More changes per Oliver 091420 pass.

poly-form.tex

@@ -123,12 +123,13 @@ Let $\numocc{G}{H}$ denote the number of occurrences of pattern $H$ in graph $G$
 For any graph $G$, the following formulas compute $\numocc{G}{H}$ for their respective patterns in $O(\numedge)$ time, with $d_i$ representing the degree of vertex $i$.
 \begin{align}
 	&\numocc{G}{\ed} = \numedge, \label{eq:1e}\\
-	&\numocc{G}{\twopath} = \sum_{i \in V} \binom{d_i}{2} \label{eq:2p}\\						&\numocc{G}{\twodis} = \sum_{(i, j) \in E}\binom{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\
+	&\numocc{G}{\twopath} = \sum_{i \in V} \binom{d_i}{2} \label{eq:2p}\\						
+	&\numocc{G}{\twodis} = \sum_{(i, j) \in E} \frac{\numedge - d_i - d_j + 1}{2}\\%\binom{\numedge - d_i - d_j + 1}{2}\label{eq:2m}\\
 	&\numocc{G}{\oneint} = \sum_{i \in V} \binom{d_i}{3}\label{eq:3s}\\
-	&\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = \sum_{(i, j) \in E} \binom{\numedge - d_i - d_j + 1}{3}\label{eq:2pd-3d}
+	&\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = \sum_{(i, j) \in E} \binom{\numedge - d_i - d_j + 1}{2}\label{eq:2pd-3d}
 \end{align}
-\AH{The formula doesn't seem to work.  This is where I left off. 091620 pm}
-A quick argument to why \cref{eq:2m} is true.  Note that for each edge connecting arbitrary vertices $i$ and $j$, we can get rid of all neighbors, and choose one distinct edge.  Thus, subtracting $(d_i - 1)$ from $m$ yields the number of edges disjoint to vertex $i$ and likewise for $j$.  Choand $d_j$ from $m$  and adding $2$ gives the number of edges disjoint to edge $(i, j)$.  We need to add two since $d_i + d_j$ double counts the shared edge, and since we keep that edge we need to give back both the edge and its double count., since this edge itself is one of the two in the two matching.    The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + \numocc{G}{\threedis}$.
+\AH{The formula ~\cref{eq:2pd-3d} doesn't seem to work.  This is where I left off. 091620 pm}
+A quick argument to why \cref{eq:2m} is true.  Note that for edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other edges in $G$ disjoin to $(i, j)$ is equivalent to finding all edges that are not connected to either vertex $i$ or $j$.  The number of such edges is $m - d_i - d_j + 1$, where we add $1$ since edge $(i, j)$ is removed twice when subtracting both $d_i$ and $d_j$.  Since the summation is iterating over all edges, division by $2$ eliminates the double counting.  Equation ~\ref{eq:2pd-3d} is true for similar reasons.  For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected.  As in ~\cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not.    The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + \numocc{G}{\threedis}$.
 
 
 Now consider the query $q_E(X_1,\ldots, X_\numvar) = \sum\limits_{(i, j) \in E} X_i \cdot X_j$. For the following discussion, set $\poly_{G}(\vct{X}) = \left(q_E(X_1,\ldots, X_\numvar)\right)^3$.