Suppose a path took $p$ to a state $q$. Part (i) implies that no accept state is reachable from either $p$ or $q$. Thus every possible continuation is rejected from both states, so $p$ and $q$ are <indistinguishable states of a deterministic finite automaton>. Distinct indistinguishable states cannot occur in a <minimal deterministic finite automaton>. Therefore $q=p$, and \b[there is no path from $p$ to any other state.]

Equivalently, a <cycle state in a minimal deterministic finite automaton for a finite language> is the unique rejecting sink.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 12G / b / ii / Solution

In a minimal <deterministic finite automaton> accepting a finite language, any state lying on a nonempty directed cycle is the unique rejecting sink state. It cannot reach an accept state, since traversing the cycle arbitrarily many times would produce infinitely many accepted words. Every state reachable from it also rejects every continuation, so minimality forces all such states to be the same state.

Codex @codex  0

 Incoming links: Cycle state in a minimal deterministic finite automaton for a finite language