Changes per Atri's suggestions 071820.
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@ -22,6 +22,7 @@
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\newcommand{\union}{\cup}
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\newcommand{\sch}{sch}
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\newcommand{\rw}{\textbf{W}}%\rw for random world
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\newcommand{\graph}[1]{G^{(#1)}}
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%PDBs
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\newcommand{\ti}{TIDB}
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144
poly-form.tex
144
poly-form.tex
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@ -193,51 +193,52 @@ $\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = $ the number of occurrences o
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\begin{proof}[Proof of \cref{lem:const-p}]
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The argument for \cref{lem:gen-p} cannot be applied to \cref{lem:const-p} since we have that $\prob$ is fixed. We have hope in the following: we assume that we can solve this problem for all graphs, and the hope would be be to solve the problem for say $G_1, G_2, G_3$, where $G_1$ is arbitrary, and relate the values of $\numocc{G}{H}$, where $H$ is a placeholder for the relevant edge combination. The hope is that these relations would result in three independent linear equations, and then we would be done.
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The argument for \cref{lem:gen-p} cannot be applied to \cref{lem:const-p} since we have that $\prob$ is fixed. We have hope in the following: we assume that we can solve this problem for all graphs, and the hope would be be to solve the problem for say $\graph{1}, \graph{2}, \graph{3}$, where $\graph{1}$ is arbitrary, and relate the values of $\numocc{G}{H}$, where $H$ is a placeholder for the relevant edge combination. The hope is that these relations would result in three independent linear equations, and then we would be done.
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The following is an option.
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\begin{enumerate}
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\item Let $G_1$ be an arbitrary graph
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\item Build $G_2$ from $G_1$, where each edge in $G_1$ gets replaced by a 2 path.
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\item Let $\graph{1}$ be an arbitrary graph
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\item Build $\graph{2}$ from $\graph{1}$, where each edge in $\graph{1}$ gets replaced by a 2 path.
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\end{enumerate}
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Then $\numocc{G_2}{\tri} = 0$, and if we can prove that
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Then $\numocc{\graph{2}}{\tri} = 0$, and if we can prove that
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\begin{itemize}
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\item $\numocc{G_2}{\threepath} = 2 \cdot \numocc{G_1}{\twopath}$
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\item $\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}$
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\item $\numocc{\graph{2}}{\threepath} = 2 \cdot \numocc{\graph{1}}{\twopath}$
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\item $\numocc{\graph{2}}{\threedis} = 8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis} + 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}$
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\end{itemize}
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we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{G}{\threedis}$, a hard problem.
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we solve our problem for $q_E^3$ based on $\graph{2}$ and we can compute $\numocc{G}{\threedis}$, a hard problem.
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\end{proof}
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\AH{Proving the above linear combination for 3-matchings in $G_2$ always holds for an arbitrary $G_1$.}
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\AH{Proving the above linear combination for 3-matchings in $\graph{2}$ always holds for an arbitrary $\graph{1}$.}
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Consider graph $G_2$, constructed from an arbitrary graph $G_1$. We wish to show that the number of 3-matchings in $G_2$ will always be the linear combination above, regardless of the construction of $G_1$.
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Consider graph $\graph{2}$, constructed from an arbitrary graph $\graph{1}$. We wish to show that the number of 3-matchings in $\graph{2}$ will always be the linear combination above, regardless of the construction of $\graph{1}$.
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\AR{I did not make a pass on the above since it looks incomplete and does not seem to have changed for a while. Also it would be good to define $G_2$ and $G_3$ outside of the proofs below.}
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\AR{I did not make a pass on the above since it looks incomplete and does not seem to have changed for a while. Also it would be good to define $\graph{2}$ and $\graph{3}$ outside of the proofs below.}
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\AH{{\Huge\bf Changes start here. 07-18-2020.}}
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\subsubsection{$f_k$ and $G_k$}
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\subsubsection{$f_k$ and $\graph{k}$}
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\AR{Sorry for doing this late but I think the notation $G_1,G_2,G_3$ is not that great because it does not immediately imply that $G_2$ and $G_3$ are related and that $G_1$ is arbitrary. I think $G^{(2)}$ and $G^{(3)}$ might be better-- we cannot use $G^2$ and $G^3$ since those represent other well-known modications.}
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\begin{Definition}\label{def:Gk}
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For $k > 1$, let graph $G_k$ be a graph generated from an arbitrary graph $G_1$, by replacing every edge $e$ of $G_1$ 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 $G_1$.
