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11 votes

Guaranteed correct digits of elementary expressions

This problem is Turing equivalent to the constant problem (see also Wikipedia-constant problem), but it is open whether this problem is decidable. The constant problem is the problem of determining wh …
Joel David Hamkins's user avatar
35 votes
Accepted

Are there any undecidability results that are not known to have a diagonal argument proof?

Let me propose a candidate: Kolmogorov complexity is not computable. That is, there is no computable procedure that, given a finite sequence $s$, produces the size of the smallest program (with respec …
Joel David Hamkins's user avatar
5 votes
Accepted

Computability of fillability of unit cube in $\mathbb{R}^n$ by $k$ $\varepsilon$-balls

The question whether $(p,q,n,s)\in R$ in any instance can be expressed as a sentence in the language of the structure $\langle\mathbb{R},+,\cdot,0,1,<\rangle$, a real-closed field, and by Tarski's the …
Joel David Hamkins's user avatar
2 votes

Deciding isomorphism between graphs which interpret in the pure set

Update. As noted in the comments, this answer applies only to definable quotients of $\mathbb{N}$, rather than $\mathbb{N}^d$, and so it doesn't answer the question. The answer is yes, your relatio …
Joel David Hamkins's user avatar
11 votes
Accepted

Show that the positive existential theory is undecidable

I like this question very much. Before answering, let me try to explain the question in my words. You are considering the structure $\mathbb{C}[t,e^{\lambda t}]_{\lambda\in\mathbb{C}}$, which is the …
Joel David Hamkins's user avatar
11 votes

Is being rational decidable?

As Timothy Chow pointed out, the OP's question is most naturally formulated as a promise problem: given a finite system of equations and inequalities in finitely many variables and the promise that it …
Joel David Hamkins's user avatar
37 votes

What do you do if you believe a problem is undecidable?

The first thing to say is that for a statement to be independent of some axioms means really two things, namely, that it is consistent with those axioms, and also that the negation of the statement is …
Joel David Hamkins's user avatar
14 votes
Accepted

Decidability of decidability

$\newcommand\Con{\text{Con}} \newcommand\Dec{\text{Dec}}$ Let $F$ be the formal system in which the proofs are to be carried out, when it comes to your formal assertions of the form $\Dec(\varphi)$. …
Joel David Hamkins's user avatar
11 votes

Quantifier elimination vs decidability

Furthermore, the procedure $\bar\varphi\mapsto R_\varphi$ is effective, regardless of the decidability of $T$. …
Joel David Hamkins's user avatar
23 votes
Accepted

Is the first-order theory (with =) of real numbers with addition and multiplication complete...

What you cannot do while remaining under the decidability result is quantify over the integers or the rational numbers. …
Joel David Hamkins's user avatar
7 votes

Is the isomorphism problem for amenable groups decidable?

So the point is that you shouldn't consider such arbitrary enumerations when asking decidability questions about finitely presented groups, but rather you want to ask about decidability questions for the …
Joel David Hamkins's user avatar
12 votes

Decidability of periodic tilings of the plane

The answer appears to be no. Consider first the case of the anchor-tile periodic tiling problem, where we insist that a particular anchor tile is used. Let's modify the usual Wang tile argument, due …
Joel David Hamkins's user avatar
2 votes

Decidability of matrix algebra

If you want to determine truth in this language with real or complex entries, then Yes. All this is expressible in the language of real-closed fields, simply by using components, and is therefore expr …
Joel David Hamkins's user avatar
22 votes
Accepted

Non-constructive proofs of decidability?

When I teach computability, I usually use the following example to illustrate the point. Let $f(n)=1$, if there are $n$ consecutive $1$s somewhere in the decimal expansion of $\pi$, and $f(n)=0$ oth …
40 votes
Accepted

Has decidability got something to do with primes?

Goedel did indeed use the Chinese remainder theorem in his proof of the Incompleteness theorem. This was used in what you describe as the `boring' part of the proof, the arithmetization of syntax. Con …
Joel David Hamkins's user avatar

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