Suppose we have a complete graph $K_{2n}$, where $n$ is an integer, and we number vertices counterclockwise as $1,2,\cdots,2n.$ Then we give weight to the edge connecting $u,v$ as $|u-v|$.
Now, delete all the edges with weight less than $n-1$, that is, keep all the edges with weight $n$ and $n-1$, we get a cubic graph $G$.
I want to show that $G$ is Hamiltonian. I got a complicated construction. I am wondering if there is some theorem that can get this result quick and clean.
There's no good general results for when a cubic graph is Hamiltonian that apply here; it's easiest to solve such problems by finding a Hamiltonian cycle. But the construction can be made simple.
If $n$ is even, then we get a Hamiltonian cycle by going from $i$ to $i + (n+1)$, starting at any vertex and of course wrapping around modulo $2n$.
If $n$ is odd, this instead finds two disjoint $n$-cycles; each one visits every other vertex around $K_{2n}$. Deleting the edges $(1,n+2)$ and $(2,n+1)$ in those cycles gives two paths of length $n-1$: one starting at $1$ and ending at $n+2$, and one starting at $2$ and ending at $n+1$. We can use the edges $(2,n+2)$ and $(1,n+1)$ to join these two paths into a Hamiltonian cycle.