8. LL(k) Parsing. Canonical LL(k) parser dual of the canonical LR(k) parser generalization of strong LL(k) parser

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1 3/8/ LL(k) Parsing LL(k) Parsing Canonical LL(k) parser The most general deterministic parsing method in which the input string is parsed (1) in a single Left-to-right scan, (2) producing a Left parse, and (3) using lookahead of length k. Canonical LL(k) parser dual of the canonical LR(k) parser generalization of strong LL(k) parser Generalization of shift-reduce pareser produce-shift parser Stack symbols valid stack string valid stack string viable prefix viable suffix LR(k) equivalence LL(k) equivalence

2 3/8/ LL(k) Parsing Viable Suffixes produce-shift parser adding lookahead adding lookback G abl : S aaab bab A cab a B SLL(k) parser LL(k) parser L(G abl ) = a{c}*{, a}ab b{c}*{, a}b. G abl is not SLL(k) for any k. First k ( Follow k (A)) = {k:ab, k:b} First k (afollow k (A)) = {k:aab, k:ab} First k ( Follow k (A)) First k (afollow k (A)) = {k:ab} produce-produce conflict for A a and A. adding lookahead A ab A a a ab A ab G abl is not SLL(k) for any k. adding lookahaed and lookback bb m A ab A a bb m a ab bab m A ab A bab m ab G abl is LL(2) A a A. A a A.

3 3/8/ LL(k) Parsing 3 3 A string is a viable stack string of pda M, if $ s xy$ * $ y$ * $ f $ in M. stack string in some accepting computation M. G abl : S aaab bab A cab a B. Viable stack string of G abl (in regular exp) {S} {baaa} {baa} {ba} {b} { } {bab} {ba} {bab n Ac n 1} {bb n Ac n 1} {bab n A n 1} {bb n A n 1} {bab n a n 1} {bb n a n 1} {bab n n 1} {bb n n 1} An action r is valid for viable stack string of M if $ y $ r $ y$ in M. k:y$ First k ( R $). Note that the action r is valid for not for. Two viable stack string belongs to the same equivalent class if they have same set of valid actions. At most + k P G k+1 distinct actions at most 2 G k+1 equivalent classes

4 3/8/ LL(k) Parsing 4 4 prev. class: valid action equi. class [S] S a baaa a [baaa] [S] S b bab b [bab] [baaa] a a [baa] [bab] b b [ba] [baa] A c BAc c [bab + Ac] [baa] A aa a aa [baa] [baa] A ab ab [ba] [baa] a a [ba] [ba] a a [b] [b] b b [ ] [ba] A c BAc c [bb + Ac] [ba] A ab a ab [ba] [ba] A b$ b$ [b] [bab + Ac] c c [bab + A] [bb + Ac] c c [bb + A] [bab + A] A c BAc c [bab + Ac] [bab + A] A aa a aa [bab + a] [bab + A] A ab ab [bab + ] [bab + a] a a [bab + ] [bab + ] B ab ab [bab + ] [bab + ] B ab ab [ba] [bb + A] A c BAc c [bb + Ac] [bb + A] A ab a ab [bb + Aa] [bb + A] A b$ b$ [bb + ] [bb + Ac] c c [bb + A] [bb + Aa] a a [bb + A] [bb + ] B b$ [bb + ] [bb + ] B b$ [b] [S]

5 3/8/ LL(k) Parsing 5 5 The equivalence should be (1) right-invariant (2) any two equivalent viable stack strings should end with the same symbol {, ba} { }, {ba} {bb n n 0} {b}, {bb n n 1} {bb n A n 1} {bab n A n 1}, {bb n A n 1}, {bab n A n 1} {baa} {bab n a n 1} {A aa a aa} {bab n a n 0} same valid actions 16 equivalent classes [ ], [S], [b], [ba[, [baa], [baaa], [ba], [bab], [bab + ], [bab + A], [bab + c], [bab * a], [bb + ], [bb + A], [bb + c], [bb + a].

6 3/8/ LL(k) Parsing 6 6 S aaab bab A cab a B a [bb + a] c [bb + Ac] B [bb + ] A [bb + A] a [ ] b B [b] a [ba] B B a [bab + ] S A A [S] [ba] [baa] b [bab] a [baaa] [bab * a] A [bab + A] c [bab + Ac]

7 3/8/ LL(k) Parsing 7 7 A simplified canonical LL(k) parser the new produce action [ A] y Y n Y 1 y, where A is viable stack string, A X 1... X n is a rule, y First k (X 1 X n R $), Y i = [ X n X i ] X i N X i X i. same shift action a a. the shift action is same as that of SLL(k) parser. Example: simplified canonical LL(2) parser of G abl (p. 202 ) [S] 2 aa ba[baa] 2 a aa, S aaab. in the SLL(2) parser A ab a ab, A ab ab. in the simplified canonical LL(2) parser [ba] 2 ab a ab, [bb + A] 2 ab a ab. [baa] 2 ab ab, [bab + A] 2 ab ab. G is not SLL(2) but simlified canonical LL(2).

8 3/8/ LL(k) Parsing 8 8 String ( ( ) R R R ) V * is a viable suffix of G, if S lm * xa lm x ( x R ) where V *, x *, and A P. is a complete viable suffix, if =. Note that the viable suffix is reversed, since the top of the stack is the rightmost symbol. Fact 8.1 Any (reversal of) viable suffix is a (suffix) prefix of some complete viable suffix. Lemma 8.2 n 0, 1, 2 V*, r (a) 1 1 r n rm 2 in G iff 1 R lm r 1R r n R 2 R in G R (b) 1 lm r 1 r n 2 in G iff 1 R rm r 1R r n R Proof induction on n. 2 R in G R r R : A R in G R whenever r: A in G. The set of viable suffixes of G coincides with the set of viable prefixes of G R.

9 3/8/ LL(k) Parsing 9 9 Lemma8.3 (a) V * is a (complete) viable prefix of G iff is a (complete) viable suffix of G R. (b) V * is a (complete) viable suffix of G iff is a (complete) viable prefix of G R. Proof. (a) (only if) If S rm * Ay r rm y (= y) holds in G, then, S * lm ( Ay) R = y R A R * lm y R ( ) R R = y R R R = y R R R holds in G R. (L 8.2) (if) (G R ) R = G. only if part of (b) = if part of (a) only if part of (a) = if part of (b)

10 3/8/ LL(k) Parsing Theorem 8.4 the set of all viable suffixes of G is a regular language over V Given a grammar G = (N,, P, S), Let G VS = (N VS, VS, P VS, [S]) where N VS = {[A] A N}, VS = N, and P VS = {[A] R A P} {[A] R [B] A B P, B N}. (G abl ) VS : G abl : [S] b ba baa baaa ba bab ba[a] b[a] S aaab bab, [A] B BA BAc [B] B[A] a A cab a, [B] B. a A b a A b b [S] c A a B [A] [B]

11 3/8/ LL(k) Parsing [ ] S S S aaab S bab [(b ba)b + B ] b [b] S aaa b A ca B S ba b A cab B A cab A A A a B B A a B a [(b ba)b + A] A S aa ab A c AB A a A cab c A [(b ba)b + Ac] A a A cab a B a A a S S S [S] A [ba] S b Ab [bab] b S bab [ba] A [baa] S a Aab [baaa] a S aaab [(bb + bab * )a] B [(b ba)b + ] A [(b ba)b + A] c S [ ] B b A [b] a a B [ba] A a [(b ba)b + Ac] [(bb + bab * )a] [S] [ba] [baa] b a [bab] [baaa]

12 3/8/ LL(k) Parsing Lemma 8.5 If S lm x = xa in G, and. Then S rm x A lm r x = x x rm x r " =, and 1: = 1:. (L6.2) If is a prefix of some nontrivially derived left sentential form not extending over the first nonterminal, then there is the derivation contains a segment that proves R to be a viable prefix, even so that the righthand side of the rule r cuts properly. S A x x = A

