How many Nerode equivalence classes does the language $xneq y$ have?
$begingroup$
I have a language $L_k$ over the alphabet $Sigma={0,1,#}$ defined as follows:
begin{equation}
L_k={x#y|xin{0,1}^k,yin {0,1}^*wedge xneq y}
end{equation}
I would like to find all the Nerode equivalence classes for this language. Since this definition is somewhat hard to google I'm including it here.
The Nerode equivalence $R_L$ on a language $L$ over an alphabet $Sigma$ is defined as follows. For $s_1,s_2in Sigma^*$ $s_1R_L s_2$ if and only if $forall tinSigma^* s_1tin Liff s_2tin L$.
If we have $s_1inSigma^h$ where $0<hleq k$ such an $s_1$ must be in it's own class. We can show this by contradiction in parts. Suppose $s_2inSigma^*$ and $s_2neq s_1$.
Case 1)
$|s_2|neq |s_1|$
Choose $t=0^{h-k}#$ then $s_1t=s_10^{h-k}# in L_k$ but $s_2t=s_20^{h-k}#rnotin L_k$ since $s_2t$ has a $#$ somewhere other than at the $k+1$'st position.
Case 2)
$|s_2|=|s_1|=h$
Define $t=0^{h-k}#s_10^{h-k}$. Then $s_1t=s_10^{h-k}#s_10^{h-k}not in L_k$ since the sides around the $#$ are equal, but $s_2t=s_20^{h-k}#s_10^{h-k} in L_k$ since $s_1neq s_2$ and so the padding stays unequal.
Now how do I proceed for $s_1$ longer than $k$? Any hints or help appreciated.
computer-science formal-languages automata
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
$endgroup$
add a comment |
$begingroup$
I have a language $L_k$ over the alphabet $Sigma={0,1,#}$ defined as follows:
begin{equation}
L_k={x#y|xin{0,1}^k,yin {0,1}^*wedge xneq y}
end{equation}
I would like to find all the Nerode equivalence classes for this language. Since this definition is somewhat hard to google I'm including it here.
The Nerode equivalence $R_L$ on a language $L$ over an alphabet $Sigma$ is defined as follows. For $s_1,s_2in Sigma^*$ $s_1R_L s_2$ if and only if $forall tinSigma^* s_1tin Liff s_2tin L$.
If we have $s_1inSigma^h$ where $0<hleq k$ such an $s_1$ must be in it's own class. We can show this by contradiction in parts. Suppose $s_2inSigma^*$ and $s_2neq s_1$.
Case 1)
$|s_2|neq |s_1|$
Choose $t=0^{h-k}#$ then $s_1t=s_10^{h-k}# in L_k$ but $s_2t=s_20^{h-k}#rnotin L_k$ since $s_2t$ has a $#$ somewhere other than at the $k+1$'st position.
Case 2)
$|s_2|=|s_1|=h$
Define $t=0^{h-k}#s_10^{h-k}$. Then $s_1t=s_10^{h-k}#s_10^{h-k}not in L_k$ since the sides around the $#$ are equal, but $s_2t=s_20^{h-k}#s_10^{h-k} in L_k$ since $s_1neq s_2$ and so the padding stays unequal.
Now how do I proceed for $s_1$ longer than $k$? Any hints or help appreciated.
computer-science formal-languages automata
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
$endgroup$
$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15
add a comment |
$begingroup$
I have a language $L_k$ over the alphabet $Sigma={0,1,#}$ defined as follows:
begin{equation}
L_k={x#y|xin{0,1}^k,yin {0,1}^*wedge xneq y}
end{equation}
I would like to find all the Nerode equivalence classes for this language. Since this definition is somewhat hard to google I'm including it here.
The Nerode equivalence $R_L$ on a language $L$ over an alphabet $Sigma$ is defined as follows. For $s_1,s_2in Sigma^*$ $s_1R_L s_2$ if and only if $forall tinSigma^* s_1tin Liff s_2tin L$.
If we have $s_1inSigma^h$ where $0<hleq k$ such an $s_1$ must be in it's own class. We can show this by contradiction in parts. Suppose $s_2inSigma^*$ and $s_2neq s_1$.
Case 1)
$|s_2|neq |s_1|$
Choose $t=0^{h-k}#$ then $s_1t=s_10^{h-k}# in L_k$ but $s_2t=s_20^{h-k}#rnotin L_k$ since $s_2t$ has a $#$ somewhere other than at the $k+1$'st position.
