Finished pass on Appendix B.
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%root: main.tex
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We use \Cref{lem:qEk-multi-p} to prove \Cref{thm:mult-p-hard-result}:
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{\Cref{lem:pdb-for-def-qk}}
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\begin{Lemma}\label{lem:pdb-for-def-qk}
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@ -10,23 +10,34 @@ Assuming that each $v \in \vset$ has degree $\geq 1$,\footnote{This is WLOG, sin
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Only two relations need be constructed, one for the set $\vset$ and one for the set $\edgeSet$. By a simple linear scan, each can be constructed in time $\bigO{\numedge + \numvar}$. Given that the degree of each $v \in \vset$ is at least $1$, we have that $m\ge \Omega(n)$,
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%$\abs{\edgeSet}$ is at least within a constant factor of $\abs{\vset}$,
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and this yields the claimed runtime.
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\qed
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\end{proof}
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\subsection{Proof of \Cref{lem:tdet-om}}
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\begin{proof}
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By the recursive defintion of $\qruntime{\cdot, \cdot}$ (see \Cref{sec:gen}), we have the following equation for our hard query $\query$ when $k = 1$.
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By the recursive defintion of $\qruntime{\cdot, \cdot}$ (see \Cref{sec:gen}), we have the following equation for our hard query $\query$ when $k = 1$, (we denote this as $\query^1$).
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\begin{equation*}
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\qruntime{\query^1, \dbbase} = \abs{\dbbase.\vset} + \abs{\dbbase.\edgeSet} + \abs{\dbbase.\vset} + \abs{\dbbase.\vset \join \dbbase.\edgeSet \join \dbbase.\vset}
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\end{equation*}
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The quantity $\abs{\dbbase.\vset \join \dbbase.\edgeSet \join \dbbase.\vset}$ is at most $\numedge$ by definition of $\query^1$. Then by the assumption of \Cref{lem:pdb-for-def-qk} (each $v \in \vset$ has degree $\geq 1$), the sum of the first three terms is $\bigO{\numedge}$. We then obtain that $\qruntime{\query^1, \dbbase} = \bigO{\numedge} + \bigO{\numedge} = \bigO{\numedge}$. For $\query^k = \query_1^1 \times\cdots\times\query_k^1$, we have the recurrence $\qruntime{\query^k, \dbbase} = \qruntime{\query_1^1, \dbbase} + \cdots +\qruntime{\query_k^1, \dbbase} + \abs{\query_1^1\join\cdots\join\query_k^1}$. Since $\query^\kElem$ (and $\query^1$) all output a count, we have $\abs{\query_1^1\join\cdots\join\query_k^1}=1$. Thus, we have
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The quantity $\abs{\dbbase.\vset \join \dbbase.\edgeSet \join \dbbase.\vset}$ is at most $\numedge$ by definition of $\query^1$. Then by the assumption of \Cref{lem:pdb-for-def-qk} (each $v \in \vset$ has degree $\geq 1$), the sum of the first three terms is $\bigO{\numedge}$. We then obtain that $\qruntime{\query^1, \dbbase} = \bigO{\numedge} + \bigO{\numedge} = \bigO{\numedge}$. For $\query^k = \query_1^1 \times\cdots\times\query_k^1$, we have the recurrence $\qruntime{\query^k, \dbbase} = \qruntime{\query_1^1, \dbbase} + \cdots +\qruntime{\query_k^1, \dbbase} + \abs{\query_1^1\join\cdots\join\query_k^1}$. Since $\query^\kElem$ outputs a count, we have $\abs{\query_1^1\join\cdots\join\query_k^1}=1$. Thus, we have
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\[\qruntime{\query^k, \dbbase} \le k\cdot O(m)+1\le O(km),\]
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as desired.
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\qed
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%The dominating term in the sum of the recursive definition is $\abs{\query_1^1\join\cdots\join\query_k^1} = \bigO{\numedge^k} = O_k(\numedge)$.
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%Since by definition, $\dbbase = \cup_{\db \in \idb}\db$, it follows that $\dbbase$ consists of the relations that contain all possible $v \in \vset$ and $e \in \edgeSet$. Because the result for $\query^1$ cannot be any larger than the relation encoding $\edgeSet$ (i.e., $\abs{\edgeSet}$), it follows that (using an efficient query evaluation strategy such as indexing) the runtime of $\qruntime{\query^1, \dbbase}$ is indeed $O(\numedge)$. When $k > 1$, since by \Cref{def:qk} $\query^k$ is simply a cross product of the original query $\query^1$, we arrive at the desired runtime of $O_k(\numedge)$.
