142 lines
14 KiB
TeX
142 lines
14 KiB
TeX
%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{Proof of Theorem~\ref{thm:mult-p-hard-result}}
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\begin{proof}
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For the sake of contradiction, let us assume we can solve our problem in $f(\kElem)\cdot m^c$ time for some absolute constant $c$. Then given a graph $G$ we can compute the query polynomial or rather, expression tree representation of $\rpoly_G^\kElem$ (in the obvious way) in $O(km)$ time. Then after we run our algorithm on $\rpoly_G^\kElem$, we get $\rpoly_{G}^\kElem(\prob_i,\ldots, \prob_i)$ for $0\leq i\leq 2\kElem$ in additional $f(\kElem)\cdot m^c$ time. \Cref{lem:qEk-multi-p} then computes the number of $k$-matchings in $G$ in $O(\kElem^3)$ time. Thus, overall we have an algorithm for computing the number of $k$-matchings in time
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\begin{align*}
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O(km) + f(\kElem)\cdot m^c + O(\kElem^3)
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&\le \inparen{O(\kElem^3) + f(\kElem)}\cdot m^{c+1} \\
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&\le \inparen{O(\kElem^3) + f(\kElem)}\cdot n^{2c+2},
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\end{align*}
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which contradicts \Cref{thm:k-match-hard}.
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\qed
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\end{proof}
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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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The proofs for \Cref{eq:1e}, \Cref{eq:2p} and \Cref{eq:3s} are immediate.
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\begin{proof}[Proof of \Cref{eq:2m}]
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For edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other edges in $G$ disjoint to $(i, j)$ is equivalent to finding all edges that are not connected to either vertex $i$ or $j$. The number of such edges is $m - d_i - d_j + 1$, where we add $1$ since edge $(i, j)$ is removed twice when subtracting both $d_i$ and $d_j$. Since the summation is iterating over all edges such that a pair $\left((i, j), (k, \ell)\right)$ will also be counted as $\left((k, \ell), (i, j)\right)$, division by $2$ then eliminates this double counting. Note that $m$ and $d_i$ for all $i \in V$ can be computed in one pass over the set of edges by simply maintaining counts for each quantity. Finally, the summation is also one traversal through the set of edges where each operation is either a lookup ($O(1)$ time) or an addition operation (also $O(1)$) time.
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\qed
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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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\qed
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\end{proof}
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\begin{proof}[Proof of \Cref{eq:3p-3tri}]
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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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%Originally the proofs in hardenss section
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\input{app_single-p-proof-defs}
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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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\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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\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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\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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\begin{Definition}[$\binom{\edgeSet}{t}$ and $\binom{\edgeSet}{\leq t}$]\label{def:ed-sub}
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Let $\binom{E}{t}$ denote the set of subsets in $E$ with exactly $t$ edges. In a similar manner, $\binom{E}{\leq t}$ is used to mean the subsets of $E$ with $t$ or fewer edges.
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\end{Definition}
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The following function $f_\ell$ is a mapping from every $3$-edge shape in $\graph{\ell}$ to its `projection' in $\graph{1}$.
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\begin{Definition}\label{def:fk}
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Let $f_\ell: \binom{\esetType{\ell}}{3} \rightarrow \binom{\esetType{1}}{\leq3}$ be defined as follows. For any element $s \in \binom{\esetType{\ell}}{3}$ such that $s = \pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}$, define:
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\[ f_\ell\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}.\]
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\end{Definition}
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\begin{Definition}[$f_\ell^{-1}$]\label{def:fk-inv}
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For an arbitrary subgraph $\sg{1}$ of $\graph{1}$ with at most $m \leq 3$ edges, the inverse function $f_\ell^{-1}: \binom{\esetType{1}}{\leq 3}\rightarrow 2^{\binom{\esetType{\ell}}{3}}$ 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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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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\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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\begin{proof}\label{subsubsec:proof-fk}
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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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\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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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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\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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\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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\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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\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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\begin{itemize}
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\item Triangle ($\tri$)
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\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.
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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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\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 $\pbrace{(e_2, 0), (e_2, 1)}$. 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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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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%e have the case that all edges in $\eset{3}$ have the property that $(e_i, b_i)$ is disjoint to $(e_j, b_j)$ for $i \neq j$. For each $e_i$, there are then $3$ choices, independent of each other, and it results that there are $3^3 = 27$ many 3-matchings in $f_3^{-1}(\eset{1})$.
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\subsubsection{Proof of \Cref{lem:tri}}
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\begin{proof}%[Proof of \Cref{lem:tri}]
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The number of triangles in $\graph{\ell}$ for $\ell \geq 2$ will always be $0$ for the simple fact that all cycles in $\graph{\ell}$ will have at least six edges.
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\qed
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\end{proof}
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\subsubsection{Proof of \Cref{lem:lin-sys}}
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\input{lin_sys} |