More tweaks on Appendix B.
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@ -30,7 +30,7 @@ For edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other e
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\end{proof}
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\begin{proof}[Proof of \Cref{eq:2pd-3d}]
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\Cref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in our argument for \Cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. Note, the factor $3$ of $\threedis$ is necessary to account for the triple counting of $3$-matchings. It is also the case that, since the two path in $\twopathdis$ is connected, that there will be no double counting by the fact that the summation automatically disconnects the current edge, meaning that a two matching at the current vertex will not be counted. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$. Note that all $d_i$ and $d_i - 3$ factorials can be computed in $O(m)$ time, and then each combination $\binom{n}{3}$ can be performed with constant time operations, yielding the claimed $O(m)$ run time.
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\Cref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in our argument for \Cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. Note, the factor $3$ of $\threedis$ is necessary to account for the triple counting of $3$-matchings, since it is indistinguishable to the closed form expression which of the remaining edges are either disjoint or connected to each of the edges in the {\emph{initial}} set of edges disjoint to the edge under consideration. Observe that the disjoint case will be counted $3$ times since each edge of a $3$-path is visited once, and the same $3$-path counted in each visitation. For the latter case however, it is true that since the two path in $\twopathdis$ is connected, there will be no multiple counting by the fact that the summation automatically disconnects the current edge, meaning that a two matching at the current vertex under consideration will not be counted. Thus, $\twopathdis$ will only be counted once, precisely when the single disjoint edge is visited in the summation. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$. Note that all $d_i$ and $d_i - 2$ factorials can be computed in $O(m)$ time, and then each combination $\binom{n}{2}$ can be performed with constant time operations, yielding the claimed $O(m)$ run time.
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\qed
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\end{proof}
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\begin{proof}[Proof of \Cref{eq:3p-3tri}]
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@ -42,11 +42,13 @@ To compute $\numocc{G}{\threepath}$, note that for an arbitrary edge $(i, j)$,
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\subsection{Proofs for \Cref{lem:3m-G2}, \Cref{lem:tri}, and \Cref{lem:lin-sys}}\label{subsec:proofs-struc-lemmas}
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Before proceeding, let us introduce a few more helpful definitions.
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\AH{Main concerns: More consistent and less-bulky (if possible) notation. Make sure types match. Remove extraneous information.}
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\begin{Definition}\label{def:ed-nota}
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For $\ell > 1$, we use $E_\ell$ to denote the set of edges in $\graph{\ell}$. For any graph $\graph{\ell}$, its edges are denoted by the a pair $(e, b)$, such that $b \in \{0,\ldots, \ell-1\}$ and $e\in E_1$, where $(e,0),\dots,(e,\ell-1)$ is the $\ell$-path that replaces the edge $e$.
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\end{Definition}
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\AH{It might be helpful to keep the subscripts the same between the above and below definition.}
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\begin{Definition}[$\eset{\ell}$]
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Given an arbitrary subgraph $\sg{1}$ of $\graph{1}$, let $\eset{1}$ denote the set of edges in $\sg{1}$. Define then $\eset{\ell}$ for $\ell > 1$ as the set of edges in the generated subgraph $\sg{\ell}$ (i.e. when we apply \Cref{def:Gk} to $\sg{1})$.
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\end{Definition}
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@ -67,20 +69,24 @@ Let $f_\ell: \binom{E_\ell}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows
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For an arbitrary subgraph $\sg{1}$ of $\graph{1}$ with at most $m \leq 3$ edges, the inverse function $f_\ell^{-1}: \binom{E_1}{\leq 3}\mapsto 2^{\binom{E_\ell}{3}}$ takes $\eset{1}$ and outputs the set of all elements $s \in \binom{\eset{\ell}}{3}$ such that
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$f_\ell(s) = \eset{1}$.
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\end{Definition}
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\AH{The above definition seems choppy. Perhaps saying ``takes the set of edges in $\sg{1}$ and outputs...''}
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Note, importantly, that when we discuss $f_\ell^{-1}$, that each \textit{edge} present in $\eset{1}$ must have an edge in $s\in f_\ell^{-1}(S)$ that projects down to it. In particular, if $|\eset{1}| = 3$, then it must be the case that each $s\in f_\ell^{-1}(S)$ consists of the following set of edges: $\{ (e_i, b), (e_j, b'), (e_m, b'') \}$, where $i,j$ and $m$ are distinct.