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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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\end{Definition}
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For any graph $G_k$, we denote its edges to be a pair $(e, b)$, such that $b \in \{0,\ldots, k-1\}$ and $e\in E_1$.
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For any graph $\graph{k}$, we denote its edges to be a pair $(e, b)$, such that $b \in \{0,\ldots, k-1\}$ and $e\in E_1$.
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Before defining $f_k$, the following notation will be useful. 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 size $\leq t$. The set of edges in $\graph{k}$ is written as $E_k$.
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\begin{Definition}\label{def:fk}
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Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$.
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Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$. For $S \in \binom{E_3}{3}$, $f_k\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}$.
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\end{Definition}
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\AR{$f_k$ is not defined!}
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The function $f_k$ is a mapping from every $3$-edge shape in $G_k$ to its `projection' in $G_1$. The notation $\binom{S}{t}$ is standard and is used to denote the set of subsets in $S$ with exactly $t$ edges. The set of edges in $G_k$ is written as $E_k$.\AR{Um, this is not {\em that} standard so I recommend that you state this definition before defining $f_k$. Also need to define $\binom{S}{\le t}$.}
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The function $f_k$ is a mapping from every $3$-edge shape in $\graph{k}$ to its `projection' in $\graph{1}$.
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\begin{Lemma}
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$f_k$ is a function.
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\end{Lemma}
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\begin{proof}
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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 $G_1$. 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.
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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.
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\end{proof}
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\qed
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@ -250,67 +251,108 @@ We wish to briefly state the possible subgraphs $S$ containing exactly three edg
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\item Disjoint Two-Path ($\twopathdis$)--this subgraph consists of a two path and a remaining disjoint edge.
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\item 3-matching ($\threedis$)--this subgraph is composed of three disjoint edges.
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\end{itemize}
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\subsection{Three Matchings in $G_2$}
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\subsection{Three Matchings in $\graph{2}$}
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\begin{Lemma}\label{lem:3m-G_2}
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The number of $3$-matchings in graph $G_2$ satisfies the following identity,
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\[\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}.\]
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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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\[\numocc{\graph{2}}{\threedis} = 8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis} + 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}.\]
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\end{Lemma}
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\begin{proof}
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\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.}
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Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible edges in $S \times \{0, 1\}$ which $f_2$ maps to $S$. Then we count the number of $3$-matchings in the $3$-edge subgraphs of $G_2$ 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 set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
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\AR{Generally a good idea is you are handling a bunch of cases to put them in a bulleted list: so please do so for each 3-edge patter below.}
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Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible edges in $S \times \{0, 1\}$ which $f_2$ maps to $S$. Then 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 set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
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Consider the $S = \threedis$ pattern. Note that edges in $f_2^{-1}(S)$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2\}$.. All subsets for $h \neq i \neq j$,\AR{Why do you need the variables $h,i,j$ here?} $b_1, b_2, b_3 \in \{0, 1\}$, $(e_h, b_1), (e_i, b_2), (e_j, b_3)$ will compose a 3-matching. One can see that we have a total of two possible choices for each edge $e$ in $G_1$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(S)$.
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\begin{itemize}
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\item $3$-matching ($\threedis$)
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\end{itemize}
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Consider the $S = \threedis$ pattern. Note that edges in $f_2^{-1}(S)$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2\}$.. 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 $e$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(S)$.
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%\AH{The comment below is an important comment.}
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%\AR{I think your argument seems to implicitly assume that $G_1$ is the subset $S$ and $G_2$ is the corresponding mapping under $f^{-1}$. This is {\bf not} correct. You should present the argument as in the outline above above. I.e. fix an $S\in\binom{E_1}{\le 3}$ in $G_1$ and then consider all possible subgraphs in $G_2$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.}
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%\AR{I think your argument seems to implicitly assume that $\graph{1}$ is the subset $S$ and $\graph{2}$ is the corresponding mapping under $f^{-1}$. This is {\bf not} correct. You should present the argument as in the outline above above. I.e. fix an $S\in\binom{E_1}{\le 3}$ in $\graph{1}$ and then consider all possible subgraphs in $\graph{2}$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.}
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For $S = \twopathdis$ edges $e_1, e_2$ form a $2$-path with $e_3$ 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 $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f^{-1}.$ Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices. Edge $e_1$ cannot produce a $2$-matching,\AR{Do we really need this comment?} and are done with $\twopathdis$.
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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 $S = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_3$ 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 $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f_2^{-1}.$ Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}$.
