Equality of exact and unique non-deterministic automatic complexity

Establish whether, for all finite words x over any finite alphabet Σ with |Σ| ≥ 2, the non-deterministic automatic complexity Ne(x)—defined as the minimum number of states in a non-deterministic finite automaton that exactly accepts x while rejecting every other word of the same length—equals Hyde’s unique non-deterministic automatic complexity N(x)—defined as the minimum number of states in a non-deterministic finite automaton that exactly accepts x via a unique accepting path and is unambiguous on Σ^{|x|}. If the equality fails, construct an explicit word x such that Ne(x) < N(x).

Background

The paper studies two related measures of automatic complexity for finite words: N(x), the unique non-deterministic automatic complexity introduced by Hyde, and Ne(x), a weaker variant defined by the authors that only requires exact acceptance but not uniqueness of the accepting path. By construction, Ne(x) ≤ N(x) for every word x.

Determining whether Ne(x) = N(x) for all words would clarify whether the requirement of unambiguity and uniqueness in Hyde’s definition strictly increases complexity relative to exact acceptance. A separation (Ne(x) < N(x)) would show that the computation path matters for non-deterministic automatic complexity; conversely, equality would simplify computation of N(x) by allowing proofs and algorithms to work with exact acceptance instead of unambiguous acceptance.