We show that the asymptotic linear complexity of a multisequence a ∈ F M q ∞ that is I := lim inf n→∞ L a (n) n and S := lim sup n→∞ L a (n) n satisfies the inequalities M M + 1 S 1 and M(1 − S) I 1 − S M , if all M sequences have nonzero discrepancy infinitely often, and all pairs (I, S) satisfying these conditions are met by 2 ℵ 0 multisequences a. This answers an Open Problem by Dai et al.
We introduce rational complexity, a new complexity measure for binary sequences. The sequence s ∈ Bω is considered as binary expansion of a real fraction $s \equiv {\sum }_{k\in \mathbb {N}}s_{k}2^{-k}\in [0,1] \subset \mathbb {R}$
s
≡
∑
k
∈
ℕ
s
k
2
−
k
∈
[
0
,
1
]
⊂
ℝ
. We compute its continued fraction expansion (CFE) by the Binary CFE Algorithm, a bitwise approximation of s by binary search in the encoding space of partial denominators, obtaining rational approximations r of s with r → s. We introduce Feedback in$\mathbb {Q}$
ℚ
Shift Registers (F$\mathbb {Q}$
ℚ
SRs) as the analogue of Linear Feedback Shift Registers (LFSRs) for the linear complexity L, and Feedback with Carry Shift Registers (FCSRs) for the 2-adic complexity A. We show that there is a substantial subset of prefixes with “typical” linear and 2-adic complexities, around n/2, but low rational complexity. Thus the three complexities sort out different sequences as non-random.
Harvey Friedman gives a comparatively short description of an "unimaginably large" number n(3) , beyond e.g. the values A(7, 184) < A(7198, 158386) < n(3) of Ackermann's function -but finite.We implement Friedman's combinatorial problem about subwords of words over a 3-letter alphabet on a family of Turing machines, which, starting on empty tape, run (more than) n(3) steps, and then halt. Examples include a (44,8) (symbol,state count) machine as well as a (276,2) and a (2,1840) one.In total, there are at most 37022 non-trivial pairs (n, m) with Busy Beaver values BB(n, m) < A(7198, 158386).We give algorithms to map any (|Q|, |E|) TM to another, where we can choose freely eitherGiven the size of n(3) and the fact that these TMs are not holdouts, but assured to stop, Friedman's combinatorial problem provides a definite upper bound on what might ever be possible to achieve in the Busy Beaver contest. We also treat n(4) > A (A( 187196)) (1).
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