Combining Sketches

analysis.tex

@@ -394,7 +394,7 @@ Substituting the Cauchy Schwarts bounds into the Chebyshev calculations gives
 \begin{align}
 &\sketchCols \leq \frac{3\norm{\genV}_2^2\left(|\pw|\right) + \norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}}\nonumber\\
 &\sketchCols \leq \frac{4\norm{\genV}_2^2\left(|\pw|\right)}{\norm{\genV}_2\sqrt{|\pw|}}\nonumber\\
-&\sketchCols \leq 4\norm{\genV}_2\sqrt{|\pw|}
+&\sketchCols \leq 4\norm{\genV}_2\sqrt{|\pw|}\label{eq:b-cauchy}
 \end{align}
 \AH{\textbf{BEGIN}: Old  Bound calculations}
 \begin{align*}

combining.tex

@@ -0,0 +1,41 @@
+% -*- root: main.tex -*-
+\section{Combining Sketches}
+\label{sec:combining}
+\subsection{Adding Sketches}
+When assuming that the variables are independent, as in the TIDB model, it is a known result that 
+\[
+\varParam{X + Y} = \varParam{X} + \varParam{Y}.
+\]
+It then immediately follows that adding $n$ base sketches results in the following variance:
+\[
+n \cdot 4\norm{\genV}_2 |\pw|^{1/2}.
+\]
+
+\subsection{Multiplying Sketches}
+For the case of multiplication it is a known result that
+\[
+\varParam{X \cdot Y} = \expect{X^2}\expect{Y^2} - (\expect{X})^2 (\expect{Y})^2.
+\]
+Assuming discreet variables the expectation of the square of a random variable is simply the sum of its weighted squares.  This yields
+\begin{align}
+&\expect{\left(\sum_{\wVec \in \pw}\sketchJParam{\sketchHashParam{\wVec}}\cdot \sketchPolarParam{\wVec}\right)^2}\\
+=& \sum_{\wVec \in \pw}\expect{\left(\sketchJParam{\sketchHashParam{\wVec}}\cdot\sketchPolarParam{\wVec}\right)^2}\\
+=& \sum_{\wVec \in \pw}\expect{\left(\sum_{\substack{\wVecPrime \in \pw \st \\
+									\sketchHashParam{\wVecPrime} = \sketchHashParam{\wVec}}} \genVParam{\wVecPrime}\sketchPolarParam{\wVecPrime}\sketchPolarParam{\wVec}\right)^2}\\
+=& \sum_{\wVec \in \pw}\expect{\left(\genVParam{\wVec}^2\sketchPolarParam{\wVec}^2 + \sum_{\substack{\wVecPrime \in \pw \st \\
+									\sketchHashParam{\wVecPrime} = \sketchHashParam{\wVec},
+									\wVecPrime \neq \wVec}} \genVParam{\wVecPrime}\sketchPolarParam{\wVecPrime}\sketchPolarParam{\wVec}\right)^2}\\
+=& \sum_{\wVec \in \pw}\expect{\genVParam{\wVec}^2}\\
+=& \sum_{\wVec \in \pw}\genVParam{\wVec}^2.
+\end{align}
+\begin{Justification}
+\hfill
+	\begin{itemize}
+		\item stuff goes here
+	\end{itemize}
+\end{Justification}
+It then follows that the muliplication of two base sketches results in
+\begin{align}
+&\sum_{\wVec \in \pw}\genV_1\paramBox{\wVec}^2\sum_{\wVec \in \pw}\genV_2\paramBox{\wVec}^2 - \left(\sum_{\wVec \in \pw} \genV_1\paramBox{\wVec}\right)^2\left(\sum_{\wVec \in \pw} \genV_2\paramBox{\wVec}\right)^2\\
+=&\norm{\genV_1}_2^2\norm{\genV_2}_2^2 - \norm{\genV_1}_1^2\norm{\genV_2}_1^2.
+\end{align}

instantiation.tex

@@ -4,11 +4,22 @@
 
 \AH{This section has been started, but needs to be completed.}
 \subsection{TIDB}
-Consider the case of a TIDB with $\numTup$ tuples, with $\prob = \frac{1}{2}$ for given tuple $t$.  The vector $\genV$ can then be defined as 
+Consider the case of a TIDB with $\numTup$ tuples, with $\prob = \frac{1}{2}$ for given tuple $t$.  Because TIDB has the property of set semantics, the vector $\genV$ can then be defined as a binary bit vector $\{0, 1\}^\numTup$, whose value represents a possible world, and, where each index represents a specific tuple $t$ id.  Under these semantics, with $w_t$ representing the index mapped to a a tuple $t$'s identity, $\genV$ can alternatively be viewed as a function
 
 \begin{equation*}
 \genV = \begin{cases}
 		1, &w_t = 1\\
 		0, &otherwise
 		\end{cases}
+\end{equation*}
+where a value of $1$ indicates that the tuple is present in a given world, and $0$ denotes that the tuple is absent in the world represented by the binary bit string.
+
+In this representation, a few properties of $\genV$ immediately stand out.  First, the length of $\genV$ is the same as the number of tuples in the TIDB, $|\genV| = \numTup$.  This combined with the assumption of $\prob = \frac{1}{2}$ implies that the L1 norm of $\genV$ is $\frac{\numTup}{2}$ and that the L2 norm of $\genV$ squared is also the same value, 
+\begin{equation*}
+|\genV| = \numTup \wedge \prob = \frac{1}{2} \Rightarrow \norm{\genV}_1 = \norm{\genV}_2^2.
+\end{equation*}
+
+By \eqref{eq:b-cauchy} this yields a bucket size of 
+\begin{equation*}
+\sketchCols \leq 4\sqrt{\frac{\numTup}{2}} \cdot 2^{(\numTup/2)}.
 \end{equation*}

main.tex

@@ -126,6 +126,7 @@
 
 \input{notation}
 \input{analysis}
+\input{combining}
 \input{instantiation}
 \input{hash_const}
 \input{exact}