% Nivio: 29/jan/06
% Time-stamp: <Monday 30 Jan 2006 04:01:40am EDT yoshi@ime.usp.br>
\enlargethispage{2\baselineskip}
\subsection{Determining~$b$}
\label{sec:determining-b}
\begin{table*}[t]
\begin{center}
{\small %\scriptsize
\begin{tabular}{|c|ccc|ccc|}
\hline
\raisebox{-0.7em}{$n$} & \multicolumn{3}{c|}{\raisebox{-1mm}{b=128}}  &
\multicolumn{3}{c|}{\raisebox{-1mm}{b=175}}\\
\cline{2-4} \cline{5-7}
 & \raisebox{-0.5mm}{Worst Case} & \raisebox{-0.5mm}{Average} &\raisebox{-0.5mm}{Eq.~(\ref{eq:maxbs})} 
 & \raisebox{-0.5mm}{Worst Case} & \raisebox{-0.5mm}{Average} &\raisebox{-0.5mm}{Eq.~(\ref{eq:maxbs})} \\
\hline
$1.0 \times 10^6$  & 177 & 172.0 & 176  & 232 & 226.6 & 230 \\
%$2.0 \times 10^6$  & 179 & 174.0 & 178  & 236 & 228.5 & 232 \\
$4.0 \times 10^6$  & 182 & 177.5 & 179  & 241 & 231.8 & 234 \\
%$8.0 \times 10^6$  & 186 & 181.6 & 181  & 238 & 234.2 & 236 \\
$1.6 \times 10^7$  & 184 & 181.6 & 183  & 241 & 236.1 & 238 \\
%$3.2 \times 10^7$  & 191 & 183.9 & 184  & 240 & 236.6 & 240 \\
$6.4 \times 10^7$  & 195 & 185.2 & 186  & 244 & 239.0 & 242 \\
%$1.28 \times 10^8$ & 193 & 187.7 & 187  & 244 & 239.7 & 244 \\
$5.12 \times 10^8$ & 196 & 191.7 & 190  & 251 & 246.3 & 247 \\
$1.0 \times 10^9$  & 197 & 191.6 & 192  & 253 & 248.9 & 249 \\
\hline
\end{tabular}
\vspace{-1mm}
}
\end{center}
\caption{Values for $\mathit{BS}_{\mathit{max}}$: worst case and average case obtained in the experiments and using Eq.~(\ref{eq:maxbs}), 
considering $b=128$ and $b=175$ for different number $n$ of keys in $S$.} 
\label{tab:comparison}
\vspace{-6mm}
\end{table*}

The partitioning step can be viewed as the well known ``balls into bins'' 
problem~\cite{ra98,dfm02} where~$n$ keys (the balls) are placed independently and 
uniformly into $\lceil n/b\rceil$ buckets (the bins). The main question related to that problem we are interested 
in is: what is the maximum number of keys in any bucket? 
In fact, we want to get the maximum value for $b$ that makes the maximum number of keys in any bucket
no greater than 256 with high probability. 
This is important, as we wish to use 8 bits per entry in the vector $g_i$ of
each $\mathrm{MPHF}_i$, 
where $0 \leq i < \lceil n/b\rceil$.      
Let $\mathit{BS}_{\mathit{max}}$ be the maximum number of keys in any bucket.

Clearly, $\BSmax$ is the maximum
of~$\lceil n/b\rceil$ random variables~$Z_i$, each with binomial
distribution~$\Bi(n,p)$ with parameters~$n$ and~$p=1/\lceil n/b\rceil$.
However, the~$Z_i$ are not independent.  Note that~$\Bi(n,p)$ has mean and
variance~$\simeq b$.  To give an upper estimate for the probability of the
event~$\BSmax\geq \gamma$, we can estimate the probability that we have~$Z_i\geq \gamma$
for a fixed~$i$, and then sum these estimates over all~$i$.
Let~$\gamma=b+\sigma\sqrt{b\ln(n/b)}$, where~$\sigma=\sqrt2$.
Approximating~$\Bi(n,p)$ by the normal distribution with mean and
variance~$b$, we obtain the
estimate~$(\sigma\sqrt{2\pi\ln(n/b)})^{-1}\times\exp(-(1/2)\sigma^2\ln(n/b))$ for
the probability that~$Z_i\geq \gamma$ occurs, which, summed over all~$i$, gives
that the probability that~$\BSmax\geq \gamma$ occurs is at
most~$1/(\sigma\sqrt{2\pi\ln(n/b)})$, which tends to~$0$ as~$n\to\infty$.
Thus, we have shown that, with high probability, 
\begin{equation}
  \label{eq:maxbs}
  \BSmax\leq b+\sqrt{2b\ln{n\over b}}.
\end{equation}

