diff --git a/hardness-app.tex b/hardness-app.tex index 02aa8f3..b034b05 100644 --- a/hardness-app.tex +++ b/hardness-app.tex @@ -200,30 +200,31 @@ To compute $\numocc{G}{\threepath}$, note that for an arbitrary edge $(i, j)$, Before proceeding, let us introduce a few more helpful definitions. \begin{Definition}\label{def:ed-nota} -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} -\color{red}I don't know that I feel comfortable with the notation following. What about $E_S^{(\ell)}$ or something similar? And maybe just $S$ for subgraph rather than $SG$.\color{black} -\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 $\graph{1}=S\graph{1}$. +For $\ell > 1$, 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} -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}[$\eset{\ell}$] +Given an arbitrary subgraph $\sg{1}$ of $\graph{1}$, let $\eset{1}$ denote the set of edges in $\sg{1}$. Define then $\eset{\ell}$ for $\ell > 1$ as the set of edges in the generated subgraph $\sg{\ell}$ (i.e. when we apply \Cref{def:Gk} to $\sg{1})$. +\end{Definition} + +For example, consider $\sg{1}$ with edges $\eset{1} = \{e_1\}$. Then the edge set of $\sg{2}$ is defined as $\eset{2} = \{(e_1, 0), (e_1, 1)\}$. \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. +Let $\binom{E}{t}$ denote the set of subsets in $E$ with exactly $t$ edges. In a similar manner, $\binom{E}{\leq t}$ is used to mean the subsets of $E$ with $t$ or fewer edges. \end{Definition} 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 $f_\ell: \binom{E_\ell}{3} \mapsto \binom{E_1}{\leq3}$ 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: +Let $f_\ell: \binom{E_\ell}{3} \mapsto \binom{E_1}{\leq3}$ be defined as follows. For any element $s \in \binom{E_\ell}{3}$ 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} \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 $h \leq \binom{\abs{s^{(1)}} \cdot \ell}{3}$. +For an arbitrary subgraph $\sg{1}$ of $\graph{1}$ with at most $m \leq 3$ edges, the inverse function $f_\ell^{-1}: \binom{E_1}{\leq 3}\mapsto 2^{\binom{E_\ell}{3}}$ takes $\eset{1}$ and outputs the set of all elements $s \in \binom{\eset{\ell}}{3}$ such that %each 3-edge set $T\in f_\ell^{-1}(S)$ is mapped to the input set $\eset{1}$ by $f_\ell$, i.e. +$f_\ell(s) = \eset{1}$. %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 $f_\ell^{-1}$, 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 $T\in f_\ell^{-1}(S)$ consists of the following set of edges: $\{ (e_i, b), (e_j, b), (e_m, b) \}$, where $i,j$ and $m$ are distinct. +Note, importantly, that when we discuss $f_\ell^{-1}$, that each \textit{edge} present in $\eset{1}$ must have an edge in $s\in f_\ell^{-1}(S)$ that `projects` down to it. In particular, if $|\eset{1}| = 3$, then it must be the case that each $s\in f_\ell^{-1}(S)$ consists of the following set of edges: $\{ (e_i, b), (e_j, b), (e_m, b) \}$, where $i,j$ and $m$ are distinct. We first note that $f_\ell$ is well-defined: \begin{Lemma}\label{lem:fk-func} @@ -243,11 +244,11 @@ We are now ready to prove the structural lemmas. Note that $f_\ell$ maps subsets \begin{proof}[Proof of \Cref{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 $s \in \{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1),$ $(e_3, 0), (e_3, 1)\}$ such that $f_\ell(s) = \{e_1, e_2, e_3\}$. -We do a case analysis based on the `shape' of $\eset{1}$: +We do a case analysis based on the subgraph $\sg{1}$ induced by $\eset{1}$ (denoted $\eset{1} \equiv \sg{1}$): \begin{itemize} \item $3$-matching ($\threedis$) \end{itemize} -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 choices for $b_i$ for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(\eset{1})$. +When $\eset{1} \equiv \threedis$, it is the case that edges in $\eset{2}$ are {\em not} disjoint only for the pairs $(e_i, 0), (e_i, 1)$ for $i\in \{1,2,3\}$. 