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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 21}
\date{September 16, 2009}
\begin{document}
\begin{frame}
  \maketitle
\end{frame}

\begin{frame}
  \frametitle{Merge sort}

  \begin{itemize}
  \item In the last class we saw quick sort.
  \item This class we will see another algorithm called merge sort.
  \end{itemize}

\end{frame}

\begin{frame}
  \frametitle{Merging sorted lists}
    
    Given two sorted lists $\{a_1,\ldots, a_n\}$ and $\{b_1,\ldots,
    b_m\}$ how fast can we merge ?

    \pause
    \begin{code}
void mergearrays(int a[], int b[], int c[], int n, int m)
{
  int i=0,j=0,k=0;
  
  while( i < n && j < m)
  {
    if(a[i] < b[j]) { c[k] = a[i]; i++}
    else if(a[i] >= b[j]){ c[k] = b[j]; j++}
    k++;
  }
  
  while(i < n){ c[k] = a[i]; i++; k++;}
  while(j < m){ c[k] = b[j]; j++; k++;}
}
    \end{code}
\end{frame}

\begin{frame}
  \frametitle{Main idea}
  \begin{itemize}
  \item If the array is of size 1 then it is already sorted.
  \item Otherwise divide the array into two equal parts say $A$ and $B$
  \item Sort $A$ and $B$ recursively.
  \item Merge $A$ and $B$ to get the sorted list.
  \end{itemize}

  \pause
  \begin{block}{How many comparisons}

  \[
  \T{n} = \T{\frac{n}{2}} + \T{\frac{n}{2}} + n = 2 \T{\frac{n}{2}} + n.
  \]
  \end{block}
 
\end{frame}

\begin{frame}
  \frametitle{Unrolling the recursion}
  \begin{eqnarray*}
    \T{n} & = & 2 \T{\frac{n}{2}} + n \\
          & = & 2 \left(2 \T{\frac{n}{4}} + \frac{n}{2}\right) + n \\
          & = & 2^2 \T{ \frac{n}{2^2} } + (n + n)\\
    \ldots & \ldots & \ldots \\
          & = & 2^k \T{\frac{n}{2^k}} + \overbrace{n + \ldots + n}^{k\textrm{ times}}
  \end{eqnarray*}
  Choosing $k = \log{n}$ we have 
  \[ \T{n} = 2^{\log{n}} \T{1} + n \log{n} = \BigOh{n \log{n}}.\]
\end{frame}

\begin{frame}

  \begin{tikzpicture}
    \frametitle{Merging array slices}
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\end{frame}

\begin{frame}
   \frametitle{Merging array slices in |C|}
  \begin{code}

int merge(int *src, int mid, int n, int *dest)
{
  int i=0, j = 0, k = 0;

  int *a = src;
  int *b = src + mid;
  while( i < mid && j < n - mid)
  {
    if(a[i] < b[j]) { dest[k] = a[i]; i++}
    else if(a[i] >= b[j]){ dest[k] = b[j]; j++}
    k++;
  }
  while(i < mid)      { dest[k] = a[i]; i++; k++;}
  while(j < n - mid)  { dest[k] = b[j]; j++; k++;}

}

    \end{code}

\end{frame}


\begin{frame}
  \frametitle{Merge-sorting array using an auxilary array}
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\end{frame}

\begin{frame}

  \frametitle{Merge sorting in |C|}
  \begin{code}

void mergeSort( int *ptr, int n, int *aux)
{
  if(n <= 1) return;
  mid = n /2;
  copy(ptr, mid, aux);
  copy(ptr + mid, n-mid, aux + mid);
  mergeSort(aux,mid, ptr);
  mergeSort(aux + mid, n - mid, ptr + mid);
  merge(aux, mid, n, a);
}

\end{code}

\end{frame}

\end{document}


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