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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 …
Steven Stadnicki's user avatar
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 …
Steven Stadnicki's user avatar
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 …
Steven Stadnicki's user avatar
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 …
Steven Stadnicki's user avatar