of points of $\mathcal{ELP}(G)$ that have rank at most $r$. Suppose now that the first $f(r)$ rows of $\mathcal{N}(G)$ are filled for some $r \geq 2$. Let $\Lambda_i$ be a graph of rank $r+1$, and let $\Lambda_j$ be a graph of rank at most $r$. Then $N_{ij}$ is computed by applying Lemma \[lem-kelly1\] with $q = r$. $$\sum_{\Lambda_k| \rho(\Lambda_k) = r}
\displaystyle\binom{\Lambda_i}{\Lambda_k} 
\displaystyle\binom{\Lambda_k}{\Lambda_j}
= \displaystyle\binom{v(\Lambda_i)-v(\Lambda_j)}{r-v(\Lambda_j)}
\displaystyle\binom{\Lambda_i}{\Lambda_j}
= (r+1-v(\Lambda_j))N_{ij}$$ On the LHS, $\displaystyle\binom{\Lambda_k}{\Lambda_j}$ are known by induction hypothesis. Since $\Lambda_k$ are the graphs covered by $\Lambda_i$, $\displaystyle\binom{\Lambda_i}{\Lambda_k}$ are the edge labels. Therefore, $N_{ij}$ can be computed. This completes the construction of $\mathcal{N}(G)$ from $\mathcal{ELP}(G)$.

To construct $\mathcal{ELP}(G)$ from $\mathcal{N}(G)$, define a partial order $\preceq $ on $\{\Lambda_1, \Lambda_2, \ldots , \Lambda_I\}$ as follows: $\Lambda_j \preceq \Lambda_i$ if $N_{ij} \neq 0$. In this poset, if $\Lambda_i$ covers $\Lambda_j$ then assign an edge label $N_{ij}$ to the edge $\Lambda_j-\Lambda_i$ of the Hasse diagram of the poset. This completes the construction of $\mathcal{ELP}(G)$ from $\mathcal{N}(G)$.

\[lem-num-vertices\] Given $\mathcal{N}(G)$, $v(\Lambda_i)$ and $e(\Lambda_i)$ can be counted for each graph in $\Lambda(G)$.

There is a unique row in $\mathcal{N}(G)$ that has only one nonzero entry (the diagonal entry 1). This row corresponds to $\Lambda_1 \cong K_2$, and 