Regular

Question

Is this regular 

L={w | w $\varepsilon$ (0,1)^*  w is of the form (0ⁱ1)ⁿ for i=1,2,3...n ,n>=0} ?

Comments

Regular expression: (0⁺1)*

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@ akash.dinkar12 here i should be greater or equal to 1 and  will be less than or equal to n..so smallest string will be 01 ..but your language also except e..and we have to count i so that n should  always be greater or equal to i...so i don't think it is regular.

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@Red_devil

If we put n=0 in language L then Epsilon will be accepted.

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@ akash.dinkar12  well e is not a problem but how will you manage to count i so than n will always exceed it or be equal to it. because 0<i<=n

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I think it is not regular because no of 0 should always be less or equal to n.

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Please put the dead state i forget to write :Dis this correct ?

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@junaid No, the DFA you have shown will accept strings like $001 \implies (0^21)^1$ with $i=2$ and $n=1$ which do not belong to language $L$

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see it is accepting 001

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Yes, it is accepting. But it should not accept because $001$ do not belong to given language.

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But how, see language is generating 001 when i=2 and n=1

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you are forgetting that i<=n