MadeEasy Test Series 2018: Theory of Computation - Identify Class Language

Question

Isn't the second one is CFL?

Comments

no..

we push 0 n times and pop out on every 1 and that iis m times now for 2 that is 2n we dont left with n from which we can compare

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ohh my mistake....got it thanks!!

Answer 1

• Simplelly we can put n in place of m  Becz n=m .  Then language become 
• L=$0^{n}1^{n} 2^{2n}$ which is csl.