1. Regular Grammars Chapter 7 1

2. Regular Grammars A regular grammar G is a quadruple (V, , R, S), where: ● V is the rule alphabet, which contains nonterminals and terminals, ●  (the set of terminals) is a subset of V, ● R (the set of rules) is a finite set of rules of the form: X  Y, X  Y, ● S (the start symbol) is a nonterminal. 2

3. Regular Grammars In a regular grammar, all rules in R must: ● have a left hand side that is a single nonterminal ● have a right hand side that is: ● , or ● a single terminal, or ● a single terminal followed by a single nonterminal. Legal: S  a, S  , and T  a S Not legal: S  a S a and a S a  T 3

4. L = {w  { a, b }* : |w| is even} Regular expression: (( aa )  ( ab )  ( ba )  ( bb ))* FSM: Regular Grammar Example 4

5. L = {w  { a, b }* : |w| is even} Regular expression: (( aa )  ( ab )  ( ba )  ( bb ))* FSM: Regular grammar:G = (V, , R, S), where V = {S, T, a, b},  = {a, b}, R = { S   S  a T S  b T T  a S T  b S } Regular Grammar Example 5

6. Regular Languages and Regular Grammars Theorem: The class of languages that can be defined with regular grammars is exactly the regular languages. Proof: By two constructions. 6

7. Regular Languages and Regular Grammars Regular grammar  FSM: grammartofsm(G = (V, , R, S)) = 1. Create in M a separate state for each nonterminal in V. 2. Start state is the state corresponding to S. 3. If there are any rules in R of the form X  w, for some w  , create a new state labeled #. 4. For each rule of the form X  w Y, add a transition from X to Y labeled w. 5. For each rule of the form X  w, add a transition from X to # labeled w. 6. For each rule of the form X  , mark state X as accepting. 7. Mark state # as accepting. FSM  Regular grammar: Similarly. 7

8. Example 1 - Even Length Strings S  S   S  aTS  aT S  bTS  bT T  aST  aS T  bST  bS 8

9. S  S   S  aTS  aT S  bTS  bT T  aST  aS T  bST  bS 9

10. Strings that End with aaaa L = {w  { a, b }* : w ends with the pattern aaaa }. S  aSS  aS S  bSS  bS S  aBS  aB B  aCB  aC C  aDC  aD D  aD  a 10

11. Strings that End with aaaa L = {w  { a, b }* : w ends with the pattern aaaa }. S  aSS  aS S  bSS  bS S  aBS  aB B  aCB  aC C  aDC  aD D  aD  a 11

12. Example 2 – One Character Missing S   A  bAC  aCS   A  bAC  aC S  aBA  cAC  bCS  aBA  cAC  bC S  aC A  C  S  aC A  C   S  bA B  aBS  bA B  aB S  bC B  cBS  bC B  cB S  cA B  S  cA B   S  cBS  cB 12

13. Example 2 – One Character Missing S   A  bAC  aCS   A  bAC  aC S  aBA  cAC  bCS  aBA  cAC  bC S  aC A  C  S  aC A  C   S  bA B  aBS  bA B  aB S  bC B  cBS  bC B  cB S  cA B  S  cA B   S  cBS  cB 13

14. Regular Languages and Regular Grammars FSM  Regular grammar: Show by construction that, for every FSM M there exists a regular grammar G such that L(G) = L(M). 1.Make M deterministic  M’ = (K, , , s, A). Construct G = (V, , R, S) from M’. 2.Create a nonterminal for each state in the M’. V = K  . 3.The start state becomes the starting nonterminal. S = s. 4.For each transition  (T, a) = U, make a rule of the form T  aU. 5.For each accepting state T, make a rule of the form T  . 14

15. Strings that End with aaaa L = {w  { a, b }* : w ends with the pattern aaaa }. 15

16. Strings that End with aaaa L = {w  { a, b }* : w ends with the pattern aaaa }. S  aSS  aS S  bSS  bS S  aBS  aB B  aCB  aC C  aDC  aD D  aD  a 16