sec 3
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Boris Glavic
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@ -4,8 +4,7 @@
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\label{sec:hard}
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\AH{The notation used here is different than in~\Cref{sec:background}, in particular~\Cref{eq:expect-q-nx}. Maybe we should decide on a notation and try to stick to it as much as possible?}
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\BG{We sometimes use $\expct_{\vct{X} \sim P}$ sometimes $\expct_{\vct{X}}$}
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In this section, we will prove that computing $\expct\limits_{\vct{X} \sim \pd}\pbox{\poly(\vct{X})}$ for a \tis-lineage polynomial $\poly(\vct{X})$ generated from a project-join query is \sharpwonehard. Note that this implies hardness for \bis and general $\semNX$-PDBs. Furthermore,
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using popular hardness conjectures in fine-grained complexity we demonstrate \Cref{sec:single-p} that the problem remains hard, even if $\pd(X_i) = p$ for all $X_i$ and some fixed valued $p$ as long as these conjectures hold.
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In this section, we will prove that computing $\expct\limits_{\vct{X} \sim \pd}\pbox{\poly(\vct{X})}$ for a \tis-lineage polynomial $\poly(\vct{X})$ generated from a project-join query is \sharpwonehard. Note that this implies hardness for \bis and general $\semNX$-PDBs. Furthermore, we demonstrate \Cref{sec:single-p} that the problem remains hard, even if $\pd(X_i) = p$ for all $X_i$ and some fixed valued $p$ as long as these conjectures hold. Finally, using popular hardness conjectures in fine-grained complexity we show that if these conjectures hold and except for the trivial choices of $p \in \{0,1\}$, the problem is hard for any given $p$.
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% We would like to argue for a compressed version of $\poly(\vct{X})$, in general $\expct\limits_{\vct{X} \sim \pd}\pbox{\poly(\vct{X})}$ even for tis, 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 $p$ 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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@ -120,7 +119,7 @@ Given that we then have $2\kElem + 1$ distinct values of $\rpoly_{G}^\kElem(\pro
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We claim that $c_{2\kElem}$ is $\kElem! \cdot \numocc{G}{\kmatch}$. This can be seen intuitively by looking at the original factorized representation
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\[\poly_{G}^\kElem(\vct{X}) = \sum_{\substack{(i_1, j_1),\\\cdots,\\(i_\kElem, j_\kElem) \in E}}X_{i_1}X_{j_1}\cdots X_{i_\kElem}X_{j_\kElem},\]
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where across each of the $\kElem$ products, an arbitrary $\kElem$-matching can be selected $\prod_{i = 1}^\kElem \kElem = \kElem!$ times. Next, note that each $\kElem$-matching $(i_1, j_1)\ldots$ $(i_k, j_k)$ in $G$ corresponds to the monomial $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ in $\poly_{G}^\kElem(\vct{X})$, with all indexes distinct. %Since each index is distinct, then each variable has an exponent $e = 1$ and this monomial survives in $\rpoly_{G}^{\kElem}(\vct{X})$ Since $\rpoly$ contains only exponents $e \leq 1$, the only degree $2\kElem$ terms that can exist in $\rpoly_{G}^\kElem$ are $\kElem$-matchings since every other monomial in $\poly_{G}^\kElem(\vct{X})$ has strictly less than $2\kElem$ distinct variables, which, as stated earlier implies that every other non-$\kElem$-matching monomial in $\rpoly_{G}^\kElem(\vct{X})$ has degree $< 2\kElem$.
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Second, the only surviving monomiasl $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ of degree exactly $2k$ in $\rpoly_{G}^{\kElem}(\vct{X})$ must have that all of $i_1,j_1,\dots,i_\kElem,j_\kElem$ are distinct in $\poly_{G}^{\kElem}(\vct{X})$. Then, by the last two statements, only monomials composed of $2k$ distinct variables in $\poly_{G}^{\kElem}(\vct{X})$ (and hence of degree $2\kElem$ in $\rpoly_{G}^{\kElem}(\vct{X})$) correspond to a $k$-matching in $G$.
