States and Heads Do Count for Unary Multi-head Finite Automata

Stateless One-Way Multi-Head Finite Automata With Pebbles

“…A first answer is already known by Example 1 and a result in [12]. Example 1 gives a stateless one-way two-head finite automaton with one pebble that accepts a non-semilinear unary language.…”

“…In [12] it is shown that, for any prime number m, there is a unary language accepted by some one-head m-state automaton that cannot be accepted by any (m − 1)-state automaton having an arbitrary number of heads.…”

“…In [12] it is shown that, for any prime number m, there is a unary regular language L accepted by some one-head m-state automaton that cannot be accepted by any (m−1)-state automaton having an arbitrary number of heads. In particular, language L is accepted by an m-state 1DHPA(1, 0) but is not accepted by any stateless 1DHPA(h, 0) and, thus, is not accepted by any stateless 1DHPA(1, p).…”

“…Example 1 gives a stateless one-way two-head finite automaton with one pebble that accepts a non-semilinear unary language. In [12] it is shown that, for any number h ≥ 1, every unary language accepted by a stateless one-way h-head finite automaton without pebbles is either finite or cofinite and, thus, semilinear. So, in this sense even an arbitrary number of heads cannot compensate for a pebble in general.…”

“…Despite the strong restriction to one state only, these devices are still very powerful which is demonstrated in [24] by showing that the emptiness problem is undecidable for stateless deterministic oneway three-head finite automata. The best result known so far for one-way devices with two heads is that automata with four states have an undecidable emptiness problem [12]. The two-way case is considered in [9], where the undecidability of the emptiness problem is obtained for stateless automata with two heads.…”