Small notational changes in lemma 3.10 and preceding definitions of App 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}

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)}}
 
 
 

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 $\numocc{\graph{1}}{S}$.
-%$%^&*(
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