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The author is referring to something called Chaitin incompleteness. https://en.wikipedia.org/wiki/Kolmogorov_complexity#Chaitin's_incompleteness_theorem https:
by srcreigh 3y ago
The author is referring to something called Chaitin incompleteness.
https://en.wikipedia.org/wiki/Kolmogorov_complexity#Chaitin's_incompleteness_theorem https://en.wikipedia.org/wiki/Kolmogorov_complexity#Chaitin'...
Of course trivially some KC can be proven, ex a language with 1 or 0 characters that is interpreted to a specific string. Or to prove KC(x) where the compressed value has length N and you can list out all the results for all strings of length less than N, and they don't equal x, proves KC(x)=N.
The interesting number paradox (Berry's paradox) is more related to Chaitin incompleteness.
Basically, given a language there’s some code which enumerates proofs that KC of a string is more than some constant L, and returns the first one it finds.
If the constant L is large enough, it becomes larger than the entire proof generating code. So the proof generating code will never find a proof of any KC larger than L.
It's interesting to think about that the language gets more complex, proofs for larger strings become possible. And what it would mean for the languages to keep getting more complex indefinitely.
it's a similar train of thought to busy beaver numbers and how systems of logic (PA,ZFC) become independent to values like BB(745), and what it could mean to have more and more advanced types of logic which don't become independent until some high target n.
- causal 3y agoThis seems to assume that KC can be infinite. That must have been proven at some point? Otherwise it may be that there is some upper-bound for L which happens to also be the KC for a KC-computer.
- srcreigh 3y agoyes, it’s a pigeonhole principle argument. A bit tricky to actually enumerate everything. Imagine KC had a limit k. Then there’s a fixed number of strings that can be compressed to k characters. so considering how there’s an infinite number of strings of length greater than k, they can’t all be compressed to be at most k. Therefore KC has no limit
- tromp 3y agoKC(x) is always finite for any finite x. What we can say instead is that KC is unbounded. I.e. there is no finite bound on the value of KC(x).