Done till proof of Lemma 3.15
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\section{Missing details from Section~\ref{sec:hard}}
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\label{app:hard}
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Blah
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\subsection{Proofs of~\cref{eq:1e}-\cref{eq:2pd-3d}}
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\label{app:easy-counts}
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\cref{eq:1e},~\cref{eq:2p} and~\cref{eq:3s} are immediate.
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A quick argument to why \cref{eq:2m} is true. Note that 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.
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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 ~\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 $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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single_p.tex
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single_p.tex
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\subsection{Single $\prob$ value}
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\label{sec:single-p}
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In this discussion, let us fix $\kElem = 3$.
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%In this discussion, let us fix $\kElem = 3$.
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While~\cref{thm:mult-p-hard-result} shows that computing $\rpoly(\prob,\dots,\prob)$ in general is hard it does not rule out the possibility that can one compute this value exactly for a {\em fixed} value of $p$. Indeed, it is easy to check that once can compute $\rpoly(\prob,\dots,\prob)$ exactly in linear time for $p\in \inset{0,1}$. In this section, we show that these two are the only possibilities:
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\AH{@atri needs to put in the result for triangles of $\numvar^{\frac{4}{3}}$ runtime.}
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\begin{Theorem}\label{th:single-p}
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If we can compute $\rpoly_{G}^3(\vct{X})$ in T(\numedge) time for $X_1 =\cdots= X_\numvar = \prob$, then we can count the number of triangles, 3-paths, and 3-matchings in $G$ in $T(\numedge) + O(\numedge)$ time.
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\begin{Theorem}\label{cor:single-p-hard}
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Fix $p\in (0,1)$. Then assuming~\cref{conj:graph} is true, then any algorithms that compute $\rpoly_{G}^3(\prob,\dots,\prob)$ exactly has to run in time $\Omega\inparen{\abs{E(G)}^{1+\eps_0}}$, where $\eps_0$ is as defined in~\cref{conj:graph}.
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\end{Theorem}
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%\begin{proof}[Proof of Corollary ~\ref{cor:single-p-gen-k}]
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%Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$. By ~\cref{th:single-p}, query $\poly''$ with $\kElem = 4$ has $\Omega(\numvar^{\frac{4}{3}})$ complexity.
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%\end{proof}
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The above shows the hardness for a very specific query polynomial but it is easy to come up with an infinite family of hard query polynomials by `embedding' $\rpoly_{G}^3$ into an infinite family of trivial query polynomials. However, unlike~\cref{thm:mult-p-hard-result} the above result does not show that computing $\rpoly_{G}^3(\prob,\dots,\prob)$ for a fixed $p\in (0,1)$ is \sharpwonehard. By contrast, in~\cref{sec:algo} we show that if we are willing to compute an approximation that this problem (and indeed solving our problem for a much more general setting) is in linear time.
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%\AH{@atri needs to put in the result for triangles of $\numvar^{\frac{4}{3}}$ runtime.}
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We will prove the above result by the following reduction:
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\begin{Theorem}\label{th:single-p}
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Fix $p\in (0,1)$. Let $G$ be a graph on $\numedge$ edges.
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If we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ exactly in $T(\numedge)$ time, then we can exactly compute $\numocc{G}{\tri}$, $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ %count the number of triangles, 3-paths, and 3-matchings in $G$
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in $O\inparen{T(\numedge) + \numedge}$ time.
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\end{Theorem}
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\begin{proof}[Proof of~\cref{cor:single-p-hard}]
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For the sake of contradiction, let us 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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Let $G$ be the input graph. It is easy to see that one can compute the expression tree for $\poly_{G}^3(\vct{X})$ in $O(m)$ time. Then by~\cref{th:single-p} we can compute $\numocc{G}{\tri}$, $\numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ in further time $o\inparen{m^{1+\eps_0}}+O(m)$. Thus, the overall, reduction takes $o\inparen{m^{1+\eps_0}}+O(m)= o\inparen{m^{1+\eps_0}}$ time, which violates~\cref{conj:graph}.
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\end{proof}
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\qed
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Before moving on to prove ~\cref{th:single-p}, let us state the results, lemmas and defintions that will be useful in the proof.
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We need to list all possible edge patterns in an arbitrary $G$ consisting of $\leq 3$ distinct edges.
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\subsubsection{Preliminaries and Notation}
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We need to list all possible edge patterns in an arbitrary $G$ consisting of at most three distinct edges. We have already seen $\tri,\threepath$ and $\threedis$, so here 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 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 three 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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%\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 compute $\numocc{G}{H}$ for their respective patterns in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$.
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For any graph $G$, the following formulas for $\numocc{G}{H}$ for their respective patterns can be used to compute them 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}{\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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\end{align}
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A quick argument to why \cref{eq:2m} is true. Note that 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.