13 3/8/ LL(k) Parsing Lemma 8.6 S lm + xa. Then R A is a viable suffix. Proof. = A =. (L 8.5) Lemma 8.7 Any (suffix) prefix of a reversal of viable suffix is a (reversal of) viable suffix. Proof. Let 1 2 is a viable suffix,then 1 2 is a viable prefix of G R.If 1 is a viable prefix of G R (L 6.4), then a viable suffix of (G R ) R = G. (L 8.3(a)) Lemma 8.8 A P, if A is a viable suffix of G, then so is R. Lemma 8.9 Let M is its SLL(k) parser for G and $ y$ $ $ in M. Then $ y$ $ z$ " $ $ in M,and = ". Proof. induction on. (L 6.6) Lemma 8.10 Any prefix of a viable stack string of an SLL(k) parser is a viable stack string.(l 8.9)

14 3/8/ LL(k) Parsing Theorem 8.11 Any viable stack string of SLL(k) parser M is either S or viable suffix of G. Conversely, any viable suffix of G is a viable stack string of M, provided that G is reduced. (T6.7) Proof. (the first part of theorem) S, r the last produce action in $S w$ $ z$ * $ $ in M $S w$ $ A yz$ r $ R yz$ = $ y R yz$ $ z$, where 1 r 2 =, w=xyz 2 is a y}-lengthed shift actions. S lm *xa R in G, A P, y R = R.(L 5.27) S lm * xa R lm x R = x( y R ) R, y R is viable suffix, so is. Q.E.D.

15 3/8/ LL(k) Parsing Valid LL(k) Items Let A P. Then [A, y] is a k-item. A is a core item, y k: * is a lookahead string. A k-item [A, y] is LL(k)-valid(or valid) for string V * if S lm * xa lm x (=x R ) and y First k ( R ). The stack string is the prediction of remained input string. Fact 8.12 If [A, y] is a LL(k) valid item for string, then is a viable suffix,[a, y] is a k- item, R is a suffix of (= R R ) and y First k ( R ) = First k ( ) First k ( Follow k (A)) Conversely, if a string is a viable suffix, then some item is LL(k)-valid for.(f6.12) Example : LL(k) valid item of G abl

16 3/8/ LL(k) Parsing Let V *. Then LL(k) G = Valid LL(k) G ( ) = {I I is an LL(k)-valid item for } = {[A, y] S lm * xa lm x =x R y First k ( R )} Valid G LL(k) Valid LL(k) Valid k Valid G LL(k) LL(k) k. String 1 is LL(k)-equivalent(equivalent) to string 2, written 1 LL(k) 2 (or 1 k 2 ), if Valid k ( 1 ) = Valid k ( 2 ). The relation k is the LL(k)-equivalence for G. [ ] k denotes an equivalent class of under k [ ] k = { k } [ ] k [ ] k [ ]. We denote [ ] k by [ ] k (or even [ ]).

17 3/8/ LL(k) Parsing Fact 8.13 From the definition of a viable suffix implies that [A 1 1 1, y 1 ], [A 2 2 2, y 2 ] k {z [A 1 1 1, z] k } = {z [A 2 2 2, z] k } = First k ( R ). Note that the viable suffix is reversed in LL parsing. Theorem 8.14 The LL(k)-equivalence k for G is of finite index, and the index of k is at most 2 G ( +1)k.(T6.13) Proof. The number of distinct subsets of k: * is at most 2 ( +1)k, each of there subsets may occur in at most 2 G sets 0. (F 8.13) Hence, 2 G 2 ( +1)k =2 G ( +1)k. Q.E.D.

18 3/8/ LL(k) Parsing Lemma 8.15 Let k l. Then k = {[A, k:y] [A,y] l }.(L6.14) Lemma 8.16 Let k l. Then LL(l)-equivalence is a refinement of LL(k)-equivalence. More specifically [ ] k = [ ] l.(l6.15) canonical collection of set of LL(k)-valid items for G = canonical LL(k) collection for G: C k. Simplified canonical LL(k) machine M (or deterministic LL(k) machine) M = (C k, V, { k X X k }, k, ) Example : the canonical LL(2) machine for G abl

19 3/8/ LL(k) Parsing Construction of the canonical LL(k) machine LL(k) : relation on LL(k) items [A B, y] LL(k) [B y] desc LL(k), LL(k) k. I 2 is an immediate LL(k)-descendant of I 1, if I 1 I 2. I 2 is an LL(k)-descendant of I 1, if I 1 * I 2. I 1 is an (immediate) LL(k)-ancestor of I 2, if I 2 is an (immediate) LL(k)-descendant of I 1. k n = {[A,y] S lm n xa lm x (=x R ), y First k ( R )} Fact 8.17 k 0, k = n=0 n k.(f6.16) Lemma 8.18 If[A B, y] k n and m v *.Then B P, [B, y] k n+1.(l6.17) Proof. S lm n xa lm x B lm m xvb lm xv Lemma 8.19 k is closed under, i.e., * ( k ) = k.(l6.18)

20 3/8/ LL(k) Parsing Lemma 8.20 If [B, y] n k and n 0. Then [A B, y] m k, n-m-1 v, m n.(l6.19) Fact 8.21 k 0 = {[S R, ] S R P}. (F6.20) [A ] is LL-essential(or essential), if inessential, otherwise. Ess LL(k) (q) = {I q I is LL-essential}. Lemma 8.22 Let I k n, k, n 0. (1) n = 0, =, I = [S, ]. (2) and I is essential. (3) n 0, and I has an immediate ancestor in some k m, m n.(l6.21) Lemma 8.23 n k * ({S, ] S P}), n k * (Ess( n k )), if.(l6.22) Lemma 8.24 n k = * ({S, ] S P}) n k = * (Ess( n k )), if.(l6.23)

21 3/8/ LL(k) Parsing X k : relation on set of LL(k) items. [A X, y] X k [A X, z], z First k (Xy) pass-x, or X for short Basis LL(k) (q, X) = {[A X, z] [A X, y] q, and z First k (Xy)} X k (q). X k : relation on set of LL(k) items. Goto LL(k) (q, X) = * k (Basis LL (q, X)) = * k ( X k (q)) X k (q). X-successor, X k for short Lemma 8.25 If [A, y] k n, then R is a viable suffix and [A, z] R n k. If [A, z] n k, then s.t. [A, y] n k = R, z First k ( y).(f6.24) Lemma 8.26 Ess( X k ) = Basis( k, X)(L6.25) Lemma 8.27 X k = Goto( k, X)(L6.26)

22 3/8/ LL(k) Parsing Algorithm for constructing the canonical LL(k) machine begin compute the relation for X V do compute the X k relation q s := * k ({[S S, ]}); Q M := {q s }; P M := ; repeat for q Q M and X V do begin p := * k ( X k (q)); Q M := Q M {p}; P M := P M {q X p} end until nothing is added to Q M and P M. end

23 3/8/ LL(k) Parsing Lemma 8.28 Let M = (Q M, V, P M, q s, F) be a canonical LL(k) machine for G. Then the followings hold (a) M is deterministic. (b) q Q M, q has unique entry symbol. (c) q is accessible upon reading, iff q = k. (d) If F = { k for a given }, L M = [ ] k. (e) If F = { k k }, L M = Set of viable suffixes of X. (f) If F = { k for all }, L M = V *.(L6.27) Theorem 8.29 (a) The LL(k) equivalence of G is the equivalence induced by the canonical LL(k) machine of G. [ 1 ] = [ k 2 ], iff k q s 1 * q s q and q s 2 * q s q. (b) The LL(k) equivalence of G is right invariance. If [ 1 ] = [ k 2 ], then [ k 1 X] = [ k 2 X]. k (c) The LL(k) equivalence of G is ends with same symbols. If [ 1 ] = [ k 2 ], then k 1 :1 = 2 :1.(T6.28)

