1 / 16

Regular Grammars

Regular Grammars. Chapter 7. 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 ,

lars
Télécharger la présentation

Regular Grammars

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Regular Grammars Chapter 7

  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, ●S (the start symbol) is a nonterminal.

  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aS Not legal: SaSa and aSaT

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

  5. Regular Grammar Example 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aT SbT TaS TbS }

  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.

  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.

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

  9. Example 1 - Even Length Strings S SaT SbT TaS TbS

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

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

  12. Example 2 – One Character Missing S AbA CaC SaB AcA CbC SaC A C SbA BaB SbC BcB ScA B ScB

  13. Example 2 – One Character Missing S AbA CaC SaB AcA CbC SaC A C SbA BaB SbC BcB ScA B ScB

  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). Make M deterministic M’ = (K, , , s, A). Construct G = (V, , R, S) from M’. Create a nonterminal for each state in the M’. V = K. The start state becomes the starting nonterminal. S = s. For each transition (T, a) = U, make a rule of the form TaU. For each accepting state T, make a rule of the form T.

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

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

More Related