【正文】 83。SbA→cdA→aSAabcI0I6I5I4I3I1S’ →SA→SbbI2I2A→SbI3S→cAdA→cdA→AScA→SbA→cdA→aS→AAdS→cAdS→bSAabcdI8I9I5I4I3I7I4S→bI5A→aI6S→AAdA→AScA→AScA→SbA→cdA→aS→AAdS→cAdS→bSAabcI11I10I5I4I3I7A→cdI8A→SbbI2I9S→cAdA→AScS→AAdS→AAdS→cAdS→bA→AScA→SbA→cdA→aSAabcdI11I10I5I4I3I14I10S→AAdA→AScS→AAdS→AAdS→cAdS→bA→AScA→SbA→cdA→aSAabcdI11I10I5I4I3I13I11A→AScA→SbbcI2I12I12A→AScI13S→AAdI14S→cAdACTIONGOTOabcdSA0s5s4s3161s2acc2r5r5r5r5r53s5s4s3s7894r3r3r3r3r35r7r7r7r7r76s5s4s37r6r6r6r6r611108s29s5s4s3s14111010s5s4s3s13111011s2s1212r4r4r4r4r413r1r1r1r1r114r2r2r2r2r2SLR(1)分析法：FOLLOW(S)={c,b} FOLLOW(A)={a,b,c,d}ACTIONGOTOabcdSA0s5s4s3161s2acc2r5r5r5r53s5s4s3s7894r3r35r7r7r7r76s5s4s37r6r6r6r611108s29s5s4s3s14111010s5s4s3s13111011s2s1212R4r4r4r413r1r114r2r2兩表的異同：兩表都用ACTION表和GOTO表反映了移進、歸約（包括接受）的狀態(tài)和動作。不同之處在于SLR(1)增加了在歸約的時候考慮向前符號a以解釋可能出現(xiàn)的“移進——歸約”沖突；SLR（1）的分析較稀疏些，原因是填寫歸約項時，并不是在一狀態(tài)對應(yīng)行上全部填寫歸約動作，而是考慮了相應(yīng)非終結(jié)符的FOLLOW集因素。另外，在分析效率上SLR（1）分析的效率更高一些，因為在發(fā)現(xiàn)錯誤方面，SLR（1）比LR（0）分析發(fā)現(xiàn)的更早些。 解：求LR(1)項目集和狀態(tài)轉(zhuǎn)換表： 狀態(tài)項目集經(jīng)過的符號到達的狀態(tài)I0S’ →SS→AA→BAA→B→aB,a,bB→b,a,bSABabI1I2I3I4I5I1S’ →SI2S→AI3A→BAA→A→BAB→aB,a,bB→b,a,bABabI6I3I4I5I4B→aB,a,bB→aB,a,bB→b,a,bBabI7I4I5I5B→b,a,bI6A→BAI7B→aB,a,b相應(yīng)的LR(1)分析表為：STATEACTIONGOTOabSAB0S4S5R31231ACC2R23S4S5R3634S4S575R5R5R56R27R4R4R4用 LR(1)分析表對輸入符號串a(chǎn)bab的分析過程：步驟狀態(tài)棧中符號余留符號分析動作下一狀態(tài)10ababS44204ababS553045ababR574047aBabR43503BabS446034BabS5570345BabR4780347BaBR439033BBR36100336BBAR2611036BAR211201Aacc 解： （1）求LR(1)項目集和狀態(tài)轉(zhuǎn)換圖：狀態(tài)項目集經(jīng)過的符號到達的狀態(tài)I0S’ →SS→AA→AB,a,bA→SAI1I2I1S’ →SI2S→AA→AB,a,bB→aB,a,bB→b,a,bBabI3I5I4I3A→AB,a,bI4B→b,a,bI5B→aB,a,bB→aB,a,bB→b,a,bBabI6I5I4I6B→aB,a,b相應(yīng)的LR(1)分析表為：STATEACTIONGOTOabSAB0R3R3R3121Acc2S5S4R13R2R2R24R5R5R55S5S466R4R4R4表中沒有多從定義的元素，所以文法是LR(1)文法。