Lemma 1 and Lemma 2 of poly write up

poly-form.tex

@@ -100,10 +100,9 @@ To this end, consider the following graph $G(V, E)$, where $|E| = m$, $|V| = \nu
 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}\label{lem:const-p}
-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.
+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-matchings) 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 $G$.
 
 \begin{Claim}
@@ -163,7 +162,29 @@ $\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of thre
 \[\numocc{\twopathdis} + \numocc{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v - 1}{2}\]  The implication in \cref{claim:four-two} follows by the above and \cref{claim:four-one}.\qed
 	\end{proof}
 \end{Claim}
+
+\begin{proof}
+\underline{Lemma 2}
+
 \cref{claim:four-two} of Claim 4 implies that if we know $\rpoly_3(\prob,\ldots, \prob)$, then we can know in O(m) additional time
-\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\]  We can think of each term in the above equation as a variable, where one can solve a linear system given 3 distinct $\prob$ values, assuming independence of the three linear equation.  In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so.
+\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\]  We can think of each term in the above equation as a variable, where one can solve a linear system given 3 distinct $\prob$ values, assuming independence of the three linear equation.  In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so, and I need to ask him for understanding why this is the case, of which I suspect that it has to do with basic result(s) in linear algebra.\qed
 \end{proof}
 
+\begin{proof}
+\underline{Lemma 1}
+
+The argument for lemma 2 cannot be applied to lemma 1 since we have that $\prob$ is fixed.  We have hope in the following:  we assume that we can solve this problem for all graphs, and the hope would be be to solve the problem for say $G_1, G_2, G_3$, where $G_1$ is arbitrary, and relate the values of $\numocc{H}$, where $H$ is a placeholder for the relevant edge combination.  The hope is that these relations would result in three independent linear equations, and then we would be done.
+
+The following is an option.
+\begin{enumerate}
+	\item Let $G_1$ be an arbitrary graph
+	\item Build $G_2$ from $G_1$, where each edge in $G_1$ gets replaced by a 2 path.
+\end{enumerate}
+
+Then $\numocc{\tri}_2 = 0$, and if we can prove that
+\begin{itemize}
+	\item $\numocc{\threepath}_2 = 2 \cdot \numocc{\twopath}_1$
+	\item $\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1$
+\end{itemize}
+we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{\threedis}$, a hard problem.
+\end{proof}