Language contex-free or not?
$begingroup$
I was wondering whether this language is context-free or not?
It's $L = { AB ~|~ |A| = |B| text{ and } A neq B }$ .
The alphabet is ${ a, b }$.
In my textbook it's written that it is Context-free, but I can't seem to find proper grammar for it.
How can I keep that the words are always different?
If it's not CF, how can I prove it with the pumping lemma?
formal-languages context-free-grammar formal-grammar
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
$endgroup$
add a comment |
$begingroup$
I was wondering whether this language is context-free or not?
It's $L = { AB ~|~ |A| = |B| text{ and } A neq B }$ .
The alphabet is ${ a, b }$.
In my textbook it's written that it is Context-free, but I can't seem to find proper grammar for it.
How can I keep that the words are always different?
If it's not CF, how can I prove it with the pumping lemma?
formal-languages context-free-grammar formal-grammar
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
$endgroup$
$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35
add a comment |
$begingroup$
I was wondering whether this language is context-free or not?
It's $L = { AB ~|~ |A| = |B| text{ and } A neq B }$ .
The alphabet is ${ a, b }$.
In my textbook it's written that it is Context-free, but I can't seem to find proper grammar for it.
How can I keep that the words are always different?
If it's not CF, how can I prove it with the pumping lemma?
formal-languages context-free-grammar formal-grammar
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
$endgroup$
I was wondering whether this language is context-free or not?
It's $L = { AB ~|~ |A| = |B| text{ and } A neq B }$ .
The alphabet is ${ a, b }$.
In my textbook it's written that it is Context-free, but I can't seem to find proper grammar for it.
How can I keep that the words are always different?
If it's not CF, how can I prove it with the pumping lemma?
formal-languages context-free-grammar formal-grammar
formal-languages context-free-grammar formal-grammar
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
edited Dec 21 '18 at 4:39
Fabio Somenzi
6,52821321
6,52821321
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
asked Dec 21 '18 at 3:48
Angeld55Angeld55
565
565
$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35
add a comment |
$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35
$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35
$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solution comes from looking at the problem with a bit of a squint. I'll just describe the trick (although you'll find it a lot more satisfying to figure it out yourself, so you might not want to read this yet.)
We know that $|A| = |B| text{ and } A ne B$. That means there is at least one $i$ for which $A_i ne B_i$. Now consider the complete sentence $W = AB$, and let $n = |A|$. Clearly $A_i$ is $W_i$ and $B_i$ is $W_{n+i}$. That means that
$$W = Sigma^{i-1}A_iSigma^{n-i}Sigma^{i-1}B_iSigma^{n-i}$$
But that's the same as
$$W = Sigma^{i-1}A_iSigma^{i-1}Sigma^{n-i}B_iSigma^{n-i}$$
In other words, $W$ is the concatenation of two odd-length substrings whose middle characters differ.
answered Dec 21 '18 at 17:57
ricirici
39829
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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votes
$begingroup$
The solution comes from looking at the problem with a bit of a squint. I'll just describe the trick (although you'll find it a lot more satisfying to figure it out yourself, so you might not want to read this yet.)
We know that $|A| = |B| text{ and } A ne B$. That means there is at least one $i$ for which $A_i ne B_i$. Now consider the complete sentence $W = AB$, and let $n = |A|$. Clearly $A_i$ is $W_i$ and $B_i$ is $W_{n+i}$. That means that
$$W = Sigma^{i-1}A_iSigma^{n-i}Sigma^{i-1}B_iSigma^{n-i}$$
But that's the same as
$$W = Sigma^{i-1}A_iSigma^{i-1}Sigma^{n-i}B_iSigma^{n-i}$$
In other words, $W$ is the concatenation of two odd-length substrings whose middle characters differ.
answered Dec 21 '18 at 17:57
ricirici
39829
$endgroup$
add a comment |
$begingroup$
The solution comes from looking at the problem with a bit of a squint. I'll just describe the trick (although you'll find it a lot more satisfying to figure it out yourself, so you might not want to read this yet.)
We know that $|A| = |B| text{ and } A ne B$. That means there is at least one $i$ for which $A_i ne B_i$. Now consider the complete sentence $W = AB$, and let $n = |A|$. Clearly $A_i$ is $W_i$ and $B_i$ is $W_{n+i}$. That means that
$$W = Sigma^{i-1}A_iSigma^{n-i}Sigma^{i-1}B_iSigma^{n-i}$$
But that's the same as
$$W = Sigma^{i-1}A_iSigma^{i-1}Sigma^{n-i}B_iSigma^{n-i}$$
In other words, $W$ is the concatenation of two odd-length substrings whose middle characters differ.
answered Dec 21 '18 at 17:57
ricirici
39829
$endgroup$
add a comment |
$begingroup$
The solution comes from looking at the problem with a bit of a squint. I'll just describe the trick (although you'll find it a lot more satisfying to figure it out yourself, so you might not want to read this yet.)
We know that $|A| = |B| text{ and } A ne B$. That means there is at least one $i$ for which $A_i ne B_i$. Now consider the complete sentence $W = AB$, and let $n = |A|$. Clearly $A_i$ is $W_i$ and $B_i$ is $W_{n+i}$. That means that
$$W = Sigma^{i-1}A_iSigma^{n-i}Sigma^{i-1}B_iSigma^{n-i}$$
But that's the same as
$$W = Sigma^{i-1}A_iSigma^{i-1}Sigma^{n-i}B_iSigma^{n-i}$$
In other words, $W$ is the concatenation of two odd-length substrings whose middle characters differ.
answered Dec 21 '18 at 17:57
ricirici
39829
$endgroup$
The solution comes from looking at the problem with a bit of a squint. I'll just describe the trick (although you'll find it a lot more satisfying to figure it out yourself, so you might not want to read this yet.)
We know that $|A| = |B| text{ and } A ne B$. That means there is at least one $i$ for which $A_i ne B_i$. Now consider the complete sentence $W = AB$, and let $n = |A|$. Clearly $A_i$ is $W_i$ and $B_i$ is $W_{n+i}$. That means that
$$W = Sigma^{i-1}A_iSigma^{n-i}Sigma^{i-1}B_iSigma^{n-i}$$
But that's the same as
$$W = Sigma^{i-1}A_iSigma^{i-1}Sigma^{n-i}B_iSigma^{n-i}$$
In other words, $W$ is the concatenation of two odd-length substrings whose middle characters differ.
answered Dec 21 '18 at 17:57
ricirici
39829
answered Dec 21 '18 at 17:57
ricirici
39829
answered Dec 21 '18 at 17:57
ricirici
39829
answered Dec 21 '18 at 17:57
ricirici
39829
39829
add a comment |
add a comment |
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$begingroup$
Your textbook is right. The language ${AA ~|~ A in {a,b}^* }$ is well-known not to be context-free, but its complement and the closely-related $L$ of your problem are CFLs.
$endgroup$
– Fabio Somenzi
Dec 21 '18 at 5:35