Done with pass on Sec 3
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lin_sys.tex
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lin_sys.tex
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%root: main.tex
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\subsubsection{Developing a Linear System}
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\subsubsection{Proof of~\cref{lem:lin-sys}}
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\begin{proof}[Proof of Lemma \ref{lem:lin-sys}]
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Our goal is to build a linear system $M \cdot (x~y~z)^T = \vct{b}$, such that, assuming an indexing starting at $1$, each $i^{th}$ row in $M$ corresponds to the RHS of ~\cref{eq:LS-subtract} for $\graph{i}$ \textit{in} terms of $\graph{1}$. The vector $\vct{b}$ analogously has the terms computable in $O(\numedge)$ time for each $\graph{i}$ at its corresponing $i^{th}$ entry for the LHS of ~\cref{eq:LS-subtract}. Lemma ~\ref{lem:qE3-exp} gives the identity for $\rpoly_{G}(\prob,\ldots, \prob)$ when $\poly_{G}(\vct{X}) = q_E(X_1,\ldots, X_\numvar)^3$, and using
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Note that our goal is to compute $\vct{b}[i]$ for $i\in [3]$ in $O(m)$such that
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~\cref{eq:LS-subtract}, $\vct{b}[1] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}\big)\prob^2$.
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\begin{align}
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\label{eq:lin-eq-1}
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&\numocc{G}{\tri} +\numocc{G}{\threepath}\cdot p - \numocc{G}{\threedis}\cdot (3p^2-p^3) =\vct{b}[1]\\
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\label{eq:lin-eq-2}
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&-2\numocc{G}{\tri}\cdot (3p^2-p^3) -4\numocc{G}{\threepath}\cdot (3p^2-p^3) + 10\cdot \numocc{G}{\threedis}\cdot (3p^2-p^3) =\vct{b}[2]\\
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\label{eq:lin-eq-3}
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&-18\numocc{G}{\tri}\cdot (3p^2-p^3) -21\numocc{G}{\threepath}\cdot (3p^2-p^3) +45 \numocc{G}{\threedis}\cdot (3p^2-p^3) =\vct{b}[3]
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\end{align}
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As previously outlined, assume graph $\graph{1}$ to be an arbitrary graph, with $\graph{2}, \graph{3}$ constructed from $\graph{1}$ as defined in \cref{def:Gk}.
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%Our goal is to build a linear system $M \cdot (x~y~z)^T = \vct{b}$, such that, assuming an indexing starting at $1$, each $i^{th}$ row in $M$ corresponds to the RHS of ~\cref{eq:LS-subtract} for $\graph{i}$ \textit{in} terms of $\graph{1}$. The vector $\vct{b}$ analogously has the terms computable in $O(\numedge)$ time for each $\graph{i}$ at its corresponing $i^{th}$ entry for the LHS of ~\cref{eq:LS-subtract}. Lemma ~\ref{lem:qE3-exp} gives the identity for $\rpoly_{G}(\prob,\ldots, \prob)$ when $\poly_{G}(\vct{X}) = q_E(X_1,\ldots, X_\numvar)^3$, and using
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\subsubsection{$\graph{2}$}
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Towards that end, we first state the values in $\vct{b}$ (where $\graph{1}=G$ while $\graph{2}$ and $\graph{3}$ follow from~\cref{def:Gk}):
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\[\vct{b}[1] = \frac{\rpoly_{\graph{1}}^3(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{\graph{1}}}{\ed}}{6\prob} - \numocc{\graph{1}}{\twopath} - \numocc{\graph{1}}{\twodis}\prob - \numocc{\graph{1}}{\oneint}\prob - \big(\numocc{\graph{1}}{\twopathdis} + 3\numocc{\graph{1}}{\threedis}\big)\prob^2\]
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\begin{align*}
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&\vct{b}[2] = \frac{\rpoly^3_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&- \numocc{\graph{2}}{\oneint}\prob- \left(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\right)\prob^2- 2\cdot \numocc{\graph{1}}{\twopath}\prob \\
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&+ \left(4\cdot\numocc{\graph{1}}{\oneint}+ 6\cdot\left(\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right)\right)\left(3\prob^2 - \prob^3\right).
