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?

How do I make it so that my answer can satisfy this: "However, for a rigorous proof of NP$-$completeness of ABOCP by reduction from HCP$,$ a generalization to arbitrary graphs and a clear demonstration that all transformations maintain the problem's constraints are necessary. The analysis should expand to include all potential configurations to ensure no exceptions invalidate the reduction. This expansion is crucial for establishing the reduction as both sound and complete."