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%A quick argument to why \cref{eq:2m} is true. Note that 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.
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Equation ~\ref{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 ~\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 $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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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 ~\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 $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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%Original lemma proving the exact coefficient terms in qE3
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsubsection{The proofs}
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Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
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\begin{Lemma}\label{lem:qE3-exp}
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When we expand $\poly_{G}^3(\vct{X})$ out and assign all exponents $e \geq 1$ a value of $1$, we have the following result,
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%When we expand $\poly_{G}^3(\vct{X})$ out and assign all exponents $e \geq 1$ a value of $1$, we have the following result,
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For any $p$, we have:
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\begin{align}
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&\rpoly_{G}^3(\prob,\ldots, \prob) = \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis} + 6\numocc{G}{\tri}\prob^3\nonumber\\
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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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\begin{proof}%[Proof of \cref{lem:qE3-exp}]
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By definition we have that
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\[\poly_{G}(\vct{X}) = \sum_{\substack{(i_1, j_1),\\ (i_2, j_2),\\ (i_3, j_3) \in E}} \prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}.\]
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Rather than list all the expressions in full detail, let us make some observations regarding the sum. Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$. Notice that each expression in the sum consists of a triple $(e_1, e_2, e_3)$. There are three forms the triple $(e_1, e_2, e_3)$ can take.
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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 $p^i$ in $\rpoly_{G}^3(\vct{X})$, where $i$ 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 $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$. Notice that each expression in the sum consists of a triple $(e_1, e_2, e_3)$. There are three forms the triple $(e_1, e_2, e_3)$ can take (and in each case, we will account for their contribution to $\rpoly_{G}^3(\vct{X})$).
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\textsc{case 1:} $e_1 = e_2 = e_3$, where all edges are the same. There are exactly $\numedge$ such triples, each with a $\prob^2$ factor in $\rpoly_{G}\left(\prob_1,\ldots, \prob_\numvar\right)$.
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\textsc{case 1:} $e_1 = e_2 = e_3$, where 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. It is the analogue for 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. All $3 + 3 = 6$ combinations of two distinct values consist of the same monomial in $\rpoly$, i.e. $(e_1, e_1, e_2)$ is the same as $(e_2, e_1, e_2)$. This case produces the following edge patterns: $\twopath, \twodis$.
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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 $p^3$ and $p^4$ respectively to $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
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\textsc{case 3:} $e_1 \neq e_2 \neq e_3$, i.e., when all edges are distinct. For this case, we have $3! = 6$ permutations of $(e_1, e_2, e_3)$. This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$.
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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 a different monomial in the SOP expansion of $\poly_{G}^3(\vct{X})$. This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $p^3,p^4,p^4,p^5$ and $p^6$ 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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Since $p$ is fixed,~\cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}, \numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to~\cref{eq:1e}-\cref{eq:2pd-3d}). However, we plan to generate two more independent linear equations in these three variables. Towards, this end we generate more graphs that are related to $G$:
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\begin{Definition}\label{def:Gk}
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For $k > 1$, let graph $\graph{k}$ be a graph generated from an arbitrary graph $\graph{1}$, by replacing every edge $e$ of $\graph{1}$ with a $k$-path, such that all $k$-path replacement edges are disjoint 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 $\ell$-path replacement edges are disjoint. % 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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Next, we relate the various sub-graph counts in $\graph{2}$ and $\graph{3}$ to their counterparts in $\graph{1}=G$.
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\begin{Lemma}\label{lem:3m-G2}
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The number of $3$-matchings in graph $\graph{2}$ satisfies the following identity,
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\begin{align*}
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\end{Lemma}
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\begin{Lemma}\label{lem:tri}
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For $k > 1$, any graph $\graph{k}$ has the property that $\numocc{\graph{k}}{\tri} = 0$.
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For $\ell > 1$, any graph $\graph{\ell}$ has the property that $\numocc{\graph{\ell}}{\tri} = 0$.
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\end{Lemma}
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Due to lack of space we defer the proof of the above results to~\cref{app:hard}.
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\AR{Need to do this move.}
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Using the results we have obtained so far, we will prove the following reduction result:
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\begin{Lemma}\label{lem:lin-sys}
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Using the identities of lemmas [\ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri}] to compute $\numocc{G}{\threedis}, \numocc{G}{\threepath}, \numocc{G}{\tri}$ for $G \in \{\graph{2}, \graph{3}\}$, there exists a linear system $\mtrix{\rpoly}\cdot (x~y~z~)^T = \vct{b}$ which can then be solved to determine the unknown quantities of $\numocc{\graph{1}}{\threedis}, \numocc{\graph{1}}{\threepath}$, and $\numocc{\graph{1}}{\tri}$.
