1. Context Free Grammar

2. Introduction Why do we want to learn about Context Free Grammars?  Used in many parsers in compilers  Yet another compiler-compiler, yacc  GNU project parser generator, bison  Web application  In describing the formats of XML (eXtensible Markup Language) documents

3. Context Free Grammar G = (V, T, P, S)  V – a set of variables, e.g. {S, A, B, C, D, E}  T – a set of terminals, e.g. {a, b, c}  P – a set of productions rules  In the form of A  , where A  V,   (V  T)* e.g. S  aB  S is a special variable called the start symbol

4. CFG – Example 1 Construct a CFG for the following language : L 1 ={0 n 1 n | n>0} Thought: If  is a string in L 1, so is 0  1, i.e. 0 k+1 1 k+1 = 0(0 k 1 k )1 S  01 | 0S1

5. CFG – Example 2 Construct a CFG for the following language : L 2 ={0 i 1 j | i  j and i, j>0} Thought: Similar to L1, but at least one more ‘1’ or at least one more ‘0’ S  AC | CB A  0A | 0 B  B1 | 1 C  0C1 | 

6. CFG – Example 3 Determine the language generated by the CFG: S  AS |  A  A1 | 0A1 | 01 S generates consecutive ‘A’s A generates 0 i 1 j where i  j Languages: each block of ‘0’s is followed by at least as many ‘1’s

7. Derivation How a string  is generated by a grammar G Consider the grammar G S  AS |  A  A1 | 0A1 | 01 S  0110011 ?

8. Derivation A S S A S A 1 0 101 A  01 S  AS  A1S  011S  011AS  0110A1S  0110011S  0110011  S  AS  AAS  AA   A0A1   A0011   A10011   0110011  Parse Tree or Derivation Tree Leftmost Derivation Rightmost Derivation

9. Ambiguity Each parse tree has one unique leftmost derivation and one unique rightmost derivation. A grammar is ambiguous if some strings in it have more than one parse trees, i.e., it has more than one leftmost derivations (or more than one rightmost derivations).

10. Ambiguity Consider the following grammar G: S  AS | a | b A  SS | ab A string generated by this grammar can have more than one parse trees. Consider the string abb: b a AS SS b S AS ab b S

11. Ambiguity As another example, consider the following grammar: E  E + E | E * E | (E) | x | y | z There are 2 leftmost derivations for x + y + z: E  E + E  x + E  x + E + E  x + y + E  x + y + z E  E + E  E + E + E  x + E + E  x + y + E  x + y + z E + E x y + z E E + E xy E z +

12. Simplification of CFG Remove useless variables  Generating Variables  Reachable Variables Remove  -productions, e.g. A   Remove unit-productions, e.g. A  B

13. A variable X is useless if: X does not generate any string of terminals, or The start symbol, S, cannot generate X. S   X    where ,   (V  T)*, X  V and   T* Useless Variables

14. Removal of Non-generating Variables 1 Mark each production of the form: X   where   T*  Repeat Mark X   where  consists of terminals or variables which are on the left hand side of some marked productions. Until no new productions is marked. 3 Remove all unmarked productions.

15. Example CFG: S  AB|CA A  a B  BC|AB C  aB|b S  CA  A  a C  b

16. Example CFG: S  aAa|aB A  aS|bD B  aBa|b C  abb|DD D  aDa S  aAa|aB  A  aS B  aBa|b C  abb

17. Removal of Non-Reachable Variables 1 Mark each production of the form: S   where   (V  T)*  Repeat Mark X   where X appears on the right hand side of some marked productions. Until no new productions is marked. 3 Remove all unmarked productions.

18. Example CFG: S  aAa|aB A  aS B  aBa|b C  abb S  aAa|aB  A  aS B  aBa|b

19. Example CFG: S  Aab|AB A  a B  bD | bB C  A | B D  a S  Aab|AB  A  a B  bD | bB D  a

20. Remove Useless Variables A  Bac|bTC|ba T  aB|TB B  aTB|bBC C  TBc|aBC|ac A  ba  C  ac  A  ba

21. Removal of  -productions Step 1: Find nullable variables Step 2: Remove all  -productions by replacing some productions

22. How to find nullable variables ? 1 Mark all variables A for which there exists a production of the form A  . 2 Repeat Mark X for which there exists X   where  V* and all symbols in  have been marked. Until no new variables is marked.

23. Removal of  -Productions We can remove all  -productions (except S   if S is nullable) by rewriting some other productions. e.g., If X 1 and X 3 are nullable, we should replace A  X 1 X 2 X 3 X 4 by: A  X 1 X 2 X 3 X 4 | X 2 X 3 X 4 | X 1 X 2 X 4 | X 2 X 4

24. Removal of  -productions Example: S  A | B | C A  aAa | B B  bB | bb C  aCaa | D D  baD | abD |  Step1: nullable variables are D, C and S

25. Removal of  -productions(3) Step 2: eliminate D   by replacing :  D  baD | abD into D  baD | abD | ba | ab eliminate C   by replacing :  C  aCaa | D into C  aCaa | D | aaa  S  A | B | C into S  A | B | C |  The new grammar:  S  A | B | C |   A  aAa | B  B  bB | bb  C  aCaa | D | aaa  D  baD | abD | ba | ab Example: S  A | B | C A  aAa | B B  bB | bb C  aCaa | D D  baD | abD | 

26. Example S  ABCD A  a B   C  ED |  D  BC E  b S  ABCD|ABC|ABD|ACD|AB|AC|AD|A A  a C  ED|E D  BC|B|C E  b

27. Removal of Unit Productions Unit production: A  B, where A and B  V 1. Detect cycles of unit productions: A1  A2  A3  …  A1 A1  A2  A3  …  A1 Replaced A 1, A 2, A 3, …, A k by any one of them 2. Detect paths of unit productions: A 1  A 2  A 3  …, A k  , where   (V  T)*/V Add productions A 1  , A 2  , …, A k   and remove all the unit productions

28. Removal of Unit Productions Example: S  A | B | C A  aa | B B  bb | C C  cc | A Cycle: A  B  C  A Remove by replace B,C by A S  A | A | A A  aa | A A  bb | A A  cc | A Becomes: S  A A  aa | bb | cc Path: S  A Remove by adding productions S  aa | bb | cc

29. Example CFG : S → A | Bb A → C | a B → aBa | b C → aSa S → Bb | aSa |a → A → a | aSa B → aBa | b C → aSa

30. Example Substitute

31. Remove Unit productions of form can be removed immediately Example

32. Substitute

33. Example Remove repeated productions Final grammar

34. Thank You