将ω写ω=ωωω,同时足ω|≤,|ωω1,ωωωωω,k*(1+j)且k≠,因此必定不是质,即ωωω盾故言.由组成a,b,c的个数相同的所有字符.明:假设言,由理,数p,当ω∈L且|≥ω=ab

c

k

≥p),将ωω=ωωω,同足|ωω|≤ωω含a和c,因下,|ω如果ω(b或c)ω=j

(b

j

)(j≤,i1时,ωω现a,b,c等如果ωa和(b和c),ωωωω中会出现a,b的c的不等盾L言是Chomsky法在ωL为ω的,与的.解:设边ω,最ω为|2

.范知Chomsky范树n中长2,因ω度示言:求的{0m1n

|m≤};解:设PDAM=(Г,,Z),Q={q,q},01f

000000000000T={,Г={Z},F={q},fδ定义如下δ(q

0

,ε,Z)={(q,Z)},010δ(qδ(qδ