1 Module 33 Non-context free languages –Intuition and Examples CFL Pumping Lemma –Comparison to...
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Transcript of 1 Module 33 Non-context free languages –Intuition and Examples CFL Pumping Lemma –Comparison to...
1
Module 33
• Non-context free languages– Intuition and Examples
• CFL Pumping Lemma– Comparison to regular language pumping
lemma– What it means– Proof overview– Applications
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Examples and Intuition
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Examples
• What are some examples of nonregular languages?
• Can we build on any of these languages to create a non context-free language?
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Intuition
• Try and prove that these languages are CFL’s and identify the stumbling blocks– Why can’t we construct a CFG to generate this language?
– Compare to similar CFL languages to try and identify differences.
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Comparison to regular language pumping lemma/condition
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What’s different about CFL’s 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 CFL’s, multiple substrings can “pump”– Consider the language {anbn | n > 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
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Modified Pumping Condition
• A language L satisfies the regular language pumping
condition if: – there exists an integer n > 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| ≤ n and
• |v| ≥ 1 and
• For all k ≥ 0, uvkw is in L
• A language L satisfies the CFL
pumping condition if: – there exists an integer n > 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| ≤ n and
• |vy| ≥ 1 and
• For all k ≥ 0, uvkwykz is in L
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Pumping Lemma
• All CFL’s satisfy the CFL pumping condition
All languages over {a,b}
CFL’s
“Pumping Languages”
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Implications
• 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
CFL
Pumping
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Pumping Lemma
What does it mean?
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Pumping Condition
• A language L satisfies the CFL pumping condition if: – there exists an integer n > 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| ≤ n and
• |vy| ≥ 1 and
• For all k ≥ 0, uvkwykz is in L
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v and y can be pumped
• 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 ≥ 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 • …
1) x in L2) x = uvwyz3) For all k ≥ 0, uvkwykz 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 > 0 such that• related to number of variables in CNF grammar G
– 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| ≤ n and
– v and y are contained within n characters of x– these are NOT necessarily the first n characters of x
• |vy| ≥ 1 and– v and y cannot both be – One of them might be , but not both
• For all k ≥ 0, uvkwykz is in L
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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?
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Pumping Lemma
Proof
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Proof
• For any CFL L, consider a CNF grammar G with m variables s.t. L(G) = L – {λ}
• Let n = 2m
• Consider any string x in L of length at least n, and let T be a parse tree for x
• How short can the longest path of T be?– Must have length at least m+1.
• From the last m+1 variables in the longest path of T, identify two variables that are identical.
• Formulate u, v, w, y, and z based on these choice.
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Illustration: k=2
u v w y z
S
A
A u v y z
S
A
A
u v y z
S
A
A
v w y
A
slide originally from Chris Umans, Cal Tech
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Illustration: k=0
u v w y z
S
A
A u z
S
A
u z
S
A
x
slide originally from Chris Umans, Cal Tech
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Pumping Lemma
Applying it to prove a specific language L is not context-free
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How we use the Pumping Lemma
• We choose a specific language L– For example, {ajbjcj | j > 0}
• We show that L does not satisfy the pumping condition
• We conclude that L is not context-free
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Showing L “does not pump”
• A language L satisfies the CFL pumping condition if: – there exists an integer n > 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| ≤ n and• |vy| ≥ 1 and• For all k ≥ 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| ≤ n and• |vy| ≥ 1
– There exists a k ≥ 0 such that uvkwykz is not in L
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Example Proof• 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| ≤ n and• |vy| ≥ 1
– There exists a k ≥ 0 such that uvkwykz is not in L
• Proof that L = {aibici | i>0} 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| ≤ n and• |vy| ≥ 1
• Argue that uvkwykz is not in L for some k ≥ 0
– Argument must apply to all possible u,v,w,y,z
– Continued on next slide
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Example Proof Continued• Proof that L = {aibici | i>0} 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| ≤ n and• |vy| ≥ 1
• Argue that uvkwykz is not in L for some k ≥ 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 a’s
• Case 2– vwy contains no c’s
• Must argue uvkwykz is not in L for both cases described above
– Can use different values of k– Continued on next slide
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Example Proof Continued
• Identify possible cases for vwy• What is impossible for vwy?
• Case 1– vwy contains no a’s
• Case 2– vwy contains no c’s
• 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 a’s– vy contains at least 1 b or c
• follows from – vwy contains no a’s and– |vy| ≥ 1
– uwz is not in L• uwz has n a’s
– follows from fact vwy contains no a’s and x originally had n a’s
• uwz has fewer than n b’s or fewer than n c’s
– follows from vy contains at least 1 b or c and x originally only had n b’s and n c’s
• Continued next slide
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Example Proof Continued
• Case 1: vwy contains no a’s– vy contains at least 1 b or c
• follows from – vwy contains no a’s and– |vy| ≥ 1
– uwz is not in L• uwz has n a’s
– follows from fact vwy contains no a’s and x originally had n a’s
• uwz has fewer than n b’s or fewer than n c’s
– follows from vy contains at least 1 b or c and x originally only had n b’s and n c’s
• Continued next column
• Case 2: vwy contains no c’s– vy contains at least
– uv2wy2z is not in L• uv2wy2z has n c’s
– follows from fact vwy contains no c’s and x originally had n c’s
• uv2wy2z has more than n a’s or more than n b’s
– follows from vy contains at least 1 a or b and x originally has n a’s and n b’s
• Continued next slide
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Example Proof Completed
• For all possible u, v, w, y, z, we have shown there exists a k ≥ 0 such that– uvkwykz is not in L
• Note, we used a different value of k for each case (though we didn’t have to)
• Therefore L does not satisfy the CFL pumping condition
• There L is not a CFL
• Case 2: vwy contains no c’s– vy contains at least
– uv2wy2z is not in L• uv2wy2z has n c’s
– follows from fact vwy contains no c’s and x originally had n c’s
• uv2wy2z has more than n a’s or more than n b’s
– follows from vy contains at least 1 a or b and x originally has n a’s and n b’s
• 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 a’s equals the number of b’s equals the number of c’s
• {aibjck | i < j < 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 k=0 or k=2– k=0
• This reduces number of occurrences of v and y
– k=2• This increases number of occurrences of v and y
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Summary
• Examples of some non-CFL’s• 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 Ogden’s 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