regular language

Question

Explain why L={(ab)^n a^k:n>k,k>=0} is not regular?

Comments

Because you have the relation between n and k

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So if we have a relation between n and k than how can we say that it is Not regular.

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if relation exist then you have to compare n and k right?

you need memorization for comparing n and k but FA doesn't have memory, therefore it is not RL

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@Shaik, Just little bit modification required.
Here you need infinite memorization (like by using a stack), but in FA we have only states to keep track of past information and FA has finite no. of states. :)

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yes mam, it should be updated with your sentence,