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\subsection{Polynomial Formulation and Equivalences}

Since we have shown that computing the expected multiplicity of a result tuple is equivalent to computing the expectation of a polynomial (for that tuple) given a probability distribution over all possible assignments of variables in the polynomial to $\{0,1\}$, we from now on focus on this problem exclusively.
Before proceeding, note that the following is assuming \bis (which subsume \tis as a special case). Thus, variables are independent of each other and each variable $X$ is associated with a probability $\vct{p}(X)$.

Let us use the expression $(x + y)^2$ for a running example in the following definitions.

\begin{Definition}[Monomial]\label{def:monomial}
A monomial is a product of a fixed set of variables, each raised to a non-negative integer power.
\end{Definition}

For the term $2xy$, by ~\cref{def:monomial} the monomial is $xy$.

\begin{Definition}[Standard Monomial Basis]\label{def:smb}
A polynomial is in standard monomial basis when it is fully expanded out such that no product of sums exist and where each unique monomial appears exactly once.
\end{Definition}

The standard monomial basis for the running example is $x^2 +2xy + y^2$.  While $x^2 + xy + xy + y^2$ is an expanded form of the expression, it is not the standard monomial basis since $xy$ appears more than once.

Throughout this paper, we also make the following \textit{assumption}.
\begin{Assumption}\label{assump:poly-smb}
All polynomials considered are in standard monomial basis, i.e., $\poly(\vct{X}) = \sum\limits_{\vct{d} \in \mathbb{N}^\numvar}q_d \cdot \prod\limits_{i = 1, d_i \geq 1}^{\numvar}X_i^{d_i}$, where $q_d$ is the coefficient for the monomial encoded in $\vct{d}$ and $d_i$ is the $i^{th}$ element of $\vct{d}$.
\end{Assumption}

While the definition of polynomial $\poly(\vct{X})$ over a $\bi$ input doesn't change, we introduce an alternative notation which will come in handy.  Given $\ell$ blocks, we write $\poly(\vct{X})$ = $\poly(X_{\block_1, 1},\ldots, X_{\block_1, \abs{\block_1}},$ $\ldots, X_{\block_\ell, \abs{\block_\ell}})$, where $\abs{\block_i}$ denotes the size of $\block_i$, and $\block_{i, j}$ denotes tuple $j$ residing in block $i$ for $j$ in $[\abs{\block_i}]$.
The number of tuples in the $\bi$ instance can be (trivially) computed as $\numvar = \sum\limits_{i = 1}^{\ell}\abs{\block_i}$ .

\begin{Definition}[Degree]\label{def:degree}
The degree of polynomial $\poly(\vct{X})$ is the maximum sum of the exponents of a monomial, over all monomials when $\poly(\vct{X})$ is in SOP form.
\end{Definition}

The degree of the running example is $2$.  In this paper we consider only finite degree polynomials.

\begin{Definition}[$\rpoly(\vct{X})$] \label{def:qtilde}
Define $\rpoly(X_1,\ldots, X_\numvar)$ as the reduced version of $\poly(X_1,\ldots, X_\numvar)$, of the form
$\rpoly(X_1,\ldots, X_\numvar) = $

\[\poly(X_1,\ldots, X_\numvar) \mod X_1^2-X_1\cdots\mod X_\numvar^2 - X_\numvar.\]
\end{Definition}

\begin{Example}\label{example:qtilde}
Consider when $\poly(x, y) = (x + y)(x + y)$.  Then 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}

Intuitively, $\rpoly(\textbf{X})$ is the SOP form of $\poly(\textbf{X})$ such that if any $X_j$ term  has an exponent $e > 1$, it is reduced to $1$, i.e. $X_j^e\mapsto X_j$ for any $e > 1$.
Alternatively, one can gain intuition for $\rpoly$ by thinking of $\rpoly$ as the resulting SOP of $\poly(\vct{X})$ with an idemptent product operator.

When considering $\bi$ input, it becomes necessary to redefine $\rpoly(\vct{X})$.

\begin{Definition}[$\rpoly$ $\bi$ Redefinition]
A polynomial $\poly(\vct{X})$ over a $\bi$ instance is reduced to $\rpoly(\vct{X})$ with the following criteria.  First, all exponents $e > 1$ are reduced to $e = 1$.  Second, all monomials sharing the same $\block$ are dropped.  Formally this is expressed as

\begin{equation*}
\rpoly(\vct{X}) = \poly(\vct{X}) \mod X_i^2 - X_i \mod X_{\block_s, t}X_{\block_s, u}
\end{equation*}
for all $i$ in $[\numvar]$ and for all $s$ in $\ell$, such that for all $t, u$ in $[\abs{block_s}]$, $t \neq u$.
\end{Definition}

The usefulness of this reduction will be seen in ~\cref{lem:exp-poly-rpoly}.

