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

I was first taken by your metaphore, but I do not think it works. Actually he is saying that he is himself a decision procedure, though he cannot describe his own internal program. Hence Justice Stewa …
answered Jun 26 '15 by babou
Designing a Turing machine is pretty much like writing a program. You have to choose a representation for the data and a corresponding code (read * transitions*) to manipulate the data. Remember how w …
answered May 18 '15 by babou
bijection $\phi: \mathbb N\to L_{\emptyset,0}$ Then given any subset $S$ of $\mathbb N$, we can get get with the bijection $\phi$ a subset $\phi(S)\subset L_{\emptyset,0}$ which has the same level of computability … Of course, other levels of computability can be obtained in the same way. …
answered Jun 20 '15 by babou
I am not certain of what you are after, so I am giving you below pointers to various related areas of research that may answer your quesry. To be true, they fit your examples more closely than the tag …
answered Jul 25 '15 by babou
There are many issues in your question. I am no longer very expert in some of them, but let's try. Basically, a standard program as you like it takes some input $I$, satisfying some property $P$, and …
answered Feb 6 '15 by babou
Take the language $L= \{a^{2^p}\mid p\geq 1\}$. Try to prove it is a counter-example.
answered May 20 '15 by babou
The case of 1-state DFA Accepting one word is unrelated to having one state. If there is only one state, it is necessarily starting and accepting state, unless there is no accepting state which make …
answered Feb 27 '15 by babou
I will not attempt to define what a computation is, which was done rather well by Luke Mathieson and Yuval Filmus. However, thinking about an exploding device as a computation lead me to an important …
answered Jun 27 '15 by babou
In computability (at least for what I know of it), everything is either finite or countably infinite. So existence of mappings is usually not an issue. The problem is whether it is a computable one. …
answered Feb 28 '15 by babou
The empty set is recursive, hence B cannot be the empty set. You are right the B-A is recursively enumerable. You cannot prove B-A is not recursive for a good reason: it may be recursive. Maybe you …
answered Dec 7 '14 by babou
Context-free languages are decidable, and decidable languages are closed under intersection. So, though the intersection of two CF languages may not be CF, it is decidable. Remarks on your example: …
answered Jul 18 '14 by babou
There is no general way to find a decider TM for $L_k$ You are correct that $L_k$ is recursive because, being a subset of the finite set $\Sigma^k$, it is also finite. You would like to rather find …
answered Jul 16 '15 by babou
One possible definition of recursively enumerable (most likely the one used by your instructor) is that it is the domain of a partial function. In Turing Machine terms, that means the following defini …
answered Apr 10 '15 by babou
I do not have a general answer as I am not sure this can be actually answered in a single proof. So your question is really too broad, involving many special cases. And I am afraid there are several …
answered Feb 17 '15 by babou
My view is that vernacular would consider that S is not empty, i.e. $\emptyset \neq S\subseteq T$, while mathematical language would consider that S can be empty, i.e. \$\emptyset\subseteq S\subseteq T …
answered Apr 3 '15 by babou

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