How many Nerode equivalence classes does the language $xneq y$ have?
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 at 11:37
DRF
4,477926
add a comment |
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 at 11:37
DRF
4,477926
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.
– Peter Leupold
Dec 1 at 18:15
add a comment |
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 at 11:37
DRF
4,477926
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 at 11:37
DRF
4,477926
asked Dec 1 at 11:37
DRF
4,477926
edited Dec 4 at 13:41
asked Dec 1 at 11:37
DRF
4,477926
asked Dec 1 at 11:37
DRF
4,477926
asked Dec 1 at 11:37
DRF
4,477926
4,477926
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.
– Peter Leupold
Dec 1 at 18:15
add a comment |
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.
– Peter Leupold
Dec 1 at 18:15
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.
– Peter Leupold
Dec 1 at 18:15
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.
– Peter Leupold
Dec 1 at 18:15
add a comment |
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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.
– Peter Leupold
Dec 1 at 18:15