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\author{Piyush P Kurur\\
  Office no: 224\\
  Dept. of Comp. Sci. and Engg.\\
  IIT Kanpur}


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\title{Fundamentals of Computing: Lecture 12}
\date{August 24, 2009}
\begin{document}
\begin{frame}
  \maketitle
\end{frame}

\begin{frame}
  \frametitle{Summary of the last class}
  \begin{spec}
    while( C )
    { /* invariant is a condition that is true here */
      S
    }
  \end{spec}
  
  \begin{block}{How to design loop?}
    \begin{itemize}
    \item Write down the desired outcome $\varphi$.
      \pause
    \item Choose a loop invariant $I$ and a loop condition $C$ 
      such that $I \land \lnot C \Rightarrow \varphi$.
      \pause
    \item $I$ is obtained by parameterisation (i.e. replacing
      constants of $\varphi$ by variables)
      \pause
    \item The condition $C$ is the negation of parameterisation.
      \pause
    \item Initialise variables so that $I$ is true in the base case
      \pause
    \item Design the body $S$ to preserve the validity of $I$.
    \end{itemize}
      
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{Finding the smallest in a sequence of $n$ numbers}

  The desired condition is 
  \[ \varphi \equiv  s = \mathrm{min}\{ a[0],\ldots, a[n-1]\}
  \textrm{ (n is a constant here).} \] 
  \pause
  Parameterising we get the invariant
    \[ I_i \equiv s = \mathrm{min}\{ a[0],\ldots, a[i-1]\}. \]
  \pause
  Note that 
  \[ I_i \land (i = {n}) \Rightarrow \varphi. \]
  So the condition C is $i \neq n$.

  Hence the loop.

  \begin{spec}
    s = a[0];
    i = 1;  
    while( i != n )
    {
      if( s > a[i] ) s = a[i];
      i++;
    }
  \end{spec}
\end{frame}



\newcommand{\Length}[1]{\mathrm{length}\left(#1\right)}
\newcommand{\SortedArray}[1]{\mathrm{SortedArray}\left(#1\right)}
\newcommand{\Sorted}[2]{\mathrm{Sorted}\left(#1,#2\right)}
\begin{frame}
  \frametitle{Sorting}
  Let us define what is sorted array 
  \[ 
  \SortedArray{a} \equiv \forall i\ 0 \leq i < \Length{a} - 1 \Rightarrow a[i] \leq a[i+1].
  \]
  \pause
    \[
  \Sorted a s \equiv \forall i \  0 \leq i < s \Rightarrow a[i] \leq a[i+1]
  \]
\end{frame}

\newcommand{\SortedP}[3]{\mathrm{Sorted}'(#1,#2,#3)}

\begin{frame}
  Choose the invariant $\Sorted{a}{i}$ for a parameter $i$.
  
  \begin{spec}
    i = 0;
    while( i != n)
    {
      S; /* Do something to restore invariant */
      i++;
    }
  \end{spec}

  \pause
  $S$ itself is a loop statement.

  \[ 
  \SortedP a  r k \equiv \forall j\  0 \leq j < r  \ a[j] \leq a[j+1] )
  \lor  j = k - 1
  \]
  
  \begin{spec}
    k = i+1;
    while( k != 0 )
    {
      if( a[k-1] > a[k] ) /* swap a[k-1] and a[k] */
      k--;
    }
  \end{spec}


\end{frame}
\end{document}