MadeEasy Subject Test: Theory of Computation - Regular Languages

Question

https://gateoverflow.in/?qa=blob&qa_blobid=9622019941645658924 

please some one explain this??

Comments

if we will take the complement of L3 it will be regular and regular is closure under complement

Answer 1

L1 = {a^{1^1} , a^{2^2}, a^{3^3}............} ={a¹,a⁴,a⁹ ...... } which is not in AP {Arithmetic progression}

L2 = {a^{1^1}, a^{4^2}, a^{9^3}..........} not in AP.

L3 ={a^{2^1}, a^{3^1}, a^{4^1.........}} put n=1 and m>n

regular expression for L3= {aa⁺}

hence, L3 is regular.

Comments

but in L3 if we change value of n in every term then it is not in AP. why u put n=1 in every term

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because {aa+} will generate all lang.