in $Z_{n}$ has at most 1 neighbor in $X_{n}$. It follows that $$\begin{aligned}
\pi(\bar{P}_{4}) &=\lim_{n\to\infty} \pi_{n}(\bar{P}_{4}) \\
           &= \lim_{n\to \infty} h_{n}(G_{n}) \\
           &\leq \lim_{n\to\infty} \frac{|X_{n}|}{\binom{n}{1}} + \frac{|X_{n}|\cdot |Y_{n}|}{\binom{n}{2}} + \frac{|Y_{n}|\cdot |Z_{n}|}{\binom{n}{2}} + \frac{\binom{|Z_{n}|}{2}}{\binom{n}{2}} + \frac{|Z_{n}|}{\binom{n}{2}} \\
       &\leq \lim_{n\to\infty} \frac{xn}{\binom{n}{1}} + \frac{xyn^{2}}{\binom{n}{2}} + \frac{y(1-x-y)n^{2}}{\binom{n}{2}} + \frac{\frac{(1-x-y)^{2}n^{2}}{2}}{\binom{n}{2}} + \frac{(1-x-y)n}{\binom{n}{2}} \\
       &\leq \max_{\substack {0\leq x\leq 1 \\ 0\leq y\leq 1-x}} x + 2xy + 2y(1-x-y)+ (1-x-y)^{2} \\
       &=\frac{9}{8}.\end{aligned}$$ The last inequality is an easy multivariate calculus exercise. One can also verify it with software, such as *Mathematica*, the syntax being:

    Maximize[{x^2-x-y^2+2*x*y+1, 0<=x<=1, 0<=y<=1-x}, {x,y}].

It may be of interest to note that the maximum value of the function is obtained when $x=\frac{3}{4}$ and $y=\frac{1}{4}$. In this case $Z_{n}$ is empty. Since our upper bound matches the lower bound, we have the desired result. $\square$

Let $H=H^{1}\cup H^{2}$. If $H^{2}$ is bipartite and $H^{2}$ does not contain a closed $2k$-path for any $k\geq 1$, then $\pi(H)=1$.

[**Proof:**]{} First, since $|R(H)|=2$ we have, trivially, that $\pi(H)\geq 1$. Since $H$ contains no path of length $2k$ for any $k\geq 1$ it must be the case that $H$ is contained in a blow-up of a chain $C^{\{1,2\}}=\{\{x\}, \{x,y\}\}$. This is most clearly seen by again, considering the 