Changes propagated according to Atri's suggestions 071620.
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@ -215,13 +215,15 @@ Consider graph $G_2$, constructed from an arbitrary graph $G_1$. We wish to sho
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\AR{I did not make a pass on the above since it looks incomplete and does not seem to have changed for a while. Also it would be good to define $G_2$ and $G_3$ outside of the proofs below.}
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\AH{Changes start here.}
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\subsubsection{$f_k$ and $G_k$}
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\begin{Definition}\label{def:Gk}
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For $k > 1$, let graph $G_k$ be a graph generated from an arbitrary graph $G_1$, by replacing every edge $e$ of $G_1$ with a $k$-path, such that all $k$-path replacement edges are disjoint in the sense that they only intersect at the original intersection endpoints as seen in $G_1$.
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\end{Definition}
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For any graph $G_k$, we denote its edges to be a pair $(e_i, b)$, such that $b \in \{0,\ldots, k\}$.
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For any graph $G_k$, we denote its edges to be a pair $(e, b)$, such that $b \in \{0,\ldots, k\}$.
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\begin{Definition}\label{def:fk}
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Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$.
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@ -232,9 +234,13 @@ The function $f_k$ is a mapping from every $3$-edge shape in $G_k$ to its genera
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\begin{Lemma}
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$f_k$ is a function.
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\end{Lemma}
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Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \leq 3$, since it has to be the case that any subset of $3$ edges in $E_k$ will map to at most 3 edges in $G_1$. All mappings are in the required range. Then, since for any $b \in \{0,\ldots, k\}$ the edge $(e, b) \mapsto e$ is a mapping for which $(e, b)$ maps to no other edge than $e$, and this implies that $f_k$ is a function.
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\begin{proof}
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Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \leq 3$, since it has to be the case that any subset of $3$ edges in $E_k$ will map to at most 3 edges in $G_1$. All mappings are in the required range. Then, since for any $b \in \{0,\ldots, k\}$ the edge $(e, b) \mapsto e$ is a mapping for which $(e, b)$ maps to no other edge than $e$, and this implies that $f_k$ is a function.
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\end{proof}
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\qed
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\subsubsection{Subgraph patterns with 3 edges}
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We wish to briefly state the possible subgraphs $S$ containing exactly three edges.
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\begin{itemize}
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\item Triangle ($\tri$)
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\item 3-path ($\threepath$)
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@ -242,7 +248,7 @@ Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \l
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\item Disjoint Two-Path ($\twopathdis$)--this subgraph consists of a two path and a remaining disjoint edge.
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\item 3-matching ($\threedis$)--this subgraph is composed of three disjoint edges.
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\end{itemize}
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\subsection{Three Matchings}
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\subsection{Three Matchings in $G_2$}
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\begin{Lemma}\label{lem:3m-G_2}
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The number of $3$-matchings in graph $G_2$ is computed by the following identity,
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@ -250,20 +256,16 @@ The number of $3$-matchings in graph $G_2$ is computed by the following identity
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\end{Lemma}
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\begin{proof}
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%Denote $3_{match}$ as the set of 3-matchings in $G_2$. \AR{I do not like the notation $3_{match}$: perhaps $\mathcal{M}_3$ would be better?} Denote $SG$ as the set of subgraphs imposed on $G_1$. \AR{The set of all edge-subgraphs of $G_1$ already has an existing notation-- $2^{E_1}$, i.e. the power set of the set of edges of $G_1$.}
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\AH{I think that all of the above can be generalized to 3-paths as well.}
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\AR{Yep, this is why I am advocating for defining $f_k$ more generally above.}
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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 edges 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 $G_2$ 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 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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Consider the $S = \threedis$ pattern. Note that edges in $f_2^{-1}$ intersect only at $(e_i, 0), (e_i, 1)$. All subsets for $h \neq i \neq j$, $b_1, b_2, b_3 \in \{0, 1\}$, $(e_h, b_1), (e_i, b_2), (e_j, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choices for each edge $e$ in $G_1$ yielding $2^3 = 8$ possible 3-matchings in $G_2$.
