Minor changes.

macros.tex

@@ -23,7 +23,7 @@
 \newcommand{\sch}{sch}
 \newcommand{\rw}{\textbf{W}}%\rw for random world
 \newcommand{\graph}[1]{G^{(#1)}}
-\newcommand{\linsys}[1]{LS(G_{#1})}
+\newcommand{\linsys}[1]{LS(\graph{#1})}
 \newcommand{\lintime}[1]{LT^{\graph{#1}}}
 \newcommand{\aug}[1]{AUG^{\graph{#1}}}
 \newcommand{\mtrix}[1]{M_{#1}}

poly-form.tex

@@ -265,9 +265,11 @@ For any graph $\graph{k}$, $\numocc{\graph{k}}{\tri} = 0$.
 \end{Lemma}
 
 \begin{Lemma}\label{lem:lin-sys}
-Using the identities for computing $\numocc{G}{\threedis}, \numocc{G}{\threepath}, \numocc{G}{\tri}$ for $G \in \{\graph{2}, \graph{3}\}$, a linear system can be developed and solved for the unknown quantities of $\numocc{\graph{1}}{\threedis}, \numocc{\graph{1}}{\threepath}$, and $\numocc{\graph{1}}{\tri}$.
+Using the identities for computing $\numocc{G}{\threedis}, \numocc{G}{\threepath}, \numocc{G}{\tri}$ for $G \in \{\graph{2}, \graph{3}\}$, there exists a linear system $\mtrix{\rpoly}\cdot (x~y~z~)^T = \vct{b}$ which can then be solved to determine the unknown quantities of $\numocc{\graph{1}}{\threedis}, \numocc{\graph{1}}{\threepath}$, and $\numocc{\graph{1}}{\tri}$.
 \end{Lemma}
 
+\AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
+
 Using \cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
 We then show that for any of the subgraphs $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$.  Then, using \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
 %$%^&*(
@@ -472,7 +474,7 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the
 \begin{proof}[Proof of Lemma \ref{lem:lin-sys}]
 
 In \cref{lem:qE3-exp} is the identity for $\rpoly_{G}(\prob,\ldots, \prob)$ when $\poly_{G}(\vct{X}) = q_E(X_1,\ldots, X_\numTup)^3$. 
-Let us maintain a vector $\vct{b}(G_1)$ to hold the entries for the terms that are computable in $O(m)$ time, for each of $G_1, G_2,$ and $G_3$.  From \cref{eq:LS-subtract}, $\vct{b}(G_1)[0] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\big)\prob^2$.
+Let us maintain a vector $\vct{b}(\graph{1})$ to hold the entries for the terms that are computable in $O(m)$ time, for each of $\graph{1}, \graph{2},$ and $\graph{3}$.  From \cref{eq:LS-subtract}, $\vct{b}(\graph{1})[0] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\big)\prob^2$.
 
  
 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}.
@@ -502,7 +504,7 @@ Note that there are terms computable in $O(m)$ time which can removed from $\lin
 \end{equation}
 and
 \begin{align*}
-\vct{b}(G_1)[1] &= \frac{\rpoly(\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 - \big(\numocc{\graph{2}}{\twopathdis} + \numocc{\graph{2}}{\threedis}\big)\prob^2 +\\
+\vct{b}(\graph{1})[1] &= \frac{\rpoly(\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 - \big(\numocc{\graph{2}}{\twopathdis} + \numocc{\graph{2}}{\threedis}\big)\prob^2 +\\
 &6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(\prob^2 - \prob^3\right) + 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right) - 2\cdot \numocc{\graph{1}}{\twopath}\prob
 \end{align*}
 
@@ -526,7 +528,7 @@ Removing terms to the other side of \cref{eq:LS-subtract}, we get
 \end{equation}
 and
 \begin{align*}
-\vct{b}(G_1)[2] =& \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 - \big(\numocc{\graph{3}}{\twopathdis} + \numocc{\graph{3}}{\threedis}\big)\prob^2 + \\
+\vct{b}(\graph{1})[2] =& \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 - \big(\numocc{\graph{3}}{\twopathdis} + \numocc{\graph{3}}{\threedis}\big)\prob^2 + \\
 & \pbrace{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath} + 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob
 \end{align*}
 
@@ -538,7 +540,7 @@ We now have a linear system consisting of three linear combinations, for $\graph
 \end{pmatrix},\]
 and the following linear equation 
 \begin{equation}
-\mtrix{\rpoly}\cdot (x~ y~ z~)^T = \vct{b}(G_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 $(0,1)$}