\begin{Lemma}\label{lem:pre-poly-rpoly}
When $\poly(X_1,\ldots, X_\numvar) = \sum\limits_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}} \cdot \prod\limits_{\substack{i = 1\\s.t. d_i\geq 1}}^{\numvar}X_i^{d_i}$, we have then that $\rpoly(X_1,\ldots, X_\numvar) = \sum\limits_{\vct{d} \in \{0,\ldots, B\}^\numvar} q_{\vct{d}}\cdot\prod\limits_{\substack{i = 1\\s.t. d_i\geq 1}}^{\numvar}X_i$.
\end{Lemma}
\begin{proof}
Follows by the construction of $\rpoly$ in \cref{def:qtilde}.
\end{proof}

\qed

Note the following fact:
\begin{Proposition}\label{proposition:q-qtilde}
\[\text{For all } (X_1,\ldots, X_\numvar) \in \{0, 1\}^\numvar, \poly(X_1,\ldots, X_\numvar) = \rpoly(X_1,\ldots, X_\numvar).\]
\end{Proposition}

\begin{proof}[Proof for Proposition ~\ref{proposition:q-qtilde}]
Note that any $\poly$ in factorized form is equivalent to its sum of product expansion.  For each term in the expanded form, further note that for all $b \in \{0, 1\}$ and all $e \geq 1$, $b^e = b$.
\end{proof}

\qed

Define all variables $X_i$ in $\poly$ to be independent.
\begin{Lemma}\label{lem:exp-poly-rpoly}
The expectation over possible worlds in $\poly(\vct{X})$ is equal to $\rpoly(\prob_1,\ldots, \prob_\numvar)$.
\begin{equation*}
\expct_{\vct{w}}\pbox{\poly(\vct{w})}  = \rpoly(\prob_1,\ldots, \prob_\numvar).
\end{equation*}
\end{Lemma}

Note that in the preceding lemma, we have assigned $\vct{p}$ (introduced in ~\cref{subsec:def-data}) to the variables $\vct{X}$.

\begin{proof}[Proof for Lemma ~\ref{lem:exp-poly-rpoly}]
%Using the fact above, we need to compute \[\sum_{(\wbit_1,\ldots, \wbit_\numvar) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numvar)\].  We therefore argue that
%\[\sum_{(\wbit_1,\ldots, \wbit_\numvar) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numvar) = 2^\numvar \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).\]

Let $\poly$ be the generalized polynomial, i.e., the polynomial of $\numvar$ variables with highest degree $= B$: %, in which every possible monomial permutation appears,
\[\poly(X_1,\ldots, X_\numvar) = \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar X_i^{d_i}\].


Then, assigning $\vct{w}$ to $\vct{X}$, for expectation we have
\begin{align}
\expct_{\vct{w}}\pbox{\poly(\vct{w})} &= \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \expct_{\vct{w}}\pbox{\prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar w_i^{d_i}}\label{p1-s1}\\
&= \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar \expct_{\vct{w}}\pbox{w_i^{d_i}}\label{p1-s2}\\
&= \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar \expct_{\vct{w}}\pbox{w_i}\label{p1-s3}\\
&= \sum_{\vct{d} \in \{0,\ldots, B\}^\numvar}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numvar \prob_i\label{p1-s4}\\
&= \rpoly(\prob_1,\ldots, \prob_\numvar)\label{p1-s5}
\end{align}

In steps \cref{p1-s1} and \cref{p1-s2}, by linearity of expectation (recall the variables are independent), the expecation can be pushed all the way inside of the product.  In \cref{p1-s3}, note that $w_i \in \{0, 1\}$ which further implies that for any exponent $e \geq 1$, $w_i^e = w_i$.  Next, in \cref{p1-s4} the expectation of a tuple is indeed its probability.

%\OK{
%	You don't need to tie this to TI-DBs if you define the variables ($X_i$) to be independent.
%	Annotations
%	Boolean expressions over uncorrelated boolean variables are sufficient to model TI-, BI-, and
%	PC-Tables.  This should still hold for arithmetic over the naturals.
%}


Finally, observe \cref{p1-s5} by construction in \cref{lem:pre-poly-rpoly}, that $\rpoly(\prob_1,\ldots, \prob_\numvar)$ is exactly the product of probabilities of each variable in each monomial across the entire sum.

\qed
\end{proof}

\begin{Corollary}\label{cor:expct-sop}
If $\poly$ is given as a sum of monomials, the expectation of $\poly$, i.e., $\expct\pbox{\poly} = \rpoly\left(\prob_1,\ldots, \prob_\numvar\right)$ can be computed in $O(|\poly|)$, where $|\poly|$ denotes the total number of multiplication/addition operators.
\end{Corollary}

\begin{proof}[Proof For Corollary ~\ref{cor:expct-sop}]
Note that \cref{lem:exp-poly-rpoly} shows that $\expct\pbox{\poly} =$ $\rpoly(\prob_1,\ldots, \prob_\numvar)$.  Therefore, if $\poly$ is already in sum of products form, one only needs to compute $\poly(\prob_1,\ldots, \prob_\numvar)$ ignoring exponent terms (note that such a polynomial is $\rpoly(\prob_1,\ldots, \prob_\numvar)$), which indeed has $O(|\poly|)$ compututations.\qed
\end{proof}