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

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

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

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 !!!

Fall 2006Costas Busch - RPI4 The Pigeonhole Principle

Fall 2006Costas Busch - RPI5 pigeons pigeonholes

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

Fall 2006Costas Busch - RPI7........... pigeons pigeonholes

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

Fall 2006Costas Busch - RPI9 The Pigeonhole Principle and DFAs

Fall 2006Costas Busch - RPI10 Consider a DFA with states

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

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

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:

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

Fall 2006Costas Busch - RPI15...... Repeated state Walk of.... Arbitrary DFA If, by the pigeonhole principle, a state is repeated in the walk In General:

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

Fall 2006Costas Busch - RPI17 The Pumping Lemma

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

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

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 :

Fall 2006Costas Busch - RPI21.... Second occurrence First occurrence One dimensional projection of walk : We can write

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

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

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

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

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

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

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

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

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

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

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)

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

Fall 2006Costas Busch - RPI34 Applications of the Pumping Lemma

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)

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

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

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

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

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

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

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

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

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

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

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

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

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