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\end{align*}
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\begin{align*}
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&\vct{b}[3] = \frac{\rpoly^3_{\graph{3}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{3}}{\ed}}{6\prob} - \numocc{\graph{3}}{\twopath} - \numocc{\graph{3}}{\twodis}\prob \nonumber\\
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& - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\\
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& - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob + \left\{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} \right. \right.\nonumber\\
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&\left.\left.+ 3\numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath}\right.\\
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&\left.+ 6 \cdot \numocc{\graph{1}}{\twodis}\right\}\cdot\left(3\prob^2 - \prob^3\right)
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\end{align*}
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Let us call the linear equation for graph $\graph{2}$ $\linsys{2}$. Using the hard to compute terms of the RHS in ~\cref{lem:qE3-exp}, let us consider the RHS,
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Further, note that by~\cref{eq:1e} to~\cref{eq:2pd-3d} (and the fact that each of those quantities can be computed in $O(m)$ time) means that $\vct{b}$ can be computed in $O(m)$ time as needed. We first verify that all of~\cref{eq:lin-eq-1} to~\cref{eq:lin-eq-3} indeed hold.
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Note that~\cref{eq:lin-eq-1} follows from~\cref{lem:qE3-exp} and the definition of $\vct{b}$. Next, we derive~\cref{eq:lin-eq-2} and~\cref{eq:lin-eq-3}.
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%As previously outlined, assume graph $\graph{1}$ to be an arbitrary graph, with $\graph{2}, \graph{3}$ constructed from $\graph{1}$ as defined in \cref{def:Gk}.
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\paragraph*{Derivation of~\cref{eq:lin-eq-2}}
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%Let us call the linear equation for graph $\graph{2}$ $\linsys{2}$. Using the hard to compute terms of the RHS in ~\cref{lem:qE3-exp}, let us consider the RHS,
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Consider the following relations:
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\begin{align}
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& \numocc{\graph{2}}{\tri} + \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)\nonumber\\
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= &\numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)\label{eq:ls-2-1}\\
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@ -25,35 +51,37 @@ Let us call the linear equation for graph $\graph{2}$ $\linsys{2}$. Using the h
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%define $\linsys{2} = \numocc{\graph{2}}{\tri} + \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)$. By \cref{claim:four-two} we can compute $\linsys{2}$ in $O(T(\numedge) + \numedge)$ time with $\numedge = |E_2|$, and more generally, $\numedge = |E_k|$ for a graph $\graph{k}$.
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Equation ~\ref{eq:ls-2-1} follows by \cref{lem:tri}. Similarly ~\cref{eq:ls-2-2} follows by both \cref{lem:3m-G2} and \cref{lem:3p-G2}. Finally, ~\cref{eq:ls-2-3} follows by a simple rearrangement of terms.
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In the above,~\cref{eq:ls-2-1} follows by \cref{lem:tri}. Similarly ~\cref{eq:ls-2-2} follows by both \cref{lem:3m-G2} and \cref{lem:3p-G2}. Finally, ~\cref{eq:ls-2-3} follows by a simple rearrangement of terms.
