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CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10

CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10 Mälardalen University 2011. Content The Pumping Lemma for CFL Applications of the Pumping Lemma for CFL Midterm Exam 2: Context-Free Languages. The Pumping Lemma for CFL’s.

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CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10

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  1. CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 10 Mälardalen University 2011

  2. ContentThe Pumping Lemma for CFLApplications of the Pumping Lemma for CFLMidterm Exam 2: Context-Free Languages

  3. The Pumping Lemma for CFL’s

  4. Comparison with the Pumping Lemma for Regular Languages

  5. ...... ...... What is the difference between Context Free Languages and Regular Languages? In regular languages, a single substring “pumps”

  6. What is the difference between Context Free Languages and Regular Languages? In CFL’s, multiple substrings can be “pumped” • Consider the language {an bn | 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 Thus, a modified pumping lemma applies.

  7. stringsof terminals String Last repeated variable repeated

  8. A language L satisfies the RLpumping condition if: there exists an integer m > 0 such that for all strings w in L of length at least m there exist strings x, y, zsuch that w = xyzand |xy| ≤ mand |y| ≥ 1and For all i ≥ 0, uvizis in L A language L satisfies theCFLpumping condition if: there exists an integer m > 0such that for all strings w in L of length at least m there exist strings u, v, x, y, z such that w = uvxyzand |vxy| ≤ mand |vy| ≥ 1and For all i ≥ 0, uvixyiz is in L Pumping Conditions for RL and CFL

  9. CFL’s “Pumping Languages” All languages over {a,b} Pumping Lemma All CFL’s satisfy the CFL pumping condition

  10. Implications CFL’s “Pumping Languages” All languages over {a,b} • We can use the pumping lemma to prove a language Lis not a CFL • Show that L does not satisfy the CFL pumping condition • We cannot use the pumping lemma to prove a language is CFL • 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 m > 0such that • for all strings w in L of length at least m • there exist strings u, v, x, y, zsuch that • w = uvxyzand • |vxy| ≤ m and • |vy| ≥ 1 and Then for all i ≥ 0, uviwyizis in L

  13. v and y can be pumped 1) w in L2) w = uvxyz3) For all i ≥ 0, uvixyiz is in L • Let w = abcdefgbe in L • Then there exist substrings v and y in w such that v and y can be repeated (pumped) and the resulting string is still in L • uvixyiz is in L for all i ≥ 0

  14. For example w =abcdefg v = cdand y = f uv0xy0z = uxz =abegis in L uv1xy1z = uvxyz = abcdefgis in L uv2xy2z = uvvxyyz = abcdcdeffgis in L uv3xy3z = uvvvxyyyz = abcdcdcdefffg is in L …

  15. What the other parts mean A language L satisfies the CFL pumping condition if: • there exists an integer m > 0 such that • for all strings w in L of length at least m • w must be in L and have sufficient length

  16. What the other parts mean • There exist strings u, v, x, y, z such that • w = uvxyz and • |vxy| ≤ m and • v and y are contained within m characters of w • Note: these are NOT necessarily the first m characters of w • |vy| ≥ 1 and • v and y cannot both be 1, • One of them might be 1, but not both • For all i ≥ 0, uvixyiz is in L

  17. Pumping Lemma Applying CFL pumping lemma to prove that a specific language L is not context-free

  18. How we use the Pumping Lemma • We choose a specific language L For example {aj bj cj | j > 0} • We show that L does not satisfy the pumping condition • We conclude that L is not context-free

  19. A language L satisfies the CFL pumping condition if: there exists an integer m > 0 such that for all strings w in L of length at least m there exist strings u, v, x, y, z such that w = uvxyz and |vxy| ≤ m and |vy| ≥ 1 and For all i ≥ 0, uvixyiz is in L A language L does not satisfy the CFL pumping condition if: for all integers m of sufficient size there exists a string w in L of length at least m such that for all strings u, v, x, y, z where w = uvxyz and |vxy| ≤ m and |vy| ≥ 1 There exists a i ≥ 0 such that uvixyiz is not in L Showing that L “does not pump”

  20. Example Languages • L = {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} • L ={aibjck | i < j < k}

  21. More Pumping Lemma Applications

  22. For infinite context-free language there exists an integer such that for any string we can write with lengths and The Pumping Lemma for CFL

  23. Let be a context free grammar. There exists an integer such that can be written with lengths and The Pumping Lemma for CFL

  24. Unrestricted grammarlanguages Non-regular languages Context-Free Languages Regular Languages

  25. Theorem The language is not context free Proof Use the Pumping Lemma for context-free languages

  26. Assume the contrary - that is context-free Since is context-free and infinite we can apply the pumping lemma

  27. Pumping Lemma gives a number such that: Pick any string of with length at least we pick:

  28. and with lengths We can write: Pumping Lemma says: for all

  29. We examine all the possible locations of string in

  30. Case 1: is within the first

  31. Case 1: is within the first

  32. Case 1: is within the first

  33. Case 1: is within the first However, from Pumping Lemma: Contradiction!

  34. Case 2: is in the first is in the first

  35. Case 2: is in the first is in the first

  36. Case 2: is in the first is in the first

  37. Case 2: is in the first is in the first However, from Pumping Lemma: Contradiction!

  38. Case 3: overlaps the first is in the first

  39. Case 3: overlaps the first is in the first

  40. Case 3: overlaps the first is in the first

  41. Case 3: overlaps the first is in the first However, from Pumping Lemma: Contradiction!

  42. Case 4: in the first Overlaps the first Analysis is similar to case 3

  43. or or Other cases: is within Analysis is similar to case 1:

  44. More cases: overlaps or Analysis is similar to cases 2,3,4:

  45. There are no other cases to consider. Since , it is impossible for to overlap: neither nor nor

  46. is not context-free In all cases we obtained a contradiction. Therefore: The original assumption that is context-free must be wrong Conclusion: END OF PROOF

  47. Unrestricted grammarlanguages Non-regular languages Context-Free Languages Regular Languages

  48. Theorem The language is not context free Proof Use the Pumping Lemma for context-free languages

  49. Since is context-free and infinite we can apply the pumping lemma Assume to the contrary that is context-free

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