ODIn/paper-BagRelationalPDBsAreHard
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Fixes to 2.6 in computing determinant.

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Aaron Huber 2020-11-04 16:31:39 -05:00
parent 30a9e6e754
commit 5391c0f3e5
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24
poly-form.tex
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@ -587,8 +587,8 @@ For the LHS we get
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,
\[ \mtrix{\rpoly} = \begin{pmatrix}
1 & \prob & -(3\prob^2 - \prob^3)\\
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3)
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3)
\end{pmatrix},\]
and the following linear equation
\begin{equation}
@ -607,18 +607,18 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a
\begin{equation*}
\begin{vmatrix}
1 & \prob & -(3\prob^2 - \prob^3)\\
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3)
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3)
\end{vmatrix}
= (3\prob^2 - \prob^3)^2 \cdot
\begin{vmatrix}
-4 & -2\\
-21 & -3
-4 & 10\\
-21 & 45
\end{vmatrix}
~ - ~ \prob(3\prob^2 - \prob^3)^2~ \cdot
\begin{vmatrix}
-2 & -2\\
-18 & -3
-2 & 10\\
-18 & 45
\end{vmatrix}
- ~(3\prob^2 - \prob^3)^3~ \cdot
\begin{vmatrix}
@ -629,11 +629,11 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a
Compute each RHS term starting with the left and working to the right,
\begin{equation}
(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot -3) - (-21 \cdot -2)\right) = (3\prob^2 - \prob^3)^2\cdot(12 - 42) = -30(3\prob^2 - \prob^3)^2.\label{eq:det-1}
(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot 45) - (-21 \cdot 10)\right) = (3\prob^2 - \prob^3)^2\cdot(-180 + 210) = 30(3\prob^2 - \prob^3)^2.\label{eq:det-1}
\end{equation}
The middle term then is
\begin{equation}
-\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot -3) - (-18 \cdot -2)\right) = -\prob(3\prob^2 - \prob^3)^2 \cdot ( 6 - 36) = 30\prob(3\prob^2 - \prob^3)^2.\label{eq:det-2}
-\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot 45) - (-18 \cdot 10)\right) = -\prob(3\prob^2 - \prob^3)^2 \cdot (-90 + 180) = -90\prob(3\prob^2 - \prob^3)^2.\label{eq:det-2}
\end{equation}
Finally, the rightmost term,
\begin{equation}
@ -642,8 +642,8 @@ Finally, the rightmost term,
Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
\begin{align}
\dtrm{\mtrix{\rpoly}} =& -30(3\prob^2 - \prob^3)^2 + 30\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3 = 30(3\prob^2 - \prob^3)^2\left(-1 + \prob + (3\prob^2 - \prob^3)\right) = 30\left(9\prob^4 - 6\prob^5 + \prob^6\right)\left(-p^3 + 3p^2 + p - 1\right)\nonumber\\
=&\left(30\prob^6 - 180\prob^5 + 270\prob^4\right)\cdot\left(-p^3 + 3p^2 + p - 1\right).\label{eq:det-final}
\dtrm{\mtrix{\rpoly}} =& 30(3\prob^2 - \prob^3)^2 - 90\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3 = 30(3\prob^2 - \prob^3)^2\left(1 - 3\prob + (3\prob^2 - \prob^3)\right) = 30\left(9\prob^4 - 6\prob^5 + \prob^6\right)\left(-\prob^3 - 3\prob^2 + 3\prob + 1\right)\nonumber\\
=&\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.}