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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 $S = \oneint$, in $f_2^{-1}$, the inner edges $(e_i, 1)$ 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 $f_2^{-1}(S)$.
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When $S =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This translates to a $6$-path in the edges of $f_2^{-1}$, where all edges from $(e_0, 0),(e_0,1),(e_2,0),(e_2, 1)$ 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: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
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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 $S =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This translates to a $6$-path in the edges of $f_2^{-1}$, where all edges from $(e_0, 0),(e_0,1),(e_2,0),(e_2, 1)$ 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 $f_2^{-1}(S)$.
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\begin{itemize}
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\item Triangle ($\tri$)
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\end{itemize}
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For $S = \tri$, note that it is the case that the edges in $f_2^{-1}$ 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 six path above, one must use caution not to consider the first and last edges as disjoint, since they are connected. This rules out both $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with two remaining edge combinations that produce a 3 matching.
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Let us also consider when $S \in \binom{E_1}{\leq 2}$. When $|S| = 2$, we have one of two possibile constraints on the output of $f_2^{-1}$. First, we could have that $e_1$ and $e_2$ are connected, forming a $2$-path, and thus all edges from $(e_1, 0)$ to $(e_3, 1)$ are connected successively. As alluded to previously, a 3-matching would require at least to alternate edges to ensure disjointedness, which requires $\geq 5$-path. Second, it could be that $e_1$ is disjoint from $e_2$, and thus $(e_i, b)$ is disjoint to $(e_j, b)$ for $i \neq j$ and $b \in \{0, 1\}$. For a $3$-matching to exist at least one of the disjoint $2$-paths must be a $3$-path in order to produce a $2$-matching, and this is not the case. When $|S| = 1$, by construction of $G_2$, it is the case there does not exist an $M \in \binom{E_2}{3}$ such that $f(M) = S$. Therefore only subgraphs $S$ of $3$ edges need to be considered.\AR{There is a much simpler argument. Just note that since we can only pick one among the two pairs $\pbrace{(e_0,0),(e_0,1)}$ and $\pbrace{(e_1,0),(e_1,1)}$, there are not enough edges to create a $3$-matching in $f_2^{-1}(S)$.}
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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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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, $\pbrace{(e_0, 0), (e_0, 1)}$ and $\pbrace{(e_1, 0), (e_1, 1)}$. 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 $\pbrace{(e_0, 0), (e_0, 1)}$, thus no $3$-matching exists in $f_2^{-1}(S)$.
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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.
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\end{proof}
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\qed
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\subsection{Three matchings in $G_3$}
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\subsection{Three matchings in $\graph{3}$}
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\begin{Lemma}\label{lem:3m-G3}
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The number of 3-matchings in $G_3$ satisfied the following identity,
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The number of 3-matchings in $\graph{3}$ satisfy the following identity,
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\begin{align*}
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\numocc{G_3}{\threedis} = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 18\pbrace{\numocc{G_1}{\tri}} + 21\pbrace{\numocc{G_1}{\threepath}}\\
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&+ 24\pbrace{\numocc{G_1}{\twopathdis}} + 20\pbrace{\numocc{G_1}{\oneint}} + 27\pbrace{\numocc{G_1}{\threedis}}.
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\numocc{\graph{3}}{\threedis} = &4\pbrace{\numocc{\graph{1}}{\twopath}} + 6\pbrace{\numocc{\graph{1}}{\twodis}} + 18\pbrace{\numocc{\graph{1}}{\tri}} + 21\pbrace{\numocc{\graph{1}}{\threepath}}\\
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&+ 24\pbrace{\numocc{\graph{1}}{\twopathdis}} + 20\pbrace{\numocc{\graph{1}}{\oneint}} + 27\pbrace{\numocc{\graph{1}}{\threedis}}.
