$A$ be the set of all singleton edges. For any $x,y\in A$, $xy$ is not a 2-edge of $G$. We have $$\begin{aligned}
h_n(G)&\leq \frac{|A|}{n} + 1-\frac{{|A|\choose 2}}{{n\choose 2}} \\
      &= 1+ \frac{|A|}{n} -\frac{|A|^2}{n^2} + O\left(\frac{1}{n}\right) \\
      &\leq 1+ \frac{1}{4} + O\left(\frac{1}{n}\right).\end{aligned}$$ In the last step, we use the fact that $f(x)=1+x-x^2$ has the maximum value $\frac{5}{4}$. Combining the upper and lower bounds we have $\pi(K^{\{1,2\}}_2)=\frac{5}{4}$.

The argument is easily generalized to the complete graph $K^{\{1,k\}}_k$ (for $k>1$). We have $$\label{eq:k1k}
 \pi(K^{1,k}_k)=1+ \frac{k-1}{k^{\frac{k}{k-1}}}.$$

Let $H$ be a hypergraph. The [*suspension*]{} of $H$, denoted by $S(H)$, is a new hypergraph with the vertex set $V(H)\cup \{\ast\}$ and the edge set $\{F\cup \{\ast\}\colon F\in E(H)\}$. Here $\ast$ is a new vertex not in $H$.

Let $H$ be a hypergraph. The $k$-degree of a vertex $x$, denoted $d_{k}(x)$, is the number of edges of size $k$ containing $x$.

[**Example 2:**]{} Consider $H:=S(K^{\{1,2\}}_2)$. The edges of $H$ are $\{1,2\}$, $\{2,3\}$, $\{1,2,3\}$. We claim $$\pi(S(K^{\{1,2\}}_2))= \frac{5}{4}.$$

The lower bound is from the following construction. Partition $[n]$ into two parts $A$ and $B$ of nearly equal size. Consider the hypergraph $G$ with the edge set $E=E_{2}\bigcup E_{3}$ where $E_{2}=\binom{A}{2}\bigcup \binom{B}{2}$ and $E_{3}=\binom{[n]}{3}\setminus \left(\binom{A}{3} \bigcup \binom{B}{3}\right)$. It is 