Incorporated all of @atri Riot 120920 suggestions.

lin_sys.tex

@@ -205,26 +205,24 @@ Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
 \begin{align}
 &\dtrm{\mtrix{\rpoly}} =  30(3\prob^2 - \prob^3)^2 - 90\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3\nonumber\\
 &= 30(3\prob^2 - \prob^3)^2\left(1 - 3\prob + (3\prob^2 - \prob^3)\right) \nonumber\\
-&= 30\left(9\prob^4 - 6\prob^5 + \prob^6\right)\left(-\prob^3 + 3\prob^2 - 3\prob + 1\right)\nonumber\\
-&=\left(30\prob^6 - 180\prob^5 + 270\prob^4\right)\cdot\left(-\prob^3 + 3\prob^2 - 3\prob + 1\right).\label{eq:det-final}
+&= 30\prob^4\left(3 - \prob\right)^2\left(-\prob^3 + 3\prob^2 - 3\prob + 1\right)\nonumber\\
+&= 30\prob^4\left(3 - \prob\right)^2\left(1 - \prob\right)^3.\label{eq:det-final}
 \end{align}
 
-\AH{It appears that the equation below has roots at p = 0 (left factor) and p = 1, with NO roots $\in (0, 1)$.}
-
-It can be shown through standard polynomial roots computation techniques \footnote{An online roots solver such as https://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php will suffice}, that $\dtrm{\mtrix{\rpoly}}$ has no roots in $(0, 1)$, ensuring independence for all $\prob$ values in $(0, 1)$, and thus  ~\cref{lem:lin-sys} follows.
+From ~\cref{eq:det-final} it can easily be seen that the roots of $\dtrm{\mtrix{\rpoly}}$ are $0, 1,$ and $3$.  Hence there are no roots in $(0, 1)$ and ~\cref{lem:lin-sys} follows.
 \end{proof}
 
 \qed
 \begin{proof}[Proof of \cref{th:single-p}]
-Thus, by ~\cref{lem:lin-sys} we have proved ~\cref{th:single-p} for fixed $p \in (0, 1)$.
+The proof follows by ~\cref{lem:lin-sys}.
 \end{proof}
 \qed
 
 \begin{Corollary}\label{cor:single-p-gen-k}
 For every value $\kElem \geq 3$, there exists a query with $\kElem$ product width that is hard.
 \end{Corollary}
-\begin{proof}[Proof of Corollary ~\cref{cor:single-p-gen-k}]
-Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$.  By ~\cref{th:single-p}, query $\poly''$ with $\kElem = 4$ is hard.
+\begin{proof}[Proof of Corollary ~\ref{cor:single-p-gen-k}]
+Consider $\poly^3_{G}$ and $\poly' = 1$ such that $\poly'' = \poly^3_{G} \cdot \poly'$.  By ~\cref{th:single-p}, query $\poly''$ with $\kElem = 4$ has $\Omega(\numvar^{\frac{4}{3}})$ complexity.
 \end{proof}
 
 \qed

mult_distinct_p.tex

@@ -5,10 +5,10 @@
 We would like to argue for a compressed version of $\poly(\vct{w})$, in general $\expct_{\vct{w}}\pbox{\poly(\vct{w})}$ cannot be computed in linear time. 
 \AR{Added the hardness result below.}
 The hardness result is based on the following hardness result:
-\begin{theorem}[\cite{k-match}]
+\begin{Theorem}[\cite{k-match}]
 \label{thm:k-match-hard}
-Given a positive integer $k$ and  an undirected graph $G$ with no self-loops of parallel edges, couting the number of $k$-matchings in $G$ is $\#W[1]$-hard.
-\end{theorem}
+Given a positive integer $k$ and  an undirected graph $G$ with no self-loops or parallel edges, counting the number of $k$-matchings in $G$ is $\#W[1]$-hard.
+\end{Theorem}
 The above result means that we cannot hope to count the number of $k$-matchings in $G=(V,E)$ in time $f(k)\cdot |V|^{O(1)}$ for any function $f$. In fact, all known algorithms to solve this problem takes time $|V|^{\Omega(k)}$.
 
 To this end, consider the following graph $G(V, E)$, where $|E| = \numedge$, $|V| = \numvar$, and $i, j \in [\numvar]$.  
@@ -45,8 +45,11 @@ By ~\cref{lem:qEk-multi-p}, the term $c_{2\kElem}$ can be exactly computed.  Add
 \end{proof}
 
 \qed
-
-\begin{Corollary}\label{cor:reduct}
-By ~\cref{lem:qEk-multi-p} and ~\cref{cor:lem-qEk} it follows that computing $\rpoly(\vct{X})$ is hard.
+\begin{Corollary}\label{cor:tilde-q-hard}
+Computing $\rpoly(\vct{X})$ is $\#W[1]$-hard.
 \end{Corollary}
 
+\begin{proof}[Proof of Corollary ~\ref{cor:tilde-q-hard}]
+The proof follows by ~\cref{thm:k-match-hard}, ~\cref{lem:qEk-multi-p} and ~\cref{cor:lem-qEk}.
+\end{proof}
+

poly-form.tex

@@ -112,7 +112,7 @@ If $\poly$ is given as a sum of monomials, the expectation of $\poly$, i.e., $\e
 \end{Corollary}
 
 \begin{proof}[Proof For Corollary ~\ref{cor:expct-sop}]
-Note that \cref{lem:exp-poly-rpoly} shows that $\expct\pbox{\poly} = \rpoly(\prob_1,\ldots, \prob_\numvar)$.  Therefore, if $\poly$ is already in sum of products form, one only needs to compute $\poly(\prob_1,\ldots, \prob_\numvar)$ ignoring exponent terms (note that such a polynomial is $\rpoly(\prob_1,\ldots, \prob_\numvar)$), which is indeed has $O(|\poly|)$ compututations.\qed
+Note that \cref{lem:exp-poly-rpoly} shows that $\expct\pbox{\poly} =$ $\rpoly(\prob_1,\ldots, \prob_\numvar)$.  Therefore, if $\poly$ is already in sum of products form, one only needs to compute $\poly(\prob_1,\ldots, \prob_\numvar)$ ignoring exponent terms (note that such a polynomial is $\rpoly(\prob_1,\ldots, \prob_\numvar)$), which indeed has $O(|\poly|)$ compututations.\qed
 \end{proof}
 
 

single_p.tex

@@ -5,6 +5,7 @@
 
 In this discussion, let us fix $\kElem = 3$.
 
+\AH{@atri needs to put in the result for triangles of $\numvar^{\frac{4}{3}}$ runtime.}
 \begin{Theorem}\label{th:single-p}
 If we can compute $\rpoly_{G}^3(\vct{X})$ in T(\numedge) time for $X_1 =\cdots= X_\numvar = \prob$, then we can count the number of triangles, 3-paths, and 3-matchings in $G$ in $T(\numedge) + O(\numedge)$ time.
 \end{Theorem}