\begin{frame}{DFAs are Deterministic} Recall that $\delta$ is a function from $Q \times \Sigma$ to $Q$. \begin{goal}{} \emph{DFAs are deterministic:} \begin{itemize} \item [] For every state $q\in Q$ and every symbol $a \in \Sigma$, the state $q$ has \emph{precisely one outgoing arrow} with label $a$. \end{itemize} \end{goal} \pause\medskip Hence, for every input word, there is precisely one path from the starting state through the transition graph. \begin{exampleblock}{} The following picture shows the path for \alert{$aaba$}: \begin{center} \begin{tikzpicture}[default,node distance=20mm,->] \node (q0) [state] {$q_0$}; \draw ($(q0) + (-10mm,0mm)$) -- (q0); \node (q2) [state,right of=q0] {$q_2$}; \node (q4) [fstate,right of=q2] {$q_4$}; \begin{scope}[node distance=15mm] \node (q1) [state,above of=q2] {$q_1$}; \node (q3) [state,below of=q2,node distance=13mm] {$q_3$}; \end{scope} \draw [red] (q0) to node [label,above left] {$a$} (q1); \draw (q0) to[bend left=10] node [label,above right] {$b$} (q3); \draw [red] (q1) to[bend left=10] node [label,right] {$a$} (q2); \draw (q1) to node [label,above right] {$b$} (q4); \draw (q2) to[bend left=10] node [label,left] {$a$} (q1); \draw [red] (q2) to node [label,above] {$b$} (q4); \draw (q3) to[bend left=10] node [label,below left] {$a$} (q0); \draw (q3) to[bend left=10] node [label,above left] {$b$} (q4); \draw [red] (q4) to[bend left=10] node [label,below right] {$a$} (q3); \draw (q4) to[rloop] node [label,right] {$b$} (q4); \end{tikzpicture} \end{center} \end{exampleblock} \end{frame}