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

The Gateway to Computer Science Excellence

0 votes

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

please some one explain this??

+1 vote

**L1 **= {a^{1}^{^1 }, a^{2^2}, a^{3^3}............} ={a^{1},a^{4},a^{9 }...... } 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.**

52,215 questions

59,987 answers

201,185 comments

94,647 users