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 choices for $b_i$ for each edge $e_i$ in $\graph{1}$ yielding $2^3 = 8$ possible 3-matchings in $f_2^{-1}(\eset{1})$. \begin{itemize} \item Disjoint Two-Path ($\twopathdis$) @@ -267,7 +268,7 @@ When $\eset{1} \equiv \oneint$, the inner edges $(e_i, 1)$ of $\eset{2}$ are all \begin{itemize} \item $3$-path ($\threepath$) \end{itemize} -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. +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. For a $3$-matching to exist in $f_2^{-1}(\eset{1})$, we cannot pick both $(e_i,0)$ and $(e_i,1)$ or both $(e_i, 1)$ and $(e_j, 0)$ where $i = j + 1$. % there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}$, $\pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}$, $\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$ $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$, a total of four 3-matchings in $f_2^{-1}(\eset{1})$. \begin{itemize} diff --git a/macros.tex b/macros.tex index f684f85..3a22547 100644 --- a/macros.tex +++ b/macros.tex @@ -33,7 +33,7 @@ \newcommand{\attr}[1]{attr\left(#1\right)} \newcommand{\rw}{\textbf{W}}%\rw for random world \newcommand{\graph}[1]{G^{(#1)}} -\newcommand{\eset}[1]{S^{(#1)}} %edge set for arbitrary subgraph +\newcommand{\eset}[1]{E^{(#1)}_S} %edge set for arbitrary subgraph \newcommand{\linsys}[1]{LS(\graph{#1})} \newcommand{\lintime}[1]{LT^{\graph{#1}}} \newcommand{\aug}[1]{AUG^{\graph{#1}}} @@ -253,6 +253,7 @@ } } \newcommand{\bsym}[1]{\boldsymbol{#1}}%b for bold; sym for symbol +\newcommand{\sg}[1]{S^{(#1)}} diff --git a/single_p.tex b/single_p.tex index 45ab202..05dccd4 100644 --- a/single_p.tex +++ b/single_p.tex @@ -107,7 +107,7 @@ For $\ell > 1$, let graph $\graph{\ell}$ be a graph generated from an arbitrary \end{Definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -Next, we relate the various sub-graph counts in $\graph{2}$ and $\graph{3}$ to their counterparts in $\graph{1}=G$. +Next, we relate the various sub-graph counts in $\graph{2}$ to $\graph{1}$ ($G$). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{Lemma}\label{lem:3m-G2} @@ -119,32 +119,6 @@ The $3$-matchings in graph $\graph{2}$ satisfy the identity: \end{Lemma} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -%\begin{Lemma}\label{lem:3m-G3} -%The 3-matchings in $\graph{3}$ satisfy the identity: -%\begin{align*} -%\numocc{\graph{3}}{\threedis} &= 4\numocc{\graph{1}}{\twopath} + 6\numocc{\graph{1}}{\twodis} + 18\numocc{\graph{1}}{\tri}\\ -%&+ 21\numocc{\graph{1}}{\threepath}+ 24\numocc{\graph{1}}{\twopathdis} + 20\numocc{\graph{1}}{\oneint}\\ -%&+ 27\numocc{\graph{1}}{\threedis}. -%\end{align*} -% -%\end{Lemma} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -%\begin{Lemma}\label{lem:3p-G2} -%The $3$-paths in $\graph{2}$ satisfy the identity: -%\[\numocc{\graph{2}}{\threepath} = 2 \cdot \numocc{\graph{1}}{\twopath}.\] -%\end{Lemma} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -%\begin{Lemma}\label{lem:3p-G3} -%The $3$-paths in $\graph{3}$ satisfy the identity: -%\[\numocc{\graph{3}}{\threepath} = \numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}.\] -%\end{Lemma} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{Lemma}\label{lem:tri} For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$. @@ -156,7 +130,6 @@ For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$. Using the results we have obtained so far, we will prove the following reduction result: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{Lemma}\label{lem:lin-sys} -%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 $\numocc{\graph{1}}{\tri}$. Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, we can compute in $O(m)$ time a vector $\vct{b}\in\mathbb{R}^3$ such that \[ \begin{pmatrix} 1 - 3p & -(3\prob^2 - \prob^3)\\ @@ -178,21 +151,9 @@ Due to lack of space we defer the proof of the above results to \Cref{subsec:pro This result immediately implies \Cref{th:single-p}: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{proof}[Proof of \Cref{th:single-p}] -We can compute $\graph{2}$ and $\graph{3}$ from $\graph{1}=G$ in $O(m)$ time (also note that these graphs also have $O(m)$ edges). Thus, -in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [3]$ and \Cref{lem:lin-sys} completes the proof. +We can compute $\graph{2}$ from $\graph{1}$ in $O(m)$ time. Additionally, if in time $O(T(m))$, we have $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\ell\in [2]$, then the theorem follows by \Cref{lem:lin-sys}. \end{proof} \qed -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -% \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.} - -%Using \Cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$. -%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 $\numocc{\graph{1}}{S}$. -%$%^&*( - - - - - %%% Local Variables: