poly

poly-form.tex

@@ -69,11 +69,11 @@ $\poly(\vct{X})$ = $\poly(X_{\block_1, 1},\ldots, X_{\block_1, \abs{\block_1}},$
 \begin{equation*}
 \rpoly(\vct{X}) = \smbOf{\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 $[\abs{block_s}]$, $t \neq u$.
+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}
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-Intuitively, in the reduced form all exponents $e > 1$ are reduced to $e = 1$ and, all monomials containing more than one variable from the same block $\block$ are dropped. Note that for the special case of \tis, there is no dropping of monomials since every block contains a single tuple.
+Intuitively, in the reduced form all exponents $e > 1$ are reduced to $e = 1$ and, all monomials containing more than one variable from the same block $\block$ are dropped (tuples from the same block are disjoint events in \bis and, thus, any world containing more than one tuple from a block has $0$ probability and can be ignored). Note that for the special case of \tis, the second step (dropping monomials with variables from the same block) is not necessary since every block contains a single tuple.
 Alternatively, one can think of $\rpoly$ as the \abbrSMB of $\poly(\vct{X})$ when the product operator is idempotent.
 
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@@ -132,7 +132,7 @@ Note that any $\poly$ in factorized form is equivalent to its \abbrSMB expansion
 
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 \begin{Lemma}\label{lem:exp-poly-rpoly}
-Let $\pxdb$ be a \bi over variables $\vct{X} = \{X_1, \ldots, X_\numvar\}$ and with probability distribution $\vct{p} = (\prob_1, \ldots, \prob_\numvar)$. For any \bi-lineage polynomial $\poly(\vct{X})$ we have
+Let $\pxdb$ be a \bi over variables $\vct{X} = \{X_1, \ldots, X_\numvar\}$ and with probability distribution $\vct{p} = (\prob_1, \ldots, \prob_\numvar)$. For any \bi-lineage polynomial $\poly(\vct{X})$ based on $\pxdb$ and some 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{X}}\pbox{\poly(\vct{X})}  = \rpoly(\vct{p}).
@@ -170,7 +170,7 @@ In steps \cref{p1-s1} and \cref{p1-s2}, by linearity of expectation (recall the
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-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.
+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}
@@ -178,11 +178,11 @@ Finally, observe \cref{p1-s5} by construction in \cref{lem:pre-poly-rpoly}, that
 
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 \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 in $\poly$.
+If $\poly$ is a \bi-lineage polynomial, then the expectation of $\poly$, i.e., $\expct\pbox{\poly} = \rpoly\left(\prob_1,\ldots, \prob_\numvar\right)$ can be computed in $O(|\smbOf{\poly}|)$, where $|\poly|$ denotes the total number of multiplication/addition operators in $\poly$.
 \end{Corollary}
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 \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
+Note that \cref{lem:exp-poly-rpoly} shows that $\expct\pbox{\poly} =$ $\rpoly(\prob_1,\ldots, \prob_\numvar)$.  Therefore, if $\poly$ is already in \abbrSMB 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 $O(\smb{|\poly|})$ computations.\qed
 \end{proof}
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