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CSC 3130: Automata theory and formal languages

Fall 2008. The Chinese University of Hong Kong. CSC 3130: Automata theory and formal languages. LR( k ) grammars. Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130. LR(0) example from last time. 4. A  aA•b. a. A. b. 2. 5. A  a•Ab A  a•b A  •aAb A  •ab. 1. A  aAb•.

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CSC 3130: Automata theory and formal languages

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  1. Fall 2008 The Chinese University of Hong Kong CSC 3130: Automata theory and formal languages LR(k) grammars Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

  2. LR(0) example from last time 4 A  aA•b a A b 2 5 A  a•Ab A  a•b A  •aAb A  •ab 1 A  aAb• a A  •aAb A •ab b 3 A  ab• A  aAb | ab

  3. LR(0) parsing example revisited S Input A Stack a 1 1 1a2 1a2a2 1a2a2b3 1a2A4 1a2A4b5 1A aabb abb bb b b   1 2 2 3 4 5 2 A  •aAb A •ab S S S R S R a A  a•Ab A  a•b A  •aAb A  •ab 3 b A  ab• A • A 5 4 b a A • • • • A  aAb• A  aA•b b a b • • A  aAb | ab A  aAb  aabb

  4. Meaning of LR(0) items eNFA transitions to: X  •g A undiscovered part shift focus to subtree rooted at X (if X is nonterminal) b a X • focus A  aX•b A  a•Xb move past subtreerooted at X

  5. Outline of LR(0) parsing algorithm • Algorithm can perform two actions: • What if: no complete itemis valid there is one valid item,and it is complete reduce (R) shift (S) some valid itemscomplete, some not more than one validcomplete item R / R conflict S / R conflict

  6. Definition of LR(0) grammar • A grammar is LR(0) if S/R, R/R conflicts never occur • LR means parsing happens left to right and produces a rightmost derivation • LR(0) grammars are unambiguous and have a fastparsing algorithm • Unfortunately, they are not “expressive” enoughto describe programming languages

  7. Hierarchy of context-free grammars context-free grammars parse using CYK algorithm (slow) LR(∞) grammars … java perl python … LR(1) grammars LR(0) grammars parse using LR(0) algorithm

  8. A grammar that is not LR(0) S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a

  9. A grammar that is not LR(0) S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a possibilities: shift (3), reduce (4)reduce (5), shift (6) S valid LR(0) items: A  a•A, A  a• B  a•, B  a•b, A  •aA, A  •a S S A A B A A A S/R, R/R conflicts! a a a a a a c • • •

  10. Lookahead S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a peek inside! S valid LR(0) items: A  a•A, A  a• B  a•, B  a•b, A  •aA, A  •a S S A A B A A A a a a a a a c • • •

  11. Lookahead S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a a peek inside! S valid LR(0) items: A  a•A, A  a• B  a•, B  a•b, A  •aA, A  •a A A … a a • action: shift parse tree must look like this

  12. Lookahead S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a a a peek inside! S valid LR(0) items: A  a•A, A  a• A  •aA, A  •a A A A … a a • action: shift parse tree must look like this

  13. Lookahead S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) input: a a a S valid LR(0) items: A  a•A, A  a• A  •aA, A  •a A A A a a a • action: reduce parse tree must look like this

  14. LR(0) items vs. LR(1) items A LR(1) A LR(0) A A b b a a • • b b a a A A A  a•Ab [A  a•Ab, b] a a b b A  aAb | ab

  15. LR(1) items • LR(1) items are of the formto represent this state in the parsing [A  a•b, x] or [A  a•b, e] A A x a b a b • •

  16. Outline of LR(1) parsing algorithm • Step 1: Build eNFA that describes valid item updates • Step 2: Convert eNFA to DFA • As in LR(0), DFA will have shift and reduce states • Step 3: Run DFA on input, using stack to remember sequence of states • Use lookahead to eliminate wrong reduce items

  17. Recall eNFA transitions for LR(0) • States of eNFA will be items (plus a start state q0) • For every item S  •a we have a transition • For every item A  •X we have a transition • For every item A  a•Cb and production C  •d e q0 S  •a X A  •X A  X• e A  •C C  •d

  18. eNFA transitions for LR(1) • For every item [S  •a, e]we have a transition • For every item A  •X we have a transition • For every item [A  a•Cb, x] and production C  dfor every y in FIRST(bx) e q0 [S  •a, e] X [A  •X, x] [A  X•, x] e [A  •C, x] [C  •d, y]

  19. FIRST sets • Example FIRST(a) is the set of terminals that occuron the left in some derivation starting from a FIRST(a) = {a} FIRST(A) = {a}FIRST(S) = {a, c} FIRST(bAc) = {b} FIRST(BA) = {a} FIRST(e) = ∅ S  A(1) | cB(2)A  aA(3) | a(4)B  a(5) | ab(6)

  20. Explaining the transitions A A x x b b a X a X • • X [A  •X, x] [A  X•, x] C b A y • d x b a C • e [A  •C, x] [C  •d, y] y ∈ FIRST(bx)

  21. Example [S  A•, e] S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) A [A  •aA, e] e [S  •A, e] [A  •a, e] e e . . . q0 [S  B•c, e] e B e [S  •Bc, e] [B  •a,c] e [B  •ab,c]

  22. Convert NFA to DFA • Each DFA state is a subset of LR(1) items, e.g. • States can contain S/R, R/R conflicts • But lookahead can always resolve such conflicts [A  a•A, ] [A  a•, ] [B  a•, c] [B  a•b, c] [A  •aA, ] [A  •a, ]

  23. Example S  A(1) | Bc(2)A  aA(3) | a(4)B  a(5) | ab(6) look ahead! input valid items A stack  a ab B Bc S abc bc c c   [S  •A, ] [S  •Bc, ] [A  •aA, ] [A  •a, ] [B  •a, c] [B  •ab, c] SS R S R [A  a•A, ] [A  a•, ] [B  a•, c] [B  a•b, c][A  •aA, ] [A  •a, ] [B  ab•, c] [S  B•c, ] [S  Bc•, ]

  24. LR(k) grammars • A context-free grammar is LR(1) if all S/R, R/Rconflicts can be resolved with one lookahead • More generally, LR(k) grammars can resolve allconflicts with k lookahead symbols • Items have the form [A  •, x1...xk] • LR(1) grammars describe the semantics of mostprogramming languages

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