Abstract
If a computer is given access to an oracle—the characteristic function of a set whose membership relation may or may not be algorithmically calculable—this may dramatically affect its ability to compress information and to determine structure in strings, which might otherwise appear random. This leads to the basic question, ‘given an oracle A , how many oracles can compress information at most as well as A ?’ This question can be formalized using Kolmogorov complexity. We say that B ≤ LK A if there exists a constant c such that K A ( σ )< K B ( σ )+ c for all strings σ , where K X denotes the prefix-free Kolmogorov complexity relative to oracle X . The formal counterpart to the previous question now is, ‘what is the cardinality of the set of ≤ LK -predecessors of A ?’ We completely determine the number of oracles that compress at most as well as any given oracle A , by answering a question of Miller ( Notre Dame J. Formal Logic , 2010, 50 , 381–391), which also appears in Nies ( Computability and randomness . Oxford, UK: Oxford University Press, 2009. Problem 8.1.13); the class of ≤ LK -predecessors of a set A is countable if and only if Chaitin's halting probability Ω is Martin-Löf random relative to A .
Highlights
Kolmogorov complexity is a fundamental notion, which has found applications in topics as diverse as combinatorics, language recognition, information distance, thermodynamics and chaos theory
The basic idea behind this approach to quantifying the degree of randomness of a finite binary string is that a string is simple or non-random if it has a short description relative to its length
Programs can be identified with binary strings
Summary
Kolmogorov complexity is a fundamental notion, which has found applications in topics as diverse as combinatorics, language recognition, information distance, thermodynamics and chaos theory. The weight of a prefix-free set S of strings, denoted wgt(S ), is defined to be the sum s∈S 2−|s|.
Published Version (Free)
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have