Removed more from S3 to appendix.

app_hardness-results.tex

@@ -37,10 +37,10 @@ For edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other e
 To  compute $\numocc{G}{\threepath}$, note that for an arbitrary edge $(i, j)$, a 3-path exists for edge pair $(i, \ell)$ and $(j, k)$ where $i, j, k, \ell$ are distinct.  Further, the quantity $(d_i - 1) \cdot (d_j - 1)$ represents the number of 3-edge subgraphs with middle edge $(i, j)$ and outer edges $(i, \ell), (j, k)$ such that $\ell \neq j$ and $k \neq i$.  When $k = \ell$, the resulting subgraph is a triangle, and when $k \neq \ell$, the subgraph is a 3-path.  Summing over all edges (i, j) gives \cref{eq:3p-3tri} by observing that each triangle is counted thrice, while each 3-path is counted just once.  For reasons similar to \cref{eq:2m}, all $d_i$ can be computed in $O(m)$ time and each summand can then be computed in $O(1)$ time, yielding an overall $O(m)$ run time.
 \qed
 \end{proof}
-\subsection{Proof for \Cref{lem:qE3-exp}}
-\input{app_hardness-exact-coefficients}
+%Originally the proofs in hardenss section
+\input{app_single-p-proof-defs}
 
-\subsection{Proofs for~\Cref{lem:3m-G2}-\Cref{lem:lin-sys}}\label{subsec:proofs-struc-lemmas}
+\subsection{Proofs for~\Cref{lem:3m-G2}, \Cref{lem:tri}, and \Cref{lem:lin-sys}}\label{subsec:proofs-struc-lemmas}
 Before proceeding, let us introduce a few more helpful definitions.
 
 \begin{Definition}\label{def:ed-nota}

app_single-p-proof-defs.tex

@@ -0,0 +1,73 @@
+%root: main.tex
+\subsection{Proof of \Cref{th:single-p-hard}}
+\begin{proof}
+For the sake of contradiction, assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
+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 $\numocc{G}{\tri}$ 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}.
+\qed
+\end{proof}
+
+\subsection{Tools to prove \Cref{lem:lin-sys}}
+
+Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{Lemma}\label{lem:qE3-exp}
+%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,
+For any $\prob$, we have:
+	{\small 
+	\begin{align}
+		\rpoly_{G}^3(\prob,\ldots, \prob) &= \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3\nonumber\\
+		&+ 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}
+	\end{align}}
+\end{Lemma}
+\subsubsection{Proof for \Cref{lem:qE3-exp}}
+\begin{proof}%[Proof of \Cref{lem:qE3-exp}]
+By definition we have that
+		\[\poly_{G}^3(\vct{X}) = \sum_{\substack{(i_1, j_1), (i_2, j_2), (i_3, j_3) \in E}}~\; \prod_{\ell = 1}^{3}X_{i_\ell}X_{j_\ell}.\]
+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 $\prob^\nu$ in $\rpoly_{G}^3(\vct{X})$, where $\nu$ is the number of distinct variables in the monomial.
+		%Rather than list all the expressions in full detail, let us make some observations regarding the sum.  
+Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$.  
+We compute $\rpoly_{G}^3(\vct{X})$ by considering each of the three forms that the triple $(e_1, e_2, e_3)$ can take.
+
+\textsc{case 1:} $e_1 = e_2 = e_3$ (all edges are the same).  There are exactly $\numedge=\numocc{G}{\ed}$ such triples, each with a $\prob^2$ factor in $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
+
+\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.  
+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 $(e_2, e_1, e_2)$.  
+Since $e\ne e'$, this case produces the following edge patterns: $\twopath, \twodis$, which contribute $6\prob^3$ and $6\prob^4$ respectively to  $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
+
+\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 the same monomial in the \textsc{SMB} representation of $\poly_{G}^3(\vct{X})$.  This case consists of the following edge patterns: $\tri, \oneint, \threepath, \twopathdis, \threedis$, which contribute $6\prob^3, 6\prob^4, 6\prob^4, 6\prob^5$ and $6\prob^6$ respectively to  $\rpoly_{G}^3\left(\prob,\ldots, \prob\right)$.
+\qed
+\end{proof}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+Since $\prob$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to equations (\ref{eq:1e})-(\ref{eq:3p-3tri})). However, we need to generate one more independent linear equation in these two variables. Towards this end we generate another graph related to $G$:
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{Definition}\label{def:Gk}
+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 inner vertexes of an $\ell$-path replacement edge are disjoint from the inner vertexes of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
+\end{Definition}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+Next, we relate the various sub-graph counts in $\graph{2}$ to $\graph{1}$ ($G$).
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{Lemma}\label{lem:3m-G2}
+The $3$-matchings in graph $\graph{2}$ satisfy the identity:
+\begin{align*}
+\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}.
+\end{align*}
+\end{Lemma}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{Lemma}\label{lem:tri}
+For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$.
+\end{Lemma}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Proof of \Cref{th:single-p}}
+\begin{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}.
+\qed
+\end{proof}
+In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.  
+

mult_distinct_p.tex

@@ -11,7 +11,7 @@ In this section, we will prove that computing $\expct\limits_{\vct{W} \sim \pd}\
 \subsection{Preliminaries}
 
