From 073304e49bae44ecbb72b5e13439f2343295ec28 Mon Sep 17 00:00:00 2001 From: Aaron Huber Date: Thu, 5 Nov 2020 10:38:21 -0500 Subject: [PATCH] Added explicit detail to the proofs where eq(10) was propagated. --- poly-form.tex | 22 +++++++++++----------- 1 file changed, 11 insertions(+), 11 deletions(-) diff --git a/poly-form.tex b/poly-form.tex index 0b7744c..7a1794e 100644 --- a/poly-form.tex +++ b/poly-form.tex @@ -165,11 +165,11 @@ By definition we have that \begin{proof}[Proof of Claim \ref{claim:four-two}] %We have shown that the following subgraph cardinalities can be computed in $O(\numedge)$ time: %\[\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis}, \numocc{G}{\oneint}, \numocc{G}{\twopathdis} + \numocc{G}{\threedis}.\] -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. +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. Using the result of \cref{lem:qE3-exp}, let us show a derivation to the identity of the consequent in \cref{claim:four-two}. -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, +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, \begin{align} &\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\\ &\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}\\ @@ -514,10 +514,10 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the \subsection{Developing a Linear System} -\AH{The changes in ~\cref{eq:2pd-3d} have been propagated 110220. Barring any errors, everything should be updated and correct.} +\AH{The changes in ~\cref{eq:2pd-3d} have been propagated 110420. Barring any errors, everything should be updated and correct.} \begin{proof}[Proof of Lemma \ref{lem:lin-sys}] -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 +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 %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 ~\cref{eq:LS-subtract}, $\vct{b}[1] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}\big)\prob^2$. @@ -558,11 +558,11 @@ For the left hand side, following the above steps, we obtain &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 \end{align*} -We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$. +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. \subsubsection{$\graph{3}$} -Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, substitute the identities into $\linsys{3}$, +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 $\linsys{3}$, \begin{align} \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\\ &\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}\\ @@ -570,9 +570,9 @@ Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem: &+ \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} \end{align} -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. +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. -Removing terms to the other side of \cref{eq:LS-subtract}, we get +Removing $O(\numedge)$ computable terms to the other side of \cref{eq:LS-subtract}, we get \begin{equation} \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'} \end{equation} @@ -584,7 +584,7 @@ For the LHS we get & \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 \end{align*} -We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. 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, +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, \[ \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)\\ @@ -620,7 +620,7 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a -2 & 10\\ -18 & 45 \end{vmatrix} -- ~(3\prob^2 - \prob^3)^3~ \cdot ++ \left(- ~(3\prob^2 - \prob^3)^3\right)~ \cdot \begin{vmatrix} -2 & -4\\ -18 & -21 @@ -646,7 +646,7 @@ Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have, =&\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} \end{align} -\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.} +\AH{It appears that the equation below has roots at p = 0 (left factor) and p = 1, with NO roots $\in (0, 1)$.} %Equation \cref{eq:det-final} has no roots in $(0, 1)$. \AH{I need to understand how lemma ~\ref{lem:lin-sys} follows.}