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

There is no such thing as "the fastest growing function". In fact, there is even no sequence of fastest growing functions. This was already shown by Hausdorff. Given two functions $f,g\colon \mathbb{N … answered Sep 22 '12 by Yuval Filmus You are looking for the recursion theorem. Let$F(e)$be the function that on input$0x$returns$P(x)$and on input$1x$returns$e$. A fixed point$\varphi_e = \varphi_{F(e)}$is what you're looking … answered Apr 16 '14 by Yuval Filmus Consider the set$S$consisting of everything, and the partially computable function$f$which on input$n$simulates the$n$th Turing machine, returning whatever the machine outputs (if it ever halts … answered Sep 27 '17 by Yuval Filmus In one word: no. It doesn't follow logically. It is also definitely not the case that no infinite co-r.e. language can have infinite recursive subsets: every infinite recursive set, say$0^*$, is also … answered Nov 11 '14 by Yuval Filmus Your problem is not so trivial, though classical. See the proof of Corollary 11.9 in lecture notes of Kevin T. Kelly. answered Jan 8 '17 by Yuval Filmus The proof of the recursion theorem is constructive: it gives you an algorithm to compute$n$from$f$. See for example these notes of Rebecca Weber. answered Feb 24 '19 by Yuval Filmus Your function is computable. Just run the input machine on the empty input for$B(10^{1000})$steps. Note that$B(10^{1000})$is just a constant which can be hardcoded into your code. answered Aug 21 '19 by Yuval Filmus Notice that$A \leq B \leq \overline{A}$. Since$B$is r.e.,$A$is r.e., hence$\overline{A}$is co-r.e. Since$A \leq \overline{A}$,$A$is also co-r.e. Hence$A$is recursive. answered Dec 11 '20 by Yuval Filmus Fix a value of$n$. For$b \in \{0,1\}^n$, consider the following algorithm$A_b$: If$x \leq n$then output$b_x$, otherwise output$0$. Clearly one of the$2^n$algorithms of the form$A_b$comput … answered Jan 17 '21 by Yuval Filmus The notation is explained in the Wikipedia entry (though regrettably after its first use): for partial functions$f,g$, we say that$f\simeq g$if for all inputs$x$,$f$halts on$x$iff$g$halts on … answered Mar 16 '14 by Yuval Filmus Let's prove a more general result using diagonalization. Suppose$L_1, L_2, \ldots$is a countably infinite list of infinite languages. Then there is an infinite language$L$that doesn't contain any … answered Nov 24 '14 by Yuval Filmus First of all, encoding is up to you — you get to decide the answer to the question you state. Let's consider a more generic situation, in which you want to serialize (convert to bits) a sequence of v … answered Mar 23 '17 by Yuval Filmus .), and is a fundamental concept in computability theory. The relation between computable languages and c.e. ones is the same as that between P and NP. … answered Nov 12 '14 by Yuval Filmus Hint: Such sets$A,B$(sans the requirement of recursive enumerability) are called recursively inseparable. According to Wikipedia "it is possible for$A$and$B$to be recursively inseparable, disjoi … answered Dec 25 '14 by Yuval Filmus Suppose$f$is a fixpoint of$A$. First, $$f(0) = f(1) = f(2) = f(3) = 3f(3),$$ and so$f(0) = f(1) = f(2) = f(3) = 0$. Second, $$f(5) = 5f(4) = 5f(5),$$ and so$f(4) = f(5) = 0$. Third, if$n = 4 …
answered Jan 8 '13 by Yuval Filmus

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