# 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
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- 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
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- Prof. Busch - LSU40 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
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- 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:
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- Prof. Busch - LSU43 From the Pumping Lemma: Thus:
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- 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