Finished Section 2 incorporating Oliver's 081120 comments.
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@ -315,10 +315,10 @@ Let $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows. For
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The function $f_k$ is a mapping from every $3$-edge shape in $\graph{k}$ to its `projection' in $\graph{1}$.
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\begin{Definition}[$f_k^{-1}$]
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The inverse function $f_k^{-1}: \binom{E_1}{\leq 3}\mapsto \binom{E_k}{3}$ takes an arbitrary $\eset{1}$ of at most $3$ edges and outputs the set of all subsets of $\binom{\eset{k}}{3}$ such that each subset of theoutput set $s^{(k)}$ is mapped to the input set $s^{(1)}$ by $f_k$, i.e. $f_k(s^{(k)}) = s^{(1)}$.
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The inverse function $f_k^{-1}: \binom{E_1}{\leq 3}\mapsto \{\binom{E_k}{3}\}^{\leq \binom{E_k}{3}}$ takes an arbitrary $\eset{1}$ of at most $3$ edges and outputs the set of all subsets of $\binom{\eset{k}}{3}$ such that each subset $s^{(k)}$ of the output set is mapped to the input set $s^{(1)}$ by $f_k$, i.e. $f_k(s^{(k)}) = s^{(1)}$.
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\end{Definition}
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Note, importantly, that when we discuss $f_k^{-1}$, that, although counterintuitive, \textit{each edge present in} $s^{(1)}$ must have an edge in $s^{(k)}$ that `projects` down to it. \textit{Meaning}, if $|s^{(1)}| = 3$, then it must be the case that $s^{(k)}$ be a set $\{ (e_i, b), (e_j, b), e_\ell, b) \}$ where $i \neq j \neq \ell$.
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Note, importantly, that when we discuss $f_k^{-1}$, that, although counterintuitive, each \textit{edge} present in $s^{(1)}$ must have an edge in $s^{(k)}$ that `projects` down to it. \textit{Meaning}, if $|s^{(1)}| = 3$, then it must be the case that $s^{(k)}$ be a set $\{ (e_i, b), (e_j, b), e_\ell, b) \}$ where $i \neq j \neq \ell$.
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\begin{Lemma}\label{lem:fk-func}
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$f_k$ is a function.
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@ -340,34 +340,34 @@ Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \l
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\subsection{Three Matchings in $\graph{2}$}
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\AR{TODO for {\em later}: I think the proof will be much easier to follow with figures: just drawing out $S\times \{0,1\}$ along with the $(e_i,b_i)$ explicity notated on the edges will make the proof much easier to follow.}
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\begin{proof}[Proof of Lemma \ref{lem:3m-G2}]
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Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible sets of $3$-edge subgraphs in $S \times \{0, 1\}$ which $f_2$ maps to $S$. Then we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$. We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is the set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
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For each edge pattern $S$, we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$. We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is the set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
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\begin{itemize}
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\item $3$-matching ($\threedis$)
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\end{itemize}
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Consider the $S = \threedis$ pattern. Note that edges in $f_2^{-1}(S)$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2\}$.. All subsets for $b_1, b_2, b_3 \in \{0, 1\}$, $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choices for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(S)$.
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Consider the $\eset{1} = \threedis$ pattern. Note that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. All subsets for $b_1, b_2, b_3 \in \{0, 1\}$, $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choices for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(S)$.
