%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}):
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}.
%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,
%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.
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] = $
%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},
%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
%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.
%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.
%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 $\cref{eq:2pd-3d}$.
%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,
%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}
To show that $\mtrix{\rpoly}$ indeed has full rank, we will show that $\dtrm{\mtrix{\rpoly}}\ne0$ 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
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.
%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.