%root: main.tex

\subsubsection{Proof of~\cref{lem:lin-sys}}

Note that our goal is to compute $\vct{b}[i]$ for $i\in [3]$ in $O(m)$such that

\begin{align}
\label{eq:lin-eq-1}
&\numocc{G}{\tri} +\numocc{G}{\threepath}\cdot p - \numocc{G}{\threedis}\cdot (3p^2-p^3)  =\vct{b}[1]\\
\label{eq:lin-eq-2}
&-2\numocc{G}{\tri}\cdot (3p^2-p^3) -4\numocc{G}{\threepath}\cdot (3p^2-p^3) \nonumber\\
&+ 10\cdot \numocc{G}{\threedis}\cdot (3p^2-p^3)  =\vct{b}[2]\\
\label{eq:lin-eq-3}
&-18\numocc{G}{\tri}\cdot (3p^2-p^3) -21\numocc{G}{\threepath}\cdot (3p^2-p^3) \nonumber\\
&+45 \numocc{G}{\threedis}\cdot (3p^2-p^3)  =\vct{b}[3]
\end{align}

%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 

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}):
\begin{align*}
&\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 \nonumber\\
&- \numocc{\graph{1}}{\oneint}\prob - \big(\numocc{\graph{1}}{\twopathdis} + 3\numocc{\graph{1}}{\threedis}\big)\prob^2
\end{align*}
\begin{align*}
&\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\\
 &- \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+ \left(4\cdot\numocc{\graph{1}}{\oneint}+ 6\cdot\left(\numocc{\graph{1}}{\twopathdis}\right.\right.\nonumber\\
&\left.\left. + 3\cdot\numocc{\graph{1}}{\threedis}\right)\right)\left(3\prob^2 - \prob^3\right).
\end{align*}
\begin{align*}
&\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\\
& - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\\
& - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob + \left\{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} \right.  \right.\nonumber\\
&\left.\left.+ 3\numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath}\right.\\
&\left.+ 6 \cdot \numocc{\graph{1}}{\twodis}\right\}\cdot\left(3\prob^2 - \prob^3\right) 
\end{align*}

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.

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}.  
%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}.

\paragraph*{Derivation of~\cref{eq:lin-eq-2}}

%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,
Consider the following relations:
\begin{align}
& \numocc{\graph{2}}{\tri} + \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)\nonumber\\
= &\numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)\label{eq:ls-2-1}\\
= &2 \cdot \numocc{\graph{1}}{\twopath}\prob - \left(8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis} \right.\nonumber\\
&\left.+ 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}\right)\left(3\prob^2 - \prob^3\right)\label{eq:ls-2-2}\\
= &\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath} - 8\cdot\numocc{\graph{1}}{\threedis}\right.\nonumber\\
&\left.- 6\cdot\numocc{\graph{1}}{\twopathdis}\right)\cdot\left(3\prob^2 - p^3\right) + 2\cdot\numocc{\graph{1}}{\twopath}\prob\nonumber \\
&- 4\cdot\numocc{\graph{1}}{\oneint}\cdot\left(3\prob^2 - \prob^3\right).\label{eq:ls-2-3}
\end{align} 

%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 $\graph{k}$.

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.

Now, rearranging the terms in the identity of~\cref{lem:qE3-exp} and recalling $\prob\ne 0$ we deduce the following identities:
\begin{align}
&\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\\
&- \numocc{\graph{2}}{\oneint}\prob - \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 \nonumber\\
&=\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath}\right.\nonumber\\
&\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\\
&- 4\cdot\numocc{\graph{1}}{\oneint}\cdot\left(3\prob^2 - \prob^3\right)\label{eq:lem3-G2-1}\\
&\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\\
&- \numocc{\graph{2}}{\oneint}\prob- \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 \nonumber\\
&- 2\cdot\numocc{\graph{1}}{\twopath}\prob+ 4\cdot\numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right)\nonumber\\
&=\left(-2\cdot\numocc{\graph{1}}{\tri} - 4\cdot\numocc{\graph{1}}{\threepath} - 8\cdot\numocc{\graph{1}}{\threedis}\right. \nonumber\\
&\left.- 6\cdot\numocc{\graph{1}}{\twopathdis}\right)\cdot\left(3\prob^2 - p^3\right)\label{eq:lem3-G2-2}\\
&\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\\
&- \numocc{\graph{2}}{\oneint}\prob - \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2\nonumber\\
&- 2\cdot\numocc{\graph{1}}{\twopath}\prob + \left(4\cdot\numocc{\graph{1}}{\oneint}+ 6\cdot\left(\numocc{\graph{1}}{\twopathdis}\right.\right. \nonumber\\
&\left.\left.+ 3\cdot\numocc{\graph{1}}{\threedis}\right)\right)\left(3\prob^2 - \prob^3\right)\nonumber\\
&=\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}
\end{align}

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.  

