$L_1 =(a^nb^n)$ and $L_2 =(a^nb^{2n})$. Is $L_1 cup L_2$ DCFL?












1












$begingroup$


I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.



Therefore, $L_1 cup L_2$ is not cfL.



Is that correct?






context-free-grammar regular-expressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '14 at 8:47












  • $begingroup$
    Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:59


















1












$begingroup$


I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.



Therefore, $L_1 cup L_2$ is not cfL.



Is that correct?






context-free-grammar regular-expressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '14 at 8:47












  • $begingroup$
    Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:59
















1












1








1





$begingroup$


I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.



Therefore, $L_1 cup L_2$ is not cfL.



Is that correct?






context-free-grammar regular-expressions







share|cite|improve this question











$endgroup$




I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.



Therefore, $L_1 cup L_2$ is not cfL.



Is that correct?






context-free-grammar regular-expressions




context-free-grammar regular-expressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '14 at 8:36







Sachin Divakar

















asked Dec 11 '14 at 7:52









Sachin DivakarSachin Divakar

1076




1076












  • $begingroup$
    Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '14 at 8:47












  • $begingroup$
    Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:59




















  • $begingroup$
    Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
    $endgroup$
    – Steven Stadnicki
    Dec 11 '14 at 8:47












  • $begingroup$
    Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:59


















$begingroup$
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
$endgroup$
– Steven Stadnicki
Dec 11 '14 at 8:47






$begingroup$
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
$endgroup$
– Steven Stadnicki
Dec 11 '14 at 8:47














$begingroup$
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
$endgroup$
– Sachin Divakar
Dec 11 '14 at 8:59






$begingroup$
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
$endgroup$
– Sachin Divakar
Dec 11 '14 at 8:59












1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint:



$L_1$ is produced by



S := $epsilon$
| aSb



$L_2$ is produced by



S := $epsilon$
| aSbb



(assuming $a^a$ is a typo, meaning $a^n$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes $a^n$ but is it dCFL
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:32








  • 1




    $begingroup$
    Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
    $endgroup$
    – Ronald
    Dec 11 '14 at 10:42











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint:



$L_1$ is produced by



S := $epsilon$
| aSb



$L_2$ is produced by



S := $epsilon$
| aSbb



(assuming $a^a$ is a typo, meaning $a^n$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes $a^n$ but is it dCFL
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:32








  • 1




    $begingroup$
    Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
    $endgroup$
    – Ronald
    Dec 11 '14 at 10:42
















0












$begingroup$

Hint:



$L_1$ is produced by



S := $epsilon$
| aSb



$L_2$ is produced by



S := $epsilon$
| aSbb



(assuming $a^a$ is a typo, meaning $a^n$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    yes $a^n$ but is it dCFL
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:32








  • 1




    $begingroup$
    Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
    $endgroup$
    – Ronald
    Dec 11 '14 at 10:42














0












0








0





$begingroup$

Hint:



$L_1$ is produced by



S := $epsilon$
| aSb



$L_2$ is produced by



S := $epsilon$
| aSbb



(assuming $a^a$ is a typo, meaning $a^n$)






share|cite|improve this answer









$endgroup$



Hint:



$L_1$ is produced by



S := $epsilon$
| aSb



$L_2$ is produced by



S := $epsilon$
| aSbb



(assuming $a^a$ is a typo, meaning $a^n$)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '14 at 8:09









RonaldRonald

673510




673510












  • $begingroup$
    yes $a^n$ but is it dCFL
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:32








  • 1




    $begingroup$
    Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
    $endgroup$
    – Ronald
    Dec 11 '14 at 10:42


















  • $begingroup$
    yes $a^n$ but is it dCFL
    $endgroup$
    – Sachin Divakar
    Dec 11 '14 at 8:32








  • 1




    $begingroup$
    Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
    $endgroup$
    – Ronald
    Dec 11 '14 at 10:42
















$begingroup$
yes $a^n$ but is it dCFL
$endgroup$
– Sachin Divakar
Dec 11 '14 at 8:32






$begingroup$
yes $a^n$ but is it dCFL
$endgroup$
– Sachin Divakar
Dec 11 '14 at 8:32






1




1




$begingroup$
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
$endgroup$
– Ronald
Dec 11 '14 at 10:42




$begingroup$
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
$endgroup$
– Ronald
Dec 11 '14 at 10:42


















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