\begin{frame}{\alt<1-11>{Termination}{\alert{Termination is undecidable!}\onslide<12>{}}} \vspace{-.5ex} \begin{goal}{} Assume there would be a \emph{program $\terminatoronitself$}\; with the behaviour: \vspace{-.5ex} \begin{itemize}\setlength{\itemsep}{0pt} \item input: a program \forestgreen{$P$} \item output: \emph{yes} if \forestgreen{$P$} terminates on input \forestgreen{$P$}, \emph{no} otherwise \end{itemize} \end{goal} \pause \begin{center}\vspace{-.25ex} \begin{tikzpicture}[p/.style={rectangle,minimum width=37mm,fill=yellow!50!orange!20,draw=black,dashed,rounded corners=2mm,inner sep=2mm,align=left}] \node (p) [p] {p = read input;\\[-.5ex]\hspace{1.1cm}\scalebox{.8}{\vdots}\\[-.5ex]print result;}; \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] ($(p.south west) + (-2mm,0)$) to node [left,xshift=-3mm] {program $\terminatoronitself$} ($(p.north west) + (-2mm,0)$); \mpause[1]{ \node (p') at (p.south) [p,anchor=north] {\dm{if} result = yes\\\dm{then}\;\, loop forever\\\dm{else}\;\; terminate}; \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] ($(p.north east) + (2mm,0)$) to node [right,xshift=3mm] {program $\diagonalisator$} ($(p'.south east) + (2mm,0)$); } \end{tikzpicture} \end{center}\vspace{-1.2ex} \pause\pause \begin{goal}{} What happens if we run $\diagonalisator$ with input $\diagonalisator$?\vspace{-.5ex} \begin{itemize}\setlength{\itemsep}{0pt} \pause \item initial part $\terminatoronitself$\, decides whether $\diagonalisator$ terminates on input $\diagonalisator$ \pause \item if the result is \alert{yes}, then $\diagonalisator$ runs forever\pause{} $\,$ \alert{$\xmark$} \pause \item if the result is \alert{no}, then $\diagonalisator$ terminates\pause{} $\,$ \alert{$\xmark$} \end{itemize} \end{goal} \pause \vspace{-.5ex} \begin{alertblock}{} Thus $\terminatoronitself$ has made a mistake!\pause{} And thus also $\terminator\,$! \pause \end{alertblock} \vspace{10cm} \end{frame}