Started texing poly reformation write up.

macros.tex

@@ -2,6 +2,13 @@
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % NOTATION
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+%Polynomial Reformulation
+\newcommand{\wbit}{w}
+%using \wVec for world bit vector notation
+\newcommand{\poly}{Q}
+\newcommand{\rpoly}{\widetilde{Q}}%r for reduced as in reduced 'Q'
+\newcommand{\out}{output}%output aggregation over the output vector
 %David's Scheme
 \newcommand{\vecform}[1]{\textbf{#1}}%auxiliary cmd to use only with macros, so that we can change the format easily if need be
 \newcommand{\sone}{\vecform{x}}
@@ -268,6 +275,7 @@
 \newtheorem{Definition}{Definition}
 \newtheorem{Lemma}{Lemma}
 \newtheorem{Proposition}{Proposition}
+\newtheorem{Property}{Property}
 \newtheorem{Corollary}{Corollary}
 \newtheorem{Example}{Example}
 \newtheorem{Axiom}{Axiom}

main.tex

@@ -125,19 +125,21 @@
 %\input{abstract}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\input{prob_def}
-\input{notation}
-\input{analysis}
-\input{est_bounds}
-\input{combining}
-
-\input{instantiation}
-\input{hash_const}
-\input{exact}
-\input{var_estj}
-\input{pos}
-\input{sop}
-\input{davidscheme}
+%\input{prob_def}
+%\input{notation}
+%\input{analysis}
+%\input{est_bounds}
+%\input{combining}
+%
+%\input{instantiation}
+%\input{hash_const}
+%\input{exact}
+%\input{var_estj}
+%\input{pos}
+%\input{sop}
+%\input{davidscheme}
+\input{ra-to-poly}
+\input{poly-form}
 %\input{}
 
 

poly-form.tex

@@ -0,0 +1,41 @@
+%root: main.tex
+\section{Polynomial Formulation}
+
+Let $\vect_1,\ldots, \vect_\numTup$ be vectors annotating $\numTup$ tuples in a TIDB such that \begin{equation}\vect_i[(\wbit_1,\ldots,\wbit_\numTup)] = 
+\begin{cases} 1	&\wbit_i = 1,\\
+0	&otherwise.\label{eq:vec-def}
+\end{cases}
+\end{equation}
+Here we define vector indexing by the $\numTup$-bit binary tuple $\wVec = (\wbit_1,\ldots,\wbit_\numTup)$ such that the possible world $\wVec$ is identified by its bit vector binary value.
+
+Futher we define the polynomial $\poly(\vect_1,\ldots,\vect_\numTup)$ as an arbitrary polynomial defined over the input vectors, whose addition and multiplication operations are defined as the traditional point-wise vector addition and multiplication.  We overload notation and denote the $i^{th}$ world of $\poly$ as $\poly[\wVec]$, where $\poly$ can be viewed as the output annotation vector, and the L-1 norm can be represented as $\poly = \sum\limits_{\wVec \in \wSet} \poly[\wVec]$.
+
+By noting that \cref{eq:vec-def} is equivalent to $\vect_i[(\wbit_1,\ldots, \wbit_\numTup)] = \wbit_i$, we can further reformulate the problem by saying that $\poly[\wVec] = \poly(\wbit_1,\ldots, \wbit_\numTup)$, where each element of the RHS input is a bit element of the $\wVec$ bit vector, and we can thus replace each variable of $\poly$ with its corresponding input bit, and solve for the particular world $\wVec$.  We can then further rewrite our L-1 norm output as $\sum\limits_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup)$.
+
+Further, define $\rpoly(\wbit_1,\ldots, \wbit_\numTup)$ as the reduced version of $\poly(\wbit_1,\ldots, \wbit_\numTup)$, formally defined as
+\[\rpoly(\wbit_1,\ldots, \wbit_\numTup) = \poly(\wbit_1,\ldots, \wbit_\numTup) \mod \wbit_1^2-\wbit\cdots\mod \wbit_\numTup^2 - \wbit_\numTup.\]  Intuitively, $\rpoly(\wbit_1,\ldots, \wbit_\numTup)$ is the expanded sum of products form of $\poly(\wbit_1,\ldots, \wbit_\numTup)$ such that if any $\wbit_j$ term  has an exponent $e > 1$, it is reduced to $1$, i.e. $\wbit_j^e\mapsto \wbit_j$ for any $e > 1$.  This usefulness of this reduction will be seen shortly.
+
+First, note the following fact:
+\[\text{For all } (\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup, \poly(\wbit_1,\ldots, \wbit_\numTup) = \rpoly(\wbit_1,\ldots, \wbit_\numTup).\]
+
+\begin{proof}
+For all $b \in \{0, 1\}$ and all $e \geq 1$, $b^e = 1$.\qed
+\end{proof}
+
+As stated previously, the aggregate output we seek to compute is the L1-norm of the output vector $\poly$, which we'll denote as $\out$.  Assuming each tuple has a probability $p = 0.5$, we note that
+\begin{Property}
+\begin{equation*}
+\out = \sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup} \poly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).
+\end{equation*}
+\end{Property}
+
+\begin{proof}
+Using the fact above, we need to compute \[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup)\].  We therefore argue that 
+\[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).\]
+
+Note that for any single monomial, this is indeed the case since 1) each tuple appears in a possible world with probability of $\frac{1}{2}$, 2) the value $2^\numTup$ is the number of worlds, and 3) the product of expectation and total number of worlds yields and exact result.
+
+The final result follows by noting that $\rpoly$ is a sum of monomials, and we can, by rearranging the sum, equivlently push the sum into the monomials.\qed
+\end{proof}
+
+

ra-to-poly.tex

@@ -0,0 +1,4 @@
+%root: main.tex
+
+\section{Query translation into polynomials}
+This section will involve the set of queries (RA+) that we are interested in, the probabilistic/incomplete models we address, and the outer aggregate functions we perform over the output \textit{annotation}.