24 3/8/ LL(k) Parsing Canonical LL(k) Parser The canonical LL(k) parser for G is a pushdown transducer M = ([G] k,,,, [ ] k [S] k, {[ ] k }) where [G] k = {[ ] k V * } = {[ ] k [ A] k y [ ] k [ X n ] k [ X n X 1 ] k y [A X 1 X n, y] X n X 1 k } (pa) {[ ] k [ a] k ay [ ] k y [A a, k:ay] a k } (sa) ([ ] k [ A] k y [ ] k [ X n ] k [ X n X 1 ] k y) = A X 1 X n, ([ ] k [ a] k ay [ ] k y) =. compare in canonical LR(k) parser = {[ ] k [ X 1 ] k [ X 1 X n ] k y [ ] k [ A] k y [A X 1 X n, y] X 1 X n k } (ra) {[ ] k ay [ ] k [ a] k y a, [A a, z] k, y First max{k-1, 0} ( z)} (sa)

25 3/8/ LL(k) Parsing [B A, x] k q [B A,y] A k q A [A X 1 X n, x] k q (= q 0 ) [A X 1 X 2 X n, x n ] X n k q n [A X 1 X n, x 1 ] X n X 1 k q 1 Fig. 8.5 Algorithm for computing the parsing action of the canonical LL(k) parser. for q do for A X 1 X n P s.t. [A X 1 X n, x] q 0 do where q i+1 = GOTO(q i, X i+1 ) for 1 i n q A = GOTO(q,A) for [A X 1 X n, y] q 1 do where q q 0 generate qq A y qq n q 1 y od od for [A a, ay] q do generate q ay y od od

26 3/8/ LL(k) Parsing Fig 8.2 The Canonical LL(1) machine of G abl First 1 (S) = {a, b} First 1 (A) = {, a, c} First 1 (B) = { } [ ] S S, S S aaab, S S, {a,b} S bab, [S] [bb + B ] b [b] [ba] S aaa b, b A A ca B, b S ba b, b S b Ab,{a,b,c} A cab, b B A cab, b A, b A, b [bab] b A a, b A a, b S bab, b B, b B, b a [bb + A [ba] A] S aa ab, a A [baa] A c AB, A a, a S a Aab, {a,b,c} {a,c} A cab, a c A, a [(b ba)b + Ac] [baaa] a A a, a A cab, c a B, a S aaab,a a B B [(bb + bab * )a] [bab + ] A ca B, a A a, a [bab + A] A cab, a a A c AB,{a,c} A, a A a, a c B, a

27 3/8/ LL(k) Parsing Fig 8.3 The Canonical LL(2) machine of G abl First 2 (S) = {aa, ac, ba, bb, bc} First 2 (A) = {, a, c, ca} First 2 (B) = { } [ ] S S, S S aaab, S S, {aa,ac,ba,bb,bc} S bab, [S] [bb + B ] b [b] [ba] S aaa b,b A A ca B, b S ba b, b S b Ab, {ab,b,ca,cb,cc} A cab, b B A cab, b A, b A, b [bab] b A a, b A a, b S bab, B, b {ba,bb,bc} B, b a [bb + A [ba] A] S aa ab,ab A [baa] A c AB, A a, ab S a Aab, {ab,b,ca,cb,cc} A cab, ab {aa,ab,ca,cc} c A, ab [bb + Ac] [baaa] a A a, ab A cab, a B, ab S aaab,a {ca,cb,cc} B [bab + {aa,ac} B [bb + a] ] A ca B,ab A a,ab [bab + A] a A cab,ab A c AB,{a,c} A, ab {aa,ab,ca,cc} [bab * a] A a, ab c B, ab [bab + Ac] A a,aa a A cab, {ca,cc}

28 3/8/ LL(k) Parsing $[ ][S] accab$ S aaab $[ ][b][ba][baa][baaa] accab$ a $[ ][b][ba][baa] ccab$ A cab $[ ][b][ba][bab][baba][babac] ccab$ c $[ ][b][ba][bab][baba] cab$ A cab $[ ][b][ba][bab][babb][babba][babbac] cab$ c $[ ][b][ba][bab][babb][babba] ab$ A $[ ][b][ba][bab][babb] ab$ B $[ ][b][ba][bab] ab$ B $[ ][b][ba] ab$ a $[ ][b] b$ b $[ ] $ $[ ][S] bccb$ S bab $[ ][b][ba][bab] bccb$ b $[ ][b][ba] ccb$ A cab $[ ][b][bb][bba][bbac] ccb$ c $[ ][b][bb][bba] cb$ A cab $[ ][b][bb][bbb][bbba][bbbac] cb$ c $[ ][b][bb][bbb][bbba] b$ A $[ ][b][bb][bbb] b$ B $[ ][b][bb] b$ B $[ ][b] b$ b $[ ] $

29 3/8/ LL(k) Parsing Example: the difference between the simplified canonical LL(k) parser and the true canonical LL(k)parser.(p. 222 ) in simplified canonical LL(k) parser, [ A] k y in true canonical LL(k)parser [ ] k [ A] k y.

30 3/8/ LL(k) Parsing The canonical LL(k) parser yields a valid left parser. LL(k) parser left parser L8.30, L8.31(L6.29, L6.30) left parser LL(k) parser L8.32, L8.33, L8.34(L6.31, L6.32, L6.33) LL(k) parser left parser T8.35(T6.34) Lemma 8.30 Let M be a LL(k) parser for G. If $[$] k [$X 1 ] k [X 1 X m ] k w$ in M. Then = $[$] k [$Y 1 ] k [$Y 1 Y n ] k y$, w=xy, X m X 1 ( ) lm xy n Y 1 in G, and = ( ) + x. Proof Induction on. (L 5.12 or L 5.27) Lemma 8.31 Let M be a canonical LL(k) parser for G. Then (1) L(M) L(G), (2) is a left parse of w in M, ( ) is a left parse of w in G, (3) Time G (w) Time M (w) - w.

31 3/8/ LL(k) Parsing Lemma 8.32 Let M be a LL(k) parserfor G. If a n a 1 is a viable suffix of G, k:y$ k First k ( R ). Then s.t. is a string of an n-length shift actions of M, $ [ ] k [ a n a 1 ] k a 1 a n y$ k $ [ ] k y$ k Lemma 8.33 Let M be a LL(k) parser for G. If X m X 1 lm xy n Y 1 in G, [A ] $X 1 X m 0, k:y$ First k (Y n Y 1 ), and either Y n Y 1 = or Y n is nonterminal, Then ( ) =, = + x, and $[$] k [$X 1 ] k [$X 1 X m ] k xy$ $[$] k [$Y 1 ] k [$Y 1 Y n ] k y$ Proof. Induction on.

32 3/8/ LL(k) Parsing Lemma 8.34 Let M be a canonical LL(k) parser for G. Then (1) L(G) L(M), (2) is a left parse of w in G, ( ) =,s.t. is a left parse of w in M, (3) Time G (w) Time M (w) + w. Theorem 8.35 Let M be a canonical LL(k) parser for G. Then (1) M is a left parser for G. (2) w L(G), M produces all left parses of w. (3) Time M (w) = Time G (w) + w.

33 3/8/ LL(k) Parsing Lewis-Sterns Canonical LL(k) Parser The canonical LL(k) parser for G is a pushdown transducer M = ([L] k,, [S, {$ k }] k, { },, ) where [L] k = {[X, R] k X V, R Follow k (X)} = {[A, R] k y [X n, R n ] k [X 1, R 1 ] k y A X 1 X n P, y First k (X 1 X n R), R n = R, R i = First k (X i+1 X n R)} (pa) {[a, R] k ay y a } (sa) ([A, R] k y [X n, R n ] k [X 1, R 1 ] k y) = A X 1 X n, ([a, R] k ay y) =.