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Now, by simple algebraic manipulations of ~\cref{lem:qE3-exp}, we deduce,
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Now, rearranging the terms in the identity of~\cref{lem:qE3-exp} and recalling $p\ne 0$ we deduce the following identities:
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\begin{align}
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&\frac{\rpoly_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&\frac{\rpoly^3_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&- \numocc{\graph{2}}{\oneint}\prob - \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 \nonumber\\
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&=\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath}\right.\nonumber\\
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&\left. - 8\cdot\numocc{\graph{1}}{\threedis} - 6\cdot\numocc{\graph{1}}{\twopathdis}\right)\cdot\left(3\prob^2 - p^3\right) + 2\cdot\numocc{\graph{1}}{\twopath}\prob\nonumber\\
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&- 4\cdot\numocc{\graph{1}}{\oneint}\cdot\left(3\prob^2 - \prob^3\right)\label{eq:lem3-G2-1}\\
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&\frac{\rpoly_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&\frac{\rpoly^3_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&- \numocc{\graph{2}}{\oneint}\prob- \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 \nonumber\\
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&- 2\cdot\numocc{\graph{1}}{\twopath}\prob+ 4\cdot\numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right)\nonumber\\
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&=\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath} - 8\cdot\numocc{\graph{1}}{\threedis}\right. \nonumber\\
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&\left.- 6\cdot\numocc{\graph{1}}{\twopathdis}\right)\cdot\left(3\prob^2 - p^3\right)\label{eq:lem3-G2-2}\\
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&\frac{\rpoly_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob\nonumber\\
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&\frac{\rpoly^3_{\graph{2}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob\nonumber\\
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&- \numocc{\graph{2}}{\oneint}\prob - \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2\nonumber\\
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&- 2\cdot\numocc{\graph{1}}{\twopath}\prob + \left(4\cdot\numocc{\graph{1}}{\oneint}+ 6\cdot\left(\numocc{\graph{1}}{\twopathdis}\right.\right. \nonumber\\
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&\left.\left.+ 3\cdot\numocc{\graph{1}}{\threedis}\right)\right)\left(3\prob^2 - \prob^3\right)\nonumber\\
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&=\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath} + 10\cdot\numocc{\graph{1}}{\threedis}\right)\cdot\left(3\prob^2 - \prob^3\right)\label{eq:lem3-G2-3}
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\end{align}
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Equation ~\ref{eq:lem3-G2-1} follows by substituting ~\cref{eq:ls-2-3} in the RHS. We then arrive with ~\cref{eq:lem3-G2-2} by adding the inverse of the last 3 terms of ~\cref{eq:ls-2-3} to both sides. Finally, we arrive at ~\cref{eq:lem3-G2-3} by adding the $O(\numedge)$ computable term (by ~\cref{eq:2pd-3d}) $6\left(\cdot\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right)$ to both sides.
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In the above,~\cref{eq:lem3-G2-1} follows by substituting ~\cref{eq:ls-2-3} in the RHS. We then arrive with ~\cref{eq:lem3-G2-2} by adding the inverse of the last 3 terms of ~\cref{eq:ls-2-3} to both sides. Finally, we arrive at ~\cref{eq:lem3-G2-3} by adding term $6\left(\cdot\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right)$ to both sides.
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Denote the matrix of the linear system as $\mtrix{\rpoly_{G}}$, where $\mtrix{\rpoly_{G}}[i]$ is the $i^{\text{th}}$ row of $\mtrix{\rpoly_{G}}$. From ~\cref{eq:lem3-G2-3} it follows that $\mtrix{\rpoly_{\graph{2}}}[2] = $
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\begin{equation*}
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\left(-2 \cdot \numocc{\graph{1}}{\tri} - 4 \cdot \numocc{\graph{1}}{\threepath} + 10 \cdot \numocc{\graph{1}}{\threedis}\right)\cdot \left(3\prob^2 - \prob^3\right)
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\end{equation*}
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Note that the LHS of~\cref{eq:lem3-G2-3} is $\vct{b}[2]$ and that~\cref{eq:lem3-G2-3} is the same as~\cref{eq:lin-eq-2}.
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and
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%Denote the matrix of the linear system as $\mtrix{\rpoly_{G}}$, where $\mtrix{\rpoly_{G}}[i]$ is the $i^{\text{th}}$ row of $\mtrix{\rpoly_{G}}$. From ~\cref{eq:lem3-G2-3} it follows that $\mtrix{\rpoly_{\graph{2}}}[2] = $
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%\begin{equation*}
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%\left(-2 \cdot \numocc{\graph{1}}{\tri} - 4 \cdot \numocc{\graph{1}}{\threepath} + 10 \cdot \numocc{\graph{1}}{\threedis}\right)\cdot \left(3\prob^2 - \prob^3\right)
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%\end{equation*}
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%and
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%By \cref{lem:tri}, the first term of $\linsys{2}$ is $0$, and then $\linsys{2} = \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)$.
