Started pass on Sec 3
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%root:main.tex
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\section{Multiple Distinct $\prob$ Values}
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\section{Hardness of exact computation}
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We would like to argue for a compressed version of $\poly(\vct{w})$, in general $\expct_{\vct{w}}\pbox{\poly(\vct{w})}$ cannot be computed in linear time.
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\AR{Added the hardness result below.}
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Our hardness result is based on the following hardness result:
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We would like to argue for a compressed version of $\poly(\vct{w})$, in general $\expct_{\vct{w}}\pbox{\poly(\vct{w})}$ even for TIDB, cannot be computed in linear time. We will argue two flavors of such a hardness result. In Section~\ref{sec:multiple-p}, we argue that computing the expected value exactly for all query polynommials $\poly(\vct{X})$ for multiple values of $p$ is \sharpwonehard. However, this does not rule out the possibility of being able to solve the problem for a any {\em fixed} value of $p$ being say even in linear time. In Section~\ref{sec:single-p}, we rule out even this possibility (based on some popular hardness conjectures in fine-grained complexity).
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\subsection{Preliminaries}
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Our hardness results are based on (exactly) counting the number of occurrences of a fixed graph $H$ as a subgraph in $G$. Let $\numocc{G}{H}$ denote the number of occurrences of pattern $H$ in graph $G$. %, where, for example, $\numocc{G}{\ed}$ means the number of single edges in $G$.
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In particular, we will consider the problems of computing the following counts (given $G$ as an input in its adjaceny list representation): $\numocc{G}{\tri}$ (the number of triangles), $\numocc{G}{\threepath}$ (the number of $3$-paths), $\numocc{G}{\threedis}$ (the number of $3$-matchings or collection of three node disjoint edges) and its generalization $\numocc{G}{\kmatch}$ (the number of $k$-matchings or collections fo $k$ node disjoint edges).
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Our hardness result in Section~\ref{sec:multiple-p} is based on the following hardness result:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Theorem}[\cite{k-match}]
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\label{thm:k-match-hard}
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Given a positive integer $k$ and an undirected graph $G$ with no self-loops or parallel edges, counting the number of $k$-matchings in $G$ is \sharpwonehard.
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Given a positive integer $k$ and an undirected graph $G$ with no self-loops or parallel edges, computing $\numocc{G}{\kmatch}$ exactly is %counting the number of $k$-matchings in $G$ is
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\sharpwonehard.
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\end{Theorem}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -23,6 +31,11 @@ Consider the query $\poly_{G}(\vct{X}) = q_E(X_1,\ldots, X_\numvar) = \sum\limit
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For the following discussion, set $\poly_{G}^\kElem(\vct{X}) = \left(q_E(X_1,\ldots, X_\numvar)\right)^\kElem$.
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\subsection{Multiple Distinct $\prob$ Values}
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\label{sec:multiple-p}
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\begin{Lemma}\label{lem:qEk-multi-p}
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Let $\prob_0,\ldots, \prob_{2\kElem}$ be distinct values in $(0, 1]$. Then given the values $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for $0\leq i\leq 2\kElem$, the number of $\kElem$-matchings in $G$ can be computed in $poly(\kElem)$ time.
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\end{Lemma}
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%root: main.tex
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\subsection{Single $\prob$ value}
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\label{sec:single-p}
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In this discussion, let us fix $\kElem = 3$.
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@ -25,7 +26,7 @@ We need to list all possible edge patterns in an arbitrary $G$ consisting of $\l
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\item 3-matching ($\threedis$)--this subgraph is composed of three disjoint edges.
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\end{itemize}
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Let $\numocc{G}{H}$ denote the number of occurrences of pattern $H$ in graph $G$, where, for example, $\numocc{G}{\ed}$ means the number of single edges in $G$.
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%Let $\numocc{G}{H}$ denote the number of occurrences of pattern $H$ in graph $G$, where, for example, $\numocc{G}{\ed}$ means the number of single edges in $G$.
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For any graph $G$, the following formulas compute $\numocc{G}{H}$ for their respective patterns in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$.
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\begin{align}
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@ -316,4 +317,4 @@ The argument follows along the same lines as in the proof of \cref{lem:3p-G2}.
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\begin{proof}[Proof of Lemma \ref{lem:tri}]
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The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the simple fact that all cycles in $\graph{k}$ will have at least six edges.
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\end{proof}
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\qed
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\qed
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