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Parse : v.

(American Heritage Dict .). Parse : v. To break (a sentence) down into its component parts of speech with an explanation of the form, function, and syntactical relationship of each part. the dog loves the cat. ×. the loves dog the cat. ×. the cat the dog loves.

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Parse : v.

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  1. (American Heritage Dict.) Parse: v. To break (a sentence) down into its component parts of speech with an explanation of the form, function, and syntactical relationship of each part. the dog loves the cat × the loves dog the cat × the cat the dog loves IT 327

  2. The most practical Parsers: Predictive parser: No back tracking. • input (token string) • Stacks, parsing table • output (syntax tree, intermediate codes) IT 327

  3. Tow kinds of predictive parsers: Top-Down The syntax tree is built up from the root Example: LL(1) parser Left to right scanning Leftmost derivations 1 symbol look-ahead • Bottom-Up: • The syntax tree is built up from the leaves • Example: LR(1) parser Left to right scanning Rightmost derivations 1 symbol look-ahead IT 327

  4. end-of-file symbol A left-most derivation LL(1) Grammar • S •  ASb • aSb • aASbb • aaSbb • aaASbbb • aaaSbbb • aaaCbbb • aaacCbbb • aaaccCbbb • aaaccbbb • S A S b • S  C • A  a • C  c C • C   LL(1) Parsing Table aaaccbbb IT 327

  5. Recursive-descent Parser all possible terminal and end-of-file symbols • S A S b • S  C • A  a • C  c C • C   S(): Switch(token) { case a: A();S();get(b); build S ASb; break; case b: C(); build S C; break; case c: C(); built S C; break; case $: C(); built S C; break; } LL(1) Parsing Table IT 327

  6. Recursive-descent Parser • S A S b • S  C • A  a • C  c C • C   A(): Switch(token) { case a: get(a); build A a; break; case b: error; break; case c: error; break; case $: error; break; } LL(1) Parsing Table IT 327

  7. Recursive-descent Parser • S A S b • S  C • A  a • C  c C • C   C(): Switch(token) { case a: error; break; case b: build C ; break; case c: get(c);C(); built C cC; break; case $: build C ; break; } LL(1) Parsing Table IT 327

  8. S A S b • S  C • A  a • C  c C • C   LL(1) Parsing aaaccbbb aaaccbbb aaaccbbb aaaccbbb aaaccbbb aaaccbbb aaaccbbb aaaccbbb aaaccbbb S(); A();S();get(b); get(a);S();get(b); S();get(b); A();S();get(b);get(b); get(a);S();get(b);get(b); S();get(b);get(b); A();S();get(b);get(b);get(b); get(a);S();get(b);get(b);get(b); S();get(b);get(b);get(b); C();get(b);get(b);get(b); get(c);C();get(b);get(b);get(b); C();get(b);get(b);get(b); get(c);C();get(b);get(b);get(b); C();get(b);get(b);get(b); get(b);get(b);get(b); get(b);get(b); get(b); • S •  ASb • aSb • aASbb • aaSbb • aaASbbb • aaaSbbb • aaaCbbb • aaacCbbb • aaaccCbbb • aaaccbbb IT 327

  9. LL(1) Grammar A grammar having an LL(1) parsing table.i.e., There is no conflict in the parsing table • S A S b • S  C • A  a • C  c C • C   LL(1) Parsing Table LL(1) Grammars allow -production. IT 327

  10. Not every CFG is an LL(1) grammar <stmt> ::= <if-stmt> | s1 | s2 <if-stmt> ::= if <expr> then <stmt> else <stmt> | if <expr> then <stmt> <expr> ::= e1 | e2 if e1 thenif e2 then s1 else s2 if (a > 2) if (b > 1) b++; else a++; if (a > 2) if (b > 1) b++; else a++; IT 327

  11. The recursive-descent parser does not work for every CFG • E E + T • ET • T  T * F • T  F • F  ( E ) • F  id E(): Switch(token) { case id: E(); ... ... ... } id+id*id Left-recursions IT 327

  12. A left-recursive grammar Remove left-recursion Left-recursions • A  A’ • A’   A’ • A’ • A A  • A A A   A’ A  A’  A   A’ A       A’  IT 327

  13. Eliminating left-recursions   • E T E’ • E’ + T E’ • E’  • T  F T’ • T’  * F T’ • T’  • F  ( E ) • F  id • E E + T • ET • T  T * F • T  F • F  ( E ) • F  id   IT 327

  14. An Algorithm for Eliminating immediate left-recursions Given a CFG G, let A be one of its non-terminal symbols such that A A  A 1. Add a new non-terminal symbol A’ to G; 2. For each production A  such that A is not the 1st symbol in  add A  A’ to G; 3. For each production A A replace it by A  A’; 4. Add A’ to G; • A  A’ • A’   A’ • A’ IT 327

