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Fixed 3-matchings linear combination analysis.

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Aaron Huber 2020-07-15 14:32:01 -04:00
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poly-form.tex
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@ -279,8 +279,8 @@ The linear combination of 3-edge $G_1$ subgraphs to compute the number of 3-matc
\begin{align*}
\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}}.
\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}}\\
&+ 24\pbrace{\numocc{G_1}{\twopathdis}} + 20\pbrace{\numocc{G_1}{\oneint}} + 27\pbrace{\numocc{G_1}{\threedis}}.
\end{align*}
\AH{Justification.}
@ -290,17 +290,14 @@ For subgragh of two disjoint edges, $\twodis$, this becomes two disjoint 3-paths
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.
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. $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 1), (e_3, 0)},\ldots, \pbox{(e_1, 0), (e_2, 2), (e_3, 1)},$\newline$\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$%,\ldots, (e_5, e_7, e_9)$ gives a total of
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.
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.
\[\sum_{i = 1}^4i+ \sum_{i = 1}^4i + \sum_{i = 1}^3i + \sum_{i = 1}^2i + 1 = 30\]
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.
matchings. 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 five other 3-matchings that couldn't be counted in the case of the Tri subgraph, namely $(e_1, e_3, e_9),\ldots, (e_1, e_7, e_9)$, thus $30$ (from the Tri analysis)$ + 5 = 35$ three-matchings.
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.
For the $\twopathdis$ subgraph, it is the case that this graph becomes a 6-path with a disjoint 3-path in $G_3$. 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.
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 $27 + 18 = 45$ three-matchings.
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.
\subsection{Three Paths}