be contained in some blow-up of $K_{2}^{\{1,2\}}$ since $H^{2}$ is bipartite, i.e. there exists some $s>2$ such that $H\subseteq K_{2}^{\{1,2\}}(s)$. So, by the squeeze theorem we have $$\frac{5}{4}=\pi(K_{2}^{\{1,2\}}) \leq \pi(H) \leq \pi(K_{2}^{\{1,2\}}(s))=\frac{5}{4}.$$ Hence $\pi(H)=\frac{5}{4}$ as claimed. $\square$

We will say that $H=H^{1}\cup H^{2}$ is a ***closed path*** (from $x_{1}$ to $x_{k}$) of length $k$ if $V(H)=\{x_{1}, x_{2},...,x_{k}\}$ and $E(H^{1})=\{\{x_{1}\}, \{x_{k}\}\}$ and $E(H^{2})=\{ \{x_{i}, x_{i+1}\}:1\leq i\leq k-1\}$. We will denote a closed path of length $k$, or a closed $k$-path, by $\bar{P}_{k}$.

Pictorially, we view a closed path of length $k$ as follows:

at (0,0) \[vertex\_closed\] (v1) \[label=below:[$x_{1}$]{}\] ; at (1,0) \[vertex\_open\] (v2) \[label=below:[$x_{2}$]{}\] ; at (2,0) \[vertex\_open\] (v3) \[label=below:[$x_{3}$]{}\] ; at (3,0) \[vertex\_open\] (v4) \[label=below:[$x_{k-2}$]{}\] ; at (4,0) \[vertex\_open\] (v5) \[label=below:[$x_{k-1}$]{}\] ; at (5,0) \[vertex\_closed\] (v6) \[label=below:[$x_{k}$]{}\] ; (v1)–(v2); (v2)–(v3); (v4)–(v5); (v5)–(v6); (v3)–(2.3, 0); (2.7, 0)–(v4); at (2.5, 0) [$\dots$]{};

Let $H=H^{1}\cup H^{2}$. If $H^{2}$ is bipartite and $H$ does not contain a copy of $K_{2}^{\{1,2\}}$ and $H$ contains a closed path of length $2k$, then $\pi(H)=\frac{9}{8}$.

[**Proof:**]{} First, we will give a construction giving us the lower bound. For any $n\in \mathbb{N}$ let $G_{n}$ have vertex set $[n]$. Partition the vertices of $G_{n}$ into two sets $X$ and $\bar{X}$ 