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

Because $D$ is minimal, every state is accessible. Choose a word $u$ taking the initial state to $p$, and let the nonempty word $v$ label the closed path $\gamma$. If a word $w$ took $p$ to an accept state, thenThese words have strictly increasing lengths because $v\ne ϵ$, contradicting the assumption that $\mathcal{L}(D)$ is finite. Hence no accept state is reachable from $p$.

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 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.