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Second, the only surviving monomials $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ of degree exactly $2k$ in $\rpoly_{G}^{\kElem}(\vct{X})$ must have that all of $i_1,j_1,\dots,i_\kElem,j_\kElem$ are distinct in $\poly_{G}^{\kElem}(\vct{X})$. Then, by the last two statements, only monomials composed of $2k$ distinct variables in $\poly_{G}^{\kElem}(\vct{X})$ (and hence of degree $2\kElem$ in $\rpoly_{G}^{\kElem}(\vct{X})$) correspond to a $k$-matching in $G$.
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%It has already been established above that a $\kElem$-matching ($\kmatch$) has coefficient $c_{2\kElem}$. As noted, a $\kElem$-matching occurs when there are $\kElem$ edges, $e_1, e_2,\ldots, e_\kElem$, such that all of them are disjoint, i.e., $e_1 \neq e_2 \neq \cdots \neq e_\kElem$. In all $\kElem$ factors of $\poly_{G}^\kElem(\vct{X})$ there are $k$ choices from the first factor to select an edge for a given $\kElem$ matching, $\kElem - 1$ choices in the second factor, and so on throughout all the factors, yielding $\kElem!$ duplicate terms for each $\kElem$ matching in the expansion of $\poly_{G}^\kElem(\vct{X})$.
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Notice that %we have $\kElem!$ duplicates of
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81
single_p.tex
81
single_p.tex
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%root: main.tex
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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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While~\cref{thm:mult-p-hard-result} shows that computing $\rpoly(\prob,\dots,\prob)$ in general is hard it does not rule out the possibility that one can compute this value exactly for a {\em fixed} value of $p$. Indeed, it is easy to check that one can compute $\rpoly(\prob,\dots,\prob)$ exactly in linear time for $p\in \inset{0,1}$. In this section, we show that these two are the only possibilities:
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While \Cref{thm:mult-p-hard-result} shows that computing $\rpoly(\prob,\dots,\prob)$ in general is hard it does not rule out the possibility that one can compute this value exactly for a {\em fixed} value of $p$. Indeed, it is easy to check that one can compute $\rpoly(\prob,\dots,\prob)$ exactly in linear time for $p\in \inset{0,1}$. In this section, we show that these two are the only possibilities:
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\begin{Theorem}\label{cor:single-p-hard}
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Fix $p\in (0,1)$. Then assuming~\cref{conj:graph} is true, then any algorithm that computes $\rpoly_{G}^3(\prob,\dots,\prob)$ exactly has to run in time $\Omega\inparen{\abs{E(G)}^{1+\eps_0}}$, where $\eps_0$ is as defined in~\cref{conj:graph}.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Theorem}\label{th:single-p-hard}
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Fix $p\in (0,1)$. Then assuming \Cref{conj:graph} is true, then any algorithm that computes $\rpoly_{G}^3(\prob,\dots,\prob)$ exactly has to run in time $\Omega\inparen{\abs{E(G)}^{1+\eps_0}}$, where $\eps_0$ is as defined in \Cref{conj:graph}.
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\end{Theorem}
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%\begin{proof}[Proof of Corollary ~\ref{cor:single-p-gen-k}]
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%Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$. By ~\cref{th:single-p}, query $\poly''$ with $\kElem = 4$ has $\Omega(\numvar^{\frac{4}{3}})$ complexity.
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%\begin{proof}[Proof of Corollary ~\ref{th:single-p-gen-k}]
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%Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$. By \Cref{th:single-p}, query $\poly''$ with $\kElem = 4$ has $\Omega(\numvar^{\frac{4}{3}})$ complexity.
