编译原理5、6

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5.1

---

(1)
$S -> (L) | a$
$L -> SL'$
$L'-> ,SL' | ℇ$
First($S$) = $\{(, a\}$
First($L$) = $\{(, a\}$
First($L'$) = $\{,\}$
Follow($S$) = $\{^{,} , )\}$
Follow($L$) = $\{ ) \}$
Follow($L'$) = $\{ ) \}$

(2)The parsing table:

-	$a$	$,$	$($	$)$	$
$S$	$S->a$		$S->(L)$
$L$	$L->SL'$		$L->SL'$
$L'$		$L'->,SL'$		$L'->\epsilon$

(3)

Step	Stack	Input	Reference	Action	Output
0	$\$S$	$(a,(a,a))\$$	$[S, (] = S->(L)$	derive	$S->(L)$
1	$\$)L($	$(a,(a,a))\$$		match
2	$\$)L$	$a,(a,a))\$$	$[L, a] = L->SL'$	derive	$L->SL'$
3	$\$)L'S$	$a,(a,a))\$$	$[S, a] = S->a$	derive	$S->a$
4	$\$)L'a$	$a,(a,a))\$$		match
5	$\$)L'$	$,(a,a))\$$	$[L', ^{,}] = L'->,SL'$	derive	$L'->,SL'$
6	$\$)L'S,$	$,(a,a))\$$		match
7	$\$)L'S$	$(a,a))\$$	$[S, (] = S->(L)$	derive	$S->(L)$
8	$\$)L')L($	$(a,a))\$$		match
9	$\$)L')L$	$a,a))\$$	$[L, a] = L->SL'$	derive	$L->SL'$
10	$\$)L')L'S$	$a,a))\$$	$[S, a] = S->a$	derive	$S->a$
11	$\$)L')L'a$	$a,a))\$$		match
12	$\$)L')L'$	$,a))\$$	$[L', ^{,}] = L'->,SL'$	derive	$L'->,SL'$
13	$\$)L')L'S,$	$,a))\$$		match
14	$\$)L')L'S$	$a))\$$	$[S, a] = S->a$	derive	$S->a$
15	$\$)L')L'a$	$a))\$$		match
16	$\$)L')L'$	$))\$$	$[L', )] = L->\epsilon$	derive	$L'->\epsilon$
17	$\$)L')$	$))\$$		match
18	$\$)L'$	$)\$$	$[L', )] = L->\epsilon$	derive	$L'->\epsilon$
19	$\$)$	$)\$$		match
20	$\$$	$\$$		accept

5.2

---

(1)
$A->BA'$
$A'->C|\epsilon$
$B->aB|\epsilon$
$C->ab$
First($A$) = First($B$) = First($A'$)= $\{a, \epsilon\}$
First($C$) = $\{a\}$
Follow($A$)=$\{\$\}$
Follow($A'$)=$\{\$\}$
Follow($B$)=$\{a, \$\}$
Follow($C$)=$\{\$\}$

(2)The parsing table:

-	$a$	$\$$
$A$	$A->BA'$	$A->\epsilon$
$A'$	$A'->C$	$A'->\epsilon$
$B$	$B->aB$|$B->\epsilon$	$B->\epsilon$
$C$	$C->ab$

Exist more than one element in $[A', a]$, so this is not a LL(1) grammar.

(3)When lookahead == 2:
First($A$) = $\{aa, ab, \epsilon\}$
First($B$) = $\{aa, \epsilon\}$
First($A'$)= $\{ab, \epsilon\}$
First($C$) = $\{ab\}$
Follow($A$)=$\{\$\}$
Follow($A'$)=$\{\$\}$
Follow($B$)=$\{ab, \$\}$
Follow($C$)=$\{\$\}$
The parsing table:

-	$aa$	$ab$	$\$$
$A$	$A->BA'$	$A->BA'$	$A->\epsilon$
$A'$		$A'->C$	$A'->\epsilon$
$B$	$B->aB$	$B->\epsilon$	$B->\epsilon$
$C$		$C->ab$

6.1

---

Step	Stack	Input	Reference	Action	Output
0	$\$$	$(a,(a,a))\$$	$\$<($	shift
1	$\$($	$a,(a,a))\$$	$(<a$	shift
2	$\$(a$	$,(a,a))\$$	$a>,$	reduce	$S->a$
3	$\$(S$	$,(a,a))\$$	$(<,$	shift
4	$\$(S,$	$(a,a))\$$	$,<($	shift
5	$\$(S,($	$a,a))\$$	$(<a$	shift
6	$\$(S,(a$	$,a))\$$	$a>,$	reduce	$S->a$
7	$\$(S,(S$	$,a))\$$	$(<,$	shift
8	$\$(S,(S,$	$a))\$$	$,<a$	shift
9	$\$(S,(S,a$	$))\$$	$a>)$	reduce	$S->a$
10	$\$(S,(S,S$	$))\$$	$,>)$	reduce	$L->S$
11	$\$(S,(S,L$	$))\$$	$,>)$	reduce	$L->L,S$
12	$\$(S,(L$	$))\$$	$(=)$	shift
13	$\$(S,(L)$	$)\$$	$)>)$	reduce	$S->(L)$
14	$\$(S,S$	$)\$$	$,>)$	reduce	$L->S$
15	$\$(S,L$	$)\$$	$,>)$	reduce	$L->L,S$
16	$\$(L$	$)\$$	$(=)$	shift
17	$\$(L)$	$\$$	$)>\$$	reduce	$S->(L)$
18	$\$)S$	$\$$		accept