178 lines
7.6 KiB
TeX
Executable File
178 lines
7.6 KiB
TeX
Executable File
% Nivio: 29/jan/06
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% Time-stamp: <Monday 30 Jan 2006 12:37:22am EST yoshi@flare>
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\vspace{-2mm}
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\subsection{Performance of the internal memory based algorithm}
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\label{sec:intern-memory-algor}
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%\begin{table*}[htb]
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%\begin{center}
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%{\scriptsize
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%\begin{tabular}{|c|c|c|c|c|c|c|c|}
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%\hline
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%$n$ (millions) & 1 & 2 & 4 & 8 & 16 & 32 \\
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%\hline
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%Average time (s)& $6.1 \pm 0.3$ & $12.2 \pm 0.6$ & $25.4 \pm 1.1$ & $51.4 \pm 2.0$ & $117.3 \pm 4.4$ & $262.2 \pm 8.7$\\
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%SD (s) & $2.6$ & $5.4$ & $9.8$ & $17.6$ & $37.3$ & $76.3$ \\
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%\hline
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%\end{tabular}
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%\vspace{-3mm}
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%}
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%\end{center}
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%\caption{Internal memory based algorithm: average time in seconds for constructing a MPHF,
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%the standard deviation (SD), and the confidence intervals considering
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%a confidence level of $95\%$.}
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%\label{tab:medias}
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%\end{table*}
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Our three-step internal memory based algorithm presented in~\cite{bkz05}
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is used for constructing a MPHF for each bucket.
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It is a randomized algorithm because it needs to generate a simple random graph
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in its first step.
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Once the graph is obtained the other two steps are deterministic.
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Thus, we can consider the runtime of the algorithm to have the form~$\alpha
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nZ$ for an input of~$n$ keys, where~$\alpha$ is some machine dependent
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constant that further depends on the length of the keys and~$Z$ is a random
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variable with geometric distribution with mean~$1/p=e^{1/c^2}$ (see
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Section~\ref{sec:mphfbucket}). All results in our experiments were obtained
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taking $c=1$; the value of~$c$, with~$c\in[0.93,1.15]$, in fact has little
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influence in the runtime, as shown in~\cite{bkz05}.
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The values chosen for $n$ were $1, 2, 4, 8, 16$ and $32$ million.
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Although we have a dataset with 1~billion URLs, on a PC with
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1~gigabyte of main memory, the algorithm is able
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to handle an input with at most 32 million keys.
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This is mainly because of the graph we need to keep in main memory.
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The algorithm requires $25n + O(1)$ bytes for constructing
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a MPHF (details about the data structures used by the algorithm can
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be found in~\texttt{http://cmph.sf.net}.
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% for the details about the data structures
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%used by the algorithm).
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In order to estimate the number of trials for each value of $n$ we use
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a statistical method for determining a suitable sample size (see, e.g.,
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\cite[Chapter 13]{j91}).
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As we obtained different values for each $n$,
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we used the maximal value obtained, namely, 300~trials in order to have
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a confidence level of $95\%$.
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% \begin{figure*}[ht]
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% \noindent
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% \begin{minipage}[b]{0.5\linewidth}
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% \centering
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% \subfigure[The previous algorithm]
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% {\scalebox{0.5}{\includegraphics{figs/bmz_temporegressao.eps}}}
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% \end{minipage}
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% \hfill
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% \begin{minipage}[b]{0.5\linewidth}
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% \centering
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% \subfigure[The new algorithm]
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% {\scalebox{0.5}{\includegraphics{figs/brz_temporegressao.eps}}}
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% \end{minipage}
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% \caption{Time versus number of keys in $S$. The solid line corresponds to
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% a linear regression model.}
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% %obtained from the experimental measurements.}
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% \label{fig:temporegressao}
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% \end{figure*}
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Table~\ref{tab:medias} presents the runtime average for each $n$,
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the respective standard deviations, and
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the respective confidence intervals given by
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the average time $\pm$ the distance from average time
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considering a confidence level of $95\%$.
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Observing the runtime averages one sees that
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the algorithm runs in expected linear time,
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as shown in~\cite{bkz05}.
