Proof of linear combination of G_1 for G_2 3-matchings.

poly-form.tex

@@ -209,7 +209,30 @@ The following is an option.
 Then $\numocc{\tri}_2 = 0$, and if we can prove that
 \begin{itemize}
 	\item $\numocc{\threepath}_2 = 2 \cdot \numocc{\twopath}_1$
-	\item $\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1$
+	\item $\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1 + 6 \cdot \numocc{\twopathdis}_1 + 4 \cdot \numocc{\oneint}_1 + 4 \cdot \numocc{\threepath}_1 + 2 \cdot \numocc{\tri}_1$
 \end{itemize}
 we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{\threedis}$, a hard problem.
 \end{proof}
+
+Consider graph $G_2$, constructed from an arbitrary graph $G_1$.  We wish to show that the number of 3-matchings in $G_2$ will always be the linear combination above, regardless of the construction of $G_1$.
+
+\begin{proof}
+Denote $3_{match}$ as the set of 3-matchings in $G_2$.  Denote $SG$ as the set of subgraphs imposed on $G_1$.  Notate each edge in $G_2$ as $(e, b)$ such that $b \in \{0, 1\}$, where $b$ identifies either the first or second edge of the two path that replaced the original $G_1$ edge.  Let $f: 3_{match} \mapsto SG$ be a function that maps a distinct 3-matching in $G_2$ to its generating subgraph.  An arbitrary $M \in 3_{match}$ in $G_2$ is then denoted by $(e_1, b_1), (e_2, b_2), (e_3, b_3)$, and $f(M) = \{e_1, e_2, e_3\}$.  Note that $f(M)$ is a set, i.e., the distinct edges of the generating subgraph in $G_1$.
+
+\underline{f is a function}:
+
+First, let us argue that $f$ is indeed a function.  To do this, first note that for $G_2$ to contain a 3-matching, it must be that the generating subgraph in $G_1$ has at least 3 distinct edges.  This is because, with only 2 distinct $G_1$ edges, there are only two subgraph patterns, i.e., disjoint and two-path, both of which when transformed into $G_2$ will not have enough disjoint edges to create a 3-matching.
+
+Consider then a minimal $G_1$, i.e. a graph with at most 3 edges.  Choose an arbitrary 3-matching $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ in the generated $G_2$ graph.  Choose an arbitrary edge $(e_i, b_i)$ and remove its corresponding generating edge in $G_1$.  Notice that this 3-matching now disappears from $G_2$.  Now replace the removed edge in $G_1$ with a new edge, placing it anywhere other than its original position.  Note that, no matter where the replacement edge lies, it cannot be that it generates the same 3-matching that the original edge helped uniquely generate, since a one-to-one correspondence exists between $(e_i, b_i)$ and $e_i$.  Thus, since any $M \in 3_{match}$ cannot be generated from more than one subgraph in $G_1$, $f$ must be a function.
+
+\underline{For any $M \in 3_{match}, | f(M) |\leq 3$}
+
+Since any arbitrary 3-matching $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ has an image of $\{e_1, e_2, e_3\}$, which is a set, it has to be the case that $|f(M)|$ can be no more than $3$.
+
+\underline{For any $G' \in SG: |s| \leq 3$, there are a fixed number of 3-matchings}
+\underline{ $M'$ s.t. $f(M') = G'$}
+
+First, note that there are a fixed number of 3-edge subgraphs that can appear on a given $G_1$.  Second, these subgraphs themselves are fixed, with the number of edges, vertices, and intersections always being the same.  Third, since each possible subgraph is fixed, it must then be the case that each possible subgraph will always generate the same number of 3-matchings in $G_2$.
+\end{proof}
+
+\qed