1. **Problem:** Do an example of finite automata. 2. A finite automaton is a simple machine used to recognize patterns in strings. A common formal description is $M=(Q,\Sigma,\delta,q_0,F)$, where $Q$ is the set of states, $\Sigma$ is the input alphabet, $\delta$ is the transition function, $q_0$ is the start state, and $F$ is the set of accepting states. 3. Example: build a finite automaton over the alphabet $\Sigma=\{0,1\}$ that accepts all strings ending in $1$. 4. Let the states be $Q=\{q_0,q_1\}$. Use $q_0$ as the start state and $F=\{q_1\}$ as the accepting state. 5. Define the transition rule: $$\delta(q_0,0)=q_0,\quad \delta(q_0,1)=q_1,\quad \delta(q_1,0)=q_0,\quad \delta(q_1,1)=q_1$$ 6. Why this works: the machine remembers only the last symbol seen. If the last symbol is $1$, it ends in the accepting state $q_1$. If the last symbol is $0$, it ends in the non-accepting state $q_0$. 7. Test a few strings: - For $101$, the machine moves $q_0 \to q_1 \to q_0 \to q_1$, so it accepts. - For $1100$, the machine moves $q_0 \to q_1 \to q_1 \to q_0 \to q_0$, so it rejects. - For $1$, the machine moves $q_0 \to q_1$, so it accepts. 8. Final answer: an example finite automaton for strings ending in $1$ is $M=(\{q_0,q_1\},\{0,1\},\delta,q_0,\{q_1\})$ with transitions $\delta(q_0,0)=q_0$, $\delta(q_0,1)=q_1$, $\delta(q_1,0)=q_0$, and $\delta(q_1,1)=q_1$.