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}.

%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] = $
%\begin{equation*}
%\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)
%\end{equation*}

%and


%By \cref{lem:tri}, the first term of $\linsys{2}$ is $0$, and then $\linsys{2} = \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)$.
%
%Replace the next term with the identity of \cref{lem:3p-G2} and the last term with the identity of \cref{lem:3m-G2},
%\begin{equation*}
%\linsys{2} = 2 \cdot \numocc{\graph{1}}{\twopath}\prob - \pbrace{8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis} + 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}}\left(3\prob^2 - \prob^3\right).
%\end{equation*}
%Rearrange terms into groups of those patterns that are 'hard' to compute and those that can be computed in $O(\numedge)$,
%\begin{equation*}
%\linsys{2} = -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + \left(8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis}\right)}\left(3\prob^2 - \prob^3\right) + \pbrace{2 \cdot \numocc{\graph{1}}{\twopath}\prob - 4 \cdot \numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right)}.
%\end{equation*}
%
%Note that there are terms computable in $O(\numedge)$ time which can be subtracted from $\linsys{2}$ and added to the other side of \cref{eq:LS-subtract}, i.e., $\vct{b}[2]$.  This leaves us with
%\begin{align}
%&\linsys{2} = \left(-2 \cdot \numocc{\graph{1}}{\tri} - 4 \cdot \numocc{\graph{1}}{\threepath} - 2 \cdot \numocc{\graph{1}}{\threedis} - 4\cdot \numocc{\graph{1}}{\twopathdis}\right) \cdot \left(3\prob^2 - \prob^3\right)\label{eq:LS-G2'}\\
%&\linsys{2} = \left(-2 \cdot \numocc{\graph{1}}{\tri} - 4 \cdot \numocc{\graph{1}}{\threepath} - 2 \cdot \numocc{\graph{1}}{\threedis} + 12 \cdot \numocc{\graph{1}}{\threedis}\right)\cdot \left(3\prob^2 - \prob^3\right)\label{eq:LS-G2'-1}\\
%&\linsys{2} = \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:LS-G2'-2}
%\end{align}
%
%Equation ~\ref{eq:LS-G2'} is the result of collecting $2\cdot\left(\numocc{\graph{1}}{\twopathdis} + 3\numocc{\graph{1}}{\threedis}\right)$ and moving them to the other side.  Then ~\cref{eq:LS-G2'-1} results from adding $4\cdot\left(\numocc{\graph{1}}{\twopathdis} + 3\numocc{\graph{1}}{\threedis}\right)$ to both sides.  Equation ~\ref{eq:LS-G2'-2} is the result of simplifying terms.
%
%For the left hand side, following the above steps, we obtain



%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.

\paragraph*{Derivation of~\cref{eq:lin-eq-3}}

%Following the same reasoning for $\graph{3}$, 
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
\begin{align}
&\numocc{\graph{3}}{\tri} + \numocc{\graph{3}}{\threepath}\prob - \numocc{\graph{3}}{\threedis}\left(3\prob^2 - \prob^3\right)\nonumber\\
=& \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\\
&\left.+ 18 \cdot \numocc{\graph{1}}{\tri} + 21 \cdot \numocc{\graph{1}}{\threepath} + 24 \cdot \numocc{\graph{1}}{\twopathdis} +\right.\nonumber\\
&\left.20 \cdot \numocc{\graph{1}}{\oneint} + 27 \cdot \numocc{\graph{1}}{\threedis}\right\}\left(3\prob^2 - \prob^3\right)\label{eq:LS-G3-sub}\\
=&\left\{ -18\numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 24 \cdot \numocc{\graph{1}}{\twopathdis}\right. \nonumber\\
&\qquad\left.- 27 \cdot \numocc{\graph{1}}{\threedis}\right\}\left(3\prob^2 - \prob^3\right) \nonumber\\
&+ \pbrace{-20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(3\prob^2 - \prob^3\right)\nonumber\\
&+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob. \label{eq:lem3-G3-1}
\end{align}