34 3/8/ LL(k) Parsing Algorithm Construction of [L] k and. [L] k := {[S, {$ k }] k }; := {}; repeat for [A, R] k [L] k do for A X 1 X n P do R n := R; for i :=n downto 1 do [L] k := [L] k [X i, R i ] k ; R i-1 := First k (X i R i ) if X i then := {[X i, R i ] k X i y y y R i } fi od; := {[A, R] k y [X n, R n ] k [X 1, R 1 ] k y y R 0 } od od until [L] k does not change

35 3/8/ LL(k) Parsing LL(k) Grammars A grammar G is an LL(k) grammar, if its canonical LL(k) parser is deterministic. K-item [A 1 1,y 1 ] and [A 2 2,y 2 ] of G exhibit a produce-produce conflict if A 1 =A 2, y 1 =y 2 but 1 2. Lemma 8.36 The canonical LL(k) parser for G is nondeterministic iff there is a viable suffix, and k- item [A 1,y] [A 2,y] exhibiting a produceproduce conflict s.t. (a) [A 1,y] 1 R k [A 2,y] 2 R k.(l6.36) Proof. ( ) Assume (a) holds, then [ ] k [ A] k y [ ] k [ X m ] k [ X m X 1 ] k y, X 1 X m = 1 [ ] k [ A] k y [ ] k [ Y n ] k [ Y n Y 1 ] k y, Y 1 Y n = 2 1 2, the parser is nondeterministic.

36 3/8/ LL(k) Parsing ( ) Assume the parser is nondeterministic Let be configuration to which r 1, r 2 are applicable. then (1) r 1 =[ ] k [ A] k y [ ] k [ X m ] k [ X m X 1 ] y, r 2 =[ ] k [ B] k z [ ] k [ Y n ] k [ Y n Y 1 ] z, (2) r 1 =[ ] k [ A] k y [ ] k [ X m ] k [ X m X 1 ] y, r 2 =[ a] k az z, (3) r 1 =[ a] k ay y, r 2 =[ b] k bz z. r 1, r 2 are both in same configuration, (2),(3) impossible. only case (1) remains. [A X 1 X m,y] X m X 1 k, [B Y 1 Y n,z] Y n Y 1 k = Y n Y 1 k A=B, y=z, but X 1 X m Y 1 Y n,otherwise r 1 =r 2. Q.E.D.

37 3/8/ LL(k) Parsing Nonterminal A N has the LL(k) property if First k ( 1 ) First k ( 2 )=. xa is left sentential forms, A 1, A 2 P (L6.37, L6.38; SLL(k) property) Theorem 8.37 The following statements are logically equivalent.(l6.39; T5.34) (a) G is an LL(k) grammar (b) [A 1,y] R 1 k, and [A 2,y] 2 R k always imply that 1 = 2. (c) A N have the LL(k) property. (d) S lm * xa lm x 1 lm * xv 1, S lm * xa lm x 2 lm * xv 2, and k:v 1 =k:v 2 always imply that 1 = 2. Proof. (a)=(b) (L 8.36) (b)=(b ) by LL(k)-validity(p. 207 ), (b ) S lm *x 1 A R lm x 1 1 R, y First k ( 1 R ), S lm *x 2 A R lm x 2 2 R, y First k ( 2 R ), always imply that 1 = 2. G G, S S, x 2 x 1 (S lm * x 1 A R, A 2 P)

38 3/8/ LL(k) Parsing (b") S lm * xa R lm x 1 R, y First k ( 1 R ), S lm * xa R lm x 2 R, y First k ( 2 R ), always imply that 1 = 2.(b")=(c)=(d). Theorem 8.38 for 1 l k, LA(k+1)LL(l) grammars LA(k+1)LL(l+1) grammars. for 0 l k, LA(k)LL(l) grammars LA(k+1)LL(l) grammars. Proof. examples(p. 231 ).

39 3/8/ LL(k) Parsing Equivalence of LL(1) and SLL(1) Lemma 8.39 Every LL(1) grammar is an SLL(1) grammar. Proof. Let G be an LL(1) grammar and consider, (1) S * lm x 1 A 1 lm x * lm x 1 v 1, S * lm x 2 A 2 lm x * lm x 2 v 2. Assume 1 * lm y 1, 1 * lm z 1, y 1 z 1 = v 1 2 lm * y 2, 2 lm * z 2, y 2 z 2 = v 2 (2) S lm * x 1 A 1 lm x lm * x 1 y 1 z 1 S * lm x 1 A 1 lm x * lm x 1 y 2 z 1, and (3) S * lm x 2 A 2 lm x * lm x 2 y 1 z 2 S lm * x 2 A 2 lm x lm * x 2 y 2 z 2 if y 1, then 1:v 1 = 1:v 2 1:y 1 = 1:y 2 z 2. (3) if y 2, then 1:v 1 = 1:v 2 1:y 1 z 1 = 1:y 2. (2) if y 1 =, y 2 =, then 1:v 1 = 1:v 2 1:z 1 = 1:z 2. If G = LL(1), (1) and 1:v 1 = 1:v 2 implies 1 = 2. G is SLL(1). (T 5.34) Q.E.D.

40 3/8/ LL(k) Parsing Lemma 8.40 k 0, any SLL(k) grammar is an LA- LL(k) grammar and any LALL(k) grammar is an LL(k) grammar. Thoerem 8.41 The classes of SLL(1), LALL(1), and LL(1) grammars coincide. Theorem 8.42 k 1, the class of SLL(k) grammar is properly contained in the class of LALL(k) grammar. Proof. counter example grammar which is LALL(k) but not SLL(k), for k 1. (p. 233 ) Properly containment of theorem 8.42 are properties of grammar only, but not properties of language. i.e. any LL(k) grammar can be transformed into an equivalent SLL(k) grammar.

41 3/8/ LL(k) Parsing Transformation of LL(k) grammar into SLL(k) T k (G) = ({[A, R] A N, R Follow k (A)},, P T(k), [S, {$ k }]), where P T(k) = [A, R] X 1...X m, and X i = [X i, First k (X i+1...x m R)], X i N, X i, X i. Nonterminal A is split into several nonterminals [A, R 1 ]...[A, R n ], each representing an occurrence of the original nonterminals in a particular context. Let h k homomorphism s.t. h k ([A, R] X 1...X m ) = A X 1...X m. G abl : LL(2) S aaab bab A cab a B. T k (G abl ): SLL(2) [S, {$ 2 }] a [A, {ab}] a b b [A, {b$}] b [A, {ab}] c [A, {ab}] [B, {ab}] a [A, {b$}] c [A, {b$}] [B, {b$}] a [B, {ab}] [B, {b$}]

42 3/8/ LL(k) Parsing Covering LL(k) grammars by SLL(k) grammars (1) T k (G) left-to-left covers G. T8.45(L8.43, L8.44) (2) T k (G) is SLL(k), iff G is LL(k). T8.47 T8.48, T8.49 Lemma 8.43 Let [A, R] lm xx 1...X n in T k (G).Then A lm h k ( ) xx 1...X n in G. where X 1 = [X i, First k (X i+1 X n R)], if X i N, X i, if X i. Proof. induction on. Lemma 8.44 Let A lm xx 1...X n in G, an either X 1...X n = or X 1 N. Then P^*. h k ( ) =, [A, R] lm xx 1...X n in T k (G). Proof. By induction on. Theorem 8.45 T k (G) left-to-left covers G w.r.t. h k. Corollary 8.46 If (M, ) is a left parser of T k (G), then (M, h k ) is a left parser of G.

