Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Bookmarks inbookmarks:mine
inbookmarks:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only user 98

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

0
votes
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
2
votes
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
1
vote
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
2
votes
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
2
votes
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
2
votes
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
2
votes
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
5
votes
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
7
votes
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
6
votes
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
3
votes
$L$ 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
3
votes
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
1
vote
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
0
votes
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
2
votes
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

1
2 3 4 5
8
15 30 50 per page