Convert this grammar to language
$begingroup$
I want to convert the following grammar to language but I am not able to think and answer.
$$S to aSa mid bSa mid ab mid ba$$
It gives me a lot of choices when I tried building its derivation tree.
automata context-free-grammar
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
$endgroup$
add a comment |
$begingroup$
I want to convert the following grammar to language but I am not able to think and answer.
$$S to aSa mid bSa mid ab mid ba$$
It gives me a lot of choices when I tried building its derivation tree.
automata context-free-grammar
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
$endgroup$
add a comment |
$begingroup$
I want to convert the following grammar to language but I am not able to think and answer.
$$S to aSa mid bSa mid ab mid ba$$
It gives me a lot of choices when I tried building its derivation tree.
automata context-free-grammar
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
$endgroup$
I want to convert the following grammar to language but I am not able to think and answer.
$$S to aSa mid bSa mid ab mid ba$$
It gives me a lot of choices when I tried building its derivation tree.
automata context-free-grammar
automata context-free-grammar
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
edited Dec 8 '18 at 20:15
Hans Hüttel
3,1972921
3,1972921
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
asked Dec 8 '18 at 19:55
Nikhil SatiNikhil Sati
31
31
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Welcome to MSE. There is basically one case:
Start either with $Srightarrow^* a^mSa^m$ or $Srightarrow^* b^nSa^n$ and alternate and finally replace $S$ either by $ab$ or $ba$. This gives the language
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ab)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}$$
union with
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ba)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}.$$
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Welcome to MSE. There is basically one case:
Start either with $Srightarrow^* a^mSa^m$ or $Srightarrow^* b^nSa^n$ and alternate and finally replace $S$ either by $ab$ or $ba$. This gives the language
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ab)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}$$
union with
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ba)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}.$$
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
$begingroup$
Welcome to MSE. There is basically one case:
Start either with $Srightarrow^* a^mSa^m$ or $Srightarrow^* b^nSa^n$ and alternate and finally replace $S$ either by $ab$ or $ba$. This gives the language
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ab)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}$$
union with
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ba)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}.$$
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
$endgroup$
add a comment |
$begingroup$
Welcome to MSE. There is basically one case:
Start either with $Srightarrow^* a^mSa^m$ or $Srightarrow^* b^nSa^n$ and alternate and finally replace $S$ either by $ab$ or $ba$. This gives the language
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ab)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}$$
union with
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ba)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}.$$
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
$endgroup$
Welcome to MSE. There is basically one case:
Start either with $Srightarrow^* a^mSa^m$ or $Srightarrow^* b^nSa^n$ and alternate and finally replace $S$ either by $ab$ or $ba$. This gives the language
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ab)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}$$
union with
$${a^{n_1} b^{n_2} ldots a^{n_{k-1}} b^{n_k} (ba)
a^{n_k} a^{n_{k-1}} ldots a^{n_2} a^{n_1}mid n_1,ldots,n_kgeq 0, kgeq 0}.$$
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
edited Dec 10 '18 at 13:36
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
answered Dec 10 '18 at 13:09
WuestenfuxWuestenfux
4,2871413
4,2871413
add a comment |
add a comment |
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