# Are all $4$-regular graphs Hamiltonian

It is easy to show that all connected $2$-regular graphs are Hamiltonian. The Petersen graph is a $3$-regular graph that is not Hamiltonian. Are there any $4$ regular graphs that are not Hamiltonian?

Thank you

• Thank you all. I up voted all answers.
– Amr
Nov 14, 2012 at 20:41

The answer is no. The graph on the picture is the smallest counterexample

You can most likely generalize this construction for every $k > 4$ and deduce that you can always find connected $k$-regular graphs that are not Hamiltonian.

Note that you didn't specify connected graph, which is certainly necessary for Hamiltonicity. More generally, a Hamilton cycle has the property that deleting any $k$ vertices breaks the cycle into at most $k$ connected components. In the case $k=1$ a Hamiltonian graph must be strongly connected.

Pick your favourite $4$-regular graph $G$, and delete one edge $e$ so it has two vertices of degree $3$. Take two copies of $G-e$, and create a new vertex that is adjacent to the four deficient vertices. The resulting graph is $4$-regular, connected, but not strongly connected.

Also, did you try Googling 4-regular non-Hamiltonian? The Meredith graph is very strongly connected, $4$-regular and still not Hamiltonian.

However, Tutte proved that a 4-connected planar graph is Hamiltonian.

• Thanks for this piece of information
– Amr
Sep 18, 2013 at 17:02