65 lines
3.3 KiB
TeX
65 lines
3.3 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 $\Omega\inparen{m^{1+\eps_0}}$, 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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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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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The following result immediately implies \Cref{th:single-p}:
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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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%
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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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