Chapter 12: Context-Free Languages and Pushdown Automata

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Chapter 12 Context-Free Languages and Pushdown
Automata

• Section 12.4 Context-Free Language Topics

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Remove ?-productions

• Algorithm. Remove ?-productions from grammars for
  langauges without ?.
• 1. Find nonterminals that derive ?.
• 2. For each production A ? w construct all
  productions A ? w where w is obtained from
• w by removing one or more occurrences of the
  nonterminals from Step 1.
• 3. Combine the original productions with those of
  step 2 and eliminate any ?-productions.

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Example

• Remove ?-productions from the grammar
• S ? ABc
• A ? aA ?
• B ? bB ?.
• Solution.
• Step 1 The nonterminals A and B derive ?.
• Step 2 From the production S ? ABc we construct
  S ? Bc Ac c.
• From the production A ? aA we construct A ? a.
• From the production B ? bB we construct B ? b.
• Step 3 S ? ABc Bc Ac c
• A ? aA a
• B ? bB b.

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Quiz

• Remove ? -productions from
• S ? ABc Ab c
• A ? ABa ?
• B ? Bbc ?.
• Solution.
• S ? ABc Ab c Bc Ac b
• A ? ABa Ba Aa a
• B ? Bbc bc.

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Chomsky Normal Form

• Productions have one of the following forms
• A ? b (b a terminal)
• A ? BC
• S ? ? (if ? is in the language).
• Advantages Parse trees are binary, which are
  easy to represent. Any string of length n gt 0 can
  be derived in 2n 1 steps.

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Transform context-free grammar to Chomsky normal
form

• Algorithm. Transform context-free grammar to
  Chomsky normal form
• 1. Remove A ? ? (if A ? S) by previous algorithm.
  (If S ? ? is removed, add it back.)
• 2. Remove unit productions (i.e., A ? B) If A ?
  B or A B, then construct productions A ? w
  where B ? w is not a unit production. Now remove
  all unit productions.
• 3. For each production whose right side has two
  or more symbols, replace all occurrences of each
  terminal a with a new nonterminal A and also add
  the new production A ? a.
• 4. Replace each production B ? C1Cn with n gt 2
  with B ? C1D where D ? C2 Cn.
• Repeat this step until all right sides have
  length two.

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Example

• Construct a Chomsky normal form for the grammar
• S ? aSb D
• D ? Dc ?.
• Solution.
• Step 1 S ? aSb ab D ?
• D ? Dc c.
• Step 2 S ? aSb ab Dc c ?
• D ? Dc c.
• Step 3 S ? ASB AB DC c ?
• D ? DC c
• A ? a
• B ? b
• C ? c.
• Step 4 Replace S ? ASB by
• S ? AE and E ? SB.

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Quiz

• Construct a Chomsky normal form for the grammar
• S ? aSbb T ?
• T ? cT d.
• Solution.
• Step 1 No change in ? -productions.
• Step 2 Remove unit production S ? T to obtain
• S ? aSbb cT d ?
• T ? cT d.
• Step 3 Transform right sides of length at least
  two into strings of nonterminals.
• S ? ASBB CT d ?
• T ? CT d
• A ? a
• B ? b
• C ? c.
• Step 4 Transform right sides into strings of
  length at most two.
• S ? AD CT d ?
• D ? SE
• E? BB
• T ? CT d

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Greibach Normal Form

• Productions have one of the following forms
• A ? b (b a terminal)
• A ? bD1Dk
• S ? ? (if ? is in the language).
• Advantage Any string of length n gt 0 can be
  derived in n steps.

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Transform context-free grammar to Greibach normal
form

• Algorithm (idea). Transform context-free grammar
  to Greibach normal form.
• 1. Remove all left-recursion.
• 2. Remove ?-productions. (If S ? ? is removed,
  add it back.)
• 3. Make substitutions to transform the grammar
  into the proper form.

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Example

• Put the following grammar into Greibach normal
  form.
• S ? AB Ac d
• A ? aA a
• B? Ab c.
• Solution Steps 1 and 2 are unnecessary for this
  grammar.
• Step 3 Replace A in S ? AB Ac d with aA a
  to obtain
• S ? aAB aB aAc ac d.
• Replace A in B? Ab c with aA a to obtain
• B? aAb ab c.
• Add the new productions C ? c and D ? b to
  obtain the proper form
• S ? aAB aB aAC aC d
• A ? aA a
• B? aAD aD c
• C ? c
• D ? b.