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%Using the identities of lemmas [\ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri}] to compute $\numocc{G}{\threedis}, \numocc{G}{\threepath}, \numocc{G}{\tri}$ for $G \in \{\graph{2}, \graph{3}\}$, there exists a linear system $\mtrix{\rpoly}\cdot (x~y~z~)^T = \vct{b}$ which can then be solved to determine the unknown quantities of $\numocc{\graph{1}}{\threedis}, \numocc{\graph{1}}{\threepath}$, and $\numocc{\graph{1}}{\tri}$.
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Fix $p\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$, we can compute in $O(m)$ time a vector $\vct{b}\in\rel^3$ such that
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\[ \begin{pmatrix}
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1 & \prob & -(3\prob^2 - \prob^3)\\
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-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\
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-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3)
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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\numocc{G}{\tri}]\\
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\numocc{G}{\threepath}\\
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\numocc{G}{\threedis}
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\end{pmatrix}
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=\vct{b},
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\]
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from which we can compute $\numocc{G}{\tri}, \numocc{G}{\threepath}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
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\end{Lemma}
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\AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
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The above result immediately imples~\cref{th:single-p}:
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\begin{proof}[Proof of~\cref{th:single-p}]
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It is easy to check that in $O(m)$ time we can compute $\graph{2}$ and $\graph{3}$ from $\graph{1}=G$ (and further note that these graphs also have $O(m)$ edges). Thus,
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in time $O(T(m))$, we can compute $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$.~\Cref{lem:lin-sys} then completes the proof.
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\end{proof}
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\qed
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%\AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
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Using \cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
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We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using ~\cref{claim:four-two}, \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
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%Using \cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
|
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%We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using ~\cref{claim:four-two}, \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
|
||||
%$%^&*(
|
||||
|
||||
|
||||
\subsubsection{More notation}
|
||||
Before proceeding, let us introduce a few more helpful definitions.
|
||||
|
||||
\begin{Definition}\label{def:ed-nota}
|
||||
For the set of edges in $\graph{k}$ we write $E_k$. For any graph $\graph{k}$, its edges are denoted by the a pair $(e, b)$, such that $b \in \{0,\ldots, k-1\}$ and $e\in E_1$.
|
||||
For the set of edges in $\graph{\ell}$ we write $E_\ell$. 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 $e$.
|
||||
\end{Definition}
|
||||
|
||||
\begin{Definition}[$\eset{k}$]
|
||||
Given an arbitrary subgraph $S\graph{1}$ of $\graph{1}$, let $\eset{1}$ denote the set of edges in $S\graph{1}$. Define then $\eset{k}$ for $k > 1$ as the set of edges in the generated subgraph $S\graph{k}$.
|
||||
\begin{Definition}[$\eset{\ell}$]
|
||||
Given an arbitrary subgraph $S\graph{1}$ of $\graph{1}$, let $\eset{1}$ denote the set of edges in $S\graph{1}$. Define then $\eset{\ell}$ for $\ell > 1$ as the set of edges in the generated subgraph $S\graph{\ell}$ (i.e. when we apply~\Cref{def:Gk} to $\graph{1}=S\graph{1}$.
|
||||
\end{Definition}
|
||||
|
||||
For example, consider $S\graph{1}$ with edges $\eset{1} = \{e_1\}$. Then the edges of $S\graph{2}$, $\eset{2} = \{(e_1, 0), (e_1, 1)\}$.
|
||||
For example, consider $S\graph{1}$ with edges $\eset{1} = \{e_1\}$. Then the edges of $S\graph{2}$, s defined as $\eset{2} = \{(e_1, 0), (e_1, 1)\}$.
|
||||
\begin{Definition}\label{def:ed-sub}
|
||||
Let $\binom{S}{t}$ denote the set of subsets in $S$ with exactly $t$ edges. In a similar manner, $\binom{S}{\leq t}$ is used to mean the subsets of $S$ with $t$ or fewer edges.
|
||||
\end{Definition}
|
||||
|
||||
The following function $f_k$ is a mapping from every $3$-edge shape in $\graph{k}$ to its `projection' in $\graph{1}$.
|
||||
The following function $f_\ell$ is a mapping from every $3$-edge shape in $\graph{\ell}$ to its `projection' in $\graph{1}$.
|
||||
\begin{Definition}\label{def:fk}
|
||||
Let $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows. For any $S \in \binom{E_k}{3}$, such that $S = \pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}$, define:
|
||||
\[ f_k\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}.\]
|
||||
Let $f_\ell: \binom{E_\ell}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows. For any $S \in \binom{E_\ell}{3}$, such that $S = \pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}$, define:
|
||||
\[ f_\ell\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}.\]
|
||||
\end{Definition}
|
||||
|
||||
\AH{Just questioning if the notation is clear in the ~\cref{def:fk-inv}. For more details, see the immediately following todo note which is commented out.}
|
||||
%\AH{Just questioning if the notation is clear in the ~\cref{def:fk-inv}. For more details, see the immediately following todo note which is commented out.}
|
||||
%\AH{I found ~\cref{def:fk-inv} a bit imprecise and bulky and have attempted to refine it.