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@ -79,17 +107,13 @@ and
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%For the left hand side, following the above steps, we obtain
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\begin{align*}
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&\vct{b}[2] = \frac{\rpoly(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob \nonumber\\
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&- \numocc{\graph{2}}{\oneint}\prob- \left(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\right)\prob^2- 2\cdot \numocc{\graph{1}}{\twopath}\prob \\
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&+ \left(4\cdot\numocc{\graph{1}}{\oneint}+ 6\cdot\left(\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right)\right)\left(3\prob^2 - \prob^3\right).
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\end{align*}
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We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$. Note that by ~\cref{eq:2pd-3d}, it is the case that any term of the form $x \cdot \left(\numocc{\graph{i}}{\twopathdis}\right.$ + $\left.3\cdot \numocc{\graph{i}}{\threedis}\right)$ is computable in linear time. By ~\cref{eq:1e}, ~\cref{eq:2p}, ~\cref{eq:2m}, and ~\cref{eq:3s} the same is true for $\numocc{\graph{i}}{\ed}$, $\numocc{\graph{i}}{\twopath}$, $\numocc{\graph{i}}{\twodis}$, and $\numocc{\graph{i}}{\oneint}$ respectively.
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%We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$. Note that by ~\cref{eq:2pd-3d}, it is the case that any term of the form $x \cdot \left(\numocc{\graph{i}}{\twopathdis}\right.$ + $\left.3\cdot \numocc{\graph{i}}{\threedis}\right)$ is computable in linear time. By ~\cref{eq:1e}, ~\cref{eq:2p}, ~\cref{eq:2m}, and ~\cref{eq:3s} the same is true for $\numocc{\graph{i}}{\ed}$, $\numocc{\graph{i}}{\twopath}$, $\numocc{\graph{i}}{\twodis}$, and $\numocc{\graph{i}}{\oneint}$ respectively.
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\subsubsection{$\graph{3}$}
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\paragraph*{Derivation of~\cref{eq:lin-eq-3}}
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Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, starting with the RHS of ~\cref{eq:LS-subtract}, we derive
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%Following the same reasoning for $\graph{3}$,
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Using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, we derive % starting with the RHS of ~\cref{eq:LS-subtract}, we derive
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\begin{align}
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&\numocc{\graph{3}}{\tri} + \numocc{\graph{3}}{\threepath}\prob - \numocc{\graph{3}}{\threedis}\left(3\prob^2 - \prob^3\right)\nonumber\\
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=& \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob - \left\{4 \cdot \numocc{\graph{1}}{\twopath} + 6 \cdot \numocc{\graph{1}}{\twodis}\right.\nonumber\\
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@ -101,7 +125,7 @@ Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:
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&+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob. \label{eq:lem3-G3-1}
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\end{align}
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Looking at ~\cref{eq:LS-subtract},
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By the identity in~\cref{lem:qE3-exp} (along with the fact that $p\ne 0$), we get:
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\begin{align}
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&\frac{\rpoly_{\graph{3}}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{3}}{\ed}}{6\prob} - \numocc{\graph{3}}{\twopath} - \numocc{\graph{3}}{\twodis}\prob \nonumber\\
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& - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\nonumber\\
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@ -117,50 +141,54 @@ Looking at ~\cref{eq:LS-subtract},
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&\cdot\left(3p^2 - p^3\right)\label{eq:lem3-G3-3}
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\end{align}
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Equation ~\ref{eq:lem3-G3-2} follows from substituting ~\cref{eq:lem3-G3-2} in for the RHS of ~\cref{eq:LS-subtract}. We derive ~\cref{eq:lem3-G3-3} by adding the inverse of all $O(\numedge)$ computable terms, and for the case of $\twopathdis$ and $\threedis$, we add the $O(\numedge)$ computable term $24\cdot\left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right)$ to both sides.
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%Equation ~\ref{eq:lem3-G3-2} follows from substituting ~\cref{eq:lem3-G3-2} in for the RHS of ~\cref{eq:LS-subtract}.
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In the above,~\cref{eq:lem3-G3-3} follows by moving terms and adding the term $24\cdot\left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right)$ to both sides.
