Context-Free Grammars

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Context-Free Grammars Chomsky Normal Form

• Lecture 16
• Section 2.1
• Wed, Sep 26, 2007

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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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Chomsky Normal Form

• Theorem Every context-free language is generated
  by a grammar in CNF.

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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.

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Chomsky Normal Form

• Proof (detailed)
• Add a new start symbol S0.
• Add the rule S0 ? S.
• Eliminate all ? rules A ? ?.
• For each rule A ? ? and each rule B ? uAv, add a
  rule B ? uv.
• Eliminate the rule A ? ?.

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Example

• Start with the grammar
• S ? SXS ?
• X ? ab ?
• Add the rule
• S0 ? S

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Example

• We now have
• S0 ? S
• S ? SXS ?
• X ? ab ?

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Example

• Apply the rules S ? ? and X ? ? to the other
  rules, creating the rules
• S ? X
• S ? SS
• S ? XS
• S ? SX
• S ? S
• Dont bother with the last rule.

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Example

• Eliminate the rules
• S ? ?
• X ? ?

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Example

• Add the rule
• S0 ? ?
• because in the original, S could be replaced by
  ?.

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Example

• We now have
• S0 ? S ?
• S ? SXS SS SX XS X
• X ? ab

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Chomsky Normal Form

• Proof (detailed)
• Eliminate all unit rules A ? B.
• If A ? B and B ? u are rules, then add the rule A
  ? u.
• Eliminate the rule A ? B.

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Chomsky Normal Form

• Add the rules
• S ? ab
• S0 ? SXS SS SX XS X ab
• Eliminate the rules
• S0 ? S
• S ? X

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Example

• We now have
• S0 ? SXS SS SX XS ab ?
• S ? SXS SS SX XS ab
• X ? ab

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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.

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Example

• Add the rules
• A ? a
• B ? b
• and rewrite the string ab as AB.

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Example

• We now have
• S0 ? SXS SS SX XS AB ?
• S ? SXS SS SX XS AB
• X ? AB
• A ? a
• B ? b

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Chomsky Normal Form

• Finally, eliminate all long rules.
• Break all strings of length ? 2 into a series of
  strings of length 2.

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Chomsky Normal Form

• Replace the rule
• A ? B1B2Bk
• with
• A ? B1C1
• C1 ? B2C2
• Ck 2 ? Bk 2Ck 2
• Ck 1 ? Bk 1Bk

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Example

• Replace
• S0 ? SXS
• S ? SXS
• with
• S0 ? SY
• S ? SY
• Y ? XS

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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

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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.

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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.

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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.

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Example

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

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