|
||||
%\par Since this an inverse function, the signature is reversed, \vari{but},
|
||||
%\par...the challenge is in quantifying the size of the set (of 3 edge subsets) that is returned...
|
||||
|
|
@ -143,97 +197,104 @@ Let $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows. For
|
|||
%\par...but the catch is that for $r \geq 3$, the set will be strictly less than $\binom{r\cdot k}{3}$ since $f_k$ does not map e.g. an input $\{(e_a, b_1), (e_a, b_2), (e_a, b_3)\}$ (where $a$ is constant and $b_1, b_2, b_3 \in \{0,\ldots, k -1\}$) to more than one edge, \textit{and} it is the case for $r \geq 3$ that $f_k^{-1}$ will not map such an input to its input of size $r$, meaning we must subtract off all such subsets of $\binom{E_k}{3}$.
|
||||
%\par My fix was to use a variable in the exponent and explain in prose. Perhaps there is a better, simpler notation/solution.}
|
||||
|
||||
\begin{Definition}[$f_k^{-1}$]\label{def:fk-inv}
|
||||
The inverse function $f_k^{-1}: \binom{E_1}{\leq 3}\mapsto \left\{\binom{E_k}{3}\right\}^{h}$ takes an arbitrary $\eset{1}$ of at most $3$ edges and outputs the set of all subsets of $\binom{\eset{k}}{3}$ such that each subset $s^{(k)}$ of the output set is mapped to the input set $s^{(1)}$ by $f_k$, i.e. $f_k(s^{(k)}) = s^{(1)}$. The set returned by $f_k^{-1}$ is of size $h$, where $h$ depends on $\abs{s^{(1)}}$, such that $h \leq \binom{\abs{s^{(1)}} \cdot k}{3}$.
|
||||
\begin{Definition}[$f_\ell^{-1}$]\label{def:fk-inv}
|
||||
The inverse function $f_\ell^{-1}: \binom{E_1}{\leq 3}\mapsto 2^{\binom{E_\ell}{3}}$ takes an arbitrary subset $S\subseteq E_1$ of at most $3$ edges and outputs the set of all subsets of $\binom{\eset{\ell}}{3}$ such that each set $T\in f_\ell^{-1}(S)$ is mapped to the input set $S$ by $f_\ell$, i.e. $f_\ell(T) = S$. %The set returned by $f_\ell^{-1}$ is of size $h$, where $h$ depends on $\abs{s^{(1)}}$, such that $h \leq \binom{\abs{s^{(1)}} \cdot \ell}{3}$.
|
||||
\end{Definition}
|
||||
|
||||
Note, importantly, that when we discuss $f_k^{-1}$, that, although potentially counterintuitive, each \textit{edge} present in $s^{(1)}$ must have an edge in $s^{(k)}$ that `projects` down to it. \textit{Meaning}, if $|s^{(1)}| = 3$, then it must be the case that each $s^{(k)}$ be a set $\{ (e_i, b), (e_j, b), e_\ell, b) \}$ where $i \neq j \neq \ell$.
|
||||
Note, importantly, that when we discuss $f_\ell^{-1}$, that each \textit{edge} present in $S$ must have an edge in $T\in f_\ell^{-1}(S)$ that `projects` down to it. In particular, if $|S| = 3$, then it must be the case that each $T\in f_\ell^{-1}(S)$ si given by $T=\{ (e_i, b), (e_j, b), (e_m, b) \}$ where $i,j$ and $m$ are distinct.
|
||||
|
||||
We first note that $f_\ell$ is well-defined:
|
||||
\begin{Lemma}\label{lem:fk-func}
|
||||
$f_k$ is a function.
|
||||
\end{Lemma}
|
||||
\begin{proof}[Proof of Lemma \ref{lem:fk-func}]
|
||||
Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \leq 3$, since it has to be the case that any subset of $3$ edges in $E_k$ will map to at most 3 edges in $\graph{1}$. All mappings are in the required range. Then, since for any $b \in \{0,\ldots, k-1\}$ the edge $(e, b) \mapsto e$ is a mapping for which $(e, b)$ maps to no other edge than $e$, and this implies that $f_k$ is a function.
|
||||
\begin{proof}%[Proof of Lemma \ref{lem:fk-func}]
|
||||
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, which %` mapping for which $(e, b)$ maps to no other edge than $e$, and this
|
||||
implies that $f_\ell$ is a function.
