122 lines
6.9 KiB
TeX
122 lines
6.9 KiB
TeX
%root: main.tex
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%!TEX 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 $\prob$. Indeed, it is easy to check that one can compute $\rpoly(\prob,\dots,\prob)$ exactly in linear time for $\prob\in \inset{0,1}$. In this section, we show that these two are the only possibilities:
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\begin{Theorem}\label{th:single-p-hard}
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Fix $\prob\in (0,1)$. Then assuming \Cref{conj:graph} is true, any algorithm that computes $\rpoly_{G}^3(\prob,\dots,\prob)$ from $G$ 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{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.
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Unlike \Cref{thm:mult-p-hard-result} the above result does not show that computing $\rpoly_{G}^3(\prob,\dots,\prob)$ for a fixed $\prob\in (0,1)$ is \sharpwonehard.
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However, 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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\begin{Theorem}\label{th:single-p}
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Fix $\prob\in (0,1)$. 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 $\numocc{G}{\tri}$ %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{th:single-p-hard}]
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For the sake of contradiction, 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 $\numocc{G}{\tri}$ 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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In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.
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%Before we move on to the proof itself, we state the results, lemmas, and defintions that will be useful in the proof.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection{The proofs}
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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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\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 $\prob$, we have:
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{\small
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\begin{align}
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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}\prob^4 + 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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Since $\prob$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to \Cref{eq:1e}-\Cref{eq:3p-3tri}). However, we need to generate one more independent linear equation in these two variables. Towards this end we generate another graph 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 inner vertexes of an $\ell$-path replacement edge are disjoint from the inner vertexes of any other $\ell$-path replacement edge. % 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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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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Next, we relate the various sub-graph counts in $\graph{2}$ to $\graph{1}$ ($G$).
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Lemma}\label{lem:3m-G2}
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The $3$-matchings in graph $\graph{2}$ satisfy the identity:
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\begin{align*}
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\numocc{\graph{2}}{\threedis} &= 8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis}\\
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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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\begin{Lemma}\label{lem:tri}
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For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$.
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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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Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, we can compute in $O(m)$ time a vector $\vct{b}\in\mathbb{R}^3$ such that
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\[ \begin{pmatrix}
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1 - 3p & -(3\prob^2 - \prob^3)\\
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10(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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\numocc{G}{\tri}]\\
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\numocc{G}{\threedis}
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\end{pmatrix}
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=\vct{b},
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\]
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allowing us to compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
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\end{Lemma}
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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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%
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This result immediately implies \Cref{th:single-p}:
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\begin{proof}[Proof of \Cref{th:single-p}]
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We can compute $\graph{2}$ from $\graph{1}$ in $O(m)$ time. Additionally, if in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, then the theorem follows by \Cref{lem:lin-sys}.
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\end{proof}
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\qed
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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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