1、N 2|4|6|8|OD 0| NS O| AOA AD | NP36-8文法:E T| E T|E TT F|T*F|T/ FF ( E)|i最左推导 :EET T T FTiTi T* Fi F* F ii* F i i *iETT*FF*Fi*Fi *(E) i*( ET)i *( TT) i*( F T)i *( iT ) i*( i F)*( i i)最右推导TEET*iF *i Ei*iT i *i F i* i i i*F*TF*( E)*( E T)F *( EF)F *( E i )F *( Ti)F *( F i )F *( ii *( i i )/*P36-9句子iiie
2、i 有两个语法树:S iSeS iSei iiSei iiieiS iS iiSeS iiSei iiieiP36-10/*S TS |TT (S)|()*/P36-11/*L1:S ACA aAb | ab C cC |L2:S ABA aA|B bBc|bcL3:A aAb|B aBb|L4:S A| BA 0A1|B 1B0| A第三章习题参考答案P64 - 7确定化:1X1,2,32,32,3,42,3,52,3,4,Y2,3,4,最小化:0,1,2,3,4,5,60,123,4,5。 1,3,5 0,1,2,3,4,51 1,2,40,1,2,3,4, 5,60,1,2,34。1,3
3、,50,1,23,4, 5,60,1,23。 1,3 0,123 1,2,40,1,2,04, $,60,1 0 1 0,1 1 1,22,30 B 门匡40,1,2,3, 4, 5, 6P64 8(1|0)*01 (1|2|3|4|5|6|7|8|9)(0|1|2|3|4|5|6|7|8|9)*(0|5)|(0|5)0 1(0|10 1) |1 0(0 |10 1)P64 - 12ab00,11给状态编号:23最小化:0,1, 2,30,1 a 1 0,1b 2 2,3a 0,3 2,3b 30,1, 2, 3已经确定化了,进行最小化0,1, 2,3,4,50,1a 1 0,1b 2,42,
4、3,4,5爲1,3,0,5 2,3,4,5b 2,3,4,52,4a 1,0 2,4b 3,53,5a 3,5 3,5b 2,40,1, 2,4, 3,50,1a2,4a3,5a1,03,50,叽2,42,4b3,5b2,4:X,1,Y1,Y21,Y0,1,2,30,10 1 0,11 22,3o 1,3 2,3i 30,1,2,3P81 - 1(1)按照T,S的顺序消除左递归S aF|(T)T STT ,ST |递归子程序:procedure S;beginif sym=a or sym=人the n abva neeelse if sym=(the n begi nadvance;T;)
5、the n adva nee; else error;endelse erroren d;procedure T;S;procedure ;,advanee;其中:sym:是输入串指针IP所指的符号advanee:是把IP调至下一个输入符号 error:是出错诊察程序FIRST(S)=a,(FIRST(T)=af,(FIRST()=,FOLLOW(S)=),#FOLLOW(T)=)FOLLOW()=)预测分析表A(J#S是LL(1)文法P81 - 2文法:E TEE E |T FTT T |F PFF *F |P (E)|a|b$FIRST(E)=(,a,bfFIRST(E)=+, FIRST
6、(T)=(,a,bFIRST(T)=(,a,b,A, FIRST(F)=(,a,b,AFIRST(F)=*, FIRST(P)=(,a,bfFOLLOW(E)=#,)FOLLOW(E)=#,)FOLLOW(T)=+,),#FOLLOW(T)=+,),#FOLLOW(F)=(,a,b,A,+,),#FOLLOW(F)=(,a,b,A,+,),#FOLLOW(P)=*,(,a,bf,+,),#考虑下列产生式:E E|T T|P (E)|A|a|bFIRST(+E) A FIRST( )=+ n = $FIRST(+E) A FOLLOW(E)=+ A #,)= $FIRST(T) A FIRST(
7、 )=(,a,b,A A FIRST(T) A FOLLOW(T)=(,a,b,A A +,),#= $FIRST(*F) A FIRST( )=* A ) A FOLLOW(F)=* A (,a,b,A,+,),#= $FIRST(E) A FIRST(a) A FIRST(b) A FIRST(A)= $ 所以,该文法式LL(1)文法.+*E TEEE ETT TPFFF *FF :P(E)P Aprocedure E;b the n beg in T; E end else errorprocedure E:+the n beg in adva nee; E endelse if sym and sym# then error end the n beg in F; Tprocedure T; the n T* then errorprocedure F; the n beg in P; Fprocedure F endprocedure P;Athe n adva nee the n E; the n adva nee else error end else errorP81 3(1)是,满足三个条件。(2)不是,对于A不满足条件3。(3)不是,A B均不满足条件3。(4)是,满足三个条件。*
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