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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 $G_1$ is the subset $S$ and $G_2$ 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 $G_1$ and then consider all possible subgraphs in $G_2$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.}
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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 $G_1$ is the subset $S$ and $G_2$ 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 $G_1$ and then consider all possible subgraphs in $G_2$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.}
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For $S = \twopathdis$ edges $e_1, e_2$ form a $2$-path with $e_3$ 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 $f^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f^{-1}.$ 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. Edge $e_1$ cannot produce a $2$-matching, and are done with $\twopathdis$.
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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 for a 3-star subgraph. \AR{Overall this argument if good. But remember to make the change that should follow from the comment in the first case on how to setup the argument.}
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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 for a 3-star subgraph.
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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 $f_2^{-1}$, where all edges from $(e_0, 0)$ to $(e_2, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. A sequence of size $6$ produces $4$ such possibilities. The following edge combinations, $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ produce four possible 3-matchings.
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@ -272,51 +274,59 @@ For $S = \tri$, note that it is the case that the edges in $f_2^{-1}$ are connec
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Let us also consider when $S \in \binom{E_1}{\leq 2}$. When $|S| = 2$, we have one of two possibile constraints on the output of $f_2^{-1}$. First, we could have that $e_1$ and $e_2$ are connected, forming a $2$-path, and thus all edges from $(e_1, 0)$ to $(e_3, 1)$ are connected successively. As alluded to previously, a 3-matching would require at least to alternate edges to ensure disjointedness, which requires $\geq 5$-path. Second, it could be that $e_1$ is disjoint from $e_2$, and thus $(e_i, b)$ is disjoint to $(e_j, b)$ for $i \neq j$ and $b \in \{0, 1\}$. For a $3$-matching to exist at least one of the disjoint $2$-paths must be a $3$-path in order to produce a $2$-matching, and this is not the case. When $|S| = 1$, by construction of $G_2$, it is the case there does not exist an $M \in \binom{E_2}{3}$ such that $f(M) = S$. Therefore only subgraphs $S$ of $3$ edges need to be considered.
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Observe that all of the arguments above focused solely on the shape/pattern of $S$. In other words, all $S$ of a given shape yield the same number of $3$-matchings, and this is why we get the required identity.
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\AR{I did not make a pass on the rest-- though see a quick comment later on. Please propagate the changes above to the rest of the argument.}
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\end{proof}
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\qed
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\subsubsection{$G_3$}
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\AH{Linear Equation computing 3-matchings in $G_3$ using all 3-edge subgraphs in $G_1$.}
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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 $G_1$.
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The linear combination of 3-edge $G_1$ subgraphs to compute the number of 3-matchings in $G_3$ follows is
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\subsection{Three matchings in $G_3$}
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\begin{Lemma}\label{lem:3m-G3}
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The number of 3-matchings in $G_3$ is computed by the following identity,
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\begin{align*}
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\numocc{G_3}{\threedis} = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 18\pbrace{\numocc{G_1}{\tri}} + 21\pbrace{\numocc{G_1}{\threepath}}\\
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&+ 24\pbrace{\numocc{G_1}{\twopathdis}} + 20\pbrace{\numocc{G_1}{\oneint}} + 27\pbrace{\numocc{G_1}{\threedis}}.
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\end{align*}
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\AH{Justification.}
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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.
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\end{Lemma}
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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 $2 * 3 = 6$ unique 3-matchings.
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\begin{proof}
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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 $f_3^{-1}(S)$. Note, that for $G_1$, we represent edges as $e_1, e_2, e_3$, and their corresponding $3$-paths in $G_3$ as $(e_1, 0), (e_1, 1), (e_1, 2),\ldots, (e_3, 2)$.
