# prof. busch - lsu1 non-regular languages (pumping lemma)

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• Slide 1
• Prof. Busch - LSU1 Non-regular languages (Pumping Lemma)
• Slide 2
• Prof. Busch - LSU2 Regular languages Non-regular languages
• Slide 3
• Prof. Busch - LSU3 How can we prove that a language is not regular? Prove that there is no DFA or NFA or RE that accepts Difficulty: this is not easy to prove ( since there is an infinite number of them) Solution: use the Pumping Lemma !!!
• Slide 4
• Prof. Busch - LSU4 The Pigeonhole Principle
• Slide 5
• Prof. Busch - LSU5 pigeons pigeonholes
• Slide 6
• Prof. Busch - LSU6 A pigeonhole must contain at least two pigeons
• Slide 7
• Prof. Busch - LSU7........... pigeons pigeonholes
• Slide 8
• Prof. Busch - LSU8 The Pigeonhole Principle........... pigeons pigeonholes There is a pigeonhole with at least 2 pigeons
• Slide 9
• Prof. Busch - LSU9 The Pigeonhole Principle and DFAs
• Slide 10
• Prof. Busch - LSU10 Consider a DFA with states
• Slide 11
• Prof. Busch - LSU11 Consider the walk of a long string: A state is repeated in the walk of (length at least 4)
• Slide 12
• Prof. Busch - LSU12 Pigeons: Nests: (Automaton states) Are more than Walk of The state is repeated as a result of the pigeonhole principle (walk states) Repeated state Repeated state
• Slide 13
• Prof. Busch - LSU13 Consider the walk of a long string: A state is repeated in the walk of (length at least 4) Due to the pigeonhole principle:
• Slide 14
• Prof. Busch - LSU14 Automaton States Pigeons: Nests: (Automaton states) Are more than Walk of The state is repeated as a result of the pigeonhole principle: (walk states) Repeated state
• Slide 15
• Prof. Busch - LSU15...... Repeated state Walk of.... Arbitrary DFA If, by the pigeonhole principle, a state is repeated in the walk In General:
• Slide 16
• Prof. Busch - LSU16 Walk of Pigeons: Nests: (Automaton states) Are more than (walk states).... A state is repeated Number of states in walk is at least
• Slide 17
• Prof. Busch - LSU17 The Pumping Lemma
• Slide 18
• Prof. Busch - LSU18 Take an infinite regular language There exists a DFA that accepts states (contains an infinite number of strings)
• Slide 19
• Prof. Busch - LSU19 (number of states of DFA) then, at least one state is repeated in the walk of...... Take string with Walk in DFA of Repeated state in DFA
• Slide 20
• Prof. Busch - LSU20 Take to be the first state repeated.... There could be many states repeated.... Second occurrence First occurrence Unique states One dimensional projection of walk :
• Slide 21
• Prof. Busch - LSU21.... Second occurrence First occurrence One dimensional projection of walk : We can write
• Slide 22
• Prof. Busch - LSU22... In DFA:... Where corresponds to substring between first and second occurrence of
• Slide 23
• Prof. Busch - LSU23 Observation:lengthnumber of states of DFA Since, in no state is repeated (except q) unique states in... Because of
• Slide 24
• Prof. Busch - LSU24 Observation:length Since there is at least one transition in loop...
• Slide 25
• Prof. Busch - LSU25 We do not care about the form of string... may actually overlap with the paths of and
• Slide 26
• Prof. Busch - LSU26 The string is accepted Additional string:... Do not follow loop...
• Slide 27
• Prof. Busch - LSU27 The string is accepted... Follow loop 2 times Additional string:...
• Slide 28
• Prof. Busch - LSU28 The string is accepted... Follow loop 3 times Additional string:...
• Slide 29
• Prof. Busch - LSU29 The string is accepted In General:... Follow loop times...
• Slide 30
• Prof. Busch - LSU30 Therefore: Language accepted by the DFA...
• Slide 31
• Prof. Busch - LSU31 In other words, we described: The Pumping Lemma !!!
• Slide 32
• Prof. Busch - LSU32 The Pumping Lemma: Given a infinite regular language there exists an integer for any string with length we can write with and such that: (critical length)
• Slide 33
• Prof. Busch - LSU33 In the book: Critical length = Pumping length
• Slide 34
• Prof. Busch - LSU34 Applications of the Pumping Lemma
• Slide 35
• Prof. Busch - LSU35 Observation: Every language of finite size has to be regular Therefore, every non-regular language has to be of infinite size (contains an infinite number of strings) (we can easily construct an NFA that accepts every string in the language)
• Slide 36
• Prof. Busch - LSU36 Suppose you want to prove that n infinite language is not regular 1. Assume the opposite: is regular 2. The pumping lemma should hold for 3. Use the pumping lemma to obtain a contradiction 4. Therefore, is not regular
• Slide 37
• Prof. Busch - LSU37 Explanation of Step 3: How to get a contradiction 2. Choose a particular string which satisfies the length condition 3. Write 4. Show thatfor some 5. This gives a contradiction, since from pumping lemma 1. Let be the critical length for
• Slide 38
• Prof. Busch - LSU38 Note: It suffices to show that only one string gives a contradiction You dont need to obtain contradiction for every
• Slide 39
• Prof. Busch - LSU39 Theorem: The language is not regular Proof: Use the Pumping Lemma Example of Pumping Lemma application
• Slide 40
• Prof. Busch - LSU40 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
• Slide 41
• Prof. Busch - LSU41 Let be the critical length for Pick a string such that: and length We pick
• Slide 42
• Prof. Busch - LSU42 with lengths From the Pumping Lemma: we can write Thus:
• Slide 43
• Prof. Busch - LSU43 From the Pumping Lemma: Thus:
• Slide 44
• Prof. Busch - LSU44 From the Pumping Lemma: Thus:
• Slide 45
• Prof. Busch - LSU45 BUT: CONTRADICTION!!!
• Slide 46
• Prof. Busch - LSU46 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF
• Slide 47
• Prof. Busch - LSU47 Regular languages Non-regular language

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