courtesy costas busch - rpi1 non-regular languages

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  • Slide 1
  • Courtesy Costas Busch - RPI1 Non-regular languages
  • Slide 2
  • Courtesy Costas Busch - RPI2 Regular languages Non-regular languages
  • Slide 3
  • Courtesy Costas Busch - RPI3 How can we prove that a language is not regular? Prove that there is no DFA that accepts Problem: this is not easy to prove Solution: the Pumping Lemma !!!
  • Slide 4
  • Courtesy Costas Busch - RPI4 The Pigeonhole Principle
  • Slide 5
  • Courtesy Costas Busch - RPI5 pigeons pigeonholes
  • Slide 6
  • Courtesy Costas Busch - RPI6 A pigeonhole must contain at least two pigeons
  • Slide 7
  • Courtesy Costas Busch - RPI7........... pigeons pigeonholes
  • Slide 8
  • Courtesy Costas Busch - RPI8 The Pigeonhole Principle........... pigeons pigeonholes There is a pigeonhole with at least 2 pigeons
  • Slide 9
  • Courtesy Costas Busch - RPI9 The Pigeonhole Principle and DFAs
  • Slide 10
  • Courtesy Costas Busch - RPI10 DFA with states
  • Slide 11
  • Courtesy Costas Busch - RPI11 In walks of strings:no state is repeated
  • Slide 12
  • Courtesy Costas Busch - RPI12 In walks of strings:a state is repeated
  • Slide 13
  • Courtesy Costas Busch - RPI13 If string has length : Thus, a state must be repeated Then the transitions of string are more than the states of the DFA
  • Slide 14
  • Courtesy Costas Busch - RPI14 In general, for any DFA: String has length number of states A state must be repeated in the walk of...... walk of Repeated state
  • Slide 15
  • Courtesy Costas Busch - RPI15 In other words for a string : transitions are pigeons states are pigeonholes...... walk of Repeated state
  • Slide 16
  • Courtesy Costas Busch - RPI16 The Pumping Lemma
  • Slide 17
  • Courtesy Costas Busch - RPI17 Take an infinite regular language There exists a DFA that accepts states
  • Slide 18
  • Courtesy Costas Busch - RPI18 Take string with There is a walk with label :......... walk
  • Slide 19
  • Courtesy Costas Busch - RPI19 If string has length (number of states of DFA) then, from the pigeonhole principle: a state is repeated in the walk...... walk
  • Slide 20
  • Courtesy Costas Busch - RPI20...... walk Let be the first state repeated in the walk of
  • Slide 21
  • Courtesy Costas Busch - RPI21 Write......
  • Slide 22
  • Courtesy Costas Busch - RPI22...... Observations:lengthnumber of states of DFA length
  • Slide 23
  • Courtesy Costas Busch - RPI23 The string is accepted Observation:......
  • Slide 24
  • Courtesy Costas Busch - RPI24 The string is accepted Observation:......
  • Slide 25
  • Courtesy Costas Busch - RPI25 The string is accepted Observation:......
  • Slide 26
  • Courtesy Costas Busch - RPI26 The string is accepted In General:......
  • Slide 27
  • Courtesy Costas Busch - RPI27 In General:...... Language accepted by the DFA
  • Slide 28
  • Courtesy Costas Busch - RPI28 In other words, we described: The Pumping Lemma !!!
  • Slide 29
  • Courtesy Costas Busch - RPI29 The Pumping Lemma: Given a infinite regular language there exists an integer for any string with length we can write with and such that:
  • Slide 30
  • Courtesy Costas Busch - RPI30 Applications of the Pumping Lemma
  • Slide 31
  • Courtesy Costas Busch - RPI31 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 32
  • Courtesy Costas Busch - RPI32 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 33
  • Courtesy Costas Busch - RPI33 Let be the integer in the Pumping Lemma Pick a string such that: length We pick
  • Slide 34
  • Courtesy Costas Busch - RPI34 it must be that length From the Pumping Lemma Write: Thus:
  • Slide 35
  • Courtesy Costas Busch - RPI35 From the Pumping Lemma: Thus:
  • Slide 36
  • Courtesy Costas Busch - RPI36 From the Pumping Lemma: Thus:
  • Slide 37
  • Courtesy Costas Busch - RPI37 BUT: CONTRADICTION!!!
  • Slide 38
  • Courtesy Costas Busch - RPI38 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:
  • Slide 39
  • Courtesy Costas Busch - RPI39 Regular languages Non-regular languages
  • Slide 40
  • Courtesy Costas Busch - RPI40 Regular languages Non-regular languages
  • Slide 41
  • Courtesy Costas Busch - RPI41 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 42
  • Courtesy Costas Busch - RPI42 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 43
  • Courtesy Costas Busch - RPI43 We pick Let be the integer in the Pumping Lemma Pick a string such that: length and
  • Slide 44
  • Courtesy Costas Busch - RPI44 Write it must be that length From the Pumping Lemma Thus:
  • Slide 45
  • Courtesy Costas Busch - RPI45 From the Pumping Lemma: Thus:
  • Slide 46
  • Courtesy Costas Busch - RPI46 From the Pumping Lemma: Thus:
  • Slide 47
  • Courtesy Costas Busch - RPI47 BUT: CONTRADICTION!!!
  • Slide 48
  • Courtesy Costas Busch - RPI48 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:
  • Slide 49
  • Courtesy Costas Busch - RPI49 Regular languages Non-regular languages
  • Slide 50
  • Courtesy Costas Busch - RPI50 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 51
  • Courtesy Costas Busch - RPI51 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 52
  • Courtesy Costas Busch - RPI52 We pick Let be the integer in the Pumping Lemma Pick a string such that: length and
  • Slide 53
  • Courtesy Costas Busch - RPI53 Write it must be that length From the Pumping Lemma Thus:
  • Slide 54
  • Courtesy Costas Busch - RPI54 From the Pumping Lemma: Thus:
  • Slide 55
  • Courtesy Costas Busch - RPI55 From the Pumping Lemma: Thus:
  • Slide 56
  • Courtesy Costas Busch - RPI56 BUT: CONTRADICTION!!!
  • Slide 57
  • Courtesy Costas Busch - RPI57 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:
  • Slide 58
  • Courtesy Costas Busch - RPI58 Regular languages Non-regular languages
  • Slide 59
  • Courtesy Costas Busch - RPI59 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 60
  • Courtesy Costas Busch - RPI60 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 61
  • Courtesy Costas Busch - RPI61 We pick Let be the integer in the Pumping Lemma Pick a string such that: length
  • Slide 62
  • Courtesy Costas Busch - RPI62 Write it must be that length From the Pumping Lemma Thus:
  • Slide 63
  • Courtesy Costas Busch - RPI63 From the Pumping Lemma: Thus:
  • Slide 64
  • Courtesy Costas Busch - RPI64 From the Pumping Lemma: Thus:
  • Slide 65
  • Courtesy Costas Busch - RPI65 Since: There must exist such that:
  • Slide 66
  • Courtesy Costas Busch - RPI66 However:for for any
  • Slide 67
  • Courtesy Costas Busch - RPI67 BUT: CONTRADICTION!!!
  • Slide 68
  • Courtesy Costas Busch - RPI68 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:
  • Slide 69
  • 69 Summary Showing regular construct DFA, NFA construct regular expression show L is the union, concatenation, intersection (regular operations) of regular languages. Showing non-regular pumping lemma assume regular, apply closure properties of regular languages and obtain a known non-regular language.

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