diff --git a/macros.tex b/macros.tex index 3e5713e..683d0d3 100644 --- a/macros.tex +++ b/macros.tex @@ -26,6 +26,8 @@ \newcommand{\linsys}[1]{LS^{\graph{#1}}} \newcommand{\lintime}[1]{LT^{\graph{#1}}} \newcommand{\aug}[1]{AUG^{\graph{#1}}} +\newcommand{\mtrix}[1]{M_{#1}} +\newcommand{\dtrm}[1]{Det\left(#1\right)} %PDBs \newcommand{\ti}{TIDB} diff --git a/poly-form.tex b/poly-form.tex index 2137405..0e95f85 100644 --- a/poly-form.tex +++ b/poly-form.tex @@ -402,6 +402,7 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the \end{proof} \qed +\AH{\LARGE \bf New material starts here.} \subsection{Developing a Linear System} In \cref{lem:qE3-exp} is the identity for $\rpoly(\prob,\ldots, \prob)$ when $\poly(\wElem_1,\ldots, \wElem_N) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$. (This lemma still needs a proof, but for now we will pretend the proof is there.) @@ -435,13 +436,13 @@ Rearrange terms into groups of those patterns that can be computed in $O(m)$ and \end{equation*} Let $\lintime{2}$ represent all the terms we wish to remove from $\linsys{2}$. Then, the following are true. -\begin{align*} -\lintime{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^2\\ -\linsys{2'} = \linsys{2} -\lintime{2} &= -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\threepath}}\left(\prob^2 - \prob^3\right)\\ -\aug{2'} = \aug{2} - \lintime{2} &= \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^2\\ -\implies \aug{2'} &= \linsys{2'} -\end{align*} +\begin{align} +\lintime{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^2\nonumber\\ +\linsys{2'} = \linsys{2} -\lintime{2} &= -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\threepath}}\left(\prob^2 - \prob^3\right)\label{eq:LS-G2'}\\ +\aug{2'} = \aug{2} - \lintime{2} &= \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 +\nonumber \\ +&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^2\nonumber\\ +\implies \aug{2'} &= \linsys{2'}\nonumber +\end{align} We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$ where $\linsys{2'}$ is the linear combination and $\aug{2}$ is the augmented side of the matrix, or the constant value, since all of the terms in $\aug{2}$ can be solved in $O(m)$ time. @@ -460,12 +461,74 @@ Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identi Collecting terms we would like to send to the other side, the following equations are true, \begin{align} \lintime{3} =& \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}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\ -\linsys{3'} =& \linsys{3} - \lintime{3} = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 3 \cdot \numocc{\graph{1}}{\threedis}}\left(p^2 - p^3\right)\nonumber \\ -\aug{3'} =& \aug{3} - \lintime{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 - \big(\numocc{\graph{3}}{\twopathdis} + \numocc{\graph{3}}{\threedis}\big)\prob^2 + \\ +\linsys{3'} =& \linsys{3} - \lintime{3} = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 3 \cdot \numocc{\graph{1}}{\threedis}}\left(p^2 - p^3\right)\label{eq:LS-G3'} \\ +\aug{3'} =& \aug{3} - \lintime{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 - \big(\numocc{\graph{3}}{\twopathdis} + \numocc{\graph{3}}{\threedis}\big)\prob^2 + \nonumber\\ & \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}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\ -&\implies \aug{3'} = \linsys{3'} +&\implies \aug{3'} = \linsys{3'}\nonumber \end{align} -We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. +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 $\linsys{3'}$, \cref{eq:LS-G2'}, \cref{eq:LS-G3'} respectively, the following matrix is obtained, +\[ \mtrix{\rpoly} = \begin{pmatrix} +x & (\prob)y & z(\prob^2 - \prob^3)\\ +-2(\prob^2 - \prob^3)x & -4(\prob^2 - \prob^3)y & -2(\prob^2 - \prob^3)z\\ +-18(\prob^2 - \prob^3)x & -21(\prob^2 - \prob^3)y & -3(\prob^2 - \prob^3)z +\end{pmatrix}.