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%\end{proof}
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The above shows the hardness for a very specific query polynomial but it is easy to come up with an infinite family of hard query polynomials by `embedding' $\rpoly_{G}^3$ into an infinite family of trivial query polynomials. However, unlike~\cref{thm:mult-p-hard-result} the above result does not show that computing $\rpoly_{G}^3(\prob,\dots,\prob)$ for a fixed $p\in (0,1)$ is \sharpwonehard. By contrast, in~\cref{sec:algo} we show that if we are willing to compute an approximation that this problem (and indeed solving our problem for a much more general setting) is in linear time.
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The above shows the hardness for a very specific query polynomial but it is easy to come up with an infinite family of hard query polynomials by `embedding' $\rpoly_{G}^3$ into an infinite family of trivial query polynomials. However, unlike \Cref{thm:mult-p-hard-result} the above result does not show that computing $\rpoly_{G}^3(\prob,\dots,\prob)$ for a fixed $p\in (0,1)$ is \sharpwonehard. By contrast, in \Cref{sec:algo} we show that if we are willing to compute an approximation that this problem (and indeed solving our problem for a much more general setting) is in linear time.
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%\AH{@atri needs to put in the result for triangles of $\numvar^{\frac{4}{3}}$ runtime.}
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We will prove the above result by the following reduction:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Theorem}\label{th:single-p}
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Fix $p\in (0,1)$. Let $G$ be a graph on $\numedge$ edges.
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If we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ exactly in $T(\numedge)$ time, then we can exactly compute $\numocc{G}{\tri}$, $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ %count the number of triangles, 3-paths, and 3-matchings in $G$
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in $O\inparen{T(\numedge) + \numedge}$ time.
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\end{Theorem}
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\begin{proof}[Proof of~\cref{cor:single-p-hard}]
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{proof}[Proof of \Cref{th:single-p}]
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For the sake of contradiction, let us assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
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Let $G$ be the input graph. It is easy to see that one can compute the expression tree for $\poly_{G}^3(\vct{X})$ in $O(m)$ time. Then by~\cref{th:single-p} we can compute $\numocc{G}{\tri}$, $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ in further time $o\inparen{m^{1+\eps_0}}+O(m)$. Thus, the overall, reduction takes $o\inparen{m^{1+\eps_0}}+O(m)= o\inparen{m^{1+\eps_0}}$ time, which violates~\cref{conj:graph}.
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Let $G$ be the input graph. It is easy to see that one can compute the expression tree for $\poly_{G}^3(\vct{X})$ in $O(m)$ time. Then by \Cref{th:single-p} we can compute $\numocc{G}{\tri}$, $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ in further time $o\inparen{m^{1+\eps_0}}+O(m)$. Thus, the overall, reduction takes $o\inparen{m^{1+\eps_0}}+O(m)= o\inparen{m^{1+\eps_0}}$ time, which violates \Cref{conj:graph}.
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\end{proof}
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\qed
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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Before moving on to prove ~\cref{th:single-p}, let us state the results, lemmas and defintions that will be useful in the proof.
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Before moving on to prove \Cref{th:single-p-hard}, let us state the results, lemmas and defintions that will be useful in the proof.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection{Preliminaries and Notation}
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We need to list all possible edge patterns in an arbitrary $G$ consisting of at most three distinct edges. We have already seen $\tri,\threepath$ and $\threedis$, so here we define the remaining patterns:
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@ -48,7 +54,7 @@ We need to list all possible edge patterns in an arbitrary $G$ consisting of at
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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 for $\numocc{G}{H}$ for their respective patterns can be used to compute them exactly in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$ (proofs are in~\cref{app:easy-counts}):
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For any graph $G$, the following formulas for $\numocc{G}{H}$ for their respective patterns can be used to compute them exactly in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$ (proofs are in \Cref{app:easy-counts}):
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\begin{align}
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&\numocc{G}{\ed} = \numedge, \label{eq:1e}\\
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&\numocc{G}{\twopath} = \sum_{i \in V} \binom{d_i}{2} \label{eq:2p}\\
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@ -57,9 +63,9 @@ For any graph $G$, the following formulas for $\numocc{G}{H}$ for their respecti
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&\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis} = \sum_{(i, j) \in E} \binom{\numedge - d_i - d_j + 1}{2}\label{eq:2pd-3d}
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\end{align}
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%A quick argument to why \cref{eq:2m} is true. Note that for edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other edges in $G$ disjoint to $(i, j)$ is equivalent to finding all edges that are not connected to either vertex $i$ or $j$. The number of such edges is $m - d_i - d_j + 1$, where we add $1$ since edge $(i, j)$ is removed twice when subtracting both $d_i$ and $d_j$. Since the summation is iterating over all edges such that a pair $\left((i, j), (k, \ell)\right)$ will also be counted as $\left((k, \ell), (i, j)\right)$, division by $2$ then eliminates this double counting.
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%A quick argument to why \Cref{eq:2m} is true. Note that for edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other edges in $G$ disjoint to $(i, j)$ is equivalent to finding all edges that are not connected to either vertex $i$ or $j$. The number of such edges is $m - d_i - d_j + 1$, where we add $1$ since edge $(i, j)$ is removed twice when subtracting both $d_i$ and $d_j$. Since the summation is iterating over all edges such that a pair $\left((i, j), (k, \ell)\right)$ will also be counted as $\left((k, \ell), (i, j)\right)$, division by $2$ then eliminates this double counting.
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%\cref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in ~\cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. Note, the factor $3$ of $\threedis$ is necessary to account for the triple counting of $3$-matchings. It is also the case that, since the two path in $\twopathdis$ is connected, that there will be no double counting by the fact that the summation automatically 'disconnects' the current edge, meaning that a two matching at the current vertex will not be counted. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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%\Cref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in \Cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. Note, the factor $3$ of $\threedis$ is necessary to account for the triple counting of $3$-matchings. It is also the case that, since the two path in $\twopathdis$ is connected, that there will be no double counting by the fact that the summation automatically 'disconnects' the current edge, meaning that a two matching at the current vertex will not be counted. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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%Original lemma proving the exact coefficient terms in qE3
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\subsubsection{The proofs}
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Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
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Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Lemma}\label{lem:qE3-exp}
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%When we expand $\poly_{G}^3(\vct{X})$ out and assign all exponents $e \geq 1$ a value of $1$, we have the following result,
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For any $p$, we have:
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&\rpoly_{G}^3(\prob,\ldots, \prob) = \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis} + 6\numocc{G}{\tri}\prob^3\nonumber\\
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&+ 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
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\end{align}
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\end{Lemma}
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\end{Lemma}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{proof}%[Proof of \cref{lem:qE3-exp}]
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{proof}%[Proof of \Cref{lem:qE3-exp}]
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By definition we have that
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\[\poly_{G}^3(\vct{X}) = \sum_{\substack{(i_1, j_1),\\ (i_2, j_2),\\ (i_3, j_3) \in E}} \prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}.\]
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Hence $\rpoly_{G}^3(\vct{X})$ has degree six. Note that the monomial $\prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}$ will contribute to the coefficient of $p^i$ in $\rpoly_{G}^3(\vct{X})$, where $i$ is the number of distinct variables in the monomial.
