paper-BagRelationalPDBsAreHard/poly-form.tex
2022-02-09 09:35:36 -05:00

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%\subsection{Reduced Polynomials and Equivalences}
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%\AH{\Cref{def:reduced-poly} replaces this.}
%\begin{Definition}[Reduced \bi Polynomials]\label{def:reduced-bi-poly}
% Let $\poly(\vct{X})$ be a \bi-lineage polynomial.
% The reduced form $\rpoly(\vct{X})$ of $\poly(\vct{X})$ is the same as \Cref{def:reduced-poly} with the added constraint that all monomials with variables $X_{\block, i}, X_{\block, j}, i\neq j$ from the same block $\block$ are omitted.
%\end{Definition}
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%
%
%Consider a $\abbrBIDB$ polynomial $\poly\inparen{\vct{X}} = X_{1, 1}X_{1, 2} + X_{1, 2}X_{2, 1}^2$. Then by \Cref{def:reduced-bi-poly}, we have that $\rpoly\inparen{\vct{X}} = X_{1, 2}X_{2, 1}$. Next, we show why the reduced form is useful for our purposes.
%%Removing this example to save space
\iffalse
\begin{Example}\label{example:qtilde}
Consider $\poly(X, Y) = (X + Y)(X + Y)$ where $X$ and $Y$ are from different blocks. The expanded derivation for $\rpoly(X, Y)$ is
\begin{align*}
(&X^2 + 2XY + Y^2 \mod X^2 - X) \mod Y^2 - Y\\
= ~&X + 2XY + Y^2 \mod Y^2 - Y\\
= ~& X + 2XY + Y
\end{align*}
\end{Example}
\fi
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%\begin{Lemma}\label{lem:exp-poly-rpoly}
%Let $\pdb$ be a \abbrBIDB over $\numvar$ input tuples such that the probability distribution $\pdassign$ over $\{0,1\}^\numvar$ (the all worlds set) is induced by the probability vector $\probAllTup = (\prob_1, \ldots, \prob_\numvar)$. As in \Cref{lem:tidb-reduce-poly} for \abbrTIDB, any \abbrBIDB-lineage polynomial $\poly(\vct{X})$ based on $\pdb$ and query $\query$ we have:
% % The expectation over possible worlds in $\poly(\vct{X})$ is equal to $\rpoly(\prob_1,\ldots, \prob_\numvar)$.
%\begin{equation*}
%\expct_{\vct{W}\sim \pdassign}\pbox{\poly(\vct{W})} = \rpoly(\probAllTup).
%\end{equation*}
%\end{Lemma}
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Let $\abs{\poly}$ be the number of operators in $\poly$.
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\begin{Corollary}\label{cor:expct-sop}
If $\poly$ is a \bi-lineage polynomial already in \abbrSMB, then the expectation of $\poly$, i.e., $\expct\pbox{\poly} = \rpoly\left(\prob_1,\ldots, \prob_\numvar\right)$ can be computed in $\bigO{\abs{\poly}}$ time.
\end{Corollary}
\secrev{
Queries over probabilistic databases are evaluated using the so-called possible world semantics. Under the possible world semantics, the result of a query $\query$ over an incomplete database $\Omega$ is the set of query answers produced by evaluating $\query$ over each possible world $\omega\in\Omega$: $\inset{\query\inparen{\omega}: \omega\in\Omega}$.
The result of a query is the pair $\inparen{\query\inparen{\omega}, \bpd'}$ where $\bpd'$ is a probability distribution that assigns to each possible query result the sum of the probabilites of the worlds that produce this answer: $\probOf\pbox{\omega\in\Omega} = \sum{\omega'\in\Omega,\\\query\inparen{\omega'}=\query\inparen{\omega}}\probOf\pbox{\omega'}$.
}
Recalling \Cref{fig:nxDBSemantics} again, which defines the lineage polynomial $\apolyqdt$ for any $\raPlus$ query. We now make a meaningful connection between possible world semantics and world assignments on the lineage polynomial.
\begin{Proposition}[Expectation of polynomials]\label{prop:expection-of-polynom}
Given a \abbrBPDB $\pdb = (\Omega,\bpd)$, $\raPlus$ query $\query$, and lineage polynomial $\apolyqdt$ for arbitrary result tuple $\tup$, %$\semNX$-\abbrPDB $\pxdb = (\idb_{\semNX}',\pd')$ where $\rmod(\pxdb) = \pdb$,
we have (denoting $\randDB$ as the random variable over $\idb$):
$ \expct_{\randDB \sim \bpd}[\query(\randDB)(t)] = \expct_{\vct{\randWorld}\sim \pdassign}\pbox{\apolyqdt\inparen{\vct{\randWorld}}}. $
\end{Proposition}
\noindent A formal proof of \Cref{prop:expection-of-polynom} is given in \Cref{subsec:expectation-of-polynom-proof}.\footnote{Although \Cref{prop:expection-of-polynom} follows, e.g., as an obvious consequence of~\cite{IL84a}'s Theorem 7.1, we are unaware of any formal proof for bag-probabilistic databases.}
We focus on the problem of computing $\expct_\pdassign\pbox{\apolyqdt\inparen{\vct{\randWorld}}}$ from now on, assume implicit $\query, \tupset, \tup$, and drop them from $\apolyqdt$ (i.e., $\poly\inparen{\vct{X}}$ will denote a polynomial).
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