On Monday, February 03, 2020, at 3:32 PM, Matt Mahoney wrote: > Less than 8.7 x 10^244 bits. That's the square of the Bekenstein bound of a > black hole with a Schwartzchild radius equal to the Hubble radius,13.8 > billion light years. Edit distance expressed as the shortest program that > inputs one string and outputs the other is equal to the size of the output > for the vast majority of random string pairs.
I would think bigger than that? This would be the set of all the combinatorial subsets with order, permutations without repetition, summing the edit distances from every element to every other element. So, 1 particular bit's edit distance to the full universe string would be something like 10^124 bits but do this for every bit in the universe and for every n bit ordered sequence to every other n bit ordered sequence. It's immense. Why does it matter? It's a measurement of the magnitude, and there might be better ones, of the edit "separation" of everything. Most of the stuff is physically non-computable so you have to take the max. Effectively though it could be compressed way down to the KC of the universe I would guess... Well, that is if KC is some sort of minimization of edit separation or Levenshtein separation which it has to be? Speaking colloquialliy here, just speculating :) ------------------------------------------ Artificial General Intelligence List: AGI Permalink: https://agi.topicbox.com/groups/agi/T65747f0622d5047f-M2483f5f88d7bbc1ea4729af9 Delivery options: https://agi.topicbox.com/groups/agi/subscription
