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

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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?






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











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















up vote
1
down vote

favorite












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























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











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













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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















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

























asked Dec 11 '14 at 7:52









Sachin Divakar

1076




1076











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











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

















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











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
















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





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













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





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











1 Answer
1






active

oldest

votes

















up vote
0
down vote













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




















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







  • 1




    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.
    – Ronald
    Dec 11 '14 at 10:42










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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













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




















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







  • 1




    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.
    – Ronald
    Dec 11 '14 at 10:42














up vote
0
down vote













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




















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







  • 1




    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.
    – Ronald
    Dec 11 '14 at 10:42












up vote
0
down vote










up vote
0
down vote









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












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









Ronald

543410




543410











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







  • 1




    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.
    – Ronald
    Dec 11 '14 at 10:42
















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







  • 1




    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.
    – Ronald
    Dec 11 '14 at 10:42















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





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





1




1




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.
– Ronald
Dec 11 '14 at 10:42




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.
– Ronald
Dec 11 '14 at 10:42

















 

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