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 Feb 24 at 12:16
NetHackerNetHacker
1455
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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 Feb 24 at 12:16
NetHackerNetHacker
1455
$endgroup$
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 Feb 24 at 12:16
NetHackerNetHacker
1455
$endgroup$
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 Feb 24 at 12:16
NetHackerNetHacker
1455
asked Feb 24 at 12:16
NetHackerNetHacker
1455
edited Feb 24 at 13:00
NetHacker
asked Feb 24 at 12:16
NetHackerNetHacker
1455
asked Feb 24 at 12:16
NetHackerNetHacker
1455
asked Feb 24 at 12:16
NetHackerNetHacker
1455
1455
add a comment |
add a comment |
1 Answer
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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 Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
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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 Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
$endgroup$
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 Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
$endgroup$
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 Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
$endgroup$
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 Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
answered Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
answered Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
answered Feb 24 at 13:36
Yuval FilmusYuval Filmus
194k14181346
194k14181346
add a comment |
add a comment |
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