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\begin{document}

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\begin{frame}[t]

\frametitle{%

  \begin{tabular}{@ {}c@ {}}

  \\

  \LARGE Security Engineering (9)\\[-3mm] 

  \end{tabular}}\bigskip\bigskip\bigskip

  \normalsize

  \begin{center}

  \begin{tabular}{ll}

  Email:  & christian.urban at kcl.ac.uk\\

  Office: & S1.27 (1st floor Strand Building)\\

  Slides: & KEATS (also homework is there)\\

  \end{tabular}

  \end{center}

\end{frame}

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  \begin{frame}[c]

  \begin{center}

  \includegraphics[scale=0.6]{../pics/bridge-limits.png}

  \end{center}

  \end{frame}

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  \begin{frame}[c]

  \frametitle{Old-Fashioned Eng.~vs.~CS}

  \begin{center}

  \begin{tabular}{@{}cc@{}}

  \begin{tabular}{@{}p{5.2cm}} 

  \includegraphics[scale=0.058]{../pics/towerbridge.jpg}\\ 

  {\bf bridges}: \\

  \raggedright\small

  engineers can ``look'' at a bridge and have a pretty good

  intuition about whether it will hold up or not\\ 

  (redundancy; predictive theory)

  \end{tabular} &

  \begin{tabular}{p{5cm}} 

  \includegraphics[scale=0.265]{../pics/code.jpg}\\

  \raggedright

  {\bf code}: \\

  \raggedright\small

  programmers have very little intuition about their code; 

  often it is too expensive to have redundancy;

  not ``continuous'' 

  \end{tabular}

  \end{tabular}

  \end{center}

  \end{frame}

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  \begin{frame}[c]

  \frametitle{Dijkstra on Testing}

  \begin{bubble}[10cm]

  ``Program testing can be a very effective way to show the

  presence of bugs, but it is hopelessly inadequate for showing

  their absence.''

  \end{bubble}\bigskip

  unfortunately attackers exploit bugs (Satan's computer vs 

  Murphy's)

  \vfill

  \small

  Dijkstra: shortest path algorithm, 

  dining philosophers problem,

  semaphores

  \end{frame}

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\begin{frame}[c]

\frametitle{\Large Proving Programs to be Correct}

\begin{bubble}[10cm]

\small

{\bf Theorem:} There are infinitely many prime 

numbers.\medskip\\

{\bf Proof} \ldots\\

\end{bubble}\bigskip

similarly\bigskip

\begin{bubble}[10cm]

\small

{\bf Theorem:} The program is doing what 

it is sup+ed to be doing.\medskip

{\bf Long, long proof} \ldots\\

\end{bubble}\bigskip\medskip

\small This can be a gigantic proof. The only hope is to have

help from the computer. `Program' is here to be understood to be

quite general (protocol, OS,\ldots).

\end{frame}

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  \begin{frame}[c]

  \frametitle{\Large{}Mars Pathfinder Mission 1997}

  \begin{center}

  \includegraphics[scale=0.15]{../pics/marspath1.png}

  \includegraphics[scale=0.16]{../pics/marspath3.png}

  \includegraphics[scale=0.3]{../pics/marsrover.png}

  \end{center}

  \begin{itemize}

  \item despite NASA's famous testing procedures, the lander crashed frequently on Mars

  \item a scheduling algorithm was not used in the OS

  \end{itemize}

  \end{frame}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  \begin{frame}[c]

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  \end{tabular}

  \end{textblock}

  \begin{textblock}{0}(2.5,13)%

  \small

  \onslide<3->{

  \begin{bubble}[8cm]%

  Scheduling: You want to avoid that a high 

  priority process is staved indefinitely.

  \end{bubble}}

  \end{textblock}

  \end{frame}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  \begin{frame}[c]

  \frametitle{\Large Priority Inheritance Scheduling}

  \begin{itemize}

  \item Let a low priority process $L$ temporarily inherit 

    the high priority of $H$ until $L$ leaves the critical 

    section unlocking the resource.\bigskip

  \item Once the resource is unlocked $L$ returns to its original 

    priority level.

  \end{itemize}

  \end{frame}

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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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    \node at (1.5,0.9) {\small $A_L$};

    \node at (2.0,0.9) {\small $B_L$};

    \node at (3.5,3.9) {\small $A_R$};

    \node at (6.7,0.9) {\small $B_R$};

    \draw[very thick,blue!100] (1.5,0) -- (1.5,0.6);

    \draw[very thick,blue!100] (2.0,0) -- (2.0,0.6);

    \draw[very thick,blue!100] (3.5,3) -- (3.5,3.6);

    \draw[very thick,blue!100] (6.7,0) -- (6.7,0.6);

    \draw[fill,blue!100] (3.5,3) rectangle (4.5,3.6);

    \node at (4,3.3) {\small $A$};

    \draw[very thick,blue!100] (7,3) rectangle (8,3.6);

    \node at (7.5,3.3) {\small $B$};