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

3
votes
This answer gives us a hint, namely that an interpreter for a language more powerful than primitive recursion, is itself not primitive recursive. For example, System F has no self-interpreter, and thu …
answered Jul 20 '17 by jmite
3
votes
Absolutely not. In fact, these are most useful when $B$ is undecidable. Diagonalization was used to show the Halting problem undecidable, but most other problems were shown to be undecidable through …
answered Aug 25 '16 by jmite
2
votes
What these slides are describing is an algorithm (Turing Machine), which looks at a DFA and asks "does this DFA accept the empty language?" So the diagrams in the slides are correct. If there is no p …
answered Jan 8 '16 by jmite
2
votes
The problem is, you haven't defined a language. You have defined a function that returns a language. What you call $S$ is really $S(M)$, for some Turing Machine $M$. What is undecidable is the functi …
answered Jul 10 '13 by jmite
1
vote
Reversal Bounded Counter Automata have decidable equivalence. They're a fascinating class, because augmenting them in any obvious way makes one of their main properties (equivalence, emptiness, etc.) …
answered Jul 17 '19 by jmite
2
votes
For recursive languages, membership is decidable by definition, so most undecidable questions with regards to them will involve quantifiers. It's decidable to check if there's a given input is accepte …
answered Oct 25 '17 by jmite
1
vote
In the general case, this is undecidable, and it turns out the answer has nothing to do with being rational or irrational. Consider a Turing Machine with state set $Q$ with a single halting state $H$. …
answered Feb 15 '21 by jmite
17
votes
Yes, every undecidable (not semi-decidable) language has this property. For example, consider the set $L = \{(x,M) \mid M \text{ does not halt on input } x \}$. Suppose we have an algorithm that can …
answered Sep 22 '17 by jmite
15
votes
The key is quantifiers, both in the theorem, and in the "mathematical statements." First, the theorem says that "there is no algorithm that can take in an arbitrary mathematical statement and prove if …
answered Feb 10 '21 by jmite
233
votes
Because a lot of really practical problems are the halting problem in disguise. A solution to them solves the halting problem. You want a compiler that finds the fastest possible machine code for a g …
answered Nov 8 '14 by jmite
57
votes
It all comes from undecidability of the halting problem. Suppose we have a "perfect" dead code function, some Turing Machine M, and some input string x, and a procedure that looks something like this: …
answered Nov 10 '15 by jmite
2
votes
It is absolutely theoretically justified. First realize that a loop is just a form of recursion: do the loop body, then either stop or do the loop body again with different variable values. System …
answered May 22 '16 by jmite
0
votes
For me, the easiest way to think of it is this: You're asking yourself "out of a bunch of candidates in a set $X$, do any of them fulfill the property $P$"? Decidable/recursive means that you only n …
answered May 8 '14 by jmite
2
votes
You've stumbled across the difference between isorecursive and equirecursive types. Equi-recursive types say "types are (possibly) infinite trees, and a recursive type is the solution to a recursive …
answered Aug 30 '19 by jmite
2
votes
Your instructor is right, and this seems like an odd choice. $O(\infty)$ arises usually as a way to say "this algorithm may never terminate, it's awful, let's never use it." Big-O and Running Time …
answered Apr 27 '16 by jmite

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