How to scale algorithm to fit in one frame and centered in the frame at same time?

Question

Originally, the algorithm is too long to fit in the frame. If I use the \scalebox{0.8}{}, it can fit in the frame but doesn't look like to be centered in the frame. Here is the code and output. First picture is for the original code, second is for the code with scalebox. Are there any methods to make the the algorithm to both fit in and be centered in the frame. (And I think the hat of A in the last line is ugly.)

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}
\begin{frame}{Hello world}
  \begin{center}
    \begin{minipage}{0.8\linewidth}
    %\scalebox{0.8}{ 
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
      %}
    \end{minipage}
  \end{center}

\end{frame}
\end{document}

Answer 1

Simply invert minipage with \scalebox, i.e. embed the minipage inside \scalebox.

MWE:

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}
\begin{frame}{Hello world}
  \begin{center}
    \scalebox{0.75}{
    \begin{minipage}{0.7\linewidth}
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}%
    }
  \end{center}

\end{frame}
\end{document} 

Output:

Answer 2

For content that contains text, choosing an appropriate font size is to be preferred over scaling the content, see Why not scale elements that contain text for some explanation.

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}


\begin{frame}{Hello world}
  \begin{center}
    \begin{minipage}{0.8\linewidth}
        \scriptsize
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}
  \end{center}
\end{frame}

\end{document}

---

However if you don't mind a sub-optimal result, you could use one of the most evil beamer options: shrink:

\documentclass{beamer}
\mode<presentation>
{
  \usetheme{CambridgeUS}
  \usecolortheme{dolphin}
  \usecolortheme{rose}
  \setbeamerfont*{title}{shape=\itshape, family=\rmfamily}
  \setbeamercovered{transparent}
}
\usepackage{algorithm2e}

\begin{document}

% This solution is evil!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
% use it at your own risk!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\begin{frame}[shrink]{Hello world}
  \begin{center}
    \begin{minipage}{0.8\linewidth}
      \begin{algorithm}[H]
    \DontPrintSemicolon
    \KwSty{type} brow : \KwSty{int}$[M+1]$\;
    \KwSty{type} bcol : \KwSty{int}$[N+1]$\;
    \KwSty{type} val : \KwSty{real}$[k]$\;
    \KwSty{type} val\_ptr : \KwSty{int}$[K+1]$\;
    \KwSty{type} ind : \KwSty{int}$[K]$\;
    \KwSty{type} ptr : \KwSty{int}$[M+1]$\;
    \everypar={\nl}
    \ForEach{block row $I$}{
      $i_0 \leftarrow brow[I]$\;
      $r \leftarrow brow[I+1]$\;
      Let $\hat{y} \leftarrow y_{i_0:(i_0+r-1)}$\;
      \For{$b=ptr[I] \KwTo ptr[I+1]$}{
        $J \leftarrow ind[b]$\;
        $j_0 \leftarrow bcol[J]$\;
        $c \leftarrow bcol[J+1]-bcol[J]$\;
        Let $\hat{x} \leftarrow x_{j_0:(j_0+c-1)}$\;
        Let $\hat{A} \leftarrow a_{i_0:(i_0+r-1),j_0:(j_0+c-1)}$\;
        Perform $r \times c$ block multiply, $\hat{y} \leftarrow \hat{y}+\hat{A} \cdot \hat{x}$\;
      }
    }
      \end{algorithm}
    \end{minipage}
  \end{center}
\end{frame}



\end{document}