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\end{align*}
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\end{Lemma}
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\begin{proof}
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For any $S \in \binom{E_1}{\leq3}$, we now consider $f_3^{-1}(S)$, which lists all possible subsets of $3$ edges in $S \times \{0, 1, 2\}$ to which $f_3$ maps to $S$. We again then count the number of $3$-matchings in $f_3^{-1}(S)$. Note, that for $G_1$, we represent edges as $e_1, e_2, e_3$,\AR{Technically $S\in\binom{E_1}{\le 2}$ cannot have $e_3$ in it.} and their corresponding $3$-paths in $G_3$ as $(e_1, 0), (e_1, 1), (e_1, 2),\ldots, (e_3, 2)$. \AR{Generally be vary about using $\ldots$ since it make sense only if everything in lex order between the two terms around the $\ldots$ appears. This is true here but not everywhere $\ldots$ are used in this proof. Please go through all uses of $\ldots$ to check this.}
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For any $S \in \binom{E_1}{\leq3}$, we now consider $f_3^{-1}(S)$, which lists all possible subsets of $3$ edges in $S \times \{0, 1, 2\}$ to which $f_3$ maps to $S$. We again then count the number of $3$-matchings in $f_3^{-1}(S)$. Let $S'$ be all the edges of $\graph{3}$, formally, $S' = \pbrace{(e_1, 0),\ldots, (e_3, 2)}$. Note, that for $\graph{1}$, when $|S| = 1$, we represent $S$ as $e_1$, when $|S| = 2$, as $e_1, e_2$, and when $|S| = 3$ we represent edges as $e_1, e_2, e_3$. The representation is symmetrical in $S'$, where, for example, when $|S| = 1$, then $S' = \pbrace{(e_1, 0), (e_1, 1), (e_1, 2)}$. \AR{Generally be vary about using $\ldots$ since it make sense only if everything in lex order between the two terms around the $\ldots$ appears. This is true here but not everywhere $\ldots$ are used in this proof. Please go through all uses of $\ldots$ to check this.}
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\AR{Bullet list of cases.}
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When $S = \ed$, $f_3^{-1}(S)$ gives one result, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(S)$ for this case.% All edges in the subset are a $3$-path, and it is the case as alluded in $G_2$ discussion that no 3-matching can exist in a single $3$-path.
|
||||
|
||||
Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S' = \{(e_1, 0),\ldots(e_1, 2), (e_2, 0),\ldots, (e_2, 2)\}.$\AR{You use the notation $S'$ in other cases as well-- I would recommend that you pull this out and explicitly define it before considering the cases.} All edges in $S'$ form a $6$-path, and as stated above\AR{This is a vague reference-- you actually meant the proof of Lemma 7 and not earlier part of this proof. So explicitly refer to proof of Lemma 7. Check for other occurences of this in the proof.}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
\begin{itemize}
|
||||
\item $1$ edge ($\ed$)
|
||||
\end{itemize}
|
||||
When $S = \ed$, $f_3^{-1}(S)$ gives one result, $(e_1, 0), (e_1, 1), (e_1, 2)$, which clearly does not contain a $3$-matching. Thus there are no $3$-matchings in $f_3^{-1}(S)$ for this case.% All edges in the subset are a $3$-path, and it is the case as alluded in $\graph{2}$ discussion that no 3-matching can exist in a single $3$-path.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-path ($\twopath$)
|
||||
\end{itemize}
|
||||
Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S'$. All edges in $S'$ form a $6$-path, and as stated above in \cref{lem:3m-G2}\AR{This is a vague reference-- you actually meant the proof of Lemma 7 and not earlier part of this proof. So explicitly refer to proof of Lemma 7. Check for other occurences of this in the proof.}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-matching ($\twodis$)
|
||||
\end{itemize}
|
||||
For $S = \twodis$, then all subsets in the output of $f_3^{-1}(S)$ 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 $(e_j, 0),\ldots, (e_j, 2)$ again for $i \neq j$. Since the selections are independent, it follows that there exist $2 \cdot 3 = 6$ $3$-matchings in $f_3^{-1}(S)$.
|
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|
||||
Now, we consider the 3-edge subgraphs of $G_1$, starting with a $S = \tri$. As discussed in the case of $G_2$, $f_3^{-1}(S)$ subsets are conditioned on the fact that all the edges in $S'$ 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 $s \in S'$, $s$ is a $3$-matching when we have that $j \geq i + 2, k \geq j + 2$\AR{Again, why do you need $i,j,k$ here? Why not just use $1,2,3$?} for the edges $(e_i, b_1), (e_j, b_2), (e_k, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$ for all $i \in \{1, 2\}$ it is the case that if $b_i = 2$ then $b_{i + 1} \neq 0$ and if $b_1 = 0$ then $b_3 \neq 2$\AR{You can state this succintly by stating that if $b_i=2$ for $i\in [3]$ then $b_{(i+1)\mod{3}+1}$.} . Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$\AR{I think this is one of the place where the $\ldots$ are not used appropriately. It would be better to list the possibilities separately for each choice of $b_1$ for $e_1$.} gives a total of 18 3-matchings in $f_3^{-1}(S)$.
