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

A TM decides a language if it enters the accepting state for word in the language and it enters the rejecting state if it is not. Thus it halts on all inputs. Note the machines defined above are not e …
answered Jun 4 '12 by Dave Clarke
The iPhone has a Turing Machine App. Therefore, if it has an infinite amount of memory, the iPhone would be Turing complete. But no general purpose computer is Turing complete because it has a finite …
answered Jun 8 '12 by Dave Clarke
I'm not sure I full get your encoding and based on the comments there seem to be some problems with it, but I'll nevertheless provide my answer (or "answer", if you prefer). A sequence of languages ( …
answered Sep 16 '15 by Dave Clarke
The time taken for a primitive computation is limited by the speed of light and the size of atoms, as far as we understand physics on this very day, 15 September, 2015. The computation unit needs to …
answered Sep 14 '15 by Dave Clarke
$L_{M_1}$ is in $R$ simply because the number of machine descriptions smaller than a given machine description is finite and any finite language is in $R$.
answered Aug 8 '12 by Dave Clarke
Goldbach's conjecture is either true or false. Do a case analysis on the two possibilities. In one case, $f(x)=x$, which is primitive recursive. In the other case, $f(x)=0$, which is also primitive re …
answered Oct 17 '12 by Dave Clarke
What's missing is the way you run the Turing Machine $M$ on strings to get the Enumerator. Rather than generate each string, run $M$, and then output this string if the $M$ accepts – which as you iden …
answered Jun 5 '12 by Dave Clarke
There are two cases to consider. Assume that $\text{P=NP}$. Then $L=\{\langle M\rangle \mid L(M)\in \emptyset\}=\emptyset$. The empty language is decidable; as no word belongs to it, it is trivial t …
answered Aug 9 '12 by Dave Clarke
In the more algebraic side of theoretical computer science, co- means dual, as in Gilles' answer, but it has a very precise interpretation. If the concept of interest (let's say a product) is formalis …
answered Apr 20 '12 by Dave Clarke
If you have all the features you mention, you don't need recursion. Using dynamic memory you can implement a stack. This can be used to save "partial computations" and the while loop can loop through …
answered Apr 15 '15 by Dave Clarke
One of the key uses of state transition systems, also known as labelled transition systems, is for modelling concurrent systems. One very nice, delightful even, book that uses labelled transition sys …
answered Apr 18 '12 by Dave Clarke
In contrast to what the nay-sayers say, there are many effective techniques for doing this. Bisimulation is one approach. See for example, Gordon's paper on Coinduction and Functional Programming. A …
answered May 25 '12 by Dave Clarke
The combinators $\mathbf{S}$ and $\mathbf{K}$ where, $(\mathbf{S}\ x\ y\ z) = (x\ z\ (y\ z))$ and $(\mathbf{K}\ x\ y) = x$ are sufficient to express any (closed) lambda term, therefore any computable …
answered Apr 19 '12 by Dave Clarke
Consider this from a different perspective. First-order logic is undecidable, that is, there is no decision procedure that determines whether arbitrary formulas are logically valid. (But the set of …
answered Aug 21 '12 by Dave Clarke