Please tell me if my next optimization can work:
I have symmetric, weighted, undirected graph, where it is knowen that the optimal tour, that visit all the vertices exactly once and return to the starting vertex, only contains edges with the weight of 1, it is also known that there is no edge with weight less than 1.
My goal is to find that tour.
Within this graph I want to look in to the case where there's a node, $B$ with only two edges with weight of 1 connected to it, let's call them, $AB$ and $BC$. I know that $AB,BC$ must be part of the optimal tour because there is no other way to get to $B$ and from $B$ with weight of 1. So I am removing the node $B$ and adding extra edge $AC$ between the endpoints of the removed node, and I am giving it the weight 0. Now when I find the optimal tour of that new graph(I assume that $AC$ will be part of it), I will replace it back to $AB,BC$, and I believe it will be the optimal tour of my original graph.
Do you think it will work?