Most programs stop quickly or never halt

Abstract

The aim of this paper is to provide a probabilistic, but non-quantum, analysis of the Halting Problem. Our approach is to have the probability space extend over both space and time and to consider the probability that a random N -bit program has halted by a random time. We postulate an a priori computable probability distribution on all possible runtimes and we prove that given an integer k > 0, we can effectively compute a time bound T such that the probability that an N -bit program will eventually halt given that it has not halted by T is smaller than 2 −k .We also show that the set of halting programs (which is computably enumerable, but not computable) can be written as a disjoint union of a computable set and a set of effectively vanishing probability.Finally, we show that "long" runtimes are effectively rare. More formally, the set of times at which an N -bit program can stop after the time 2 N + constant has effectively zero density.

Anytime Algorithms for Non-Ending Computations

“…The algorithmic complexity is similar to the complexities studied in [10,6,5,4,11]; the plain Kolmogorov complexity is about the logarithm of the algorithmic complexity. While the Kolmogorov complexity is optimal up to an additive constant, the optimality of Ò is up to a multiplicative constant.…”

Information: The Algorithmic Paradigm

“…A time t will be called algorithmically random if bin(t) is algorithmically random. In [12] one proves the following result:…”

“…Then, we continue by halving the length of the interval. For example, to guess the number 10 we need 8 questions (corresponding to i = 1, 2, 4, 8,16,12,9,10). In general, to guess the number n we have to ask the first 2 log n + 1 questions (here log n is the integer part of log 2 n, the base-2 logarithm of n).…”

“…In [12] a relation between incompleteness and uncertainty is established. To present it we will use the natural complexity ∇ = ∇ U induced by a universal self-delimiting Turing machine U ; recall that H = H U .…”

“…Of course, this is a formal approach and much more is required to check its "physical" base (see more in [12]). …”