Given a regular language L, prove or disprove L' is regular
$begingroup$
Given $NFA$ $N$ , $L(N)$ regular language and two words $w1$,$w2$ $in$ $sum^*$ such that $w1$ $neq$ $w2$.
I have to prove or disprove that
$L'=$ {$zin sum^*|exists$ $w1,w2$ :$w1z$ $in$ $L(N)$ $wedge$ $w2z$ $notin$ $L(N)$} is regular.
I believe this is correct but I'm having a hard time proving it.
Any help will do.
Thanks in advance.
computer-science formal-languages automata
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
$endgroup$
add a comment |
$begingroup$
Given $NFA$ $N$ , $L(N)$ regular language and two words $w1$,$w2$ $in$ $sum^*$ such that $w1$ $neq$ $w2$.
I have to prove or disprove that
$L'=$ {$zin sum^*|exists$ $w1,w2$ :$w1z$ $in$ $L(N)$ $wedge$ $w2z$ $notin$ $L(N)$} is regular.
I believe this is correct but I'm having a hard time proving it.
Any help will do.
Thanks in advance.
computer-science formal-languages automata
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
$endgroup$
$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18
add a comment |
$begingroup$
Given $NFA$ $N$ , $L(N)$ regular language and two words $w1$,$w2$ $in$ $sum^*$ such that $w1$ $neq$ $w2$.
I have to prove or disprove that
$L'=$ {$zin sum^*|exists$ $w1,w2$ :$w1z$ $in$ $L(N)$ $wedge$ $w2z$ $notin$ $L(N)$} is regular.
I believe this is correct but I'm having a hard time proving it.
Any help will do.
Thanks in advance.
computer-science formal-languages automata
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
$endgroup$
Given $NFA$ $N$ , $L(N)$ regular language and two words $w1$,$w2$ $in$ $sum^*$ such that $w1$ $neq$ $w2$.
I have to prove or disprove that
$L'=$ {$zin sum^*|exists$ $w1,w2$ :$w1z$ $in$ $L(N)$ $wedge$ $w2z$ $notin$ $L(N)$} is regular.
I believe this is correct but I'm having a hard time proving it.
Any help will do.
Thanks in advance.
computer-science formal-languages automata
computer-science formal-languages automata
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
asked Nov 30 '18 at 17:29
Avishai YanivAvishai Yaniv
173
173
$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18
add a comment |
$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18
$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18
add a comment |
1 Answer
1
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$begingroup$
The automaton $A$ for $L'$ can work like this:
- Without reading any input, it guesses a $w_1$ and simulates $N$ on this string. The guessing is done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_1$ is remembered.
$A$ guesses a $w_2$ and simulates $N$ on this string. The guessing is again done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_2$ is remembered.- Now $A$ reads the input $z$ and simulates two cmputations of $N$ in parallel: that of $w_1z$ and that of $w_2z$. If the former ends in some final state while the latter does not, $A$ accepts.
For fixed $w_1$ and $w_2$ you could do the same without the guessing. However, it would be more efficient to do the simulations of $N$ once on paper and let $A$ start from there directly.
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
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add a comment |
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1 Answer
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$begingroup$
The automaton $A$ for $L'$ can work like this:
- Without reading any input, it guesses a $w_1$ and simulates $N$ on this string. The guessing is done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_1$ is remembered.
$A$ guesses a $w_2$ and simulates $N$ on this string. The guessing is again done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_2$ is remembered.- Now $A$ reads the input $z$ and simulates two cmputations of $N$ in parallel: that of $w_1z$ and that of $w_2z$. If the former ends in some final state while the latter does not, $A$ accepts.
For fixed $w_1$ and $w_2$ you could do the same without the guessing. However, it would be more efficient to do the simulations of $N$ once on paper and let $A$ start from there directly.
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
$endgroup$
add a comment |
$begingroup$
The automaton $A$ for $L'$ can work like this:
- Without reading any input, it guesses a $w_1$ and simulates $N$ on this string. The guessing is done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_1$ is remembered.
$A$ guesses a $w_2$ and simulates $N$ on this string. The guessing is again done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_2$ is remembered.- Now $A$ reads the input $z$ and simulates two cmputations of $N$ in parallel: that of $w_1z$ and that of $w_2z$. If the former ends in some final state while the latter does not, $A$ accepts.
For fixed $w_1$ and $w_2$ you could do the same without the guessing. However, it would be more efficient to do the simulations of $N$ once on paper and let $A$ start from there directly.
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
$endgroup$
add a comment |
$begingroup$
The automaton $A$ for $L'$ can work like this:
- Without reading any input, it guesses a $w_1$ and simulates $N$ on this string. The guessing is done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_1$ is remembered.
$A$ guesses a $w_2$ and simulates $N$ on this string. The guessing is again done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_2$ is remembered.- Now $A$ reads the input $z$ and simulates two cmputations of $N$ in parallel: that of $w_1z$ and that of $w_2z$. If the former ends in some final state while the latter does not, $A$ accepts.
For fixed $w_1$ and $w_2$ you could do the same without the guessing. However, it would be more efficient to do the simulations of $N$ once on paper and let $A$ start from there directly.
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
$endgroup$
The automaton $A$ for $L'$ can work like this:
- Without reading any input, it guesses a $w_1$ and simulates $N$ on this string. The guessing is done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_1$ is remembered.
$A$ guesses a $w_2$ and simulates $N$ on this string. The guessing is again done letter by letter with $lambda$-transitions. The states in which $N$ ends after reading $w_2$ is remembered.- Now $A$ reads the input $z$ and simulates two cmputations of $N$ in parallel: that of $w_1z$ and that of $w_2z$. If the former ends in some final state while the latter does not, $A$ accepts.
For fixed $w_1$ and $w_2$ you could do the same without the guessing. However, it would be more efficient to do the simulations of $N$ once on paper and let $A$ start from there directly.
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
answered Dec 1 '18 at 18:00
Peter LeupoldPeter Leupold
56526
56526
add a comment |
add a comment |
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$begingroup$
Are $w_1$ and $w_2$ fixed words? The description of $L’$ seems to use any string, presumably in $Sigma^*$.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 18:02
$begingroup$
what do you mean by "fixed words"?
$endgroup$
– Avishai Yaniv
Nov 30 '18 at 20:13
$begingroup$
Are we to solve the problem for some given two words $w_1$ and $w_2$? Because the definition of $L’$ allows the words to change depending on the value of $z$. I’m asking if the words can vary based on $z$ or if they’re “fixed”.
$endgroup$
– Joey Kilpatrick
Nov 30 '18 at 20:18