Case 2)
$|s_2|=|s_1|=h$
Define $t=0^{h-k}#s_10^{h-k}$. Then $s_1t=s_10^{h-k}#s_10^{h-k}not in L_k$ since the sides around the $#$ are equal, but $s_2t=s_20^{h-k}#s_10^{h-k} in L_k$ since $s_1neq s_2$ and so the padding stays unequal.
Now how do I proceed for $s_1$ longer than $k$? Any hints or help appreciated.
computer-science formal-languages automata
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
$endgroup$
I have a language $L_k$ over the alphabet $Sigma={0,1,#}$ defined as follows:
begin{equation}
L_k={x#y|xin{0,1}^k,yin {0,1}^*wedge xneq y}
end{equation}
I would like to find all the Nerode equivalence classes for this language. Since this definition is somewhat hard to google I'm including it here.
The Nerode equivalence $R_L$ on a language $L$ over an alphabet $Sigma$ is defined as follows. For $s_1,s_2in Sigma^*$ $s_1R_L s_2$ if and only if $forall tinSigma^* s_1tin Liff s_2tin L$.
If we have $s_1inSigma^h$ where $0<hleq k$ such an $s_1$ must be in it's own class. We can show this by contradiction in parts. Suppose $s_2inSigma^*$ and $s_2neq s_1$.
Case 1)
$|s_2|neq |s_1|$
Choose $t=0^{h-k}#$ then $s_1t=s_10^{h-k}# in L_k$ but $s_2t=s_20^{h-k}#rnotin L_k$ since $s_2t$ has a $#$ somewhere other than at the $k+1$'st position.
Case 2)
$|s_2|=|s_1|=h$
Define $t=0^{h-k}#s_10^{h-k}$. Then $s_1t=s_10^{h-k}#s_10^{h-k}not in L_k$ since the sides around the $#$ are equal, but $s_2t=s_20^{h-k}#s_10^{h-k} in L_k$ since $s_1neq s_2$ and so the padding stays unequal.
Now how do I proceed for $s_1$ longer than $k$? Any hints or help appreciated.
computer-science formal-languages automata
computer-science formal-languages automata
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
edited Dec 4 '18 at 13:41
DRF
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
asked Dec 1 '18 at 11:37
DRFDRF
4,497926
4,497926
$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15
add a comment |
$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15
$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15
$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15
add a comment |
1 Answer
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The equivalence classes are as follows:
- Class $B$: All words with (1) $#$ not on position $k+1$, (2) $#$ missing from position $k+1$, (3) more than $2k+1$ letters, or of the form (4) $x#x$, where $|x|=k$.
- Class $G_ell$ for $1 leq ell leq k$: All words of the form $x#y$, where $|x|=k$, $|y| = ell$, and $x_i neq y_i$ for some $i leq ell$.
- Every other word is its own equivalence class. These are words of the form $x$ for $|x| leq k$ and words of the form $x # y$, where $|x| = k$, $|y| < k$, and $y$ is a prefix of $x$ (possibly $y = epsilon$).
All words in $B$ and all of their extensions are not in $L_k$. In contrast, all other words extend to a word in $L_k$. This shows that $B$ is an equivalence class. All words in $G_ell$ can be extended to a word in $L_k$ by any string of length exactly $k-ell$. In contrast, words of the singleton classes don't have these properties. This shows that the $G_ell$ are equivalence classes.
Finally, to show that the singletons are equivalence classes, it suffices to consider two words of the same length and show that they can be extended so that one of them is in $L_k$ and the other aren't. There are a few cases to consider, but they are all pretty simple. One also needs to check that the equivalence classes together cover all words.
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
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$begingroup$
The equivalence classes are as follows:
- Class $B$: All words with (1) $#$ not on position $k+1$, (2) $#$ missing from position $k+1$, (3) more than $2k+1$ letters, or of the form (4) $x#x$, where $|x|=k$.
- Class $G_ell$ for $1 leq ell leq k$: All words of the form $x#y$, where $|x|=k$, $|y| = ell$, and $x_i neq y_i$ for some $i leq ell$.
- Every other word is its own equivalence class. These are words of the form $x$ for $|x| leq k$ and words of the form $x # y$, where $|x| = k$, $|y| < k$, and $y$ is a prefix of $x$ (possibly $y = epsilon$).
All words in $B$ and all of their extensions are not in $L_k$. In contrast, all other words extend to a word in $L_k$. This shows that $B$ is an equivalence class. All words in $G_ell$ can be extended to a word in $L_k$ by any string of length exactly $k-ell$. In contrast, words of the singleton classes don't have these properties. This shows that the $G_ell$ are equivalence classes.
Finally, to show that the singletons are equivalence classes, it suffices to consider two words of the same length and show that they can be extended so that one of them is in $L_k$ and the other aren't. There are a few cases to consider, but they are all pretty simple. One also needs to check that the equivalence classes together cover all words.