By the identity in~\cref{lem:qE3-exp} (along with the fact that $\prob\ne 0$), we get:
\begin{align}
&\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\\
& - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\nonumber\\
&= \left\{ -18\numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 24 \cdot \numocc{\graph{1}}{\twopathdis}\right. \nonumber\\
&\left.- 27 \cdot \numocc{\graph{1}}{\threedis}\right\}\left(3\prob^2 - \prob^3\right) \nonumber\\
&+ \pbrace{-20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(3\prob^2 - \prob^3\right)\nonumber\\
&+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob. \label{eq:lem3-G3-2}\\
&\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\\
&- \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2 - \left(\numocc{\graph{1}}{\ed}\right.\nonumber\\
&\left.+ \numocc{\graph{1}}{\twopath}\right)\prob+ \left(24\left(\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right) \right.\nonumber\\
&\left.+ 20\cdot\numocc{\graph{1}}{\oneint} + 4\cdot\numocc{\graph{1}}{\twopath}+ 6\cdot\numocc{\graph{1}}{\twodis}\right)\left(3\prob^2 - \prob^3\right)\nonumber\\
&= \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} + 45 \cdot \numocc{\graph{1}}{\threedis}}\nonumber\\
&\cdot\left(3p^2 - p^3\right)\label{eq:lem3-G3-3} 
\end{align}

%Equation ~\ref{eq:lem3-G3-2} follows from substituting ~\cref{eq:lem3-G3-2} in for the RHS of ~\cref{eq:LS-subtract}.  
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.

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}.

%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.

%It then follows that
%Removing $O(\numedge)$ computable terms to the other side of \cref{eq:LS-subtract}, we get
%\begin{align}
%&\mtrix{\rpoly_{G}}[3] = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} + 45 \cdot \numocc{\graph{1}}{\threedis}}\nonumber\\
%&\cdot\left(3p^2 - p^3\right)\label{eq:LS-G3'} 
%\end{align}
%and

%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 $\cref{eq:2pd-3d}$.
%
%For the LHS we get

\paragraph*{Wrapping it up.}

%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,
For notational convenience, define $x = \numocc{\graph{1}}{\tri}$, $y = \numocc{\graph{1}}{\threepath}, z = \numocc{\graph{1}}{\threedis}$. 
% Using $\linsys{2}$ and $\linsys{3}$, the following matrix is obtained,
If we denote
\[ \mtrix{\rpoly} = \begin{pmatrix}
1					& 	\prob	 				&	-(3\prob^2 - \prob^3)\\
-2(3\prob^2 - \prob^3) 	&	-4(3\prob^2 - \prob^3) 	&	10(3\prob^2 - \prob^3)\\
-18(3\prob^2 - \prob^3) 	&	-21(3\prob^2 - \prob^3) 	&	45(3\prob^2 - \prob^3)
\end{pmatrix},\]
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}),
\end{equation}
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}

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 $\prob\in (0,1)$. Towards that end, we will show that $\dtrm{\mtrix{\rpoly}}$ as a polynomial in $\prob$ does not have any root in $(0,1)$.

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)\\
-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{vmatrix}
= \\
&(3\prob^2 - \prob^3)^2 \cdot
\begin{vmatrix}
-4	&	10\\
-21	&	45
\end{vmatrix}
- \prob(3\prob^2 - \prob^3)^2~ \cdot
\begin{vmatrix}
-2	&	10\\
-18	&	45
\end{vmatrix}\\
&+ \left(- ~(3\prob^2 - \prob^3)^3\right)~ \cdot
\begin{vmatrix}
-2	&	-4\\
-18	&	-21
\end{vmatrix}.
\end{align*}

Compute each RHS term starting with the left and working to the right, 
\begin{align}
&(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot 45) - (-21 \cdot 10)\right) = (3\prob^2 - \prob^3)^2\cdot(-180 + 210)\nonumber\\
&= 30(3\prob^2 - \prob^3)^2.\label{eq:det-1}
\end{align}
The middle term then is
\begin{align}
&-\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot 45) - (-18 \cdot 10)\right) \nonumber\\
&= -\prob(3\prob^2 - \prob^3)^2 \cdot (-90 + 180) = -90\prob(3\prob^2 - \prob^3)^2.\label{eq:det-2}
\end{align}
Finally, the rightmost term,
\begin{align}
&-\left(3\prob^2 - \prob^3\right)^3 \cdot \left((-2 \cdot -21) - (-18 \cdot -4)\right) \nonumber\\
&= -\left(3\prob^2 - \prob^3\right)^3 \cdot  (42 - 72) = 30\left(3\prob^2 - \prob^3\right)^3.\label{eq:det-3}
\end{align}

Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
\begin{align}
&\dtrm{\mtrix{\rpoly}} =  30(3\prob^2 - \prob^3)^2 - 90\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3\nonumber\\
&= 30(3\prob^2 - \prob^3)^2\left(1 - 3\prob + (3\prob^2 - \prob^3)\right) \nonumber\\
&= 30\prob^4\left(3 - \prob\right)^2\left(-\prob^3 + 3\prob^2 - 3\prob + 1\right)\nonumber\\
&= 30\prob^4\left(3 - \prob\right)^2\left(1 - \prob\right)^3.\label{eq:det-final}
\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}

%\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}

%\qed