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Quiz

• Put the following grammar into Greibach normal
  form.
• S ? BaS B
• B? cSd a.
• Solution Steps 1 and 2 are unnecessary for this
  grammar.
• Step 3 Replace B in S ? BaS B with cSd a to
  obtain
• S ? cSdaS aaS cSd a.
• Add the new productions A ? a and D ? d to
  obtain the proper form
• S ? cSDAS aAS cSD a
• A ? a
• D ? d
• B? cSD a (Not needed in this example).

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Properties of Context-Free Languages

• When we know some properties of context-free
  languages they can help us argue, BWOC, that
  certain languages are not context-free.

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The Pumping Lemma

• If L is an infinite context-free language, then
  any grammar for L must be recursive, so there
  must be derivations of the following form where
  u, v, w, x, and y are terminal strings.
• S ? uNy
• N ? vNx (where v and x are not both ?)
• N ? w.
• These derivations lead to derivations like
• S ? uNy ? uvNxy ? uv2Nx2y ? uvkNxky ?
  uvkwxky ? L for all k ? N.
• This is the basis for the Pumping Lemma
• There is an integer m gt 0 such that if z ? L and
  z m, then z has the form z uvwxy where
• 1 vx vwx m and uvkwxky ? L for all
  k ? N.
• Note The number m depends on the grammar as
  well see in the following example.

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Example

• Suppose we have the following grammar for ?,
  bbc ? abcnd n ? N.
• S ? aNd bbc ?
• N? Nc b.
• Here are a few derivations
• S ? aNd ? abd
• S ? aNd ? aNcd ? abcd
• S ? aNd ? aNcd ? aNccd ? abccd
• S ? abckd for any k in N.
• For this grammar m 4 can be used in the pumping
  lemma because any derivation of a string z with
  z 4 must use the nonterminal N. For example,
  if z 8 and z abcccccd,
• then the pumping lemma factors z abcccccd
  uvwxy where 1 vx vwx 4 and
• uvkwxky ? L for all k ? N. In this case let u
  a, v ?, w b, x c, and y ccccd.

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Example

• The language L anbncnk k, n ? N is not
  context-free.
• Proof Assume, BWOC, that L is context-free. L is
  infinite, so pumping lemma applies.
• Choose z ambmcm where m is the positive
  integer from the lemma.
• Then z ambmcm uvwxy where 1 vx vwx
  m and uvkwxky ? L for all k ? N.
• Observe neither v nor x can contain distinct
  letters. For example, if v ab, then
• v2 abab, which cant appear as a
  substring of any string in L.
• So v and x must be strings of repeated
  occurrences of a single letter.
• Now since vwx m, there are two possible
  places in ambmcm where v and x must occur
• (1) v and x occur in ambm.
• (2) v and x occur in bmcm.
• But we obtain the following contradictions
  because v and x are not both ?.
• (1) Let k 2 to obtain uv2wx2y amibmjcm,
  where i gt 0 or j gt 0. So uv2wx2y ? L
• (2) Let k 0 to obtain uwy ambm-icmj, where
  i gt 0 or j gt 0. So we have uwy ? L.
• These contradictions imply that L is not
  context-free. QED.

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Example/Quiz

• Prove that the language L ss s ? a, b is
  not context-free.
• Proof Assume, BWOC, that L is context-free. L is
  infinite, so pumping lemma applies.
• Choose z ambmambm where m is the positive
  integer from the lemma.
• Then z ambmambm uvwxy where 1 vx
  vwx m and uvkwxky ? L for all k ? N.
• Now since vwx m, there are three possible
  places in ambmambm where v and x must occur
• (1) v and x occur in ambm (on the left of z).
• (2) v and x occur in bmam (in the center of z).
• (3) v and x occur in ambm (on the right of z).
• Notice that v and x can consist only of
  repetitions a single letter. For example, in case
  (1) suppose v aibj for some i gt 0 and j gt 0 and
  x bn for some n 0. Then, letting k 0, we
  would obtain uwy amibmjnambm, which cannot
  be in L. The argument is similar for the other
  cases. So v and x must consist only of
  repetitions of a single letter.

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Example/Quiz Contd

• We need to find a contradiction in each of the
  three cases. Well do it by using k 0.
• This tells us that uwy ? L. But we obtain the
  following contradictions because v and x are not
  both ?.
• (1) uwy amibmjambm where either i gt 0 or j gt
  0 So uwy ? L,
• (2) uwy ambmiamjbm where either i gt 0 or j gt
  0. So uwy ? L.
• (3) uwy ambmamibmj where either i gt 0 or j gt
  0. So uwy ? L.
• Therefore L is not context-free. QED.
• Remark Be careful that the choice of z is not in
  a context-free sublanguage of L. For example, if
  we chose z (ab)m(ab)m in the preceding example,
  we would not get any contradictions.

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The End of Chapter 12 - 4