Looking for Pairs that Hard to Separate: A Quantum Approach

Abstract

Abstract. Determining the minimum number of states required by a deterministic finite automaton to separate a given pair of different words (to accept one word and to reject the other) is an important challenge. In this paper, we ask the same question for quantum finite automata (QFAs). We classify such pairs as easy and hard ones. We show that 2-state QFAs with real amplitudes can separate any easy pair with zero-error but cannot separate some hard pairs even in nondeterministic acceptance mode. When using complex amplitudes, 2-state QFAs can separate any pair in nondeterministic acceptance mode, and here we conjecture that they can separate any pair also with zero-error. Then, we focus on (a more general problem) separating a pair of two disjoint finite set of words. We show that QFAs can separate them efficiently in nondeterministic acceptance mode, i.e., the number of states is two to the power of the size of the small set.

New Results on Vector and Homing Vector Automata

“…In this section we investigate the string separation problem for vector and homing vector automata. Recently, the same question has been investigated in [1,2] for different models such as probabilistic, quantum, and affine finite automata (respectively, PFA, QFA, AfA). (We refer the reader to [6,24] for details of these models.)…”

“…where the first entry is set to v 0 [1] and then the first block is set to v 0 since V starts its computation in q 1 , and, all other blocks are set to zeros.…”

“…Proof. Let w ∈ L and suppose that the string w = w [1] w [2] · · · w [n] is accepted by H. Let A 1 A 2 · · · A n be the product of the matrices labeling the computation such that…”

“…where v 0 is the initial vector of H. Since the matrices are commutative, then for any permutation τ , This leads to the acceptance of the string w = w [τ (1)…”

“…where the first entry is set to v 0 [1] and then the first block is set to v 0 since V starts its computation in q 1 , and, all other blocks are set to zeros. Since V is stateless, it has a single (1+nk)×(1+nk) matrix for every (σ, τ ) pair.…”