|
||||
\end{proof}
|
||||
\qed
|
||||
|
||||
\AR{TODO for {\em later}: I think the proof will be much easier to follow with figures: just drawing out $S\times \{0,1\}$ along with the $(e_i,b_i)$ explicity notated on the edges will make the proof much easier to follow.}
|
||||
|
||||
\subsubsection{Three Matchings in $\graph{2}$}
|
||||
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 $f_\ell^{-1}(S)$.
|
||||
|
||||
\begin{proof}[Proof of Lemma \ref{lem:3m-G2}]
|
||||
For each edge pattern $S$, we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$. We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is the set of all $3$-edge subsets of the set
|
||||
%\subsubsection{Three Matchings in $\graph{2}$}
|
||||
|
||||
\subsubsection{Proof of Lemma \ref{lem:3m-G2}}
|
||||
|
||||
For each subset $\eset{1}\in \binom{E_1}{\le 3}$, we count the number of $3$-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 of the set
|
||||
\begin{equation*}
|
||||
\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}.
|
||||
\end{equation*}
|
||||
|
||||
We do a case analysis based on the `shape' of $\eset{1}$:
|
||||
\begin{itemize}
|
||||
\item $3$-matching ($\threedis$)
|
||||
\end{itemize}
|
||||
Consider the $\eset{1} = \threedis$ pattern. Note that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. All subsets for $b_1, b_2, b_3 \in \{0, 1\}$, $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choices for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(S)$.
|
||||
When $\eset{1} \equiv \threedis$, that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. All choices for $b_1, b_2, b_3 \in \{0, 1\}$, $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choicesi for $b_i$ for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item Disjoint Two-Path ($\twopathdis$)
|
||||
\end{itemize}
|
||||
For $\eset{1} = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick either $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from $e_2$ and $e_3$. Note that a four path allows there to be 3 possible 2 matchings, specifically,
|
||||
For $\eset{1} \equiv \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_1$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick either $(e_1, 0)$ or $(e_1, 1)$ for $f_2^{-1}(\eset{1})$, and then we need to pick a $2$-matching from $e_2$ and $e_3$. Note that the four path allows there to be 3 possible 2 matchings, specifically,
|
||||
\begin{equation*}
|
||||
\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}.
|
||||
\end{equation*}
|
||||
Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
|
||||
Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-star ($\oneint$)
|
||||
\end{itemize}
|
||||
When $\eset{1} = \oneint$, the inner edges $(e_i, 1)$ of $\eset{2}$ 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. When exactly one inner edge is chosen, there are 3 such possibilities. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are $3 + 1 = 4$ 3-matchings in $f_2^{-1}(S)$.
|
||||
When $\eset{1} \equiv \oneint$, the inner edges $(e_i, 1)$ of $\eset{2}$ 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. When exactly one inner edge is chosen, there are 3 such possibilities. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are $3 + 1 = 4$ many 3-matchings in $f_2^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-path ($\threepath$)
|
||||
\end{itemize}
|
||||
When $\eset{1} =\threepath$ 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 in the edges of $f_2^{-1}(S)$, where all edges from $(e_1, 0),\ldots,(e_3, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
|
||||
When $\eset{1} \equiv\threepath$ 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 in the edges of $f_2^{-1}(\eset{1})$, where all edges from $(e_1, 0),\ldots,(e_3, 1)$ are successively connected. For a $3$-matching to exist in $f_2^{-1}(\eset{1})$, we cannot pick both $(e_i,0)$ and $(e_i,1)$. % there must be at least one edge separating edges picked from a sequence.
|
||||
There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
|
||||
For $\eset{1} = \tri$, 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 $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with $2$ remaining edge combinations that produce a 3 matching.
|
||||
For $\eset{1} \equiv \tri$, 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 $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with $2$ remaining edge combinations that produce a 3 matching.
|
||||
|
||||
Let us now consider when $S \in \binom{E_1}{\leq 2}$, i.e. patterns among
|
||||
\begin{itemize}
|
||||
\item $2$-matching ($\twodis$), $2$-path ($\twopath$), $1$ edge ($\ed$)
|
||||
\end{itemize}
|
||||
Let us also consider when $S \in \binom{E_1}{\leq 2}$. When $|S| = 2$, 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 $f_2^{-1}(S)$. The same argument holds for $|S| = 1$, where we can only pick one edge from the pair $\pbrace{(e_1, 0), (e_1, 1)}$, thus no $3$-matching exists in $f_2^{-1}(S)$.
|
||||
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 $\pbrace{(e_1, 0), (e_1, 1)}$, thus no $3$-matching exists in $f_2^{-1}(\eset{1})$.