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Now for the 3-edge subgraphs, starting with a triangle. Note that we are considering 3-matchings $M$ $s.t.$ $f(M) = G'$, where $G'$ is a 3-edge subgraph. In other words, the function $f$ will produce 3 \textit{distinct} edge outputs. This then disalllows double counting of 3-matchings from a 3-edge subgraph using only two of the edges.
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When $S = \ed$, $f_3^{-1}(S)$ gives one result, $(e_1, 0), (e_1, 1), (e_1, 2)$. All edges in the subset are a $3$-path, and it is the case as alluded in $G_2$ discussion that no 3-matching can exist in a single $3$-path.
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Note that for the case of $G_3$, edges are denoted as $(e_i, b_i)$ where $b \in \{0, 1, 2\}$. 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 $(e_1, 0)$ and $(e_3, 2)$ cannot be in the same 3-matching set together. For this $G_3$ it is also the case that for all $i \in \{0, 1, 2\}$ $(e_i, 2)$ and $(e_{i + 1}, 0)$ are neighbors and cannot share a 3-matching. Iterating through all possible combinations producing 3-matchings, i.e. $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 1), (e_3, 0)},$\newline$\ldots, \pbox{(e_1, 0), (e_2, 2), (e_3, 1)},\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$ gives a total of 18 3-matchings.
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Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S' = \{(e_1, 0),\ldots(e_1, 2), (e_2, 0),\ldots, (e_2, 2)\}.$ All edges in S' form a $6$-path, and as stated above, this forms $4$ $3$-matchings.
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Consider next a 3-path in $G_1$, where the resulting subgraph in $G_3$ is a 9-path. In this case, because the endpoints are disconnected, we have 3 other 3-matchings that couldn't be counted in the case of the Tri subgraph, namely $\pbox{(e_1, 0), (e_2, 0), (e_3, 2)},\pbox{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbox{(e_1, 0), (e_2, 2), (e_3, 2)}$, thus $18$ (from the Tri analysis)$ + 3 = 21$ three-matchings.
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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 $i \neq j$ 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.
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For the $\twopathdis$ subgraph, it is the case that this graph becomes a 6-path with a disjoint 3-path in $G_3$. Starting with the 6-path, there are 8 distinct two matchings in the form of $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 2)}, \ldots \pbox{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$. Notice again that the edge pattern $\pbox{(e_1, 2), (e_2, 0)\ldots}$ is forbidden. Each of these 8 2-matchings can be paired with one of the 3 edges in the disjoint 3-path, yielding $8 * 3 = 24$ 3-matchings.%We have to consider 3 possibilities of 3-matchings. First, the 6-path produces 4. Second, it is the case that the 6-path produces 10 two-matchings, which can be paired with any one of the three edges in the disjoint 3-path, producing $10 * 3$ 3-matchings. Third, the disjoint 3-path can produce one 3-matching which can be paired with a third edge from any one of the edges in the 6-path, giving $1 * 6 = 6$ three-matchings, for a total of $4 + 30 + 6 = 40$ three-matchings.
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Now for the 3-edge subgraphs of $G_1$, starting with a $S = \tri$. As discussed in the case of $G_2$, $f_3^{-1}(S)$ 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 $s \in S'$, $s$ is a $3$-matching when we have that $j \geq i + 2, k \geq j + 2$ for the edges $(e_i, b_1), (e_j, b_2), (e_k, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$ for all $i \in \{1, 2\}$ it is the case that if $b_i = 2$ then $b_{i + 1} \neq 0$ and if $b_1 = 0$ then $b_3 \neq 2$. Iterating through all possible combinations producing 3-matchings, i.e. $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 1), (e_3, 0)},\ldots, \pbox{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbox{(e_1, 1), (e_2, 0), (e_3, 0)},\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$ gives a total of 18 3-matchings.
