More poly-formulation.

macros.tex

@@ -9,6 +9,16 @@
 \newcommand{\poly}{Q}
 \newcommand{\rpoly}{\widetilde{Q}}%r for reduced as in reduced 'Q'
 \newcommand{\out}{output}%output aggregation over the output vector
+\newcommand{\numocc}[1]{\#\left(G, #1\right)}
+	%Graph Symbols
+	\newcommand{\ed}{|}
+	\newcommand{\twodis}{\|}
+	\newcommand{\twopath}{\land}
+	\newcommand{\threedis}{| | |}
+	\newcommand{\tri}{\triangle}
+	\newcommand{\twopathdis}{| \land}
+	\newcommand{\threepath}{\sqcap}
+	\newcommand{\oneint}{\Upilson}
 %David's Scheme
 \newcommand{\vecform}[1]{\textbf{#1}}%auxiliary cmd to use only with macros, so that we can change the format easily if need be
 \newcommand{\sone}{\vecform{x}}

poly-form.tex

@@ -26,7 +26,7 @@ First, note the following fact:
 For all $b \in \{0, 1\}$ and all $e \geq 1$, $b^e = 1$.\qed
 \end{proof}
 
-Assuming each tuple has a probability $\prob = 0.5$, we note that
+Assuming each tuple has a probability $\prob = \frac{1}{2}$, we note that
 \begin{Property}\label{prop:l1-rpoly-numTup}
 The L-1 norm of Q is equal to $\rpoly$ times the number of possible worlds, $|\wSet| = 2^\numTup$.
 \begin{equation*}
@@ -38,12 +38,12 @@ The L-1 norm of Q is equal to $\rpoly$ times the number of possible worlds, $|\w
 Using the fact above, we need to compute \[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup)\].  We therefore argue that 
 \[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).\]
 
-Note that for any single monomial, this is indeed the case since 1) each tuple appears in a possible world with probability of $\frac{1}{2}$, 2) the value $2^\numTup$ is the number of worlds, and 3) the product of expectation and total number of worlds yields and exact result.
+Note that for any single monomial, this is indeed the case since the variables in a single monomial are independent and their joint probability equals the product of the probabilities of each variable in the monomial, i.e., for monomial $M$, $\prob[M] = \prod_{x_i \in M}\prob[x_i].$  This is equivalent to the sum of all probabilities of worlds where each variable in $M$ is a $1$.  Since $1$ is the identity element, it is also the case  that $\prod_{x_i \in M}\prob[x_i] = \ex{M}$. (Note all other terms in the expectation will not contribute since $M$ will equal $0$, and a product containing a factor of $0$ always equals $0$.)  It follows then that $\ex{M} = \rpoly(\wElem_1,\ldots, \wElem_\numTup)$.  Next, observe that the value $2^\numTup$ is the number of worlds, and finally, that the product of expectation and total number of worlds yields the exact sum.
 
-The final result follows by noting that $\rpoly$ is a sum of monomials, and we can, by rearranging the sum, equivlently push the sum into the monomials.\qed
+The final result follows by the fact that $\rpoly$ is a sum of monomials, and we can, by linearity of expectation, equivlently push the expectation through the sum and into the monomials.\qed
 \end{proof}
 
-\begin{Property}
+\begin{Property}\label{prop:exp-rpoly}
 For the case of general $\prob$, where each tuple in the TIDB is present with probability $\prob$, the expectation of polynomial $\poly$ is equal to $\rpoly(\prob,\ldots, \prob).$
 \end{Property}
 
@@ -64,7 +64,28 @@ For any $M_i$,
 &\implies \ex{M_1} +\cdots+\ex{M_t} = \prod_{i_1 \in \{j | x_j \in M_1\}}\prob_{i_1} +\cdots+\prod_{i_v \in \{j | x_j \in M_v\}} \prob_{i_v}\\
 &=\rpoly(\prob_1,\ldots, \prob_\numTup).\qed
 \end{align*}
-I just realized, that I could have saved a lot of time by noting that for the case of TIDB, all monomial variables in $M_i$ are independent, and then using linearity of expectation to conclude the proof.  
+\AH{I just realized, that I could have saved a lot of time by noting that for the case of TIDB, all monomial variables in $M_i$ are independent, and then using linearity of expectation to conclude the proof.} 
 \end{proof}
 
+\begin{Corollary}
+If $\poly$ is given to us in a sum of monomials form, the expectation of $\poly$ ($\ex{\poly}$) can be computed in $O(|\poly|)$, where $|\poly|$ denotes the total number of multiplication/addition operators.
+\end{Corollary}
+
+The corollary follows by \cref{prop:l1-rpoly-numTup} and \cref{prop:exp-rpoly}, and by the fact that the total number of operations in sum of monomials form is exactly the number of addition/multiplication operations.
+
+\subsection{When $\poly$ is not in sum of monomials form}
+We would like to argue that in the general case there is no computation of expectation in linear time.
+
+To this end, consider the follow graph $G(V, E)$, where $|E| = m$, $|V| = \numTup$, and $i, j \in [\numTup]$.  Consider the query $q_E(\wElem_1,\ldots, \wElem_\numTup) = \sum\limits_{(i, j) \in E} \wElem_i \cdot \wElem_j$.
+
+\begin{Lemma}
+If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for fixed $\prob$, then we can count the number of triangles in $G$ in T(m) + O(m) time.
+\end{Lemma}
+\begin{Lemma}
+If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for O(1) distinct values of $\prob$ then we can count the number of triangles (and the number of 3-paths, the number of 3-mathcings) in $G$ in O(T(m) + m) time.
+\end{Lemma}
+
+\begin{proof}
+First, let us do a warm-up by computing $\rpoly(\wElem_1,\dots, \wElem_\numTup)$ when $\poly = q_E(\wElem_1,\ldots, \wElem_\numTup)$.  Before doing so, we introduce a notation.  Let $\numocc{H}$ denote the number of occurrences that $H$ occurs in $G$.  So, e.g., $\numocc{\ed}$ is the number of edges ($m$) in $G$.
+\end{proof}