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Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identities in ~\cref{lem:3m-G3}, ~\cref{lem:3p-G3}, and ~\cref{lem:tri}. We then get \cref{eq:LS-G3-rearrange} by simply rearranging the operands.
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Note that the LHS of~\cref{eq:lem3-G3-3} is $\vct{b}[3]$ and that~\cref{eq:lem3-G3-3} is indeed~\cref{eq:lin-eq-3}.
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It then follows that
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%Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identities in ~\cref{lem:3m-G3}, ~\cref{lem:3p-G3}, and ~\cref{lem:tri}. We then get \cref{eq:LS-G3-rearrange} by simply rearranging the operands.
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%It then follows that
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%Removing $O(\numedge)$ computable terms to the other side of \cref{eq:LS-subtract}, we get
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\begin{align}
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&\mtrix{\rpoly_{G}}[3] = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} + 45 \cdot \numocc{\graph{1}}{\threedis}}\nonumber\\
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&\cdot\left(3p^2 - p^3\right)\label{eq:LS-G3'}
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\end{align}
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and
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%\begin{align}
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%&\mtrix{\rpoly_{G}}[3] = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} + 45 \cdot \numocc{\graph{1}}{\threedis}}\nonumber\\
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%&\cdot\left(3p^2 - p^3\right)\label{eq:LS-G3'}
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%\end{align}
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%and
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%The same justification for the derivation of $\linsys{2}$ applies to the derivation above of $\linsys{3}$. To arrive at ~\cref{eq:LS-G3'}, we move $O(\numedge)$ computable terms to the left hand side. For the term $-24\cdot\numocc{\graph{1}}{\twopathdis}$ we need to add the inverse to both sides AND $72\cdot\numocc{\graph{1}}{\threedis}$ to both sides, in order to satisfy the constraint of $\cref{eq:2pd-3d}$.
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%
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%For the LHS we get
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\begin{align*}
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&\vct{b}[3] = \frac{\rpoly(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{3}}{\ed}}{6\prob} - \numocc{\graph{3}}{\twopath} - \numocc{\graph{3}}{\twodis}\prob \nonumber\\
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& - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\\
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& - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob + \left\{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} \right. \right.\nonumber\\
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&\left.\left.+ 3\numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath}\right.\\
|
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&\left.+ 6 \cdot \numocc{\graph{1}}{\twodis}\right\}\cdot\left(3\prob^2 - \prob^3\right)
|
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\end{align*}
|
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\paragraph*{Wrapping it up.}
|
||||
|
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We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. Note that the constants for $\graph{1}$ follow the RHS of ~\cref{eq:LS-subtract}. To make it easier, use the following variable representations: $x = \numocc{\graph{1}}{\tri}, y = \numocc{\graph{1}}{\threepath}, z = \numocc{\graph{1}}{\threedis}$. Using $\linsys{2}$ and $\linsys{3}$, the following matrix is obtained,
|
||||
%We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. Note that the constants for $\graph{1}$ follow the RHS of ~\cref{eq:LS-subtract}. To make it easier,
|
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For notational convenience, define $x = \numocc{\graph{1}}{\tri}, y = \numocc{\graph{1}}{\threepath}, z = \numocc{\graph{1}}{\threedis}$.
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% Using $\linsys{2}$ and $\linsys{3}$, the following matrix is obtained,
|
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If we denote
|
||||
\[ \mtrix{\rpoly} = \begin{pmatrix}
|
||||
1 & \prob & -(3\prob^2 - \prob^3)\\
|
||||
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\
|
||||
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3)
|
||||
\end{pmatrix},\]
|
||||
and the following linear equation
|
||||
then~\cref{eq:lin-eq-1} to~\cref{eq:lin-eq-3} implies that
|
||||
%and the following linear equation
|
||||
\begin{equation}
|
||||
\mtrix{\rpoly}\cdot (x~ y~ z~)^T = \vct{b}(\graph{1}).