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\end{proof}
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\subsection{\Cref{lem:qEk-multi-p}}
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\noindent The following lemma reduces the problem of counting $\kElem$-matchings in a graph to our problem (and proves \Cref{thm:mult-p-hard-result}):
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\begin{Lemma}\label{lem:qEk-multi-p}
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Let $\prob_0,\ldots, \prob_{2\kElem}$ be distinct values in $(0, 1]$. Then given the values $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for $0\leq i\leq 2\kElem$, the number of $\kElem$-matchings in $G$ can be computed in $\bigO{\kElem^3}$ time.
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\end{Lemma}
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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{Proof of Theorem~\ref{thm:mult-p-hard-result}}
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\begin{proof}
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For the sake of contradiction, assume we can solve our problem in $\littleo{\kmatchtime}$ time. Given a graph $G$ by \Cref{lem:pdb-for-def-qk} we can compute the \abbrPDB encoding in $\bigO{\numedge}$ time. Then after we run our algorithm on $\rpoly_G^\kElem$, we get $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for every $0\leq i\leq 2\kElem$ in additional $\bigO{k}\cdot \littleo{\kmatchtime}$ time. \Cref{lem:qEk-multi-p}\AR{Broken ref} then computes the number of $k$-matchings in $G$ in $O(\kElem^3)$ time. Adding the runtime of all of these steps, we have an algorithm for computing the number of $k$-matchings that runs in time
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For the sake of contradiction, assume we can solve our problem in $\littleo{\kmatchtime}$ time. Given a graph $G$ by \Cref{lem:pdb-for-def-qk} we can compute the \abbrPDB encoding in $\bigO{\numedge}$ time. Then after we run our algorithm on $\rpoly_G^\kElem$, we get $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for every $0\leq i\leq 2\kElem$ in additional $\bigO{k}\cdot \littleo{\kmatchtime}$ time. \Cref{lem:qEk-multi-p} then computes the number of $k$-matchings in $G$ in $O(\kElem^3)$ time. Adding the runtime of all of these steps, we have an algorithm for computing the number of $k$-matchings that runs in time
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\begin{align}
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%&\bigO{\numedge} + \bigO{k}\cdot \littleo{\kmatchtime + \numedge} + O(\kElem^3)\label{eq:proof-omega-kmatch1}\\
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&\bigO{\numedge} + \bigO{k}\cdot \littleo{\kmatchtime} + O(\kElem^3)\label{eq:proof-omega-kmatch2}\\
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@ -38,15 +49,6 @@ Thus we obtain the contradiction that we can achieve a runtime $\littleo{\kmatch
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\qed
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\end{proof}
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\subsection{\Cref{lem:qEk-multi-p}}
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\noindent The following lemma reduces the problem of counting $\kElem$-matchings in a graph to our problem (and proves \Cref{thm:mult-p-hard-result}):
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\begin{Lemma}\label{lem:qEk-multi-p}
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Let $\prob_0,\ldots, \prob_{2\kElem}$ be distinct values in $(0, 1]$. Then given the values $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for $0\leq i\leq 2\kElem$, the number of $\kElem$-matchings in $G$ can be computed in $\bigO{\kElem^3}$ time.
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\end{Lemma}
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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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@ -60,7 +62,7 @@ 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, 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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\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 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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@ -74,11 +76,11 @@ To compute $\numocc{G}{\threepath}$, note that for an arbitrary edge $(i, j)$,
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Before proceeding, let us introduce a few more helpful definitions.
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\begin{Definition}[$\esetType{\ell}$]\label{def:ed-nota}
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For $\ell > 1$, we use $\esetType{\ell}$ 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 $e\in \esetType{1}$, where $(e,0),\dots,(e,\ell-1)$ is the $\ell$-path that replaces the edge $e$.
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For $\ell > 1$, we use $\esetType{\ell}$ 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 $b \in \{0,\ldots, \ell-1\}$ where $(e,0),\dots,(e,\ell-1)$ is the $\ell$-path that replaces the edge $e$ for $e\in \esetType{1}$.
