Changed to one column.

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Aaron Huber 2020-07-14 11:45:57 -04:00
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%root: main.tex
%!TEX root = ./main.tex
%\onecolumn
\section{Polynomial Formulation}
We can think of $\poly(\vct{w})$ as a function whose input are the variables $X_1,\ldots, X_M$ as in $\poly(X_1,\ldots, X_M)$. Denote the sum of products expansion of $\poly(X_1,\ldots, X_\numTup)$ as $\poly(X_1,\ldots, X_\numTup)_{\Sigma}$
@ -206,10 +206,10 @@ The following is an option.
\item Build $G_2$ from $G_1$, where each edge in $G_1$ gets replaced by a 2 path.
\end{enumerate}
Then $\numocc{G}{\tri}_2 = 0$, and if we can prove that
Then $\numocc{G_2}{\tri} = 0$, and if we can prove that
\begin{itemize}
\item $\numocc{G}{\threepath}_2 = 2 \cdot \numocc{G}{\twopath}_1$
\item $\numocc{G}{\threedis}_2 = 8 \cdot \numocc{G}{\threedis}_1 + 6 \cdot \numocc{G}{\twopathdis}_1 + 4 \cdot \numocc{G}{\oneint}_1 + 4 \cdot \numocc{G}{\threepath}_1 + 2 \cdot \numocc{G}{\tri}_1$
\item $\numocc{G_2}{\threepath} = 2 \cdot \numocc{G_1}{\twopath}$
\item $\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}$
\end{itemize}
we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{G}{\threedis}$, a hard problem.
\end{proof}
@ -248,7 +248,7 @@ The fact that there is a \textit{fixed} number of possible subgraphs that can be
Earlier, we claimed the following.
\[\numocc{G}{\threedis}_2 = 8 \cdot \numocc{G}{\threedis}_1 + 6 \cdot \numocc{G}{\twopathdis}_1 + 4 \cdot \numocc{G}{\oneint}_1 + 4 \cdot \numocc{G}{\threepath}_1 + 2 \cdot \numocc{G}{\tri}_1\]
\[\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}\]
Beginning with the leftmost of RHS terms and proceeding to the consecutive rightmost terms, let us show this to be the case.
@ -268,19 +268,19 @@ For Tri, note that it is the case that the graph $G_2$ is a 'triangle of two pat
\end{proof}
\qed
\AH{Linear Equation computing 3-matchings in $G_3$ using all 3-edge subgraphs in $G_1$.}
In a similar way we can count the number of 3-matchings in graph $G_3$, where each edge in a given $G_1$ gets replaced with a disjoint 3-path, disjoint meaining that no other 3-path intersects another 3-path, except at its endpoints as in the original graph. Because of $G_3$ construction, we now need to also account for two paths in $G_1$.
The linear combination for 3-matchings in $G_3$ follows.
The linear combination of 3-edge $G_1$ subgraphs to compute the number of 3-matchings in $G_3$ follows is
\begin{align*}
\numocc{G}{\threedis}_3 = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 30\pbrace{\numocc{G_1}{\tri}} + 35\pbrace{\numocc{G_1}{\threepath}}\\
&+ 40\pbrace{\numocc{G_1}{\twopathdis}} + 32\pbrace{\numocc{G_1}{\oneint}} + 45\pbrace{\numocc{G_1}{\threedis}}
\numocc{G_3}{\threedis} = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 30\pbrace{\numocc{G_1}{\tri}} + 35\pbrace{\numocc{G_1}{\threepath}}\\
&+ 40\pbrace{\numocc{G_1}{\twopathdis}} + 32\pbrace{\numocc{G_1}{\oneint}} + 45\pbrace{\numocc{G_1}{\threedis}}.
\end{align*}
\AH{Justification next.}
\AH{Justification.}
Enumerate through the RHS in a similar fashion. Beginning with a two path $\twopath$ in $G_1$, it is the case that in $G_3$ this becomes a six-path. As discussed previously, this yields four three matching subgraphs. For subgragh of two disjoint edges, $\twodis$, this becomes two disjoint 3-paths. It is the case in one 3-path, that we have one subgraph of two disjoint edges, where a third disjoint edge can be picked from any of the three edges in the remaining disjoint 3-path. The process can be repeated starting with the alternative 3-path, giving $2 * 3$ unique 3-matchings.
Now for the 3-edge subgraphs, starting with a triangle. When a triangle in $G_1$ is transformed into $G_3$, it becomes a 'triangle' where each leg is a three-path. This is very similar to a 9-path, with the caveat that the first and last edge cannot be in the same 3-matching set together. Iterating through all possible combinations producing 3-matchings, i.e. $(e_1, e_3, e_5),\ldots, (e_1, e_3, e_8), (e_1, e_4, e_6),\ldots, (e_1, e_4, e_8),$\newline$\ldots, (e_1, e_6, e_8),\ldots, (e_5, e_7, e_9)$ gives a total of
@ -293,4 +293,4 @@ matchings. Consider next a 3-path in $G_1$, where the resulting subgraph in $G_
Given the $Fan$ subgraph, where 3 distinct edges are connected at one common endpoint, occurring in $G_1$. In $G_3$, this becomes 3 distinct 9-paths, with each 9-path intersecting the others at one common and shared endpoint. If we consider the outermost non-intersecting edges along with the middle non-intersecting edges, we have $2 * 2 * 2 = 8$ possible 3-matchings. Considering the inner, intersecting edges, we have the condition that only one can appear at a time in a 3-matching set. When we pick an arbitrary inner edge, we have one of two possibilities, we can pick the outer edge of the same 3-path the inner edge is located on, while picking any of the other 4 remaining edges in the middle and outer edges of the other two 3-paths. This gives $4 * 3$ more unique 3-matchings. The remaining possibility exists in combining the arbitrary inner edge with any of the 4 combinations of the middle and outer edges of the other 3-paths. This yields again $3 * 4 = 12$ unique three-matchings, together which make $8 + 12 + 12 = 32$ three-matchings.
Given the $\threedis$ subgraph occurring in $G_1$, the resulting graph consists of three disjoint 3-paths in $G_3$. There are two considerations. First, if we pull one edge from each disjoint 3-path, we have three choices from each path, which is $3^3 = 27$ three-matchings. The second consideration is that we can pull a two matching from any of the given disjoint 3-paths, matching it with a third disjoint edge from any of the other edges in the other 2 three-paths, giving $3 * 6 = 18$ more unique 3-matchings for a total of $27 + 18 = 45$ 3-matchings.
Given the $\threedis$ subgraph occurring in $G_1$, the resulting graph consists of three disjoint 3-paths in $G_3$. There are two considerations. First, if we pull one edge from each disjoint 3-path, we have three choices from each path, which is $3^3 = 27$ three-matchings. The second consideration is that we can pull a two matching from any of the given disjoint 3-paths, matching it with a third disjoint edge from any of the other edges in the other 2 three-paths, giving $3 * 6 = 18$ more unique 3-matchings for a total of $27 + 18 = 45$ three-matchings.

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%root: main.tex
%!TEX root=./main.tex
\onecolumn
\section{Query translation into polynomials}
%\AH{This section will involve the set of queries (RA+) that we are interested in, the probabilistic/incomplete models we address, and the outer aggregate functions we perform over the output \textit{annotation}
%1) RA notation