Finished determinant calculation for linear system in proving lemma 3.
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@ -26,6 +26,8 @@
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\newcommand{\linsys}[1]{LS^{\graph{#1}}}
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\newcommand{\lintime}[1]{LT^{\graph{#1}}}
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\newcommand{\aug}[1]{AUG^{\graph{#1}}}
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\newcommand{\mtrix}[1]{M_{#1}}
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\newcommand{\dtrm}[1]{Det\left(#1\right)}
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%PDBs
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\newcommand{\ti}{TIDB}
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@ -402,6 +402,7 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the
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\end{proof}
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\qed
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\AH{\LARGE \bf New material starts here.}
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\subsection{Developing a Linear System}
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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.)
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@ -435,13 +436,13 @@ Rearrange terms into groups of those patterns that can be computed in $O(m)$ and
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\end{equation*}
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Let $\lintime{2}$ represent all the terms we wish to remove from $\linsys{2}$. Then, the following are true.
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\begin{align*}
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\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\\
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\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)\\
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\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 + \\
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&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\\
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\implies \aug{2'} &= \linsys{2'}
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\end{align*}
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\begin{align}
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\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\\
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\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'}\\
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\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 \\
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&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\\
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\implies \aug{2'} &= \linsys{2'}\nonumber
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\end{align}
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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.
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@ -460,12 +461,74 @@ Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identi
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Collecting terms we would like to send to the other side, the following equations are true,
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\begin{align}
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\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\\
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\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 \\
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\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 + \\
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\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'} \\
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\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\\
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& \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\\
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&\implies \aug{3'} = \linsys{3'}
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&\implies \aug{3'} = \linsys{3'}\nonumber
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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 $\graph{1}$.
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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 $\linsys{3'}$, \cref{eq:LS-G2'}, \cref{eq:LS-G3'} respectively, the following matrix is obtained,
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\[ \mtrix{\rpoly} = \begin{pmatrix}
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x & (\prob)y & z(\prob^2 - \prob^3)\\
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-2(\prob^2 - \prob^3)x & -4(\prob^2 - \prob^3)y & -2(\prob^2 - \prob^3)z\\
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-18(\prob^2 - \prob^3)x & -21(\prob^2 - \prob^3)y & -3(\prob^2 - \prob^3)z
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\end{pmatrix}.\]
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Now we seek to show that all rows of the system are indeed independent.
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The method of minors can be used to compute the determinant, $\dtrm{\mtrix{\rpoly}}$, giving
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\begin{equation*}
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\begin{vmatrix}
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x & (\prob)y & z(\prob^2 - \prob^3)\\
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-2(\prob^2 - \prob^3)x & -4(\prob^2 - \prob^3)y & -2(\prob^2 - \prob^3)z\\
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-18(\prob^2 - \prob^3)x & -21(\prob^2 - \prob^3)y & -3(\prob^2 - \prob^3)z
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\end{vmatrix}
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=
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\begin{vmatrix}
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-4(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\
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-21(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3)
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\end{vmatrix}
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~ - ~ \prob~ \cdot
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\begin{vmatrix}
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-2(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\
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-18(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3)
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\end{vmatrix}
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+ ~(\prob^2 - \prob^3)~ \cdot
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\begin{vmatrix}
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-2(\prob^2 - \prob^3) & -4(\prob^2 - \prob^3)\\
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-18(\prob^2 - \prob^3) & -21(\prob^2 - \prob^3)
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\end{vmatrix}.
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\end{equation*}
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Compute each RHS term starting with the left and working to the right,
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\begin{equation}
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-4(\prob^2 - \prob^3)\cdot -3(\prob^2 - \prob^3) - \left(-21(\prob^2 - \prob^3) \cdot -2(\prob^2 - \prob^3)\right) =
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12(\prob^2 - \prob^3)^2 - 42(\prob^2 - \prob^3)^2 = -30(\prob^2 - \prob^3)^2.\label{eq:det-1}
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\end{equation}
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The middle term then is
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\begin{equation}
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-\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) =
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-\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}
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\end{equation}
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Finally, the rightmost term,
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\begin{equation}
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\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}
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\end{equation}
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Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
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\[\dtrm{\mtrix{\rpoly}} = -30(\prob^2 - \prob^3)^2-\prob\left(-30(\prob^2 - \prob^3)^2\right) + \left(\prob^2 - \prob^3\right)\left(-30(\prob^2 - \prob^3)^2\right)\]
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Expanding out $\left(\prob^2 - \prob^3\right)^2$ gives
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\begin{equation*}
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\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).
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\end{equation*}
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Further algebraic manipulations result in
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\begin{align}
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\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}\\
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&=\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}\\
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&= \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}.
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\end{align}
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\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.
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The roots for \cref{eq:det-combine} are $p = 0, p = 1$, and $p = i$. Thus, we have proved the lemma for fixed $p \in (0, 1)$.
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