Asymptotic growth of a non-computable sequence
Dennis E. Hamilton
dennis.hamilton at acm.org
Thu Aug 12 18:55:16 EDT 2021
Cabellero wrote:
> Kreinovich, Vladik wrote:
> > Since a(n) has n binary digits, it is between 2^n and 2^{n+1} = 2*2^n, so
this is a clear asymptotic
> where a(n) = the smallest number having n binary digits, which can't be
> compressed concerning a fixed Turing machine
> I suspect that a(n)/2^n converges to 1 as n goes to infinity.
[orcmid responds]
The TL:DR: There is something more meant about the fixed Turing machine and
what it means to have uncompressible cases. That needs to be expressed.
A thought experiment,
Not to be overly fussy, we can presume that the leading bit is always 1 (to
avoid trivial compressibility) and in that case, the range of candidate binary
numbers is from 2^(n-1) to 2^n-1 and there are 2^(n-1) of them. Cabellero's
suspicion can then be stated that as n approaches infinity, a(n)/2^(n-1)
approaches 1. In effect, the proportion compressible by any fixed TM
approaches 0.
We are left to deal with what it means to be uncompressible. Maybe
information theory shows us enough about that without consideration of Turing
machines at all.
Presume, first, that lossless compression is intended: From the number
representing a compressed value, we can recover the original.
I contrived an easy way to do that by assuming the a(n) candidates to all have
the first of the n bits being 1. Suppose then that the number corresponding
to a compressed value first bit of 0 and fewer than n of them. If we allow
exactly n all we have to do is flip the leading bit to 0 or even flip all the
bits. We know how to decompress all of those with a trivial algorithm. But
there's no compression and we have one wasted bit, the leading 1/0.
If there must be fewer than n bits in the compression, there are at most n-1
of them, including the leading 0 bit, and that means there are fewer than
2^(n-2) of them. So, without trying, we know as many as half of the originals
cannot be compressed by any method that is lossless and for which the
compression is shorter than the original. There's probably a tighter proof
by induction. We don't need that to see the bind.
We are now up against a well-known characteristic of compression algorithms.
The more successful the compression is for many cases, there must also be
cases where the compression result is worse -- has more bits than the
original. In practical schemes, the cost of expanded cases is minimized (like
that fixed initial 1 bit taken to mean not compressed). And in practical
schemes, there is something that is known about the nature of the strings of
bits that has the algorithm provide successful compressions in practice, with
pessimistic cases producing tolerably-larger results.
I suspect that the fixed Turing machine case is around a different question,
albeit constrained in at least the manner I have sketched.
- Dennis
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