 Our hardness results are based on (exactly) counting the number of occurrences of a subgraph $H$ in $G$. Let $\numocc{G}{H}$ denote the number of occurrences of $H$ in graph $G$.  We can think of $H$ as being of constant size and $G$ as growing.  In query processing, $H$ can be viewed as the query while $G$ as the database instance.
-In particular, we will consider the problems of computing the following counts (given $G$ as an input and its adjacency list representation): $\numocc{G}{\tri}$ (the number of triangles), $\numocc{G}{\threepath}$ (the number of $3$-paths),  $\numocc{G}{\threedis}$ (the number of $3$-matchings) and its generalization $\numocc{G}{\kmatch}$ (the number of $k$-matchings).  Our hardness result in \Cref{sec:multiple-p} is based on the following result:
+In particular, we will consider the problems of computing the following counts (given $G$ as an input and its adjacency list representation): $\numocc{G}{\tri}$ (the number of triangles), $\numocc{G}{\threedis}$ (the number of $3$-matchings), and the latter's generalization $\numocc{G}{\kmatch}$ (the number of $k$-matchings).  Our hardness result in \Cref{sec:multiple-p} is based on the following result:
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{Theorem}[\cite{k-match}]
@@ -20,7 +20,6 @@ Given a positive integer $k$ and  an undirected graph $G$ with no self-loops or
 \sharpwonehard (parameterization is in $k$).
 \end{Theorem}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
 The above result means that we cannot hope to count the number of $k$-matchings in $G=(V,E)$ in time $f(k)\cdot |V|^{c}$ for any function $f$ and constant $c$ independent of $k$. In fact, all known algorithms to solve this problem take time $|V|^{\Omega(k)}$.
 %
 Our hardness result in Section~\ref{sec:single-p} is based on the following conjectured hardness result:

single_p.tex

@@ -29,64 +29,13 @@ in $O\inparen{T(\numedge) + \numedge}$ time.
 \end{Theorem}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{proof}[Proof of \Cref{th:single-p-hard}]
-For the sake of contradiction, assume that for any $G$, we can compute $\rpoly_{G}^3(\prob,\dots,\prob)$ in $o\inparen{m^{1+\eps_0}}$ time.
-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 $\numocc{G}{\tri}$ 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}.
-\end{proof}
-\qed
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
-In other words, if \Cref{th:single-p} holds, then so must \Cref{th:single-p-hard}.  
+
 %Before we move on to the proof itself, we state the results, lemmas, and defintions that will be useful in the proof.
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 
-
-\subsubsection{The proofs}
-
-Note that $\rpoly_{G}^3(\prob,\ldots, \prob)$ as a polynomial in $\prob$ has degree at most six. Next, we figure out the exact coefficients since this would be useful in our arguments:
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{Lemma}\label{lem:qE3-exp}
-%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,
-For any $\prob$, we have:
-	{\small 
-	\begin{align}
-		\rpoly_{G}^3(\prob,\ldots, \prob) &= \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3\nonumber\\
-		&+ 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}
-	\end{align}}
-\end{Lemma}
-
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-Since $\prob$ is fixed, \Cref{lem:qE3-exp} gives us one linear equation in $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ (we can handle the other counts due to \Cref{eq:1e}-\Cref{eq:3p-3tri}). However, we need to generate one more independent linear equation in these two variables. Towards this end we generate another graph related to $G$:
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{Definition}\label{def:Gk}
-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 inner vertexes of an $\ell$-path replacement edge are disjoint from the inner vertexes of any other $\ell$-path replacement edge. % in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
-\end{Definition}
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-Next, we relate the various sub-graph counts in $\graph{2}$ to $\graph{1}$ ($G$).
-
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{Lemma}\label{lem:3m-G2}
-The $3$-matchings in graph $\graph{2}$ satisfy the identity:
-\begin{align*}
-\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}.
-\end{align*}
-\end{Lemma}
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{Lemma}\label{lem:tri}
-For $\ell > 1$ and any graph $\graph{\ell}$, $\numocc{\graph{\ell}}{\tri} = 0$.
-\end{Lemma}
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-
-
-Using the results we have obtained so far, we will prove the following reduction result:
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{Lemma}\label{lem:lin-sys}
 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
@@ -104,15 +53,9 @@ Fix $\prob\in (0,1)$. Given $\rpoly_{\graph{\ell}}^3(\prob,\dots,\prob)$ for $\e
 allowing us to compute $\numocc{G}{\tri}$ and $\numocc{G}{\threedis}$ in $O(1)$ time.
 \end{Lemma}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-Due to lack of space we defer the proof of the above results to \Cref{subsec:proofs-struc-lemmas}.
 %
 This result immediately implies \Cref{th:single-p}:
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\begin{proof}[Proof of \Cref{th:single-p}]
-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
 
 
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