43 3/8/ LL(k) Parsing Theorem 8.47 T k (G) is SLL(k) whenever G is LL(k). G is LL(k) whenever T k (G) is so. Proof. (proof of the first part of theorem) Let [A, R] X 1 X m, [A, R] Y 1 Y n P Tk, 0 m p, 0 n q, s.t. (1) [S, {$}] * lm x 1 [A, R] X m+1 X p lm x 1 X 1 X m X m+1 X p lm * x 1 y 1, [S, {$}] lm * x 2 [A, R]Y n+1 Y q lm x 2 Y 1 Y n Y n+1 Y q lm * x 2 y 2, k: y 1 = k: y 2 implies X 1 X? m = Y 1 Y n. (4) X 1 X m * lm u 1, X m+1 X p * lm v 1, u 1 v 1 = y 1. Y 1 Y n lm * u 2, Y n+1 Y q lm * v 2, u 2 v 2 = y 2. (5) X 1 X m lm * u 1, X m+1 X p lm * v 1. Y 1 Y n lm * u 2, Y n+1 Y q lm * v 2. (6) First k (X m+1 X p $) = R = First k (Y n+1 Y q $) (7) S lm * x 1 AX m+1 X p.(l 8.43)

44 3/8/ LL(k) Parsing (8) X m+1 X p * lm v 2 s.t. k:v 2 = k:v 2, so k: u 1 v 1 = k: y 1 = k: y 2 = k: u 2 v 2 = k: u 2 v 2 then S lm *x 1 AX m+1...x p lm x 1 X 1...X m X m+1...x p lm *x 1 y 1, S lm *x 1 AX m+1...x p lm x 1 Y 1...Y n X m+1...x p lm *x 1 y 2, k: y 1 = k: y 2 where y 2 = u2v2 Thus if G is LL(k) (i.e. X 1 X m = Y 1 Y n ) then Tk(G) is SLL(k) (i.e. X 1 X m = Y 1 Y n ). (proof of the LL(k)-ness of T k (G) implies the LL(k)- ness of G.) Assume that (8) is true, then (9) [S, {$}] lm *x 1 [A, R]X m+1...x p (10) [S, {$}] * lm x 1 [A, R]X m+1 X p * lm x 1 X 1 X m X m+1 X p * lm x 1 y 1, [S, {$}] * lm x 1 [A, R]X m+1 X p * lm x 1 Y 1 Y n X m+1 X p * lm x 1 y 2, k: y 1 = k: y 2. If T k (G) is LL(k)(i.e. X 1 X m = Y 1 Y n ), then G is LL(k)(i.e. X 1 X m = Y 1 Y n ). Q.E.D.

45 3/8/ LL(k) Parsing Theorem 8.47 T k (G) is SLL(k) whenever G is LL(k). G is LL(k) whenever T k (G) is so. Proof. Consider a derivation in G S * lm xax m+1 X p. Then [S, $ k ] * lm x[a, R]X m+1 X p. X i = [X i, First k (X i+1 X n R)], if X i N, = X i, if X i. First k (X m+1 X p ) = First k (X m+1 X p ) Assume G is LL(k). A N, First k ( 1 ) First k ( 2 ) =. A 1 A 2, S * lm xa. [S, $ k ] * lm x[a, R] where k ( ) =. where k ([A, First k (

46 3/8/ LL(k) Parsing Theorem 8.48 k 1, any grammar can be transformed into an equivalent grammar that left to left covers the original grammar, and is SLL(k) iff the original is LL(k). In fact, the transformed grammar is structurally equivalent to the original grammar. Theorem 8.49 For any k, the families of LL(k) language, LALL(k) languages, and SLL(k) languages coincide. Theorem 8.50 Any LL(k) grammar is unambiguous. Proof. Let S lm w and S lm w. Assume the contrary 1 = r 1 1, 2 = r 2 2, r 1 r 2 s.t. S lm xa lm r 1 1 xy, S r lm xa 2 2 lm xy, where xy = w. As G is LL(k), r 1 = r 2. (T 8.37) Thus 1 = 2 and G is unambiguous.q.e.d. Theorem 8.51 A reduced left recursive grammar is not LL(k), k 0. Proof. (T5.40, C5.41).

47 3/8/ LL(k) Parsing Comparison on the classes of LL(k) and LR(k) grammars. S * xaz x z * xyz LR(k) parser reduce A after scanning xy and seeing k:z LL(k) parser produce A after scanning x and seeing k:yz LL(k) grammars LR(k) grammar rightmost derivation leftmost derivation (viable suffix) leftmost derivation rightmost derivation (viable prefix)

48 3/8/ LL(k) Parsing Lemma8.52 Let G is LL(k), S 1 lm x 1 A 1 r 1 lm x 1 1 1, 1 1 lm v 1, 1 * lm y 1, S 2 lm x 2 A 2 r 2 lm x 2 2 2, 2 2 lm v 2, 2 * lm y 2, x 2 v 2 = x 1 v 1 v and k:y 1 = k:vy 2. Then either 1 r 1 1 is a prefix of 2 r 2 2 or vice versa. Proof. For the sake of contradiction 1 r 1 1 and 2 r 2 2 are not prefix of each other. (case1) 1 = 2 and r 1 r 2. (case2) 1 = r 1 " 1, 2 = r 2 " 2, where r 1 r 2 (case3) 1 r 1 = 2 r 2, 1 = r 1 " 1, and 2 = r 2 " 2, where r 1 r 2 Each case is in contradiction to the assumption that G is LL(k). Lemma 8.53 Let LL(k) grammar G has S * lm x 1 A 1 1 lm x 1 1 1, 1 * lm v 1, 1 * lm y 1, S * lm x 2 A 2 2 lm x 2 2 2, 2 * lm v 2, 2 * lm y 2, x 2 v 2 = x 1 v 1 v and k: y 1 = k: vy 2. Then either 1 * v 2 or v 2 * 1. Proof. (L 8.52).

49 3/8/ LL(k) Parsing S + rm Ay vs. S + lm xa R.( * x and R * y) viable prefix induced by S + lm xa R viable suffix R induced by S rm + Ay Let be a viable prefix of G, P *. Then is a viable prefix induced by, if (p1) = and = or (p2) =, = 1 r 2,. is a viable prefix induced by 1 and S 1 lm xa R r lm x B R 2 u. lm Let be a viable suffix of G, ˆ P * then is a viable suffix induced by ˆ if (s1) = and ˆ = or (s2) = R, ˆ = ˆ1r ˆ2 s.t. is a viable suffix induced by ˆ1 and S ˆ 1 rm Ay r rm B y, ˆ 2 rm v. (refer to figures in p )

50 3/8/ LL(k) Parsing Lemma 8.54 No rule string in P * induces more than one viable prefix (resp. viable suffix). Proof. Assume induces two distinct viable prefixes 1 and 2. S lm 1 x 1 A 1 1 R lm r 1 x 1 1 B R, 1 lm 1 u 1, S lm 2 x 2 A 2 2 R lm r 2 x 2 2 B R, 2 lm 2 u 2 x 1 u 1 B R x 2 u 2 B R 1 r r 2 2 therefore x 1 u 1 x 2 u 2, B 1 B 2, 1 1 R 2 2 R. If 1 2, then 1 r r 2 2 imply that 1 r 1 is a prefix of 2 or 2 r 2 is a prefix of 1. Assume 1 r 1 is a prefix of 2, 2 1 r 1, r 2 2 1, then S 1 r 1 lm x 1 1 B 1 1 R 1 R lm x 2 A 2 2 r 2 lm x 2 2 B 2 2 R 2 2 R lm x 2 u 2 B B 1 1 R 1 can not be involved in the derivation that uses r 2 2. B 1 1 R 1 can not be involved in the derivation x 1 1 B 1 1 R 1 lm x 2 A 2 R 2 r lm x 2 2 B 2 2 R 2.