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\end{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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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 $S$ to generate $\sg{\ell}$).
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\end{Definition}
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For example, consider $\sg{1}$ with edges $\eset{1} = \{e_1\}$. Then the edge set of $\sg{2}$ is defined as $\eset{2} = \{(e_1, 0), (e_1, 1)\}$.
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@ -98,7 +100,7 @@ For an arbitrary subgraph $\sg{1}$ of $\graph{1}$ with at most $m \leq 3$ edges,
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$f_\ell(s) = \eset{1}$.
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\end{Definition}
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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 $s\in f_\ell^{-1}(\eset{1})$ 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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Note, importantly, that when we discuss $f_\ell^{-1}$, that each \textit{edge} present in $\eset{1}$ must have an edge in $s\in f_\ell^{-1}(\eset{1})$ that projects down to it. In particular, if $|\eset{1}| = 3$, then it must be the case that each $s\in f_\ell^{-1}(\eset{1})$ 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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%\begin{Lemma}\label{lem:fk-func}
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%$f_\ell$ is a function.
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@ -108,50 +110,55 @@ Note, importantly, that when we discuss $f_\ell^{-1}$, that each \textit{edge} p
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%For any $b \in \{0,\ldots, \ell-1\}$, the map $(e, b) \mapsto e$ is a function since it has exactly one mapping. It then follows 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 $\esetType{1}$. To prove the structural lemmas, we will use the map $f_\ell^{-1}$. In particular, to count the number of occurrences of $\tri$ and $\threedis$ 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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We are now ready to prove the structural lemmas.
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%Note that $f_\ell$ maps subsets of three edges in $\graph{\ell}$ to a subset of at most three edges in $\esetType{1}$.
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To prove the structural lemmas, we will
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%use the map $f_\ell^{-1}$. In particular, to
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count the number of occurrences of $\tri$ and $\threedis$ in $\graph{\ell}$ we count for each $S\in\binom{E_1}{\le 3}$, how many $\threedis$ and $\tri$ subgraphs appear in $f_\ell^{-1}(\eset{1})$.
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\subsubsection{Proof of Lemma \ref{lem:3m-G2}}
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\begin{proof}%[Proof of \Cref{lem:3m-G2}]
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For each subset $\eset{1}\in \binom{E_1}{\le 3}$, we count the number of {\emph{$3$-matchings }}in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(\eset{1})$. We first consider the case of $\eset{1} \in \binom{E_1}{3}$, where $\eset{1}$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(\eset{1})$ is the set of all $3$-edge subsets $s \in \{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1),$ $(e_3, 0), (e_3, 1)\}$ such that $f_\ell(s) = \{e_1, e_2, e_3\}$. Note that the size of the output denoted $\abs{f_2^{-1}(\esetType{1})}$ is $8$ in such a case, where each set of edges of the form $\{(e_1, b_1), (e_2, b_2), (e_3, b_3)\}$ for $b_i \in [2], i \in [3]$ is present. We count the number of $3$-matchings from the set $f_2^{-1}(\eset{1})$.
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For each subset $\eset{1}\in \binom{E_1}{\le 3}$, we count the number of {\emph{$3$-matchings }}in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(\eset{1})$. We first consider the case of $\eset{1} \in \binom{E_1}{3}$, where $\eset{1}$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(\eset{1})$ is the set of all $3$-edge subsets $s \in \{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1),$ $(e_3, 0), (e_3, 1)\}$ such that $f_\ell(s) = \{e_1, e_2, e_3\}$. The size of the output is denoted $\abs{f_2^{-1}(\esetType{1})}$. For the case where each set of edges of the form $\{(e_1, b_1), (e_2, b_2), (e_3, b_3)\}$ for $b_i \in [2], i \in [3]$ is present, we have $\abs{f_2^{-1}(\esetType{1})} = 8$. We count the number of $3$-matchings from the set $f_2^{-1}(\eset{1})$.
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We do a case analysis based on the subgraph $\sg{1}$ induced by $\eset{1}$. %(denoted $\eset{1} \equiv \sg{1}$):
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\begin{itemize}
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\item $3$-matching ($\threedis$)
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\end{itemize}
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When $\sg{1}$ is isomorphic to $\threedis$, it is the case that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. By definition, each set of edges is a three matching. One can see that we have a total of two possible choices for $b_i$ for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(\eset{1})$.