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%\AH{The comment below is an important comment.}
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%\AR{I think your argument seems to implicitly assume that $\graph{1}$ is the subset $S$ and $\graph{2}$ is the corresponding mapping under $f^{-1}$. This is {\bf not} correct. You should present the argument as in the outline above above. I.e. fix an $S\in\binom{E_1}{\le 3}$ in $\graph{1}$ and then consider all possible subgraphs in $\graph{2}$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.}
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\begin{itemize}
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\item Disjoint Two-Path ($\twopathdis$)
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\end{itemize}
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For $S = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f_2^{-1}(S).$ Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
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For $\eset{1} = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick either $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from $e_2$ and $e_3$. Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
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\begin{itemize}
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\item $3$-star ($\oneint$)
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\end{itemize}
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When $S = \oneint$, in $f_2^{-1}$, the inner edges $(e_i, 1)$ are all connected, and the outer edges $(e_i, 0)$ are all disjoint. Note that for a valid 3 matching it must be the case that at most one inner edge can be part of the set of disjoint edges. When exactly one inner edge is chosen, there are 3 such possibilities. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are $3 + 1 = 4$ 3-matchings in $f_2^{-1}(S)$.
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When $\eset{1} = \oneint$, the inner edges $(e_i, 1)$ of $\eset{2}$ are all connected, and the outer edges $(e_i, 0)$ are all disjoint. Note that for a valid 3 matching it must be the case that at most one inner edge can be part of the set of disjoint edges. When exactly one inner edge is chosen, there are 3 such possibilities. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are $3 + 1 = 4$ 3-matchings in $f_2^{-1}(S)$.
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\begin{itemize}
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\item $3$-path ($\threepath$)
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\end{itemize}
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When $S =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This translates to a $6$-path in the edges of $f_2^{-1}(S)$, where all edges from $(e_1, 0),\ldots,(e_3, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
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When $\eset{1} =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This means that the edges of $\eset{2}$ form a $6$-path in the edges of $f_2^{-1}(S)$, where all edges from $(e_1, 0),\ldots,(e_3, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
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\begin{itemize}
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\item Triangle ($\tri$)
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\end{itemize}
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For $S = \tri$, note that it is the case that the edges in $f_2^{-1}(S)$ are connected in a successive manner, but this time in a cycle, such that $(e_1, 0)$ and $(e_3, 1)$ are also connected. While this is similar to the discussion of the three path above, one must use caution not to consider the first and last edges as disjoint, since they are connected. This rules out both subsets of $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with $2$ remaining edge combinations that produce a 3 matching.
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For $\eset{1} = \tri$, note that it is the case that the edges in $\eset{2}$ are connected in a successive manner, but this time in a cycle, such that $(e_1, 0)$ and $(e_3, 1)$ are also connected. While this is similar to the discussion of the three path above, the first and last edges are not disjoint, since they are connected. This rules out both subsets of $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with $2$ remaining edge combinations that produce a 3 matching.
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\begin{itemize}
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\item $2$-matching ($\twodis$), $2$-path ($\twopath$), $1$ edge ($\ed$)
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@ -380,32 +380,32 @@ Observe that all of the arguments above focused solely on the shape/pattern of $
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\subsection{Three matchings in $\graph{3}$}
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Let $S'$ be all the edges of $\graph{k}$ which 'project' down to any set of edges $S$ in $\graph{1}$, formally, for $|S| = 1$, then $S = \{e_1\}$ and $S' = \pbrace{(e_1, 0),\ldots, (e_1, k-1)}$. Similarly, when $|S| = 2$, then $S = \pbrace{e_1, e_2}$ and $S' = \pbrace{(e_1, 0),\ldots, (e_2, k -1)}$, and so on for $|S| = 3$.
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% Let $S'$ be all the edges of $\graph{k}$ which 'project' down to any set of edges $S$ in $\graph{1}$, formally, for $|S| = 1$, then $S = \{e_1\}$ and $S' = \pbrace{(e_1, 0),\ldots, (e_1, k-1)}$. Similarly, when $|S| = 2$, then $S = \pbrace{e_1, e_2}$ and $S' = \pbrace{(e_1, 0),\ldots, (e_2, k -1)}$, and so on for $|S| = 3$.
