Finished Atri's suggestions 071820.

poly-form.tex

@@ -215,7 +215,7 @@ Consider graph $\graph{2}$, constructed from an arbitrary graph $\graph{1}$.  We
 
 \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 $\graph{2}$ and $\graph{3}$ outside of the proofs below.}
 
-\AH{{\Huge\bf Changes start here. 07-18-2020.}}
+\AH{{\Huge\bf Changes start here. 07-20-2020.}}
 
 \subsubsection{$f_k$ and $\graph{k}$}
 
@@ -228,7 +228,7 @@ For any graph $\graph{k}$, we denote its edges to be a pair $(e, b)$, such that
 
 Before defining $f_k$, the following notation will be useful.  Let $\binom{S}{t}$ denote the set of subsets in $S$ with exactly $t$ edges.  In a similar manner, $\binom{S}{\leq t}$ is used to mean the subsets of $S$ with size $\leq t$.  The set of edges in $\graph{k}$ is written as $E_k$.
 \begin{Definition}\label{def:fk}
-Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$.  For $S \in \binom{E_3}{3}$, $f_k\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}$.
+Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$.  For $S \in \binom{E_3}{3}$, such that $S = \pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}$, $f_k\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}$.
 \end{Definition}
 
 
@@ -261,7 +261,7 @@ The number of $3$-matchings in graph $\graph{2}$ satisfies the following identit
 \begin{proof}
 \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.}
 
-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 $\graph{2}$ 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 set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
+Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible sets of 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 $\graph{2}$ 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 set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
 
 \begin{itemize}
 	\item $3$-matching ($\threedis$)
@@ -311,7 +311,7 @@ The number of 3-matchings in $\graph{3}$ satisfy the following identity,
 \end{Lemma}
 
 \begin{proof}
-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 $f_3^{-1}(S)$.  Let $S'$ be all the edges of $\graph{3}$, formally, $S' = \pbrace{(e_1, 0),\ldots, (e_3, 2)}$.  Note, that for $\graph{1}$, when $|S| = 1$, we represent $S$ as $e_1$, when $|S| = 2$, as $e_1, e_2$, and when $|S| = 3$ we represent edges as $e_1, e_2, e_3$.  The representation is symmetrical in $S'$, where, for example, when $|S| = 1$, then $S' = \pbrace{(e_1, 0), (e_1, 1), (e_1, 2)}$. \AR{Generally be vary about using $\ldots$ since it make sense only if everything in lex order between the two terms around the $\ldots$ appears. This is true here but not everywhere $\ldots$ are used in this proof. Please go through all uses of $\ldots$ to check this.}
+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 $f_3^{-1}(S)$.  Let $S'$ be all the edges of $\graph{3}$ which 'project' down to any edge in $S$, formally, for $|S| = 1$, then $S = \{e_1\}$ and $S' = \pbrace{(e_1, 0), (e_1, 1), (e_1, 2)}$.  Similarly, when $|S| = 2$, then $S = \pbrace{e_1, e_2}$ and $S' = \pbrace{(e_1, 0),\ldots, (e_2, 2)}$, and so on for $|S| = 3$.
 
 
 
@@ -323,7 +323,7 @@ When $S = \ed$, $f_3^{-1}(S)$ gives one result, $(e_1, 0), (e_1, 1), (e_1, 2)$,
 \begin{itemize}
 	\item $2$-path ($\twopath$)
 \end{itemize}
-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 as stated above in \cref{lem:3m-G2}\AR{This is a vague reference-- you actually meant the proof of Lemma 7 and not earlier part of this proof. So explicitly refer to proof of Lemma 7. Check for other occurences of this in the proof.}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
+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 as stated above in \cref{lem:3m-G2}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
 
