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2021 | OriginalPaper | Chapter

State Complexity of the Set of Synchronizing Words for Circular Automata and Automata over Binary Alphabets

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Abstract

Most slowly synchronizing automata over binary alphabets are circular, i.e., containing a letter permuting the states in a single cycle, and their set of synchronizing words has maximal state complexity, which also implies complete reachability. Here, we take a closer look at generalized circular and completely reachable automata. We derive that over a binary alphabet every completely reachable automaton must be circular, a consequence of a structural result stating that completely reachable automata over strictly less letters than states always contain permutational letters. We state sufficient conditions for the state complexity of the set of synchronizing words of a generalized circular automaton to be maximal. We apply our main criteria to the family \(\mathscr {K}_n\) of automata that was previously only conjectured to have this property.

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Footnotes
1
The circular automata are a proper subfamily of the generalized circular automata, as shown by \(\mathscr {A} = (\{a,b\}, [3], \delta )\) with \(\delta (0, a) = 1, \delta (1, a) = 0, \delta (2,a) = 2\) and \(\delta (0, b) = 0, \delta (1, b) = 2, \delta (2, b) = 1\). The word ba cyclically permutes the states.
 
2
For, if we choose a finite number of words and build the automaton by identifying these words with new letters, distinguishability or reachability of states (or subsets of states) of this new automaton is inherited to the original automaton. Hence, all results are also valid when stated with words instead of letters, but otherwise the same conditions.
 
3
Note that \(0< m < n\) implies \(\delta (q, b^m) \ne q\). Also note that we added n on the right hand side to account for values \(d > 1\). In Proposition 10 we only subtract one from the exponent of b, which is always non-zero and strictly smaller than n, and so we do not needed this “correction for the b-cycle” in case of resulting negative exponents.
 
4
Note that here, even if the bounds for m from Theorem 8 do not include this case, \(\delta (q, w) = \delta (q, b^nw) = \delta (q, wb^{n-1})\), which is equivalent with \(\delta (q, w) = \delta (q, b)\).
 
5
I slightly changed the numbering of the states with respect to the action of the letter a compared to [13].
 
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Metadata
Title
State Complexity of the Set of Synchronizing Words for Circular Automata and Automata over Binary Alphabets
Author
Stefan Hoffmann
Copyright Year
2021
DOI
https://doi.org/10.1007/978-3-030-68195-1_25

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