Why augment the grammar?
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- Sybil Stokes
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1 Why augment the grammar? Consider -> F + F FOLLOW F -> i :->.F+ ->.F F->.i i F 2:F->i. Action Goto + i $ F s2 1 1 s3? 2 r3 r3 3 s ? 1:->F.+ ->F. i 4:->F+. i + F 3:->F+. ->.F+ ->.F F->.i? Reduce ->F or Accept $ F $, +? Reduce -> F + or Accept
2 S -> S S -> L = R S -> R L -> R L -> id R -> L Follow S $ L =, $ R =, $ L -> id. id xample not SLR(1) S -> R. id S ->.S S ->.L=R S ->.R L ->.R L ->.id R ->.L R R L ->.R S S ->S. L S -> L.=R R -> L. R -> L. L L ->.R R ->.L L ->.R L ->.id id = L S -> L=.R R ->.L L ->.R L ->.id R S -> L=R.
3 Another xample not SLR(1) S -> G G -> = -> + -> FOLLOW G $ $,+,== $,+,=, 10: ->. 11: ->. G 8: ->.+ G -> =. 9: -> +. ->. 0:S ->.G G ->. G ->.= ->. ->.+ ->. ->. 1:S ->.G + 2:G ->. ->. 3:G ->.= ->.+ 4: ->. ->. = 7:G -> =. ->. ->.+ ->. ->. + 5: -> +. ->. ->. 6: ->.
4 G 17:S ->G.,$ (1) 0:S ->.G,$ G->.=,$ G->.,$ ->.,=,+ ->.,=,+, ->.,=,+, ->.+,=,+ (0) LR(1) Parser (or same grammar) 2:->.,=,+ ->.,=,+, (4) 3:G->.,$ ->.,=,+, (2) = 15:->.,$,+, (10) 1:G->.=,$ ->.+,=,+ (3) 4:G->=.,$ ->.,$,+ ->.,$,+, ->.,$,+, ->.+,$,+ (7) 8: ->.,$,+, (6) + 5:->+.,=,+ ->.,=,+, ->.,=,+, (5) 6:G->=.,$ ->.+, $,+ (8) 7: ->.,$,+ ->.,$,+, (4) 14:->+.,$,+ ->.,$,+, (9) 16: ->.,$,+, (11) 9:->.,=,+, (6) 10: ->+.,=,+ ->.,=,+, (9) + 11:->.,=,+, (10) 13: ->+.,$,+ ->.,$,+, ->.,$,+, (5) 12:->.,=,+, (11)
5 LR(1) Parser (or same grammar) LR(1) Item [A -> α.β, c] A->αβ is a production c is the lookahead
6 Constructing DFA or LR(1) Parser Basic Idea o closure: For item [A -> α.xβ, c], where X is nonterminal and production X->δ exists, there exists a rightmost derivation G => ϕauw => ϕαxβuw item [X ->. δ, v] is valid or viable preix ϕα with v in FIRS(βuw) I β => the empty string, then v = u. Closure o set o Items I: cset = I; repeat or each item [A-> α.xβ, a] in cset where X is a nonterminal, Add all items [X->. δ,b] or all b in FIRS(βa) to cset (i not already in cset) until no more items added; GOO(I,X): Assume [A-> α.xβ, a] in I. hen GOO(I,X) = closure o items [A-> αx.β, a].
7 Computing Closure o LR(1) items S -> G G -> = -> + -> In LR(1) machine: [S ->.G,$] In LR(0) machine: S ->.G G ->.= G ->. ->. ->.+ ->. ->. S: [G->.=,$] [G->.,$]
8 S0: S ->.G,$ G->.=,$ G->.,$ ->.,=,+ ->.,=,+, ->.,=,+, ->.+,=,+ Computing GOO(I,X) in LR(1) machine GOO(S0,G) GOO(S0,) GOO(S0,) GOO(S0,)
9 More xtensive xample o GOO(I,X) in LR(1) 1: G->.=,$ ->.+,=,+ GOO(1,+) 5: GOO(1,=) 4:
10 LR(1) able Construction Partial xample 4: G->=.,$ ->.,$,+ ->.,$,+, ->.,$,+, ->.+,$, : ->.,$,+ ->.,$,+, 15 State Action GOO = + $ G 4 7
11 G 17:S ->G.,$ (1) 0:S ->.G,$ G->.=,$ G->.,$ ->.,=,+ ->.,=,+, ->.,=,+, ->.+,=,+ (0) LR(1) Parser (or same grammar) 2:->.,=,+ ->.,=,+, (4) 3:G->.,$ ->.,=,+, (2) = 15:->.,$,+, (10) 1:G->.=,$ ->.+,=,+ (3) 4:G->=.,$ ->.,$,+ ->.,$,+, ->.,$,+, ->.+,$,+ (7) 8: ->.,$,+, (6) + 5:->+.,=,+ ->.,=,+, ->.,=,+, (5) 6:G->=.,$ ->.+, $,+ (8) 7: ->.,$,+ ->.,$,+, (4) 14:->+.,$,+ ->.,$,+, (9) 16: ->.,$,+, (11) 9:->.,=,+, (6) 10: ->+.,=,+ ->.,=,+, (9) + 11:->.,=,+, (10) 13: ->+.,$,+ ->.,$,+, ->.,$,+, (5) 12:->.,=,+, (11)
12 LALR(1) Grammars and Parsing Characteristic: Same number o states as SLR(1) with more power due to lookahead in states. But, less power than canonical LR(1) because less states. 2 Approaches to able Construction: Construct LR(1) sets o items (DFA) and merge states with same core. Construct LR(0) sets o items and generate lookahead inormation or each o those states. Properties o LALR we will see: May perorm RDUC rather than RROR like LR(1), but will catch error beore any more input is processed. LALR derived rom LR with no shit-reduce conlict will also have no shit-reduce conlict (Shit-reduce conlicts arise rom core, not lookahead thereore merging has no eect.) LALR merging can create reduce-reduce conlicts not in LR rom which LALR derived.
