In the book embedding of a graph $G$ , each vertex of $G$ is placed on the spine and each edge is placed in the pages without crossing each other edge. If vertices have degree at most one in each page, the book embedding is matching . The minimum number of pages in which a graph can be matching book embedded is called matching book thickness. For Convenience, we denote the matching book thickness of a graph $G$ by $\mathrm{mbt}(G)$.

For the Wheel graph $W_n$ with $n$ vertices $O,1,2,3,...,n-1$, I want to know $\mathrm{mbt}(W_n)$. For the case $n$ is odd, it is not hard to know that $\mathrm{mbt}(W_n)=\Delta(W_n)=n-1.$ For the case $n$ is even, I guess $\mathrm{mbt}(W_n)=n$. For example, when $n=6$, I have tried some matching book embeddings of $W_6$ with different orderings $\omega$ of the vertices on the spine. And I always get the $mbt(W_6,\omega)=6$. But I have no idea about the proof.

I will appreciate it if someone could give any suggestions.

In the book embedding of a graph $G$ , each vertex of $G$ is placed on the spine and each edge is placed in the pages without crossing each other edge. If vertices have degree at most one in each page, the book embedding is matching . The minimum number of pages in which a graph can be matching book embedded is called matching book thickness. For Convenience, we denote the matching book thickness of a graph $G$ by $\mathrm{mbt}(G)$.

For the Wheel graph $W_n$ with $n$ vertices $O,1,2,3,...,n-1$, I want to know $\mathrm{mbt}(W_n)$. For the case $n$ is odd, it is not hard to know that $\mathrm{mbt}(W_n)=\Delta(W_n)=n-1.$ For the case $n$ is even, I guess $\mathrm{mbt}(W_n)=n$. For example, when $n=6$, I have tried some matching book embeddings of $W_6$ with different orderings $\omega$ of the vertices on the spine. And I always get a matching book embedding on 6-page. For $W_4=K_4$, $mbt(W_4)=4.$ But I was wondering whether the equality holds for any even $n$,i.e.$\mathrm{mbt}(W_n)=n$ .

I will appreciate it if someone could give any suggestions.