Given a total recursive function, can you always compute its fixed-point?Clarification of Theorem 1.11 in the...
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Given a total recursive function, can you always compute its fixed-point?
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We know from Kleene's recursion theorem that if $f$ is total recursive, there must be an integer $n$ for which $varphi_n=varphi_{f(n)}$. My question is: for every $f$ total recursive, is there a computable (total) procedure that computes such a point?
ie: $exists g.forall f.varphi_{g(n)}=varphi_{f(g(n))}$.
computability
asked 1 hour ago
NetHackerNetHacker
1314
$endgroup$
add a comment |
$begingroup$
We know from Kleene's recursion theorem that if $f$ is total recursive, there must be an integer $n$ for which $varphi_n=varphi_{f(n)}$. My question is: for every $f$ total recursive, is there a computable (total) procedure that computes such a point?
ie: $exists g.forall f.varphi_{g(n)}=varphi_{f(g(n))}$.
computability
asked 1 hour ago
NetHackerNetHacker
1314
$endgroup$
add a comment |
$begingroup$
We know from Kleene's recursion theorem that if $f$ is total recursive, there must be an integer $n$ for which $varphi_n=varphi_{f(n)}$. My question is: for every $f$ total recursive, is there a computable (total) procedure that computes such a point?
ie: $exists g.forall f.varphi_{g(n)}=varphi_{f(g(n))}$.
computability
asked 1 hour ago
NetHackerNetHacker
1314
$endgroup$
We know from Kleene's recursion theorem that if $f$ is total recursive, there must be an integer $n$ for which $varphi_n=varphi_{f(n)}$. My question is: for every $f$ total recursive, is there a computable (total) procedure that computes such a point?
ie: $exists g.forall f.varphi_{g(n)}=varphi_{f(g(n))}$.
computability
computability
asked 1 hour ago
NetHackerNetHacker
1314
asked 1 hour ago
NetHackerNetHacker
1314
edited 1 hour ago
NetHacker
asked 1 hour ago
NetHackerNetHacker
1314
asked 1 hour ago
NetHackerNetHacker
1314
asked 1 hour ago
NetHackerNetHacker
1314
1314
add a comment |
add a comment |
1 Answer
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$begingroup$
The proof of the recursion theorem is constructive: it gives you an algorithm to compute $n$ from $f$. See for example these notes of Rebecca Weber.
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The proof of the recursion theorem is constructive: it gives you an algorithm to compute $n$ from $f$. See for example these notes of Rebecca Weber.
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
$endgroup$
add a comment |
$begingroup$
The proof of the recursion theorem is constructive: it gives you an algorithm to compute $n$ from $f$. See for example these notes of Rebecca Weber.
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
$endgroup$
add a comment |
$begingroup$
The proof of the recursion theorem is constructive: it gives you an algorithm to compute $n$ from $f$. See for example these notes of Rebecca Weber.
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
$endgroup$
The proof of the recursion theorem is constructive: it gives you an algorithm to compute $n$ from $f$. See for example these notes of Rebecca Weber.
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
answered 26 mins ago
Yuval FilmusYuval Filmus
193k14181345
193k14181345
add a comment |
add a comment |
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