I need to prove that determining whether a graph has a relaxed-Hamiltonian cycle (definition given ahead) is NP-hard.
A relaxed-Hamiltonian cycle in $G$ is a closed walk $C$ that visits every vertex of $G$ exactly once, except for at most one vertex that $C$ visits more than once (i.e. that vertex may repeated twice or even more times).
Note: In a closed walk we can visit a vertex or edge multiple times, but first and last vertex are same. In the relaxed-Hamiltonian cycle this first and last vertex being the same is not counted as repetition.
I am thinking reducing to relaxed-Hamiltonian cycle from Hamiltonian Cycle.
Claim: Construct graph $G'$ given $G$ such that $G$ has Hamiltonian cycle iff $G'$ had relaxed-Hamiltonian cycle.
My idea is that if a graph $G$ has one or more articulation points then it cannot have a Hamiltonian Cycle so simply give a $G'$ which does not have a relaxed-Hamiltonian cycle.
The problem is when graph $G$ does not have any articulation points. In this case the graph $G$ may or may not have an Hamiltonian Cycle. I cannot think of a construction for this case such that my claim holds.
Is my approach correct? If yes please suggest how I could take care of the mentioned case? Otherwise, please hint me towards the correct direction of thinking to get the right construction.