51 3/8/ LL(k) Parsing B R is a suffix of 2 R, then contradiction because 1 1 R 2 2 R. Q.E.D.

52 3/8/ LL(k) Parsing Lemma 8.55 Let G has at least one rule with left handside S, and (a) S lm xa R and R * y. Then induces some viable prefix, and is a viable suffix induced by ˆ P *, s.t. (b) S ˆ rm Ay and * x Proof. Base case, if =, immediate by definition. If and = 1 r 2, r=a 1 A then (1) S 1 lm x 1 A 1 R r lm x 1 A R x 1 2 lm x (2) S lm 1 x 1 A 1 R, R * y 1 implies(by I.H.) 1 induces some viable prefix, is a viable suffix induced by ˆ1 P*, and ˆ (3) S 1 A rm 1 y 1 and * x 1. ˆ (4) S 1 A rm 1 y 1 r ˆ rm A y 1 and 2v rm (5) S rm ˆ Avy 1 = Ay, = * x 1 u=x (6) S lm 1 x 1 A 1 R lm r x 1 A R, and lm 2 u in (4) = R, is a viable suffix induced by ˆ1 is a viable suffix induced by ˆ= ˆ1r ˆ2 in (6) =, is a viable suffix induced by 1 is a viable suffix induced by = 1 r 2 Q.E.D.

53 3/8/ LL(k) Parsing Lemma 8.56 Let G has at least one rule with left handside S, and ˆ (a) S rm Ay and * x. Then ˆ reduces some viable suffix, and is a viable prefix induced by P *, and (b) S lm xa R and R * y.

54 3/8/ LL(k) Parsing Theorem 8.57 k 0, any reduced LL(k) grammar is an LR(k) grammar. Proof. S + S impossible. Let (1) S ˆ 1 rm 1 A 1 y 1 r 1 rm 1 1 y 1 = y 1 S rm ˆ 2 2 A 2 y 2 rm r y 2 = vy 2 and k:y 1 =k:vy 2 (2) S rm ˆ 1 1 A 1 y 1, 1 * x 1 (G is reduced) S rm ˆ 2 2 A 2 y 2, 2 * x 2 (3) S lm 1x1 A 1 1 R, 1 R * y 1 (L 8.56) S lm 2x2 A 2 2 R, 2 R * y 2 where 1, 2 are viable suffix induced by ˆ1 and ˆ2 respectively, 1, 2 are viable prefix induced by 1 and 2 (4) S lm 1x1 A 1 1 R lm r 1 x R, 1 * v 1, 1 R * y 1 S lm 2x2 A 2 2 R lm r 1 x R, 2 * v 2, 2 R * y 2 x 2 v 2 =x 1 v 1 v and k:y 1 =k:vy 2 (from 3, 1) If G is LL(k) grammar. Then either 1 R derives v 2 R or v 2 R derives 1 R ( L 8.53).

55 3/8/ LL(k) Parsing (case 1) 1 R derives v 2 R (5) S lm 1x1 A 1 1 R and 1 R * v 2 R * vy 2 (6) S rm * 1 A 1 vy 2 rm 1 1 vy 2 = vy 2 S rm * 2 A 2 y 2 rm 2 2 y 2 = vy 2 G is LL(k) and hence unambiguous, then 1 A 1 vy 2 = 2 A 2 y 2, hence 1 = 2, A 1 =A 2, v=, 1 1 = = v= 2 2 = 1 2 implies 1 = 2 (1) implies 1 = 2, A 1 =A 2, and 1 = 2, so G is LR(k). (case 2) v 2 R derives 1 R the proof is analogous to that of case 1. Q.E.D.

56 3/8/ LL(k) Parsing Theorem 8.58 k 0, the class of reduced LL(k) grammars is properly contained in the class of reduced LR(k) grammars Proof. By Theorem 8.57 and LR(0) grammar may be left-recursive. Proposition 8.59 k 0, the family of LL(k) languages is properly contained in the family of LR(1) languages Theorem 8.60 k 1, the class of SLL(k) grammars is incompatible with the class of SLR(k) grammars, and the class of LALL(k) grammars is incompatible with the class of LALR(k) grammars Proof. example grammars(p. 248 ).

57 3/8/ LL(k) Parsing Construction of LL(k) Parsers Construction of canonical LL(0) machine. Let G = (V,, P, S). M 0 = (I 0 {q s }, V, q s, I f, 0 ) is the nondeterministic LL(0) machine for G, where I 0 : set of 0-items, V: input alphabet, q s I 0 : initial state, I f = I 0 : set of final states, 0 : (I 0 {q s }) (V { }) 2 I 0 of the form; q s [S ], [A X ]X [A X ], X V, and [A B ] [B ] Lemma 8.61 The set of viable suffixes of any G is the language accepted by M 0 of G, and for any viable suffix, 0 = {q q I 0 : q 0 '(q s, )} where 0 '(q, X) = {p V *, X V, r 0 '(q, ): p 0 (r, X)}.

58 3/8/ LL(k) Parsing Theorem 8.62 Deterministic LL(0) machine for any grammar G can be computed in time O(2 G +2log G ). (T 3.30) Let G = (V,, P, S). M 1 = (I 1 {q s }, V, q s, I f, 1 ) is the nondeterministic LL(1) machine for G where I 1 : set of 1-items, V: input alphabet, q s I 1 : initial state, I f = I 1 : set of final states, 1 : (I 1 {q s }) (V { }) 2 I 1 of the form; q s [S, ], [A X, y]x [A X, z], X V, z First 1 (Xy). [A B, y] [B, y] Lemma 8.63 The set of viable suffixes of any grammar G is the language accepted by M 1 of G, and for any viable suffix, 1 = {q q I 1 : q 1 '(q s, )} where 1 '(q, X) = {p V *, X V, r 1 '(q, ): p 1 (r, X)}.

59 3/8/ LL(k) Parsing Theorem 8.64 The cannonical LL(1) machine for any grammar G can be constructed in time O(2 2 G +3log G ). Theorem 8.65 The canonical LL(1) parser of any grammar G can be constructed in time O(2 2 G +3log G ).

60 3/8/ LL(k) Parsing Construction of LALL(1) Parsers To construct LALL(1) parser of a grammar G construct canonical LL(0) machine of its $-augmented grammar G' of G and lookahead symbols of states. For state q in the canonical LL(0) machine of G' Lall(q) = {b {$} [A, b] [ ] 0 1 } where Lall(q) is Lall(1) lookahead set of q. Fact 8.66 The LALL(1) parser has a produce action of the form [ ] 0 [ A] 0 b [ ] 0 [ X n ] 0... [ X n... X 1 ] 0 b iff [ X 1... X n ] X n... X 1 0 and b Lall( X n... X 1 0 ) Lemma 8.67 For any state q in the canonical LL(0) machine, Lall(q)={b First 1 ( R ) $V *, 0 = q} In other words, Lall(q) is the union of all sets First 1 ( R ). Proof. (F 8.13)

61 3/8/ LL(k) Parsing Computation of Lall(1) lookahead set q contains-item A, if A q. (q, A) goes-to Goto(q, A), if Goto(q, A) ø q has-null-transition (q, B), if Goto(q, B) ø and B is nullable. has-lall-lookahead = (goes-to -1 has-null-transition -1 ) * contains-item points first-of -1 where A X points X a first-of X if and only if a First 1 (X) X. q '.. q Portion of a canonical LL(0) machine. If is nullable and a First 1 (X) then q (goes-to -1 has-null-tranisiton -1 ) *q contains-item A X points X first-of -1 a that is, q has-lall-lookahead a

62 3/8/ LL(k) Parsing Theorem 8.68 Let G be a reduced grammar. a, a Lall(q) iff q has-lall-lookahead a. Proof. ( ) q has-lall-lookahead a = q (goes-to -1 has-null-transition -1 ) * q' contains-item points first-of -1 a for some q If q = 0 for a viable suffix, then q = 0, *. q contains-item points first-of -1 a implies that a First 1 ( R ). a Lall(q),since a First 1 ( R R ), is nullable. ( )Assume that a Lall(q), of G such that 0 =q and a First 1 ( R ). a First 1 ( R ) implies that R = R X R where R *, a First 1 (X). q (goes-to -1 has-null-transition -1 ) * q' contains-item points first-of -1 a where q = X 0. q has-lall-lookahead a. Q.E.D.

