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Context-Free Grammars – Chomsky Normal Form

Context-Free Grammars – Chomsky Normal Form. Lecture 16 Section 2.1 Wed, Sep 26, 2007. Chomsky Normal Form. A context-free grammar is in Chomsky Normal Form (CNF) if each rule is of the form A  BC , or A  a , where B and C are not S . Furthermore, the rule S   is allowed.

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Context-Free Grammars – Chomsky Normal Form

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  1. Context-Free Grammars – Chomsky Normal Form Lecture 16 Section 2.1 Wed, Sep 26, 2007

  2. Chomsky Normal Form • A context-free grammar is in Chomsky Normal Form (CNF) if each rule is of the form • ABC, or • Aa, where B and C are not S. • Furthermore, the rule S is allowed.

  3. Chomsky Normal Form • Theorem: Every context-free language is generated by a grammar in CNF.

  4. Chomsky Normal Form • Constructive proof (outline): • Add a new start symbol S0. • Eliminate all  rules A. • Eliminate all unit rules AB. • Convert all remaining rules to the proper form.

  5. Chomsky Normal Form • Proof (detailed): • Add a new start symbol S0. • Add the rule S0S. • Eliminate all  rulesA. • For each rule A and each rule B  uAv, add a rule B  uv. • Eliminate the rule A.

  6. Example • Start with the grammar • SSXS |  • X  ab |  • Add the rule • S0S

  7. Example • We now have • S0S • SSXS |  • X  ab | 

  8. Example • Apply the rules S and X  to the other rules, creating the rules • S  X • S  SS • S  XS • S  SX • S  S • Don’t bother with the last rule.

  9. Example • Eliminate the rules • S • X  

  10. Example • Add the rule • S0 because in the original, S could be replaced by .

  11. Example • We now have • S0S |  • S  SXS | SS | SX | XS | X • X  ab

  12. Chomsky Normal Form • Proof (detailed): • Eliminate all unit rulesAB. • If AB and Bu are rules, then add the rule Au. • Eliminate the rule AB.

  13. Chomsky Normal Form • Add the rules • S  ab • S0SXS | SS | SX | XS | X | ab • Eliminate the rules • S0S • S  X

  14. Example • We now have • S0SXS | SS | SX | XS | ab |  • S  SXS | SS | SX | XS | ab • X  ab

  15. Chomsky Normal Form • Eliminate all mixed rules. • Add rules • Aa for all terminals appearing in strings of length  2. • Then replace a with A in those strings.

  16. Example • Add the rules • A a • B b and rewrite the string ab as AB.

  17. Example • We now have • S0SXS | SS | SX | XS | AB |  • S  SXS | SS | SX | XS | AB • X  AB • A a • B b

  18. Chomsky Normal Form • Finally, eliminate all long rules. • Break all strings of length  2 into a series of strings of length 2.

  19. Chomsky Normal Form • Replace the rule • A B1B2…Bk with • A  B1C1 • C1  B2C2 • … • Ck – 2  Bk – 2Ck – 2 • Ck – 1  Bk – 1Bk

  20. Example • Replace • S0SXS • S  SXS • with • S0SY • S  SY • Y XS

  21. Example • The final result is • S0SY | SS | SX | XS | AB |  • S  SY | SS | SX | XS | AB • X  AB • Y XS • A a • B b

  22. A Derivation in CNF • Use this grammar in CNF to derive the string ababab. • S0 SY  SXS  ABXS  ABABS  ABABAB  aBABAB  abABAB  abaBAB  ababAB ababaB  ababab.

  23. CNF Derivations • Theorem: If a grammar G is in CNF and a string w in L(G) has length n, then w will be derived from G in exactly 2n – 1 steps.

  24. The Membership Problem • This theorem allows us to determine whether a given string is derivable from a given grammar. • This is called the Membership Problem.

  25. Example • Show that the string abba is not derivable from the grammar of the previous example.

  26. A Tree of all Possible Derivations S0  SY SS AB SY SS SY SS AB SY a b XS AB AB SS etc.

  27. Example • Put the grammar EE + E | E * E | (E) | a | b | c into CNF. • Then show that the string c++ is not derivable from it.

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