Search Results
| 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" |
| Saves | in:saves |
| 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 turing-machines
Search options not deleted
user 72757
Questions about Turing machines, a theoretical model of mechanical computation capable of simulating any computer program.
1
vote
2
answers
526
views
Is $L=\{\langle M \rangle|M$ makes more than 10 steps on some input$\}$ decidable?
I want to show that it satisfies Rice's theorem:
(i) $L$ is non-trivial, since $\langle M\rangle \notin L$ where $L(M)=\emptyset$, as $M$ halts/rejects any input string in the first step.
(ii) if $L …
- 173
0
votes
1
answer
378
views
Prove that there exists a machine which decides an infinite subset of halting problem
We already know that $H:=\{\langle M,w\rangle | M$ halts on $w\}$ is undecidable, then how can there possibly be a machine that decides any infinite subset of $H$?
- 173