More touch up on the 2-col format.

approx_alg.tex

@@ -340,8 +340,8 @@ Consider when $\etree$ encodes the expression $(x_1 + x_2)(x_1 - x_2) + x_2^2$.
 
 \begin{figure}[h!]
 	\begin{tikzpicture}[thick, every tree node/.style={default_node, thick, draw=black, black, circle, text width=0.3cm, font=\bfseries, minimum size=0.65cm}, every child/.style={black}, edge from parent/.style={draw, thick},
-level 1/.style={sibling distance=1.25cm},
-level 2/.style={sibling distance=1.0cm},
+level 1/.style={sibling distance=0.95cm},
+level 2/.style={sibling distance=0.7cm},
 %level 2+/.style={sibling distance=0.625cm}
 %level distance = 1.25cm,
 %sibling distance = 1cm,

lin_sys.tex

@@ -29,8 +29,8 @@ Equation ~\ref{eq:ls-2-1} follows by \cref{lem:tri}.  Similarly ~\cref{eq:ls-2-2
 
 Now, by simple algebraic manipulations of ~\cref{lem:qE3-exp}, we deduce,
 \begin{align}
-&\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 - \numocc{\graph{2}}{\oneint}\prob\nonumber\\
-& - \big(\numocc{\graph{2}}{\twopathdis} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 \nonumber\\
+&\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\\
+&- \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}\\
@@ -103,17 +103,18 @@ Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:
 
 Looking at ~\cref{eq:LS-subtract}, 
 \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 - \numocc{\graph{3}}{\oneint}\prob \nonumber\\
-&- \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2\nonumber\\
+&\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 - \numocc{\graph{3}}{\oneint}\prob\nonumber\\
-&- \big(\numocc{\graph{3}}{\twopathdis} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2 - \left(\numocc{\graph{1}}{\ed} + \numocc{\graph{1}}{\twopath}\right)\prob\nonumber\\
-&+ \left(24\left(\numocc{\graph{1}}{\twopathdis} + 3\cdot\numocc{\graph{1}}{\threedis}\right) + 20\cdot\numocc{\graph{1}}{\oneint} + 4\cdot\numocc{\graph{1}}{\twopath}\right.\nonumber\\
-&\left.+ 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}}\left(3p^2 - p^3\right)\label{eq:lem3-G3-3} 
+&\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}.  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.
@@ -122,9 +123,10 @@ Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identi
 
 It then follows that
 %Removing $O(\numedge)$ computable terms to the other side of \cref{eq:LS-subtract}, we get
-\begin{equation}
-\mtrix{\rpoly_{G}}[3] = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} + 45 \cdot \numocc{\graph{1}}{\threedis}}\left(3p^2 - p^3\right)\label{eq:LS-G3'} 
-\end{equation}
+\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 $\cref{eq:2pd-3d}$.
@@ -132,10 +134,11 @@ and
 %For the LHS we get
 
 \begin{align*}
-&\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 - \numocc{\graph{3}}{\oneint}\prob \nonumber\\
-&- \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} + 3\numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath} \right.\nonumber\\
-&\left.+ 6 \cdot \numocc{\graph{1}}{\twodis}\right\}\left(3\prob^2 - \prob^3\right) 
+&\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\\
+& - \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*}
 
 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,

single_p.tex

@@ -155,9 +155,10 @@ Note that $f_k$ is properly defined.  For any $S \in \binom{E_k}{3}$, $|f(S)| \l
 \end{proof}
 \qed   
 
+\AR{TODO for {\em later}: I think the proof will be much easier to follow with figures: just drawing out $S\times \{0,1\}$ along with the $(e_i,b_i)$ explicity notated on the edges will make the proof much easier to follow.}
 
 \subsubsection{Three Matchings in $\graph{2}$}
-\AR{TODO for {\em later}: I think the proof will be much easier to follow with figures: just drawing out $S\times \{0,1\}$ along with the $(e_i,b_i)$ explicity notated on the edges will make the proof much easier to follow.}
+
 \begin{proof}[Proof of Lemma \ref{lem:3m-G2}]
 For each edge pattern $S$, we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$.  We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is the set of all $3$-edge subsets of the set 
 \begin{equation*}
@@ -172,7 +173,11 @@ Consider the $\eset{1} = \threedis$ pattern.  Note that edges in $\eset{2}$ are
 \begin{itemize}
 	\item Disjoint Two-Path ($\twopathdis$)
 \end{itemize}
-For $\eset{1} = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint.  This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path.  We can only pick either $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from $e_2$ and $e_3$.  Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$.  Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
+For $\eset{1} = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint.  This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path.  We can only pick either $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from $e_2$ and $e_3$.  Note that a four path allows there to be 3 possible 2 matchings, specifically, 
+\begin{equation*}
+\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}.
+\end{equation*}
+Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
 
 \begin{itemize}
 	\item $3$-star ($\oneint$)