Suppose $G$ is a simple graph and $V(G)=V(C)\bigcup \{u\_1,...,u\_n\}$,where $C$ is a $2n$-cycle in $G$ and $V(C)=\{a\_1,...,a\_n,b\_1,...,b\_n\}$ such that 


$(1)V(C)\bigcap \{u\_1,...,u\_n\}=\varnothing$;


$(2)E(G)=\{a\_1u\_1,u\_1b\_1,a\_2u\_2,u\_2b\_2,...,a\_nu\_n,u\_nb\_n\}\bigcup E(C)$.


I think there must exists a proper subset $S$ of $\{u\_1,...,u\_n\}$ such that $G-S$ is a Hamiltonian graph,is it ture?



a conjecture about Hamiltonian graph
The conjecture is right.It is equivalent to the claim below:


Let G be a simple graph which is a $2n$-cycle equipped with $n$ chords such that $G$ is $3$-regular,in other words,the set of the $n$ chords is a perfect matching of $G$(that is,every vertex of $G$ is matched).Then there must exist at least two different $2n$-cycles in $G$.


The proof is given by Tony Huynh,see my another question:[Does this graph contain at least two Hamiltonian cycles?](https://mathoverflow.net/questions/145531/an-interesting-conjecture-about-even-cycle).


