does recursive (decidable) languages closed under division (Quotient) with any language?
up vote
3
down vote
favorite
I need to prove or disprove that R languages are closed under divison.
I have managed to prove thet CFL are't closed under division. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. I would really apreciate any help.
computational-complexity computability-theory computer-science
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
3
down vote
favorite
I need to prove or disprove that R languages are closed under divison.
I have managed to prove thet CFL are't closed under division. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. I would really apreciate any help.
computational-complexity computability-theory computer-science
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I need to prove or disprove that R languages are closed under divison.
I have managed to prove thet CFL are't closed under division. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. I would really apreciate any help.
computational-complexity computability-theory computer-science
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need to prove or disprove that R languages are closed under divison.
I have managed to prove thet CFL are't closed under division. I read in wikipedia that RE languages are closed, but I didn't find any proof and also I didn't find anything about R languages. I would really apreciate any help.
computational-complexity computability-theory computer-science
computational-complexity computability-theory computer-science
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
oren harlev
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
oren harlev
162
asked 18 hours ago
oren harlev
162
162
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
oren harlev is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
8
down vote
The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xyin L$ for some $yin R$.
If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $x$ and $y$ such that $xyin L$ and $yin R$, and when found, output $x$. This will enumerate the quotient $L/R$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $L$ be the sets of strings consisting of strings of the form $xy$, where $x$ codes a Turing machine program $p$ and input $n$ (with a suitable end-of-code marker) and $y$ codes the halting computation of $p$ on $n$, provided that it does halt. And let $R$ be the string with just the strings $y$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $L/R$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.
The big-picture perspective here is that the quotient construction allows you to perform an existential quantifier, and one does not expect such an operation to preserve decidability, although it will preserve computable enumerability.
answered 16 hours ago
Joel David Hamkins
163k25500860
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
oren harlev is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317586%2fdoes-recursive-decidable-languages-closed-under-division-quotient-with-any-l%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xyin L$ for some $yin R$.
If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $x$ and $y$ such that $xyin L$ and $yin R$, and when found, output $x$. This will enumerate the quotient $L/R$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $L$ be the sets of strings consisting of strings of the form $xy$, where $x$ codes a Turing machine program $p$ and input $n$ (with a suitable end-of-code marker) and $y$ codes the halting computation of $p$ on $n$, provided that it does halt. And let $R$ be the string with just the strings $y$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $L/R$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.
The big-picture perspective here is that the quotient construction allows you to perform an existential quantifier, and one does not expect such an operation to preserve decidability, although it will preserve computable enumerability.
answered 16 hours ago
Joel David Hamkins
163k25500860
add a comment |
up vote
8
down vote
The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xyin L$ for some $yin R$.
If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $x$ and $y$ such that $xyin L$ and $yin R$, and when found, output $x$. This will enumerate the quotient $L/R$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $L$ be the sets of strings consisting of strings of the form $xy$, where $x$ codes a Turing machine program $p$ and input $n$ (with a suitable end-of-code marker) and $y$ codes the halting computation of $p$ on $n$, provided that it does halt. And let $R$ be the string with just the strings $y$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $L/R$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.
The big-picture perspective here is that the quotient construction allows you to perform an existential quantifier, and one does not expect such an operation to preserve decidability, although it will preserve computable enumerability.
answered 16 hours ago
Joel David Hamkins
163k25500860
add a comment |
up vote
8
down vote
up vote
8
down vote
The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xyin L$ for some $yin R$.
If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $x$ and $y$ such that $xyin L$ and $yin R$, and when found, output $x$. This will enumerate the quotient $L/R$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $L$ be the sets of strings consisting of strings of the form $xy$, where $x$ codes a Turing machine program $p$ and input $n$ (with a suitable end-of-code marker) and $y$ codes the halting computation of $p$ on $n$, provided that it does halt. And let $R$ be the string with just the strings $y$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $L/R$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.
The big-picture perspective here is that the quotient construction allows you to perform an existential quantifier, and one does not expect such an operation to preserve decidability, although it will preserve computable enumerability.
answered 16 hours ago
Joel David Hamkins
163k25500860
The quotient of one language $L$ by another $R$ is the set of strings $x$ such that $xyin L$ for some $yin R$.
If both $L$ and $R$ are computably enumerable (what you call RE), then the quotient is clearly enumerable, since we can simply search for all strings $x$ and $y$ such that $xyin L$ and $yin R$, and when found, output $x$. This will enumerate the quotient $L/R$.
But in the case of decidable sets (what you call R), it is not true that the quotient is necessarily decidable. To see this, let $L$ be the sets of strings consisting of strings of the form $xy$, where $x$ codes a Turing machine program $p$ and input $n$ (with a suitable end-of-code marker) and $y$ codes the halting computation of $p$ on $n$, provided that it does halt. And let $R$ be the string with just the strings $y$ coding the halting computations. These are each decidable, since we can look at a string and easily decide if it codes the information or not.
But the quotient $L/R$ will consist of strings coding the halting TM program and input pairs that halt. That is, it is the halting problem, and this is not decidable.
The big-picture perspective here is that the quotient construction allows you to perform an existential quantifier, and one does not expect such an operation to preserve decidability, although it will preserve computable enumerability.
answered 16 hours ago
Joel David Hamkins
163k25500860
edited 16 hours ago
answered 16 hours ago
Joel David Hamkins
163k25500860
answered 16 hours ago
Joel David Hamkins
163k25500860
answered 16 hours ago
Joel David Hamkins
163k25500860
163k25500860
add a comment |
add a comment |
oren harlev is a new contributor. Be nice, and check out our Code of Conduct.
oren harlev is a new contributor. Be nice, and check out our Code of Conduct.
oren harlev is a new contributor. Be nice, and check out our Code of Conduct.
oren harlev is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317586%2fdoes-recursive-decidable-languages-closed-under-division-quotient-with-any-l%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
