So this may be a question that is born of my inability to properly express my intentions to Google.
In a fully connected non-directional graph such that any three points can be properly represented in a triangle in Euclidean geometry it would seem that a greedy first algorithm which chooses the shortest edge would find the shortest path from a starting point to visit all other points.
I can't find a condition under which this is false, but one of my neighbors (another programmer) insists that I am wrong, though still cannot come up with a condition where I am wrong.
Edit:
I am not sure this question is an instance of of the general TSP, which I know is NP hard, or if it is I thought that all of the constraints on the allowed graphs would have impacted the difficulty.
From wiki: TSP is the shortest possible route that visits each city exactly once and returns to the origin city. In this problem there is no requirement to return to the origin city. Further, we don't care about selecting the optimal origin city since the origin city is set in our specific example that we were discussing.
Additionally, the graph in this problem must conform to the constraints of euclidean distances in either 2 or 3 dimensions so a three node graph such as below (key is the node, val is the distances between other nodes) could not exist since the edge from A to B would make the edges from C to A and B infeasible.
A:{B:80, C:10}, B:{A:80, C:5}, C:{A:10, B:5} Also by fully connected I meant all nodes have edges between them and every other node.
Sorry if the description came off as confusing, but I am wondering if given all these constraints, does it become solvable in a more simplistic manner.