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\begin{proof}[Proof of Lemma \ref{lem:3m-G3}]
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For any $S \in \binom{E_1}{\leq3}$, we now consider $f_3^{-1}(S)$, which lists all possible subsets of $3$ edges in $S \times \{0, 1, 2\}$ to which $f_3$ maps to $S$. We again then count the number of $3$-matchings in $f_3^{-1}(S)$.
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For any $S \in \binom{E_1}{\leq3}$, we again then count the number of $3$-matchings in $f_3^{-1}(S)$.
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\begin{itemize}
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\item $1$ edge ($\ed$)
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\end{itemize}
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When $S = \ed$, $f_3^{-1}(S)$ has one subset, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(S)$ for this case.% All edges in the subset are a $3$-path, and it is the case as alluded in $\graph{2}$ discussion that no 3-matching can exist in a single $3$-path.
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When $\eset{1} = \ed$, $f_3^{-1}(\eset{1})$ has one subset, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(\eset{1})$ for this case.% All edges in the subset are a $3$-path, and it is the case as alluded in $\graph{2}$ discussion that no 3-matching can exist in a single $3$-path.
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\begin{itemize}
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\item $2$-path ($\twopath$)
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\end{itemize}
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Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S'$. All edges in $S'$ form a $6$-path, and similar to the discussion in the proof of \cref{lem:3m-G2} (when $S = \threepath$ in $\graph{2}$), this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
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Fix then $\eset{1} = \twopath$ and now we have all edges in $\eset{3}$ form a $6$-path, and similar to the discussion in the proof of \cref{lem:3m-G2} (when $eset{1} = \threepath$ in $\graph{2}$), this leads to $4$ $3$-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item $2$-matching ($\twodis$)
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\end{itemize}
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For $S = \twodis$, then all subsets in the output of $f_3^{-1}(S)$ are predicated on the fact that $(e_i, b)$ is disjoint with $(e_j, b)$ for $i \neq j\in \{1,2\}$ and $b \in \{0, 1, 2\}$. Pick an aribitrary $e_i$ and note, that $(e_i, 0), (e_i, 2)$ is a $2$-matching, which can combine with any of the $3$ edges in $(e_j, 0),\ldots, (e_j, 2)$ again for $i \neq j$. Since the selections are independent, it follows that there exist $2 \cdot 3 = 6$ $3$-matchings in $f_3^{-1}(S)$.
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For $\eset{1} = \twodis$, all edges of $\eset{3}$ are predicated on the fact that $(e_i, b)$ is disjoint with $(e_j, b)$ for $i \neq j\in \{1,2\}$ and $b \in \{0, 1, 2\}$. Pick an aribitrary $e_i$ and note, that $(e_i, 0), (e_i, 2)$ is a $2$-matching, which can combine with any of the $3$ edges in $(e_j, 0), (e_j, 1), (e_j, 2)$ again for $i \neq j$. Since the selections are independent, it follows that there exist $2 \cdot 3 = 6$ $3$-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item Triangle ($\tri$)
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\end{itemize}
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Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $S = \tri$. As discussed in proof of \cref{lem:3m-G2} for the case of $\tri$, $f_3^{-1}(S)$ subsets are conditioned on the fact that all the edges in $S'$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $s \in f_3^{-1}(S)$, $s$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i \mod{3} + 1} \neq 0$. Iterating through all possible combinations, we have
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Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $\eset{1} = \tri$. As discussed in proof of \cref{lem:3m-G2} for the case of $\tri$, the edges of $\eset{3}$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $s \in f_3^{-1}(S)$, $s$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i \mod{3} + 1} \neq 0$. Iterating through all possible combinations, we have
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\begin{itemize}
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\item \textsc{$(e_1, 0)$}
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\begin{itemize}
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@ -430,27 +430,27 @@ Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $S = \tri$.
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\item $\pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$
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\end{itemize}
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\end{itemize}
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for a total of 18 3-matchings in $f_3^{-1}(S)$.
