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

I am reading between the lines. If you want to ask this: Given semi-decidable languages $L_1$ and $L_2$, is the language $L = L_1 \setminus L_2$ also semi-decidable? In this case, the answer is: …
answered Jun 15 '13 by Raphael
We will show that any Turing-complete "programming language" has such a pair of programs by using classic recursion theory, that is the notions of Gödel numberings with s-m-n property/theorem (smn) …
answered Dec 4 '14 by Raphael
Depends on what exactly you have in mind. If the problem is to find an element with decidable (!) properties among a recursively enumerable set $A$ (of finitely represented elements), then yes: if $… answered Nov 5 '17 by Raphael Let's review what an index set is. Given a Gödel numbering$\phi$and a set$P \subseteq \mathcal{P}$of (partially) recursive functions, we call$\qquad\displaystyle I_P := \{ i \mid \phi_i …
answered Apr 24 '15 by Raphael
Your "proof sketch" contains almost no information; you need to note the special reduction partner. Usually, the halting problem is reduced to deciding the given, arbitrary non-trivial encoding set $L … answered Jun 21 '12 by Raphael In fact, given sufficient background in computability, it is clear that such a function can not exist. Assume any computable, total function$\mathtt{alter} : \mathbb{N} \to \mathbb{N}$. … answered May 4 '15 by Raphael It's easy to encode the Halting problem as unary language:$\qquad\displaystyle L_H = \{ a^{\langle M \rangle} \mid M \text{ is a TM}, M(\langle M \rangle) \text{ halts}\}$is decidable if and only … answered Nov 8 '17 by Raphael My favorite related example is the computability of whether$\pi$contains$0^k$. Try to understand why that's the case, then revisit this question. … answered Sep 5 '17 by Raphael Let's do this step by step. "Z(A,b) where {A,b!=Z}" I understand this to mean "just forbid input Z for algorithm Z". You can not do that. A machine that is supposed to solve the halting problem has t … answered Jan 9 '14 by Raphael You can construct a recognizer following the same principle used for the recognizer for HALT. The only extra bit is how you check "all" inputs without getting stuck in a non-terminating computation. … answered Mar 22 '14 by Raphael$L$can indeed be "everything". Let$L$the set of Turing machines that always move right, never write to the tape, and halt when they read the blank symbol. Clearly, we can decide these critera … answered Jun 22 '15 by Raphael No, that is not possible. There is an extended version of Rice's theorem¹ to prove an index set is not recursively enumerable. In your notation, the theorem states that if a (non-trivial)$C$contain … answered Jun 8 '12 by Raphael What you are proposing can be used to implement an algorithm to solve this problem: Given a runtime bound$\Theta(f)$and a procedure proc, decide whether proc runs in that time (in the worst case … answered Jul 10 '13 by Raphael Yes; every computable function has infinitey many Turing machines that compute it. That is a fundamental property of all Turing-complete models/formalisms. It is a consequence of the s-m-n theorem an … answered May 19 '16 by Raphael I'm going to assume that you mean$\qquad L = \{ \langle M \rangle \mid |L_{\leq 10}(M)| = \infty \}$with$L_{\leq m}(M)$the set of words that$M$accepts after at most$m\$ steps. In particular, i …
answered Apr 15 '17 by Raphael

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