Finished addressing Atri's suggestions for 3-matchings 071620.

poly-form.tex

@@ -250,31 +250,26 @@ The number of $3$-matchings in graph $G_2$ is computed by the following identity
 \end{Lemma}
 
 \begin{proof}
-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 $G_1$.}  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.  
-
-In $G$, the following possible subgraphs that can be composed of three edges.
+%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 $G_1$.}  
 
 \AH{I think that all of the above can be generalized to 3-paths as well.}
 \AR{Yep, this is why I am advocating for defining $f_k$ more generally above.}
 
-Given any $S \in \binom{E_1}{\leq3}$, we consider $f^{-1}(S)$, which is the set of all possible edges in $S \times \{0, 1\}$ which $f$ maps to $S$.  Then we count the number of $3$-matchings in the $3$-edge subgraphs of $G_2$ in $f^{-1}(S)$.
-Beginning with the leftmost of RHS terms and proceeding to the consecutive rightmost terms, let us show \cref{lem:3m-G_2} to be true.
+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 $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
 
-\AR{I think it would be good to give an overview of the argument below. Something along these lines. Given any $S\in\binom{E_1}{\le 3}$, we consider $f^{-1}(S)$ (which is the set of all possible subsets of three edges in $S\times \{0,1\}$ that $f$ maps to $S$). Then in $f^{-1}(S)$, we count how many of the corresponding three edge subgraphs in $G_2$ is a 3-matching.}
-
-Consider the $\threedis$ pattern.  For $G_2$, this gives us three disjoint two paths.
+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 $G_2$.
 \AH{The comment below is an important comment.}
-\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.}  One can see that we have a total of two possible choices for each disjoint two path, which yields $2^3 = 8$ possible 3-matchings in $G_2$.
+\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.}  
 
-For the $\twopathdis$, i.e. DT pattern, when we convert to $G_2$, we have a four path and disjoint two path.  Let the generating subgraph in $G_1$ be $S = \{e_0, e_1, e_2\}$, and $e_1, e_2$ form the $2$-path.  We can only pick $(e_0, 0)$ or $(e_0, 1)$ from $f^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f^{-1}.$  The corresponding $4$-path in $G_2$ is $(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1).$  A four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_1, 0), (e_2, 0)}, \pbrace{(e_1, 0), (e_2, 1)}, \pbrace{(e_1, 1), (e_2, 1)}$.  Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices.  Edge $e_0$ cannot produce a $2$-matching, and are done with $\twopathdis$.
+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 $\twopathdis$.
 
-For the $\oneint$, i.e. 3-star pattern, the resulting $G_2$ graph produces 3 two paths, each of which has one and only one endpoint that intersects with one and only one endpoint of the other two paths.  Call the intersecting edges of the two paths inner edges.  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.}
+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.}
 
-When $S \in G_1$ is $\threepath$ it is the case that $S$ becomes a six path.  Let $S = (e_0, e_1, e_2)$ and the six path be $(e_0, 0), (e_0, 1), (e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1)$.  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.
+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.
 
-For $S = \tri$, note that it is the case that the generated subgraph in $G_2$ is a 'triangle of two paths', where each trianglular edge is a two path.  While this is similar to the discussion of the six path above, one must use caution not to consider the first and last edges as disjoint, since they are connected.  This rules out both $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with two remaining edge combinations that produce a 3 matching.
+For $S = \tri$, note that it is the case that the edges in $f_2^{-1}$ 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 six path above, one must use caution not to consider the first and last edges as disjoint, since they are connected.  This rules out both $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with two remaining edge combinations that produce a 3 matching.
 
-Consider when $S$ consists of at most $2$ distinct edges.  When $S$ has exactly $2$ edges, the generated subgraphs would either be a $4$-path or $2$ disjoint $2$-paths.  Neither of these can produce a $3$-matching.  For the former, a 3-matching would require at least to alternate edges to ensure disjointedness, which requires $\geq 5$-path.  The latter requires that at least one of the disjoint $2$-paths be a $3$-path in order to produce a $2$-matching.  When $S$ has $< 2$ edges, it is the case we have a generated $2$-path, which as stated above cannot produce a $3$-matching.  Therefore only subgraphs $S$ of $3$ edges need to be considered.
+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.
 
 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.