Chaitin's constant is normal?

Question

According to this source, Chaitin's constant $\Omega$ is normal.

Each halting probability is a normal and transcendental real number that is not computable, which means that there is no algorithm to compute its digits. Indeed, each halting probability is Martin-Löf random, meaning there is not even any algorithm which can reliably guess its digits.

Source (Wikipedia)

Furthermore, the definition of normal is that each digit occurs with equal probability $1/b$. And that each duets of digits occur with $1/b^2$ probability and every triplets occurs with probability $1/b^3$ and so on.

Chaitin's omega is calculated via

$\Omega = \sum_{p \in halts} 2^{-|p|}$

Writing $\Omega$ in binary, we obtain a list of 0 and 1. For example,

2^-1=0.1 +
2^-2=0.01 +
2^-3=0.001 +
~skip 2^-4 as it does not halt
2^-5=0.00001 +
...
=\Omega
=0.11101...

Clearly, we can see that the position of each bits corresponds to the halting state of the program of length corresponding to the bit.

Here is what I am struggling with

If $\Omega$ is indeed normal, then it means that exactly 50% of programs halt and exactly 50% do not. This seems very counter intuitive.

For example, suppose I generate java programs by randomly concatenating single characters. The majority of them, I would guess more than 99.99% would not even compile. Would this not imply that at least 99.99% of them will not halt? How do we justify that exactly half will halt and exactly half will not, by virtue of $\Omega$ being normal.

Or is wikipedia incorrect about $\Omega$ being normal?

Answer 1

In contrast to your example, Chaitin's constant is not defined as follows:

$$ \Omega = \sum_{n\colon \text{$n$th program halts}} 2^{-n}. $$

Instead, there is a set $\Pi \subseteq \{0,1\}^*$ of allowed programs which is prefix-free (no string is a prefix of another string). Each of the programs in $\Pi$ is legal (this negates your Java example). If the programs are encoding in unary then it is indeed the case that the $n$th program has length $n$, and then your definition of $\Omega$ works. But for other encodings, the definition of $\Omega$ is

$$ \Omega = \sum_{ p \in \Pi\colon p \text{ halts} } 2^{-|p|}, $$ where $|p|$ is the length of the binary string $p$. Kraft's inequality shows that $\sum_{p \in \Pi} 2^{-|p|} \leq 1$.

Chaitin's constant is algorithmically random: the (prefix) Kolmogorov complexity of the first $n$ bits is $n - O(1)$. To show this, note first that $\Omega_n$, the first $n$ bits of $\Omega$, suffice to determine whether a program of length $n$ (under the encoding $\Pi$) halts or not. Indeed, as a fraction, $\Omega_n \leq \Omega < \Omega_n + 2^{-n}$. Run all programs in parallel, and whenever $p$ stops, add $2^{-|p|}$ to some counter $C$ (initialized at zero). Eventually $C \geq \Omega_n$ (since $C \to \Omega$ from below). At this point, if the input program of length $n$ didn't halt, then you know that it doesn't halt, since otherwise $\Omega \geq C + 2^{-n} \geq \Omega_n + 2^{-n}$.

Given this, suppose that for some $K>0$ and infinitely many $n$, you could compute $\Omega_n$ using $n - K$ bits. For each such $n$, you can find a string whose Kolmogorov complexity is larger than $n$, by considering the output of all halting programs of length at most $n$. For large enough $K$, the result is a program of length at most $n$ for computing the a string whose Kolmogorov complexity is more than $n$. This contradiction proves that for some $K$, the Kolmogorov complexity of $\Omega_n$ is at least $n-K$.

Algorithmic randomness of $\Omega$ implies, in particular, that the frequency of 0s and 1s in its binary expansion tends to 1/2. Indeed, suppose that for some (rational) $\epsilon > 0$ there exist infinitely many $n$ such that the fraction of 1s in $\Omega_n$ is at most $1/2-\epsilon$. Since there are only at most $2^{h(1/2-\epsilon)n}$ strings with at most $1/2-\epsilon$ many 1s, we can compress $\Omega_n$ to size $h(1/2-\epsilon)n + 2\log n + C_\epsilon$ (the constant $C_\epsilon$ depends on $\epsilon$ since the program needs to know $\epsilon$). However, this is $n - \omega(1)$, contradicting the algorithmic randomness of $\Omega$.

Answer 2

Your mistake is on the following line:

If $\Omega$ is indeed normal, then it means that exactly 50% of programs halt and exactly 50% do not.

Nope. That's not what "normal" means. Or, to say it another way: Define $d_0(n)$ to be the number of bits that are a 0, in the first $n$ bits of the base-2 expansion of $\Omega$. Saying that $\Omega$ is normal implies that

$$\lim_{n \to \infty} {d_0(n) \over n} = 1/2.$$

In other words, "normal" is an asymptotic notion. It means that as you look far enough into the bits of $\Omega$, on average about half of them will be 0's. (Similarly, about half with be 1's.)

However, this says nothing about the first few bits of $\Omega$. For instance, there is no implication that the binary expansion $\Omega$ starts out as 0.01... or at anything else -- and there is no implication that $\Omega$ is close to 1/2. $\Omega$ might be close to 0, or close to 1, or anywhere in between; that doesn't contradict being normal, and being normal doesn't imply anything about how large $\Omega$ is.

Answer 3

Yeah I also found it weird when looking through Wikipedia. after reading the most upvoted response here, I looked through all references on Wikipedia, none of them mention Chaitin's constant being normal. Seems like someone misunderstood something and added that by mistake, and unfortunately Wikipedia is done by people so such things happen... Fixed the Wikipedia article.