Can a transcendental number like $e$ or $\pi$ be compressed as not algorithmically random?

Question

The related and interesting fields of Information Theory, Turing Computability, Kolmogorov Complexity and Algorithmic Information Theory, give definitions of algorithmically random numbers.

An algorithmically random number is a number (in some encoding, usually binary) for which the shortest program (e.g using a Turing Machine) to generate the number, has the same length (number of bits) as the number itself.

In this sense numbers like $\sqrt{e}$ or $\pi$ are not random since well known (mathematical) relations exist which in effect function as algorithms for these numbers.

However, especially for $e$ and $\pi$ (which are transcendental numbers) it is known that they are defined by infinite power series.

For example $e = \sum_{n=0}^\infty \frac{1}{n!}$

So even though a number, which is the binary representation of $\sqrt{e}$, is not alg. random, a program would (still?) need the description of the (infinite) bits of the (transcendental) number $e$ itself.

Can transcendental numbers (really) be compressed?

Where is this argument wrong?

UPDATE:

Also note the fact that for almost all transcendental numbers, and irrational numbers in general, the frequency of digits is uniform (much like a random sequence). So its Shannon entropy should be equal to a random string, however the Kolmogorov Complexity, which is related to Shannon Entropy, would be different (as not alg. random)

Thank you

Answer 1

The problem is in your poor definition of "algorithmically random number" as applied to irrational numbers. In particular:

has the same length (number of bits) as the number itself.

has no meaning if the number is of unbounded length.

Your Wikipedia link gives better definitions, which don't have this problem. For example (and paraphrasing formatting):

Kolmogorov complexity [...] can be thought of as a lower bound on the algorithmic compressibility of a finite sequence (of characters or binary digits). It assigns to each such sequence $w$ a natural number $K(w)$ that, intuitively, measures the minimum length of a computer program (written in some fixed programming language) that takes no input and will output $w$ when run. Given a natural number $c$ and a sequence $w$, we say that $w$ is $c$-incompressible if $K(w) \geq |w| - c$.

An infinite sequence $S$ is Martin-Löf random if and only if there is a constant $c$ such that all of $S$'s finite prefixes are $c$-incompressible.

This is a test passed by $\sqrt{e}$ by setting $c$ a bit larger than the program to generate $\sqrt{e}$ and including in it the length to generate.

Answer 2

Surprisingly, the Kolmogorov complexity of some arbitrary number is known to be uncomputable. So the general form of your question, is an arbitrary number algorithmically compressible is undecidable (i.e. the problem of computing the algorithmic complexity of a sequence). This can be proven by reduction to the Halting problem.

Problems that are known to be related to "computable analysis" (the theory of computable reals) (cf. K. Weihrauch, Computable Analysis - An Introduction) are the following:

1. To effectively enumerate all digits of a real number, one needs an infinite time Turing machine
2. Distinguishing two Turing machines that compute two reals is uncomputable (i.e. the machine equivalence problem is undecidable)

A consequence of 1 and 2 is that the equality operator on Turing machines is not defined (at least in terms of a Choice Axiom).

In terms of undecidability, we are typically referring to Omega-like numbers (cf. Chatin's constant for which C. Calude et al. surprisingly provided "a" computation - Exact approximations of omega numbers).

Now, $e$ and $\pi$ are computable reals. The Borwein-Bailey-Plouffe formula can even compute the $n$-th digit of $\pi$.

Computability, compressibility and decidability theories refer to metamathematics. Compressibility refers to the program description of a sequence.

So for your example, $e$ would be described by 9 symbols ($\sum$,$n$,$=$,$0$,$\infty$,$\ldots$). We are assuming that the program is defined on some Universal Turing Machine, in this case, say, a symbolic algebra system. So in this case, $9 + O(1)$ symbols since the Universal Turing Machine is considered to be $O(1)$.

The problem of enumerating all digits of $e$ is a problem distinct from the problem of representing $e$ algorithmically.

Furthermore, in computer science, we only have pseudo-randomness since any random number generator that is fully described by an algorithm is by definition pseudo-random (in contrast to a hardware random number generator).

Answer 3

Your statement

Also note the fact that for transcendental numbers, and irrational numbers in general, the frequency of digits is uniform

is wrong. Look at Liouville's number. It is a transcendental number, containing almost all zeros, with a sparse amount of ones added (the nth 1 is at the n! decimal position). The distribution of digits to this number is completely one sided containing only 2 different digits, and one with a dominating frequency.