Novikovian Computation (2/2)

(continued from previous post)

Let's simplify and generalize: if the communications channel c has `b` bits, and there are `a` non-paradoxical answers, then (absent NSP) the probability of receiving a valid answer, `P(A_v)`, is `(1-epsilon) a/2^b`; the probability of an invalid answer, `P(A_{stackrel ~ v})`, is `(1-epsilon) {:(2^b - a):}/2^b`, and the probability of external failure, `P(_|_)`, is still `epsilon`. Note that, as required, `P(A_v) + P(A_{stackrel ~ v}) + P(_|_) = 1`, as these three exclusive possibilities make up the entirety of the event space.

In the presence of the NSP, however, the probability function `P_N(*)` is different: `P_N(A_{stackrel ~ v}) = 0`. For convenience, let `bar{:A_{stackrel ~ v}:} = A_v uu _|_` be denoted by `C` (for consistent).

The NSP essentially gives us that, for any event-set `E sube C`, `P_N(E) = P(E|C)`. We can therefore compute `P_N(E)` by Bayes' theorem: `P_N(E) = P(E|C) = {:P(C|E)P(E):}/{:P(C):}`. Of course, `P(C|E) = 1`; so after some mild algebra, we have`P_N(A_v) = {:a (1-epsilon):}/{:a (1-epsilon) + 2^b epsilon:}`, `P_N(_|_) = {:2^b epsilon:}/{:a (1-epsilon) + 2^b epsilon:}`.
If we have a reasonably reliable system, such that `epsilon ≪ 1`, these will basically be `{:a:}/{:a + 2^b epsilon:}` and `{:2^b epsilon:}/{:a + 2^b epsilon:}` respectively. Note that the probability of failure goes up exponentially with the number of bits being communicated and tested (assuming `a` remains constant as `b` increases). For Novikovian computation to be useful, we must have `a > 2^b epsilon`. (Recall that `a` is a small integer, probably on the order of 1 or 2, and definitely bounded above by `2^b`.)

Now suppose multiple time-loop-logic computational apparatuses exist (say, `n` of them), and are all executing problems concurrently. These are not discrete systems, because the probability that two apparatuses `c_1` and `c_2` will fail is not quite independent: for example, a relatively nearby star could suddenly go Type Ia supernova, flooding the solar system with ionizing radiation and thereby destroying all the apparatuses (and quite incidentally all life on Earth).

For the single-system case, divide `_|_` into `_|_ _ L` (local failure, independent) and `_|_ _ G` (global failure), with NSP-less probabilities `epsilon_L` and `epsilon_G` respectively (`epsilon_L + epsilon_G > epsilon`, since there is some overlap). Then, our consistent set `C` is `_|_ _ G + prod_{:i=0:}^n(_|_ _ {:L_i:} + A_{:v_i:})`, so we have`{:(P(C),=,epsilon _ G + (1-epsilon_G)prod_{:i=0:}^n(epsilon_{:L_i:} + (1-epsilon_{:L_i:})a/2^b)),(,=,epsilon _ G + (1-epsilon_G)1/2^{:bn:}prod_{:i=0:}^n(2^b epsilon_{:L_i:} + (1-epsilon_{:L_i:})a)):}`and therefore`P_N(_|_ _ G) = P(_|_ _ G|C) = epsilon _ G / {: epsilon _ G + (1-epsilon_G)1/2^{:bn:}prod_{:i=0:}^n(2^b epsilon_{:L_i:} + (1-epsilon_{:L_i:})a) :}`.The term on the right of the numerator is bounded above by `1/2^{:bn:}prod_{:i=0:}^n(2^b epsilon_{:L_i:} + a)`, which in the case of useful computation is itself bounded above by `1/2^{:bn:}prod_{:i=0:}^n(a + a) = 2a^n/2^{:bn:} = 2(a/2^b)^n`. This clearly goes to zero in the limit as `n` increases.

Therefore we have `lim_{:n rarr infty:} P_N(_|_ _ G) = epsilon _ G / epsilon _ G = 1`, or, in plain English:

Theorem. As the number of useful Novikovian computers in the world increases, the probability of apocalypse nears certainty.

See also the Fermi paradox, and excuse me while I go make an eschaton tag.

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Edit 2009-03-13: Fixed typo in Bayesian restatement; adjusted formatting; eliminated references to `F`.