Prove that function is totally computable
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The problem that I'm working on is
Graph of function $f : mathbb{N}rightarrow mathbb{N}$ is set {$(x, f(x))$, $x in mathbb{N}$ and $f(x)$ $neq perp$} $subseteq mathbb{N}^{2}$. Prove that function $f$ is totally computable when $f(x)$ is defined for every $x $ $in $ $mathbb{N}$ and his graph is recursively enumerable set.
Do you have any suggestions how to prove it?
I have recently started learning theory of computability so some easy to understand answers would be appreciated.
computability
asked yesterday
Bedřich Kletorný
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The problem that I'm working on is
Graph of function $f : mathbb{N}rightarrow mathbb{N}$ is set {$(x, f(x))$, $x in mathbb{N}$ and $f(x)$ $neq perp$} $subseteq mathbb{N}^{2}$. Prove that function $f$ is totally computable when $f(x)$ is defined for every $x $ $in $ $mathbb{N}$ and his graph is recursively enumerable set.
Do you have any suggestions how to prove it?
I have recently started learning theory of computability so some easy to understand answers would be appreciated.
computability
asked yesterday
Bedřich Kletorný
102
New contributor
Bedřich Kletorný 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 |
The problem that I'm working on is
Graph of function $f : mathbb{N}rightarrow mathbb{N}$ is set {$(x, f(x))$, $x in mathbb{N}$ and $f(x)$ $neq perp$} $subseteq mathbb{N}^{2}$. Prove that function $f$ is totally computable when $f(x)$ is defined for every $x $ $in $ $mathbb{N}$ and his graph is recursively enumerable set.
Do you have any suggestions how to prove it?
I have recently started learning theory of computability so some easy to understand answers would be appreciated.
computability
asked yesterday
Bedřich Kletorný
102
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The problem that I'm working on is
Graph of function $f : mathbb{N}rightarrow mathbb{N}$ is set {$(x, f(x))$, $x in mathbb{N}$ and $f(x)$ $neq perp$} $subseteq mathbb{N}^{2}$. Prove that function $f$ is totally computable when $f(x)$ is defined for every $x $ $in $ $mathbb{N}$ and his graph is recursively enumerable set.
Do you have any suggestions how to prove it?
I have recently started learning theory of computability so some easy to understand answers would be appreciated.
computability
computability
asked yesterday
Bedřich Kletorný
102
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Bedřich Kletorný
102
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Bedřich Kletorný
102
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Bedřich Kletorný
102
asked yesterday
Bedřich Kletorný
102
102
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bedřich Kletorný is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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Fix a computable enumeration of the graph of the function. On input $n$, wait for a pair $(n,y)$ to appear in the graph. When you find such a pair, output $y$. This is a computable procedure that computes $f$.
The totality of the function is irrelevant to this argument. What the procedure shows is that a function $f$ is computable if and only if the graph of $f$ is computably enumerable.
But if you also know (by your hypothesis) that the function is total, then you've got a total computable function.
answered yesterday
JDH
32.4k679145
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1 Answer
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1 Answer
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active
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votes
Fix a computable enumeration of the graph of the function. On input $n$, wait for a pair $(n,y)$ to appear in the graph. When you find such a pair, output $y$. This is a computable procedure that computes $f$.
The totality of the function is irrelevant to this argument. What the procedure shows is that a function $f$ is computable if and only if the graph of $f$ is computably enumerable.
But if you also know (by your hypothesis) that the function is total, then you've got a total computable function.
answered yesterday
JDH
32.4k679145
add a comment |
Fix a computable enumeration of the graph of the function. On input $n$, wait for a pair $(n,y)$ to appear in the graph. When you find such a pair, output $y$. This is a computable procedure that computes $f$.
The totality of the function is irrelevant to this argument. What the procedure shows is that a function $f$ is computable if and only if the graph of $f$ is computably enumerable.
But if you also know (by your hypothesis) that the function is total, then you've got a total computable function.
answered yesterday
JDH
32.4k679145
add a comment |
Fix a computable enumeration of the graph of the function. On input $n$, wait for a pair $(n,y)$ to appear in the graph. When you find such a pair, output $y$. This is a computable procedure that computes $f$.
The totality of the function is irrelevant to this argument. What the procedure shows is that a function $f$ is computable if and only if the graph of $f$ is computably enumerable.
But if you also know (by your hypothesis) that the function is total, then you've got a total computable function.
answered yesterday
JDH
32.4k679145
Fix a computable enumeration of the graph of the function. On input $n$, wait for a pair $(n,y)$ to appear in the graph. When you find such a pair, output $y$. This is a computable procedure that computes $f$.
The totality of the function is irrelevant to this argument. What the procedure shows is that a function $f$ is computable if and only if the graph of $f$ is computably enumerable.
But if you also know (by your hypothesis) that the function is total, then you've got a total computable function.
answered yesterday
JDH
32.4k679145
answered yesterday
JDH
32.4k679145
answered yesterday
JDH
32.4k679145
answered yesterday
JDH
32.4k679145
32.4k679145
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Bedřich Kletorný is a new contributor. Be nice, and check out our Code of Conduct.
Bedřich Kletorný is a new contributor. Be nice, and check out our Code of Conduct.
Bedřich Kletorný is a new contributor. Be nice, and check out our Code of Conduct.
Bedřich Kletorný is a new contributor. Be nice, and check out our Code of Conduct.
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