Reducing the Hamiltonian cycle to the travelling salesman problem and self loops If this is my adjacency matrix for the hamiltonian cycle:
$$\begin{pmatrix}0&amp;1&amp;0&amp;1\\ 1&amp;0&amp;1&amp;0\\ 0&amp;1&amp;0&amp;1\\ 1&amp;0&amp;1&amp;0\end{pmatrix}$$
Then a reduction algorithm to reduce this to a TSP problem is to introduce 1 anywhere an edge is missing and let pre-existing edges have a cost of zero (according to this), applying this yields:
$$\begin{pmatrix}1&amp;0&amp;1&amp;0\\ 0&amp;1&amp;0&amp;1\\ 1&amp;0&amp;1&amp;0\\ 0&amp;1&amp;0&amp;1\end{pmatrix}$$
Now I see that I have also introduced weighted-loops to my undirected graph, would asking whether this has a zero-cost salesman tour be equivalent to asking whether the unreduced version has a hamiltonian cycle, or should I maybe have something else on the diagonal of the reduced matrix?
Suppose your original graph $G$ has a Hamiltonian cycle $C$.
Then the cost of the tour induced by $C$ in the new graph $G'$ you defined is indeed $0$ and conversely, any TSP tour of cost $0$ can only use edges of cost $0$ as you did not introduce any edges with a negative cost.
Adding the loops does not change this, so the reduction works out if we are willing to allow non-simple graphs which isn't done usually as the definition of the TSP asks for a tour visiting each vertex exactly once, i.e. no valid tour would include loop edges anyways.
Given this, we also see that you can remove the loop edges in $G'$ to make it a simple graph to obtain an equally valid reduction.
From a technical standpoint this doesn't make much of a difference but since people like simple graphs, I would personally prefer the reduction yielding a simple graph.