Module 36

1
Module 36

• Non context-free languages
• Examples and Intuition
• Pumping lemma for CFLs
• Pumping condition
• No proof of pumping lemma
• Applying pumping lemma to prove that some
  languages are not CFLs

2
Examples and Intuition
3
Examples

• What are some examples of nonregular languages?
• Can we build on any of these languages to create
  a non context-free language?

4
Intuition

• Try and prove that these languages are CFLs and
  identify the stumbling blocks
• Why cant we construct a CFG to generate this
  language?
• Why cant we construct a PDA to accept this
  language?
• Compare to similar CFL languages to try and
  identify differences.

5
Pumping Lemma for CFLs
6
Comparison to regular language pumping
lemma/condition
7
Whats different about CFLs than regular
languages?

• In regular languages, a single substring pumps
• Consider the language of even length strings over
  a,b
• We can identify a single substring which can be
  pumped
• In CFLs, multiple substrings can pump
• Consider the language anbn n gt 0
• No single substring can be pumped and allow us to
  stay in the language
• However, there do exist pairs of substrings which
  can be pumped resulting in strings which stay in
  the language
• This results in a modified pumping condition

8
Modified Pumping Condition

• A language L satisfies the regular language
  pumping condition if
• there exists an integer n gt 0 such that
• for all strings x in L of length at least n
• there exist strings u, v, w such that
• x uvw and
• uv lt n and
• v gt 1 and
• For all k gt 0, uvkw is in L

• A language L satisfies the CFL pumping condition
  if
• there exists an integer n gt 0 such that
• for all strings x in L of length at least n
• there exist strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• vy gt 1 and
• For all k gt 0, uvkwykz is in L

9
Pumping Lemma

• All CFLs satisfy the CFL pumping condition

CFLs
10
Implications
Pumping
CFL

• We can use the pumping lemma to prove a language
  L is not a CFL
• Show L does not satisfy the CFL pumping condition
• We cannot use the pumping lemma to prove a
  language is context-free
• Showing L satisfies the pumping condition does
  not guarantee that L is context-free

11
Pumping Lemma

• What does it mean?

12
Pumping Condition

• A language L satisfies the CFL pumping condition
  if
• there exists an integer n gt 0 such that
• for all strings x in L of length at least n
• there exist strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• vy gt 1 and
• For all k gt 0, uvkwykz is in L

13
v and y can be pumped
1) x in L2) x uvwyz3) For all k gt 0, uvkwykz
is in L

• Let x abcdefg be in L
• Then there exist 2 substrings v and y in x such
  that v and y can be repeated (pumped) in place
  any number of times and the resulting string is
  still in L
• uvkwykz is in L for all k gt 0
• For example
• v cd and y f
• uv0wy0z uwz abeg is in L
• uv1wy1z uvwyz abcdefg is in L
• uv2wy2z uvvwyyz abcdcdeffg is in L
• uv3wy3z uvvvwyyyz abcdcdcdefffg is in L

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What the other parts mean

• A language L satisfies the CFL pumping condition
  if
• there exists an integer n gt 0 such that
• Since we skip this proof, we will not see what n
  really means
• for all strings x in L of length at least n
• x must be in L and have sufficient length
• there exist strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• v and y are contained within n characters of x
• Note these are NOT necessarily the first n
  characters of x
• vy gt 1 and
• v and y cannot both be l,
• One of them might be l, but not both
• For all k gt 0, uvkwykz is in L

15
Example

• Let L be the set of palindromes over a,b
• Let x aabaa
• Let n 3
• What are the possibilities for v and y ignoring
  the pumping constraint?
• Which ones satisfy the pumping lemma?

16
Pumping Lemma

• Applying it to prove a specific language L is not
  context-free

17
How we use the Pumping Lemma

• We choose a specific language L
• For example, ajbjcj j gt 0
• We show that L does not satisfy the pumping
  condition
• We conclude that L is not context-free

18
Showing L does not pump

• A language L satisfies the CFL pumping condition
  if
• there exists an integer n gt 0 such that
• for all strings x in L of length at least n
• there exist strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• vy gt 1 and
• For all k gt 0, uvkwykz is in L

• A language L does not satisfy the CFL pumping
  condition if
• for all integers n of sufficient size
• there exists a string x in L of length at least n
  such that
• for all strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• vy gt 1
• There exists a k gt 0 such that uvkwykz is not in
  L

