Constructive proof to show the quotient of two regular languages is regular
$begingroup$
I have a question regarding the quotient of two regular languages, $R$ and $L$.
I saw the answers to this question: are regular languages closed under division
and the proof sketch is not constructive, because $L$ can be any language.
I'm thinking about a constructive proof when $L$ is regular: how can we construct that $F2$ group in the case when $L$ is regular?
We can't go over every word $x$ in $L$ and check if $delta(q,x)in F1$ in the case when $L$ is infinite...
automata regular-languages finite-automata
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
$endgroup$
add a comment |
$begingroup$
I have a question regarding the quotient of two regular languages, $R$ and $L$.
I saw the answers to this question: are regular languages closed under division
and the proof sketch is not constructive, because $L$ can be any language.
I'm thinking about a constructive proof when $L$ is regular: how can we construct that $F2$ group in the case when $L$ is regular?
We can't go over every word $x$ in $L$ and check if $delta(q,x)in F1$ in the case when $L$ is infinite...
automata regular-languages finite-automata
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
$endgroup$
add a comment |
$begingroup$
I have a question regarding the quotient of two regular languages, $R$ and $L$.
I saw the answers to this question: are regular languages closed under division
and the proof sketch is not constructive, because $L$ can be any language.
I'm thinking about a constructive proof when $L$ is regular: how can we construct that $F2$ group in the case when $L$ is regular?
We can't go over every word $x$ in $L$ and check if $delta(q,x)in F1$ in the case when $L$ is infinite...
automata regular-languages finite-automata
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
$endgroup$
I have a question regarding the quotient of two regular languages, $R$ and $L$.
I saw the answers to this question: are regular languages closed under division
and the proof sketch is not constructive, because $L$ can be any language.
I'm thinking about a constructive proof when $L$ is regular: how can we construct that $F2$ group in the case when $L$ is regular?
We can't go over every word $x$ in $L$ and check if $delta(q,x)in F1$ in the case when $L$ is infinite...
automata regular-languages finite-automata
automata regular-languages finite-automata
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
edited Dec 25 '18 at 17:02
Apass.Jack
10.3k1939
10.3k1939
asked Dec 25 '18 at 13:44
A.GA.G
283
asked Dec 25 '18 at 13:44
A.GA.G
283
asked Dec 25 '18 at 13:44
A.GA.G
283
283
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
In the question you link to, the automaton for the operation "division" (usually known as "quotient") $L_1/L_2$ is obtained from a FSA $M_1 = (Q,Sigma,delta,q_0,F_1)$ for $L_1$ by changing the final states. The new states are given as $F_2={qin Q mid delta(q,x)in F_1 text{ for some } xin L_2}$.
The quotient is regular for regular $L_1$ even when $L_2$ is arbitrarily complex.
We can't go over every word $x$ in $L_2$ and check if $delta(q,x)in F_1$
in the case when $L$ is infinite...
You are right that in general the construction is not effective. We just know how the final states are chosen, but we cannot definitely say which states are final. But in case $L_2$ is regular, we can determine which states are final, as regular languages are closed under intersection.
For any state $q$ change $M_1$ simply by setting $q$ as its new initial state (instead of $q_0$): $M_q = (Q,Sigma,delta,q,F_1)$. Now $q$ is final (in $F_2$) iff the intersection $L(M_q) cap L_2$ is nonempty. Because any $x$ is the intersection is precisely an $xin L_2$ for which $delta(q,x)in F_1$ as required.
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
$endgroup$
add a comment |
$begingroup$
The quotient of two languages $R$ and $L$ is
$$R/L = left{xin Sigma^{*} : exists y in L. xy in Rright}$$
Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.
Let $(Q_1,Sigma,delta_1,s_1,F_1)$ and $(Q_2,Sigma,delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA
$$D=(Q_1 times {s_2} times {1} cup Q_1 times Q_2times {2}, Sigma, delta, (s_1,s_2, 1), F_1 times F_2times {2})$$
where the transitions are:
$delta((q_1,s_2,1),sigma) = { (delta_1(q_1,sigma),s_2, 1)}$.
$delta((q_1,s_2,1),epsilon) = { (q_1,s_2, 2)}$.
$delta((q_1,q_2,2),sigma) = { (delta_1(q_1,sigma),delta_2(q_2,sigma), 2)}$.