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\vspace{-2mm}
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\begin{table*}[htb]
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\begin{center}
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{\scriptsize
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\begin{tabular}{|c|c|c|c|c|c|c|c|}
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\hline
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$n$ (millions) & 1 & 2 & 4 & 8 & 16 & 32 \\
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\hline
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Average time (s)& $6.1 \pm 0.3$ & $12.2 \pm 0.6$ & $25.4 \pm 1.1$ & $51.4 \pm 2.0$ & $117.3 \pm 4.4$ & $262.2 \pm 8.7$\\
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SD (s) & $2.6$ & $5.4$ & $9.8$ & $17.6$ & $37.3$ & $76.3$ \\
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\hline
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\end{tabular}
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\vspace{-1mm}
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}
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\end{center}
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\caption{Internal memory based algorithm: average time in seconds for constructing a MPHF,
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the standard deviation (SD), and the confidence intervals considering
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a confidence level of $95\%$.}
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\label{tab:medias}
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\vspace{-4mm}
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\end{table*}
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% \enlargethispage{\baselineskip}
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% \begin{table*}[htb]
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% \begin{center}
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% {\scriptsize
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% (a)
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% \begin{tabular}{|c|c|c|c|c|c|c|c|}
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% \hline
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% $n$ (millions) & 1 & 2 & 4 & 8 & 16 & 32 \\
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% \hline
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% Average time (s)& $6.119 \pm 0.300$ & $12.190 \pm 0.615$ & $25.359 \pm 1.109$ & $51.408 \pm 2.003$ & $117.343 \pm 4.364$ & $262.215 \pm 8.724$\\
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% SD (s) & $2.644$ & $5.414$ & $9.757$ & $17.627$ & $37.333$ & $76.271$ \\
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% \hline
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% \end{tabular}
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% \\[5mm] (b)
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% \begin{tabular}{|l|c|c|c|c|c|}
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% \hline
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% $n$ (millions) & 1 & 2 & 4 & 8 & 16 \\
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% \hline % Part. 16 \% 16 \% 16 \% 18 \% 20\%
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% Average time (s) & $6.927 \pm 0.309$ & $13.828 \pm 0.175$ & $31.936 \pm 0.663$ & $69.902 \pm 1.084$ & $140.617 \pm 2.502$ \\
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% SD & $0.431$ & $0.245$ & $0.926$ & $1.515$ & $3.498$ \\
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% \hline
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% \hline
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% $n$ (millions) & 32 & 64 & 128 & 512 & 1000 \\
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% \hline % Part. 20 \% 20\% 20\% 18\% 18\%
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% Average time (s) & $284.330 \pm 1.135$ & $587.880 \pm 3.945$ & $1223.581 \pm 4.864$ & $5966.402 \pm 9.465$ & $13229.540 \pm 12.670$ \\
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% SD & $1.587$ & $5.514$ & $6.800$ & $13.232$ & $18.577$ \\
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% \hline
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% \end{tabular}
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% }
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% \end{center}
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% \caption{The runtime averages in seconds,
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% the standard deviation (SD), and
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% the confidence intervals given by the average time $\pm$
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% the distance from average time considering
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% a confidence level of $95\%$.}
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% \label{tab:medias}
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% \end{table*}
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\enlargethispage{2\baselineskip}
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Figure~\ref{fig:bmz_temporegressao}
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presents the runtime for each trial. In addition,
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the solid line corresponds to a linear regression model
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obtained from the experimental measurements.
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As we can see, the runtime for a given $n$ has a considerable
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fluctuation. However, the fluctuation also grows linearly with $n$.
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\begin{figure}[htb]
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\vspace{-2mm}
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\begin{center}
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\scalebox{0.4}{\includegraphics{figs/bmz_temporegressao.eps}}
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\caption{Time versus number of keys in $S$ for the internal memory based algorithm.
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The solid line corresponds to a linear regression model.}
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\label{fig:bmz_temporegressao}
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\end{center}
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\vspace{-6mm}
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\end{figure}
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The observed fluctuation in the runtimes is as expected; recall that this
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runtime has the form~$\alpha nZ$ with~$Z$ a geometric random variable with
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mean~$1/p=e$. Thus, the runtime has mean~$\alpha n/p=\alpha en$ and standard
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deviation~$\alpha n\sqrt{(1-p)/p^2}=\alpha n\sqrt{e(e-1)}$.
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Therefore, the standard deviation also grows
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linearly with $n$, as experimentally verified
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in Table~\ref{tab:medias} and in Figure~\ref{fig:bmz_temporegressao}.
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%\noindent-------------------------------------------------------------------------\\
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%Comentario para Yoshi: Nao consegui reproduzir bem o que discutimos
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%no paragrafo acima, acho que vc conseguira justificar melhor :-). \\
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%-------------------------------------------------------------------------\\
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