Added explicit detail to the proofs where eq(10) was propagated.
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@ -165,11 +165,11 @@ By definition we have that
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\begin{proof}[Proof of Claim \ref{claim:four-two}]
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%We have shown that the following subgraph cardinalities can be computed in $O(\numedge)$ time:
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%\[\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis}, \numocc{G}{\oneint}, \numocc{G}{\twopathdis} + \numocc{G}{\threedis}.\]
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It has already been shown previously that $\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis},$ and $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$ can be computed in O(\numedge) time.
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It has already been shown previously that $\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis},$ and $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$ can be computed in $O(\numedge)$ time.
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Using the result of \cref{lem:qE3-exp}, let us show a derivation to the identity of the consequent in \cref{claim:four-two}.
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All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d} show that we can compute the respective edge patterns in $O(\numedge)$ time. Rearrange $\rpoly_{G}$ with all linear time computations on one side, leaving only the hard computations,
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All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d} show that we can compute the respective edge patterns in $O(\numedge)$ time. Rearrange ~\cref{claim:four-one}, $\rpoly_{G}$, with all linear time computations on one side, leaving only the hard computations,
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\begin{align}
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&\rpoly_{G}(\prob,\ldots, \prob) = \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\tri}\prob^3 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6\nonumber\\
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&\rpoly_{G}(\prob,\ldots, \prob) - \numocc{G}{\ed}\prob^2 - 6\numocc{G}{\twopath}\prob^3 - 6\numocc{G}{\twodis}\prob^4 - 6\numocc{G}{\oneint}\prob^4 = 6\numocc{G}{\tri}\prob^3 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6\label{eq:LS-rearrange}\\
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@ -514,10 +514,10 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the
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\subsection{Developing a Linear System}
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\AH{The changes in ~\cref{eq:2pd-3d} have been propagated 110220. Barring any errors, everything should be updated and correct.}
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\AH{The changes in ~\cref{eq:2pd-3d} have been propagated 110420. Barring any errors, everything should be updated and correct.}
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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. 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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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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%Let us maintain a vector $\vct{b}$ to hold the entries for the terms that are computable in $O(\numedge)$ time, for each of $\graph{1}, \graph{2},$ and $\graph{3}$. From
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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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@ -558,11 +558,11 @@ For the left hand side, following the above steps, we obtain
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&6\cdot\left(\numocc{\graph{1}}{\twopathdis}+ 3\numocc{\graph{1}}{\threedis}\right)\left(3\prob^2 - \prob^3\right) + 4 \cdot \numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right) - 2\cdot \numocc{\graph{1}}{\twopath}\prob
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\end{align*}
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We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$.
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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 $x \cdot \left(\numocc{\graph{i}}{\twopathdis} + 3\cdot \numocc{\graph{i}}{\threedis}\right)$ 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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Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, substitute the identities into $\linsys{3}$,
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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} $\linsys{3} = \numocc{\graph{3}}{\tri} + \numocc{\graph{3}}{\threepath}\prob - \numocc{\graph{3}}{\threedis}\left(3\prob^2 - \prob^3\right)$ , substitute the identities into $\linsys{3}$,
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\begin{align}
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\linsys{3} =& \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} + 18 \cdot \numocc{\graph{1}}{\tri} + 21 \cdot \numocc{\graph{1}}{\threepath} + 24 \cdot \numocc{\graph{1}}{\twopathdis} +\right.\nonumber\\
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&\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}\\
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@ -570,9 +570,9 @@ Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:
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&+ \left\{\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)+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob\right\}. \label{eq:LS-G3-rearrange}
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\end{align}
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Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identities. We then get \cref{eq:LS-G3-rearrange} by simply rearranging the operands into a formation that is more organized for our purposes.
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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 into a formation that is more organized for our purposes.
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Removing terms to the other side of \cref{eq:LS-subtract}, we get
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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{equation}
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\linsys{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'}
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\end{equation}
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@ -584,7 +584,7 @@ For the LHS we get
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& \pbrace{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} + 6 \cdot \numocc{\graph{1}}{\twodis}}\left(3\prob^2 - \prob^3\right) - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob
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\end{align*}
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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 $\graph{1}$. 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,
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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,
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\[ \mtrix{\rpoly} = \begin{pmatrix}
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1 & \prob & -(3\prob^2 - \prob^3)\\
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-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\
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@ -620,7 +620,7 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a
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-2 & 10\\
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-18 & 45
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\end{vmatrix}
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- ~(3\prob^2 - \prob^3)^3~ \cdot
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+ \left(- ~(3\prob^2 - \prob^3)^3\right)~ \cdot
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\begin{vmatrix}
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-2 & -4\\
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-18 & -21
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@ -646,7 +646,7 @@ Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
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=&\left(30\prob^6 - 180\prob^5 + 270\prob^4\right)\cdot\left(-\prob^3 - 3\prob^2 + 3\prob + 1\right).\label{eq:det-final}
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\end{align}
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\AH{It appears that the equation below has roots at p = 0 (left factor) and p = 0.4608 (right factor), \textit{UNLESS} I made a mistake somewhere.}
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\AH{It appears that the equation below has roots at p = 0 (left factor) and p = 1, with NO roots $\in (0, 1)$.}
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%Equation \cref{eq:det-final} has no roots in $(0, 1)$.
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\AH{I need to understand how lemma ~\ref{lem:lin-sys} follows.}
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