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Fall 2006Costas Busch - RPI1 Non-regular languages (Pumping Lemma)

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Fall 2006Costas Busch - RPI2 Regular languages Non-regular languages

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Fall 2006Costas Busch - RPI3 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 !!!

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Fall 2006Costas Busch - RPI4 The Pigeonhole Principle

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Fall 2006Costas Busch - RPI5 pigeons pigeonholes

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Fall 2006Costas Busch - RPI6 A pigeonhole must contain at least two pigeons

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Fall 2006Costas Busch - RPI pigeons pigeonholes

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Fall 2006Costas Busch - RPI8 The Pigeonhole Principle pigeons pigeonholes There is a pigeonhole with at least 2 pigeons

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Fall 2006Costas Busch - RPI9 The Pigeonhole Principle and DFAs

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Fall 2006Costas Busch - RPI10 Consider a DFA with states

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Fall 2006Costas Busch - RPI11 Consider the walk of a “long’’ string: A state is repeated in the walk of (length at least 4)

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Fall 2006Costas Busch - RPI12 Pigeons: Nests: (Automaton states) Are more than Walk of The state is repeated as a result of the pigeonhole principle (walk states) Repeated state

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Fall 2006Costas Busch - RPI13 Consider the walk of a “long’’ string: A state is repeated in the walk of (length at least 4) Due to the pigeonhole principle:

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Fall 2006Costas Busch - RPI14 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

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Fall 2006Costas Busch - RPI Repeated state Walk of.... Arbitrary DFA If, by the pigeonhole principle, a state is repeated in the walk In General:

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Fall 2006Costas Busch - RPI16 Walk of Pigeons: Nests: (Automaton states) Are more than (walk states).... A state is repeated

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Fall 2006Costas Busch - RPI17 The Pumping Lemma

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Fall 2006Costas Busch - RPI18 Take an infinite regular language There exists a DFA that accepts states (contains an infinite number of strings)

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Fall 2006Costas Busch - RPI19 (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

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Fall 2006Costas Busch - RPI20 Take to be the first state repeated.... There could be many states repeated.... Second occurrence First occurrence Unique states One dimensional projection of walk :

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Fall 2006Costas Busch - RPI Second occurrence First occurrence One dimensional projection of walk : We can write

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Fall 2006Costas Busch - RPI22... In DFA:... contains only first occurrence of

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Fall 2006Costas Busch - RPI23 Observation:lengthnumber of states of DFA Since, in no state is repeated (except q) Unique States...

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Fall 2006Costas Busch - RPI24 Observation:length Since there is at least one transition in loop...

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Fall 2006Costas Busch - RPI25 We do not care about the form of string... may actually overlap with the paths of and

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Fall 2006Costas Busch - RPI26 The string is accepted Additional string:... Do not follow loop...

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Fall 2006Costas Busch - RPI27 The string is accepted... Follow loop 2 times Additional string:...

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Fall 2006Costas Busch - RPI28 The string is accepted... Follow loop 3 times Additional string:...

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Fall 2006Costas Busch - RPI29 The string is accepted In General:... Follow loop times...

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Fall 2006Costas Busch - RPI30 Therefore: Language accepted by the DFA...

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Fall 2006Costas Busch - RPI31 In other words, we described: The Pumping Lemma !!!

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Fall 2006Costas Busch - RPI32 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)

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Fall 2006Costas Busch - RPI33 In the book: Critical length = Pumping length

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Fall 2006Costas Busch - RPI34 Applications of the Pumping Lemma

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Fall 2006Costas Busch - RPI35 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)

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Fall 2006Costas Busch - RPI36 Suppose you want to prove that An 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

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Fall 2006Costas Busch - RPI37 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

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Fall 2006Costas Busch - RPI38 Note: It suffices to show that only one string gives a contradiction You don’t need to obtain contradiction for every

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Fall 2006Costas Busch - RPI39 Theorem: The language is not regular Proof: Use the Pumping Lemma Example of Pumping Lemma application

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Fall 2006Costas Busch - RPI40 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

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Fall 2006Costas Busch - RPI41 Let be the critical length for Pick a string such that: and length We pick

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Fall 2006Costas Busch - RPI42 with lengths From the Pumping Lemma: we can write Thus:

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Fall 2006Costas Busch - RPI43 From the Pumping Lemma: Thus:

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Fall 2006Costas Busch - RPI44 From the Pumping Lemma: Thus:

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Fall 2006Costas Busch - RPI45 BUT: CONTRADICTION!!!

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Fall 2006Costas Busch - RPI46 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF

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Fall 2006Costas Busch - RPI47 Regular languages Non-regular language

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