6/4/2016IT 3271 The most practical Parsers: Predictive parser: 1.input (token string) 2.Stacks, parsing table 3.output (syntax tree, intermediate codes) No back tracking.

6/4/2016IT 3272 Tow kinds of predictive parsers: 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 Left to right scanning Leftmost derivations 1 symbol look-ahead Top-Down The syntax tree is built up from the root Example: LL(1) parser

6/4/2016IT 3273 LL(1) Grammar 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 LL(1) Parsing Table aaaccbbb S  ASb  aSb  aASbb  aaSbb  aaASbbb  aaaSbbb  aaaCbbb  aaacCbbb  aaaccCbbb  aaaccbbb A left-most derivation end-of-file symbol

6/4/2016IT 3274 Recursive-descent Parser 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 LL(1) Parsing Table 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; } all possible terminal and end-of-file symbols

6/4/2016IT 3275 Recursive-descent Parser A():Switch(token) { case a: get(a); build A  a; break; case b: error; break; case c: error; break; case $: error; break; } 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 LL(1) Parsing Table

6/4/2016IT 3276 Recursive-descent Parser C():Switch(token) { case a: error; break; case b: build C  ; break; case c: get(c);C(); built C  cC; break; case $: build C  ; break; } 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 LL(1) Parsing Table

IT 3277 LL(1) Parsing 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 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); aaaccbbb S  ASb  aSb  aASbb  aaSbb  aaASbbb  aaaSbbb  aaaCbbb  aaacCbbb  aaaccCbbb  aaaccbbb

6/4/2016IT 3278 LL(1) Grammar A grammar having an LL(1) parsing table. i.e., There is no conflict in the parsing table LL(1) Grammars allow -production. 1.S  A S b 2.S  C 3.A  a 4.C  c C 5.C  abc$ S1222 A3 C545 LL(1) Parsing Table

6/4/2016IT 3279 Not every CFG is an LL(1) grammar ::= | s1 | s2 ::= if then else | if then ::= e1 | e2 if e1 then if e2 then s1 else s2 if (a > 2) if (b > 1) b++; else a++; if (a > 2) if (b > 1) b++; else a++;

6/4/2016IT The recursive-descent parser does not work for every CFG 1.E  E + T 2.E  T 3.T  T * F 4.T  F 5.F  ( E ) 6.F  id E():Switch(token) { case id: E();... } Left-recursions id+id*id

6/4/2016IT Left-recursions 1.A  A  2.A   A  A  A  A   A A’    1.A   A’ 2.A’   A’ 3.A’  A left-recursive grammarRemove left-recursion 

6/4/2016IT Eliminating left-recursions 1.E  E + T 2.E  T 3.T  T * F 4.T  F 5.F  ( E ) 6.F  id 1.E  T E’ 2.E’  + T E’ 3.E’  4.T  F T’ 5.T’  * F T’ 6.T’  7.F  ( E ) 8.F  id    

6/4/2016IT An Algorithm for Eliminating immediate left-recursions Given a CFG G, let A be one of its non-terminal symbols such that 1. Add a new non-terminal symbol A’ to G ; 2. For each production A   such that A is not the 1 st 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   1.A   A’ 2.A’   A’ 3.A’ 

6/4/2016IT Indirect left-recursions 1.S  A a 2.S  b 3.A  S d 4.A  e S a A d S a A d S b bdada

6/4/2016IT Indirect left-recursions 1.S  A a 2.S  b 3.A  SdA’ 4.A  eA’ 5.A’  cA’ 6.A’  find all immediate left recursions A  A  A   A   A’ A’   A’ A’  1.S  A a 2.S  b 3.A  A c 4.A  S d 5.A  e   1.S  SdA’ a 2.S  eA’a 3.S  b 4.A’  cA’ 5.A’  if any, remove the last non-terminal symbol Z with rule Z  X… find all immediate left recursions   1.S  eA’aS’ 2.S  bS’ 3.S’  dA’aS’ 4.S’  5.A’  cA’ 6.A’  repeat 

6/4/2016IT An Algorithm for Eliminating left-recursions Given a CFG G, let A 1, A 2,..... A n, 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 A i  A j ω do { For each production A j  , add A i   ω to the grammar; Remove A i  A j ω by } } // end for j Eliminate the immediate left-recursion caused by A i } // end for i

6/4/2016IT A Grammar for if statements 1.S  iCtSE 2.S  a 3.E  eS 4.E  5.C  b abeit$ S21 E3,44 C5 Is it an LL(1) grammar? Is there an LL(1) parsing table for it? No!

