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Is this grammar {a^n b^2n | n >= 0} regular or not?

原文 2016-01-10 22:35:22 0 1 computer-science/ regular-language

I don't understand the pumping lemma for this statment, can someone help me?

{a^nb^2n | n >= 0}

This is regular or not? If yes why, if no why?

Thanks to everybody!!

1 answers

Language is not regular if the value of n is changed the number of states in DFA will also change.In the definition of DFA to accept regular language, the set of states is a finite set. Of course the other way to prove it as non regular ,use pumping lemma for regular languages.

Is this grammar {a^2n b^n | n >= 0} context free or not?

Is L = {a^n a^n b^m |m, n ≥ 0} a regular or irregular language?

Are L1 = {a^n b^n | n < 4 } and L2 = {a^n b^n | n < 10^10^10 }, regular languages?

Why is {a^nb^n | n >= 0} not regular?

If L = {0^n 1^n | n > 0 } is L^c (complement of L) regular?

Is a^n, where n>1000 a regular language?

Why is {a^nb^n | n <= 10} regular?

Is n log n = O(n) and is n log n = Omega(n)

pda to accept the language L={a^n b^m | n<m}

Calculating Reciprocals: n**(-1) or (1/n)?

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Related Question Is this grammar {a^2n b^n | n >= 0} context free or not? Is L = {a^n a^n b^m |m, n ≥ 0} a regular or irregular language? Are L1 = {a^n b^n | n < 4 } and L2 = {a^n b^n | n < 10^10^10 }, regular languages? Why is {a^nb^n | n >= 0} not regular? If L = {0^n 1^n | n > 0 } is L^c (complement of L) regular? Is a^n, where n>1000 a regular language? Why is {a^nb^n | n <= 10} regular? Is n log n = O(n) and is n log n = Omega(n) pda to accept the language L={a^n b^m | n<m} Calculating Reciprocals: n**(-1) or (1/n)?

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Is this grammar {a^n b^2n | n >= 0} regular or not?

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1 answers

solution1 0 2016-02-22 11:14:08

solution1
0 2016-02-22 11:14:08