What would be the complement of $L ={a^{n}b^{m}a^{n}b^{m} | n,m geq 1}$
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I understand the complement of L would be all the strings not in L, but I'm having a hard time writing down the structure of all the strings not in L.
formal-languages context-free-grammar
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
$endgroup$
add a comment |
$begingroup$
I understand the complement of L would be all the strings not in L, but I'm having a hard time writing down the structure of all the strings not in L.
formal-languages context-free-grammar
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
$endgroup$
$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06
add a comment |
$begingroup$
I understand the complement of L would be all the strings not in L, but I'm having a hard time writing down the structure of all the strings not in L.
formal-languages context-free-grammar
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
$endgroup$
I understand the complement of L would be all the strings not in L, but I'm having a hard time writing down the structure of all the strings not in L.
formal-languages context-free-grammar
formal-languages context-free-grammar
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
edited Dec 16 '18 at 21:28
Lord Shark the Unknown
104k1160132
104k1160132
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
asked Dec 16 '18 at 21:04
sharprabbitzsharprabbitz
31
31
$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06
add a comment |
$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06
$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06
$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06
add a comment |
1 Answer
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As well as all sentences which are not in the form $a^ib^ja^kb^l$, there's the set ${a^ib^ja^kb^lmid ine k lor j ne l}$. The two inequalities are not mutually exclusive, but it should be clear that only if both inequalities are false (i.e. both equalities would be true) would the sentences be in $L$.
answered Dec 16 '18 at 22:54
ricirici
39829
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1 Answer
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$begingroup$
As well as all sentences which are not in the form $a^ib^ja^kb^l$, there's the set ${a^ib^ja^kb^lmid ine k lor j ne l}$. The two inequalities are not mutually exclusive, but it should be clear that only if both inequalities are false (i.e. both equalities would be true) would the sentences be in $L$.
answered Dec 16 '18 at 22:54
ricirici
39829
$endgroup$
add a comment |
$begingroup$
As well as all sentences which are not in the form $a^ib^ja^kb^l$, there's the set ${a^ib^ja^kb^lmid ine k lor j ne l}$. The two inequalities are not mutually exclusive, but it should be clear that only if both inequalities are false (i.e. both equalities would be true) would the sentences be in $L$.
answered Dec 16 '18 at 22:54
ricirici
39829
$endgroup$
add a comment |
$begingroup$
As well as all sentences which are not in the form $a^ib^ja^kb^l$, there's the set ${a^ib^ja^kb^lmid ine k lor j ne l}$. The two inequalities are not mutually exclusive, but it should be clear that only if both inequalities are false (i.e. both equalities would be true) would the sentences be in $L$.
answered Dec 16 '18 at 22:54
ricirici
39829
$endgroup$
As well as all sentences which are not in the form $a^ib^ja^kb^l$, there's the set ${a^ib^ja^kb^lmid ine k lor j ne l}$. The two inequalities are not mutually exclusive, but it should be clear that only if both inequalities are false (i.e. both equalities would be true) would the sentences be in $L$.
answered Dec 16 '18 at 22:54
ricirici
39829
answered Dec 16 '18 at 22:54
ricirici
39829
answered Dec 16 '18 at 22:54
ricirici
39829
answered Dec 16 '18 at 22:54
ricirici
39829
39829
add a comment |
add a comment |
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$begingroup$
Are you sure this language is context-free? I haven't worked through the pumping lemma rigorously, but I don't think this language is context-free --- see this wiki section and how it suggests a language containing $a^nb^na^nb^n$ is not context-free.
$endgroup$
– apnorton
Dec 17 '18 at 7:06