Find an LL(2) grammar for the following language
The question asks to find both an LL(1) and an LL(2) grammar for the following language
{𝑎^𝑚 𝑏^𝑛 𝑐^𝑚+𝑛 | m,n ϵ N}
I have an LL(1) grammar like so
S --> aSc | T
T --> bTc | lambda
But what is the LL(2) grammar? I have this, but I don't feel confident in it
LL(2)
S --> aaScc | aSc | T
T --> bbTcc | bTc | $Lambda$
logic automata regular-language context-free-grammar
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
The question asks to find both an LL(1) and an LL(2) grammar for the following language
{𝑎^𝑚 𝑏^𝑛 𝑐^𝑚+𝑛 | m,n ϵ N}
I have an LL(1) grammar like so
S --> aSc | T
T --> bTc | lambda
But what is the LL(2) grammar? I have this, but I don't feel confident in it
LL(2)
S --> aaScc | aSc | T
T --> bbTcc | bTc | $Lambda$
logic automata regular-language context-free-grammar
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08
add a comment |
The question asks to find both an LL(1) and an LL(2) grammar for the following language
{𝑎^𝑚 𝑏^𝑛 𝑐^𝑚+𝑛 | m,n ϵ N}
I have an LL(1) grammar like so
S --> aSc | T
T --> bTc | lambda
But what is the LL(2) grammar? I have this, but I don't feel confident in it
LL(2)
S --> aaScc | aSc | T
T --> bbTcc | bTc | $Lambda$
logic automata regular-language context-free-grammar
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
The question asks to find both an LL(1) and an LL(2) grammar for the following language
{𝑎^𝑚 𝑏^𝑛 𝑐^𝑚+𝑛 | m,n ϵ N}
I have an LL(1) grammar like so
S --> aSc | T
T --> bTc | lambda
But what is the LL(2) grammar? I have this, but I don't feel confident in it
LL(2)
S --> aaScc | aSc | T
T --> bbTcc | bTc | $Lambda$
logic automata regular-language context-free-grammar
logic automata regular-language context-free-grammar
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
asked Apr 18 '16 at 14:09
Kendall WeiheKendall Weihe
1213
1213
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08
add a comment |
yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08
yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08
add a comment |
1 Answer
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The question doesn't make sense: the $n$ in $LL(n)$ is a property of a language not a grammar.
An $LL(n)$ language is one that can be parsed by an $LL$ parser that uses at most $n$ look-ahead tokens. So any $LL(1)$ language is $LL(2)$ (because a parser can just ignore the second look-ahead token).
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
add a comment |
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1 Answer
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1 Answer
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The question doesn't make sense: the $n$ in $LL(n)$ is a property of a language not a grammar.
An $LL(n)$ language is one that can be parsed by an $LL$ parser that uses at most $n$ look-ahead tokens. So any $LL(1)$ language is $LL(2)$ (because a parser can just ignore the second look-ahead token).
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
add a comment |
The question doesn't make sense: the $n$ in $LL(n)$ is a property of a language not a grammar.
An $LL(n)$ language is one that can be parsed by an $LL$ parser that uses at most $n$ look-ahead tokens. So any $LL(1)$ language is $LL(2)$ (because a parser can just ignore the second look-ahead token).
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
add a comment |
The question doesn't make sense: the $n$ in $LL(n)$ is a property of a language not a grammar.
An $LL(n)$ language is one that can be parsed by an $LL$ parser that uses at most $n$ look-ahead tokens. So any $LL(1)$ language is $LL(2)$ (because a parser can just ignore the second look-ahead token).
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
The question doesn't make sense: the $n$ in $LL(n)$ is a property of a language not a grammar.
An $LL(n)$ language is one that can be parsed by an $LL$ parser that uses at most $n$ look-ahead tokens. So any $LL(1)$ language is $LL(2)$ (because a parser can just ignore the second look-ahead token).
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
edited Apr 19 '16 at 0:13
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
answered Apr 18 '16 at 22:04
Rob ArthanRob Arthan
29.1k42866
29.1k42866
add a comment |
add a comment |
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yes I'd say it is $LL(2)$ since (during parsing) it has to check $2$ tokens ahead to see if one should apply the rule $S to aaScc$ or the rule $S to aSc$, and it doesn't need backtracking. and honestly I don't see what could be the answer if it is not that grammar
– reuns
Apr 18 '16 at 14:58
maybe you could do something : the rule $S to a S c$ you can replace it by $S to a T c$, and the rule $T to b T c$ replaced by $T to bc$
– reuns
Apr 18 '16 at 15:08