Computability Theory (ii): uncomputable numbers.

October 20, 2022 · 4 mins · 832 words

During my current incursion in computability theory I learnt about uncomputable numbers, this is, numbers that can’t be computed with arbitrary precision. This means that even given all the computational power in the universe you could not compute these numbers. Even if God himself came from Heaven, he could not compute these numbers. Uncomputable numbers are numbers that can’t be computed because of maths limitations. Just the idea that maths has some inherent limits amazed me. In my last post, I talked about the halting problem, which is a limitation of maths, but the halting problem is a little bit abstract in my opinion. In this post, instead, I’ll show that there are numbers for which we can know an approximate value but can’t be known with arbitrary precision.

Busy Beaver Numbers

Let’s start by defining the Busy Beaver $n$-game. Its rules are the following

Let’s define the Busy Beaver Number $\Sigma(n)$ as the number of prints that the absolute winner of the Busy Beaver $n$-game does. This is, $\Sigma(n)$ is the maximum number of prints you can do with a program with $n$ characters that eventually halts.

These numbers were first defined by Tibor Radó in the paper On Non-Computable Functions. For a more precise and mathematical definition of the game please read the original article.

Relation with halting problem

What I find fascinating about these numbers, is that they can’t be computed for all $n$. We can prove that by contradiction by assuming that we have a method busy_beaver_number() that returns the value of $\Sigma(n)$ for any $n$. If that was true we could then define the following halts method

def busy_beaver_number(n: int) -> int:
def halts(program) -> bool:
	n_steps = 0
	bb = busy_beaver_number(len(program))
	while n_steps < bb:
		n_steps += 1
	if has_halted(program):
		return True
	return False

This is, given a program of length $n$ run the program for busy_beaver_number(n) steps. If after these steps the program hasn’t halted it means it’ll never halt (by the definition of the $\Sigma(n)$). Since we know that the halting problem can’t be solved we must conclude that our initial assumption is wrong, and therefore busy beaver numbers can’t be computed for any $n$.

However, we can compute $\Sigma(n)$ for some $n$. Currently (2022) the known busy beaver numbers are

We can say that $\Sigma(n)$ grows too fast to be computable.

Computable numbers are real numbers that can be computed within any desired precision by a finite, terminating algorithm.

Small uncomputable numbers

In the last section, I showed that $\Sigma(n)$ grows incredibly fast, and then it can’t be computed. However, in this section, I want to show you that not all uncomputable numbers are enormous. Let’s define $\Omega$ constant as

\[\Omega = \sum_{i=1}^{\infty} 2^{-\Sigma(i)}\]

Since $\Sigma(n)$ grows faster than any other standard function we know that the sum converges to some value. However, since we don’t know $\Sigma(n)$ we can’t know the exact value of $\Omega$. But, what’s interesting is that, even if we can’t know the exact value, we can compute its first digits

\[\Omega = 2^{-1} + 2^{-6} + 2^{-21} + 2^{-107} + ... = 0.515625476837158...\]

So we have a number for which we can compute some of its digits but we can’t know with arbitrary precision. That’s amazing!

Let me emphasize some things about the above result by comparing $\Omega$ and $\pi$. For both cases we can give upper and lower bounds, ie: $\pi \in (3.14, 3.15)$ and $\Omega \in (0.51, 0.52)$. In the case of $\pi$, we don’t know all of its digits, however, we can compute it with arbitrary precision. For example, using the Chudnovsky formula

\[\frac{1}{\pi} =\frac{\sqrt {10005}}{4270934400} \sum_{k=0}^{\infty }\frac{(6k)!(13591409+545140134k)}{(3k)!\,k!^{3}(-640320)^{3k}}\]

Currently, the number of known digits of $\pi$ is 100 trillion. On the other hand, we can’t compute $\Omega$ with an arbitrary level of precision. We know it exists, and we know more or less its value, but we will never have an algorithm that can calculate it as precisely as we want.

Maybe this is a known fact by most mathematicians, but for me (a simple physicist) this has been a revelation that has changed my perception about maths (and physics as well). Hope you have enjoyed this post as much as I have enjoyed this discovery.