Identify the Nature of the given Language
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If I am true then following two languages are not equal:-
$L_1 = {(a^nb^m)^l / n,m,l geq 1 }$
$L_2 = {(a^*b^*)^*}$
And I think $L_1$ is not $CFL$, because suppose a case where I put $n=2$, $m = 3 $ and $ l = 100$, then each and every time when $aabbb$ occurs we have to check whether $n_a = 2 text{ and } n_b = 3$, which I think is not possible with one stack.
formal-languages regular-language
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
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add a comment |
$begingroup$
If I am true then following two languages are not equal:-
$L_1 = {(a^nb^m)^l / n,m,l geq 1 }$
$L_2 = {(a^*b^*)^*}$
And I think $L_1$ is not $CFL$, because suppose a case where I put $n=2$, $m = 3 $ and $ l = 100$, then each and every time when $aabbb$ occurs we have to check whether $n_a = 2 text{ and } n_b = 3$, which I think is not possible with one stack.
formal-languages regular-language
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
$endgroup$
1
$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
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Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48
add a comment |
$begingroup$
If I am true then following two languages are not equal:-
$L_1 = {(a^nb^m)^l / n,m,l geq 1 }$
$L_2 = {(a^*b^*)^*}$
And I think $L_1$ is not $CFL$, because suppose a case where I put $n=2$, $m = 3 $ and $ l = 100$, then each and every time when $aabbb$ occurs we have to check whether $n_a = 2 text{ and } n_b = 3$, which I think is not possible with one stack.
formal-languages regular-language
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
$endgroup$
If I am true then following two languages are not equal:-
$L_1 = {(a^nb^m)^l / n,m,l geq 1 }$
$L_2 = {(a^*b^*)^*}$
And I think $L_1$ is not $CFL$, because suppose a case where I put $n=2$, $m = 3 $ and $ l = 100$, then each and every time when $aabbb$ occurs we have to check whether $n_a = 2 text{ and } n_b = 3$, which I think is not possible with one stack.
formal-languages regular-language
formal-languages regular-language
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
edited Nov 30 '18 at 17:06
Shubhanshu
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
asked Nov 30 '18 at 16:59
ShubhanshuShubhanshu
134
134
1
$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
$begingroup$
Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48
add a comment |
1
$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
$begingroup$
Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48
1
1
$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
$begingroup$
Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48
add a comment |
1 Answer
1
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$begingroup$
Your intuition about $L_1$ is correct, but your argumentation is not. Any fixed string by itself is a regular language. So the string you provide can be recognized by a finite automaton. The problem here is that there is an infinite variety of strings of that nature.
The most common way to prove that $L_1$ is not context-free is probably the pumping lemma. If the pumping constant for $L_1$ is $k$, you can use the word $(a^kb^k)^2$.
And since $L_2$ is by definition regular, this also shows that the two languages are not equal.
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
$endgroup$
add a comment |
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$begingroup$
Your intuition about $L_1$ is correct, but your argumentation is not. Any fixed string by itself is a regular language. So the string you provide can be recognized by a finite automaton. The problem here is that there is an infinite variety of strings of that nature.
The most common way to prove that $L_1$ is not context-free is probably the pumping lemma. If the pumping constant for $L_1$ is $k$, you can use the word $(a^kb^k)^2$.
And since $L_2$ is by definition regular, this also shows that the two languages are not equal.
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
$endgroup$
add a comment |
$begingroup$
Your intuition about $L_1$ is correct, but your argumentation is not. Any fixed string by itself is a regular language. So the string you provide can be recognized by a finite automaton. The problem here is that there is an infinite variety of strings of that nature.
The most common way to prove that $L_1$ is not context-free is probably the pumping lemma. If the pumping constant for $L_1$ is $k$, you can use the word $(a^kb^k)^2$.
And since $L_2$ is by definition regular, this also shows that the two languages are not equal.
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
$endgroup$
add a comment |
$begingroup$
Your intuition about $L_1$ is correct, but your argumentation is not. Any fixed string by itself is a regular language. So the string you provide can be recognized by a finite automaton. The problem here is that there is an infinite variety of strings of that nature.
The most common way to prove that $L_1$ is not context-free is probably the pumping lemma. If the pumping constant for $L_1$ is $k$, you can use the word $(a^kb^k)^2$.
And since $L_2$ is by definition regular, this also shows that the two languages are not equal.
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
$endgroup$
Your intuition about $L_1$ is correct, but your argumentation is not. Any fixed string by itself is a regular language. So the string you provide can be recognized by a finite automaton. The problem here is that there is an infinite variety of strings of that nature.
The most common way to prove that $L_1$ is not context-free is probably the pumping lemma. If the pumping constant for $L_1$ is $k$, you can use the word $(a^kb^k)^2$.
And since $L_2$ is by definition regular, this also shows that the two languages are not equal.
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
answered Dec 1 '18 at 18:06
Peter LeupoldPeter Leupold
56826
56826
add a comment |
add a comment |
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$begingroup$
If you're asking how to prove that the languages are not the same, note that, for example, $aabbab$ is in $L_2$ but not $L_1$.
$endgroup$
– Connor Harris
Nov 30 '18 at 17:11
$begingroup$
Yeah, that's clear my doubt little bit. It means that if we put n = 2 and m = 3 then in for each value of l they must stick to their initial values.
$endgroup$
– Shubhanshu
Nov 30 '18 at 17:17
$begingroup$
@ConnorHarris Please post your comment as an answer so that the quesiton will be marked as answered (and so I can upvote).
$endgroup$
– Joey Kilpatrick
Dec 1 '18 at 2:37
$begingroup$
Language $L_1$ is not $CFL$. Is it true?
$endgroup$
– Shubhanshu
Dec 1 '18 at 2:48