Prove that function is totally computable
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
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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
New contributor
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
New contributor
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
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New contributor
New contributor
asked yesterday
Bedřich Kletorný
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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.
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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.
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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.
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.
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
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