|
||||
|
||||
Observe that all of the arguments above focused solely on the shape/pattern of $S$. In other words, all $S$ of a given shape yield the same number of $3$-matchings, and this is why we get the required identity.
|
||||
\end{proof}
|
||||
\qed
|
||||
Observe that all of the arguments above focused solely on the shape/pattern of $S$. In other words, all $S$ 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.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
||||
\subsubsection{Three matchings in $\graph{3}$}
|
||||
\subsubsection{Proof of~\cref{lem:3m-G3}}
|
||||
|
||||
|
||||
\begin{proof}[Proof of Lemma \ref{lem:3m-G3}]
|
||||
For any $S \in \binom{E_1}{\leq3}$, we again then count the number of $3$-matchings in $f_3^{-1}(S)$.
|
||||
For any $\eset{1} \in \binom{E_1}{\leq3}$, we again then count the number of $3$-matchings in $f_3^{-1}(\eset{1})$ via a case analysis:
|
||||
|
||||
|
||||
|
||||
\begin{itemize}
|
||||
\item $1$ edge ($\ed$)
|
||||
\end{itemize}
|
||||
When $\eset{1} = \ed$, $f_3^{-1}(\eset{1})$ has one subset, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(\eset{1})$ for this case.
|
||||
When $\eset{1} \equiv \ed$, $f_3^{-1}(\eset{1})$ has one subset, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(\eset{1})$ for this case.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-path ($\twopath$)
|
||||
\end{itemize}
|
||||
Fix then $\eset{1} = \twopath$ and now we have all edges in $\eset{3}$ form a $6$-path, and similar to the discussion in the proof of \cref{lem:3m-G2} (when $eset{1} = \threepath$ in $\graph{2}$), this leads to $4$ $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
When $\eset{1} \equiv \twopath$ and now we have all edges in $\eset{3}$ form a $6$-path, and similar to the discussion in the proof of \cref{lem:3m-G2} (when $\eset{1} \equiv \threepath$ in $\graph{2}$), this leads to four $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-matching ($\twodis$)
|
||||
\end{itemize}
|
||||
For $\eset{1} = \twodis$, all edges of $\eset{3}$ are predicated on the fact that $(e_i, b)$ is disjoint with $(e_j, b)$ for $i \neq j\in \{1,2\}$ and $b \in \{0, 1, 2\}$. Pick an aribitrary $e_i$ and note, that $(e_i, 0), (e_i, 2)$ is a $2$-matching, which can combine with any of the $3$ edges in $(e_j, 0), (e_j, 1), (e_j, 2)$ again for $i \neq j$. Since the selections are independent, it follows that there exist $2 \cdot 3 = 6$ $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
For $\eset{1} \equiv \twodis$, all edges of $\eset{3}$ are predicated on the fact that $(e_i, b)$ is disjoint with $(e_j, b)$ for $i \neq j\in \{1,2\}$ and $b \in \{0, 1, 2\}$. Pick an aribitrary $e_i$ and note, that $(e_i, 0), (e_i, 2)$ is a $2$-matching, which can combine with any of the $3$ edges in $(e_j, 0), (e_j, 1), (e_j, 2)$ again for $i \neq j$. Since the selections are independent, it follows that there exist $2 \cdot 3 = 6$ many $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
|
||||
Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $\eset{1} = \tri$.
|
||||
\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
|
||||
Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $\eset{1} = \tri$. As discussed in proof of \cref{lem:3m-G2} for the case of $\tri$, the edges of $\eset{3}$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $s \in f_3^{-1}(S)$, $s$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i \mod{3} + 1} \neq 0$. Iterating through all possible combinations, we have
|
||||
As discussed in proof of \cref{lem:3m-G2} for the case of $\tri$, the edges of $\eset{3}$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $T \in f_3^{-1}(\eset{1})$, $T$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i \mod{3} + 1} \neq 0$. Iterating through all possible choices for $e_1$, we have
|
||||
\begin{itemize}
|
||||
\item \textsc{$(e_1, 0)$}
|
||||
\item For \textsc{$(e_1, 0)$}, there are five possibilities:
|
||||
\begin{itemize}
|
||||
\item $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}$
|
||||
\item $\pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}$
|
||||
|
|
@ -241,13 +302,13 @@ Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $\eset{1} =
|
|||
\item $\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}$
|
||||
\item $\pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}$
|
||||
\end{itemize}
|
||||
\item \textsc{$(e_1, 1)$}
|
||||
\item For \textsc{$(e_1, 1)$}, there are eight possibilities:
|
||||
\begin{itemize}
|
||||
\item $\pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}, \ldots\pbrace{(e_1, 1), (e_2, 1), (e_3, 2)}$
|
||||
\item $\pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}$
|
||||
\item $\pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}$
|
||||
\end{itemize}
|
||||
\item \textsc{$(e_1, 2)$}
|
||||
\item For \textsc{$(e_1, 2)$}, there are five possibilities:
|
||||
\begin{itemize}
|
||||
\item $\pbrace{(e_1, 2), (e_2, 1), (e_3, 0)}$
|
||||
\item $\pbrace{(e_1, 2), (e_2, 1), (e_3, 1)}$
|
||||
|
|
@ -256,12 +317,12 @@ Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $\eset{1} =
|
|||
\item $\pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$
|
||||
\end{itemize}
|
||||
\end{itemize}
|
||||
for a total of 18 3-matchings in $f_3^{-1}(\eset{1})$.