 \begin{itemize}
 	\item $2$-matching ($\twodis$)
@@ -333,17 +333,17 @@ For $S = \twodis$, then all subsets in the output of $f_3^{-1}(S)$ are predicate
 \begin{itemize}
 	\item Triangle ($\tri$)
 \end{itemize}
-Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with a $S = \tri$.   As discussed in the case of $\graph{2}$, $f_3^{-1}(S)$ 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 $s \subseteq S'$, $s$ 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 $b_{i + 1 \mod{4}} = 0$.  Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)},  \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}$\newline$\ldots, \pbrace{(e_1, 1), (e_2, 1), (e_3, 2)} \pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}, \pbrace{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$, giving a total of 18 3-matchings in $f_3^{-1}(S)$.
+Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with a $S = \tri$.   As discussed in \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 $b_{i + 1 \mod{4}} = 0$.  Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)},  \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}$\newline$\ldots, \pbrace{(e_1, 1), (e_2, 1), (e_3, 2)} \pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}, \pbrace{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$, giving a total of 18 3-matchings in $f_3^{-1}(S)$.
 
 \begin{itemize}
 	\item $3$-path ($\threepath$)
 \end{itemize}
-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.
+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 $18 + 3 = 21$ three-matchings.
 
 \begin{itemize}
 	\item Disjoint Two-Path ($\twopathdis$)
 \end{itemize}
-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 (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 $\pbox{(e_2, 0), (e_3, 0)},\ldots, \pbox{(e_2, 1), (e_3, 2)}, \pbox{(e_2, 2), (e_3, 1)}, \pbox{(e_2, 2), (e_3, 2)}$. 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 in $f_3^{-1}(S)$.
+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 (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 $f_3^{-1}(S)$.
 
 \begin{itemize}
 	\item $3$-star ($\oneint$)
@@ -368,7 +368,7 @@ The number of $3$-paths in $\graph{2}$ is computed by the following linear combi
 \end{Lemma}
 
 \begin{proof}
-For $s \subseteq S'$ such that $s$ is a $3$-path, it \textit{must} be the case that all edges in $f_2(s)$ have at least one mapping from an edge in $s$. \AR{Here you mean to talk about edges in $E_2$ but the states talking about edges in $E_1$}\AH{Not sure if I have understood this comment, but hopefully fixed the problem.}  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 $f_2(s) = S$.\AR{Technically the argument for $\ed$ is different from the others in that $f_2^{-1}(\ed)$ has not have enough edges in total.}\AH{I'm slightly confused by the above comment...intuitively I understand, but doesn't $f_2^{-1}(S) = \emptyset$ technically?}  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 $f_2^{-1}(S)$.\AH{\LARGE Here where I left off.}\AR{Explicitly state the paths.}
+For $s \subseteq S'$ such that $s$ is a $3$-path, it \textit{must} be the case by definition of $f$ that all edges in $f_2(s)$ have at least one mapping from an edge in $s$. \AR{Here you mean to talk about edges in $E_2$ but the states talking about edges in $E_1$}\AH{Not sure if I have understood this comment, but hopefully fixed the problem.}  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 $f_2(s) = S$.\AR{Technically the argument for $\ed$ is different from the others in that $f_2^{-1}(\ed)$ has not have enough edges in total.}\AH{I'm slightly confused by the above comment...intuitively I understand, but doesn't $f_2^{-1}(S) = \emptyset$ technically?}  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 $f_2^{-1}(S), \pbrace{(e1, 0), (e_1, 1), (e_2, 0)} \text{ and }\pbrace{(e_1, 1), (e_2, 0), (e_2, 1)}$.\AH{\LARGE Here where I left off.}\AR{Explicitly state the paths.}
 \end{proof}
 \qed
 %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,
@@ -381,7 +381,7 @@ The number of $3$-paths in $\graph{3}$ is computed by the following linear combi
 \end{Lemma}
 
 \begin{proof}
-The argument follows along the same lines as above.  Given $M = \threepath \in \graph{3}$, it \textit{must} be that there is a $3$-path across all edges of $S = f_3(M)$\AR{Same comment as in the $\graph{2}$ case.}.  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 $e_2$.  For the latter, it it trivial to see that a $1$-path in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.  
+The argument follows along the same lines as above.  Given $s \subseteq S'$, it \textit{must} be that every edge in $f_3(s)$ has at least one edge in $s$ mapped to it.  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 a $1$-path in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.  
 \end{proof}
 \qed