I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
In the question you link to, the automaton for the operation "division" (usually known as "quotient") $L_1/L_2$ is obtained from a FSA $M_1 = (Q,Sigma,delta,q_0,F_1)$ for $L_1$ by changing the final states. The new states are given as $F_2={qin Q mid delta(q,x)in F_1 text{ for some } xin L_2}$.
The quotient is regular for regular $L_1$ even when $L_2$ is arbitrarily complex.
We can't go over every word $x$ in $L_2$ and check if $delta(q,x)in F_1$
in the case when $L$ is infinite...
You are right that in general the construction is not effective. We just know how the final states are chosen, but we cannot definitely say which states are final. But in case $L_2$ is regular, we can determine which states are final, as regular languages are closed under intersection.
For any state $q$ change $M_1$ simply by setting $q$ as its new initial state (instead of $q_0$): $M_q = (Q,Sigma,delta,q,F_1)$. Now $q$ is final (in $F_2$) iff the intersection $L(M_q) cap L_2$ is nonempty. Because any $x$ is the intersection is precisely an $xin L_2$ for which $delta(q,x)in F_1$ as required.
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
$endgroup$
add a comment |
$begingroup$
In the question you link to, the automaton for the operation "division" (usually known as "quotient") $L_1/L_2$ is obtained from a FSA $M_1 = (Q,Sigma,delta,q_0,F_1)$ for $L_1$ by changing the final states. The new states are given as $F_2={qin Q mid delta(q,x)in F_1 text{ for some } xin L_2}$.
The quotient is regular for regular $L_1$ even when $L_2$ is arbitrarily complex.
We can't go over every word $x$ in $L_2$ and check if $delta(q,x)in F_1$
in the case when $L$ is infinite...
You are right that in general the construction is not effective. We just know how the final states are chosen, but we cannot definitely say which states are final. But in case $L_2$ is regular, we can determine which states are final, as regular languages are closed under intersection.
For any state $q$ change $M_1$ simply by setting $q$ as its new initial state (instead of $q_0$): $M_q = (Q,Sigma,delta,q,F_1)$. Now $q$ is final (in $F_2$) iff the intersection $L(M_q) cap L_2$ is nonempty. Because any $x$ is the intersection is precisely an $xin L_2$ for which $delta(q,x)in F_1$ as required.
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
$endgroup$
add a comment |
$begingroup$
In the question you link to, the automaton for the operation "division" (usually known as "quotient") $L_1/L_2$ is obtained from a FSA $M_1 = (Q,Sigma,delta,q_0,F_1)$ for $L_1$ by changing the final states. The new states are given as $F_2={qin Q mid delta(q,x)in F_1 text{ for some } xin L_2}$.
The quotient is regular for regular $L_1$ even when $L_2$ is arbitrarily complex.
We can't go over every word $x$ in $L_2$ and check if $delta(q,x)in F_1$
in the case when $L$ is infinite...
You are right that in general the construction is not effective. We just know how the final states are chosen, but we cannot definitely say which states are final. But in case $L_2$ is regular, we can determine which states are final, as regular languages are closed under intersection.
For any state $q$ change $M_1$ simply by setting $q$ as its new initial state (instead of $q_0$): $M_q = (Q,Sigma,delta,q,F_1)$. Now $q$ is final (in $F_2$) iff the intersection $L(M_q) cap L_2$ is nonempty. Because any $x$ is the intersection is precisely an $xin L_2$ for which $delta(q,x)in F_1$ as required.
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
$endgroup$
In the question you link to, the automaton for the operation "division" (usually known as "quotient") $L_1/L_2$ is obtained from a FSA $M_1 = (Q,Sigma,delta,q_0,F_1)$ for $L_1$ by changing the final states. The new states are given as $F_2={qin Q mid delta(q,x)in F_1 text{ for some } xin L_2}$.
The quotient is regular for regular $L_1$ even when $L_2$ is arbitrarily complex.
We can't go over every word $x$ in $L_2$ and check if $delta(q,x)in F_1$
in the case when $L$ is infinite...
You are right that in general the construction is not effective. We just know how the final states are chosen, but we cannot definitely say which states are final. But in case $L_2$ is regular, we can determine which states are final, as regular languages are closed under intersection.
