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1、作业题:2.1(a,c,d), 2.8(a,c,d), 2.12, 3.3, 3.4, 4.8, 4.125.8(a,b,c), 5.12, 6.7, 6.8, 6.137.4, 7.15第二章作业：21 (a) a | aa-z*a(c) 1-90-9*(d) 0-9*024682.8（a）Start a a b-z a b-z 正则表达式中丢了单独 a 的情况的比较多，另外有些同学在有 NFA 到 DFA 的转化过程中，不能够正确的确定最终的状态，造成有两个终结状态的情况。（c）Start 1-90-9 （d）问题比较多，建议同学画出 DFA 图生成的全部过程。2.1234 （c）.rew
2、rite this grammar to establish the correct precedences for the operator.rexp - rexp “|” rexp1 | rexp1rexp1 - rexp1 rexp2 | rexp2rexp2 - rexp3 * | rexp3rexp3 - (rexp) | letter左结合48 （a） 消除左递归lexp - atom | list atom - number | identifierlist - (lexp-seq)lexp-seq - lexp lexp-seqlexp-seq - lexp lexp-seq
3、| first (lexp)= number ,identifier,( first (atom)= number, identifier first (list) = ( first (lexp-seq)= number ,identifier,( first (lexp-seq)= number ,identifier,( , a abaaba bba aaba bbfollow(lexp)= $,),number,identifier,(follow (atom)= $,), number,identifier,( follow (list) = $,), number,identifi
4、er,(follow (lexp-seq)= )follow (lexp-seq)= )MN,T number identifier ( ) $lexp lexp-atom lexp-atom lexp-listatom atom-number atom-identifierlist list - (lexp-seq)lexp-seq lexp-seq - lexp lexp-seqlexp-seq - lexp lexp-seqlexp-seq - lexp lexp-seqlexp-seq lexp-seq - lexp lexp-seqlexp-seq - lexp lexp-seqle
5、xp-seq - lexp lexp-seqlexp-seq - 4.12(a) LL（1）文法不会是二义的，因为它的分析表的每个入口是唯一的。就不会是二义文法了。(b) 二义文法也不可能是 LL（1）的，否则在 LL（1）的 parsing table 中会产生各种冲突。非二义的文法不一定是 LL（1）的，因为二义文法是非 LL（1）文法的一个因素，但不是唯一的因素。如一些带有左递归的文法也不是 LL（1）文法。第五章作业512 s ssa A d | b B d | a B e | b A eAcBc5.8 Consider the following grammardeclaration
6、 - type var-listtype - int | floatvar-list - identifier, var-list | identifier (a) declaration - type var-listtype - int | floatvar-list - var-list, identifier | identifier s s,$sa A d,$ s b B d ,$s a B e,$ s b A e,$absaA d,$ saB e,$ Ac, dBc, es bB d ,$s bA e,$Ac, eBc, ds s,$saAd,$AsaAd,$saBe,$BsaBe
7、,$As bAe,$sbAe,$BsbBd,$sbBd,$Ac, eBc, dAc, dBc, ec c(b) decl - type vlist type - inttype -float 7vlist - vlist, idvlist - id061 243 5follow(decl)= $ , follow(type)= id follow(vlist)= $, , (c ) construct the SLR(1) parsing table for the rewritten grammarstate input gotoint float id , $ type Decl vlis
8、t0 6 7 11 3 22 4 r13 r5 r54 55 r4 r46 r27 r36.7 ,Decl- . type vlist type-.inttype -.floattypedecl - type. vlistvlist - . vlist , idvlist - .id vlistdecl - type vlist.vlist - vlist. , idvlist - vlist, .idvlist - vlist, id.vlist - id.idtype-int. type - float.iddecl var-list: type var-list.dtype = type
9、.dtypevsr-list11 var-list2, id var-list2.dtype = var-list1.dtypeid.dtype = var-list1.dtypetype integer type.dtype = inttype real type.dtype = float 6.8 decl id var-list id.dtype = var-list.dtypevar-list1 , id var-list2 var-list1.dtype = var-list2.dtypeid.dtype = var-list2.dtype var-list : type var-l
10、ist.dtype = type.dtypetype integer type.dtype = inttype real type.dtype = float6.13 Solution:a.u S vu A v u B v u C vabcOne correct order can be:C.v = 1B.u = S.uB.v = B.uA.u = B.v + C.vA.v = 2 * A.uS.v = A.vb.8.c.u S vu A v u B v u C vabcThe value is not calculable due to the circularity of the depe
11、ndency graph. The value of C.u depends on that of A.v, which in turn depends on that of A.u, which in its own turn depends on C.v, which finally depends on that of C.u, which is to be decided by the calculation.6.13 s.v = -2, 存在环，不是一个拓扑排序。74 main procedure a procedureb procedurec procedureb procedurec procedure715A、1，1B、3，1C、2，1D、2，2control linkreturn addressx: access link control linkreturn addressaccess link control linkreturn addressaccess linkcontrol linkreturn addressaccess link control linkreturn addressfpsp