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\textsc{case 3:} All $e_1,e_2$ and $e_3$ are distinct. For this case, we have $3! = 6$ permutations of $(e_1, e_2, e_3)$, each of which contribute to a different monomial in the SOP expansion of $\poly_{G}^3(\vct{X})$. This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $p^3,p^4,p^4,p^5$ and $p^6$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\end{proof}
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\qed
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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Since $p$ is fixed,~\cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri},$ $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to~\cref{eq:1e}-\cref{eq:2pd-3d}). However, we plan to generate two more independent linear equations in these three variables. Towards, this end we generate more graphs that are related to $G$:
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Since $p$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri},$ $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to \Cref{eq:1e}-\Cref{eq:2pd-3d}). However, we plan to generate two more independent linear equations in these three variables. Towards, this end we generate more graphs that are related to $G$:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Definition}\label{def:Gk}
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For $\ell > 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary graph $\graph{1}$, by replacing every edge $e$ of $\graph{1}$ with a $\ell$-path, such that all $\ell$-path replacement edges are disjoint. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
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\end{Definition}
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Next, we relate the various sub-graph counts in $\graph{2}$ and $\graph{3}$ to their counterparts in $\graph{1}=G$.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Lemma}\label{lem:3m-G2}
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The number of $3$-matchings in graph $\graph{2}$ satisfies the following identity,
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\begin{align*}
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&+ 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}.
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\end{align*}
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\end{Lemma}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Lemma}\label{lem:3m-G3}
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The number of 3-matchings in $\graph{3}$ satisfy the following identity,
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\begin{align*}
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\end{align*}
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\end{Lemma}
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\begin{Lemma}\label{lem:3p-G2}
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The number of $3$-paths in $\graph{2}$ satisfies the following identity,
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\[\numocc{\graph{2}}{\threepath} = 2 \cdot \numocc{\graph{1}}{\twopath}.\]
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\end{Lemma}
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\begin{Lemma}\label{lem:3p-G3}
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The number of $3$-paths in $\graph{3}$ satisfies the following identity,
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\[\numocc{\graph{3}}{\threepath} = \numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}.\]
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\end{Lemma}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Lemma}\label{lem:tri}
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For $\ell > 1$, any graph $\graph{\ell}$ has the property that $\numocc{\graph{\ell}}{\tri} = 0$.
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\end{Lemma}
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Using the results we have obtained so far, we will prove the following reduction result:
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\begin{Lemma}\label{lem:lin-sys}
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%Using the identities of lemmas [\ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri}] to compute $\numocc{G}{\threedis}, \numocc{G}{\threepath}, \numocc{G}{\tri}$ for $G \in \{\graph{2}, \graph{3}\}$, there exists a linear system $\mtrix{\rpoly}\cdot (x~y~z~)^T = \vct{b}$ which can then be solved to determine the unknown quantities of $\numocc{\graph{1}}{\threedis}, \numocc{\graph{1}}{\threepath}$, and $\numocc{\graph{1}}{\tri}$.
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Fix $p\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$, we can compute in $O(m)$ time a vector $\vct{b}\in\rel^3$ such that
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\]
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from which we can compute $\numocc{G}{\tri}, \numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
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\end{Lemma}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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Due to lack of space we defer the proof of the above results to~\Cref{subsec:proofs-struc-lemmas}.
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Due to lack of space we defer the proof of the above results to \Cref{subsec:proofs-struc-lemmas}.
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The above result immediately implies~\cref{th:single-p}:
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\begin{proof}[Proof of~\cref{th:single-p}]
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The above result immediately implies \Cref{th:single-p-hard}:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{proof}[Proof of \Cref{th:single-p-hard}]
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It is easy to check that in $O(m)$ time we can compute $\graph{2}$ and $\graph{3}$ from $\graph{1}=G$ (and further note that these graphs also have $O(m)$ edges). Thus,
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in time $O(T(m))$, we can compute $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$.~\Cref{lem:lin-sys} then completes the proof.
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in time $O(T(m))$, we can compute $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$. \Cref{lem:lin-sys} then completes the proof.
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\end{proof}
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\qed
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%\AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% \AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
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%Using \cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
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%We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using ~\cref{claim:four-two}, \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
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%Using \Cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
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%We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using \Cref{claim:four-two}, \Cref{lem:qE3-exp} and \Cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
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%$%^&*(
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@ -174,3 +200,8 @@ in time $O(T(m))$, we can compute $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ f
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "main"
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%%% End:
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Reference in a new issue