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\AH{The type for the input seems inconsistent. Does it take in a subgraph, or a set of edges? We need to be precise and consistent with this.}
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We first note that $f_\ell$ is well-defined:
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\begin{Lemma}\label{lem:fk-func}
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$f_\ell$ is a function.
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\end{Lemma}
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\AH{I think that only the third sentence is necessary to prove the lemma's claim.}
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\begin{proof}\label{subsubsec:proof-fk}
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Note that $f_\ell$ is properly defined. For any $S \in \binom{E_\ell}{3}$, $|f(S)| \leq 3$, since it has to be the case that any subset of $3$ edges in $E_\ell$ will map to at most three edges in $E_1$. All mappings are in the required range. Then, since for any $b \in \{0,\ldots, \ell-1\}$ the map $(e, b) \mapsto e$ is a function and has exactly one mapping, which %` mapping for which $(e, b)$ maps to no other edge than $e$, and this
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implies that $f_\ell$ is a function.\qed
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\end{proof}
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We are now ready to prove the structural lemmas. Note that $f_\ell$ maps subsets of three edges in $\graph{\ell}$ to a subset of at most three edges in $E_1$. To prove the structural lemmas, we will use the map $f_\ell^{-1}$. In particular, to count the number of occurrences of $\tri,\threepath,\threedis$ in $\graph{\ell}$ we count for each $S\in\binom{E_1}{\le 3}$, how many of $\tri/\threepath/\threedis$ subgraphs appear in $f_\ell^{-1}(S)$.
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We are now ready to prove the structural lemmas. Note that $f_\ell$ maps subsets of three edges in $\graph{\ell}$ to a subset of at most three edges in $E_1$. To prove the structural lemmas, we will use the map $f_\ell^{-1}$. In particular, to count the number of occurrences of $\tri,\threepath,\threedis$ in $\graph{\ell}$ we count for each $S\in\binom{E_1}{\le 3}$, how many $\threedis$ and $\tri$ subgraphs appear in $f_\ell^{-1}(S)$.
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\subsubsection{Proof of Lemma \ref{lem:3m-G2}}
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@ -123,9 +129,9 @@ Let us now consider when $\eset{1} \in \binom{E_1}{\leq 2}$, i.e. patterns among
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\begin{itemize}
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\item $2$-matching ($\twodis$), $2$-path ($\twopath$), $1$ edge ($\ed$)
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\end{itemize}
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When $|\eset{1}| = 2$, we can only pick one from each of two pairs, $\pbrace{(e_1, 0), (e_1, 1)}$ and $\pbrace{(e_2, 0), (e_2, 1)}$. This implies that a $3$-matching cannot exist in $f_2^{-1}(\eset{1})$. The same argument holds for $|\eset{1}| = 1$, where we can only pick one edge from the pair $\pbrace{(e_1, 0), (e_1, 1)}$. Trivially, no $3$-matching exists in $f_2^{-1}(\eset{1})$.
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When $|\eset{1}| = 2$, we can only pick one from each of two pairs, $\pbrace{(e_1, 0), (e_1, 1)}$ and $\pbrace{(e_2, 0), (e_2, 1)}$. This implies that a $3$-matching cannot exist in $f_2^{-1}(\eset{1})$. The same argument holds for $|\eset{1}| = 1$, where we can only pick one edge from the pair $\pbrace{(e_1, 0), (e_1, 1)}$. Trivially, no $3$-matching exists in $f_2^{-1}(\eset{1})$ either.
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Observe that all of the arguments above focused solely on the subgraph $\sg{1}$ is isomorphmic. In other words, all $\eset{1}$ of a given ``shape'' yield the same number of $3$-matchings in $f_2^{-1}(\eset{1})$, and this is why we get the required identity using the above case analysis.
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Observe that all of the arguments above focused solely on the property of subgraph $\sg{1}$ being isomorphmic. In other words, all $\eset{1}$ of a given ``shape'' yield the same number of $3$-matchings in $f_2^{-1}(\eset{1})$, and this is why we get the required identity using the above case analysis.
