Set that is recursively enumerable but is NOT decidable
$begingroup$
I was trying to find a set that is recursively enumerable
i.e
$$
exists f; text{computable and a program $P$ that computes}; f
$$ such that
$$
A = { x; :; P(x)downarrow }.
$$
But it is not decidable so it's characteristic function is not computable.
I tried with using the Halting problem and taking $A$ as
$$
A = { x; :; text{ the $x$-th computable function stops }}
$$
but I don't think that A is recursively enumerable.
Thank you.
computability computational-mathematics
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
$endgroup$
add a comment |
$begingroup$
I was trying to find a set that is recursively enumerable
i.e
$$
exists f; text{computable and a program $P$ that computes}; f
$$ such that
$$
A = { x; :; P(x)downarrow }.
$$
But it is not decidable so it's characteristic function is not computable.
I tried with using the Halting problem and taking $A$ as
$$
A = { x; :; text{ the $x$-th computable function stops }}
$$
but I don't think that A is recursively enumerable.
Thank you.
computability computational-mathematics
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
$endgroup$
add a comment |
$begingroup$
I was trying to find a set that is recursively enumerable
i.e
$$
exists f; text{computable and a program $P$ that computes}; f
$$ such that
$$
A = { x; :; P(x)downarrow }.
$$
But it is not decidable so it's characteristic function is not computable.
I tried with using the Halting problem and taking $A$ as
$$
A = { x; :; text{ the $x$-th computable function stops }}
$$
but I don't think that A is recursively enumerable.
Thank you.
computability computational-mathematics
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
$endgroup$
I was trying to find a set that is recursively enumerable
i.e
$$
exists f; text{computable and a program $P$ that computes}; f
$$ such that
$$
A = { x; :; P(x)downarrow }.
$$
But it is not decidable so it's characteristic function is not computable.
I tried with using the Halting problem and taking $A$ as
$$
A = { x; :; text{ the $x$-th computable function stops }}
$$
but I don't think that A is recursively enumerable.
Thank you.
computability computational-mathematics
computability computational-mathematics
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,2322622
2,2322622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
575
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The set $S_h={ (x,y) | text{program x halts when run on input y} }$ is recursively enumerable since a universal program that takes input $(x,y)$ and emulates program $x$ running on input $y$ will halt for exactly the inputs in $S_h$. However, as the halting problem shows, $S_h$ is not decideable because its complement is not recursively enumerable.
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
$endgroup$
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: "69"
};
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',
autoActivateHeartbeat: false,
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
});
}
});
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%2fmath.stackexchange.com%2fquestions%2f3029676%2fset-that-is-recursively-enumerable-but-is-not-decidable%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
$begingroup$
The set $S_h={ (x,y) | text{program x halts when run on input y} }$ is recursively enumerable since a universal program that takes input $(x,y)$ and emulates program $x$ running on input $y$ will halt for exactly the inputs in $S_h$. However, as the halting problem shows, $S_h$ is not decideable because its complement is not recursively enumerable.
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
$endgroup$
add a comment |
$begingroup$
The set $S_h={ (x,y) | text{program x halts when run on input y} }$ is recursively enumerable since a universal program that takes input $(x,y)$ and emulates program $x$ running on input $y$ will halt for exactly the inputs in $S_h$. However, as the halting problem shows, $S_h$ is not decideable because its complement is not recursively enumerable.
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
$endgroup$
add a comment |
$begingroup$
The set $S_h={ (x,y) | text{program x halts when run on input y} }$ is recursively enumerable since a universal program that takes input $(x,y)$ and emulates program $x$ running on input $y$ will halt for exactly the inputs in $S_h$. However, as the halting problem shows, $S_h$ is not decideable because its complement is not recursively enumerable.
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
$endgroup$
The set $S_h={ (x,y) | text{program x halts when run on input y} }$ is recursively enumerable since a universal program that takes input $(x,y)$ and emulates program $x$ running on input $y$ will halt for exactly the inputs in $S_h$. However, as the halting problem shows, $S_h$ is not decideable because its complement is not recursively enumerable.
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,689725
8,689725
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3029676%2fset-that-is-recursively-enumerable-but-is-not-decidable%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