（2）LR(1)分析法：狀態(tài)項目集經(jīng)過的符號到達的狀態(tài)I0E’ →EE→E+T/+E→T/+T→(E)/+T→a/+ET(aI1I1I5I4I1E’ →EE→E+T/++I2I2E→E+T/+T→(E)/+T→a/+T(aI3I5I4I3E→E+T/TI4T→a/+I5T→(E)/+E→E+T+/)E→T)/+T→(E)+/)T→a+/)ECTaI7I8I9I12I6E→T/+I7T→(E)/+E→E+T+/))+I10I11I8T→(E))/+E→E+T+/)E→T)/+T→(E)+/)T→a+/)(aETI8I12I14I9I9E→T+/)I10T→(E) /+I11E→E+T+/)T→(E)+/)T→a+/)(TaI8I13I12I12T→a+/)I13E→E+T)/+I14T→(E)+/)E→E+T+/)+)I11I15I15T→(E) +/)LALR(1)分析：（合并同心集）狀態(tài)項目集經(jīng)過的符號到達的狀態(tài)I0E’ →EE→E+T/+E→T/+T→(E)/+T→a/+ETaI1I6/I9I4/I12I1E’ →EE→E+T/++I2/I11I2/I11E→E+T/+/）T→(E)/+/）T→a/+/）T(aI3/I13I5/I8I4/I12I3/I13E→E+T)/+/I4/I12T→a+/)/I5/I8T→(E)/+/）E→E+T+/)E→T)/+T→(E)+/)T→a+/)T(EaI6/I9I5/I8I7/I14I4/I12I6/I9E→T+/)/I7/I14T→(E)+/)/E→E+T+/)+)I2/I11I10/I15I10/I15T→(E) +/)/LR(1)ACTIONGOTO+()aET0S5S4161S2ACC2S5S433R1R14R4R45S8S12796R2R27S11S108S8S121499R2R210R3R311S8S121312R4R413R1R114S11S1515R3R3可以看出，表中無沖突項，所以是LR(1)文法；LALR（1）分析表：LR(1)ACTIONGOTO+()aET0S5/S8S4/S1216/91S2/S11ACC2/11S5/S8S4/S123/133/13R1R1R14/12R4R4R45/8S5/S8S4/S127/146/96/9R2R2R210/15R3R3R37/14S2/S11S10/S15 解： （1）求LR(1)項目集和狀態(tài)轉(zhuǎn)換圖：狀態(tài)項目集經(jīng)過的符號到達的狀態(tài)I0E’ →EE→E+E,+,*E→E*E,+,*E→i,+,*EiI1I2I1E’ →EE→E+E,+,*E→E*E,+,*+*I3I4I2E→i,+,*I3E→E+E,+,*E→E+E,+,*E→E*E,+,*E→i,+,*EiI5I2I4E→E*E,+,*E→E+E,+,*E→E*E,+,*E→i,+,*EiI6I2I5E→E+E,+,*E→E+E,+,*E→E*E,+,*+*I3I4I6E→E*E,+,*E→E+E,+,*E→E*E,+,*+*I3I4依據(jù)以上圖求出該文法的LR(1)分析表知道由于項目I5,I6導(dǎo)致了有多重定義的元素，所以不是LR(1)文法。事實上，從文法本身可以看出它是二義性的，因此不可能是LR（1）文法。等價的LR(1)文法為：E→E+T|TF→T*F|FF→i。另外，對原文法的沖突項來說，若考慮算術(shù)運算符的運算優(yōu)先級別，以及結(jié)合方式，上述沖突是可解決的。例如，假設(shè)表達式運算滿足左結(jié)合律（即a+b+c=(a+b)+c而不是右結(jié)合律