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Questions related to computability theory, a.k.a. recursion theory

To show that the language is $\Pi^0_2$ complete you have to show two things: The language is $\Pi^0_2$ Every $\Pi^0_2$ language is many-one reducible to your language To show that your language is …
answered Aug 9 '12 by Carl Mummert
The thing you are trying to prove is called the "separation principle" for $\Pi^0_1$ sets (remember co-r.e. sets are $\Pi^0_1$ and r.e. sets are $\Sigma^0_1$ in the arithmetical hierarchy). The dual …
answered Feb 17 '13 by Carl Mummert
One famous open question about the poset $(D, \leq_T)$ of Turing degrees is whether it has any non-trivial automorphisms. That is, does there exist a non-identity bijection $f\colon D \to D$ such that …
answered Mar 16 '12 by Carl Mummert
If you take any noncomputable set of natural numbers, the characteristic function of the set takes only the values $\{0,1\}$ and is noncomputable. So it is not the case that every noncomputable functi …
answered Jul 28 '12 by Carl Mummert
Turing machines can handle interaction just fine, using oracle tapes. It works are follows: from the point of view of a computer that handles interaction, the input of the operator is simply another s …
answered Mar 18 '12 by Carl Mummert
A general theorem that partially covers the example given is that any $\Sigma^0_1$-hard property of the machine will be undecidable. The halting problem is $m$-reducible to the state-reachability prob …
answered Mar 6 '12 by Carl Mummert
Stronger systems that have universal functions, such as Turing computability, simply must have partial functions in order to allow the universal function to exist. …
answered Mar 13 '12 by Carl Mummert
Let's look at the definition of the set $S = \{ x : f_x \not = h\}$. There are two ways that a given $f_x$ could differ from $h$: 1) For some $n$, $f_x(n) \downarrow \not = h(n)$ 2) $f_x(n) \uparro … answered Feb 12 '14 by Carl Mummert In practice, we don't usually prove just that a language is r.e. or not r.e.. If the language is r.e., we want to know whether it is recursive. If it is not r.e., we want to know what sort of Turing d … answered May 3 '12 by Carl Mummert You cannot show, for an arbitrary unrecognizable language$C$, that$\overline{A_{TM}} \leq_m C$. If$\overline{A_{TM}} \leq_m C$then in particular the Turing degree of$C$is greater than or equal … answered May 3 '12 by Carl Mummert There is a general fact that every noncomputable Turing degree contains infinitely many distinct$m$-degrees. (This result follows at least from results of Jockusch, "Relationships between reducibil … answered May 3 '12 by Carl Mummert I think the question is how to show that$L$is not r.e. One way to do that is to reduce the complement of the halting problem to$L\$, because the complement of the halting problem is not r.e. Here …
answered May 22 '12 by Carl Mummert
Yes, this is decidable, because you can do an exhaustive search of all possible paths. There is no need to look at any paths that repeat a vertex, since the "detour" could be skipped. But the length o …
answered Mar 9 '12 by Carl Mummert
There are various single instructions that lead to Turing complete languages. The typical example is "subtract and branch if zero". These are well known in the context of assembly language programming …
answered Apr 19 '12 by Carl Mummert