|
||||
\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
|
||||
Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with a $S = \tri$. As discussed in the case of $\graph{2}$, $f_3^{-1}(S)$ subsets are conditioned on the fact that all the edges in $S'$ 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 $s \subseteq S'$, $s$ 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 $b_{i + 1 \mod{4}} = 0$. Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}$\newline$\ldots, \pbrace{(e_1, 1), (e_2, 1), (e_3, 2)} \pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}, \pbrace{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$, giving a total of 18 3-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-path ($\threepath$)
|
||||
\end{itemize}
|
||||
Consider when $S = \threepath$ and $f_3^{-1}(S)$ has the constraint that all edges 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. The relaxation yields 3 other 3-matchings that couldn't be counted in the case of the $S = \tri$, namely $\pbox{(e_1, 0), (e_2, 0), (e_3, 2)},\pbox{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbox{(e_1, 0), (e_2, 2), (e_3, 2)}$. There are therefore $18 + 3 = 21$ three-matchings.
|
||||
|
||||
Assume $S = \twopathdis$, then $f_3^{-1}$ has successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $S'$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbox{(e_2, 0), (e_3, 0)}, \pbox{(e_2, 0), (e_3, 1)}, \pbox{(e_2, 0), (e_3, 2)},\pbox{(e_2, 1), (e_3, 0)},\ldots, \pbox{(e_2, 1), (e_3, 2)}, \pbox{(e_2, 2), (e_3, 1)}, \pbox{(e_2, 2), (e_3, 2)}$.\AR{Again this use of $\ldots$ might not be appropriate} 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 $f_3^{-1}(S)$.
|
||||
\begin{itemize}
|
||||
\item Disjoint Two-Path ($\twopathdis$)
|
||||
\end{itemize}
|
||||
Assume $S = \twopathdis$, then $f_3^{-1}$ has successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $S'$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbox{(e_2, 0), (e_3, 0)},\ldots, \pbox{(e_2, 1), (e_3, 2)}, \pbox{(e_2, 2), (e_3, 1)}, \pbox{(e_2, 2), (e_3, 2)}$. 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 $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-star ($\oneint$)
|
||||
\end{itemize}
|
||||
Given $S = \oneint$, the subsets of $f_3^{-1}(S)$ 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 discussed previously, 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 $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-matching ($\threedis$)
|
||||
\end{itemize}
|
||||
Given $S = \threedis$ subgraph, we have the case that all subsets in $f_3^{-1}(S)$ 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 $f_3^{-1}(S)$.
|
||||
|
||||
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.
|
||||
|
|
@ -318,28 +360,28 @@ All of the observations above focused only on the shape of $S$, and since we see
|
|||
\qed
|
||||
|
||||
\subsection{Three Paths}
|
||||
Computing the number of 3-paths in $G_2$ and $G_3$ consists of much simpler linear combinations.
|
||||
\subsubsection{$G_2$}
|
||||
Computing the number of 3-paths in $\graph{2}$ and $\graph{3}$ consists of much simpler linear combinations.
|
||||
\subsubsection{$\graph{2}$}
|
||||
\begin{Lemma}
|
||||
The number of $3$-paths in $G_2$ is computed by the following linear combination,
|
||||
\[\numocc{G_2}{\threepath} = 2 \cdot \numocc{G_1}{\twopath}.\]
|
||||
The number of $3$-paths in $\graph{2}$ is computed by the following linear combination,
|
||||
\[\numocc{\graph{2}}{\threepath} = 2 \cdot \numocc{\graph{1}}{\twopath}.\]
|
||||
\end{Lemma}
|
||||
|
||||
\begin{proof}
|
||||
For a $M = \threepath \in G_2$, it \textit{must} be the case that there is successive connectivity for $3$ edges across $f_2(M) = S$\AR{Here you mean to talk about edges in $E_2$ but the states talking about edges in $E_1$}. This constraint rules out every pattern $S \in G_1$ consisting of $3$ edges, as well as when $S = \twodis$ and for $S = \ed$.\AR{Technically the argument for $\ed$ is different from the others in that $f_2^{-1}(\ed)$ has not have enough edges in total.} The only surviving pattern is $S = \twopath$, where it can be seen in $f_2^{-1}(S)$ that each subset has successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then $2$ $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(S)$.\AR{Explicitly state the paths.}