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for a total of 18 3-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item $3$-path ($\threepath$)
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\end{itemize}
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Consider when $S = \threepath$ and $f_3^{-1}(S)$ has the constraint that all edges are successively connected to form a $9$-path. Since $(e_1, 0)$ is disjoint to $(e_3, 2)$, both of these edges can exist in a $3$-matching. This relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $S = \tri$, namely $\pbrace{(e_1, 0), (e_2, 0), (e_3, 2)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 2)}$. There are therefore $18 + 3 = 21$ $3$-matchings in $f_3^{-1}(S)$.
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Consider when $\eset{1} = \threepath$ and all edges in $\eset{3}$ are successively connected to form a $9$-path. Since $(e_1, 0)$ is disjoint to $(e_3, 2)$, both of these edges can exist in a $3$-matching. This relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $\eset{1} = \tri$, namely $\pbrace{(e_1, 0), (e_2, 0), (e_3, 2)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 2)}$. There are therefore $18 + 3 = 21$ $3$-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item Disjoint Two-Path ($\twopathdis$)
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\end{itemize}
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Assume $S = \twopathdis$, then $f_3^{-1}(S)$ has successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $S'$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbrace{(e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_2, 1), (e_3, 2)}, \pbrace{(e_2, 2), (e_3, 1)}, \pbrace{(e_2, 2), (e_3, 2)}$. These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ 3-matchings in $f_3^{-1}(S)$.
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Assume $\eset{1} = \twopathdis$, then the edges of $\eset{3}$ have successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $\eset{3}$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbrace{(e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_2, 1), (e_3, 2)}, \pbrace{(e_2, 2), (e_3, 1)}, \pbrace{(e_2, 2), (e_3, 2)}$. These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ 3-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item $3$-star ($\oneint$)
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\end{itemize}
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Given $S = \oneint$, the subsets of $f_3^{-1}(S)$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As previously mentioned in the proof of \cref{lem:3m-G2}, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_k$, where $i \neq j \neq k$. These choices are independent and we have $4 \cdot 3 = 12$ 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ $3$-matchings in $f_3^{-1}(S)$.
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Given $\eset{1} = \oneint$, the edges of $\eset{3}$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As previously mentioned in the proof of \cref{lem:3m-G2}, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_k$, where $i \neq j \neq k$. These choices are independent and we have $4 \cdot 3 = 12$ 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ $3$-matchings in $f_3^{-1}(\eset{1})$.
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\begin{itemize}
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\item $3$-matching ($\threedis$)
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\end{itemize}
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Given $S = \threedis$ subgraph, we have the case that all subsets in $f_3^{-1}(S)$ have the property that $(e_i, b)$ is disjoint to $(e_j, b)$ for $i \neq j$. For each $e_i$, there are then $3$ choices, independent of each other, and it results that there are $3^3 = 27$ 3-matchings in $f_3^{-1}(S)$.
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Given $\eset{1} = \threedis$ subgraph, we have the case that all edges in $\eset{3}$ have the property that $(e_i, b)$ is disjoint to $(e_j, b)$ for $i \neq j$. For each $e_i$, there are then $3$ choices, independent of each other, and it results that there are $3^3 = 27$ 3-matchings in $f_3^{-1}(\eset{1})$.
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All of the observations above focused only on the shape of $S$, and since we see that for fixed $S$, we have a fixed number of $3$-matchings, this implies the identity.
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\end{proof}
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@ -462,7 +462,7 @@ Computing the number of 3-paths in $\graph{2}$ and $\graph{3}$ consists of much
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\begin{proof}[Proof of Lemma \ref{lem:3p-G2}]
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For $\mathcal{P} \subseteq S'$ such that $\mathcal{P} $ is a $3$-path, it \textit{must} be the case by definition of $f$ that all edges in $f_2(\mathcal{P} )$ have at least one mapping from an edge in $\mathcal{P} $ (and recall that $\mathcal{P} $ is connected). This constraint rules out every pattern $S \in \graph{1}$ consisting of $3$ edges, as well as when $S = \twodis$. For $S = \ed$, note that $S$ doesn't have enough edges to have any output in $f_2^{-1}(S)$, i.e., there exists no $s \in \binom{E_2}{3}$ such that $f_2(\mathcal{P} ) = S$. The only surviving pattern is $S = \twopath$, where it can be seen in $f_2^{-1}(S)$ that each subset has successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then $2$ $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(S), \pbrace{(e_1, 0), (e_1, 1), (e_2, 0)} \text{ and }\pbrace{(e_1, 1), (e_2, 0), (e_2, 1)}$.
