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 |
computable sets and functions, Turing degrees, c.e. degrees, models of computability, primitive recursion, oracle computation, models of computability, decision problems, undecidability, Turing jump, halting problem, notions of computable randomness, computable model theory, computable equivalence relation theory, arithmetic and hyperarithmetic hierarchy, infinitary computability, $\alpha$-recursion, complexity theory.
1
vote
Completeness, easiest, hardest problems
It sounds like what you're talking about is the very basics of recursion theory, particularly the Turing degrees and even more specifically the degree 0' that contains the complete set K (defined as t …
6
votes
What do we know about the computable surreal numbers?
I did indeed find the (or at least a) definition of $\sqrt{x}$ for surreal $x$ in On Numbers And Games (page 22 of the second edition). It looks like this:
$$\sqrt{x}=y=\left\{\sqrt{x^L}, \dfrac{x+y^L …
2
votes
Finding the largest integer describable with a string of symbols of predefined length
Building in the exposition above
Spencer observes that, between
experts, this game is not funny and
reduces to claims of legitimacy (over
the validity of the axioms they are
supposed to u …
10
votes
Undecidable infinite analogs of NP-complete problems?
There's at least one notable counter-example: solving the mate-in-$poly(n)$ problem for chess on an $n\times n$ board is PSPACE-complete and thus NP-hard (per https://arxiv.org/abs/2010.09271 ), but t …
18
votes
Accepted
Why don't Zeilberger and Gosper's algorithms contradict Richardson's theorem?
(Comment turned into an answer:)
It's as simple as "the composition of hypergeometric terms is not hypergeometric". $f(n)=2^n$ is a hypergeometric term because $\frac{f(n+1)}{f(n)}=2$ is a rational te …