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When $\sg{1}$ is isomorphic to $\threedis$, it is the case that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. By definition, each set of edges in $f_2^{-1}\inparen{\eset{1}}$ is a three matching and $\abs{f_2^{-1}\inparen{\eset{1}}} = 8$ possible 3-matchings.%. One can see that we have a total of two possible choices for $b_i$ for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(\eset{1})$.
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\begin{itemize}
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\item Disjoint Two-Path ($\twopathdis$)
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\end{itemize}
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For $\sg{1}$ isomorphic to $\twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that in $\sg{2}$ edges $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can pick either $(e_1, 0)$ or $(e_1, 1)$ for the first edge in the $3$-path, while it is necessary to have a $2$-matching from path $(e_2, 0),\ldots(e_3, 1)$. Note that the $4$-path allows for three possible $2$-matchings, specifically,
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For $\sg{1}$ isomorphic to $\twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that in $\sg{2}$ edges $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can pick either $(e_1, 0)$ or $(e_1, 1)$ for the first edge in the $3$-matching, while it is necessary to have a $2$-matching from path $(e_2, 0),\ldots(e_3, 1)$. Note that the $4$-path allows for three possible $2$-matchings, specifically,
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\begin{equation*}
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\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}.
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\end{equation*}
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Since these two selections can be made independently, there are $2 \cdot 3 = 6$ \emph{distinct} $3$-matchings in $f_2^{-1}(\eset{1})$.
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Since these two selections can be made independently, $\abs{f_2^{-1}\inparen{\eset{1}}} = 2 \cdot 3 = 6$ \emph{distinct} $3$-matchings in $f_2^{-1}(\eset{1})$.
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\begin{itemize}
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\item $3$-star ($\oneint$)
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\end{itemize}
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When $\sg{1}$ is isomorphic to $\oneint$, the inner edges $(e_i, 1)$ of $\sg{2}$ 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. For the case of when exactly one inner edge is chosen, there exist $3$ possiblities, based on which inner edge is chosen. Note that if $(e_i, 1)$ is chosen, the matching has to choose $(e_j, 0)$ for $j \neq i$ and $(e_{j'}, 0)$ for $j' \neq i, j' \neq j$. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are four 3-matchings in $f_2^{-1}(\eset{1})$.
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When $\sg{1}$ is isomorphic to $\oneint$, the inner edges $(e_i, 1)$ of $\sg{2}$ 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. For the case of when exactly one inner edge is chosen, there exist $3$ possiblities, based on which inner edge is chosen. Note that if $(e_i, 1)$ is chosen, the matching has to choose $(e_j, 0)$ for $j \neq i$ and $(e_{j'}, 0)$ for $j' \neq i, j' \neq j$. The remaining possible 3-matching occurs when all 3 outer edges are chosen, and $\abs{f_2^{-1}\inparen{\eset{1}}} = 4$. %Thus, there are four 3-matchings in $f_2^{-1}(\eset{1})$.
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\begin{itemize}
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\item $3$-path ($\threepath$)
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\end{itemize}
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When $\sg{1}$ is isomorphic to $\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This means that the edges of $\eset{2}$ form a $6$-path. For a $3$-matching to exist in $f_2^{-1}(\eset{1})$, we cannot pick both $(e_i,0)$ and $(e_i,1)$ or both $(e_i, 1)$ and $(e_j, 0)$ where $j = i + 1$. % there must be at least one edge separating edges picked from a sequence.
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There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}$, $\pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}$, $\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$ $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$, a total of four 3-matchings in $f_2^{-1}(\eset{1})$.
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There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}$, $\pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}$, $\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$ $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ and $\abs{f^{-1}_2\inparen{\eset{1}}} = 4.$%a total of four 3-matchings in $f_2^{-1}(\eset{1})$.