  15. Indirect left-recursions S a • S A a • S b • A  S d • A  e A d S a A d S b bdada IT 327

  16. repeat Indirect left-recursions if any, remove the last non-terminal symbol Z with rule ZX… find all immediate left recursions find all immediate left recursions  • S A a • S b • ASdA’ • A eA’ • A’ cA’ • A’  • S eA’aS’ • S bS’ • S’ dA’aS’ • S’  • A’ cA’ • A’  • S A a • S b • A  A c • A  S d • A  e • S SdA’ a • S eA’a • S b • A’ cA’ • A’     A  A’ A’   A’ A’ A A  A  IT 327

  17. An Algorithm for Eliminating left-recursions Given a CFG G, let A1, A2, ..... An, be its nonterminal symbols for i:= n down to 1 do { for j := 1 to i-1 do { // find one level of indiretion For each production Ai Aj ωdo { For each production Aj ,add Ai ω to the grammar; Remove Ai Ajωby } } // end for j Eliminate the immediate left-recursion caused by Ai } // end for i IT 327

  18. A Grammar for if statements • S iCtSE • S a • E  eS • E  • C  b Is it an LL(1) grammar? Is there an LL(1) parsing table for it? No! IT 327

  19. ibtibtae…… A Grammar for if statements • S iCtSE • S a • E  eS • E  • C  b Why there is a conflict? • S ... • i b t S E… ... • i b t ibtSEE… • i b t ibtaEE… • i b t ibta E… • i b t ibta eS… • S ... • i b t S E… ... • i b t ibtSEE… • i b t ibtaEE… • i b t ibtaeSE… 4: 3: IT 327

  20. A Grammar for if statements • S iCtSE • S a • E  eS • E  • C  b Can we have an unambiguous equivalent grammar for this grammar? Yes! In general, No! Some inherently ambiguous languages exist. Can we write a program to test whether a given grammar is ambiguous? No! Can we write a program to get an unambiguous equivalent grammar from any grammar of a language that is known to be not inherently ambiguous? No! IT 327

  21. Is there an LL(2) Grammar ? Yes! We need to look two symbols ahead in order to determine which rule should be used. { ambnc | m ≥ 1 and n ≥ 0 } • S A B • A  aA • A  a • B  b B • B  c LL(2) Parsing Table a a a a a b b b b c IT 327

  22. LL(2) Parsing Table • S A B • A  aA • A  a • B  b B • B  c LL(2) Parsing S(); a a a b c A();B(); a a a b c get(a);A();B(); a a a b c A();B(); a a b c get(a);A();B(); a a b c A();B(); a b c get(a);B() a b c B(); b c get(b);B(); b c B(); c get(c); c IT 327

  23. Is there an LL(1) grammar equivalent to the following LL(2) grammar? Yes { ambnc | m ≥ 1 and n ≥ 0 } • S a A B • A  aA • A   • B  b B • B  c • S A B • A  aA • A  a • B  b B • B  c a a a a a b b b b c IT 327

  24. Every left-recursive grammar is not an LL(k) grammar But we can effectively find an equivalent one • S aS’ • S’  AS’ • S’   • A  b • S S A • S  a • A  b • S • SA • SAA • SAAA • SAAAA • aAAAAA • .... • abbbbbb • E T E’ • E’ + T E’ • E’  • T  F T’ • T’  * F T’ • T’  • F  ( E ) • F  id • E E + T • ET • T  T * F • T  F • F  ( E ) • F  id Are we happy with this? IT 327

  25. Does any LL(2) grammar always has an equivalent LL(1) grammar? No LL(2) grammar LL(k) grammar, k  2 • S a SA • S   • A  ak-1bS • A  c • S a SA • S   • A  abS • A  c no equivalent LL(k-1) grammar no equivalent LL(1) grammar KuriKi-Sunoi [1969] LL(1)  LL(2)  LL(3)  ..... LL(k)  LL(k+1)  ... IT 327

  26. LL(k) grammar, k  2 (KuriKi-Sunoi [1969]) There exists DCFL that is not LL(k) • S a SA • S   • A  ak-1bS • A  c This grammar is inherently ambiguous. Is there an unambiguous CFG that is not an LL(k) grammar? Yes -- Stearns [1970] { an | n ≥ 0 }  { anbn | n ≥ 0 } IT 327

  27. LL(1) Parser Implementation • E T E’ • E’ + T E’ • E’  • T  F T’ • T’  * F T’ • T’  • F  ( E ) • F  n • p.s. Let n be any positive integer less than 32767 Programming Assignment Details will be announced later. IT 327

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