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For $\mathcal{P} \subseteq \eset{2}$ such that $\mathcal{P} $ is a $3$-path, it \textit{must} be the case by definition of $f$ that all edges in $f_2(\mathcal{P} )$ have at least one mapping from an edge in $\mathcal{P} $ (and recall that $\mathcal{P} $ is connected). This constraint rules out every pattern $\eset{1}$ consisting of $3$ edges, as well as when $\eset{1} = \twodis$. For $\eset{1} = \ed$, note that $\eset{1}$ doesn't have enough edges to have any output in $f_2^{-1}(\eset{1})$, i.e., there exists no $s \in \binom{E_2}{3}$ such that $f_2(\mathcal{P} ) = \eset{1}$. The only surviving pattern is $\eset{1} = \twopath$, where the edges of $\eset{2}$ have successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then $2$ $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(\eset{1}), \pbrace{(e_1, 0), (e_1, 1), (e_2, 0)} \text{ and }\pbrace{(e_1, 1), (e_2, 0), (e_2, 1)}$.
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\end{proof}
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\qed
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%we have two 3-paths generated: $\pbox{(e_1, 0), (e_1, 1), (e_2, 0)}$ and $\pbox{(e_1, 1), (e_2, 0), (e_2, 1)}$. Thus,
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@ -472,7 +472,7 @@ For $\mathcal{P} \subseteq S'$ such that $\mathcal{P} $ is a $3$-path, it \texti
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\begin{proof}[Proof of Lemma \ref{lem:3p-G3}]
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The argument follows along the same lines as in the proof of \cref{lem:3p-G2}. Given $\mathcal{P} \subseteq S'$, it \textit{must} be that every edge in $f_3(\mathcal{P})$ has at least one edge in $\mathcal{P}$ mapped to it (and $\mathcal{P}$ is connected). Notice again that this cannot be the case for any $S \in \binom{E_1}{3}$, nor is it the case when $S = \twodis$. This leaves us with two patterns, $S = \twopath$ and $S = \ed$. For the former, it is the case that we have $2$ $3$-paths across $e_1$ and $e_2$, $\pbrace{(e_1, 1), (e_1, 2), (e_2, 0)}$ and $\pbrace{(e_1, 2), (e_2, 0), (e_2, 1)}$. For the latter pattern $\ed$, it it trivial to see that an edge in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.
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The argument follows along the same lines as in the proof of \cref{lem:3p-G2}. Given $\mathcal{P} \subseteq \eset{3}$, it \textit{must} be that every edge in $f_3(\mathcal{P})$ has at least one edge in $\mathcal{P}$ mapped to it (and $\mathcal{P}$ is connected). Notice again that this cannot be the case for any $\eset{1} \in \binom{E_1}{3}$, nor is it the case when $\eset{1} = \twodis$. This leaves us with two patterns, $\eset{1} = \twopath$ and $\eset{1} = \ed$. For the former, it is the case that we have $2$ $3$-paths across $e_1$ and $e_2$, $\pbrace{(e_1, 1), (e_1, 2), (e_2, 0)}$ and $\pbrace{(e_1, 2), (e_2, 0), (e_2, 1)}$. For the latter pattern $\ed$, it it trivial to see that an edge in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.
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\end{proof}
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\qed
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Reference in a new issue