|
||||
for a total of 18 many 3-matchings in $f_3^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-path ($\threepath$)
|
||||
\end{itemize}
|
||||
Consider when $\eset{1} = \threepath$ and all edges in $\eset{3}$ are successively connected to form a $9$-path. Since $(e_1, 0)$ is disjoint to $(e_3, 2)$, both of these edges can exist in a $3$-matching. This relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $\eset{1} = \tri$, namely
|
||||
When $\eset{1} \equiv \threepath$ and all edges in $\eset{3}$ are successively connected to form a $9$-path. Since $(e_1, 0)$ is disjoint to $(e_3, 2)$, both of these edges can exist in a $3$-matching. This relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $\eset{1} = \tri$, namely
|
||||
\begin{equation*}
|
||||
\pbrace{(e_1, 0), (e_2, 0), (e_3, 2)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 2)}.
|
||||
\end{equation*}
|
||||
|
|
@ -274,47 +335,42 @@ Assume $\eset{1} = \twopathdis$, then the edges of $\eset{3}$ have successive co
|
|||
\begin{equation*}
|
||||
\pbrace{(e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_2, 1), (e_3, 2)}, \pbrace{(e_2, 2), (e_3, 1)}, \pbrace{(e_2, 2), (e_3, 2)}.
|
||||
\end{equation*}
|
||||
These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ 3-matchings in $f_3^{-1}(\eset{1})$.
|
||||
These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ many 3-matchings in $f_3^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-star ($\oneint$)
|
||||
\end{itemize}
|
||||
Given $\eset{1} = \oneint$, the edges of $\eset{3}$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As previously mentioned in the proof of \cref{lem:3m-G2}, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_k$, where $i \neq j \neq k$. These choices are independent and we have $4 \cdot 3 = 12$ 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
When $\eset{1} \equiv \oneint$, the edges of $\eset{3}$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As previously mentioned in the proof of \cref{lem:3m-G2}, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_m$, where $i \neq j \neq m$. These choices are independent and we have $4 \cdot 3 = 12$ many 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ many $3$-matchings in $f_3^{-1}(\eset{1})$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-matching ($\threedis$)
|
||||
\end{itemize}
|
||||
Given $\eset{1} = \threedis$ subgraph, we have the case that all edges in $\eset{3}$ have the property that $(e_i, b)$ is disjoint to $(e_j, b)$ for $i \neq j$. For each $e_i$, there are then $3$ choices, independent of each other, and it results that there are $3^3 = 27$ 3-matchings in $f_3^{-1}(\eset{1})$.
|
||||
When $\eset{1} \equiv \threedis$ subgraph, we 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 $f_3^{-1}(\eset{1})$.
|
||||
|
||||
All of the observations above focused only on the shape of $S$, and since we see that for fixed $S$, we have a fixed number of $3$-matchings, this implies the identity.
|
||||
\end{proof}
|
||||
\qed
|
||||
All of the observations above focused only on the shape of $\eset{1}$, and since we see that for fixed $\eset{1}$, we have a fixed number of $3$-matchings, this implies the identity.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
||||
\subsubsection{Three Paths in $\graph{2}$}
|
||||
Computing the number of 3-paths in $\graph{2}$ and $\graph{3}$ consists of much simpler linear combinations.
|
||||
\subsubsection{Proof of~\cref{lem:3p-G2}}
|
||||
|
||||
For $\mathcal{P} \in f_2^{-1}\inparen{ \eset{2}}$ such that $\mathcal{P} $ is a $3$-path, it \textit{must} be the case by definition of $f_2$ that (i)eall edges in $f_2(\mathcal{P} )$ have at least one mapping from an edge in $\mathcal{P} $ and recall that (ii) $\mathcal{P} $ is connected. These constraint rules out every pattern $\eset{1}$ consisting of $3$ edges (it can be verified that in each three-edge pattern at least one of (i) or (ii) is violated), as well as when $\eset{1} = \twodis$. For $\eset{1} = \ed$, note that $\eset{1}$ doesn't have enough edges to have any output in $f_2^{-1}(\eset{1})$, i.e., there exists no $\eset{1} \in \binom{E_2}{3}$ such that $f_2(\mathcal{P} ) = \eset{1}$. The only surviving pattern is $\eset{1} \equiv \twopath$, where the edges of $\eset{2}$ have successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then two $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(\eset{1}), \pbrace{(e_1, 0), (e_1, 1), (e_2, 0)} \text{ and }\pbrace{(e_1, 1), (e_2, 0), (e_2, 1)}$.
