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L= { $a^{n}b^{m}$ | $n<=m<=2n$ }

a) DCFL

b) CFL but not DCFL

c) Not CFL
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I am still unclear with the solution.

Can anyone elaborate it ?

We are doing two comparisons here n<=m and m<=2n.

+1 vote
Option b) cfl but not dcfl
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grammar will be

$S\rightarrow aSb|aSbb|\epsilon$

why not DCFL?
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Here push and pop are not cleare.i.e if a comes then push into the stack and b comes then we pop with b or bb. So we need to create two copy .
you have to keep two copy of m one is m=n and other is m=2n which can't do dpda but npda can do ,so it will be cfl not dcfl
L=$a^{n}b^{n}\cup a^{n}b^{2n}$.so we can easily say that above language is CFL  not DCFL.