= 27d567bc2d48d9f40e9ccc20ea370b15)
gRR comments on [SEQ RERUN] The Dilemma: Science or Bayes? - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (26)
Is it known what is the highest complexity, beyond which the Chaitin's Incompleteness applies? If it is relatively large, it is possible that all hypotheses interesting for humans have complexity lower than that...
In particular I wish to extract from that paper the following very simple (albeit non-constructive) proof of Chaitin's Incompleteness:
Given a (reasonable, sound) formal theory F, we know that F cannot prove all true sentences of the form "Program P never halts" (the reason is that if it could, we could solve the halting problem by searching over all possible proofs in F for the proof of P either halting with a particular run, or never halting, being sure our search will finish in finite time). Consider the shortest program P such that P never halts but F cannot prove that fact. Let L(P) be its length. Claim: F can never prove that Kolmogorov complexity of anything can be greater than L(P). Proof: given any output X, F can never refute the possibility that P might yet halt at some future time and output exactly X. Therefore L(P) must remain a candidate for the Kolmogorov complexity of X as far as F is concerned.
Edit: nevermind this. I've realized the proof is wrong. It's only true that "F can never refute the possibility that P might yet halt at some time and output some Y", but it is not true that "F can never refute the possibility that P might yet halt and output this specific X". It's conceivable (albeit unusual) that P doesn't halt, F is unable to prove that, but is able to prove that should P halt, its output will not be X. For example, think of P as a Turing machine with one halt state, which is easily "backwards-traceable" to a sequence of actions that erases the entire tape so far and writes out "123". Then F can easily be strong enough to be able to prove that if P halts at all, it outputs "123" and not anything else.
I emailed the article's author and he replied acknowledging the problem, which has been raised by a bunch of people before, and giving me links to a few paywalled articles with the correct exposition. However, this correct exposition is nowhere as succint and attractive as the short and faulty proof above.
I don't know much about it, but searching leads to this pretty interesting-looking paper which argues that the bound is largely incidental: http://www.mv.helsinki.fi/home/praatika/chaitinJPL.pdf
Thanks! The paper does give an answer, obvious in hindsight: the threshold constant for a formal system F is determined by the shortest Turing Machine that does not halt, but that fact is not provable in F.
Unfortunately, this turns out to be subtly wrong - see my update in a sibling comment.
That's a pity. Still, given any non-halting TM for which F cannot prove that, it is easy (costs very little additional constant complexity) to build a TM for which F also cannot prove that it does not return any x. And this still answers my original question (whether the threshold is very high) in the negative. So, bad for K-complexity, we need something better.