The busy beaver function $\text{BB}(n)$ describes the maximum number of steps that an $n$-state Turing machine can execute before it halts (assuming it halts at all). It is not a computable function because computing it allows you to solve the halting problem.

Are functions like $\text{BB}(n) \bmod 2$, or more generally $\text{BB}(n) \bmod m$ for a modulus $m$, computable? Computing these functions doesn't solve the halting problem, so the above argument doesn't apply.

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    $\begingroup$ This seems like it might well depend sensitively on the details of your machine setup. $\endgroup$ Jan 7, 2013 at 0:01
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    $\begingroup$ Some discussion on this question: scottaaronson.com/blog/?p=46 $\endgroup$ Jan 7, 2013 at 0:44
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    $\begingroup$ A variation: can it be shown that $\text{BB}(n)$ is composite infinitely often? This version is seemingly less sensitive to the encoding. $\endgroup$ Jan 7, 2013 at 4:02
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    $\begingroup$ 1-D BB Turing machines are hard to visualize, so I made a page for 2-D Turing Machine BBs.. Once a 1-D Turing machine becomes predictable, it can be classified as halting or infinite. Thus, the point of predictability is the important point. This rarely happens elegantly. The champions tend to be machines that can be extended forward as they get into temporary predictable behaviors. $\endgroup$
    – Ed Pegg
    Feb 26, 2013 at 15:51
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    $\begingroup$ Related. $\endgroup$ Jul 23, 2013 at 16:21

1 Answer 1


Define $\text{BB}(n)$ as the largest natural number whose Kolmogorov complexity (in a prefix-free binary language) is less than or equal to $n$ bits.

Consider $\text{BB}(n) \space \text{mod} \space 4^n$. This number has a Kolmogorov complexity less than $n + o(n)$, since it can be computed from $\text{BB}(n)$, and $K(\text{BB}(n)) \le n$.

Also consider $\lfloor \Omega \cdot 4^n \rfloor$ where $\Omega$ is Chaitin's constant. This number's Kolmogorov complexity is at least $2 \cdot n - o(n)$ bits (by definition of algorithmic randomness).


$\text{BB}(n) \space \text{mod} \space 4^n \stackrel{?}{=} \lfloor \Omega \cdot 4^n \rfloor$

is computable since it is false for all but finitely many $n$.

Given the first $n$ bits of $\Omega$ it is possible to compute not just $\text{BB}(n)$ but all the $\text{BB}(i)$ for $i$ up to $n$. We can use this to turn the above statement sideways and say something about only the lower bits of each busy beaver number:

$K(\sum_{i \le n}{[4^i \cdot (\text{BB}(i) \space \text{mod} \space 4)]}) < n + o(n)$

implying that

$\sum_{i}{\frac{\text{BB}(i) \space \text{mod} \space 4}{4^i}}$

is not algorithmically random, and in particular,

$\Omega \ne \sum_{i}{\frac{\text{BB}(i) \space \text{mod} \space 4}{4^i}}$ .

A couple more observations:

There is a total computable function $\text{CC}:\mathbb{N}\rightarrow\mathbb{N}$ that inverts $\text{BB}$, i.e. $\text{CC}(\text{BB}(n)) = n$ for all $n \in \mathbb{N}$. It works like this: on input $k$, run every TM with $k$ or fewer states for $k$ steps, and return the fewest number of states of any that halted on the last step. For all $k$ there is a $k$-state machine that terminates in exactly $k$ steps, so there will be a smallest one. This implies immediately that Busy Beaver numbers have some computable properties, for example if $f$ is any computable function, then there is another computable function $g$ such that $f(n) = g(\text{BB}(n))$, namely $g(k) = f(\text{CC}(k))$. But also, we can make $f$ and $g$ be the same function: $\text{CC}$ is non-increasing so it has no cycles and at least one fixed point, call the computable function that finds it $\text{CC}^*$. So, $\text{CC}^*(\text{BB}(n)) = \text{CC}^*(n)$. For $\text{CC}^*$ to be non-trivial there need to be at least two fixed points, surely there always are, but if not just redefine $\text{CC}(k) = k$ on some particular $k$ which is not a $\text{BB}$ number.

On the other extreme, I believe there exists a total computable function $g$ such that $\sum{\frac{g(\text{BB}(n))}{2^n}}$ is algorithmically random: $g(k)$ computes the $\text{CC}(k)^{\text{th}}$ bit of $\Omega$ using the assumption that $\text{BB}(\text{CC}(k)) = k$. I think it should work to to count all programs shorter than $\text{CC}(k)$ that terminate in at most $k$ steps (but more care is needed to describe this and prove that it is total).


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