13 Constructing LALR rom LR(1) 1. Construct LR(1) DFA. 2. Identiy all sets o states with same core. In our example: (2,7), (5,13), (8,9), (10,14), (11,15), (12,16) 3. Merge the states with the same core into a single state in LALR: 1. create single state with that core 2. merge lookaheads rom all LR(1) states with that core xample: LALR(1) state 2: rom LR(1) states 2 and 7: ->., ->., Add edges to LALR machine based on edges o LR machine.
14 17:S ->G.,$ G 0:S ->.G,$ G->.=,$ G->.,$ ->.,=,+ ->.,=,+, ->.,=,+, ->.+,=,+ (0) 2:->.,=,+,$ ->.,=,+,,$ (2/7) LALR(1) Parser (or same grammar) 3:G->.,$ ->.,=,+, = 1:G->.=,$ ->.+,=,+ Final LALR machine. 4:G->=.,$ ->.,$,+ ->.,$,+, ->.,$,+, ->.+,$,+ 5:->+.,=,+,$ ->.,=,+,,$ + ->.,=,+,,$ (5/13) 8: ->.,$,+,,= (8/9) 6:G->=.,$ ->.+, $,+ 10: ->+.,=,+,$ ->.,=,+,,$ (10/14) 11:->.,=,+,,$ (11/15) 12:->.,=,+,,$ (12/16)
15 Is a grammar LR(1)? LALR(1)? Construct LR(1) (or LALR(1)) parse table using lookahead inormation. I there exists any multideined entries, then the grammar is NO LR(1) (LALR(1)). or Construct LR(1) (or LALR(1)) DFA. I there exists any inadequate states or which lookahead does not resolve the local ambiguity as below, then the grammar is NO LR(1) (LALR(1)). I or all states including A -> α., {a1,a2,,an} B -> β., {b1,b2, bm} {a1,a2, an} {b1,b2,,bm} = 0 AND I or all states including A -> α., {a1,a2,,an} B -> β.aδ, {b1,b2, bm} {a1,a2,,an} {a} = 0 then the grammar is LR(1) (LALR(1)).
16 Property 1: May reduce beore error. Consider string: + LR(1): Stack Input 0 +$ 03 +$ 02 +$ 01 +$ 01+5 $ $ RROR! LALR(1): 0 +$ 03 +$ 02 +$ 01 +$ 01+5 $ $ $ 01 $ RROR! - 2 extra reductions.
17 Property 2: No shit-reduce in LR => no shit-reduce conlict in LALR. Assume when merge, we get a shit-reduce conlict in some state: LALR state: A -> α., a = > reduce on a B -> β.aδ,b = > shit on a his implies that some set Si rom LR(1) machine has the item A -> α., a to be included in this merge. Since the cores o all states merged together are the same, Si must also contain B -> β.aδ,c or some lookahead c. his implies that the shit-reduce conlict also must exist in Si within the LR(1) machine. Contradiction. KY: Merging states cannot cause shit-reduce conlicts in LALR.
18 Property 3: Merging states or LALR(1) can produce reduce-reduce conlicts. Consider: S1->S S -> aad bbd abe bae A -> c B -> c o create LALR, merge: A->c., d,e B->c., e,d S ->.S,$ S->.aAd, $ S->.bBd, $ S ->.abe,$ S ->.bae,$ a b S->b.Bd,$ S->b.Ae,$ B->.c,d A->.c,e S ->a.ad,$ S ->a.be,$ A->.c,d B->.c,e c c B->c.,d A->c.,e A->c.,d B->c.,e Reduce-reduce conlict!! hereore, not LALR grammar
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