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For $A, B ⊆ Σ^*$, define
$A/B = \{x ∈ Σ^* | ∃y ∈ B, xy ∈ A\}$
If $L$ is a $CFL$ and $R$ is regular, then $L/R$ is

  1. Regular
  2. $CFL$ but not regular
  3. Recursive but not $CFL$
  4. None of the above
asked in Theory of Computation by (115 points)
edited by | 55 views

1 Answer

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Best answer

Answer : D

$L/R$ where $L$ is CFL and $R$ is Regular, then $L/R$ will definitely be CFL because CFL's are closed under right Quotient operation with Regular languages. (See proof here if interested :  https://cs.stackexchange.com/questions/20090/closure-of-cfl-against-right-quotient-with-regular-languages )

Though, CFL's are closed under right Quotient operation with Regular languages, But this only means that CFL under right Quotient operation with Regular languages will be CFL. It doesn't mean that It can't be Regular.

For instance, If the given CFL itself is Regular then result will also be regular.

Or Even if $L$ is Non-regular CFL, then also $L/R$ can be Regular.

For example, Consider this :

$L = \left \{ a^nb^n \,\,|\,\,n \geq 1 \right \}$  and $R = \left \{ b^n \,\,|\,\,n \geq 1 \right \}$

then $L/R = \left \{ a^n \,\,|\,\,n \geq 1 \right \}$ Which is Regular.

So, the correct answer for the asked question will be "$L/R \,\,is\,\, CFL$. Hence, Option D is correct.

answered by Boss (18.7k points)
selected by
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For example, Consider this :

L={anbn|n≥1}L={anbn|n≥1}  and R={bn|n≥1}R={bn|n≥1}

then L/R={an|n≥1}L/R={an|n≥1} Which is Regular.

 

Deepakk Poonia (Dee)

plz explain this part!



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