|
||||
|
||||
All of the observations above focused only on the shape of $\eset{1}$, and since we see that for fixed $\eset{1}$, we have a fixed number of $3$-paths, this implies the identity.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
||||
|
||||
\subsubsection{Proof of~\cref{lem:3p-G3}}
|
||||
The argument follows along the same lines as in the proof of \cref{lem:3p-G2}. Given $\mathcal{P} \in f_3^{-1}\inparen{\eset{1}}$, it \textit{must} be that every edge in $f_3(\mathcal{P})$ has at least one edge in $\mathcal{P}$ mapped to it (and $\mathcal{P}$ is connected). Notice again that this cannot be the case for any $\eset{1} \in \binom{E_1}{3}$, nor is it the case when $\eset{1} = \twodis$. This leaves us with two patterns, $\eset{1} = \twopath$ and $\eset{1} = \ed$. For the former, it is the case that we have two $3$-paths across $e_1$ and $e_2$, $\pbrace{(e_1, 1), (e_1, 2), (e_2, 0)}$ and $\pbrace{(e_1, 2), (e_2, 0), (e_2, 1)}$. For the latter pattern $\ed$, it it trivial to see that an edge in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.
|
||||
|
||||
All of the observations above focused only on the shape of $\eset{1}$, and since we see that for fixed $\eset{1}$, we have a fixed number of $3$-paths, this implies the identity.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
||||
|
||||
|
||||
\begin{proof}[Proof of Lemma \ref{lem:3p-G2}]
|
||||
For $\mathcal{P} \subseteq \eset{2}$ such that $\mathcal{P} $ is a $3$-path, it \textit{must} be the case by definition of $f$ that all edges in $f_2(\mathcal{P} )$ have at least one mapping from an edge in $\mathcal{P} $ (and recall that $\mathcal{P} $ is connected). This constraint rules out every pattern $\eset{1}$ consisting of $3$ edges, as well as when $\eset{1} = \twodis$. For $\eset{1} = \ed$, note that $\eset{1}$ doesn't have enough edges to have any output in $f_2^{-1}(\eset{1})$, i.e., there exists no $s \in \binom{E_2}{3}$ such that $f_2(\mathcal{P} ) = \eset{1}$. The only surviving pattern is $\eset{1} = \twopath$, where the edges of $\eset{2}$ have successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then $2$ $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(\eset{1}), \pbrace{(e_1, 0), (e_1, 1), (e_2, 0)} \text{ and }\pbrace{(e_1, 1), (e_2, 0), (e_2, 1)}$.
|
||||
\end{proof}
|
||||
\qed
|
||||
\subsubsection{Proof of~\cref{lem:tri}}
|
||||
|
||||
|
||||
\subsubsection{Three Paths in $\graph{3}$}
|
||||
|
||||
|
||||
\begin{proof}[Proof of Lemma \ref{lem:3p-G3}]
|
||||
The argument follows along the same lines as in the proof of \cref{lem:3p-G2}. Given $\mathcal{P} \subseteq \eset{3}$, it \textit{must} be that every edge in $f_3(\mathcal{P})$ has at least one edge in $\mathcal{P}$ mapped to it (and $\mathcal{P}$ is connected). Notice again that this cannot be the case for any $\eset{1} \in \binom{E_1}{3}$, nor is it the case when $\eset{1} = \twodis$. This leaves us with two patterns, $\eset{1} = \twopath$ and $\eset{1} = \ed$. For the former, it is the case that we have $2$ $3$-paths across $e_1$ and $e_2$, $\pbrace{(e_1, 1), (e_1, 2), (e_2, 0)}$ and $\pbrace{(e_1, 2), (e_2, 0), (e_2, 1)}$. For the latter pattern $\ed$, it it trivial to see that an edge in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.
|
||||
\end{proof}
|
||||
\qed
|
||||
|
||||
|
||||
|
||||
\subsubsection{Triangles}
|
||||
|
||||
|
||||
\begin{proof}[Proof of Lemma \ref{lem:tri}]
|
||||
The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the simple fact that all cycles in $\graph{k}$ will have at least six edges.
|
||||
\end{proof}
|
||||
\qed
|
||||
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.
|
||||
%\end{proof}
|
||||
%\qed
|
||||
|
|
|
|||
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Reference in a new issue