% The traditional approach used to estimate $\mathit{BS}_{\mathit{max}}$ with high probability is 
% to consider $\mathit{BS}_{\mathit{max}}$ as a random variable that follows a binomial distribution 
% that can be approximated by a poisson distribution. This yields a good approximation
% when the number of balls is lower than or equal to the number of bins~\cite{g81}. In our case,
% the number of balls is greater than the number of buckets.
% % and that is why we have used more recent works to estimate $\mathit{BS}_{\mathit{max}}$. 
% As $b > \ln (n/b)$, we can use the result by Raab and Steger~\cite{ra98} to estimate
% $\mathit{BS}_{\mathit{max}}$ with high probability. 
% The following equation gives the estimation, where $\sigma=\sqrt{2}$:  
% \begin{eqnarray} \label{eq:maxbs}
% \mathit{BS}_{\mathit{max}} = b + O \left( \sqrt{b\ln\frac{n}{b}} \right) = b + \sigma \times \left(\sqrt{b\ln\frac{n}{b}} \right)
% \end{eqnarray}

% In order to estimate the suitable constant $\sigma$ we did a linear 
% regression suppressing the constant term. 
% We use the equation $BS_{max} - b = \sigma \times \sqrt{b\ln (n/b)}$ 
% in the linear regression considering $y=BS_{max} - b$ and $x=\sqrt{b\ln (n/b)}$. 
% In order to obtain data to be used in the linear regression we set 
% b=128 and ran the new algorithm ten times 
% for n equal to 1, 2, 4, 8, 16, 32, 64, 128, 512, 1000 million keys.  
% Taking a confidence level equal to 95\% we got 
% $\sigma = 2.11 \pm 0.03$. 
% The coefficient of determination was $99.6\%$, which means that the linear 
% regression explains $99.6\%$ of the data variation and only $0.4\%$ 
% is due to experimental errors.
% Therefore, Eq.~(\ref{eq:maxbs}) with $\sigma = 2.11 \pm 0.03$ and $b=128$ 
% makes a very good estimation of the maximum number of keys in any bucket.

% Repeating the same experiments for $b$ equals to $175$ and  
% a confidence level of $95\%$ we got $\sigma = 2.07 \pm 0.03$.
% Again we verified that Eq.~(\ref{eq:maxbs}) with $\sigma = 2.07 \pm 0.03$ and $b=175$ is 
% a very good estimation of the maximum number of keys in any bucket once the
% coefficient of determination obtained was $99.7 \%$ and $\sigma$ is in a very narrow range.

In our algorithm the maximum number of keys in any bucket must be at most 256. 
Table~\ref{tab:comparison} presents the values for $\mathit{BS}_{\mathit{max}}$
obtained experimentally and using Eq.~(\ref{eq:maxbs}).
The table presents the worst case and the average case, 
considering $b=128$,  $b=175$ and Eq.~(\ref{eq:maxbs}),
for several numbers~$n$ of keys in $S$. 
The estimation given by Eq.~(\ref{eq:maxbs}) is very close to the experimental
results.

Now we estimate the biggest problem our algorithm is able to solve for 
a given $b$. 
Using Eq.~(\ref{eq:maxbs}) considering $b=128$, $b=175$ and imposing
that~$\mathit{BS}_{\mathit{max}}\leq256$, 
the sizes of the biggest key set our algorithm 
can deal with are $10^{30}$ keys and $10^{10}$ keys, respectively. 
%It is also important to have $b$ as big as possible, once its value is
%related to the space required to store the resultant MPHF, as shown later on.
%Table~\ref{tab:bp} shows the biggest problem the algorithm can solve.
% The values were obtained from Eq.~(\ref{eq:maxbs}), 
% considering $b=128$ and~$b=175$ and imposing
% that~$\mathit{BS}_{\mathit{max}}\leq256$. 

% We set $\sigma=2.14$ because it was the greatest value obtained for $\sigma$
% in the two linear regression we did.
% \vspace{-3mm} 
% \begin{table}[htb]
% \begin{center}
% {\small %\scriptsize
% \begin{tabular}{|c|c|}
% \hline
% b    & Problem size ($n$) \\
% \hline
% 128  & $10^{30}$ keys  \\
% 175  & $10^{10}$ keys  \\
% \hline
% \end{tabular}
% \vspace{-1mm}
% }
% \end{center}
% \caption{Using Eq.~(\ref{eq:maxbs}) to estimate the biggest problem our algorithm can solve.}
% %considering $\sigma=\sqrt{2}$.}
% \label{tab:bp}
% \vspace{-14mm}
% \end{table}