1

Let BB(n) denote the largest number of 1's a halting Turing Machine with n states can print. Clearly this is uncomputable.

It BB(n) mod 10 computable?

I suspect it may not be, but I cannot prove it.

Jacob Urick
  • 53

1 Answers1

0

I strongly suspect it isn't. Say we could determine $\Sigma(n)$ modulo $m$ for any $m$ if we do it for a sufficent large $m$, we have $\Sigma(n)$, which we know is impossible to compute. Now perhaps 10 is one of the finite set of $m$ where it can be computed...

vonbrand
  • 27,812
  • 2
    That only holds if we can do that uniformly in $m$. There's no reason $\Sigma(n)$ mod $m$ couldn't be computable in each $m$ but not uniformly computable. In fact, I suspect we can whip up an acceptable enumeration of the partial computable functions with this property (each such function would be eventually constant, but we wouldn't know when it becomes constant). – Noah Schweber Feb 25 '20 at 01:33
  • In fact this is easy to do: just take the usual model of computation and "slow down" the $k$th computation by a factor of $k!$. Of course this results in an artificial system, but it shows that the general heuristic is inapplicable. – Noah Schweber Feb 25 '20 at 04:21
  • @Noah Schweber could you elaborate about $\Sigma (n)$ mod $m$ being computable for each $m$, but nor being uniformly computable? – Jacob Urick Feb 25 '20 at 04:30
  • 1
    @JacobUrick Suppose I have a sequence of functions $(f_n)_{n\in\mathbb{N}}$. It may be the case that each $f_n$ is a computable function but the binary function $(x,y)\mapsto f_x(y)$ is not computable. For example, let $f_n(a)=1$ iff $n$ is in the halting problem. Then for each $n$ the function $f_n$ is computable (since it's constant), but we don't have "uniform computability." – Noah Schweber Feb 25 '20 at 04:33
  • In this case, my $f_n$ is $\Sigma(k)$ mod $n$. Vonbrand is correct that if the binary function $\alpha: (k,n)\mapsto \Sigma(k)$ mod $n$ were computable, then $\Sigma$ itself would be computable, and so $\alpha$ can't be computable. However, this doesn't rule out the possibility that each "row" of $\alpha$ (= each function of the form $k\mapsto \Sigma(k)$ mod $n$ for some $n$) is computable, just "non-uniformly." – Noah Schweber Feb 25 '20 at 04:34
  • 1
    @JacobUrick As a relevant concrete example, let $g$ be any function and define $$h(n)=n!g(n).$$ For each $m$ the function $$h_m: n\mapsto h(n)\mbox{ mod }m$$ is eventually constant, hence computable, but $h$ itself is Turing equivalent to $g$ (and so won't be computable unless $g$ is) - the issue being exactly that the sequence of functions $(h_m)_{m\in\mathbb{N}}$ is not uniformly computable. – Noah Schweber Feb 27 '20 at 21:35