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\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
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For $\sg{1}$ isomorphic to $\tri$, note that it is the case that the edges in $\eset{2}$ 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 three path above, the first and last edges are not disjoint, since they are connected. This rules out both subsets of $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$, yielding two 3-matchings.
|
||||
For $\sg{1}$ isomorphic to $\tri$, note that it is the case that the edges in $\eset{2}$ 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 three path above, the first and last edges are not disjoint.%, since they are connected.
|
||||
This rules out both subsets of $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$, so that $\abs{f_2^{-1}\inparen{\eset{1}}} = 2$.%yielding two 3-matchings.
|
||||
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||||
Let us now consider when $\eset{1} \in \binom{E_1}{\leq 2}$, i.e. fixed subgraphs among
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\noindent Let us now consider when $\eset{1} \in \binom{E_1}{\leq 2}$, i.e. fixed subgraphs among
|
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\begin{itemize}
|
||||
\item $2$-matching ($\twodis$), $2$-path ($\twopath$), $1$ edge ($\ed$)
|
||||
\end{itemize}
|
||||
When $|\eset{1}| = 2$, we can only pick one from each of two pairs, $\pbrace{(e_1, 0), (e_1, 1)}$ and $\pbrace{(e_2, 0), (e_2, 1)}$. The third edge for any set of edges in $f_2^{-1}(\esetType{1})$ would break the disjoint property of a $3$-path. Thus, a $3$-matching cannot exist in $f_2^{-1}(\eset{1})$. A similar argument holds for $|\eset{1}| = 1$, where the output of $f_2^{-1}$ is $\{\emptyset\}$ since there are not enough edges in the input to produce any other output. %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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When $|\eset{1}| = 2$, we can only pick one from each of two pairs, $\pbrace{(e_1, 0), (e_1, 1)}$ and $\pbrace{(e_2, 0), (e_2, 1)}$. The third edge choice in $\eset{2}$ will break the disjoint property of a $3$-matching. Thus, a $3$-matching cannot exist in $f_2^{-1}(\eset{1})$. A similar argument holds for $|\eset{1}| = 1$, where the output of $f_2^{-1}$ is $\{\emptyset\}$ since there are not enough edges in the input to produce any other output. %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 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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|
|
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|
@ -19,7 +19,7 @@ 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 $e_1 = (i_1, j_1), e_2 = (i_2, j_2), and e_3 = (i_3, j_3)$.
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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.
|
||||
|
||||
\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 $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
|
||||
|
|
@ -36,7 +36,7 @@ Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \two
|
|||
|
||||
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 $G$:
|
||||
\begin{Definition}\label{def:Gk}
|
||||
For $\ell \geq 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary graph $G$, by replacing every edge $e$ of $G$ with a $\ell$-path, such that all inner vertexes of an $\ell$-path replacement edge are disjoint from all other vertexes.\footnote{Note that $G\equiv \graph{1}$.}% of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
|
||||
For $\ell \geq 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary graph $G$, by replacing every edge $e$ of $G$ with an $\ell$-path, such that all inner vertexes of an $\ell$-path replacement edge are disjoint from all other vertexes.\footnote{Note that $G\equiv \graph{1}$.}% of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
|
||||
\end{Definition}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
We will prove \Cref{th:single-p-hard} by the following reduction:
|
||||
|
|
@ -77,10 +77,5 @@ Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\e
|
|||
allowing us to compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
|
||||
\end{Lemma}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\subsection{Proof of \Cref{th:single-p}}
|
||||
\begin{proof}
|
||||
We can compute $\graph{2}$ from $\graph{1}$ in $O(m)$ time. Additionally, if in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, then the theorem follows by \Cref{lem:lin-sys}.
|
||||
\qed
|
||||
\end{proof}
|
||||
In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.
|
||||
|
||||
|
||||
|
|
|
|||
|
|
@ -9,7 +9,8 @@ We note that $c_i$ is {\em exactly} the number of monomials in the SMB expansion
|
|||
|
||||
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.
|
||||
Thus, after $O(k^3)$ work, we know $\vct{c}$ and in particular, $c_{2k}$ exactly.
|
||||
Next, we show why we can compute $\numocc{G}{\kmatch}$ from $c_{2k}$ in $O(1)$ additional time.
|
||||
|
||||
Next, we show why we can compute $\numocc{G}{\kmatch}$ from $c_{2k}$ in $O(1)$ additional time.