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Example Proof

• Proof that L aibici igt0 does not satisfy
  the CFL pumping condition
• Let n be the integer from the pumping lemma
• Choose x anbncn
• Consider all strings u, v, w, y, z s.t.
• x uvwyz and
• vwy lt n and
• vy gt 1
• Argue that uvkwykz is not in L for some k gt 0
• Argument must apply to all possible u,v,w,y,z
• Continued on next slide

• A language L does not satisfy the CFL pumping
  condition if
• for all integers n of sufficient size
• there exists a string x in L of length at least n
  such that
• for all strings u, v, w, y, z such that
• x uvwyz and
• vwy lt n and
• vy gt 1
• There exists a k gt 0 such that uvkwykz is not in
  L

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Example Proof Continued

• Proof that L aibici igt0 does not satisfy
  the CFL pumping condition
• Let n be the integer from the pumping lemma
• Choose x anbncn
• Consider all strings u, v, w, y, z s.t.
• x uvwyz and
• vwy lt n and
• vy gt 1
• Argue that uvkwykz is not in L for some k gt 0
• Argument must apply to all possible u,v,w,y,z
• Continued next column

• Identify possible cases for vwy
• What is impossible for vwy?
• Case 1
• vwy contains no as
• Case 2
• vwy contains no cs
• Must argue uvkwykz is not in L for both cases
  described above
• Can use different values of k
• Continued on next slide

21
Example Proof Continued

• Identify possible cases for vwy
• What is impossible for vwy?
• Case 1
• vwy contains no as
• Case 2
• vwy contains no cs
• Must argue uvkwykz is not in L for both cases
  described above
• Can use different values of k
• Continued next column

• Case 1 vwy contains no as
• vy contains at least 1 b or c
• follows from
• vwy contains no as and
• vy gt 1
• uwz is not in L
• uwz has n as
• follows from fact vwy contains no as and x
  originally had n as
• uwz has fewer than n bs or fewer than n cs
• follows from vy contains at least 1 b or c and x
  originally only had n bs and n cs
• Continued next slide

22
Example Proof Continued

• Case 1 vwy contains no as
• vy contains at least 1 b or c
• follows from
• vwy contains no as and
• vy gt 1
• uwz is not in L
• uwz has n as
• follows from fact vwy contains no as and x
  originally had n as
• uwz has fewer than n bs or fewer than n cs
• follows from vy contains at least 1 b or c and x
  originally only had n bs and n cs
• Continued next column

• Case 2 vwy contains no cs
• vy contains at least
• uv2wy2z is not in L
• uv2wy2z has n cs
• follows from fact vwy contains no cs and x
  originally had n cs
• uv2wy2z has more than n as or more than n bs
• follows from vy contains at least 1 a or b and x
  originally has n as and n bs
• Continued next slide

23
Example Proof Completed

• For all possible u, v, w, y, z, we have shown
  there exists a kgt0 such that
• uvkwykz is not in L
• Note, we used a different value of k for each
  case (though we didnt have to)
• Therefore L does not satisfy the CFL pumping
  condition
• There L is not a CFL

• Case 2 vwy contains no cs
• vy contains at least
• uv2wy2z is not in L
• uv2wy2z has n cs
• follows from fact vwy contains no cs and x
  originally had n cs
• uv2wy2z has more than n as or more than n bs
• follows from vy contains at least 1 a or b and x
  originally has n as and n bs
• Continued next column

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Other example languages

• TWOCOPIES ww w is in a,b
• abbabb is in TWOCOPIES but abaabb is not
• EQUAL3 the set of strings over a, b, c such
  that the number of as equals the number of bs
  equals the number of cs
• aibjck i lt j lt k

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Pumping Lemma

• Two rules of thumb

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Two Rules of Thumb

• Try to use blocks of at least n characters in x
• For TWOCOPIES, choose x anbnanbn rather than
  anbanb
• Guarantees v and y cannot be in more than 2
  blocks of x
• Try k0 or k2
• k0
• This reduces number of occurrences of v and y
• k2
• This increases number of occurrences of v and y

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Summary

• We use the Pumping Lemma to prove a language is
  not a CFL
• Note, does not work for all non CFL languages
• Can be strengthened to Ogdens Lemma
• In book
• Choosing a good string x is first key step
• Choosing a good k is second key step
• Typically have several cases for v, w, y