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1: Regular

2: Regular

3: DCFL

@ Ankit Srivastava 7   How 3rd one is DCFL and not CFL?

If it is DCFL then obviously it is CFL

I am not saying it is not CFL

DCFL is subset of CFL

if you want most accurate answer then it will be DCFL

see L1={a^n b^m | n=m} is complement of given language

and this language is DCFL and DCFL's are closed under complement...so this language is also DCFL

Ankit Srivastava 7  Thanks bro!

A)$L_{1}=\left \{ a^{n}\,|\, n> 0 \right \} L_{2}=\left \{ b^{n}\,|\, n> 0 \right \}$

Regular language are closed under Concatenation.Hence regular

$L_{1}.L_{2}=\left \{ a^{x}b^{y}\,\,|\,\,x,y > 0 \right \}$

Regular expression-:$a^{+}b^{+}$

B)$L_{2}=\left \{ a^{n}b^{m}\,\,|\,\,n,m \geq 0 \right \}$

Regular Expression-:$a^{*}b^{*}$

Hence regular

C) Not regular need a stack to remember $n$,Thus can't be solved by FSM

Hence not regular

answered by Veteran (13.3k points) 16 55 115
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@sourav Whoa! Thanks, that was quick! Keep up, mate. Thanks again.

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