|
||||
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
|
||||
\[\poly_{G}^\kElem(\vct{X}) = \sum_{\substack{(i_1, j_1),\cdots,(i_\kElem, j_\kElem) \in E}}X_{i_1}X_{j_1}\cdots X_{i_\kElem}X_{j_\kElem},\]
|
||||
where a unique $\kElem$-matching in the multi-set of product terms can be selected $\prod_{i = 1}^\kElem i = \kElem!$ times.
|
||||
|
|
|
|||
14
lin_sys.tex
14
lin_sys.tex
|
|
@ -12,7 +12,7 @@ The lemma claims that for $\vct{M} =
|
|||
\numocc{G}{\tri}]\\
|
||||
\numocc{G}{\threedis}
|
||||
\end{pmatrix}$
|
||||
satisfies the system $\vct{M} \cdot \vct{x} = \vct{b}$.
|
||||
satisfies the linear system $\vct{M} \cdot \vct{x} = \vct{b}$.
|
||||
|
||||
To prove the first step, we use \Cref{lem:qE3-exp} to derive the following equality (dropping the superscript and referring to $G^{(1)}$ as $G$):
|
||||
\begin{align}
|
||||
|
|
@ -30,7 +30,7 @@ Note that the LHS of \Cref{eq:b1-alg-2} is obtained using \cref{eq:2pd-3d} and \
|
|||
|
||||
We follow the same process in deriving an equality for $G^{(2)}$. Replacing occurrences of $G$ with $G^{(2)}$, we obtain an equation (below) of the form of \cref{eq:b1-alg-2} for $G^{(2)}$. Substituting identities from \cref{lem:3m-G2} and \Cref{lem:tri} we obtain
|
||||
\begin{align}
|
||||
0-\left(8\numocc{G}{\threedis}\right.&\left.+6\numocc{G}{\twopathdis}+4\numocc{G}{\oneint}+4\numocc{G}{\threepath}+2\numocc{G}{\tri}(3\prob^2 -\prob^3)\right)=\nonumber\\
|
||||
0-\left(8\numocc{G}{\threedis}\right.&\left.+6\numocc{G}{\twopathdis}+4\numocc{G}{\oneint}+4\numocc{G}{\threepath}+2\numocc{G}{\tri}\right)(3\prob^2 -\prob^3)=\nonumber\\
|
||||
&\frac{\rpoly_{\graph{2}}^3(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath}-\numocc{\graph{2}}{\twodis}\prob-\numocc{\graph{2}}{\oneint}\prob\nonumber\\
|
||||
&-\left[\numocc{\graph{2}}{\twopathdis}\prob^2+3\numocc{\graph{2}}{\threedis}\prob^2\right]-\left[\numocc{\graph{2}}{\threepath}\prob + 3\numocc{\graph{2}}{\tri}\prob\right]\label{eq:b2-sub-lem}\\
|
||||
(10\numocc{G}{\tri} &+ 10{G}{\threedis})(3\prob^2 -\prob^3) = \nonumber\\
|
||||
|
|
@ -42,7 +42,7 @@ The steps to obtaining \cref{eq:b2-final} are analogous to the derivation immedi
|
|||
|
||||
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.
|
||||
|
||||
To show that $\vct{M}$ indeed has full rank, we show derive in what follows that $\dtrm{\vct{M}}\ne 0$ for every $\prob\in (0,1)$.
|
||||
To show that $\vct{M}$ indeed has full rank, we show in what follows that $\dtrm{\vct{M}}\ne 0$ for every $\prob\in (0,1)$.
|
||||
$\dtrm{\vct{M}} = $
|
||||
\begin{align}
|
||||
&\begin{vmatrix}
|
||||
|
|
@ -59,6 +59,14 @@ From \Cref{eq:det-final} it can easily be seen that the roots of $\dtrm{\vct{M}
|
|||
\qed
|
||||
\end{proof}
|
||||
|
||||
\subsection{Proof of \Cref{th:single-p}}
|
||||
\begin{proof}
|
||||
We can compute $\graph{2}$ from $\graph{1}$ in $O(m)$ time. Additionally, if in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, then the theorem follows by \Cref{lem:lin-sys}.
|
||||
\qed
|
||||
\end{proof}
|
||||
|
||||
In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.
|
||||
|
||||
\subsection{Proof of \Cref{th:single-p-hard}}
|
||||
\begin{proof}
|
||||
For the sake of contradiction, assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
|
||||
|
|
|
|||
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Reference in New Issue