|
||||
For $s \subseteq S'$ such that $s$ is a $3$-path, it \textit{must} be the case that all edges in $f_2(s)$ have at least one mapping from an edge in $s$. \AR{Here you mean to talk about edges in $E_2$ but the states talking about edges in $E_1$}\AH{Not sure if I have understood this comment, but hopefully fixed the problem.} This constraint rules out every pattern $S \in \graph{1}$ consisting of $3$ edges, as well as when $S = \twodis$. For $S = \ed$, note that $S$ doesn't have enough edges to have any output in $f_2^{-1}(S)$, i.e., there exists no $s \in \binom{E_2}{3}$ such that $f_2(s) = S$.\AR{Technically the argument for $\ed$ is different from the others in that $f_2^{-1}(\ed)$ has not have enough edges in total.}\AH{I'm slightly confused by the above comment...intuitively I understand, but doesn't $f_2^{-1}(S) = \emptyset$ technically?} The only surviving pattern is $S = \twopath$, where it can be seen in $f_2^{-1}(S)$ that each subset has successive connectivity from $(e_1, 0)$ to $(e_2, 1)$. There are then $2$ $3$-paths sharing edges $e_1$ and $e_2$ in $f_2^{-1}(S)$.\AH{\LARGE Here where I left off.}\AR{Explicitly state the paths.}
|
||||
\end{proof}
|
||||
\qed
|
||||
%we have two 3-paths generated: $\pbox{(e_1, 0), (e_1, 1), (e_2, 0)}$ and $\pbox{(e_1, 1), (e_2, 0), (e_2, 1)}$. Thus,
|
||||
|
||||
|
||||
\subsubsection{$G_3$}
|
||||
\subsubsection{$\graph{3}$}
|
||||
\begin{Lemma}
|
||||
The number of $3$-paths in $G_3$ is computed by the following linear combination,
|
||||
\[\numocc{G_3}{\threepath} = \numocc{G_1}{\ed} + 2 \cdot \numocc{G_1}{\twopath}.\]
|
||||
The number of $3$-paths in $\graph{3}$ is computed by the following linear combination,
|
||||
\[\numocc{\graph{3}}{\threepath} = \numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}.\]
|
||||
\end{Lemma}
|
||||
|
||||
\begin{proof}
|
||||
The argument follows along the same lines as above. Given $M = \threepath \in G_3$, it \textit{must} be that there is a $3$-path across all edges of $S = f_3(M)$\AR{Same comment as in the $G_2$ case.}. Notice again that this cannot be the case for any $S \in \binom{E_1}{3}$, nor is it the case when $S = \twodis$. This leaves us with two patterns, $S = \twopath$ and $S = \ed$. For the former, it is the case that we have $2$ $3$-paths across $e_1$ and $e_2$. For the latter, it it trivial to see that a $1$-path in $G_1$ becomes a $3$-path in $G_3$, and this proves the identity.
|
||||
The argument follows along the same lines as above. Given $M = \threepath \in \graph{3}$, it \textit{must} be that there is a $3$-path across all edges of $S = f_3(M)$\AR{Same comment as in the $\graph{2}$ case.}. Notice again that this cannot be the case for any $S \in \binom{E_1}{3}$, nor is it the case when $S = \twodis$. This leaves us with two patterns, $S = \twopath$ and $S = \ed$. For the former, it is the case that we have $2$ $3$-paths across $e_1$ and $e_2$. For the latter, it it trivial to see that a $1$-path in $\graph{1}$ becomes a $3$-path in $\graph{3}$, and this proves the identity.
|
||||
\end{proof}
|
||||
\qed
|
||||
|
||||
|
|
@ -347,12 +389,12 @@ The argument follows along the same lines as above. Given $M = \threepath \in G
|
|||
|
||||
\subsection{Triangle}
|
||||
\begin{Lemma}
|
||||
For any graph $G_k$, $\numocc{G_k}{\tri} = 0$.
|
||||
For any graph $\graph{k}$, $\numocc{\graph{k}}{\tri} = 0$.
|
||||
\end{Lemma}
|
||||
|
||||
\begin{proof}
|
||||
The number of triangles in $G_k$ for $k \geq 2$ will always be $0$ for the simple fact that when we replace a single edge with $\geq 2$-path, the possibility of a triangle of single edge sides disappears, since the only way a single edged triangle could exist is if it existed in $G_1$ and then was passed to $G_2$ or $G_3$ without replacing each single edge with $\geq 2$-paths.\AR{I think this confusing. It is easier to just state that all cycles in $G_2$ and $G_3$ have size at least six.}
|
||||
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
|
||||
|
||||
\AH{Changes propagated up to this point.}
|
||||
|
||||
|
|
|
|||
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Reference in a new issue