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\qed
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\end{proof}
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@ -25,10 +25,10 @@ By definition we have that
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\[\poly_{G}^3(\vct{X}) = \sum_{\substack{(i_1, j_1), (i_2, j_2), (i_3, j_3) \in E}}~\; \prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}.\]
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Hence $\rpoly_{G}^3(\vct{X})$ has degree six. Note that the monomial $\prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}$ will contribute to the coefficient of $\prob^\nu$ in $\rpoly_{G}^3(\vct{X})$, where $\nu$ is the number of distinct variables in the monomial.
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%Rather than list all the expressions in full detail, let us make some observations regarding the sum.
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Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$.
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Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), and e_3 = (i_3, j_3)$.
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We compute $\rpoly_{G}^3(\vct{X})$ by considering each of the three forms that the triple $(e_1, e_2, e_3)$ can take.
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\textsc{case 1:} $e_1 = e_2 = e_3$ (all edges are the same). There are exactly $\numedge=\numocc{G}{\ed}$ such triples, each with a $\prob^2$ factor in $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\textsc{case 1:} $e_1 = e_2 = e_3$ (all edges are the same). When we have that $e_1 = e_2 = e_3$, then the monomial corresponds to $\numocc{G}{\ed}$. There are exactly $\numedge$ such triples, each with a $\prob^2$ factor in $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\textsc{case 2:} This case occurs when there are two distinct edges of the three, call them $e$ and $e'$. When there are two distinct edges, there is then the occurence when $2$ variables in the triple $(e_1, e_2, e_3)$ are bound to $e$. There are three combinations for this occurrence in $\poly_{G}^3(\vct{X})$. Analogusly, there are three such occurrences in $\poly_{G}^3(\vct{X})$ when there is only one occurrence of $e$, i.e. $2$ of the variables in $(e_1, e_2, e_3)$ are $e'$. %Again, there are three combinations for this.
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This implies that all $3 + 3 = 6$ combinations of two distinct edges $e$ and $e'$ contribute to the same monomial in $\rpoly_{G}^3$. % consist of the same monomial in $\rpoly$, i.e. $(e_1, e_1, e_2)$ is the same as $(e_2, e_1, e_2)$.
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@ -43,7 +43,7 @@ Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \two
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Since $\prob$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to equations (\ref{eq:1e})-(\ref{eq:3p-3tri})). However, we need to generate one more independent linear equation in these two variables. Towards this end we generate another graph related to $G$:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{Definition}\label{def:Gk}
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For $\ell > 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary graph $\graph{1}$, by replacing every edge $e$ of $\graph{1}$ with a $\ell$-path, such that all inner vertexes of an $\ell$-path replacement edge are disjoint from the inner vertexes of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
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For $\ell > 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary graph $\graph{1}$, by replacing every edge $e$ of $\graph{1}$ with a $\ell$-path, such that all inner vertexes of an $\ell$-path replacement edge are disjoint from all other vertexes of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
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\end{Definition}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -10,13 +10,13 @@ We note that $c_i$ is {\em exactly} the number of monomials in the SMB expansion
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Given that we then have $2\kElem + 1$ distinct values of $\rpoly_{G}^\kElem(\prob,\ldots, \prob)$ for $0\leq i\leq2\kElem$, it follows that we have a linear system of the form $\vct{M} \cdot \vct{c} = \vct{b}$ where the $i$th row of $\vct{M}$ is $\inparen{\prob_i^0\ldots\prob_i^{2\kElem}}$, $\vct{c}$ is the coefficient vector $\inparen{c_0,\ldots, c_{2\kElem}}$, and $\vct{b}$ is the vector such that $\vct{b}[i] = \rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$. In other words, matrix $\vct{M}$ is the Vandermonde matrix, from which it follows that we have a matrix with full rank (the $p_i$'s are distinct), and we can solve the linear system in $O(k^3)$ time (e.g., using Gaussian Elimination) to determine $\vct{c}$ exactly.
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Thus, after $O(k^3)$ work, we know $\vct{c}$ and in particular, $c_{2k}$ exactly.
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Next, we show why we can compute $\numocc{G}{\kmatch}$ from $c_{2k}$ in $O(1)$ additional time.