\] -Now we seek to show that all rows of the system are indeed independent. \ No newline at end of file +Now we seek to show that all rows of the system are indeed independent. + +The method of minors can be used to compute the determinant, $\dtrm{\mtrix{\rpoly}}$, giving +\begin{equation*} +\begin{vmatrix} +x & (\prob)y & z(\prob^2 - \prob^3)\\ +-2(\prob^2 - \prob^3)x & -4(\prob^2 - \prob^3)y & -2(\prob^2 - \prob^3)z\\ +-18(\prob^2 - \prob^3)x & -21(\prob^2 - \prob^3)y & -3(\prob^2 - \prob^3)z +\end{vmatrix} += +\begin{vmatrix} +-4(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\ +-21(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3) +\end{vmatrix} +~ - ~ \prob~ \cdot +\begin{vmatrix} +-2(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\ +-18(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3) +\end{vmatrix} ++ ~(\prob^2 - \prob^3)~ \cdot +\begin{vmatrix} +-2(\prob^2 - \prob^3) & -4(\prob^2 - \prob^3)\\ +-18(\prob^2 - \prob^3) & -21(\prob^2 - \prob^3) +\end{vmatrix}. +\end{equation*} + +Compute each RHS term starting with the left and working to the right, +\begin{equation} +-4(\prob^2 - \prob^3)\cdot -3(\prob^2 - \prob^3) - \left(-21(\prob^2 - \prob^3) \cdot -2(\prob^2 - \prob^3)\right) = + 12(\prob^2 - \prob^3)^2 - 42(\prob^2 - \prob^3)^2 = -30(\prob^2 - \prob^3)^2.\label{eq:det-1} +\end{equation} +The middle term then is +\begin{equation} +-\prob\left(-2(\prob^2 - \prob^3)\cdot -3(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -2(\prob^2 - \prob^3)\right) = + -\prob\left(6\cdot (\prob^2 - \prob^3)^2 - 36(\prob^2 - \prob^3)^2\right) = -\prob\left(-30(\prob^2 - \prob^3)^2\right).\label{eq:det-2} +\end{equation} +Finally, the rightmost term, +\begin{equation} +\left(\prob^2 - \prob^3\right) \left(-2(\prob^2 - \prob^3)\cdot -21(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -4(\prob^2 - \prob^3)\right) = \left(\prob^2 - \prob^3\right)\left(42\cdot (\prob^2 - \prob^3)^2 - 72(\prob^2 - \prob^3)^2\right) = \left(\prob^2 - \prob^3\right)\left(-30(\prob^2 - \prob^3)^2\right).\label{eq:det-3} +\end{equation} + +Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have, +\[\dtrm{\mtrix{\rpoly}} = -30(\prob^2 - \prob^3)^2-\prob\left(-30(\prob^2 - \prob^3)^2\right) + \left(\prob^2 - \prob^3\right)\left(-30(\prob^2 - \prob^3)^2\right)\] + +Expanding out $\left(\prob^2 - \prob^3\right)^2$ gives +\begin{equation*} +\dtrm{\mtrix{\rpoly}} = -30\left(\prob^4 - 2\prob^5 + \prob^6\right)-\prob\left(-30\left(\prob^4 - 2\prob^5 + \prob^6\right)\right) + \left(\prob^2 - \prob^3\right)\left(-30\left(\prob^4 - 2\prob^5 + \prob^6\right)\right). +\end{equation*} +Further algebraic manipulations result in +\begin{align} +\dtrm{\mtrix{\rpoly}} &= \left(30\prob^4\right)\left(-1 + 2\prob - \prob^2\right) -\prob\left(-1\left(1 - 2\prob + \prob^2\right)\right) + \left(\prob^2 - \prob^3\right)\left(-1\left(1 - 2\prob + \prob^2\right)\right)\label{eq:det-factor}\\ +&=\left(30\prob^4\right)\left(\left(-1 + 2\prob - \prob^2\right) + \left(\prob - 2\prob^2 + \prob^3\right) + \left( - \prob^2 + 2\prob^3 - \prob^4 + \prob^3 - 2\prob^4 + \prob^5\right)\right)\label{eq:det-mult}\\ +&= \left(30\prob^4\right)\left(\prob^5 - 3\prob^4 + 4\prob^3 - 4\prob^2 + 3\prob - 1\right)\label{eq:det-combine}. +\end{align} + +\cref{eq:det-factor} results from factoring out common terms. \cref{eq:det-mult} is the case when multiplying terms in the right hand factor. We arrive at \cref{eq:det-combine} through a simple rearranging and combining of like terms. + +The roots for \cref{eq:det-combine} are $p = 0, p = 1$, and $p = i$. Thus, we have proved the lemma for fixed $p \in (0, 1)$. \ No newline at end of file