# costas busch - rpi1 standard representations of regular languages regular languages dfas nfas...

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• Slide 1
• Costas Busch - RPI1 Standard Representations of Regular Languages Regular Languages DFAs NFAs Regular Expressions Regular Grammars
• Slide 2
• Costas Busch - RPI2 When we say: We are given a Regular Language We mean:Language is in a standard representation
• Slide 3
• Costas Busch - RPI3 Elementary Questions about Regular Languages
• Slide 4
• Costas Busch - RPI4 Membership Question Question:Given regular language and string how can we check if ? Answer:Take the DFA that accepts and check if is accepted
• Slide 5
• Costas Busch - RPI5 DFA
• Slide 6
• Costas Busch - RPI6 Given regular language how can we check if is empty: ? Take the DFA that accepts Check if there is any path from the initial state to a final state Question: Answer:
• Slide 7
• Costas Busch - RPI7 DFA
• Slide 8
• Costas Busch - RPI8 Given regular language how can we check if is finite? Take the DFA that accepts Check if there is a walk with cycle from the initial state to a final state Question: Answer:
• Slide 9
• Costas Busch - RPI9 DFA is infinite DFA is finite
• Slide 10
• Costas Busch - RPI10 Given regular languages and how can we check if ? Question: Find if Answer:
• Slide 11
• Costas Busch - RPI11 and
• Slide 12
• Costas Busch - RPI12 or
• Slide 13
• Costas Busch - RPI13 Non-regular languages
• Slide 14
• Costas Busch - RPI14 Regular languages Non-regular languages
• Slide 15
• Costas Busch - RPI15 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 16
• Costas Busch - RPI16 The Pigeonhole Principle
• Slide 17
• Costas Busch - RPI17 pigeons pigeonholes
• Slide 18
• Costas Busch - RPI18 A pigeonhole must contain at least two pigeons
• Slide 19
• Costas Busch - RPI19........... pigeons pigeonholes
• Slide 20
• Costas Busch - RPI20 The Pigeonhole Principle........... pigeons pigeonholes There is a pigeonhole with at least 2 pigeons
• Slide 21
• Costas Busch - RPI21 The Pigeonhole Principle and DFAs
• Slide 22
• Costas Busch - RPI22 DFA with states
• Slide 23
• Costas Busch - RPI23 In walks of strings:no state is repeated
• Slide 24
• Costas Busch - RPI24 In walks of strings:a state is repeated
• Slide 25
• Costas Busch - RPI25 If string has length : Thus, a state must be repeated Then the transitions of string are more than the states of the DFA
• Slide 26
• Costas Busch - RPI26 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 27
• Costas Busch - RPI27 In other words for a string : transitions are pigeons states are pigeonholes...... walk of Repeated state
• Slide 28
• Costas Busch - RPI28 The Pumping Lemma
• Slide 29
• Costas Busch - RPI29 Take an infinite regular language There exists a DFA that accepts states
• Slide 30
• Costas Busch - RPI30 Take string with There is a walk with label :......... walk
• Slide 31
• Costas Busch - RPI31 If string has length (number of states of DFA) then, from the pigeonhole principle: a state is repeated in the walk...... walk
• Slide 32
• Costas Busch - RPI32...... walk Let be the first state repeated in the walk of
• Slide 33
• Costas Busch - RPI33 Write......
• Slide 34
• Costas Busch - RPI34...... Observations:lengthnumber of states of DFA length
• Slide 35
• Costas Busch - RPI35 The string is accepted Observation:......
• Slide 36
• Costas Busch - RPI36 The string is accepted Observation:......
• Slide 37
• Costas Busch - RPI37 The string is accepted Observation:......
• Slide 38
• Costas Busch - RPI38 The string is accepted In General:......
• Slide 39
• Costas Busch - RPI39 In General:...... Language accepted by the DFA
• Slide 40
• Costas Busch - RPI40 In other words, we described: The Pumping Lemma !!!
• Slide 41
• Costas Busch - RPI41 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 42
• Costas Busch - RPI42 Applications of the Pumping Lemma
• Slide 43
• Costas Busch - RPI43 Theorem: The language is not regular Proof: Use the Pumping Lemma
• Slide 44
• Costas Busch - RPI44 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
• Slide 45
• Costas Busch - RPI45 Let be the integer in the Pumping Lemma Pick a string such that: length We pick
• Slide 46
• Costas Busch - RPI46 it must be that length From the Pumping Lemma Write: Thus:
• Slide 47
• Costas Busch - RPI47 From the Pumping Lemma: Thus:
• Slide 48
• Costas Busch - RPI48 From the Pumping Lemma: Thus:
• Slide 49
• Costas Busch - RPI49 BUT: CONTRADICTION!!!
• Slide 50
• Costas Busch - RPI50 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:
• Slide 51
• Costas Busch - RPI51 Regular languages Non-regular languages

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