Reworked example after Problem 1.6 to be simpler.
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@ -214,35 +214,34 @@ Given one circuit $\circuit$ that encodes $\apolyqdt$ for all result tuples $\tu
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\end{Problem}
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For an upper bound on approximating the expected count, it is easy to check that if all the probabilties are constant then (with an additive adjustment) $\poly\left(\prob_1,\dots, \prob_n\right)$ (i.e. evaluating the original lineage polynomial over the probability values) is a constant factor approximation.
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This is illustrated in the following example using $\query_1^2$ from earlier. To aid in presentation we assume $\bound = 2$ for variable $X$ and $\bound = 1$ for all other variables. Let $\prob_A$ denote $\probOf\pbox{A = 1}$.
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This is illustrated in the following example using $\query_1^2$ from earlier. To aid in presentation we again limit our focus to $\refpoly{1, }^{\inparen{ABX}^2}$, assume $\bound = 2$ for variable $X$ and $\bound = 1$ for all other variables. Let $\prob_A$ denote $\probOf\pbox{A = 1}$.
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In computing $\rpoly$, we have some cancellations to deal with:
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\begin{footnotesize}
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\begin{align*}
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\refpoly{1, }^2\inparen{\vct{X}} &= A^2\inparen{X_1^2 + 4X_1X_2 + 4X_2^2}B^2 + B^2Y^2E^2 + B^2Z^2C^2 + 2AX_1B^2YE \\
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&\qquad+ 2AX_2B^2YE + 2AX_1B^2ZC + 2AX_2B^2ZC + 2B^2YEZC\\
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\end{align*}
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\begin{equation*}
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\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{X}} = A^2\inparen{X_1^2 + 4X_1X_2 + 4X_2^2}B^2 =A^2X_1^2B^2 + 4A^2X_1X_2B^2+4A^2X_2^2B^2
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%&\qquad+ 2AX_2B^2YE + 2AX_1B^2ZC + 2AX_2B^2ZC + 2B^2YEZC\\
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\end{equation*}
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\end{footnotesize}
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This then implies
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\begin{footnotesize}
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\begin{align*}
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\rpoly^2\inparen{\vct{X}} &= AX_1B+4AX_2B+BYE+BZC+2AX_1BYE+2AX_2BYE+2AX_1BZC\\
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&\qquad+2AX_2BZC+2BYEZC\\
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\end{align*}
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\end{footnotesize}
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Substituting $\vct{\prob}$ for $\vct{X}$, $\refpoly{1, }^2\inparen{\probAllTup} =$
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%\begin{footnotesize}
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%\begin{equation*}
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$\rpoly_1^{\inparen{ABX}^2}\inparen{\vct{X}} = AX_1B+4AX_2B$.
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%\end{equation*}
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%\end{footnotesize}
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Substituting $\vct{\prob}$ for $\vct{X}$,
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\begin{footnotesize}
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\begin{align*}
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\hspace*{-3mm}
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&\prob_A^2\prob_{X_1}^2\prob_B^2 + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2 + 4\prob_A^2\prob_{X_2}^2\prob_B^2 + \prob_B^2\prob_Y^2\prob_E^2 + \prob_B^2\prob_Z^2\prob_C^2 + 2\prob_A\prob_{X_1}\prob_B^2\prob_Y\prob_E\\
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&\qquad+ 2\prob_A\prob_{X_2}\prob_B^2\prob_Y\prob_E + 2\prob_A\prob_{X_1}\prob_B^2\prob_Z\prob_C + 2\prob_A\prob_{X_2}\prob_B^2\prob_Z\prob_C+ 2\prob_B^2\prob_Y\prob_E\prob_Z\prob_C\\
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&\leq\prob_A\prob_{X_1}\prob_B + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2 + 4\prob_A\prob_{X_2}\prob_b + \prob_B\prob_Y\prob_E + \prob_B\prob_Z\prob_C + 2\prob_A\prob_{X_1}\prob_B\prob_Y\prob_E \\
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&\qquad + 2\prob_A\prob_{X_2}\prob_B\prob_Y\prob_E+ 2\prob_A\prob_{X_1}\prob_B\prob_Z\prob_C + 2\prob_A\prob_{X_2}\prob_B\prob_Z\prob_C + 2\prob_B\prob_Y\prob_E\prob_Z\prob_C\\
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&= \rpoly_1^2\inparen{\vct{p}} + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2.
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\refpoly{1, }^{\inparen{ABX}^2}\inparen{\probAllTup} &=\prob_A^2\prob_{X_1}^2\prob_B^2 + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2 + 4\prob_A^2\prob_{X_2}^2\prob_B^2\\% + \prob_B^2\prob_Y^2\prob_E^2 + \prob_B^2\prob_Z^2\prob_C^2 + 2\prob_A\prob_{X_1}\prob_B^2\prob_Y\prob_E\\
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%&\qquad+ 2\prob_A\prob_{X_2}\prob_B^2\prob_Y\prob_E + 2\prob_A\prob_{X_1}\prob_B^2\prob_Z\prob_C + 2\prob_A\prob_{X_2}\prob_B^2\prob_Z\prob_C+ 2\prob_B^2\prob_Y\prob_E\prob_Z\prob_C\\
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&\leq\prob_A\prob_{X_1}\prob_B + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2 + 4\prob_A\prob_{X_2}\prob_b\\% + \prob_B\prob_Y\prob_E + \prob_B\prob_Z\prob_C + 2\prob_A\prob_{X_1}\prob_B\prob_Y\prob_E \\
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%&\qquad + 2\prob_A\prob_{X_2}\prob_B\prob_Y\prob_E+ 2\prob_A\prob_{X_1}\prob_B\prob_Z\prob_C + 2\prob_A\prob_{X_2}\prob_B\prob_Z\prob_C + 2\prob_B\prob_Y\prob_E\prob_Z\prob_C\\
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&= \rpoly_1^{\inparen{ABX}^2}\inparen{\vct{p}} + 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2.
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\end{align*}
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\end{footnotesize}
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If we assume that all probability values are at least $p_0>0$, then given access to $\refpoly{1, }^2\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$
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we get that $\refpoly{1, }^2\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$ is in the range $\left(\inparen{p_0}^3\cdot\inparen{\rpoly^2_1\vct{\prob}}, \rpoly_1^2\inparen{\vct{\prob}}\right]$.
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If we assume that all probability values are at least $p_0>0$, then given access to $\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$
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we get that $\refpoly{1, }^{\inparen{ABX}^2}\inparen{\vct{\prob}} - 4\prob_A^2\prob_{X_1}\prob_{X_2}\prob_B^2$ is in the range $\pbox{p_0^3\cdot\rpoly^{\inparen{ABX}^2}_1\inparen{\vct{\prob}}, \rpoly_1^{\inparen{ABX}^2}\inparen{\vct{\prob}}}$.
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%We can simulate sampling from $\refpoly{1, }^2\inparen{\vct{X}}$ by sampling monomials from $\refpoly{1, }^2$ while ignoring any samples $A^2X_1X_2B^2$.
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Note however, that this is \emph{not a tight approximation}.
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In~\cref{sec:algo} we demonstrate that a $(1\pm\epsilon)$ (multiplicative) approximation with competitive performance is achievable.
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