Simplification of Context-Free Grammars

1. Simplification of Context-Free Grammars

2. Simplification of Context-Free Grammars Some useful substitution rules. • Removing useless productions. • Removing -productions. • Removing unit-productions. Removing Left Recursion

3. Some Useful Substitution Rule G = (V, T, S, P) A  x1Bx2 P B y1 | y2 | ... | yn  P L(G) = L(G^) G^ = (V, T, S, P^) A  x1y1x2 | x1y2x2 | ... | x1ynx2 P^

4. Example G = ({A, B}, {a, b}, A, P) A  a | aaA | abBc B b G^ = (V, T, S, P^) A  a| aaA| abbc

5. Removing Useless Productions I. S  aSb |  | A A  aA S  A is redundant as A cannot be transformed into a terminal string. II. S  A A aA | λ B  bA B is useless as it will never be reached from S

6. Removing Useless Productions G = (V, T, S, P) A  V is useful iff there is w  L(G) if it is: • generating ie. A * w • reachable ie. S * xAy So in the process of removing useless productions, we first check that a symbol is generating or not and then check for reachability.

7. Example G = ({S, A, B, C}, {a, b}, S, P) S  aS | A | C A  a B  aa C  aCb S  aS | A A  a S  aS | A A  a B  aa Here C is not generating And B is not reachable

8. Removing -Productions • Any production of a context-free grammar of the form: A   is called a -production. • Any variable A for which the derivation: A *  is possible is called nullable.

9. Example 1 S  aS1b S  aS1b | ab S1  aS1b |  S1 aS1b | ab

10. Example 2 • S  ABC • A  aAA |l • B  bBB |l • Cc Here A & B are nullable So S ABC|AC|BC|C A aAA | aA | a B  bBB | bB |b Cc

11. Example S  ABaC S  ABaC | BaC | AaC | ABa | aC | Aa |Ba | a A  BC A  B | C | BC B  b |  B  b C  D |  C  D D  d D  d VN = {A, B, C}

12. Removing Unit-Productions Any production of a context-free grammar of the form: A  B is called a unit-production.

13. Example S  Aa | B B  A | bb A  a | bc | B

14. Example S  Aa A  a | bc B  bb S  Aa | B B  A | bb A  a | bc | B S * A S * B A * B B * A S  a | bc | bb A  bb B  a | bc

15. Example Simplify the following grammar : S aA | aBB A aaA | l B bB | bbC C B Obtain its Language also.

16. Solution • B & C are not genarating remove thm • Remove the l production of A. • We get S aA |a A  aaA |aa L(G) = (a(aa)*))