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@ -223,7 +223,7 @@ In the remainder of this work, we demonstrate that a $(1\pm\epsilon)$ (multiplic
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\input{two-step-model}
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Like set-probabilistic databases, our approach adopts the two-step intensional model of query evaluation, as illustrated in \Cref{fig:two-step}:
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We adopt the two-step intensional model of query evaluation used in set-\abbrPDB\xplural, as illustrated in \Cref{fig:two-step}:
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(i) \termStepOne (\abbrStepOne): Given input $\dbbase$ and $\query$, output every tuple $\tup$ that possibly satisfies $\query$, annotated with its lineage polynomial ($\poly(\vct{X})=\apolyqdt\inparen{\vct{X}}$);
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(ii) \termStepTwo (\abbrStepTwo): Given $\poly(\vct{X})$ for each tuple, compute $\expct\pbox{\poly(\vct{\randWorld})}$.
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Let $\timeOf{\abbrStepOne}(Q,\dbbase,\circuit)$ denote the runtime of \abbrStepOne when it outputs $\circuit$ (which is a representation of $\poly$ as an arithmetic circuit --- more on this representation shortly).
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@ -30,7 +30,7 @@ There exists a constant $\eps_0>0$ such that given an undirected graph $G=(\vset
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\end{hypo}
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%
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Based on the so called {\em Triangle detection hypothesis} (cf.~\cite{triang-hard}), which states that detection of whether $G$ has a triangle or not takes time $\Omega\inparen{|\edgeSet|^{4/3}}$, implies that in Conjecture~\ref{conj:graph} we can take $\eps_0\ge \frac 13$.
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The so called {\em Triangle detection hypothesis} (cf.~\cite{triang-hard}), which states that detecting the presence of triangles in $G$ takes time $\Omega\inparen{|\edgeSet|^{4/3}}$, implies that in Conjecture~\ref{conj:graph} we can take $\eps_0\ge \frac 13$.
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All of our hardness results rely on a simple lineage polynomial encoding of the edges of a graph.
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To prove our hardness result, consider a graph $G=(\vset, \edgeSet)$, where $|\edgeSet| = m$, $\vset = [\numvar]$. Our lineage polynomial has a variable $X_i$ for every $i$ in $[\numvar]$.
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