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Given the 3-star subgraph, where 3 distinct edges are connected at one common endpoint, occurring in $G_1$. In $G_3$, this becomes 3 3-paths joined at one common endpoint. Note the non-intersecting outer and middle edges. For each 3-path, there are 2 possible choices to create a 3-matching, yielding $2^3 = 8$ 3-matchings. Finally, consider the joined inner edges, recalling that at most one of them can be in a 3-matching. Pick an arbitrary edge, and there are four possible combinations that result from the middle and outer edges of the other 2 3-paths. This gives $3 * 4 = 12$ more 3-matchings, a total of $8 + 12 = 20$.% 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.
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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. The relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $S = \tri$, namely $\pbox{(e_1, 0), (e_2, 0), (e_3, 2)},\pbox{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbox{(e_1, 0), (e_2, 2), (e_3, 2)}$. There are therefore $18 + 3 = 21$ three-matchings.
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Given the $\threedis$ subgraph occurring in $G_1$, the resulting graph consists of three disjoint 3-paths in $G_3$. Since only one edge can be used at a time from each 3-path, it results that there are $3^3 = 27$ 3-matchings.%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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Assume $S = \twopathdis$, then $f_3^{-1}$ 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 in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbox{(e_2, 0), (e_3, 0)}, \pbox{(e_2, 0), (e_3, 1)}, \pbox{(e_2, 0), (e_3, 2)},\pbox{(e_2, 1), (e_3, 0)},\ldots, \pbox{(e_2, 1), (e_3, 2)}, \pbox{(e_2, 2), (e_3, 1)}, \pbox{(e_2, 2), (e_3, 2)}$. 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.
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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 discussed previously, 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 $8 + 12 = 20$
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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.
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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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\qed
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\subsection{Three Paths}
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Computing the number of 3-paths in $G_2$ and $G_3$ consists of much simpler linear combinations.
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\subsubsection{$G_2$}
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It can be easily verified that a single edge in $G_1$ becomes a 2-path in $G_2$, and generates $0$ 3-paths. Similarly, it is the case for 2 disjoint edges. When we look at output of $f$ that has 3 edges, we see the same result for the case of 3 disjoint edges in $G_1$. When we have a 2-path and 1 disjoint edge, there is no way to create a 3-path out of all three edges. The same holds for the remaining subgraphs consisting of 3 edges, the 3-star, triangle, and 3-path.\AR{For each pattern here, you should {\em explicitly} argue why they give rise to zero 3-paths. It does not have to be a separate argument for each shape-- you can make a general argument that rules out everything except the 2-path. Whatever is easier.}
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\begin{Lemma}
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The number of $3$-paths in $G_2$ is computed by the following linear combination,
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\[\numocc{G_2}{\threepath} = 2 \cdot \numocc{G_1}{\twopath}.\]
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\end{Lemma}
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All the above to say that there exists only one subgraph in $G_1$ that will produce 3-paths, namely the 2-path. This subgraph becomes a 4-path in $G_2$, and using both edges (as required by definition of $f$), 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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\begin{proof}
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For a $M = \threepath \in G_2$, it \textit{must} be the case that there is successive connectivity for $3$ edges across $f_2(M) = S$. This constraint rules out every pattern $S \in G_1$ consisting of $3$ edges, as well as when $S = \twodis$ and for $S = \ed$. 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 $e_2$.
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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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\[\numocc{G_2}{\threepath} = 2 * \numocc{G_1}{\twopath}.\]
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\AH{Changes propagated up to this point.}
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\subsubsection{$G_3$}
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In a similar fashion, enumerate through the various subgraphs in $G_1$ with $\leq 3$ edges, starting with the smallest. Note, that one edge in $G_1$ generates one 3-path in $G_3$. Moving on to a 2-path, again we see that we have 2 2-paths that consist of both $G_1$ generating edges. For the subgraph of 2 disjoint edges, as in the case of $G_2$, there is no way to make a 3-path out of disjoint edges, and this rolls over into the subgraph consisting of 3 disjoint edges, the subgraph made of a 2 path and disjoint edge, 3-star, triangle, and 3-path. All of these subgraphs provide no way to create a 3-path from all edges $e_1, e_2, e_3$ in $G_1$. The combination is then
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Reference in a new issue