Context-Free Grammars – Chomsky Normal Form

Lecture 16Section 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.

Chomsky Normal Form

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

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.

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 .

Example

• Start with the grammar• S SXS | • X ab |

• Add the rule • S0 S

Example

• We now have• S0 S

• S SXS | • X ab |

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.

Example

• Eliminate the rules• S • X

Example

• Add the rule• S0

because in the original, S could be replaced by .

Example

• We now have• S0 S |

• S SXS | SS | SX | XS | X• X ab

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.

Chomsky Normal Form

• Add the rules• S ab

• S0 SXS | SS | SX | XS | X | ab

• Eliminate the rules• S0 S

• S X

Example

• We now have• S0 SXS | SS | SX | XS | ab |

• S SXS | SS | SX | XS | ab• X ab

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.

Example

• Add the rules• A a• B b

and rewrite the string ab as AB.

Example

• We now have• S0 SXS | SS | SX | XS | AB |

• S SXS | SS | SX | XS | AB• X AB• A a• B b

Chomsky Normal Form

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

a series of strings of length 2.

Chomsky Normal Form

• Replace the rule• A B1B2…Bk

with• A B1C1

• C1 B2C2

• …• Ck – 2 Bk – 2Ck – 2

• Ck – 1 Bk – 1Bk

Example

• Replace• S0 SXS

• S SXS

• with• S0 SY

• S SY• Y XS

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

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.

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.

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.

Example

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

A Tree of all Possible Derivations

S0

SY SS AB

a bSY SS AB SY SS ABXS AB SS SY

etc.

Example

• Put the grammarE E + E | E * E | (E) | a | b | c

into CNF.• Then show that the string c++ is

not derivable from it.