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We claim that $c_{2\kElem}$ is $\kElem! \cdot \numocc{G}{\kmatch}$. This can be seen intuitively by looking at the original factorized representation
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We claim that $c_{2\kElem}$ is $\kElem! \cdot \numocc{G}{\kmatch}$. This can be seen intuitively by looking at the expansion of the original factorized representation
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\[\poly_{G}^\kElem(\vct{X}) = \sum_{\substack{(i_1, j_1),\cdots,(i_\kElem, j_\kElem) \in E}}X_{i_1}X_{j_1}\cdots X_{i_\kElem}X_{j_\kElem},\]
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where across each of the $\kElem$ products, an arbitrary $\kElem$-matching can be selected $\prod_{i = 1}^\kElem i = \kElem!$ times.
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Indeed, note that each $\kElem$-matching $(i_1, j_1)\ldots$ $(i_k, j_k)$ in $G$ corresponds to the monomial $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ in $\poly_{G}^\kElem(\vct{X})$, with distinct indexes. Second, the only surviving monomials $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ of degree exactly $2k$ in $\rpoly_{G}^{\kElem}(\vct{X})$ must have that all of $i_1,j_1,\dots,i_\kElem,j_\kElem$ are distinct in $\poly_{G}^{\kElem}(\vct{X})$.
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where a unique $\kElem$-matching in the multi-set of product terms can be selected $\prod_{i = 1}^\kElem i = \kElem!$ times.
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Indeed, note that each $\kElem$-matching $(i_1, j_1)\ldots$ $(i_k, j_k)$ in $G$ corresponds to the monomial $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ in $\poly_{G}^\kElem(\vct{X})$, with distinct indexes, and this implies that each distinct $\kElem$-matching appears the exact number of permutations that exist for the set of its edges, or $k!$. Second, the only surviving monomials $\prod_{\ell = 1}^\kElem X_{i_\ell}X_{j_\ell}$ of degree exactly $2k$ in $\rpoly_{G}^{\kElem}(\vct{X})$ must have that all of $i_1,j_1,\dots,i_\kElem,j_\kElem$ are distinct in $\poly_{G}^{\kElem}(\vct{X})$.
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By the last two statements, only monomials composed of $2k$ distinct variables in $\poly_{G}^{\kElem}(\vct{X})$ (and hence of degree $2\kElem$ in $\rpoly_{G}^{\kElem}(\vct{X})$) correspond to a $k$-matching in $G$.
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Notice that each of the $k!$ permutations of an arbitrary monomial maps to the same distinct $\kElem$-matching in $G$, and this implies a $\kElem!$ to $1$ mapping between degree $2\kElem$ monomials in $\rpoly_{G}^{\kElem}(\vct{X})$ and $\kElem$-matchings in $G$.
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As noted above, each of the $k!$ permutations of an arbitrary monomial maps to the same distinct $\kElem$-matching in $G$, and this implies a $\kElem!$ to $1$ mapping between degree $2\kElem$ monomials in $\rpoly_{G}^{\kElem}(\vct{X})$ and $\kElem$-matchings in $G$.
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It then follows that $c_{2\kElem}= \kElem! \cdot \numocc{G}{\kmatch}$.
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Thus, simply dividing $c_{2\kElem}$ by $\kElem!$ gives us $\numocc{G}{\kmatch}$, as needed. \qed
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\end{proof}
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@ -1,5 +1,5 @@
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%root: main.tex
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\AH{Just wanted to go over the math calculation in this before moving on.}
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\begin{proof}%[Proof of \Cref{lem:lin-sys}]
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The proof consists of two parts. First we need to show that a vector $\vct{b}$ satisfying the linear system exists and further can be computed in $O(m)$ time. Second we need to show that $\numocc{G}{\tri}, \numocc{G}{\threedis}$ can indeed be computed in time $O(1)$.
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@ -42,8 +42,8 @@ As in the previous equality derivation for $G$, note that the LHS of \Cref{eq:b2
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Note that if $\vct{M}$ has full rank then one can compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ using Gaussian elimination.
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To show that $\vct{M}$ indeed has full rank, we will show that $\dtrm{\vct{M}}\ne 0$ for every $\prob\in (0,1)$.
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Let $\vct{M} = $
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To show that $\vct{M}$ indeed has full rank, we show derive in what follows that $\dtrm{\vct{M}}\ne 0$ for every $\prob\in (0,1)$.
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$\dtrm{\vct{M}} = $
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\begin{align}
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&\begin{vmatrix}
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1-3\prob &-(3\prob^2 - \prob^3)\\
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