63 3/8/ LL(k) Parsing Non-Left-Recursive Grammatical Covers A reduced left-recursive grammar cannot be an LL(k) grammar. (T 8.51) Nonterminal A is directly left-recursive if the grammar has a rule A A Transformations of left-recursion to right-recursion G: A A Consider A * lm A n lm n * lm y n * lm yxn i = w L (w) = (A A ) n (A ) L (y) L (x i ) n. A rm A * rm Ax n * rm A(x i ) n n rm (x i ) n * rm yx n i. R (w) = R (y)(a ) ( R (x i )(A A )) n. G^:A A A A A lm A * lm ya lm y A * lm y n * lm yxn i = w ^L (w) = (A A ) L (y) ((A A ) L (x i )) n (A ). A rm A rm A * lm n A rm n * rm yx n i. ^R (w) = R (y) ( R (x i ))n (A ) (A A ) n (A A ).

64 3/8/ LL(k) Parsing No way to make a homomorphism that produce a cover. G : A A R 1 R 2 R 1, R 2 G^ : A R 2 A A R 1 A R 1, R 2 h(r 1 ) = A A h(r 2 ) = A h(a R 2 A ) = h(a R 1 A ) = ^L (w)=(a R 2 A ) L (y)(r 2 )((A R 1 A ) L (x i ) (R 1 )) n (A ). h( ^L (w)) = R (y)(a )( R (x i )(A A ))n = R (w) G : A R 1 A R 2 R 1, R 2

65 3/8/ LL(k) Parsing Removing of left recursion from a grammar such that a cover is obtained A reduced grammar, which is -free and does not contain any nonterminal A such that A + A,called a proper grammar Lemma 8.71 Any proper grammar G can be transformed in time O( G ) into a grammar Ĝ =(Vˆ,T,Pˆ,S) such that the following statements hold (1) Ĝ is not directly left-recursive (2) V Vˆ (3) L Ĝ (X) = L G (X) for all X V Proof. A A 1 A n 1 m, n 1, m 1 A 1 A m A, A 1 A n A Q.E.D.

66 3/8/ LL(k) Parsing Theorem 8.72 Any proper grammar G can be transformed in time O( G 2 ) into a grammar Ĝ =(Vˆ,,Pˆ,S) such that the following statements hold (1) Ĝ is not left-recursive (2) V Vˆ (3) L Ĝ (X) = L G (X) for all X V Proof. Algorithm for removing left recursion from a proper grammar G=({A 1,,A n },, P, S}. Pˆ := P; for i := 2 to n do begin for j := 1 to i-1 do if A i A j for some is in Pˆ then replace A i A j by the set of rules end {A i j A j j Pˆ }; eliminate the direst left recursion possibly caused by nnterminal A i (use the method described in the proof of L 8.71).

67 3/8/ LL(k) Parsing A non left-recursive covering grammar, G T, for a given any proper grammar G G T =(V P, P,P T,S},where P T = {A (A, ) (A, ) P}. Lemma 8.73 Let a string w ( P)*, n>0, such that (a) X n w in G T Then, (b) w = y 1 r 1 y 2 r 2 y n r n, for some y i * and r X n r 1 rm y 1 y 2 y n in G. Proof. Base n=1, immediate. Case n>0, follows from construction of G T and I.H. (a) implies that (L 4.1) X X 1 X l (X,X 1 X l ) P T. X X 1 X l (X,X 1 X l ), X i n i w i, i=1,,l, w 1 w l (X,X 1 X l )=w, and n 1 + +n l = n-1 r n r 1 = r n (X X 1 X l ) and r n-1 r 1 Q.E.D. (from I.H.)

68 3/8/ LL(k) Parsing Lemma 8.74 Let n 0, such that (a) X r n r 1 rm y in G Then, (b) y = y 1 y 2 y n, and X n y 1 r 1 y 2 r 2 y n r n in G T. Let Ĝ T =(Vˆ, P,Pˆ T,S) be a grammar obtained by removing left recursion from G T by means of theorem Let Ĝ N =(Vˆ,,PˆN,S) where PˆN = Pˆ T { r r P} Then Ĝ N left-to-right covers G w.r.t. h. where, h : PˆN * P* by: (1) h(r ) = r, where r P (2) h(a ) =, where A Pˆ T Ĝ N interprets each rule in P as a nonterminal.

69 3/8/ LL(k) Parsing Lemma 8.75 Let y be a sentence in L(Ĝ T ) and ˆ a left parse of y in Ĝ T.Then h( ˆ ) is right parse of y in G. Proof. As ˆ is left parse, then ˆ = ˆ 1(r 1 ) ˆ 2(r 2 ) ˆ n(r n ) ˆ n+1 S lm ˆ 1 y 1 r 1 1 lm y 1 1 lm ˆ 2 y 1 y 2 r 2 2 lm y 1 y 2 2 lm ˆ n y 1 y 2 y n r n n lm y 1 y 2 y n n lm ˆ n+1 y 1 y 2 y n y n+1 = y, where ˆ n+1 =, y n+1 = = n (rule in PˆT do not derive * ) S ˆ 1 ˆ n lm y 1 r 1 y n r n in Ĝ T S + lm y 1 r 1 y n r n in G T. (L 8.72) S r n r 1 rm y 1 y n in G. ( L 8.73) h( ˆ )=r 1 r n Q.E.D.

70 3/8/ LL(k) Parsing Lemma 8.76 Let y be a sentence in L(G) and a right parse of y in G.Then y has in Ĝ N a left parse ˆ such that h( ˆ )=. Proof. = r 1 r n, S rm r n r 1 y in G, is right parse, S n y 1 r 1 y n r n in G T, y 1 y n =y, (L 8.74) S + y 1 r 1 y n r n in Ĝ T, (L 8.72) S lm y 1 r 1 y n r n in G T, (L 4.2)where PˆT *. Get ˆ a left parse of Ĝ N from by inserting r i at that place in where r i become the first nonterminal in the derivation. ˆ = ˆ 1(r 1 ) ˆ n(r n ), where ˆ 1 ˆ n = then ˆ S lm y 1 y n = y in Ĝ N with h( ˆ )= Q.E.D. Theorem 8.77 For any proper grammar G, there is a non-left-recursive grammar Ĝ and a homorphism h such that (Ĝ,h) is a left-to-right cover to G.

71 3/8/ LL(k) Parsing Predictive LR(k) Grammars Converting a proper grammar into LL(k) form. A L X if A P, X V { }, 1: = X -augmented grammar G G =(N {S } { P {S S}, S The LL-transformed grammar T LL (G) for G T LL (G)=(V t,,p t,[s, ]) If A Z X P,, V *, X, Z V. Then (1) [A, Z ] a[a, Z a], where X L * a, a. (2) [A, Z Y] [B,Y][A, Z B] where X L * B, B L Y, Y V { }. (3) [A, Z X ].