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
$endgroup$
add a comment |
$begingroup$
The equivalence classes are as follows:
- Class $B$: All words with (1) $#$ not on position $k+1$, (2) $#$ missing from position $k+1$, (3) more than $2k+1$ letters, or of the form (4) $x#x$, where $|x|=k$.
- Class $G_ell$ for $1 leq ell leq k$: All words of the form $x#y$, where $|x|=k$, $|y| = ell$, and $x_i neq y_i$ for some $i leq ell$.
- Every other word is its own equivalence class. These are words of the form $x$ for $|x| leq k$ and words of the form $x # y$, where $|x| = k$, $|y| < k$, and $y$ is a prefix of $x$ (possibly $y = epsilon$).
All words in $B$ and all of their extensions are not in $L_k$. In contrast, all other words extend to a word in $L_k$. This shows that $B$ is an equivalence class. All words in $G_ell$ can be extended to a word in $L_k$ by any string of length exactly $k-ell$. In contrast, words of the singleton classes don't have these properties. This shows that the $G_ell$ are equivalence classes.
Finally, to show that the singletons are equivalence classes, it suffices to consider two words of the same length and show that they can be extended so that one of them is in $L_k$ and the other aren't. There are a few cases to consider, but they are all pretty simple. One also needs to check that the equivalence classes together cover all words.
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
$endgroup$
add a comment |
$begingroup$
The equivalence classes are as follows:
- Class $B$: All words with (1) $#$ not on position $k+1$, (2) $#$ missing from position $k+1$, (3) more than $2k+1$ letters, or of the form (4) $x#x$, where $|x|=k$.
- Class $G_ell$ for $1 leq ell leq k$: All words of the form $x#y$, where $|x|=k$, $|y| = ell$, and $x_i neq y_i$ for some $i leq ell$.
- Every other word is its own equivalence class. These are words of the form $x$ for $|x| leq k$ and words of the form $x # y$, where $|x| = k$, $|y| < k$, and $y$ is a prefix of $x$ (possibly $y = epsilon$).
All words in $B$ and all of their extensions are not in $L_k$. In contrast, all other words extend to a word in $L_k$. This shows that $B$ is an equivalence class. All words in $G_ell$ can be extended to a word in $L_k$ by any string of length exactly $k-ell$. In contrast, words of the singleton classes don't have these properties. This shows that the $G_ell$ are equivalence classes.
Finally, to show that the singletons are equivalence classes, it suffices to consider two words of the same length and show that they can be extended so that one of them is in $L_k$ and the other aren't. There are a few cases to consider, but they are all pretty simple. One also needs to check that the equivalence classes together cover all words.
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
$endgroup$
The equivalence classes are as follows:
- Class $B$: All words with (1) $#$ not on position $k+1$, (2) $#$ missing from position $k+1$, (3) more than $2k+1$ letters, or of the form (4) $x#x$, where $|x|=k$.
- Class $G_ell$ for $1 leq ell leq k$: All words of the form $x#y$, where $|x|=k$, $|y| = ell$, and $x_i neq y_i$ for some $i leq ell$.
- Every other word is its own equivalence class. These are words of the form $x$ for $|x| leq k$ and words of the form $x # y$, where $|x| = k$, $|y| < k$, and $y$ is a prefix of $x$ (possibly $y = epsilon$).
All words in $B$ and all of their extensions are not in $L_k$. In contrast, all other words extend to a word in $L_k$. This shows that $B$ is an equivalence class. All words in $G_ell$ can be extended to a word in $L_k$ by any string of length exactly $k-ell$. In contrast, words of the singleton classes don't have these properties. This shows that the $G_ell$ are equivalence classes.
Finally, to show that the singletons are equivalence classes, it suffices to consider two words of the same length and show that they can be extended so that one of them is in $L_k$ and the other aren't. There are a few cases to consider, but they are all pretty simple. One also needs to check that the equivalence classes together cover all words.
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
answered Jan 16 at 22:13
Yuval FilmusYuval Filmus
48.5k471144
48.5k471144
add a comment |
add a comment |
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$begingroup$
I think that the equivalence classes of the syntactical correspond exactly to the reachable states of a minimized DFA. So if you are able to construct some automaton, you could determinize and minimize it and kind of "read" the classes from there. Your relation takes only the left context, sou your equivalence classes should be unions of classes of the syntactical congruence. This would be an almost mechanic approach, but maybe not a very efficient one.
$endgroup$
– Peter Leupold
Dec 1 '18 at 18:15