\begin{frame}{The Halting Problem is Undecidable}\onslide<11>{} \vspace{-.5ex} \begin{goal}{}\vspace{-.5ex} Assume there would be a program $T$ with the behaviour: \begin{itemize}\setlength{\itemsep}{0pt} \item input: a program $M$ \item output: \textit{yes} if $M$ terminates on input $M$, \textit{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},scale=.95,nodes={scale=.95}] \node (p) [p] {p = read input;\\[-.5ex]\hspace{1.1cm}\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 $T$} ($(p.north west) + (-2mm,0)$); \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 $T'$} ($(p'.south east) + (2mm,0)$); } \end{tikzpicture} \end{center}\vspace{-1.2ex} \pause\pause \begin{goal}{}\vspace{-.5ex} What happens if we run $T'$ with input $T'$? \begin{itemize}\setlength{\itemsep}{0pt} \pause \item initial part $T$ decides whether $T'$ terminates on input $T'$ \pause \item if the result is \alert{yes}, then $T'$ runs forever\pause{} $\,$ \alert{Contradiction} \pause \item if the result is \alert{no}, then $T'$ terminates\pause{} $\,$ \alert{Contradiction} \end{itemize} \end{goal} \pause \vspace{-.5ex} \begin{alertblock}{} Thus $T$ cannot exist!\pause{} The halting problem is undecidable! \end{alertblock} \vspace{10cm} \end{frame}