I don't find the above answer to be helpful. This is my solution, please let me know if this is correct: so if my graph G has four nodes and is connected with edges to make it look like a square with four nodes, plus an additional edge connecting two nodes diagonally, that is a yes instance of the HCP$.$ By reducing this yes instance to be a yes instance of ABOCP through taking away one node that has two edges connected to it $($that would remove one node, while still maintaining a graph with a cycle$)$ and demonstrating that this new graph is in NP$,$ would this would prove that ABOCP is NP complete through this polynomial time reduction?