A370489 Deterministic complexity of binary strings, in shortlex order.

1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 4, 3, 3, 3, 2, 3, 2, 2, 3, 2, 3, 3, 3, 4, 1, 1, 5, 4, 4, 3, 3, 4, 3, 3, 3, 2, 4, 4, 3, 4, 2, 2, 4, 3, 4, 4, 2, 3, 3, 3, 4, 3, 3, 4, 4, 5, 1, 1, 6, 5, 5, 4, 4, 5, 4, 4, 3, 3, 5, 4, 4, 5, 3, 3, 4, 3, 5, 5, 2

Offset: 1

Author

Lucas B. Vieira, Mar 02 2024

Keywords

• nonn
• tabf

Comments

a(n) = DC(w[n]), where w[n] is the n-th binary string in shortlex order: 0, 1, 00, 01, 10, 11, 000, 001, .... (row n-1 of A076478).

The Deterministic Complexity (DC) of a string w is the minimum number of states in a deterministic Mealy automaton outputting the string w. Alternatively, DC(w) = |u|+|v| for the smallest (possibly empty) u and primitive word v, such that w = trunc(uv^k,|w|) for some k >= 1, where trunc(x,L) truncates the word x to length L.

1..L each appear at least twice in row L since DC(0^i 1^(L-i)) = DC(1^i 0^(L-i)) = i+1 for i = 0..L-1. - Michael S. Branicky, Mar 03 2024

Examples

			For n = 1, w[1] = 0, and DC(w[1]) = 1. The minimal deterministic Mealy automaton outputting w[1] has a single state, transitioning to itself and outputting 0.
For n = 36, w[36] = 00101, and DC(w[36]) = 3. The minimal deterministic Mealy automaton has 3 states: transitions 1->2 and 2->3 output 0, and 3->2 outputs 1.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Lucas B. Vieira and Costantino Budroni, Temporal correlations in the simplest measurement sequences, Quantum 6 p. 623 (2022).

Crossrefs

Cf. A059412, A076478, A370821.

Programs

Mathematica

(* e.g. DC[{0,0,1,0,1}] = 3 *)
DC[seq_] := With[{L = Length[seq]}, Do[If[With[{c = dc - t}, Take[Flatten[{Take[seq, t], ConstantArray[Take[Drop[seq, t], c], Ceiling[(L - t)/c]]}], L]] == seq, Return[dc]], {dc, 0, L}, {t, 0, dc - 1}]];
a[n_] := DC[Rest[IntegerDigits[n + 1, 2]]];
(* Table for all sequences up to length 10 *)
With[{L = 10}, Table[a[n], {n, 2*(2^L-1)}]]

Python

def DC(w): return next(dc for dc in range(1, len(w)+1) for i in range(dc) if w == (w[:i]+w[i:dc]*(1+(len(w)-i)//(dc-i)))[:len(w)])
def a(n): return DC(bin(n+1)[3:])
print([a(n) for n in range(1, 88)]) # Michael S. Branicky, Mar 03 2024