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Having the function:

$$f(y) = \begin{cases} \ 1 &\text{if }\forall n \Phi_y(n)=n\lor \Phi_y(n) \!\uparrow\\ \ 0 &\text{otherwise.} \end{cases}$$

By the rule of thumb it should not be computable. How to prove it formally? If it's computable, I assumed a function like this exists: $$g(y) = \begin{cases} \ 1 &\text{if $f$(y) = 1 }\\ \ 0 &\text{if $f$(y) = 0} \end{cases}$$

I assumed $g(y)$ is computable. Now there exists a function $\Phi_{x_0}(x_0)$ which simulates $g(y)$. This means: $\Phi_{x_0}(x_0) \downarrow$ iff $\Phi_{x_0} (x_0) \uparrow$, which is a contradiction. Is this sufficient as a proof?

TechCrap
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  • Welcome to Computer Science! The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Jan 22 '17 at 14:55
  • Our reference question may be of help. – Raphael Jan 22 '17 at 14:55
  • I don't quite understand your proof. Try a reduction from the halting problem. – Yuval Filmus Jan 22 '17 at 15:41
  • So it's enough to show that we can map it onto the halting problem? I.e. having a set of instances of (A, D), where A is $\Phi_y(n)$ and D is some x from $N$, we get pairs in S such as ($\Phi_y$, 1), ($\Phi_y$, 2), ...., now H halts if $\Phi_y(n) $ = n. Thus we can map it onto the halting problem, therefore it's not computable? – TechCrap Jan 22 '17 at 16:00
  • Could you please provide a sample solution? So I could see the right way to go for exercises like this one. – TechCrap Jan 22 '17 at 16:03

2 Answers2

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Since I assume this is a homework question, I won't give a fully detailed proof but rather a sketch with some gaps to fill.

Consider the function $$g(x)= \begin{cases} x+1 & \text{ if } \varphi_x(x)\downarrow\\ \uparrow & \text{ otherwise}.\end{cases}$$ This function is computable (convince yourself of that!), so it has an index $e$.

Now consider the function $f$ on $e$, then computing $f(e)$ comes down to verifying whether $\forall n\ \varphi_e(n)\uparrow$. Can you see why this contradicts that the halting problem is not computable?

Dino Rossegger
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  • Thansk for the answer! Just to make it clear, this is not a homework assignment, it's already the exam preparation :). I see that the g(x) is computable, as we can construct a turing machine for each $\Phi_x(x)$ so that if it halts it gives back the answer. However if the $\Phi_e$ halts and gives back the answer, it also won't diverge as it will halt, right? Then f(e) gives back 0. However if $\Phi_e$ diverges we still assume f(x) to give back 1 which we can never decide? – TechCrap Jan 23 '17 at 21:15
  • To your first question, a computation either halts (converges) or it does not halt (diverges). So of course, if it halts, it won't diverge. – Dino Rossegger Jan 23 '17 at 21:34
  • And is the second question correct contradiction? :) – TechCrap Jan 23 '17 at 21:38
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    To your second question, $f(e)=1 \text{ if } \forall n\ \varphi_e(n)\uparrow$. This i ridiculous though, as it converges if another function diverges. – Dino Rossegger Jan 23 '17 at 21:43
  • One could also define $g$ more in the light of Yuval Filmus' answer, by $g(x)={x+1 \text{ if } \varphi_{x_0}(x_0)\downarrow}$ for some natural number $x_0$. $g$ is again computable, say by $e$. Then we have solved the Halting problem as $f(e)=1$ if $\varphi_{x_0}(x_0)\uparrow$ and $f(e)=0$ if $\varphi_{x_0}(x_0)\downarrow$.

    I would accept both answers, although the one using this definition is maybe a little better as it shows better how to reduce the Halting problem to $f$.

     – Dino Rossegger Jan 23 '17 at 21:54
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Here's how to solve the halting problem using $f$. Suppose that we are given a Turing machine $T$, and we want to know whether it halts on the empty input. Construct a Turing machine $T'$ which on input $n$ runs $T$, and if it ever halts, it outputs $n+1$. So $T$ halts on the empty input iff $f(\langle T' \rangle) = 0$.

Conversely, we can calculate $f$ using an oracle for the halting problem. Given a Turing machine $T$, construct a Turing machine $T'$ which runs $T$ on all possible inputs in parallel, and stops if it finds out that $T(n) \neq n$ for some $n$. Then $f(\langle T \rangle) = 0$ iff $T'$ halts.

What we have shown is that $f$ (more exactly, the set of 1-inputs of $f$) is $\Pi_1$-complete.

Yuval Filmus
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    Thank you for your answer! Unfortunately I can only accept only the single answer as a correct one! – TechCrap Jan 23 '17 at 22:59