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

6
votes
In fact, by your first argument, the restriction of a function to a finite domain is always computable, regardless of the computability of the original function. ¹ A disjoint union is the union of the …
answered Jun 21 '21 by Gilles 'SO- stop being evil'
1
vote
A language is infinite if and only if it contains arbitrarily long strings. In other word, a language is finite if and only if there is an upper bound on the length of its strings. (Prove it; it's eas …
answered Sep 3 '13 by Gilles 'SO- stop being evil'
5
votes
First, Apple may wish to restrict applications, but in practice, all versions of the iPhone can be jailbroken: you can then upload arbitrary machine executables. Whether the phone vendor likes or disl …
answered Jun 8 '12 by Gilles 'SO- stop being evil'
5
votes
To answer the literal question that you asked, $\mathsf{HALT}$ is a boolean function, i.e. a function whose values are in the set $\{\mathsf{false}, \mathsf{true}\}$. The negation of $\mathsf{HALT}$ i …
answered Oct 23 '12 by Gilles 'SO- stop being evil'
5
votes
The problem statement is indeed incomplete, but when you see this, you can safely assume that “representing integers in decimal notation” or “representing integers in binary notation” was meant. So …
answered Mar 21 '12 by Gilles 'SO- stop being evil'
3
votes
Meta-knowledge: you want to find a non-decidable language which nonetheless has some computational property. An arbitrary non-decidable language is probably not going to lead you very far. But a semi- …
answered Mar 13 '12 by Gilles 'SO- stop being evil'
4
votes
Russel's paradox does not apply to the lambda calculus. Intuitively speaking, Russel's paradox comes up when there is a level system (e.g. sets that are members of “bigger” sets one level up), and the …
answered Aug 29 '16 by Gilles 'SO- stop being evil'
4
votes
Indeed, the theoretical undecidability of the halting problem does not apply to any concrete computer with finite memory. That's a finite system, so for any program, it's mathematically possible to de …
answered Sep 1 '16 by Gilles 'SO- stop being evil'
48
votes
You can use the algorithm which detects whether a linked list loops to implement the Halting Function with space complexity of O(1). To do that, you need to store at least two copies of the parti …
answered Nov 9 '14 by Gilles 'SO- stop being evil'
4
votes
Here's an approach to this problem that doesn't involve imagination. $\{f(w) \mid w \in L\}$ is strongly reminiscent of the definition of a recursively enumerable set — specifically, a non-empty set $ …
answered Aug 19 '14 by Gilles 'SO- stop being evil'
5
votes
A simple way to demonstrate this kind of property without getting bogged into details is to use the following lemma: Lemma: For any compiler C for a Turing-complete language, there exists a function …
answered Nov 11 '15 by Gilles 'SO- stop being evil'
6
votes
I'm not knowledgeable in that field either, but I think I can provide a non-constructive answer. The first-order theory of real closed fields is decidable. Your problem can be stated as a system of al …
answered Oct 15 '12 by Gilles 'SO- stop being evil'
4
votes
Every procedure has a functional equivalent, but not a structural equivalent, going by contextual definitions of functional and structural equivalent based on your question. That every procedure has …
answered Jun 12 '16 by Gilles 'SO- stop being evil'
38
votes
Yes, sure. Many typed lambda calculi accept only strongly normalizing terms, by design, so they cannot express arbitrary computations. But a type system can be anything you like; make it broad enough, …
answered Jul 6 '12 by Gilles 'SO- stop being evil'
54
votes
It's possible to replace recursion by iteration plus unbounded memory. If you only have iteration (say, while loops) and a finite amount of memory, then all you have is a finite automaton. With a fin …
answered Dec 26 '16 by Gilles 'SO- stop being evil'

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