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
2,1972622
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
2,1972622
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
2,1972622
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
2,1972622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,1972622
asked Dec 7 '18 at 9:01
Melissa Robles CarmonaMelissa Robles Carmona
575
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,1972622
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,1972622
edited Dec 7 '18 at 9:43
Daniele Tampieri
2,1972622
2,1972622
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
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$begingroup$
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
8,554725
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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
8,554725
$endgroup$
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
8,554725
$endgroup$
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
8,554725
$endgroup$
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
8,554725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,554725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,554725
answered Dec 7 '18 at 9:43
gandalf61gandalf61
8,554725
8,554725
add a comment |
add a comment |
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