Removed proofs from S3
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%root: main.tex
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\begin{proof}%[Proof of \Cref{lem:qE3-exp}]
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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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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 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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Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \twodis$, which contribute $6\prob^3$ and $6\prob^4$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\textsc{case 3:} All $e_1,e_2$ and $e_3$ are distinct. For this case, we have $3! = 6$ permutations of $(e_1, e_2, e_3)$, each of which contribute to the same monomial in the \textsc{SMB} representation of $\poly_{G}^3(\vct{X})$. This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $6\prob^3, 6\prob^4, 6\prob^4, 6\prob^5$ and $6\prob^6$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\qed
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\end{proof}
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@ -0,0 +1,21 @@
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%root: main.tex
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We need all the possible edge patterns in an arbitrary $G$ with at most three distinct edges. We have already seen $\tri,\threepath$ and $\threedis$, so we define the remaining patterns:
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\begin{itemize}
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\item Single Edge $\left(\ed\right)$
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\item 2-path ($\twopath$)
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\item 2-matching ($\twodis$)
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\item 3-star ($\oneint$)--this is the graph that results when all three edges share exactly one common endpoint. The remaining endpoint for each edge is disconnected from any endpoint of the remaining two edges.
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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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\end{itemize}
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For any graph $G$, the following formulas for $\numocc{G}{H}$ compute their respective patterns exactly in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$ (proofs are in \Cref{app:easy-counts}):
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\begin{align}
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&\numocc{G}{\ed} = \numedge, \label{eq:1e}\\
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&\numocc{G}{\twopath} = \sum_{i \in V} \binom{d_i}{2} \label{eq:2p}\\
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&\numocc{G}{\twodis} = \sum_{(i, j) \in E} \frac{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\%\binom{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\
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&\numocc{G}{\oneint} = \sum_{i \in V} \binom{d_i}{3}\label{eq:3s}\\
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&\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis} = \sum_{(i, j) \in E} \binom{\numedge - d_i - d_j + 1}{2}\label{eq:2pd-3d}\\
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&\numocc{G}{\threepath} + 3\numocc{G}{\tri} = \sum_{(i, j) \in E} (d_i - 1) \cdot (d_j - 1)\label{eq:3p-3tri}
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\end{align}
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@ -17,7 +17,8 @@ which contradicts \Cref{thm:k-match-hard}.
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\subsection{Proof of Lemma~\ref{lem:qEk-multi-p}}
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\input{lem_mult-p}
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\subsection{Subgraph Notation and $O(1)$ Closed Formulas}
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\input{app_hardness-notation-easy-counts}
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\subsection{Proofs of~\cref{eq:1e}-\cref{eq:3p-3tri}}
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\label{app:easy-counts}
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@ -36,6 +37,8 @@ For edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other e
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To compute $\numocc{G}{\threepath}$, note that for an arbitrary edge $(i, j)$, a 3-path exists for edge pair $(i, \ell)$ and $(j, k)$ where $i, j, k, \ell$ are distinct. Further, the quantity $(d_i - 1) \cdot (d_j - 1)$ represents the number of 3-edge subgraphs with middle edge $(i, j)$ and outer edges $(i, \ell), (j, k)$ such that $\ell \neq j$ and $k \neq i$. When $k = \ell$, the resulting subgraph is a triangle, and when $k \neq \ell$, the subgraph is a 3-path. Summing over all edges (i, j) gives \cref{eq:3p-3tri} by observing that each triangle is counted thrice, while each 3-path is counted just once. For reasons similar to \cref{eq:2m}, all $d_i$ can be computed in $O(m)$ time and each summand can then be computed in $O(1)$ time, yielding an overall $O(m)$ run time.
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\qed
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\end{proof}
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\subsection{Proof for \Cref{lem:qE3-exp}}
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\input{app_hardness-exact-coefficients}
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\subsection{Proofs for~\Cref{lem:3m-G2}-\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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41
single_p.tex
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single_p.tex
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@ -40,32 +40,7 @@ In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard
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%Before we move on to the proof itself, we state the results, lemmas, and defintions that will be useful in the proof.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection{Preliminaries and Notation}
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We need all the possible edge patterns in an arbitrary $G$ with at most three distinct edges. We have already seen $\tri,\threepath$ and $\threedis$, so we define the remaining patterns:
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\begin{itemize}
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\item Single Edge $\left(\ed\right)$
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\item 2-path ($\twopath$)
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\item 2-matching ($\twodis$)
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%\item Triangle ($\tri$)
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%\item 3-path ($\threepath$)
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\item 3-star ($\oneint$)--this is the graph that results when all three edges share exactly one common endpoint. The remaining endpoint for each edge is disconnected from any endpoint of the remaining two edges.
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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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%Let $\numocc{G}{H}$ denote the number of occurrences of pattern $H$ in graph $G$, where, for example, $\numocc{G}{\ed}$ means the number of single edges in $G$.
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For any graph $G$, the following formulas for $\numocc{G}{H}$ compute their respective patterns exactly in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$ (proofs are in \Cref{app:easy-counts}):
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\begin{align}
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&\numocc{G}{\ed} = \numedge, \label{eq:1e}\\
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&\numocc{G}{\twopath} = \sum_{i \in V} \binom{d_i}{2} \label{eq:2p}\\
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&\numocc{G}{\twodis} = \sum_{(i, j) \in E} \frac{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\%\binom{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\
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&\numocc{G}{\oneint} = \sum_{i \in V} \binom{d_i}{3}\label{eq:3s}\\
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&\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis} = \sum_{(i, j) \in E} \binom{\numedge - d_i - d_j + 1}{2}\label{eq:2pd-3d}\\
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&\numocc{G}{\threepath} + 3\numocc{G}{\tri} = \sum_{(i, j) \in E} (d_i - 1) \cdot (d_j - 1)\label{eq:3p-3tri}
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\end{align}
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\subsubsection{The proofs}
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@ -81,23 +56,7 @@ For any $\prob$, we have:
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&+ 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
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\end{align}}
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\end{Lemma}
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\begin{proof}[Proof of \Cref{lem:qE3-exp}]
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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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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 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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Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \twodis$, which contribute $6\prob^3$ and $6\prob^4$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\textsc{case 3:} All $e_1,e_2$ and $e_3$ are distinct. For this case, we have $3! = 6$ permutations of $(e_1, e_2, e_3)$, each of which contribute to the same monomial in the \textsc{SMB} representation of $\poly_{G}^3(\vct{X})$. This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $6\prob^3, 6\prob^4, 6\prob^4, 6\prob^5$ and $6\prob^6$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\end{proof}
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\qed
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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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 \Cref{eq:1e}-\Cref{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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