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

9
votes
This is a classical exercise in computability. Since $A,B\in coRE$, there exist TMs $M,N$ that recognize $\overline{A}$ and $\overline{B}$ respectively. …
answered Feb 17 '13 by Shaull
2
votes
You are confusing languages with algorithms. An algorithms is not "computable" or "non-computable". It can halt or not halt, but the latter doesn't mean that the language it defines is non-computable …
answered May 14 '13 by Shaull
2
votes
This is quite a fundamental point in computability theory, which has to do with the difference between being able to find and algorithm and knowing that an algorithm exists. …
answered Jan 10 '15 by Shaull
3
votes
Explaining intuitively why this language is recursive is easy: it's almost trivial to write a computer program in a high-level programming language (e.g. Java, C) that decides this language. Since TMs …
answered Jun 21 '13 by Shaull
2
votes
A simple example would be a reduction from e.g. $SAT$ to it's complement $\overline{SAT}$, which works as follows: given a formula $\varphi$, you can decide whether $\varphi$ is satisfiable by decidin …
answered Nov 22 '14 by Shaull
2
votes
Note that in your model, you only ever have a single item $x$ on the stack, namely the current number of inputs seen. Thus, you're not really using the stack. This is sometimes called a register, and …
answered Oct 27 '19 by Shaull
3
votes
It would be great, wouldn't it? And like most problems regarding Turing Machines (or "algorithms"), there is no such algorithm, and there never will be: Consider for example the language $$\{M: M\tex …
answered Jul 10 '13 by Shaull
3
votes
The condition about $B$ can be transformed into a condition about $B\setminus A$, as follows. If $B\setminus A$ is decidable, then $B$ is unrecognizable. Indeed, if you could recognize $B$, you could …
answered Feb 4 '15 by Shaull
2
votes
Rice's theorem says that any nontrivial semantic property of TMs is undecidable. The property in question is clearly nontrivial, but let's see if it's semantic. A semantic property of TMs is a set of …
answered May 15 '17 by Shaull
5
votes
This problem is not in RE nor in coRE. First, your intuition that it's not in RE is correct, although note that very crucially - it's just an intuition, and not a direction for a proof! However, your …
answered Mar 6 '18 by Shaull
4
votes
The question is not well defined, so the answer can be either decidable or undecidable. Here are two extreme (and naive) examples of this issue. Suppose your input consists of a single DFA $A$ (i.e …
answered Dec 2 '17 by Shaull
16
votes
A property $P$ is a set of Turing machines. The property is trivial if it contains every TM, or if it is empty. Essentially, in order to check if a property is trivial, just check if there is a TM th …
answered Feb 27 '14 by Shaull
1
vote
Proving the correctness and computability of this procedure is not hard. …
answered Jul 7 '16 by Shaull
1
vote
Technical hint: assume that every string is a legal encoding of a TM. Can you solve this now? Intuitive hint: the language $C$ can contain many other encodings as well as those that are equivalent to …
answered Sep 29 '13 by Shaull
2
votes
First, there is a small typo in (1) - if $\alpha$ is not a legal encoding, then you should return something that is in $ALL_{TM}$ (since you are reducing to the complement). For your question: the ke …
answered Apr 19 '13 by Shaull

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