72 3/8/ LL(k) Parsing Assume X=B 0 L B 1 L B 2 L L B n L a (1)[A, Z ] a[a, Z a] (2)[A, Z a] [B n, a] [A, Z B n ] [A, Z B n ] [B n-1, B n ] [A, Z B n-1 ] [A, Z B 2 ] [B 1, B 2 ] [A, Z B 1 ] [A, Z B 1 ] [B 0, B 1 ] [A, Z B 0 ] A Z B 0 * xb 0 xb 1 1 xb * xb n n 1 xa n+1 n 1 * xay n+1 n 1 * xay n+1 y n y 1 * xay n+1 y n y 1 z [A, Z ] a[a, Z a] a[b n, a] [A, Z B n ] * ay n+1 [A, Z B n ] ay n+1 [B n-1, B n ] [A, Z B n-1 ] * ay n+1 y n [A, Z B n-1 ] * ay n+1 y n y 2 [B 0, B 1 ] [A, Z B 0 ] * ay n+1 y n y 2 y 1 [A, Z B 0 ] * ay n+1 y n y 2 y 1 z

73 3/8/ LL(k) Parsing Lemma 8.78 T LL (G) is of size O( G 2 ), can be constructed from G in time O( G 2 ). G assign : S S S i A i B S L i A A * P P A L A, A L P B A = A B L A P i (A) P L i, P L ( T LL (G assign ) is: [S, ] i[s, i] [S, i] [S,i][S, S] [S, S] [S,i] [S,i ] [S,i ] ([S,i (] i[s,i i] [S,i (] [P,(][S,i P] [S,i i] [P,i][S,i P] [S,i P] [A,P][S,i A] [S,i A] [A,A][S,i A] [B,A][S,i B] [S,i B]

74 3/8/ LL(k) Parsing E E E E + T T T T * F F F a ( E ) [E, ] a [E, a] ( [E, (] a ( [E, a] [F, a] [E, F] [E, (] [F, (], [E, F] [E, F] [T, F] [E, T] [E, T] [T, T] [E, T] [E, T] [E, E] * +, $ [E, E] [E, E] [E, E] + $ [E, E] + [E, E+] [E, E+] a [E, E+a] ( [E, E+(] a ( [E, E+a] [F, a] [E, E+F] [E, E+(] [F, (] [E, E+F] [E, E+F] [T, F] [E, E+T] [E, E+T] [T, T] [E, E+T] * +, $ [E, T] [T, T] * [T, T*] [T, T*] a [T, T*a] ( [T, T*(] a ( [T, T*a] [F, a] [T, T*F] [T, T*(] [F, (] [T, T*F] [T, T*F] [T, F] [F, (] a [F, (a] ( [F, ((] a ( [F, (a] [F, a] [F, (F] [F, ((] [F, (] [F, (F] [F, (F] [T, F] [F, (T] [F, (T] [T, T] [F, (T] [E, T] [F, (E] * +, $ [F, (E] [E, E] [E, (E] ) [F, (E)] + ) [F, (E)] [F, a]

75 3/8/ LL(k) Parsing [E, ] a [E, a] a [F, a] [E, F] a [E, F] (F a) a [T, F] [E, T] a [E, T] (T F) a [E, T] [E, E] a [E, E] (E T) a [E, E] [E, E] a + [E, E+] [E, E] a+ a [E, E+a] [E, E] a+a [F, a] [E, E+F] [E, E] a+a [E, E+F] [E, E] (F a) a+a [T, F] [E, E+T] [E, E] a+a [E, E+T] [E, E] (T F) a+a [T, T] [E, E+T] [E, E] a+a * [T, T*] [E, E+T] [E, E] a+a* a [T, T*a] [E, E+T] [E, E] a+a*a [F, a] [T, T*F] [E, E+T] [E, E] a+a*a [T, T*F] [E, E+T] [E, E] (F a) a+a*a [E, E+T] [E, E] (T T*F) a+a*a [E, E] (E E+T) a+a*a (E E)

76 3/8/ LL(k) Parsing Let homomorphism, h T from the rule strings of T LL (G) to the rule strings of G h T ([A, ] ) = A, = and A P; =, otherwise. Then T LL (G) left-to-right covers G w.r.t. homomorphism h T. Lemma 8.79 Let G be -free, T a rule string of T LL (G), r T a rule of T LL (G) and (a) [A, Y] T r T lm x in T LL (G), Y V { } Then (b) [A, Y] T r lm x[a, X ] Tx lm in TLL (G), X rm h T ( T ) R Yx in G Proof. induction on T Base case, if T =, r T =[A, Y] x, x *, so x= and A Y in G, X=Y, =,so X Y. immediately (a) implies (b).

77 3/8/ LL(k) Parsing (case 1) the first rule in T is of type 1 let r T = [A, Y] a[a, Ya] T =r T T, x=ax where [A, Ya] T r T x lm in TLL (G) by I.H. [A, Ya] T lm x [A, YX ] r T x lm in TLL (G) X h T ( T ) R rm ax in G then [A, Y] T lm ax [A, YX ] r T lm ax in T LL (G) YX h T ( T ) R rm Yax = Yx in G observe that h T ( T ) R =h T (r T T ) R =h T ( T ) R because h T (r T )= for type 1 (b) holds when we choose X=Y and =X

78 3/8/ LL(k) Parsing case 2: the first rule in T is of type 2 let r T = [A, Y] [B,Y][A, B] T r T =r T T r" T " T r T, x=x x" where [B,Y] T r" T lm x and [A, B] " T r T lm x" in T LL (G) by I.H. [B,Y] T lm x [B,X ] r" T lm x in T LL (G) X rm h T ( T ) R Yx in G [A, B] lm " T x"[a, X ] lm r T x" in T LL (G) h X T ( " T ) R rm Bx" in G then [A, Y] T lm x x"[a, X ] r T lm x x" = x in T LL (G) X rm h T ( " T ) R Bx" rm r" X x" rm h T ( T ) R Yx x" = Yx in G observe that r"=b X, h T (r" T )=r" h T ( " T ) R r"h T (r T ) R = (h T (r T )h T ( T )h T (r" T )h T ( " T )) R = h T ( T ) R where h T (r T ) = (r T is of type 2). Q.E.D.

79 3/8/ LL(k) Parsing Lemma 8.80 L(T LL (G)) L(G) moreover left parse of sentence w in T LL (G), h T ( ) is a right parse of w in G. Proof. in Lemma 8.79, let A=S, =, Y=, x=w, T r T = s.t. [S, ] lm Tw[S,X ] lm r T w in T LL (G) and h X T ( T ) R rm w in G. Then S h T ( T ) R rm w in G h T ( T ) R = h T ( T r T ) R = h T ( ) R, because h T (r T )=, r T =[S, S] Q.E.D. Lemma 8.81 Let G be an -free grammar A rm X and X rm Yx in G. Then, [A, Y] lm Tx[A, X ] in TLL (G) and h T ( T )= R Proof. induction on. Lemma 8.82 L(G) L(T LL (G)),and right parse of sentence w in G, left parse of w in T LL (G) such that h T ( ) =.

80 3/8/ LL(k) Parsing Theorem 8.83 (T LL (G), h T ) is a left-to-right cover of G. A grammar G is a Predictive LR(k) grammar (PLR(k)), k 1,if the transformed grammar T LL (G) is LL(k) Proposition 8.84 k 1, LL(k) grammar PLR(k) grammar LR(k) grammar Intuitive characterization PLR(k) grammar Let A X S * waz, X * x, * y (S * wxyz) LL(k) parser emits the rule A X after scanning w, seeing k:xyz PLR(k) parser recorgnize the rule A X after scanning wx, seeing k:yz LR(k) parser recognize the rule A X after scanning wxy, seeing k:z

81 3/8/ LL(k) Parsing Proposition 8.85 If G is proper, then T LL (G) is nonleft-recursive. A grammar is left-factored if it has no two distinct rules A 1 and A 2. Fact 8.86 For any G, T LL (G) is left-factored. Theorem 8.87 Left-factoring cannot produce a PLR(k) grammar from a non-plr(k) grammar, and a PLR(k) grammar cannot be converted into a non- PLR(k) gramma by left-factoring. Proof. Process of constructing T LL (G) also involves the process of left-factoring.