|
||||
\mtrix{\rpoly}\cdot (x~ y~ z~)^T = \vct{b}(\graph{1}),
|
||||
\end{equation}
|
||||
\AR{
|
||||
Also the top right entry should be $-(p^2-p^3)$-- the negative sign is missing. This changes the rest of the calculations and has to be propagated. If my calculations are correct the final polynomial should be $-30p^2(1-p)^2(1-p-p^2+p^3)$. This still has no root in $(0,1)$}
|
||||
which proves the first part of the lemma.
|
||||
%\AR{
|
||||
%Also the top right entry should be $-(p^2-p^3)$-- the negative sign is missing. This changes the rest of the calculations and has to be propagated. If my calculations are correct the final polynomial should be $-30p^2(1-p)^2(1-p-p^2+p^3)$. This still has no root in $(0,1)$}
|
||||
|
||||
\AH{While propagating changes in ~\cref{eq:2pd-3d}, I noticed and corrected some errors, most notably, that for pulling out the \textbf{$a^2$} factor as described next, I hadn't squared it. That has been addressed. 110220}
|
||||
%\AH{While propagating changes in ~\cref{eq:2pd-3d}, I noticed and corrected some errors, most notably, that for pulling out the \textbf{$a^2$} factor as described next, I hadn't squared it. That has been addressed. 110220}
|
||||
|
||||
Now we seek to show that all rows of the system are indeed independent.
|
||||
Note that if $\mtrix{\rpoly}$ has full rank then one can compute $x,y,z$ in $O(1)$ using Gaussian elimination.
|
||||
%Now we seek to show that all rows of the system are indeed independent.
|
||||
%
|
||||
%The method of minors can be used to compute the determinant,
|
||||
To show that $\mtrix{\rpoly}$ indeed has full rank, we will show that $\dtrm{\mtrix{\rpoly}}\ne 0$ for every $p\in (0,1)$. Towards that end, we will show that $\dtrm{\mtrix{\rpoly}}$ as a polynomial in $p$ does not have any root in $(0,1)$.
|
||||
|
||||
The method of minors can be used to compute the determinant, $\dtrm{\mtrix{\rpoly}}$.
|
||||
|
||||
We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $ae$, the determinant is $a^2be - a^2cd = a^2(be - cd)$.
|
||||
We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $ae$, the determinant is $a^2be - a^2cd = a^2\cdot\begin{vmatrix} b&c \\d &e\end{vmatrix}$. We have $\dtrm{\mtrix{\rpoly}}$ is
|
||||
\begin{align*}
|
||||
&\begin{vmatrix}
|
||||
1 & \prob & -(3\prob^2 - \prob^3)\\
|
||||
|
|
@ -210,19 +238,19 @@ Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
|
|||
\end{align}
|
||||
|
||||
From ~\cref{eq:det-final} it can easily be seen that the roots of $\dtrm{\mtrix{\rpoly}}$ are $0, 1,$ and $3$. Hence there are no roots in $(0, 1)$ and ~\cref{lem:lin-sys} follows.
|
||||
\end{proof}
|
||||
%\end{proof}
|
||||
|
||||
\qed
|
||||
\begin{proof}[Proof of \cref{th:single-p}]
|
||||
The proof follows by ~\cref{lem:lin-sys}.
|
||||
\end{proof}
|
||||
\qed
|
||||
%\qed
|
||||
%\begin{proof}[Proof of \cref{th:single-p}]
|
||||
%The proof follows by ~\cref{lem:lin-sys}.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
||||
\begin{Corollary}\label{cor:single-p-gen-k}
|
||||
For every value $\kElem \geq 3$, there exists a query with $\kElem$ product width that is hard.
|
||||
\end{Corollary}
|
||||
\begin{proof}[Proof of Corollary ~\ref{cor:single-p-gen-k}]
|
||||
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.
|
||||
\end{proof}
|
||||
%\begin{Corollary}\label{cor:single-p-gen-k}
|
||||
%For every value $\kElem \geq 3$, there exists a query with $\kElem$ product width that is hard.
|
||||
%\end{Corollary}
|
||||
%\begin{proof}[Proof of Corollary ~\ref{cor:single-p-gen-k}]
|
||||
%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.
|
||||
%\end{proof}
|
||||
|
||||
\qed
|
||||
%\qed
|
||||
|
|
|
|||
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Reference in a new issue