6/4/2016 IT A Grammar for if statements 1.S  iCtSE 2.S  a 3.E  eS 4.E  5.C  b abeit$ S 21 E 3,44 C 5 Why there is a conflict? S ...  i b t S E…...  i b t ibtSE E…  i b t ibtaE E…  i b t ibta E…  i b t ibta eS… S ...  i b t S E…...  i b t ibtSE E…  i b t ibtaE E…  i b t ibtaeS E… 4: 3: ibtibtae……

6/4/2016IT A Grammar for if statements 1.S  iCtSE 2.S  a 3.E  eS 4.E  5.C  b abeit$ S21 E3,44 C5 Can we have an unambiguous equivalent grammar for this grammar? Yes! No! In general, No! Some inherently ambiguous languages exist. 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? Can we write a program to test whether a given grammar is ambiguous? No!

6/4/2016IT Is there an LL(2) Grammar ? Yes! 1.S  A B 2.A  a A 3.A  a 4.B  b B 5.B  c abc S1 Aabc 233 B45 LL(2) Parsing Table We need to look two symbols ahead in order to determine which rule should be used. { a m b n c | m ≥ 1 and n ≥ 0 } a a a a a b b b b c

6/4/2016IT LL(2) Parsing 1.S  A B 2.A  a A 3.A  a 4.B  b B 5.B  c LL(2) Parsing Table 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 abc S1 Aabc 233 B45

6/4/2016IT Is there an LL(1) grammar equivalent to the following LL(2) grammar? 1.S  a A B 2.A  a A 3.A  4.B  b B 5.B  c { a m b n c | m ≥ 1 and n ≥ 0 } a a a a a b b b b c 1.S  A B 2.A  a A 3.A  a 4.B  b B 5.B  c Yes

6/4/2016IT Every left-recursive grammar is not an LL(k) grammar 1.S  a S’ 2.S’  AS’ 3.S’  4.A  b 1.S  S A 2.S  a 3.A  b S  SA  SAA  SAAA  SAAAA  a AAAAA ....  abbbbbb 1.E  E + T 2.E  T 3.T  T * F 4.T  F 5.F  ( E ) 6.F  id 1.E  T E’ 2.E’  + T E’ 3.E’  4.T  F T’ 5.T’  * F T’ 6.T’  7.F  ( E ) 8.F  id But we can effectively find an equivalent one Are we happy with this?

6/4/2016IT Does any LL(2) grammar always has an equivalent LL(1) grammar? 1.S  a S A 2.S  3.A  abS 4.A  c No no equivalent LL(1) grammar 1.S  a S A 2.S  3.A  a k-1 bS 4.A  c no equivalent LL(k-1) grammar LL(2) grammar LL(k) grammar, k  2 LL(1)  LL(2)  LL(3) ..... LL(k)  LL(k+1) ... KuriKi-Sunoi [1969]

6/4/2016IT There exists DCFL that is not LL(k) -- Stearns [1970] { a n | n ≥ 0 }  { a n b n | n ≥ 0 } 1.S  a S A 2.S  3.A  a k-1 bS 4.A  c LL(k) grammar, k  2 This grammar is inherently ambiguous. (KuriKi-Sunoi [1969 ]) Is there an unambiguous CFG that is not an LL(k) grammar? Yes

6/4/2016IT LL(1) Parser Implementation 1.E  T E’ 2.E’  + T E’ 3.E’  4.T  F T’ 5.T’  * F T’ 6.T’  7.F  ( E ) 8.F  n p.s. Let n be any positive integer less than n+*()$ E11 E’233 T44 T’6566 F87 Programming Assignment Details will be announced later.