For any state $q$ change $M_1$ simply by setting $q$ as its new initial state (instead of $q_0$): $M_q = (Q,Sigma,delta,q,F_1)$. Now $q$ is final (in $F_2$) iff the intersection $L(M_q) cap L_2$ is nonempty. Because any $x$ is the intersection is precisely an $xin L_2$ for which $delta(q,x)in F_1$ as required.
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
answered Dec 25 '18 at 23:19
Hendrik JanHendrik Jan
21.2k2667
21.2k2667
add a comment |
add a comment |
$begingroup$
The quotient of two languages $R$ and $L$ is
$$R/L = left{xin Sigma^{*} : exists y in L. xy in Rright}$$
Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.
Let $(Q_1,Sigma,delta_1,s_1,F_1)$ and $(Q_2,Sigma,delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA
$$D=(Q_1 times {s_2} times {1} cup Q_1 times Q_2times {2}, Sigma, delta, (s_1,s_2, 1), F_1 times F_2times {2})$$
where the transitions are:
$delta((q_1,s_2,1),sigma) = { (delta_1(q_1,sigma),s_2, 1)}$.
$delta((q_1,s_2,1),epsilon) = { (q_1,s_2, 2)}$.
$delta((q_1,q_2,2),sigma) = { (delta_1(q_1,sigma),delta_2(q_2,sigma), 2)}$.
I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
$endgroup$
add a comment |
$begingroup$
The quotient of two languages $R$ and $L$ is
$$R/L = left{xin Sigma^{*} : exists y in L. xy in Rright}$$
Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.
Let $(Q_1,Sigma,delta_1,s_1,F_1)$ and $(Q_2,Sigma,delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA
$$D=(Q_1 times {s_2} times {1} cup Q_1 times Q_2times {2}, Sigma, delta, (s_1,s_2, 1), F_1 times F_2times {2})$$
where the transitions are:
$delta((q_1,s_2,1),sigma) = { (delta_1(q_1,sigma),s_2, 1)}$.
$delta((q_1,s_2,1),epsilon) = { (q_1,s_2, 2)}$.
$delta((q_1,q_2,2),sigma) = { (delta_1(q_1,sigma),delta_2(q_2,sigma), 2)}$.
I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
$endgroup$
add a comment |
$begingroup$
The quotient of two languages $R$ and $L$ is
$$R/L = left{xin Sigma^{*} : exists y in L. xy in Rright}$$
Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.
Let $(Q_1,Sigma,delta_1,s_1,F_1)$ and $(Q_2,Sigma,delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA
$$D=(Q_1 times {s_2} times {1} cup Q_1 times Q_2times {2}, Sigma, delta, (s_1,s_2, 1), F_1 times F_2times {2})$$
where the transitions are:
$delta((q_1,s_2,1),sigma) = { (delta_1(q_1,sigma),s_2, 1)}$.
$delta((q_1,s_2,1),epsilon) = { (q_1,s_2, 2)}$.
$delta((q_1,q_2,2),sigma) = { (delta_1(q_1,sigma),delta_2(q_2,sigma), 2)}$.
I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
$endgroup$
The quotient of two languages $R$ and $L$ is
$$R/L = left{xin Sigma^{*} : exists y in L. xy in Rright}$$
Here is a constructive solution using a non-deterministic finite automaton (NFA) to show $R/L$ is regular when both $R$ and $L$ are regular.
Let $(Q_1,Sigma,delta_1,s_1,F_1)$ and $(Q_2,Sigma,delta_2,s_2,F_2)$ be deterministic finite automata for $R$ and $L$ respectively. Define a NFA
$$D=(Q_1 times {s_2} times {1} cup Q_1 times Q_2times {2}, Sigma, delta, (s_1,s_2, 1), F_1 times F_2times {2})$$
where the transitions are:
$delta((q_1,s_2,1),sigma) = { (delta_1(q_1,sigma),s_2, 1)}$.
$delta((q_1,s_2,1),epsilon) = { (q_1,s_2, 2)}$.
$delta((q_1,q_2,2),sigma) = { (delta_1(q_1,sigma),delta_2(q_2,sigma), 2)}$.
I'll leave it as an exercise for you to verify the language accepted by $D$ is $R/L$.
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
answered Dec 25 '18 at 22:00
Apass.JackApass.Jack
10.3k1939
10.3k1939
add a comment |
add a comment |
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